Engineering Mathematics Transforms and Partial Differential Equations [3, 2 ed.]

Table of contents :
1. PARTIAL DIFFERENTIAL EQUATIONS

Formation – Solutions of standard types of first order equations –
Lagrange’s Equation - Linear partial differential equations of second
and higher order with constant coefficients.

2. Fourier Series
Dirichlet’s conditions – General Fourier series – Half range Sine and
cosine series – Parseval’s identity – Harmonic Analysis.

3. Boundary value problems
Classification of second order linear partial differential equations – So-
lutions of one – dimensional wave equation, one-dimensional heat equa-
tion – Steady state solution of twodimensional heat equation – Fourier
series solution in Cartesian coordinates.

4. Laplace Transforms
Transforms of simple functions – Basic operational properties – Trans-
forms of derivatives and integrals – Initial and final value theorems –
Inverse transforms – Consvolution theorem – Periodic function – Ap-
plications of Laplace transforms of solving linear ordinary differential
equations upto second order with constant coefficients and simultaneous
equat ions of first order with constant ecoefficients.

5. Fourier Transform
Statement of Fourier integral theorem – Fourier transform pairs – Fou-
rier Sine and Consine transforms – Properties – Transforms of simple
functions – Convolution theorem – Parseval’s identity.

Citation preview

ENGINEERING MATHEMATICS VOLUME III (Third Semester)

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ENGINEERING MATHEMATICS VOLUME III [For B.E., B. Tech. & B.Sc. (Applied Sciences)]

P. KANDASAMY

K. THILAGAVATHY

M.A., Ph.D.

Professor of Mathematics (Retd.) P.S.G. College of Technology Coimbatore Visiting Professor Amrita Institute of Technology and Science Coimbatore

M.Sc., M.Phil, Ph.D.

Reader in Mathematics Kongunadu Arts and Science College Coimbatore

K. GUNAVATHY M.E., Ph.D.

Asst. Professor, Department of Electronics and Communication Engineering P.S.G. College of Technology, Coimbatore

2002 S. CHAND & COMPANY LTD. RAM NAGAR, NEW DELHI-110 055

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PREFACE TO THE SECOND EDITION This revised edition is according to the syllabus of Anna University for third semester from the year 2002 onwards. The chapter on Laplace Transform is shifted from volume II to this volume III. AUTHORS

PREFACE TO THE FIRST EDITION The existing Third Volume of our series of textbooks on Engineering Mathematics for students of B.E., B. Tech and B.Sc. (Applied Sciences) has been now split into two volumes, to cater to the needs of the syllabus semester-wise. This volume caters to the syllabus of fourth semester. Many worked examples are added in each chapter and a large number of problems are included in the Exercises. In addition, short Answer Questions, Model and University Question Papers have been added at the end. We are confident that with all these modifications/additions etc., this present edition with its new format will prove to be even more useful and our esteemed readers will receive it well. We request the readers to feel free to write to us for the improvement of the book. AUTHORS Author’s address for contact : 66, Thirty Feet Road, Krishnasamy Nagar, Ramanathapuram Coimbatore 641045

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SYLLABUS ANNA UNIVERSITY

MA231 MATHEMATICS III L T P C 3 1 0 4 1.

PARTIAL DIFFERENTIAL EQUATIONS 9 Formation – Solutions of standard types of first order equations – Lagrange’s Equation - Linear partial differential equations of second and higher order with constant coefficients. 2. Fourier Series Dirichlet’s conditions – General Fourier series – Half range Sine and cosine series – Parseval’s identity – Harmonic Analysis. 3. Boundary value problems (9) Classification of second order linear partial differential equations – Solutions of one – dimensional wave equation, one-dimensional heat equation – Steady state solution of twodimensional heat equation – Fourier series solution in Cartesian coordinates. 4. Laplace Transforms (9) Transforms of simple functions – Basic operational properties – Transforms of derivatives and integrals – Initial and final value theorems – Inverse transforms – Consvolution theorem – Periodic function – Applications of Laplace transforms of solving linear ordinary differential equations upto second order with constant coefficients and simultaneous equat ions of first order with constant ecoefficients. 5. Fourier Transform (9) Statement of Fourier integral theorem – Fourier transform pairs – Fourier Sine and Consine transforms – Properties – Transforms of simple functions – Convolution theorem – Parseval’s identity. L : 45 + T : 15 = 60 Text Books : 1. Kreyszig, E., “Advanced Engineering Mathematics“ (8th Edition), John Wiley and sons, (Asia) Pte Ltd., Singapore 2000. 2. Grewal, B.S., “Higher Engineering Mathematics” (35th Edition), Khanna Publishers, Delhi 2000.

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Books for Reference : 3. Kandasamy, P., Thilagavathy, K., and Gunavathy, K., “Engineering Mathematics”, Volumes II & III (4th Revised Edition), S. Chand & Co. New Delhi, 2001. 4. Narayanan, S., Manicavachagom Pillay, T.K., Ramanaiah, G., “Engineering Mathematics for Engineering Students”, Volumes II & III (2nd Edition), S. Viswanathan (Printers & Publishers, Pvt, Ltd.) 1992. 5. Venkataraman, M.K. “Engineering Mathematics” Volumes III – A & B, 13th Edition National Publishing Company, Chennai, 1998. 6. Shanmugam, T.N. : http://www.annauniv.edu/shan/trans.htm

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CONTENTS Chapter I.

Fourier Series

II.

Partial Differential Equations

III.

Applications of Partial Differential Equations Boundary Value Problems Vibration of Strings Heat Flow—One Dimension Heat Flow—Two Dimension (Cartesian) Heat Flow—Polar Equations

IV.

Fourier Transforms Short Answer Questions Answers

V.

The Z-Transforms Answers

V.

.

Linear Algerba Answers Model Question Papers Solved Question Oapers Examination Question Paper

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1

fourier series

Periodic functions. Let f (x) be a function of x, defined as follows. For each positive integer x, f (x) is the remainder obtained when x is divided by 3. We can easily make a list of the first few values of f (x): f (1) = 1, f (2) = 2, f (3) = 0; f (4) = 1, f (5) = 2, f (6) = 0; f (7) = 1, f (8) = 2, f (9) = 0; f (10) = 1, f (11) = 2, f (12) = 0. From an examination of this table, it is found that for any positive integer x, f (x + 3) = f (x); further, if y is any positive integer less than 3, f (x + y) ≠ f (x). We say that f (x) is a periodic function with the period 3. Definition. A function f (x) is said to be periodic, if and only if (x + p) = f (x) is true for some value of p and every value of x. The Smalles value of p for which this equation is true for every value of x will be called the period of the function. When n is any integer, sin (x + 2np) = sin x, for all x; hence sin x is periodic. Taking n = 1, sin (x + 2p) = sin x for every x. However, if a is any positive number smaller than 2p, it can be shown that sin (x + a) = sin x is not true for all values of x. Therefore, 2p is the smallest number p such that sin (x + p) = sin x for every x. Accordingly, 2p is the period of sin x. Similarly, cos x is a periodic function with the period 2p. Now tan (x + p) = tan x, for all x. Hence tan x is periodic; its period is p. The graph of a periodic function of period p is obtained by periodic repetition of its graph in any interval of length p. Note that f (x) = k, where k is a constant, is a periodic function in the sense of the definition, since f (x + p) = f (x). Again, if f (x) and F (x) have the same period p, then the function f (x) = af (x) + bF (x) has the period p. Exercise 1 (a) 1. Find the smallest period p of the following functions: (i) sin 2x, (ii) cos 3x, (iii) sin nx, (iv) cos 2px, (v) sin (2px/k). 2. Plot graphs of the following functions: p 1 + cos x, (iii) sin x + sin 3x, (iv) sin 2px, (i) sin x, (ii) 4 3 (v) f (x) = x, when – p < x < p and f (x + 2p) = f (x), (vi) f (x) = x2, when – p < x < p and f (x + 2p) = f (x),

{ (viii) f (x) = {0 when –p < x < 0 and f (x + 2p) = f (x) x when 0 < x < p

(vii) f (x) = – 1 when –p < x < 0 and f (x + 2p) = f (x) + 1 when 0 < x < p

3. If p is a period of f (x), show that np is also a period, where n is any integer, (positive or negative) 4. Show that the function f (x) = constant, is a periodic function of period p, for any value of p. 5. If f (x) and F (x) have the period p, show that f = af + bF where a and b are constants, has the same period p.

AVC3/S Chand/1st Proof: 9/7/2014; 2nd Proof: 14/7/2014

2

Engineering Mathematics-III (Third Semester )

Bounds of a function. If for all values of x in a given interval, f (x) is never greater than some fixed number M, the number M is called an upper bound for f (x) in that interval; if f (x) is never less than some number m, then m is called a lower bound. One or both these bounds may not exist. 1 Thus, in the interval (0, 2p), y = sin x has an upper bound +1 and a lower bound –1. y = + cos 2 3 1 1 x, has an upper bound and a lower bound – . The function y = in the interval (0, 1) has a 2 2 x 1 lower bound unity, but has no upper bound. The function y = x – has neither an upper nor a x lower bound in the interval (0, ∞). If a function has an upper bound and a lower bound in a given interval, then that function is said to be bounded. We can put the idea differently: for all values of x in a given interval, if | f (x) | ≤ a, where a is some fixed number, then the function is bounded in that interval. Continuity of a function. It is sometimes important to know the behaviour of a function f (x) when x → a in such a way that x always remains greater than a, that is when x is allowed to approach a from the right side only. In such a case, the following notation is adopted: we write x → a+, and the limit is called the right-hand limit. If the limit is b, we give the symbol lim f (x) = b, or f (a +) = b. x → a+



The left-hand limit is defined similarly, and is denoted by either

lim f (x), or f (a–). x → a–

Let f (x) =

3 x−a . + 2 2| x−a|

Here f (a –) = 1, and f (a +) = 2. The symbols f (a–) and F (a+) should not be confused with the symbol F (a), which means the value of f (x) calculated at the point x = a. In the present 3 1 0 example, f (a) = + × ; hence f (a) does not exist. 2 2 0 1



Or again, if f (x) = 2 x −1 , f (1 –) = 0, f (1 +) = ∞, but f (1) is not defined.

If f (a –) = f (a +) = b, then we call b the limit of f (x) as x → a and write lim f (x) = b. x→a

If f (x) =

2

2

x −a 0 , f (a –) = 2a and f (a +) = 2a; thus, lim f (x) = 2a; note that f (a) = and x−a q x→a

hence is not defined.

We have already studied that a function f (x) is said to be continuous at x = a, if lim f (x) = f (a). x→a

With our present knowledge, we can say that f (x) is continuous at x = a, if f (a –) = f (a) = f (a +). x2 − a2 , when x ≠ a, and f (x) = 2a, when x = a is continuous at x = a, even though Thus, f (x) = x−a x2 − a2 as such is not continuous at x = a. x−a From the above, it is clear that f (x) is discontinuous at x = a, if f (a –) ≠ f (a +). The discontinuity at x = a is called infinite whenever f (x) becomes infinite as x → a. If the function is discontinuous at point x = a but is bounded in its vicinity, the discontinuity is called finite or ordinary. If f (a +) the function F (x) =

3

Fourier Series

and f (a –) exist and are unequal, their difference f (a + ) – f (a –) is called the jump of the function 3 x−a has a jump of one unit at x = a. On the other hand, the at x = a. Thus the function + 2 2| x−a| 1

function 2 x − 1 has an infinite discontinuity at x = 1. A function which is continuous everywhere except for a finite number of jumps in a given interval is called piece-wise continuous or sectionally continuous, in that interval. Exercise 1(b) 1. If f (x) = x2, find f (3 –), f (3 +). What is lim x3 ? x→3

2. What is lim

x → p /3

cos x? cos

3. Given that f (x) =

( p3 + ) ? cos ( p3 – ) ?

1 , find f (1 +) and f (1 –) x −1

4. Find f (0 +) and f (0 –), if f (x) = 5. If f (x) =

e1/ x ; what is lim f (x) ? x→0 1 + e1/ x

1 , find f (0 +) and f (0 –). What is f (0) ? 1 − e −1/ x

Is the function continuous at x = 0? What is the jump of the function at x = 0? Find f (0). 6. f (x) = 1, 0 < x < p f (x) = 2, p < x < 2p. Find f (p –) and f (p +). Evaluate the jumps of the function at x = p.

Fourier Series. Periodic functions occur frequently in engineering problems. Such periodic functions are often complicated; it is therefore desirable to represent these in terms of the simple periodic functions of sine and cosine. A development of given periodic function into a series of sines and cosines was effected by the French physicist and mathematician Joseph Fourier (1768–1830); the series of sines and cosines in known after him. Below we prove a few results which we need in deriving Fourier series. Given that m and n are positive integers or zero: 1. If n ≠ 0,

c + 2p

∫



c

 cos nx  sin nx dx =  − n  

= Note. if n = 0,

c + 2p

∫

= −

cos n (c + 2p) cos nc + n n

cos nc – cos (nc + 2np) cos nc − cos nc = = 0. n n

c + 2p

sin = nx dx

c

2. If n ≠ 0,

c + 2p

c + 2p

∫

= 0·dx 0. c

c + 2p

 sin nx  cos nx dx = 0. =  n  c

∫ c

3.

c + 2p

∫

sin mx cos nx dx

c

=

1 2

c + 2p

∫ c

[sin (m + n) x + sin (m − n) x] dx = 0, as in 1.

4

4. If m ≠ n,

Engineering Mathematics-III (Third Semester ) c + 2p

∫

sin mx sin nx dx

c

= 5. If m ≠ n,

1 2

c + 2p

∫

[cos (m − n) x − cos (m + n) x]dx = 0, as in (2.)

c

c + 2p

∫

cos mx cos nx dx

c

= 6. If n ≠ 0,

c + 2p

∫

2 sin= nx dx

c

1 2

∫

∫ c

∫

[cos (m + n) x + cos (m − n) x] dx = 0, as in (2)

c

(1 − cos 2nx)dx

c

= c + 2p

c + 2p

c + 2p

=

7. If n ≠ 0,

1 2

1 2

c + 2p

∫

dx −

c

1 2

c + 2p

∫

cos 2= nx dx

c

1 c + 2p 1 [ x] − ×0 2 c 2

1 1 {(c + 2p) – c} = × 2p = p. 2 2

c + 2p

1 cos 2 nx dx = (1 + cos 2nx) dx = p. 2

∫ c

Theorem 1. If f (x) is a periodic function with period 2p and if f (x) can be represented by a trigonometric series in (– ∞, ∞). f (x) =



∞ ao + (a cos nx + bn sin nx), …(i) 2 n =1 n

∑

1 an = p

then

c + 2p

∫ c

1 f ( x) cos nx dx, and bn = p

c + 2p

∫

f ( x) sin nx dx.

c

Proof. First of all, let us determine a0. Integrating both the sides of (i) from c to c + 2p, we get c + 2p c + 2p ∞  a0  f ( x) dx = (an cos nx + bn sin nx)  dx.  + 2  n =1 c c  

∫

∑

∫

We assume that term by term integration of the series is allowed. [This is possible, if the series is uniformly convergent.] Then we have c + 2p



∫ c

f ( x) dx =

a0 2

c + 2p

∫ c

dx +

 c + 2 p cos nx dx + bn an  n =1 c ∞

∑

∫

c + 2p

∫ c

 sin nx dx  . 

a0 a [ x]cc + 2 p= 0 × 2p= a0 p. Using formulae 2 and 1 on p. 4, the other 2 2 integrals on the right become zero each. Thus, The first term on the right is c + 2p



∫ c

f ( x) dx = a0p,  giving  a0 =

1 p

c + 2p

∫ c

f ( x) dx.

5

Fourier Series

We now determine a1, a2, a3, … by a similar method. Multiplying (i) by cos mx, where m is any fixed positive integer, and integrating from c to c + 2p, we get c + 2p

∫



c + 2p

∫

f ( x) cos mx dx =

c



c

∞  a0  + (an cos nx + bn sin nx)  cos mx dx …(ii) 2 n =1  

∑

Using term by term integration, the right hand side becomes c + 2p

c + 2p

∞ ∞ a0 cos mx dx + cos nx cos mx dx + an bn 2 = n 1= n 1

∑

∫ c

∑

∫ c

c + 2p

∫

sin nx cos mx dx.

c

By formula 2 on p. 4, the first of these integrals is zero. Again by formula 3 on p. 4, all the integrals of the last S are zero each. The remaining integrals are c + 2p

a1

∫

c + 2p

cos x cos mx dx + a2

c

∫

c + 2p

cos 2 x cos mx dx + … + am

c

∫

cos 2 mx dx + …

c

c + 2p

+ an



∫

cos nx cos mx dx + …

c

c + 2p



∫

Of these, am

cos 2 mx= dx am · p, using formula 7, p. 5.

c

Lastly by formula 5, p. 5 each of the remaining integrals is zero. Substituting the above results in (ii) above, we have c + 2p



∫ c

1 p

f ( x) cos mx dx = am. p, giving am =

Writing n in the place of m, we get an =

1 p

c + 2p

∫

f ( x) cos mx dx .

c

c + 2p

∫

f ( x) cos nx = dx, n 1, 2, 3, …

c

Finally multiplying (i) by sin mx, where m is any fixed positive integer, and integrating from c to c + 2p, we have c + 2p



∫

c + 2p

∫

f ( x)sin mx dx =

c

c

∞  a0 (an cos nx + bn sin nx) sin mx dx …(iii) 2 + n =1 

∑

c + 2p

Integrating the R.H.S. term by term, we find that every one of the integrals except

∫

bm sin 2 mx dx

c

is zero; and by formula 6 on p. 5, this integral is bm p. Hence from (iii) above c + 2p



∫ c

f ( x)sin mx dx = bmp, giving bm =

Writing n in the place of m,

bn =

1 p

1 p

c + 2p

∫

f ( x)sin mx dx.

c

c + 2p

∫

f ( x)sin nx = dx, n 1, 2, 3, …

c

Thus we have found the values of the coefficients a0, an and bn. These relations are known as Euler formulas. Thus, the Euler formulas are:

6

Engineering Mathematics-III (Third Semester )



a0 =

1 p

bn =

1 p

c + 2p

∫ c

f ( x) dx; an =

1 p

c + 2p

∫

f ( x) cos nx dx, and

c

c + 2p

∫

f ( x)sin nx dx= n 1, 2,…

c

They are sometimes called Euler-Fourier formulas. The R.H.S. of (i), when the coefficients there are given by the Euler formulas, is known as the Fourier series of f (x). In most applications, the interval over which f (x) is to be expanded is either (–p, p) or (0, 2p), so that the value of c in the Euler formulas is either –p or 0. Actually, the formula for a0 need not be listed, for it can be obtained from the general expression for an by putting n = 0; it was to achieve this that we wrote a the constant term in the original expansion as 0 , (and not a0 as usual). 2 Note that we have not proved that any given function f (x) has a Fourier expansion that converges to f (x). What we have shown is that if a function f (x) has an expansion of the form (i) for which term by term integration is valid, then the coefficients in that series will be given by the Euler formulas. Questions concerning the convergence of the Fourier series, and if they converge, the conditions under which they will represent the functions which generated them are many and difficult. But, all the functions that you will have to deal with will be covered by the famous theorem of Dirichlet, (the great German mathematician 1805–1859). Theorem 2. If f (x) is a bounded function of period 2p which in any one period has at most a finite number of maxima and minima and a finite number of points of discontinuity, then the Fourier series of f (x) converges to f (x) at all points where f (x) is continuous; also, the series converges to the average value of the right and left-hand limits of f (x) at each point where f (x) is discontinuous. The conditions given in this theorem are known as the Dirichlet condiditons. (The proof of this theorem is not expected of you; you have merely to memorise the statement). These conditions are not necessary. Dirichlet’s conditions If a function f (x) is defined in c ≤ x ≤ c + 2p, it can be expanded as a Fourier series of the ∞ a form 0 + (a cos nx + bn sin nx), provided the following DIRICHLET’S conditions are 2 n =1 n

∑

satisfied. (i) f (x) is single valued and finite in (c, c + 2p) (ii) f (x) is continuous or piece-wise continuous with finite number of finite discontinuities in (c, c + 2p) (iii) f (x) has a finite number of maxima or minima in (c, c + 2p) Note 1. These conditions are not necessary but only sufficient for the existence of Fourier series. Note 2. If f (x) satisfies Dirichlet’s conditions and f (x) is defined in (– ∞, ∞), then f (x) has to be periodic of periodicity 2p for the existence of Fourier series of period 2p. Note 3. If f (x) satisfies Dirichlet’s conditions and f (x) is defined in (c, c + 2p), then f (x) need not be periodic for the existence of Fourier series of period 2p. Note 4. If x = a is a point of discontinuity of f (x), then the value of the Fourier series at 1 x = a is [f (a +) + f (a –)]. 2

7

Fourier Series

The Dirichlet conditions make it clear that a function need not be continuous in order to possess a valid Fourier expansion. This means that the graph of a function may consist of a number of dis-jointed arcs of different curves, each defined by a different equation, and still be representable by a Fourier series. In using the Euler formulas to find the coefficients in the expansion of such a function, it will be necessary to break up the range of integration (c, c + 2p) to correspond to the various segments of the function. Thus, in the above function, f (x) is defined by three expressions f1 (x), f2 (x) and f3 (x) on successive portions of the period interval c ≤ x ≤ c + 2p. Hence it is necessary to write the Euler formulas as an =



1 p

c + 2p

∫

f ( x) cos nx dx

c

1  =  f ( x) cos nx dx + p c a

b

c + 2p

∫

a

b

c + 2p

a

b

1 =  f1 ( x) cos nx dx + p c

∫

a

∫

 f ( x) cos nx dx  . 

∫

f ( x) cos nx dx +

b

∫ f2 ( x) cos nx dx + ∫

 f3 ( x) cos nx dx  . 

A similar break up is necessary for evaluating bn. According to the above theorem, the Fourier series of the function shown in the figure will converge to the average values, indicated by dots, at the discontinuities at c, a and b, regardless of the definition (of lack of definition) of the function at these points. The following two standard integrals which you have studied in lower classes occur in almost all problems in this chapter. It is better to remember these results to apply them in problems. 1.

∫e

ax

sin bx dx =

∫

e ax cos bx dx 2. =

e ax [a sin bx – b cos bx]. a + b2 2

e ax [a cos bx + b sin bx]. a 2 + b2

Bernouli’s Generalised Formula of Integration by Parts

This formula is only an extension of the formula of integration by parts.



We know that



∫ u dv = uv – ∫ v du.

But in many problems, the integral will be of the form ∫ uv dx and not ∫ u dv. Now, we will find a formula to evaluate ∫uvdx.

In what follows, we write

and

du a 2u = u ′′, etc. = u′, dx dx 2 ∫ u dx = v1, ∫ v1 dx = v2 etc

Hence,



∫ uv dx = ∫ ud (v1)

= uv1 – ∫ v1 du, using integration by parts

8

Engineering Mathematics-III (Third Semester )

= uv1 – ∫ v1u′dx

= uv1 – ∫ u′ d (v2)

= uv1 – u′v2 + ∫v2u″ dx

= uv1 – u′v2 + ∫ u″ d (v3)

= uv1 – u′v2 + u″v3 – ∫ v3u′′′ dx







……………………………………….







……………………………………….

= uv1 – u′ v2 + u″v3 – u′′′v4 + … Thus,

∫ uv dx = uv1 – u′ v2 + u″ v3 – … is called Bernoulli’s extented formula of integration by parts.

Note 1. Whenever there is a polynomial as a factor in the integrand, take that polynomial as u and the remaining factor as v, since after a certain stage, the dervatives of u will vanish.

To illustrate the advantage of this formula, consider

∫ (x3 + 2x2 + 5) cos 2x dx

Here take x3 + 2x2 + 5 = u and cos 2x = v.

∫ (x3 + 2x2 + 5) cos 2x dx = ∫ uv dx  sin 2 x  = (x3 + 2x2 + 5)   – (3x2 + 4x)  2 

 cos 2 x  −  + (6x + 4) 4  

 sin 2 x   cos 2 x  –  – (6)  +C 8    16 



Note 2. In the above problem, we take the given integral as ∫ uv dx and not as ∫u dv.



Example 1. Obtain the Fourier series of periodicity 2p for f (x) = e–x in the interval 0 < x >

2p. Hence deduce the value of

∞

(–1) n

∑ 1+ n2 . Further, derive a series for cosech p.

n= 2

Let

e–x =

∞ a0 + (a cos nx + bn sin nx) …(i) 2 n =1 n

∑

Taking c = 0 in the Euler formulas, we have,



a0 =

2p

2p

0

0

1 1 1 [ – x ]2 p = f ( x) dx = e − x dx e 0 p p p

∫

∫

1 1 − e −2 p =  −e −2 p + 1 = p p For n = 1, 2, 3, …

an =

1 p

2p

∫ f ( x) cos nx dx 0

2p

1 = e − x cos nx dx p

∫ 0

2p

 1 e− x {(−1) cos nx + n sin nx} , =  p  (−1) 2 + n 2 0 

using the result of the standard integral.

9

Fourier Series

1 e −2 p (−1) − (−1)  = p(1 + n 2 ) 1 (1 − e −2 p ) = p (1 + n 2 )

Similarly, bn =



1 p

2p

∫e

−x

sin nx dx

0

2p

 1  e− x {(−1) sin nx − n cos nx} =  2 p 1 + n 0 1 = [−ne −2 p + n] 2 p(1 + n ) n (1 − e −2 p ) = p(1 + n 2 )

Substituting the values of a0, an, bn in (i), and remembering that the constant term is

a0 , (and 2

not a0), we get that in the range (0, 2p), e–x =



1 − e −2 p p

1 ∞ 1  (cos nx + n sin nx)  …(ii)  + 2 2 n = 11 + n  

∑

In (0, 2p), e–x is continuous. Therefore, at x = p, the value of the Fourier series equals the value of the function e–x.

Hence replacing x by p in (ii), e–p =



\



When n = 1,



Hence

1 − e −2 p p

 1 ∞ cos np   + 2   2 n = 1 1 + n 

∑

∞ cos np 1 pe −p p p + = = = −2 p p −p 2 n = 1 1 + n2 p 2 sinh 1− e e −e

∑

cos np 1 = ; further cos np = (–1)n 2 1 + n2 ∞

(–1) n

∑ 1 + n2 =

p cosech p 2

cosech p =

(–1) 2 . p n = 11 + n2

n =1

∞

∑

n



\



Example 2. If f (x) = −k , when – p < x < 0 and f (x + 2p) = f (x) for all x, derive the Fourier k , when 0 < x < p

{

series for f (x). Deduce that

p 1 1 1 = 1 – + + + … to infinity. 3 5 7 4

10

Engineering Mathematics-III (Third Semester )

f (x) =

∞ a0 + (a cos nx + bn sin nx) (i) 2 n =1 n

∑



Let



Taking c = –p in the Euler formulas we have



a0 =

1 p

p

∫

f ( x) dx

−p

0

1   f ( x) dx + p  −p

∫

p

∫ 0

 f ( x) dx  . 

Now using the hypothesis for the value of f (x), we get



0 p  1 0 p 1  a0 =  (−k ) dx + kdx  = ( −kx ) – p + ( kx )0 p p 0 –p 

∫

{

∫

}

1 = {(0 – kp) + (kp – 0)} p Thus a0 = 0. Again for n = 1, 2, 3,…

an =

1 p

p

∫

f ( x) cos nx dx

−p

0 p  1  =  f ( x) cos nx dx + f ( x) cos nx dx  . p  0 – p

∫



∫

Substituting the values supplied for f (x), we have



0 p  1  an =  (−k ) cos nx dx + k cos nx dx  p 0 –p 

∫

∫

0 p 1  sin nx   sin nx   =  −k  + k   p  n −p  n 0 



Since sin 0, sin (–np) and sin np are all zero, we get an = 0.



bn =

1 p

p

∫

f ( x) sin nx dx

−p

0 p  1  =  f ( x) sin nx dx + f ( x) sin nx dx  p 0 – p 

∫

∫

0 p  1  (– k ) cos nx dx =  + k cos nx dx  p  0 –p

∫

∫

0 p 1   cos nx   cos nx    =   k + − k p  n  −p  n  0 

{

}{

}

1 k k k k  =  cos 0 − cos (−np) + − cos np + cos 0  p n n n n 

But cos (–a) = cos a, giving cos (–np) = cos np; further, cos 0 = 1.



Hence

bn =

k k [{1 – con np} + {– cos np + 1}] = (2 – 2cos np) np np

11

Fourier Series



\



Hence b1 =



bn =

4k 4k 4k 4k ; b2 = 0; b3 = ; b = 0; b5 = ; b = 0; b7 = … p 3p 4 5p 6 7p Using the values of an and bn in (i) we obtain

4k 1 1 1 {sin x + sin 3x + sin 5x + sin 7x + … to ∞} p 3 5 7 In the above equation putting x = p/2, we get f (x) =



−1, for odd n 2k (1 − cos np). Now cos np = +1, for even n np (−1) n , for any integer n.



But, by hypothesis, f



Hence

f

( p2 ) = 4pk {1 − 13 + 15 − 71 + …to ∞}

( p2 ) = k.

k =

{

}

4k 1 1 1 1 − + + + … to ∞ p 3 5 7

p , we have 4k 1 1 1 p = 1 – + − + … to ∞. 4 3 5 7 Note. Functions of the type given in this example occur as external forces acting on mechanical systems, electromotive forces in electric circuits etc.



Multiplying both the sides by



Example 3. Obtain the Fourier series of the periodic function defined by f (x) = –p, if –p < x 1 1 1 p2 (MS. 1986A) < 0 and f (x) = x, if 0 < x < p. Deduce that 2 + 2 + 2 + … to ∞ = .  8 1 3 5



Let



Then



f (x) =

∞ a0 + (a cos nx + bn sin nx) …(i) 2 n =1 n

∑

p 0 p  1 1 f ( x) dx a0 = =  f ( x) dx + f ( x) dx  p p −p 0 –p 

∫

a0 =

∫

∫

0 p  1  (−p) dx + x dx  p  0  −p

∫

∫

1 p2  =  −p2 +  = – p/2 2 p

an =

1 p

p

∫

f ( x) cos nx dx

−p

0 p  1 =  (−p) cos nx dx + x cos nx dx  p  0  −p

∫

{

∫

}

1  sin nx  =  (−p)   p   n 

0

p   sin nx   cos nx    + ( x )   − (1)  − 2     n  n  0   −p 

12

Engineering Mathematics-III (Third Semester )

1 1 =  2 (cos np − 1)  p n  1 = 2 [(−1) n − 1] pn bn =



1 p

p

∫

f ( x) sin nx dx

−p

p p  1 ( ) sin nx dx =  −p + x sin nx dx  p  0  −p

∫

∫

{

}

1  cos nx  =  (−p)   p   n 

0

p   cos nx   sin nx    + ( x )   − (1)  − 2     n  n  0   −p 

1 p p =  (1 − cos np) − cos np  n p n  1 = (1 − 2 cos np) n Substituting the values of the coefficients in (i), cos 3 x cos 5 x p 2  f (x) = − −  cos x + + + … 4 p 32 52 



sin 2 x 3 sin 3 x sin 4 x 3 sin 5 x + − + − …(ii) 2 3 4 5 Zero is a point of discontinuity of f (x). But f (0 –) = – p, and f (0 +) = 0. Hence f (0−) + f (0+) p = − . 2 2 By theorem 2 on p. 8, this is the sum of the Fourier series in (ii) for the value 0 of x. Hence + 3 sin x –



−

p 2 1 1 p = − − 1 + 2 + 2 + … 4 p 2 3 5 



−

2 1 1 p = − 1 + 2 + 2 + … 4 p 3 5 

\

p2 1 1 = 1 + 2 + 2 +… 8 3 5 Example 4. What is the Fourier expansion of the periodic function, whose definition in one 1 1 1 1 period is f (x) = 0, when − p < x < 0 ? Evaluate − + − + … to ∞. sin x, when 0 < x < p 1.3 3.5 5.7 7.9



Hence

{



Let

f (x) =

∞ a0 + (a cos nx + bn sin nx) …(i) 2 n =1 n

∑

[In the great majority of the examples, we can calculate an and bn for n = 1 to ∞. But, this is a special example, where a1 and b1 are to be calculated separately; but a0 can be got from an.]

0 p  1  a1 =  0. cos x dx + sin x cos x dx  p 0  −p 

∫

∫

13

Fourier Series p 1 1 = sin 2 x  0 = × 0 = 0. 2p p For n = 0, 2, 3, 4,… 0 p  1  an =  0. cos nx dx + sin x cos nx dx  p  0  −p

∫



∫

p

1 = {sin (1 + n) x + sin (1 − n) x} dx 2p

∫ 0

p

1  cos (1 + n) x cos (1 − n) x  =  − −  ,  2p  1+ n 1− n 0

for n ≠ 1

{

}

1   cos (p + np) x cos (p − np) x  1 1  =  − −  + 1+ n + 1− n  2p   1+ n 1− n   1  cos np cos np 1 1  =  − + + 2p  1 + n 1− n 1 + n 1 − n  1 1 + cos np 1 + cos np  1 + cos np + = =  2p  1 + n 1 − n  p(1 − n 2 )

i.e.,

an = −

1 + cos np p(n − 1)(n + 1)

0 p  1  b1 =  0. sin x dx + sin x sin x dx  p 0 –p 

∫



∫

p

p

{2

}

1 1 1 1 1 p 1 2 = ∫ sin x dx = ∫ − cos 2 x dx = × = p



p

0

0

2

p

2

2

n = 2, 3, 4, 5,…

For

0 p  1  bn =  0. sin nx dx + sin x sin nx dx  p 0  −p 

∫



∫

p

1 = {cos (n − 1) x − cos (n + 1) x} dx 2p

∫ 0

p

1  sin (n − 1) x sin (n + 1) x  1 =  − = × 0 = 0. 2p  n −1 n + 1  0 2p

Substituting the values of the coefficients in (i), f (x) =



Putting x =



p , 2 f

()

{

}

1 sin x 2 cos 2 x cos 4 x cos 6 x cos 8 x + – + + + +… to ∞ p 2 p 3 15 35 63

( p2 ) = p1 + 12 − p2 {− 13 + 151 − 351 + 631 − … to ∞}

p p But, by hypothesis f= sin = 1. Hence 2 2

14

Engineering Mathematics-III (Third Semester )

{



\



i.e.,

}

1 1 2 1 1 1 1 − + – … to ∞ 1 = + − − + p 2 p 1·3 3·5 5·7 7·9

1−

{ } 1 1 2 1 1 1 1 − = { − + − + … to ∞} 2 p p 1·3 3·5 5·7 7·9 1 1 2 1 1 1 1 − + − – … to ∞ − = p 1·3 3·5 5·7 7·9 p 2

p−2 1 1 1 1 = – … to ∞. − + − 4 1·3 3·5 5·7 7·9 Note. When k is a positive integer, in the Fourier expansion of any function containing either sin kx or cos kx, the coefficients ak and bk are to be calculated separately. This is necessary in ex. 5 also.

\



Example 5. Expand in Fourier series of periodicity 2p of f (x) = x sin x, for 0 < x < 2p. (MS. 1991 A, N)



Let

f (x) =



Then

a0 = 2p



Let us evaluate

∫ x sin rx dx,

∞ a0 + (a cos nx + bn sin nx) …(i) 2 n =1 n

∑

1 p

2p

∫

f ( x) dx =

0

1 p

2p

∫ x sin x dx …(ii) 0

when r is an integer;

0

2p



∫ 0

2p

  cos rx   sin rx   x sin rx dx = ( x)  −  − (1)  − 2   r   r 0  

2p 2p = − cos 2pr =− …(iii) r r 1 2p Using this result in (ii), we get a0 = − = –2 p 1

{ }



Hence by (iii), a1 =



For n = 2, 3, 4,…



\ Now using (iii),

2p

2p

2p

0

0

0

1 1 1 f ( x) cos x dx = x sin x cos x dx a1 == 2p p p

∫

{ }

∫

∫ x sin 2 x dx

1 2p 1 − = − . 2p 2 2

an =

1 p

2p

∫

x sin x cos nx dx =

0

1 2p

2p

∫ x(2 sin x cos nx) dx 0

2p

an =

1 2p

an =

1 2p 2p 1 1 − + = − + 2p n + 1 n − 1 n +1 n −1

∫ x{sin (n + 1) x – sin (n − 1) x} dx. 0

{

}

−(n − 1) + (n + 1) 2 = 2 = (n + 1)(n − 1) n −1

15

Fourier Series 2p

2p

2p

0

0

0

1 1 1 f ( x) sin x dx = x sin 2 x dx b1 = = 2p p p

∫



∫



But cos 2x = 1 – 2 sin2 x, giving 2 sin2 x = 1 – cos 2x.



Substituting this, b1 =



When r is an integer, 2p



∫ 0



1 2p

2

x) dx

2p

∫ x(1 − cos 2 x) dx …(iv) 0

2p

 sin rx  x cos rx dx =  x − r  0 

2p

∫ 0

2p

sin rx  cos rx  dx = 0+ 2  r  r 0

= 0

…(v)

Using this in (iv),



∫ x(2 sin

b1 =

2p  2 (2p) 2 2p 1   x 2  − 0 = × = = p.   2p   2  0 2 2  2p

bn =

1 p

Lastly for n = 2, 3, 4,…



2p

∫

x sin x sin nx dx =

0

1 = 2p

2p

∫ x(2 sin x sin nx) dx 0

2p

∫ x{cos (n − 1) x − cos (n + 1) x} dx= 0



Thus, for n = 2, 3, 4, …, bn = 0.



Substituting the values of the coefficients in (i), we get



1 2p

1 {0 − 0}, using (v). 2p

 ∞ cos nx  1 f (x) = –1 − cos x + 2   + p sin x. 2 2  n = 2 n − 1 

∑

p , we derive that 2 p−2 1 1 1 . − + −… to ∞ = 1.3 3.5 5.7 4 Example 6. Derive the Fourier series of f (x) = k, where k is a constant, the periodicity being 2p. Note. Putting x =



a0 = an =

1 p 1 p

c + 2p

∫ c

f ( x) dx=

1 p

c + 2p

∫

k c + 2p k [ x] = × 2p= 2k . p c p

k dx=

c

c + 2p

c + 2p

∫

f ( x) cos nx dx =

c

c + 2p

1 k  sin nx  k cos nx dx = p p  n  c

∫ c



bn =



= 1 p

c + 2p

∫ c

f ( x) sin nx dx =

1 p

c + 2p

∫

k sin nx dx =

c

=

k ×0= 0. p c + 2p

k  cos nx  − n  c p  k ×0= 0 p

16

Engineering Mathematics-III (Third Semester )

Thus the first term (constant term) in the Fourier expansion is k; all the other coefficients are zero each. So, no Fourier expansion, (other than what is supplied), is available for k.

Apparently, the labour involved in this example is a waste.



Note. k being any constant, the result of this example is valid for any constant number like p.

Theorem 3. The Fourier coefficients of a sum f1 + f2 are the sums of the corresponding Fourier coefficients of f1 and f2.

Theorem 4. If c is a constant, the Fourier coefficients of cf are the corresponding Fourier coefficients of f multiplied by c.

Example 7. Derive the Fourier expansion for



Let

{ f (x) = {−k , when − p < x < 0 + k , when 0 < x < p

F (x) = 0, when − p < x < 0 and f (x + 2p) = f (x), for all x. 2k , when 0 < x < p 1

and f2 (x) = k, for all x.

Then we have developed the Fourier expansions of f1 and f2 in worked examples 2 and 6 respectively. But the given F (x) = f1(x) + f2(x).

Hence by theorem 3,



f (x) = k +

{

}

4k 1 1 1 sin x + sin 3 x + sin 5 x + sin 7 x + … to ∞ . p 3 5 7

Note. One can directly work out this problem as in example 2 without splitting into f1 and f2.



Example 8. Derive the Fourier series for the periodic function, whose definition in one period is





− 1 p, when − p < x < 0  F (x) =  4 1  p + sin x, when 0 < x < p 4

− 1 p, when − p < x < 0  Let f1(x) =  4 1  p, when 0 < x < p. 4 Then working exactly as in ex. 2,



1 1 sin 3x + sin 5x + … to ∞. 3 5 Now F (x) = f (x) + f1 (x), where f (x) is the one supplied in ex. 4.



Therefore using Th. 3,





f1(x) = sin x +

F (x) =

1 2 − p p +

 cos 2 x cos 4 x cos 6 x cos 8 x  + + + + ... to ∞   15 35 63  3 

3 1 1 1  sin x +  sin 3 x + sin 5 x + sin 7 x + ... to ∞  . 2 5 7 3 

Note. Work out this example directly as in example 2. Even and odd functions. f (x) is said to be even, if f (– x) = f (x); x2, cos kx, 2, cosh x, x2 – 3 cos x + 1 are all even functions. The graph of an even function will be symmetric about the y-axis. (Refer Fig. 1.)

17

Fourier Series

Fig. 1             Fig.2

The function f (x) is odd, if f (– x) = – f (x). For example, x, sin x, 3x – 4 sin x are odd functions of x. The graph of an odd function will be symmetric about the origin. Refer Fig. 2.

There are functions which are neither odd nor even; 1 + x and ex and such functions.



Results to Remember. (from integral calculus)



a

∫

−a

and

a

∫

a

f ( x) dx = 2 ∫ f ( x) dx if f (x) is even 0

f ( x) dx = 0 if f (x) is odd.

−a

Theorem 5. When an even function f (x) is expanded in a Fourier series over the interval – p to p, the series will have cosine terms only, and the coefficients of the series are given by p 2 an = ∫ f ( x) cos nx dx; and bn = 0 p0

Proof. We have already seen that,



an =

1 p

p

∫

f ( x) cos nx dx

−p p

2 = ∫ f ( x) cos nx dx since the integrand is even p0

bn =

1 p

p

∫

f ( x) sin nx dx = 0  integrand is odd

−p

{f (x) is even and sin n x is odd}

a0 =

2 p

p

∫ f ( x) dx 0

Theorem 5 (a). When an odd function f (x) is developed in a Fourier series in the interval from – p to p, p 2 an = 0,  and  bn = ∫ f ( x) sin nx dx. p0

(The student can write out the proof.)

Note that the Fourier series of an odd function will contain terms in sines only. The function in worked example 2 is an odd function; and, the expansion had terms in sines only. Example 9. Find the Fourier series of f (x) = x + x2 in (– p, p) of periodicity 2p. Hence deduce p2 1 (MS. 1993 April) ∑ n2 = 6 



18

Engineering Mathematics-III (Third Semester )



Let

f (x) =



Then

a0 =

∞ a0 + ∑ (an cos nx + bn sin nx) …(1) 2 n =1

1 p

p

∫ (x + x

2

)dx

−p p

1 1 = ∫ xdx + p −p p

p

∫ x dx 2

−p

p

0+ =

p

2 2 x dx p ∫0

Since the first integral vanishes because integer and x is odd and the second integral is

2 2 x dx since x2 is even. p

∫ 0

we use the theorem,

a

∫

f ( x) dx = 0 if f (x) is odd

−a

and

a

∫

−a

a

f ( x) dx = 2 ∫ f ( x) dx if f (x) is even. 0

p



Hence



2  x3  a0 =   = p  3 0 an =

1 p

p

2 2 p 3

∫ (x + x

2

) cos nx dx

−p p

1 1 = ∫ x cos nx dx + p −p p

p

∫x

2

cos nx dx

−p

p

0+ =

2 2 x cos nx dx,  p ∫0

since x cos nx is odd p

2  sin nx   cos nx   sin nx   = ( x 2 )   − (2 x)  −  + (2)  −  2 p n  n3   0  n    2  2p  =  2 cos np = p n 

bn =

1 p

p

∫ (x + x

2

4 (− 1) n n2

) sin nx dx

−p p

1 1 = ∫ x sin nx dx + p −p p

p

∫x

2

sin nx dx.

−p

p

2 = ∫ x sin nx dx + 0 p0 p

2   − cos nx   − sin nx   = ( x)   − (1)   n p    n 2   0

19

Fourier Series

2 − p 2  =  cos np  = − (−1) n p n n  Substituting the values of a0, an, bn, in (1), we get x + x2 =



∞ p2 2 4  + ∑ (− 1) n  2 cos nx − sin nx  3 n =1 n n 

Deduction: x = – p and x = p are end points of the range.

Therefore the value of Fourier series at x = p is the average of the values of f (x) at x = p and x=–p

Hence, put x = p in Fourier series



∞ p2 4 f (p) + f (−p) (p + p2 ) + (−p + p2 ) + ∑ (−1) n 2 cos np = = 3 n =1 2 2 n ∞ ∞ p2 1 1 2 + 4 ∑ 2 = p2  \  4 ∑ 2 = p2 3 3 n =1 n 1 n ∞ 1



1

∑n

\

2

=

p2 1 1 1 + + + ∞ = ..... to 6 12 22 32

Example 10. In – p < x < p, Express sinh ax and cosh ax in Fourier series of periodicity 2p. (MS. 1989A; 1988 Nov.)

 Let

f (x) = sinh ax = p

1 f ( x) dx a0 = = p −∫p



Here,



 the integrand f (x) is odd.



a0 + ∑ (an cos nx + bn sin nx) 2 n =1

an =

1 p

bn =

1 p

∞

p

1 sinh ax dx = p −∫p

0

p

∫ sinh ax cos nx dx

= 0    the integrand is odd

−p p

∫ sinh ax sin nx dx

−p p

=

2 sinh ax sin nx dx since the integrand is even. p ∫0

=

2 e ax − e − ax sin nx dx 2 p ∫0

=

1 ax 1 e sin nx dx − ∫ e − ax sin nx dx p ∫0 p0

=

 1  e ax e − ax {a sin nx − n cos nx} − 2 {− a sin nx − n cos nx} ⋅ 2 2 x a + n a + n2 0

=

1 1 [− ne ap (−1) n + n + ne − ap (−1) n − n] ⋅ 2 2 p a +n

p

p

p

p

20

Engineering Mathematics-III (Third Semester )

=

1 1 (−1) n − 1 ⋅ n(e ap − e − ap ) ⋅ 2 p a + n2

=

2n (−1) n − 1 ⋅ sinh ap p a 2 + n2



Hence

(ii) Let

sinh ax =

∞ 2 (−1) n − 1 n sinh ap ∑ 2 sin nx. 2 p n =1 a + n

cosh ax =

∞ a0 + ∑ (an cos nx + bn sin nx) 2 n =1

a0 =

1 p

p

∫

f ( x) dx =

−p

1 p

p

∫ cosh ax dx

−p

p

=

2 cosh ax dx p ∫0

=

2  sinh ax    p  a 0

=

2 sinh ap pa

p



an =

1 p

p

∫ cosh ax cos nx dx

−p p

=

2 cosh ax cos nx dx p ∫0

=

2 p

=

p

∫ 0

e ax + e − ax ⋅ cos nx dx 2

p p  1  ax cos e nx dx + e − ax cos nx dx  ∫ ∫ p 0 0 

1 = p

 e ax e − ax {a cos nx + n sin nx} + 2  2 2 a + n2 a + n

=

1 1 e ap ⋅ a (−1) n − a − e − ap ⋅ a (−1) n + a  ⋅ 2 p a + n2

=

1 1 ⋅ 2 a (−1) n [e ap − e − ap ] p a + n2

=

2 1 a (−1) n ⋅ sinh ap 2 p a + n2



bn =

1 p

p

− a cos nx    + n sin nx  0

p

∫ cosh ax sin nx dx

= 0 ( integrand is odd)

−p

f (x) = cosh ax =

∞ 1 2 1 sinh ap + ∑ ⋅ ⋅ 2 a(–1)n sinh ap cos nx. pa p a + n2 1

21

Fourier Series

=

∞ 2a 1  1  sinh ap  2 + ∑ 2 (−1) n ⋅ cos nx  2 p 1 a + n  2a 



Example 11. If f (x) = sin x in 0 ≤ x ≤ p





= 0  in p ≤ x ≤ 2p

find a Fourier series of periodicity 2p and hence evaluate Let

f (x) = a0 =

1 1 1 + + + ⋅⋅⋅ to ∞. 1.3 3.5 5.7

∞ a0 + ∑ (an cos nx + bn sin nx) …(1) 2 n =1

1 p

2p

∫

f ( x) dx

0

=

p 2p  1 sin x dx 0 ⋅ dx  + ∫ ∫ p 0 p 

=

1 2 (− cos x)0p = …(2) p p



an =

p

1 ⋅ sin x cos nx dx p ∫0 p

1 2p

∫ [sin (n + 1) x − sin (n − 1) x] dx

=

1 2p

 cos (n + 1) x cos (n − 1) x  + −  if n ≠ 1 n +1 n −1  0

=

1 2p

1  1  n n −1 − 1}  n + 1 {(−1) + 1} + n − 1 {(−1)  

=

1 1 [(n − 1) {1 + (−1) n } − (n + 1) {(−1) n + 1}] 2p n 2 − 1

=

1 {(−1) n (−2) − 2} 2p (n 2 − 1)

=

−1 [1 + (−1) n ] p (n 2 − 1)

=

0

p

= 0 if n is odd =

a1 =

−2 if n is even, n ≠ 1. p (n 2 − 1) p

1 sin x cos x dx p ∫0 p

=

1 sin 2 x dx 2p ∫0

=

1  1   − cos 2 x  2p  2 0

p

…(3)

22

Engineering Mathematics-III (Third Semester )

1 (1 − 1) = 0 4p

= −

bn =

p

1 sin x sin nx dx p ∫0 p

=

1 [cos (n − 1) x − cos (n + 1) x] dx 2p ∫0

=

1  sin (n − 1) x sin (n + 1) x  − if n ≠ 1 2p  n − 1 n + 1  0

p

= 0 if n ≠ 1 p

b1 =

1 sin x sin x dx p ∫0

=

1 1 − cos 2 x dx p ∫0 2

=

1 2p



p



\



f (x) =

p

1 1   x − sin 2 x  =  2 0 2

1 1 2 + sin x − p 2 p

1 cos nx 2 ( − 1) n n = 2, 4, 6 ∞

∑

x = 0 is an end point in the range



\ The value of the Fourier series at x = 0 is equal to



Putting x = 0 in the Fourier series,



1 2 − p p



1 1 = 2 n = 2, 4, 6 ( n − 1) ( n + 1)

i.e.,

∞

∑

n = 2, 4, 6

f (0) + f (2p) 2

1 0+0 = 0 = 2 n −1 2

∞

∑

1 1 1 1 + + + ⋅⋅⋅⋅⋅ to ∞ = 1.3 3.5 5.7 2

Example 12. Express f (x) = (p – x)2 as a Fourier series of periodicity 2p in 0 < x < 2p and ∞ 1 hence deduce the sum ∑ 2  (MS. 1989 April) n n=1 ∞ a Let f (x) = 0 + ∑ (an cos nx + bn sin nx) 2 n =1



a0 =

1 p

2p

=

1 p

2p



∫

f ( x) dx

0

∫ (p − x)

2

dx

0

2p

=

1  − ( p − x )3    p 3 0

23

Fourier Series

= −

1 [− p3 − p3 ]= 3p

2 2 p 3

an =

1 p

2p

=

1 p

  cos nx   sin nx   2  sin nx  (p − x)  n  + 2 (p − x)  − n 2  + 2  − n3    0

=

1  2p 2p  4 + = p  n 2 n 2  n2



bn =



1 p

1 = p =

1 p

∫ (p − x)

cos nx dx

0

2p

2p

∫ (p − x)

2

sin nx dx

0

2p

 cos nx   sin nx   cos nx   2  (p − x)  − n  + 2 (p − x)  − n 2  + 2  n3    0  p2 p2  2 + + (1 − 1)  = 0 − n n3  n 

f (x) = (p – x)2 =

\

2

∞ p2 1 + 4 ∑ 2 cos nx 3 n =1 n

x = 0 is an end point. The value of Fourier series at x = 0 is

\



\

∞ p2 1 p2 + p2 + 4 ∑ 2 = = 3 2 n =1 n ∞ 1



1

∑n

2

=

f (0) + f (2p) 2

p2

p2 . 6

Example 13. Find the Fourier series of f (x) = ex in (– p, p), of periodicity 2p (MS. 1991 Nov.)

Let where

f (x) = a0 =

∞ a0 + ∑ (an cos nx + bn sin nx) …(1) 2 n =1

1 p

p

∫

e x dx

−p

=

1 x p (e ) − p p

=

1 p (e – e– p) p

=

2 sinh p p



an =

1 p

p

∫e

x

cos nx dx

−p

p

 1  ex (cos nx + n sin nx)  =  2 p 1 + n − p

24

Engineering Mathematics-III (Third Semester )

=

1 [e p (−1) n − e −p (−1) n ] p (1 + n 2 )

=

2 (−1) n sinh p p (1 + h 2 )



bn =

1 p

p

∫

e x sin nx dx

−p p

 1  ex (sin nx − n cos nx  =  2 p 1 + n − p =

1 [e p (−n) (−1) n + e − p n (−1) n ] p (1 + n 2 )

=

−2 (−1) n ⋅ n sinh p p (1 + n 2 )

ex =

sinh p p

∞   2 (−1) n 1 (cos nx − n sin nx)  +  ∑ 2  n =1 1 + n 



\



Example 14. Find the Fourier series of periodicity 2p



for

Let where

 x when − p < x < 0  p  0 when 0 < x < f (x) =   2  p p  x − when < x < p 2 2  f (x) = a0 =

∞ a0 + ∑ (an cos nx + bn sin nx) …(1) 2 n =1

1 p

p

∫

f ( x) dx

−p

p 0 p 2 1 =  ∫ x dx + ∫ 0 dx − ∫ p −p p 0  2

1 = p =

1 p

an =

1 p



(Kerala University)

 p    x −  dx  2  

2 p   x 2  0  1   p    +   x −    2   p   2  − p  2    2

 p2 p2  3p +  = − − 8 8  2 p

∫

−p

f ( x) cos nx dx

25

Fourier Series p 0 p 2 1 =  ∫ x cos nx dx + ∫ 0 ⋅ cos nx dx + ∫ p −p p 0  2

1 = p

 p   x cos nx dx −    2  

0    sin nx   cos nx   ( x ) (1) − −      n 2  − p     n  p  p   sin nx   cos nx    +  x −   − (1) −     2 n  n2  p    2



=

11 1  np  {1 − (−1) n } + 2 (−1) n − cos   2 n  2  p n

=

11 p  n 2



bn =

np    1 − cos  2   

0 p 1  ∫ x sin nx dx + ∫ p − p p 2 

1 = p

0    cos nx   sin nx    ( x )  −  − (1)  −  n  n 2  − p     p  p   cos nx   sin nx    +  x −   − − (1) −     n  2 n2  p    2

 =

1 p

= −

 p   x −  sin nx dx  2   

\

p 1 np   p  − n cos np − 2n cos np − n 2 sin 2  1 p

1 np   3p n  2n (−1) + n 2 sin 2 

− 3p 1 ∞ 1  np  1 ∞  3p (−1) n + ∑ 2 1 − cos  cos nx − ∑  16 2  p n 1= p n 1  2n n  = f (x) =



+



Example 15. Find the Fourier series of periodicity 2p



for

in (0, p)  x f (x) =  2 p − x in (p, 2 p)

and hence deduce

1 1 1 p2 + + + = ..... 8 12 32 5 2

Let

f (x) =



a0 =

∞ a0 + ∑ (an cos nx + bn sin nx) 2 n =1 p 2p  1  ∫ x dx + ∫ (2p − x) dx  p 0 p 

np  1 sin  sin nx 2 2  n

26

Engineering Mathematics-III (Third Semester )

1 = p =

an =

2p  x 2  p  1     −  (2p − x) 2   p   2 0  2

1  p2 p2  p  +  = p2 2 p 2p  1  ∫ x cos nx dx + ∫ (2p − x) cos nx dx  p 0 p 

1 = p

p    sin nx   cos nx    ( x )   − (1)  −  n 2  0  n     2p

  sin nx   cos nx   + (2p − x)   − (−1)  −  n 2  p  n   

 =

11 1  {(−1) n − 1} − 2 {1 − (−1) n } p  n 2 n 

=

2 [(−1) n − 1] pn 2



bn =

=

p 2p  1  ∫ x sin nx dx + ∫ (2p − x) sin nx dx  p 0 p 

1 p

p    cos nx   sin nx    ( x )  −   − (1)  −  n  n 2  0     2p   − cos nx   sin nx    + (2p − x)  − ( − 1) −     n n 2  p     

 =

Hence

f (x) =

1 − p p  (−1) n + (−1) n  = 0 p  n n  p 2 + 2 p

∞

∑

n =1

1 [(−1) n − 1] cos nx n2

p 4 1 1  − cos x + 2 cos 3 x + 2 cos 5 x + ..... to ∞  2 p 3 5  x = 0 is an end point of the range. =

The value of Fourier series at an end is equal to the average of [f (0) + f (2p)]

\ At x = 0, p 41 1 1 1  −  2 + 2 + 2 + ...  = [f (0) + f (2p)] 2 p 1 2 3 5  1 0 = [0 + 0] = 2 41 1 1 p  + + ...  =  + p  12 32 52 2  p2 1 1 1 + 2 + 2 + ... = . 2 8 1 3 5

  

27

Fourier Series

Example 16. In (– p, p), find the Fourier series of periodicity 2p for in 0 < x < p  1+ x f (x) =  −1 + x in − p < x < 0



The function f (x) is odd evidently; for



Let 0 < a < p; Than f (a) = 1 + a

Let

f (–a) = –1 – a = – (1 + a) = – f (a) ∞ a f (x) = 0 + ∑ (an cos nx + bn sin nx) 2 n =1 a0 =

1 p

an =

1 p

bn =

1 p

=

2 p

where

p

∫

f ( x) dx = 0

−p p

∫

f ( x) cos nx dx = 0 ( integrand is o)

−p p

∫

f ( x) sin nx dx

−p p

∫ f ( x) sin nx dx 0

p

2 = ∫ (1 + x) sin nx dx p0 p

=

2 p

  cos nx   sin nx   (1 + x)  − n  − (1)  − n 2    0

=

2 p

  cos np  1  (1 + p)  − n  + n   

2 [1 – (1 + p) (–1)n] np 2 ∞ 1 f (x) = ∑ [1 − (1 + p) (−1) n ] sin nx. p n =1 n

=

\

Example 17. Write down the analytic expressions of the following functions given in the graph. Hence find Fourier series of f (x). (a) f (x) = x,  0 < x < p    = 0  p < x < 2p (b) From the figure on page 36, Slop of BC =

1− 0 −1 = p 0−p

equation of BC is y = − Slop of AC is

1 (x – p) p

1− 0 1 = 0+p p

Equation of AC is y =

1 (x + p) p

28

Engineering Mathematics-III (Third Semester )

f (x) =

\

1 (x + p),  – p < x < 0 p

= −

1 (x – p),  if  0 < x < p p

{ f (x) is an even function} (c) The two segments are parallel to x axis \ f (x) = –1,  – p < x < 0 = 1,  0 < x < p (d) Refer to figure Slope of OA =

1 p

Equation of OA is y = Slope of AB =

1 x p

1− 0 −1 = p − 2p p

Equation of AB is y = −

1  (x – 2p) p x f (x) = ,   0 < x < p p

\

=

1 (2p – x), p < x < 2p p

Fourier series of the above functions can be got in the usual way. Example 18. If f (x) is defined in (– p, p) and if

f (x) = x + 1 in (0, p) find f (x) in (– p, 0) if

(i) f (x) is odd   (ii) f (x) is even.

Ans: If f (x) is even then,



f (x) = – x + 1  in (– p, 0)

If f (x) is odd, then

f (x) = x – 1  in (– p, 0) Exercise 1(c)

Expand each of the functions in 1 to 17 in a Fourier series. It is given that each of the functions has the period 2p. 1.  f (x) = ex, – p < x < p. Derive a series for

p . sinh p

2. f1 (x) = x, – p < x < p. Deduce the value of 3. F (x) = x + ex, – p < x < p. 4. f (x) = x2, – p < x < p. Deduce the relations

p . 4

29

Fourier Series

1+

p2 1 1 1 + 2 + 2 + .... to ∞ = ; 2 6 2 3 4



1−

p2 1 1 1 + 2 − 2 + .... to ∞ = ; 2 12 2 3 4



1+

p2 1 1 1 + 2 + 2 + .... to ∞ = . 2 8 3 5 7

5. f (x) = x3, – p < x < p. − p , when − p < x < 0  6. f (x) =  4 p  , when 0 < x < p,  4 and f (– p) = f (0) = f (p) = 0, and f (x + 2p) = f (x) for all x. p 1 1 1 = 1 − + − + .... to ∞.  Deduce that 4 3 5 7

(Madras, 72 Engg.)

7. f (x) = 0, if – p ≤ x ≤ 0; and f (x) = p, if 0 ≤ x < p. Show how this can be got from example 6. 8. F (x) = – x, if – p ≤ x ≤ 0; and f (x) = p – x, if 0 < x < p. Show how this can be got from example 2 and example 7. 1 1 9. f (x) = p, if – p < x ≤ p; and f (x) = 0, if p < x ≤ p. 2 2 10. f (x) = 1, if −

1 1 1 3 p < x < p; and f (x) = –1, if p < x < p. 2 2 2 2

11. f (x) = 1, if 0 < x
0.

146



Engineering Mathematics-III (Third Semester)

Example 3. Classify the equations given below

(i)

∂ 2u ∂ 2u ∂ 2u ∂u ∂u +4 + 4 2 − 12 + + 7u = x2 + y2 2 ∂x∂y ∂x ∂y ∂x ∂y

(ii) (1 + x2)fxx + (5 + 2x2) fxy + (4 + x2) fyv = 2 sin (x + y) Solution. (i) Here, A = 1, B = 4, C = 4 B2 – 4AC = 16 – 4 (1) (4) = 0



\



Hence, the given equation is parabolic for all x, y.



Example 4. Classify the equations:

(i) The Laplace equation

∂ 2u ∂ 2u + = 0 ∂x 2 ∂y 2

(ii) The Poisson equation

∂ 2u ∂ 2u + = f ( x, y ) ∂x 2 ∂y 2

(iii) One dimensional heat equation a 2

∂ 2u ∂u = ∂x 2 ∂t

∂ 2u ∂ 2u = ∂x 2 ∂t 2 Here A = 1, B = 0, C = 1

(iv) One dimensional wave equation a 2

Solution. (i)



\

B2 – 4AC = – 4 < 0 Hence the equation is elliptic

(ii) Here also B2 – 4AC = – 4 < 0. Hence it is elliptic (iii) Here A = a2, B = 0, C = 0 \ B2 – 4AC = 0

Hence it is parabolic

(iv) Here A = a2, B = 0, C = –1 Hence, B2 – 4AC = 4a2 > 0

Therefore, the equation is hyperbolic. Exercise 3 (b)

1. Prove fxx + 2fxy + 4 fyy = 0 is elliptic. 2. Prove fxx = 2fxy and fxy = fx are hyperbolic 3. Prove fxx – 2fxy + fyy = 0 and uxx = ut are parabolic. 4. Classify : (1 – x2) fxx – 2xy fxy + (1 – y2) fyy = 0

3.2  Transverse Vibrations of a Stretched Elastic String Let us consider small transverse vibrations of an elastic string of length l, which is stretched and then fixed at its two ends. Now we will study the transverse vibration of the string when no external forces act on it. Take an end of the string as the origin and the string in the equilibrium position as the x-axis and the line through the origin and perpendicular to the x-axis as the y-axis. We make the following assumptions: (1) The motion takes place entirely in one plane. This plane is chosen as the xy plane. (2) In this plane, each particle of the string moves in a direction perpendicular to the equilibrium position of the string.

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Applications of Partial Differential Equations

(3) The tension T caused by stretching the string before fixing it at the end points is constant at all times at all points of the deflected string. (4) The tension T is very large compared with the weight of the string and hence the gravitational force may be neglected. (5) The effect of friction is negligible. (6) The string is perfectly flexible. It can transmit only tension but not bending or shearing forces. (7) The slope of the deflection curve is small at all points and at all times.

When the string is in motion in the xy plane, the displacement y of any point of the string is a function of x and time t. Let P(x, y) and Q(x + dx, y + dy)> be two neighbouring points on the string. Let y and y + dy be the inclinations made by the tangents at P and Q respectively with the x-axis. Let m be the mass per unit length of the string which is homogeneous. Consider the motion of the infinitesimal element PQ of the string. The vertical component of the force to which this element is subjected to is

T sin (y + dy) – T sin y

= T [(y + dy) – y], since sin y = y (approximately). = T dy, (approximately).

The acceleration of the element in the oy direction is

∂2 y . If the length of PQ is ds, then the ∂t 2

mass of PQ is m ⋅ ds.

Hence by the second law of Newton, the equation of motion becomes,



⋅ m ⋅ ds ⋅

∂2 y = T⋅dy ∂t 2

∂2 y T ∂y = m ∂s ∂t 2 As Q → P, ds → 0. Therefore, taking limit as ds → 0, the above equation becomes,



i.e.,





∂2 y T ∂y , where = m ∂s ∂t 2 ∂2 y ∂x 2

∂y = curvature at P of the deflection curve = 3/2 ∂s   ∂y 2  + 1     ∂x     2

∂2 y  ∂y  = 2 , approximately since   is negligible.  ∂x  ∂x

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Engineering Mathematics-III (Third Semester)

∂2 y T ∂2 y . = m ∂x 2 ∂t 2 T = a2, (positive), Putting m the displacement y (x, t) is given by the equation

Hence

∂2 y ∂2 y = a 2 2 . 2 ∂t ∂x Note. The partial differential equation is known as one-dimensional wave equation.



2 ∂2 y 2∂ y = a . ∂t 2 ∂x 2 Method 1. (By the method of separation of variables).

3.3  Solution of the wave equation

∂2 y ∂2 y = a 2 2 . …(1) 2 ∂t ∂x Let y = X (x) ⋅T(t) be a solution of (1), where X (x) is a function of x only and T (t) is a function of t only.



∂2 y ∂2 y = X ′′T , = XT″ and ∂x 2 ∂t 2

d2X d 2T and T″ = . dx 2 dt 2 Hence (1) becomes, XT″ = a2 X″T X ′′ T ′′ = 2 …(2) i.e., X aT The L.H.S. of (2) is a function of x only whereas the R.H.S. is a function of time t only. But x and t are independent variables. Hence (2) is true only if each is equal to a constant. X ′′ T ′′ \ = 2 = k (say) where k is any constant. X aT Hence X″ – kX = 0 and T″– a2 kT = 0  …(3)

where

X″ =



Solutions of these equations depend upon the nature of the value of k.



Case 1. Let k = l2, a positive value.



Now the equation (3) are X″ – l2 X = 0 and T″ – a2l2 T = 0.



Solving the ordinary differential equations we get,



X = A1elx + B1e–lx

T = C1elat + D1e–lat



and



Case 2. Let k = l2, a negative number.



Then the equation (3) are X″ + l2 X = 0 and T″ + a2l2 T = 0.



Solving, we get



and

X = A2 coslx + B2 sin lx

T = C2 cos lat + D2 sin l at.

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Applications of Partial Differential Equations



Case 3. Let k = 0. a negative number.



Now the equation (3) are X″ = 0 and T″ = 0.



Then integrating,

X = A3x + B3

T = C3t + D3

and

Thus the various possible solution of the wave equation are



y = ( A1elx + B1e −lx )(C1elat + D1e −lat ) …(i)



y = ( A2 cos lx + B2 sin lx)(C2 cos lat + D2 sin lat ) …(ii)



y = (A3x + B3) (C3t + D3)…(iii)

Out of these solutions, we have to select that particular solution which suits the physical nature of the problem and the given boundary conditions. In the case of vibration of string, it is evident that y must be a periodic function of x and t. Hence we select the solution II as the probable solution of the wave equation. The constants are determined by using the boundary conditions in the problem. In doing problems, we shall select the solution II directly. 3.4  D’ Alembert’s Solution of the Wave Equation

(Method 2)



The one-dimensional wave equation is

∂2 y ∂2 y = a 2 2 . …(1) 2 ∂t ∂x



∂ ∂ = D and = D′, the equation becomes, (D2 – a2D′2)y = 0 ∂t ∂x The auxiliary equation is m2 – a2 = 0. i.e., m = ± a



Hence the general solution of the wave equation is

Writing

y = f(x + at) + g(a – at)…(2),

where f, g are arbitrary functions.

Suppose we assume that the displacement and the velocity of the string at t = 0 are



y = h(x) and

∂y = v(x)…(3) ∂t

Then substituting t = 0 in (2),we get



h(x) = f(x) + g(x)…(4)

and

n(x) = af′(x) – ag′(x)…(5)



Integrating the last equation, we get, x



f(x) – g(x) =

1 n(θ) d θ, where c is arbitrary  a

∫

…(6)

c

x



From (4) and (6),

1 1 f(x) = h( x) + n(θ) d θ, and 2 2a

∫ c

x



1 1 g(x) = h( x) − n(θ) d θ 2 2a

∫ c



Hence

1 1 y = [h( x + at ) + h ( x − at )] + 2 2a

x + at

∫

x − at

n(θ) d θ.

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Engineering Mathematics-III (Third Semester)



This solution is called as D’ Alembert’s solution of the one-dimensional wave equation. If the 1 string is at rest at t = 0, i.e., v = 0, then y = [h(a + at ) + h ( x − at )]. …(7) 2 Example 1. A tightly stretched string with fixed end points x = 0 and x = 6 is initially in the ∂y position y = f (x). It is set vibrating by giving to each of its points a velocity = g (x) at t = 0. ∂t Find y (x, t) in the form of Fourier series.

The displacement y (x, t) is governed by ∂2 y ∂2 y = a 2 2 …(1) 2 dt ∂x The boundary conditions under which (1) is to be solved are



(i) y (0, t) = 0 for t ≥ 0 (ii) y (l, t) = 0 for t ≥ 0 (iii) y (x, 0) = f (x), for 0 < x < l  ∂y  = g (x), for 0 < x < l. (iv)    ∂t t = 0

Solving (1) by the method of separation of variables, we get,



y (x, t) = (A1 elx + B1 e–lx)(C1 elat + D1 e–lat)…(I)



y = (A2 cos lx + B2 sin lx)(C2 cos lat + D2 sin lat)…(II)



y = (A3x + B3)(C3t + D3).…(III)

Since the solution should be periodic in t, we reject solutions (I) and (III) and select (II) to suit the boundary conditions (i), (ii), (iii) and (iv).

\

y (x, t) = (A cos lx + B sin lx)(C cos lat + D sin lat)…(2),

where A, B, C, D are arbitrary constants.

Using boundary condition (i) in (ii),

A (C cos lat + D sin lat) = 0 for all t ≥ 0. A = 0.



\



Applying the boundary condition (ii) in (2),



B sin ll (C cos lat + D sin lat) = 0, for all t ≥ 0.

If B = 0, the solution becomes y = 0 which is not true.

sin ll = 0, B ≠ 0



i.e., ll = np, where n is any integer.



\



\

i.e.,

l =

np l

y (x, t) = B B sin y (x, t) = sin

(

npx npat npat C cos + D sin l l l

(

)

)

npx npat npat Cn cos + Dn sin …(3), l l l

where BC = Cn and BD = Dn.

Since the wave equation is linear and homogeneous, the most general solution of it is

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Applications of Partial Differential Equations



y (x, t) =

∑ ( Cn cos nplat + Dn sin nplat ) sin npl x …(4). ∞

n =1

This satisfies boundary condition (i) and (ii). To find Cn and Dn we make use of the initial conditions (iii) and (iv)

y (x, 0) =

∞

∑ Cn sin npl x = f ( x) …(5)

n =1

 ∂y    =  ∂t t = 0

and

∑ npl a Dn sin npl x = g ( x) …(6)

The left hand sides of (5) and (6) are Fourier sine series of the right hand side functions.

Hence

Cn =

l

2 npx f ( x) sin dx …(7) l l

∫ 0 l

npa 2 npx g ( x)sin dx …(8) Dn = l l l

∫

and

0

Substituting the values of Cn and Dn in (4), we get the solution of the wave equation satisfying the given boundary conditions.

Example 2. A tightly stretched string with fixed end points x = 0 and x = l is initially in a px position given by y (x, 0) = y0 sin3 . If it is released from rest from this position, find the l

( )

(BR. 1995 Ap.)

displacement y at any time and at any distance from the end x = 0.

The displacement y of the particle at a distance x from the end x = 0 at time t is governed by 2 ∂2 y 2∂ y = a …(1) dt 2 ∂x 2 The boundary, conditions are



y (0, t) = 0,

for all t ≥ 0

(i)

y (l, t) = 0,

for all t ≥ 0,

(ii)

 ∂y    = 0,  ∂t 

for 0 ≤ x ≤ l (iii)

t =0

y (x, 0) = y0 sin3

( plx ) ,

for 0 ≤ x ≤ l (iv)

Now solving (1) and selecting the proper solution to suit the physical nature of the problem and making use of the boundary conditions (i) and (ii) as in the pervious problem, we get

y (x, t) = B sin

Again using the boundary condition (iii),



(

)

npx npat npat C cos + D sin …(2) l l l

(

)

npx npa  ∂y  D· .   = 0 = B sin l l  ∂t t = 0

If B = 0, (2) takes the form y (x, t) = 0. Hence B cannot be zero.

\ D = 0.

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Engineering Mathematics-III (Third Semester)



Hence (2) becomes,



y (x, t) = Bn



npx npat cos , where n is any integer and Bn is any constant. l l The most general solution satisfying (1) and the boundary conditions (i), (ii) and (iii) is



y (x, t) =

∞

∑ Bn sin npl x cos nplat …(3).

n =1



To find Bn use the boundary condition (iv).



y (x, 0) =

∑ Bn sin npl x y0 sin3 ( plx ) ∞

n =1

(

y0 px 3px 3 sin = − sin l l 4

)

y 3 y0 , B3 = − 0 and Bn = 0, for n ≠ 1, 3. 4 4 Using these values in (3), the solution of the equation is

This is only if B1 =



y (x, t) =

3 y0 px pat y0 3px 3pat − sin cos sin cos 4 l l 4 l l

Example 3. A tightly stretched with fixed end points x = 0 and x = l is initially at rest in its equilibrium position. If it is set vibrating giving each point a velocity 3x (l – x), find the displacement. (Madras, 1991 Ap.)

The equation to be solved is

∂2 y ∂2 y = a 2 2 …(1) 2 ∂t ∂x

The ends are fixed at x = 0 and x = l. The boundary conditions are y (0, t) = 0,

for t ≥ 0

(i);

y (l, t) = 0,

for t ≥ 0,

(ii);

y (x, 0) = 0,

for 0 ≤ x ≤ l (iii);

 ∂y    = 0,  ∂t 

for 0 ≤ x ≤ l (iv);

t =0

Solving (1) and selecting the suitable solution and making use of the boundary conditions (i) and (ii) as before,

y (x, t) = B sin

(

Again using (iii),



y (x, t) = BC sin

npx = 0, for all 0 ≤ x < l. l

Since B ≠ 0, C = 0. npx npat sin , l l where Bn is an arbitrary constant, and n is any integer. Therefore,

)

npx npat npat C cos + D sin …(2) l l l

y (x, t) = Bn sin

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Applications of Partial Differential Equations



Since (1) is homogeneous and linear, the most general solution of (1) is



y (x, t) =

∞

∑ Bn sin npl x sin nplat …(3)

n =1



Making use of (iv),



∞

npa npx  ∂y  Bn sin = 3 x (l − x) for 0 ≤ x ≤ l.   = l l  ∂t t = 0 n =1

∑

If Fourier sine series of 3x (l – x) in (0, l) is



3x (l – x) =

∞

∑ bn sin npl x ,

then

n +1



Bn

l

npa 2 npx = = bn 3 x(l − x) sin dx l l l

∫ 0

l

 npx  npx    cos npx      sin 6   cos l   6 2  l = (lx − x )  − − (l − 2 x)  − 2 2  + (−2)  3 3   np  l     n 2p   n 3p   l   l    l   0   6  −2l 3 =  3 3 (cos np − 1)  l n p  12l 2 = 3 3 [1 − (−1) n ] np = 0, if n is even, and 24l 2 = 3 3 , if n is odd np 24l 3 , if n is odd an 4 p4 = 0, if n is even. Bn =



\



Substituting the value of Bn in (3), we get,



y (x, t) =

24l 3 p4= an

24l 3 = 3 pa

∞

1 npx npat sin sin . 4 l l n 1, 3, 5,…

∑

∞

∑ (2r1− 1) sin

r =1

(2r − 1)px (2r − 1)pat sin . l l

Example 4. The points of trisection of a tightly stretched string of length l with fixed ends are pulled aside through a distance d on opposite sides of the position of equilibrium, and the string is released from rest. Obtain an expression of the displacement of the string at any subsequent time and show that the mid point of the string always remains at test. (Madras, 65 B.E.) Let B and C be the points of trisection of the string OA. The initial position of the string is shown by lines ODEA, where

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Engineering Mathematics-III (Third Semester)

BD = CE = d.



The displacement y (x, t) is governed by 2 ∂2 y 2∂ y = a …(1) ∂t 2 ∂x 2 The boundary conditions here are



y (0, t) = 0,

for t ≥ 0

(i);

y (l, t) = 0,

for t ≥ 0,

(ii);

 ∂y  and   = 0,  ∂t 

for 0 ≤ x ≤ l (iii);

t =0



To find the initial position of the string, we require the equation of ODEA.



The equation of OD is



The equation of DE is

y =

d 3dx . x= l/3 l

y – d = −

d ( x − l/3) (l/6)

3d (l − 2 x). l 3d y = ( x − l ). l

i.e.,

y =



The equation of EA is



The fourth initial condition is



 3dx  l  3d y (x, 0) =  (l − 2 x)  l  3d ( x − l )  l

for 0 ≤ x ≤ l/3 1 2l ≤ x ≤ …(iv) 3 3 2l for ≤ x ≤ l 3

for

Solving (1) and selecting the suitable solution and using the boundary conditions (i), (ii) and (iii) as in example 2, we get

y (x, t) =

∞

∑ Bn sin npl x cos nplat …(2)

n =1



Using the initial condition (iv), we get, ∞

∑

Bn sin n =1

npx 3dx = y (x, 0) = for 0 ≤ x ≤ l/3 l l

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Applications of Partial Differential Equations

3d 1 2l = (l − 2 x), for ≤ x ≤ l 3 3 3d 2l = ( x − 1), for ≤ x ≤ l. l 3 Finding Fourier sine series of y (x, 0) in (0, l) we get in the usual

way y (x, 0) =

∞

∑ bn sin npl x .

n =1



\

\

l

2 npx y ( x, 0) sin dx l l

∫

Bn = bn =

0

1/3

2 3dx npx sin dx + Bn =  l l l 0

∫

2l /3

∫

l /3

3d npx (l − 2 x) sin dx + l l

l

3d npx  ( x − l ) sin dx  l l  2l /3

∫

l /3

  npx   sin npx   cos   6d   l l  = x −  − (1)  − 2 2   2   n p l     n 2p   l  l    0  

2l /3

  cos npx   sin npx       6d  l + 2 (l − 2 x)  − − (−2)  − 2 l2   np  p l  n     l   l 2   l /3  



l

  cos npx   sin npx      6d  l l  + 2 ( x − 1)  −  − (1)  − 2 2   n p l     n 2p   l    l    2l /3



18d = n 2 p2

sin np − sin 2np   3 3 

18d = n 2 p2

np   np sin 3 − sin np − 3 

(

)

18d  np np sin + cos np · sin  = 3 3 n 2 p2  18d np sin [1 + (−1) n ] = 3 n 2 p2 = 0 if n is odd. 36d np = sin if n is even. 2 2 3 n p Hence, y (x, t) =

i.e., y (x, t) =

36d p2 = n 9d p2

∞

∞

1 np npx npat sin sin cos 2 3 l l n 2,4,6…

∑

∑ n12 sin 2n3p sin 2nlpx cos 2nlpat .

n =1

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Engineering Mathematics-III (Third Semester)



By putting x = l/2, we get the displacement of the midpoint.



\ y

( 2l , t ) = 0, since sin 2nlpx become sin np = 0 when x = l/2.

Example 5. A string is stretched between two fixed points at a distance 2l apart and the points of the string are given initial velocities v where cx , in 0 < x < l l c = (2l – x), in l < x < 2l, l x being the distance from an end point. Find the displacement of the string at any subsequent time.  (Madras, 1988 N.) v =



Take the origin at the end referred to. Let y be the displacement of any point at a distance x from the origin. 2 ∂2 y 2∂ y = a …(1) ∂t 2 ∂x 2 The boundary conditions are

Then

y (0, t) = 0,  for t ≥ 0

…(i)

y (2l, t) = 0,  for t ≥ 0,

…(ii)

y (x, 0) = 0,  for 0 ≤ x ≤ 2l…(iii) cx  ∂y    = , in 0 ≤ x ≤ l l  ∂t t = 0 c = (2l − x), in l < x < 2l…(iv). l As in the pervious examples, using boundary conditions (i) and (ii), we get

y (x, t) = sin

npx  npat npat  C cos + Dn sin 2l  n 2l 2l 

y (x, t) = sin

npx  npat npat  C cos + Dn sin 2l  n 2l 2l 



Using (iii), Cn = 0.



\



The most general solution of the equation (1) is y (x, t) =



∞

∑ Dn sin n2plx + sin np2lat …(2)

n =1



( )

∞ ∂y npa npx npat Dn ( x, t ) = sin cos . 2 l 2 l 2l ∂t n =1

∑

Using (iv),



∑ Dn ( n2pla ) sin n2plx = v = ∞

n =1

cx , in 0 < x < 1. l

c = (2l − x), in l < x < 2l. l

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Applications of Partial Differential Equations

Expanding v in Fourier sine series, we get



\

Dn .

l 2l 2 c npx c npx  npa =  x sin dx + (2l − x) sin dx  2l  l 2l l 2l 2l  l  0

∫

∫

l   npx   sin npx   cos  2c   2l  − (1)  − 2l    x − Dn =  2 2  np  npal    n p      2l     4l 2  0  

  cos npx   sin npx        2 l  + (2l − x)  − − (−1)  − 2 22l   np     n p2    2l   4l    



2l 

l

2

2

2

    

2

2c  −2l np 4l np 2l np 4l np  cos + sin + cos + sin  =  npal  np 2 n 2 p2 2 np 2 n 2 p2 2 2c 8t 2 np = · 2 2 sin 2 npal n p 16lc np = sin 3 3 2 npa Substituting this value of Dn in (2),

y (x, t) =

∞

16cl 1 np npx npat sin sin sin 2 2l 2l ap3 n = 1 n3

∑

Example 6. If a string of length of length l is initially at rest in equilibrium position and each point of it is given the velocity px  ∂y  = v0 sin3 , 0 < x < l, determine the transverse displacement y (x, t). (Ms. 1990 Ap.)   l  ∂t t = 0

The displacement y (x, t) is governed by 2 ∂2 y 2∂ y = a …(1) ∂t 2 ∂x 2 Solving this, by the method of separation of variables,



we get,

y (x, t) = (A1 elx + B1 e–lx) (C1 elat + D1 e–lat)…I

= (A2 cos lx + B2 sin lx) (C2 cos lat + D2 sin lat)…II

= (A3x + B3) (C3t + D)…III

The boundary conditions are



y (0, t) = 0,  for t ≥ 0

(i)



y (l, t) = 0,  for t ≥ 0,

(ii)



y (x, 0) = 0,  for 0 ≤ x ≤ l(iii)



px  ∂y  for 0 ≤ x ≤ l(iv)   = v0, sin3 l  ∂t t = 0



Selecting the solution II, and using boundary conditions (i) and (ii)

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Engineering Mathematics-III (Third Semester)

y (x, t) = B sin



We get



using (iii), C = 0



Therefore,



The most general solution is

y (x, t) = Bn sin

y (x, t) =



(

npx npat npat C cos + D sin l l l

)

npx npat sin , n any integer l l

∞

∑ Bn sin npl x sin nplat …(3)

n =1

∞ ∂y npa npx npat Bn . sin cos …IV = l l l ∂t n =1

∑



 ∂y    =  ∂t t = 0



∞

∑ Bn sin npl a sin npl x = v0 sin3 plx 1



Comparing both sides



B1 .

\



3 3px  px 1 = v0  sin − sin l 4 l  4

3v v pa 3pa = 0 and B3 . = − 0 and Bn = 0 for n ≠ 1, n ≠ 3 l 4 l 4 3lv B1 = 0 4pa −v l B3 = 0 12pa Bn = 0, n ≠ 1, n ≠ 3.

Using in (3),

vl 3lv0 3px 3pat px pat sin sin − 0 sin sin 4pa l l 12pa l l Example 7. A tightly stretched string with end points x = 0 and x = l is initially in a position px given by y (x, 0) = y0 sin . If it is released from rest from this position, find the displacement y (x, l t) at any point of the string. y (x, t) =



The boundary conditions are

y (0, t) = 0 for all r ≥ 0

…(i)



y (l, t) = 0 for all t ≥ 0

…(ii)



px for 0 ≤ x ≤ l…(iv) l Looking into example 2, and using the first three conditions, we have



 ∂y    = for 0 ≤ x ≤ l…(iii)  ∂t t = 0



y (x, 0) = y0 sin

y (x, t) =

∞

∑ Bn sin npl x cos nplat …(IV)

n =1

159

Applications of Partial Differential Equations

Using boundary condition (iv), in IV, npx px = y0 sin l l Comparing both sides, B1 = y0 and Bn = 0 n ≠ 1



y (x, 0) = S Bn sin

Using B1 in IV,

y (x, t) = y0 sin

px pat cos . l l

Example 8. A string is stretched and fastened to two points l apart. Motion is started by displacing the string into the form y = k(lx – x2) from which it is released at time t = 0. Find the displacement of any point of the string at a distance x from one end at any time t. (M.S. 1990 Nov.)



The boundary conditions are



y (0, t) = 0  t ≥ 0



y (l, t) = 0  t ≥ 0…(ii)



 ∂y    = 0  0 ≤ x ≤ l…(iii)  ∂t t = 0



y (x, 0) = k(lx – x2), 0 ≤ x ≤ l…(iv)

…(i)

Using boundary conditions (i), (ii) and (iii) as in example 2, we get

y (x, t) =

∞

∑ Bn sin npl x . cos nplat …(IV)

n =1



using boundary condition (iv),



y (x, 0) =

∞

px ∑ Bn sin n= l

k (lx − x 2 )

n =1

This shows that this is the half range Fourier sine series of k(lx – x2). Using the formula for Fourier coefficients,

Bn = bn =

l

2 npx k (lx − x 2 ) sin dx l l

∫ 0

l

 npx  npx    cos npx      sin l   cox l   2k  2  l = (lx − x )  − − (1 − 2 x)  − 2 2  + (−2)  3 3   np  l     n 2p   n 3p   l    l    l   0  2k  −2t 3 =  3 3 {(−1) n − 1} l n p  4kl 2  n = 1 − (−1)  n3p3 = 0 if n is even 8kl 2 if n is odd = n3p3

160



Engineering Mathematics-III (Third Semester)

Substituting in IV,



y (x, 0) =

∞

(2n − 1)px (2n − 1)pat 8kt 2 1 sin cos 3 3 6 l p n = 1 (2n − 1)

∑

Example 9. A taut string of length 2l is fastened at both ends. The mid point of the string is taken to a height b and then released from the rest in that position. Find the displacement of the string. (MS. 1992 Ap.)

Taking an end as origin, the boundary conditions are



y (x, 0) = 0, t ≥ 0

…(i)



y (2l, t) = 0, t ≥ 0

…(ii)



 ∂y    = 0, 0 ≤ x ≤ 2l…(iii)  ∂t t = 0

b ( x − 2l ), l ≤ x ≤ l l b = (x − 2l ), l ≤ x ≤ 2l l

y (x, 0) =



y−0 b−0 b  since, equation of OA is y x and equation of AB is = l x − 2l l − 2l  



Starting with the solution



using the first boundary condition,



y (x, t) = (A cos lx + B cos lx)(C cos lat + D sin lat)…I

\

using

y (o, t) = A(C cos lat + D sin lab) = 0 A = 0, y (2l, t) = 0 we get

  B sin 2ll (C cos lat + D sin lat) = 0 Using

\

B ≠ 0; ,ll = np; l =

np 2l

 ∂y    ; = D = 0.  ∂t t = 0 y (x, t) =

∞

∑ Bn sin n2plx cos np2lat …IV

n =1



Using boundary condition (iv) in IV,



y (x, 0) =

∞

x ∑ Bn sin n2p= l 1

b x, 0 ≤ x ≤ l l

b = − ( x − 2l ), l ≤ x ≤ 2l l This is half-range Fourier sine series

\

Bn =

2 2l

2l

∫ f ( x) sin 0

npx dx 2l

161

Applications of Partial Differential Equations l 2l 1 b npx b npx  dx − ( x − 2l ) sin dx  =  x sin l l l 2l 2l  l 0

∫

∫

l 2l   npx   npx     cos npx   sin npx    cos sin  b    2l  −  − 2l   − ( x − 2l )  − 2l  −  − 2l     ( x)  − =           2  2 2 2 2 np np l     n p      n p   2 2l   2 l         4l 4l 2  l   0  

(

)

b  2t 2 np 4t 2 np 2l 2 np 4l 2 np  cos + 2 2 sin cos + 2 2 sin  = − + 2  np 2 n p 2 np 2 n p 2 l  b  8t 2 np  sin  = 2  2 2 2 l n p 8b np = 2 2 sin 2 n p = 0 for n even 8b np for odd n. = 2 2 sin 2 n p Substituting in IV, y (x, 0) =

∞

(2n − 1)px (2n − 1)pat 8b 1 p sin (2n − 1) · sin cos 2 2l 2l p2 n = 1 (2n − 1) 2

∑

∞ (−1) n − 1 (2n − 1)px (2n − 1)pat 8b sin cos = 2 2l 2l p n = 1 (2n − 1) 2

∑

Example 10. Show that the solution of the equation of a vibrating string of length l, satisfying the conditions y (0,t) = y (l, t) = 0 and

∂y y = f (x), (0) at t 0 is = g= ∂t



y =

∞

∑ an sin npl x cos nplat 1

where

an =

l

2 npx f ( x) sin dx l l

∫ 0



The four boundary and initial conditions are given in the problem itself



As in problem 2, using the first two conditions,



y (x, t) = B sin



Again using the boundary condition



We get,

Since,

(

)

npx npat npat C cos + D sin …IV l l l

∂y ( x,0) = 0 ∂t

(

)

∂y npx npa ( x, 0) = B sin · D =0 l l ∂t B ≠ 0; D = 0

162



Engineering Mathematics-III (Third Semester)

Therefor, the most general solution is ∞



y (x, t) = Sum Bn sin

Using

y (x, 0) = f (x), we get



y (x, 0) = S Bn sin

n =1

∞ 1

Bn =

Hence,



npx f ( x) = l

l

2 npx f ( x) sin dx …(4) l l

∫ 0

Hence the solution is



npx npat · cos …(3) l l

∞

y (x, t) = S Bn sin 1

npx npat · cos where l l

Bn is given by (4)

Example 11. Solve the problem of the vibrating string for the following boundary conditions.

(i) y (0, t) = 0, (ii) y (l, t) = 0 ∂y (iii) ( x, 0) = x(x – l), 0 ≤ x ≤ l ∂t (iv) y (x, 0) = x in 0 ≤ x ≤ l/2 l < x < l. = l – x in 2 This is same as the problem (Example 1) where f (x) and g (x) are clearly specified. g (x) = x(x – l)

(MI. 1978)

f (x) = x in 0 < x < l/2

= l – x in l/2 < x < l Hence

Cn =

l

2 npx f ( x) sin dx l l

∫ 0

l /2 l 2 npx npx  d n + (l − x) sin dx  =  x sin l l l  l /2 0

∫

∫

l /2   npx   npx   cos sin  2   l −− l   =  ( x)  −   2 2  n p l   n p    2 l    l   0  



npx  npx        cos l   sin l    + (l − x)  − − (−1)  − 2 2   np     n 2p    l   l    

2  l2 np l2 np l2 np l2 np  =  − cos cos + 2 2 sin + + 2 2 sin  l  2 np 2 n p 2 2 np 2 l p 2 2  2l 2 np  =  2 2 sin  2 l n p

163

Applications of Partial Differential Equations

= 0 if n is even 4l np sin for n odd = 2 2 2 n p Again, Dn =

l

2 npx x( x − l ) sin dx npa l

∫ 0

l

 npx  npx  npx       cos l   sin l   cos l   2  2 = ( x − lx)  − − (2 x − l )  − 2 2  + (2)  3 3   np  npa     n 2p   n 3p   l   l    l   0   2  2t 3 =  3 3 ( (−1) n − 1)  npa  n p  = 0 if n is even −8t 3 if n is odd = n 4 p4 a Hence, using the result of the Example 1, Page 180, y (x, t) =

∞

∞

4l np npx npat 8l 3 npx npat − − sin sin · cos sin · sin 2 2 4 4 l l l l 2 n n a . p p n= n= 1,3,5… 1,3,5…

∑

∞

∑

= sin n =1

∑

(2n − l )px  (1−) n −1.4l (2n − 1)pat (2n − 1)pat  8t 3 cos · sin −   2 2 4 4 l l l (2n − 1) p a  (2n − 1) p  Exercises 3(b)

1. (a) A tightly stretched string with fixed end points x = 0 and x = l is initially in a position given by 2px . If it is released from rest from this position, find the displacement y at any y(x, 0) = y0 sin l distance x from one end at any time t. px 2px  − Sin (b) In the above problem, if y(x, 0) = k  Sin  find y(x, t) l l   2. If a string of length l is initially at rest in equilibrium position and each of its points is given the velocity px  ∂y  , 0 < x < l , determine the displacement y(x, t) . = v0 sin  ∂t  l  t = 0

(Madras, 63 B.E.)

3. A string is stretched tightly between x = 0 and x = l and both ends are given displacement y = a sin pt perpendicular to the string. If the string satisfies the differential equation ∂2 y ∂2 y = c 2 2 , show that the oscillations of the string are given by ∂t2 ∂x y = a sec

pl  px pl  cos  −  sin pt 2c  c 2c 

4. A string is stretched and fastened to two points l apart. Motion is started by displacing the string into the form y = 3(lx – x2) from which it is released at time t = 0. Find the displacement of any point on the string at a distance of x from one end at any time t. (Madras, 67 B.E. 75, 78) 5. A tightly stretched string with fixed end points x = 0 and x = l is initially at rest in equilibrium position. If it is set vibrating giving each point a velocity lx(l – x), show that

164 y(x, t) =

Engineering Mathematics-III (Third Semester) ∞

(2n − 1)px (2n − 1)px 8ll 3 1 sin sin  4 4 l l ap n = 1 (2n − 1)

∑

(MS. 1989 A.)

6. A uniform elastic string of length 60 cms. is subjected to a constant tension of 2 kg. If the ends are fixed and the initial displacement y(x, 0) = 60x – x2 for 0 < x < 60, –2 while the initial velocity is zero, find the displacement function y(x, t). (Madras, 64 ; 66 B.E.) 7. A taut string of length 2l is fastened at both ends. The mid point of the string is taken to a height b and then released from rest in that position. Show that the displacement is y(x, t) =

∞

(2n −1)px (2 −1)pat (−1) n − 1 8b sin cos . 2 2l 2l p n = 1 (2n − 1) 2

∑

8. If a string of length l is initially at rest in its equilibrium position and each of its points is given a velocity v such that v = cx, for 0 < x ≤ l / 2 l = c(l – x), for < x ≤ l, show that the displacement y(x, t) at any time t is given by 2 y(x, t) =

px pat 1 4l 2c  3px 3pat sin sin sin − 3 sin + ...  l l l l 3 p3a  

(S.V.U. 67 B.E.)

9. Find the displacement if a string of length a is vibrating between fixed end points with initial velocity zero and initial displacement given by y(x, 0) =

2 px for 0 < x < a/2 a

   = 2p –

2 px for a/2 < x < a. a

(Madras, 64 B.E.)

10. An elastic string is stretched between two points at a distance l apart. One end is taken as the origin and 2l from this end, the string is displaced a distance d transversely and is released from rest at a distance 3 when it is in this position. Find the equation of the subsequent motion. 11. A uniform string of line density r is stretched to tension ra2 and executes a small transverse vibration in a plane through the undisturbed line of the string. The ends x = 0 and x = l of the string are fixed. The string is at rest, with the point x = b drawn aside through a small distance d and released at time t = 0. Show that at any subsequent time t the transverse displacement y is given by y(x, t) =

∞

2dt 2 1 npb npx npat sin sin cos . 2 2 l l l p b(l − b) n = 1 n

∑

12. A uniform string of density r is stretched to tension rc2 and executes a small transverse vibration in a plane through the undisturbed line of the string. The ends x = 0, l of the string are fixed. The string is 4∈ released from rest in the position y = 2 x(l – x). Show that the motion is given by l

y =

∞

(2 + 1)px (2n + 1)pct 32 ∈ 1 sin cos . l l 3 n = 0 (2n + 1)3

∑

(Madras, 72 B.E.)

Also find the total energy of the string in terms of Fourier coefficients. (Madras, 65 B.E.) 13. A taut string of length l, fastened at both ends, is disturbed from its position of equilibrium by imparting to each of its points an initial velocity of magnitude f (x). Show that the solution of the problem is

y =

2 pa

∞



l



=1



0



d θ ∑  1n sin npl x sin nplat ∫ f (θ) sin npθ l 

165

Applications of Partial Differential Equations

14. If the string is released from rest in the position y = f (x), show that the total energy of the string is p2T 4l

∞

∑n

n =1

2

bn2 where

l

bn =

2 npx f ( x) sin dx. l l

∫ 0

15. The differential equation of a vibrating string that is viscously damped is

∂2 y ∂2 y ∂y = a 2 2 − 2b . 2 ∂t ∂t ∂x

If the string is of length l and is fastened at both ends and has initial displacement f (x) when the initial velocity is zero, show that,

y =

∞

∑a

n

n=0

e −bt sin

 npx  b cos a nt + sin a nt  an t  

l

where an2 =

n 2 p2 a 2 2 npx dx. . − b 2 and a n = f ( x) sin l l l2 0

∫

Discuss the case when l = p. (Madras, 63 B.E.) 16. Show that the total energy of a string which is fixed at the points x = 0, x = l and is executing small transverse vibrations is

l 2 2 1  ∂y  1  ∂y   W = T   + 2    dx. 2  ∂x  a  ∂t   0

∫

Show that if y = f (x – at) , 0 ≤ x ≤ l, then the energy of the wave is equally divided between potential and kinetic energy.

3.5.  One Dimensional Heat Flow In this article, we shall consider the flow of heat and the accompanying variation of temperature with position and with time in conducting solids. The following empirical laws are taken as the basis of investigation.

1. Heat flows from a higher to lower temperature.



2. The amount of heat required to produce a given temperature change in a body is proportional to the mass of the body and to the temperature change. This constant of proportionality is known as the specific heat (c) of the conducting material.



3. The rate at which heat flows through an area is proportional to the area and to the temperature gradient normal to the area. This constant of proportionality is known as the thermal conductivity (k) of the material.

Consider a bar or rod of homogeneous material of density r(gr./cm3) and having a constant cross–sectional area A (cm2). We suppose that the sides of the bar are insulated so that the streamlines of heat flow are all parallel and perpendicular to the area A. Take an end of the bar as the origin and the direction of heat flow as the positive x-axis.

166

Engineering Mathematics-III (Third Semester)

Let c be the specific heat and k the thermal conductivity of the material. Consider an element got between two parallel sections BDEF and LMNP at distances x and x + dx from the origin O, the sections being perpendicular to the x-axis. The mass of the element = Ardx. Let u(x, t) be the temperature at a distance x at time t. By the second law enunciated above, ∂u the rate of increase of heat in the element = Ardxc . If R1 and R2 are respectively the rates ∂t (cal./sec.) of inflow and outflow, for the sections x = x and x = x + dx, then ∂u R1 = −kA    ∂x l



∂u , the negative sign being due to the fact that heat flows from higher to and R2 = −kA    ∂x  x + dx lower temperature.

 i.e., ∂u is negative    ∂x  



Equating the rates of increase of heat from the two empirical laws,



Arcdx .

∂u = R1 – R2 ∂t

 ∂u ∂u  = kA   −    x ∂   x + dx  ∂x  x    ∂u  ∂u  −      ∂u k   ∂x  x + dx  ∂x  x  = ∂t rc  dx 



\



Taking the limit as dx → 0 i.e. when x + dx → x.



  ∂u  ∂u  −      ∂x  ∂x  x  x + dx ∂u k  = Lt  ∂t rc dx → 0  dx 

k ∂ ∂u =   rc ∂x  ∂x  i.e.,

2  ∂u  = k ∂ u   rc ∂x 2  ∂t 

k is called the diffusivity (cm2/sec.) of the substance. If we denote it by a2, the above equation pc takes the form

∂u ∂ 2u = a 2 2 . ∂t ∂x

3.6.  Solution of Heat Equation by the Method of Separation of Variables

We have to solve the equation



∂u ∂ 2u = a 2 2 …(1) ∂t ∂x

167

Applications of Partial Differential Equations



Assume a solution of the form



u(x, t) = X(x) . T(t),

where X is a function of x alone and T is a function of t along.

Then (1) becomes,



XT′ = a2 X′′T,

d2X dT and T ′ = 2 dt dx X ′′ T′ = 2 …(2) i.e., X aT The left hand side is a function of x alone and the right hand side is a function of t alone when x and t are independent variables. (2) can be true only if each expression is equal to a constant. X ′′ T′ = 2 = k (constant) \ Let X aT …(3) \ X ″– kX = 0, and T ′ – a2kT = 0 where

X″ =



The nature of solutions of (3) depends upon the values of k.



Case 1. Let k = l2, a positive number.



Then (3) becomes,



X ″ – l2X = 0, and T′ – a2l2T = 0.

Solving, we get

2 2

X = A1elx + B1elx and T = C1ea l t . .

Case 2. Let k = – l2, a negative number. Then (3) becomes X″ + l2X = 0, and T′ + a2l2 T = 0.



Solving, we obtain

2 2

X = A2 cos lx + B2 sin lx, and T = C2e −a l t . .

Case 3. Let k = 0 .





Then X ″ = 0 and T ′ = 0 .

Solving, we arrive at,



X = A3x + B3 and T = C3 .

Hence the possible solutions of (1) are 2 2



u(x, t) = ( A1elx + B1e −lx ) C1ea l t …(I)



u(x, t) = ( A2 cos lx + B2 sin lx) C2e −a l t …(II)



u(x, t) = (A3x + B3) C3 …(III)

2 2

Out of these three possible solutions, we have to select that solution which will suit the physical nature of the problem and the given boundary conditions. As we are concerned with heat conduction, u(x, t) must decrease with increase of time. Therefore, out of the three solutions, we select the second solution to suit the physical nature of the problem. In the steady-state conditions, when the temperature no longer varies with time, the solution of the diffusion equation (1) will be the last solution (III). Example 12. A rod l cm. with insulated lateral surface is initially at temperature f (x) at an inner point distant x cm. from one end. If both the ends are kept at zero temperature, find the temperature at any point of the rod at any subsequent time. (Calicut, 72 Engg.)

168

Engineering Mathematics-III (Third Semester)

Let u(x, t) be the temperature at any point distant x from one end at any time t seconds. Then u satisfies the partial differential equation ∂ 2u 1 ∂u = 2 …(1) ∂x 2 a ∂t The boundary conditions, here, are





u(0, t) = 0 for all t ≥ 0

…(i)



u(l, t) = 0 for all t ≥ 0

…(ii)

and the initial condition is



u(x, 0) = f (x), for 0 < x < l …(iii)

Solving the equation (1) by the method of separation of variables and selecting the suitable solution to suit the physical nature of the problem as explained in the method §3.6, we get

2 2

u(x, t) = (A cos lx + B sin lx) e −a l t …(2)

Substituting the boundary condition (i) in (2), we get,



2 2

u(0, t) = Ae −a l t = 0, for all t ≥ 0 A = 0



\



Employing the boundary condition (ii) in (2), we obtain,

i.e.,

2 2

u(l, t) = B sin ll e −a l t = 0, for all t ≥ 0 B sin ll = 0.

If B = 0, (2) will be a trivial solution. Hence





\

i.e.,

sin ll = 0 ll = np, where n is any integer. l =

np , where n is any integer. l

Then (2) reduces to

npx − e u(x, t) = Bn sin k

a 2 n 2 p2 l2

…(3),

where Bn is any constant. Since the equation (1) is linear, its most general solution is obtained by a linear combination of solutions given by (3).

Hence the most general solution is

u(x, t) =

∞

∑

n =1

npx − Bn sin e l



(4) should satisfy the initial condition (iii).



Using (iii) in (4),

a 2 n 2 p2t l2

…(4).

169

Applications of Partial Differential Equations



u(x, 0) =

∞

∑B

n

sin

n =1

npx = f (x), for 0 < x < 1 (given) l

…(5).

If u(x, 0), for 0 < x < l, is expressed in a half-range Fourier sine series in 0 < x < l , we know that

u(x, 0) = f (x) =

∞

∑b

n

sin

n =1

bn =



∫ f ( x) sin 0

npx dx. l

Comparing this with (5), we get Bn = bn =



2 l

l

npx , where l

2 l

l

∫ f ( x) sin 0

npx dx. l

Therefore the temperature function u (x, t), is a 2 l npx  npx −  f ( x) sin dx sin e u(x, t) = l l l  n = 1 0  ∞



∑ ∫

2 2 2

n pt l2

Example 13. Find the temperature u(x, t) in a silver bar (of length 10 cm, constant cross– section of 1 cm2 area, density 10.6 gm/cm3, thermal conductivity 1.04 cal/cm deg. sec; specific heat 0.056 cal/gm.deg.) which is perfectly insulated laterally, if the ends are kept at italic 0°C. and if, initially, the temperature is italic 5°C. at the centre of the bar and falls uniformly to zero at its ends. a2 =

k 1.04 = = 1.75. rc (10.6)(0.056)



In this problem,



Initial temperature distribution is given by the graph given below.



The equation to OB is u = x.



The equation to BA is u = – (x – 10) = 10 – x.



Therefore, the initial temperature





in 0 < x < 5  x, f (x) =  10 − x, in 5 < x < 10.

We can directly use the result of the previous example 12, for l = 10.

Bn =

2 10

10

∫ f ( x) sin 0

npx dx 10

5 10 1 npx npx  dx + (10 − x) sin dx =  x sin 5 10 10  5 0

∫

1  200 np  =  2 2 sin 5 n p 2 

∫

170

Engineering Mathematics-III (Third Semester)

40 np = sin 2 n 2 p2 \ Bn = 0, if n is even

40 = , for n = 1, 5, 9, … n 2 p2 40 − 2 2 , for n = 3, 7, 11, … = n p Hence the temperature u(x, t) =

40  1 np − 0.0175p2t 1 3px − 0.0175(3p)2 t  sin e e − 2 sin + ... 2  2 10 10 3 p 1 

Example 14. A rod, 30 cm. long, has its ends A and B kept at 20°C. and 80°C., respectively, until steady state conditions prevail. The temperature at each end is then suddenly reduced to 0°C. and kept so. Find the resulting temperature function u(x, t)) taking x = 0 at A.

The P.D.E. of one dimensional heat flow is



∂u ∂ 2u = a 2 2 …(1) ∂t ∂x

In steady-state conditions, the temperature at any particular point does not vary with time. That is, u depends only on x and not on time t.

Hence the P.D.E. (1) in steady-state becomes



d 2u = 0 dx 2

…(2)

Solving (2), we get,



u = ax + b



u = 20, when x = 0;

and

u = 80, when x = 30.



Using these conditions in (3), we obtain,



…(3)

The initial conditions, in steady-state, are

\

b = 20, a = 2. u(x) = 2x + 20

…(4)

When the temperatures at A and B are reduced to zero, the temperature distribution changes and the state is no more steady-state. For this transient state, the boundary conditions are

u(0, t) = 0 " t ≥ 0

…(i)



u(30, t) = 0 " t ≥ 0

…(ii).

The initial temperature of this state is the temperature in the previous steady-state. Hence the initial condition is

u(x, 0) = 2x + 20 for 0 < x < 30

…(iii)

Now, we have to find u(x, t) satisfying the conditions (i), (ii) and (iii) and the P.D.E. (1). The suitable solution of (1) is of the form

2 2

u(x, t) = (A cos lx + B sin lx) e −a l t …(5)

Using (i) in (5), we get that A = 0.

171

Applications of Partial Differential Equations



Using (ii) in (5),

Since B ≠ 0,

B sin 30l = 0. sin 30l = 0.

i.e.,

30l = np

i.e.,

l =



np , where n is any integer. 30

Therefore (5) reduces to a 2 n 2 p2t

npx − 900 u(x, t) = Bn sin …(6) e 30 The most general solution of (1) is obtained by a linear combination of terms given by (6).

\

u(x, t) =

∞

∑

n =1



npx − Bn sin e 30

a 2 n 2 p2t 900

…(7)

Using (iii), u(x, 0) =



∞

∑B

n

n =1

∞

∑B

n

n =1

sin

sin

npx = 2x + 20, for 0 < x < 30 . 30

npx is the Fourier sine series for f (x) = 2x + 20, in 0 < x < 30. 30 30

Hence

2 npx (2 x + 20) sin dx Bn = 30 30

∫ 0

30

  npx    cos npx   sin    1  30 − (2)  − 30   = (2 x + 20)  − 2 2 np  15   n p      30    302   0  40 = [1 − 4(−1) n ] np Substituting in (7),

u(x, t) =

∞

∑

n =1

40  npx − 1 − 4 (−1) n  sin e np  30

a 2 n 2 p2t 900

degrees.

Example 15. A bar, 10 cm. long, with insulated sides, has its ends A and B kept at 20° and 40°C., respectively, until steady-state conditions prevail, that is, until the temperature at any interior point no longer changes with time. The temperature at A is then suddenly raised to 50°C. and at the same instant that at B is lowered to 10°C. Find the subsequent temperature function u(x, t) at any time. (MS 1990 Nov.)

The partial differential equation satisfied by u(x, t) is



∂u ∂ 2u = a 2 2 …(1) ∂t ∂x

In steady-state, this equation reduces to



d 2u = 0. dx 2

172



Engineering Mathematics-III (Third Semester)

Solving this, u = ax + b, where a and b are arbitrary constants.



u = 20 at x = 0, and u = 40 at x = 10.



b = 20 and 40 = 10a + b.

\

Solving, a = 2.

Thus the temperature function in steady-state is



u(x) = 2x + 20

…(2)

When the temperatures at A and B are changed, the state is no longer steady. Then the temperature function u(x, t) satisfies (1).

The boundary conditions in the second state are



u(0, t) = 50 " t > 0

…(i)



u(10, t) = 10 " t > 0

…(ii)

and the initial condition is

u(x, 0) = 2x + 20 for 0 < x < 10

…(iii),

as explained in the previous problem. Now we have non-zero boundary values and the procedure adopted in the previous problem has to be modified.

So we break up the required function u(x, t) into two parts and write



u(x, t) = us (x) + ut (x, t)

…(3)

where us (x) is a solution of (1) involving x only and satisfying the boundary conditions (i) and (ii) ut (x, t) is a function defined by (3) and satisfying the equation (1). Thus us (x) is a steady-state solution of (1) and ut (x, t) may then be regarded as a transient solution which decreases with increase of t; us (x) satisfies (1).

\

Solving, and Hence a = – 4 . Thus

d 2u s = 0, where us (0) = 50 and us (10) = 10. dx 2 us (x) = ax + b.

us (0) = b = 50 , using (i);

us (10) = 10a + 50 = 10, using (ii). us (x) = 50 – 4x …(4)

Consequently, and i.e.,

ut (0, t) = u(0, t) – us (0) = 50 – 50 = 0

ut (10, t) = u(10, t) – us (10) = 10 – 10 = 0

(iv) (v)

ut (x, 0) = u(x, 0) – us (x) = (2x + 20) – (50 – 4x) ut (x, 0) = 6x – 30

(vi)

Now, ut (x, t) also satisfies (1) and (iv), (v), (vi).

Solving (1) and selecting a suitable solution



2 2

ut (x, t) = (A cos lx + B sin lx) e −a l t …(5)

Using the boundary condition (iv) in (5),

173

Applications of Partial Differential Equations 2 2

ut (0, t) = Ae −a l t = 0 " t > 0



\ A = 0



Using (v) in (5),







2 2

B sin 10l . e −a l t = 0, for " t > 0

Since B ≠ 0,

sin 10l = 0.

Hence

10l = np

i.e.,

l =

np , where n is any integer. 10

Therefore, (5) becomes,

npx − ut (x, t) = Bn sin e 10 The most general solution of (1) is,

∑

n =1

npx − Bn sin .e 10

a 2 n 2 p2t 100

…(6)

Using the initial condition (vi) in (6),



ut (x, t) =

∞

a 2 n 2 p2t 100

ut (x, 0) =

∞

∑B

n

sin

n =1

npx = 6x – 30, for 0 < x < 10. 10

Thus is itself Fourier half-range sine series for 6x – 30 in 0 < x < 10 if 10



Bn =

2 npx (6 x − 30) sin dx 10 10

∫ 0

10

  npx    cos npx  sin      1 10 − (6)  − 10   = (6 x − 30)  − 2 2 np  5  n p      10    102   0  1 300 300  cos np − =  − 5  np np  − 60 = [1 + (−1) n ] np −120 = , for n even np = 0, for n roman odd.

Substituting this value of Bn in (6), ∞



 −120  sin npx e − ut (x, t) =   10  np  n = 2, 4, 6...

∑

60 − = p

∞

∑

n =1

1 npx − sin e n 5

a 2 n 2 p2t 25

a 2 n 2 p2t 100

174



Engineering Mathematics-III (Third Semester)

u(x, t) = us (x) + ut (x, t)

\

60 50 − 4 x − = p

∞

∑

n =1

1 npx − sin e 5 n

a 2 n 2 p2t 25

This series converges for 0 < x < 10 roman and t > 0. We note that u(x, t) tends to us (x) as t → ∞. Example 16. Solve subject to the conditions



(i) u is not infinite as t → ∞ (ii) u = 0 for x = 0 and x = p for all t (iii) u = px – x2 t = 0 in (0, p) Solving the differential equation by the method of separation of variables, we get

u(x, t) = (Aelx + Be–lx)e …(I)



2l2t

= (A cos lx + B sin lx) ea

…(II)

= (Ax + B) …(III) as three possible solutions As t → ∞, u → ∞, in solution I. Hence we reject solution as u is not infinite as t → ∞, inf as per condition (1). Further solution III is independent of t (steady state solution). Hence we select solution II as possible one.

Let



2l2t

u(x, t) = (A cos lx + B sin lx) ea u(0, t) = 0. using this, 2l2t

A . ea

= 0  \  A = 0

u(p, t) = 0. using in II, 2l2t

B sin lp . ea

= 0

B ≠ 0 (otherwise u ≡ 0).

Hence,

sin lp = 0  \  l = n, any integer



\

u(x, t) = Bn sin nxea



Hence, the most general solution is

…(II)

2l2t

u(x, t) =

, n any integer.

∞

∑ Bn sin nxe−a n t …(IV) 2 2

n =1



Now , we will use the initial condition to find Bn.



u(x, 0) =

∞

∑ Bn sin nx =px − x2

n =1



This is half range Fourier sine series for px – x2. p



Bn = bn =

2 (px − x 2 ) sin nx dx p

∫ 0

p



sin nx 2  cos nx   cos nx   = (px − x 2 )  −  − (p − 2 x) (− 2 ) + (−2)   n  p  n  n3   0

175

Applications of Partial Differential Equations

2 2 =  3 (1 − (−1) n )  p n  4 = 3 1 − (−1) n  pn = 0 if n is even 8 = 3 if n is odd pn Substituting Bn value in IV,

\



Example 17. Solve

u(x, t) =

∞

2 2 8 1 sin(2n − 1) x . e −a (2 n − 1) r p n = 1 (2n − 1)3

∑

∂u ∂ 2u = a 2 2 subject to ∂t ∂x



u(o, t) = 0 (i) for t ≥ 0



u(l, t) = 0 (ii) for t ≥ 0 

(M.S. 1990 Ap.)

u(x, 0) = x for 0 ≤ x ≤ l/2 l ≤ x ≤ l(iii) = l – x for 2



Solving the given differential equation by the method of separation of variables we get three possible types of solutions, out of which we neglect the solution I and III for reasons explained in the previous question. Hence, we select.

2l2t

u(x, t) = (A cos lx + B sin lx) e–a



using u(0, t) = 0 in II, we get A = 0



using u(l, t) = 0 in II, B sin ll = 0



B ≠ 0;  \  ll = np



l =

np , n any integer. l

npx − Hence, u(x, t) = Bn sin .e l Hence, the most general solution is

…II

∞

a 2 n 2 p2 l2

npx − u(x, t) = Bn sin .e l n =1

∑

n any integer.

a 2 n 2 p2t l2

…IV

Now we use the third initial condition in IV



u(x, 0) =

∞

∑ Bn sin npl x

= given function of x.

1



\

l /2 l 2 npx npx  dn + (l − x) sin dx  Bn =  x sin l l l l /2 o 

∫

∫

176

Engineering Mathematics-III (Third Semester)

l /2   npx   sin npx   cos  2   l  − (1)  − l   =  ( x)  −  2 2  np  l n p      l    l 2  0    l

     l /2 

npx  npx        cos l   sin l   + (1 − x)  − − (−1)  − 2 2   np     n 2p    l    l   



2  l2 np l2 np l2 np l2 np  cos + 2 2 sin cos + + 2 2 sin  =  − 2 n p 2 2 np 2 n p 2 l  2 np 4l np = 2 2 sin 2 n p \ using the value of Bn in IV,

∞

4l 1 np npx − sin sin e u(x, t) = 2 2 2 l p n =1 n

∑

a 2 n 2 p2t l2

∂u ∂ 2u = a 2 2 subject to the conditions. ∂t ∂x (i) u is not infinite as t → ∞



Example 18. Solve

∂u = 0 for x = 0 and x = l (two ends thermally insulated) ∂x (iii) u = lx – x2 for t = 0 , 0 0 ∂q ( x, t ) = 0 at x = 1, t > 0 ∂x q(x, 0) = x for 0 < x < 1



In this problem, the end x = 1 is insulated whereas the end x = 0 is not.



In the differential equation a2 = 0



Hence, starting with the solution (II type)

178

Engineering Mathematics-III (Third Semester)



q(x, t) = (A cos lx + B sin lx) e–l t …(II)

Since

q(0, t) = 0,  A = 0





∂q = Bl cos lxe–l ∂x

∂q (1, t ) = 0  \  Bl cos l = 0; B ≠ 0. ∂x p cos l = 0  \  l = (2n – 1) 2

(2n − 1)px − e 2 The most general solution is

Hence q(x, t) = Bn sin



t

(2 n − 1)2 p2t 4 ,

n any integer

∞

(2n − 1)px − Bn sin e q(x, t) = 2 n =1

∑

(2 n − 1)2 p2 t 4

…(IV)

At t = 0,



∞

q(x, 0) =

∑ Bn sin

n =1



Bn =

(2n − 1)px =x 2

1

(2n − 1)px 2 x sin dx 1 2

∫ 0

1

  (2n − 1)px   (2n − 1)px     cos   sin  2 2 = 2 ( x)  − − − (2n − 1)p   (2n − 12 )p2        2     4   0 8(−1) n − 1 = (2n − 1) 2 p2 (2 n − 1)2 p2

∞ t (−1) n − 1 (2n − 1)px − 8 4 sin q(x, t) = 2 e (2 1) − p n 2 p n =1 2 Example 20. A uniform rod of length l whose surface is thermally insulated is initially at a constant temperature q = q0. At time t = 0, one end is suddenly cooled to zero temperature while the other end remains thermally insulated. This state is maintained subsequently. Find the temperature at the end x = l. (BR. 1995 Ap.)

∑



The boundary conditions are

(i) q(o, t) = o for t ≥ 0 ∂q (l, t) = 0 for t ≥ 0 since insulated (ii) ∂x (iii) q(x, 0) = q0 for 0 < x < l. (iv) q is finite. Since q is finite, we start with the solution

q(x, t) = (A cos lx + B sin lx) e–a

l t…II

179

Applications of Partial Differential Equations



Using boundary condition (1), A = 0 ∂q (l , t ) = 0 implies ∂x Bl cos ll = 0



B ≠ 0, cos ll = 0; l =



q(x, t) = Bn sin



\



Hence, most general solution is q(x, t) =



∞

∑ 1



At



\

(2n − 1)p 2l

(2n − 1)px −a 2l 2t e 2l

(2n − 1)px − Bn sin e 2l

a 2 (2 n − 1)2 p2 4l 2

t = 0, q = qo ∞

∑ Bn sin 1

(2n − 1)px dx = q0 2l Bn =



l

(2n − 1)px 2 dx q sin 2l l o

∫ o

l

(2n − 1)px   dx  2qo  cos 2l =  −  (2n − 1)p l   2l  0 4q = o (2n − 1)p u(x, t) =



u(l, t) =



4qo p

∞

∑ (2n1− 1) sin

n =1

 −a 2 (2n − 1) 2 p2t  (2n − 1)px Exp.   2l 4t 2  

 −a 2 (2n − 1) 2 p2t  4q0 ∞ (−1) n − 1 . Exp.   p n = 1 (2n − 1) 4l 2  

∑

Example 21. An insulated rod of length l has its end A and B kept at a degree centigrade and b degree centigrade until steady state conditions prevail. The temperature at each end in suddenly reduced to zero degree centigrade and kept so. Find the resulting temperature at any point of the rod taking the end A as origin.

Let the temperature at any point P of the rod at time t be u where AP = x.

Then u satisfies

∂u ∂ 2u k = a 2 2 . where a2 = . ∂t r c ∂x

∂u d 2u = 0. Therefore =0 ∂t dx 2



In steady,



i.e.,

u = Dx + E



At

x = 0, u = a



At

x = l, u = b

180



Engineering Mathematics-III (Third Semester)

E = a

\



b = Dl + E Dl + a

b−a l Hence, the steady state temperature is D =

\

b−a x + a …(2) l This temperature is the initial temperature of the next unsteady state. u(x) =



Let u(x, t) be the temperature in unsteady state. The boundary conditions are

u(o, t) = 0 for t > o…(i)



u(l, t) = 0 for t > o…(ii)



u(x, o) =

Now, in the unsteady state,



b−a x + a for o < x < l…(iii) l

u(x, t) = (A cosl x + B sin lx)e–a

l t

…II

Using the first two boundaries conditions,

A = 0; l =

np , n any integer l

npx − Hence, u(x, t) = Bn sin .e l The most general solution is

a 2 n 2 p2 t l2

∞

npx – u(x, t) = Bn sin .e l n =1

∑

a 2 n 2 p2 t l2

…IV

Using boundary condition (iii),



u(x, o) =

∞

npx ∑ Bn sin= l

b−a x + a …(3) l

n =1



This is half-range Fourier sine series



\

Bn =

2 l

∫( l

0

)

b−a npx x + a sin dx l l

 2 b−a =  x+a l l  

(

)

 cos npx   l  b−a  − np  − l   l  

( )

l

 sin npx    l   − 2 2  n p   l 2   0 

2 bl al cos np +  =  − l  np np  2 =  a − b(−1) n  np Hence

u(x, t) =

200 p

∞



2 2 2



∑ 1n (−1)n + 1 sin npl x Exp.  − a nl 2 p t  1

181

Applications of Partial Differential Equations

Note: In many problems, a, b and l only will very. The result can be verified by putting the values of a, b, l in the result of the above problem.

For example, if a = 0°C, b = 100°C and l = l



u(x, t) =

200 p

∞



2 2 2



∑ 1n (−1)n + 1 sin npl x Exp.  − a nl 2 p t  1

Example 22. A rod of length l cm long, with insulated sides, has its ends A and B kept at a° centigrade and b° centigrade respectively until steady state conditions prevail. The temperature at A is then suddenly raised to c° centigrade and at the same that at B is lowered to d° centigrade. Find the temperature distribution u (x, t) subsequently.

i.e.,

d 2u =0 dx 2 u = Dx + E



At

x = 0, u = a



At

x = l, u = b



\

E = a



In the steady state, u satisfies



…(1)

Dl + a = b

b−a l b−a …(2) \ u(x) = x + a in steady state  l When the temperature at A and B are changed, the state is no longer steady. It becomes transient. Let u(x, t) be the temperature in the next state. The boundary conditions are

D =



u(o, t ) = c, t > o…(i)



u(l, t) = d, t > o…(ii)

b−a x + a, for o < x < l…(iii) l Now, the boundary conditions are not zeros. This transient state, after a long time, will again be steady. The boundary conditions both in the transient state and subsequent steady state will be u(o, t) = c, u(l, t) = d and

u(x, o) =



In other words u(x, t) consists of two portions



us(x) and ut(x, t) where us(x) is the steady state function and us + ut the transient state function.



\

i.e., which implies,

u(x, t) = us(x) + ut(x, t)…(3) us(o) = c and u(o, t) = us(o) + ut(o, t) = c us(l) = d  u(l, t) = us(l) + ut(l, t) = d

ut(o, t) = 0, ut(l, t) = 0

us(x), the steady state function satisfies,



Now use



\

d 2u s = 0 i.e., us = ax + b…(4) dx 2 us(o) = c, us(l) = d b = c and al + b = d

182



Engineering Mathematics-III (Third Semester)

d −c l d −c x + c …(5) us(x) = l a =

\

i.e.,



∂u ∂ 2u = a 2 2 ∂t ∂x

Now Ut satisfies

ut = (A cos lx + B sin lx ) e–a

\

l t

…II

ut(o, t) = o…(iv)

subject to

ut(l, t) = o…(v)



ut(x, o) = u(x, o) –us(x)



(

)(

)

b−a d −c x+c = x + a − l l b+c−a−d = x + (a − c) …(vi) l Now find A, B, in II using (iv), (v), (vi)

ut(o, t) = 0 A = 0



ut(l, t) = 0 l =



npx − \ ut(x, t) = Bn sin .e l The most general solution is ∞

npx − ut(x, t) = Bn sin .e l n =1

∑

Now ut(x, o) =

np . l

\

Bn =

∞

npx n sin ∑ B= l 1

a 2 a 2 p 2t l2

a 2 a 2 p 2t l2

…(IV)

…(IV)

b+ x−a−d x + (a − c). l

l

2 b + c − a − d npx x + a − c  sin dx l  l l 

∫ 0

 2 b+c−a−d =  x+a−c l l  

(

)

l

npx    cos npx     b + c − a − d  sin l   l ×  − 2 2   − np  − l    n 2p   l   l    0

(

)

l (a − c)  2  (b − d )l (−1) n + =  − l np np  2 = (a − c) − (b − d )(−1) n  pn using Bn in IV and then using (3), we get u(x, t) = us(x) + ut(x, t)

(

)

∞

d −c 2 1 npx x+c + (a − c) − (b − d )(−1) n  sin u(x, t) = ×e l l p n =1 n 

∑

−a 2 n 2 p2t l2

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Applications of Partial Differential Equations



Note 1. Putting a = 20, b = 40, c = 50, d = 10, l = 30 cm, we get the worked example 15.



Note 2. Worked examples 21, and 22 give more general problem. Exercises 3 (c)

1. Solve the one dimensional diffusion equation for the function q(x, t) in the region 0 ≤ x ≤ p, t ≥ 0 when (i) q remains finite as t → ∞ (ii) q = 0 if x = 0 or p for all values of t > 0; and (iii) at t = 0, q = x for 0 ≤ x ≤ p/2 and p ≤ x ≤ p. (Madras, ’65 B.E.) q = p – x for 2 2. Find the solution to the equation

∂u ∂ 2u = a 2 2 that satisfies the conditions ∂t ∂x

u(0, t) = 0, u(l, t) = 0 for t > 0 and u(x, 0) = x for 0 ≤ x ≤ l/2    = l – x for l/2 < x < l 3. Obtain permissible product solutions of the heat-flow equation

(Madras, ’69 B.E.; Kerala, ’68 B.E.)

∂ 2T 1 ∂T 2 = 2 which satisfy the conditions ∂x a ∂t ∂T ∂T = (0, t ) 0= and (l , t ) 0.  ∂x ∂t

(Madras, ’69 B.E.)

4. Determine the solution of one dimensional heat equation ∂q ∂ 2q = a 2 2 under the boundary conditions ∂t ∂x q(0, t) = q(l, t) = 0 for t > 0 and the initial condition q(x, 0 ) = x for 0 < x < l, l being the length of the bar. 5. Solve the equation

∂u ∂ 2u = a 2 2 subject to the conditions given below: ∂t ∂x

(i) u is not infinite as t → ∞ (ii) u = 0 for x = 0 or x = p for all values of t (iii) u = px – x2 if t = 0 in (0, p). 6. Show that the solution of the equation ∂q ∂ 2q = k 2 subject to the conditions ∂t ∂x (i) q is finite as t → ∞ ∂q = 0 when x = 0 for all t > 0 (ii) < ∂x (iii) q = 0 when x = l for all t > 0 (iv) q = q0 when t = 0 for 0 < x < l is q q =4 0 p

∞

(−1) n (2n + 1)px − cos ·e (2 n 1) 2l + n=0

∑

k (2 n + 1)3 · p2t 4l 2



7. Solve

∂u ∂ 2u = a 2 2 with adiabatic boundary conditions: ∂t ∂x

(i)

∂u ∂ (0, t ) = 0, (ii) u (l , t ) = 0, ( ) u(x, 0) = x. ∂x ∂x

(MS. 1989 Nov.)

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Engineering Mathematics-III (Third Semester)

8. Solve

∂u ∂ 2u = a 2 2 for the conduction of heat along a rod, without radiation, subject to the following ∂t ∂x

conditions. (i) u is not infinite as t → ∞ ∂u = 0 for x = 0 or x = l (ii) ∂x (iii) u = lx – x2 for t = 0 for 0 < x < l. 9. Solve

(Madras, ’65 B.E.)

2

∂u ∂u = a2 2 = 0 for 0 < x < p, t > 0 ∂t ∂x

ux(0, t) = 0, ux(p, t) = 0 and u(x, 0) = sin x. 10. Show that the solution of

∂u ∂ 2u = a 2 2 subject to the conditions ∂t ∂x

∂u = 0 when x = 0  ∂x   for all t > 0 ∂u (ii) = 0 when x = l  ∂x

(i)

(iii) u = f (x) when t = 0 is u=

∞ a0 npx + Exp.[−a 2 n 2 p2t/l 2 ] a cos 2 n =1 n l

∑

l

where an =

2 npx f ( x) cos dx. l l

∫ 0

11. A bar with insulated sides is initially at temperature 0°C. throughout. The end x = 0 is kept at 0° symbol C. and the heat is suddenly applied at the end x = l at a constant rate, so that ∂u = K for x = l, K being a constant. Show that ∂x u(x, t) = Kx +

∞

(−1)1 (2n − 1)px 8 Kl sin . Exp [−(2n − 1) 2 p2a 2t/4l 2 ]  2 2 2l p n = 1 (2n − 1)

∑

(MS. 1987 Nov.)

12. The temperatures at the ends x = 0 and x = 100 of a rod 100 cm. in length are held at 0° and 100°C respectively until steady-state conditions prevail. Then at t = 0, the two ends are suddenly insulated. Find the resultant temperature distribution in the rod. (Madras, 69 B.E.) 13. (a) A rod of length l has its ends A and B kept at 0°C. and 100°C. respectively until steady-state conditions prevail. If the temperature at B is reduced suddenly to 0°C. and kept so, while that of A is maintained, find the temperature u(x, t) at a distance x from A and at time t. (Madras, B.E. 78.) (b) Solve the above problem, if the change consists of raising the temperature of A to 25°C. and reducing that of B to 75°C. (O.U., 65 B.E.) 14. Solve the heat equation ∂q ∂ 2q = k 2 in the range 0 ≤ x ≤ 2p, t > 0 subject to the conditions ∂t ∂x    q(x, 0) = sin3x, 0 ≤ x ≤ 2p and q(x, 0) = q(2p, t) = 0 for t > 0. (Calicut, 71 B.Sc. Engg.) 15. In a bar of 10 cm. long with its sides impervious to heat, the heat flow is one-dimensional and the two ends A and B are kept at 50°C. 100°C. respectively, until steady-state conditions prevail. The temperature at A is then suddenly raised to 90°C. and at the same instant that at B is lowered to 60°C.; these end temperatures are maintained thereafter. Find an expression for the temperature at a distance x from A and at any time t subsequent to the changes in temperatures at the ends. (S.V.U., 66 B.E.)

185

Applications of Partial Differential Equations

16. A rod AB of length 10 cm. has the ends A and B kept at temperatures 30°C. and 100°C, respectively until the steady-state conditions prevail. At sometime later, the temperature at A is lowered to 20°C, and that at B to 40°C, and then these temperatures are maintained. Find the subsequent temperature distribution.  (64 B.E.) 17. A bar, 10 cm. long, with insulated sides, has its ends A and B kept at 20°C, and 40°C, respectively, until steady-state conditions prevail. The temperature at A is then suddenly raised to 50°C, and at the same time that at B is lowered to 10°C. Find the subsequent temperature distribution u(x, t) and show that the temperature at the middle point of the bar remains unaltered for all time, regardless of the material of the bar. (Madras, 63 B.E.) 18. A bar 40 cm. long has originally a temperature of 0°C, along all its length. At time t = 0 sec., the temperature at the end x = 0 is raised to 50°C, while that at the other end is raised 100°C. Determine the resulting temperature distribution. (Madras, 63 B.E.) 19. A rod of length l has its ends A and B kept at 0°C. and 100°C. respectively until steady-state conditions prevail. If the temperature of A is suddenly raised to 50°C. and that of B to 150°C., find the temperature distribution at any point of the rod and at any time. (Madras, 71 B.E.) 20. A metal bar 50 cms. long whose surface is insulated is at temperature 60°C. At t = 0, a temperature of 30°C. is applied at one end and a temperature of 80°C. to the other end and these temperature are maintained. Determine the temperature of the bar at any time assuming the diffusivity K = 0.15 cgs unit.  (Madras, 70 B.E.) 21. The ends A and B of a rod of 30 cms. long have their temperatures kept at 20°C. and the other at 80°C., until steady-state conditions prevail. The temperature of the end B is suddenly reduced to 60°C. and kept so while the end A is raised to 40°C. Find the temperature distribution in the rod after time t.  (67 B.E.) 22. The temperature at one end of a bar, 50 cm. long and with insulated sides, is kept at 0°C. and the other end is kept at 100°C. until steady-state conditions prevail. The two ends are then suddenly insulated, so that the temperature gradient is zero at each end thereafter. Find the temperature distribution.  (’63, ’66 B.E.) 23. A uniform rod of length a whose surface is thermally insulated, is initially at temperature q = q0. At time t = 0, one end is suddenly cooled to temperature q = 0 and subsequently maintained at this temperature. The other end remains thermally insulated; show that the temperature at this end at time t is given by  (2n + 1) 2 K p2t  4q ∞ (−1) n 0 exp  −  where K is the thermometric conductivity. p n = 0 (2n + 1) 4a 2  

∑

24. The equation for the conduction of heat along the bar of length l is

(BR. 1995 April)

∂q ∂ 2q = k 2 , neglecting radiation. ∂t ∂x

Find an expression for q, if the ends of the bar are maintained at zero temperature and if, initially, the temperature is T at the centre of the bar and falls uniformly to zero at its ends. 25. A rod of length l is at a uniform temperature of 50°C. Suddenly, the end x = 0 is cooled to 0°C. by an application of ice, and the end x = l is heated to 100°C. by an application of steam and these two temperatures are maintained at the ends. Also the rod is insulated along its length so that no transfer of heat can occur from the sides. Find the temperature distribution in the rod at any subsequent time. 26. An insulated metal rod of length 100 cm has one end A kept at 0°C., and the other end B 100°C. until steady-state conditions prevail. At t = 0, the temperature at A is then suddenly raised to 50°C. and thereafter maintained while at the same time t = 0 the end B is insulated. Find the temperature at any point of the rod at any subsequent time. (S.V.U. ’67 B.E.)

3.7  Temperature in a Slab with Faces at Zero Temperature. Consider a slab of homogeneous material bounded by two parallel planes x = 0 and x = l, having an initial temperature u = f (x) which varies only with the distance from the faces, the two faces being kept at temperature zero. Since the heat-flow is one dimensional, the temperature is a

186

Engineering Mathematics-III (Third Semester)

function of x and t only. Hence the function u(x, t) is the function representing the temperature at any interior point of the slab and satisfying the one-dimensional feat-flow equation ∂ 2u 1 ∂u = 2 . 2 ∂x a ∂t Example 23. The faces of a slab of width l are kept at temperature zero. Its initial temperature is given by u = f (x), where





f (x) = A in 0 < x < l/2

= 0 in l/2 < x < l.

Find the temperature distribution.

The temperature function u(x, t) satisfies ∂u ∂ 2u = a 2 2 …(1) ∂t ∂x The boundary conditions are



u(0, t) = 0 for all t > 0

(i)



u(l, t) = 0 for all t > 0

(ii)



The initial conditions is



u(x, 0) = A, for 0 < x < l/2

= 0, for l/2 < x < l(iii).

Solving (1) and selecting the suitable solution as usual,



Using (i), we get



Using (ii),

u(x, t) = (A cos lx + B sin lx )e–a

l t …(2)

A = 0. np l = , where n is any integer. l a 2 n 2 p2t

npx − l 2 Hence u(x, t) = Bn sin e l The most general solution of (1) is a 3 n 2 p 2t ∞ npx − l 2 u(x, t) = Bn sin e …(3) l n =1

∑



Using (iii),



u(x, 0) =

∞

∑ Bn sin npl x …(4)

n =1

Expressing u(x, 0) as a half-range Fourier sine series in (0, l) and comparing it with (4), we find

l

2 npx u ( x, 0) sin dx l l

∫

Bn = bn =

0

l /2

2 npx dx, using (iii) = A sin l l

∫ 0

l /2

 cos npx  2A  l  =  − np  l   l 0

Applications of Partial Differential Equations

187

2A l np = . 1 − cos  l np  2 4A = sin 2 np/4 np Substituting in (3), u(x, ) =



∞  a 2 n 2 p 2t  4A 1 2 np npx sin sin exp  −  l 4 p n =1 n l2  

∑

Exercises 3(d) 1. The faces x = 0 and x = c of a slab which is initially at temperature f (x) are kept at temperature

zero. Derive the temperature formula

u(x, t) =

c ∞ 2  npq  npx f (q) sin d q sin exp (− n 2 p2a 2t/c 2 ).  c n = 1 c c 0 

∑∫

2. If f (x) = sin

px in problem 1, show that c

u(x, t) = sin

px exp (a 2 p2t/c 2 ). c

3. Two slabs of iron, each of 20 cm. thick, one at temperature 100°C. and the other at 0°C. throughout, are placed face to face in perfect contact, and their outer faces are kept at 0°C. Given that a2 = 0.15 centimetre-gram-second unit, show that the temperature at their common face 10 minutes after the contact was made is approximately 37°C. 4. If the slabs in question (3) above are made of material whose a2 = 0.005 cgs unit, show that five hours are necessary for the common face to reach the temperature 36°C. 5. A homogeneous slabs of conducting material is bounded by the planes x = 0 and x = p which are kept at temperature zero. If initially when t = 0, the slab is raised to temperature pkx in 0 ≤ x ≤ p/2 u(x, 0) = 4 =

pk (p − x) in p/2 ≤ x ≤ p 4

where k is the constant in the heat equation ∂u ∂ 2u = k 2 , show that the temperature at any subsequent time is ∂t ∂x 2 2 1 1 k  e − kt sin x − 2 e −3 kt sin 3 x + 2 e −5 kt sin 5 x −… . 3 5  

6. Two slabs of the same material, one 2 feet thick and the other 1 foot thick, are placed side by side. The thicker slab is initially at constant temperature A, the thinner one initially at zero. The outer faces are held at temperature zero for t > 0. Show that the temperature at the centre of the two-feet slab is ∞

4A 1 2 np sin exp (−a 2 n 2 p2 · t/9). p n =1 n 3

∑

7. The faces x = 0 and x = l of a slab which is initially at temperature f (x) are insulated. Derive the formula u(x, t) =

∞

1 npx a + a cos exp ( − n 2 p2a 2t/l 2 ) 2 0 n =1 n l

∑

l

where an =

2 npx f ( x) cos dx. l l

∫ 0

188

Engineering Mathematics-III (Third Semester)

3.8.  Two Dimensional Heat-Flow When the heat-flow is along curves instead of along straight lines, all the curves lying in parallel planes, then the flow is called two-dimensional. Let us consider now the flow of heat in a metal-plate in the xoy plane. Let the plate be of uniform thickness h, density r, thermal conductivity k and the specific heat c. Since the flow is two dimensional, the temperature at any point of the plate is independent of the z-coordinate. The heat-flow lies in the xoy plane and is zero along the direction normal to the xoy plane.

Now, consider a rectangular element ABCD of the plate with sides dx and dy, the edges being parallel to the coordinates axes, as shown in the figure. Then the quantity of heat entering the element ABCD per sec. through the surface AB is ∂u = −k   dx · h.  ∂y  y

Similarly the quantity of heat entering the element ABCD per sec. through the surface AD is

( ∂∂ux )

= −k

y

dy · h.

The amount of heat which flows out through the surfaces BC and CD are

−k

( ∂∂ux )

∂u dx · h respectively. · dx · h and − k   x + dx  ∂y  y + dy



Therefore the total gain of heat by the rectangular element ABCD



per sec. = inflow–outflow

( )

  ∂u kh   = ∂x 

x + dx

( )

 ∂u  ∂x = khdx·dy  

−

( ∂∂ux )  dy +  ∂∂uy  x

x + dx

−

( ∂∂ux )

dx

x

∂u  −   dx  y ∂   y + dy y 

∂u   ∂u  −      ∂y  x + dy  ∂y  y  +  …(1) dy 

The rate of gain of heat by the element ABCD is also given by

∂u …(2) ∂t Equating the two-expressions for gain of heat per sec. from (1) and (2), we have, rdx · dy · h · c ·

( )

 ∂u  ∂x ∂u rdx · dy · h · c · = h k dx dy  ∂t 

x + dx

−

dx

( ∂∂ux )

x

∂u   ∂u  −      ∂y  x + dy  ∂y  y  + dy 

189

Applications of Partial Differential Equations

( )

 ∂u  k  ∂x ∂u = rc  ∂t

x + dx

−

( ∂∂ux )

x



i.e.,



Taking the limit as dx → 0, dy → 0, the above reduces to ∂u k  ∂ 2u ∂ 2u  =  2 + 2  ∂t rc  ∂x ∂y 

Putting a2 =

dx

∂u   ∂u  −      ∂y  x + dy  ∂y  y  + dy 

k as before, the equation becomes, rc  ∂ 2u ∂ 2u  ∂u = a 2  2 + 2  …(3) ∂t ∂y   ∂x



The equation (3) gives the temperature distribution of the plate in the transient state.



In the steady-state, u is independent of t, so that

of the plate in the steady-state is

∂u = 0. Hence the temperature distribution ∂t

∂ 2u ∂ 2u + = 0. ∂x 2 ∂y 2

i.e., ∇2u = 0, which is known as Laplace’s Equation in two-dimensions.

∂u of the ∂t temperature in the direction of the y-axis will be zero. Then the heat-flow equation reduces to ∂u ∂ 2u = a 2 2 which is the heat-flow equation in one-dimension. ∂x ∂y



Corollary. If the stream lines are parallel to the x-axis, then the rate of change

3.9  Solution of the Equation

∂ 2u ∂ 2u + = 0. ∂x 2 ∂y 2

∂ 2u ∂ 2u + = 0 ∂x 2 ∂y 2



The equation is



Assume the solution u(x, y) = X (x) · Y (y),

…(1)

where X is a function of x alone and Y a function of y alone. 2 ∂ 2u ′′ Y , and ∂ u2 XY ′′. = \ X= 2 ∂x ∂y

The Laplace equation ∇2u = 0 becomes X″Y + Y″X = 0 X ′′ −Y ′′ = …(2) i.e., X Y The left hand side of (2) is a function of x alone and the right hand side is a function of y alone. Also x and y are independent variables. Hence, this is possible only if each quantity is equal to a constant k. X ′′ −Y ′′ \ Let = = k …(3) X Y i.e., X″ – kX = 0, and Y″ + kY = 0. …(4)

190



Engineering Mathematics-III (Third Semester)

Case 1. Let k =

Then X″ –

l2X

l2,

a positive number.

= 0, and Y″ + l2 Y = 0.

Solving, X = A1 elx + B1 e–lx and Y = 1 cos ly + D1 sin ly.

Case 2. Let k = – l2, a negative number.



Then (4) becomes X″ + l2X = 0 and Y″ – l2Y = 0.



Solving these equations, we have, A2 cos lx + B2 sin lx and Y = C2ely + D2e–ly.

X =

Case 3. Let k = 0. Then (4) reduces to

X = 0 and Y = 0.



On solving these equations, X = A3x + B3 and Y = C3y + D3.



Therefore, the possible solutions of (1) are



u(x, y) = (A1elx + B1e–lx) (C1 cos ly + D1 sin ly) …(I)

u(x, y) = (A2 cos lx + B2 sin lx) (C2ely + D2e–ly) …(II)

u(x, y) = (A3x + B3) (C3y + D3) …(III)

In problems where the boundary conditions are given, we have to select a suitable solution or a linear combination of solutions to satisfy (1) and the boundary conditions. Example 24. An infinitely long plane uniform plate is bounded by two parallel edges x = 0 and x = l, and an end at right angles to them. The breadth of this edge y = 0 is l and is maintained at a temperature f (x). All the other three edges are at temperature zero. Find the steadystate temperature at any interior point of the plate. Let u(x, y) be the temperature at any point (x, y) of the plate. Then u satisfies

∂ 2u ∂ 2u + = 0 ∂x 2 ∂y 2

…(1)

The boundary conditions are



u(0, y) = 0,

for 0 ≤ y ≤ ∞ …(i)



u(l, y) = 0,

for 0 ≤ y ≤ ∞ …(ii)



u(x, ∞) = 0,

for 0 ≤ x ≤ l …(iii)



u(x, 0) = f (x), for 0 < x < l …(iv)



Solving (1), we get,



u(x, y) = (A1elx + B1e–lx) (C1 cos ly + D1 sin ly) …(I) u(x, y) = (A2 cos lx + B2 sin lx) (C2ely + D2e–ly)

…(II)

u(x, y) = (A3x + B3) (C3y + D3) …(III)

Of these solutions, we have to select a solution to suit the boundary conditions.

Since u = 0 as y → ∞, we select the solution (II) as a possible solution, (rejecting the other two).

\

u(x, y) = (A cos lx + B sin lx) (Cely + De–ly) …(2)

191

Applications of Partial Differential Equations



Using the boundary condition (1), u(0, y) = A(Cely + De–ly) = 0, for 0 ≤ y ≤ ∞.   \  A = 0



Using the boundary condition (ii) in (2),



u(l, y) = B sin ll (Cely + De–ly) = 0, for 0 ≤ y ≤ ∞.

Since B ≠ 0,

sin ll = 0. Hence ll = np

np , where n is any integer. l As y → ∞, u → 0, from (iii).  \  C = 0. i.e.,

l =

npx − e l Therefore the most general solution of (1) is u(x, y) = Bn sin

Hence

u(x, y) =



∞

∑

Bn sin

n =1



npy l

npx − e l

, where BD = Bn. npy l

…(3)

Using the boundary condition (iv) in (3), u(x, 0) =



∞

∑B

sin

n

n =1

npx = f (x), roman in 0 < x < l l

…(4)

Expressing f (x) as a half-range Fourier sine series in (0, l), we have f (x) =



∞

∑b

n

sin

1

where bn =

2 l

l

∫ f ( x) sin 0

npx …(5), l

npx dx. . l

Comparing (4) and (5),

Bn = bn =

2 l

l

∫ f ( x) sin 0

npx dx. l

Therefore the solution is npy 2 l npx  npx − l  f ( x) sin dx sin .e u(x, y) = l l l  n = 1 0  ∞



∑ ∫

Note. If f (x) is given explicitly in any problem, evaluate the value of Bn from the integral and substitute. Example 25. The vertices of a thin square plate are (0, 0), (l, 0), (l, l) and (0, l). The upper edge of the square is maintained at an arbitary temperature given by u(x, l) = f (x). The other three edges are kept at temperature zero. Find the steady-state temperature at any point within the plate. Explain how you will find the steady-state temperature if an arbitary temperature distribution existed along each edge. Suppose that u(x, y) is the temperature at any point (x, y) of the plate in steady-state.

192

Engineering Mathematics-III (Third Semester)

∂ 2u ∂ 2u + = 0 ∂x 2 ∂y 2



Then



The boundary conditions are





u(0, y) = 0,

for 0 < y < l …(i)



u(l, y) = 0,

for 0 < y < l

…(ii)



u(x, 0) = 0,

for 0 < x < l

…(iii)



u(x, l) = f (x), for 0 < x < l



…(1)

…(iv)

Solving (1), we get the three possible solutions, u(x, y) = (Aelx + Be–lx) (C cos ly + D sin ly)



(Cely

De–ly)

…(I)



u(x, y) = (A cos lx + B sin lx)



u(x, y) = (Ax + B) (Cy + D) …(III)

+



…(II)

where A, B, C, D are different arbitary constants in each solution.

Now we shall select the solution II. u(x, y) = (A cos lx + B sin lx) (Cely + De–ly)

i.e.,

Using the boundary condition (i) in (II),



A(Cely + De–ly) = 0, for 0 ≤ y < l.  \  A = 0

Using the condition (ii) in (II),



u(l, y) = B sin ll (Cely + De–ly) = 0. But B ≠ 0; sin ll = 0

i.e.,

ll = np

i.e.,

l =



…(II)

np where nn roman is any integer. l

Using (iii) in II,



u(x, 0) = (C + D) (B sin lx) = 0, for 0 ≤ x ≤ l.

B ≠ 0 Hence C + D = 0.   \  D = – C.

Hence (II) reduces to,

(

npx

npy

)

− npx e l −e l l npy npx where Bn = 2BC. = sin Bn sinh l l Hence the most general solution is



u(x, y) = BC sin



u(x, y) =

∞

∑B

n

sin

npy npx sinh …(2) l l

sin

npx sin np = f (x) for 0 < x < l l

n =1



Using the boundary condition (iv) in (2),



u(x, l) =

∞

∑B

n

1

…(3)

Expressing f (x) as a half range Fourier sine series in (0, l), we get

f (x) =

∞

∑b

n

1

sin

npx …(4) l

193

Applications of Partial Differential Equations

2 where bn = l

l

∫ f ( x) sin 0

npx dx l



Comparing (3) and (4), Bn sinh np = bn



\



Hence putting this value in (2),

Bn =

u(x, y) =



bn 2 = sinh np l sinh np

l

∫ f ( x) sin 0

npx dx. l

l  py npx  npx 2  f ( x) sin dx sin sin n l sinh np l l l   n = 1 0  ∞

∑

∫

Suppose that arbitrary temperature distribution exists along each edge, we can get four solutions each corresponding to a problem in which zero temperatures are prescribed along three of the four edges. Then by superimposing these four solutions, we get the solution of the given problem. In this case, note that you are doing four different problems of the type worked out in the above example. Example 26. A rectangular plate is bounded by the lines x = 0, y = 0, x = a and y = b. Its surfaces are insulated and the temperature along two adjacent edges are kept at 100°C. while the temperature along the other two edges are at 0°C. Find the steady-state temperature at any point in the plate. Also find the steady-state temperature at any point of a square plate of side a if two adjacent edges are kept at 100°C. and the others at 0°C. (Madras B.E.) If u(x, y) be the temperature at any interior point of the rectangular plate, then

∂ 2u ∂ 2u + = 0 ∂x 2 ∂y 2

…(1)

The boundary conditions are



u(0, y) = 0,

for 0 < y < b

…(i)



u(x, 0) = 0,

for 0 < x < a

…(ii)



u(a, y) = 100, for 0 < y < b

…(iii)



u(x, b) = 100, for 0 < x < a

…(iv)



Now we will split the solution into two solutions.

i.e., Let

u(x, y) = u1(x, y) + u2(x, y)

…(2)

where u1(x, y) and u2(x, y) are solutions of (1) and further u1(x, y) is the temperature at any point P with the edge BC maintained at 100°C. and the other three edges at 0°C. while u2(x, y) is the temperature at P with the edge AB maintained at 100°C. and the other three edges at 0°C. (Refer to figure on page 234)

Therefore the boundary conditions for the functions u1(x, y) and u2(x, y) are as follows.

u1(0, y) = 0

u1(a, y) = 0

u2(0, y) = 0

u2(a, y) = 100

u1, u2 each satisfies (1).

(v) (vii) (ix) (xi)

u1(x, 0) = 0

u1(x, b) = 100 u2(x, 0) = 0 u2(x, b) = 0

(vi) (viii) (x) (xii)

194

Engineering Mathematics-III (Third Semester)

Therefore u1 will have three possible values when the equation (1) is solved for u. i.e.,



u1(x, y) = (Aelx + Be–lx) (C cos ly + D sin ly)

u1(x, y) = (A cos lx + B sin lx)

(Cely

+

De–ly)



…I …II

u1(x, y) = (Ax + B) (Cy + D) …III

Select II solution and make use of the boundary conditions (v), (vi), (vii) as in example 25. We get,

u1(x, y) =

∞

∑ B sin napx sinh n

n =1

npy …(3) a

Using boundary condition (viii) in (3)



u1(x, b) =

∞

∑ B sin napx sinh npab = 100 …(4) n

n =1

Expressing 100 as a half–range Fourier sine series, ∞



100 =

∑ b sin napx …(5) n

1

a



where

bn =

2 npx 200 100sin dx = (1 − cos np) a a np

∫ 0

npb Comparing (5) and (4) Bn sinh = bn . a 200 \ Bn = (1 − cos np) = 0 if n is even npb np sinh a 400 = if n is odd npb np sinh a Substituting this value in (3),



∞

npy 400 1 npx ⋅ sin ⋅ sinh …(6) np sinh npb a a n = 1, 3, 5... a Similarly considering the boundary conditions of u2(x, y), we can easily obtain



Since

u1(x, y) =

∑

∞

npy 1 1 npx sinh sinh …(7) p n a n b b n = 1, 3, 5... sinh b u(x, y) = u1(x, y) + u2(x, y),

u2(x, y) =

400 p

400 u(x, y) = p

∑

(2n − 1)py  sin(2n − 1)px sinh 1  a a (2n − 1)pb 2n −1  n =1 sinh  a 

∑

sin(2n − 1)py (2n − 1)px  .sinh  b b +  (2n − 1)pa sinh  b 

195

Applications of Partial Differential Equations

Putting b = a, the rectangular plate becomes a square plate.

Hence the temperature at any point (x, y) in the square plate of side a is u(x, y) =



400 p

∞

1 ∑ (2n − 1)sinh(2 n − 1)p n =1

(2n − 1)py (2n − 1)py (2n − 1)px   (2n − 1)px sinh sinh × sin + sin  a a a a  Note. While selecting the solution for u2(x, y), we choose solution I and not II.

Example 27. An infinite long plate is bounded by two parallel edges and an end at right angles to them. The breadth is p . This end is maintained at a constant temperature u0 at all points and the other edges are at zero temperature. Find the steady–state temperature at any point (x,y) of the plate. If u(x, y) is the temperature at any point (x, y), then

∂ 2u ∂ 2u + = 0 (steady–state) ∂x 2 ∂y 2

…(1)

Solving, we get, three solutions u(x, y) = (Aelx + Be–lx) (C cos ly + D sin ly)



= (A cos lx + B sin lx)



(Cely

+

De–ly)



…I …II

= (Ax + B) (Cy + D) …III where A, B, C, D are different arbitrary constants.

The boundary conditions are u(0, y) = 0 u(p, y) = 0

 (i)  (ii) 

0 a



=

1 p  p + 0 for= |x| a  = 22  4

x=0

Setting ∞



sin as p ds = s 2 0

∫

∞

∴

∫ 0

sin as p dx = x 2

Setting ax = θ, we get, ∞



∫ 0

sin q p dq = q 2

Convolution Theorem or Faltung Theorem Def. The convolution of two functions f (x) and g(x) is defined as f *g =

1 2p

∞

∫

f (t ) g ( x − t ) dt

−∞

Theorem. The Fourier transform of the convolution of f(x) and g(x) is the product of their Fourier transforms. That is, F f (x) * (x)} = F(s).G(s) = F{f(x)} . F{g(x)}

230

Engineering Mathematics-III (Third Semester) ∞

1

F { f * g} =



2p

∫ ( f * g) e

∞

∞

 1  ∫ 2p −∞  2p ∞

1 2p

=

dx

−∞

1

=

ixs

∫

−∞

 f (t ) g ( x − t )dt  eixs dx  

∞

  f (t )  ∫ g ( x − t )eixs dx  dt    −∞ 

∫

−∞

by changing the order of integration 1

=

2p =



1 2p

∞

∫

∞

f (t ) eits G ( s ) dt , using shifting theorem

∫

−∞

1

= G ( s ).



f (t ) F {g ( x − t )} dt

−∞

2p



= G(s). F(s)



= F(s).G(s)

∞

f (t ) eits dt

∫

−∞

By inversion, F–1 {F (s) G (s)} = f * g = F–1 {F (s) * F–1 {G (s)} Parseval’s identity. If F(s) is the Fourier transform of f (x). ∞

then,

2

∫ | f ( x) |

−∞

dx =

∞

2

∫ | F ( s) |

ds

−∞

Proof. Firstly let us prove F f (− x)} = F ( s ) where F ( s ) indicates the complex conjugate of F(s) F (s) =



=

1 2p 1

∞

f ( x) e −isx dx

∫

−∞ ∞

∫

2 p −∞

f (−v) e isv dv putting x = – v

= F{ f (−v)} = F{ f (− x)} , changing the dummy variable By convolution Theorem,

F{f(x)} * g(x)} = F(s) G(s)



f * g = F–1 {F(s) G(s)} 1

∞

∫

2 p −∞

f (t) g( x − t)dt =

1

∞

∫ F(s) G(s) e

2 p −∞

− isx

ds

... (1)

231

Fourier Transforms

Putting x = 0, we get ∝



∫

−∞

∞

f (t) g(−t) dt = ∫ F(s) G(s) ds

... (2)

−∞

Since it is true for all g(t), take g(t) = f (−t) g(– t) = f (t)

\

G(s) = F{g (t)} = F { f (−t)} = F( s) using (1). Use this in (2). ∞

∫

∞

∫ F(s)F(s) ds

f (t) f (t) dt =

−∞

−∞

∞

∞

−∞

−∞

2 ∫ | f (t)| dt =

\

2

∫ | F(s)|

ds. 2

∞

p  sin t  Example 6. Using Parseval’s identity, prove ∫   dt = . t  2 0 1for| x|< a In example 5, if f(x) =  0 for| x|> a > 0 We have proved F(s) =

2 sin as . p s

Using Parseval’s identity. ∞

∝

−∞

−∞

2 ∫ | f (t)| dt =



a

∫ 1 dt =

\

−a

2a =

∫ | F(s)|

2

2

∞

2  sin as  ∫ p  p  ds. −∞ 2 p

∞

2

∞

\

\

 sin t  ∫  t  dt = p −∞ ∞

2

∞

2

2

 sin as  ∫  s  ds. −∞

Setting as = t, we get

ds

 sin t  2∫   dt = p t  0  sin t  ∫  t  dt = p/2 0

232

Engineering Mathematics-III (Third Semester)

Example 7. Find the Fourier transform of f(x) = 1 – | x | if | x | < 1 and hence find the value sin 4 t (Anna Ap. 2005) ∫ t 4 dt. = 0 for | x | > 1 0

∞



F{f(x)} =



=



=



1

∞

∫ (1−| x|) e

2 p −∞ 1

isx

dx

1

∫ (1−| x|)(cos sx + i sin sx) dx

2 p −1 1 2p

1

⋅ 2 ∫ (1 − x )cos sx dx since (1 – | x |) is even 0

2  1 − cos s    p  s2 

=

Using Parseval’s identity. ∞

2 (1 − cos s) ∫ s4 ds = p −∞ ∞

1

∫ (1−| x| )dx 2

−1

1

2

4 (1 − cos s) ds =2 ∫ (1 − x )2 dx =2/ 3 4 p ∫0 s 0 ∞

16 sin 4 s / 2 ds = 2/3 p ∫0 s4 s = x , we get 2

Setting ∞

∫ 0

sin 4 x x4

dx =

p 3

Infinite Fourier cosine transform and sine transform: Infinite Fourier cosine transform Let f(x) be defined for all x ≥ 0. Now we shall extend the function f(x) to the negative side of x axis. That is, let the extended function F(x) be defined as

F(x) = f(x) for x ≥ 0 = f(– x) for x < 0 so that F(x) is even in (–∞, ∞). Now, F{F(x)} =



=



=

1

∞

∫ F(x)e

2 p −∞ 1

isx

dx

∞

∫ F(x)[cos sx + i sin sx] dx

2 p −∞ 1 2p

∞

× 2 ∫ F( x ) cos sx dx since F(x) is even 0

233

Fourier Transforms



∞

2 p

=

∫ f (x) cos sx dx since F(x) = f(x) for x > 0 0

The R.H.S. is defined as the infinite Fourier cosine transform of f(x) denote by Fc(s) \ Infinite Fourier cosine transform of f(x) is defined to be 2 p

Fc(s) =

∞

∫ f (x)cos sx dx

... (1)

0

Then by inversion theorem, F(x) =

=



=

∞

1

∫ Fc (s) e

− isx

2 p −∞

ds

∞

1

∫ FC (s){cos sx − i sin sx} ds

2 p −∞

∞

1

× 2 ∫ Fc ( s) cos sx dx since Fc(s) is even

2p

0

Since F(x) = f(x) in (0, ∞),

f (x) =

2 p

∞

∫ Fc (s)cos sx ds

... (2)

0

Equation (1) defines the Fourier cosine transform of f(x) and (2) gives the Inversion Theorem for Fourier cosine transform. Infinite Fourier Sine Transform Let f(x) be defined for x ≥ 0. We extend f (x) to the negative side of x axis. Thus, let,

= –f (– x) for x < 0

so that F(x) is odd in (– ∞, ∞)

F{F(x)} =



=



=



= i

∞

1

∫ F( x ) e

isx

2 p −∞ 1

dx

∞

∫ F(x) [cos sx + i sin sx]dx

2 p −∞ i

∞

∫ F(x) sin sx dx since F(x) is odd

2 p −∞ 2 p

∞

∫ f (x)sin sx dx 0

Now, we define Fourier Sine transform of f(x) as Fs(s) = Imaginary point of F {F(x)} Fs {f(x)} = Fs(s) =

2 p

∞

∫ f (x)sin sx dx 0

... (3)

234

Engineering Mathematics-III (Third Semester)

By inversion theorm, F(x) = F–1 {iFs (s)}

1

= i

∞

∫ Fs (s)e

2 p −∞

isx

ds

∞

2 Fs ( s)sin sx ds p ∫0

f (x) =

... (4)

since Fs(s) is odd function of s and F(x) = f(x) in (0, ∞) Equation (3) defines infinite Fourier sine transform of f(x) and (4) gives the inversion theorem for Fourier sine transform. Hence the transform pairs are Fc{f(x)} = Fs{f(x)} =

2 p

∞

2 p

∞

∫

f ( x )cos sx dx ; f(x) =

0

∫ f (x)sin sx dx ;

f(x) =

0

2 p

∞

2 p

∞

∫ Fc (s)cos sx ds 0

∫ Fs (s)sin sx ds 0

Properties regarding cosine and sine transforms

1. Cosine and sine transforms are linear.

That is, Fc{af(x) + bg (x) = aFc {f(x)} + bFc [g(x)} The proof is obvious.

2. Fs [af (x) + bg(x)] = aFs {f(x)}+ bFs {g(x)}



3. Fs [f(x) sin ax] =



4. Fs [ f(x) cos ax] =



1 [F (s – a) – Fc (s + a)] 2 c

1 [Fs (s + a) + Fs (s – a)] 2 1 5. Fc [f(x) sin ax] = [Fs (a + s) + Fs (a – s)] 2

6. Fe [f(x) cos ax] =

1 [Fc (s + a) + Fc (s – a)] 2

Proof. Fs [ f(x)] sin ax] =

= = Fs [f(x) cos ax] =

2 p

∞

∫ f (x) sin sx dx 0

1 2 2 p

∞

∫ f (x) ⋅ [cos (s – a)x – cos (a + s)x] dx 0

1 [Fc(s – a) – Fc (s + a)] 2 2 p

∞

∫ f (x)cos ax sin sx dx 0

235

Fourier Transforms

1 2 2 p

∞

∫ f (x) . [sin (s + a)x + sin (s – a)]dx



=



1 = [Fs ( s + a) + Fs ( s − a)] 2

0

Similarly, we can prove the results (5) and (6).

7. Fc{f (ax)} =

1 s Fc   a a



8. Fs{f (ax)} =

1 s Fs   a a

Hint. put ax = u and proceed. Identities If Fc (s), Gc (s) are the Fourier cosine transforms and Fs(s), Gs (s) are the Fourier sine transforms of f(x) and g(x) respectively, then ∞

1.

∫

∞

∞

0

2.

∞

f ( x ) g( x ) dx = ∫ fc ( s)Gc ( s)ds 0

∫ f (x)g(x)dx = ∫ Fs (s) Gs (s) ds 0

3.

0

∞

∞

∞

0

0

0

| f ( x )|2 dx ∫=

Proof.

| fc ( s)|2 ds ∫=

∝

∞  ( ) ( ) F s G s ds = s s ∫ ∫ Fs (s)  0 0  ∞  = ∫ g( x )   0

∫| Fs (s)|

2 p

2

∞

ds. 

∫ g(x)sin sx dx  ds 

  2 Fs ( s)sin sx ds dx ∫ p 0  0

∝

∞



=

∫ g(x) f (x) dx 0

Similarly, we can prove the first identity and the third follows by setting g(x) = f(x). Example 8. Find Fourier cosine and sine transforms of e–ax, a > 0 and hence deduce the inversion formula.

Fs(e–ax) =

2 p

∞

∫e

− ax

sin sx dx

0

∞



=

 2  e − ax (− a sin sx − s cos sx )  p  a 2 + s2 0

=

2 s , if a > 0 2 p a + s2

... (1)

236

Engineering Mathematics-III (Third Semester)

2 p

Fc (e − ax ) =

∞

∫e

− ax

cos sx dx

0

∞

=

 2  e − ax  2 2 (− a cos sx + s sin sx ) p a + s 0

=

2 a if a > 0 ⋅ p a 2 + s2

... (2)

By inversion formula of (1). ∞

\

2 p

e–ax =

∞

 2  s ⋅ sin sx  ds 2 2 p a +s  0

∫ 

=

p − ax e , a>0 2

dx =

p −αa e ,a > 0 2

s

∫ a2 + s2 sin sx ds 0

Changing the variables, ∞



∫ 0

x sin αx 2

a +x

2

... (3)

Again, by inversion formula of (2)

e–ax = ∞

\

cos sx

∫ a2 + s2 ds

=

0

2 p

∞

 2 a 2 p a + x2 0

∫ 

  cos sx ds 

p − ax e 2a

Changing the variables, ∞

cos αx

dx ∫ a2= + x2 0

p −αa e ,a > 0 2a

Example 9. Find Fourier sine transform of

 x  Fs  2  =  a + x2 

2 p

∞

... (4)

x 2

a +x

2

and Fourier cosine transform of

x

∫ a2 + x 2 sin sx dx 0

=

2  p − as  e  using (3) of Example 8. p  2 

=

p − as e 2



 1  Fc  2  =  a + x2 

=

2 p

∞

1

∫ a2 + x 2 cos sx dx 0

2  p − as  e  using (4) of Example 8. p  2 a 

1 2

a + x2

.

237

Fourier Transforms

p 1 − as ⋅ e . 2 a

=

Example 10. Using Parseval’s identity evaluate ∞

dx

∫ ( a 2 + x 2 )2



∞

x2

∫ (a2 + x 2 )2 dx

and

0

2 s p a 2 + s2

By example 8. If f(x) = e–ax, Fs (s) = 2 a p a 2 + s2

Fc (s) =

and

if a > 0.

0

Using Parseval’s identity. ∞

2

s2

∞

−2 ax

dx



 e −2 ax =   −2 a

  0



=

1 2a

=

p ⋅ if a > 0 4a



∫ p (a2 + s2 )2 ds

∫e

=

0

0

∞

\

x2

∫ (a2 + x 2 )2 dx 0

∞

Also

2 a2 ∫ p (a2 + s2 )2 ds = 0 ∞

dx

∫ ( a 2 + x 2 )2

=

0

∞

∫e

∞

−2 ax

dx =

0

p

1 2a

if a > 0

4a 3

∞

dx

∫ (a2 + x 2 )(b2 + x 2 )

Example 11. Evaluate

using transform methods.

0

Let

f(x) = e–ax, g (x) = e–bx Fc(s) =



=

similary,

Gc (s) =

2 p

∞

∫e

− ax

cos sxdx

0

2 a by example 8 p a 2 + s2 2 b p b 2 + s2

∞

\

Using,

∫ Fc (s) Gc (s)ds 0

∞

=

∫ f (x) g(x) dx 0

238

Engineering Mathematics-III (Third Semester) ∞

2 ab ds = ∫ 2 2 p 0 ( a + s )(b 2 + s2 ) ∞

dx

∫ (a2 + x 2 )(b2 + x 2 )

∞

∫e

−( a + b ) x

dx

0

=

1 a+b

=

p , if a, b > 0. 2 ab( a + b)

0

Example 12. Find Fourier cosine transform of

cos x in 0 < x < a f(x) =  x≥a 0 2 p

∞

2 p

a

Fc(s) =

=

∫ f (x)cos sx dx 0

∫ cos x cos sx dx 0

a



=

21 [cos( s + 1)x + cos( s − 1)x] dx p 2 ∫0



=

1  sin( s + 1)a sin( s − 1)a  + if s ≠ 1, – 1 s − 1  2 p  s + 1

Example 13. Find Fourier sine transform of 1 Fs   = x



2 p

∞

∞

1 x

(Anna Ap. 2005)

sin sx dx x 0

∫

sin q dq, putting sx = q q 0



=

2 p



=

2 p × p 2



=

p 2

∫

Example 14. Find Fourier sine and cosine transform of xn–1. ∞



We know, Γ(n) =

∫e

− x n −1

x

0



∞

=

∫e

dx , n > 0

− ax

( ax )n −1 adx setting x as (as ax, a > 0)

0

∞

\

∫e 0

− ax n −1

x

dx =

Γ(n) an

⋅ n > 0, a > 0

239

Fourier Transforms

We can prove the above result even if a is complex. Setting a = is, ∞

∫e



− isx n −1

x

dx =

0





Γ(n) (is)n (−i )a Γ(n)

=

sn e

=

p − ni 2

sn

Γ(n)since − i =3

p − i 2

Equating real and imaginary parts on both sides, we get ∞

∫x



n −1

Γ(n)

cos sx dx =

s

0

∞

∫x

n −1

sin sx dx =

Γ(n)

0

\

Taking n = 1/2.

n

s

n

cos

np 2

sin

np 2

Fc(xn–1) =

2 Γ(n) np cos 2 p sn

Fs (xn–1) =

2 Γ(n) np sin n 2 p s

1  1  1 Fc  since Γ   =p  = s x 2   1  1  Fs   = s  x

Note:

1 x

is self reciprocal under Fourier sine and cosine transforms.

Example 15. Show that

(i) Fs[xf(x)] = −



(ii) Fc[xf(x)] =



Fc(s) =

d Fc ( s) ds

d Fs ( s) and hence find Fourier cosine and sine transform of xe–ax. ds 2 p

d 2 Fc ( s) = − p ds

∞

∫ f (x)cos sx dx 0

∞

∫ xf (x)sin sx dx 0

= – Fs {xf (x)}

240

Engineering Mathematics-III (Third Semester)

Similarly,

Fs(s) =

d Fs ( s) = ds

2 p

∞

2 p

∞

∫ f (x)sin sx dx 0

∫ xf (x)cos sx dx 0

= Fc {xf(x)}



d Fs (e − ax ) ds

Fc(xe–ax) =

=

d 2 s  2 ds  p a + s2



=

2 a 2 − s2 p ( a 2 + s 2 )2

Fs(xe–ax) = −

= −



=

  using example 8 

d Fc (e − ax ) ds d 2 a    (Refer example 8) 2 ds  p a + s2 

2 2 as 2 p ( a + s 2 )2

Example 16. Find Fourier cosine transform of e − a transform of x e Fc( e − a

2 2

x

− a2 x 2 ∞



(



2 p

= Real part of

Fs xe − a

2 − a2 x 2 .cos sx dx e p ∫0

)=

= 2 2

x

1 a 2

e

−

s2 4 a2

c

= =

a 2

e

−

s 2 a3 2

∫e

− a2 x 2 + isx

dx

0

(Refer example 4)

) = − dsd F (e 1

∞

− a2 x 2

s2 4 a2

e

−

)

s 2 a2 s 4 a2

2 2

x

and hence evaluate, Fourier sine

241

Fourier Transforms

Example 17. Solve for f(x) from the integral equation ∞



−α

e ∫ f ( x ) cos αx dx = 0

Proof. Multiplying by

2 p

∞

∫f

2 , we get p

( x) cos αx =

0

2 −α e , α parameter p

{ f ( x)} Fc=



2 −α e p

−1  2 −α  f ( x) = Fc  e   p 

∴

=



=



2 p

∞

∫e

−α

cos αx d α

0

2 1 on integration ⋅ p 1 + x2

Example 18. Solve for f(x) from the integral equationz

∞

∫ 0

1for 0 ≤ s < 1  f ( x )sin sx dx = 2 for 1 ≤ s < 2 0 for 2 ≥ 2 

Poof. Multiplying by



\

2 , both sides, p  2 for 0 ≤ s < 1   p  2 for 1 ≤ s < 2 Fs (f(x)) = 2  p 0 for s≥2   f(x) = Fs−1 (R.H.S) 1

2



=

2 4 sin sx ds + ∫ sin sx ds p ∫0 p1



=

2  1 − cos x  4  cos x − cos 2 x   +   p x x  p 

f (x) =

2 (1 + cos x − 2 cos 2 x ) px

242

Engineering Mathematics-III (Third Semester) 2

Example 19. Find the Fourier cosine transform of e − x (second method). 2

I = Fc (e − x ) =



∞

2 − x 2 cos sx dx e p ∫0



∞ 2 dl 2 = − xe − x sin sx dx ∫ ds p0



∞  e − x2 2 sin sx d  = +  2 p ∫0 

   

∞ 2   ∞  2 2  e − x 1  sin sx  − ∫ se − x cos sx dx    20 p  2 0  



=



s = − I 2 dI s = − ds I 2

\

log I = −

\

−

s2 4



I = ce

when

s = 0, I =

s2 + log c 4



... (1)

∞

2 − x2 1 e dx = ∫ p0 2

Using in (1),

\

1 2

=c

I = Fc (e

− x2

)=

1 2

e

−

s2 4

Example 20. Find the complex Fourier transform of dirac delta function d(t–a).

F{d(t – a)} =



=



=



=

1

∞

∫e

ist

2 p −∞ 1 2p

d(t − a)dt

a+ h

Lt

h →0

∫ a

1 ist e dt h

1  e ist  2 p h →0 h  is

1

Lt

a+ h

  a

 e ish − 1  Lt e isa   2 p h →0  ish 

1

243

Fourier Transforms



=

e isa 2p

eq − 1 =1 q→0 q

since Lt

Note. Dirac delta function d(t – a) is defined as d(t – a) = Lt I(h, t – a) where h→0

I(h, t – a) =

1 for a < t < a + h h

= 0 for t < a and t > a + h



Example 21. Find the function if its sine transform is e − as s

Let

Fs (f(x)) =

Then,

f(x) =

2 e − ax sin sx ds p ∫0 s

\

df = dx

2 p

∞

∞

∫e

− as

... (1)

cos sx ds

0

= f(x) =

\

e − as . s

=

2 a ⋅ 2 p a + x2 2 dx ⋅a p ∫ a2 + x 2 2 x tan −1 + c a p

... (2)

At x = 0, f(0) using (1) Using this in (2), c = 0 Hence,

f(x) =

2 x tan −1 . p a

1 Fs−1   = s

2 p p ⋅ = p 2 2

setting a = 0.

Example 22. Prove (i) F{xn f(x)} = (– i)n

d n F ( s) dsn

and (ii) F{f n(x)} = (– is)n F(s) (iii) Hence

solve for f(x) if ∞



∫

−∞

f (t) e −|x −t| dt = f( x ) where f(x) is known.

244

Engineering Mathematics-III (Third Semester)

Proof. (i)

1

F(s) = dn ds (−i )n

n

dn ds

n

∞

∫e

isx

f ( x ) dx

2 p −∞ 1

F ( s) =

∞

∫ (ix) e

n isx

2 p −∞ 1

F ( s) =

∞

∫x

n isx

e

2 p −∞

f ( x ) dx

f ( x ) dx

= F{xn f(x)}

(ii) Similarly,

1

F{f n(x)} =

∞

∫e

isx

dn dx n

2 p −∞

f ( x ) ⋅ dx

= (is)n F(s),



Using integration by parts successively and making assuptions that f, f 1, ..., f → ± ∞. (iii)

1 2p

1

f( x ) =

∞

∫

2 p −∞

f (t)e −|x −t| dt , from the given the equation

= f(x) * e–| x |



By convolution theorem,





1 2p

2 1 p 1 + s2

f( s) = F(s) . F(s) =

1 (1 + s2 ) f ( s) 2

1 [f( s) − (−is)2 f( s)] 2 1 1 f(x) = f( x ) − f′′ ( x ) using the result derived in (ii). 2 2



=

\

Exercises 4 (a)

1. Show that the Fourier transform of



f(x) = a – | x | for | x | < a = 0 for | x | > a > 0

is

2 1 − cos as . Hence show that p s2

∞

2. Show that the Fourier transform of





2

 sin t  ∫  t  dt = p / 2 0

f(x) = 0 for x < a



= 1 for a < x < b



= 0 for x > b

(n–1)

→ 0 as x

245

Fourier Transforms

is

1  e ibs − e iαs  is 2 p 

  

3. Show that the Fourier transform of  2p sin sa , for| x|≤ a  f(x) =  2 a is sa 0 for| x|> a 





4. Find the Fourier and cosine transforms of e–x and hence using the inversion formulae, ∞ ∞ x sin αx p −α cos αx e dx show that ∫ dx = = ∫ 2 2 2 0 1+ x 0 1+ x



5. Show that the Fourier sine transform of for 0 < x < 1 x  f(x) = 2 − x for 1 < x < 2 0 for x > 2 



is

2 2 p

(1987 Ap. Barathadasan)

sin s (1 – cos s) / s2.



6. Find the Fourier sine and cosine transform of cosh x – sinh x (same as question 4).



7. Find the Fourier sine and cosine transform of ae–ax + be–bx, a, b > 0. 1for| x|≤ a 8. Find the Fourier transform of f(x) =  0 for| x|> a 9. Show that the Fourier cosine transform of 10. Show that the Fourier sine transform of 11. Show that the Fourier transform of e 12. Find Fourier transform of e

–| x |

13. Find Fourier transform of

−

−x 2

1 1+ x x

1+ x

2

2

is

is

p −s e 2 p −s e 2

2

is self-reciprocal.

if a > 0.

1 | x|.

Fourier Transform of Derivatives We have already seen (refer Example 22) that, F{f n(x)} = (– is)n F(s)

 ∂2u  (i) \ F  2  = (– is)2 F{u(x)} = – s2 u where u is Fourier transform of u w.r.t. x.  ∂x 

(ii) Fc{f ′(x)} = −

2 f (0) + sFs ( s) p

246

Engineering Mathematics-III (Third Semester)

L.H.S. =

=

2 p

∞

2 p

∞

′

∫ f (x) ⋅ cos sx dx 0

∫ cos sx d{ f (x)} 0

∞  2 ∞ { f ( x )cos sx}0 + s ∫ f ( x )sin sx dx  p  0 



=



= sFs ( s) −

2 f (0) assuming f(x) → 0 as x → ∞. p

∞

(iii) Fs{f ′(x)} =

2 sin sx d [ f ( x )] p ∫0



=

∞  2 ∞ ( f ( x )sin sx )0 − s ∫ f ( x )cos sx dx  p  0 



= – sFc (s)

(iv) Fc(F′′(x)) =

∞

2 cos sx d [ f ′( x )] p ∫0 ∞  ∞ 2 { f ( x )cos sx } s f ′( x)sin sx dx  ′ +  ∫ 0 p  0 



=



= −



= − s 2 Fc ( s ) −

2 f ′(0) + sFs { f ′( x )} p 2 f ′ (0) p

assuming f(x), f ′(x) → 0 as x → ∞

=

∞  2  ∫ sin sx d [ f ′( x )] p  0 



=

∞  2 ∞ ( f ′( x )sin sx )0 − s ∫ f ′( x )cos sx dx  p   0



= – sFc {f ′(x)}



  2 = − s  sFs ( s) − f (0) p  



= –s2 Fs(s) +

assuming f(x), f′(x) → 0 as x → ∞.

2 sf (0) p

247

Fourier Transforms

Relationship between Fourier and Laplace Transforms Consider e − xt g(t)for t > 0 f(t) =  for t < 0 0



... (1)

Then the Fourier transform of f(t) is given by

F{f(t)} =



=



=



=



=

∞

1

∫e

ist

2 p −∞ ∞

1 2p

∫e

2p

∫e

2p 1 2p

( is − x )t

g(t) dt

0

∞

1

f (t) dt

0

∞

1

ist

f (t) dt

∫e

− pt

g(t)dt where p = x – is

0

L{ g(t)}

\ Fourier transform of f(t) =

1 2p

× Laplace transform of g(t) defined by (1).

APPLICATIONS TO BOUNDARY VALUE PROBLEMS ∂u ∂2u = K 2 , −∞ < x < ∞ , t > 0 with the conditions, Example 1. Solve the diffusion equation ∂t ∂x ∂u u(x, 0) = f(x), and , u tend to zero as x tend to ± ∞. ∂x Proof. Here, u = u(x, t), – ∞ < x < ∞, t > 0 Defining Fourier transform of u(x, t) as u (s, t) = i.e.,

1

∞

∫ u(x , t)e

2 p −∞

isx

dx , take Fourier transform of the given differential equation.

2  ∂ u   ∂u  KF  2  = F    ∂t   ∂x 

K (− s2 u( s, t)) =

1

∞

∂u

∫ e 2 p ∂t 0

d {F(u)} dt du = dt =

isx

dx

248

Engineering Mathematics-III (Third Semester)

du 2 + s ⋅ ku = 0 where u = Fourier transform of u dt

\

Solving (1), u (s, t) = ce − s

2

... (1)

kt

Since u(x, 0) = f(x), taking transform w.r.t. x. u (s, o) = F(s)



Using (3) in (2), we get, c = F(s) u (s, t) = F(s) e − s



2

kt

Taking inverse transform, −1 −s u(x, t) = F {F( s)e



2

kt

= f(x) * F–1 (e s = f(x) * 

(since

s2

F–1  e 4 a2   

−

kt

)

x2 4 e kt

2 kt

  = 2 ae − a2 x 2 by example 4)   

x−q 2 kt

4 pkt

∫

f (q) e

−

( x −q)2 4 kt dq

−∞

= f, we get q = x – 2 ktf u(x, t) =

\ Example 2. Solve

∞

1

=

Putting,

2

}

∂2u ∂t

2

= α2

∂2u ∂x 2

1

∞

∫

p −∞

2

f ( x − 2 kt f) e −f df

, – ∞ < x < ∞, t ≥ 0 with conditions u(x, 0) = f(x),

∂u ∂u ( x , 0) = g( x ) and assuming u, → 0 as x → ± ∞. ∂t ∂x Taking Fourier transform on both sides of the differential equation, d2u dt

2

=α 2 (− s2 u ) where u is Fourier transform of u with respect to x. d2u dt 2

+ α 2 s2 u = 0

Auxiliary equation is m2 + a2s2 = 0 m = ± ias \

u (s, t) = Aeiast + Be – iast

... (1)

249

Fourier Transforms

Since u(x, 0) = f(x) and u (s, 0) = F(s) and

∂u ( x , 0) = g(x), ∂t

du ( s, o) = G(s) on taking transforms. dt

Using these conditions in (1).

u (s, 0) = A + B = F(s) du ( s, 0) = ias(A – B) = G(s) dt

Solving

A=

G( s)  1 F ( s) +  iαs  2



B=

G( s)  1 F ( s) −  iαs  2

... (2) ... (3)

Using these values in (1), u (s, t) =

1 G( s)  iαst 1  G( s)  iαst F ( s) + e e +  F ( s) − iαs  iαs  2  2

... (4)

By inversion theorem, (4) reduce to,

u(x, t) =

1 1  f ( x − αt ) − 2  α

x −αt

∫

α

x +αt   1 1 ( ) g(q) dq  +  f ( x + αt) + g q d q  2  α α∫  

Using the result x  F ( s) F  ∫ f (t)dt  =   (−is) α  Boundary Value Problems using sine and cosine Transforms Example 3. Solve

∂u ∂ 2 u = for x > 0, t > 0 given that ∂t ∂x 2

(i) u(0, t) = 0 for t > 0 1 for 0 < x < 1 (ii) u(x, 0) =  and 0 for x ≥ 1 (iii) u(x, t) is bounded. ∂u (Note. If u at x = 0 is given, take Fourier sine transform and if at x = 0 is given, take ∂x Fourier cosine transform.) In this problem, u at x = 0 is given. Therefore, take fourier (infinite) sine transform.  ∂2u   ∂u  Fs   = Fs  2   ∂t   ∂x  ∞



2 ∂u sin sx dx = p ∫0 ∂t

∞

2 ∂2u sin sx dx p ∫0 ∂x 2

250

i.e.,

Engineering Mathematics-III (Third Semester)

du 2 = – s2 u + su (0, t) where u stands for Fourier sine transform of u. p dt

Using boundary condition (i), we get du 2 +s u = 0 dt

Solving

2

u( s, t) = ce − s t 1for 0 < x < 1 Now, u(x, 0) =  0 for x ≥ 1

... (iv)

Taking Fourier sine transform of this initial condition, 1

u (s, 0) =

2 sin sx dx p ∫0



2  1 − cos s    p s 

=

... (v)

Using (v) in (iv)

c=

2  1 − cos s    s p 

Hence (iv) reduces to, u (s, t) =

2  1 − cos s  − s2 t  ⋅e s p 

By iunversion theorem, this becomes, u(x, t) = Example 4. Solve

∞

2  1 − cos s  − s2 t ⋅ sin sx ds.  e s p ∫0  

∂u ∂2u = K 2 for 0 ≤ x < ∞, t > 0 given the conditions ∂t ∂x

(i) u(x, 0) = 0 for x ≥ 0

∂u (ii) (0, t) = – a constant ∂x

(iii) u(x, t) is bounded. In this problem,

∂u at x = 0 is given. Hence, take Fourier cosine transform on both sides of ∂x

the given equation.

 ∂2u   ∂u  Fc   = Fc  K 2   ∂t   ∂x 

251

Fourier Transforms

  du 2 ∂u ⋅ (0, t)  = K  − s2 u dt p ∂x  



2 = −ks2 u =ka using condition (ii) p du + ks2 u = dt



2 ka p

This is linear in u . Therefore, solving ue ks



2

t

=

2 ks2 t kae dt p

∫

2



u (s, t) =



2 e ks t ka +c p ks2

=

2 2 a + ce − ks t 2 ps

... (iv)

Since u(x, 0) = 0 for x ≥ 0.

u (s, 0) = 0 for x ≥ 0,



u (s,0) = 0.

Using this in (iv), we get

u (s,0) = c + c= −

∴

2 a 0 = p s2 2 a p s2

Substituting this in (iv) u (s, t) =

(

s 2 a 1 − e−k t p s2

)

By inversion theorem, ∞

u(x, t) =

Example 5. Solve

∂u ∂2u = k 2 for x ≥ 0, t ≥ 0 under the given conditions u = u0 at x = 0, t > dt ∂x

0 with initial condition u(x, 0) = 0, x ≥ 0 Taking Fourier sine transforms

2

2 1 − e ks t ⋅ a∫ cos sx ds. p 0 s2

 ∂2u   ∂u  Fs   = Fs  k 2   ∂t   ∂x 

252

Engineering Mathematics-III (Third Semester)

  d 2 u = k  − s2 u + su(0, t) du p  

= – ks2 u +



2 ksu0 where u is the Fourier sine transform of u. p du + ks2 u = dt



2 ksu0 p

This is linear in u . ue ks

\

2

t



2 2 ku0 ∫ s e kx t dt p

=

2 u0 ks2 t e +c p s

=

... (1)

Since, u(x, 0) = 0, u (s, 0) = 0. Using this in (1) 2 u0 +c p s

0 = \ c = − 2



e ks t u =

\

u =

2 u0 p s

( ) 2u 1− e ) p s ( 2 u0 ks2 t e −1 p s 0

ks2 t

By inversion theorem, 2 ∞ 2u0  1 − e − ks t  u(x, t) = s p ∫0  



  sin sx ds.  

Exercises 4(b)

1. Show that the solution of

∂u ∂2u =2 2. ∂t ∂x

subject to u(0, t) = 0, for t > 0 and u(x, 0) = e–x for x > 0 and u(x, t) is bounded, is 2

∞

2 se −2 s t sin sx ds p ∫ 1 + s2

2. Solve



(i)

∂u ∂ 2 u = if ∂t ∂x 2

∂u (0, t) = 0 for t > 0. ∂t

 x for 0 ≤ x ≤ 1 (ii) u(x, 0) =  0 for x > 1

253

Fourier Transforms



(iii) and u(x, t) is bounded for x > 0, t > 0.



∞   2  sin s cos s − 1  − s2 t u( x , t) cos sxds  = + Ans.  e ∫ 2 p 0 s s   

FINITE FOURIER TRANSFORMS Let f(x) denotes a function which is sectionally continous over the range (0, l). Then the finite Fourier sine transform of f(x) on this interval is defined as l

Fs(p) = f s ( p) = ∫ f ( x )sin 0

ppx dx l

where p is an integer (instead of s, we take p as a parameter). Inversion formula for sine transform It fs(p) = Fs(p) is the finite Fourier sine transform of f(x) in (0, t) then the inversion formula for sine transform is

f(x) =

ppx 2 ∞ f s ( p)sin ∑ l p =1 l

Proof. For the given function f(x) in (0, l), if we find the half range Fourier sine series, we get, f(x) =

∞

∑ bn sin

n =1

npx l

... (1)

l

where bn =

2 npx f ( x )sin dx l ∫0 l

\ bp =

ppx 2 f ( x )sin dx ∫ l0 l

=

2 f s ( p) by definition l

l

Substituting in (1), \ f(x) =

ppx 2 ∞ ∑ fs (p)sin l l p =1

Finite Fourier Cosine Transform Let f(x) denote a sectionally contionous function in (0, l). Then the Finite Fourier cosine transform of f(x) over (0, l) is defined as

fc(p) = fc ( p) =

l

∫ f (x)cos 0

ppx dx where p is an integer. l

Inversion Formula for Cosine Transform If f s ( p) is the finite Fourier cosine transform of f(x) in (0, l), then the inversion formula for cosine transform is

254

Engineering Mathematics-III (Third Semester)



f(x) =

where

fc (0) =

ppx 1 2 ∞ fc (0) + ∑ fc ( p)cos l l p =l l l

∫ f (x) dx. 0

Proof. If we find half Fourier cosine series for f(x) in (0, l), we obtain,

f(x) =

a0 ∞ npx + ∑ an cos 2 n =1 l



an =

2 npx f ( x )cos dx ∫ l0 l

\

ap =

2 fc ( p) l



a0 =

=

l

l

2 f ( x ) dx l ∫0 2 fc (0) . l

Substituting in (2), we get,

f(x) =

ppx 1 2 ∞ fc (0) + ∑ fc ( p)cos l l p =1 l

Example 1. Find the finite Fourier sine and cosine transforms of

(i) f(x) = 1 in (0, p)



(ii) f(x) = x in (0, l)



(iii) f(x) = x2 in (0, l)



(iv) f(x) = 1 in 0, < x < p/2



= – 1 in p/2 < x < p



(v) f(x) = x3 in (0, l)



(vi) f(x) = eax in (0, l)

(i) f s (p) = Fs(1) =

p

∫ 1 ⋅ sin 0

ppx dx p

p



 cos px  = −  p 0 



=

1 − cos pp if p ≠ 0 p p



fc ( p) =

∫ 1 ⋅ cos px dx 0

... (2) where

255

Fourier Transforms p



 sin px  1 (0 − 0)= 0 =   = p  0 p l

∫ x sin

(ii) f= s ( p) F= s ( p)

0

ppx dx l l



  ppx   l cos − = ( x )    pp   l 



=

−1 (1cos pp) pp



=

−l 2 (−1)p if p ≠ 0 pp

fc ( p) = Fc(x) =

1

 ppx     sin   l  − (1)  − 2 2    p p      l2  0

∫ x cos 0

ppx dx l l





   ppx  ppx     sin  cos   l − (1)  − l  = ( x )     pp   p 2 p2        l  l2  0  =

l2 2 2

p p

[(−1)p − 1] if p ≠ 0

(iii) f s (p) = Fs(x2) =

l

∫x

2

0

sin

ppx dx l l



   ppx  ppx   ppx     cos  sin   cos    l  l − (2 x )  − l   + (2)  = ( x 2 )  −  pp     p 3 p3   p 2 p2           l   l2    l3 0 



= fc ( p) =

−l 3 2l 3 [(−1)p ] + 3 3 [(−1)p − 1] if p ≠ 0 pp p p l

∫ (x 0

2

)cos

ppx dx l

256

Engineering Mathematics-III (Third Semester)

   ppx  ppx  ppx     sin  cos   sin    l − (2 x )  − l  + (2)  l  = ( x 2 )      p 3 p3   p 2 p2   pp       l   l2    l3  





2l 3

=

2 2

p p p/2

∫

(iv) Fs {f(x)} =

p

∫

sin px dx +

(−1)sin px dx

p/2 p/2

p



 cos px  = −  p 0 



pp  1 1 − 1  + (cos pp − cos pp / 2) = −  cos 2 p  p



=

p/2

∫

p

cos px dx − p/2

 sin px  =    p 0

∫x 0

3

sin

∫

cos px dx

p/2

0

1

 cos px  +   p 2 x

pp  1 cos pp − −2 cos + 1  if p ≠ 0  p 2 

Fc(f(x)) =

(v) Fs(x3) =

0

[(−1)p ] if p ≠ 0

0



l

p

 sin px  pp 2 sin − = if p ≠ 0  p p 2  p/2

ppx dx l l





    ppx  ppx   ppx  ppx      cos sin cos sin        l  l  − (3 x 2 )  − l   + (6 x )  l  –(6)  − = ( x 3 )  −  pp     p 3 p3   p 4 p4   p 2 p2             l   l2 l4    l3   0  = −

Fc(x3) =

l

l4  6l 4 p n ( − 1) +  p 3 p3 (−1) if p ≠ 0 pp 

∫x 0

3

cos

ppx dx l

   ppx  ppx  ppx  ppx      cos   sin   cos   3  sin l  2  l  + 6( x )  l  −(6)  l  − − (3 ) x = ( x )     p 3 p3   p 4 p4   p 2 p2    pp         l   l2    l3   l4 

l

0

257

Fourier Transforms

=

(vi) Fs (eax) =

l

∫e

3l 4 p2 p

ax

(−1)p − 2

sin

0

ppx dx l



  e ax  =  2 2  a2 + p p  l2



=

l

 ppx pp ppx     a sin l − l cos l     0

e ax

1  pp   pp  ;  − (−1)p  +  2 2  p a  l  2 p p  l  a2 + 2 a + 2 l l 2 2

  e ax  Fc(eax) =  2 2  a2 + p p  l2

6l 4  (−1)p − 1 if p ≠ 0 4 4  p p

e al

=

a2 +

2 2

p p

l

 ppx pp ppx     a cos l + l sin l     0 1

a(−1)p − a2 +

l2

p 2 p2

( a)

l2

Note. In all the above problems, if p = 0, do the integration separately Example 2. Find f(x) if its finite Fourier sine transform is

2p p3

(−1) p −1 for p = 1, 2, ...., 0 < x < π.

By inversion Theorem, f(x) =

2 ∞ 2p ∑ (−1)p−1 sin px p p =1 p 3 ∞

= 4 ∑

p =1

(−1)p −1 p3

sin px

Example 3. Find f(x) if its finitie Fourier sine transform is given by (i) Fs(p) = (ii) Fs(p) =

1 − cos pp p 2 p2 16(−1)p −1 p3

for p = 1, 2, 3, ...... and 0 < x < p for p = 1, 2, 3, ..... where 0 < x < 8

2 pp 3 for p = 1, 2, 3, ..... and 0 < x < 1. (iii) Fs(p) = (2 p + 1)2 cos

258

Engineering Mathematics-III (Third Semester)

Solution. By inversion theorem (i) f(x) =

2 ∞  1 − cos pp   ⋅ sin px ∑ p p =1  p 2 p2 



=

 1 − cos pp   ⋅ sin px p2 p p =1  

(ii) f(x) =

2 ∞  ppx  Fs ( p)sin  ∑  l p =1  l 



=

2 ∞ 16(−1)p −1  ppx  sin  ∑  since l = 8 3 8 p =1 p  8 



= 4∑

p =1



∑ 

3

∞

(iii) f(x) =

∞

2

(−1)p −1 p

3

 ppx  sin    8 

2 ∞  ppx  Fs ( p)sin  ∑  l p =1  l 

 2 pp  cos    3  sin (ppx) since l = 1 = 2∑ 2 p =1 (2 p + 1) ∞

Example 4. Find f(x) if its finite Fourier cosine transform is (i) Fc(p) =

1  pp  sin   for p = 1, 2, 3..... 2p  2 



p for p = 0 given 0 < x < 2p 4

=

(ii) Fc(p) =

=

pp − cos pp 2 for p = 1, 2, 3, ..... (2 p + 1)p

6 sin

2 for p = 0 given 0 < x < 4 p

 2 pp  cos    3  for p = 1, 2, 3, ..... (iii) Fc(p) = (2 p + 1)2

= 1 for p = 0 given 0 < x < 1

Solution: By inversion theorem, (i) Here

f(x) =

ppx 1 2 ∞ . Fc (0) + ∑ Fc ( p) ⋅ cos l l p =1 l

Fc(0) = p/4 and l = 2p

259

Fourier Transforms



=

(ii) Here, Fc(0) =



1 p 2 ∞ 1  pp   ppx  ∑ sin  2  cos  2p   + 2 p  4  2 p p =1 2 p

f(x) =

\

1 1 ∞ 1  pp   px  sin   cos   . + ∑ 8 2 p p =1 p  2   2 

2 and l = 4 p

pp   6 sin − cos pp   12 2 2  cos  ppx  f(x) =   + ∑   4  4  p  4 p =1 (2 p + 1)p   ∞



pp   6 sin − cos pp   1 1 2   ⋅ cos  ppx  + = ∑   2 p 2 p p =1 (2 p + 1)  4  ∞



(iii) Here Fc(0) = 1, l = 1 \ f(x) =

1 2 ∞ 1  2 pp  + ∑ cos   ⋅ cos( ppx) 1 1 p =1 (2 p + 1)2  3 

 2 pp  cos    3  cos( ppx) = 1 + 2∑ 2 p =1 (2 p + 1) ∞



Example 5. Find the finite Fourier sine transform of in f(x) = 1 (0, p). Use the inversion theorem and find Fourier sine series for f(x) = 1 in (0, p) Hence prove (i) 1 −

1 1 1 + − + .... = p/ 4 3 5 7

Solution. Fs(1) = f s ( p) =

p

(ii)

1 2

1

+

1 3

2

+

1 52

+ .... = p2 / 8

 ppx  dx p 

∫0 1 ⋅ sin 

1 − cos pp if p ≠ 0 p

By inversion theorem, f(x) =

ppx 2 ∞ Fs ( p)sin ∑ l p =1 l



1=

2 ∞ 1 − (−1)p ∑ p ⋅ sin px since l = p p p =1



1=

4 1 1  sin x + sin 3 x + sin 5 x + ..... 3 5 p  

This is the half range Fourier sine series for f(x) = 1 in (0, p) getting x = p/2.

4 1 1 1  1 − + − + ..... = 1 p  3 5 7 

260

Engineering Mathematics-III (Third Semester)

\

1−

1 1 1 + − + ..... = p/4 3 5 7

In the half Fourier sine series ln =

4 1 ⋅ for n odd p n

By using Parseval’s Theorem.

1  (range)  ∑ bn2  = 2  

p

∫0 (1)

2

dx

 1 16 ∞ 1 p ⋅ 2 ∑ =p 2  2 p n =1,3,5... n  i.e.,

∞

∑

n =1,3,5

1 n2

=

p2 8

Example 6. Find f(x) if its finite Fourier cosine transform is

and is

2l 3 2 2

p p

(−1)p for p = 1, 2, 3, ....

l3 for p = 0; 0 < x < l. 3

Proof: By inversion formula,

f(x) =

1 2 ∞  ppx  fc (0) + ∑ fc ( p)cos   l l p =1  l 

=

ppx 1 l 3 2 ∞ 2l 3 (−1)p cos ⋅ + ∑ 2 2 l 3 l p =1 p p l

=

l 2 4l 2 + 3 p2

∞

∑

(−1)p

p =1

p

2

 ppx  cos    l 

Exercise 4 (c) Find the finite Fourier sine and cosine transforms of

1. f(x) = 2x in (0, 4) 2. f(x) = x in (0, p)



3. f(x) = cos ax in (0, p) 4. f(x) = 1 –



 x in (0, p / 2) 5. f(x) =  6. f(x) = e–ax in (0, l) p − x in (p / 2, p)



x  7. Find finite Fourier cosine transform of  1 −  , 0 < x < p p 



x in (0, p) p

2

 pp  sin    2  , p = 1, 2, 3..... 8. Find f(x) if fc ( p) = 2p

(BR 1995 April)

261

Fourier Transforms



and =

p if p = 0 given 0 < x < 2p. 4

9. Find finite Fourier cosine and sine transform of f(x) =

p x2 −x+ . 3 2p

(BR 1995 April)

Finite Fourier sine and cosine transform of derivatives Using the definition and the integration by parts, we can easily prove the following results. For 0 ≤ x ≤ 1. Fs {f ′′(x)} = − Fc{f′′(x)} = − Fs {f′(x)} = −

p 2 p2 l

f s ( p) +

2

2 2

p p l2

pp  f (0) − (−1)p f (l)  l 

fc ( p) + f ′ (l)(−1)p − f ′(0)

pp fc ( p) l

Fc {f′(x)} = f(l) (– 1)p – f(0) + Proof: (i) Fs {f′(x)} =

l

∫ f ′(x)sin 0



=

l

∫ sin 0

pp f s ( p) l

ppx dx l

ppx ⋅ d[ f ( x )] l l

l



ppx  ppx pp  − ∫ f ( x ) ⋅ cos ⋅ dx =  f ( x )sin  l 0 0 l l 



= −

(ii) Fc {f ′(x)} =

l

pp fc ( p) l

∫ f ′(x)cos 0

ppx dx l l

l



ppx  pp ppx  − ∫ f ( x ) ⋅ sin dx =  f ( x )cos  l 0 0 l l 



= (– 1)p f(l) – f(0) +

(iii) Fs {f′′(x)} =

l

∫ sin 0

ppx d[ f ′ ( x )] l l



pp f s ( p) l

ppx  pp l ′ ppx  − f ( x )cos dx =  f ′ ( x )sin  ∫ l 0 l 0 l 

262

Engineering Mathematics-III (Third Semester)



= −



= −

(iv) Fc{f′′(x)} =

pp  pp  (−1)p f (l) − f (0) + f s ( p) l  l  p 2 p2 l

2

l

∫ cos 0

f s ( p) +

pp  f (0) − (−1)n f (1)  l 

ppx d[ f ′( x )] l l

ppx  pp  + =  f ′ ( x )cos  l 0 l 



= (– 1)p f′(l) – f′(0) +



= −

p 2 p2 l

2

l

∫ f ′(x)sin 0

ppx dx l

pp  pp  fc ( p) l  l 

fc ( p) + f ′(1)(−1)n − f ′(0)

Note. If u = u(x, t), then pp  ∂u  Fs   = − Fc (u) l  ∂x 



pp  ∂u  Fc   = Fs(u) – u(0, t) + (– 1)p u(l, t) x ∂ l    ∂2u  p 2 p2 pp Fs  2  = − 2 Fs (u) + [u(0, t) − (−1)p u(l , t)] l l  ∂x   ∂2u  p 2 p2 ∂u ∂u Fc  2  = − 2 Fc (u) + (l , t)cos pp − (0, t) ∂x ∂x l  ∂x  Example 1. Using finite Fourier transform, solve ∂u ∂ 2 u = given u(0, t) = 0 and u(4, t) = 0 ∂t ∂x 2 and u(x,0) = 2x where 0 < x < 4, t > 0 Proof. Since u(0, t) is given, take finite fourier sine transform. 4

4

ppx ppx ∂u ∂2u ∫ ∂t sin 4 dx = ∫ ∂x 2 sin 4 dx 0 0

 ∂2u  d us = Fs  2  dt  ∂x 



= −

p 2 p2 pp us + [u(0, t) − (−1)p u(4, t)] 16 4



= −

p 2 p2 us using u(0, t) = 0, u(4, t) = 0 16

263

Fourier Transforms

p 2 p2 dus dt = − 16 us

Integrating

log us =

p 2 p2 t+c 16

us = Ae



−

x 2 p2 t 16

since u(x, 0) = 2x

us (p, 0) =

4

 ppx  dx 4 

∫ (2 x)sin  0



= −

32 cos pp pp

... (2)

Using (2) in (1), us = (p, 0) = A = −



32 cos pp. pp

Substituting in (1), 2 2

p p 32 n − us = − (−1) e 16 np

\

t

By inversion theorem, − 2 ∞ 32 u(x, t) = ∑ (−1)p +1 e 4 p =1 p p



Example 2. Solve Proof. Since



6

 ppx  sin    4 

∂u ∂u ∂u ∂ 2 u = , 0 < x < 6, t > 0, given (0, t) = 0, (6, t) = 0 and u(x, 0) = 2x ∂x ∂t ∂x 2 ∂x

∂u (0, t) is given, use finite Fourier cosine transform. ∂x

ppx ∂u dx = cos ∂t 6 0

∫

p 2 p2 t 16

6

∂2u

∫ ∂x 2 cos 0

ppx dx 6

p 2 p2 d ∂u ∂u uc = − uc + (6, t)cos pp − (0, t) dt 36 ∂x ∂x = −

p 2 p2 uc 36

p 2 p2 duc dt = − 36 uc

264

Engineering Mathematics-III (Third Semester)

log uc = −

p 2 p2 t+c 36

uc = Ae



−

p 2 p2 t 36



... (1)

u(x, 0) = 2x. t=0

\ At

uc (p, 0) =

6

∫ (2 x)cos 0



=

72

ppx dx 6

(cos pp − 1)

p 2 p2

... (2)

Using this in (1), 72

uc (p, 0) = A =

p 2 p2

(cos pp − 1)

Substituting in (1), uc (p, t) =

72 p 2 p2

(cos pp − 1)e

−

p 2 p2 t 36

By inversion theorem, u(x, t) =

1 2 ∞  ppx  fc (0) + ∑ fc ( p)cos   l l p =1  l  6

1 2 = ∫ (2 x ) dx + 60 6



= 6+



Example 3. Solve

24 x

2

∞

∑

p =1

∞

∑

p =1

72 2 2

p p

(cos pp − 1) p

2

e

−

(cos pp − 1)e p 2 p2 t 36

−

p 2 p2 t 36

 ppx  ⋅ cos    6 

 ppx  cos  .  6 

∂u ∂2u = 2 2 , 0 < x < 4, t > 0 given u(0, t) = 0, u(4, t) = 0; u(x, 0) = 3 sin ∂t ∂x

px –2 sin 5px. Proof. Since u(0, t) is given, take finite Fourier sine transform. The equation becomes (as in example 1)  p 2 p2  pp d uc = 2  − us + {u(0, t) − (−1)p u(4, t)}  dt 4  16 







= −

p 2 p2 u, 8

265

Fourier Transforms

Solving we get, us = Ae



−

p 2 p2 t 8

u(x, 0) = 3 sin px – 2 sin 5px Taking sine transform, us (p, 0) =

4

∫ (3 sin px − 2 sin 5px)sin 0

ppx dx 4

= 0 if p ≠ 4, p ≠ 20.

If

p = 4, us (4, 0) = 6

If

p = 20, us (20, 0) = – 4 u(x, t) =

2 ∞  ppx  ∑ us (p, t)sin  4  4 p =1

p 2 p2  p 2 p2  − t 1 − 8 t 6e sin px − 4 e 8 sin 5 px  =  2  



where p in the first term is 4 and p in the second term is 20 2

2

= 3 e 2 p t sin px − 2 e −50 p t sin 5 px.

Example 4. Solve

∂u ∂ 2 u = , 0 < x < p, t > 0 given u(0, t) = u (x, t) = 0 for t > 0 and u(x, 0) ∂t ∂x 2

= sin3 x. Solution. Since u(0, t) and u(p, t) are given, take finite Fourier sine transforms on both sides of the given equation with respect to x in (0, p). p ∂u

∫0

∂t

sin

ppx dx = p

p∂

2

u

∫0 ∂x 2 sin

ppx dx p

 ∂2u  d us = Fs  2  dt  ∂x 

= – p2 us + p[u(0, t) – (– 1)p u(p, t)]



= – p2 us using u(0, t) = u(p, t) = 0.



dus = – p2dt us

Integrating \

log( us ) = – p2t + k us (p, t) = ce − pt

... (1)

266

Engineering Mathematics-III (Third Semester)

Since, u(x, 0) = sin3 x

p

∫0 sin

us (p, 0) =

p

3

x sin px dx

3

1



∫0  4 sin x − 4 sin 3 x  sin px dx



=



= 0 for p ≠ 1, p ≠ 3 p = 1, us (1, 0) =

when

p./2



= 2∫



3 1  = 2  ⋅ ⋅ p/ 2  4 2  



=

when

p = 3, us (3, 0) =

0

p

∫0 sin

4

xdx

sin 4 x dx

3 p 8 p

... (2) 1

∫0  − 4 sin

2

 3 x  dx 

1 p 1 − cos 6 x dx 4 ∫0 2



= −



1 sin 6 x  = − x − 8 6  0



= –p/8

p

Using (2) and (3) in (1),

... (3) 3p 8

when

p = 1, us (1, 0) = C =

when

p = 3, us (3, 0) = C = – p/8

For all other values of p, c = 0 By inversion formula, u(x, t) = i.e.,

u(x, t) =

Example 5. Solve

=

2 ∞ ∑ us (p, t)sin nx p p =1 2  3p −t p  e sin x − e −9t sin 3 x   8 p 8  3 −t 1 3 sin x − e −9t sin 3 x 4 4

∂u ∂ 2 u = , 0 < x < 10 given u(0, t) = u(10, t) = 0 for t > 0 and u(x, 0) = ∂t ∂x 2

10 x – x2 for 0 < x < 10. Solution. As the previous problem, take finite Fourier sine transform on both sides of the heat equation with respect to x in (0, 10). \

p 2 p2 pp d us + [u(0, t) − (−1)p u(10, t)] us = − 100 10 dt

267

Fourier Transforms



= −

p 2 p2 u , using u(0, t) = u(10, t) = 0 100

Solving for us , we get us = C e

− p 2 p2 t 100

... (1)

Since u(x, 0) = 10x – x2,

10

∫0

us (p, 0) =

(10 x − x 2 )sin

ppx dx 10 10



   cos ppx   ppx  ppx   cos sin −        10  − (10 − 2 x )  − 10  + (−2)  10   = (10 x − x 2 )    pp  p 2 p2   p 3 p3        1000      10 100  0



= −



 4000 if p is odd  =  p 3 p3 0 if p is even 

2000 3 3

D p

[(−1)p − 1]

Using (1), us (p, 0) = C.. using this again in (1), C is eliminated. Hence, taking inversion formula, use (1) u(x, t) =

=

ppx 2 us ⋅ sin ∑ 10 10 800 p3

∞

ppx ∑ p3 sin 10 e p =1,3,5 1

− p2 p2 t 100

Exercises 4 (d)

1. Solve

2

∂u ∂ u = , 0 < x < 6, t > 0 given that ∂t ∂x 2

1for 0 < x < 3  u(0, t) = u(6, t) and u(x, 0) =   0 for 3 < x < 6    2 Ans. u( x , t) = p  

2. Solve

pp  2 2     1 − cos 2  − p p t p p x   ∑  p  e 36 sin  6    p =1      p

∂y ∂ 2 y = subject to conditions v(0, t) = 1, v(p, t) = 3 ∂t ∂x 2

268

Engineering Mathematics-III (Third Semester)

 4 ∞ cos pp − p2 t 2x  = e sin px + 1 + . Ans. v( x , t) v(x, 0) = 1 for 0 < x < p, t > 0. ∑ p p =1 p p    ∂q ∂ 2 q = 3. Solve given ∂t ∂x 2 q(0, t) = 0, q(p, t) = 0, q(x, 0) = 2x for 0 < x < p, t > 0

4. Solve

∂u ∂u ∂u ∂ 2 u (1, t ) = 0 for t > 0 and u(x, = 2 for 0 < x < l, t > 0 given (0, t) = 0 ∂x ∂t ∂x ∂x  l2 l2 Ans. u( x , t= ) −  6 p2 

0) = 1x – x for 0 < x < l 2



5. Solve

6. Solve



2 ppx ∑ p2 cos l e 1

−4 p 2 p2 t l2

   

∂u ∂2u = 2 2 for 0 < x < 1, t > 0 given u(0, t) = u (l, t) = 0 for t > 0 and u(x, 0) ∂t ∂x 2 2 Ans.= u( x , t) e −2 p t sin px + e −18 p t sin 3 px   

= sin 3px + sin px.

[Refer example 3]

∂u ∂2u = α 2 2 subject to ∂t ∂x

u(0, t) = u (1, t)0 for t > 0 and u(x, 0) = x for 0 ≤ x ≤ l/2 = l – x for l/2 ≤ x = < 1  Ans. u( x , t) = 4l  p2 



pp ppx ∑ p2 sin 2 sin l e 1

−α 2 p 2 p2 t t2

   

SHORT ANSWER QUESTIONS

1. Define Periodic function.



2. If f(x) is expressed in Fourier series of period 2p in (c, c + 2p) write down the Fourier series and formulae for Euler’s Coefficients.



3. If f(x) is expressed in Fourier Series of period 2l in (c, c + 2l) write down the Fourier series and forumulae for Euler’s Coefficients.



4. If f(x) is expressed in half range Fourier cosine series or sine series in (0, l) of periodicity 2l, write down two series and the corresponding integrals for coefficients.



5. State Dirichlet’s conditions.



6. Draw the graph of f(x) = | x |



7. Define | x | in (– ∞, ∞)



8. Find half range Fourier Cosine series or sine series in (0, p) of periodicity 2p if f(x) = k.



9. Define odd and even functions.



10. State any property you know regarding

l

∫ f (x) dx

−l

if f(x) is odd or even.

269

Fourier Transforms



11. Evaluate

1

∫ (1−| x|)

cos n p x dx, n any integer.

−1



12. Find the half range since series in 0 < x < p if f(x) = x2.



13. Find the half range cosine series in 0 < x < p if f(x) = x + 1.



14. Write down the analytic expression for f(x) if y = f(x) has the graph. Y

P (a, d)

O

X (I, o)



15. Define root mean square value of f(x) over the range (a, b)



16. Using the R.M.S. value fill up the blanks in terms of Fourier coefficients



17. Write down the complex Fourier series, stating the formula for coefficient Cn. Write true or false, with reason, if possible, (18 to 19).



18. In (–l, l), f(x) = (1 – | x |) sin nx is odd.

c+2p

∫

[ f ( x )]2 dx = .....

c

19.

c+2p

∫

sin mx cos nx dx = 0, for all integral values of m, n.

c

20. Find the Fourier series of periodicity of 2p for f(x) = x in – p < x < p. 21. f(x) = 1 +

=1–

2x in p < x < 0 p 2x in 0 < x < p is even. State true or false. p



22. If fourier series is found out for f(x) = x2 in – 1 < x < 1, the series will contain only cosine terms. State true or false.



23. If f(x) = x3 in – p < x < p, find the constant terms of its Fourier series.



24. If f(x) = sin ax, in – p < x < p, a not an integer, find the constant terms of its Fourier expansion.



25. If f(x) = x2 in (–l, l) is expressed as a Fourier series of periodity 2l, find the constant term of the series.

By eliminating the arbitrary constants, form the differential equations (26 to 30).

26. z = (x2 + a) (y2 + b)



27.



28. z = ax + by + ab



29. (x – a)2 + (y – b)2 + z2 = 1

x y z + + = 1 a a c

270



30.

Engineering Mathematics-III (Third Semester)

x2 a2

+

y2 b2

+

z2 c2

=1

By eliminating the arbitrary functions from the partial differential equation (31 to 37).

31. z = F(x2 + y2)



32. z = f(x2 + y2 + z2)



33. xyz = f(x + y + z)



 xy  34. z = f    z 



35. f(x2 + y2 + z2, xyz) = 0



36. z = ey f(x + y)



37. z = f(2x + 3y) + F(y + 2x)



38. Find the partial differential equation of all spheres whose centres lie on the z-axis.



39. Find the partial differential equation of all spheres of radius K with centres on the xy plane.



40. Define complete solution of a.p.d.e.



41. If



42. Solve



43. Solve for z ⋅

∂2 z = sin x find z ∂x∂y ∂2 z ∂x 2

= xy ∂z ∂z = 3x − y; = cos y – x. ∂x ∂y

Solve for z. (44 to 54)

44.



45. z = px + qy +



46. p = 2qx



47. p + q = x + y



48. p + q = pq



49. q + sin p = 0



50. (z – px – qy) (p + q) = 1



51. (y – z)p + (z – x)q = x – y



52. px2 + qy2 = (x + y)z



53. x(y – z)p + y(z – x)q = z(x – y)



54. px + qy = nz

p+ q = 1 1 + p2 + q2

Find the complimentary function of the following Partial differential equation.

55. (D2 – 6DD′ + 5D′2)z = 0

271

Fourier Transforms



56. (D3 – 7DD′2 – 6D′3)z = x2y



57. (D2 – DD′)z = cos x cos 2y



58. (D2 –D′2)z = ex+2y

Find the Particular integrals of the following

59. (D2 – D′2)z = ex+2y



60. (D2 – D′2)z = ex + y



61. Evaluate



62. Solve: (2D2 + 5DD′ + 2D′2) z = 0

ex+2y 3

D − 3D 2 D′ + 4D′ 3

Write True or false

63. If (D – D′) (D + D′)z = 0, the general solution is

z = f1(y + x) + f2(y – x)

64. If (D2 – 2DD′ + D′2)z = 0, its general solution is z = f1(y + x) + xf2(y + x).



65. Solve 2xp – 3yq = 0 by the method of separation of variables.



66. Write down one dimensional wave equation.



67. What are the various solutions of



68. a tighly stretched string with fixed end points x = 0 and x = l is initially at rest in equilibrium position. If it is set vibrating giving each point a velocity 3x(l – x). Write down the boundary conditions of the problem.



69. Write the analytic expression for the functions given by graph y

∂2 y ∂t 2

= a2

∂2 y ∂x 2

.

l , d 3

D B

A

C

(l, 0)

x

E

2l , – d 3



70. Write down the partial differential equation that governs the temperature flow in two dimension (cartesian)



71. Write down the steady-state differential equation that governs the temperature function. Write also the various types of solutions you get in solving it.



72. A rectangular plate is bounded by x = 0, y = 0, x = a and y = b. Its surfaces are insulated and temperature along two edges x = a and y = b are 100°C while the temperature along the other edges are 0°C. Write down the Boundary conditions in mathematical form.



73. Express

∂2u ∂x 2

+

∂2u ∂y 2

in its equilvalent in polar form.

272

Engineering Mathematics-III (Third Semester)



74. What are the solutions of the equation ∇2u = 0 in polar form?



75. Evaluate



76. A semi-circular plate of radius a is kept at temperature u0 along the bounding diameter and at u1 along the circumference. In solving the above problem for u(r, q) in steadystate; write down the boundary conditions in u. Further, making a transformation u(r, q) = v(r, q) + u0 what are the boundary conditions for v(r, q)



77. Define steady-state solution.



78. An infinitely long plane uniform plate is bounded by edges x = 0, x = l and an end at right angles to them. The breadth of this edge y = 0 is l and is maintained at f(x). All the other three edges, are kept at 0°C. Write down the boundary conditions in mathematical form.



79. In the definition of Fourier transform, F(s) =



80. Define infinite complex Fourier transform of f(x) and write its inversion formula also.



81. State Fourier integral theorem.



82. State inversion theorem for complex Fourier Transform.



83. Prove Fourier Transform is linear.



84. If F{f(x)} = F(s), F{f(x – a)} = .....



85. If F{f(x)} = F(s), F{f(ax)} = .....



86. F{eiax f(x)} = ......



87. F{f(x) cos ax} =



88. F{(xn f(x)} = .....



89. F{f′(x)} = ..... if f(x) – > 0 as x → ± ∞.



90. Find Fourier transform of

p

400 (pq − q2 ) sin n q dq; n positive integer. p 0

∫

1

∞

∫

2 p −∞

1 [ + ] (modulation theorem). 2

1if| x|< a f (x) =  0 if| x|> a > 0.

91. Define convolution of two functions of f(x) and g(x).



92. Prove convolution of two functions is commutative.



93. F–1{F(s) G(s)} = .....



94. State convolution Theorem.



95. State parseval’s identity.



96. Define infinite Fourier sine and cosine Transform.



97. Write the corresponding inversion forms.



98. Find Fourier sine and cosine transform of e–x, x ≥ 0,



99. Find Fourier sine transform of

1 . x

f ( x ) e isx dx what is the kernel?

273

Fourier Transforms



∞

∫ f (x) cos a x = e–a.

100. Solve f(x) if

0

1 − x2 2 e



101. The Fourier transform of



102. Define Finite Fourier sine and cosine transform of f (x).



103. Write down the corresponding inversion results.



104. Find fourier sine and cosine transforms of f ′(x).



105. Find Fourier sine and cosine transforms of f ′′(x).



106. Find finite Fourier sine and cosine transform of

is.....

(i) f(x) = x in (0 , p) (ii) f(x) = x2 in (0, l) (iii) f(x) = 1 −

x in (0, p) p

(iv) f(x) = 1 in (0, p)



107. If Fs(p) =

2p p

3

(−1)p −1 for p = 1, 2, 3... 0 ≤ x ≤ p find f(x) pp 2 if p = 1, 2, 3.... 2p

sin

108. Find f(x), if Fc(p) =



p if p = 0 in 0 < x < 2p 4

=

pp fc ( p) (finite) l



109. Prove Fs{f ′(x)} = −



110. Prove Fc (f ′′(x)) = −



111. Prove Fc{f′(x)} = f(l) (–1)p – f(0) +



112. Prove Fs(f ′′ (x)) =



113. Find Fourier cosine and sine transform e–x of



114. Fc{f(ax)} = .....



115. Fs{f(ax)} = .....



116. If F{f(x)} = F(s) than F [f(x – a)] = .....



117. F [e–|x|] (i) does not exist.

p 2 p2 l2

− p 2 p2 l

2

fc ( p) + f ′(l)(−1)p − f ′(0)

f s (p) +

pp f s ( p) (finite). l pp [ f (0) − (−1)p f (1)] l

274

Engineering Mathematics-III (Third Semester)

(ii) is (iv) is

2 1+ s

(iii) is

2

(v) is –

s 1+ s

1

2

1 + s2 2 1 + s2

State true or false:

118. Dirchlet’s conditions are necessary for the uniform convergence of Fourier series.



119. When the number of independent variables is equal to the number of arbitary constants, elimination of the latter leads to first order partial differential equation?



120. One dimensional heat conduction equation is

∂u ∂ 2 u + = 0. ∂t ∂x 2

121. Two dimensional Laplace’s equation in Cartesian coordinate is





∂2u ∂x 2

∂2u

+

= 1.

∂y 2

122. F{f * g} =

∞

∫

f (u) g(u − t)du.

−∞

Fill in the blanks:

123. Euler’s formula for the Fourier coefficients in the half-range sine series of f(x) in (0, 2l) are .....



124. The general solution in terms of arbitrary functions for the partial differential equation 2p + 3q = 1 is ......



125. The steady state temperature of a rod of length l whose ends are kept at 30° and 40° is ......



126. A separable solution of Laplace equation in polar coordinates suitable for a circular disc, which are periodic in q is ......



127. If F [f(t)] = f (s), then F [f(at)] = .....

Pick the correct answer:

128. If the complex form of the Fourier series of f(x) is ∞

∑ cn einx in (0, 2p), then Cn is −∞



(a)

(c)

1 p

2p

∫

f ( x )e − inx dx.    (b)

0

1 2p

2p

∫ 0

f ( x )e inx dx.   

2 p

2p

∫ 0

p

(d)

f ( x )e − inx dx.

2 f ( x )e − inx dx. p ∫0

275

Fourier Transforms

(e)



1 2p

2p

∫

f ( x )e inx dx.

0

129. The general solution of

∂2 z ∂x 2

= 0 is

(a) z = f (x) + g(y) .         

(b) z = x f(y) + g(y).

(c) z = yf (y) + g(y) .        

(d) z = f (x) + g(x).

(e) z = yf (x) + g(x).



130. The general solution y(x, t) of vibratory motion of a string of length l with fixed end points and zero initial velocity is (a) Sl n sin

npx npat . npx npat . sin cos    (b) Sl n cos l l l l

(c) Sl n sin

npx npat . npx npat . cos sin      (d) Sl n cos l l l l

(e) Sl n sin

npx . l

131. The equation of steady-state conduction in a plate is

(a) r2urr + rur +uqq =0.       

(b) r2urr + rur + uqq = 1.

(c) rurr + r2ur + uqq =0 .      

(d) r2urr + rur + quqq = 0.

(e) r2urr + 2rur + uqq = 0.

132. If the Fourier series of f(x) and g(x) are

a0 + S[an cos nx + bn sin nx] and A0 + S[2An cos nx + 2Bn sin nx] over [– p, p]. Write 2 down the Fourier series of 2f (x) + g(x) over [–p, p].

133. Find the complete integral of the partial differential equation z = px + qy + pq . 134. Express the boundary conditions in respect of insulated ends of a bar of length a and also the initial temperature distribution.



135. Write down a suitable separable solution for the partial differential equation governing steady state heat conduction in a rectangular plate when the edges y = 0, y = b of a rectangular plate bounded by x = 0, x = a, y = 0, y = b are insulated.



136. Find the finite cosine transform of x when 0 < x < p. Questions from University Papers

State true or false.

137. Fourier series of period 2 for | x | in (0, 2) contains only cosine terms.



138. A solution of a p.d.e. involving arbitrary functions, is called complete integral.



139. The p.d.e. governing the transverse vibrations of an elastic string is

∂u ∂2u = C2 2 . ∂t ∂x

276

Engineering Mathematics-III (Third Semester)



140. A separable solution of Laplace equation in polar coordinates, suitable for a circular plate, is u(r q) = r–p (A cos pq + B sin p q).



141. The inverse finite sine transform of F(s) is

∞

∫ F ( s)

sin sxds.

0

Fill in the blanks:

142. The cosine series for f (x) = sin2 x in (0, p) is __________



143. The p.d.e. whose solution is z = f(x2 + y2) (f, using an arbitrary function) is __________



144. A uniform rod of length a is initially at temperature q = q0. At time t = 0, one end is suddenly cooled to temperature q = 0 and subsequently maintained at this temperature. The other end remains thermally insulated. The mathematical formulation of this problem is __________.



145. The two-dimensional steady state heat conduction equation is __________.



146. The Fourier transform of f(x) = 1, | x | < a = 0, | x | > a is __________

Choose the correct answer.

147. Sum of the Fourier series for

 f ( x )= x , 0 < x < 1   at x = 1 is = 21 < x < 2   (a) 0       (b) 1       (c) 2      (d)

148. The degree of the p.d.e.

3

∂ 2 u  ∂u  =   is ∂x∂y  ∂z 

(a) 3        (b) 1       (c) 2      (d)

149. A separable solution of

1 3     (e) . 2 2

∂2 y ∂t 2

= C2

∂2 y ∂x 2

1     (e) not defined 2

satisfying the end conditions y(0, t) = y(1, t) = 0 is − c 2 n 2 p2 t e t

 npct   npx  (a) sin   sin   l    l 

(b)

 npCt  (c) x(x – l) sin    l 

npx  npCt  (d) sin   cos l  l 



 npx  sin    l 

150. A suitable expression for the steady state temperature in a rectrangular plate when its edges x = 0 and x = a are kept at zero temperature is

(a) (C1 cos px + C2 sin px) epy

(b) (C1 epx + C2e–px) sin py

(c) x(x – a) sin py

 npy   npx  (d) sin   sin    a   b 

(e)

px − py a

eae

277

Fourier Transforms



151. If f (s) is Fourier transform of f(t), then the Fourier transform of f(t – a) is :

(a) f (s – a)   (b) f (s + a)   (c) eias f (s)

(d) e–ias f (s)

(e) f (s)



152. Write down the complex form of the Fourier Series series for f(x) in a given interval.



153. Solve:



154. Write down one-dimensional heat e4 equation and a separable solution for the same.



155. In the case of flow of heat in a metal plate in the x-y plane, how do you calculate the rate of gain of heaf by the plate.



156. State inversion theorem for a complex Fourier transform.

∂2 z

= sin xy.

∂y 2

State True or False:

157. The application of Fourier series is not restricted to the expansion of periodic functions of period of 2p.



158. The Fourier series for f(x) in the interval c ≤ x ≤ c + 2p is given by



f(x) =

where a0 = an =



a0 ∞ + ∑ an cos nx 2 n =1 1 p

2p

1 p

c+2p

∫

f ( x ) dx

0

∫

f ( x ) cos nx dx.

c

159. z = px – qy, where p =

∂z ∂z ,q= is the partial differential equation, obtained by elimi∂x ∂y

nating arbitary constants a and b from the equation z = ax + by. ∂2u



160.



161. f(x) =

∂2u

is the Laplace’s equation, which arises in the conduction of heat in plate in ∂x 2 ∂y 2 steady state. +

g(w) =

1

∞

∫e

iwt

2 p −∞ 1

∞

∫e

2 p −∞

g(w) dw where

− iws

f ( s) ds

does not constitute the symmetrical fourier transform. Fill in the blanks:

162. At a point x = x0, where f(x) has finite discontinuity, the sum of the fourier series is ...........



163. For half-range cosine series in 0 < x < l the co-efficient an is given by an ...........



164. Pp + Qq = R is known as ........... where P, Q, R are functions of x, y, z.

278

Engineering Mathematics-III (Third Semester)



165. The quantity of heat required to produce a given temperature change in a body is proportional to the mass of the body and to the ............



166. f(t) =

∞∞

2 f ( s) cos ws. cos wt ds dw is called the fourier cosine integral and it is similar t ∫0 ∫0

to ..............

Pick the correct answer;

∞

npx in (0, l), then an is l

∑ an cos

167. If f(x) =

n =1

(a) (c)

l

l

1 npx 2 npx f ( x )cos dx    (b) ∫ f ( x )cos dx ∫ l −l l l −1 l l

∫ f (x)cos

−l

(e)

npx dx        (d) Zero l

l

1 npx f ( x )cos dx. l ∫0 l

168. The complete integral of the P.D.E. of the form f(z, p, q) is

(a) g(z, a) = x + ay + j(a)      (b) g(z, a) = x + ay + b (c) g′(z, a) = y + j′(a)

(d) g(z, a) = ax + by + j(a)

(e) g(z, a) = ax + y + j′(a)

169. The complementary function of P.D.E. (Dx + Dy)2 z = ex–y is

(a) z = j1(y + x) + xj2 (y – x)   (b) z = j1(y – x) + xj2 (y – x) (c) z = j1(y + x) + j2(y – 2x)   

(d) z = j1(y + x) + j2(y – 2x)

(e) z = j1(y – 2x) + j2(y + x).

170. The equation of the two dimensional heat flow is (a)

(c)

∂2u ∂t 2 u ∂2u ∂x

2

∂2

= a2

+

∂x 2

∂2u ∂y

2



+

(b)

∂2u = 0 ∂x∂y

(d)

∂2u ∂x 2

+

∂2u ∂y 2

= 0

∂u ∂2u = α2 2 ∂t ∂x

171. The Fourier integral in its exponential or complex form is

(a) f (t) =

(c) f (t) =

∞

1 ∫ 2 p −∞ ∞ ∞

∞

∫ f ( s) e

− iw( s − t )

ds dw (b) f(t) =

0

1 − iw( s − t ) ds dw ∫ f (s)e 2 p ∫0 −∞

(d) f(t) =

∞

1 ∫ 2 p −∞ −∞

1 ∫ 2 p −∞

∞

∫

f ( s)e − iw( s −t ) ds dw

−∞ ∞

∫

f ( s)e iw( s + i ) ds dw

−∞

(e) None

172. Examine whether the function f (x) = sin (1/x) in the interval [–p, p] can be expanded in a Fourier series.

279

Fourier Transforms



173. Find the complete integral of the P.D.E pq = k, where p =



174. Obtain the general solution of the equation

tan x

∂z ∂z , q= . ∂y ∂x

∂z ∂z + tan y = tan z. ∂x ∂y



175. Solve (D2 + 2DD′ + D′2) z = 0



176. Write down the various possible solutions of the wave equation

∂2 y ∂t 2

= a2

∂2 y ∂x 2

.

State True or False:

177. One of the Dirichlets conditions for convergence is that a function may have finite number of maxima or minima in any one period.



178. The number of arbitary constants will be the order of the partial differential equation on their elimination to form the partial differential equation.



179. In one-dimensional heat flow equation, if the temperature function u is independent of time, then the solution is u = ax + b.



180. One of the possible solutions to two-dimensional heat flow equation in cartesian system of co-ordinates is u =( c1x + c2) (c3y + c4).



181. Is F [f (x)] =

∞

1 − sx ∫ e f (x) dx. ? 2 p −∞

Fill in the blanks:

182. If a periodic function f (x) is even (–l, l) then an an is _________.



183. The Lagranges auxilary equation to x



184. If y = X(x). T(t) is the solution of



185. In two dimensional heat flow, the temperature at any point (of the xy plane) is independent of _________ coordinate.



186. The fourier cosine transform of f (x) can be defined as _________.

∂2 y ∂t

2

∂z ∂z ∂z +y +t = xyt are _________. ∂x ∂y ∂t = c2

∂2 y ∂x

2

, then

X′′ = _________. X

Choose the correct answer:

187. In harmonic analysis upto the fundamental harmonic, the given function

(a) f (x) = a0 + a1 cos x.

(b) f (x) =

a0 a sin x + b1 cos x. 2 1

(c) f (x) = a0 + a1 cos x + b1 sin x. (d) a1 cos x + b1 sin x +

188. The solution to

dx x2 + a

=

dy is y

a0 = f(x). 2

280

Engineering Mathematics-III (Third Semester)

 x  (a) sin–1   = loge y + b.  a

 x  (b) cos–1   = loge y + b.  a

 x  (c) sinh–1   = loge y + b.  a

 x  (d) cosh–1   = loge y + b.  a



189. When the ends of a rod is non-zero for one-dimensional heat flow equation, the temperature function u(x, t) is modified as the sum of steady state and transient state temperatures. The transient part of the solution which

(a) increases with increase of time. (b) decrease, with increase of time (c) increase with decrease of time. (d) decrease, with decrease of time

190. In two dimensional heat flow, the rate of heat flow across an area is proportional to the

(a) area and to the temperature gradient parallel to the area. (b) area and to the temperature normal to the area. (c) area and to the temperature gradient normal to the area. (d) area and to the temperature parallel to the area.

191. If F [f(x)] = f (l), then F [f(x + a)] is

(a) eialf(l) .    

(b) e–ial f (l).

(c) eial f (il).

(d) e–ial f(il).



192. What is the equivalent Fourier constant to the expression 2X mean value of f (x) in (l, l + 2p).



193. Form the partial differential equation by eliminating f from u = f(x3 – y3).



194. The ends of a rod length 5 units are maintained at 15°C and 5°C respectively until steady state prevails. Determine the temperature of that state.



195. If u(r, q) =



equation of a thin semicircular plate of radius a units, then determine u (r, θ) when r = a. and u (a, q) = k. 196. Find the Fourier transform of second derivative of a function f(t). State True or False.



197. Is the function



∞

n

r ∑ bn  a  sin nq is the solution of two dimensional steady state heat flow 1

 2x 1 + p , −p + x < 0 f (x) =  odd? 1 − 2 x , 0 < x < p p 



198. If the number of constants to be eliminated is equal to the number of independent variables, then the P.D.E’ is of first order.



199. One dimensional heat flow equation is



200. Two dimensional Laplace’s equation in polar co-ordinates is r 2

∂u ∂ 2 u − = 0. ∂t ∂x 2

∂2u ∂r 2

+r

∂u ∂ 2 u + = 0. ∂r ∂q2

281

Fourier Transforms



201. The Fourier transform of the convolution of two functions f(x) and g(x) is ∞

1 ∫ f (t) g( x − t)dt. 2 p −∞ Fill in the blanks:

202. Euler’s formulae for the Fourier co-efficients in the half-range cosine series of f(x) in (0, 1) are __________.



203. The complete integral of the equation z = px + qy + f (p, q) is __________.



204. D’Alembert’s solution of the one dimensional wave equation is __________.



205. In steady-state, two dimensional heat-flow equation in Cartesian co-ordinates is __________.



206. The Fourier transform of

x

∫ f (x) dx

is __________.

a

Pick the correct answer:



207. The exponential form of the Fourier series of f(x) in (–1, 1) is given by f(x) =

l

− 2 f ( x) e ∫ l −l

inpx l dx.



(b)

l

inpx 2 − (c) ∫ f ( x ) e l dx. l0

(e)



l

1 f ( x )e 2l ∫0

∑

n = −∞

where cn is (a)

∞

l

inpx l dx.

l

inpx l dx.

− 2 f (x) e ∫ 2l − l

− 1 f ( x )e (d) ∫ 2l 0

inpx l dx.

208. The general integral of

y 2 z ∂z ∂z + xz = y 2 is x ∂x ∂y

(a) F(x3 – y3, x2 – z2) = 0.     (b) F(x3 – y2, x – z) = 0. (c) F(x2 – z3, x2 – y2) = 0.     (d) F(y3 – z3, x2 – y2) = 0. (e) F(x – y, x2 – z2) = 0.

209. The P.D.E. of one dimensional heat flow in steady state is given by

(a)  

(c)

∂u = 0.   ∂t ∂2u ∂x

2

−



∂u = 0. ∂t

(b) (d)

∂2u ∂t 2 ∂2u ∂x

2

= 0. = 0.

(e)

du = 0. dt

cn e

inpx l

282



Engineering Mathematics-III (Third Semester)

210. A semicircular plate of radius ‘a’ is kept at temperature zero along the bounding diameter and at f (q), 0 < q < p along the circumference. The steady-state temperature u(r, q) at any point of the plate is given by

n

∞

r ∑  a  sin nq where bn is n =1

∞

(a)

1 f (q)sin nq dq. p ∫0

(c)

1 f (q)sin qdq. p ∫0

(e)

2 f (q)cos nq dq. p ∫0

p

(b)

2 f (q)sin nq dq. p ∫0

(d)

1 f (q)cos q dq. p ∫0

p

p

p

211. The Fourier sine transform of xf (x) is

(a) Fc′ ( s) (b) Fs′ ( s) (c) − Fc′ ( s) (d) − Fs′ ( s)

(e) Fs′′ ( s)

212. Find the Fourier series for

 x − 1, −p < x < 0 f (x) =   x + 1, 0 < x < p  xy  213. Find the P.D.E. by eliminating the arbitrary function in z = f   .  z 



214. A tightly stretched string with fixed end points x = 0 and x = 1 is initially at rest in equilibrium position. If it set vibrating giving each point a velocity lx(1 – x), write down the boundary conditions of the problem.



215. Write down a suitable separable solution for the P.D.E. governining steady state heat conduction in a square plate when the edges y = 0, y = 10 of a square plate bounded by X = 0, X = 10, Y = 0, Y = 10 are insulated.



216. Find the Fourier Transform of e–a| x |, a > 0.

State True or False:

217. If a function f (x) has a finite discontinuity at x = a, then the sum of the Fourier series is the arithmetic mean of the left and right-hand limits of f(x) at x = a.



218. The singular integral is not contained in the complete integral whereas the particular integral is obtained from the complete integral.



219. One dimensional wave equation is



220. The temperature distribution of the plate in the steady-state is ∇2u = 0



221. The Fourier transform of e–ax f (x) in terms of Fourier transform F(s) of f (x) is given by F(s – a).

∂2u ∂t 2

+

∂2u ∂x 2

= 0

283

Fourier Transforms



222. If a periodic function f (x) is even, its Fourier expansion contains only _________ .



223. The particular integral of (D2 – 2DD′ + D′2) z = ex+2y is _________.



224. In the steady-state condition, when the temperature no longer varies with time, the solution of the diffusion equation is _________.



225. Two dimensional heat flow equation in polar co-ordinates is _________.



226. The Fourier sine transform of f(x) sin ax in terms of Fourier transform F(s) of f(x) is _________.

Pick the correct answer :

a0 ∞ npx + ∑ an cos is the half-range consine series of f(x) of period 2l in (0, l), then 2 n =1 l

227. If 1

∫ [ f ( x )]

2

dx is equal to

0

 a02

∞   a2 ∞  l  a2 ∞  l  a2 ∞  + ∑ an2  . (b)  0 + ∑ an2  . (c)  0 + ∑ an2  . (d) l  0 + ∑ an2  . 2  2 n =1  2  2 n =1   2 n =1   2 n =1 

(a) 2l 



2 (e) 2l  a0 +





∞



∑ a2  . n

n =1



228. The complete integral of the P.D.E of the form F(p, q) = 0 given by

(a) Z = ax + f′ (a) y + b.

(b) Z = ax + f(a) y + b.

(c) Z = a + f(a) (a)x.

(d) Z = ax + f(a).

(e) Z = ax + f′ (a)y.



229. The general solution u(x, t) of vibratory motion of string of length 2l with fixed end points and zero initial velocity is ∞

(a)

∑ Bn sin

n =1 ∞

(c)

∑ Bn sin

n =1 ∞

(e)

∑ Bn sin

n =1

npx npat . sin (b) l l npx npat . cos (d) 2l 2l

∞

∑ Bn sin

n =1 ∞

∑ Bn sin

n =1

npx npat . cos l l npx npat . sin 2l 2l

npx npat . cos 2l 2l

∞

230. ∑ sin x is of period n =1

(a) 0. (e) 2p.

(b) p/2.

(c) p. (d)

3p . 2

284



Engineering Mathematics-III (Third Semester)

231. If f(x) → 0 as x → ∞, then Fourier cosine transform of first derivative of f(x) is given by.

(a) sFs(s).

(b) –s Fc (s).



(e) −  sFs ( s) +





(c) sFs(s) –

 2 f (0) . p 

2 f (0) . p

(d) sFc(s) –

2 f (0) . p

232. Find the Fourier series for f(x) = | x | in (–p, p).

State true or false.

233. At a point x = x0, where f(x) has a finite discontinuity, the sum of the Fourier series is 1 lim [hf(x0 – h) + hf(x0 + h)]. 2 h→0



234. If the number of constants to be eliminated is just equal to the number of independent variables, the process of elimination will give rise to partial differential equation of the second order.



235. In one dimensional heat flow equation, the temperature funtion u is to decrease with increase of time.



236. One of the possible solutions to two dimensional heat flow equation in polar co-ordinate is

u(r, q) = (C1 log r + C2) (C3 log q + c)

237. Fourier sine transform of f(x) is equal to

1 2p

∞

∫ sin sx f (x) dx. 0

Fill in the blanks:

238. If a periodic function f (x) is an odd function in (–l, l), then bn in the Fourier series is ________.



239. A solution which contains the maximum possible number of arbitrary function is called ________.



240. The one dimensional wave equation is ________.



241. Two dimensional Laplace equation in polar co-ordinate is ________.



242. The Fourier inverese cosine transform is defined as ________.



243. The root-mean square value of a function y = f (x) over an interval from x = a to x = b is defined as (a) y =

1

b

2 ∫ y dx. (b) y =

1 2 (b − a) a

1

b

∫ f (x) dx.

1 2 (b − a) a

1

1

b  1 b 2 2 1  2 2  ∫ y dx  . (d) y =  (c) y = ∫ y dx  .  (b − a)  a   (b − 1) a 



244. A rod of length l has its ends A and B kept at 0°C and 100°C respectively, until steadystate conditions prevail. Then the initial condition is given by

(a) u(x, 0) = ax + b + 100l (b) u(x, 0) =

100x l

285

Fourier Transforms (c) u(x, 0) = 100 xl



(d) u(x, 0) = (x + l)100

245. The solution of the Lagrange’s linear partial differential equation Pp + Qq = R is of the form.

(a) f(u + v) = 0 where f is an arbitrary function (b) u = f(v) where f is an arbitrary function (c) uv = f (v) where f is an arbitrary function (d) v = f(u + v) where f is an arbitrary function

246. The solution of one dimensional wave equation is

(a) y(x, t) = (C1 cosh px + C2 sinh px) (C3 cos pat + C4 sin pat) (b) y(x, t) = (C1 cosh px + C2 sinh px) (C3 epat + C4 e–pat) (c) y(x, t) = (C1 cos px + C2 sin px) (C3 cos pat + C4 sin pat) (d) y(x, t) = (C1epx + C2 e–px) (C3 epy + C4 e–py)

x  247. F  ∫ f ( x ) dx  is equal to   a 

(a)

F ( s) F ( s) F ( s) F ( s) (b) (c) (d) e ia −is is s −is



248. If f(x) = | x | when – p < x < p, then obtain Fourier series expansion for f(x).



249. Form the partial differential equation by eliminating f and f and Z = f(x + ct) + f( x – ct).



250. A rod 30 cm long, has its end A and B kept at 20°C and 80°C respectively, unitl steady state conditions prevail. Determine the temperature at that state.



251. For the steady steady state heat flow in a circular plate, is it possible to have the general solution u(r, q) as (A log r + B) = (Cq + D)? If not, why?



252. If f (x) = e–ax, a > 0, find Fourier sine transform of f(x).



253. Write the Fourier sine series of K in the interval (0, p).



254. Fill in the blanks

The fourier series of f (x) (0, l) is f (x) =

∞ 1 npx a0 + ∑ an cos , then a0 = ________ and 2 l n =1

an = ________.

255. What is the transform of f (ax) if fs(s) is the Fourier sine transform of f(x)?



256. State Parseval’s identity on Fourier transform.



257. Write true or false.

The non-linear equation f (p, q) = 0 has singular integral.

258. Fill in the blanks

The general solution in terms of arbitrary functions of the PDE 2p + 3q = 1 is ________.

259. Pick the correct answer

The general solution of

∂2 z ∂x 2

= 0 is

286

Engineering Mathematics-III (Third Semester)

(a) z = f (x) = +g(y)

(b) z = xf (y) + g(y)

(c) z = yf (x) + g(x)

(d) z = f (x) + g(x)



260. Solve: (D – 4DD′ + 5D′ ) = 0.



261. Write the three possible solutions of the equation



262. Choose the correct answer

2

2

∂2u ∂t 2

= C2

∂2u ∂x 2

.

The one dimensional heat equation in steady-state is (a)

(c)



∂2u ∂x 2 d2u dx

2

+

∂2u ∂y 2

= 0 (b)

= 0 (d)

∂u ∂2u = c2 2 ∂t ∂x ∂2u ∂t

2

= c2

∂2u ∂x 2

263. Analyse the possible solution of steady state heat flow in two dimension for the following boundary conditons. (a) u(0, y) = 0, 0 ≤ y ≤ ∞ (b) u(l, y) = 0, 0 ≤ y ≤ ∞ (c) u(x, ∞) = 0, 0 ≤ x ≤ l (d) u(x, 0) = f(x), 0 < x < l



264. Express ∇2u in polar form.



265. Write down the boundary conditions for the following boundary value problem.

If a string of length l is initially at rest in its equilibrium position and each of its points is px  ∂y  given the velocity   = v0 sin 3 , 0 < x < l, determine the displacement y(x, t). l  ∂t t =0 State true of false

266. Fourier series of period 2 for x sin x in the internal (–1, 1) contains only sine terms.



267. The difference between the partial differential equation and the ordinary differential equation is that order of the differential equation equals the number of arbitrary constants eliminated.



268. In one dimensional heat flow equation, if the temperature function u is independent of time, then solution is u = ax + b.



∂u ∂ 2 u ∂ 2 u + 269. The temperature distribution of the plate in the transient state is given by = . ∂t ∂x 2 ∂y 2 ∞ 1 isx 270. Is F{f(x)} = ∫ e f (x) dx ? 2 p −∞

Fill in the blanks:

271. The half-range sine series for f(x) = x in (0, 1) is _________.



272. The PDE whose solution on is z = ax + by, where a and b are arbitrary constants, is given by _________.



273. The steady state temperature of a rod of length 2l whose ends are kept at 20°C and 70°C is _________.

287

Fourier Transforms



274. The two-dimensional steady state heat conduction equation is _________.



 x  275. If F{f(x)} = F(s). then F  ∫ f ( x )dx  = _______.  a 

Choose the correct Answer.

276. The Fourier series of f(x) = C, where C is a constant, the periodicity being 2p is

(a) 0



(c) 2p (d) C

(b) 2C

277. The complete integral of the equation p = 2qx is

(a) z = ax2 + ay + c (c) z = ay2 + ax + c





(b) z = ax + c

(d) z = ay + c.

278. The general solution y(x, t) of vibratory motion of a string of length l with fixed end points and zero initial velocity is (a)

∑ ln sin

npx npat sin (b) l l

∑ ln sin

npx npat cos l l

(c)

∑ ln cos

npx npat cos (d) l l

∑ ln cos

npx npat sin l l

279. In two dimensional heat flow, the rate of heat flow across an area is proportional to the (a) area and to the temperature gradient parallel to the area (b) area and to the temperature normal to the area (c) area and to the temperature gradient normal to the area (d) area and to the temperature parallel to the area



280. Fourier sine transform of

(a) p (b)



1 is x

p (c) 2

p / 2 (d) 2p



281. Find the r.m.s. value of f (x) = x – x2 in the internal.



282. Obtain partial differential eqution by eliminating the arbitrary functions f and g from

z = f (2x + y) + g(3x – y).

283. The ends of a rod of length 20 cm are maintained at 10°C and 30°C respectively until steady-state prevails. Find the temperature at that state.



284. Write down the various possible separable solutions of Laplace’s equation in two dimensions.



285. Find the Fourier transform of





1−| x|if| x|< 1 f (x) =  if| x|> 1. 0



286. Find the Fourier constant an for x cos x in (– p, p), n = 2, 3, ... to ∞.



287. Find the sine series of f(x) = K in (0, l)



288. What is known as harmonic analysis?

288

Engineering Mathematics-III (Third Semester)



289. Form the p.d.c by eliminating ‘f’ from z = f(y/x)



290. Find the solution of px2 + qy2 = z2.



291. Find the complete integral of p + q = x + y.



292. Find the particular integral of

(D2 – 4DD′ + 3D′2)Z = ex+y.

293. Write down the one dimensional wave equation.



294. Write different possible solutions of one dimensional wave equation.



295. What is meant by steady state condition in one-dimensional heat flow?



296. If the ends x = 0 and x = l are insulated in one-dimensional problems, write the boundary conditions.



297. Write down the unsteady two-dimensional heat flow equation in Cartesian co-ordinates.



298. Write down the different solutions of Laplace equation in Cartesian co-ordinates.



299. Write the most general solution to find the steady-state temperature over annulus.



300. A quadrant plate or radius ‘a’ cms has its temperature at 0°C on the bounding radii and 60°C on its circumference. Write the corresponding boundary conditions.



301. If the Fourier transform of f(x) is f (s), what is the Fourier transform of f(x – a)?



302. Find the Fourier sine transform of e–2x.



303. State the Parseval’s identity on Fourier transform.



304. Find the finite Fourier Cosine transform of f (x) = x in (0, p).

Say True or False (305 to 309):

1 − x in 0 < x < p 305. f(x) =  1 + x in p < x < 2 p

is an even function and hence the Fourier series of f(x) contains only cosine terms.

306. When the number of independent variables is equal to the number of arbitary constants, the partial differential equation obtained by eliminating the arbitrary constants is of first order.



307. The one dimensional heat conduction equation is



308. e − x



309. Fc[xf(x)] = Fs′ [f(x)].



310. The Fourier sine series of f(x) in 0 < x < p is f (x) =

2

/2

∂2u ∂t

2

= α2

∂2u ∂x 2

.

is self reciprocal under Fourier transform. ∞

∑ bn 1

sin nx, where bn = ________.



311. The complete integral of the partial differential equation z = px + qy + pq is ________.



312. The general solution of

∂2 z ∂x 2

−

∂2 z ∂y 2

= 0 is ________.

289

Fourier Transforms



313. The steady state temperature distribution in a rod of length 10 cm whose ends x = 0 x = 10 are kept at 20°C and 50°C respectively, is ________.



314. If F [f (x)] = F(s), then F [eix f (x)] = ________.

Choose the correct answer (315 to 319):

315. The sum of the Fourier series of f(x) = x + x2 in – p < x < p at x = p is

(a) p (b) p2 (c)



p p2 (d) 2 2

316. The general solution of the partial differential equation (4D2 – 4DD′ + D′2)z = 0 is  

1  1   x  + j1  y + x  2  2  

 

1  1   x  + xj 2  y + x  . 2  2  

(a) z = j1(y + 2x) + xj2(y + 2x) (b) z = j1  y +



(c) z = j1(x + 2y) + j2 (x + 2y)

317. The solution y(x, t) of the transverse vibrations of a stretched string with fixed end points and zero initial velocity is  npx   npat   sin   (b) l   l 

∑ Bn cos 

 npx   npat   cos   l   l 

∑ Bn cos 

(a)

∑ Bn sin 

(c)

∑ Bn sin 



(d)

 npx   npat   sin   l   l   npx   npat   cos   l   l 

318. The two dimensional heat equation in steady state is

(a)



(d) z = j1  y +

∂2u ∂t 2

= α2

∂2u ∂x 2

(b)

 ∂2u ∂2u  ∂u ∂2u ∂u = α 2 2 (c) = α 2  2 + 2  ∂t ∂t ∂x ∂x   ∂x

(d)

∂2u ∂x 2

+

∂2u ∂y 2

= 0

319. If [f (x)] = f (s), then F [f (x + a)] =

(a) F(s – a) (b) F(s + a) (c) eias F(s) (d) e–ias F(s).



320. Write down the Parsevals formula corresponding to the Fourier cosine series of f(x) in 0 < x < p.



321. Form the partial differential equation by eliminating the arbitrary constants from z = ax2 + by2.



322. List all the possible solutions of the one dimensional wave equation (obtained by the method of separation of variables) and state the proper solution.



323. @



324. @



325. If f(x) = x2 in the interval –2 ≤ x ≤ 2, what is the value of bn?



326. Find the value of an in the cosine series expansion of f(x) = k in (0, 10).



327. What is known as harmonic analysis?



328. Form the partial differential equationby eliminating the arbitrary function f from z = y f  . x



329. Write the complete integral of p + q = x + y.

290

Engineering Mathematics-III (Third Semester)



330. Find the solution of p x + q y = z.



331. Write the particular integral of (D2 + 3DD′ + 4D′2)z = ex–y.



332. Write any two solution of the transverse vibration of a string.



333. If the ends of a string of length l are fixed and the mid point of the string is drawn aside through a height h and the string is released from rest, write the initial conditions.



334. Write down steady state one dimensional heat flow equation.



335. The upper end of a rod of length l is always kept insulated. Write this condition mathematically.



336. Write the Laplace’s equation in Cartesian co-ordination.



337. A rectangular plate is bounded by x = 0, y = 0, x = a and y = b. Its surfaces are insulated and temperature along two edges x = a and y = b are 100°C while the temperature along the other edges are 0°C. Write down the boundary conditions for the above problem.



338 Write down any two solution of Laplace’s equation in polar coordinates.



339 A circular plate of radius 10 cm has the upper half of its circumference at 0°C and the lower half of 100°C. Write the corresponding boundary conditions.



340 Define Fourier transform of f(x).



341. Define self-reciprocal function w.r.t. Fourier Cosine transform.



342. Find the Fourier sine transform of



343. Find the finite fourier sine transform of f(x) = x in (0, l).

1 . x

Say True or False (344 – 348)

344. The fourier series of f(x) = x cos x in – p < x < p contains only cosine terms.



345. The general solution of



346. Two dimensional steady-state heat equation in polar coordinates is r 2



347. If f(x) is an even functions of x, then its Fourier transform F(s) is also an even function of s.



348. e − x

2

/2

∂2 z = 0 is z = f(x) + g(y). ∂x∂y ∂2u ∂r 2

+

∂u ∂2u +r 2 = 0 ∂r ∂q

is self-reciprocal under Fourier sine transform.

Fill in the blanks (349 to 353)

349. The value of a0 in the Fourier series expansions of f(x) = p – x in 0 < x < 2p is _______.



350. The complete solution of the partial differential equation pq = 1 is _______.



351. The general solution of



352. The steady-state temperature distribution in a rod of length 10 cm whose ends are kept at 10°C and 30°C is _______.



353. If F [f (x) = F(s), then F [f (2x)] = _______.

∂2 z ∂x 2

−2

∂2 z ∂2 z + = 0 is _______. ∂x∂y ∂y 2

291

Fourier Transforms

Choose the correct answer (354-358)

354. The general solution of

∂2 z ∂y 2

= 0 is

(a) z = f (x) + g(y) (b) z = f (x) = yg (x) (c) z = f (y) + xg (y) (d) z = f (y) + xg (y)



355. The one dimensional heat equation in steady state is ∂2u

(a)



∂2u

+

∂x 2

∂y 2

= 0 (b)

∂ 2 u ∂u ∂u d2u = 0 (c) − = 0 (d) =0 ∂t ∂x 2 ∂t dx 2

356. The general solution for the displacement y(x, t) of string of length l vibrating between fixed end points with initial velocity zero and intial displacement f(x) is  npx   npat   npx   npat   cos   (b) SBn sin   sin    l   l   l   l 

(a) SBn sin 

 npx   npx   npat   sin   (d) SBn sin    l   l   l 

(c) SBn cos 



357. If f (s) is the fourier transform of f(x), then the Fourier transform of f(x – a) is

(a) F(s + a) (b) F(s – a) (c) eias F(s) (d) e–ias F(s)



358. Write down the complex form of Fourier series of f (x) defined in –p < x < p.



359. Find the particular integral of



360. Write down the three possible solutions of the one dimensional heat equation and state the suitable solution.



361. If F(s) is the Fourier transform of f (x), what is the Fourier transform of f′′(x)?



362. Using Parseval’s theorem prove (i)

(iii)

p

∫ −p cos p

4

∫ −p cos



∂2 z ∂y 2

= cos (x – y)

3p (ii) 4

xdx =

35 p 64

8

p

∫ −p cos

1 1 + cos 2 x 2 2

a0 1 1 = , a2 = 2 2 2





∂x 2

+

xdx =

Hint: (i) f(x) = cos2 x =



∂2 z

 a02 1 2 2 2  2 [ f ( x )] dx = p  + ∑ ( an + bn ) ∫−p  4 2  p  1 1  1  4 ∫−p cos x dx = 2p  4 + 2  4  p

=

3p 4

6

x dx =

5p 8

292

Engineering Mathematics-III (Third Semester)

(ii) Take f (x) = cos3 x =

3 1 cos x + cos 3 x. 4 4

a1 = 3/4; a3 = 1/4 p

∫ −p cos



6

1  1 9 xdx = 2 p   +    2  16 16   =

5p 8

 1 + cos 2 x  (iii) Take f (x) = cos4 x =   2  

2

3 1 1 = + cos 2 x + cos 4 x 8 2 8



3 1 1 a0 = = , a2 = , a4 4 2 8



ANSWERS (FOURIER SERIES) Exercises 1(a) Page 1

1. (i) p. (ii)

2 p. 3

(iii) 2p/ p. (iv) k.

Exercises 1(b) Page 3 1 1 1 ; ; . 2 2 2



1. 9; 9; 9. 2.



3. +∞; – ∞. 4. 1, 0 does not exist.



5. 1, 0; f(x) is not continous at x = 0; jump at 0 is 1 : f(0) = 1.



6. 1, 2 ; 1. Exercises 1(c) Page 37  e 2 p − 1  1 ∞ (−1)n (cos nx − n sin nx ) . +∑ 2 p  pe  2 n −1 n + 1 



1. f(x) =



1 1 1 1 1  p 2. f1(x) = 2  sin x − sin 2 x + sin 3 x − ...  =1 − + − + ... 2 3 3 5 7  4



3. F(x) = f(x) + f1(x), where f(x) and f1(x) are given above.



4. f(x) =

∞ p2 (−1)n + 4 ∑ 2 cos nx. 3 n =1 n

293

Fourier Transforms



2   x2 6   p2 6  6   x − 3  sin x −  − 3  sin 2 x +  − 3  sin 3 x − ... . 5. f(x) = 2   2 2   3 3   1 1  



1 1 6. f(x) = sin x + sin 3 x + sin 5 x + ...to ∞. 3 5



7. f(x) =

1 1 1   p + 2  sin x + sin 3 x + sin 5 x + ...to ∞  . 2 3 5  



8. f(x) =

1 1 1 1  p + 2  sin 2 x + sin 4 x + sin 6 x + ...to ∞  . 2 4 6 2 



9. f(x) =

3 p ∞ (−1)n −1 cos(2n − 1)x − sin(2n − 1)x + sin 2(2n − 1)x +∑ 4 n =1 2n − 1



10. f(x) =



11. a0 =



1, 2, 4, 5, 7, 8,... 3/ np, n = 12. a0 = 2; an = 0, when n ≠ 0; bn =  0, n = 3, 6, 9,...



13. a0 =



14. f (x) =



15. a0 =



16. an = 0; bn =



17.

4 cos 3 x cos 5 x  + − ...to ∞  .  cos x − 3 5 p 

1, 3, 5,... 0, n − 2, 4, 6,... 1/ np, n = 1   ; a= 1/ p , = 1, 5, 9 p = n n b = 2/ n , n 2, 6,10,...   n n 2 −1/ np, n =  3, 7,11...  0, n = 4, 8,12...

p −2 1 ; an = 2 for odd n, and = 0 for even n; bn = ± , as n is odd or even. 2 n n p 2 1 2 1 2 sin x + sin 2 x − sin 3 x − sin 4 x + sin 5 x + ...to ∞. 2 9p 4 25 p p

p2 (−1)n 2 p 4 p ; an = , for n ≠ 0; bn = − 3 for odd n, and = − for even n. 2 3 n pn n n

cos x 4 + p 2

(−1)n +1 8n p(4n2 − 1)

∞

∑

n =1

n sin 2nx 4n 2 − 1

2i  cos 2q cos 4q cos 6q  − − − ... 1 − 1⋅ 3 3⋅5 5 ⋅7 p 



18. i =



19. (i) f (x) =

4 1 sin nx for n odd. ∑ p n

(ii) f (x) =

3 2 1 1  − sin x + sin 3 x + sin 5 x + ... 2 p 3 5 

(iii) f (x) =

2 1 1  sin x + sin 3 x + sin 5 x + ... 3 5 p  

294

Engineering Mathematics-III (Third Semester)

1 2 1 1  +  sin x + sin 3 x + sin 3 x + ...  2 p 3 5 

(v) f (x) = (vi) f(x) =

p2  cos x cos 2 x cos 3 x  + 4 2 + + + ...  3 22 32  1  Exercises 1(d) Page 42

2  p 2 4 −  p  1 12

  p2 4 p2 − 2  sin x − sin 2 x +  2   3 3

  p2  sin 3 x − sin 4 x + ... 4  



1.



 p 2 6    p2 6   p2 6  2. 2  − 3  sin x −  − 3  sin 2 x +  − 3  sin 3 x − ...  1 1    3 2   3 3 



1 8 ∞ r sin 2r p. 3. − sin x − ∑ 2 p p r =1 4 r − 1



4.

sin pα ∞ 2α sin pα + ∑ (−1)n cos nx. pα p(α 2 − n2 ) r =1



5.

2 sin pa  sin x 2 sin 2 x 3 sin 3 x  ... − 2 +  2 2 p 1 − α 2 − α2 32 − α2  Exercises 1(e) Page 51



1. (i)

2 p sin 2 x (p + 2)sin 3 x p sin 4 x  + − + ... ; ( x + 2)sin x − p 2 3 4  4 cos 3 x cos 5 x p  + 1 − cos x + + + ... 2 2 2 p 3 5 

(ii)

2 ∞ n{1 − (−1)n e p } e p − 1 2 ∞ 1 − (−1)n e p sin nx ; − ∑ cos nx ∑ n2 + 1 p p n =1 n 2 + 1 p n =1

(iii) The same as exercise 1(d), ex. 1; cos 2 x cos 3 x cos 4 x p2   − 4 cos x − + − + ... . 3 22 32 42   cos 6 x cos10 x cos 2(2r + 1)x p 2   − cos 2 x + + + ... + + ... . 2 2 2 4 p  3 5 (2r + 1) 



2.



3. The sine series for sin x is sin x.This answer is consistent with the theorem. If two Fourier series in sines alone (or cosines alone) converge to the same sum for all values of x, then the corresponding coefficients are equal. Exercises 1(f) Page 60



1. (a) The same as in 1(b), with k = 1.

(b)

k 2k  px 1 3 px 1 5 px  + + sin + sin + ... . sin 2 p  2 3 2 5 2 

295

Fourier Transforms



2.

4 ∞ 1 (2n + 1)px sin . ∑ 2 p 0 2n + 1



3.

1 2 px 1 3 px 1 5 px  + cos − cos + cos − ... . 2 p 2 3 2 5 2 



4.

2 sin 2 px sin 3 px  + − ... sin px − 2 3 p 



5.

∞ 1 ∞ 4 cos(2n − 1)px cos 2npx 4 −∑ 6. ∑ 4n 2 p 2 2 n =1 p2 (2n − 1)2 n =1



 4  n2 p2 , n = 1, 3, 5,...  1 0, n 4, 8,12,... 7. bn = 0; a0 = = , an = 2  8  2 2 , n = 2, 6,10,...  n p



8.

1 4  cos 2 px cos 3 px cos 4 px  − 2 cos px − + − + ... 3 p  4 6 8 



9.

2 4  cos 2 x cos 4 x cos 6 x cos 4 px  −  + + − + ... 3.5 5.6 7.9 p p  1.3 



10. 1 + (2p sin 1) 11. a0 =

an =

∞

(−1)n +1 n sin npx

n =1

n 2 p2 − 1

∑

2 1 and a3 = . For n = 1, 2, 4, 5, 6, 7, ... 3 3 18 2

(n − 9)np

sin

np ; bn = 0, for all n. 3

E E 2E  cos 2ωt cos 4ωt cos 6ωt  + sin ωt + + + − ...  3⋅5 5⋅7 p 2 p  1⋅ 3 



12.



13. a0 =

2 kn 1 (3 – e–2); an = where kn = 1 – 1(– 1)n e–2, ln = 4 + n2p2. 2 ln

For even n, bn =

npkn npk a 2 − . and for odd n, bn = ln np ln



15.

1 ∞ (−1)n (2n − 1)x cos . −∑ 2 n =1 2 n − 1 2



16.

1 4 − 6 p2

 cos 2 pt cos 4 pt cos 6 pt cos 8 pt  + + + + ... ;  16 9 64  4 

8  sin pt sin 3 pt sin 5 pt  + + + ...  27 125 p2  1 

296

Engineering Mathematics-III (Third Semester)

k 16 k  1 2 pt 1 6 pt pt 1 3 pt 1 5 pt  8k   − + 2 cos + ... ; 2 sin − 2 sin + 2 sin − ...  cos 2 p2  2 2 l l l 3 l l 6 5  p  



17.



16  4 1, 5, 9,...  n 2 p2 − n 3 p3 , n =  − 16 ⋅ n = 2, 6,10,...  n2 p2 5 18. a0 = ; an =  6  4 + 16 , n = 3, 7,11,...  n 2 p2 n 3 p3  8  , n = 4, 8,12,...  n2 p2



19. −



20.



21. For even n, bn = 0 and for odd n, bn =



22.

8  sin x 2 sin 2 x 3 sin 3 x 4 sin 4 x  + + + + ...  3.5 5.7 7.9 p  1.3 



23.

2 4 1 2 px 1 4 px 1 6 px  cos cos cos ... . −  + + 3.5 5.7 p p  1.3 l l l 



24.

∑ 1 + n 2 p2

32  px p2 2 px 1 3 px p2 4 px  sin sin sin + + 3 sin + + ... 3  2 8 2 2 16 2 3 p  

  p2 4  p2 p2 2  p2 4  sin sin 2 sin 3 sin 4 px + ... − p x + p x + − p x +      3  3  3   2 4 p  1 1    3 3 

∞

np

4n 2

(n − 4)p

.

(1 – e cos np) sin npx.

n =1

Exercises 1(h) Page 90

1. y = 0.73 + 0.359 cos x – 0.085 cos 2x – 0.117 cos 3x + ...



+ 0.466 sin x – 0.289 sin 2x – 0.090 sin 3x + ...

2. 1.45 – 0.367 cos x – 0.1 cos 2x + 0.033 cos 3x + 0.173 sin x – 0.0067 sin 2x + 0(sin 3x) + ...



3. 5.73 cos q + 30.26 sin q – 5.10 cos 3q + 3.17 sin 3q



4. 1.26 + 0.04 cos x + 0.53 cos 2x – 0.1 cos 3x – 0.63 sin x – 0.23 sin 2x + 0.085 sin 3x.



5. 7850 sin q + 1500 sin 2q; T = 8332



6. 1.071



8. y = 2.10 – 0.283 cos x – 0.18 cos 2x + 1.6 sin x – 0.49 sin 2x

10. y = 6.44 – 1.65 cos x – 0.34 cos 2x + 0.71 sin x – 0.69 sin 2x Exercises 1(i) Page 97



1. eax =

sinh ap sinh ap ∞ (−)n 2 a cos nx 2 sinh ap ∞ (−1)n n sin nx + ∑ a 2 + n 2 − p2 ∑ a 2 + n 2 ap p2 1 1

297

Fourier Transforms

sin αp  1 ∞ (−1)n 2α cos nx   +∑  p  α 1 α 2 − n2 



2. (i)



3. e–x =

∞

(1 − inp)

∑ (−1)n 1 + n2 p2 einpx

n = −∞



(ii)

2 sin αp ∞ (−1)n n sin nx ∑ α 2 − n2 . p 1

 cos px cos 2 px cos 3 px  = sinh 1 – 2 sin h 1  − + − ...  1 + p2 1 + 4 p 2 1 + 9p 2 

 sin px 2 sin 2 px 3 sin 3 px  – 2p sin h 1  − + − ... 2 2 2 1 + 4p 1 + 9p 1+ p 

2  2  − inp   1 + inp  e 3 − 1 3    ∞ sin ap n sinh p ∞ 1 + in inx n inx x (−1) 2 (−1)n e 8. (i) sin ax = e (ii) e = ∑ ∑ 2 p −∞ p 1 + n2 a −n −∞

1 3 4. C0 = = ; Cn 6 4n 2 p2

(Partial Differential Equations) Exercise 2(a) Page 107

1. (i) px + qy = 3z, (ii) z = px + qy + pq, (iii) p = q,

(iv) z2(p2 + q2 + 1) (v) z(px + qy) = z2 – 1, (iv) px + qy = q2, (vii) z = pq.

2. (i) px + qy = nz (ii) z2 = x2 + y2 + 2z(px + qy)

(iii) p + x = qy (iv) p + q = px + qy (v) p2 + q2 = 16 p2 q2 (x + y) (vi) p2 + q2 = tan2 a (vii) z = px + qy +

b −q q

3. (i) py = qx, (ii) py = qx, (iii) px(y – z) + qy(z – x) = z(x – y),

(iv) xy

∂2 z ∂2 z ∂2 z ∂2 z = px + qy = z, (v) + − 6 = 0 , (vi) py – qx = y2 – x2, ∂x∂y ∂x 2 ∂y∂x ∂y 2

2 (vii) px + qy = 0, (viii) px = qy, (ix) px – qy = x – y, (x) a

∂2 z ∂y

2

− 2 ab

∂2 z ∂2 z + b2 2 = 0, ∂x∂y ∂x

∂ z ∂ z (xi) (x + iy)  2 + 2  = 2(p + iq), (xii) p(x + y) (x + 2z) – q(x + y) (y + 2z) = z(x – y),  ∂x ∂y   2

2

(xiii) q – p = z, (xiv) qz – pz = x – y, (xv) (z – y) p + (x – z) q = y – x, (xvi) z = px + qy – xys – y2t, (xvii)

∂4 z ∂x 4

−2

∂4 z ∂x 2 ∂y 2

+

∂4 z ∂y 4

= 0 , (xviii) px2 + qy = 2y2,

(xix) (x – 2z)p + (2z – y)q = y – x, (xx) 3r – 8s + 4t = 0,

4. py – qx = 0     5. z2 (p2 + q2 + 1) = k2     9. z = px + qy + k(1 + p2 + q2)

298

Engineering Mathematics-III (Third Semester)

Exercise 2(b) Page 114

1. z = f(y) 2. z = f(x) + yf(x)



3. z = – cos x + xf(y) + f(y) 4. z = 6xy –



5. z = −



6. z =



7. z = − x −



9. z = e2yj(x) + e3y f(x) + 2y +

1 a2 b

1 2 x log y + F(y) + f(y) + j(x) 2

sin (ax + by) + xF(y) + f(y) + j(x)

1 cos (2x – y) – x3y3 + xF(y) + f(x) + j(y) 4 y + x 3 f ( y ) 8. z = (1 + cos x) cos y. 3 5 3



10. z =

x3 y y3 x + F(y) + f(x) 11. z = ev cosh x + e–y sinh x. + 3 3



12. z =

3x2 − xy + sin y + k. 13. u = j(x + iy) + f(x – iy) 2



14. z = f(2x + y) + f(3x + y) 15. z = axy + bx + cy + d. Exercise 2(c) Page 132 a 1 y + b 2. z = ax + y+b a −1 a



1. z = ax +



3. z = ax + y (m2 − a 2 ) + b 4. z = b2x + by + c.



5. z = ax +



7. z = (2 + 3b – b2) x + by + c. 8. z = 3(x tan a + y sec a) + b.



9. z = ax – y sin a + b. 10. z = ax +

ay 1 − 2a y + c, 6. z = ax + [n ± (n2 − 4)] + b. 3 2 5 − a3 y + b. 3 − 2a



11. z = ax + by + a2 – b2. 12. z = ax + by + 3ab.



13. z = ax + by – 4a2b2. 14. (a + b) (z – ax – by) = 1



15. z = ax + by +



17. 4(1 + a2)z = (x + ay + b)2.



18. az (1 + a 2 z) − log az + (1 + a 2 z 2 ) = 2a(ax + y + b).



19. 4(az – a2 – 1) = (x + ay + b)2. 20. 3z3 = (a2 – 9)x + 3ay + 3b.



21. (a2 + 1) (1 – z2) = (x + ay + b)2. 22. z2 = (x + ay + b)2 + a2.

( a 2 + b 2 ) 16. a(log z)2 = (x + ay + b)2.

299

Fourier Transforms

2 2 ( x + a)3/2 − ( a − y )3/2 + b . 3 3 2 y + b . 26. 2z = ax2 + + b. a



23. az – 1 = bx + ay 24. z =



25. z = a 2 ( x + y ) +



x 2 4 a 3/2 − x 2 3 2 2 27. z = ( x + a)3/2 + ( y − a)3/2 + b , 28. 2z = (x + a)2 + (y – a)2 + b 3 3



29. z = a(x – y) – (cos x + cos y) + b. 30. (2z + ay2 – 2b)2 = 4a(1 – x2).



31. z = – cos x + ax +



33. z3/2 = (x + a)3/2 + (y + a)3/2 + b.



35. z =



36. (4 + a2) (log z)2 = (x2 a log y + b)2.



37. z =



38. z = 2ax ev + 2a2e2y + b. 39.



40. z 2 1 + a 2 = ± 2(ay + log x) + b. 41. z3/2 + (x + a)3/2 + (y + a)3/2 + bf(xy + yz + zx)

sin y + b . a

32. z = a log x – a log sec y – y + b. 34. z = a ( x + y ) + (1 − a 2 ) ( x − y ) + b .

y 1 a log( x 2 + y 2 ) + (1 − a 2 ) tan −1 + b. 2 x y a2 x x a2 x sin h −1 + ( a2 + x 2 ) + ( y 2 − a 2 ) − cosh −1 + b. 2 2 2 a 2 a (1 + a)z=

( ax ) + y + b.

Exercises 2(d) Page 145

 x − y xy  1. F  ,  = 0 2. F(xy, x – log z) = 0 z   z



3. log (x2 + y2 + z2 + 2xy) – 2x = F(x + y)



  1 1 1 1 x 4. F  xy − z 2 ,  = 0 . 5. f  − , −  = 0 y  x y y z

 a−x b−y 6. F(x – z, y – z) = 0 7. F  ,  = 0. b−y c−z  8.

x − y= f ( y − z ) 9. (x – y) = f(x – z)

y 10. z = x n f   . 11. F(x3 – y3, x2 – z2) = 0 x 12. xyz = F(x2 + y2 – 2z) 13.

 1 1 1 1 + = f  + . x y y z

 cos y cos z  14. x2 + y2 = f(yz – y2). 15. f  , =0.  cos x cos y  x+y+2 16. y − z= f ( x − y ) . 17. z(y – x) = f   z  

300

Engineering Mathematics-III (Third Semester)

18. x2 – y2 – z2 = F(y2 – x2 + 2xy) x+y+z 2 2 2 23. x2 – z2 = f   . 24. x + y + z = f(xyz) y   25. log (x2 + y2 + z2 + 2xy) – zx = f(x + y) Exercises 2(e) Page 164 1   1. z = Q1(y – 2x) + Q2  y − x  . 2. z = f1(y – 2x) + f2(y + x). 2   3. z = Q1(y – 3x) + x Q2 (y – 3x). 4. z = j1(y + x) + xj2(y + x) + x2j3 (y + x). 5. z = f1(y + 2x) + xf2 (y + 2x) + f3(y – 3x). 6. y = f(x – at) + j(x + at). x2 . 4

7. z = j1(y – x) + j2(x + 2y) +

8. z = j1(y + x) + xj2 (y + x) + j3 (y + 2x) –

9. z = j1(2y – 3x) + xj2(2y – 3x) +

x 2 y + x 1 y −2 x e − e + xe y + 2 x . 2 36

x 2 ex − 2 y e . 8

1 x+2y e 3



10. z = f1(y + x) + f2(y – x) –



11. z = j1(y + x) + xj2(y + x) + ex + 2y.



12. z = f1(y) + f2(y + 2x) +



13. z = f1(y) + xf2(y) + f3(y + 2x) +



14. z = j1(y + x) + j2(y – 5x) +

x3 y4 − . 6 60



15. z = j1(y – x) + xj2 (y – x) +

x4 y x5 − . 12 30



16. z = f1(y – 2x) + f2(y + 3x) +

x4 x3 y + . 24 6



17. z = f1 (y – 2x) + f2 (y – x) +

x2 y x3 − 2 3



18. z = f (y – x) + f (y – 2x) + F(y + 3x) +



19. z = f1(y + x) + f2(y – x) + xf3 (y – x) +

1 2x x4 y x5 e + + 4 12 30 1 2x 1 5 1 e + x y + x6 . 4 20 60

x 5 1 1 4 2 x2 y3 x y + cos( x − y ) + x 6 + x 5 (1 + 21y ) + 4 72 60 24 6 1 2x+y 1 e − x cos( x + y ) . 9 4

301

Fourier Transforms

1 1 cos( x + 2 y ) − cos (x – 2y) 2 6



20. z = f1(y) + f2(y + x) +



21. z = f1(y + 2x) + xf2 (y + 2x) +



22. z = f1(y + 3x) + xf2(y + 3x) + x2(3x + y).



23. z = f1(y + 2x) + f2 (y – 3x) +



24. z = f1 (y + x) + xf2(y + x) – sin x.



25. z = f1(y) + f2(x + y) +



26. z = f(y – 2x) + f(y + 2x) + F(y + 3x) +



27. z = f1(y + 2x) + f2(y – x) + yex.



28. z = f1(y – 5x) + f2(y – x) –



x 2 2x+y e . 2

x sin (2x + y). 5

1 (sin x cos 2y + 2 cos x sin 2y) 3

1 2 1 xy 2 − y 2 − y 4 . 10 75 60 x 1 29. z = f1(y + 2x) + f2(y – 2x) – cos (2x + y) + sin (2x + y) 4 6



30. z = f1(y + x) + xf2(y – x) + f3(y + x) –



31. u =



3x sin x + y. 4

2

y 1  [log (x2 + y2)]2 + 2  tan −1  + f(x + iy) + g(x – iy) x 2  3x 32. z = f(y – x) + xf(y – x) + F(y + x) – sin (x + y). 4



33. z = f1(y – x) + f2(y + 3x) + f3(y – 2x) +



34. z = f1(x + y) + xf2(x + y) +



1. z = exf(y – x) + e–x f(y) –



1 sin (2x + y). 4

x x 5 x 4 y 2 x3 y3 7 x5 y 5x6 + + + + cos( y − x ) + 4 60 24 6 20 72

x2 x+y e [1 + cos (x + y)] 2 Exercise 2(f) Page 170

x –x e . 2 x 1  x3 y x2 y x x2 x4  2. z = f1(y – x) + e2x f2(y – x) – ex – y –  + + + − . 2  3 2 2 2 12  2



3. z = f1(y) + ex f2 (y – x) + e2x f3(y – 3x) +



4. z = f1(x) + e3y f2(x – 2y) +

1 2 x + 3 y 1  x 2 y 3 xy 5x2 (e )+  + + 7x + 72 2  2 2 4

3 [4 cos (3x – 2y) + 3 sin (3x – 2y)]. 50

  . 

302

Engineering Mathematics-III (Third Semester)



5. z = f1(x – 2y) + e3y f2(y + x).



6. z = f1(y – x) + e–2x f2(y + 2x) –

x 1 − cos (2 x + y ) + (6 xy − 2 x 2 + 9 x − 6 y − 12). 6 24



7. z = exf1(y) + e–xf2(y + x) +



x 8. z = f1(xy) + f2   y



1 2x+3y e 10

1 sin (x + 2y) – xey. 2 9. z = f(y) + e–xf(x + y) +

1 3 x + xy2 – x2 + 6x 3

1 sin( x + 2 y ) 2 1 y–2x 11. z = exf1(x + y) + e2x f2(x + y) + e . 20

10. z = exf(y) + e–x f(x + y) – xey +

1 1 y–x sin (x + 2y) e . 2 2



12. z = exf(y) + e–xf(y + x) +



13. Z = exf1(y) + e–x f2(x + y) +

1 sin (x + 2y) 2

(BOUNDARY VALUzE PROBLEMS) Exercise 3(a) Page 175

{



1. z = Ae[1+



2. z =



3. z = ke 4 cx ⋅ e −3 cy .



4. z = ( Ae x

1 2

(1+ k )]x

+ Be[1−

(1+ k )x ]

}e

− ky

.

sinh x 2 + e −3 y sin x . 4

3

k

+ Be − x

z = (A sin x

k

)e 2 ky if k > 0.

−k + B cos x

−k e2ky if k < 0

z = Ax + B if k = 0.

6. z = 6e–3x – 2y    7. u = 4e–3x – 2t

8. u = 4 e

3 −x+ y 2

Exercises 3(b) Page 200

1. (a) y = y0 sin



2. y(x, t) =

px pat 2 px 2 pat 2 px 2 pat − k sin cos . (b) y = k sin cos cos l l l l l l

v0 l px pat sin sin . pa l l

303

Fourier Transforms



4. y (x, t) = 6. y(x, t) =

∞

24l 2 p 8l p

1

∑ (2n − 1)3 sin

3

n =1

2 ∞

3

1

∑ (2n − 1)3 sin

n =1

(2n − 1)px (2n − 1)pat cos . l l

(2n − 1)px (2n − 1)pat cos l l

where l = 60 and a2m = 2000 and m = mass per unit length of the string.

9. y(x, t) =

∞

(−1)n +1

∑ (2n − 1)2 p2 sin

n =1



10. y(x, t) =

2(n − 1)px (2n − l)pct cos a a

9d 3  px pat 1 2 px 2 pat 1 4 px 4 pat  sin cos cos cos − 2 sin + 2 sin − ... . 3  l l l l l l 2p  2 4  Exercise 3(c) Page 225 2 2 2 2 4  − a2 t 1 1  sin x − 2 e −3 a t sin 3 x + 2 e −5 a t sin 5 x − ... . e  p 3 5 



1. q(x, t) =



a 2 p2 t 3 2 p2 a 2 t   px − t2 4l  1 1 3 px − t2 2. u(x, t) = 2 2 sin e e − 2 sin + ... .  1 l 3 p 1  



3. T(x, t) =



a2 n2 p2 t/t 2  2l  ∞ (−1)n +1 npx − t2 sin e 4. q(x, t) = ∑ l p  n =1 n 

ao ∞ npx + ∑ an cos exp(– a2 n2 p2 t/t2). 2 n =1 l  .  

2 2 8 ∞ 1 8 ∞ 1 ( , ) sin(2n − 1)xe −α (2 n −1) t 5. u(x, t) = = u x t ∑ ∑ 3 3 p p (2n − 1) n 1= n 1 (2n − 1) =





l 4l 7. u(x, t) = − 2 2 p



8. u(x, t) =



9. u(x, t) =

∞

(2n + 1)px − ∑ (2n + 1)2 cos l ⋅ e n=0

l2 l2 − 6 p2

1

∞

1

∑ n2 cos

n =1

a2 (2 n +1)2 p2 t t2

2npx Exp [– 4a2n2p2t/l2]. l

∞

2 2 2 4 1 cos 2nx ⋅ e −4 n a t . − ∑ 2 p p n =1 (4n − 1)

∞



(2n − 1)px − cos e 12. u(x, t) = 50 − 2 ∑ 2 100 p n =1 (2n − 1)



13. (a) u(x, t) =

400

1

a2 p2 (2 n −1)2 t 100 2

200 ∞ npx ∑ (−1)n+1 sin l Exp [– n2p2a2t/l2] p n =1

(b) u(x, t) = 25 +

50 x 50 ∞ 1 2npx − ∑ sin l Exp [– 4n2 p2 a2t/l2] p n =1 n l

304

Engineering Mathematics-III (Third Semester)

3 − kt 1 3 sin x − e −9 kt sin 3x. 4 4



14. u(x, t) =



80 ∞ 1 npx − sin e 15. u(x, t) = 90 – 30x – ∑ 5 p n =1 n



20 16. u(x, t) = 20x + 20 + p



60 ∞ 1 npx − sin e 17. u(x, t) = 50 – 4x – ∑ 5 p n =1 n



18. u(x, t) = −



19. u(x, t) = 50 +



20. u(x, t) = 30 + x +



2x 40 ∞ 1 npx − + 40 − sin e 21. u(x, t) = ∑ p n =1 n 3 5



(2n − 1)px − cos e 22. u(x, t) = 50 − 2 ∑ 2 50 p 1 (2n − 1)



(2n + 1)px − sin e 24. q(x, t) = 2 ∑ 2 l p n =1 (2n + 1)



100 ∞ 1 2npx − sin e 25. u(x, t) = ∑ l p n =1 n



 (−1)n +1 800 200  (2n − 1)px − − e 26. u(x, t) = 50 + ∑   sin 2 2 (2n − 1)p  2l 1  (2n − 1) p

n 2 p2 a 2 t 25

∞

1 npx − ∑ n (1 − 6 cos np)sin 10 e 1

α 2 n 2 p2 t 100

α 2 n 2 p2 t 25

5x 100 ∞ 1 npx − − 50 + (2 cos n p − 1)sin e ∑ 4 p 1 n 40

n 2 p2 a 2 t 1600

100 x 200 ∞ 1 (2n − 1)px −α 2 (2n − 1)2 p2 t − sin e ∑ l p n =1 (2n − 1) l l2  0 ⋅ 15n2 p2 t  20 ∞ 1 npx (3 + 2 cos np) ⋅ sin ⋅ exp  − . ∑ n 50 2500  p n =1 

400

8T

.

∞

∞

α 2 n 2 p2 t 225

1

(−1)n

α 2 p2 (2 n −1)2 t 2500

(2 n +1)2 p2 kt l2

α 2 4 n 2 p2 t t2

.

∞

(2 n −1)2 α 2 n2 t 4t2

.

Exercise 3(e) Page 249



400 ∞ 5. u(x, t) = ∑ p n =1



sin

u (50, 50) =

(2n − 1)py (2n − 1)px sinh 100 100 (2n − 1)sinh ( 2n − 1) p 400 p

∞

∑

n =1

1 (2n − 1)p (2n − 1)p . sin sin h (2n − 1) sin h (2n − 1) p 2 2

305

Fourier Transforms



6. T(x, y) =

∞

3200

1

∑ (2n − 1)3 sin

p3

n =1

(2n − 1)px − e 10

∞



7. T(x, y) =

160 1 (2n − 1)px − sin e ∑ 30 p n =1 (2n − 1)



9. u(x, y) =

4u0 p



10. u(x, y) = 11. u(x, y) =

.

sin(2n − 1)px /100 ⋅ sinh(2n − 1)py /100 (2n − 1)sinh(2n − 1)p n =1

∑

∞

3200

∑

p3 ∞

(2 n −1)py 30

∞

u (50, 50)=



(2 n −1)p 10

2u0 p

sin

n =1



∑  Cn cosh

n =1

3p 1 5p p 1   sec h 2 − 3 sec h 2 + 5 sec h 2 − ... .  

( 2n − 1)py (2n − 1)px sinh 20 20 (2n − 1)3 sinh(2n − 1)p npy npy  npx + Dn sinh sin a a  a

a

where Cn =

2 npx npb f ( x )sin dx and Dn = – Cn tanh . a ∫0 a a

npy npy npx  npb  200 ∞ 1 sinh tanh . ∑ (1 − cos np)sin a  cos a = a a  p n =1 n



12. u(x, y) =



13. u(x, y) = 3 sin



16. u(x, y) =

8 a2 p3

3 x( b − y ) 4 p(b − y ) 3 px 3 pb 4 px 4 pb sinh cosech + 5 sin sinh cosech . a a a a a a ∞

∑

n =1

sin

(2n − 1)p(b − y ) (2n − 1)px sinh a a (2n − 1)pb 3 (2n − 1) sinh a npy 5



18. u(x, y) =

30 ∞ (−1)n +1 npx − sin e ∑ n 5 p 1



21. u(x, y) =

∞ (−1)n sinh n(p − y )cos nx p (p − y) + 4 ∑ 3 n2 sinh np n =1

.

Exercises 3(f) Page 266

2. u(r, q) =

3 5  800  r  1  r  1  r  sin 3 sin q − q +     sin 5q − ... . 2  2  2  3  10  5  10  p  10  



3. u(r, q) =

200 ∞ (−1)n +1  r  ∑   p n =1 (2n − 1)2  10 



8k 4. u(r, q) = p

∞

∑

n =1

r 3   (2n − 1)  a  1

2 n −1

2 n −1

sin(2n − 1)q

sin(2n − 1)q .

306



Engineering Mathematics-III (Third Semester)

5. u(r, q) =

400 p2

∞

1

∑ n2 sin

n =1

n

np  r    sin nq . 2 a

200 ∞ 1 r ∑   p n =1 (2n − 1)  a 



6. u(r, q) = 50 +



7. u(r, q) =



800 1 r 8. u(r, q) = ∑ 3   9p (2n − 1)  a 



9. V =



2 ∞ 1 a ∑   p n =1,3,5... n3  r 

2n

2 n −1

sin(2n − 1)q .

 r 4n − b 4n  4 n 4n  a −b 2 n −1 3

  sin 2nq . 

 2n − 1  ⋅ sin  q .  3 

n

r ∑  a  Cn cos nq . n

10. u(r, q) =

400 p

∞

1  r   3  n =1,3,5... n  10 

∑

2n

sin 2n q; 45.7°.

Exercises 3(g) Page 272

1  400  100   1. T = 2  r − − r  cos q .  sin q +  2 r r    



1  1  2. u(r, q) = 4  r −  sin q + 4  r +  cos q . r r   

5

THE Z-TRANSFORMS

The Z-transform plays an important role in the communication engineering and control engineering. It is used in the analysis and representation of discreate-time linear shift invariance system. The application of Z-transform in discrete-time systems is similar to that of the Laplace Transform in continuous-time systems. When continuous signals are sampled, discrete-time functions arise. We shall see something about the sampler and holding devices. Sampler and holding devices: The important element of a discrete-time system is the sampler. In a sampler, a switch closes to admit an imput signal in every T seconds. A sampler is a conversion device which converts a continuous signal into a sequence of pulses occuring at the sampling instants 0, T, 2T, .....; here T is called the sampling period. If two signals give equal values at the sampling instants, then they give the same sampled signal. A holding device is one which converts the sampled signal into a continuous signal, which is very close to the signal applied to the sampler. The simplest holding device is one which converts the sampled signal into one which is constant between two consecutive sampling instants. Definition of the Z-Transform Let {x(n)} be a sequence defined for n = 0, ±1, ±2, ±3, ... . Then the two sided Z-transform of the sequence x(n) is defined as

Z {x (n)} = X (z) =

∞

∑

n = −∞

x ( n) ⋅ z − n

... (1)

where z is a complex variable in general. If {x(n)} is a causel sequence, i.e., x(n) = 0 for n < 0, then the Z-transform reduces to one-sided Z-transform and its definition is

Z {x(n)} = X(z) =

∞

∑

n=0

x(n) ⋅ z − n

... (2)

The infinite series on the right hand side of (1) or (2) will be convergent only for certain values of z depending on the sequence x(n). The inverse z-transform of Z {x (n)} = X(z) is defined as Z–1 [X (z)] = {x(N)} Symbols: Sometimes we write Z{x(n)} = Z[{x(n)}] and {x(n)} as x(n) (which is a sequence and not a single data). The double braces { } are used for a sequence. Unit Sample Sequence and Unit Step Sequence The unit sample sequence d(n) is defined as the sequence with values 1 for n = 0 d (n) =  0 for n ≠ 0 The unit step sequence u (n) has values 1 for n ≥ 0 u (n) =   0 for n < 0 307

... (3)

... (4)

308

Engineering Mathematics-III (Third Semester)

The above two sequences are related as

u (n) =

n

∑

m = −∞

d ( m)

... (5)

and d (n) = u(n) – u(n – 1) we have d (n – k) = 1 if k = n = 0 if k ≠ n and u (n – k) = 1 for (n – k) ≥ 0 = 0 for (n – k) < 0 Also,

x (n) =

∞

∑

... (6) ... (7) ... (8)

x(m) d (n − m)

m = −∞

In discrete-time systems, the input signal f(t) is sampled at discrete instants 0, T, 2T, ...nT, ... where T is the sampling period. For such functions, the Z-transform (one sided) becomes

Z [f (t)] = F(z) =

∞

∑

n=0

f (nT )z − n

... (9)

Def. Therefore, if f(t) is a function defined for discrete values of t where t = nT, n = 0, 1, 2, 3, ... to ∞, T being the sampling period, then Z-transform of f(t) is defined as

∞

−n Z [f(t)] = ∑ f (t)z =

∞

∑

= n 0= n 0

f (nT )z − n

... (10)

The continuous function f(t) and discrete function f(t) = f(nT), n = 0, 1, 2, ... ∞ are given below for understanding by way of graph. Notation: Z [f(t)] = F(z) Convergence of series

The series of one-sided z-transform, namely, Lt if

n →∞

∞

∑

n=0

x(n) z − n is convergent, by Cauchy’s ratio  test,

x(n + 1) ⋅ z − n −1 x ( n) z − n

1. z − 1 1 − z −1

Proof:

∞

∞

( n) z − n Z {u(n)} = ∑ u= = n 0

=

∞

= z ∑ ∑ −n

= n 0= n 0

1   z

n

1 z = if |z–1| < 1 i.e., |z| > 1 1 − z −1 z − 1

(Here u(n) is the unit step sequence already defined) z Theorem 4. Z[an f(t)] = F   a or Proof:

z Z[{an x (n)}] = X   a ∞

−n n Z[an f(t)] = ∑ a f (nT ) z n=0

=

∞

∑ n=0

z f (nT ) ⋅   a

−n

z = F  a

Or

∞

−n n Z[{an x(n)}] = ∑ a x(n) z n=0 ∞

z x ( n)   a

−n

=

∑

Theorem 5.

z if |z| < a. z−a

Proof:

Z{an u (n)} =

n=0

z =X  a

∞

n −n Z {an u (n)} = ∑ a z n=0 ∞

1 if |az–1| > 1 1 − az −1

=

∑

=

z if |z| > |a| z−a

Theorem 6.

(az −1 ) n =

n=0

Z[n f (t)] = − z

sF ( z ) dz

311

The Z-transforms

Proof: \ Theorem 7.

F(z) = Z[f (t)] =

∞

∑

f (nT ) z − n

n=0

dF ( z ) − n −1 = ∑ − n f (nT ) z dz n=0 ∞

z

∞ dF −n − Z [nf (t )] = −∑ n f (nT ) z = dz n=0

Z [n f (t)] = − z

dF ( z ) dz

Z [f(t + kT)] = Z[ f(n + k)T ]

f (1.T ) f (2 T ) f ((1 − 1)T )  k  − − = z  F ( z ) − f (0.T ) −  z z2 z k −1   Proof:

∞

−n Z [f(k + n) T] = ∑ f ((n + k )T ) z n=0

∞

−(n+ k ) k = z ∑ f ((n + k )T ) z n=0 ∞

−m k = z ∑ f (mT ) z where m = n + k m=k

∞ k −1 −m −m  k  = z  ∑ f (mT ) z − ∑ f (mT ) z  =  m 0=m 0 



f (T ) f (2T ) f (3T ) f ((k − 1)T )  k  − − − ... − = z  F ( Z ) − f (0) −  z z2 z3 z k −1  

Note: If f ((n + k)T) is denoted by fn + k, then Z [f(t + kT)] = Z [fn + k] f  f f k  = z  F ( z ) − f 0 − 1 − 22 − ... − kk −−11  z z z   Z-Transform of Standard Functions Find the Z-Transform of sequence {x (n)} or {fn} where x (n) is given by (i) x (n) = k (ii) x (n) = (–1)n (iii) x (n) = an (iv) x(n) = n (v) x (n) = nan (vi) x(n) = sin nq (vii) x (n) = cos nq (viii) x (n) = rn sin nq (ix) x (x) = rn cos nq (x) x(n) = n(n – 1) (ix) x(n) = n2 ∞  1 1  −n Solution. (i) Z (k) = ∑ kz = k 1 + + 2 + ...to ∞  z z   n=0 1 kz if | z | > 1 k = = 1 z −1 1− z

312

Engineering Mathematics-III (Third Semester)

Cor.

Z (1) =

z if |z| > 1. z −1 ∞

∞

−n n −n Z {(–1)n} = ∑ (−1) z =∑ (− z )

(ii)

= n 0= n 0

=

1

1 1+ z

if | z | >1 =

z if | z | > 1 z +1

∞ ∞ a n −n Z {an} = ∑ a z = ∑   = n 0= n 0  z 

(iii)

=

1

a 1− z ∞

z|>|a| if | =

−n Z {(n)} = ∑ nz = − z

(iv)

n=0

= − z z {nan} = − z

(v)

n

z if | z | > | a | z−a

d [ Z (1)] dz

d  z  z  = dz  z − 1  ( z − 1) 2 d  z  az =   dz  z − a  ( z − a ) 2

(vi), (vii) Z {sin nq} and Z {cos nq} z if |z| > |a| z−a

We have proved

Z {an} =

Taking

a = eiq, z Z [einq] = z − eiq



Z {cos nq + i sin nq} =

=

z ( z − cos q + i sin q) ( z − cos q − i sin q) ( z − cos q + i sin q) z ( z − cos q + i sin q) ( z − cos q) 2 + sin 2 q

Equating real and imaginary parts, and

z ( z − cos q) if | z | > 1 z − 2 z cos q + 1 z sin q Z {sin nq} = 2 if | z | > 1 z − 2 z cos q + 1 Z (cos nq) =

2

(viii)

Z {rn cos nq} =

z ( z − r cos q ) if | z | > | r | z − 2 zr cos q + r 2

(ix)

Z {rn sin nq} =

zr sin q if | z | > | r | z 2 − 2 zr cos q + r 2

2

313

The Z-transforms

a = reiq in Z {an} z Z {rn ein q} = z − reiq  z  z  z    − cos q + i sin q    r r r       Z {rn (cos nq + i sin nq)} = = 2  z  iq  z   2 e −    r  − cos q  + sin q r    Taking

Equating real and imaginary parts,

Z {rn cos nq} =

z ( z − r cos q) z − 2 zr cos q + r 2



Z {rn sin nq} =

zr sin q if | z | > | r | z − 2 zr cos q + r 2

2

2

(x)

Z (n2) = Z (n n) = − z

(xi)

Z [{n (n – 1)}] = Z [{n2 – n}]

d  z  z ( z + 1)   = dz  ( z − 1) 2  ( z − 1)3 z ( z + 1) z − ( z − 1)3 ( z − 1) 2

2 ) = Z {n } − z (n=

=

z 2 + z − ( z − 1) ( z − 1)

3

=

2z ( z − 1)3

Standard Results Find the Z-transform of f (t) where f (t) is is given by (i) t (ii) e–at (iii) eat (iv) sin wt (v) cos wt (vi) tk (i)

Z (t) =

∞

∑

n 0= n 0 =

= T ⋅ (− z ) = (ii)

d d  z  [ Z {1}] =−Tz   dz dz  z − 1 

Tz ( z − 1) 2 ∞

∞

− anT − n − aT n − n Z (e–at) = ∑ e z = ∑ (e ) z

= n 0= n 0

= (iii)

∞

(nT ) z − n = T ∑ nz − n

z using the result of Z {an} if |n| > |e–aT| z − e − aT ∞

an T − n Z (eat) = ∑ e z =

∑

(e at ) z − n

= n 0= n 0

=

z if | z | > |eaT| z − e aT

314

(iv)

Engineering Mathematics-III (Third Semester) ∞

Z [sin wt] = ∑ sin n wT. z–n n=0

= Z {sin nq} where q = wT z sin wT = 2 if |z| > 1 z − 2 z cos wT + 1 (v)

∞

−n Z[cos wt] = ∑ cos n wT ⋅ z n=0

= Z {cos nq} where q = wT z ( z − cos wT ) = 2 z − 2 z cos wT + 1  e i wt + e − i wt  Z [cos wt] = Z   2   1 z z  + =  i wT − i wT  2 z −e z −e  i wT − i wT 1  2 z − (e + e )  = ⋅ z   2  ( z − eiwT )( z − e − iwT )  or

=

 1  2 z − 2 cos wT ⋅z  2  z 2 − ( e i wT + e − i wT ) z + 1 

=

z ( z − cos wT ) z − 2 z cos wT + 1

Similarly,

2

 e i wt − e − i wt  Z [sin wt] = Z .   2i  

1 z z  − i wT − i wT   2i  z − e z −e  i wT − i wT  z  (e − e ) =   2i  z 2 − z (eiwT + e − iwT ) + 1  =

=

z sin wT z − 2 z cos wT + 1 2

d [ Z (t k −1 )] dz



Z (tk) = − Tz

(vi)

k −n Z (tk) = ∑ (nT ) z

∞

n=0

∞

∑

= Tz

n k T k −1 z − ( n +1)

n=0 ∞

− ( n +1) k −1 = Tz ∑ (nT ) ⋅ nz n=0

Similarly,

∞

k −1 − n Z (tk – 1) = ∑ (nT ) z n=0

... (1)

315

The Z-transforms



∞ d [ Z (t k −1 )] = ∑ (nT ) k −1 (−n) z − ( n +1) dz n=0 ∞

− ( n +1) k −1 = −∑ (nT ) n ⋅ z

... (2)

n=0

Using (2) in (1), we get

Z (tk) = −Tz

d [ Z (t k −1 )] dz

... (3)

Setting k = 1, 2, 3, .... we get Z (t), Z (t2), .... d Z (t) = −Tz [ z (1)] dz = −Tz

d  z    dz  z − 1 

Tz ( z − 1) 2



Z (t) =



Z(t2) = −Tz



Z (t2) =

... (4)

d d  Tz  Z (t ) = −Tz   dz dz  ( z − 1) 2 

T 2 z ( z + 1) ( z − 1)3

... (5)

Shifting Theorem I If Proof:

Z [f(t)] = F(z), then Z [e–at f(t)] = F [zeaT] ∞

− anT f (nT ) z − n Z[e–atf(t)] = ∑ e n=0

=

∞

∑

f (nT )( ze aT ) − n

n=0

= F [zeaT] since F (z) = i.e.,

∞

∑

f (nT ) z − n

n=0

Z [e–at f(t)] = z [ f (t )]z → zeaT

= F (z) where z → zeaT Deductions: Using shifting theorem, derive the following results (which are already proved by using definition). 1. Z (e–et) = Z (e–at . 1) = [ Z (1)]z → zeaT ze aT  z  = =   aT  z − 1  z → zeaT ze − 1

Z (e–at) =

z z − e − aT

316

2.

Engineering Mathematics-III (Third Semester)

Z[e–at . t] = [ z (t )]z → zeaT

 Tz  =  2   ( z − 1)  z → zeaT =

Tz e aT ( ze aT − 1) 2

Z (te–at) =

Tz e − aT ( z − e − aT ) 2

3.

Z [e–iat] = Z [e–at . 1] Z [cos at – i sin at] = [z (1)]z → zeiaT  z  =    z − 1  z → zeinT =

zeiaT z = zeiaT − 1 z − e − iaT

=

z ( z − eiaT ) ( z − e − iaT )( z − eiaT )

=

z[ z − (cos aT + i sin aT )] z 2 − z (eiaT + e − iaT ) + 1

=

z[( z − cos aT ) − i sin aT ] z 2 − 2 z cos aT + 1



Equating real and imaginary parts, z ( z − cos aT ) Z [cos at] = 2 z − 2 z cos aT + 1 and 4.

Z [sin at] =

z sin aT z 2 − 2 z cos aT + 1

Z [e–at cos bt] = Z (cos bt)z → zeaT

 z ( z − cos bT )  =  2   z − 2 z cos bT + 1  z → zeaT = 5.

ze aT [ ze aT − cos bT ] z 2 e 2 aT − 2 z e aT cos bT + 1

Z [e–at sin bt] = [Z (sin bt)]z → zeaT

=

2

z e

2 aT

ze aT sin bT − 2 z e aT cos bT + 1

Initial Value Theorem If Z [f (t)] = F (z), then f (0) = Lt F ( z ) n →∞

317

The Z-transforms

Proof:

F(z) = Z [f(t)] =

∞

∑

f (nT ) z − n

n=0

f (1.T ) f (2.T ) + + ...to ∞ = f(0.T) + z z2 = f (0) +

1 1 f (T ) + 2 f (2T ) + ... to ∞ z z

Taking limit as z → ∞ Lt F ( z ) = f (0) z →∞

z ( z − cos aT ) , find f (0). z 2 − 2 z cos aT + 1

Example. If

F(z) =



f (0) = Lt F ( z ) = 1 by L’Hospital’s rule. n →∞

Shifting Theorem II If

Z [f(t)] = F (z) then Z [f(t + T)] = z [F(z) – f (0)] ∞

−n Z [f(t + T)] = ∑ f (mT + T ) z

Proof:

n=0

=

∞

∑

f ((n + 1)T ) ⋅ z − n

n=0

∞

− ( n +1) = z ∑ f [(n + 1)T ]z n=0 ∞

−k = z ∑ f (kT ) z k =1

∞  −k = z  ∑ f (kT ) z − f (0)   k =1  = z [F(z) – f (0)] Final Value Theorem f (t ) Lt ( z − 1) F ( z ) Z [f(t)] = F(z) then Lt =

If

t →∞

z →1

∞

−n Z [f (t + T) – f(t)] = ∑ [ f (nT + T ) − f (nT )]z

Proof:

n=0 ∞



−n Z [f(t + T) – Z [f (t)]] = ∑ [ f (nT + T ) − f (nT )]z



−n zF (z) – zf (0) – F (z) = ∑ [ f (nT + T ) − f (nT )]z

n=0 ∞

n=0

Taking limit as z → 1,

∞

Lt ( z − 1) F ( z ) − f (0) = Lt ∑ [ f (nT + T ) − f (nT )]z − n

z →1

=

z →1 ∞

∑ n=0

n=0

[ f (nT + T ) − f (nT )]

318

Engineering Mathematics-III (Third Semester)

= Lt [ f (T ) − f (0) + f (2T ) − f (T ) + ... + f ((n + 1)T ) − f (nT )] n →∞

= Lt f ((n + 1)T ) − f (0) n →∞

= f(∞) – f(0) (t ) Lt ( z − 1) F ( z ) f (∞) = Lt f= t →∞

t →1

z , find Lt f (t ). Example. If F (z) = t →∞ z − e −T Solution.

∞ Lt z ( z − 1) ⋅ LT f (t ) = ∑ −T t →∞ z →1 z −e n=0

0 = Lt f (t )



t →∞

10 z , find f (0). Example. If F (z ) = ( z − 1) ( z − 2) = f (0) Solution.

Lt = F ( z)

z →∞

10 Lt= 0 2z − 3

z →∞

Example 1. Find the Z-transform of 1  (i)   n

np   (ii) cos  2 

Solution. (i)

 1  (iii)  , n ≥ 1  n(n + 1) 

 1  ∞ 1 −n Z =∑ z  n  n =1 n

=

1 1 1 + + + ... to ∞ z 2 z 2 3z 3

 1 1 1  z −1  (ii)

∞ np  np  Z cos  = ∑ cos ⋅ z − n 2 2  n=0

= 1 −

1 1 + − ... to ∞ z2 z4 −1

1  z2  if |z| > 1 = 1 + 2  = 2 z +1  z  (iii) Let

1 A B = + n(n + 1) n n +1

=

1 1 − n n +1

∞   1  1 −n ∞ 1 ⋅ z −n Z   = ∑ ⋅ z − ∑ n n + n ( n + 1) 1 = n 1 = n 1    

319

The Z-transforms

z 1 1  − + + ... z − 1  2 z 3 z 2  2  1  1  1  1 3  z − z    +   + ... = log z −1  2  z  3  z   z  1 + z log 1 −  + 1 = log z −1  z  z −1  = ( z − 1) log   +1  z  Shift Property. (Shift of a causal sequence to the right) Z {x (n – m)} = z–m X (z) where x (n) is a causal sequence and m is a positive integer. = log

Proof:

∞

−n Z {x(n – m)} = ∑ x(n − m) z n=0

= =

∞

∑

x( p) z − ( m + p )

p= −m ∞

∑

x( p) z − p ⋅ z − m

p =0

= z–m X (z) \ Z {x (n – m)} = z–m X (z) Cor. Z–1 [z–m X(z)] = {x(n – m)} = [Z–1 X(x)]n → n – m Example 2. Find Z–1

   1    z− 1 2 

Solution.

      −1  z    1  −1    = Z  z    z − 1   z− 1   2 2   

Z–1

   z  = Z   1  z −  2  n → n −1  −1

1 =   2

n −1

1 or   2

n −1

u (n − 1)

z   1  – 1  n Example 3. Evaluate Z – 1   given Z    = (–1) .  z +1  z +1 Solution.

z   1  −1  −1 Z – 1   = Z  z ⋅ z + 1  z + 1    

320

Engineering Mathematics-III (Third Semester)

z  −1  = Z    z + 1  n → n −1 = (–1)n–1, n = 1, 2, 3, ... z ( z − a cos q) Example 4. Prove Z {an cos nqu (n)} = 2 z − 2az cos q + a 2 Z {an sin n qu (n)} =

az sin q . z 2 − 2az cos q + a 2

Solution. These are same as Z{rn cos nq} and Z {rn sin nq} (which are worked out earlier) Example 5. Find the Z-transform of np u ( n) (i) u (n – 1) (ii) 3n d(n – 1) (iii) cos (iv) d(n – k) 2 Solution. (i)

Z [{u (n – 1)}] =

∞

∑

1⋅ z − n

n =1

1 1 1 + + + ... to ∞ z z 2 z3   1 1  1 1 z −1 ∞

∑ n=0

3 3n d(n − 1) z − n = z

np   (iii) This problem is same as Z cos  2  See Example 1. (iv)

∞

−n Z [d (n – k)] = ∑ d(n − k ) z n=0

= \

Z [d(n – 1)] =

1 , if k is a positive integer. zk 1 z

Example 6. Find the Z-transform of (i) an cos np (ii) e–t t2 (iii) et sin 2t (iv) e3t cos t (v) e–2t t2 (vi) e3t + 7 (vii) eat + g Solution. (i)

∞

−n n Z (an cos np) = ∑ a cos np ⋅ z n=0

321

The Z-transforms 2

= 1 −

a a a 3 +   −   h + .... to ∞ z z z −1

a  a |a| z+a

Z (e–t t2) = [Z(t2)]z → zeT

 T 2 z ( z + 1)  =  3   ( z − 1)  z → zer = (iii)

T 2 ⋅ zeT ( zeT + 1) ( zeT − 1)3

Z [et sin 2t] = Z [sin 2t ]z → ze−T

  z sin 2T –T =  2  where z → ze z 2 z cos 2 T 1 − +   = (iv)

z 2 e −2T

ze −T sin 2T − 2 ze −T cos 2T + 1

Z [e3t cos t] = [ z (cos t )]z → ze − 3T

 z ( z − cos T )  –3T =  2  where z → ze  z − 2 z cos T + 1  = (v)

ze −3T ( ze −3T − cos T ) z e −6T − 2 ze −3T cos T + 1 2

3 Z [e–2t t3] = [ z (t )]z → ze2T

 T 3 z (1 + 4 z + z 2 )  =   ( z − 1) 2   z → ze2T = (vi)

T 3 ze 2T [1 + 4 ze 2T + z 2 e 4T ] ( ze 2T − 1) 4

Z[e3t + 7] = e7 Z (e3t) = e7 .

z z − e3T

Z [eat + b] = eb . Z (eat) z b = e ⋅ if |z| > |eaT| z − e aT (vii)

Example 7. Find the Z-transforms of (i) e2(t + T) (ii) sin (t + T) (iii) (t + T) e–(t + T)

322

Engineering Mathematics-III (Third Semester)

Solution. (i) Z [e2(t + T)] = Z [f(t + T)] where f (t) = e2t = z [F (z) – f(0)]  z  − 1 = z  2T z −e  =

z ⋅ e 2T , using shifting theorem II. z − e 2T

Z[e2 (i + T)] = e2T Z(e2t) z 2T = e z − e 2T Or

(ii) Z [sin (t + T)] = Z [f(t + T)] where f(t) = sin t = z [F (z) – f(0)]   z sin T − 0 = z  2  z − 2 z cos T + 1  = Or

z 2 sin T z 2 − 2 z cos T + 1

Z [sin (t + T)] = Z [sin t cos T + cos t sin T]

   z ( z − cos T )  z sin T = cos T  2  + sin T  2  2 cos 1 z z T − +    z − 2 z cos T + 1  =

z 2 sin T z − 2 z cos T + 1 2

(iii) Z[(t + T) e–(t + T)] = e–T Z[(t + T) e–t] = e–T [Z (te–t) + TZ (e–t)] TzeT Tz  −T  + = e  T  2 z − e −T   ( ze − 1) TzeT TeT z  −T  + T = e  T  2 ze − 1   ( ze − 1) zeT  Tz 2 eT −T T  = = e ⋅ Tze  T 2  T 2  ( ze − 1)  ( ze − 1) Or Z [(t + T) e ] = Z [f(t + T)] where f (t) = te–t = z [F(z) – f(0)] –(t + T)

 TzeT  − 0 = z  T 2  ( ze − 1)  =

Tz 2 − eT ( zeT − 1) 2

Example 8. Find the Z-transforms of (i) abn, (a ≠ 0, b ≠ 0) (ii) x (n) = n, n ≥ 0 = 0, n < 0

323

The Z-transforms

(iii)

(n + 1) (n + 2) 2

Solution. (i)

(iv) x (n) = 1 for n = k = 0 otherwise. ∞

n −n Z [{abn}] = ∑ ab z n=0

∞ b = a ∑   n=0  z 

n

 b  b 2  = a 1 + +   + ... to ∞   z  z   b if | b | z −b = a ⋅

1

∞

−n z [x(n)] = ∑ n ⋅ z

(ii)

n=0

=

1 2 3 + + + ... to ∞ z z 2 z3

=

2  1 1 1 1 + 2   + 3   + ...... z  z z 

1 1 = 1 −  z z

−2

1  z −1    z z 

−2

= =

z if | z | > 1 ( z − 1) 2

if

1 1 2  ( z − 1) 2 ( z − 1) 2 ( z − 1)  (iii)

∞

−n Z [{x(n)}] = ∑ x(n) z

(iv)

n=0

−k = z =

Example 9. Find the Z-transform of (i) x (n) = 0 if n > 0 = 1 if n < 0 (ii) x (n) =

an for n > 0 n!

= 0 otherwise.

1 zk

324

Engineering Mathematics-III (Third Semester)

Solution. (i)

Z [{x (n)}] =

∞

∑

x ( n) z − n =

n = −∞

0

∑

z −n

n = −∞

= 1 + z + z + ... to zn + ... ∞ 1 = if |z| < 1 1− z 2

(ii)

∞

a a n − n ∞ (az −1 ) n az −1 z z e e = = = ∑ n! n! 0= n 0

Z[{x (n)}] = ∑ = n

Convolution of Sequences The convolution of two sequences {x(n)} and {y(n)} is defined as

∞

w(n) = ∑ x(k ) y (n − k ) k = −∞

[Note: If it is one sided (right sided) sequence, take x (k) = 0, y (k) = 0 for k < 0] Convolution Theorem If w (n) is the convolution of two sequences x (n) and y (n), then Z [w(n)] = W(z) = Z[x(n)] . Z[y (n)] = X(z) Y (z) Proof: W(z) = Z [w (n)]  ∞  = Z  ∑ x(k ) y (n − k )   k = −∞  = =

∞

∑

n = −∞ ∞

∑

k = −∞

 ∞  −n  ∑ x(k ) y (n − k )  z  k = −∞   ∞  x(k )  ∑ (n − k ) z − n   n = −∞  by change of order of summation

= = =

∞

∑

k = −∞

  x(k )  ∑ y (m) z − ( m + k )  putting n – k = m  m = −∞  ∞

∑

 ∞  x ( k ) z − k  ∑ y ( m) z − m   m = −∞ 

∑

x(k ) z − k ⋅ Y ( z )

∞

k = −∞ ∞ k = −∞

∞

−k = Y ( z ) ∑ x(k ) z k = −∞

= Y (z) X (z) = X (z) Y (z) Note: The result is true only for those values of z inside the region of convergence. Another Form of Convolution Theorem If F(z) and G(z) are one sided Z-transforms of f(t) and g (t) respectively, then the Z-transform of

n

∑ k =0

f (kT) g (nT – kT) is F (z) . G (z).

325

The Z-transforms

Proof: (We are dealing here with one sided Z-transform) ∞



−i F (z) = ∑ f (iT ) z



−n G (z) = ∑ g (nT ) z

\

i =0 ∞

n=0

∞  ∞ −i   −n  F(z) G (z) =  ∑ f (iT ) z   ∑ g (nT ) z  =  i 0=  n 0  ∞

∞

=

∑ ∑

=

∑  ∑

[ f (iT ) g (nT ) z − i ⋅ z − n ]

= n 0=i 0



∞

n

= n 0= k 0

 f (kT ) g ((n − k )T )  z − n 

Collecting the coefficient of z in double sigma summation. –n

=

∞

∑ n=0

C (n) z − n where C(n) =

n

∑

f (kT ) g ((n − k )T )

k =0

= Z (C (n)) = Z-transform of Note: By symmetry, Z-transform of

n

∑ k =0

n

∑

f (kT ) g ((n − k )T )

k =0

g(kT) f((n – k)T) is also the same value F(z) . G(z).

Example 10. Find the Z-transform of the convolution of x(n) = u (n) and y (n) = an u(n). ∞ z −n Solution. X (z) = ∑ z = if | z | > 1 z −1 n=0 1 z = if | z | > | a | a − z a n=0 1− z The Z-transform of the convolution of x (n) and y (n) is

∞

n −n z Y(z) = ∑ a=

z z z2 ⋅ = if | z | > max (| a |, 1) z − 1 z − a ( z − 1) ( z − a ) Example 11. Find the Z-transform of the convolution of x(n) = an u(n), and y(n) = bn u (n). z Solution. X (z) = if | z | > | a | z−a z Y (z) = if | z | > | b | z −b

W(z) =



W (z) = Z-transform of convolution of x (n) and y (n).

= Note:

z2 if | z | > max (| a |, | b |) ( z − a ) ( z − b) ∞

−n Z [x (n) y (n)] = ∑ x(n) y (n) z n=0

326

Engineering Mathematics-III (Third Semester)

=

∞

∑

(ab) n ⋅= z −n

n=0

1 z = if | z | > | ab | −1 z − ab 1 − abz

The convolution of two causal sequences x (n) and y (n), we define

n

{x (n)} * {y (n)} = ∑ x(n − k ) y (k ) k =0

Convolution Theorem Ff f(n) and g(n) are two causal sequences, Z[{f(n) * g(n)}] = Z {f(n)} . Z {g (n)} = F (z) . G(z) Proof:

∞ ∞ −n   −n  F (z) . G (z) =  ∑ f (n) z   ∑ g (n) z  =  n 0=  n 0  ∞



n

=

∑ ∑

=

∑ ∑

= n 0= k 0

 f (n − k ) g (k )  z − n 

... (1)

∞  n  −n Z [f(n) * g(n)] = ∑  ∑ f (n − k ) g (k )  z = n 0= k 0 

.... (2)

∞



 f (k ) g (n − k )  z − n 

n

= n 0= k 0

By definition,

From (1) and (2), Z [f (n) * g(n)] = F (z) G(z) = Z [f(n)] . Z [g (n)] –1 Cor. Z [F (z) . G (z)] = f (n) * g (n) =

n

∑

f (n − k ) g (k ) = Z −1 ( F ( z )) * Z −1 (G ( z ))

k =0

Example 12. Find the inverse Z-transform of X(Z) = theorem. Solution.   Z–1

z2 using convolution 1  1  z − z −    2  4 

      2    z  −1  z  z −1   = Z  *Z   1    z − 1  z − 1    z − 1   z −      2 4   2  4    n

n

n 1 1 1 =   *   = ∑   2 4 k =0  2  k

n−k

1   4

k

1 n k   1 n 4 1 n 1 = =   ∑   ∑   k  2  k 0= 2 k 0 2 = 1   2 n

327

The Z-transforms n 2 n 1  1 1 1  =   1 + +   + ...     2   2  2   2  

  1  n +1  n 1 −    1  2  =    2   1− 1   2  

1 x (n) =   2

n −1

  1  n +1   1  n −1  1  2 n 1 −    =  −   2   2    2 

[Note: We will do the same problem by partial fraction method later.] Example 13. Find the inverse Z-transform of Solution.

z2 using convolution theorem. ( z − a)2

 z2  z   z Z −1  ⋅ Z–1  2  =  z −a z −a  ( z − a) 

z  −1  z  −1  n n = Z  *Z   = a *a z a z a − −     =

n

∑

n

a n − k ⋅ a k = ∑ a n = (n + 1) a n

k 0= k 0 =

or = (n + 1) an u (n) Table of Z-Transforms x (n) 1

X(z)

(–1)n

z z +1

an u (n) u (n – m)

z z −1

z if |z| > |a| z−a z–m .

z z −1

n

z ( z − 1) 2

n2

z2 + z ( z − 1)3

n (n – 1)

2z ( z − 1)3

328

Engineering Mathematics-III (Third Semester)

x (n) n(k)

X(z) k !z ( z − 1) k +1

an u(n–1)

a z−a

u (n –1)

1 z−n

n an

az ( z − a)2

u (n) cos nq

z ( z − cos q) z 2 − 2 z cos q + 1

u (n) sin nq

z sin q z − 2 z cos q + 1 2

rn cos nq

z ( z − r cos q) z − 2 zr cos q + r 2 2

rn sin nq

rz sin q z 2 − 2 zr cos q + r 2

an x (n)

z X  a

n x (n)

1 dx ( z ) z dz −1

x(n) or f(t)

X(z) of F(z)

1 n d (n) d (n – k)

 z  log   if | z | > 1  z −1  1 1 zk

an u (n) an cos nq . u (n)

z z−a z ( z − a cos q) z − 2az cos q + a 2 2

an sin nq . u (n)

az sin q z − 2az cos q + a 2 2

n an u (n)

az ( z − a)2

329

The Z-transforms

(n + 1) an u (n)

z2 ( z − a)2

n (n – 1) an u (n)

2a 2 z ( z − a )3 aX (z) + bY (z) z–m X (z) X(z) Y(z) z if |z| > 1 z −1

ax (n) + by (n) x (n – m) x (n) * y (n) u (n) Z (tk)

−Tz

d [ Z (t k −1 )] dz

Z (t)

Tz ( z − 1) 2

Z (t2)

T 2 z ( z + 1) ( z − 1)3

Z (t3)

T 3 z (1 + 4 z − z 2 ) ( z − 1) 4

np 2

z2 z + a2

np 2 an f (t)

az z2 + a2 z F  a

a n cos

2

a n sin

n f (nT) = n f (t)

−z

k

kz if |z| > 1 z −1

e–at

z if | z | >|e − aT | z − e − aT z if | z | > | e aT | z − e aT

eat sin wt

d F ( z) dz

z sin wT if |z| < 1 z − 2 z cos wT + 1 2

cos wt

z ( z − cos wT ) z − 2 z cos wT + 1 2

Theorems.

Z [e–at f (t)] = F(zeeT) Z[f ((n + 1)T)] = Z [f (t + T)] = z [F (z) – f (0)] f (T ) f (2T ) k  − Z [f (t + kT)] = Z [fn + k] = z  F ( z ) − f (0) − z z2 

330

Engineering Mathematics-III (Third Semester)

−

Z [fn

+ 2] = z

2

f (3T ) z3

− ...

f ((k − 1)T )   z k −1 

f (T )    F ( z ) − f (0) − z   

Table of Inverse Z-Transform X (z) z z −1

x (n) = Z–1 [X(z)] 1

z z +1

(–1)n

z z−a

an u (n)

z −m ⋅

z z −1

u(n – m)

z ( z − 1) 2

n

z2 + z ( z − 1)3

n2

2z ( z − 1) 2

n (n – 1)

1 z −1

u (n – 1)

az ( z − a)2

nan

z X  a

an x (n)

z ( z − r cos q) z 2 − 2rz cos q + r 2

rn cos nq

rz sin q z 2 − 2 zr cos q + r 2

rn sin nq

z2 z2 + a2 az z + a2 1 z−a 2

a n cos

np 2

np 2 u (n – 1)

a n sin an –1

etc. See the other table also for more functions.

331

The Z-transforms

Inverse Z-Transform Method I. Long Division Method: ∞

∑

Since Z-transform is defined by the series X (z) =

n=0

x(n) z − n (one sided), to find the inverse

Z-transform [x(n) = z–1[X (z)] of X (z), expand X(z) in the proper power series and collect the coefficient of z–n to get x(n). Long division method is useful to get x (n). Example 1. Find inverse Z-transform of (i)

10 z ( z − 1) ( z − 2)

Solution.= (i) X ( z )

(ii)

( z + 2) z

(iii)

2

z + 2z + 4

2 z ( z 2 − 1) 2

( z + 1)

2

(iv)

z 2

z + 7 z + 10

10 z 10 z 10 z −1 = = 2 ( z − 1) ( z − 2) z − 3 z + 2 1 − 3 z −1 + 2 z −2

By actual division, 10z–1 + 30z–2 + 70z–3 + ... 10 – 3z–1 + 2z–2 ) 10z–1 10z–1 – 30z–2 + 20z–3 30z–2 – 20z–3 30z–2 – 90z–3 + 30z–4 70z–3 – 60z–4 70z–3 – 210z–4 + 40z–5 Comparing the quotient with

∞

∑

+ 150z–4 – 140z–5 x ( n) z − n

n =0

= x (0) + x (1)z–1 + x(2) z – 2 + ... We get the sequence x (n), x (0) = 0, x (1) = 10, x(2) = 30, x (3) = 70, .... i.e., we can get x (n) = 10 (2n – 1), n = 0, 1, 2, .... (ii) X(z) =

z2 + 2z z2 + 2z + 4

=

1 + 2 z −1 1 + 2 z −1 + 4 z −2

By actual division, 1 – 4z–2 + 8z–3 – 32z–5 + ...

1 + 2z–1 + 4z–2 ) 1 + 2z–1

1 + 2z–1 + 4z–2

–4z–2



–4z–2 – 8z–3 – 16z–4



8z–3 + 16z–4



8z–3 + 16z–4 + 32z–5



–32z–5

332

Engineering Mathematics-III (Third Semester)

Therefore x (0) = 1, x (1) = 0, x(2) = – 4 x (3) = 8, x (4) = 0, x (5) = –32, ....



The sequence is 1, 0, –4, 8, 0, –32, ... 2 z ( z 2 − 1) 2z3 − 2z 2 z −1 − 2 z −3 = (iii) X(z) = = z 4 + 2 z 2 + 1 1 + 2 z −2 + z −4 ( z 2 + 1) 2 By actual division, this is equal to

X(z) = 2z–1 – 6z–3 + 10z–5 – 14z–7 + ...

\

x(0) = 0, x (1) = 2, x (2) = 0, x (3) = – 6,



x(4) = 0, x (5) = 10, x(6) = 0.

np , n = 0, 1, 2, .... 2 z z −1 = (iv) X (z) = 2 z + 7 z + 10 1 + 7 z −1 + 10 z −2 In general x (n) = 2n sin

By actual division, X (z) = z–1 – 7z–2 + 39z–3 – 203z–4 + ....



x(0) = 0, x (1) = 1, x (2) = –7, x(3) = 39, x (4) = – 203, ...

\

Example 2. Find the Z-inverse of X (z) = Solution. X (z) =

(1 − e − aT ) z ( z − 1) ( z − e − aT )

=

(1 − e − aT ) z ( z − 1)( z − e − aT ) (1 − e − aT ) z −1

1 − (1 + e − aT ) z −1 + e − aT z −2

By actual division, X (z) = (1 – e–aT) z–1 + (1 – e–2aT) z–2 + [(1 – e–aT) + (1 + e–aT) (1 – e–2 aT)] z–3 + ... \

x (0) = 0, x (1) = 1 – e–aT



x (2) = (1 – e–2aT)



x (3) = (1 – e–aT) + (1 + e–aT) (1 – e–2aT)



...................................................................

Method 2. (By Partial Fraction Method) Example 3. Find the inverse Z-transform of z z2 + z (i) 2 (ii) z + 7 z + 10 ( z − 1) ( z 2 + 1) (iv)

z2 1  1   z −  z −  2 4  

Solution. (i) Let X(z) =

(v)

z ( z − 1) ( z + 1)

z 2

2

z + 7 z + 10

(iii)

kTz ( z − 1 + kT ) ( z − 1)

333

The Z-transforms

\ \

X ( z) 1 A B 1 1 1 1 = + =− ⋅ + ⋅ = z ( z + 5) ( z + 2) z + 5 z + 2 3 z +5 3 z +2 1 z z  − X (z) =  3  z + 2 z + 5  z  −1  1  z − x (n) = Z    z z + + 3 2 5    

1 z [(−2) n − (−5) n ], n 0,1, 2,... since Z ( a n ) . = = = 3 z−a z ( z + 1) (ii) Let X (z) = ( z − 1) ( z 2 + 1) X ( z) z +1 z +1 = = 2 z ( z − 1) ( z + 1) ( z − 1) ( z − i ) ( z + i ) A B C 1 1 1 1 1 + + = − ⋅ − ⋅ = z −1 z − i z + i z −1 2 z − i 2 z + i

\ (iii)

X (z) =

z 1 z 1 z − ⋅ − ⋅ z −1 2 z − i 2 z + i

1 n 1 n x(n) = 1 − (i ) − (−i ) . 2 2 kTz X (z) = ( z − 1 + kT ) ( z − 1) X ( z) kT A B = + = z ( z + kT − 1) ( z − 1) ( z + kT − 1) z − 1

=

1 1 − z − 1 z − (1 − kT )

\

X (z) =

z z − z − 1 z − (1 − kT )

\

x (n) = Z–1 [RHS]

z −1  z  − Z −1 = Z   z − (1 − kT )  z − 1 (iv) Let





x (n) = 1 – (1 – kT)n, n = 0, 1, 2, .... z2 1 1  z− z−  2 4  z A B X ( z) 2 1 = + = − = 1 1 1 1 1  1 z   z −  z −  z − 2 z − 4 z − 2 z − 4 2  4  z z − X (z) = 2 ⋅ 1 1 z− z− 2 4 X (z) =

334

Engineering Mathematics-III (Third Semester)



X (n) = 2 Z

−1

    n  z  1 −1  z  2⋅  −Z  =    2 z− 1 z− 1   2 4

  1 n =  2     2  (v) Let



X (z) =

1 −  4

n

 u ( n) 

z 2

( z − 1) ( z + 1)

1 1 1 X ( z) 1 = 4 − 4 + 2 = z ( z − 1) 2 ( z + 1) z + 1 z − 1 ( z − 1) 2



1 z 1 z 1 z − ⋅ + ⋅ X (z) = ⋅ 4 z + 1 4 z − 1 2 ( z − 1) 2



x (n) = Z–1 [RHS]

\

n

1 0,1, 2,.... −  , n = 4

1 1 1 (−1) n − + n, n = 0, 1, 2, .... 4 4 2

=

Method 3. (Inverse of Z-transform by Inverse integral method) By using the theory of complex variables, it can be shown that the inverse Z-transform is given by

x (n) =

1 X ( z ) ⋅ z n −1dz 2pi ∫

where c is the circle (may be even closed contour) which contains all the isolated singularities of X(z) and containing the origin of the z-plane, in the region of convergence. By Cauchy’s Residue theorem, x(n) = SR where SR is the sum of the residues of X(z) . zn–1. at the isolated singularities of X(z) . zn – 1. Example 4. Find the inverse Z-transform of (i)

z ( z − 1) ( z − 2)

Solution. (i) Let

(ii)

z ( z 2 − 1) ( z 2 + 1) 2

using residues.

X (z) =

z ( z − 1) ( z − 2)

X (z) . zn – 1 =

zn ( z − 1) ( z − 2)

has simple poles at z = 1 and z = 2. z = 0 is not a singularity of X (z) zn 1 since n = 0, 1, 2, ...

335

The Z-transforms

Residue of X (z) . zn – 1 at z = 1 } = Lt ( z − 1) ⋅

zn =−1 ( z − 1) ( z − 2)

Residue of X (z) . zn – 1 at z = 2} = Lt ( z − 2) ⋅

zn = 2n ( z − 1) ( z − 2)

n →1

z →2

\

x (n) = SR = 2n – 1, n = 0, 1, 2, ...

(ii) Let

X (z) =



z ( z 2 − 1) ( z 2 + 1) 2

f(z) = X (z) . zn–1 =

z n ( z 2 − 1) ( z 2 + 1) 2

poles of X(z) . zn–1 are z = ± i, each pole of order 2; z = 0 is not a singularity.

R1 = Residue of f (z) at z = i

= Lt

z →i

d  ( z − i ) 2 ⋅ X ( z ) ⋅ z n −1  dz 

d  z n ( z 2 − 1)   2  z →i dz  ( z + i) 

= Lt

= Lt

( z + i ) 2 [ z n ⋅ 2 z + nz n −1 ( z 2 − 1)] − z n ( z 2 − 1) ⋅ 2( z + i )

z →i

( z + i)4

n n −1 t 2 R2 = Residue of f (z) at z = – i

= Similarly,

n (−i ) n −1 2

= \

n n −1 n −1 x(n) = [(i ) + (−i ) ], n =0,1, 2,... 2

z   Example 5. Find Z–1  2  by Residue method.  z − 2z + 2  z Solution. X(z) = 2 ( z − 2 z + 2) zn f(z) = X(z) . zn – 1 = 2 z − 2z + 2 The poles of f (z) are z = 1 ± i, each simple pole. R1 = Residue to f(z) at z = 1 + i =

P(a) (1 + i ) n 1 = = (1 + i ) n Q ′(a ) 2(1 + i ) − 2 2i

R2 = Residue of f(z) at z = 1 – i = –

1 (1 − i ) n 2i

(pole of order 2)

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Engineering Mathematics-III (Third Semester)

x(n) =

\

1 [(1 + i ) n − (1 − i ) n ] 2i

1 np np  np np     ( 2) n cos + i sin  − ( 2) n cos − i sin   2i  4 4 4 4    np , n = 0, 1, 2, .... x (n) = ( 2 )n sin 4  z 2 − 3z     ( z + 2) ( z − 5) 

= Example 6. Find Z–1 Solution. Let

X (z) =

z 2 − 3z ( z + 2) ( z − 5)

f(z) = X (z) . zn – 1 are z = –2, 5.

The simple poles of Residue of f(z) at z = – 2 is

5 (−2) n 7

Residue of f(z) at z = 5 is

2 n 5 7

\

x (n) =

1 [2(5)n + 5(–2)n], n = 0, 1, 2, .... 7

Example 7. Find Z–1 (X(z)) where X (z) = Solution.

4z2 − 2z z3 − 5z 2 + 8z − 4

f (z) = X(z) . zn – 1 has poles given by z3 – 5z2 + 8z – 4 = 0

i.e.,

z = 2 is a simple pole and



z = 2 is a pole of order 2.

Residue of f(z) at

z = 1 is Lt ( z − 1) f ( z ) z →1

= Lt

(4 z 2 − 2 z ) z n −1

z →1

( z − 2) 2

=2

Residue of f (z) at z = 2 is 2n (3n – 2) \

x (n) = 2 + 2n (3n – 2), n = 0, 1, 2, ...

Example 8. Find the inverse Z-transform of Solution. Let

z ( z + 1) ( z − 1)3

f(z) = X (z) . z­n – 1 =

.

z n ( z + 1) ( z − 1)3

f(z) has only one pole at z = 1 of order 3. Residue of f (z) at z = 1 is =

1 d2 Lt 2 [ z ( z + 1) ⋅ z n −1 ] 2! z →1 dz

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The Z-transforms

=

1 Lt [(n + 1)nz n −1 + n(n − 1) z n − 2 ] 2 z →1

1 2 [ n + n + n 2 − n] = n2 2 x (n) = n2, for n = 0, 1, 2, ...

= \

 1 Example 9. If X (z) = (z– 1) log 1 −  + 1, find x (n)  z x (n) = Z

Solution.

−1

  1  ( z − 1) log  1 − z  + 1    

1 1  1   −1  = Z ( z − 1)  − − 2 − 3 ....  + 1 z 2z 3z     1 1 1  1   −1  = Z  −1 − − 2 − ...  +  + 2 + ...  + 1 2 z 3z   z 2z    1 1 1 1 1  −1  = Z 1 −  +  −  2 + ... 2 z 2 3   z    1 1 1 1 −1  1 1 + + ... ⋅ n + ... = Z  2 n(n + 1) z 1.2 z 2.3 z  ∞

−1 = Z ∑

n =1

1 1 ⋅ n(n + 1) z n

if n = 0  0  =  1  n(n + 1) , if n ≥ 1  Differentiation Let

Z [{x (n)}] = X (z).

An infinite series can be differentiated term by term within its region of convergence. X (z) may be treated as a function of z–1.

∞

∞

−n X (z) = ∑ x(n) ⋅ z = ∑ x(n) ( z −1 )n

= n 0= n 0

Differentiate both sides w.r.t. z–1. dX ( z )



dz −1 dX ( z )

∞

−1 n −1 = ∑ n x (n) ⋅ ( z ) n =0 ∞

−n = ∑ n( x(n)) z = Z [nx (n)]



z −1



Z [nx (n)] = z

dz −1

... (1)

n =0

−1

dX dz −1



... (2)

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Engineering Mathematics-III (Third Semester)

Differentiating (1) w.r.t. to z–1 again, d 2 X ( z)



d ( z −1 ) 2 z −2



d 2 X ( z) d ( z −1 ) 2

∞

−1 n − 2 = ∑ n(n − 1) x (n) ( z ) n =0 ∞

−n = ∑ n(n − 1) x (n) n n =0

= Z [n (n – 1) x (n)] Z [n (n – 1) x (n)] = z

\

– 2

d 2 X (z ) d ( z – 1 )2

... (3)

Example 10. Find the Z-transform of (i) ndn u (n)

(ii) n (n – 1) an u (n)

Solution. Take x (n) = an u (n) in (2) and (3)  z    d (z )  z − a  d −1 [1 − az −1 ]−1 = z −1 d (z ) Z [nan u (n)] = z

(i)

−1 = z

(ii)

Z [n (n – 1) an u (n)] = z

d

−1

−2

−1

a −1 2

(1 − az )

=

az −1 (1 − az −1 ) 2

d 2 [1 − az −1 ]−1 d ( z −1 ) 2

=

2a 2 z 2 (1 − az −1 )3

Setting a = 1,

Z [nu (n)] =



Z [n (n – 1) u (n)] =

z −1 (1 − z −1 ) 2 2 z −2 (1 − z −1 ) 2

Example 11. If X (z) = (1 – az–1)–2, find x (n). Solution.

X (z) =

= =

1 (1 − az −1 ) 2 1 −1

(1 − az )

+

=

1 − az −1 + az −1 (1 − az −1 ) 2 az −1

(1 − az −1 ) 2

z az + z − a ( z − a)2

x (n) = Z–1 [RHS]

= an u(n) + nan u (n) = (n + 1)an u (n) \

Z [(n + 1) an u (n)] =

z2 ( z − a)

= (1 − az −1 ) −2

2

339

The Z-transforms

Applications to Solving Finite Difference Equations Z-transform can be employed in solving difference equation remembering Z[x(n – m)] = z–m X(z) and the expansions of Z[fn + k], Z[fn + 1], Z [fn + 2] etc.

Z [y (n + 2)] = z2 Y (z) – z2 y (0) – zy (1)



Z [y (n + 1)] = zY (z) – zy (0) where Y (z) = Z [y (n)]

Example 1. Solve : yn +2 – 4yb + 1 + 4yn = 0 given y0 = 1 and y1 = 0. Solution. Taking Z-transform on both sides of the difference equation, we get Z (yn + 2) – 4Z (yn + 1) + 4Z (yn) = Z (0) y   z 2 Y ( z ) − y0 − 1  − 4[ z (Y ( z ) − y0 )] + 4Y ( z ) = 0 z  (z2 – 4z + 4) Y (z) = z2 – 4z = Y ( z)

z2 − 4z z2 − 4z = 2 z − 4 z + 4 ( z − 2) 2

Y (z) has pole at z = 2 (of order 2) \ y(n) = R where R is residue of Y (z) . zn – 1 d [( z 2 − 4 z ) z n −1 ] z → 2 dz

= Lt

n n −1 = Lt [(n + 1) z − 4nz ] n→2

= (n + 1)2n – 4n 2n – 1 = 2n [n + 1 – 2n] y (n) = 2n (1 – n), n = 0, 1, 2, .... Example 2. Solve yn + 2 + 6yn + 1 + 9yn = 2n given y0 = y1 = 0. Solution. Taking Z-transform on both sides of the difference equation, Z [yn + 2] + 6Z [yn + 1] + 9Z [yn] = Z (2n) [z2 Y (z) – z2y0 – zy1] + 6[z(Y(z) – y0)] + 9Y (z) = [z2 + 6z + 9] Y(z) = \

Y(z) =

z z−2

z ( z − 2)( z + 3) 2

Y ( z) 1 A B C = + + = z ( z − 2)( z + 3) 2 z − 2 z + 3 ( z + 3) 2

=

1 1 1 1 1 1 ⋅ − ⋅ − ⋅ 25 z − 2 25 z + 3 5 ( z + 3) 2

Y(z) =

1 z 1 z 1 z ⋅ − ⋅ − ⋅ 25 z − 2 25 z + 3 5 ( z + 3)3



z z−2

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Engineering Mathematics-III (Third Semester)

Taking inverse Z-transform,

y(n) =

Since

Z[n(–3)n] =



y(n) =

1 n 1 1 ⋅ 2 − (−3) n − n(−3) n −1 25 25 5 −3 z ( z + 3) 2 1  n 5  2 − (−3) n + n(−3) n   25  3 

Example 3. Solve x(n + 2) – 3x(n + 1) + 2x(n) = 0, given x(0) = 0, x(1) = 1. Solution. Taking Z-transform of the equation given, Z[x(n + 2)] – Z[3x(n + 1)] + 2Z[x(n)] = 0 z2X(z) – z2 x(0) – zx(1) – 3[z(X (z) – x(0))] + 2X(z) = 0 (z2 – 3z + 2) X(z) = z X ( z) 1 1 1 = − = ( z − 1)( z − 2) z − 2 z − 1 z

\

z z − z − 2 z −1

X(z) =

\

Taking inverse Z-transform, x(n) = 2n – 1, where n = 0, 1, 2, ... Example 4. Solve x(k + 2) – 3x(k + 1) + 2x(k) = u(k) given x(k) = 0 for k ≤ 0 and u(0) = 1, u(k) = 0 for k < 0 and k > 0. Solution. Taking Z-transform on both sides of the given equation, Z[x (k + 2)] – 3Z[x(k + 1)] + 2Z[x(k)] = Z[u(k)] [z2X(z) – z2 x (0) – zx(1)] – 3[zX(z) – zx(0)] + 2X(z) = Z[u(k)] ... (1) x(0) = 0, x(–1) = 0; putting k = –1 in the given equation,

x(1) – 3x(0) + 2x(–1) = u(–1) x(1) = 0 since u(–1) = 0

\

∞

Z[u(k)] = ∑ u (k )z



−k

=1

k =0

Equation (1) reduces to, (z2 – 3z + 2) X(z) = 1



X(z) =



1 ( z − 1)( z − 2)

Poles of X(z).zk–1 are z = 1, z = 2, for k = 1, 2, ... Residue of f(z) = X(z)zk–1 at z = 1 is

Lt ( z − 1)

z →1

z k −1 = –1 ( z − 1)( z − 2)

341

The Z-transforms

Residue of f(z) at z = 2 is

Lt ( z − 2)

z →2

z k −1 = 2k–1 ( z − 1)( z − 2)

\ x(k) = ΣR = 2k–1 –1, for k = 1, 2, 3,... Example 5. Solve : y(n) – ay(n – 1) = u(n). Solution. Taking Z-transform on both sides Z[y(n)] – aZ[y(n – 1)] = Z[u(n)]

z z −1 z (1 – az–1) Y(z) = z −1

Y(z) – a.z–1 Y(z) =



Y(z) =

z2 ( z − 1)( z − a )



Y(z) =

1  z az  −  1 − a  z − 1 z − a 

Taking inverse Z-transform,

y(n) =

1 [1 − a n +1u (n)] (1 − a )

Example 6. Solve : y(n) – y(n – 1) = u(n) + u(n – 1). Solution. Taking Z-transform on both sides, Z[y(n)] – Z[y(n – 1)] = Z[u(n)] + Z[u(n – 1)]

z z + z −1 ⋅ z −1 z −1 z + 1 (1 – z–1)Y(z) = z −1 z ( z + 1) Y(z) = ( z − 1) 2 z +1 z −1+ 2 Y ( z) 1 2 = = + = 2 2 z − 1 ( z − 1) 2 z ( z − 1) ( z − 1)

Y(z) – z–1 Y(z) =

Y(z) =

z 2z + z − 1 ( z − 1) 2

y(n) = Z–1 [RHS] = 1 + 2n Example 7. Solve x(n + 1) – 2x(n) = 1, given x(0) = 0. Solution. Taking Z-transform on both sides, Z[x(n + 1)] – 2Z[x(n)] = Z[1]

z z −1 z (z – 2) X(z) = z −1

z[X(z) – x(0)] – 2.X(z) =

342

\

Engineering Mathematics-III (Third Semester)

X(z) =

z ( z − 1)( z − 2)

X ( z) 1 1 1 = − = ( z − 1)( z − 2) z − 2 z − 1 z z z − ( z − 2) ( z − 1)

\

X(z) =



x(n) = Z–1 [RHS]

x(n) = 2n – 1 Example 8. Solve yn+2 + yn = 2, given y0 = y1 = 0. Solution. Z[yn + 2] + Z[yn] = Z(2) z2Y(z) – z2y(0) – zy(1) + Y(z) = 2 ⋅

z z −1



(z2 + 1) Y(z) = 2 ⋅



Y(z) = 2 ⋅



y(n) = Z

z z −1 z z2 + z = − ( z − 1)( z 2 + 1) z − 1 z 2 + 1 z

z  −1  z ( z + 1)   −Z  2   z −1   z +1 

−1 

np np   + sin  u (n) = 1 − cos 2 2   Y ( z) A Bz + C z +1 2 1 = = + 2 = − 2  Hint. 2 z ( z − 1) ( z + 1) z − 1 z + 1 z − 1 z + 1  Hence Y= ( z)

z ( z 2 + z)  − 2  z −1 z +1 

Example 9. Solve yn+2 – 4yn = 0 using Z-transform. Solution. Here, the conditions y0 and y1 are not given. \ Take y0 = A, y1 = B \ Z[yn + 2] – 4Z[yn] = 0 z2Y(z) – z2y(0) – zy(1) – 4Y(z) = 0 (z2 – 4) Y(z) = Az2 + Bz

Y(z) =

Az 2 2

z −4

+

Bz 2

z −4

A z z  B z z  + +  −   2  z − 2 z + 2  4  z − 2 z + 2  A n B n n n y(n) = [2 + (−2) ] + [2 − (−2) ] 2 4  A B n  A B n =  +  2 +  −  (−2) 2 4 2 4 =



y(n) = C.2 + D(–2)n

343

The Z-transforms

Example 10. Using Z-transform, solve the simultaneous equations, xn+1 – yn = 1 yn+1 + xn = 0 given x0 = 0, y0 = – 1 Solution. Taking Z-transforms

z z −1 z zX(z) – Y(z) = z −1

z[X(z) – x(0)] – Y(z) =



... (1)

Similarly, z[Y(z) – y(0)] + X(z) = 0 zY(z) + X(z) = – z



... (2)

Solving for X(z) and Y(z) from (1) and (2), z



X(z) =

\

x(n) = Z–1 [RHS]

=

( z − 1) ( z 2 + 1) np np  1 1 − cos − sin  (Refer Example 8)  2 2 2

From given equation, = − xn yn + 1 =



1 np np  cos + sin − 1  2 2 2 

1 p p  yn = cos(n − 1) + sin(n − 1) − 1 2 2 2  1 np p np p = cos cos + sin sin 2 2 2 2 2 \

+ sin

p p  np np cos − cos sin − 1 2 2 2 2 



np  1  np − cos − 1 yn = sin 2 2 2 



np np  1 − sin  xn = 1 − cos 2 2 2

EXERCISE 1 1. Find the Z-transform (one sided) of the sequences {x(n)} where x(n) is n

n

1 1 n (i)   u (n) (ii) d(n – 2) (iii) u(n – 3) +   + 2 2 3 np np (iv) (–1)n u(n) (v) 2n sin (vi) 3n cos 2 2 (vii) n.2n (viii) 2n u(n) (ix) n u(n), n2 u(n) np np (x) u(n – 2) (xi) u(n) cos (xii) u(n) sin 4 4

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Engineering Mathematics-III (Third Semester)

(xiii) an2 + bn + c; (Give region of convergence) (xiv) 2nu(n – 1) (xv) u(n – k) 2. Find the Z-transform of

(xvi) 2n u(n – k)

(i) e2t + 3 (ii) e–3t – 7 (iii) sin 2t (iv) cos 3t (v) sin2 3t (vi) cos2 t (vii) sin3 t (viii) cos3 t (ix) n.2n (x) n sin nq (xi) n cos nq 3. Find the inverse Z-transform of (assume x(n) is a causal sequence) by any method; if possible do by more than one method. (i) (iv)

(vii)

z2 + z ( z − 1) 2 z2 + z ( z − 1)

3

(ii) (v)

1 + 2 z −1 1 − z −1

(iii)

z2 + 2z 4 z −1 (vi) ( z − 1)( z − 2)( z − 3) (1 − z −1 ) 2

z2 z z2 − z (viii) (ix) ( z − a )( z − b) ( z − 1)( z − 2)( z − 3) ( z + 1) 2

11 −1 z + 7 z −2 2 (xii)  1 −1  −1 −2 1 − z  (1 + 2 z + 4 z )  2  z2 z 2z2 + 4z (xiii) (xiv) (xv) ( z + 2)( x 2 + 4) z 2 + 11z + 30 ( z − 2)3 z2

4 − 8 z −1 + 6 z −2 (x) (xi) 2 ( z − 1) (1 − 2 z −1 ) 2 (1 + z −1 )

z ( z − 1) ( z − 2)

(xvi)

3+

5z 5z z ( z 2 − z + 2) (xvii) (xviii) (2 − z )(3 z − 1) (2 z − 1) ( z − 3) ( z + 1) ( z − 1) 2

4. Find the Z-transforms of (i) e–3t.t (ii) e–t.t2 (iii) e–at sin wt (iv) e–t cos 2t (v) at2 + bt + c (vi) at2 + bt + c 5. Find the Z-transforms of the convolutions of x(n) * y(n) (i) 2n * n (ii) n(n – 1) * 3n (iii) 3n * cos nq np np (iv) 2n * sin nq (v) cos *sin 2 2 6. Using convolution theorem, find the inverse Z-transform of

(i)

z2 1 (ii) ( z − a ) ( z − b)  1 −1  1 −1  1 − z 1 − z   2  4 

7. Find (i) an u(n) * an u(n) (ii) 3n * 3n 8. Find the inverse Z-transform of 5 3 − z −1 1 6 if |Z| > 3  1 −1   1 −1  1 − z   z  4 3   

345

The Z-transforms

9. Find the Z-transform of f(n) = – u(–n – 1) where u(n) is the unit step sequence. 10. Solve the difference equations, using Z-transforms: x(n + 2) – 3x (n + 1) – 10x(n) = 0, given x(0) = 1, x(1) = 0 y(k + 2) – 5y (k + 1) + 6y (k) = 6k if y(0) = y(1) = 0 y(n + 2) – 5y(n + 1) + 6y (n) = 4n given y(0) = 0, y (1) = 1 yn + 2 + 3yn + 1 + 2yn = 0 if y0 = 0, y1 = 1 f(n) + 3f(n – 1) – 4f(n – 2) = 0, n ≥ 2 given that f(0) = 3 and f(1) = –2 −5 19 (vi) f(n) – 2f(n –1) – 3f(n – 2) = 0, n ≥ 0 given that f(–1) = and f(–2) = 3 9 (vii) x(k + 2) + 2x (k + 1) + x(k) = u(k) where x(0) = 0, x(1) = 0 and u(k) = k for k = 0, 1, 2, ... (viii) yn + 2 – 4yn + 1 + 3yn = 0 given y0 = 2, y1 = 4 (ix) yn + 2 – 4yn + 1 + 4yn = p, if y0 = y1 = 0 (x) yn + 1 – 3yn = 0 where y0 = 1 (xi) yn + 3 + yn + 2 – 8yn + 1 – 12yn = 0 given y0 = 1, y1 = y2= 0 (xii) ux + 2 + ux = 5.2x given u0 = 1, u1 = 0 (xiii) yx + 2 + yx = xax (xiv) yn + 2 – 7yn+1 + 12yn = 2n given y0 = y1 = 0 (xv) yn + 3 – 3yn + 1 + 2yn = 0 given y1 = 0, y2 = 8 and y3 = – 8 (xvi) yx + 2 – 4yx + 1 + 4yx = 0 if y0 = 1, y1 = 2 (xvii) yn + 2 – 4yn = 2n given y0 = 0, y1 = 0 (xviii) x(n + 2) – 5x (n + 1) + 6x(n) = 5n given x(0) = x(1) = 0 (xix) y(n) – ay(n – 1) = u(n) (xx) yn + 2 – 5yn + 1 + 6yn = 36 given y0 = y1 = 0 (xxi) xn + 1 = 7xn + 10yn yn + 1 = xn + 4yn given x0 = 3, y0 = 2 (xxii) an + 1 = an + abn bn + 1 = bn + aan given a0 = 0, b0 = 1 a Find n as a function of a and n. bn (xxiii) yn + 2 – 2 cos a yn + 1 + ayn = 0 given y0 = 1, y1 = cos a (xxiv) yn + 2 – 2 cos a yn + 1 + yn = 0 if y0 = 0, y1 = 1 (xxv) yn + 2 – yn = 2n if y0 = 0, y1 = 1 (xxvi) y(n + 2) = y(n + 1) + y(n) if y(0) = 0, y(1) = 1 (xxvii) yn + 1 + yn = 1 if y0 = 1 (xxviii) yn + 1 – 3yn = 2n if y0 = 1 (xxix) yn + 2 + 2yn + 1 + yn = n if y0 = y1 = 0 (xxx) yn + 2 – 4yn + 1 + 4yn = 0 given y0 = 1, y1 = 0 (i) (ii) (iii) (iv) (v)

EXERCISE 2 (Short Answer Questions) 1. u(n) – u(n – 1) is ................ .

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Engineering Mathematics-III (Third Semester)

2. Radius of convergence of

∞

∑ x ( n) z − n

n =0

is ................ .

3. Z[an u(n)] exists only if ................ . 4. Z-transform of nx (n) is ................ . 5. Z-transform of n(n – 1) u(n) is ................ . 6. Define the sampler. 7. Define a holding device. 8. Define two sided Z-transform. 9. Define one sided Z-transform. 10. Define unit step sequence and unit sample sequence. 11. Define u(n – k) 12. Z[nf(t)] = ................ . n 13. Z{(– 1) } = ................ . 14. If x(n) = n, then Z [x, (n)] is ................ . Write Ture or False z2 np   16. Z cos  = 2 2 z +1  1  z  18. Z   = log   if |z| > 1 n  z −1 

z  np  15. Z sin  = 2 2 z +1   1 17. d(n – k) = k z −1  z  19. Z   if |z| > 1 is ................ .  z −1 

20. State convolution theorem for casual sequences.  z −1  21. Z   is ................ .  ( z − 3) ( z − 4)  2  −1  z 23. Z  2  is ................ .  z + 1

22. If Z[x(n)] = X(z) then Z[nx(n)] ................ . −1  z  24. Z  2  is ................ .  z + 1

25. Solve : yn + 1 – 2yn = 1 given y0 = 0 27. Find the Residue of

z n ( z − 4) ( z − 2) 2

26. Find the Residue of

z n ( z + 1) ( z − 1)3

28. State initial value theorem.



29. State final value theorem.

30. Z[n(n – 1)x (n)] is ................ . ANSWERS (Z-Transforms) EXERCISE 1

1. (i)

3z 1 1 2z z + + (ii) 2 (iii) 2 − − 3z − 1 2 z 1 z 2 z z ( z − 1)

(iv)

z2 z 2z (v) 2 (vi) 2 z +1 z +4 z +9

(vii) (x)

2z ( z − 2) 2

(viii)

1 (xi) z ( z − 1)

z z z ( z + 1) , (ix) z−2 ( z − 1) 2 ( z − 1)3 z ( 2 z − 1) 2

2[ z − 2 z + 1]

(xii)

z 2

2[ z − 2 z + 1]

347

The Z-transforms

(xiii)

az ( z + 1) ( z − 1)

3

+

bz ( z − 1)

2

+

cz 2 (xiv) z −1 z−2 −k

−k (xv) z

3 2. (i) e ⋅

(iv)

z z   ⋅  z 2 2 ⋅ (xvi) z −1 z   −1 2 z z −e

2T

−7 (ii) e ⋅

z ( z − cos 3T ) 2

z − 2 z cos 3T + 1

(v)

z z −e

−3T

(iii)

1 z 1  z ( z − cos 2T )  + 2 z − 1 2  z 2 − 2 z cos 2T + 1 

(vii)

z sin T z sin 3T 3  1  −  2   2 4  z − 2 z cos T + 1  4  z − 2 z cos 3T + 1 

(viii)

3  z ( z − cos T )  1  z ( z − cos 3T )  + 4  z 2 − 2 z cos T + 1  4  z 2 − 2 z cos 3T + 1  2z ( z − 2) 2

−1 (xi) z

( x) z −1

 sin q ⋅ z −1 d   −1 −1 −2  dz 1 − 2 z cos q + z 

 d  1 − cos q ⋅ z −1 −1  −1 −2  dz 1 − 2 cos q ⋅ z + z 

3. (i) 2n + 1

(ii) 1 + 2u(n – 1)

2

t (v) T2 1 n +1 n +1 [a − b ] (viii) (vii) ab (iv)

z − 2 z cos 2T + 1

1 z 1  z ( z − cos 6T )  ⋅ −  2 2 z − 1 2  z − 2 z cos 6T + 1 

(vi)

(ix)

z sin 2T 2

(iii) (2n – 1)u(n)

3 5 − 4.2n + ⋅ 3n (vi) 2n(n – 1) u(n) 2 2 1 1 n 1 n − 2 + ⋅ 3 (ix) (2n + 1) (– 1)n 2 2 2

(x) (n + 1) u(n) (xi) 2(– 1)n + 2n(n + 2) n

2pn 1 2pn  1 n sin + (xii) 2   + 2 cos 3 3  3 2  (xiii)

(−2) n +1 (2i ) n +1 (2i ) n +1 + − (xiv) (–5)n – (–6)n 8 8i (1 + i ) 2i (1 − i ) n



1 n n (xvi)   − 2 (xvii) 3 3

(xv) n2 2n

(xviii) (–1)n + 4. (i)

1 (2n – 1), n = 0, 1, 2,... 2

T ⋅ ze3T ( ze

3T

− 1)

2

(ii)

T 2 ⋅ zeT ( zeT + 1) T

( ze − 1)

3

(iii)

1 −  2

n

Ze aT sin wt 2 2 aT

z e

− 2 ze aT cos wT + 1

348

Engineering Mathematics-III (Third Semester)

(iv) 5. (i) (iii) (v)

zeT ( zeT − cos 2T ) 2 2T

T

z e

− 2 ze cos 2T + 1

z

2

( z − 2)( z − 1)

2

(v)

(ii)

z 2 ( z − cos q) 2

( z − 3)( z − 2 z cos q + 1)

(iv)

aT 2 z ( z + 1) ( z − 1) 2z

3

+

2

( z − 1) 2 ( z − 3)

bTz ( z − 1)

+

cz z −1



z 2 sin q ( z − 2)( z 2 − 2 z cos q + 1)

z3 ( z 2 + 1) 2 n

6. (i)

2

1 21 1 1 [a n +1 − b n +1 ] (ii)   +  −  a −b 3 2 3 4

n

7. (i) (1 + n) an (ii) (n + 1)3n n

8.

n

1 1   − 2  4   3

9.

z if |z| < 1 z −1

10. (i) x(n) =

2 5 . (5)n + (–2)n, n = 0, 1, 2, .... 7 7

(ii) y(k) =

1 k 1 k 1 k 6 − ⋅ 3 + 2 , k = 0, 1, 2, ... 12 3 4

1 n (4 − 2n ), n = 0, 1, 2, ... (iv) yn = (– 1)n (– 2)n . n = 0, 1, 2, ... 2 1 n 14 n (v) f(n) = ⋅ 4 + (−1) (vi) f(n) = 3n + 2(–1)n 5 5 (iii) y(n) =

(vii) x(k) =

k + 1 (−1) k + (1 – k), k = 0, 1, 2, ... 4 4

  nn (viii) yn = 1 + 3n (ix) yn = p 1 + 2  − 1   2  

(x) y(n) = 3n (xi) yn =

(xii) ux = 2x – 2 sin (xiii) yx = A cos

px 2

px px ax  2a 2 x + B sin + −  2 2 1 + a 2  1 + a2

  

8 4 + (−2) n 3 3 1 (xvi) yx = 2x (xvii) yn = [(–2)n – 2n + n . 2n + 1] 16 1 n 1 n 5n 1 (1 − a n +1 ) (xviii) x(n) = ⋅ 2 − ⋅ 3 + (xix) y(n) = 1− a 3 2 6 (xiv) yn = (– 1) 3n +

1 n . 4 + 2n – 1 2

4 1 ⋅ 3 + (21 − 15n) (−2) n 25 25

(xx) yn = 18.3n – 36 . 2n + 18

(xv) yn =

(xxi) xn = 5.9n – 2.2n; yn = 9n + 2n

349

The Z-transforms

(xxii) tan (n tan–1 a) (xxiii) yn = cos n a sin na 1 (xxiv) yn = (xxv) yn = [2n – (– 1)n] sin a 3 n n 1− 5   1  5 + 1   −    2   5  2      1 (xxvii) yn = [1 – (–1)n] (xxviii) yn = 2.3n – 2n 2 n −1 [1 − (−1) n ] (xxix) yn = (xxx) yn = 2n (1 – n) 4

(xxvi) yn =

EXERCISE 20000 1. d(n) 4. z–1 12. − z

2.

dX dz

−1



dF ( z ) dz

5. 13.

x(n + 1) x ( n)

Lt

n →∞

2 z −1 −1 3

(1 − z )



z if | z | > 1 z +1

15. True

18. True

21. 4n – 3n

−1 22. z

np 24. sin 2

25. 2n – 1

27. 2n (1 – n)

30. z–2

= if n k  1= 11.  =  0 if n ≠ k 14.

z ( z − 1) 2

19. u(n)

dX dz

3. |z| > |a|

−1



d2X d ( z −1 ) 2

23. cos 26. n2

np 2

VI

LINEAR ALGEBRA

Revision: THE GROUP Definition. A group is a non-empty set G of elements together with a binary operation * satisfying the following axions or postulates. P1: Closure property. For all a, b ∈ G, a * b ∈ G. ie.., the set G is closed with respect to the composition *. P2: Associative law. For all a, b, c ∈ G, (a * b) * (c) = a *(b * c) i.e., the operation * obeys associative law. P3: Existence of identity element. There exists an element e ∈ G such that

a * e = e * a = a all for a ∈ G.

i.e., the binary composition admits identity element. P4: Existence of inverse element: For every a ∈ G, there exists an element a–1 ∈ G such that

a * a–1 = a–1 * a = e

where e is identity element or unit element of G. Definition: Abelian group. A group (G, *) is said to be an Abelian group or commutative group if the group operation * satisfies the commutative law. That is, for every a, b, ∈ G, a * b = b * a. Note. 1. If the first postulate only is satisfied then G is called a groupoid or an algebraic structure or quasi group.

2. If the first two postulates are satisfied, then G is called a semi group.



3. If the first three postulates are satisfied, then G is called a monoid.



4. If all the four postulates are satisfied, then G is called a group. Properties of groups (without proof) Let (G, *) be a group. Then,



1. The identity element e is unique



2. For every element a ∈ G, its inverse a–1 unique



3. For all a, b, c ∈ G, a * b = a * c ⇒ b = c b * a = c * a ⇒ b = c



4. For a, b ∈ G, the equations a * x = b and y * a = b have unique solution in G for x and y.



5. For a, b ∈ G,

(a * b)–1 = b–1 * a–1 350

351

Linear Algebra

THE RING A non empty set R is said to be an associative ring if, in R, there are two defined operations ‘+’ and ‘.’ such that for all a, b, c ∈ R. I : P1 : a + b ∈ R. P2 : a + b = b + a P3 : (a + b) + c = a + ( b + c) P4 : There exists an element 0 in R, such that 0 + a = a + 0 = a for every a in R P5 : There exists an element – a in R such that a + (– a) = (– a) + a = 0 – a, b in R, a. b ∈ R II : P6 : ∨ – a, b, c ∈ R, (a. b) . c = a . (b . c) P7 : ∨ III : P8 : Distributive laws.

For all a, b, c ∈ R,

a . (b + c) = a . b + a . c and (b + c) . a = b . a + c. a

(dot is distributive over plus-both left and right)

These eight postulates must be satisfied in R, called associative Ring. Note: 1. Axions P1 to P5 state that R is an abelian group under the operation ‘+’, (addition)

2. Axions P6 and P7 insist that R must be closed under an associative operation ‘.’. (multiplication)

3. Axiom P8 serves to interrelate the two operations. That is, . is distributive over +, both right and left. – a ∈ R, then R Unit element : If there exists an element 1 ∈ R such that 1 . a = a . 1 = a ∨ is a ring with unity or unit element 1. – a, b ∈ R, than R is said Comumutative ring : If the operation ‘.’ is such that a . b = b . a ∨ to be a commutative ring.

Zero division : If R is a commutative ring then a ≠ 0 and a ∈ R, is said to be a zero divisor if there exists an element b ∈ R, b ≠ 0 such that a . b = 0. Integral Domain : A commutative ring is an integral domain if it has no zero divisors. Division ring : A ring is said to be a division ring if its non zero elements form a group under multiplication. Notation. A ring is denoted by (R, +, .) THE FIELD A field is a commutative division ring. A finite integral domain is a field. In other words. Definition. A ring (R, +, .) which has at least two element is called a field if (i) it is a commutative ring with unity and (ii) all the non-zero elements are invertical w.r.t the multiplication. – a ≠ 0 and a ∈ R, i.e., ∨

352

Engineering Mathematics-III (Third Semester)

There exists b ∈ R such that a . b = b . a = 1 Stating in other words (R has at least two elements)

(1) (R, +) is an abelian group



(2) (R – {0}, .) is an abelian group



(3) distributive law of. over + holds good (both left and right) Examples: (Prove yourself)



1. The set of integers with ordinary addition (+) and multiplication (.) as the compositions forms a ring’.



2. The set of numbers of the form a + b 3 where a, b are integers forms a ring under ‘+’ and ‘–’.



3. The set of 2 × 2 matrices is a ring w.r.t addition and multiplication of matrices. Note: 1. In a field 0 and 1 are distinct elements. 2. A subfield S of a field (F, + , .) is a subset of F which is also field under + and . (same laws of compositions)



Fields (Examples) (Prove yourself)

1. The set of all real numbers R is a field w.r.t ‘+’ and ‘.’



2. The set of all rational number Q is a field w.r.t. and.



3. (C, + , .) is a field where C is the set of all complex numbers.



4. The set of positive integers with addition and multiplication modulo p, p being prime, is a finite field with p elements.

VECTOR SPACE Definition: Let (F, + , .) be a given field and V be a non-empty set having defined in it one internal and one exteral composition ⊕ and *. Then the given set V is said to be a vector space or a linear space over the field F, denoted by V(F), if the following postulates are satisfied. V1 : The algebraic structure (V, ⊕) is an abelian group.

That is, for all x, y, z ∈ V.

(1) x ⊕ y ∈V

(2) (x ⊕ y) ⊕ z = x ⊕ (y ⊕ z)

– x ∈ V, x ⊕ 0 = 0 ⊕ x = x (3) ∃ an element 0 ∈ V, such that ∨ ∨ – x ∈V, ∃ an element – x ∈ V, such that x ⊕ (– x) = (– x) ⊕ x = 0 (4) – x, y, z ∈ V (5) x⊕y=y⊕x ∨ V2 : (6) Scale multiplication is closed. – a ∈ F, x ∈ V, a * x ∈ V (a * x is a vector) i.e., ∨

(7) The scalar multiplication is associative.

i.e., For a, b ∈ F, and x ∈ V (a . b) * x = a * (b * x) (8) For unit scalar 1 ∈ F, – x∈V 1 * x = x ∨

Linear Algebra

353

V3 : (9) Scalar multiplication is distribution over vector addition. a * (x ⊕ y) = (a * x) ⊕ (a * y) for a ∈ F, x, y ∈V. (10) Multiplication by vector is distributive over scalar addition. (a + b) * x = (a * x) ⊕ (b * x), a, b ∈ F and x∈V. Note: 1. The elements of V are called vectors.

2. The elements of the field F are called scalars.

3. V1 defines the additive structure of the system. 4. V2 defines the multiplicative structure of the system. 5. V3 express the connection between the two structures. Null space: The vector space which has only one element 0 , the additive identity of the vector space, is called a null space or zero vector space and is denoted by { 0 }. Note: 1. 1 does not belong to V but it belong to F. 2. Normally, we take (R, +, .) or (C, +, .) as fields. In that case, real number or complex numbers are scalars. In our course, mostly we adopt this. 3. We use a, b, g ....... for scalars and xyz for vectors. Examples 1. A field F is a vector space over F or over any subfield S of F. i.e., F(F) or F(S) is a vector space. 2. C(R), C(C), R(R), are vector spaces. (where C is the complex field and R is the real field). Let us prove one of the above namely C(R) is a vector space. To Prove V1 : Here + and . of the field are ordinary + and. The ⊕ and * of the space C are also ordinary + and . of C. (C, ⊕) = (C, +) is an abelian group This is true from group ideas. We accept this is true V2 : a, b ∈ F means a . b ∈ R x, y ∈ V means x, y ∈ C. i.e., x, y, z are complex numbers and a, b are real numbers a * x = a . x = (real no) . (complex no) = complex no ∈ C. So a * x ∈ C (a . b) * x = (a . b) . x = (b . x) = a . (b . x). is true. Also 1 ∈ R. Hence – x∈C 1 * x = 1 . x = x is true ∨ V3 : a * (x ⊕ y) = a . (x + y) = a . x + a . y = (a * x) ⊕ (a * y) is true (a + b) * x = (a + b) . x = a . x + b . x = a * x ⊕ b * x All the ten axioms of the vector space are verified to be true. Hence C(R) is vector space (C is space and R is the field here) Example 2. Is R(C) a vector space? Here R is the space and C is the field. a, b ∈ C and x, y ∈ R. V1 is true here. (you can try). a * x ≠ (complex no) . (real no) = complex no ∉ R

354

Engineering Mathematics-III (Third Semester)

a * x ∈ field and not to space This postulates is violated. Hence it is not a vector space. Example 3. Show that V, the set of all ordered n-tuples of the elements of any field F for a fixed n ∈ N, is vector space. Proof : We will verify the axioms of a vector space. (i) Let x = (x1, x2 , ..... xn), y = (y1, y2, ...... yn) be two vectors of V. Here xi, yi ∈ F We define x ⊕ y = (x1, x2, ..... xn) ⊕ (y1, y2, ..... yn) = (x1 + y1, x2 + y2, ..... xn + yn)



xi + yi ∈ F since xi, yi ∈ F.

Hence x ⊕ y ∈ V.

Addition of vectors is binary. i.e., (ii) Associativity: x ⊕ ( y ⊕ z ) = (x1, x2, ..... xn) ⊕ (y1, y2, ..... yn) ⊕ (z1, z2, ..... zn)]

= (x1, x2, ..... xn) ⊕ (y1 + z1, y2 + z2, ..... yn + zn)



= (x1 + y1 + z1, x2 + y2 + z2, ..... xn + yn + zn)

Similarly ( x ⊕ y ) ⊕ z = (x1 + y1 + z1 x2 + y2 + z2 + ..... + xn + yn + zn)

Hence x ⊕ ( y ⊕ z ) = ( x ⊕ ( y ) ⊕ z

(iii) V x = (x1, x2, ..... xn), ∃ 0 = (0, 0, ..... 0) ∈ V Such that x ⊕ 0 = (x1 + 0, x2 + 0, ..... xn + 0) = (x1, x2, ..... xn) = x Also 0 ⊕ x = x – x = (x1, x2, .... xn), – x = (– x1, x2, ..... – xn) ∈ V (iv) ∨ and x + (– x ) = 0 and (– x ) + x = 0 (v) Also x ⊕ y = y ⊕ x is true Hence (V, ⊕) is an abelian group. (vi) Let a, b ∈ F Define a * x = a . (x1, x2, ... xn) = (ax1, ax2, ...... axn) ∈ V since a1, x1 ∈ F and so axi ∈ F \ a ∈ F, x ∈ v ⇒ a * x ∈ v (vii) a * (b * x) = a * (bx1, bx2 ..... bxn)

= (abx1, abx2 ...... abxn)



= (a . b) * (x1, x2 ..... xn)



= (a . b) * x

– x ∈ V, and 1 ∈ F (viii) ∨

1 * x = 1 * (x1, x2 ..... xn)

355

Linear Algebra



= (1. x1, x2 ..... 1. xn)



= (x1, x2 ..... xn)



= x

– x , y ∈ V, a ∈ F, (ix) ∨ a* ( x ⊕ y ) = a * (x1 + y1 + x2 + y2, ..... xn + yn)





= (a x1 + a y1, a x2, + a y2, ..... a xn + a yn) = (a x1, a x2 ...... a xn) ⊕ (a y1, a y2 ..... a yn)



=a* x ⊕a* y

(x) (a + b) * x = ((a + b)x1, (a + b)x2, ..... (a + b)xn)

= (ax1 + bx1, ax2 + bx2, ..... axn + bxn)



= (ax1, ax2, ..... axn) + (bx1, bx2, ...... bxn)



=a* x ⊕b* y

Hence V(F) is a vector space. EXERCISE 6 (a) 1. Let F be a field and let V be the set of all ordered pairs (x, y) where x, y ∈ F. Define (x1, y1) ⊕ (x2, y2) = (x1 + x2, y1 + y2) and a(x, y) = (ax, y) where a ∈ F. Prove V(F) is not a vector space. 2. V = {(a, b) : a, b ∈ R} Define (a, b) ⊕ (c, d) = (0, b + d) a * (a, b) = (aa, ab) Prove V(R) is not a vector space

3. In problem 2, define a * (a, b) = (aa, b), a ∈ R. Prove V(R) is not a vector space.



4. Prove C(C), C(R), R(R) are vector spaces, where addition and scalar multiplication of vectors are ordinary addition and multiplication.



5. Show that the set of all m × n matrices with elements as real numbers is a vector space over the field R with usual operations of matrices.



6. If A is the set of all real valued continuous functions defined in [0, 1], show that A is a vector space over R with addition and scalar multiplication defined as follows. – f, g ∈ A (f + g) (x) = f(x) + g(x) ∨

– f ∈ A, a ∈ R. (af) (x) = af (x), ∨



7. Let P(x) denotes the set of all polynomials in one variable x over a field F. Then prove P (x) is a vector space over F with addition defined as addition of polynomials and scalar multiplication defined as the product of the polynomials and scalar multiplication defined as the product of the polynomials by an element of F.



8. Rn is the set of n-tuples of real numbers. Prove Rn is a real space.



9. Cn is the set of all n-tuples of complex numbers. Define x ⊕ y = (x1 + y1, x2 + y2, ..... xn + yn) and a * x = (x1, x2, ...... axn) as usual. Prove Cn is a vector space over C.



10. Taking n = 1 in problem 9, prove C(C) is a vector space.



11. V consists of only zero vector with scalar multiplication defined by a * 0 = 0 for all a ∈ R. Prove V(R) is a vector space.

356

Engineering Mathematics-III (Third Semester)

LINEAR INDEPENDENCE AND DEPENDENCE Definition: If V be a vector space over the field F, then the vectors x1, x2, ..... xn ∈V are said to be linearly dependent over F, if there exist elements a1, a2 ..... an ∈F, not all of them zero, that is at least one of the ai′ s is not equal to zero, such that

n

0 ∑ a i xi = i =1

Otherwise, the vectors are said to be linearly independent. n

In other words, if

0 implies each a = 0 ∑ a i xi = 1 i =1

Then x1, x2, ...... xn are linearly independent and at least one a1, is not zero, then x1. x2 ..... xn are linearly dependent. Note. 1. The set S = {x1, x2, ..... xn} is linearly independent or linearly dependent according as the vectors are linearly independent or linearly dependent.

2. Any subset of a linearly independent set is linearly independent.

Proof. Let x1, x2, ..... xn be linearly independent.

Therefore,

n

∑ a i xi i =1

= 0 ⇒ a1 = a2 = .... an = 0

... (1)



To prove x1, x2 ..... xm are linearly independent, m ≤ n.



The order of the elements is immaterial. m

m

i= 1

i= 1

∑ a i xi =0 ⇒ ∑ a1 x1 + i.e.,

n

∑

i= m +1

0 ⋅ xi =0

n

0 where ai = 0, for i = m = 1, .... n ∑ a i xi = i =1



By (1), each a1 = 0

\

a1, a2, ..... am = 0

Hence x1, x2, ..... xm are linearly independent.

3. Any superset of a linearly dependent set is linearly dependent.



4. Any subset of a vector space is either linearly independent or linearly dependent.



5. A set containing zero vector only is linearly dependent.



6. Any set of a vector space containing only a non-zero vector is linearly independent.



7. A set containing one of the vectors as zero vector is linearly dependent.



8. The set of non-zero vectors x1, x2, ...... xn is linearly dependent iff some xk, 2 ≤ k ≤ n, is a linear combination of the preceding ones.



9. If x is a linear combination of x1, x2, ..... xn then (x, x1, x2, ..... xn) is linearly dependent. Example 1. Determine whether the following vectors are linearly dependent or not.

(i) x = (4, 3, –1), y = (1, 1, 5) in R3 (ii) x = (1, 2, 3, 4); y = (2, 4, 6, 8) in R4  1 −2 4   2 −4 8  (iii) x =  ,y =   0 −2   3 0 −1   6 (iv) x = t3 + 3t + 4, y = t3 + 4t + 3

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Linear Algebra

Proof. When two vectors are given, they are linearly dependent if one is scalar multiple of other. Hence (i) x ≠ ky ∴ x, y are linearly independent (ii) y = 2x ∴ x, y are linearly dependent (iii) y = 2x; x, y are linearly dependent (iv) x, y ae linearly independent. Example 2. Determine whether the following vectors in V3(R) or R3 are linearly dependent or not. (i) x = (1, 2, 3), y = (4, 1, 5), z = (– 4, 6, 2) (ii) x = (1, 2, – 3), (1, –3, 2), (2, –1, 5) (iii) x = (3, 0, –3), (–1, 1, 2), (4, 2, –2) (2, 1, 1) Proof. (i) Let a, b, g ∈ R (scalars) ax + by + gz = 0 ⇒ a(1, 2, 3) + b(4, 1, 5) + g(–4, 6, 2) = (0, 0, 0) \ a + 4b – 4g = 0 ... (1) (1) + (2) gives (3) 2a + b + 6g = 0 ... (2) Solve a + 4b – 4g = 0

3a + 5b + 2g = 0 ... (3) 2a + b + 6g = 0

Hence a = – 4, b = 2 , g = 1

Hence the vectors are linearly dependent.

(iii) Write the vectors as row vectors of a matrix.

 1 2 −3    = A  1 −3 2   2 −1 5   



 1 2 −3    −  0 −5 5   0 −5 11   



 1 2 −3    −  0 −5 5  0 0 6   R(A) = 3 and three vectors are linear independent (iii) There are 4 vectors in R3. Hence they are linearly dependent. Example 3. Prove that the vectors (1, 0, 0), (0, 1, 0), (0, 0, 1) and (1, 1, 1) are linearly dependent. Also prove any three vectors are linearly independent. Proof. There are 4 vectors in R3. Hence they are linearly dependent.

OR a (1, 0, 0) + b (0, 1, 0) + g (0, 0, 1) + d (1, 1, 1) = (0, 0, 0)

⇒

a + d = 0, b + d = 0, g + d = 0

⇒

a = b = g = 1, d = – 1

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Engineering Mathematics-III (Third Semester)

Hence the vectors are linearly dependent. Also (1, 0, 0), (0, 1, 0), (0, 0, 1) are linearly independent. Example 4. If x, y, z are linearly independent vectors in V(F) prove (i) x + y, y + z, z + x are linearly independent and (ii) x – y, y – z, z – x are linearly dependent. Proof: a(x + y) + b(y + z) + g(z + x) = 0 ⇒ (a + g)x + (a + b)y + (b + g)z = 0 ⇒ a + g = 0, a + b = 0, b + g = 0 Adding 2 (a + b + g) = 0 \ a+b+g=0 using (1), a = 0, b = 0, g = 0 Hence x + y, y + z, z + x are linearly independent.

... (1)

(ii) a(x – y) + b(y – z) + g(z – x) = 0 ⇒ a = 1, b = 1, g = 1 Hence x – y, y – z, z – x are linearly dependent. Example 5. If V(R) is a vector space of 2 × 3 matrices over R, show that

1 −1  1 −3   2  1  4 A = =  , B = , C  5  3 −2 4   −2 0  1



in V(R) are linearly independent.

−1 −2

2  3

Proof. aA + bB + gC = 0 ⇒

 2 a  3

1 −1   1  + b −2 4   −2

1 0

−3   4 + g 5  1

−1 −2

2  =0 3

\ 2a + b + 4g = 0

...(1)

a + b – g = 0

... (2)



– a – 3b + 2g = 0

... (3)

3a – 2b + g = 0

... (4)

– 2a . – 2g = 0

... (5)



... (6)

4a + 5b + 3g = 0

Solving we get the only solution a = 0, b = 0, g = 0 Hence they are linearly independent. Example 6. Prove (1, 0, 0) + b(0, 1, 0) + g(0, 0, 1) = 0 are linearly independent in R3. Find three sets of three linearly independent vectors from your answer. ⇒

a(1, 0, 0) + b(0, 1, 0) + g (0, 0, 1) = 0 a = 0, b = 0, g = 0

Hence the vectors are linearly independent. x + y, y + z, z + x are also linearly independent. Hence (1, 0, 0) + (0, 1, 0), (0, 1, 0) + (0, 0, 1), (0, 0, 1) + (1, 0, 0) i.e., (1, 1, 0), (0, 1, 1) and (1, 0, 1) are linearly independent. (1, 2, 1), (1, 1, 2) and (2, 1, 1) are linearly independent. Also (2, 3, 3), (3, 2, 3) and (3, 3, 2) are linearly independent.

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Linear Algebra

Example 7. Show that the vectors (1, 2, 1, 0), (1, 3, 1, 2) (4, 2, 1, 0) and (6, 1, 0, 1) are linearly independent in R4. Proof. Write vectors as row vectors of a matrix A. 1  1 A =  4  6

2 3 2 1

1 1 1 0

  ~    

1 0 0 0

2 1 −6 −11

  ~    

1 0

2 1

0 0

0 0

0  2 0  1 1 0 −3 −6

0 2 0 1

     

1 0

0  2 −3 12   −6 23 

1 0  1 2   0 1 0 2 ~   0 0 −3 12    0 0 −1   0 R(A) = 4. Hence the four vectors are linearly independent. Example 8. Express (2, –1, 4) as a linear combination of (4, 0, 12) and (0, 1, 2). Proof. Hence

Let (2, –1, 4) = a(4, 0, 12) + b(0, 1, 2) 2a = 2, b = –1, 12a + 2b = 4 a = 1/2, b = –1 satisfy the three equations.

\

(2, –1, 4) =

1 (4, 0, 12) – (0, 1, 2). 2 EXERCISE 6 (b)



1. Show that the following vectors are linearly dependent. Find their relationship in each case. (i) (1, 2, 0), (2, 3, 0) and (8, 13, 0) of R3 (ii) (2, 3, –1, –1), (1, –1, –2, –4), (3, 1, 3, –2) and (6, 3, 0, –7) of R4 (iii) (1, 2, –1, 3), (0, –2, 1, –1), and (2, 2, –1, 5) of R4 (iv) (1, 2, 3), (3, –2, 1) and (1, –6, –5) of V3 (R)



2. Show the the vectors are linearly independent in each case (i) (1, 1, 1, 1 ), (1, –1, 1, 1), (1, 1, –1, 1) and (1, 1, 1, –1) of R4 (ii) (1, 2, 3) and (2, –2, 0) of R3 (iii) (1, 0, 0, 0), (0, 1, 0, 0), (0, 0, 1, 0) and (0, 0, 0, 1) of V4 (R). (iv) (0, 2, 1, –1), (1, –1, –1, 3) and (7, 2, –7, 4) of R4.

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3. If x, y, z are linearly independent, prove x + y, x – y, x – 2y + z are also linearly independent.



4. Express (x, y, z) as a linear combination of the vectors (1, 1, 0), (0, 1, 1), (1, 0, 1).

BASES AND DIMENSION Definition: A basis in a vector space V(F) is a set B of linearly independent vectors of V(F) such that every vector in V is a linear combination of elements of B. In other words, a basis for a vector space V over a field F, is a linearly independent set of vectors from V which spans V. If a basis for V(F) contains only a finite number of vectors, then V(F) is finite dimensional. Definition: The number of elements in any basis of a finite dimensional vector space V(F) is called the dimensions of the vector space and is demand as dim V. Theorem. If {x1, x2, ..... xn} is a basis of V(F), then any vector of V(F) can be expresssed uniquely as a combination of the vectors of the basis Let y =

n

∑ a i xi i =1

Now we have to prove the uniqueness of the expression. let y =

So,

n

∑ bi i−1

\

n

∑ a 1 xi

=

i =1

Hence

xi, if not unique.

n

∑ b i xi i −1

n

∑ ai − bi x1 = 0 i =1

Hence ai – bi = 0 for i = 1, 2, ...... n since (x1, x2, ..... xn) is linearly independent. So,

ai = bi

Hence

y=

n

∑ a i xi

is unique.

i =1

Note 2. A set of vectors having zero vector as one of the vectors cannot be a basis. n

Note 3. If {x1 , x2, ... xn} is a basis and if x ∈ V(F) fand x =

∑ ai xi , then the set of scalars i =1

{a1, a2, ...... an} can be taken as coordinates of the vector x corresponding to the basis set as coordinate system. Note 4. There exists a basis for each finite dimensional vector space. Theorem: Each set of (n + 1) or more vectors in an n-dimensional vector space V(F) is linearly dependent. Cor: In an n-dimensional vector space V(F), any linearly independent set of n vectors is a basis of V(F). Theorem: If V(F) is finite dimensional vector space of dimension n and if W is any subspace of V then W is also finite dimensional and dim W ≤ dim V.

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Linear Algebra

Example 1. Show that S = {(1, 0, 0, 0), (0, 1, 0, 0), (0, 0, 1, 0), (0, 0, 0, 1) is a basis of V4 (R) or R4. Firstly, let us prove S is linearly independent.

a1 (1, 0, 0, 0) + a2 (0, 1, 0, 0) + a3(0, 0, 1, 0) + a4 (0, 0, 0, 1) = 0

⇒

(a1, a2, a3, a4) (0, 0, 0, 0)

Hence

a1 = 0, a2 = 0, a3 = 0, a4 = 0

Therefore S is linearly by independent. Next, we should prove any vector of R4 is a linear combination of vectors of S. Let

(x1, x2, x3, x4) ∈ R4

Then (x1, x2, x3, x4) = x1(1, 0, 0, 0) + x2 (0, 1, 0, 0) + x3 (0, 0, 1, 0) + x4 (0, 0, 0, 1) Therefore S spans R4. Hense S is a basis of R4. Example 2. If V(R) = Pn, be vector space of polynomials in t over the field of real, of degree ≤ n, show that S = {1, t, t2, ...... tn} is a basis of V(R). Proof. V(R) = Pn = {P(t) : p(t) = a0 + a1t + a2t2 + ..... + antn} for a1∈ R Let ai’ s be scalars such that

a0 . 1 + a1 . t + a2 . t2 + ..... + an . tn = 0 (zero polynomial) a0 = 0, a1 = 0, ..... an = 0.

⇔ So, S in linearly independent. To Prove: S spans V(F).

Let b0 + b1 t + b2t2 + ..... + bn tn ∈V(F) Then b0 + b1t + .... + bntn = b1. 1 + b2.t + b3 . t2 + ..... + bn . tn = a linear combination of S.

Hence S is a basis of Pn.

Example 3. Prove S = {(1, 0, 0), (1, 1, 0), (1, 1, 1)} is a basis of R3. Hence find the coordinates of the vector (6, 5, 3) w.r.t. this basis. What is the dimensions of the space? Proof. a1(1, 0, 0) + a2 (1, 1, 0) + a3 (1, 1, 1) = 0 ⇒

a1 + a2 + a3 = 0



a2 + a3 = 0



a3 = 0

Going backwards, a3 0, a2 = 0, a1 = 0 Hence S is linearly independent in R3. Therefore S spans R3. S is a basis of R3.

(6, 5, 3) = a1 (1, 0, 0) + a2 (1, 1, 0) + a3 (1, 1, 11)

\

a1 + a2 + a3 = 6



a2 + a3 = 5



a3 = 3

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Engineering Mathematics-III (Third Semester)

a2 = 2 and a1 = 1

Hence

Coordinates of the vector (6, 5, 3) are (1, 2, 3); the dimension of R3 is three. Example 4. Show that B = {(1, 0, 0)}, (1, 1, 0), (1, 1, 1)} is a basis of V3 (C). Hence find the coordinates of the vector (3 + 4i, 6i, 3 + 7i) in V3(C) w.r.t. B. Proof. a1(1, 0, 0) + a2(1, 1, 0) + a3(1, 1, 1) = (0, 0, 0) implies

a1 + a2 + a3 = 0



a2 + a3 = 0



a3 = 0.

Hence, solving,

a3 = 0, a2 = 0, a1 = 0.

Therefore B is linearly independent. V3(C) is of dimension 3 and B has 3 vectors. Hence B is a basis of V3 (C). (3 + 4i, 6i, 3 + 7i) = a1 (1, 1, 0) + a2 (1, 1, 0) + a3 (1, 1, 1) ⇒

a1 + a2 + a3 = 3 + 4i a2 + a3 = 6i



a3 = 3 + 7i

\

a3 = 3 + 7i, a2 = – 3 – i, a1 = 3 – 2i

Coordinates are (3 – 2i, –3 –i, 3+ 7i) Example 5. Show that the vector a1 = (1, 0, –1), a2 = (1, 2, 1) and a3 = (0, – 3, 2) from a basis of R3. Express each of the standard basis vector as a linear combination of a1, a2, a3. Proof. We can easily prove {a1, a2, a3} is linearly independent and they form a basis. Let Then

(1, 0, 0) = a (1, 0, –1) + b(1, 2, 1) + r(0, –3, 2) a=

7 3 1 ,b ,r = = 10 10 5

Coordinates of (1, 0, 0) w.r.t the basis {a1, a2, a3} are  7 3 1  , ,   10 10 5  Similarly, we can find the coordinates of (0, 1, 0) and (0, 0, 1). Example 6. Prove B = {(1, 0, 0), (0, 1, 0), (0, 0, 1)} is a standard basis of R3. Find the coordinates of the vector (6, 5, 3) w.r.t. this standard basis. Proof. Let us prove B is linearly independent Hence

a1(1, 0, 0) + a2(0, 1, 0) + a3(0, 0, 1) = 0 implies (a1, a2, a3) = (0, 0, 0) a1 = 0, a2 = 0, a3= 0

Further, B is linearly independent set of three vectors in a three diamensional vector space R3. Hence B is a basis of R3 (called standard basis of R3)

(6, 5, 3) = 6(1, 0, 0) + 5(0, 1, 0) + 3(0, 0, 1)

Hence the coordinates of the vector (6, 5, 3) are (6, 5, 3) Example 7. (a, b) and (c, d) are two vectors which form a basis of R2. Prove ad – bc ≠ 0.

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Linear Algebra

a b Proof. Since (a, b) and (c, d) are linearly independent, the rank of   is 2 and hence c d a b ≠ 0 i.e., ad – bc ≠ 0 c d EXERCISE 6(C)

1. Prove that the following sets of vectors are basis for V3(R).

(a) {(5, 5, 6), (6, 5, 5), (5, 6, 5)} (b) {(2, 3, 3), (3, 2, 2), (3, 3, 2)} (c) {(1, 2, 1), (1, 1, 2), (2, 1, 1)} (d) {(2, –3, 1), (0, 1, 2), (1, 1, –2)}

2. Prove {(1, 0, 0, 0), (0, 1, 0, 0), (0, 0, 1, 0) (0,0, 0, 1)} constitute a basis for R4. Also find the coordinates of the vector (a, b, c, d) with respect to this basis. What is the dimension this space?

INNER PRODUCT SPACES General Definition: If F be either the field of real numbers or the field of the complex number and V(F) be a vector space, then an inner product in V(F) is a numerical valued function which assigns to each ordered pair of vectors x, y in V, a scalar (x, y) in F, such that I1 : ( x , y ) = (y, x) (conjugate symmetry) I2 : (ax + by, z) = a(x, z) + b(y, z) where x, y, z ∈ V and a, b ∈ F (linearity) I3 : (x, x) ≥ 0 and (x, x) = 0 ⇔ 0

(The above is the general definition of an inner product.)

The vector space V(F) with an inner product is called an inner product space. Norm: We define norm of vector x or length of vector x as || x || =

( x , x) .

We read || x || as norm of x. Unit vector. A vector x is a unit vector if || x || = 1 Note 1 : In the case of vector space over real field, ( x , y ) = (x, y) = (y, x) Note 2. (Sai ) = S ai Note 3. ab = a b Note 4. ( a ) = a Note 5. If the field is the complex field, the inner product is complex inner product. Note 6. If the field is the real field, the inner product is real inner product. Note 7. || x || is real and ≥ 0. Example 1. In V3 (C), define an inner product

(x, y) =

n1

∑ ai bi i =1

if x = (a1, a2, ... an) and

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Engineering Mathematics-III (Third Semester)



y = (b1, b2, ... bn)

and a1, bi ∈ C. Prove this is really an inner product in V3(C). Proof. (x, y) =

n

∑ ai bi i =1

is a complex number ∈ C.

This is a scalar valued function. We will prove the three condition I1, I2, I3. Sac bi = Sbi ai = ( y , x) (i) ( x , y ) = Sai , bi = (ii) (ax, + by, z) = S(aai + bbi) ci

= aSai ci + bSbi ci



= a(x, z) + b(y, z)

(iii)

(x, x) = Sai ai = S| a1 |2 ≥ 0

If (x, x) = 0, then S | ai |2 = 0 and | ai | = 0 for each i and hence ai = 0 and then x = 0. Conversely, if x = 0, the (x, x) = (0, 0) = 0 The axions of inner product are verified to be true. Hence with this definition as inner product, V3(C) is an inner product space. This is called standard inner product in V3(C). Note 1. If x = (a1, a2, ...... an), y = (b1, b2 ... bn) in V3 (R), then the definition (x, y) = Sai bi gives an inner product. This is standard inner product in V3(R). Definition. A real inner product space is called an Euclidean space and a complex inner product space is called a unitary space. Note 2. If x = (x1, x2, ... xn) is a complex vector in Vn(C), then || x || = | x1 |2 +| x2 |2 + .....+| xn |2 Example 2. Prove every finite dimensional vector space over the field of reals or complex numbers is an inner product space with respect to the standard definition of inner product. Proof. Follow the pattern of the proof in example 1. Example 3. If one of the vectors x or y is zero. The inner product is zero. That is ( 0 , y) = 0 or (x, 0 ) = 0. Also prove (x, ay) = a (x, y), and ( z, ax, by) = a (z, x) + b (z, y) Proof. ( 0 , y) = (0. 0 , y) = 0 ( 0 , y) = 0 since (ax, y) = a (x, y)

(x, ay) = (ay , x ) = a( y , x) = a( y , x) = a( x , y)



(z, ax + by) = ( ax , by , z) = a( x , z) + b( y , z)



= a( x , z) + b( y , z)



= a( z, x ) + b( z, y )

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Linear Algebra

NORMED VECTOR SPACE Definition: A vector space V(F) which assigns to each vector x in V a real number || x || (read norm x) such that (i) || x || ≥ 0 and || x || = 0 ⇔ x = 0 for each x of V (ii) || a x || = | a | || x || for a ∈ F, x ∈ V and (iii) || x + y || ≤ || x || + || y || for x, y ∈ V is called a normed vector space. Theorem: Every inner product space is a normed vector space. Proof. Let V(F) be an inner product space. ( x , x) (i) || x || = =

a non − negative number = a non-negative number.

\

|| x || ≥ 0.





(x, x) = 0 ⇒ x = 0



|| x ||2 = 0 ⇒ x = 0

\

|| x || = 0 ⇒ x = 0



x = 0, then (x, x) = 0

If

⇒ || x || = 0 i.e., || x || = 0 2

|ax|2 = (ax, ax) = a a (x, x) = | a |2 || x ||2

(ii) \

|| ax || = | a | || x ||

(iii)

|| x + y2 || = (x + y, x + y)



= (x, x + y) + (x + y, y)



= ( x + y , x) + ( x + y , y)



= ( x , x) + ( y , x) + ( x , y) + ( y , y)



= (x, x) + (x, y) + ( x , y ) + (y, y)



= || x ||2 + 2 Real (x, y) + || y ||2



= ≤ || x ||2 + 2 | (x, y) | + || y ||2 since Real (x, y) ≤ | (x, y) |





≤ || x ||2 + 2 || x || || y || + || y ||2 by Schwarz inequality ≤ || x || + || y ||2 (see next pages)

\ || x + y || ≤ || x || + || y || The inner product space satisfies all the conditions of normed vector space. Hence, it is a normed vector space. Note 1. Combining the above theorem and worked example 2, we get that: “Every finite dimensional vector space is a normed vector space”. The converse is not true. Example 4. If a1, b ∈ F and x, y, z ∈ V where V(F) is a normed vector space, prove the following. (i) || ax + by ||2 = | a |2 || x ||2 + a b (y, x) + a b(y, x) + | b |2 || y ||2 (ii) || x + y ||2 + || x – y ||2 = 2( || x ||2 + || y ||2) (iii) (x, y) = 0 ⇔ ||ax + by||2 = | a |2 || x ||2 + | b |2 || y ||2

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(iv) || x || = || y || ⇒ (x – y, x + y) = 0 in real inner product space (v) If (x, y) = 0, then || x + y ||2 = || x ||2 + || y ||2 (vi) 4(x, y) = || x + y ||2 – || x – y ||2 + i || x + iy ||2 – i|| x – iy ||2. Proof. (i) || ax + by ||2 = (ax + by, ax + by)

= a(x, ax + by) + b(y, ax by)



= a (ax + b y , x ) + b (ax , b y , y )



= aa( x , x ) + ab( y , x ) + ba( x , y ) + bb( y , y )



= aa( x , x ) + ab( x , y ) + ba( y , x ) + bb( y , y )



= |a|2 || x ||2 + a b (x, y) + a b(y, x) + | b |2 || y ||2

(ii) Putting a = 1, b = 1 in the above result || x + y ||2 = || x ||2 + (x, y) + (y, x) + || y ||2 Put

a = 1, b = –1 in the above result, || x – y ||2 = || x ||2 – (x, y) – (y, x) + || y ||2



Hence || x + y ||2 + || x – y ||2 = 2(|| x ||2 + || y ||2) (iii) In the result of (i), put (x, y) = 0 (y, x) = 0

|| ax + by ||2 = | a |2 || x ||2 + | b |2 . || y ||2

(iv) (x – y, x + y) = (x, x + y) – (y, x + y)

= (x, x) + (x, y) – (y, x) – (y, y)



= || x ||2 – || y ||2 Q (x, y) = (y, x) in real product space



= 0 since || x || = || y ||

(v) If (x, y) = 0 use the result in (ii). Then

|| x + y |2 = || x ||2 + || x ||2

(vi) || x + iy |2 = i || x ||2 + i (x, y) + i (y, x) + i. i || y ||2 i|| x + iy ||2 = i|| x ||2 + (x, y) – (y, x) + i || y ||2 Similarly, –i || x + iy |2 – i || x ||2 + (x, y) – (y, x) – i|| y ||2 i|| x + iy ||2 – i|| x + iy |2 = 2(x, y) – 2(y, x) ...(i) From the result in (ii),

|| x + y ||2 – || x – y ||2 = 2(x, y) + 2(y, x)

... (ii)

Adding (i) and (ii)

4(x, y) = || x + y ||2 – || x – y ||2 + i|| x + iy ||2 – i|| x – iy ||2

Example 5. If V(C) be the vector space of all continous complex valued functions on the real interval a ≤ t ≤ b, and (f. g) =

b

∫a

f (t) g(t) dt, f(t). g(t) ∈ V. Prove this is an inner product in V.

Proof. From the definition of (f.g) it is scalar valued. (i) ( f , g ) =

b

∫a

f (t) g ,(t) dt

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Linear Algebra b

∫a



=



= (g, f)

f (t) g(t) dt

(ii) af(t) + b g(t) ∈ V. (af + bg.h) =

b

∫a (af + bg )h dt b

b

a

a



= a ∫ fh dt + b∫ gh dt , h(t) ∈ V



= a(f, h) + b(g, h)

(iii)

(f.f) =



=

If

b

∫a

f (t) f (t) dt

b

∫a| f (t)

2

dt ≥ 0

f = 0, (f, f) = 0 and the converse

The above definition of (f, g) is really an inner product. Example 6. x, y ∈ V2 (C) and x = (a1, a2), y = (b1, b2) z = (c1, c2). Verify whether the following defines an inner product. (i) (x, y) = 2a1b1 + a1b2 + a2 b1 + a2 b2 (ii) (x, y) = a1b1 + ( a1 + a2 )(b1 + b2 ) (iii) (x, y) = a1 b1 – a2 b1 – a1 b2 + 4a2 b2. This is left as an exercise to the reader. Theorem: Schwarz’s inequality: If x, y are vectors in an inner product space, then prove | (x, y) || x || || y || Proof. Let l be real and x, y ∈ V(F) ; l ≠ 0

|| x + ly||2 ≥ 0



|| x ||2 + l2 || y ||2 + l[(x, y) + (y, x)] ≥ 0



l2 || y ||2 + l[(x, y) + ( x , y ) + || x ||2 ≥ 0

i.e.,

l2 || y ||2 + l.[Real part of (x, y)] + || x ||2 ≥ 0

i.e.,

l2 || y ||2 + 2l |(x, y)| + || x ||2 ≥ 0 since Rl (x, y) ≤ |(x, y)|

This is a quadratic expression in l and l2 > 0 \

“b2 – 4ac” ≤ 0

i.e.,

4|(x, y)|2 – 4 || y ||2 || x ||2 ≤ 0

i.e.,

| (x, y) |2 ≤ || x ||2 || y ||2

Hence,

| (x, y) | ≤ || x || || y ||

Cauchy’s inequality: If Vn(C) is a unitary space, for any two vectors x = (a1, a2, ... an), y = (b1, b2, ... bn) define (x, y) =

n

∑ ai bi ; then by the above inequality. i =1

2



|Sai bi | ≤ (Sai ai )(Sbi bi )

i.e.,

|Sai bi |2 ≤ (S| ai |2 )(S|bi |2 )

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Engineering Mathematics-III (Third Semester)

In the case of Euclidean space (real field) (a1b1 + a2b2 + anbn)2 ≤ ( a12 + a22 + ..... + an3 )(b12 + b22 + ..... + bn2 ) ORTHOGONALITY Definition: If x and y are two vectors in inner product space V(F), then x and y are orthogonal if (x, y) = 0 i.e., x is orthogonal to y if (x, y) = 0 If x is orthogonal to y, then y is orthogonal to x because of symmetry

(x, y) = 0 ⇔ (y, x)

– x ∈V Note 1. Every vector is orthogonal to zero vector since (x, 0 ) = 0 ∨

2. Zero vector is orthogonal to itself



3. The angle between two vectors, q, is given by ( x , y) x y



cos q =



(x, y) = 0, q = 90°

If

Orthogonal set : If S be a set of vectors in an inner product space V(F), then S is called an orthogonal set if any two distinct in S are orthogonal i.e., (xi, xj) = 0, i ≠ j. Orthonormal set : In an orthogonal set S if every vector is a unit vector, then the set is an orthonormal set. In other words, S is orthonormal if (i) (x, y) = 0 for x ≠ y (ii) (x, y) = 1 for x = y when x, y ∈ S. Theorem: An orthogonal set of non-zero vectors in an inner product space V(F) is linearly independent. Proof: Let S = {x1, x2, ..... xm} be an orthogonal set of non-zero vectors \ (xi , xj ) = 0 for i ≠ j m

0 we have to prove each ai = 0 ∑ a i xi =

Let

i =1

i.e.,

m   ∑ a i xi , x j  = (0, xj) = 0  i =1  m

∑ ai (xi , x j ) = 0 i =1

i.e.,

aj (xj, xj) = 0 Q (xe, xj) = 0, i ≠ j

\

aj = 0 since xj ≠ 0, j = 1, 2, ...m.

\ S is linearly independent Note 1: Let x ≠ 0 and x ∈ V(F), inner product space; || x || ≠ 0,  x x  orthonormal singleton set for  = ,  x   x

x = ( x , x) 1 2 x

 x  x ∈ V A =   is an x  x 

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Linear Algebra

Example 7. In R3, prove B = {1, 0, 0} , (0, 1, 0), (0, 0 , 1) is an orthonormal set under usual standard inner product. Also find the angle between (1, 2, 3) and (0, 2, 5). Proof. We define (x, y) = Σaibi as inner product in R3. Let the vector be denoted by x, y, z.

(x, y) = 1.0 + 0.1 + 0.0 = 0



(y, z) = 0.0 + 1.0 + 0.1 = 0



(x, z) = 0.1 + 0.1 + 0.1 = 0



|| x ||2 = (x, x) = 1.1 = 1 \ || x || = 1

Similarly

|| y || = 1, || z || = 1

Hence B is an orthonormal set

cos q =



=



=



( x , y) ; (x, y) = Σaibi = 1.0 + 2.2 + 3.5 = 19 x y 19 1 + 4 + 9 4 + 25

sin || x || =

Sai 2

19 14 29

 19  q = cos −1    14 29 

Example 8. Find the norm of the vector n = (2, –3, 4) and normalize it. Proof.

(n, n) = || n ||2 = 22 + (– 3)2 + 42 = 29 || n || =

Normalized vector is =

29 n  2 −3 4  = , ,  n  29 29 29 

GRAM-SCHMIDT ORTHOGONALIZATION PROCESS Theorem: Every finite dimensional inner product space has an orthonormal basis. Proof. Let V be an inner product space and {b1, b2, ..... bn} be a basis for V. From this, we shall obtain an orthogonal basis (a1, a2, ...... an) by means of construction known as Gram-Schmidt orthogonalization process. First, Let

Let a1 = b1. Then a1 ≠ 0. a2 = b2 –

(b2 , a1 ) a1

2

⋅ a1

Since, b1, b2 are linearly independent, a2 ≠ 0, for If Also

a2 = 0, then b2 depends upon a1 = b1 which contradicts. (a2, a1) = 0 by direct computation, for (a2, a1) = (b2, b1) –



(b2 , a1 ) a1

2

(a1 , a1 )

= (b2, a1) – (b2, a1) = 0.

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Engineering Mathematics-III (Third Semester)

\ {a1, a2} is an orthogonal set Next, let

a3 = b3 –

(b3 , a1 ) a1

2

a1

(b3 , a 2 ) 2

a2

⋅ a2

Again a3 ≠ 0; otherwise, if a3 = 0, then b3 is a linear combination of b1 and b2 which is untrue. Further,

(a3, a1) = (b3, a1) –

(b3 , a1 ) 2

(a1 , a1 ) −



a1 = (b3, a1) – (b3, a1) – 0



=0

Similarly

(b3 , a 2 ) a2

2

(a 2 , a1 )

(a3, a2) = 0

Hence {a1, a2, a3} is an orthogonal set. Now, suppose we have constructed non-zero orthogonal vectors a1, a2, ..... ak in such a way that aj is bj minus some linear combination of b1, b2 ..... bj – 1 for 2 ≤ j ≤ k. Let

k

ak+1 = bk +1 − ∑

(bk +1 , a j )

j −1

Then

aj

2

aj

ak+1 ≠ 0 and (ak + 1, a2) = (bk + 1, ai) –

k

∑

(bk +1 , a j ) aj

j =1



= (bk+1, a1) – (bk+1, ai)



= 0 for 1 ≤ i ≤ k.

2

(a j , a i )

Thus ak+1 is orthogonal to each of the vectors a1, a2, ..... ak. Suppose ak+1 = 0 Then bk+1 is a linear combination of a1, a2, ..... ak and hence of b1, b2, ..... bk which is not true. Therefore ak+1 ≠ 0. Continuing like this, ultimately, we obtain a non zero orthogonal set (a1, a2, ..... an) containing n district vectors. Since the set is orthogonal, it is independent and has n vectors and the dimensions of V is n. Hence it is an orthogonal basis. To obtain orthonormal basis, divide each vector by its norm. Also

 a a   i , j  = 0, i ≠ j  ai a j   

and

 ai ai  , =    a1 a c 

1 ac

2

(a i ,= ai ) 1

a  a  a Therefore, the set  1 , 2 ,..... n  is the orthonormal set. a n   a1 a 2 Example 9. B = {(1, 1, 1), (0, 1, 1), (0, 0, 1)} is a linear basis of V3 (R). Transform B into an orthonormal basis of V3 (R) by Gram-Schmidt orthogonalization process (Standard inner product). Solution. Let {a1, a2, a3} be the required basis and {b1, b2, b3} be the given linear basis.

371

Linear Algebra

Take

a1 = b1 = (1, 1, 1).

Select

a2 = b2 −



(b2 , a1 ) a1

= (0, 1, 1) –

2

⋅ a1

(1.0 + 1.1 + 1.1) (1,1,1) (1 + 1 + 1) 2 (1, 1, 1) 3

= (0, 1, 1) –

 −2 1 1  =  , ,   3 3 3 Select a3 = b3 −



(b3 , a 2 ) a2

2

⋅ a2 −

(b3 , a1 ) a1

2

a1

 1  1⋅  (1.1) 3  2 1 1 − , , − ⋅ (1,1,1) = (0, 0,1) −    4 1 1   3 3 3  (1 + 1 + 1) + +   9 9 9

1 1 1 1 1 1 = (0, 0,1) +  , − , −  −  , ,  3 6 6 3 3 3 1 1  =  0, − ,  2 2 

|| a1 || =

3 , || a2 || =

 a a2 a3 Hence, orthonormal basis is  1 , ,  a1 a 2 a 3 i.e.,

2 1 , || a3 || = 3 2

  

 1 1 1    2 1 1   1 1  , , , , ,   ,  − ,0 −  6 6 6  2 2   3 3 3   

Example 10. Apply Gram-Schmidt orthogonalization process to the vectors b1 = (1, 0, 1), b2 = (1, 0 , –1), b3 = (03, 3 , 4) so that the linear basis {b1, b2, b3} becomes an orthonormal basis for V3 (R) with the standard inner product. Solution. Let (a1, a2, a3) be the required orthonormal basis. We take (x, y) = Σaibi as inner product. Take

a1 = b1 = (1, 0, 1)

Let

a2 = b2 −



(b2 , a1 ) a 12

= (1, 0, –1) –

a1

(1.1 − 1.1) a1

2

(1, 0, −1) a= 1

372

Engineering Mathematics-III (Third Semester)

a3 = b3 −



(b3, a1 ) a1

2

a1 −

(b3 ⋅ a 2 ) a2

2

a2

... (1)

(b3, a1) = (b3, b1) = 1.0 + 0.3 + 1.4 = 4

(b3, a2) = 0.1 + 3.0 + 4. (– 1) = – 4



|| a1 ||2 = 1 + 0 + 1 = 2; || a1 || =

2



|| a2 ||2 = 1 + 0 + 1 = 2; || a2 || =

2

Now use in (1). a3 = (0, 3, 4) –



4 (−4) (1, 0, 1) – . (1, 0, –1) 2 2

= (0, 3, 4) – (2, 0, 2) + (2, 0 –2) = (0, 3, 0)



|| a3 || = 3



a1 1  a2 1   1  1 , 0, ; , 0, − =  =    a1 2  a2 2  2  2



a3 a3

= (0, 1, 0)

Hence, the orthonormal basis is  1  1   1 1  , 0, , 0, − B =  ,  ,(0,1, 0) 2  2 2  2 



Example 11. The vector space P2 containing polynomials of at most second degree in t has the inner product (p(t), q(t) =

1

∫0 p(t) q(t) dt for p, q ∈ P2. Taking B = {1, t, t2} as a basis, convert

to orthonormal basis. Let {a1, a2, a3} be the required basis

b1 = 1, b2 = t, b3 = t2.

Let

a1 = b1 = 1.

and

a2 = b2 −

\

(b2, a1) =

(b2 − a1 ) a1

1

2

a1

1

∫0 t ⋅ 1dt = 2

||a1||2 = (a1, a1) =

1

dt ∫0 1=

|| a1 || = 1

Use in (1).

... (1)

1    2  ⋅1 = t − 1 a2 = t − 1 2

(t= )10 1

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Linear Algebra



a3 = b3 −

(b3 , a1 ) a1

a1 −

1 2

1 ⋅ 1 dt = 3

1 2

 1  t −  dt  2



(b3, a1) =

∫0 t



(b3, a2) =

∫0 t



=

1 3

∫0  t

=

2

(b3 , a 2 ) a2

2

a2

1  − t 2  dt 2  1 1 1 − = 4 6 12

|| a2 ||2 = (a2, a2) =

2

1 ∫0  t − 2  dt 1

1

 t3 t2 1  =  − + t   3 2 4 0 = Now use in (2).

1 1 1 1 − + = 3 2 4 12 1

|| a2|| =

12

1  1      3 12  1   ×1−   ⋅t −  a3 = t − 1  2 1 12 2

= t 2 −

1 1 1 − t + = t2 − t + 3 2 6

Hence {a1, a2, a3} is

1 2 1  1, t − , t − t +  2 6  

||a3||2 = (a3, a3) 1 2 t 0 

2

1 − t +  dt 6

=

∫

=

∫0  t

=

1 1 1 1 1 1 + + − − + 5 3 36 2 6 9

1 4

+ t2 +

1 1  1 − 2t 3 − t + t 2  dt 36 3 3 

... (2)

374

Engineering Mathematics-III (Third Semester)

=

36 + 60 + 5 − 90 − 30 + 20 180

=

1 180



|| a3 || =

1 180

=

1 3 20

Hence, orthonormal basis is  a1 a 2 a 3  , ,    a1 a 2 a 3 



1    1  2 1, 2  t −  , 3 20  t − t +   2 6     

i.e.,

Example 12. If x, y are orthogonal unit vectors what is the distance between x and y? Solution. d(x, y) = || x – y || || x – y ||2 = (x – y, x – y)



= (x, x) – (x, y) – (y, x) + (y, y) = || x ||2 – 0 – 0 + || y ||2 since x, y are orthogonal = 1 + 1 = 2 since x, y are unit vectors δ(x, y) =

\

2 EXERCISE 6(d)



1. Obtain an orthonormal basis of R3 with standard inner product starting from the linear basis.

(a) B = {(1, –1, 3), (0, 1, –1), (0, 3, –2)} (b) B = {(1, 1, 1), (1, 2, 3), (2, 3, 8)} (c) B = {(1, 2, 1), (1, 1, 2), (2, 1, 1}

2. Let V be a space of polynomials in a variable x over the reals of degree 2 or less. We define an inner product.

(p, q) =

1

∫−1 P(x) q(x)dx , p, q ∈ V.

Find an orthonormal basis starting from the basis (1, t, t2).

3. V is a vector space of R(x) of polynomials of degree at most 3. Define an inner product.

(f.g) =

1

∫0 f (t) g(t) dt. Apply Gram-Smidt orthogonalization process to the basis (1, t, t2,

t3).

4. If x and y are linearly dependent, prove

| (x, y) | = || x || || y ||.

375

Linear Algebra

ANSWERS 4(d) 1 1  1  (1, −1, 3), (4, 7,1), (−2,1,1) 1. (a)  66 6  11   1 1 1   1 1   1 −2 1   (b)  , , , 0, , , ,−  , 2 2   6 6 6   3 3 3    1 2 1  6  1 −4 7  11  12 −4 −4   , , (c)   , , ,  , ,  ,  6 6 6  11  6 6 6  176  11 11 11   2.

 1 3 45  2 1   , ⋅ t,   t −  8  3    2 2

MODEL QUESTION PAPER - I Time: 3 Hrs.

Max. Marks : 100

Answer all Questions PART-A (10 × 2 = 20) 1. Form a partial differential equation by eliminating the arbitrary function f from z = (x + y) f (x2 – y2). 2. Find the complete integral of q = 2px. 3. Find the half range sine series for f(x) = 2 in 0 < x < 4. 4. If the cosine series for f(x) = x sin x for 0 < x < p is given by x sine x = 1 –



x 1 (−1)n cos x − 2 ∑ 2 cos nx, show that 2 n=2 n − 1

1 1 p  1  1 + 2 − + − .......... = 2 1⋅ 3 3 ⋅ 5 5 ⋅ 7 

5. Classify the partial differential equation. (1 – x2)Zxx – 2xyzxy + (1 – y2)zyy + xzx + 3x2yzy – 2z = 0. 6. The steady state temperature distribution is considered in a square plate with sides x = 0, y = 0, x = a, and y = a. The edge y = 0 is kept at a constant temperature T and the other three edges are insulated. The same state is continued subsequently. Express the problem mathematically. 7. Define the sampler. 8. Find Z-transform of {n}. 9. If Fourier transform of f(x) is F(s), prove the Fourier transform of f(x) cos ax is

1 [F(s 2

– a) + F(s + a)].

1, 0 < x < 1 10. Find the Fourier cosine integral representation of f(x) =  x >1 0 PART-B (5 × 16 = 80) Question No. 11 has no choice; Questions 12 to 15 have one choice (either or type) each. 11. (i) Expand in Fourier series of periodicity if 0 < x < p x 2n of f(x) =  (8) 2 p − x if p < x < 2 p (ii) Find the half-range cosine series for the functions f(x) = x, 0 < x < p and hence de∞ 1 duce the sum of the series ∑ (8) 4 n = 0 (2n − 1) 376

377

Model Question Papers

12. (a) (i) Find the complete solution and singular solution of z = px + qy + p2 – q2. (ii) Find the general solution of x(z2 – y2) p + y (x2 – z2)q = z(y2 – x2)

(8)

OR (b) (i) Solve : (D2 – 4D′2)z = cos 2x coa 3y. (8) (ii) Solve : (D + D′ – 1) (D + 2D′ – 3)z = 4 + 3x + 6y + 3x + y.

(8)

13. (a) A taut string of length L is fastened at both ends. The midpoint of the string is taken to a height of b and then released from rest in this position. Find the displacement of the string at any time t. (16) OR (b) A rod 30 cm long, has its ends A and at 20°C and 80°C respectively, until steady state conditions prevail. The temperature at the end B is then suddenly reduced to 60°C and the end A is raised to 40°C and maintained so. Find the resulting temperature u(x, t). (16) 14. (a) (i) Find z-transform of t and eat. (8) (ii) Find inverse z-transform of

4 z2 − 2 z 23 − 5 z2 + 8 z − 4

by Residue method.

(8)

OR (b) (i) Find z-transform of {a }. (8) n

(ii) Solve : yn+2 – 5yn+1 + 6yn = 36

(8)

given y0 = y1 = 0



1−| x| for| x|≤ 1 15. (a) (i) Find the Fourier transform of f(x) =  for| x|> 1 0 (ii)

(iii)

Hence evaluate the following integral: ∞

2

∞

4

 sin x  ∫  x  dx 0  sin x  ∫  x  dx 0

(5)

OR

(b) (i) Find the Fourier sine and cosine transform of e–2x. Hence find the value of the following integrals : ∞

(ii)

(6)

dx

∫ (x 2 + 4)2 0

∞

(iii) ∫ 0

x 2 dx ( x 2 + 4)2



(5)

378

Engineering Mathematics-III (Third Semester)

MODEL QUESTION PAPER - II Time: 3 Hrs

Max Marks : 100 Answer all Questions PART-A (10 × 2 = 20 Marks)

1. Find the partial differential equation of the family of spheres having their centres on the line x = y = z. 2. Solve

∂3 z ∂x

2

−2

∂2 z 2

∂x ∂y

2

−4

∂2 z ∂x∂y

+8

2

∂3 z ∂y 3

= 0,

3. Find the constant term in the Fourier series corresponding to f(x) = cos2x expressed in the interval (–p, p). 4. To which value, the half range sine series corresponding to f(x) = x2 expressed in the interval (0, 2) converges at x = 2? 5. Classify the following partial differential equations: (a)

∂2u ∂x

2

=

∂2u

∂ 2 u  ∂u   ∂u  =     + xy (b) ∂xdy  ∂x   ∂y  ∂y 2

6. Write any two solutions of the Laplace equation uxx + uyy = 0 involving exponential terms in x or y. 7. State initial value theorem on z-transform. 8. Find z-transform of eat+b. 9. Write the Fourier transform pair. 10. If Fc(s) is the Fourier cosine transform of f(x) prove that the Fourier cosine transform of 1 s f(ax), is Fc   . a a PART-B (5 × 16 = 80 Marks) 11. (i) Find the Fourier series expansion of the periodic function f(x) of period 2l defined by f(x) = l + x, – l ≤ x ≤ 0 = 1 – x, 0 ≤ x ≤ 1 Deduce that

∞

1

∑ (2n − 1)2 1

=

p2 8

(ii) Find the half range sine series of f(x) = x cos x in (0, p). 12. (a) (i) Form the differential equation by eliminating the arbitrary functions f and g in z = f(x3 + 2y). (ii) Solve : (x + y) p + (yz – x)q = (x + y) (x – y). OR (b) (i) Find the singular solutions of z = px + qy +

p 2 + q 2 + 16

(ii) Solve: (D2 – DD′ – 30 D′2)z = xy + e6x + y.

13. (a) A tightly stretched string of length l has its ends fastened at x = 0 and x = 1. The mid point of the string is then taken to a height h and then released from rest in that position. Obtain an expression for the displacement of the string at any subsequent time.

379

Model Question Papers

OR (b) The boundary value problem governing the steady-state temperature distribution in a flat, thin, square plate is given by

∂2u ∂x 2

+

∂2u ∂y 2

= 0, 0 < x < a, 0 < y < a.

 px  u(x, 0) = 0, u(x, a) = 4 sin3   , 0 < x < a.  a 



Find the steady-state temperature distribution in the plate. 14. (a) (i) Find the inverse z-transform of

z2 1  1   z −  z −  2  4 

(ii) Solve x(n + 1) – 2x(n) = 1 if x(0) = 0 OR np   (b) (i) Find the inverse z-transform of cos  2   (ii) Find the inverse z-transform of

z2 + z ( z − 1)( z 2 + 1)

.

15. (a) (i) Find the Fourier transform of f(x) = 1 for | x | < 1 = 0 otherwise. ∞

Hence prove that

sin x ∫ x dx = 0

∞

∫

sin 2 x x

0

2

dx =

p 2

(ii) Find the Fourier sine transform f(x) = sin x, 0 < x ≤ p = 0, p ≤ x < ∞



OR (b) (i) Find the Fourier cosine transform of e–4x. Dedute that ∞

cos 2 x p −8 ∫ x 2 + 16 dx = 8 e and 0

∞

x sin 2 x

p

∫ x 2 + 16 dx = 2 e

−8

0

(ii) State and prove convolution theorem for Fourier transforms.

380

Engineering Mathematics-III (Third Semester)

MODEL QUESTION PAPER - III Time: 3 Hrs

Max Marks : 100

Answer all Questions PART-A (10 × 2 = 20 Marks) 1. Find the complete integral of p + q = pq where p = 2. Solve (D3 – 3DD′2 + 2D′)z = 0

∂z ∂z and q = . ∂x ∂y

3. If f(x) = x2 + x is expressed as a Fourier series in the interval (–2, 2) to which value this series convergesm at x = 2? 4. If the fourier series corresponding to f(x) = x in the interval (0, 2p) is a0 ∞ + ∑ {an cos nx + bbn sin nx}, without finding the values of a0an, bn find the value of 2 1 a02 ∞ 2 + ∑ an + bn2 . 2 1

(

)

5. Classify the following second order partial differential equations: (a) 4 (b)

∂2u ∂x 2

∂2u ∂x 2

+

+4

∂2u ∂2u ∂u ∂u + 2 − 6 − 8 − 16u = 0 ∂x∂y ∂y ∂x ∂y

2 ∂ 2 u  ∂u   ∂u  =   +  ∂y 2  ∂x   ∂y 

2

6. Write any two solutions of the Laplace equation obtained by the method of separation of variables. 7. Find z-transform of {(– 1)n}. 8. State Final value theorem on z-Transform. 9. If F(s) is the Fourier transform of f(x), write the formula for the Fourier transform of f(x) cos (ax) in terms in F. 10. State the convolution of f(x) cos (ax) in terms in F. PART-B (5 × 16 = 80 Marks) 11. (i) Find inverse of z-transform of

z2 ( z − a )2

using convolution theorem.

(ii) Using z transform

Solve : y(n) – ay(n – 1) = u(n)

12. (a) (i) Solve : z = 1 + p2 + q2. (ii) Solve : (y – z)p – (2x + y)q = 2x + z. OR

(b) (i) Form the partial differential equation by eliminating the arbitrary functions f and g in z = x2f(y) + y2g(x).

13. (a) (i) Obtain the Fourier series for f(x) = 1 + x + x 2 in (–p, p). Deduce that 1 1 1 p2 ...... + + + ∞ = . 6 12 2 2 3 2

381

Model Question Papers

(ii) Obtain the constant term and the first harmonic in the Fourier series expansion for f(x) where f(x) is given in the following table : x:

0

1

2

3

4

5

6

7

8

9

f(x)

18.0

18.7

17.6

15.0

11.6

8.3

6.0

5.3

6.4

9.0

10

11

12.4 15.7

OR

(b) (i) Expand the function f(x) = x sin x as a Fourier series in the interval – p ≤ x ≤ p.

(ii) Obtain the half range cosine series for f(x) = (x – 2)2 in the interval 0 < x = 2. Deduce ∞ 1 p2 = that ∑ . 2 8 1 (2n − 1) 14. (a) A tightly stretched flexible string has its ends fixed at x = 0 and x = l. At time t = 0, the string is given a shape defined by f(x) = kx2 (l – x), where k is a constant, and then released from rest. Find the displacement of any point x of the string at any time t > 0. OR (b) The ends A and B of a rod l cm long have the temperature 40°C and 90°C until steady state prevails. The temperature at A is suddenly raised to 90°C and the same time that at B is lowered to 40°C. Find the temperature distribution in the rod at time t. Also show that the temperature at the mid point of the rod remains unaltered for all time, regardless of the material of the rod. 15. (a) (i) Find the Fourier transform of e–a|x| if a > 0. Dedude that

∞ 0



(ii) Find the Fourier sine transform of xe–x–2/2. OR

(b) (i) Find the Fourier cosine transform of f(x) = 1 – x2, 0 < x < 1

= 0, otherwise. ∞

Hence prove that

∫ 0

sin x − x cos x

3

3p x cos   dx = . 16 2

(ii) Derive the Parseval’s identity for Fourier transforms.

1

p

∫ (x 2 + a2 ) dx = 4 a3

if a > 0.

382

Engineering Mathematics-III (Third Semester)

MODEL QUESTION PAPER - IV Time: Three hours

Max Marks : 100

Answer all Questions PART-A (10 × 2 = 20 Marks) 1. Obtain partial differential equation by eliminating arbitrary constants a and b from (x – a)2 + (y – b)2 + z2 = 1. ∂2 z ∂2 z ∂2 z 2. Find the general solution of 4 2 − 12 + 9 2 = 0. ∂x∂y ∂x ∂y 3. State Dirichlet’s conditions for a given function to expand in Fourier series. 4. If the Fourier series of the function f(x) = x + x2 in the interval – p < x p is p2 ∞ 2  4  + ∑ (−1)n  2 cos nx − sin nx  , then find the value of the infinite series 3 n =1 n n  1 2

1

+

1 2

2

+

1 32

+ ....

5. Classify the following partial differential equations (a) y2uxx – 2xyuxy + x2uyy + 2ux – 3u = 0 (b) y2uxx + uyy + ux2 + uy2 + 7 = 0 6. An insulated rod of length 60 cm has its end at A and B maintained at 20°C and 80°C respectively. Find the steady state solution of rod. 7. State convolution theorem on Fourier Transform. 8. Find the z-transform of u(n). 1 9. Find the z-transform of   . n  10. If Fs(s) is the Fourier sine transform of f(x), show that Fs[f(x) cos ax] =

1 [Fs(s + a) + 2

Fs(s – a)]. PART-B (5 × 16 = 80 Marks) 11. Obtain Fourier series for f(x) of period 2l and defined as follows f(x) = l – x in 0 < 0 ≤ t

= 0 in l ≤ x ≤ 2l

Hence deduce that 1 1 1 p 1 − + − + ..... = 3 5 7 4 1 1 1 p2 ..... + + + = . (16) 8 12 3 2 5 2 12. (a) (i) Find the singular integral of the partial differential equation z = px + qy + p2 – q2. (6) (ii) Solve: (D2 + 4DD′ – 5D′2)z = 3e2x – y + sin(x – 2y).

... (10)

383

Model Question Papers

OR

(b) (i) Find the general solution of



(3z – 4y)p + (4x – 2z)q = 2y – 3x

(ii) Solve : (D2 – 2DD′ + D′2 – 3D + 3D′ + 2)z = (e3x + 2e–2y)2

(10)

13. (a) A tightly stretched string with fixed end points x = 0 and x = l is initially at rest in its equilibrium position. If it is set vibtrating by giving each point a velocity kx(l – x). Find the displacement of the string at any time. (16) OR 13. (b) A rectangular plate with insulated surface is 10 cm wide so long compared to its width that it may be considered infinite length. If the temperature along short edge, are kept at 0°C, find the steady state temperature function u(x, y). (16) 14. (a) Find the Fourier transform of f(x) given by f(x) = 1 for | x | < 2





= 0 for | x | > 2 ∞



sin x dx and and hence evaluate ∫ x 0

2

∞

 sin x  ∫  x  dx . (16) 0

OR (b) Find Fourier sine and cosine transform of e–x and hence find the Fourier sine transform x 1 of and Fourier cosine transform of . (16) 2 1+ x 1 + x2 15. (a) (i) Find the inverse z-transform of e–x and hence find the Fourier sine transform of 1 x and Fourier cosine transform of . 1 + x2 1 + x2 OR (b) (i) Find the inverse transform of

z 2

z + 7 z + 10

.

(ii) Solve : yn + 2 + 6yn + 1 + 9yn = 2n if y0 = y1 = 0

(16)

384

Engineering Mathematics-III (Third Semester)

MODEL QUESTION PAPER - V Time: Three hours

Max Marks : 100

Answer all Questions PART-A (10 × 2 = 20 Marks)  xy  1. Eliminate the arbitrary function f from z = f   and form the partial differential equation.  z  2. Find the complete integral of p + q = pq. 3. Find a Fourier sine series for the function f(x) = 1; 0 < x < p. 4. If the Fourier series for the function f(x) = 0; 0 < x < p

= sin x; p < x < 2p



is f(x) = −



deduce that

1 2  cos 2 x cos 4 x cos 6 x  1 +  + + + ... + sin x 3.5 5.7 p p  1.3  2

1 1 1 p−2 − + − ...∞ = . 1.3 3.5 5.7 4

5. Classify the partial different equation uxx + xuyy = 0. 6. A rod 30 cm long has its ends A and B kept at 20°C and 80°C respectively until steady state conditions prevail. Find the steady state temperature in the rod. 7. Define the sampler. 8. Define unit step sequence and unit sample sequence. 9. State change of scale property on Fourier Transform. 10. Find the Fourier transform of f(x) if

1;| x|< a f(x) =  0;| x|> a > 0

11. (i) Solve x(y – z)p + y(z – x)q = z(x – y) (ii) Solve (D3 – 7DD′2 – 6D′3)z = sin(x + 2y) + e2x + y. 12. (a) (i) Determine the Fourier expansion of f(x) = x in the interval – p < x < p.

(ii) Find the half range cosine series for sin x in (0, p). OR

(b) (i) Obtain the Fourier series for the function

f(x) = px; 0 ≤ x ≤ 1 = p(2 – x); 1 ≤ x ≤ 2.

(ii) Find the Fourier series of period 2p for the function

1in (0, p) f(x) =  2 in (p, 2 p)



and hence find the sum of the series

1 2

1

+

1 3

2

+

1 52

+ ...∞.

385

Model Question Papers

13. (a) A string is stretched between two fixed points at a distance 2l apart and the points of the string are given initial velocities v where

v=



=

cx in 0 < x < 1 l c (2l − x ) in l < x < 2l l

x being the distance from one end point. Find the displacement of the string at any subsequent time. OR (b) Find the solution of the one dimensional diffusion equation satisfying the boundary conditions:  ∂u  (ii)   = 0 for all t  ∂x  x =0

(i) u is bounded as t

 ∂u  (iii)   = 0 for all t  ∂x  x = a

(iv) u(x, 0) = x(a – x), 0 < x < a.

14. (a) (i) Find the z-transform of te–3t. (ii) Using convolution theorem, find inverse of z-transform of

z2 ( z − a)( z − b)

OR

(b) (i) Solve: yn+2 – 4yn+1 + 4y = p if y0 = y1 = 0

(ii) Find the z-Transform of nanu(n). 15. (a) (i) Show that the Fourier transform of  a 2 − x 2 ;| x|< a f(x) =  | x|> a > 0 0;

is 2

2  sin la − l a cos la    p l3  ∞

Hence deduce

∫

sin t − cos t

0

t

3

dt =

p . 4

(ii) Find the Fourier sine and cosine transform of 0< x2  OR (b) (i) If f (l) is the Fourier transform of f(x), find the Fourier transform of f(x – a) and f(ax). (ii) Verify Parseval’s theorem of Fourier transform for the function

0; x < 0 f(x) =  − x e ; x > 0.

386

Engineering Mathematics-III (Third Semester)

SOLVED QUESTION PAPERS B.E./B.Tech. Degreee Examination, April/May 2011 Third Semester Civil Engineering MA 231-Mathematics-III (Common to all Other Branches) (Regulation 2001) Time: Three hours

Maximum Marks : 100

Answer all Questions PART-A (10 × 2 = 20 Marks) 1. Find the complete integral of pq + p + q = 0. 2. Form the partial differential equation by eliminating the arbitrary function form z = f(x2 + y2 + z2). 3. State Dirichlet conditions for the existence of the Fourier series of f(x). 4. State the Parsevel’s theorem for the Fourier series. 5. Prove that F[f(x – a)] = eias F(s). 6. Find the Fourier sine transform of f(x) = e–x, x > 0. 7. Classify the partial differential equation fxx + 2fxy + 4fyy = 0, x > 0, y > 0. 8. Write down all possible solutions of one dimensional wave equation. 9. Find the Laplace transform of e–t sin t cos t. 10. Find the inverse Laplace transform of

s 2

s + 2s + 3

.

PART-B (5 × 16 = 80 Marks) 11. (a) (i) Find the singular integral of z = px + qy + p2 + pq + q2

(ii) Solve (D2 – DD′ – 30D′2)z = xy + e6x + y. OR

(b) (i) Find the general solution of

x(z2 – y2) p + y(x2 – z2) q = z(y2 – x2).

(ii) Find the general solution of p2 + q2 = x2 + y2. 12. (a) (i) Find the Fourier series of f(x) = (p – x)2 in (0, 2p). (ii) Find the half range cosine series of f(x) = x(l – x), 0 < x < l. OR (b) (i) Find the Fourier series of f(x) = x 2 , –p < x < p. Hence show that p4 1 1 1 1 .. + + + + = 90 14 2 4 3 4 4 4 (ii) Find the Fourier series of f(x) as far as the second harmonics from the given data : x

0

p/6

p/3

p/2

2p/3

5p/6

p

f(x)

2.34

2.20

1.60

0.83

0.51

0.88

1.19

387

Model Question Papers

1 − x 2 , if | x|≤ 1 13. (a) Find the Fourier transform of f(x) =  . Hence evaluate if | x|> 1 0, ∞

x sin x − x cos x cos dx and ∫ 3 2 x 0

2

∞

 sin t − t cos t   dt . ∫  t3  0 OR

(b) Find the Fourier sine transforms of f(x) = e–ax, x > 0, a > 0 and g(x) = e–bx, x > 0, ∞ x 2 dx b > 0. Hence evaluate ∫ 2 . 2 2 2 0 ( x + a )( x + b ) 14. (a) A tightly stretched string of length l has its ends fastened at x = 0 and x – l. At time t = 0 the string was released from rest with the displacement f(x) = kx(l – x). Find the displacement at a distance x from one end at any time t. OR 14. (b) A square plate is bounded by the lines x = 0, y = 0, x = 10, y = 10. Its faces are insulated. The temperature along the upper horizontal edge is given by u(x, 10) = x(10 – x), 0 < x < 10, while the other edges are kept at 0°C. Find the temperature distribution in the plate. T   E, 0 ≤ 1 15. (a) (i) Find the Laplace transform of f(t) =  and f(t + T) = f(t). 2 −E , T / 2 ≤ t ≤ T (ii) Using Laplace transform, solve y′′ = 3y′ + 2y = e3x, where y(0) = 0, y′ (0) = 0. OR (b) (i) Find the inverse Laplace transform of

s2 − s + 2 s( s + 2)( s − 3)

(ii) Using convolution theorem, find the inverse Laplace theorem of

s2 ( s 2 + a 2 )2

.

388

Engineering Mathematics-III (Third Semester)

Answers for MA 231 – Mathematics - III Third Semester – May 2011 PART-A (10 × 2 = 20 Marks) 1. This is of the form F(p, q) = 0. Hence complete integral is z = ax + by + c where ab + a + b = 0 \b=

−a 1+ a

Hence, z = ax –

a y+c 1+ a

2. z = f(x2 + y2 + z2) ∂z = p = f ′(x2 + y2 + z2) × (2x + 2zp) ∂x ∂z = q = f ′ (x2 + y2 + z2) × (2y + 2zq) ∂y Dividing,

p x + pz = q y + qz

py + pqz = qx + pqz

\ py – qx = 0 is the required answer.



3. Dirichlet’s conditions (i) f(x) is single valued and finite in (c, c + 2p) (ii) f(x) is continous or piece-wise continuous with finite number of finite discontinuities in (c, c + 2p) (iii) f(x) has a finite number of maxima or minima in (c + c + 2p)



4. Parseval’s theorem

If f(x) =

a0 ∞ + ∑ ( an cos nx + bn sin nx ) 2 n =1

 a2  1 Then (2 p)  0 + ∑ ( an2 + bn2 ) = 2  4  5. F[f(x – a)] =

1

∞

2p 1

∫−∞ f (x − a) e ∞

∫ f (t ) e 2 p −∞



=



= e ias ⋅

1

isx

is( a + b )

∞

∫ f (t ) e 2 p −∞

c+2p

∫

[ f ( x )]2 dx .

c

dx

dt , where t = x – a

ist

dt = eias F[f(x)] = eiasF(s)

6. Fs(e–x) =

2 p



 2  e−x 2 s (− sin sx − s cos sx ) = ⋅  2 p 1 + s p 1 + s2 0

∞ −x

∫0

e

sin sx dx ∞

=

389

Model Question Papers

7. fxx + 2fxy + 4fyy = 0 A = 1, B = 2, C = 4





B2 – 4AC = 4 – 16 = – 12 = – ve

Hence, it is elliptic at every point of the domain. 2 ∂2 y 2 ∂ y 8. = wave equation. a ∂t 2 ∂x 2 The solutions are



y(x, t) = (Aelx + Be–lx) (Celat +De–lat )

Or

y(x, t) = (A cos lx + B sin lx) (C cos lat + D sin lat)

Or

y(x, t) = (Ax + B) (Ct + D)

9. L[e–t sin t cos t] =

=

1 1 2  L  e −1 sin 2t  =  2 2  2  s + 4  s→ s +1 1 ( s + 1)2 + 4

s ( s + 1) − 1   L−1  2 10. = L−1  ( s + 1)2 + ( 2)2  s + 2s + 3  s−1 = e −1 L−1 2 s + ( 2)2 1   sin 2t  = e − t cos 2t − 2  



PART-B (5 × 16 = 80 Marks) 11. (a) (i) z = px + qy + p + pq + q2 2

This is of the form z = px + qy + f(p, q) Hence complete integral is z = ax + by + a2 + ab + b2

... (1)

where a, b are arbitrary constants. To find, singular integral, differentiate (1) w.r.t. a and b partially.

0 = x + 2a + b



0 = y + 2b = a

i.e., 2a + b = – x a + 2b = – y

\





3a = y – 2x a=

y − 2x 3

x 2y 2 b = – x – 2a = − x − ( y − 2 x ) = − 3 3 3

Substituting (1)

2

2

( x − 2 y) ( y − 2 x)( x − 2 y)  y − 2x   x − 2 y  ( y − 2 x) z=  + + x + 3 + 3 9 9 9    

390

Engineering Mathematics-III (Third Semester)

9z = 3xy – 3x2 – 3y2

\ 3z = xy – x2 – y2 is the singular solution.

(ii) (D2 – DD′ – 30D′2)z = xy + e6x–y (printing mistake in question paper) Auxiliary equation is

m2 – m – 30 = 0



(m – 6) (m + 5) = 0



\

m = 6, – 5

C.F. is f1(y + 6x) + f2(y – 5x)

... (1) xy

PI1 =

2

D − DD′ − 30D

′2

xy

1

=

2

D  D′ 30D′2  1− −  D D2   −1



=

1   D′ D′2 1 − + 30   D 2   D D2



=

1  D′ 30D′2  D′ 30D′2 1+ + +  + D D2  D2 D2 D 



=

1  D′  1 + + ... ( xy ) D D 2  



D′   1 =  2 + 3  ( xy ) D  D



=



e6 x+ y D 2 − DD′ − 30D′2

=

e6 x+ y (D − 6D′)(D + 5D′)

1 e6 x+ y 1 = ⋅ xe 6 x + y 11 D − 6D′ 11

Hence general solution is

2    + ... ( xy)   

x3 y x4 + 6 24

PI2 = =

   ( xy )  

z = f1(y + 6x) + f2(y – 5x) +

11. (b) (i) x(z2 – y2) p + y (x2 – z2)q = z(y2 – x2)

x3 y x4 1 6x+y + + xe 6 24 11 (Text P. 136)

The subsidiary equations are dy dz dx = = 2 2 2 x( z − y ) y( x − z 2 ) z( y 2 − x 2 ) Taking the two sets of multipliers x, y, z and 1 1 1 dx + dy + dz xdx + ydx + zdz x y z = 2 2 2 2 2 Σ( z − y ) Σx ( z − y )

... (1) 1 1 1 , , each ratio of (1) equals x y z

391

Model Question Papers

This implies xdx + ydy + zdz = 0 and

dx dy dz + + =0 x y z

Integrating, x2 + y2 + z2 = a and log x + log y + log z = b i.e., xyz = k Hence, the general solution is f(x2 + y2 + z2, xyz) = 0. (ii) p2 + q2 = x2 + y2 (Text P. 125) p2 – x2 = y2 – q2 = a2 (say)

p=

\

a2 + x 2 , q =

y 2 − a2

dz = pdx + qdy



=

a 2 + x 2 dx + y 2 − a 2 dy

Integrating y 2 y a2 x x 2 a2 a + x2 + y − a 2 − cosh −1 + b sinh −1 + 2 2 2 a 2 a This is the complete solution and not general solution. (Question is wrongly worded) z=



12. (a) (i) let f(x) = (p – x)2 = 1 p

where a0 =

2p

a0 ∞ + ∑ ( an cos nx + bn sin nx ) 2 n =1

(Text P.28)

∫ (p − x) dx 2

0

2p



1  ( p − x )3  =   p  (−3)  0



= −

1 2x (p − x )2 cos n x dx p ∫0

an =



1 2 p2  −p3 − p3  =  3p  3

2p



=

1  cos nx   sin nx    sin nx  ( p − x )2   + 2(p − x )  −  + 2  p  n2   n    n3   0



=

1  2p 2p  4 + = p  n2 n2  n2

bn =



1 p

2p

∫ (p − x)

2

sin nx dx

0

2p



=

1  sin nx   cos nx    cos nx  ( p − x )2  −  + 2(p − x )  −  + 2   2 n  p n     n3   0



=

 1  p2 p2 2 + + 3 (1 − 1) = 0 − n n p n 



392



Engineering Mathematics-III (Third Semester) ∞ p2 1 + 4 ∑ 2 cos nx 3 n =1 n

Hence, f(x) = (p – x)2 =



(ii) f(x) = (lx – x2), 0 < x < l. Let f(x) =

a0 ∞ npx + ∑ an cos 2 n =1 e 1

2 1 2  lx 2 x 3  a0 = ∫ (lx − x 2 ) dx =  −  l 0 l 2 3 0



=



an =



2  l3  l2  = l 6 3 2 1 npx (1x − x 2 )cos dx ∫ 0 l l

   npx  npx  npx    sin   cos   sin    2 l − (l − 2 x )  − l  + (−2)  − l  (lx − x 2 )  =    n 3 p3   l n 2 p2   np       l   l2 l3      2 l l  −2l  − 2 2 cos np − 2 2  = 2 2 l n p n p  n p [1 + (−1)n ] 3

3



=



= 0 if n is odd



=

\ f(x) =

−4l 2 n 2 p2

2

if n is even

l 2 4l 2 − 6 p2

∞

∑

1

n − 2,4,6.... n

2

cos

npx l

12. (b) (i) f(x) = x , –p < x < p 2

f(x) is even in (–p, p)



Hence bn = 0 in its Fourier series f(x) =

a0 ∞ + ∑ an cos nx 2 1



a0 =

2 p 2 2 p2 x dx = ∫ p 0 3



an =

2 p 2 x cos nx dx p ∫0



\

p



=

2  2  sin nx   cos nx   sin nx   (x )   − (2 x )  −  + (2)  −  p  n2  n3   0  n   



=

2  2p  4 (−1)n  = 2 (−1)n p  n2  n

l

0

393

Model Question Papers



\

f(x) =

Here a0 =

∞ ∞ p2 (−1)n + 4 ∑ 4 ∑ 2 cos nx 3 =n 1=n 1 n

2 p2 4(−1)n , an = 3 n2

Using Parsival’s theorem  a2 1  (range)  0 + ∑ an2  =  4 2 

p

∫−p ( f (x))

2

p

dx = ∫ x 4 dx −p

 p4 1 16  p5 2p  + Σ 4  = 2 5  9 2 n  p4 1 p4 + 8Σ 4 = 9 5 n



p4 p4 4 − = p4 4 5 9 45 n ∞ 1 p4 ∑ 4 = 90 . n =1 n



8Σ

\

1

=

(ii) We can omit the last value of f(x), since f(x) is given at x = 0 Let q = 2x; As x varies from 0 to p, q varies from 0 to 2p with an increase of Let f(x) = F(q) =

q

y = F(q)

0

2.34

1

p 3

2.20

2p 3 p

2p p = . 6 3

a0 + a1 cos q + a2 cos 2q + b1 sin q + b2 sin 2q 2

sin q

sin 2 q

y cos q

y cos 2 q

y sin q

y sin 2 q

1

0

0

2.34

2.34

0

0

0.5

–0.5

0.866

0.866

1.10

–1.1

1.91

1.91

1.60

–0.5

– 0.5

0.866

–0.866

–0.80

– 0.8

1.39

–1.39

0.83

–1

1

0

0

–0.83

0.83

0

0

4p 3

0.51

–0.5

–0.5

–0.866

0.866

–0.25

–0.25

–0.43

0.43

5p 3 Σ

0.88

0.5

– 0.5

– 0.866

–0.866

0.44

–0.44

–0.76

–0.76

2.00

0.58

2.11

0.19

8.36

cos q cos 2q

394

y =

Engineering Mathematics-III (Third Semester)

a0 + a1 cos q + a2 cos 2q + b1 sin q + b2 sin 2 q. 2

2 8 ⋅ 36 a0 = Σf ( x ) = = 2.79 6 3

b1 =

2 (2, 11) = 0.70 6

2 2 a1 = (2.0) = 0.67 b2 = (0.19) = 0.06 6 6 2 a2 = (0 ⋅ 58) = 0.19 6

\ f(x) = F(q) = 1.39 + 0.67 cos q + 0.19 cos 2q + 0.7 sin q + 0.06 sin 2q where q = 2x. 1

∞

1

1

∫ e 2 p −∞

13. (a) F{f(x)} =

∫ e 2 p −1

isx

isx

f ( x ) dx

(1 − x 2 ) dx



=



=

isx 1  2 e (1 − x )  2 p   is



=

1  2 is − is −2i is − is  − ( e + e ) 3 ( e − e ) 2 p  s2 s 



= −



= −

2 s

3

1

⋅

2p

  e isx  − (−2 x )  2 2  i s

  e isx  + (−2)  3 3  i s

1

     −1

[2s cos s – 2 sin s]

4  4 cos s − sin s   2 p  s3

... (1)

Using inversion formula

f(x) =

2p ∫

1 −4  s cos− sin s  isx  e ds 2p  s3 

∞

−∞



=

−2 ∞ s cos s − sin s − isx e ds p ∫−∞ s3 ∞

∫−∞

s cos s − sin x s

3

(cos sx − i sin sx )ds = −

p (1 – x2) if | x | < 1 2

= 0 if | x | > 1

... (2)

Put x = 1/2 in (2) and equate real parts,

s p 1  −3 p cos ds = −  1 −  = 2 2 4 8 s3 ∞ 5 cos s − sin s s 3p 2∫ cos ds = − 3 0 2 8 s ∞

∫−∞

∞

∫0

s cos s − sin s

s cos s − sin s s

3

s 3p cos ds = − 2 16

(Q Integrand is even)

395

Model Question Papers ∞

∫0



sin s − s cos s s

3

s 3p cos ds = − 2 16

Put x = 0 in (2), we get ∞



∫−∞

\

∫0

∞

s cos s − sin s s3 sin s − s cos s s3

ds = −

p 2

p (change s to t) 4

ds =

13. (b) Fs(e–ax) =

2 ∞ − ax e sin sx dx p ∫0



=

 2  e − ax  2 2 (− a sin sx − s cos sx ) p a + s 0



=

2 s , a>0 ⋅ p a 2 + s2

Fs(e–bx) =

2 s ⋅ 2 2 ,b>0 p b +s

∞



Using

f ( x ) g( x )dx =

∞ − ax e 0

∫



∞

∫0

⋅ e − bx dx =

∞ −( a + b ) x e dx 0

∫

\

=

∞

∫0

Fs ( s)Gs ( s) ds

2 ∞ s2 ds p ∫0 ( a 2 + s2 )(b 2 + s2 ) 2 ∞ x2 dx p ∫0 ( a 2 + x 2 )(b 2 + x 2 )

∞

 e( a + b ) x  2 ∞ x2 dx   = ∫0 2 p ( a + x 2 )(b 2 + x 2 )  −( a + b)  0

\

=

2 ∞ x2 1 dx = p ∫0 ( a 2 + x 2 )(b 2 + x 2 ) a+b x2

∞

∫0

2

2

2

2

( a + x )(b + x )

dx =

p 2( a + b)

14. (a) Refer Text P 194.

Also Refer to Question 14 (a) under MA 2211 question paper.

(b) Boundary conditions are

y

u (0, y) = 0; 0 ≤ y ≤10       ... (i)

y = 10 P

u (x, 0) = 0; 0 ≤ x ≤10        ... (ii)

x = 10

u (10, y) = 0; 0 ≤ y ≤10       ... (iii) u (x, 10) = 10x – x2; 0 ≤ x ≤ 10   

... (iv)

Q

O

R

x

Take u(x, y) = (A cos lx + B sin lx) (Cely + De–ly) as solution of Laplace equation (Steady state)

396

Engineering Mathematics-III (Third Semester)

Using u(0, y) = 0, we get A = 0 Using u(10, y) = 0, we get B sin 10l (variable) = 0 np , n = 1, 2, 3,..... 10

B ≠ 0, l =

\

Using u(x, 0) = 0, we get

B sin lx (C + D) = 0 D=–C

\ Hence I becomes,

u(x, y) = B sin



npx × c(e ly − e −ly ) 10



= Bn sin

npx sinh ly 10



= Bn sin

npy npx sinh 10 10

Most general solution is ∞

∑ Bn sin

u(x, y) =



n =1

npy npx sinh 10 10

... (II)

Using boundary condition (iv) in II, u(x, 10) =



∞

∑ Bn sin 1

\



Bn sinh= np = bn

2 10 npx (10 x − x 2 )sin dx 10 ∫0 10

   npx  npx  npx    − cos − sin cos        1 10 − (10 − 2 x )  10  + (−2)  10   = (10 x − x 2 )   np  n 3 p3   5  n 2 p2         10    100   10 3    1  −2000  400 (−1)n = [1 − (−1)n ] −1  3 3 5  n3 p3  n p

(



=



= 0 for n even

)

800

for n odd n 3 p3 0 for n even  800 Hence, Bn =  for n odd, put in II  n3 p3 sin hnp 



npx ⋅ sinh np = 10 x – x2. 10

=

u(x, t) =

∞

∑

n =1,3,5,... n

800 3 3

p sin hnp

⋅ sin

npy npx sin h 10 10

10

0

397

Model Question Papers

15. (a) (i) Since f(t) is periodic of period T,

1

(Text P. 781)

T − st e 0

f (t) dt ∫ 1 − e − sT T E  T/2 − st e dt − ∫ e − st dt  = ∫ − sT   T 0 /2  1− e

L[f(t)] =

E

 − st  e  − s  

T T   2  e − st    −    0  − s  T  2



=



=



sT  − 2  e = 1 − s(1 − e − sT ) 

1 − e − sT

− sT − sT  − sT  2 − e 2 + 1 e − e  − sT s(1 − e )  

E

E

   

2

(ii) y′′ – 3y′ + 2y = e3x, y(0) = 0, y′(0) = 0

(Text P. 740)

Taking Laplace Transform,

L(y′′) – 3L(y′) + 2L(y) = L(e3x)



s2 y – sy(0) – y′(0) – 3[s y – y(0) + 2 y =

1 s−3

Using y(0) = 0, y′(0) = 0 we get

y (s2 – 3s + 2) =



y =



1 s−3

1 A B C = + + ( s − 1)( s − 2)( s − 3) s − 1 s − 2 s − 3

1 1 1 = 2 − + 2 s−1 s− 2 s− 3

\

y = L–1 (RHS)



y=

1 x 2x 1 3x e −e + e 2 2

 s2 − s + 2  15. (b) (i) L−1   = ? (Text P. 725)  s( s + 2)( s − 3)  Let

s2 − s + 2 A B C = + + s( s + 2)( s − 3) s s+2 s−3



 s2 − s + 2  2 −1 A =   = =  ( s + 2)( s − 3)  x =0 −6 3



 s2 − s + 2  8 4 B =   = =  s( s − 3)  s = −2 10 5

398

Engineering Mathematics-III (Third Semester)







\

 s2 − s + 2  8 C =   =  s( s + 2)  s = 3 15 L−1

(ii) L−1

s2 − s + 2  −1/ 3 4/ 5 8/15  = L−1  + + s( s + 2)( s − 3) s + 2 s − 3   s −1 4 −2 t 8 3 t = + e + e 3 5 15 s2

( s + a 2 )2



2

L−1

= ?

s2 2

2 2

(s + a )

= L−1

(Text P. 787)

s 2

s +a

2

* L−1

s 2

s + a2



= cos at * cos at



=

∫0 cos au ⋅ cos a(t − u) du



=

1 t [cos at + cos(2 au − at)] du 2 ∫0



=

1 sin(2 au − at)  y cos at +   2 2a 0



=

1 1  t cos at + (sin at + sin at) 2  2a 



=

1 (sin at + at cos at) 2a

t

t

399

Model Question Papers

B.E./B.Tech. Degree Examination, April/May 2011 Regulation 2008 Third Semester Common to all branches MA 2211 Transforms and Partial Differential Equations Time: Three Hours

Maximum : 100 Marks

Answer All Questions PART-A (10 × 2 = 20 Marks) 1. Give the expression for the Fourier Series coefficient bn for the function f(x) defined in (–2, 2). 2. Without finding the values of a0, an and bn, the Fourier coefficients of Fourier series, for  a2 ∞  the function F(x) = x2 in the interval (0, p) find the value of  0 + ∑ ( an2 + bn2 ) .  2 n =1  3. State and prove the change of scale property of Fourier Transform. 4. If FC(s) is the Fourier cosine transform of f(x), prove that the Fourier cosine transform of 1 s f(ax) is FC   . a a 5. Form the partial differential equation by eliminating the arbitrary constants a and b from z = (x2 + a) (y2 + b) 6. Solve the equation (D – D′)3 z = 0. 7. A rod 40 cm long with insulated sides has its ends A and B kept at 20°C and 60°C respectively. Find the steady state temperature at a location 15 cm from A. 8. Write down the three possible solutions of Laplace equation in two dimensions. 9. Find the Z-transform of an. 10. What advantage is gained when z-transforms is used to solve difference equation? PART-B (5 × 16 = 80 Marks) 11. (a) (i) Expand f(x) = x(2p – p) as Fourier series in (0, 2p) and hence deduce that the sum 1 1 1 1 of 2 + 2 + 2 + 2 + ..... 1 2 3 4 (b) Obtain the Fourier series for the function f(x) given by

1 − x , −p < x < 0 f(x) =  1 + x , 0 < x < p Hence deduce that

1 12

+

1 32

+

1 52

+ ....

p2 8

OR l   x in 0 ≤ x ≤ 2 11. (b) (i) Obtain the sine series for f(x) =  (l − x )in l ≤ x ≤ l  2

400

Engineering Mathematics-III (Third Semester)

(ii) Find the Fourier series up to second harmonic for y = f(x) from the following values. x

0

p/3

2p/3

p

4p/3

5p/3

2p

y

1.0

1.4

1.9

1.7

1.5

1.2

1.0

1 − x 2 if| x|< 1 12. (a) (i) Find the Fourier transform of f(x) =  0 if| x|> 1 x cos x − sin x  x  cos dx . 3 2 x  (ii) Find the Fourier transform of f(x) given by ∞

∫0 



Hence evaluate



1 for| x|< a f(x) =  0 for| x|> a > 0

and using Parseval’s identity prove that

∫0 

2

p  dt = . t  2

∞  sin t 

OR

0< x| a|

z if| z|>| a| z−a

10. The difference equation is reduced to rational integral functions and its inverse z transform becomes simple. Arbitrary constants do not occur in the solution. This is the main advantage in using the transform. PART-B (5 × 16 = 80 marks) 11. (a) (i) Let f(x) =

a0 ∞ + ∑ ( an cos nx + bn sin nx ) 2 n =1 2p



1 2p 1  2 x3  where d0 = ∫ (2 px − x 2 )dx=  px −  3 0 p 0 p



=



an =



=

1  3 8 p3  4 2  4p − = p p 3  3 1 2p (2 px − x 2 ) ⋅ cos nx dx p ∫0 2p







\

1  − sin nx    − cos nx   sin nx  (2 px − x 2 )   − (2 p − 2 x)   + (– 2)   3 p   n   n2   n 0 1  1 1  4 =  2p  − 2 − 2   = 2 p  n n  n bn =

2p

1 2p  − cos nx   − sin nx   cos nx  (2 px − x 2 )   − (2 p − 2 x)   + (−2)   ∫ 2 2 0 p  n   n   n 3 0

= −

2 pn3

(1 − 1) = 0

f(x) = x(2p – x) =

4 2 ∞ 4 p − ∑ 2 cos nx 6 n =1 n

403

Model Question Papers

At x = 0, value of Fourier series is =

f (0) + f (2 p) 2

2 2 ∞ 4 0+0 p −∑ 2 = =0 3 2 n =1 n

\

∞

1

∑ n2

\

=

1

p2 6

1 − x , − p < x < 0 (ii) f(x) =  1 + x , 0 < x < p f(x) is evidently even in (–p, p)

bn = 0 in its Fourier series.

\

a0 ∞ + ∑ an cos nx 2 n =1

Let f(x) =

p



x2  1 p 2 p 2 f ( x ) dx = ∫ (1 + x ) dx =  x +  = 2 + p ∫ 2 0 p −p p 0 p

where a0 =

1 p 2 p f ( x )cos nx = dx (1 + x )cos nx dx ∫ −p p p ∫0

an =



p

2 2 1  − cos nx     sin nx  (1 + x )  (−1)n − 1   − (1)  =  p   n   n2   0 p  n2 



=



0 if n even  =  −4  2 if n odd  pn



f(x) =

\

∞ 2+p 4 − ∑ cos nx 2 2 n =1,3,5 pn

Putting x = 0 ∞ 2+p 4 − ∑ = f(0) = 1 2 2 n =1,3,5 pn

p 4 1 1 1  + 1 −  2 + 2 + 2 + ... = 1 2 p 1 3 5 

\

1 12

+

1 32

+

1 52

+ ...to ∞ =

p2 8

 x in 0 ≤ x ≤ l / 2 11. (b) (i) f(x) =  l − x in l / 2 ≤ x ≤ 1 Let

f(x) =

∞

∑ bn sin

n =1



npx l

2 1 npx dx bn = ∫ f ( x )sin l 0 l

(

)

404

Engineering Mathematics-III (Third Semester)



l 2  l/2 npx npx  x sin dx + ∫ (1 − x)sin dx  ∫  l/2 l 0 l l 

=

l/2

 npx   npx      − cos  − sin   2 =  l (1)  l   ( x )   np  n2 l 2   l          l   l2  0 



1/2 

  npx  npx    − cos   − sin     l (−1)  l  + (10 − x )    np  n2 l 2        l    l2   0





=



=

np l2 np l 2 np l2 np  2  l2 cos cos + 2 2 sin + + 2 2 sin  − l  2 np 2 n p 2 2 np 2 n p 2  4l 2 2

n p

Hence, f(x) =

4l

sin ∞

np 2 1

∑ n2 sin p2 n =1



=

np npx sin 2 l

4l  1 np 1 3 px 1 5 px  sin ....to ∞  − 2 sin + 2 sin 2  2 l 3 l l 5 p 1 

(ii) (See Text P.86) We will exclude the last point x = 2p x

f(x)

cos x

sin x

cos 2x

sin 2x

0

1.0

1

0

1

0

p 3

1.4

0.5

0.866

-0.5

0.866

2p 3

1.9

-0.5

0.866

-0.5

-0.866

p

1.7

–1

0

1

0

4p 3

1.5

–0.5

–0.866

–0.5

0.866

5p 3

1.2

0.5

– 0.866

–0.5

– 0.866



a0 =

2 1 ∑ f (x=) 3 (1.0 + 1.4 + 1.9 + 1.7 + 1.5 + 1.2) = 2.9 6



a1 =

2 1 f ( x )cos x = (1 + 0.7 – 0.95 – 1.7 – 0.75 + 0.6) = – 0.37 ∑ 6 3

    

405

Model Question Papers

2 Σf ( x )cos 2 x = – 0.1 6

a2 =

b1 =

2 Σf ( x )sin x = 0.17 6



b2 =

2 Σf ( x )sin 2 x = – 0.06 6



f(x) = 1.45 – 0.37 cos x – 0.1 cos 2x + 0.17 sin x – 0.06 sin 2x

\

12. (a) (i) (See Text P.28) 1

∞

1

e f ( x ) dx ∫= 2 p −∞

F[f(x)] =

4

∫ e 2 p −1

isx

isx

(1 − x 2 ) dx

=

isx 1  2 e (1 − x )  2 p   is

=

1  −2 is is 2i is − is  ( e + e ) − 3 ( e + e ) 2 p  s2 s 

=

−2 s3

⋅

1 2p

  e isx  − (−2 x )  2 2  i s

1

∞

∫ 2 p −∞

−4  s cos s − sin s  isx   e ds 2p  s3 

=

−2 ∞ s cos s − sin s isx e dx p ∫−∞ s3



∫−∞ 

−p  s cos s − sin s  (1 – x2) if | x | < 1  (cos sx – i sin sx)ds = 3 2 s 

= 0 if | x | > 1



Put x =

1 and equate real parts 2

∞ s cos s − sin x s −p  1  −3 p cos ds = 1−  = ∫−∞ 3 2 2 4 8 s 



∞

s cos s − sin s

∞

3

\ 2 ∫ \

∞

s

s −p  3  cos ds =   2 2 4

s −3 p  s cos s − sin x  .  cos ds = 2 16 s3 

∫−∞ 

Change s to x to get the result. (ii)

1

     ∞

−4  5 cos s − sin s  [2 s cos s − 2 sin s] =  2 p  s3

Using inversion, f(x) =

∞

  e isx  + (−2)  3 3  t s

1 for| x|< a f(x) =  . 0 for| x|> a > 0 F[f(x)] =

1

∞

∫ f (x)e 2 p −∞

isx

dx

... (1)

406

Engineering Mathematics-III (Third Semester)

1

     =

2p

      =

1

a

∫− a

e isx dx =

(

1  e isx  2 p  is

a

  − a

)

1 isa − isa 1 1 e −e = ⋅ (2i sin a) 2 p is 2 p is ⋅

2 sin as ⋅ p s

=

... (1)

Using parseval’s identity.

∞

∫−∞

f ( x )2 dx = a

∫− a

\



∞

∫−∞

1dx =

2 sin 2 as ∫−∞ p s2 ds

2a =

2 ∞  sin as    ds p ∫−∞  s 

∞

Putting as = t, ds =

2

| F( s)| ds (Refere Text P. 286)

2

dt a

2

 sin t  ∫−∞  t  dt = p ∞

2

∞  sin t  dt = p 2 ∫  0  t  



\

∞

∫0

2

p  sin t    dt = 2  t 

 x for 0 < x < 1  12. (b) (i) f(x) = 2 − x for 1 < x < 2 0 forx > 2 

Fs(f(x)) =

(Text P.303)

2 ∞ f ( x )sin sx dx p ∫0



=

2 2 1 x sin sx dx + ∫ (2 − x )sin sx dx  ∫   0 1 p



=

2    − cos sx   − sin sx    ( x )   − (1)   s p     s 2  0

1

2   − sin sx     − cos sx  ( 1) + (2 − x )  − −     s    s2  1  



=

2 1 1 1 sin 2 x sin s  − cos s + 2 sin s + cos s − 2 + 2   s p s s s s 

407

Model Question Papers



22 sin 2 s  2 2 sin s sin s − 2  = (1 − cos s)  2 p s ps2 s 

=

dx

∞

(ii)

∫0



Fc(e–ax) =

2

2

( a + x )( x 2 + b 2 )



(Text P. 294)

2 ∞ − ax e cos sx dx p ∫0 ∞



=

 2  e − ax  2 2 (− a cos sx + s sin sx ) p a + s 0



=

2 sin as ⋅ p s

... (1)

Using parseval’s identity, ∞

2 ∫ | f (x)| dx = −∞



a

∫− a

\



1 dx = 2a =

∞

∫−∞| F(s)|

ds

(Refer Text P. 286)

2 sin 2 as ∫−∞ p s2 ds ∞

2

2 ∞  sin as    ds p ∫−∞  s 

Putting as = t, ds =

2

dt a

2

 sin t  ∫−∞  t  dt = p ∞

2

∞  sin t  dt = p 2 ∫  0  t  



\

∫0

2

p   dt = 2  t 

∞  sin t 

 x for 0 < x < 1  12. (b) (i) f(x) = 2 − x for 1 < x < 2 (Text P. 303) 0 for x > 2 

Fs(f(x)) =

2 ∞ f ( x )sin sx ax p ∫0



=

2 2 1 x sin sx dx + ∫ (2 − x )sin sx dx  ∫   0 1 p



=

2    − cos sx   − sin sx    ( x )   − (1)   s p     s 2  0  1



2   − sin sx     − cos sx  ( 1) + (2 − x )  − −      s    s2  1  

408

Engineering Mathematics-III (Third Semester)

  =

2 1 1 1 sin 2 s sin s  − cos s + 2 sin s + cos s − 2 + 2   s p s s s s 



22 sin 2 s  2 2 sin s sin s − 2 = (1 − cos s)  2 p s s  ps2

(ii)

=

dx

∞

∫0

2

2

(

(a + x ) x2 + b2

)



(Text P. 294)

Fc(e–ax) =

2 ∞ − ax e cos sx dx p ∫0



=

 2  e − ax  2 2 (− a cos sx + sin sx) p a + s 0



=

2 a , if a > 0 2 p a + s2

∞

Similarly, Fc (e–bx) = ∞

Using

∫0



∫0

2 b , if b > 0 2 p b + s2

f ( x ) g( x ) dx =

∞ − ax

e

⋅ e − bx dx =

∞

∫0

Fc ( s) Gc ( s) ds

2 ∞ ab ds p ∫0 ( a 2 + s2 )(b 2 + s2 )

∞

 e −( a + b ) x  2 ab dx   = 2 2 p ( a + x )(b 2 + x 2 )  −( a + b)  0



1 = R.H.S a+b



\

dx

∞

∫0

2

2

2

2

( a + x )(b + x )

=

p , a > 0, b > 0. 2 ab( a + b)

(Test P. 102)

13. (a) (i) z = f(x + ct) + f(x – ct)

∂z = p = f ′(x + ct) + f′(x – ct) ∂x ∂2 z

= f′′(x + ct) + f′′(x – ct) ∂x 2 ∂z = cf ′ (x + ct) – cf′ (x – ct) ∂t ∂2 z ∂t 2

= c2[f′′(x + ct) + f′′(x – ct)]



= c2f′′ (x + ct) + c2 f′′(x – ct)

∂2 z ∂t 2

= c2

∂2 z ∂x 2

... (1)

409

Model Question Papers

(ii) (n.z – ny)p + (nx – lz)q = ly – mx

(Text P.135)

Subsidiary equations are dy dz dx = = nx − lz ly − mz mz − ny



... (1)



Using two sets of multipliers x, y, z and l, m, n each of the ratios in (1) equals



xdy + ydy + zdz ldx + mdy + ndz = x(mz − ny ) + y(nx − lz) + z(ly − mx ) l(mz − ny ) + m(nx − lx ) + n(ly − mx )



xdx + ydy + zdz ldx + mdy + ndz = 0 0 This implies, xdx + ydx + zdz = 0 and ldx + mdy + ndz = 0



Integrating, x2 + y2 + z2 = a and lx + my + mz = b



Hence, general solution is



f(x2 + y2 + z2, lx + my + xz)





i.e.,

13. (b) (i) (D2 – D′2)Z = ex–y sin (2x + 3y) Auxiliary equation is m2 – 1 = 0, m = ±1 C.F. is f(y + x) + f(y – x) P.I. =

e x − y sin(2 x + 3 y ) D 2 − D ′2

= e x − y ×

= e x − y ×

= e x − y × = e x − y × = e x − y

sin(2 x + 3 y ) (D + 1)2 − (D′ − 1)2 sin(2 x + 3 y ) (D + 1)2 − (D′ − 1)2 sin(2 x + 3 y ) 2

D − D′2 + 2D + 2D′ sin(2 x + 3 y ) −4 − (−9) + 2(D + D′)

sin(2 x + 3 y ) 5 + 2(D + D′)

= ex–y(5 – 2D – 2D′) = ex–y(5 – 2D – 2D′)

= ex–y (5 – 2D – 2D′)

= ex–y (5 – 2D – 2D′) = ex–y (5 – 2D – 2D′)

sin(2 x + 3 y ) 25 − 4(D + D′)2 sin(2 x + 3 y ) 25 − 4(D 2 + D′2 + 2DD′) sin(2 x + 3 y ) 25 − 4(D + D′)2 sin(2 x + 3 y ) 25 − 4(D 2 + D′2 + 2DD′ ) sin(2 x + 3 y ) 25 − 4(−4 − 9 − 12)

410

Engineering Mathematics-III (Third Semester)

   =

ex−y (5 − 2D − 2D′ )sin(2 x + 3 y ) 125

   =

ex−y [5 sin(2 x + 3 y ) − 4 cos(2 x + 3 y ) − 6 cos(2 x + 3 y )] 125

     =

ex−y [sin (2x + 3y) – 2 cos (2x + 3y)] 25

Hence, general solution is

ex−y [sin (2x + 3y) – 2 cos (2x + 3y)] 25 (ii) D2 – 3DD′ + 2D′2 + 2D – 2D′) z = x + y + sin (2x + y) Z = f(y + x) + f(y – x) +



i.e., (D – 2D′ + 2) (D – D′)z = x + y + sin (2x + y)



Here, m1 = 2, c1 = –2, m2 = 1, c2 = 0



Hence C.F. is e–2x f(y + 2x) + f(y + x)



PI1 =

x+y = (D − 2D′ + 2)(D − D′) −1

x+y D − 2D′   1 − D′   D  (2)  1 +  D 2     −1



=

1  D′   D − 2D′  1+  1+  ( x + y) 2D  D  2 



=

  D − 2D′  D − 2D′ 2  1  D′ D′2 1 ... + + + +   1 −  ... ( x + y ) 2D  2 2 D D2     



=

 D  D 2 4DD′ 1  D′ D′2 − + D′2  ( x − y ) 1 + + 2 + ... 1 − + D′ + D D 2D  2 4 4  



=

1  1 1 1 D′ D D′  + + ⋅ + − D′ + 2  ( x − y ) 2  D 2 2 D 4 D 



=

1  1 1 D′ D D′  − + + − D′ + 2  ( x + y )  2  D 2 2D 4 D 



=

x 2 xy y 3 + − − 2 2 4 8



PI2 = =

sin(2 x + y ) D 2 − 3DD′ + 2D′2 + 2D − 2D′ sin(2 x + y ) −4 + 6 − 2 + 2D − 2D′

Replacing D2 by – a2, D′2 by – b2, DD′ by – ab.

sin(2 x + y) 1 (D + D′) = sin(2 x + y ) = 2(D − D′) 2 D 2 − D ′2 =

1 (D + D′)sin(2 x + y ) ⋅ −4 + 1 2

411

Model Question Papers



1 = − [2 cos (2x + y) + cos (x + y)] 6



1 = − cos (2x + y) 2



\ General solution is



Z = e–2x f(y + 2x) + f(y + x) +



x 2 xy y 3 1 + − − − cos (2x + y) 2 2 4 8 2

14. (a) (Refer Text P. 194) Boundary conditions are

y(0, t) = 0   t ≥ 0 ... (1, 2) y(l , t) = 0 



  ∂y    ∂t  = 0  t = 0  0 ≤ x ≤ l y( x = , 0) kx(l − x )



The differential equation governing the displacement is



Solving and selecting the probable solution.



y(x, t) = (A cos lx + B sin lx) (C cos lat + D sin lat )



and using Boundary condition (1) and (2) we get A = 0 and l =



... (3, 4) ∂2 y ∂t 2

= a2

∂2 y ∂x 2 ... (I)

np , n = 1, 2, 3...... l



∂y = (B sin lx) (– Cla sin lat + Dla cos lat) ∂l





Using



\

∂y = 0 at t = 0, we get D = 0. ∂l

y(x, t) = Bnsin

npx npat ⋅ cos , n = 1, 2, 3..... l l

Most general solution is y(x, t) =



∞

∑ Bn sin

n =1

npx npat ⋅ cos l l

... II

Using boundary condition (4) in

∞

∑ Bn sin

n =1

npx = kx (1 – x) l

This is Fourier sine series. \

Bn = bn =

2 l np 2 2 k ( lx − x )sin dx l ∫0 l

... III

412

Engineering Mathematics-III (Third Semester)

   npx  npx  npx    − cos − sin cos       2k  l  − (l − 2 x )  l  + (−2)  − l  (1x − x 2 )     =  np  n 2 p2   l  n 3 p3          2 l   l l3     

=



=

4kl 2 n 3 p3 8kl 2 n 3 p3

l

0

[1 – (–1)n] = 0 if n even if n is odd

Hence, using II, we get

y(x, t) =

8 kl 2 p

3

∞

1

∑ (2n − 1)3 sin

n =1

(2n − 1)px (2n − 1)pat ⋅ cos l l

14. (b) Take u(x, y) = (Aelx + Be–lx) (C cos ly + B sin ly) as the probable solution of Laplace equation. Boundary conditions are

u( x , 0) = 0  0 < x < ∞ u( x , 20) = 0 

...1 ...2

0 ≤ y ≤ 10 10 y , u(0, y) =  10(20 − y ),10 ≤ y ≤ 20

Using (1) in 1, we get, C = 0 Using (2) sin 20l = 0 \ = l =

...(3) ...(4)

np , n = 1, 2, 3..... 20

20

Using (3) in 1, we get A = 0; otherwise u → ∞. Hence, u(x, y) = Bn e

−

npx 20

sin

npy , n = 1, 2, 3 ... 20

Most general solution is u(x, y) =

∞

∑ Bn e

n =1

− npy 20

sin

0

npy 20

X

... II

Using boundary conditions (4) in II

u(0, y) =

∞

∑ Bn sin

n =1





\

Bn = bn =

if 0 ≤ y ≤ 10 10 y npy =  20 10(20 − y ) if 10 ≤ y ≤ 20

10 20 npy npy  2  dy + ∫ 10(20 − y)sin dy   ∫ 10 y ⋅ sin 20  0 20 20 10 

10 20       npy  npy   npy  npy     − cos    − cos   − sin   − sin   20  − (1)  20   + (20 − y )  20  − (−1)  20    =  ( y )        np np    n 2 p2    n 2 p2                 20 20     400  0  400  10   

413

Model Question Papers



=



=



\

u(x, y) =

−200 np 400 cos + np 2 n 2 p2 800 2 2

n p

800 p2

np 2

⋅ sin ∞

∑

npx

1

n =1 n

np  200 np 400 np  cos + ⋅ sin  sin + 2  np 2 n 2 p2 2 

2

sin

npy np − 20 ⋅e sin 2 20

15. (a) (i) Find inverse z transform of

Z2 , by covolution (Z − 1)(Z − 3)

  z2 −1 ′z –1  z  Z−1   =Z  *Z z−3  z −1  ( z − 1)( z − 3) 

= 1n * 3n if | z | > 3



=

n

∑ 1n−k 3k

= 1 + 3 + 32 + .. 3n =

k =0

(ii) Z(an) =

z if | z | > | a | z−a

Take

a = reiq

Z {rneinq} =

3n+1 − 1 2

z z − re iq z r

Z{rn cos nq + irn sin nq} =

z   − cos q − i sin q  r 

z  − cos q + i sin q  z  r  = 2 r z  2  − cos q  − sin q r 



Taking real Also,

Z(rn cos nq) =

z( z − r cos q) 2

z − 2 zr cos q + r 2

, if | z | > | r |

Z{e–at sin bt} = Z(sin bt) z → zeat =

ze aT sin bT z 2 e 2 aT − 2 ze aT cos bt + 1

15. (b) (i) yn+3 – 3yn+1 + 2yn = 0

Z(yn+3) – 3Z (yn + 1) + 2Z(yn) = 0

[z3y(z) – 32y (0) – zy(1)] – 3 [zy(z) – zy (0)] + 2y (z) = 0 Using y(0) = 4, y(1) = 0 and y(2) = 8, we get

414

Engineering Mathematics-III (Third Semester)

z3 y(z) – 4z2 – 3zy(3) + 12z + 2y(z) = 0 (z3 – 3z + 2) y(z) = 4z2 – 12z

4 z 2 − 12 z

y(z) = =

z3 − 3 z + 2

=

4 z 2 − 12 z ( z − 1)( z 2 + z − 2)

4 z 2 − 12 z 2

( z − 1) ( z + 2)

=z

4 z − 12 ( z − 1)2 ( z + 2)

 A B C  = z  + +  2 3 + 2  z − 1 ( z − 1)

 20/ 9 8 1 20 1  = z − ⋅ − ⋅  2 9 z + 2  z − 1 3 ( z − 1)

y(z) =

20  z  8  z  20 z −  − ⋅ 9  z − 1  3  ( z − 1)2  9 z + 2

Taking inverse transform, y(n) =

20 8 20 − n − ⋅ (−2)n 9 3 9

15. (b) (ii) yn = (A + Bn) (– 3)n

\

yn + 1 = (A + Bn + B) (–3)n+1



\

yn+2 = (A + Bn + B) (–3)n+2 yn+2 + 6yn+1 + 9yn



= (– 3)n + 2 [A + Bn + 2B – 2(A + Bn + B)]



= (– 3)n+2 (0)



yn+2 + 6yn+1 + 9yn = 0

415

Model Question Papers

B.Tech. (PT) DEGREE EXAMINATION, DECEMBER 2013 First Semester PMA201-MATHEMATICS - III Time: Three hours

Max Marks: 100

Answer All Questions PART – A (10 × 2 = 20 Marks) 1. Determine the value of an in the Fourier series expansion of f(x) = x3 in –p < x < p. 2. Expand f(x) = 1 in a sine series in 0 < x < p. 3. Find the particular integral of (D2 – DD′) = sin (x + y). 4. Find the complete integral of q = 2px. 5. The ends A and B of a rod of length 10 cm long have their temperature kept at 20°C and 70°C. Find the steady state temperature distribution on the rod. 6. In the wave equation

∂2 y ∂t 2

= C2

∂2 y ∂x 2

, what does C2 stand for?

7. Find the Fourier cosine transform of cos x , 0 < x < a f(x) =  x≥a 0,



8. If Fourier transform of f(x) = F(s), then what is Fourier transform of f(ax)? 9. Write any two applications of Y2 test. 10. Define Correlation and Regression. PART – B (5 × 16 = 80 Marks) 11. (a) Find Fourier series to represent f(x) = x – x2 from x = – p to x = p and hence deduce that p2 1 1 1 1 ... . − + − + = 12 12 2 2 3 2 4 2 OR (b) Table below gives the value of f(x) at the specified values of x. Find the first three harmonic to represent f(x) as a Fourier series.

x y

0° 1.98

60° 1.30

120° 1.05

180° 1.30

240° –0.88

300° –0.25

12. (a) (i) Solve (D2 – 2DD′)z = e2x + x3y.

(ii) Find the general solution of the following linear partial differential equation x2(y – z)p + y2(z – x)q = z2(x – y). OR

(b) (i) Solve p(1 + q) = qz.

(ii) Solve (D2 + DD′ – 6D′2)z = cos(2x + y)

416

Engineering Mathematics-III (Third Semester)

13. (a) A tightly stretched string with fixed end point x = 0 and x = l is initially at rest in its equilibrium position. If it is set to vibrate by giving to each of its position a velocity lx(l – x). Find the displacement of the string at any distance x from one end at any time t. OR (b) The ends A and B of a rod 20 cm long have the temperature kept at 0°C and 20°C respectively until steady state conditions prevail. The temperature of the end B is then suddently raised to 60°C and kept so, while that of the end A is kept at 0°C. Find the temperature u(x, t). x

14. (i) Find the Fourier sine transform of ∞

(ii) Evaluate

dx

∫ (x 2 + a2 )(x 2 + b2 )

2

x + a2

.

using Fourier transform method.

0

OR (b) (i) Find the Fourier transform of e − x

2

/2

(ii) Find the Fourier cosine transform of f(x) = sin x in 0 < x < p.

15. (a) (i) To veryfy whether a course in accounting improved performance, a similar test was given to 12 participants both before and after the course. The marks are:

Before 44 After 53

40 38

61 69

52 57

32 46

44 39

70 73

41 48

67 73

72 74

53 60

72 78

Whether the course is useful or not? (ii) The number of automobile accidents per week in a certain community are as follows : 12, 8, 20, 2, 14, 10, 15, 6, 9, 4. Are these frequencies in agreement with the belief that accident conditions were the same during the 10 week period? OR (b) (i) The means of two large samples 1000 and 2000 members are 67.5 inches respec tively. Can the samples the regarded as drawn from the same population of standard deviation 2.5 inches. (ii) Obtain two regression equations to the following data:

x y

65 67

66 68

67 65

67 68

68 72

69 72

79 69

72 71

417

Model Question Papers

B.Tech. (PT) DEGREE EXAMINATION, DECEMBER 2012 First Semester PMA211-MATHEMATICS - III Time: Three hours

Max Marks: 100

Answer All Questions PART – A (10 × 2 = 20 Marks) 1. Write a0, an in the expansion of x + x3 as a Fourier series in (–p, p). 2. Find the half range sine series for f(x) = k in 0 < x < p. 3. Form the P.D.E. by eliminating the arbitrary function from z = f(x2 – y2) 4. Solve (4D2 – D′2)z = 0. 5. In wave equation

∂2 y ∂t

2

= c2

∂2 y ∂x 2

, what does c2 stands for?

6. A rod length 20 cm whose one end is kept at 30°C and the other end is kept at 70°C is maintained so until steady state prevails. Find the steady state temperature. 7. Write any two solution of the Laplace equation uxx + uyy = involving exponential terms in x or y. 8. State two dimensional heat equation. 9. Find the Fourier sine transform of 3e–2x. 10. Prove that Fc[f(ax)] =

1 s Fc   . a a PART – B (5 × 16 = 80 Marks)

11. (a) Find the half range cosine series f(x) = (p – x2) in the interval (0, p).

Hence find the sum of the series

1 4

1

+

1 2

4

+

1 34

+ ... + ∞.

OR (b) Find the Fourier series upto 3 harmonic from the table. rd

x

0

p 3

2p 3

p

4p 3

5p 3

y

1.0

1.4

1.9

1.7

1.5

1.2

2p

12. (a) Solve: (D2 + DD′ – 6D′2)z = x2y + e3x + y OR (b) Solve: (3z – 4y)p + (4x – 2z)q = (2y – 3x) 13. (a) A string is strectched and fastened to two points l apart. Motion is started by displacing the string into the form y = k(lx – x2) from which it is released at time t = 0. Find the displacement of any point of the string at a distance x from one end at any time time ‘t’.

418

Engineering Mathematics-III (Third Semester)

14. (a) A rectangular plate with insulated surface is 10 cm wide so long compared to its width that it may be considered infinite length without introducing appreciable error. If the temperature at short edge y = 0 is given by

      

for 0 ≤ 5 20 x u=  20(10 − x ) for 5 ≤ 10 

and all the other three edges are kept 0°C. Find the u(x, y).

OR

(b) An infinte long metal plate in the form of an area is enclosed between the line y = 0 and y = p for x > 0. The temperature is zero along the edges y = 0 and y = p and at infinity. If the edge x = 0 is kept at a constant temperature T°C. Find u(x, y). ∞



dx

∫ (x 2 + 1)(x 2 + 4) . 0

15. (a) Use Fourier transform method to evaluate ∞



dx

∫ (x 2 + 1)(x 2 + 4). 0

OR (b) Find the Fourier transform of the function f(x) is defined by



1 − x 2 if | x|< 1 f(x) =  . Hence prove that if | x|≥ 1 0

∞

∫

sin s − s cos s

0

s

3

3p s cos   ds = . 16 2

B.Tech. DEGREE EXAMINATION, NOVEMBER 2013 Third Semester MA0201-MATHEMATICS - III (For the candidates admitted from the academic year 2007-2008 to 2012-2013) (Use of statistical table is permitted) Time: Three hours

Max Marks: 100

Answer All Questions PART – A (10 × 2 = 20 Marks) 1. State the Dirichlet’s condition for a function f(x) to be explained as a Fourier series. 2. Find an for f(x) = x cos x in (–p, p). 3. Form the partial differential equation by eliminating the arbitrary constants a and b from z = ax + by + ab. 4. Find the particular integral of the equation (D2 + 2DD′ + D′2)z = ex–y. 5. Classify the following partial differential equation: 6. Write down the three mathematically possible solutions of one dimensional heat equation. 7. Prove that F{eiax. f(x)} = F(s + a)

419

Model Question Papers

8. State Parseval’s identity in Fourier transform. 9. The first four central moments of a distribution are 0, 2.5, 0.7 and 18.75. Find the Kutosis. 10. Define Null and Alternative hypothesis. PART – B(5 × 16 = 80 Marks) 11. (a) (i) Expand f(x) = x 2 in – l < x < l in a Fourier series and hence deduce p2 1 1 1 ... . + + + ∞ = (10 Marks) 6 12 2 2 3 2 (ii) Find the half range sine series of the function f(x) = p – x in (0, p) and hence deduce 1 1 1 1 − + − + ...∞. (6 Marks) 3 5 7 (b) Find the first three harmones of the Fourier series of y = f(x) from the following data:

x°(edge) y

0° 0

60° 9.2

120° 14.4

180° 17.8

240° 17.3

300° 11.7

360° 0

12. (a) (i) Solve: Z = px + qy + p2 + pq + q2. Find also singular solution.

(ii) Form the partial differential equation by eliminating the arbitrary function f(x2 + y2 + z2, ax + by + cz) = 0. OR

(b) (i) Solve: x(z – y )p + y(x – z )q = z(y2 – x2). 2



2

2

2

(ii) Solve: (D2 + 4DD′ – 5D′2)z = xy + sin(2x + 3y).

13. (a) A string is stretched and fastened to two points ‘l’ apart. Motion is started by displacing the string into the form y = k(lx – x2) and then released it from this position at time t = 0. Find the displacement of the point of the string at a distance ‘x’ from one end at any time ‘t’.

OR

(b) A rectangular plate with insulated surface is 10 cm wide and so long compared to its width that it may be considered infinite in length. The temperature at the short edge y = 0 is given by 0≤x≤5 20 x , u =  and all other three edges are kept at 0°C. Find the steady 20(10 − x ), 5 ≤ x ≤ 10  state temperature at any point in the plate. 2 1 − x 14. (a) Find the Fourier transform of f(x) =  0 ∞  x cos x − sin x  x  cos   dx. ∫  x3 2  0

if | x|< 1 . Hence evaluate if | x|> 1

OR  e − ax  (b) Find the Fourier sine transform of e–ax (a > 0). Hence find Fs{x.e–ax} and Fs    x  ∞ sin sx dx. Also deduce the value of ∫ x 0

420

Engineering Mathematics-III (Third Semester)

15. (a) Marks obtained by 10 students in Mathematics (x) and Statistics (y) are given below:

x y

60 75

34 32

40 33

50 40

45 45

40 33

22 12

43 30

42 34

64 51

Find the correlation coefficient and the two regression lines. Also find y when x = 55. OR

(b) (i) The average marks scored by 32 boys is 72 with a standard deviation of 8, while that for 36 girls is 70 with a standard deviation of 6. Test at 1% level of signifi cance, whether the boys perform better than girls. (ii) The theory predicts the proportion of beans in the four groups A, B, C and D should be 9 : 3 : 3 : 1. In an experiment among 1600 beans, the numbers in the four groups were 882, 313, 287 and 118. Does the experimental results support the theory?