Differential Equations: Theory and Applications 1773614037, 9781773614038

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Differential Equations: Theory and Applications
 1773614037, 9781773614038

Table of contents :
Cover
Half Title Page
Title Page
Copyright Page
About the Author
Table of Contents
Preface
Chapter 1 Basic Concepts of Differential Equations
1.1. Introduction
1.2. The Bernoulli Equation
1.3. Differential Equations of Higher Order
1.4. The Wronskian
Chapter 2 Fundamental Concepts of Partial Differential Equations
2.1. Introduction
2.2. Classification of Second Order PDE
2.3. Summary and Discussion
2.4. Classification of Second Order PDE
Chapter 3 Application of Differential Equations In Mechanics
3.1. Introduction
3.2. Projectile Motion
3.3. Summary and Discussion
Chapter 4 Elliptic Differential Equation
4.1. Introduction
4.2. Boundary Value Problem (BVPs)
4.3. Some Important Mathematical Tools
4.4. Properties Of Harmonic Functions
4.5. Separation Of Variables
4.6. Dirichlet Problem For A Rectangle
4.7. The Neumann Problem For A Rectangle
4.8. Interior Dirichlet Problem For A Circle
4.9. Exterior Dirichlet Problem For A Circle
4.10. Interior Neumann Problem For A Circle
4.11. Solution Of Laplace Equation In Cylindrical Coordinates
4.12. Solution Of Laplace Equation In Spherical Coordinates
4.13. Miscellaneous Example
4.14. Summary And Discussions
Chapter 5 Hyperbolic Differential Equation
5.1 Introduction
5.2. Solution Of One-Dimensional Wave Equation by
Canonical Reduction
5.3. The Initial Value Problem; D’alembert’s Solution
5.4. Summary And Discussion
Chapter 6 Parabolic Differential Equations
6.1. Introduction
6.2. Boundary Conditions
6.3. Elementary Solutions Of The Diffusion Equation
6.4. Dirac Delta Function
6.5. Separation Of Variables Method
6.6. Maximum-Minimum Principle and Consequences
6.7. Miscellaneous Example
6.8. Boundary Conditions
Chapter 7 Laplace Transform Methods
7.1. Introduction
7.2. Transform Of Some Elementary Functions
7.3. Properties Of Laplace Transform
7.4. Transform Of A Periodic Function
7.5. Transform Of Error Function
7.6. Transform Of Bessel’s Function
7.7. Transform Of Dirac Delta Function
7.8. Convolution Theorem (Faltung Theorem)
Chapter 8 Green’s Function
8.1. Introduction
8.2. The Eigenfunction Method
8.3. Summary and Discussion
References
Index

Citation preview

DIFFERENTIAL EQUATIONS: THEORY AND APPLICATIONS

DIFFERENTIAL EQUATIONS: THEORY AND APPLICATIONS

Maria Catherine Borres

ARCLER

P

r

e

s

s

www.arclerpress.com

Differential Equations: Theory and Applications Maria Catherine Borres

Arcler Press 2010 Winston Park Drive, 2nd Floor Oakville, ON L6H 5R7 Canada www.arclerpress.com Tel: 001-289-291-7705         001-905-616-2116 Fax: 001-289-291-7601 Email: [email protected] e-book Edition 2019 ISBN: 978-1-77361-596-7 (e-book) This book contains information obtained from highly regarded resources. Reprinted material sources are indicated and copyright remains with the original owners. Copyright for images and other graphics remains with the original owners as indicated. A Wide variety of references are listed. Reasonable efforts have been made to publish reliable data. Authors or Editors or Publishers are not responsible for the accuracy of the information in the published chapters or consequences of their use. The publisher assumes no responsibility for any damage or grievance to the persons or property arising out of the use of any materials, instructions, methods or thoughts in the book. The authors or editors and the publisher have attempted to trace the copyright holders of all material reproduced in this publication and apologize to copyright holders if permission has not been obtained. If any copyright holder has not been acknowledged, please write to us so we may rectify.

Notice: Registered trademark of products or corporate names are used only for explanation and identification without intent of infringement. © 2019 Arcler Press ISBN: 978-1-77361-403-8 (Hardcover) Arcler Press publishes wide variety of books and eBooks. For more information about Arcler Press and its products, visit our website at www.arclerpress.com

ABOUT THE AUTHOR

Catherine is currently taking up Master of Arts in Education Major in Mathematics in Philippine Normal University – Manila. She is currently working as a Content Developer for Mathematics at the Affordable Private Education Center (APEC Schools).

TABLE OF CONTENTS

Preface...........................................................................................................xi Chapter 1

Basic Concepts of Differential Equations.................................................... 1 1.1. Introduction......................................................................................... 2 1.2. The Bernoulli Equation....................................................................... 31 1.3. Differential Equations of Higher Order............................................... 50 1.4. The Wronskian................................................................................... 85

Chapter 2

Fundamental Concepts of Partial Differential Equations........................ 111 2.1. Introduction..................................................................................... 112 2.2. Classification of Second Order PDE ................................................ 112 2.3. Summary and Discussion................................................................. 141 2.4. Classification of Second Order PDE ................................................ 141

Chapter 3

Application of Differential Equations In Mechanics............................... 143 3.1. Introduction..................................................................................... 144 3.2. Projectile Motion ............................................................................ 161 3.3. Summary and Discussion................................................................. 186

Chapter 4

Elliptic Differential Equation.................................................................. 191 4.1. Introduction..................................................................................... 192 4.2. Boundary Value Problem (BVPs)...................................................... 195 4.3. Some Important Mathematical Tools................................................ 197 4.4. Properties Of Harmonic Functions................................................... 199 4.5. Separation Of Variables ................................................................... 210 4.6. Dirichlet Problem For A Rectangle................................................... 212 4.7. The Neumann Problem For A Rectangle .......................................... 215 4.8. Interior Dirichlet Problem For A Circle ............................................ 217

4.9. Exterior Dirichlet Problem For A Circle ........................................... 222 4.10. Interior Neumann Problem For A Circle......................................... 227 4.11. Solution Of Laplace Equation In Cylindrical Coordinates .............. 229 4.12. Solution Of Laplace Equation In Spherical Coordinates................. 238 4.13. Miscellaneous Example................................................................. 247 4.14. Summary And Discussions............................................................. 276 Chapter 5

Hyperbolic Differential Equation........................................................... 279 5.1 Introduction...................................................................................... 280 5.2. Solution Of One-Dimensional Wave Equation by Canonical Reduction .................................................................... 284 5.3. The Initial Value Problem; D’alembert’s Solution.............................. 288 5.4. Summary And Discussion................................................................ 297

Chapter 6

Parabolic Differential Equations............................................................. 301 6.1. Introduction..................................................................................... 302 6.2. Boundary Conditions....................................................................... 304 6.3. Elementary Solutions Of The Diffusion Equation ............................. 305 6.4. Dirac Delta Function....................................................................... 310 6.5. Separation Of Variables Method....................................................... 316 6.6. Maximum-Minimum Principle and Consequences........................... 340 6.7. Miscellaneous Example. ................................................................. 343 6.8. Boundary Conditions....................................................................... 352

Chapter 7

Laplace Transform Methods................................................................... 357 7.1. Introduction..................................................................................... 358 7.2. Transform Of Some Elementary Functions........................................ 362 7.3. Properties Of Laplace Transform...................................................... 364 7.4. Transform Of A Periodic Function.................................................... 370 7.5. Transform Of Error Function............................................................. 372 7.6. Transform Of Bessel’s Function........................................................ 374 7.7. Transform Of Dirac Delta Function.................................................. 376 7.8. Convolution Theorem (Faltung Theorem).......................................... 382

Chapter 8

Green’s Function.................................................................................... 389 8.1. Introduction..................................................................................... 390 8.2. The Eigenfunction Method............................................................... 403

viii

8.3. Summary and Discussion................................................................. 406 References.............................................................................................. 408 Index...................................................................................................... 409

ix

PREFACE

In many respects this book is perfect book for beginners. It combines a lighthearted approach, well rounded explanations and plenty of practice opportunities. It makes an ideal companion for those students who are commencing a course in differential equation at undergraduate or at graduate level. The study of differential equations is a wide region in pure and applied mathematics, physics, and engineering. All of these disciplines are anxious with the properties of differential equations of different types. Pure mathematics emphasis on the life and incomparability of solutions, while applied number emphasizes the rigorous occasion of the way for approximating solutions. Differential equations is an important legislature in modeling virtually every physical, technical, or biological process, from celestial motion, to bridge design, to interactions between neurons. Differential equations such as those used to solve real-life problems may not necessarily be directly solvable, i.e. do not have closed example solutions. Instead, solution tins be approximated using numerical methods. Many fundamental traditions of physics and courtesy tins be formulated as differential equations. In biology and economics, differential equations are used to patterns the allotment of complex systems. The mathematical opinion of differential equations first developed together with the adeptness where the equations had originated and where the results found application. However, diverse problems, sometimes originating in quite distinct scientific fields, may give rise to identical differential equations. Whenever this happens, mathematical opinion seat the equations tins be viewed as a unifying outlook sitting diverse phenomena. As an example, consider assembly of enlightenment and sound in the atmosphere, and of shaft on the surface of a pond.

CHAPTER

1

BASIC CONCEPTS OF DIFFERENTIAL EQUATIONS

CONTENTS 1.1. Introduction......................................................................................... 2 1.2. The Bernoulli Equation....................................................................... 31 1.3. Differential Equations of Higher Order............................................... 50 1.4. The Wronskian................................................................................... 85

2

Differential Equations: Theory and Applications

1.1. INTRODUCTION Mathematical models for real world phenomena often take the form of equations relating various quantities and their rates of change (derivatives), For example, the motion of a particle involves the distances covered in time t and velocity v or acceleration a. Now the rate of change

ds of s dt

dv with respect to t is the velocity v and rate of change of velocity with respect to t is the acceleration a . A particle moving in adtstraight line has an equation of motion as s = f ( t ) , where t is in seconds and s is in meters. Its velocity satisfies the equation

v=

ds = 4t 2 + 5t − 3 dt This leads us to the definition of a differential equation (D.E).

A differential equation is a mathematical equation that relates some function with its derivatives. In applications, the functions typically represent physical quantities, the by – product represent their rates of change, and the equation defines an association between the two. Because such relations are exceptionally common, differential equations amusement a prominent role in many disciplines including engineering, physics, economics, and biology. In pure mathematics, differential equations are studied from several different perspectives, mostly concerned with their solutions-the set of functions that satisfy the equation. Only the simplest differential equations are solvable by explicit formulas; however, some ownership of solutions of a given differential equation may be determined without discovery their exact form.

1.1.1. Differential Equation and their Classification An equation involving one dependent variable and its derivatives with respect to one or more independent variables is called differential equation. For example

dy + y cos x = sin x dx

Basic Concepts of Differential Equations

3

2

d2y  dy  + xy   = 0 2 dx  dx  3

  dy 2  2 d 2 y 1+    = dx 2   dx  

1.1.2. Ordinary Differential Equation (O. D. E) A differential equation, in which ordinary derivatives of the dependent variable with respect to a single independent variable occur, is called an ordinary differential equation (O.D.E). Above equations (i) , (ii) and (iii) are examples of Ordinary Differential Equation.

Partial Differential Equation A differential equation involving partial derivatives of the dependent variable with respect to more than one independent variable is called a partial differential equation.

x

∂z ∂z + y = nx ∂x ∂y

∂ 2u ∂ 2u ∂ 2u + 2 + 2 = 0 ∂x 2 ∂y ∂z

Order of Differential Equation The order of a differential is the order of the highest derivative that occurs in the equation.

Degree of a Differential Equation The degree of a differential equation is the greatest exponent of the highest order derivative that appears in the equation. (The dependent variable and its derivatives should be expressed in a form free from radicals and fractions). •

dy + y cos x = sin x dx (order 1, degree 1)



d2y  dy  + xy   = 0 dx 2  dx  (order

2

2, degree 1)

Differential Equations: Theory and Applications

4

3



  dy 2  2 d 2 y 1+    = dx 2   dx   2

exponent of the equation •

x

d y dx 2

(order 2, degree 2)

is 2 after removing the radical by squaring both sides of

∂z ∂z + y = nx ∂x ∂y 2

2

(order 1, degree 1)

2

∂u ∂u ∂u + + 2= 0 ∂x 2 ∂y 2 ∂z

• (order 2, degree 1) In this chapter, we shall study ordinary differential equation only. Recall that a function T : U → V , where U ,V , are vector spaces over the same field is called linear if, for α , β ∈ F , x, y ∈U , T (α x + β y=) α T ( x ) + β T ( y ) .

1.1.3. Linear Differential Equation  dy d 2 y dny  F x , y , , 2 ,…, n  = 0 dx dx dx   is said

An ordinary differential equation to be d y d y , ,…, linear if F is a linear function of the variables x , dy . A similar dx dx dx definition applies to partial differential equations. Thus, the general linear ordinary differential equation of order n is 2

n

2

a0 ( x )

n

dny d n −1 y dy + a1 ( x ) +… + an −1 ( x ) + an ( x ) y = F ( x ) n dx dx n −1 dx

Where

a0 ( x )

The equation dy y . dx

is not identically zero. 2 d3y dy x d y e x3 + 2 +y = 3 2 is dx dx dx

not linear because of the term

It should be carefully noted that in a linear ordinary differential equation • The dependent variable y and its derivative are all of degree one. •

No product of y or any of its derivative appear. No transcendental function of y and / or its derivative occur.

• A differential equation that is not linear is called a nonlinear differential

Basic Concepts of Differential Equations

5

equation. Differential equations occur in the mathematical formulation of many problems in science and engineering. Some such problems are: • • • •

Determining the motion of projectile, rocket, satellite or planet. Finding the charge or current in an electric circuit. Study of chemical reactions. Determination of curves with given geometrical properties.

Solution of a Differential Equation A solution (or integral) of a differential equation is a relation between the variables, not containing derivatives, such that this relation and the derivatives obtained from it satisfy the given differential equation identically. dy = −µy −µx For example, the equation dx has a solution y = c.e , where c is an arbitrary constant.

d2y +y= 0 dx 2 The differential equation has solution = y A= cos x, y B sin x and = y A cos x + B sin x, where Aand B are arbitrary constants. A solution of a differential equation which contains as many arbitrary constants as the order of the equation is called a general solution (or integral) of the differential equation. A solution obtained from the general solution by giving particular values to the constants is called a particular solution. The graph of a particular integral is called an integral curve of the differential equation.

Formation of a Differential Equation Given a relation

f ( x , y , c1 , c2 , … , cn ) = 0

(1)

c , c , …, cn between variables x, y and containing n constants 1 2 it is always possible to form a differential equation of order n such that the given relation (1) is the general solution of the equation. This is done by differentiating (1) n times thereby obtaining equations and then eliminating the n constants from the original relation and n derived equations. The method is illustrated by means of examples. Example: The equation y= x + a (1)

6

Differential Equations: Theory and Applications

Represents a family of parallel straight lines for different values of a. Elimination of one constant ‘a’ requires two equations. The second equation is obtained by differentiating (1).

dy =1 Thus dx is the differential equation of the relation (1), with a eliminated. Example: From the differential equation by eliminating the two constants y A sin x + B cos x. A and B from the relation = (1) Solution: It is clear that three equations are required to eliminate two unknowns A and B. We obtain two other needed equations by successive differentiation of (1). Thus from (1), we have

dy = A cos x − B sin x dx (2) d2y = − A sin x − B cos x = −y dx 2 , using (1)

(3)

2

d y + y= 0 2 So, dx (4) is the required differential equation and (1) is its general solution. Example: Find the differential equation of all parabolas whose axes are y − axis. parallel to the Solution: General equation of a parabola whose axis is parallel to the y − axis is

y = ax 2 + bx + c. (1) In order to obtain its differential equation, we have to eliminate a, b, c from (1). For that, we need three more equations. Differentiating (1) successively, we have

dy = 2ax + b dx d2y = 2a, dx 2

Basic Concepts of Differential Equations

7

d3y = 0. dx 3 The last equation does not contain any of the constants a, b and c. Thus, d3y =0 dx 3 is the differential equation of all parabolas whose axes are parallel y − axis. to the

Initial and Boundary Condition We have observed that general solution of a differential equation contains the same number of arbitrary constants, as is the order of the equation. Sometimes we need to find the solutions of differential equations subject to supplementary conditions. Two types of conditions will be often encountered.

Initial Condition It is often required to find the solution of a differential equation subject to certain conditions. If the conditions relate to one value of the independent y = y0 at x = x0 (written as y ( x0 ) = y0 ) and variable such as dy = y′ ( x0 ) x = x0 , where x0 belongs to some interval ]α , β [ then dx at

x

they are called initial conditions (or one- point boundary conditions) and 0 is called the initial point. An initial value problem consists of a differential equation (of any order) together with a collection of initial conditions that must be satisfied by the solution of the differential equation and derivatives at the initial point.

1.1.4. Boundary Conditions The problem of finding the solution of a differential equation such that all the associated constraints relate to two different values of the independent variable is called a two-point boundary value problem (or simply a boundary value problem). The associated supplementary boundary conditions are called two-point boundary conditions.

dy = 2x Example: Solve dx (1)

Differential Equations: Theory and Applications

8

Such that

y (1) = 4.

2 Solution: This is an initial value problem. We note that y= x + c, c y (1) = 4, being arbitrary constant, is the general solution of (1). Since we have

y (1)= 12 + c.

1+ c or c = 3. Therefore, 4 =

y x 2 + 3 is the solution of the initial value problem (1). Thus, = Note that the general solution represents a family of parabolas for

y x 2 + 3 is a particular member of the different values of c. The solution = (1 ,4 ). family that passes through d2y +y= 0 2 Example: Solve dx (1)

π  π  1 = y = 2, y '   . Subject to the conditions  4  2 4 Solution: Since both the conditions relate to one value of x, namely

π

, 4 this is an initial value problem. We have already noted in previous example that = y A sin x + B cos x (2) x=

is the general solution of (1). Differentiating (2) w. r. t x, we get

= y′ A cos x − B sin x. (3) By the given conditions, we have respectively from (2) and (3) π y 4

A B π π  A sin + B cos =+ 2= = 4 4 2 2 

π y′  4

B π π A  1 A cos − B sin =− .  == 4 4 2 2 2 

2 and A − B = 1 Hence A + B =

Basic Concepts of Differential Equations

9

Solving for Aand B, we find that

= A

3 1 = ,B 2 2 With these values of Aand B, the particular solution of (1) is

= y

3 1 sin x + cos x 2 2

y = c cos x

y = c sin x

1 2 Example: Verify that and are solutions of 2 d y + y= 0. dx 2 Find a particular solution of the equation satisfying the boundary conditions

π  = y ( 0 ) 1= , y   2. 2 Solution: We have

y = c1 cos x.

Differentiating twice, we get

dy = −c1 sin x dx d2y d2y = c1 cos x = − y +y= 0. 2 dx 2 or dx d2y +y= 0 dx 2 (1)

y = c1 cos x is a solution of y = c2 sin x is also a solution of (1) Similarly, it can be checked that Thus,

General solution of (1) is

= y c1 cos x + c2 sin x (2) Applying the boundary conditions, we obtain from (2)

y ( 0 ) = 1 = c1 + 0

and

π  y  = 2= 0 + c2 . 2

10

Differential Equations: Theory and Applications

c =1

c = 2. Hence the particular solution of (1) satisfying

Thus 1 and 2 the given conditions is

= y cos x + 2sin x Example: Solve the d y = + y 0, y= ( 0 ) 1, y= (π ) 5. dx 2 2

boundary

value

problem

Solution: Applying the boundary conditions to the general solution = y A sin x + B cos x of the given equation, we have

y ( 0 ) = 1= B y (π ) = 5 = − B Thus, we obtain two values of B and we are unable to determine any definite value of A. Hence, the boundary value problem has no solution. It follows that a boundary value problem need not always have a solution. In above section we classified the differential equations and saw how a differential equation can be formed by eliminating constants from a given functional equation. The problem of finding general solution (integral) of a given differential may or may not exist. Even if the integral of a given equation exists, it may not be easy to find. We shall only discuss methods of solutions of special types of differential equations.

Separable Equations Definition. A differential equation of the type

F ( x ) G ( x ) dx + f ( x ) g ( y ) dy = 0 (1) Is called an equation with separable variables or simply a separable equation. Equation (1) may be written as

F ( x) f ( x)

dx +

g ( y)

G ( y)

dy = 0

which can be easily integrated.

Basic Concepts of Differential Equations

dy x 2 = dx y (1) Example: Solve Solution: Equation (1) can be written as

y dy = x 2 dx. Integrating both the sides, we get

y 2 x3 = + c1 2 3 or

3= y 2 2x2 + c which is the required solution of (1).

dy 1 = . Example: Solve dx x tan y Solution: The given equation is separable and can be written as

dx = tan y dy x Integrating, we have

ln x = − ln cos y + c

or

ln x + ln cos y = c i.e.,

ln x cosy = c

or

x cos y= e=c a ( say ) x cos y = a because a > 0. Example: Solve

(

)

x sin y dx + x 2 +1 cos y dy = 0

Solution: Dividing (1) by

(x

2

)

+1 sin y

, we have

x dx + cot y dy = 0 x +1 2

Which is a separable equation. Therefore,

(1)

11

Differential Equations: Theory and Applications

12



x dx + ∫ cot y dy= c1= ln c2 x2 + 1

1 ln x 2 +1 + ln sin y = ln c2 2 i.e.,

(

or

(

)

)

ln x 2 +1 + 2 ln sin y = 2 ln c2 . 2= ln sin y ln = sin y ln = ( sin y ) ln sin 2 y 2

Now, we can write 2

2 2

2 2

2 ln c2 = ln c2 = ln c = ln c, where c = c > 0 2 2 c is the required solution. Hence ln( x + 1) sin y =

Example: Solve the initial value problem

dy 2x = , y ( 0 ) = −2. dx y + x 2 y Solution: We have

dy 2x = dx y 1 + x 2

(

y dy =

) or

2x dx 1+ x 2

integrating, we obtain

y2 = ln(1 + x 2 ) + ln c, 2 c being a constant. y2 = ln c 1 + x 2 2

((

))

(

)

2

= y 2 ln( c 2 1 + x 2 ) Now setting 2

2

(i)

y ( 0 ) = −2 4

= 4 ln= c or c e . So (i) becomes

into (1), we have

2

and

Basic Concepts of Differential Equations

(

(

= y 2 ln e 4 1 + x 2

13

)) 2

which is the required solution.

1.1.4. Homogenous Equations f ( x, y )

Definition. A function f ( tx , ty ) = t n f ( x, y ) ,

is called homogeneous of degree n if

xy ,

x x10 + y10 and sin   2 2 x + y  y  are

where t is a nonzero real number. Thus homogeneous function of degree 1,8 and 0 respectively. A first order differential equation

dy = f ( x, y ) dx (1) is said to be homogeneous function of any degree. If (1) is written in the form M ( x, y ) dx + N ( x, y ) dy = 0 . Then it is called homogeneous functions of the same degree. y 1+ 1 dy x +y x = ln x − ln y + = ln + y y dx x− y 1− x x is homogenous, but The equation 2

dy y 3 + 2 xy  y  y = = y   + 2   is not homogeneous. 2 dx x x x dy  y =g  Theorem. A homogeneous equation dx (1)  x  can be transformed into a separable equation ( in the variables v and x) by the substitution y = vx. dy dv = v+ x Proof: Put y = vx into (1). Then and (1) becomes dx dx xdv v+ = g (v) dx or

v − g (v) + x

dv = 0 dx

Differential Equations: Theory and Applications

14

 v − g ( v )  dx + x dv = 0. or  This equation is separable and can be solved as in the previous section. dy x3 + y 3 = 2 2 Example: Solve dx x y + xy

Solution: We have

dy = dx   

 1+  y  + x

3

y  x 2  y   x

So that the equation is homogeneous. Setting y = vx into (1), we obtain

v+ x

dv 1 + v3 1+ v3 v 2 − v +1 = = = dx v + v 2 v ( 1 + v ) v

dv v 2 – v +1 1− v = x = –v dx v v

v dx dv = x or 1 − v

dx 1    −1 +  dv = x 1−v or 

integrating, we get

− v − ln 1 –= v ln x + ln c or

ln cx ( 1 – v ) + v = 0 Replacing

ln c ( x – y ) +

v by

y x in the above equation, we have

y = 0 x

which is the required solution. Example: Solve the initial value problem y ( 2 ) = 6.

(x

2

)

+ 3 y 2 dx − 2 xy dy = 0 ,

Basic Concepts of Differential Equations 2

15

 y 3   +1 3 2 dy x + 3 y x = = dx 2 xy  y 2   x  (1) Solution: Here Which shows that the equation is homogeneous. Putting y = vx into (1),

we have

v+ x

dv 3v 2 + 1 = dx 2v

dv 3v 2 +1 v 2 +1 x= . = −v dx 2v 2v

2v dx dv = 2 x or 1 + v on integrating, we get

(

)

ln 1 + v 2 = ln x + ln c or

v by

replacing y2 1+ 2 = cx x or

1 + v2 = cx

y , x we obtain

x2 + y 2 = cx x 2 Now if x ≥ 0, we can write this as

x2 + y 2 = cx3 2 3 2 or y= cx − x

± cx 3 − x 2 or y = Applying the initial condition, we get

y ( 2) = 6= ± 8c − 4

16

Differential Equations: Theory and Applications

c − 4 36 = i.e., c 5 or 8= 5 x3 − x 2 is the required solution. We take the plus sign in y ( 2) the radical since is positive. = y Hence

1.1.5. Differential Equation Reducible to Homogeneous Form Some differential equations are not homogeneous. But they can be reduced to homogeneous form.

0 ( a1 x + b1 y + c1 ) dx + ( a2 x + b2 y + c2 ) dy = The differential equation is not homogeneous. But it can be reduced to a homogeneous form as illustrated below:

a1 b ≠ 1 b2 , then make the transformations Case 1. If a2 x =X + h, y = Y + k. The given equation becomes

0 ( a1 X + b1Y + a1h + b1k + c1 ) dX + ( a2 X + b2Y + a2 h + b2 k + c2 ) dY = (1) Let h and k be the solution of the system of equations

a1h + b1k + c1 = 0  a2 h + b2 k + c2 = 0 Then for these values of h and k , (1) reduces to the homogenous equation

0 ( a1 X + b1Y ) dX + ( a2 X + b2Y ) dY = in the variables X and Y .

a1 b1 = , z a1 x + b1 y and the given equation will a b2 then put= 2 Case II. If reduce to a separable equation in the variable x and z. dy 2 y – x + 5 = Example: Solve dx 2 x – y − 4 (1)

Basic Concepts of Differential Equations

X + h, y =+ Y k Solution: Putting x = So the given equation becomes

dY 2Y – X + 2 K − h + 5 = dX 2 X –Y + 2hX k − 4 (2) The solution of the system of equations −h + 2k + 5 =0   2h − k − 4 = 0  is h = 1, k = −2. For these values of h and k , (2) reduce to

dY 2Y − X = dX 2 X − Y (3) which is homogeneous. Putting Y = vX into (3), we have v+ X

dv 2v −1 = dX 2 − v which is separable.

dX  2−v  .  2  dv = v − 1 X   Therefore, Integrating, we obtain ln

v −1 1 − ln v 2 = −1 ln X + ln c0 v +1 2 2

or

ln

v −1 2 2 = ln X + ln c0 + ln v 2 −1 v+ 1 2

{

v −1 2 = ln ln c0 X . v 2 −1 v +1 2

}

v −1 2 = c X . v −1 . v +1 , where= c c02 v +1 2

= v –1 c X . v +1 or

3

17

Differential Equations: Theory and Applications

18

Replacing

v by

Y , X this becomes

Y−X Y+X 2 =c X . X X or

Y –X = c Y+X

3

3

(4)

But X= x –1 , Y = y + 2 . Hence (4) takes the form

y – x + 3= c y + x +1

3

which is the required solution. ( 2 x + y +1 ) dx + ( 4 x + 2 y − 1 ) dy =0 Example: Solve (1) a1 2 b1 1 = = = , a 4 b 2 z. then (1) becomes 2 Solution: Since 2 we put 2 x + y =

0 ( z +1 ) dx + ( 2 z –1 )( dz − 2dx ) = 0 ( z +1 – 4 z +2 ) dx + ( 2 z –1 ) dz = 3 ( 1 – z ) dx + ( 2 z –1 ) dz = 0 i.e., which is separable. or

Dividing by 1 − z , we have

3 dx +

2 z −1 dz = 0 1− z

1  3.dx +  −2 + 1− z  or integrating, we obtain

  dz =0 

3 x − 2 z – ln 1 – z = c0 Replacing z by 2 x + y, it reduces to

3 x − 2 ( 2 x + y ) − ln 1 − 2 x − y = c0 or

– x − 2 y − ln 2 x + y −1 = c0

Basic Concepts of Differential Equations

19

ln 2 x + y −1 + x + 2 y =c, where c =−c0 or is the required solution.

Exact Equation Definition. The expression

M ( x, y ) dx + N ( x, y ) dy

(1)

is called an exact differential if there exists a continuously differentiable f ( x, y ) function of two real variables x and y such that the expression equals the total differential df . We know from calculus that

= df

∂f ∂f dx + dy. ∂x ∂y

Thus, if (1) is exact then

M ( x , y= )

∂f ∂f y ) = fy. = f x and N ( x , = ∂x ∂y

If (1) is an exact differential then the differential equation

M ( x , y ) dx + N ( x , y ) dy = 0 is called an exact equation. Theorem. The differential equation

M ( x , y ) dx + N ( x , y ) dy = 0

(1)

∂M ∂N = , ∂ y ∂ x is an exact equation8.U&hKŽ§ if and only if where the funcM ( x, y ) N ( x, y ) tions and have continuous first order partial derivatives. Proof. Suppose that the equation (1) is exact so that M dx + Mdy is an exact differential. By definition, there exists a function ∂f ∂f M ( x, = y ) = f x and N ( x, = y) = fy ∂x ∂y

Then,

N=x

M= y

∂M ∂2 f = f= xy ∂y ∂y ∂x

∂N ∂2 f = f= yx ∂x ∂x ∂y

f ( x, y )

such that

Differential Equations: Theory and Applications

20

Since M and N possess continuous first order partial derivatives, we ∂M ∂N = f xy = f yx have and, therefore ∂y ∂x as desired.

Solution of an Exact Equation If

M ( x , y ) dx + N ( x , y ) dy = 0 (1)

is an exact equation, then there exists a function

df =

f ( x, y )

such that

∂f ∂f dx + dy = M ( x, y ) dx + N ( x, y ) dy ∂x ∂y

∂f ∂f = M= and N ∂y Therefore ∂x . ∂f =M Integrating ∂x with respect to x, we have

f ( x, y ) = ∫ M dx + h ( y ) .

(2)

h( y) The constant of integration is an arbitrary function of y since it must vanish under differentiation w.r.t x. Differentiating (2) w.r.t. y , we get ∂f ∂ = ∫ Mdx + h′ ( y ) ∂y ∂y

(

)

∂f ∂ ∫ Mdx + h′ ( y ) =N = ∂ y y ∂ i.e., ∂ = h′ ( y ) N − ∫ Mdx ∂ y so

(

)

(

)

f ( x, y ) = c integrating the above equation w.r.t y, we obtain h and hence is the required solution of (1).

( 3x Example: Solve

2

)

(

)

y + 2 dx + x3 + y dy = 0 2

3

M 3 x y + 2 and N= x + y Solution: Here=

(1)

Basic Concepts of Differential Equations

21

∂M ∂N = 3= x2 , 3x 2 ∂y ∂x ∂M ∂N = ∂ y ∂x and so the equation is exact. Thus To find the solution of (1), we note that the left hand side of the equation f ( x, y ) is an exact differential. Therefore, there exists a function such that ∂f = 3x 2 y + 2 ∂x (2) ∂f = x3 + y And ∂y (3) w . Integrating (2) r.t x , we have

f ( x, y ) = x 3 y + 2 x + h ( y ) , h( y) where is the constant of integration. Differentiating the above equation w.r.t y and using (3), we obtain ∂f =+ x3 h′ ( y ) =+ x3 y ∂y or

h′ ( y ) = y. Integrating, we have

h( y) =

y2 2

Thus,

f ( x, y ) = x 3 y + 2 x +

y2 2

Hence the general solution of (1) is

x3 y + 2 x +

y2 = c 2

22

Differential Equations: Theory and Applications

Alternative Method Integrating (2) and (3) w.r.t x and y respectively, we have

f ( x, y ) = x 3 y + 2 y + h ( y ) and

f ( x, y ) = x 3 y +

y2 + g ( x). 2

y2 2 and g ( x ) = 2 x . Thus, So, the general solution is h( y) =

x3 y + 2 x +

y2 = c. 2

Example: Solve the initial value problem

( 2 y sin x cos x + y

2

)

(

)

sin x dx + sin 2 x − 2 y cos x dy = 0, y ( 0 ) = 3

2 = Solution: Here M 2 y sin x cos x + y sin x 2

= N sin x − 2 y cos x and ∂M = 2sin x cos x + 2 y sin x ∂y ∂N = 2sin x cos x + 2 y sin x ∂x ∂M ∂N = Thus ∂y ∂x showing that the given equation is exact. Hence there exists a function

f ( x, y )

such that

∂f = 2 y sin x cos x + y 2 sin x ∂x . (1) ∂f = sin 2 x − 2 y cos x ∂y (2) Integrating (1) w.r.t . x . , we have erentiating this equation w.r.t y and using (2), we get

Basic Concepts of Differential Equations

23

sin 2 x − 2 y cos x + h' ( y ) = sin 2 x − 2 y cos x = h′ ( y ) 0= and so h ( y ) c1. i.e., Hence the general solution of the given equation is f ( x, y ) = c2 i.e.,

y sin 2 x − y 2 cos x + c1 = c2

y.sin 2 x − y 2 .cos x = c2 − c1 = c

x 0,= y 3, we have −9 =c . Applying the initial condition that when= 2 2 9 is the required solution. Hence y cos x − y sin x =

1.1.6. Integrating Factors Definition. If the differential equation

M ( x, y ) dx + N ( x, y ) dy = 0 (1)

µ ( x, y ) is not exact but when it is multiplied by a function and the resulting µ ( x , y ) M ( x , y ) dx + µ ( x , y ) N ( x , y ) dy = 0 equation is exact, then µ(x ,y) is called an integrating factor (I.F) of the differential equation (1). The number of integrating factors of an equation may be infinite. We list below (without proofs) some rules to find the integrating factors of equations of special types.

M ( x , y ) dx + N ( x , y ) dy = 0 (1) M y − Nx = P, N is not exact and where P is a function of x only then (1) µ ( x) µ ( x) has an integrating factor which also depends on x. is solution of the differential equation dµ = Pµ dx Rule 1. If

i.e.,

x) (=

exp ∫ P dx

24

Differential Equations: Theory and Applications

∂M ∂N = , Nx . ∂y ∂x

My = Note that

Nx − M y Rule II. If

= Q,

M

where Q is a function of y only, then

0 has an integrating factor the differential equation Mdx + Ndy = µ (= y ) exp ∫ Q dy. 0 (1) Rule III. If M dx + Ndy = 1 is homogeneous and xM + yN ≠ 0, then xM + yN is an I.F of (1).

0 0 is of the form y f ( x, y ) dx + x g ( x, y ) dy = Rule IV. If M dx + Ndy = 1 xM − yN ≠ 0, then xM − yN is an I.F of (1). and The following differential formulas are useful in the calculation of certain exact equations:

 y  x dy − y dx d = x2 x

 x  y dx − x dy d = y2  y d ( xy = ) x dy + y dx

(

)

d x 2 + y 2= 2 ( x dx + y dy )  x  y dx − x dy d  ln  = xy  y  x  y dx − x dy d  arc tan  = 2 y  x + y2  Example: Solve

(

)

y dx + x 2 y − x dy = 0

Basic Concepts of Differential Equations

25

Solution. The equation is not exact. Rearranging the equation, we have

y dx − x dy + x 2 y dy = 0 y dx − x dy + ydy = 0 x2 or

  0. Now it is an exact equation and may be written as − d  x  + y dy = Integrating, we have y

y y2 − + = c x 2 2 cx is the general solution. or xy − 2 y = x 2 − 2 x + 2 y 2 dx +2 xy dy = 0 Example: Solve

(

)

(1)

M y − Nx 4 y − 2 y 1 = = N 2 xy x Solution. Here

dµ µ = Therefore, I.F is the solution of dx x or µ = x is an I.F. Multiplying the equation by x, we have

µ ( x, y )

(x

3

)

− 2 x 2 + 2 xy 2 dx + 2 x 2 y = 0.

This equation is exact. So we can easily find that its solution is x 2 x3 − + x2 y 2 = c0 4 3 2 2 c is the solution of (1). 4 3 or 3 x − 8 x + 12 x y = 4

 x  dx +  − sin y  dy = 0  y  Example: Solve . Solution. Here, by rule II, 1 −0 Nx − M y y 1 = = M 1 y dy ln y is a function of y only. Therefore, µ ( y ) = exp ∫ y = e = y is an I.F. Multiplying the equation by y, we have

26

Differential Equations: Theory and Applications

y dx + ( x − y sin y ) dy = 0

0 or y dx + x dy − y sin y dy = d ( xy ) − y sin y dy = 0. or

c which is the required Integrating, we get xy + y cos y − sin y = solution.

(x

2

)

(

)

y − 2 xy 2 dx − x3 − 3 x 2 y dy = 0 (1) Solution: The equation is homogeneous but not exact. We have Example: Solve

xM + yN = x 3 y − 2 x 2 y 2 − x 3 y + 3 x 2 y 2 = x 2 y 2 ≠ 0 1 1 = 2 2 So, using Rule III, xM + yN x y is an I.F. Therefore, multiplying 1 , 2 2 (1) by x y we obtain

1 2  x 3 0  −  dx −  2 −  dy = y  y x  y This equation is exact. Integrating, we get

x − 2 ln x + 3ln y = c y is the required solution.

x , y Note that, here the term involving both x and y, has been taken only once. Example: Solve

(

)

(

)

y x . y + 2 x 2 y 2 + x xy − x 2 y 2 dy = 0

Solution: The equation is of the form

y f ( xy ) dx + x g ( xy ) dy = 0

2 2 3 3 2 2 3 3 3 3 Now, xM − yN = x y + 2 x y − x y + x y = 3 x y ≠ 0.

1 3 Therefore, x y is an I.F. 3

(1)

Basic Concepts of Differential Equations

27

1 , 3 Multiplying (1) by x y we have  1  1 2  1 0.  2 +  dx +  2 −  dy = y x y x   xy 3

This is an exact equation. Integrating, we get



1 + 2 ln x − ln y = c xy

as the required solution.

Linear Equations A first order ordinary differential equation (ODE) is linear in the dependent variable y and the independent variable x if it is or can be written in the form

dy + P ( x) y = Q ( x), dx (1) where P and Q are functions of x. Solution of a Linear Equation.

dy + P( x ) = Q( x ) The linear equation dx may be written as  P ( x ) y − Q ( x )  dx + dy = 0 (2) M dx + N dy = 0, where which is of the form M= P ( x ) y − Q ( x ) and N = 1.

∂M ∂N = P= 0 ( x ) and ∂x Now ∂y

P ( x) = 0 Thus (2) is not exact unless in which case (1) is separable. However, an integrating factor (depending only on x) of (2) may be easily µ ( x) µ ( x), found. Let be an I.F. of (2). Then multiplying (2) by we get

Differential Equations: Theory and Applications

28

 µ ( x ) P ( x ) y − µ ( x ) Q ( x )  dx + µ ( x ) dy = 0 (3) Now (3) is an exact equation if and only if ∂ ∂  µ ( x ) P ( x ) y − µ ( x ) Q ( x )  =  µ ( x )  ∂y ∂x 

This condition reduces to i.e.,

µ P ( x) =

dµ or µ

µ ( x) P ( x) =

d  µ ( x )  dx 

dµ dx

= P ( x ) dx,

Integrating, we obtain

ln µ = ∫ P ( x ) dx or

µ =∫ exp[ P ( x ) dx ] > 0. Thus

e ∫ P dx or [exp ∫ P ( x ) dx ]

is an I.F. of the linear equation (1). dy ∫ P dx e ∫ P dx P ( x) y = Q ( x ) e ∫ P dx +e ∫ P dx e , dx Multiplying (1) by we have

d  ∫ P dx  y e dx  = Q ( x ) e ∫ P dx  or dx Integrating this we obtain the solution of (1) in the form y e ∫ P dx = ∫[ Q ( x ) e ∫ P dx ] dx + c

dy 2 + 4 ( x − 1) y =+ x 1 dx Example: Solve Solution: We write the equation in the standard form dy x +1 4 y =3 + dx x − 1 ( x − 1)

( x − 1)

Here

P ( x) =

3

4 . x − 1 Therefore, an I.F. of (1) is

(1)

Basic Concepts of Differential Equations

exp[ ∫

29

4 dx 4 4 ] =exp ln ( x − 1)  =( x − 1)   x −1

Multiplying (1) by this I.F., we get

( x − 1)

4

dy 3 + 4 ( x − 1) y =x 2 − 1 dx

d  4 ±( ℵ1)  2 1   or dx integrating, we obtain y ( x − 1) = 4

x3 −x+c 3

which is the required solution.

= y ( x + 2 y ) dy dx 3

Example: Solve

dy y = 3 Solution: We have dx x + 2 y

This equation is clearly not linear in y. But in the first order differential

equation, the roles of x and y are interchangeable in the sense that either variable may be regarded as dependent variable. Let us regard x as dependent variable and y as independent variable. The equation may be written as

dx 1 − x= 2 y3 ± (1)

  1   1 1 I .F =. exp  ∫  −  dy=  exp ln =   y y .   y  which is linear in x with 1 , Multiplying (1) by y we get 1 dx 1 − x= 2y y dy y 2

d x   =2y dx  y or

Differential Equations: Theory and Applications

30

x = y2 + c = x y y2 + c Integrating, we have y or solution. Example: Solve the initial value problem

(

dr π  = + r tan θ cos 2 θ , = r  1 dθ 4 Solution. The equation is linear in r with

I .F . = exp  ∫ tan θ dθ  = exp [ ln sec θ ] = sec θ , Multiplying the given equation by sec θ , we have

sec θ

dr + r sec θ tan θ = cos θ dθ

d [ r secθ ] = cos θ dθ Integrating, we obtain r sec = θ sin θ + c Applying the initial condition, we have

π  π  = 1.sec   sin   + c 4 4 or

c= 2−

1 1 = , 2 2

Therefore,

r sec = θ sin θ +

= r sin θ cos θ +

= or 2r sin 2θ +

cos θ 2

1 2

2 cos θ is the required solution.

)

is the required

Basic Concepts of Differential Equations

31

1.2. THE BERNOULLI EQUATION dy + P ( x) y = Q ( x ) y n (1) Definition. An equation of the form dx is called the Bernoulli differential equation. This equation is linear if n = 0 or 1. If n is not zero or 1, then (1) is reducible to a linear equation. n Dividing by y , (1) becomes

y−n

dy Q ( x) + P ( x) y1− n = dx (2)

1− n In (2), put v = y then it reduces to

dv + (1 –n ) P ( x) v = (1 –n )Q ( x) dx which is linear in v. Note. Consider the equation

f ′( y)

dy + P ( x) f ( y) =Q ( x) dx

Letting in v.

v = f ( y),

dv

Q ( x ) which is linear this equation becomes dx + P ( x ) v =

1 dy xy 2 + = xy 2 dx 1 − y Example: Solve (1)

1 2

Solution: Dividing by y , (1) becomes

y



1 2

1 dy x + y2 = x dx 1 − x 2 (2) 1

2 Put y = v

1 − 12 dy dv y = dx dx or 2 then (2) reduces to dv x x + .v= 2 dx 2 1 − v 2

(

)

(3)

Differential Equations: Theory and Applications

32

This is linear in v.

 x I .F . = exp  ∫  2 1 − x 2

)

  1  dx  = exp  − ln 1 − x 2  =− 1 x2  4  

Multiplying (3) by

(

1− x 2

(

( 1− x ) 2



1 4

dv + dx

(

x

(

2 1− x 2

)

5 4

)



1 4

x . v= 2 1− x 2

(

(

v 1− x 2

or

v =c 1 2

)

)

1 − 4

(

(

1 1− x = − 3 4 4

( 1− x ) 2

1 4



2

)

3 4

)

1 4

, we get

1 −  d  1  2 4 v= −  − 2 x 1− x 2  1− x dx  4   or Integrating, we have

(

) (



)

)



1 4

1 4

  

+c

1− x 2 3

1 4

1 − x2 y =− c 1 x − 3 or is the required solution of (1).

(

2

)



1.2.1. Equation Solvable for p 2 2 2 0. (1) Example: Solve x p + xp − y – y =

(x

Solution: we factorize the left hand side of (1) to obtain 2

)

p 2 − y 2 += ( xp – y )

+ y +1 ) ( xp – y )( xp=

0

Therefore, either

xp – y = 0 (2) or xp + y + 1 =0 (3)

Basic Concepts of Differential Equations

dy =y (2) gives dx or y = cx or x

From (3), we have

33

dy dx = dx x . (4) xp = − ( y +1 )

dy dx dy = − = − ( y +1 ) x or dx or y + 1 x ( y +1 ) = c. which yields (5) Combining (4) and (5), the required solution of y – 0. ( cx )( xy + x – c ) =

x

Example: Solve

(

)

(

)

(1)

xp 3 − x 2 + x + y p 2 + x 2 + xy + y p – xy = 0

is

(1)

Solution: By inspection, we find that p − 1 is a factor of left hand side of (1). Thus, the given equation is

( p –1 )  or

(

)

xp 2 − x 2 + y p + xy  = 0

( p –1 )( xp – y )( p – x ) = 0. Therefore, either

p –1 = 0 (2) or xp – y = 0 (3) or p – x = 0 (4) From (2), we have

dy = 1 or y = x + c dx (5) From (3), we get dy = x y= or y cx dx (6) From (4), we obtain

dy =x dx

Differential Equations: Theory and Applications

34

x2 y= +c or x 2 + 2 ( y – c ) = 0. 2 or (7) The general solution of (1) is obtained by combining (5), (6) and (7). y – x – c )( y – cx ) x 2 − 2 y + 2c = 0 ( Thus is the general solution of (1).

(

)

1.2.2. Equations solvable for y y p 2 x + p (1) Example: Solve= Solution. Differentiating (1) w.r.t.x, we have dy dp dp = p= p 2 + 2 xp + dx dx dx dp 0 ( 2 px + 1) + p 2 − p = or dx dp − p ( p –1 ) = dx 2 px +1

dx −2 px − 1 = dp p ( p − 1) dx 2 x 1 + = − dp p –1 p ( p –1)

i.e.,

which is linear in x.

(2)

 2  2 2 I .F . =exp  ∫ dp  =exp ln ( p − 1)  =( p − 1) .    p −1  Multiplying (2) by

( p − 1)

x ( p − 1) = c − p + ln p

2

and integrating, we get

2

x=

c − p + ln p

( p − 1)

2

(3)

.

Substituting this value of x into (1), we have

Basic Concepts of Differential Equations

= y p2.

c − p + ln p

( p − 1)

2

35

+p

y ( p − 1)= p 2 ( c − p + ln p ) + p 2

(4) Thus, (3) and (4) constitute the required solution of (1) with p as a parameter.

p 2 x 4 (1) Example: Solve y + px = Solution. From (1), we get

= y p 2 x 4 − px Differentiating the above equation w.r.t.x, we have

dy dp dp = p = 4 x3 p 2 + 2 px 4 − p−x dx dx dx 2 p − 4 x3 p 2 − 2 x 4 p

(

dp dp +x = 0 dx dx

) (

2 p 1 − 2 px3 + x 1 − 2 px3

= 0 ) dp dx

 0 (1 − 2 px )  2 p + x dp = dx  2

3 0 Hence, either 1 − 2 px =

dp = 0 dx or (2) The equation (2) gives dp dx +2 = 0 p x 2p + x

ln c or ln p + 2 ln x = 2

or px = c or

p=

c x2

Differential Equations: Theory and Applications

36

Substituting this value of p into (1), we obtain xy − c 2 x + c = 0

= y c2 −

c x or

which is the required solution.

We have yet to deal with relation (2). If we eliminate p from (1) and (2), we get

1 4 x 2 (3) It is easy to check that (3) is also a solution of (1) and this solution does not involve any constant. y=−

Equation Solvable for x 3 Example: Solve xp = 1 + p (1)

1 +p p Solution: We have (2) w . r . t . y , Differentiating (1) we have dx 1 1 dp dp = = − 2 + dy p p dy dx x=

1  1  dp = 1 − 2  p  dy or p   1  dp = 1  p−  p  dy  or

Which is a separable equation. Therefore,

 1 dy =  p −  dp p  Integrating, we get

c+ y=

p2 − ln p 2

p 2 − 2 ln p − 2c (3) i.e., 2 y = Thus, (2) and (3) constitute the solution of (1).

Basic Concepts of Differential Equations

37

1.2.3. Clairaut’s Equation Definition. The equation

= y xp + f ( p ) ,

(1)

is known as Clairaut’s equation. Differentiating both sides of (1), w.r.t.x, we get

p =p + xp′ + f ′ ( p ) . p′ Cancelling like terms, we have

xp′ + p′f ′ ( p ) = 0 or

p′  x + f ′ ( p )  = 0

Since one of the factors must be zero, two different solution arise. • If p′ = 0, then p = c and substitution of this value into (1) yields the general solution

= y cx + f ( c ) •

x + f ′( p) = 0, x = − f ′( p) If then and (1) can be rewritten as ′ y= − pf ( p ) + f ( p )

Thus x and y are both expressed as functions of p and we obtain the parametric equations

 x = − f ′( p)  = y f ( p ) − p f ′ ( p ) 

(2)

of a curve, representing a solution of (1). The solution (2) is called the singular solution. This solution is not deducible from the general solution p may be eliminated between the two equations in (2) to get a relation in

x and y involving no constant. Example: Find the general solution and singular solution of 1 = y xp + p 4 4 (1)

1 y cx + c 4 . Solution. The general solution of the equation is = 4 Differentiating (1) w.r.t.x, we have

Differential Equations: Theory and Applications

38

p =p + xp′ + p 3 p′ 3 or x = − p (2) We eliminate p from (1) and (2) and get 1

y= x ( − x )3 +

4 1 1 3 − x ( − x )3 ( – x)3 = 4 4

3 4 0 is the singular solution. or 64 y + 27 x =

x 2 ( y − px ) = yp 2 (1) Solution: This is not Clairaut’s equation. Let us write Example: Solve 2 x 2 = u and y = v

Then, 2 x dx = du and 2 y dy = dv.

y dy dv = Hence x dx du dy x dv = p = dx y du Substituting this value of p into (1), we get

 x 2 dv  x 2  dv  x y− =   y du  y  du  

2

2

2

or

y 2 − x2

dv  dv  =   du  du  2

dv  dv  v −u =   du  du  or 2 dv  dv  = v u +  du  du  (2) or This is Clairaut’s equation. Hence its general solution is

= v cu + c 2

Basic Concepts of Differential Equations

39

y 2 cx 2 + c is the general solution of (1). Differentiating (2) w.r.t u , or = we get dv d 2 v dv dv d 2 v = u 2+ +2 du du du du du 2

d 2v  dv  u+2 = 0, 2  v cu + c 2 , which is the general solution. If du du   or then = dv u+2 = 0, du then

u = −2

dv du (3) 2

dv  dv  = v u +  du  du  (4) And give the singular solution. dv Eliminating du from (3) and (4), we obtain u2 v= − 4 2 0 is the singular solution. or u + 4v =

2 2 Replacing u , v by x and y respectively, we have



2 y= cx 2 + c 2 as the general solution.



x4 + 4 y 2 = 0 as the general solution.

1.2.4. Envelope f ( x, y , c ) = 0 Let be a one-parameter family of curves. Suppose all members of the family of curves are drawn for various values of the parameter c arranged in order of magnitude. The two curves that correspond to two consecutive values of c will be designated as neighboring curves. The

40

Differential Equations: Theory and Applications

locus of the ultimate points of intersection of neighboring curves is called f ( x, y, c ) = 0. the envelope of the family Let A, B , C represents three neighboring intersecting members of the

family. Let P be the point of intersection of Aand B and Q be the point of intersection of B and C . By definition P and Q are points on the envelope. Thus the curve B and the envelope have two contiguous points common, and therefore, have ultimately a common tangent. Hence B and the envelope touch each other. In the same way, we may show that the envelope touches any other member of the family. Thus, a curve E which, at each of its points, touches someone of the curves of the family is the envelope of the family. For the family of circles with centre on the x − axis and radius 1 , that

( x − 1) is, the family of circles with equation

2

+ y2 = 1,

The pair of lines y = ±1 is the envelope. It is easy to see that the lines

y = ±1 touch each member of the family of the circles.

Equation of the Envelope To find an equation of the envelope of the family two neighboring curves and

f ( x, y, c ) = 0,

f ( x, y, c ) = 0   0  f ( x, y , c + h ) =

consider

(1) Find the intersection of these two curves and let h → 0. The point of intersection then must approach the point of contact of the curve f ( x, y , c ) = 0 with the envelope. At the point of intersection of the equation

f ( x, y , c + h ) − f ( x, y , c ) h

=0

(2)

is true as well as (1). Letting h → 0, we have from (2)

f c ( x, y , c ) = 0 and from (1) f ( x, y , c ) = 0

(3)

(4)

Basic Concepts of Differential Equations

41

If we eliminate c from (3) and (4), we obtain an equation of the envelope. f ( x, y , c ) = 0 The eliminant is called the c - discriminant of the family . The c-discriminant may contain loci other than the envelope. Example: The family of parabolas envelope.

( x − c)

2

− 2y = 0 has the x-axis as

Singular Solution Let

f ( x, y , p ) = 0

(1)

be a nonlinear first order differential equation in which the left-hand side is a polynomial in p. The general solution of this differential equation will be a one parameter family

f ( x, y , c ) = 0

(2) The envelope E of (2) is a curve which, at each of its points, touches some one member of the family (2). At a point of contact P of the envelope and a member of (2), the values of x, y, p are the same. But the values of x, y, p for the curve at P satisfy (1). Thus, the envelope of family (2) is a solution of the differential equation (1). 2 0. Example: Consider p − xp + y =

This is Clairaut’s equation and its general solution (replacing simply p by c) is

= y cx + c 2 (1) Now we find the envelope of the family (1)

f ( x, y, c ) ≡ y − cx + c 2 = 0 (2) ∂f f c ≡ =− x + 2c =0 ∂c (3) Substituting the value of c from (3) into (2), we have 2

x x y − x  +   = 0 2 2

Differential Equations: Theory and Applications

42

i.e.,

y−

x2 x2 + = 0 2 4 2

or 4 y = x . (4) is an equation of the envelope of the family (1). We check whether (4) is a solution of the given differential equation. dy = 4 2= x or 2 p x. w . r . t . x , Differentiating (4) we get dx (5) 2 Substituting (4) and (5) into p − px + y, we have

x2 x2 x2 − + = 0 4 2 4 Thus (4) is a solution of the given differential equation. This solution does not involve any constant and it cannot be derived from the general solution by giving particular values to c, such a solution is called a singular solution (S.S). Definition A solution of a differential equation a singular solution (S.S) if

f ( x, y , p ) = 0

is called



It is not derived from the general solution by giving any particular value to the arbitrary constants • At each of its points, it is tangent to some member of the one parameter family of curves represented by the general solution. We have seen that the envelope, if any, of the one parameter family of curves represented by the general solution of a differential equation is a singular solution. Thus the c- discriminant equation may contain singular solution, if any The singular solution may also be obtained from the differential equation directly without finding the general solution. Since at ultimate point of intersection of neighboring curves the p ' s for the intersecting curves

become equal, and thus the locus of the points where p′s have equal roots will include the envelope. If we eliminate p from

∂f f (= x, y, p ) 0= and 0 ∂p

Basic Concepts of Differential Equations

43

The resulting equation is called the p-discriminant. The p-discriminant f ( x, y , p ) = 0 represents the locus for each point of which has equal roots.

f ( x, y , p ) = 0 If envelope of the general solution of exists, it will be contained in the p-discriminant. Thus, the p-discriminant equation may contain (i) singular solution (ii) Solutions that are not singular and (iii) such loci that are not solutions at all. 2 0 (1) Example: Solve xp − 2 yp + 4 x =

yp xp 2 + 4 x Solution: 2= 2= y xp +

4x p

Differentiating w.r.t.x, we get

2 p= x

p− or

dp 4 4 x dp + p+ − 2 dx p p dx

4  4 x  dp = x− 2  p  p  dx

(

)

2 p 2 − 4 x p − 4 dp = p p2 dx

dp dx = x or p or p = cx (2) Eliminating p from (1) and (2), we obtain

(

)

x c 2 x 2 − 2 ycx + 4 x = 0 2 2 0 (3) or c x − 2 yc + 4 = From (3), we obtain the c - discriminant as

4 y 2 − 16 x 2 = 0 or y 2 = 4 x 2 The p-discriminant is from (1)

4 y 2 = 16 x 2 or y 2 = 4 x 2

Differential Equations: Theory and Applications

44

2

Since the c-discriminant and p-discriminant are the same viz., y = 4 x and it satisfies the given differential equation, it is the singular solution.

x 2 p 2 + yp ( 2 x + y ) + y 2 = 0 (1) = y u= , xy v and find the singular solutions. By making substitution Example: Solve

y= u , x=

Solution: Let

dy dx = 1,= du Thus du = p

u.

v v = y u

dv −v du u2

dy dy du u2 = . = dx du dx u. dv − v du Substituting into (1), we have 2

   v2  u 2 u2 2 v u + 2 + .   dv u 2  u. dv − v  u. − v du  du 

(

)

2

 dv   dv  v + 2v + u  u − v  +  u. − v  = 0 du du     or 2

(

2

)

2

dv dv dv  dv  v + 2uv − 2v 2 + u 3 . − u 2 v + u 2   − 2uv + v2 = 0 du du du  du  2

dv  dv  = v u +  du  du 

2

which is a Clairaut’s equation. Its solution is

= v cu + c 2

= cy + c 2 (2) i.e., xy is the solution of (1). From (2), the c-discriminant is

2

Basic Concepts of Differential Equations

45

0 y 2 + 4 xy = 0 i.e., y ( y + 4 x ) =

Therefore, y = 0, y + 4 x = 0 (3)

Differentiating (3), we have p = −4. Substituting p = −4 into the lefthand member of (1) and using (3), we obtain

16 x 2 − 4 x ( −4 )( −2 x ) + ( −4 x= x2 0 ) 32 x 2 − 32= 2

0 is also a singular solution. Thus y + 4 x =

(x

(x 2y =

2

)

−1 p2 − x2 xp

2

)

− 1 p 2 − 2 xyp − x 2 = 0 (1) Moreover, find the singular solution, if any. Solution: Solving the equation for y, we have Example: Solve

= xp −

p x − . x p

Differentiating w.r.t.x, we get xp′ − p p − xp′ dp 2 p = xp′ + p − − , p′ = 2 2 x p dx where

 1 x  p 1 = p p′  x − + 2  + 2 − x p  x p  or   1 1  1 1  p 1 − 2 + 2 = xp′  1 − 2 + 2  x p  x p   or  dp dx = = or p cx. p x Therefore, Substituting this value of p into (1), we get

(x

2

)

− 1 c 2 x 2 − 2 x 2 yc − x = 0

(x or

2

)

− 1 c 2 − 2 yc − 1 =0

(2)

as the general solution of (1). From (2), the c-discriminant is

(

)

4 y 2 + 4 x2 −1 = 0

Differential Equations: Theory and Applications

46

2 2 1 i.e., x + y =

which is also envelope of the family (2). We also note that p − discriminat is

(

)

4x2 y 2 + 4x2 x2 −1 = 0 2 2 1 i.e., x + y = Since the envelope and the p-discriminant are same, the singular solution is

x2 + y 2 = 1 Example: Find the singular solution of 3

x p 2 + x 2 yp + a 3 = 0 (1) Solution: The p − discriminant of (1) is

(

)

x 4 y 2 − 4= x 3 a 3 0 or x 3 y 2 x − = 4a 3 0

(2)

The parts of (2) that satisfy (1) are singular solutions of the given equation. From (2), we have 3 = x 0 or y 2 x − 4a= 0 (3)

We rewrite (1) as 2

 dx  dx x + x y + a3   = 0 dy  dy  (4) Clearly, x = 0 satisfies (4). 3

2

Thus x = 0 is a singular solution of (1). 2 3 0 w.r.t x, we have Differentiating y x − 4a =

2 xyp = + y 2 0 or y ( 2 xp = + y) 0

y . 2x i.e., Putting this value of p into the left-hand member of (1) and using (3), we obtain p=−

Basic Concepts of Differential Equations 2

47

y  3 xy 2  y  2  x−  + x y−  + a = − + a3 = − xy 2 − 4a 3 = 0 4  2x   2x 

(

)

2 3 0 and its derivative satisfy (1) and so it is a solution Thus xy − 4a =

of (1). Hence

= x 0 and xy 2 − = 4a 3 0 are singular solutions of (1).

1.2.5. The Ricatti Equation Definition. We have already studied first order linear differential equation

y′ + P ( x ) y = R ( x )

(1) q ( x) y2 If we add the term to the left-hand member of (1), we obtain a nonlinear differential equation

y′ + P ( x ) y + Q ( x ) y 2 = R ( x) (2) (2) is called the Ricatti equation. Note: In many cases, the solution of (2) cannot be expressed in terms of 2 elementary functions. However, the Ricatti equation y′ + Py + Qy = R (1) 1 y= y1 + , u where y1 can be reduced to a linear equation by the substitution is a particular solution of (1) and u is an unknown nonzero function of x . 1 y= y1 + u be as given. Differentiating w.r.t.x, we have Proof: Let dy dy1 1 du = y′ = − dx dx u 2 dx Substituting for y and y′ into (1), we get 2y dy1 1 du 1 1    − 2 + P  y1 +  + Q  y12 + 1 + 2  = R dx u dx u u u   

dy1 1  du  + Py1 + Qy12 − R − 2  − Pu − 2Qy1u − Q  = 0 dx u dx   or (2) Since

y1 is a solution of (1), we have

Differential Equations: Theory and Applications

48

dy1 + Py1 + Qy12 − R = 0 dx du − ( P + 2Qy1 ) u = Q and so (2) reduces to dx which is a linear equation. Thus if

a particular solution of (1) is known then its general solution can be found. dy − y2 = −1, y ( 0 ) = 3, dx

Example: Solve solution of the given equation.

given that y1 = 1 is a particular

0, Q = −1, R = −1. Solution: Here P = 1 y = 1+ , u the given equation reduces to Writing du − ( 0 + 2 ( −1)(1) ) = −1 dx

du + 2u = −1, i.e., dx (1)

which is a linear equation. ∫ 2 dx = e2 x . I.F. of (1) is e

Multiplying (1) by the I.F., we get

du 2 x e + 2u e 2 x = −e 2 x dx d ue 2 x = −e 2 x or dx Integrating, we obtain

(

)

u e 2 x =− ∫ e 2 x dx + c =−

e2 x +c 2

1 c u= − + 2x 2 e (2) or 1 y = 1+ , u we get by the initial condition, From

1 1 y ( 0) = 3= 1+ or u ( 0 ) = u (0) 2

Basic Concepts of Differential Equations

49

Hence from (2), we have

1 1 u (0) = = − + c or c = 1 2 2 1 1 2 – e2 x u= − + 2x = 2x 2 e 2e Thus 1 2e 2 x 2 + e2 x y =1 + =1 + = u 2 − e2 x 2 − e2 x . Required solution is dy y − − x3 y 2 = − x5 , Example: Solve dx x (1) by finding a particular solution.

1 P= − ,Q = − x3 and R = − x5 . x Solution: It is a Ricatti equation with 1 y= x + y1 = x. u into (1), An obvious solution of (1) is Substituting we have du  1  −  − + 2 − x 3 x  u =− x 3 dx  x 

(

)

du  1  +  + 2x4  u = − x3 ,  or dx  x (2) which is a linear equation. 1  ∫  + 2 x 4  dx x 

ln x

2 5 x 5

= I .F . e = e= .e xe Its

2 5 x 5

Multiplying (2) by the I.F. and integrating, we have 2

ux e 5

x5

2 5 x 1 2 x5 = − ∫ x 4 .e 5 = dx + c′ − e 5 + c′ 2

2

1 x5 c′ − e 5 2 u= 2

x5

x e5 or Hence the general solution of (1) is

Differential Equations: Theory and Applications

50

2 5 x 5

xe y= x+ 2 1 x5 c′ − e 5 2

2  1 5 x5  ′ x c + e  2   = 2 5 x 1 c′ − e 5 2

2

1 x5 c′ + e 5 y 2 = 2 x 1 5 x5 c′ − e 2 or 2

x5

2 5 x y − x e5 5 = = ce or y + x 2c′ is the required solution.

1.3. DIFFERENTIAL EQUATIONS OF HIGHER ORDER 1.3.1. Introduction In previous sections, we studied methods of solving special types of linear and nonlinear differential equations of the first order. In this chapter, we shall consider systematic methods for the solution of certain classes of differential equations of order more than one.

1.3.2. Linear Differential Equations Definition. A linear differential equation of order n in the dependent variable y and the independent variable x is of the form

a0 ( x )

dny d n −1 y dy a x + +…+ an −1 + an ( x ) y = F ( x ) , ( ) 1 n n −1 dx dx dx

Where

a0 ( x ) , a1 ( x ) ,…, an −1 ( x ) , an ( x ) and F ( x )

(1)

are functions of

a ( x) the independent variable x only and 0 is not identically zero. Using primes, (1) is also written as a0 y n + a1 y n −1 +…+ an −1 y′ + an y = F ( x) where

(2)

a0 , a1 , …, an −1 , an are real constants.

Basic Concepts of Differential Equations

51

Equation (1) is with variable coefficients while (2) is with constant coefficients. In order to solve (2), we shall first consider the equation

a0 y n + a1 y n −1 +… + an −1 y′ + an y = 0 (3)

a.

n

The coefficient of y may be made 1 by dividing throughout by 0 The differential equation (3) is called homogeneous linear differential equation of order n. The use of the word homogeneous here is quite different from F ( x) the one already mentioned in previous chapter. If is not identically zero then (2) is called non - homogeneous and (3) is called the associated homogeneous equation of (2). Definition. If

y1 ( x ) , y2 ( x ) , …, ym ( x )

are m functions of an

c , c , …, cm independent variable x and 1 2 are constants, then the expression c1 y1 ( x ) + c2 y2 ( x ) +…+ cm ym ( x ) is called a linear combination of y1 ( x ) , y2 ( x ) , …, ym ( x ) . y , y , … , ym We usually write 1 2 instead of y1 ( x ) , y2 ( x ) ,…, ym ( x ) y , y , … , ym when it is clear from the context that 1 2 are functions of x. As in vector spaces, the m nonzero functions

y1 , y2 , …, ym

linearly dependent if and only if, there exist constants one of which is nonzero, such that

are called

c1 , c2 , …, cm

at least

c1 y1 + c2 y2 +…+ cm ym = 0. The functions

y1 , y2 , …, ym

are called linear independent if and only if,

they are not linearly dependent, i.e., if and only if Implies

c1 y1 + c2 y2 +…+ cm ym = 0

c1 = c2 = … = cm = 0.

Now before investigating a solution of (2), we state the following facts: i. Every homogeneous linear nth − order differential equation

a0 y n + a1 y n −1 +…+ an −1 y′ + an y = 0 has n linearly independent solutions y1 , y2 , …, yn ii.

If

y1 , y2 ,…, y _ m

are n linealy independent solutions of (3), then

Differential Equations: Theory and Applications

52

= yc c1 y1 + c2 y2 +…+ cn yn of y1 , y2 , …, yn c , c ,…, cm being arbitrary constants. is the general solution of (3), 1 2 any linear combination

y be any particular solution of (2). i.e., p does not contain y + yp any constant, then c is the general solution of (2). (ii) and (iii) can be easily checked by actual substitutions into (3) and (2). Thus, to find the general solution of (2), we have to find a linearly independent y y set of n solutions of (3). So as to determine c and a particular solution p = y yc + y p of (2) and then obtain (4) iii.

Let

yp

y

as the general solution of (2). In the general solution (4), c is called the y Complementary Function (C.F) and p is called the Particular Integral (P.I) of (2). These statements are also true for the equation (1) with variable coefficients.

1.3.3. Homogeneous Linear Equation Consider the equation

a0

dny d n −1 y dy a + +…+ an −1 + an y, = 0 1 n n −1 dx dx dx (1)

a , a ,…, a

n −1 are real constants. To find a solution of (1) we shall try where 0 1 a judicious guess. The differential equation (1) requires a function y with the property that if y and its successive derivatives are each multiplied by a , j= n, n − 1,…,1, 0, constants j the resulting products are then added; the sum should equal zero. This can only happen if a function is such that its various derivatives are constant multiples of itself. The exponential function

y = e mx , m being a constant, has such properties. Here we have dy d2y dny = me mx , 2 = m 2 e mx , …, n = m n e mx . dx dx dx Substituting into (1) we have

a0 m n e mx + a1m m −1e mx +…+ an −1m e mx + an e mx = 0 or

e mx (a0 m n + a1m n −1e mx +…+ an −1m e mx + an e mx = 0

Basic Concepts of Differential Equations

Since e ≠ 0, we have a0 m n + a1m m −1 +…+ an −1m + an = 0

53

mx

(2)

Thus, y = e is a solution of (1) if and only if m is a solution of (2). Equation (2) is called the characteristics (or auxiliary) equation of the given differential equation (1). Observe that (2) can be obtained from (1) m k ( k= n, n − 1, …, 3, 2,1) . by merely replacing the kth derivative in (1) by Three cases arise according as the roots of (2) are mx

i. ii. iii.

Real and distinct Real and repeated Complex

Case I. Distinct Real Roots m , m , …, m

, e ,…, e n be n distinct real roots of (2). Then e Let 1 2 are n distinct solutions of (1). These n solutions are linearly independent. = y c1e m1x + c2 e m2 x +…+ cn e mn x , where Hence the general solution of (2) is m1 x

m2 x

mn x

c1 , c2 , …, cn are arbitrary constants.

Case II. Repeated Real Roots In equation (1), writing

(a D 0

n

D≡

d , dx we have

)

+ a1 D n −1 +…+ an −1 D + an y = 0

where

 f ( D )  y = 0, or 

f= ( D ) a0 D n + a1D n−1 +…+ an−1D + an

Note that if we write D for m in the characterstic equation (2) then it f ( D ) = 0. m1 , m2 , …, mn f ( D ) = 0, is the same as If are the roots of then (1) may be written as

0. ( D − m1 )( D − m2 )…( D − mn ) y = Let the root

be repeated twice say

m2 = m1.

The part of the general solution of (1) corresponding to the twice repeated

0 ( D − m1 ) y = m root 1 of (2) is solution of 2

Differential Equations: Theory and Applications

54

0 ( D − m1 )( D − m1 ) y = (3) V. ( D − m1 ) y = Let

i.e., of

Then (3) becomes

0 ( D − m1 )V =

dV − m1V = 0 or dx (4) Separating the variables, we have dV = m1dx. V

= V m1 x + k , where k is a constant.

Therefore, in

V = c2 e m1x , where c2 is a constant. Replacing V in (4), we obtain

c2 e m x ( D − m1 ) y = 1

dy − m1 y = c2 e m1x dx (5)

(

)

= exp ∫ (−m1 dx = ) e − m1x Which is a linear equation of order one. Its I.F. −m1 x Multiplying (5) by e , we get

d ye − m1x = c2 dx

(

)

1x y e − m= c2 x + c1

= y

( c1 + c2 x ) em x 1

twice repeated

= y

is the part of the general solution corresponding to the

m1. The general solution of (1) is

( c1 + c2 x ) em x + c3e m x +…+ cn em x 1

3

n

Basic Concepts of Differential Equations

55

In the same manner, if the characteristic equation (2) has the triple repeated root the solution of

( D − m1 )

3

m1 ,

the corresponding part of the general solution of (1) is

y= 0

Proceeding as before, we can easily find

(

)

y = c1 + c2 x + c3 x 2 e m1x As the part of the general solution corresponding to this triple repeated

m.

m

root 1 If the characteristics equation (2) has the real root 1 occurring k times then the part of the general solution of (1) corresponding to the

k − fold repeated root m1 is

= y

(c + c x + c x 1

2

3

2

)

+…+ ck x k −1 e m1x

Case III. Complex Roots Suppose the characteristic equation has the complex number a + ib as a non- repeated root. Since coefficients of (1) are real, the conjugate complex number a − ib is also a non- repeated root. The corresponding part of the general solution is

= y k1e( a + ib ) x + k2 e( a −ib ) x , where k1 and k2 are arbitrary constants.

y e ax  k1eibx + k2 e − ibx  = = e ax  k1 ( cos bx + i sin bx ) + k2 ( cos bx − i sin bx )  = e ax ( k1 + k2 ) cos bx + i ( k1 − k2 ) sin bx  = e ax [ c1 sin bx + c2 cos bx ] where

c1 = i ( k1 − k2 ) , c2 = k1 + k2

are two arbitrary constants.

If a + ib and a − ib are conjugate complex roots, each repeated k times, the the corresponding part of the general solution of (1) may be written as

Differential Equations: Theory and Applications

56

(

)

(

)

= y  c1 + c2 x + c3 x 2 +…+ ck x k −1 sin bx + ck +1 + ck + 2 x +…+ c2 k x k −1 cos bx 

In the examples given below we use these concepts to solve homogeneous linear differential equations. Example: Solve

(D

2

)

+ 4D + 3 y = 0

2 0 with roots Solution: The characteristic equation is D + 4 D + 3 =

D= −1, −3.

= y c1e − x + c2 e −3 x . Hence the general solution of the given equation is

(D Example: Solve

3

)

− 5D 2 + 7 D − 3 y = 0. 3

2

0 (1) Solution. The characteristic equation is D − 5 D + 7 D − 3 = By inspection, D = 1 is a solution of this equation. The other two roots

D − 1) ( D 2 − 4 D + 3) = 0 ( can be found from the quadratic factor of Hence all roots of (1) are D = 1,1,3. The general solution is

(D Example. Solve

3

y =+ ( c1 c2 x ) e x + c3e3 x .

)

− D2 + D −1 y = 0.

3 2 0. Solution. The characteristic equation is D − D + D − 1 = By inspection, D = 1 is a root of this equation.

Now,

D 3 − D 2 + D − 1=

( D − 1) ( D 2 + 1)=

0.

Hence the other two roots are D =±i =a + ib with a =0, b =1. The general solution is

y= c1e x + e0 x ( c2 sin x + c3 cos x ) y= c1e x + c2 sin x + c3 cos x.

(

)

= D 2 + D − 12 y 0, where = y ( 2 ) = 2, y′ ( 2 ) 0. Example: Solve

Basic Concepts of Differential Equations

57

2 0 with roots Solution. The characteristic equation is D + D − 12 =

D= 3, −4.

y c1e3 x + c2 e −4 x (1) Hence the general solution of the given equation is=

c1 , c2 in (1) from the initial conditions as follows. Since the initial conditions are given at x = 2, We now determine the co-efficient (constants)

we rewrite the general solution in the form

= y k1e3( x − 2) + k2 e −4( x − 2)

(2)

6

= k c= e , k2 c2 e −8 . where 1 1 Applying initial condition, ( i.e. replacing 6 k1 e , k2 e −8 respectively in (1), we get

c1 , c2 by the new constants

y ( 2 )= 2= k1 + k2

(3) Differentiating (2) w.r.t.x, we have

y′ =

dy = 3k1e3( x − 2) − 4k2 e −4( x − 2) dx Therefore,

= y′ ( 2 ) 0 = 3k1 − 4k2

(4) 8 6 = k1 = , k2 . 7 7 Solving (3) and (4), we obtain

Hence the solution (2) of the differential equation satisfying the given conditions is

= y

8 3( x − 2) 6 −4( x − 2) . e + e 7 7 Example. Solve

(D

2

)

+ 4 D += 5 0, y (= 0 ) 1, y′ ( 0 ) = 0.

2 0 which has Solution. The characteristic equation is D + 4 D + 5 = roots D =−2 ±i.

= y e −2 x ( c1 sin x + c2 cos x ) . The general solution is (1)

Differential Equations: Theory and Applications

58

Applying the given conditions, we have from (1)

y ( 0 )= 1= c2 Differentiating (1) w.r.t.x, we get

y′ = −2 e −2 x ( c1 sin x + c2 cos x ) + e −2 x ( c1 cos x − c2 sin x ) Therefore,

y′ ( 0 ) = 0 = −2c2 + c1 ,

giving

Substituting the values of

= y e

−2 x

c1 = 2.

c1 and c2

into (1), the required solution is

( 2sin x + cos x ) .

(D

Example. Solve

3

)

− 3D 2 + 4 y= 0, y ( 0= −4. ) 1, y′ ( 0=) −8, y′′ ( 0 ) =

3 2 0. Solution. The characteristic equation is D − 3D + 4 =

0. ( D + 1) ( D 2 − 4 D + 4 ) = Therefore, D = −1, 2, 2 The general solution is Now, from (1)

y =c1e − x + e 2 x ( c2 + c3 x )



(1)

y′ = −c1e − x + 2e 2 x ( c2 + c3 x ) + c3e 2 x

y′′ = c1e − x + 4e 2 x ( c2 + c3 x ) + 4c3e 2 x Applying initial conditions, we get from the above three equations

y ( 0 ) = 1 = c1 + c2

(2)

y′ ( 0 ) =−8 =−c1 + 2c2 + c3

(3)

y′′ ( 0 ) =−4 =c1 + 4c2 + 4c3

(4) Multiplying (3) by -4 and adding to (4), we have

28 = 5c1 − 4c2

(5)

Multiplying (2) by 4 and adding to (5), we obtain

= 9c1 32 = or c1

32 . 9

Basic Concepts of Differential Equations

Therefore,

c2 = 1−

59

32 23 6 = − and c3 =. 9 9 9

The required solution is

32 − x 2 x  23 6  1 32 e − x − 23 e 2 x + 6 x e 2 x . e + e  − + x=  9  9 9  9

(

= y

)

Differential Operators Let T :V →V , where V is a vector space over a field F , be a linear transformation (or an operator) from V toV . Linear transformation S ,T , U defined on V have the following properties. i.

S + T = T + S where ( S + T ) v = S ( v ) + T ( v ) , v ∈ V , ( S + T ) +U = S + ( T +U ) S ( T + U ) =ST + SU



( ST )U = S (TU )

ii.

ST ≠ TS in general.

( ST )( v ) = S (Tv ) Here ST : V → V is a linear transformation defined by: for all v ∈ V . Let X be the vector space of all real (or complex) valued real (or complex)

= k 1, 2,3, … For each functions possessing a kth order derivative for every k g ∈ X , y = g ( x ) ∈ R ( or ∈ C ) .D which associates with each y = g ( x ) dk y D k y as k . d x Since , its kth derivative, is a linear operator on X . We write the sum and product of the linear operator and scalar multiple of a linear operator are also linear operators, the linear combination f= ( D ) a0 D n + a1D n−1 +…+ an−1D + an I

(1)

n n −1 Of D , D , …, D, I is also a linear operator. Here, I is the identity = I ( y ) y= for all y g ( x ) a , a , …, an are linear operator defined by and 0 1

Differential Equations: Theory and Applications

60

scalars. The image of a

y = g ( x)

f= ( D ) y a0 D y + a1D n

= a0

n.

n −1

under

f ( D)

is written as

y +…+ an −1 Dy + an y

dny d n −1 y dy a + +…+ an −1 + an y. 1 n n −1 dx dx dx

The equation

f ( D ) y = 0.

is then a linear differential equation of order

D, D 2 , …, D n and f ( D ) ,

The operator

given by (1), are called d d2 D for , D 2 for 2 dx dx and so on differential operators. We have written n 2 d dny dy 2 d y n D n for n . = Dy = ,D y D y = . dx Thus dx dx 2 and dx n

= A f (= D ) a0 D n + a1 D n −1 +…+ an −1 D + an I The expression is called a differential operator of order n. This may be thought of as an operator which, when applied to any function y , result in Ay f = = ( D ) y a0

a dy dny d n − 1y a + +…+ n −1 + an Iy. 1 n n −1 dx dx dx

Nonhomogeneous linear Equations In this section we discuss the solution of nonhomogeneous linear equation a0 y n + a1 y n −1 +…+ an −1 y′ + an y = F ( x) .

Solution of the equation f ( D) y = F ( x) where



f= ( D ) a0 D + a1D n

n −1

(1)

+…+ an −1 D + an .

We have already discussed that the general solution of (1) consist of two parts, namely i. ii.

Complementary Function (C.F) Particular Solution (P.I)

Basic Concepts of Differential Equations

61

f ( D) y = 0 The C.F is the solution of the homogeneous equation and P . I we have described different cases of its solution. Now to find the of (1), we can write y=

1 F ( x) f ( D)

And try to evaluate

y=

1 F ( x). f ( D)

f ( D) f ( m ) = 0. A real number m is said to be a zero of if Suppose Then

f ( D)

has n distinct zeros

m1 , m2 ,…, mn .

1 1 F ( x) = f ( D) ( D − m1 )( D − m2 ) …( D − mn

Suppose

)

F ( x)

.

1 F ( x ) = y* . D − mn

dy F ( x) − mn y* = D − mn ) y = F ( x) ( Then or dx which is a linear I .F = exp ∫ ( −mn ) dx = exp ( − mn x ) . equation of the first order. Its Hence d y e − mn x = F ( x ) e − mn x dx *

(

(

)

)

= y* e mn x ∫ F ( x ) e − mn x dx, Omitting the constant of integration, since we are looking for a particular solution, Therefore

1 F ( x) f ( D)

1 e mn x ∫ e − mn x F ( x ) dx. ( D − m1 )…( D − mn−1 )

1 e mn x ∫ e − mn x F ( x ) dx – D m n −1 Next, we evaluate As before and continue the process. This is the required P.I of the equation (1).

Differential Equations: Theory and Applications

62

Alternatively, we may also write

1 1 F ( x) = F ( x) f ( D) ( D − m1 )( D − m2 )…( D − mn )

 A1 An  A2 =  + +…+  F ( x) D − mn   D − m1 D − m2 after resolving

1 f ( D)

into partial fractions

= A1 e m1x ∫ e − m1x F ( x ) dx + A2 e m2 x ∫ e − m2 x F ( x ) dx +… An e mn x ∫ e − mn x F ( x ) dx

,

which is the required P.I of the equation (1).

(D

Example: Solve

3

)

−D y = ex .

3 0 with roots 0 ,1 , −1 Solution. The characteristic equation is D − D =

C.F = yc = c1 + c2 e x + c3e − x .

P.I =

yp =

ex D ( D +1 )( D –1 )

1 x e = e x ∫ e − x .e x dx = x ex , 1 D − Now by the method already described Thus 1 1 ex = xe x D ( D +1 )( D –1 ) D ( D +1 ) Again,

1 xe x e x x x x −x xe = e ∫ e .xe dx =− D +1 2 4

(

)

Hence

1 1  x e x e x  xe x ex xe x 3 x xe x =  −  =∫ dx − ∫ dx = − e D ( D +1 ) D 2 4  2 4 2 4

( )

Therefore, the general solution of the equation is

Basic Concepts of Differential Equations

y =yc + y p =c1 + c2 e x + c3 e − x +

63

xe x 3 x − e . 2 4

3 xe x  = c1 +  c2 −  e x + c3e − x + 4 2  c1 + c2' e x + c3e − x + =

xe x 2

1.3.4. Alternative Method Consider

yp =

f ( D)

−1

as the inverse of the operator

f ( D),

so that

1 ex D ( D +1 )( D –1 )

1 1  x  1 1 1 + . y p = − + . e .  D 2 1+ D 2 D −1 

1 1 y p =− ∫ e x dx + e − x ∫ e x .e x dx + e x ∫ e − x .e x dx 2 2 1 1 xe x y p =− e x + e x + xe x = , 4 2 2 x x since e already occurs in the C.F and so the terms involving e are omitted.

1.3.5. Working Rules for finding P.I. The method of evaluating

1 F ( x) f ( D)

involves successive integrations and is F ( x) quite unwieldy. We list below short methods for finding the P.I. when is a function of a particular type. 1 e ax , f (D)

D ) a0 D n + a1 D n −1 +…+ an −1 D + an . where f (= We know that D k eax = a k eax , where k is a positive integer.

To evaluate Hence

Differential Equations: Theory and Applications

64

(a D

f ( D= ) eax

0

n

)

+ a1 D n −1 +…+ an −1 D + an e ax

= a0 .D n e ax + a1.D n −1e ax +…+ an −1.D e ax + an .e ax = a0 .a n e ax + a1.a n −1e ax +…+ an −1. a e ax + an .e ax .

= f ( a ) e ax .

(1)

Applying

f ( D) f ( D)

e ax =

e ax = f ( a ) .

1 f ( D)

to both sides of (1), we obtain

1 f ( a ) e ax f ( D) 1 e ax f ( D)

1 e ax ax = e , if f ( a ) ≠ 0 f (D) f (a) i.e., f ( a ) = 0, f ( D ) = 0. if then a is a root of k f ( D ) = 0. ( D – a ) is a factor of Let a be a k-fold root of so that f ( D ). Then,

f (= D)

( D –a )

k

∅ ( D ) , where ∅ ( a ) ≠ 0.

We first check the effect of p ( x) in x. We have

( D – a ) ( eax p ( x ) =)

(

( D –a )

k

on e ax p ( x ) , for a polynomial

)

D e ax p ( x ) − a e ax p ( x )= e ax Dp ( x )

D – a ) ( e Ax p ( x ) ) (= D – a ) ( e ax D p ( x ) ) (= 2

e ax D 2 p ( x ) .

Basic Concepts of Differential Equations

65

Continuing in this way, we get

( D –a )

k

(e

Setting

( D –a )

k

Ax

)

p ( x ) = e ax D k p ( x ) .

p ( x) = x

xe ) (= k ax

k

(3)

in (3), we are led to

ax k k e= D x k !e ax

(

)

∅ ( D ) ( D – a ) n x k e ax = ∅ ( D ) k !e ax k

= k !∅ ( D ) e ax = k !∅ ( a ) e ax ,

by (1)

1

Operating on both sides by

x k e ax

1

∅( D ) ( D −a )

1

∅ ( D ) ( D –a )

e ax =

k

k

∅( D ) ( D –a )

k

, we obtain

k !∅ ( a ) e ax

x k e ax k! ∅( a )

(4)

1.3.6. Principle of Superposition f (= D) y

(a D 0

n

)

+ a1 D n −1 +…+ an −1 D + = an y F ( x )



(1)

F= ( x ) F1 ( x ) + F2 ( x ) , Be a linear differential equation of order n. if then particular integral of (1) is the sum of particular integrals of F ( D ) y = F1 ( x ) and

(2)

F ( D) y

Proof: Let

F ( x)

y1 and y2

Then

f ( D ) y1 = F1 ( x ) and

f ( D ) y2 = F2 ( x )

(3)

be particular integrals of (2) and (3) respectively.

Differential Equations: Theory and Applications

66

y= y1 + y2 . Setting this value of y into (1), we obtain

Suppose

= f ( D ) y f ( D )( y1 + y2 ) = f ( D ) y1 + f ( D ) y2 = F1 ( x ) + F2 ( x ) = F ( x ) Showing that y is a solution of (1). Example: Solve

(D

3

)

+ 1 y =+ 1 e− x + e2 x .

Solution. The characteristic equation is 3

D + 1 =0 with roots Therefore, C.F is

−1 ,

1± i 3 2

x  3 3  yc = c1e − x + e 2  c2 sin x + c3 cos x . 2 2  

The P/I is given by

1 1+ e− x + e2 x D +1

= yp =

3

(

)

1 1 1 e0 x + 3 e 2 x + e− x 2 D +1 D +1 ( D +1 ) D – D +1

1+ =

3

e2 x 1 e− x , + 9 3 ( D +1 )

(

)

1 1 ax e ax = e , f ( D) f (a)

using 1 e − x , with D = −1 when f ( −1) ≠ 0 2 for D – D +1 e2 x 1 − x 1+ = + xe , 9 3

Basic Concepts of Differential Equations

67

Hence the general solution of the equation is

= y yc + y p x  3 3  e2 x 1 − x x + c3 cos x  + 1 + = c1e − x + e 2  c2 sin + x. e . 2 2 9 3  

F ( x ) = sin ax or cos ax.

To Find the P.I when

Here we have to evaluate

1 1 sin ax and cos ax. f ( D) f ( D)

From Euler’s theorem, we have

= e cos ax + i sin ax iax

Thus parts of

1 1 sin ax and cos ax f ( D) f ( D) 1 eiax . f ( D)

are respectively imaginary and real

If f ( D ) contains only even powers of 1  sin ax =  f D 2 cos ax that

( )

Example: Solve

( )

D, say f ( D ) = f D 2 ,

 sin ax 1 , provided − a 2 is not a zero of f D 2 . 2  f −a cos ax

(

(D

( )

)

2

)

–5 D + 6 y = sin 3 x

Solution: The roots of the characteristic equation 2

D − 5D + 6 = 0 are 2 and 3. C.F = y= c1e 2 x + c2 e3 x c

P.= I y= p

1 sin 3 x D − 5D + 6 2

1

which is imaginary part of Now

it is easy to see

( D – 2 )( D – 3 )

e3ix

Differential Equations: Theory and Applications

68

1

( D – 2 )( D – 3 ) =

e3ix =

e3ix , ( 3i – 2 )( 3i – 3 )

−1 + 5i 3ix e 78

 1 5  = − + i  ( cos 3 x + i sin 3 x )  78 78 

Its imaginary part is



1 5 sin3 x + cos 3 x = yp 78 78

Hence the required general solution is

y =c1e 2 x + c2 e3 x −

1 5 sin 3 x + cos 3 x 78 78

Theorem. If a is not zero of

f ( D)

, then

1 1 e ax F ( x ) = e ax F ( x). f ( D –a ) f ( D)

This replacing of D by D + a is known as exponential shift. Proof.

We have already shown in previous section that

( D –a )

k

(

)

. e ax p ( x ) = e ax D k p ( x )

Using linearity of differential operators, we conclude that, when is a polynomial (with constant coefficients), then

(

)

f ( D – a ) e ax p ( x ) = e ax f ( D ) p ( x ) Suppose

p ( x) =

f ( D) p ( x) = F ( x)

(1)

, then

1 F ( x) f ( D)

(2) From (1) and (2), we have

f ( D – a ) e ax

1 F ( x ) = e ax F ( x ) f ( D)

f ( D)

Basic Concepts of Differential Equations

Operating on both sides by

e ax

or

1 , f ( D –a )

69

we get

1 1 F ( x) = e ax F ( x ) f ( D) f ( D –a ) 1 1 e ax F ( x ) = e ax F ( x), f ( D –a ) f (D)

f ( D ). provided that a is not a zero of Example: Solve

(D

3

)

+ D2 − 4D − 4 y = e 2 x cos 3 x.

3 2 0. Solution. The characteristics equation is D + D − 4 D − 4 = By inspection D = 2 is a root of this equation. The other roots are

D= −2, −1

C.F . = yc = c1e − x + c2 e −2 x + c3e 2 x

.I y= P= p = e2 x

= e2 x

1 e 2 x cos 3 x ( D – 2 )( D + 2 )( D +1 )

1 cos 3 x, by the exponential shift D( D+4 ) ( D+3)

(

1 cos 3 x D 3 + 7 D 2 +12 D

)

= e2 x

1 cos 3 x, putting D 2 = −32 − 9 D – 63 +12 D

= e2 x

1 cos 3 x 3 ( D – 21 )

= e2 x

( D + 21)

(

3 D 2 – 441

)

cos 3 x

Differential Equations: Theory and Applications

70

−1 [ −3 sin 3x + 21cos 3x ] 3× 450

= e2 x .

e2 x = ( sin3x – 7 cos 3x ) 450 1 cos 3 x, D ( D + 4 )( D + 3 )

Alternative Method. Here, for follow:

(

cos 3 x 1 = Re e3ix , D 3 + 7 D 2 +12 D D 3 + 7 D 2 +12 D

)

(

= Re

1 ( − 27i − 63 + 36i )

= Re

e 3 ix 9( i − 7)

= Re = Re

)

( i + 7 ) e3ix −450

( i + 7 ) ( cos 3x + i sin 3x) −450

7 1 = − cos 3 x + sin 3 x. 450 450 Therefore,

e2 x ( sin 3x − 7 cos 3x ) 450 as before. y= yc + y p The general solution is 7 1 cos3 x + sin 3 x. = c1e − x + c2 e −2 x + c3e 2 x − 450 450

yp =

−x

= c1e + c2 e

−2 x

e2 x + c3e + 450 2x

(

sin 3 x – 7 cos 3 x ) .

we proceed as

Basic Concepts of Differential Equations

71

1.3.7. Series Expansion of Polynomials or Expressions involving Operators. This method is useful when

F ( x)

is a polynomial in x .

f ( D) y = F ( x) F ( x) Let be such that is a polynomial in x . To evaluate the particular integral

yp =

1 F ( x) f ( D) 1 f ( D)

It is often useful to expand Theorem for negative exponent so that

y p=

( 1+ a D + a D 1

2

2

in a series in D by the Binomial

)

+ a3 D 3 +… F ( x )

The derivatives on the right will vanish after certain stage since

= D n x r 0 if n > r. The following binomial expansions will be useful in this connection:

1 = 1 + x + x 2 + x 3 +…. 1– x 1 = 1 − x + x 2 − x3 +…. 1+ x Example: Solve

(D

3

)

− 2 D + 1 y = 2 x3 − 3x 2 + 4 x + 5

3 0 Solution. The characteristic equation is D − 2 D + 1 =

0 ( D –1 ) ( D 2 + D –1 ) = D =1 ,

−1 + 5 −1 − 5 , 2 2

yc = c1e x + c2 e

−1 + 2

5

x

+ c3 e

–1 − 2

5

x

Differential Equations: Theory and Applications

72

1 = 2 x3 − 3 x 2 + 4 x +5 yp 3 1− 2 D + D Next, −1 1 =1 − 2 D − D 3  3 1− 2 D + D Now

(

(

(

) (

= 1 + 2 D − D3 + 2 D − D3

)

)

) + ( 2D − D ) 2

3

3

+…

=+ 1 2 D − D3 + 4 D 2 + 8D3 , 4 Neglecting D and higher powers of D in view of the degree 3 of the p ( x) polynomial in x therefore

(

y p = 1+ 2 D + 4 D 2 + 7 D3

(

)( 2 x

3

− 3x 2 + 4 x + 5

)

)

= 2 x3 − 3 x 2 + 4 x + 5 + 2 6 x 2 − 6 x + 4 + 4 ( 12 x − 6 ) + 7 (12 )

= 2 x3 + 9 x 2 + 40 x + 73. The general solution is = c1e x + c2 e

−1 + 2

5

x

+ c3e

−1 − 5 x 2

= y yc + y p + 2 x3 + 9 x 2 + 40 x +73.

1.3.8. The Method of Undetermined Coefficients ( U. C. Method) We have studied some special cases in which particular integral can be evaluated by the inverse operator. Now we consider the method of undetermined coefficients which can prove simpler in finding the particular f ( D) y = F ( x) F ( x) integral of the equation when is ax

a. An exponential function: e b0 x n + b1 x n −1 +…+ bn b. A polynomial: c. Sinusoidal function : sinax or cos bx F ( x) d. The more general case in which is sum of a product of terms

(

)

 sin ax  = F ( x ) e ax b0 x n + b1 x n −1 + …+ bn    cos bx  of the above types, such as

(

)

Basic Concepts of Differential Equations

The P.I, i.e.,

yp

73

will be constructed according to the following table:

F ( x ) is of the form

Take y p as

1 .a

A. x k

2. ax n ( nis a + veinteger )

xk

(

A0 x n + A1 x n −1 +…+ An

)

3. a x n e rx ( nis a + veinteger ) x k ( A0 x n + A1 x n −1 +…+ An −1 x + An ) erx x k ( A cos ax + B sin ax )

4.Cx n cos ax 5.Cx n sin ax

(

)

6.C.x n . e rx cos ax

x k [ A0 x n + A1 x n −1 +…+ An −1 x + An e rx cos ax

7.C x n e rx sin ax

+

(B x 0

n

)

+ B1 x n −1 +…+ Bn −1 x + Bn e rx sin ax

k In x , k is the smallest nonnegative integer which will ensure that no y terms in p is already in the C.F.

F ( x)

y is sum of several terms, write p for each term individually y and then add up all of them. The p and its derivatives will be substituted f ( D) y = F ( x) into the equation and coefficients of like terms on the left hand and right-hand sides will be equated to determine the If

U .C. A0 , A1 , …, An , B0 , B1 , …, Bn .

follows.

The method is illustrated by examples as

′′ ′ 2 x 3 − 9 x 2 + 6 x Example: Solve y − 3 y + 2 y = Solution: C.F. is easily found as

(1)

= yc c1e x + c2 e 2 x

For a particular solution, we assume

(

)

= y p x k Ax 3 + Bx 2 + Cx + D . Since no term of the C.F. is present in

y p = Ax 3 + Bx 2 + Cx + D

yp ,

we take k = 0 and so

Differential Equations: Theory and Applications

74

y 'p = 3 Ax 2 + 2 Bx + C = y"p 6 Ax + 2 B Substituting for

y p , y 'p and y"p

into (1), we have

( 6 Ax + 2 B ) − 3 ( 3 Ax 2 + 2 Bx + C ) + 2 ( Ax3 + Bx 2 + Cx + D ) = 2 x3 − 9 x 2 + 6 x or

2 Ax 3 + x 2 ( 2 B − 9 A ) + x ( 6 A − 6 B + 2C ) + 2 B − 3C + 2 D

= 2 x3 − 9 x 2 + 6 x. Equating coefficients of like terms, we obtain

x3 : 2 A 2= or A 1. Coeff. Of= 2 −9 or 2 B = 9 A − 9 giving B = 0 Coeff. Of x : 2 B − 9 A = + 2c 6 = or C 0. Coeff. Of x : 6 A − 6 B= 0 2D 0 Coeff. Of x : 2 B − 3C +=

So,

= or D 0.

y p = x3

The required general solution is

y =c1e x + c2 e 2 x + x 3 .

′′ ′ x 2 e x . (1) Example: Solve y − 3 y + 2 y = Solution. The C.F. is

= yc c1e x + c2 e 2 x .

To construct a particular solution, we use (3) of the table

(

)

= y p x k Ax 2 + Bx + C e x . x Since e is already in C.F ., we take k = 1 so that no terms of C.F. is in y p = Ax 3 + Bx 2 + Cx e x yp. Hence the modified P.I. is

(

y 'p = y"p =

( Ax ( Ax

) ( + Cx ) e + 2 ( 3 Ax

)

)

3

+ Bx 2 + Cx e x + 3 Ax 2 + 2 Bx + C e x

3

+ Bx 2

x

2

)

+ 2 Bx + C e x + ( 6 Ax + 2 B ) e x .

Basic Concepts of Differential Equations

y p , y 'p and y"p

Substituting for

75

into (1), we have

( Ax + Bx + Cx ) e + 2 ( 3 Ax + 2Bx + C ) e + ( 6 Ax + 2B ) e + ( 6 Ax + 2B ) e − 3 ( Ax + Bx + Cx ) e − 3 ( 3 Ax + 2 Bx + C ) e + 2 ( Ax + Bx + Cx ) e = xe 3

2

3

2

x

2

x

x

2

x

3

x

2

x

x

3 x

 −3 Ax 2 + ( 6 A − 2 B ) x + C  e x = x 3e x

Equating coefficients of like terms, we obtain

1 Coeff .of x3 : −3 A = 1 or A = − 3

Coeff .of x : 6 A − 2 B = 0 or B = −1

Coeff .of x 0 : C = 0 1 − x 3e x − x 2 e x . yp = 3 So The required solution is 1 y =c1e x + c2 e 2 x − x3e x − x 2 e x . 3

′′ xe x + x sin 2 x (1) Example: Solve y + 4 y = = yc c1 sin 2 x + c2 cos 2 x.

Solution. The C.F., as readily found above, is

′′ xe x For the P.I of (1), we find the P.I. of y + 4 y =

(2)

x sin 2 x (3) And y′′ + 4 y = Separately. Their sum will be the P.I of (1) ( by the Principle of superposition). For a particular solution of (2), we have

= y p x k ( Ax + B ) e x . Since no term of C.F. is in the

= yp

( Ax + B ) e x

y 'p =( Ax + B ) e x + Ae x

yp ,

we must take k = 0. Therefore,

Differential Equations: Theory and Applications

76

y"p =( Ax + B ) e x + 2 Ae x . Substituting for

y p , y 'p and y"p

into (2), we have

xe x ( Ax + B ) e x + 2 Ae x + 4 ( Ax + B ) e x = 5 Axe x + ( 2 A + 5 B ) e x = xe x .

or

Equating coefficients of like terms, we obtain

5 A 1= Coefficient of xe x : = or A

1 5

2 0 or B = − Coefficient of e x : 2 A + 5 B = 25 1 x 2 x = yp xe − e . 5 25 So, P.I of (2) is k y= x ( Cx + D ) sin 2 x + ( Ex + F ) cos 2 x  . p Since sin 2 x and cos 2 x are already in the C.F., we must take k = 1, so that the P.I. is

( = ( Cx

) ( ) y + Dx ) ( 2 cos 2 x ) + ( 2Cx + D ) sin 2 x + ( Ex + Fx ) ( −2sin 2 x ) + ( 2 E ( x ) + F ) cos 2 x y = ( Cx + Dx ) ( −4sin 2 x ) + ( 2Cx + D )( 2 cos 2 x ) + ( 2Cx + D )( 2 cos 2 x ) + ( 2Cx + D )( 2 cos 2 x ) + 2C sin 2 x + ( Ex + Fx ) ( −4 cos 2 x ) y p = Cx 2 + Dx sin 2 x + Ex 2 + Fx cos 2 x ' p

2

2

" p

2

2

+ ( 2 Ex + F )( −2sin 2 x ) + ( 2 Ex + F )( −2sin 2 x ) + 2 E cos 2 x.

( Cx

Substituting into (3), we have 2

)

+ Dx ( −4sin 2 x ) + 4 ( 2Cx + D ) cos 2 x + 2C sin 2 x + ( Ex + Fx )( −4 cos 2 x )

Basic Concepts of Differential Equations

(

)

(

77

)

−4 ( 2 Ex + F ) sin 2 x + 2 E cos 2 x + 4 Cx 2 + Dx sin 2 x + 4 Ex 2 + Fx cos 2 x = x sin 2 x.

Equating the coefficients of like terms, we get 1 −8 E = 1 or E = −

8 Coeff. Of x sin 2 x : : 8C 0= or C 0 Coeff. Of x cos 2 x=

= 0 or F = 0 Coeff. Of sin 2 x : 2C − 4 F Coeff. Of

cos 2 x : 2 E + 4 D = 0 or D =

1 . 16

Hence P.I of (3) is

1 1 x sin 2 x − x 2 cos 2 x. 16 8 (5)

= yp

Sum of (4) and (5) is the required P.I of (I), i.e., yp =

1 x 2 x 1 1 xe − e + x sin 2 x – x 2 cos 2 x 5 25 16 8 is the particular solution of (1).

Hence the general solution of (1) is 1 2 1 1 y= c1 sin 2 x + c2 cos 2 x + xe x − e x + x sin 2 x − x 2 cos 2 x. 5 25 16 8

1.3.9. The Cauchy-Euler Equation An equation of the form

a0 x n

n −1 dny y dy n −1 d a x + +…+ an −1 x + an y = F ( x ) 1 n n −1 dx dx dx

a , a , …, a , a

n −1 n are is called Cauchy-Euler (or equidimensional) equation. 0 1 real constants. The equation can be reduced to a linear differential equation t = x e= or t ln x with constant coefficients by the transformation Then

dy dy dt 1 dy = = . dx dt dx x dt

78

Differential Equations: Theory and Applications

d 2 y d  1 dy  1 d  dy  dy d  1  = = .      + dx 2 dx  x dt  x dx  dt  dt dx  x 

1 d  dy  dt dy  1  1  d 2 y dy  −    . +  − =  x dt  dt  dx dt  x 2  x 2  dt 2 dt 

=

d 3 y d  1 d 2 y 1 dy  = −   dx3 dx  x 2 dt 2 x 2 dt  1 d  d 2 y  d 2 y d  1  1 d  dy  dy d  2 + 2 .  2 − 2  + . 2 x dx  dt  dt dx  x  x dx  dt  dt  1  dx  − 2   x 

=

1 d  d 2 y  dt 2 d 2 y 1 d  dy  dt 2 dy −  . −  . + x 2 dt  dt 2  dx x 3 dt 2 x 2 dt  dt  dx x 3 dt

= =

1 d 3 y 3 d 2 y 2 dy − + x3 dt 3 x3 dt 2 x3 dt

1  d3y d2y dy  − 3 +2  3  3 2 x  dt dt dt 

=

d d = D and = ∆ , then dx dt If we write xD = ∆ x 2 D 2 = ∆ 2 − ∆ = ∆ ( ∆ − 1) x 3 D 3 =∆ 3 − 3∆ 2 + 2∆ =∆ ( ∆ − 1)( ∆ − 2 )







x n D n =∆ ( ∆ − 1)( ∆ − 2 ) …( ∆ − n + 1) . 2

2

Substituting these values of xD, x D , … into (1), we obtain an equation of nth order with constant coefficients having t as the independendent variable. This equation can be solved by the previous methods.

Basic Concepts of Differential Equations

Example: Solve

x2

79

d2y dy − 2x + 2 y = x3 . 2 dx dx (1)

t Solution: Let x = e . Then we have t = ln x xD = ∆.

x 2 D 2 =∆ ( ∆ − 1) . Substituting into (1), we get

 ∆ ( ∆ − 1) − 2∆ + 2  y = e3t  ∆ 2 − 3∆ + 2  y = e3t . (2) The characteristic equation has roots 2 and 1. Therefore, C.F. of (2) is

= yc c1et + c2 e 2t = yp

P.I. of equation (2) is

1 1 3t = e 3t e . 2 ( ∆ − 1)( ∆ − 2 )

1 y =c1et + c2 e 2t + e3t . 2 The general solution of (2) is

(

)

ln x or et by x , Replacing t by we have

y =c1 x + c2 x 2 +

1 3 x 2 as the general solution of (1).

Example: Solve

x4

d4y d3y d2y dy 2 + 6 x 3 3 + 9 x 2 2 + 3 x + y = (1 + ln x ) . 4 dx dx dx dx

Solution: We let x = e so that t = ln x xD = ∆ t

x 2 D 2 =∆ ( ∆ − 1) = ∆ 2 − ∆. x 3 D 3 =∆ ( ∆ − 1)( ∆ − 2 ) =∆ 3 − 3∆ 2 + 2∆ x 4 D 4 =∆ ( ∆ − 1)( ∆ − 2 )( ∆ − 3) =∆ 4 − 6∆ 3 + 11∆ 2 − 6∆.

(1)

Differential Equations: Theory and Applications

80

Substituting into (1), we get

(

) (

)

 ∆ 4 − 6∆ 3 + 11∆ 2 − 6∆ + 6 ∆ 3 − 3∆ 2 + 2∆ + 9 ∆ 2 − ∆ + 3∆ + 1 y = (1 + t )2  

(∆

4

)

+ 2∆ 2 + 1 y = 1 + 2t + t 2

(2) The roots of the characteristic equation of (2) are ±i, ±i. Therefore, C.F. of (2) is

yc = ( c1 + c2t ) sin t + ( c3 + c4t ) cos t.

To find a particular integral of (2), we use the method of U .C. Let the particular integral of (2) be

y p = At 2 + Bt + C = y 'p 2 At + B y"p = 2 A. Substituting into (2), we obtain

4 A + At 2 + Bt + C =1 + 2t + t 2 At 2 + Bt + ( 4 A + C ) =t 2 + 2t + 1. Equating coefficients of like terms, we have

A= 1, B = 2, 4 A + C = 1or C = −3. Therefore, a particular integral of (2) is

y p = t 2 + 2t − 3, Hence the general solution of (2) is

y=

( c1 + c2t ) sin t + ( c3 + c4t ) cos t + t 2 + 2t − 3 (3)

Replacing t by ln x in (3), we get

y = ( c1 + c2 ln x ) sin ( ln x ) + ( c3 + c4 ln x ) cos(ln x) + ( ln x ) + 2 ln x − 3 2

as the general solution of (1).

Basic Concepts of Differential Equations

81

Reduction of Order If one solution of the second order linear equation

d2y dy + P + Qy = 0 (1) dx 2 dx

(where P, Q are not necessarily constants and may be functions of x ) is known, then we can use it to find the general solution of d2y dy + P + Qy = F ( x) 2 dx dx (2) The procedure, due to D’ Alembert, is known as the method reduction of order. Suppose it is known that

y = y1 is a solution of (1). We assume that

y = vy1 (3) is a solution of (2), where v is a function of x to be determined. From (3), we get

dy dy dv = v 1 + y1 , dx dx dx d 2 y1 d2y dv dy1 d 2v = v 2 +2 + y1 2 dx 2 dx dx dx dx Substituting into (2), we have 2

d y1 dy dv dy1 d 2v dv v 2 +2 + y1 2 + Pv 1 + Py1 + Qvy1 = F ( x) dx dx dx dx dx dx 2  dy d 2 v  dy1  dv  d y1 + Py1  +  2 + P 1 + Q y1  v = y1 2 +  2 F ( x) dx  dx dx  dx  dx 



(4)

y=y

1 is a solution of (1) Since 2 d y1 dy + P 1 + Qy1 = 0 2 dx dx

and, therefore, equation (4) reduces to

d 2 v  dy1  dv + Py1  = y1 2 +  2 F ( x) dx  dx  dx (5)

Differential Equations: Theory and Applications

82

dv =u Setting dx in (5), we obtain du  dy1  + 2 + Py1  u = y1 F ( x) dx  dx  Which is a linear equation of the first order in u and can be solved dv = u, y = vy1. The for From dx we determine v and hence the solution method is illustrated by means of example.

d2y +y= csc x. 2 Example: Solve dx (1)

= yc c1 sin x + c2 cos x.

Solution. The C.F, of (1) is

yc as y1. Let us take y1 to be the value y when = c1 1= and c2 0 . Then assume that y = v sin x of c (2) We may take any special value of

is a solution of (1). From (2), we get

dy dv = sin x + v cos x dx dx d2y d 2v dv = sin x + 2 cos x − v sin x. 2 2 dx dx dx Substituting into (1), we obtain

sin x

d 2v dv + 2 cos x − v sin x + v sin x = csc x 2 dx dx

d 2v dv + 2 cot x = csc 2 x. 2 dx dx Setting

u=

dv , dx the above equation becomes

du + 2 cot x u = csc 2 x dx (3) Which is linear equation of first order. An integrating factor of (3) is

Basic Concepts of Differential Equations

83

 exp ln sin 2 = exp  ∫ 2 cot xdx = x  sin 2 x 2 Multiplying (3) by sin x and integrating, we have

u sin 2 x = x, Neglecting constant of integration since we seek only a particular solution.

u = x csc 2 x dv = x csc 2 x dx v= ∫ x csc 2 x dx = − x cot x + ln sin x



(on integrating by parts)

Hence a particular solution of (1) is

y = v sin x = − x cos x + sin x (ln sin x ). The general solution of (1) is

y = c1 sin x + c2 cos x − x cos x + sin x(ln sin x ). Example: Solve

( x + 2).

d2y dy − ( 2 x + 5 ) . + 2 y =( x + 1) e x 2 dx dx

(1)

2x Solution: We note that y = e makes the left-hand side of (1) zero. Therefore, we put

y = v e2 x dy  dv  =  + 2v  e 2 x dx  dx 

 d 2 y  d 2v dv =  2 + 4 + 4v  e 2 x 2 dx dx  dx  Substituting into (1), we obtain  d 2v  dv  dv  + 4 + 4v  e 2 x − ( 2 x + 5 )  + 2v  e 2 x + 2v e 2 x =( x + 1) e x . 2 dx  dx   dx 

( x + 2) 

Differential Equations: Theory and Applications

84

( x + 2)

d 2v 2 x dv e +  4 ( x + 2 ) − ( 2 x + 5 )  e 2 x +  4 ( x + 2 ) − 2 ( 2 x + 5 ) + 2  v e 2 x = ( x + 1) e x dx 2 dx

( x + 2)

d 2v dv + ( 2 x + 3) =( x + 1) e − x . 2 dx dx

Writing

( x + 2)

u=

dv , dx the above equation becomes

du + ( 2 x + 3) u =( x + 1) e − x dx

du 3 x + 3 x +1 −x + u= e dx x + 2 x + 2 (2) which is linear equation of first order. An integrating factor of (2) is

  1    2x + 3  exp  ∫ dx = exp  ∫  2 −  dx x + 2    x+2    e2 x  = exp  2 x − ln ( x + 2= ) x+2 e2 x Multiplying (2) by x + 2 , we have d  e2 x  x +1 .e x u = dx  x + 2  ( x + 2 )2

u

e2 x x +1 x = ∫ .e dx + c1 x + 2 ( x + 2 )2

ex ex =∫ dx − ∫ dx + c1 2 x+2 ( x + 2) =

ex + c1 x+2

u =e − x + c1 ( x + 2 ) e −2 x

Basic Concepts of Differential Equations

85

dv =e − x + c1 ( x + 2 ) e −2 x . Therefore, dx Integrating, we get 1 v= −e − x − c1 ( 2 x + 5 ) e −2 x + c2 4 1 −e x − c1 ( 2 x + 5 ) + c2 e 2 x . y= ve 2 x = 4 Hence is the general solution of (1).

0 is satisfied by y = e ± x . Note. It is easy to see that y′′ + Py′ + Qy = 0 and by y = x if P + Qx = 0. These can be used to find a If 1 ± P + Q = 0 by inspection. solution y′′ + Py′ + Q =

1.4. THE WRONSKIAN

I = [ a ,b] are two differentiable functions of x on W = W [ y1 , y2 ] , then their Wronskian, denoted by is obtained by Definition. If

y1 , y2

W = W [ y1 , y2 ] = y1 y2' − y1' y2 =

y1 y1'

y2 y2'

Similarly, the Wronskian of three differentiable functions I =[ a ,b ] is defined by

= W W= [ y1 , y2 , y3 ]

y1 y1' y1"

y2 y2' y2"

y1 , y2 , y3

on

y3 y3' y3"

The definition can be extended in a similar manner for the Wronskian of I = [ a, b ] . n differentiable functions on Theorem. Let

y1 , y2

be two solutions of the homogeneous linear equation

d2y dy + P + Qy = 0, 2 dx dx (1)

Differential Equations: Theory and Applications

86

I = [ a, b ] . Where P, Q are functions of x and are continuous on Then their W = W [ y1 , y2 ] Wronskian is either identically zero or is never zero on I . = W y1 y2' − y1' y2 Proof. We have Differentiating w.r.t.x, we get W ′ = y1 y2" + y1' y2' − y1' y2' − y1" y2 = y1 y2" − y1" y2 . Since

y1 , y2 are solutions of (1), we have

y1" + Py1' + Qy1 = 0 (2) and

y2" + Py2' + Qy2 = 0 (3)

Multiply (2) by

y2 and (3) by y1 to have an equivalent system

y1" y2 + P y1' y2 + Qy1 y2 = 0 (4) y1 y2" + P y2' y1 + Qy1 y2 = 0 (5) Subtracting (4) from (5), we obtain

(y y

" 1 2

)

(

)

− y2 y1" + P y1 y2' − y1' y2 = 0.

dW + PW = 0. dx The general solution of the above equation is readily found as

= W c exp(− ∫ P dx ).

Since the exponential function is never zero, W is identically zero if c = 0 or is never zero when c ≠ 0.

d2y dy + P + Qy = 0, 2 y ,y dx Theorem. If 1 2 are two solutions of dx

= W W= [ y1 , y2 ] 0 if and only if y1 , y2 are Then their Wronskian linearly dependent.

Basic Concepts of Differential Equations

87

Proof. If one of the two solutions is identically zero, then the theorem is obviously true. Assume that linearly dependent. Then

y1 , y2

are both nonzero and let

y1 , y2

y1 = cy2 , where c is a constant.

be

y1 =c y i.e., 2 d  y1  y2 y1' − y1 y2' dc = = 0  =  dx  y2  y22 dx or y y ' − y1' y2 = 0 i.e., 1 2 W [ y1 , y2 ] = 0.

d  y1  W  =  = 0. W [ y1 , y2 ] = 0, dx  y2  y22 Conversely, if then y1 = c, y2 where c is a constant. Thus

y1

Corollary.

cy2

and so

y1 , y2

W [ y1 , y2 ] ≠ 0

are linearly dependent.

if and only if

y1 , y2

are linearly independent.

Above theorems can only be applied to check the linear dependence or linear independence of differentiable functions which are solutions of a linear homogeneous O.D.E. Arbitrary differentiable functions exist with their Wronskian zero but they are linearly independent. 3 = y1 x= , y2 x 3

Consider the functions on W ( y1 , y2 ) = 0 y and y2 but 1 are linearly independent.

]−∞, ∞[ .

Here

1.4.1. VARIATION OF PARAMETERS. d2y dy + P + Qy = F ( x). 2 dx We found the solution of the equation dx (1)

Where P, Q are functions of x, in the previous section by reduction of order of (1). The solution of (1) can be determined by a procedure known as the method of variation of parameters. This method can be applied even to equations of higher order.

Differential Equations: Theory and Applications

88

d2y dy ℵüP Q dx 2 dx

Suppose that linearly independent solutions of Are given by function of (1) is

y = y1 ( x )

y = y2 ( x ) .

and

(2)

Then the complementary

= yc c1 y1 + c2 y2 c1 and c2 are arbitrary constants. We replace the arbitrary c and c2 by unknown functions u1 ( x ) and u2 ( x ) and require that constants 1 = y p u1 y1 + u2 y2 Where

(3)

be a particular solution of (1). In order to determine the two functions

u1 and u2 we need two conditions. One condition is that (3) must satisfy (1).

A second condition can be imposed arbitrarily. Differentiating (3) w.r.t.x, we have

(u y + u y ) + u y + u y .

y 'p =

' 1 1

' 2

' 1 1

2

2

' 2

If we differentiate the above equation, avoid second derivatives of

y"p

u1 and u2 . we put

will contain

u1" and u2" . To

u1' y1 + u2' y2 = 0. (4) With this condition, we have So that

= y 'p u1 y1' + u2 y2'

= y"p u1 y1' + u2 y2' y"p = u1' y1' + u2' y2' + u1 y1" + u2 y2" .

So that

y p , y 'p , y"p

Substituting for

into equation (1), we get

F ( x) (u y + u y + u y + u y ) + P (u y + u y ) + Q (u y + u y ) = u ( y + Py + Qy ) + u ( y + Py + Qy ) + u y = F ( x). or ' ' 1 1

1

' 2

" 1

' 2

" 1 1

' 1

2

1

" 2

2

' 1 1

" 2

' 2

2

' 2

2

1 1

2

2

' ' 1 2

Expressions within the parenthesis are zero since of (2). Hence

y1 and y2 are solutions

Basic Concepts of Differential Equations

u1' y1' + u2' y2' = F ( x)

89

(5)

Taking (4) and (5) together, we have two equations in the two unknowns u and u2' ; ' 1

u1' y1 + u2' y = 0

u ' y ' + u2' y2' = F ( x), and 1 1 Solving these, we have − y2 F ( x )  u1' =  y1 y2' − y1' y2   y1 F ( x )  ' u2 = y1 y2' − y1' y2  (6) W ( y1 , y2 ) = y1 y2' − y1' y2 ≠ 0, y ,y In (6), the Wronskian of 1 2 namely

y1 , y2 are linearly independent solutions of (2). Integrating (6), we u and u2 as find 1 − y2 F ( x ) − y2 F ( x ) u1 = ∫ dx = ∫ dx y1 y2' − y1' y2 W (7) since

y F ( x) y F ( x) u2 = ∫ 1' dx = ∫ 1 dx ' y1 y2 − y1 y2 W (8) Where W is the Wronskian of

y1 , y2 .

y Thus p is completely determined. In numerical problems, instead of performing the complete process, formulas (7) will be directly applied to evaluate

u1 and u2 .

d2y dy x e x ln x −2 + y = 2 dx Example. Find the general solution of dx (1) Solution. The C.F. of (1) is = yc c1e x + c2 x e x Let

90

Differential Equations: Theory and Applications

= y p u1e x + c2 xe x Here x = y1 e= , y2 x e x , F= ( x ) x e x ln x

(

)

W = W ( y1 , y2 ) = y1 y2' − y1' y2 = e x e x + xe x − xe x .e x = e 2 x . By the formulas of (7) and (8), we have

− y F ( x) − xe x .xe x ln x ∫ 2 ∫ u1 = dx = dx W e2 x

y F ( x) e x . xe x ln x ∫ 1 ∫ u2 = dx = dx W e2 x and i.e.,

u1 =− ∫ x 2 ln xdx =−(ln x).

u2 = ∫ x ln x dx =

x3 x3 x3 x3 + ∫ dx = − ln x 3 3 9 3

x2 x x2 x2 ln x − ∫ dx = − + ln x. 2 2 4 2

 x3 x3  x  x2 x2  x y= x − ln   e +  − + ln x  x e p  9 3   4 2   x3 x3  x3 x3 = e x  − ln x − + ln x  4 2  9 3   5 x3 x3  = ex  − + ln x  .  36 6  The general solution of (1) is

 5 x3 x3  y = yc + y p = c1e + c2 x e = e  − + ln x  .  36 6  x

x

x

Example. Find the general solution of

d2y dy x − x ( x + 2) + ( x + 2) y = x3 2 dx dx 2

(1)

Basic Concepts of Differential Equations

91

x Given that y = xe is a solution of the associated homogeneous equation

x2

d2y dy − x ( x + 2) + ( x + 2) y = 0 2 dx dx (2)

Solution. The given equation d 2 y x + 2 dy x + 2 − + 2 y= x. dx 2 x dx x

in

the

standard

form

is

x+2 x+2 P + Qx = − +x 2 = 0, y = x x x Since is also a solution of (2). x y = xe y = x The two solutions, and are linearly independent. The complementary function of (1) is

y= c1 x + c2 xe x . c We assume that

y= u1 x + u2 xe x p

(3)

is a particular solution of (1).

= y1 x= , y2 xe x , F = ( x) x Here W = W ( y1 , y2 ) = y1 y2' − y1' y2 = x xe x + e x − xe x = x 2 e x

(

)

By the formulas of (7) and (8), we have

u1 = ∫

− y2 F ( x ) W

dx

and

u2 = ∫

y1 F ( x ) W

dx

x 2e x x2 dx x u dx = = − = ∫ −e − x . 2 2 x 2 x xe xe i.e., and u ,u Substituting for 1 2 into (3), a particular solution of (1) is u1 = − ∫

y = yc + y p = c1 x + c2 x e x − x 2 − x. Example. Explain how the method of variation of parameters can be applied to find a particular solution of a non-homogeneous linear third order differential equation whose complementary function is known. Apply the method to find a particular solution of

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92

d 3 y dy + = cos x. dx 3 dx Solution. Suppose the C.F. of a linear third order differential equation

d3y d2y dy + P + Q + Ry = F ( x) 3 2 dx dx dx (1)

y =c y + c y + c y ; y , y , y

1 1 2 2 3 3 1 2 3 being linearly independent is known to be c solutions of the associated homogeneous equation of (1).

y =u1 y1 + u2 y2 =u3 y3 , u ,u ,u We assume that p where 1 2 3 are functions of x. y 'p = u1 y1' + u3 y3' + u1' y1 + u2' y2 + u3' y3 . We set

u1' y1 + u2' y2 + u3' y3 = 0.

Then ' p

y =u1 y1' + u2 y2' + u3 y3' y"p =u1 y1" + u2 y2" + u3 y3" y '''p = u1 y1''' + u2 y2''' + u3 y3''' + u1' y1" + u2' y2'' + u3' y3" . Substituting for

y p , y 'p , y"p , y '''p

into equation (1), we have

( u y + u y + u y + u y + u y + u y ) + P (u y + u y + u y ) +Q( u y + u y + u y ) + R( u y + u y + u y ) = F ( x) u ( y + Py + Qy + Ry ) + u ( y + Py + Qy + Ry ) ''' 1 1

2

''' 2

' 1 1

''' 1

1

(

''' 3 3

2

' 2

' '' 1 1

' 2

' 3 3

'' 1

' 1

1

)

2

(

'' 2

' '' 3 3

1 1

2

''' 2

'' 1 1

2

'' 2

2

'' 2

'' 3 3

3 3 ' 2

)

2

+ u2 y + Py + Qy + Ry2 + u3 y + Py + Qy + Ry3 + u y + u2' y2'' + u3' y3'' = F ( x) ''' 2

'' 2

' 2

''' 3

'' 3

' 3

' '' 1 1

From the system of equations (2), (3) and (4), we find on integration we get

u1 , u2 and u3 .

' 1

' 2

(4) ' 3

u , u and u and

d 3 y dy + = csc x. 3 Now we apply this method to find P.I. of dx dx

Basic Concepts of Differential Equations

93

The C.F. of the equation is obtained by solving the equation D D +1 = 0. Thus

(

)

yc = c1 + c2 cos x + c3 sin x with For the particular integral, let

y p =u1 y1 + u2 y2 + u3 y3 ,

= u1 + u2 cos x + u3 sin x. Substituting for

y1 , y2 , y3

and their derivatives into equations (2), (3) and u' ,u' ,u ' (4) above, we have the following system of linear equations in 1 2 3 .

u1' + u2' cos x + u3' sin x = 0 u1' .0 − u2' sin x + u3' cos x = 0 u1' .0 − u2' cos x − u3' sin x = csc x. We solve this system by the Gaussian Elimination method. Augmented matrix of the system is

1 cos x sin x 0    0 − sin x cos x 0  0 − cos x − sin x csc x  1 0 0 csc x    0 − sin x cos x 0  by R1 + R3 0 − cos x − sin x csc x  1 0  0 − sin x cos x 0 − cos x sin x

0

csc x   by ( cos x ) R2 cos 2 x 0  and ( sin x ) R3 − sin 2 x 1 

1 0 0 csc x    2 0 − sin x cos x cos x 0  −1 0 1  0

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94

Therefore, ' 3

−u = 1or u3' = −1, from R3

0 ( − sin x cos x ) u2' + ( cos 2 x ) u3' = R from 2 or

u2' = − cot x.

u1' = csc x, Hence

from

R1.

u1 = ∫ csc x dx = ln csc x − cot x

and

u3 =∫ − dx =− x.

yp = u1 + u2 cos x + u3 sin x.

= ln csc x − cot x − ( cos x ) ln sin x − x sin x is the required particular solution.

1.4.2. Application of Differential Equations Differential equations can be used in the solution of many problems in physical, biological and social sciences. In this section, we shall study such problems which can be formulated in terms of first and second order differential equations. The solution of the differential equation will lead to the solution of the given problem. The following examples illustrate the techniques. Example. A stone weighting 8 lbs. falls from rest toward the earth from a great height. As it falls it is acted upon by air resistance that is numerically 1 v ( in pounds ) 2 . Where v is the velocity (in feet per second). Find the velocity and distance fallen at time t. Solution. We use Newton’s second law F = ma to formulate mathematical model of this problem. The forces acting on the body are

Basic Concepts of Differential Equations

i. ii.

95

F1 , the weight of the stone acting downward and hence is positive. 1 F2 , the air resistance, numerically equal to 2 v, which acts upward

and is therefore negative.

dv m = F1 + F3 Newton’s second law becomes dt 1 dv 1 = 9 − v, taking g = 32. 2 i.e., 4 dt dv = 32 − 2v dt dv + 2v = 32 dt which is a linear equation. We now solve this differential equation. Here ∫ 2 dt = I .F e= e 2t .

And so,

d v e 2t = 32 e 2t dt

(

)

Integrating, we get 2t

ve = 16 e 2t + c or v =+ 16 ce −2t Now

v ( 0 ) = 0.

16 + c or c = −16 Hence 0 =

{

= v 16 1 − e −2t Therefore,

} (1)

is the velocity after time t. (1) may be written as

dx = 16 1 − e −2t . dt (2)

(

)

From (2), we have

16t + 8 e −2t + k . Or x =

(

)

= dx 16 1 − e −2t dt

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96

Applying the initial condition

x ( 0 ) = 0,

we find k = −8.

16t + 8 e −2t − 8 is the distance fallen after time t. Hence x = Example. A body of constant mass is projected upward from the earth’s

v.

surface with an initial velocity 0 Assuming there is no air resistance, but taking into consideration the variation of the earth’s gravitational field with altitude, find the smallest initial velocity for which the body will not return to the earth (This is the so-called escape velocity).

w( x) Solution. The general expression for the weight of a body of mass m is obtained from Newton’s inverse- square law of gravitational attraction. If R is the radius of the earth and x is the altitude above sea level, then k w( x) = ,k 2 ( R + x) x 0,= w mg , hence k = mgR 2 being constant. At= mg R 2 w( x) = . 2 R+ x) ( and The only force acting on the body is its weight dv − mg R 2 m = dt ( R + x )2 which acts downward. Thus, the equation of motion is dv mg R 2 v = − . 2 dt R + x) ( or Separating the variables and integrating, we have

1 2 gR 2 = v + c. 2 R+x

= v v= 0 i.e., at x = 0. 0 at t 1 2 = c v0 − gR 2 Therefore, But

2 gR 2 v =v − 2 gR + R + x (1) And so The escape velocity is found by requiring that v given by (1) remains positive for all (positive) value of x. Thus we must have 2

2 0

Basic Concepts of Differential Equations

97

v02 ≥ n 2 g R. Hence the escape velocity is

= v0

= 2 gR 6.9miles / sec.taking = R 4000 miles

1.4.3. Summary and Discussion In this chapter, we discussed some elementary aspects of differential equation. A very important topic in real life problem. The study of differential equations is a wide sphere in pure and applied mathematics, physics, and engineering. All of these disciplines are concerned with the estate of differential equations of various types. Pure sum focuses on the life and singularity of solutions, while applied tally emphasizes the rigorous occasion of the methods for approximating solutions. Differential equations pleasure an important role in modeling virtually every physical, technical, or biological process, from celestial motion, to bridge design, to interactions between neurons. Differential equations such as those used to solve real-life problems may not necessarily be directly solvable, i.e. do not have closed pattern solutions. Instead, solutions can be approximated using numerical methods. Applications. Many fundamental jurisprudences of physics and chemistry can be formulated as differential equations. In biology and economics, differential equations are used to mold the behavior of complex systems. The mathematical opinion of differential equations first developed together with the sciences where the equations had originated and where the results found application. However, diverse problems, sometimes originating in quite distinct scientific fields, may give rise to identical differential equations. Whenever this happens, mathematical plan seats the equations can be viewed as a unifying principle seat diverse phenomena. As an example, consider dissemination of light and sound in the atmosphere, and of waves on the surface of a pond. All of them may be described by the same secondorder partial differential equation, the breaker equation, which allows ourselves to think of brightening and sound as forms of waves, scads like familiar cascade in the water. Conduction of heat, the theory of which was developed by Joseph Fourier, is governed by another second-order partial differential equation, the heat equation. It turns out that many scattering processes, while seemingly different, are described by the same equation;

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98

the Black-Scholes equation in finance is, for instance, related to the heat equation. Euler-Lagrange equation in classical mechanics, Hamilton’s equations in classical mechanics, Radioactive decay in nuclear physics Newton’s statute of cooling in thermodynamics, The breaker equation , The heat equation in thermodynamics, Laplace’s equation, which defines harmonic functions, Poisson’s equation, The geodesic equation, The Navier-Stokes equations in fluid dynamics, The Diffusion equation in stochastic processes, The Convection-diffusion equation in fluid dynamics, The Cauchy-Riemann equations in complex analysis, The Poisson-Boltzmann equation in molecular dynamics, The shallow water equations and Universal differential equation. Differential Equation. An equation involving one dependent variable and its derivatives with respect to one or more independent variables is called differential equation. For example

dy + y cos x = sin x dx Ordinary Differential Equation (O. D. E). A differential equation, in which ordinary derivatives of the dependent variable with respect to a single independent variable occur, is called an ordinary differential equation (O. D. E). Partial Differential Equation. A differential equation involving partial derivatives of the dependent variable with respect to more than one independent variable is called a partial differential equation.

x

∂z ∂z + y = nx ∂x ∂y

Order of Differential Equation .The order of a differential is the order of the highest derivative that occurs in the equation. Degree of a Differential Equation. The degree of a differential equation is the greatest exponent of the highest order derivative that appears in the equation. (The dependent variable and its derivatives should be expressed in a form free from radicals and fractions). •

dy + y cos x = sin x dx

(order 1, degree 1)

Basic Concepts of Differential Equations

99

2



d2y  dy  + xy   = 0 2 dx  dx  (order 2, degree 1) 3

  dy 2  2 d 2 y 1+    = dx 2   dx   • (order 2, degree 2) Linear Differential Equation. An ordinary differential equation  dy d 2 y dny  F x , y , , ,…, n  = 0 dx dx 2 dx   is said to be linear if F is a linear 2 dy d y dny x , , ,…, n dx dx 2 dx function of the variables It should be carefully noted that in a linear ordinary differential equation • The dependent variable y and its derivative are all of degree one. •

No product of y or any of its derivative appear. No transcendental function of y and / or its derivative occur.

• Nonlinear Differential Equation. A differential equation that is not linear is called a nonlinear differential equation. Differential equations occur in the mathematical formulation of many problems in science and engineering. Some such problems are • Determining the motion of projectile, rocket, satellite or planet. • Finding the charge or current in an electric circuit. • Study of chemical reactions. • Determination of curves with given geometrical properties. Solution of a Differential Equation A solution (or integral) of a differential equation is a relation between the variables, not containing derivatives, such that this relation and the derivatives obtained from it satisfy the given differential equation identically.

1.4.4. Initial Condition It is often required to find the solution of a differential equation subject to certain conditions. If the conditions relate to one value of the independent dy y x = y ) ( 0 ) 0 and dx = y′ ( x0 ) y = y0 x = x0 variable such as at (written as

100

at

Differential Equations: Theory and Applications

x = x0 ,

where

x0

belongs to some interval

]α , β [ then they are called

initial conditions (or one- point boundary conditions) and initial point.

x0 is called the

1.4.5. Boundary Conditions The problem of finding the solution of a differential equation such that all the associated constraints relate to two different values of the independent variable is called a two-point boundary value problem (or simply a boundary value problem). The associated supplementary boundary conditions are called two-point boundary conditions.

1.4.6. Separable Equations Definition. A differential equation of the type F ( x ) G ( x ) dx + f ( x ) g ( y ) dy = 0 is called an equation with separable variables or simply a separable equation. Equation (1) may be written as

F ( x) f ( x)

dx +

g ( y)

G ( y)

dy = 0

which can be easily integrated.

1.4.7. Homogenous Equations f ( x, y ) Definition. A function is called homogeneous of degree n n f ( tx , ty ) = t f ( x, y ) , if Where t is a nonzero real number. Thus 10 10 x x + y xy , 2 and sin   2 x + y  y  are homogeneous function of degree 1,8 and 0 respectively.

M ( x, y ) dx + N ( x, y ) dy Exact Equation. The expression is called an exact differential if there exists a continuously differentiable function f ( x, y ) of two real variables x and y such that the expression equals the ∂f ∂f = df dx + dy. ∂x ∂y total differential df . We know from calculus that

Basic Concepts of Differential Equations

Thus, if it is exact then

M ( x , y= )

101

∂f ∂f y ) = fy. = f x and N ( x , = ∂x ∂y

Integrating Factors. If the differential equation M ( x, y ) dx + N ( x, y ) dy = 0 is not exact but when it is multiplied by a µ ( x, y ) function and the resulting equation

µ ( x , y ) M ( x , y ) dx + µ ( x , y ) N ( x , y ) dy = 0 µ(x ,y)

is exact, then is called an integrating factor (I.F) of the differential equation. The number of integrating factors of an equation may be infinite. We list below (without proofs) some rules to find the integrating factors of equations of special types.

M ( x , y ) dx + N ( x , y ) dy = 0 Rule 1. If is not exact and

My

− Nx

N

= P,

µ ( x) where P is a function of x only then (1) has an integrating factor µ ( x) which also depends on x. is solution of the differential equation dµ = Pµ dx i.e.,

µ (= x ) exp ∫ P dx

My = Note that

∂M ∂N = , Nx . ∂y ∂x Nx − M y

M Rule II. If differential equation

= Q,

where Q is a function of y only, then the

Mdx + Ndy = 0 has an integrating factor

µ (= y ) exp ∫ Q dy. 0 (1) Rule III. If M dx + Ndy = 1 is homogeneous and xM + yN ≠ 0, then xM + yN is an I.F of (1).

102

Differential Equations: Theory and Applications

0 0 is of the form y f ( x, y ) dx + x g ( x, y ) dy = Rule IV. If M dx + Ndy = 1 xM − yN ≠ 0, then xM − yN is an I.F of (1). and The following differential formulas are useful in the calculation of certain exact equations:

 y  x dy − y dx d = x2 x

 x  y dx − x dy d = y2  y d ( xy = ) x dy + y dx

(

)

d x 2 + y 2= 2 ( x dx + y dy )  x  y dx − x dy d  ln  = xy  y  x  y dx − x dy d  arc tan  = 2 y  x + y2 

Linear Equations A first order ordinary differential equation (ODE) is linear in the dependent variable y and the independent variable x if it is or can be written in the dy + P ( x) y = Q ( x), form dx where P and Q are functions of x.

The Bernoulli Equation dy + P ( x) y = Q ( x) yn Definition. An equation of the form dx (1) is called the Bernoulli differential equation. This equation is linear if n = 0 or 1. If n is not zero or 1, then (1) is reducible to a linear equation.

Basic Concepts of Differential Equations

103

Clairaut’s Equation Definition. The equation

= y xp + f ( p ) ,

is known as Clairaut’s equation.

Envelope

f ( x, y , c ) = 0 Let be a one-parameter family of curves. Suppose all members of the family of curves are drawn for various values of the parameter c arranged in order of magnitude. The two curves that correspond to two consecutive values of c will be designated as neighboring curves. The locus of the ultimate points of intersection of neighboring curves is called f ( x, y, c ) = 0. the envelope of the family Singular Solution: A solution of a differential equation is called a singular solution (S.S) if i. ii.

f ( x, y , p ) = 0

It is not derived from the general solution by giving any particular value to the arbitrary constants At each of its points, it is tangent to some member of the one parameter family of curves represented by the general solution.

The Ricatti Equation Definition. We have already studied first order linear differential equation

y′ + P ( x ) y = R ( x )

(1) q ( x) y2 If we add the term to the left-hand member of (1), we obtain a nonlinear differential equation

y′ + P ( x ) y + Q ( x ) y 2 = R ( x) (2) (2) is called the Ricatti equation.

1.4.8. Linear Differential Equations Definition. A linear differential equation of order n in the dependent variable y and the independent variable x is of the form

dny d n −1 y dy a x + +…+ an −1 + an ( x ) y = F ( x ) , ( ) 1 n n −1 dx dx dx (1) a ( x ) , a1 ( x ) ,…, an −1 ( x ) , an ( x ) and F ( x ) where 0 are functions of the indea0 ( x )

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104

a ( x) pendent variable x only and 0 is not identically zero. Using primes, (1) is also written as a0 y n + a1 y n −1 +…+ an −1 y′ + an y = F ( x) (2) where

a0 , a1 , …, an −1 , an are real constants.

Definition. If

y1 ( x ) , y2 ( x ) , …, ym ( x )

are m functions of an

c , c , …, cm are constants, then the expression independent variable x and 1 2 c1 y1 ( x ) + c2 y2 ( x ) +…+ cm ym ( x ) is called a linear combination of y1 ( x ) , y2 ( x ) , …, ym ( x ) . i. Every homogeneous linear nth − order differential equation a0 y n + a1 y n −1 +…+ an −1 y′ + an y = 0 has n linearly independent solutions

y1 , y2 , …, yn

y1 , y2 ,…, y _ m are n linealy independednt solutions of (3), then = y c1 y1 + c2 y2 +…+ cn yn of y1 , y2 , …, yn any linear combination c

ii.

If

c , c ,…, c

m being arbitrary constants. is the general solution of (3), 1 2 y y Let p be any particular solution of (2). i.e., p does not contain y + yp any constant, then c is the general solution of (2).

iii.

1.4.9. Homogeneous Linear Equation Consider the equation where

a0

dny d n −1 y dy a + +…+ an −1 + an y, = 0 1 n n −1 dx dx dx (1)

a0 , a1 ,…, an −1 are real constants.

Case I. Distinct Real Roots m , m , …, m

, e ,…, e n be n distinct real roots of (2). Then e Let 1 2 are n distinct solutions of (1). These n solutions are linearly independent. = y c1e m1x + c2 e m2 x +…+ cn e mn x , Where Hence the general solution of (2) is c1 , c2 , …, cn are arbitrary constants.

m1 x

m2 x

mn x

Basic Concepts of Differential Equations

105

Case II. Repeated Real Roots

= y

( c1 + c2 x ) em x + c3e m x +…+ cn em x 3

1

n

In the same manner, if the characteristic equation (2) has the triple repeated root the solution of

( D − m1 )

3

m1 , the corresponding part of the general solution of (1) is

y= 0

Proceeding as before, we can easily find

(

)

y = c1 + c2 x + c3 x 2 e m1x Case III. Complex Roots Suppose the characteristic equation has the complex number a + ib as a non- repeated root. Since coefficients of (1) are real, the conjugate complex number a − ib is also a non- repeated root. The corresponding part of the general solution is

= y k1e( a + ib ) x + k2 e( a −ib ) x , where k1 and k2 are arbitrary constants. y e ax  k1eibx + k2 e − ibx  =

= e ax  k1 ( cos bx + i sin bx ) + k2 ( cos bx − i sin bx )  = e ax ( k1 + k2 ) cos bx + i ( k1 − k2 ) sin bx  = e ax [ c1 sin bx + c2 cos bx ] c = i ( k1 − k2 ) , c2 = k1 + k2 Where 1 are two arbitrary constants. If a + ib and a − ib are conjugate complex roots, each repeated k times, the the corresponding part of the general solution of (1) may be written as

(

)

(

)

= y  c1 + c2 x + c3 x 2 +…+ ck x k −1 sin bx + ck +1 + ck + 2 x +…+ c2 k x k −1 cos bx 

Solution of the equation • •

Complementary Function (C.F) Particular Solution (P.I)

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106

The C.F is the solution of the homogeneous equation we have described different cases of its solution. Now to find the P.I of (1), we can write

y=

f ( D) y = 0

and

1 F ( x) f ( D) y=

and try to evaluate

1 F ( x). f ( D)

To Find the P.I when

F ( x ) sin ax or cos ax.

Here we have to evaluate

1 1 sin ax and cos ax. f ( D) f ( D)

From Euler’s theorem, we have

= e cos ax + i sin ax iax

Thus

1 1 sin ax and cos ax f ( D) f ( D)

real parts of

1 eiax . f ( D)

f ( D)

are respectively imaginary and

If contains only even powers of easy to see that 1  sin ax =  f D 2 cos ax

( )

( )

D, say f ( D ) = f D 2 ,

it is

 sin ax 1 , provided − a 2 is not a zero of f D 2 . 2  f −a cos ax

(

( )

)

1.4.10. Series Expansion of Polynomials or Expressions involving Operators. This method is useful when

F ( x)

is a polynomial in x .

f ( D) y = F ( x) F ( x) Let be such that is a polynomial in x . To evaluate the particular integral

Basic Concepts of Differential Equations

yp =

1 F ( x) f ( D) 1 f ( D)

It is often useful to expand Theorem for negative exponent so that

y p=

107

( 1+ a D + a D 1

2

2

in a series in D by the Binomial

)

+ a3 D 3 +… F ( x )

The derivatives on the right will vanish after certain stage since

= D x r 0 if n > r. n

The following binomial expansions will be useful in this connection:

1 = 1 + x + x 2 + x 3 +…. 1– x 1 = 1 − x + x 2 − x3 +…. 1+ x

1.4.11. The Method of Undetermined Coefficients (U. C. Method) We have studied some special cases in which particular integral can be evaluated by the inverse operator. Now we consider the method of undetermined coefficients which can prove simpler in finding the particular f ( D) y = F ( x) F ( x) integral of the equation when is ax

a. An exponential function: e b0 x n + b1 x n −1 +…+ bn b. A polynomial: c. Sinusoidal function : sinax or cos bx F ( x) The more general case in which is sum of a product of terms of  sin ax  = F ( x ) e ax b0 x n + b1 x n −1 + …+ bn   cos bx   the above types, such as

(

)

(

The Cauchy-Euler Equation An equation of the form

)

108

Differential Equations: Theory and Applications

a0 x n

n −1 dny y dy n −1 d a x + +…+ an −1 x + an y = F ( x ) 1 n n −1 dx dx dx

a , a , …, a , a

n −1 n are is called Cauchy-Euler(or equidimensional) equation. 0 1 real constants. The equation can be reduced to a linear differential equation t = x e= or t ln x with constant coefficients by the transformation

Reduction of Order If one solution of the second order linear equation

d2y dy + P + Qy = 0 dx 2 dx (1)

(where P, Q are not necessarily constants and may be functions of x ) is known, then we can use it to find the general solution of d2y dy + P + Qy = F ( x) 2 dx dx (2) The procedure, due to D’ Alembert, is known as the method reduction of order.

The Wronskian If

y1 , y2 are two differentiable functions of x on I = [ a , b ] then their W = W [ y1 , y2 ] ,

Wronskian, denoted by

is obtained by

W = W [ y1 , y2 ] = y1 y2' − y1' y2 =

y1 y1'

y2 y2'

Similarly, the Wronskian of three differentiable functions I =[ a ,b ] is defined by

= W W= [ y1 , y2 , y3 ]

y1 y1' y1"

y2 y2' y2"

y1 , y2 , y3 on

y3 y3' y3"

The definition can be extended in a similar manner for the Wronskian of I = [ a, b ] . n differentiable functions on

Basic Concepts of Differential Equations

109

d2y dy + P + Qy = 0, 2 y ,y dx Theorem. If 1 2 are two solutions of dx Then their = W W= [ y1 , y2 ] 0 if and only if y1 , y2 are linearly dependent. Wronskian Corollary.

W [ y1 , y2 ] ≠ 0

if and only if

y1 , y2 are linearly independent.

Variation of Parameters. d2y dy + P + Qy = F ( x). 2 dx We found the solution of the equation dx (1) where P, Q are functions of x, in the previous section by reduction of

order of (1). The solution of (1) can be determined by a procedure known as the method of variation of parameters. This method can be applied even to equations of higher order.

− y2 F ( x ) − y F ( x) u1 = ∫ dx = ∫ 2 dx ' ' y1 y2 − y1 y2 W y1 F ( x ) y1 F ( x ) u2 = ∫ dx = ∫ dx y1 y2' − y1' y2 W

where W is the Wronskian of

y1 , y2 .

CHAPTER

2

FUNDAMENTAL CONCEPTS OF PARTIAL DIFFERENTIAL EQUATIONS CONTENTS 2.1. Introduction..................................................................................... 112 2.2. Classification of Second Order PDE ................................................ 112 2.3. Summary and Discussion................................................................. 141 2.4. Classification of Second Order PDE ................................................ 141

Differential Equations: Theory and Applications

112

2.1. INTRODUCTION Many practical problems in science and engineering , when formulated mathematically, give partial differential equations (often referred to as PDE ) , In order to understand the physical behavior of the mathematical model, it is necessary to have some knowledge about the mathematical character properties, and the solution of the governing PDE, An equation which ( x, y, z, t ,…) , involves several independent variables usually denoted by a dependent function u of these variables , and the partial derivatives of the dependent function u with respect to the independent variables such as

F ( x , y , z , t … . , u x , u y , u z , ut ,……u xx , u yy , ….u xy , …) = 0



(1.1)

is called a partial differential equation. A few well-known examples are:

( i ) ut=

k ( u xx + u yy + u zz )

0 ( ii ) uxx + u yy + uzz = 0 ( iii ) u xx + u yy + u zz = u x µ u xx ( iv= ) ut u=



[linear three-dimensional heat equation]

[Laplace equation in three dimension ] [linear three-dimensional wave equation]

[one linear one-dimensional Burger equation ] In all these examples, u is the dependent function and the subscripts denote b partial differential with respect to these variable. Definition 1. The order of the partial differential equation is the order of the highest derivative occurring in the equation. Thus the above examples

= ut uu xxx + sin x

are partial differential equation of second order, whereas is an example for third order partial differential equation.

2.2. CLASSIFICATION OF SECOND ORDER PDE The most general linear second order PDE, with one dependent function u X ℵ( x1 , x2 , ) , n 1, on a domain Ù of points is n

∑a

n

u x + ∑b u xi + F ( u ) = G

ij xi j i = i , j 1 =i 1

The classification of PDE depend only on the highest order derivative present.

Fundamental Concepts of Partial Differential Equations

113

The classification of PDE is motivated by the classification of the quadratic equation of the form

Ax 2 + Bxy +Cy 2 + Dx + Ey + F = 0 (3) which is elliptic parabolic or hyperbolic according as the discriminant

B 2 − 4 AC is negative , zero or positive . Thus we have the following second order linear PDE in two variables x and y Au xx + Bu xy + Cu yy + Du x + Eu y + Fu = G (4) where the coefficients A, B, C ,… may be functions of x and y, however,

for the sake of simplicity we assume then to be constant . Equation is elliptic, (x , y ) parabolic of hyperbolic at a point 0 0 according as the discriminant.

B 2 ( x0 , y0 ) − 4 A ( x0 , y0 ) C ( x0 , y0 )

is negative, zero or positive. If this Ù, is true at all points in a domain then is said to be elliptic, parabolic in that domain. If the number of independent variables is two or three, a transformation can always be found to reduce the given PDE to a canonical form ( also called normal form). In general, when the number of independent variables is greater than 3, it is not always possible to find such a transformation excepts in certain special cases. the idea of reducing the given PDE to a canonical form is that the transformed equation assumes a simple form so that the subsequent analysis of solving the equation is made easy.

2.2.1. Canonical Forms Consider the most general transformation of the independent variables

x and y of to new variables ξ ,η , where

ξ= ξ ( x , y ) ,= η= η ( x , y )

(5)

Such that the function ξ and η are continuously differentiable and the Jacobian ∂ ( ξ ,η ) ξ x ξ y j= = = ξ xη y − ξ y η x ≠ 0 ∂ ( x, y ) η x η y

(

)

In the domain Ù Eq. 4 holds, using the chain rule of partial differential . the partial derivative becomes

Differential Equations: Theory and Applications

114

= u x u ξ ξ x + uη η x = u y u ξ ξ y + uηη y

u xx = u ξξ ξ x2 + 2 u ξη ξ xη x + uηη η x2 + uξ ξ xx + uη η xx

u xy= uξξ ξ xξ y + uξη ( ξ xη y + ξ yη x ) + uηη η xη y + uξ ξ xy + uη η xy u yy = uξξ ξ y2 + 2 uξη ξ y η y + uηηη y2 + uξ ξ yy + uη η yy Substituting these expression into the original different equation , we get

Auξξ + Buξη + Cuηη + Duξ + Euη + Fu = G where

A= Aξ x2 + B ξ x ξ y + Cξ y2

B= 2 Aξ xη x + B ( ξ xη y + ξ yη x ) + 2 C ξ y η y C= Aη x2 + Bη xη y + Cη y2 D = Aξ xx + Bξ xy + Cξ yy + Dξ x + Eξ y

= F F= , G G (9)

It may be noted that the transformed equation has the same form as that of the original equation under the general transformation Since the classification of Eq. depends on the coefficient A, B and C , we can also rewrite the equation in the form

Au xx + Bu XY + Cu yy = H ( x , y , u, ux , u y )

(10)

It can be shown easily that under the transformation (5) Eq. (10) takes one of the following three canonical form :

∅ ( ξ ,η , u , uξ , uη ) ( i ) uξξ − uηη = or

uξη = ∅, (ξ ,η , u , uξ , uη )

(11a)

in the hyperbolic case

Fundamental Concepts of Partial Differential Equations

∅ (ξ ,η , u , uξ , uη ) ( ii ) uξξ + uηη = in the elliptic case ( iii ) uξξ = ∅ (ξ ,η , u, uξ , uη ) u = ∅ (ξ ,η , u , uξ , uη ) or ηη in the parabolic case

115

(11b) (11c)

We shall discuss in detail each of these cases separately . Using Eq. (9) it can also be verified that

(

B 2 − 4 AC = (ξ xη y − ξ yη x ) B 2 − 4 AC 2

)

and therefore we conclude that the transformation of the independent variables does not modify the type of PDE.

2.2.2. Canonical Form for Hyperbolic Equation 2 Since the discriminant B − 4 AC > 0 for the hyperbolic case, we set

= A 0= and C 0 in Eq. (9) which will gave us the coordinates ξ and η that reduce the given PDE to a canonical form in which the coefficient of uξξ , uηη are zero. Thus we have A = Aξ x2 + Bξ xξ y + C ξ y2 = 0 C = Aη x2 + Bη xη y + Cη y2 = 0 which, in rewriting , become 2

ξx  ξx  A 0  + B  +C = ξ y  ξ y  η A x ηy 

2

 η  0  + B  x  + C =  ηy 

Solving these equation for

ξ x − B + B 2 − 4 AC = ξy 2A η x − B − B 2 − 4 AC = ηy 2A



x

/ ξ y ) and (η x / η y )

, we get

116

Differential Equations: Theory and Applications

2 The condition B − 4 Ac implies that the slopes of the curves 2 = ξ ( x, y ) C= C2 1 ,η ( x, y ) are real . Thus, if B > 4 AC , then at any ( x, y ) there exist two real directions given by the two roots (12) point along which the PDE (4) reduces of the canonical form. These are called characteristics equation. Though there are two solutions for each quadratic, we have considered only one solution for each. Otherwise we will end up ξ ( x, y ) = c1 , we have with the same two coordinates. Along the curve

= d ξ ξ= 0 x dx ξ y dy Hence,

ξ  dy = − x   ξy  dx   Similarly, along the curve

η ( x, y ) = c2 ,

we have

η  dy = − x   ηy  dx   Integrating Eq. (13) and (14), we obtain the equations of family = ξ ( x, y ) c= c2 , 1 and η ( x, y ) of characteristics which are called the characteristic of the PDE.(4). Now to obtain the canonical form for the given PDE, we substitute the expressions of ξ and η into Eq. (8) which reduce to Eq.(11a) Though there are two solutions for each quadratic , we have considered only solution for each . Otherwise we will end up with the same two coordinates . To make the ideas clearer, let us consider the following examples

3u x +10u xy + 3u yy = 0 Comparing

with 2

the

standard

PDE

(4),

we

have

A= 3, B = 10, C = 3, B − 4 AC = 64 > 0. Hence the given equation is a

hyperbolic PDE. The corresponding characteristics are :

 ξ   − B + B 2 − 4 AC dy = − x  =   ξy   dx 2A   

 1 =  3 

Fundamental Concepts of Partial Differential Equations

 η   − B + B 2 − 4 AC dy = − x  =  ηy   dx 2A   

117

 3 =  

To find ξ and η , we first solve for y by integrating the above equations. Thus. we get

1 y =3 x + c1 , y = x + c2 3 Which give the constant as

c1 = y − 3 x, c2 = y−x/3 Therefore,

1 3

ξ =y − 3x =c1 ,η =y, − x =c2 These are the characteristics lines for the given hyperbolic equation. In this example, the characteristics are found to be straight lines in the ( x, y ) − plane along which the initial data, impulse will propagate. To find the canonical equation, we substitute the expression for

ξ and η into Eq. (9) to get

A = Aξ x2 + Bξ xξ y + Cξ y2 = 3 ( −3) +10 ( −3)(1) + 3 = 0 2

B = 2 Aξ xη x + B (ξ xη y + ξ y η x ) + 2Cξ y η y   1  1  = 2 ( 3)( −3)  −  +10 ( −3)(1) +1 −   + 2 ( 3)(1)(1)  3  3   100 64  10  =6 =10  −  + 6 =12 − =− 3 3  3 = C 0,= D 0,= E 0= F 0 Hence, the required canonical form is

64 = uξη 0= or uξη 0 3 On integration, we obtain

Differential Equations: Theory and Applications

118

u ( ξ= ,η ) f ( ξ ) + g ( η ) where f and g are arbitrary. Going back to the original variables , the general solution is

u=

( x, y ) =

f ( y −3 x ) + g ( y − x / 3

)

2.2.3. Canonical Form for parabolic Equation 2 0, which can be true For the parabolic equation, the discriminant B −4 AC = A = 0 in Eq. ( 9 ) . if B = 0 and A or C is equal to zero n. Suppose we set first then we obtain

A = Aξ x2 + Bξ x ξ y + Cξ y2 = 0 ξ A x  ξy 

 ξ  + B  x  ξy

 0  + C = 

which gives

ξ x  − B ± B 2 − 4 AC = 2A ξ y 

   

Using the condition for parabolic case, we get

ξx B =− ξy 2A

(15)

Hence, to find the function

ξ = ξ ( x, y )

which satisfies Eq. (15) . we set

ξ dy B = − x = dx ξy 2A and get the implicit solution

ξ ( x , y ) = C1 In fact, one can verify that A = 0 implies B = 0 as follows :

B= 2 Aξ xη x + B (ξ xη y + ξ y η x ) + 2 C ξ y η x

Fundamental Concepts of Partial Differential Equations

119

2 0 the above relation reduces to Since B − 4 AC =

B= 2 Aξ xη x + 2 AC ( ξ xη y + ξ y η x ) + 2C ξ y η x

(

= 2 Aξ x + Cξ y

)(

Aη x + Cη y

)

However,

ξx B 2 AC C = − = − = − ξy 2A 2A A We therefore choose ξ in such a way that both A and B are zero. Then

η can be chosen in any way we like as long as it is not parallel to the ξ coordinates . In other words. We choose η such that the Jacobian of the

transformation is not zero. Thus we can write the canonical equation for ξ and η into Eq. ( 8 ) parabolic case by simply substituting which reduces to either of the forms (11c) . To illustrate the procedure, we consider the following example:

x 2u xx − 2 xyu xy + y 2u yy = ex 2

2

2

2

2

The discriminant B − 4 AC = 4 x y − 4 x y = 0, and hence the given PDE is parabolic everywhere. The characteristics equation is

ξ dy B 2 xy y = − x = = − 2 = − dx 2x x ξy 2A On integration, we have And hence ξ = xy will satisfy the characterstics eawuation and we can chosen η = y . To finds the canonical equation , we substitute the expression ξ and η into Eq. ( 9 ) to get for

A = A y 2 + Bxy + cx 2 = x 2 y 2 − 2 x 2 y 2 + y 2 x 2 = 0 B = 0. C = y 2 , D = −2 xy

= E 0,= F 0,= G ex Hence, the transformed equation is

Differential Equations: Theory and Applications

120

η= uηη 2ξ uξ + e y uηη − 2 xyuξ = e Or The canonical form is, therefore 2ξ 1 = uηη u + 2 eξ /η 2 ξ 2

x

η

2

ξ η

η

2.2.4. Canonical Form for Elliptic Equation 2 Since the discriminant B − 4 AC < 0, for elliptic case, characteristics equation

dy B − B 2 − 4 AC − dx 2A dy B + B 2 − 4 AC − dx 2A Give us complex conjugate coordinates, say ξ and η . Now , we make (ξ ,η ) to (α , β ) so that another transformation from

= α

ξ +η

= ,β 2

ξ −η 2i

which give us the required canonical equation in the form (11b). To illustrate the procedure, we consider the following example :

u xx + x 2u yy = 0 2 −4 x 2 < 0. Hence the given PDE is elliptic The discriminant B − 4 AC = . The characteristics equations are

dy B − B 2 − 4 AC −4 x 2 = = − = −ix dx 2A 2 Integration of these equation yields

x2 x2 iy + =c1 , −iy + =c2 2 2 Hence, we may assume that

Fundamental Concepts of Partial Differential Equations

ξ=

1 2 1 x + iy ,η = x 2 − iy 2 2 Now, introducing the second transformation

= α

ξ +η

= .β 2

ξ +η 2i

We obtain

= α

x2 = ,β y 2 The canonical form can now be obtained by computing

A = Aα x2 + β α x α y + cα y2 = x 2

B= 2 Aα x β x + B ( ax β y + α y β x ) + 2c (α y β y )= 0 C = Aβ x2 + Bβ x β y + cβ y2 = x 2 D = Aaxx + Baxy + ca yy + Dax + Ea y = 1 E = Aβ xx + Bβ xy + cβ xy + Dβ x + E β y = 0

= F 0,= G 0 Thus required canonical equation is 2

x uaa + x 2uββ + uα = 0

u uαα + uββ = − α 2α or Example 1. Classify and reduce the relation y 2u xx − 2 xyu xy + x 2u yy =

y2 x2 ux + u y x y

to a canonical form and solve it. Solution. The discriminant of the given PDE is 2

B − 4 AC = 4 x 2 y 2 − 4 x 2 y 2 = 0

121

Differential Equations: Theory and Applications

122

Hence the given equation is of a parabolic type, The characteristics equation is

ξ dy B −2 xy x = − x = =2 = − dx y ξy 2A 2y x 2 + y 2 = c1.Therefore,ξ = x 2 + y 2 satisfies the Integration gives characteristic equation. The η − coordinates can be chosen arbitrary so that is not parallel to ξ , i.e. the Jacobian of the transformation is not zero . Thus, we chosen

ξ= x 2 + y 2 ,η = y2 To find the canonical equation we compute

A Aξ x2 + B ξ xξ y + Gξ y2 = 4 x 2 y 2 − 8 x 2 y 2 + 4 x 2 y 2 = 0 B= 0 , C= 4 x 2 y 2 , D= E= F= G= 0 Hence, the required canonical equation is 2

4x = y 2uηη 0= or uηη 0 To solve this equation, we integrate functions of ξ . Now , going back to the original independent variables , the required solution is

(

) (

= u y 2 f x2 + y 2 + g x2 + y 2

)

Example. 2 Reduce the following equation to a canonical form:

(1 + x ) u + ( 1 + y ) u 2

2

xx

yy

+ xu x + yu y = 0

Solution. The discriminant of the given PDE is

(

)(

)

B 2 − 4 AC = −4 1 + x 2 1 + y 2 < 0 Hence the given PDE is an elliptic type. The characteristics equation are

(

)(

)

−4 1 + x 2 1 + y 2 dy B − B 2 − 4 AC 1+ y2 = =−= = − i dx 2A 1 + x2 2 1 + x2

dy B − B 2 − 4 AC 1+ y2 = = i dx 2A 1 + x2

(

)

Fundamental Concepts of Partial Differential Equations

123

On integrating, we get

ξ= In ( x + x 2 + 1) – i In ( y + y 2 + 1)= c1

η= In ( x + x 2 + 1) – i In ( y + y 2 + 1)= c2 Introducing the second transformation

= α

ξ +η

η −ξ

= ,β 2

2i

we obtain

α = In ( x +

x 2 + 1)

β = In (y +

y 2 + 1)

Then the canonical equation for the given PDE is

uaa + uββ = 0 Example. Reduce the following equation top a canonical form and hence solve it.

u xx − 2 sin xu xy − cos 2 xu yy − cos xu y = 0 Solution. Computing with the general second order PDE (4) ,we have

A= 1, B= −2sin x, C = − cos 2 x, D= 0, E = − cos x, F = 0, G = 0 2

2

2

4(sin x + cos x) = 4 > 0. Hence the The discriminant B − 4 AC = given PDE is hyperbolic. The relevant characteristics equations are dy B − B 2 − 4 AC = = − sin x −1 dx 2A dy B + B 2 − 4 AC = =−1 − sin x dx 2A On integration, we get

= y cos x − x + c1 , = y cos x + x + c2 Thus, we choose the characteristics line as

124

Differential Equations: Theory and Applications

ξ =+ x y – cos x =c1 , η = − x + y − cos x =c2 In order to find the canonical equation ,we compute

A = Aξ x2 + Bξ x ξ y + Cξ 2y = 0

B = 2 Aξ xη x + B ( ξ xη y + ξ y η x ) + 2Cξ yη y = 2 ( sin x +1 )( sinx − 1 ) − 4sin 2 x − 2C ξ yη y

= C 0,= D 0,= E 0,= F 0,= G 0 Thus, required canonical equation is

uξη = 0 Integrating with respect to ξ , we obtain

uη = f (η ) where f is arbitrary. Integrating once again with respect to η , we have

u= ∫ f (η ) dη + g (ξ ) = u ø ( ç)+ g( î or

)

g (ξ ) where is another arbitrary function. Returning to the old variables x, y the solution of the Given PDE is u ( x, y= ) ψ ( y − x − cosx ) g ( y + x − cos x) Example. Find the characteristics of the equation

u xx + 2u xy + sin 2 ( x ) u yy + u y = 0 when it is of hyperbolic type. 2 4 − 4sin 2 x = 4 cos 2 x . Hence Solution. The discriminant B − 4 AC = x ≠ ( 2π − 1 ) π / 2 , for all the given PDE is of hyperbolic type. The characteristics equation are

dy B B 2 4 AC = = dx 2A

cos

On integration, we get

Fundamental Concepts of Partial Differential Equations

125

y= x – sin x + sin x , y = x + sin x + c2 Thus, the characteristics equation are

ξ = y − x + sin x, η = y − x − sin x Example. Reduce the following equation to a canonical form and hence solve it.

yu xx + ( x + y ) u xy + xu yy = 0 Solution. The discriminant

B 2 − 4 AC =( x + y ) − 4 xy =( x − y ) > 0 2

2

Hence the given PDE is hyperbolic everywhere except along the line y = x ; whereas on the line v = x. it is parabolic . when y ≠ x, the characteristics equations are

dy B  B 2 −4 AC = = dx 2A

( x+ y )( x – y ) 2y

Therefore,

dy dy x = 1= , dx dx y On integrating, we obtain

y= x + c1 , y 2 = x 2 + c2 Hence, the characteristics equation are

ξ =y − x,η =y 2 − x 2 These are straight lines and rectangular hyperbolas. The canonical form can be obtained by computing

A =Aξ x2 + Bξ xξ y + Cξ y2 =y − x – y + x =0 , B =− 2 ( x − y ) C= 0, D= 0 , E =− 2( x y ) , F = G= 0 Thus, the canonical equation for the given PDE is

−2 ( x − y ) uξη + 2 ( x − y ) uη = 0 2

2

Differential Equations: Theory and Applications

126

−2ξ 2 uξη + 2 ( −ξ ) uη = 0

or

∂ ∂ξ

uξη + uη ξ=

or

 ∂u  ξ =  0  ∂η 

Integration yields

∂u = f (η ) ∂η

ξ

Again integrating with respect to η , we obtain

1 u= ∫ f ( η ) dη + g ( ξ )

ξ

Hence,

u=

1 ∫ f y 2 − x2 d y 2 − x2 + g ( y − x ) y−x

(

) (

)

is then general solution. Example. Classify and transformation the following equation to a canonical form:

sin 2 ( x ) u xx + sin ( 2 x ) u xy + cos 2 ( x ) u yy = x Solution. The discriminant of the given PDE is 2

B − 4 AC=

sin 2 2 x − 4sin 2 x cos 2 x= 0

Hence, the given equation is of parabolic type. The characteristics equation

dy B = = cot x dx 2 A Integration gives

= y In sin x + c1 Hence, the characteristic equation are :

ξ= y − In sin x,η = y η is chosen in such a way that the Jacobian of the transformation is nonzero.

Now the canonical from can be obtained by computing

Fundamental Concepts of Partial Differential Equations

127

= A 0= , B 0= , C cos 2 x ,= D 1= , E 0 ,= F 0= ,G x Hence, the canonical equation is

cos 2 ( x ) uηη + uξ = x

(

)

1 −e 2(η – ξ )  uηη = sin −1 eη − ξ − uξ   or Example. Show that the equation 2N 1 u xx + ux = utt x a2 where N and a are constants, is hyperbolic and obtain its canonical form. Solution. Comparing with the general PDE (4) and replacing y by t , A 1,= B 0, C = −1 / a 2 , D = 2 N / x , and E= F= G= 0 . The we have=

B 2 − 4 AC = 4 / a 2 > 0 . Hence the given PDE is hyperbolic. discriminant The characteristics equation are dt B  B 2 − 4 AC 4 /a 2 1 = = =  dx 2A 2 a Therefore,

dt 1 dt 1 = − , = dx a dx a On integration, we get

x x t= − + c1 , t = + c 2 a a Hence, the characteristic equations are

ξ= x + at , η = x − at The canonical form can be obtained by computing

A = Aξ x2 + Bξ x ξt + Cξt2 = 0 ,

B = 2 Aξ xη x + B ( ξ xηt + ξtη x ) + 2Cξt η= 4, t

C = 0 , D = Dξ x + Eξt =

2N 2N , E = Dη x + Eηt = x x

Differential Equations: Theory and Applications

128

Thus, the canonical equation for the given PDE is

4uξη +

2N 0 ( uξ + uη ) = x

Example. Transform the following differentiate equation to a canonical form:

u xx + 2u xy + 4u yy + 2u x + 3u y = 0 2

−12 < 0. Hence, the given PDE Solution. The discriminant B − 4 AC = is elliptic. The characteristics equations are dy B − B 2 − 4 AC = =+1 − i 3 2A dx dy B + B 2 − 4 AC = =+1 + i 3 2A dx Integration of these equations yields

(

)

(

)

y= + 1 − i 3 x +c1 , y = + 1 + i 3 x + c2 Hence, we may take the characteristics equations in the form

(

)

(

)

ξ = y − 1 − i 3 x ,η = y − 1+ i 3 x In order to avoid calculations with complex variables , we introduced the second transformation

= a

ξ +η

= ,β 2

ξ −η 2i

Therefore.

a= y − x ,β = ξ +η The canonical form can now be obtained by computing

A = Aax2 + Bax a y + Ca y2 = 3

B = 2 Aax β x + B ( ax β y + a y β x ) + 2Ca y β= 0 y C = Aβ x2 + B β x β y + C β y2 = 3

Fundamental Concepts of Partial Differential Equations

129

D = Aaxx + Baxy + Ca yy + Dax + Ea y = 1

E = Aβ xx + B β xy + C β yy + Dβ x + E β y = 2 3 Thus the required canonical form is

3uaa + 3uββ + ua + 2 3 uβ = 0

1 uaa + uββ = − ( ua + 2 3uβ = 0 3 or

2.2.5. Adjoint Operators Let Lu = ∅ (16) where L is a differentiate operator given by

dn d n −1 L a0 ( x ) n + a1 ( x ) n −1 +…+ an ( x ) = dx dx * One way of introducing the adjoint differential operators L associated with L is to from the product vLu and integrate it over the interval of interest . Let B

dx ∫vLu=

[

A

B

.] BA + ∫uL * v dx A

(17)

* which is obtained after repeated integration by parts. Here L is the operator adjoint to L, where the function u and v are completely arbitrary except

* that Lu and L v should exist.

Example. Let

(

Lu = a ( x ) d 2 u / dx 2

* construct its adjoint L .

) + b ( x )( du / dx ) + c ( x ) u ;

Solution. Consider the equation

  d 2u du vLu = dx v a x + b x + c x u ( ) ( ) ( )   dx ∫A ∫A  dx 2 dx 

B

B

Differential Equations: Theory and Applications

130 B

∫ ( av ) A

d 2u du dx + ∫ ( bv ) dx ∫ ( cv ) u dx 2 dx dx A A B

B

d 2u d ' ∫A ( av ) dx 2 du = ∫A ( av ) dx u dx

B

B

( )

B

=  u va  − ∫ ( av ) u ' dx A '

B

'

A

B

B

' '' u av  −  u ( av )  + ∫u ( av ) dx = A  A '

B

A

B

B

B du ' ) dx  u ( bv )  A − ∫u ( bv ) dx ∫A ( bv= dx A B

B

A

A

∫ ( cv ) u dx = ∫u ( cv ) dx Therefore, B

B

B

' '' ' ∫ vLu dx= u′ ( av ) − u ( av ) u ( bv ) A + ∫u ( av ) − ( bv ) + ( cv ) dx A

A

Comparing this equation with Eq. (17), we get

L* v =( av ) − ( bv ) + ( cv ) =av′′ + ( 2a′ − b ) v′ + ( a′′ − b′ + c ) v ''

'

Therefore,

L* = a

d2 d + ( 2a′ − b ) + ( a′′ − b′ + c ) 2 dx dx

Consider the partial differential equation

L ( u ) = Au xx + 2 Bu xy + Cu xy + Du x + Eu y + Fu= φ which is valid in a region S of the xy − plane , where A, B, C ,….φ are

functions of x and y .In addition. Linear boundary conditions of the general form

Fundamental Concepts of Partial Differential Equations

131

au + β u x = f * are prescribed over the boundary curve ∂S , the boundary of S , then L is called the adjoint operator. In general, a second order linear partial differential operator L is denoted by n ∂ 2u ∂u + ∑Bi + Cu ∂xi ∂x j i 1 ∂x1 = i, j 1 =

L (= u)

n

∑ Aij

(19)

It’s adjoint operator is defined by n ∂2 ∂ − A v ( Bi v ) + Cv ( ) ∑ ∑ ij ∂xi ∂x j = i, j 1 = i 1 ∂xi

L* (= v)

n

Here it is assumed that

Aij ò C and Bi ò C1. 2

u , v ò C 2 , it can be shown that ∂  n  ∂u ∂v −u  ∑Aij  v ∑ ∂x  j 1  ∂x j ∂x j =i 1 =

vLu (= u ) − uL* ( v )

n

(20) For any pair of functions

n ∂A   ij  + uv  Bi − ∑  =j 1 ∂x j

     (21)

This is known as Lagrange’s identity. Example. Construct a adjoint to the Laplace operator given by

L ( u= ) uxx + u yy

(22)

Solution. Comparing Eq. (22) with the general linear PDE (19) , we = A11 1,= A22 1. From Eq. ( 20 ) the adjoint of (22) is given by have

∂2 ∂2 L* ( v ) = v + vxx = v yy ( ) 2 (v) = ∂x 2 ∂y Therefore,

L* ( u= ) uxx + u yy Hence, the Laplace operator is a self-adjoint operator Example. Find the adjoint of the differentiate operator

L ( u= ) uxx + ui

(23)

132

Differential Equations: Theory and Applications

Solution. Comparing Eq. (23) with the general linear PDE (19), we have

= A11 1,= A22 1. L* ( v )

From Eq (20), the adjoint of (22) is given by

∂2 ∂2 v + vxx + v yy ( ) (v) = ∂x 2 ∂y 2

Therefore,

L ( u= ) uxx + ut *

It may be noted that the diffusion operator is not a self-adjoint operator.

2.2.6. Riemann’s Method We have noted with interest that a linear second order PDE

L ( u ) = G ( x, y ) 2 is classified as hyperbolic if B > 4 AC , and it has two families of real

characteristics curves in the xy − plane whose equations are Here

(ξ ,η )

are the natural coordinates for the hyperbolic system. In the = ξ ( x, y ) c1= and η ( x, y ) c2 xy − plane , the curves are the characteristics

of the given PDE as shown in Fig 1(a). while in the ξη − plane , the curves

= ξ c= c2 a and η in Fig 1(b)

are families of straight lines parallel to the axes as shown

A linear second order partial differential equation in two variables , once classified as a hyperbolic equation, can always be reduce to the canonical form

∂ 2u = G ( x, y , u , u x , u y ) ∂x∂y In particular, consider an equation which is already reduced to its canonical form in the variables x, y :

L ( u )=

∂ 2u ∂u ∂u = a + b + cu= F ( x, y ) ∂x∂y ∂x ∂y (24)

Fundamental Concepts of Partial Differential Equations

133

where L is a linear differential operator and a, b c, F are functions of

x and y only and are differentiable in some domain IR. Let

v ( x, y )

be an arbitrary function having continuous second order * partial derivatives. Let us consider the adjoint operator L of L defined by

∂ 2v ∂ ∂ L* ( v ) = − ( av ) − ( bv ) + cv ∂x ∂y ∂x ∂y (25) Now we introduce ∂v ∂u = M auv – u ,= N buv + v ∂y ∂x (26) Then

M x + N= u x ( av ) + u ( av ) x − u x v y − uvxy + u y ( bv ) + u ( bv ) y + v y u x + vu xy y Adding and subtracting cuv, we get  ∂ 2v   ∂ 2u  ∂ ∂ ∂u ∂u Mx + Ny = −u  − ( av ) − ( bv ) + cv  + v  + a + b + cu  ∂y ∂x ∂y  ∂x∂y ∂x   ∂x∂y 

i.e, vLu − ul *v = M x + N y

(27)

This is knows as Lagrange identity which will be used in the subsequent * discussion . The operator L is a self – adjoint if and only if L = L . Now we shall attempt to solve Cauchy’s problem which is described as follows . Let

L ( u ) = F ( x, y )

(28)

with the condition (Cauchy data) a curve I n the xy − plane;

This is u , and its normal derivatives are prescribed on a curve à which is not a characteristics line. Let à be a smooth initial curve which is also continuous as shown in Fig 2 since Eq. (24) is in canonical form , x and y are the characteristics

134

Differential Equations: Theory and Applications

coordinates. We also assume that line tangent to Ãis nowhere parallel to the coordinates axes. Let

P (ξ ,η )

be a point at which the solution to the Cauchy problem is sought. Let us draw the characteristics PQ and PR through P to meet the u, ux u y curve Ãat Q and R. We assume that are prescribed along Ã. Let ∂IR be a closed contour PQRP bounding IR. Since Eq. (28) is already in canonical form, the characteristics are lines parallel to x and y axes . using Green’s theorem ,we have .

.

∫ ∫ ( M x + N y ) dxdy=

∮ ( M dy − N dx )

(29)

∂IR

R

where ∂IR is the boundary of IR. Applying this theorem to the surface integral of Eq.(27). We obtain .

.

∂IR

IR

* ∫ ( M dy − N dx=) ∫ vL ( u ) − uL ( v ) dx dy



(30)

In other words, .

.

.

Ã

RP

R

∫ ∫ vL ( v ) − uL ( v )  dx dy ∫ ( Mdy − N dx ) + ∫ ( M dy − N dx ) = *

= on PQ and dx 0 on PR, we have Now using the fact that dy 0= .

.

.

.

Ã

RP

PQ

IR

∫ ∫ v L ( u ) − uL ( v )  dx dy ∫ ( M dy − Ndx ) + ∫ M dy − ∫ N dx = *



(31)

From Eq. (26), we find that .

Q

Q

PQ

P

P

N dx ∫=

∫buv dx + ∫vux dx

Integrating by parts the second term on the right –hand side and grouping. the above equation becomes .

Q

∫ N dx = [ nv ] + ∫u ( bv − v ) dx Q

x

P

PQ

p

Fundamental Concepts of Partial Differential Equations

135

Substituting this result into Eq. (31), we obtain .

.

PQ

RP

[uv ] p =[uv ]Q + ∫ u ( bv − vx ) dx − ∫ u ( av − v y ) dy .

.

Ã

IR

− ∫ ( M dy − N dx ) + ∫ ∫ vL ( u ) − uL* ( v )  dx dy Let us choose

L* ( v ) = 0

v ( x, y;, ξ ,η )



(32)

to be a solution of the adjoint equation

(33)

And at the same time satisfy the following condition :

( i ) vx bv= = whenu η , i.e , on PQ ( ii )

v y = av

( iii= )v

(34 a)

when x = ξ , i.e., on PR (34 b)

1 when = x ξ= ,y η

We call this function

(34 c)

v ( x, y; ξ ,η )

as the Riemann function or the L ( u ) = F , Eq. ( 32 ) Riemann- Green function. Since reduces to .

.

Ã

IR

u ] p [uv ]Q − ∫ u ( av − v y ) dy − v ( bu + u x ) dx  + ∫ ∫ ( vF ) dx dy [=

(35)

This is called the Riemann-Green solution for the Cauchy problem described by Eq.(28) when also be written as

u and u x are prescribed on Ã. Equation (35) can

.

.

.

Ã

Ã

IR

[u = ] p [uv ]Q − Ã∫uv ( a dy − b dx ) + ∫ ( uv y dy − vux dx ) + ∫ ∫ ( vF ) dx dy

(36)

u and u x are This relation gives us the value of at a point p when u and u x are prescribed on Ã, we obtain prescribed on à . But when u]p [=

.

.

.

Ã

Ã

IR

[uv ]R − ∫uv ( a dy − b dx ) − ∫ ( uvx dx +vu y dy ) + ∫ ∫ ( vF ) dx dy

(37)

Differential Equations: Theory and Applications

136

u and u x By adding Eq. (36) and (37), the value of u at a point p when

u and u x are prescribed on Ã, we obtain are prescribed on Ã. But when = [u ] p

{

.

}

.

.

.

1 1 1 [uv ]Q + [uv ]R − ∫uv ( a dy − b dx ) − ∫u ( vx dx − v y dy ) + ∫v ( ux dx − u y dy ) + ∫ ∫ ( vF ) dx dy 2 2 2Ã Ã Ã IR

(38) Thus, we can see that the solution to the Cauchy problem at a point (ξ ,η ) depends only on the Cauchy data on Ã. the Knowledge of the Riemann-Green function therefore enables us to solve Eq. (28) with the Cauchy data prescribed on a non characteristics curve. Thus, we can see (ξ ,η ) depends only that the solution to the Cauchy problem at a point on the Cauchy data on Ã. the Knowledge of the Riemann-Green function therefore enables us to solve Eq. (28) with the Cauchy data prescribed on a non characteristics curve.

(

Lu = a ( x ) d 2u / dx 2

Example. Let

* construct its adjoint L .

) + b ( x )( du / dx )

Solution. Consider the equation

  d 2u du v a x ( ) ∫A  dx 2 + b ( x ) dx + c ( x ) u  dx

B

B

dx ∫vLu= A

B

∫ ( av ) A

B

∫ ( av ) A

d 2u du dx + ∫ ( bv ) dx ∫ ( cv ) u dx 2 dx dx A A B

B

d 2u d du = ∫ ( av ) ( u ′ ) dx 2 dx dx A B

B

= [u ′va ] A − ∫ ( av ) u ′ dx B

'

A

B

B

' ' = [u′av ]A − u ( av )  A + ∫u ( av ) 'dx B

A

+ c ( x ) u;

Fundamental Concepts of Partial Differential Equations B

137

B

B du '   bv dx u bv = ( ) ( )  A − ∫u ( bv ) dx ∫A dx  A B

B

A

A

∫ ( cv ) u dx = ∫u ( cv ) dx Therefore, B

B

B

' '' ' ∫ vLu dx= u′ ( av ) − u ( av ) u ( bv ) A + ∫u ( av ) − ( bv ) + ( cv ) dx A

A

Comparing this equation with we get

L* v =( av ) − ( bv ) + ( cv ) =av′′ + ( 2a′ − b ) v′ + ( a′′ − b′ + c ) v ''

'

Therefore,

d2 d L = a 2 + ( 2a′ − b ) + ( a′′ − b′ + c ) dx dx *

Consider the partial differential equation

L ( u ) = Au xx + 2 Bu xy + Cu xy + Du x + Eu y + Fu= φ which is valid in a region S of the xy − plane , where A, B, C ,….φ are

functions of x and y .In addition . Linear boundary conditions of the general form

au + β u x = f Are prescribed over the boundary curve ∂S , the boundary of S , then

L* is called the adjoint operator. In general , a second order linear partial differential operator L is denoted by n ∂ 2u ∂u L (= u ) ∑ Aij + ∑Bi + Cu ∂xi ∂x j i 1 ∂x1 = i, j 1 = n

It’s adjoint operator is defined by



n ∂2 ∂ − A v ( Bi v ) + Cv ( ) ∑ ∑ ij ∂xi ∂x j = i, j 1 = i 1 ∂xi

L* (= v)

n



Differential Equations: Theory and Applications

138

Here it is assumed that

Aij ò C 2 and Bi ò C1.

2

u , v ò C , it can be shown that

For any pair of functions

∂  n  ∂u ∂v −u vLu (= u ) − uL ( v ) ∑  ∑Aij  v  ∂x  j 1  ∂x j ∂x j =i 1 =

n ∂A   ij  + uv  Bi − ∑  =j 1 ∂x j Example. Verify that the Green function for the equation *

n

    

∂ 2u 2  ∂u ∂u  + 0  + = ∂x∂y x + y  ∂x ∂y 

x 2 on y x, is given by u / ∂x 3= Subject to u = 0 , ∂= v ( x, y; ξ ,η ) =

( x + y ) {2 xy + (ξ − η )( x − y ) + 2ξη} 3 (ξ + η )

and obtain the solution of the equation in the form .

(

u = ( x − y ) 2 x 2 − xy + 2 y 2

)

Solution. In the given problem,

∂ 2u 2 ∂u 2 ∂u L (u ) = + + = 0 ∂x∂y x + y ∂x x + y ∂y (39) Comparing this equation with the standard canonical hyperbolic equation (24), we have 2 a= b= . C= 0 , F= 0 x+ y Its adjoint equation is

L* ( v ) = 0

2

where

∂v ∂  2v  ∂  2 v  L* ( v ) = −  −   ∂x∂y ∂x  x + y  ∂y  x + y  (40) Such that

( i ) L * v = 0 thought the ( ii )

∂v ∂x

xy − plane

2 = v on PQ, i.e, om y η x+ y

Fundamental Concepts of Partial Differential Equations

2 = v on PR, i.e., on x ξ x+ y (41) ( iv ) v = 1 at P (ξ ,η ) ∂v iii ( ) ∂y

If v defined by

x+ y

v (= x , y ; ξ ,η )

(ξ + η )

3

 2 xy + (ξ − η ( x − y ) + 2ξη 



(42)

Then

∂v = ∂x

x+ y

(ξ + η )

∂v = ∂y

or

3

 2 xy + (ξ − η )( x − y ) + 2ξη   2 y + (ξ − η )  +   3   (ξ + η )

1

(ξ + η )

3

 4 xy = 2 x 2 − 2 y (ξ − η ) + 2ξη 

And

∂ v 4( x + y) = ∂x∂y (ξ + η )3



(43)

2

∂v = ∂y

1

(ξ + η )

3

(44)

 4 xy = 2 x 2 − 2 y (ξ − η ) + 2ξη 



Using the results described becomes

L* ( v ) =

∂ 2v 2 − ∂x∂y x + y

4( x + y)

(ξ + η )

3

L* ( v ) =

or



 ∂v ∂v  4v  + + 2  ∂x ∂y  ( x + y ) 2

( x + y )(ξ + η )

4( x + y)

(ξ + η )

3



Hence condition ∂v = ∂x

1

(ξ + η )

3

3

4( x + y)

(ξ + η )

3

(

)

 4 xy + 2 x 2 + y 2    =0

( i ) is satisfied . Also, on

 4 xη + 2η 2 + 2 x (ξ − η ) + 2ξη 

y = η.



(45)

139

Differential Equations: Theory and Applications

140

However, p

 2uv ∂u  ∫Q  x + y + v ∂x  dx=

2uv ∂v P ∫Q x + y dx + ( uv )Q − Q∫u ∂x dx P

P

= u 0= on y x, becomes Now, using the condition .

Q

 2uv ∂u  ∫ ∫ vL ( u ) − uL ( v )  dx dy =∫  − v  dx x+ y ∂x  IR P *

Q

 2uv 2uv ∂u  −∫ dx − ( uv ) p + ( uv )Q + ∫  + v  dx x+ y x+ y ∂x  Q R P

2uv  ∂u  ∫Q x + y dy − ∫P  u ∂x  dy P

-

R

Also, using conditions (ii) –(iv) of the above equation simplifies to

= (u ) p

Q

( uv )Q − ∫v R

∂u dx ∂x

Now using the given condition, viz.

∂u = 3 x 2 on RQ ∂x We obtain

 2 x  2 x 2 + 2ξη     dx = ( u ) p ( uv )Q − 3 ∫x   3  (ξ + η )  R Q

2

η

12 = − x 5 + x 3 ξη dx 3 ∫ (ξ + η ) ξ

= −

12

(ξ + η )

=

3

(

)

1 1 6 6 4 4   6 η − ξ + 4 ξη η − ξ 

(

)

(

)

ξ 2 −η 2  2 ξ 4 + ξ 2η 2 + η 4 ) + 3ξη (ξ 2 + η 4  3  ( (ξ + η )

Fundamental Concepts of Partial Differential Equations

(

= (ξ − η ) 2ξ 2 − ξη + 2η 2 Therefore,

141

)

(

u ( x, y ) = ( x − y ) 2 x 2 − xy + 2 y 2

)

hence the result.

2.3. SUMMARY AND DISCUSSION Many practical problems in science and engineering, when formulated mathematically, give partial differential equations (often referred to as PDE ). An equation, which involves several independent variables usually, ( x, y, z, t ,…) , a dependent function u . of these variables, denoted by and the partial derivatives of the dependent function u with respect to the independent variables such as

F ( x , y , z , t … . , u x , u y , u z , ut ,……u xx , u yy , ….u xy , …) = 0

is called a partial differential equation. A few well-known examples are:

( i ) ut=

k ( u xx + u yy + u zz )

( ii ) uxx + u yy + uzz

= 0



[linear three-dimensional heat equation]

[Laplace equation in three dimension ]

Order of PDE. The order of the partial differential equation is the order of the highest derivative occurring in the equation. Thus the above examples

= ut uu xxx + sin x

are partial differential equation of second order, whereas is an example for third order partial differential equation.

2.4. CLASSIFICATION OF SECOND ORDER PDE The classification of PDE depend only on the highest order derivative present. The classification of PDE is motivated by the classification of the quadratic equation of the form

Ax 2 + Bxy +Cy 2 + Dx + Ey + F = 0 which is elliptic parabolic or hyperbolic according as the discriminant

B 2 − 4 AC is negative , zero or positive . Thus we have the following second order linear PDE in two variables x and y

142

Differential Equations: Theory and Applications

Au xx + Bu xy + Cu yy + Du x + Eu y + Fu = G A , B , C ,… where the coefficients may be functions of x and y, however, for the sake of simplicity we assume then to be constant . Equation is elliptic (x , y ) , parabolic of hyperbolic at a point 0 0 according as the discriminant.

B 2 ( x0 , y0 ) − 4 A ( x0 , y0 ) C ( x0 , y0 )

is negative , zero or positive. If this Ù, is true at all points in a domain then is said to be elliptic, parabolic in that domain . if the number of independent variables is two or three , a transformation can always be found to reduce the given PDE to a canonical form ( also called normal form).

Adjoint Operators Let Lu = ∅ where L is a differentiate operator given by

L a0 ( x ) =

dn d n −1 a x + +…+ an ( x ) ( ) 1 dx n dx n −1

* One way of introducing the adjoint differential operators L associated with L is to from the product vLu and integrate it over the interval of interest . Let B

B

dx [ .] + ∫uL ∫vLu= B A

A

*

A

v dx

* which is obtained after repeated integration by parts. Here L is the operator adjoint to L, where the function u and v are completely arbitrary except

* that Lu and L v should exist.

∂  n  ∂u ∂v −u vLu (= u ) − uL ( v ) ∑  ∑Aij  v  ∂x  j 1  ∂x j ∂x j =i 1 = *

n

This is known as Lagrange’s identity

n ∂A   ij  + uv  Bi − ∑  =j 1 ∂x j

    

CHAPTER

3

APPLICATION OF DIFFERENTIAL EQUATIONS IN MECHANICS CONTENTS 3.1. Introduction..................................................................................... 144 3.2. Projectile Motion ............................................................................ 161 3.3. Summary and Discussion................................................................. 186

Differential Equations: Theory and Applications

144

3.1. INTRODUCTION The motion of a particle along a straight line is called rectilinear motion. Let the particle start from point O along a line. We take the line as x − axis . Let after time t particle be at point P at a distance x from O. Let r be the position vector of point p w.r.t. origin O. then

= r OP = xiˆ = v

dr dx ˆ = i dt dt v a = a , then Let v = and

v=

dx dt

d 2x a= 2 dt Also,

= a

dv = dt

dv dx dv dv = v =v dx dt dx dx

3.1.1. Motion with Constant Acceleration Let the particle start form O with velocity u at time t = 0 with constant acceleration a. Let the particle be at distance x from O after time t. Then

dv dt where a is constant. Integration will yield v= at + A a=

Since velocity v = u at t = 0, therefore, A = u , thus v = at + u = u + at

Application of Differential Equations in Mechanics

Now, since

v=

145

dx , dt therefore eq. (1) gives

dx = u + at dt By integrating

 t2  x= ut + a   + B 2

= x 0,= at t 0, we get, B = 0, hence Using at 2 x= ut + 2 Also, ax = v dv

a =v

dv , dx where a is constant, then

Integrating

v2 = ac + C 2 u2 , Since v = u, at x = 0, therefore, C = 2 and hence

v2 − u 2 = 2ax The equation and are called equation of motion. If the particle moves with constant retardation the acceleration is taken as negative.

3.1.2. Vertical Motion under Gravity For a failing body, the acceleration is constant. It is called acceleration due

to gravity and is denoted by ‘g’. In FPS system g = 32 ft / sec .In CGS 2 2 system g =981 cm / sec and in MKS system g = 9.81 m / sec . If the body is projected vertically upward then acceleration g is negative. For falling bodies equation of motion is v= u + gt 2

146

Differential Equations: Theory and Applications

x= ut +

1 2 gt 2

v2 − u 2 = 2 gx Example

2gh A particle is projected vertically upward with velocity and another is let fall from a height h at the same time. Find the height of the point where they meet each other. Solution. Let both particles meet at point P at height x. Then for the first particle

x= ut −

1 2 gt 2

u = 2 gh , then 1 = x 2 ght − gt 2 2 (1) For the second particle Put

x= ut +

gt

x= ut +

1 2 gt 2

Put u= 0 , x= h − x, then

h− x =0+

1 2 gt 2

(1) and ( 2 ) , we get Using eqs. u = 2 ght or

t = 2 ght Thus equation (1) gives



(2)

Application of Differential Equations in Mechanics

 h x = 2 ght   2 gh 

147

 1  h2  h 3h  − g   =h − = 4 4  2  2 gh 

Example A particle is projected vertically upward. After time t another particle is sent up from the same point with the same velocity and meets the first at height h during the downward flight of the first. Find the velocity of the projection. Solution Let u be the velocity of the projection and v be the velocity at height h, then

v2 − u 2 = −2 gh = v or

u 2 − 2 gh

Now, the time taken by the first particle from height h to the highest point and back to the height h is t, therefore, the time taken from the highest point is t / 2 .

Velocity at highest point = o Velocity at height h = v Here we use the formula v= u − gt

v 0,= u v= , t t/2, then Put = 0= v −

gt 2

From (1) and (2) gt u 2 − 2 gh = 2

g 2t 2 u 2 − 2 gh = 4 = u 2 2 gh +

g 2t 2 4

148

Differential Equations: Theory and Applications

8 gh + g 2t 2 u= 4

3.1.3. Distance Traveled in nth Second x and x b

2 Let 1 e the distance travelled in the first n and n – 1 seconds respectively, then from the equation of motion

1 x= ut + at 2 2 1 x= un + an 2 1 2 Then 1 2 x2= u ( n − 1) + a ( n − a ) 2 And Distance travelled in the nth second =

x1 − x2 = un +

x1 − x2

a 2  a 2 n − u ( n − a ) + ( n − 1)  2 2  

a 2  a  n −  un − u + n 2 − 2n + 1  2 2   = 1 u + a ( 2n − 1) = 2 un +

(

)

3.1.4. Motion with variable Acceleration

( i ) Time Dependent Acceleration As the acceleration is time dependent, then it is function of time Therefore we can write:

a = f (t )

dv = f (t ) or dt or dv = f ( t ) dt

Application of Differential Equations in Mechanics

149

Integrating

∫ dv = ∫ f ( t ) dt v g (t ) + A or= ∫ f ( t ) dt , where g(t) = and A is constant of integration, which gives the velocity. Also since

v=

dx , dt Therefore

dx = g (t ) + A dt = dx g ( t ) dt + Adt or By integrating, we get x= ∫ g ( t ) dt + At + B where B is another constant which gives the distance travelled by the particle . By using the initial conditions we can find the constants Aand B . Example 3 Discuss the motion of the particle moving in a straight line if it stars from rest t = 0 and its acceleration is equal to accost + bsint. Solution Give acceleration is = a accost + b sint By integrating, we have v = asint − b cost + A

= v 0= at t 0 , therefore A = b, thus Since = C asint – bcost + b

dx = asint − bcost + b Also, v = dt Again integrating

150

Differential Equations: Theory and Applications

x= −acost – bsint +bt + B where B is constant of integration. Since x = 0 , at t = 0, therefore B = a. Thus x= −acost – bsint + bt + a

= a (1 − cost ) + b ( t − sint ) (ii) Velocity Dependent Acceleration Let the acceleration a is function of velocity v, then

a = f (v Since

v

a =v

dv , therefore dx

dv = dx f (v) Integrating

dv = ∫ dx f (v)

∫v

x g (v) + A or =

g (v) = ∫ v

dv and f (v)

where A is constant of integration which gives the distance travelled by the particles. Again consider a = f(v). since

a=

dv , dt

Therefore

dv = dt or f (v) ∫ or

dv = ∫ dt f (v)

Application of Differential Equations in Mechanics

= t h (v) + B or where

h (v) = ∫

dv and B is f (v)

151

constant of integration and gives the time . Example: A particle moves in a straight line with an acceleration kv’ . If its initial velocity is u, find the velocity and the time spend when the particle has travelled a distance of x. Solution. Given that

a = kv3

Since

v

a=v

dv , dx therefore

dv = kv3 dx

or

v −2 dv = k dx

or ∫ v −2 dv = kdx 1 or v −= kx + A

Given

v = u , x = 0 at t = 0, therefore A = −

1 1 u = − kx or c = 1 − kux Thus v u dx v = , then dt Again since dx u = dt 1 − kux or

u dt , thus ( 1 − kux ) dx =

∫ ( 1 − kux ) dx =u ∫ dt

 x − ku  

 ut + B = 

1 u

Differential Equations: Theory and Applications

152

= x 0,= at t 0, we get = B 0. Hence Using  x2  x x − ku   = ut or t = ( 2 − kux ) 2u  2  Example. A Particle starts with a velocity u and move in a straight line. If the suffers a retardation equal to the square of velocity, find the distance travelled by the particle in time t. 2 Solution. Given retardation = v , therefore

a = −vt 2 We know that

v

dv = a , there fore dx

dv = −v dx or

dv = −dx v

or



or

In v =− x + A

dv =− ∫ dx v

where a is constant of integration . Using

A = Inu , thus u u u x or e= or v = ln   x= v ex v Since

v=

dx , dt therefore

dx u = dt e x

or e x = u dt , then

Application of Differential Equations in Mechanics x e= ut + B

where B is constant of integration .

= x 0= at t 0, we get B =1 . Hence Using ex = ut + 1 or x = in ( ut + 1) (iii) Distance Dependent Acceleration Let the acceleration be a function of distance x . Then a = f(x). Since

v

dv = f ( x ) dx dx

or ∫ vdv = ∫ f ( x ) dx v2 = g ( x) + A 2 where g(x)

∫ f ( x ) dx and A

is the constant of integration

or v = ± 2g ( x) + 2 A We know that

dx = v, therefore dt dx = ± 2g ( x) + 2 A dt dt = ±

dx 2g ( x) + 2 A

Integrating, we get

t =± ∫

dx 2g ( x) + 2 A

+B

where B is constant of integration.

153

Differential Equations: Theory and Applications

154

Example Discuss the motion of the particle moving in a straight line if it starts at a distance ‘a’ from a point O and move with an acceleration equal to times its distance from the O.

Solution Let x be the distance of particle from O, then Acceleration = a= µ x Since

or

∫ vdv =µ ∫ xdx

v2 x2 = µ +A 2 2

v 0,= x a at = t 0, so that Given= A = −µ

a2 , thus 2

 x2 a2  v2 = µ −  2  2 2  or v = ± µ x2 − a2

dx Take v =µ x 2 − a 2 , since v = , then dt ∫ or

dx 2

x − a2

= µ ∫ dt

x cosh −1  =  a

µ ∫ dt

x a at = t 0,= B 0, thus Using= x cosh −1   = µ t a

Application of Differential Equations in Mechanics

or

155

x = a cosh( µ t )

dx v= − µ x 2 − a 2 , since v = , then dt If we take dx ∫ = − µ ∫ dt + C 2 x − a2 x cosh −1   = − µ t +C a

x a=t 0, C= 0, thus Using = x cosh −1   = − µ t a

(

)

or = a cosh − µ t = a cosh

(

µt

)

3.1.5. Graphical Solution of Rectilinear Problem Consider the graph of the curve

(a) (b ) (c)

y = f ( x)

in the xy − plane, then

dy = dx slope of the tangent at any point Area bounded by the curve= y f ( x ) , from = x a to = x bs = is

b

∫ ydx a

(i) Velocity- Time Curve

Consider the graph of velocity time graph v = v(t), then acceleration = a =

156

Differential Equations: Theory and Applications

dv dt or

a = slope of velocity – time curve = tan θ

We know that dx = vdt

v=

dx , dt thus

=t t2=t t2

or

∫ dx = ∫ vdt

=t t1=t t1

t = t2

t =t

∫ vdt

or

x t = t2

or

x ( t1 ) − x ( t2 ) = ∫ vdt

1

=

t =t1

t = t2 t =t1

Change in distance = area bounded by velocity – time curve, t-axis

= t t1= and t t2

(ii) Space – Time curve

Application of Differential Equations in Mechanics

Consider the space- time curve

velocity= v=

x = x (t ) ,

as

dx dt

velocity = slope of the space-time curve = tanθ (ii) Acceleration-time graph Consider the acceleration – time graph

a

157

a = a (t ) ,

then, since

dv = or dv adt dt

=t t2=t t2

or

∫ dv = ∫ adt

=t t1=t t1

t = t2

or

v t =t = 1

t = t2

∫ adt

t =t1

t = t2

or

v ( t1 ) − v ( t2 ) = ∫ adt t =t1

Change in velocity = area bounded be the acceleration – time curve,

t − axis t= t1 and t = t2

(iv) Velocity-space curve

Differential Equations: Theory and Applications

158

Consider the velocity- space curve

a=v

v = v ( x).

Now since

dv dx

where

dv Slope of tangent = dx Therefore dv tanθ = dx In triangle PNG

tan = θ

GN GN = GP v

From (1) and (2), we have

dv GN dv = v = GN dx v or dx Therefore Acceleration= a= Lenght of subnormal at P Example. A particle starts from rest with a constant acceleration a. when its velocity acquires a certain value v, it moves uniformly and then its velocity start decreasing with constants retardation 2a till it comes to rest .

Application of Differential Equations in Mechanics

159

Find the distance traveled by the particle. If the time taken from rest to rest is t. Solution

t , and t and t

2 3 be the times for accelerated, uniform motion and Let 1 retorted motion respectively , then

t = t1 + t2 + t3 = Acceleration = a = slope of OA or

t1 =

v a

Similarly retardation = or

t3 =

2a =

v t3

v 2a

Thus

=

v t1

t = t1 + t2 + t3

v v + t2 + a 2a This implies that

t2 = t −

3v 2a

Distance = area under the velocity- time curve = area of trapezium OABC

160

= = = =

Differential Equations: Theory and Applications

1 ( OC + AB )( AD ) 2 1 ( t1 + t2 + t3 + t2 ) v 2 1 ( t + t2 ) v 2 1 3v  3v 2 t + t − v = vt −   2 2a  4a Example

A particle moving along a straight-line stars from rest and is accelerated uniformly till it attains the velocity v. The motion is the retorted and the particle come to rest after traversing a total distance x. if the acceleration is f, find the retardation and the total time take y the particle from rest to rest. Solution Let then

t1 and t2 be the time for acceleration and retorted motion respectively,

Acceleration = f

= slope of OA v v = f = or t1 t1 f Or Let g be the retardation, then g = slope of AB v 2 = t

Application of Differential Equations in Mechanics

or

t2 =

161

v g

1 = ( t1 + t2 ) v Distance = x = area of triangle OAB 2 2x t1 + t2 = v or 2x Thus total time = v = t1 Put

x=

v v = and t2 in (1) , f g then

1 v v  1 1 1  2  + v =  + v 2 f g  2 f g 

2x 1 1 = + 2 v f g or

1 2 x 1 2 xf − V 2 = 2 − = f fv 2 or g v or

g=

fv 2 2 xf − v 2

3.2. PROJECTILE MOTION An object thrown into the space with certain velocity from a gun or dropped from a moving plane under the action of gravity is called a projectile. Thus, a projectile move with a constant horizontal velocity and at the same falls freely under the action of gravity. The path of projectile is called the trajectory.

3.2.1. Equation of the Trajectory of a projectile v

Let the particle m be projected from point O with an initial velocity o making an angle α with the horizontal. Take O as the origin and the horizontal and vertical lines through O as x − axis and y − axis respectively.

Differential Equations: Theory and Applications

162

Suppose that after time t, the particle is at point vector is r . i.e r= xiˆ + y ˆj

p ( x, y )

whose position

ÿ  ˆ + y ˆj r = xi

¨

¨

¨

= r x iˆ + y ˆj The gravitational force F acting on the particle at P is F = mgjˆ ¨

m r = −mgjˆ ¨

ˆ or r = gj ¨

¨

x iˆ + y ˆj = − gjˆ, Then ¨

x= 0

¨

y = −g

and

Integrating, we get

x = A

ÿ

and y = − gt + B

at t = 0, the particle is at O, therefore

at = t 0,= x v0 cosα= , y vo sinα

Application of Differential Equations in Mechanics

163

Therefore

= A v= vo sinα o cosα , B Hence

x = v0 cos α , y = − gt + vo sinα which gives the horizontal and vertical component of velocity at any time t. again integrate

x =( v0 cos α ) t , and y =− g

t2 +t + D 2

at= t 0,= x 0,= y 0, therforeC = 0,= D 0, Hence,

x = ( v0 cos α ) t

= y v0 sinα −

1 2 gt 2

The eqs (1) and (2) are parametric equation of trajectory. To find the Cartesian equation, eliminate t from the eqs. (1) and (2), we get which is the Cartesian equation of trajectory . from

( 3) , we have

gx 2 sinα −x = −y 2 2 cosα 2vo cos α

x2 − x or

2v 2 cos 2 α sinα  2vo2 cos 2 α  −y o  = cosα  g g 

2 x v02 sinα cos α vo2 cos 2 α x − −2 y g g or = 2 v02 sinα cos α  v02 sinα cos α  2 x − 2x +  g g   or 2 v 2 cos 2 α  v02 sinα cos α  = −2 y o +  g g   2

164

Differential Equations: Theory and Applications 2

 v02 sinα cos α  2vo2 cos 2 α  vo2 sin 2 α  − = − x y     g g 2g    or  Now (i) Vertex 2  v0 sinα cos α vo2 sin 2 α  ,   g 2g   is the verte3x of the (ii) Latus Rectum

2vo2 cos 2 α 4a = g Latus Rectum = (iii) Maximum Height Maximum height attained above x- axis

H = Maximum height attained above x – axis = y − Coordinate of vertex

=

vo2 sin 2 α 2g

dy =0 This equation can also be obtained by putting dx in the equation of trajectory. (iv) Equation of Directrix Height of Directrix above x – axis 1 = H + ( latus rectum ) 4

Application of Differential Equations in Mechanics

165

vo2 sin 2 α vo2 cos 2 α + 2g 2g

=

vo2 v2 sin 2 α + cos 2 α= o 2g 2g

(

=

)

vo2 y= 2g Equation of Directrix is (v) Focus x – Coordinate of focus = x – coordinate of vertex =

v02 sinα cos α g

1 ( latus rectum ) y – Coordinate of focus = H - 4 =

vo2 sin 2 α vo2 cos 2 α + 2g 2g

=

vo2 sin 2 α + cos 2 α 2g

(

)

vo2 =− cos 2α 2g Thus, the focus

( vi )

 v 2 sinα cos α   v2  v2 v2 = 0 , − o cos 2α  =  o sin 2α − o cos 2α  g 2g 2g    2g 

Timeof Flight

The time taken by the particle to reach the horizontal plane through the point of projection is ca;;ed its time of flight. In this case put y = 0, in the equation

y g =

t2 + (vo sin α ) t 2

Differential Equations: Theory and Applications

166

t2 0= − g + (vo sin α ) t 2 Then sincet ≠ 0, therefore

2vo sinα g or is the time of flight. (vii) Range (a) Horizontal Range The horizontal range of the projectile is the horizontal distance described by the particle during the time of flight. Therefore Range R = (horizontal velocity) (time of flight) t=

 2v sinα  vo cos α  o  g   vo2 2 ( sin α cosα ) 2vo2 sin 2α = g g This range can also be obtained by putting y = 0 in the Cartesian equation of trajectory i.e putting y = 0 in the equation

y xtanα − =

gx 2 sec 2 α 2 2vo

Also, it is notable that this range will remain same if we replace

α by α=

π

2

−α

(b) Maximum Range

vo2 sin 2α g Range R =

dr =0 For maximum range dα

Application of Differential Equations in Mechanics

167

2vo2 cos 2 α = 0 g

⇒ ⇒

cos 2α = 0



π 2α = 2



π α= 4 Therefore

Rmax

vo2 = g

3.2.2. Range in inclined plane Let β the inclination of the inclined plane. Let a particle is projected from point O with velocity

vo by making an angle α to the horizontal where P ( x, y )

β < α . Let the projectile meet the inclined plane at

called the range of the inclined plane.

= x rcos = β , y rsinβ P ( x, y ) = ( rcos β , rsinβ ) Equation of trajectory is

= y tan α −

1 x2 g 2 sec 2 α 2 vo

then OP = r is

Differential Equations: Theory and Applications

168

The point

p ( rcos β , rsinβ )

lies on it, then

1 r cos 2 β = r sinβ r β tanα − g 2 2 vo cos 2 α grcos 2 β sinα = cos β − sinβ 2 2 v α cos α 2 cos o or sinα cos β − cosα sinβ = cosα =

sin (α − β ) cosα Thus

r =

=

2vo2 cos 2 α sin (α − β ) × g cos 2 β cosα

2vo2 cos 2 α sin (α − β ) g cos 2 β For range down the inclined plane, change β to − β .

Maximum Range Equation (1) can be written as, by using

sin (α + β ) − sin (α − β ) = 2 cos α sinβ

= r

vo2 (sin = ( 2α − β ) 1 , sothat gcos 2 β

2α − β =

π 2

or α =

β 2

+

Thus

= rmax

vo2 ( 1 − sin β ) gcos 2 β

π 4

Application of Differential Equations in Mechanics

vo2 ( 1 − sinβ ) g (1 − sin 2 β )

=

vo2 g (1 − sin β )

dr Or for maximum range put dx or

cos ( 2α − β )

π

169

0

=0

or 2α − β = 2

Time of Flight Horizontol distance Time of flight = Horizontol velocity x rcos β = = vo cos α vo cos α Using (1), we get

2vo2 cos 2 α sin (α − β ) cos β g cos 2 β vo cosα Time of flight = =

2vo2 sin (α − β ) g cos β For time of flight down the plane, change β to − β .

Example: Show that the speed at a point P of the projectile is the same as attained by a particle of same mass at P falling freely from the point vertically above P on the Directrix of the trajectory.

Let vo be The Velocity of projection and v be the velocity at point p ( x, y )

of trajectory, then  ˆ + yj ˆ v= xi

v=

( v0 cosα ) iˆ + ( vo sinα − gt ) ˆj

(

)

v= v = v0 cos 2 α + ( vo sin α − gt )

2

170

v=

Differential Equations: Theory and Applications

vo2 cos 2 α + vo2 sin 2 α − 2 gtvo sinα + g 2t 2

1   = v 2 − 2 g  vot sinα − gt 2  2   Since

= y vo t sin α −

1 2 gt 2

Therefore

= v

v02 − 2 gy

Let N be the point on the Directrix vertically above MN = MP + PN

P ( x, y )

. From fig.

V02 = y + PN 2g Let v’ be the velocity of the particle in failing freely a distance NP. Using the equation of motion

v 2 −U 2 = 2 gx where u is initial velocity and v is final velocity

u 0,= x PN= , v v’, then Here= v '2 − 0 = 2 g ( PN )

 v2  = v '2 2 g  o − y   2g  2 v′= vo2 − 2 gy

= v′

vo2 − 2 gy

From v′ = v

`

(1) and ( 2 ) ,

we have

Application of Differential Equations in Mechanics

171

3.2.3. Parabola if Safety As the equation of the projectile is

1 x2 = y x tanα − g 2 sec 2 α 2 v which shows that for different values of α we have different trajectories. So, we define that the locus of all the points on different trajectories which lie at maximum distance from the point of projection O is called the parabola of safety. All the trajectories from O must touch the parabola of safety at some point. Thus, it is envelope of the family of all the trajectories from with same velocity of protestation of trajectory is

vo ,α being regarded as parameter. Equation

1 x2 = y x tan α − g 2 sec 2 α 2 vo 1 x2 = y x tan α − g 2 2 v

 gx 2 gx 2  2 tan α − x tan α +  α + 2  = 0 2vo2 2vo  

It is quadratic equation in tanα . For the equation of envelope , put discriminate = 0, i.e

 gx 2   gx 2  x2 − 4  2   y + 2  = 0 vo   vo   2 Since x ≠ 0, dividing it by x , we get

Differential Equations: Theory and Applications

172

 g  gx 2  − 2  2  y + 2  = 0 v v o   o  −

2 gy g 2 x 2 − 2 = 0 vo2 vo

g 2 x 2 2 gy = +1 vo2 vo2

vo2  vo2  x = −2  y −  g 2g  2

which is the equation of the parabola of safety having vertex

 vo2   0,   2g  2vo2 Latus Rectum = g Equation of Directrix is = y

vo2 1 + ( latus rectum ) 2g 4

= y

vo2 1  2vo2  +   2g 4  g 

y=

vo2 vo2 vo2 + = 2g 2g g

x − coordinate of focus = x − coordinate of vertex = 0 y − coordinate of focus = y − coordinate of vertex = 0 −

1 ( latus rectum ) 4

Application of Differential Equations in Mechanics

173

v 2 1 2vo2 =o − = 0 2g 4 g Therefore, focus

3.2.4. Projection with Air Resistance Let a particle of mass m be projected with an initial velocity angle α with the horizontal as shown in the figure.

vo making an

Here we restrict ourselves to the simple case of air resistance proportional to the first power of velocity. We suppose that after some time t the particle is at P(x, Y). The equations of motion are, with mass canceled, ¨

x = −kx ¨

y= − g − ky From (1)

dx = −kx dt dx = −kdt x Integrate

In ( x ) = −kt + c1 At

= t 0= x vo cos α , therefore c= in ( vo cosα ) 1

Differential Equations: Theory and Applications

174

In(x)

= −kt + in ( vo cosα )

x = vo cos α e − kt



(3)



(4)

v x= − o cosα e − kt + c2 k

=t 0= , x 0, there fore At c2 = = x

vo cosα , thus k

vo v cos α e − kt + o cosα k k

v − o 1 − e − kt cosα x= k or Now, consider equation (2) dy = − g − ky dt

(

)

dy = −dt or g + ky Integrate

in ( g + ky ) = −kt + c3

= t 0,= y vo sina, that implies that = c3 in ( g + kvo sin α ) At

In

−kt + In ( g + kvo sinα ) ( g + ky ) =

In In

g + ky = −kt g + kvo sin α

g + ky = ( g + kvo sin α ) e − kt

g g  y= − +  + kvo sinα  e − kt k k  Again integrating



(5)

Application of Differential Equations in Mechanics

175

g 1g  − −  + vo sinα  e − ky + c4 y= k kk 

at= t 0,= y 0, this implies that 1g   + vo sinα  kk 

= c4

Thus

g 1g 1g   − −  + vo sinα  e − kt +  + vo sinα  y= k kk kk   g 1g  − t −  + vo sinα  1 − e − kt y= k kk 

(

)



(6)

The equation of the path is obtained by eliminating t from equation ( 4 ) and ( 6 ) as

 xcosα  xcosα  g  g y= −   + vo sinα  + 2 In 1 − vo  k vo   K  Example A shell burst on contact with the ground and pieces from it fly in all directions with all speed up to 80 ft. / sec. Prove that a man 100 ft. is in 5

2 seconds.

danger for Solution

We know that the range of the particle is

R=

vo2 sin 2α g

= R 100 = ft., vo 80 ft / sec., g = 32 ft / sec 2 ., therefore Here 100 =

or

6400 sin 2α 32

sin 2α =

1 2

Differential Equations: Theory and Applications

176

o o or α = 15 , 75

For the range 100 ft. there are two directions of projection. Let be the times of flight respectively, then

t1 and t2

2vo sin15o g

t1 =

t2 =

and

2vo sin75o g

The man is in danger for time =

t2 − t1

2vo sin15o 2vo sin75o − g g

=

2vo sin15o − sin75o g

(

=

)

160   75 + 15   75 − 15    sin    2 cos  32   2   2 

=

(

)

5= 2cos 45o sin30o Example

5 sec. 2

Find the range of a rifle bullet when α

v

is the

elvation of projection and o is speed. show that, if the rifle is field with the same elevation and speed from a car travelling with 2vvo sinα g v towards the target, the range will be increase by Solution

We know that the range is given by

R = ( horizontal velocity )( time of flight )

R

 2vo sinα  2v02 sin α cosα v= cos α   o g g  

Application of Differential Equations in Mechanics

177

When the shell is fired from a car moving with speed v towards the target the horizontal velocity is increased by v. Horizontal velocity =

vo cosα + v

New range + R’

= ( newhorizontal time of flight )

=

 2vo sinα   g  

( vo cosα + v ) 

2v02 sin α cosα  2vo v sinα  = R +  g g   '

 2vo v sinα  r′ − R =   g   Increase in range = Example A battleship is steaming ahead with speed v and a gun is mounted on a battleship so as to point straight backward and is set at an angle of elevation α if vo is the speed of projection (relative to gun), show that the range is  2vo sinα    ( vo cosα − v ) g  

Prove also that the angle of elevation for maximum range is  v + v 2 + 8v 2  o  cos −1    4vo   Solution The ship is moving with speed v away from the target, so the horizontal velocity of the projectile will be decreased by v. Actual horizontal velocity =

vo cos α − v

Actual range = R = (horizontal velocity) (time of flight)

 2vo sinα   g  

( vo cosα − v ) 

Differential Equations: Theory and Applications

178

 2vo sinα    ( vo cosα − v ) g   Since

 2v sinα  R  o  ( vo cosα − v ) g   vo (vo sin 2α − 2v sinα ) g Therefore, for maximum range dR 2vo α) 0 = = ( vo cos 2α − v cos dα g

vo cos 2 α − vcosα = 0 vo (2 cos 2 α − 1) − vcosα = 0 vo (2 cos 2 α − 1) − v cosα = 0 2v0 cos 2 α − vcosα − v0 = 0

cosα =

v ± v 2 + 8vo2 4Vo

d 2 R 2vo = ( −2vo sin 2α + vsinα ) dα 2 g

 2v sinα  = − o  4vo cosα − g  

when

cos α =

v + v 2 + 8vo2 4Vo

dR 2vo sinα = ( v 2 + 8v02 ) = −ve dα g

Application of Differential Equations in Mechanics

179

For 0 < α < 90, sinα is positive . Thus, R is maximum when

cosα =

v + v 2 + 8vo2 4Vo

 v + v 2 + 8v 2  o  α = cos    4Vo   or 2 2   −1 v + v + 8vo  α = cos    4Vo   or Example Prove that the speed required to project a particle from a height h to fall a horizontal distance a from the point of projection is at least −1

g ( a 2 + h 2 − h)

Solution Equation of trajectory is

y xtan α − = Point

gx 2 sec 2 α 2vo2

( a, −h ) lies on it, then

= −h a tan α −

ga 2 sec 2 α 2v02

Differential Equations: Theory and Applications

180

= −h a tan α −

ga 2 (1 + tan 2 α ) 2 2v0

ga 2 tan 2 α − 2avo tanα + ga 2 − 2hvo2 ) ≥ 0 Since tanα is real therefore, discriminant ≥ 0, i.e.

(

)

4a 2 vo4 − 4 ga 2 ga 2 − 2hvo2 ≥ 0

(v

2 0

)

(

2

+ gh ≥ g 2 a 2 + h 2

)

v02 + gh ≥ g a 2 + h 2 v02 ≥ g

(

vo ≥ g

a 2 + h2 − h

(

)

a 2 + h2 − h

)

v Therefore, least value of o is

g

(

a 2 + h2 − h

)

Example A particle is launched at an angle a from a cliff of height H above sea

level. If it falls into the sea at a distance D from the base of the cliff , prove that the maximum height above the sea level is

Application of Differential Equations in Mechanics

181

Solution Equation of the trajectory is

Point (D, -H) lies on it, therefore

H D tan α − − = H + D tan α − vo2 =

gD 2 sec 2 α 2 2vo

gD 2 2vo2 cos 2 α

gD 2 2 cos 2 α ( H + D tan α )

= Now, h height above x − axis, so that vo2sin 2α h= 2g Putting the value of

h=

=

vo2 ,then

gD 2 sin 2 α 2 cos 2 α ( H + D tanα ) 2 g

D 2 tan 2 α 4 ( H + D tanα ) Height above sea level= H + h

D 2 tan 2 α = H+ 4 ( H + D tanα ) Example A front and a slip are both armed with guns which give their projectile 2 gk and guns in the fort are at height h above the guns a muzzle velocity

Differential Equations: Theory and Applications

182

d

d

in the ship. If 1 and 2 are the greatest horizontal ranges at which the fort and ship respectively can engage, prove that

d1 k +h = d2 k −h Solution Let S be the ship and F be the fort.

β and SF = r Let ∠XSF = d 2 is greatest horizontal range for gun in ship, so r is maximum range on inclined plane with inclination β . Now, r=

vo2 g ( 1 + sinβ )

vo = 2 gk in the above equation, then 2 gk 2k = r = g ( 1 + sinβ ) 1 + sin β Put

r + r sin β = 2k Since form the figure h = r sin , therefore r+h= 2k Also from right angle triangle SAF, we have 2 r= h 2 + d 22

So that

Application of Differential Equations in Mechanics

h 2 + d 22 =

( 2k − h )

2

183

= 4k 2 − 4kh + h 2

d 22 = 4k 2 − 4kh = 4k ( k − h )



(1)

For gun in fort, change h to – h, then

= d12 4k ( K + h )



(2)

Dividing (2) by (1), then

d1 k +h = d2 k −h Example The horizontal rang of a ball for two different projection is same and

h and h

2 are the greatest height in the two parts for which equal to R. if 1 this is possible prove that

R 2 = 16h1h2 Solution

v Let R be the range, when α is angle of the projection and o is velocity of projection , then R=

2vo2 sinα cosα g

π Let R’ be the range when angle of projectile is 2

−α,

then

π  π  2vo2 sin  − α  cos  − α  2  2  R′ g

or

R′ =

2vo2 sinα cosα g

π −α which show that R is the same when angle of projection are α and 2 . h and h2 be the maximum height in these cases respectively, therefore Let 1

Differential Equations: Theory and Applications

184

h1 =

vo2 sin 2 α 2g

π  vo2 sin 2  − α  2 2 2  vo cos α h2 = = 2g 2g Hence

16h1h2 = 16

vo2 sin 2 α  vo2 cos 2 α    2g  2g 

vo4 sin 2 α cos 2 α 4= R2 g i.e R 2 = 16 h1h2 Example Prove that if time of flight of a bullet over a horizontal range R is T seconds then the inclination of direction to the horizontal is

 gt 2  tan −1    2r  Solution We know that 2 sin α T= o g

R=

gT 2 2R

2vo2 sinα cosα g

4vo2 sin 2 α g tanα = 2 g 4vo sinα cosα

Hence

Application of Differential Equations in Mechanics

185

 gT 2    2R 

α = tan −1 

Example A cricket ball thrown from height of 6 feet at an angle of 30 with the horizontal speed of 60 ft. / sec . is caught by another player at height 2 feet from the ground. How far apart were the two men? o

Solution Let two men are x feet apart. As the ball is caught at a height of 2 feet from the ground, so the distance it falls vertically is 6-2 = 4ft. The point ( x, −4 ) lies on the trajectory. Now

= y

( vo sinα ) t −

1 2 gt 2

y= −4 , v0 = 6 − ft / sec., α= 3= 0 , t= ? 1 −4 ==( 60 sin30 ) t − ( 32 ) t 2 2 Here

8t 2 −15t − 2 = 0 or

(t-2) (8t+1) = 0

t = 2 or t = −

1 8

1 t ≠− , 8 therefore t = 2. Since x = Horzontal distance = ( Horizontal velocity )( time )

=

vo cosα ) t (= x = 60 3 ft.

( 60 cos30 ) ( 2 ) 0

Differential Equations: Theory and Applications

186

3.3. SUMMARY AND DISCUSSION Rectilinear Motion. The motion of a particle along a straight line is called rectilinear motion.

Vertical Motion under Gravity For a failing body, the acceleration is constant. It is called acceleration due

to gravity and is denoted by ‘g’. In FPS system g = 32 ft / sec .In CGS 2 2 system g =981 cm / sec and in MKS system g = 9.81 m / sec . If the body is projected vertically upward then acceleration g is negative. For falling bodies equation of motion is v= u + gt 2

x= ut +

1 2 gt 2

v2 − u 2 = 2 gx

Distance traveled in nth Second x and x b

2 Let 1 e the distance travelled in the first n and n – 1 seconds respectively, then from the equation of motion

1 x= ut + at 2 2 1 x= un + an 2 1 2 then 1 2 x2= u ( n − 1) + a ( n − a ) 2 and Distance travelled in the nth second =

x1 − x2 = un +

=

un +

x1 − x2

a 2  a 2 n − u ( n − a ) + ( n − 1)  2 2  

a 2  a  n −  un − u + n 2 − 2n + 1  2 2  

(

)

Application of Differential Equations in Mechanics

187

1 u + a ( 2n − 1) = 2 Motion with variable Acceleration (i) Time Dependent Acceleration (ii) Velocity Dependent Acceleration (iii) Distance Dependent Acceleration

Projectile An object thrown into the space with certain velocity from a gun or dropped from a moving plane under the action of gravity is called a projectile. Thus, a projectile move with a constant horizontal velocity and at the same falls freely under the action of gravity. The path of projectile is called the trajectory. Horizontal and vertical component of velocity at any time t are , are the parametric equation of trajectory. (i) Vertex

 v sinα cos α vo2 sin 2 α  ,   g 2g   2 0

(ii)

Latus Rectum

2vo2 cos 2 α 4a = g Latus Rectum = (iii) Maximum Height =

vo2 sin 2 α 2g (iv) Equation of Directrix

y=

vo2 2g (v) Focus



Differential Equations: Theory and Applications

188

 v 2 sinα cos α   v2  v2 v2 = 0 , − o cos 2α  =  o sin 2α − o cos 2α  g 2g 2g    2g  The time taken by the particle to reach the horizontal plane through the point of projection is called its time of flight. 2v sinα t= o g is the time of flight. (vii) Range (a) Horizontal Range The horizontal range of the projectile is the horizontal distance described by the particle during the time of flight. Therefore Range R = (horizontal velocity) (time of flight)

 2v sinα  vo cos α  o  g   vo2 2 ( sin α cosα ) 2vo2 sin 2α = g g This range can also be obtained by putting y = 0 in the Cartesian equation of trajectory i.e putting y = 0 in the equation

Also, it is notable that this range will remain same if we replace

α by α=

π

2

−α

(b) Maximum Range

vo2 sin 2α g Range R =

dr =0 For maximum range dα

Application of Differential Equations in Mechanics



2vo2 cos 2 α = 0 g



cos 2α = 0



2α = 2



189

π

π α= 4

Therefore

Rmax =

vo2 g

Time of Flight Horizontol distance Time of flight = Horizontol velocity x rcos β = = vo cos α vo cos α Using (1), we get

2vo2 cos 2 α sin (α − β ) cos β g cos 2 β vo cosα Time of flight = =

2vo2 sin (α − β ) g cos β For time of flight down the plane, change β to − β .

Parabola of Safety As the equation of the projectile is = y x tanα −

1 x2 g 2 sec 2 α 2 v

which shows that for different values of α we have different trajectories. So we define that the locus of all the points on different trajectories which lie at maximum distance from the point of projection O is called the parabola of safety.

CHAPTER

4

ELLIPTIC DIFFERENTIAL EQUATION

CONTENTS 4.1. Introduction..................................................................................... 192 4.2. Boundary Value Problem (BVPs)...................................................... 195 4.3. Some Important Mathematical Tools................................................ 197 4.4. Properties Of Harmonic Functions................................................... 199 4.5. Separation Of Variables ................................................................... 210 4.6. Dirichlet Problem For A Rectangle................................................... 212 4.7. The Neumann Problem For A Rectangle .......................................... 215 4.8. Interior Dirichlet Problem For A Circle ............................................ 217 4.9. Exterior Dirichlet Problem For A Circle ........................................... 222 4.10. Interior Neumann Problem For A Circle......................................... 227 4.11. Solution Of Laplace Equation In Cylindrical Coordinates .............. 229 4.12. Solution Of Laplace Equation In Spherical Coordinates................. 238 4.13. Miscellaneous Example................................................................. 247 4.14. Summary And Discussions............................................................. 276

192

Differential Equations: Theory and Applications

4.1. INTRODUCTION In previous topics, we have seen the classification of second order partial differential equation into elliptic, parabolic and hyperbolic types. In this chapter we shall consider various properties and technique for solving Laplace and Poisson equations which are elliptic in nature. Various physical phenomena are governed by the well-known Laplace and poison equations. A few of them, frequently encountered in applications are: study heat condition, magnetic potential, torsion of prismatic shaft, bending of prismatic beams, distribution of gravitational potential, etc. In the following two sub-section, we shall give the derivation of Laplace and Poisson equations in relation to the most frequently occurring physical situation , namely the gravitational potential.

4.1.1. Derivation of Laplace Equation m and m1 Consider two particles of masses situated at Q and P separated by a distance r as shown in . According to Newton’s universal law of gravitational, the magnitude of the force, proportional to the product of their masses and inversely proportional to the square b of the distance. Between is given mm1 r 2 (1)  where G is the gravitational constant. It r represent the vector PQ, assuming unit mass at Q and G = 1, the force at Q due to the mass at p F =G

is given by

mr m  F= − 13 = ∇ 1  r  r  (2) which is called the intensity of the gravitational force. Suppose a particle of m unit mass moves under the attraction of a particle of mass 1 at P infinity up to Q; then the work done by the force F is r

r

m1 m  ∫∞F .dr = ∫∞∇  r1  .dr = r (3)

Elliptic Differential Equation

193

This is defined as the potential V at Q due to a particle at P and is denoted by

V=

m1 r (4)

Figure 1

From Eq. 2, the intensity of the force at P is F = −∇V (5) Now, if we consider a system of particles of masses which are at distances

m1 , m2 , …., mn

r1 , r2 , …, rn respectively, then the force of attraction

per unit mass at Q due to the system is n

n mi m = ∇∑ i ri =i 1 = i 1 ri (6) The work done by the force acting on the particle is

= ∑∇

r

n

mi = −V i =1 ri (7) Therefore,

∫F .dr = ∑



mi n 2 mi = ∑∇ = 0, ri ≠ 0 ri i 1 ri =i 1 = (8) n

∇2 V = − ∇2 ∑ where

194

Differential Equations: Theory and Applications

∂2 ∂2 ∂2 + + ∂x 2 ∂y 2 ∂z 2 is called the Laplace operator. In the case of continuous distribution of matter of density p in a volume r , we have

2 ∇= div ∇=

.

V ( x, y, z ) =∫∫ ∫

p (ξ ,η , ζ ) r

r

{

dr (9)

r = ( x − ξ ) + ( y −η ) + ( z − ζ ) 2

2

where can be verified that

}

2 1/2

and Q is outside the body. It

∇ 2V = 0 (10) which is called the Laplace equationA2I.

4.1.2. Derivation of Poisson Equation m , m , … mn . Consider a closed surface S consisiting of particle of masses 1 2 n

mi = M ∑ Q S . i = 1 Let be any point on Let be the total mass inside S , and let g1 , g 2 , …, g n be the gravity field at Q due to the presence of m1 , m2 …, mn n

gt = g , ∑ S i = 1 respectively within . Also, let the entire gravity field at Q. Then, according to Gauss law, we have .

∫ ∫ g.dS = − 4π GM S

(11)

.

M =∫ ∫ ∫ pdτ , p

r where is the mass density function and τ is the volume in which the masses are distributed through. Since the gravity field is conservative. We have

g = ∇V (12) where V is a scalar potential. But the Gauss divergence theorem states that

Elliptic Differential Equation .

195

.

∫ ∫g .dS = ∫ ∫ ∫ ∇.g dτ S

(13) Also. Eq. (11) gives τ

.

.

S

τ

∫ ∫g ô. dS =−4π G ∫ ∫ ∫ pd

(14)

Combining Eq. (13) and (14), we have .

∫ ∫ ∫ (∇.g + 4π Gp ) dτ =0 τ

Implying

∇.g =−4π Gp =∇. ∇ v Therefore,

∇2 V = − 4π Gp (15) This equation is known as poisson’s equation .

4.2. BOUNDARY VALUE PROBLEM (BVPS) The function V , whose analytical form we seek for the problems stated in section in addition to satisfying the Laplace and Poisson equation in a 3

bounded region IR in R , should also satisfy certain boiundary conditions on the boundary ∂IR of this region Such problems are referred to as boundary value problems (BVPs) for Laplace and Poisson equations. We shall denote the set of all boundary points of IR by ∂IR . By the closure of IR, we mean the set of all interior points of IR together with its boundary

points and is denoted by IR, we mean the set of all interior point together

IR IRU ∂IR. Symbolically, = If a function are continuous.

then all its derivative of order n

0 If it belong to c , then we mean f is continuous.

196

Differential Equations: Theory and Applications

There are mainly three types of boundary value problems for Laplace and is specified on the boundary ∂IR of some finite region 2 0 IR , the problem of determining a function ψ ( x, y, z ) suchthat ∇ ψ = ψ f on ∂IR is called the boundary value within IR and satisfying = problem of first kind, or the Dirichlet problem. For example, finding the steady state temperature within the region IR when no heat sources or sinks equation. If

are present and when the temperature is prescribed on the boundary ∂IR, is Dirichlet problem. Another example would be to find the potential inside the region IR when the potential is specified on the boundary ∂IR.These two examples correspond to the interior Dirichlet problem.

and is prescribed on the boundary ∂IR of a finite ψ ( x, y , z ) simply connected region IR, determinanting function which Similarly, if

0 outside IR and is such that ψ = f on ∂IR, is called an satisfies ∇ ψ = exterior Dirichlet problem. For example, determination of the distribution of the potential outside a body whose surface potential is prescribed , is an exterior Dirichlet problem. The second type of BVP is associated with ψ ( x, y , z ) Von Neumann. The problem is to determine the function 2

2 = so that ∇ ψ 0 within IR while ∂ψ∂n. is specified at every point of ∂IR, where ∂ψ / ∂n denotes the normal derivative of the field variable ψ .

This problem is called the Neumann problem. If ψ is the temperature,

∂ψ / ∂n is the heat flux representing the amount of heat crossing per unit volume per unit time along the normal direction, which is zero when insulated . The third type of BVP is concerned with the determination of the function 2 ψ ( x, y , z ) 0 within IR, while a boundary condition of such that ∇ ψ = the form ∂ψ / ∂n + hψ = f , where h ≥ 0 is specified at every point of ∂IR. This is called a mixed BVP or Churchill’s problem. If we assume

Newton’s law of cooling, the heat lost is hψ , where ψ is the temperature difference from the surrounding medium and h > 0 is a constant depending on the medium. The heat f supplied at a point of the boundary is partly conducted into the medium and partly lost by radiation to the surrounding. Equating these amounts, we get the third boundary condition. The third

Elliptic Differential Equation

197

type of BVP is concerned with the determination of the function ψ ( x, y, z )

2 0 within IR, while a boundary condition of the form such that ∇ ψ = ∂ψ / ∂n + hψ = f , where h ≥ 0 is specified at every point of ∂IR. This is called a mixed BVP or Churchill’s problem. If we assume Newton’s law

of cooling, the heat lost is hψ , where ψ is the temperature difference from the surrounding medium and h > 0 is a constant depending on the medium.

The heat f supplied at a point of the boundary is partly conducted into the medium and partly lost by radiation to the surrounding. Equating these amounts, we get the third boundary condition

4.3. SOME IMPORTANT MATHEMATICAL TOOLS Among the mathematical tools, we employ in deriving many important results. The Gauss divergence theorem plays a vital role, which can be stated as follows : Let ∂IR be a closed surface in the xyz − space and IR denote the bounded region enclosed by ∂IR in which F is a vector belonging to c1 in IR and continuous on IR. Then . .

.

∂ IR

IR

∫∫ F . nˆ dS = ∫ ∫ ∫ ∇.F dV where dV is an element of volume, dS is an element of surface area . and nˆ the outward drawn normal. Green’s identities which follow form divergence theorem are also useful and they can be derived as follows : Let F = f φ , where f is a vector function of position and φ is a scalar function of position. Then .

. .

IR

∂ IR

∫ ∫ ∫ ∇. ( f φ ) dV =∫ ∫ nˆ. f φ dS Using the vector identity

∇. ( f φ ) = f . ∇φ + φ∇. f we have

198

Differential Equations: Theory and Applications

.

∫ ∫ ∫ f= .∇.φ dV IR

. .

.

∂ IR

IR

∫ ∫ n. f φ dS ∫ ∫ ∫ φ ∇. f dV

If we choose f = ∇ψ , the above equation yields .

∫ ∫∇φ .∇ψ= dV IR

. .

.

∂ IR

IR

2 ∫∫φ nˆ.∇ψ dS − ∫ ∫ ∫ φ ∇ ψ dV

( 17 )



Noting that nˆ.∇ψ is trhe derivative of ψ in the direction of nˆ , we introduced the notation

nˆ .∇ψ = ∂ψ / ∂n Into Eq. (17) to get

( 18 a )



This equation is known as Green’s first identity. Of course, it is assumed

that both φ and ψ possess continuous second derivatives.

Interchanging the role of φ and ψ , we obtain form relation (18 a) the equation ( 18 b )

Now, subtracting Eq. (18b) from Eq. (18) we get

( 19 )

This is known as Green’ second identity. If we set φ = ψ in Eq. (18a) we get .

∫ ∫ ∫ (= ∇φ ) dV IR

2

. .

∫ ∫φ ∂ IR

.

∂φ dS − ∫ ∫ ∫ φ∇ 2 φ dV ∂n IR



which is a special case of Green’s first identity.

( 20 )

Elliptic Differential Equation

199

4.4. PROPERTIES OF HARMONIC FUNCTIONS Solutions of Laplace equation are called harmonic functions which possess a number of interesting properties, and they are presented in the following theorems. Theorems. If a harmonic function vanishes everywhere on the boundary , then it is identically zero everywhere. 2 0 in IR . Also, if φ = 0 Proof. If φ is a harmonic function, then ∇ φ = φ 0 in= IR IRU ∂IR. Recalling Green’s first on ∂IR, we shall show that= ( 20 ) identity, i.e., Eq. .

. .

∫ ∫ ∫ ( ∇= φ ) dV 2

∫∫ φ ∂ IR

IR

.

∂φ dS − ∫ ∫ ∫ φ ∇ 2 φ dV ∂n IR



And using the above facts we have, at once, the relation .

∫ ∫ ∫ ( ∇φ ) dV =0 2

IR

( ∇φ ) Since

2

is positive, it follows that the integral will be satisfied only ∇ φ = 0. if This implies that φ is a constant in IR . Since φ is continuous in IR and φ is zero on ∂IR, it follows that φ = 0 in IR.

0 on ∂IR, Theorem. If φ is a harmonic function in IR and ∂φ / ∂n =

then φ is a constant in IR .

Proof. Using Green’s first identity and the data of theorem, we arrive at .

∫ ∫ ∫ ( ∇φ ) dV =0 2

IR

0, i.e.,φ is a constant in IR. Since the value of φ is Implying ∇φ =

0, it is implied that φ is a not known on the boundary ∂IR while ∂φ / ∂n = constant on ∂IR and hence on IR . Theorem. If the Dirichlet problem for a bounded region has a solution, then it is unique .

200

Differential Equations: Theory and Applications

Proof. If then

φ1 and φ2 are two solutions of the interior Dirichlet problem,

∇ 2φ1 = 0 in IR; φ1 = f on ∂IR ∇ 2φ2= 0 in IR; φ2 = f on ∂IR

ψ= φ1 − φ2 . Then ∇ 2φ =∇ 2φ1 −∇ 2φ2 = 0 in IR; Let

ψ = φ1 − φ2 = f − f = 0 on ∂IR; Therefore,

∇ ψ = 0 in IR, ψ = 0 on ∂IR 2

φ = φ2 . Hence Now using, we obtain φ = 0 on IR, which implies that 1 the solution of the Dirichlet problem is unique. Theorem. If the Neumann problem for a bounded region has a solution, then it is either unique or it differ from one another by a constant only. Proof. Let Then we have

φ1 and φ2 be two distinct solution of the Neumann problem.

∇ 2φ1 = 0 in IR;

∂φ1 = f on ∂IR, ∂n

∇ 2φ2 = 0 in IR;

∂φ2 = f on ∂IR, ∂n

ψ= φ1 − φ2 . Then ∇ 2ψ = ∇ 2φ1 −∇ 2φ2 = 0 in IR Let

∂φ ∂φ1 ∂φ2 = − = 0 on ∂IR ∂n ∂n ∂n IR,i.e.,φ1 − φ2 = Hence from theorem ψ is a constant on constant. Therefore, the solution of the Neumann problem is not unique. Thus, the

Elliptic Differential Equation

201

solutions of a certain Neumann problem can differ from one another by a constant only.

4.4.1. The Spherical Mean p ( x, y , z ) Let IR be a region bounded by ∂IR and let be any point in IR. S ( p, r ) Also, let represent a sphere with centre at P and radius r such that it lies entirely within the domain IR as depicted in. Let u be a continuous function in IR. Then the spherical mean of u denote byu u is defined as 1 u (r ) = 4π r 2

.

.

∫ ∫ u ( Q ) dS S ( p ,r )



( 21 )

Figure 2

Q (ξ ,η , ζ ) S ( p, r ) where is any variable point on the surface of the sphere and dS is the surface element of integration. For a fixed radius r , the value u (r ) is the average of the value of u taken over the spehere S 9 p, r ) , and hence it is called the spherical mean. Taking the origin at p, in terms of spherical polar coordinates, we have

ξ= x + r sin θ cosφ

Differential Equations: Theory and Applications

202

η= y + r sin θ sinφ ζ = z + r cos θ Then the spherical mean can be written as u ( r=)



π

1 u ( x + r sin θ cos φ , y + r sinθ sinφ , z + r cosθ ) r 2 sin θ dθ dφ 4π r 2 =θ ∫0=θ ∫0

S ( p, r ) , u too Also since u is a continuous on is a continuous function

of r on some interval 0 < r ≤ R, which can be verified as follows.

u (r )

2π π u ( Q ) 2π π 1 sin θ d θ dφ u ( Q ) u ( Q ) sinθ dθ dφ = = 4π r 2 =φ∫0=θ ∫0 4π ∫0 ∫0

Now, taking the limit as r → 0, Q → p, we have

Ltr →0 u ( r ) = u ( p )

Hence, u is continuous in 0 ≤ r ≤ R.

( 22 )

4.4.2. Mean Value Theorem for harmonic Functions p ( x, y , z ) Theorem. Let u be harmonic in a region IR. Also, let be a S ( p, r ) given point in IR and be a sphere with centre at p such that S ( P, R ) is completely contained in the domain of harmonicas of u. Then u= ( p ) u= (r )

1 4π r 2

.

∬ u ( Q ) dS

s( p ,r )

u (r ) Proof. Since u is harmonic in IR, it spherical mean is continuous IR in and is given by u (r )

.

1 1 = u Q dS 2 ∬ ( ) 4π r s( p ,r ) 4π r 2

Therefore,

2π π

∫ ∫ u (ξ ,η , ζ ) r 0 0

2

sinθ dθ \ dφ

Elliptic Differential Equation

du ( r ) = dr =

1 4π r 2

ππ

∫ ∫ ( uξ ξ

r

203

+ uηηr + uζ ζ r ) sinθ dθ dφ

0 0

2π π

∫ ∫ ( uξ

sinθ cosφ +uτ sinθ sinφ + uζ cosθ ) sin θ d θ dφ

0 0

( 23 )

Nothing that sin θ cosφ ,sin θ sin φ and cos θ are the direction cosines of nˆ on S ( p , r ) , the normal

∇u= iuξ + juη + kuξ , nˆ=

( in1 , jn2 , kn3 ) ,

The expression within the parenthesis of the integrand of eq. (23) can be written as ∇. nˆ . Thus

du ( r ) 1 = 4π r 2 dr 1 = 4π r 2 =

.

ˆ ∬ ∇u. nr

2

sin θ dθ dφ

s( p ,r )

.

∬∇u. nˆ dS

s( p ,r ) .

1 ∬ ∇.∇u dV ( by divergencetheorem ) 4π r 2 V ( p ,r ) .

1 = ∇ 2u dV = 0 ( sinceu is harmonic ) 2 ∬ 4π r V ( p ,r )

du = 0, Therefore, dr implying u is constant. Now the continuity of u at r = 0 gives, from Eq. (22), the relation .

1 u (r ) −u ( p ) = ∬ u (Q ) d S 4π r 2 V ( p ,r )



( 24 )

4.4.3. Maximum-Minimum Principal and Consequences Theorem. Let IR be a region bounded ∂IR. Also. Let u be a function which is continuous in a closed region IR and satisfies the Laplace equation

204

Differential Equations: Theory and Applications

∇2 u = 0 in the interior of IR . Further, if u is not constant everywhere on IR , then the maximum and minimum values of u must occur only on the

boundary ∂IR .

Proof. Suppose u is a harmonic function but not constant everywhere on IR. if possible , let u attain its maximum value M at some interior point p in IR.Since M is the maximum of u which is not a constant , there

S ( p, r ) should exist a spehere about p such that some of the values of u S ( p ,r ) , on must be less than M . But by the mean value property, the u on S ( p , r ) , value of u at P is the average of the values of and hence it u = M at p. Thus x u is less than M . This contradicts the assumption that must be constant over the entire sphere

S ( p, r ) .

Let Q be any other point inside IR which can be connected to P by an arc lying entirely within the domain IR. By convergence this arc with spheres and suing the Henie-Borel Theorem to choose a finite number of covering spheres and repeating the argument given above , we can arrive at the conclusion that u will have the same constant value at Q as at P . Thus u cannot attain a maximum value at any point inside the region IR. Therefore, u can attain its maximum value only on the boundary ∂IR. By convergence this arc with spheres and suing the Henie-Borel Theorem to choose a finite number of covering spheres and repeating the argument given above , we can arrive at the conclusion that u will have the same constant value at Q as at P . Thus u cannot attain a maximum value at any point inside the region IR. Therefore, u can attain its maximum value only on the boundary ∂IR. Some important consequence of the maximum-minimum principle are given in the following theorems. Theorem.

u1 and u2 be two solutions of the Dirichlet problem and let f1 and f 2 be a given continuous functions on the boundary ∂IR such that Proof. Let

∇ 2u1= 0 in IR ; u1 = f1 on ∂IR ,

Elliptic Differential Equation

205

∇ 2 u2 = 0 in IR ; u 2= f 2 on ∂IR , Let

u= u1 − u2 . Then,

f1 − f 2 on ∂IR ∇ 2 u =∇ 2 u1 −∇ 2 u 2 = 0 in IR ; u = Hence, u is a solution of the Dirichlet problem with boundary condition

u= f1 − f 2 on ∂IR. By the maximum-minimum principle , u attains the maximum and minimum values on ∂IR. Thus at any interior point in IR, we shall have , for a given ε > 0, − ε < u min ≤ u ≤ u max < ε Therefore,

u < ε . in IR, implying u1 − u2 < ε on IR Thus, small changes in the initial data bring about an arbitrary small change in the solution. This completes the proof of the theorem. Theorem. Let

{ fn }

be a sequence of the functions, each of which is {f } continuous on IR and harmonic on IR. if the sequence n convergence uniformly on ∂IR, then it converges uniformly on IR .

{f } Proof. Since the sequence n converges uniformly on ∂IR, for a given ε > 0 , we find an integer N such that f n − f m ε for n, m N that

Hence, form stability theorem, for all n, m > N it follows immediately

f n − f m < ε In IR Therefore,

{ f n } convergence uniformly on

IR.

Example. Show that if the two-dimensional Laplace equationc¬§X ∇2 u = 0 is transformed by introducing plane polar coordinates r , θ defined = x r= cos θ , y r sin θ , it takes the form by the relations

206

Differential Equations: Theory and Applications

∂ 2u 1 ∂u 1 ∂ 2u + + = 0 ∂ r 2 r ∂r r 2 ∂ θ 2 Solution. In many practical problems, it is necessary to write the Laplace equation‑«  X in the various coordinates systems. For instance, if the boundary of the region ∂IR is circle, then it is natural to use polar

= cosθ , y r sin θ . Therefore, coordinates defined by x r=  y r2 = x2 + y 2 , θ = tan −1   . x

rx = cosθ , ry = sinθ , θ x = -

sin θ cosθ θ r , y= r

Since

sinθ   u =u ( r , θ ) u x =ur rx + uθ θ x = ur cos θ − uθ  r   Similarly,

cosθ   uy = ur ry + uθ θ y =  ur sinθ + uθ  r   Now for the second order derivatives, sinθ  sinθ   sinθ    u xx = ( ux )x = ( ux )r rx + ( ux )θ θ x =  ur cosθ −uθ  cos θ +  ur cos θ − uθ  −  r r r θ  r   

Therefore,

sinθ sinθ   u xx=  urr cos θ − uθ r + uθ 2  cosθ r r   sinθ cosθ   sinθ   +  urθ cosθ − ur sinθ − uθθ − uθ  −  r r  r  (25)  Similarly, we can show that cosθ cosθ   u yy =  urr sin θ + urθ − uθ 2  sinθ r r   cosθ sinθ   cosθ   +  urθ sinθ + ur cosθ + uθθ − uθ   r r  r  

( 26 )

Elliptic Differential Equation

207

By adding Eq. (25) and (26) and equating to zero, we get

1 1 u xx + u yy = urr + ur + uθθ = 0 ( 27 ) r r2 which is the Laplace equation in polar coordinates. One can observe that the Laplace equation in the Cartesian coordinates has constant coefficients only, whereas in polar coordinates , it has variable coefficients. Example. Show that in cylindrical coordinates r , θ , z defined by the = x r= cosθ , y r= sin θ , z z , the Laplace equation ∇ 2u = 0 takes relations the form ∂ 2u 1 ∂u 1 ∂ 2u ∂ 2u + + + = 0 ∂ r 2 r ∂r r 2 ∂θ ∂z 2 Solution. The Laplace equation in Cartesian coordinates is

∇2 u =

∂ 2u ∂ 2u ∂ 2u + + =0 ∂x 2 ∂y 2 ∂z 2

The relations between Cartesian and cylindrical coordinates gives

r2 = x 2 + y 2 ,θ = tan −1 ( y / x ) , z = z Since

u = u ( r ,θ , z )

u x = u x rx + uθ θ x + u z z x = ur cos θ − u

 sin θ  θ   r 

 cos θ  u y = ur ry + uθ θ y + u z z y = ur sin θ − uθ    r  u z = ur rz + uθ θ z + u z= u z For the second order derivatives, we find

u xx = ( ux )x = ( ux )r rx + ( ux )θ θ x + ( ux ) z z x

  sinθ = ur cosθ − uθ   r 

   sinθ   cos θ + ur cosθ − uθ   r  r 

sinθ sinθ   =  urr cosθ − urθ + uθ 2  cosθ r r  

   sinθ   −  r   θ 

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Differential Equations: Theory and Applications

sinθ cosθ  sinθ   +  urθ cosθ −ur sinθ −uθθ − uθ  −  r r  r   Similarly

(28)

u yy = ( u y ) y =( u y ) ry + ( u y ) θ y + ( u y ) z y r

θ

z

cosθ cosθ   =  ur sinθ + uθ − uθ 2  sin θ r r   cosθ sinθ  cos θ  +  urθ sinθ + +ur cosθ + uθθ − uθ  r r  r 

   (29)

u zz = u zz (30) Adding Eq. (28)-(30), we obtain

1 1 ∇ 2u = urr + ur + 2 uθθ + u zz r r (31) Example. Show that in spherical polar coordinates r , θ , φ defines by = θ cos φ , y r= sinθ sin φ , z r cos θ , , the Laplace the = relations x r sin 2 0 takes the form equations ∇ u =

∂ ∂r

1 ∂  ∂u  1 ∂ 2u  2 ∂u  r + sin + = 0 θ     ∂θ  sin 2 θ ∂φ 2  ∂r  sinθ ∂θ  Solution. In Cartesian coordinates , the Laplace equation is

2

∇ u = u xx + u yy + u xx + u zz = 0 In spherical coordinates,

u = u ( r , θ , φ ) , r 2 = x 2 + y 2 + z 2 , cos θ =

z , tan φ = y / x r .

It can be easily verified that

θx = φx =

cosθ cosφ cosθ sinφ sin θ , θy = ,θ z = − r r r sinφ cosφ φz 0 = ,φ y = r sinφ r sinφ

Elliptic Differential Equation

Now,

u x = u x rx + uθ θ x + uφ φx = ur sin θ cosφ + uθ

cos θ cos φ sinφ − uφ r rsinφ

u y = u y ry + uθ θ y + uφφ y = ur sin θ sin φ + uθ

cos θ sinφ uφ sinφ + r r sinφ

 sinθ  u z = ur rz + uθ θ y + uφ φ y = ur cos θ + uθ  −  r   For the second order derivatives,

u xx = ( u x ) r rx + ( u x ) θ θ x + ( u x )φ φx cosθ cosφ sinφ   =  ur sin θ cos φ + uθ − uφ  (sin θ cosφ ) r r sin θ  r  cosθ cosφ sinφ   sin θ cos φ   =  ur sin θ cos φ + uθ − uφ    r r sin θ  θ  r  

cosθ cos φ sinφ   sin θ   =  ur sin θ cos φ + uθ − uφ  −  r r sin θ  φ  rsin θ  

(

= sin θ cos φ 2

2

)

u rr

cos 2 θ cos 2 φ sin 2 φ + uθθ + 2 2 uφφ r2 r sin θ

 2sin θ cos θ cos 2 φ   2sinφ cos φ  + urθ   + urφ  −  r r      cos 2 θ cos 2 φ sin 2 φ   2cos θ cos φ sinφ  + uθφ  − +   + ur  − r 2 sin θ r r      sin φ cos φ cos 2 φ cos φ sinφ sin φ cos φ  + uφ  + + 2 2  r2 r 2 sin 2 θ r sin θ  

u yy =( u y ) ry + ( u y ) θ θ y + ( u y ) φ y r

φ

cosθ sin φ cos φ   =  ur sin θ sin φ + uθ + uφ , r r sin θ  

(32)

209

210

Differential Equations: Theory and Applications

cos θ sin φ cos φ  cos θ sin φ  +  ur sin θ sin φ + uθ + uφ  sin θ θ r r 

cos θ sin φ cos φ  cos φ  +  ur sin θ sin φ + uθ + uφ  sin θ φ r sin θ r 

(

)

= sin 2 θ sin 2 φ urr +

cos 2 θ sin 2 φ cos 2 φ + u uφφ θθ r2 r 2 sin 2 θ

 2sin θ cos θ sin 2 φ   2 cos φ sin φ  = + u rθ   + urφ   r r      2 cos θ cos φ sin φ +uθφ  r 2 sin θ 

 cos 2 θ sin 2 φ cos 2 φ   + u +   r r r   

 sin φ cos φ sin φ cos φ cos 2 θ sin φ cos φ  + uθ  − − 2 2 −  r2 r sin θ r 2 sin θ   Similarly,

u zz = ( u z )r rz + ( u z )θ θ z + ( uz )φ φz sinθ  sinθ  sin θ   =  ur cos θ − uθ  ( cos θ ) +  ur cos θ − uθ  — r r r θ r   = urr cos 2 θ − urθ

2sin θ cosθ sin 2 θ sin 2θ cosθ sin θ + uθθ + u + uθ r 2 r r r r2 (34)

Adding Eq. (32) - (34), we obtain

∇ 2u = urr +

1 1 2 cos θ u + 2 2 uφφ + ur + 2 uθ = 0 2 θθ r r sin θ r r sinθ

which can be rewritten as

= ∇ 2u

∂  2 ∂u  1 ∂  ∂u  1 ∂ 2u r + sin + = 0 θ     ∂r  ∂r  sin θ ∂θ  ∂θ  sin 2 θ ∂φ 2

(35)

4.5. SEPARATION OF VARIABLES The method of Separation of variables is applicable to a large number of classified linear homogeneous equations. The choice of the coordinate

Elliptic Differential Equation

211

system in general depends on the shape of the boundary For example, consider a two –dimensional Laplace equation in Cartesian coordinates .

∇ 2u = u xx + u yy = 0

(36)

We assume the solution in the form

u ( x, y ) = X ( x ) Y ( y )

(37)

Subtracting in Eq. (36), we get X ′′Y + Y ′′X = 0 i.e. X ′′ Y ′′ = = k X Y where k is separation constant. Three cases arise. 2 Case I Let k = p , p is real. Then

d2X 2 d 2Y p = X 0 and += p 2Y 0 2 2 dx dy whose solution is given by

= X c1e px + c2 e − px and

= Y c3 cos py +c4 sin py Thus the solution is

(

)

u ( x, y ) = c1e px + c2 e − px ( c3 cos py +c4 sin py ) Case II Let k = 0. Then

d2X d 2Y = 0= and 2 0 dx 2 dy Integrating twice, we get

= X c5 x + c6 and

= Y c7 y + c8



(38)

Differential Equations: Theory and Applications

212

The solution is therefore,

u ( x, y ) = ( c5 x + c6 )( c7 y + c8 )

(39)

2 Case III let k = − p . proceeding as in Case I, we obtain

= X c9 cos px + c10 sin px py

− py

= Y c11e + c12e

Hence, the solution in this case is

(

u ( x, y ) = ( c9 cos px + c10 sin px ) c11e + c12e py

− py

) (40)

c ( i = 1, 2,..,12 ) In all these cases, 1 refer to integration constants, which are calculated by using the boundary conditions.

4.6. DIRICHLET PROBLEM FOR A RECTANGLE The Dirichlet problem for a rectangle is defined as follows: 2 u 0 , 0 ≤ x ≤ a ,0 ≤ y ≤ b PDE: ∇=

BCs:

u= ( x, b ) u= ( a , y ) 0, u= ( 0 , y ) 0 , u= ( x ,0 ) f ( x )

(41)

This is an interior Dirichlet problem. The general solution of the governing PDE, using the method of variables separable, is discussed in previous section. The various solution of the Laplace equation are given by Eq. (38-40). Of three solutions, we have to choose that solution which is consistent with the physical nature of the problem and the given boundary conditions as despaired .

Figure 3: Dirichlet Boundary Conditions

Elliptic Differential Equation

213

Consider the solution given by Eq. (38)

(

)

u ( x, y ) = c1e px + c 2e px ( c3 cos py +c4 sin py ) Using the boundary condition :

u ( 0 , y ) = 0, we get

0 ( c1 + c2 )( c3 cos py + c4 sin py ) = which means that either

= c1 + c2 0 or c3 cos py += c4 sin py 0. But

c3 cos py + c4

sin py ≠ 0;

Therefore

c1 + c2 = 0

(42)

Again, using the BC;

(c

1eap

+ c2e − ap

u ( a , y ) = 0 , Eq. ( 38 )

) ( c cos py +c 3

4

gives

sin py ) = 0

Implying thereby

c1e + c2 e − ap = 0 (43) ap

c ,c ,

To determine the constants 1 2 we have to solve Eq. (42) and ( 43), being homogeneous, the determinant

11 =0 e e − ap ap

For the exercise of non-trivial solution, which is not the case. Hence ±( , ) = 0 only trivial solution is possible.

u ( x, y ) = ( c5 x + c6 )( c7 y + c8 ) , If we consider the solution given by Eq. (39) u= ( 0, y ) u= ( a, y ) 0 again yields a trivial the boundary conditions: solution. Hence, the possible solutions given by Eq. (38) and (39) are ruled out. Therefore, the only possible solutions given by Eq. (38) and (39) are rule out Therefore, the only possible solutions obtained from Eq.(40) is

214

Differential Equations: Theory and Applications

(

u( x , y ) = ( c9 cos px + c10 sin px ) c11e py + c12 e− py Using the BC : u ( a, y ) = 0 yeilds

u ( 0, y ) = 0,

(

we get

)

c9 = 0. Also, the other BC:

)

c10 sin pa c11e py + c12 e − py = 0 For non trivial solution, 10 cannot be zero , implying sin pa = 0 , which π / a., n 1 , 2 , 3, ... therefore, the possible n π or p n= is possible if pa = =

c

non-trivial solution after using the superposition principle is

u ( x , y) =



∑ sin n =1

nπ x  an exp ( nπ y / a ) + bn exp ( −nπ y / a )  a  (44)

Now, using the BC :

sin

u ( x ,b ) = 0 ,

we get

nπ x  an exp ( nπ b / a ) +bn exp ( −nπ b / a )  = 0 a  Implying thereby

an exp ( n π b / a ) + bn exp ( −nπ b / a ) = 0 which gives

bn = −an

exp ( nπ b / a )

exp ( − nπ b / a ) = , n 1 , 2 ,…, ∞ The solution (44) now becomes 2an sin ( nπ x / a )  exp {nπ ( y − b ) / a} − exp {−nπ ( y − b ) / a}    2 n =1 exp ( − nπ b / a )    ∞

u(x , y ) =∑ ∞

2an

∑ exp ( −nπ b / a ) sin ( nπ x / a ) sin h {nπ ( y − b ) / a} n =1

Let form

  nπ x   2an / exp    = An . Then the solution can be written in the  a  

Elliptic Differential Equation

215



∑ A sin ( nπ x / a ) sin h {−nπ ( y − b ) / a}

= u ( x , y)

n =1

n

Finally , using the u ( x, y ) = f ( x ) , we get

non-homogenous

boundary

condition:



f ( x) ∑ A sin ( nπ x / a ) sin h ( −nπ x / b ) = n

n =1

which is half-range Fourier series. Therefore ,

2 An sin h ( −nπ x / b ) = f ( x ) sin ( nπ x / b ) dx a ∫0 a

Thus, required solution for the given Dirichlet problem is ∞

u ( x, y ) = ∑An sin ( nπ x / b ) sin h nπ y − b / a { ( ) } n =1



(47)

where

An =

2 1 a sin h ( −nπ x / b )

a

∫ f ( x ) sin ( nπ x / b ) dx 0

4.7. THE NEUMANN PROBLEM FOR A RECTANGLE The Neumann problem for a rectangle is defined as follows

PDE : ∇ 2 u= 0, 0 ≤ x ≤ a, 0 ≤ y ≤ b BCs : u= u= 0, u= 0 , u= f ( x) x ( 0, y ) x ( a, y ) y ( x, y ) y ( x, b )

(48)

The general solution of the Laplace equation using the method of variables separated is found to be

(

u ( x , y) = ( c1 cos px +c2 sin px ) c3e py + c4e− py The BC :

u x ( a, y ) = 0

(

gives

0= −c2 p c3e py + c4 e − py

)

)

Differential Equations: Theory and Applications

216

Implying

c2 = 0. Therefore,

(

= u ( x , y ) c1 cos px c3e py + c4 e − py The BC:

u x ( a, y ) = 0

(

) (49)

gives

0 c1 p sin pa c3e py + c4 e − py For non-trivial solution,

)

c1 ≠ 0, implying

nπ = (π 0,1, 2,…) a

π, sin = pa 0, = pa n=

Thus the possible solution is

u =

x ,y) (=

cos

nπ x Ae nπ y / a + Be − nπ y / a a (50)

(

Now, using the BC :

0 cos

)

u y ( x, 0 ) = 0,

we get

nπ x  nπ nπ  −B A  a  a a 

Implying B = A. Thus, the solution is

u ( x , y ) A cos =

2 A cos

nπ x   nπ y   nπ y   exp   + exp  −   a   a   a 

nπ x nπ y cosh a a

Using the superposition principle and defining ∞

u = ∑An cos n =0

nπ x nπ y cosh a a

Finally, using the BC: ∞

f ( x ) = ∑An cos n =1

2 A = An , we get

u y ( x, b ) = f ( x ) ,

we get

nπ x nπ nπ b sin h a a a

which is the half-range Fourier cosine series. Therefore,

Elliptic Differential Equation

217

nπ nπ b 2 nπ x sinh = ∫ f ( x ) cos dx a a a0 a a

An

Hence, the required solution is ∞

u A0 + ∑ An cos = n =1

nπ x nπ y cosh a a (52)

where

2 1 nπ x f ( x ) cos dx ∫ nπ sinh ( nπ b / a ) 0 a a

An =

4.8. INTERIOR DIRICHLET PROBLEM FOR A CIRCLE The Dirichlet problem for the circle is defined as follows: 2 PDE: ∇ u= 0, 0 ≤ r ≤ a, 0 ≤ 0 ≤ 2π

BC:

u ( a= , θ ) f ( θ ) , 0 ≤ θ ≤ 2π

f (θ ) where is a continuous function on ∂IR. The task is to find the value of u at any point in the interior of the circle IR in terms of its value on ∂IR such that u is single valued and continuous on IR. In view of circular geometry , it us natural to choose polar coordinates to solve this problem and then use the variables separable method. The requirement of single-valuedness of u in IR implies the periodicity condition, i.e.,

u ( r ,θ = + 2 π ) u ( r , θ ) , 0 ≤ r ≤ a,

(54)

2

0 which in polar coordinates can be written as From Eq. (27) , ∇ u = 1 1 0 urr + ur + 2 u θθ = r r If

u ( r ,θ )

R ( r ) H (θ ) ,

the above equations reduces to

Differential Equations: Theory and Applications

218

1 1 R′′H + R′H + 2 RH ′′ = 0 r r This equation can be rewritten as

r 2 R′′ + rR′ H ′′ = −= k R H (55) which means that a function of r is equal to a function of θ and, therefore, each must be equal to a constant k ( a separation constant ). 2 Case I Let k = λ . Then

r 2 R '' + rR ' − λ 2 R = 0 (56) z which is Euler type of equation and can be solved by setting r = e . Its solution is

R= c1eλ z + c2 e – λ z = c1r λ + c2 r – λ Also,

H ′′ + λ 2 H = 0 whose solution is

= H c3 cos λθ + c4 sin λθ Therefore, u ( r ,θ ) = c1r λ + c2 r – λ

(

) ( c cos λθ + c 3

4

sin λθ )

Case II Let k = −λ 2 . Then

(57)

r 2 R′′ + rR′ + λ 2 R = 0 , H ′′ − λ 2 H= 0 Their respective solutions are

= R c1 cos ( λ in r ) + c2 sin ( λ in r ) = H c3 e λθ + c4 e − λθ Thus

u ( r ,θ )

(

c1 cos ( λ in r ) c2 sin ( λ in r )  c3 e λθ +c4 e – λθ

Case III Let k = 0 , then we have

)

(58)

Elliptic Differential Equation

219

r R '' + R ' = 0 Setting

r

R' ( r ) = V ( r ) ,

we obtain

dV dV dr +V = 0, + = 0 dr r i.e. , V Integrating , we get in

V=

Vr

in c . Therefore,

c1 dR = r dr On integration ,

= R c1 In r + c2 Also, H ′′ = 0 After integrating twice , we get

= H c3θ + c4 Thus,

u ( r ,θ ) = ( c1 in r + c2

)( c3θ + c4 )

(59) Now, for the interior problem, r = o is a point in the domain IR and since in r = 0 is a point in the domain IR and since In r is not defined

at r = 0, the solutions (58) and (59) are not acceptable. Thus the required solution is obtained from Eq. (57). The periodicity condition in θ implies

c3 cos λ + c4 sin ( λ (θ + 2π ) ) + c4 sin ( λ (θ + 2 π ) )

i.e.,

c3 cos λθ − cos ( λθ + 2λπ )  + c4 [sin ( λ (θ + 2π ) )] = 0

or 2sin λ π c3 sin ( λθ + λπ ) − c4 cos ( λθ + λπ )  = 0

= λπ 0 ,= λπ nπ = , λ n= ( n 0 ,1 , 2 , ,…) . Using the Implying sin principle of superposition and renaming the constants, the acceptable general solution can be written as

Differential Equations: Theory and Applications

220



(

)

u ( r ,θ ) = ∑ cn r n + dn r −9 ( an cos nθ + bn sin nθ ) n =0



At r = o, the solution should be finite, which requires appropriate solution assumes the form ∞

= u ( r ,θ )

∑r ( A cos nθ + B n

n

n =0

n

(60)

d n = 0. Thus the

sin nθ )

For n − 0, let the constant A0 be A0 / 2. Then the solution is A0 ∞ n + ∑ r ( a n cos n θ + Bn sin nθ ) u ( r ,θ ) = 2 n =1

(61)

which is a full-range Fourier series. Now we have to determine u ( a, θ ) = f (θ ) so that the BC: is satisfied.

An and Bn

i.e.,

= f (θ )



∑a

n

n =1

( An cos n θ + Bn sin nθ )

Hence,

A0 =

1



∫ f (θ ) dθ

π

0

a n An = 1π



∫ f (θ ) cos θ dθ 0

n a= Bn

1

π



(62)

θ ,n ∫ f (θ ) sin nθ d=

1 , 2 , ,…

0

In Eq. we replace the dummy variables θ by φ to distinguish this variable from the current variables θ in Eq. Substituting Eq. into Eq., we obtain the relation

u ( r ,θ ) =

1 2π



∞  n r cos nθ f φ d φ + ( )  n ∑ ∫0 π n =1  a



∫ cos ( nφ ) f (φ ) dφ 0

Elliptic Differential Equation

r n sin nθ + n a π

221



 sin n φ f φ d φ ( ) ( )  ∫0 

Interchanging the order of summation and integration, we get 2π

1 1 u ( r ,θ ) = f (φ ) d φ + ∫ 2π 0 π



n

∞ r f φ ( ) ∑   {cos nφ cos nθ + sin nφ sin nθ } dφ ∫0 n =1  a 

2π  1 ∞  r n  1 =∫ f (φ )  + ∑   cos n (φ − θ )  dφ π 0  2 n =1  a   (63)



2



n

r c ∑   cos n (φ − θ ) n =1  a  r s ∑   cos n (φ − θ ) n =1  a  So that ∞

 r i(φ −θ )  c + is = ∑  e  n =1  a

n

r r < a,   < 1 and e1(φ −θ ≤ 1, a Since

(r / a) r  r  i(φ −θ )  = c + is ∑ =  e    r  i(φ −θ )  n =1  a   1 −  a  e      i (φ −θ )

n



{

)}

 r  i(φ −θ ) − r 2 / a2   e a   =   r  i(φ −θ )    r  i(φ −θ )  1 −  a  e  1 −  a  e        

(

Equating the real part on both sides, we get

222

Differential Equations: Theory and Applications

 r   r 2    cos (φ − θ ) −  2   a2 − r 2  a   a  =   2r  2  a 2 − 2ar cos (φ − θ ) + r 2  2 2  1 cos / ) φ θ r a − − + ( )   a      

c

Thus , the required solution takes the form

1 u ( r ,θ ) = 2π

(a



∫ a 0



2

2

)

− r 2 f (φ )

− 2ar cos (φ − θ ) + r 2 

dφ (64)

This is known as Poisson’s integral formula for a circle, which gives a unique solution for the Dirichlet problem. The solution can be interpreted physically in many ways: it can be thought of as finding the potential u ( r , θ ) as f (φ ) a weighted average of the boundary potential weighted by the Poisson kernel p, given by

p=

a2 − r 2  a 2 − 2ar cos (φ − θ ) + r 2 

u ( r ,θ ) It can also be thought of as a steady temperature distribution in a circular disc, when the temperature u on its boundary ∂IR is given by u = f (φ ) which is independent of time.

4.9. EXTERIOR DIRICHLET PROBLEM FOR A CIRCLE The exterior Dirichlet problem is described by

PDE : ∇ 2 u = 0

BC : u ( a, θ ) = f (θ )

(65) u must be bounded as r → ∞ 2

By the method of separation of variables, the general solution of

∇ u= 0 in polar coordinates can be written as ∞

(

u ( r ,θ ) = ∑ cn r n + dn r − n n =0

)( a

n

cos nθ + bn sin nθ )

Elliptic Differential Equation

223

c = 0. Now as , r → ∞ , we require u tobe bounded, and therefore, n After adjusting the constants, the general solution now reads ∞

= u ( r ,θ )

∑( A

n

n =0

cos n θ + Bn sin nθ )

With no loss of generality, it can also be written as

A0 ∞ − n + ∑r ( An cos nθ + Bn sin nθ ) u ( r ,θ ) = 2 n =1 Using the BC:

u ( a,θ ) = f (θ ) ,

(66)

we obtain



A0 + ∑a – n ( An cos n θ + Bn sin nθ ) f (θ ) = 2 n =1 This is a full range Fourier series in

A0 = a

−n

1

f (θ ) ,

where



π

∫ f ( θ ) dθ 0

An =

a − n Bn =

1

π 1

π



∫ f ( θ ) cos nθ



0

(67)



∫ f ( θ ) sin nθ dθ 0

In Eq. (67) we replace the dummy variables θ by φ so as to distinguish it from the current variable θ . We then introduced the changed variables into solution (66) which becomes

u ( r ,θ ) = r −n an +

π

1 2π



∫ 0

sin nθ

2π ∞  −n r an f (φ )d φ + ∑  cos n θ ∫ cos ( nφ ) f ( φ ) d φ π n =1  0 2π

 sin n f d θ φ φ ( ) ( )  ∫0  or

Differential Equations: Theory and Applications

224

2π  1 ∞  a n  1 u ( r , θ ) =∫ f (φ )  + ∑   cos n (φ − θ )  dφ π 0  2 n =1  r   (68) Let



n



n

a C ∑   cos n (φ − θ ) n =1  r  a  S = ∑  n (φ − θ ) n =1  n  sin Then,

 a  i ( φ −θ )  C + iS = ∑   e  n =1  r   ∞

n

Since

a < 1 , e i (φ − θ ) ≤ 1 r we have

 a   i(φ −θ )    e  a e r = = C + iS r   a  i (φ – θ )    a  i (φ – θ )    a 1 r  e  1 –  r  e  1 −  r          i (φ −θ

)

 – i (φ – θ )  e   

Hence,

 a   a2      cos ( φ − θ ) −  2    r   r  C=    2a   a2  φ θ 1 cos − − + ( )    2    r    r  Thus the quantity in the square bracket on the right-hand side of Eq. (68) becomes

Elliptic Differential Equation

225

 a   a2   cos φ – θ ( )    2  1 r2 +a2  r   r  + = 2   2a  2  r 2 − 2ar cos ( φ −θ ) + a 2  2 2  2 1−  cos – a / r φ θ + ( )     r  

(

)

Therefore, the solution of the exterior Dirichlet problem reduce to that of an integral equation of the form

1 u ( r ,θ ) = 2π

(r



∫ 0

2

)

− a 2 f (φ )

 r − 2ar cos ( φ − θ ) + a 2  2



(69)

Example. Find the steady state temperature distribution in a semicircular plate of radius a. insulated on both the faces with its curved boundary kept at a constant temperature at zero temperature .

U0

and its bounding diameter kept

Solution The Governing heat flow equation

ut = ∇ 2 u

0,

In the steady state, the temperature is independent of time ; hence and the temperature satisfies the Laplace equation . The problem can now be stated as follows : To solve

PDE : ∇ 2 u ( u , θ ) = u rr +

1 1 + uθ θ = 0 ur r 2

= u ( a, θ ) U 0 = , u ( r , 0 ) 0,= u ( r,π ) 0 BCs: The acceptable general solution is

(

u ( r ,θ ) = cr λ + Dr − λ From the BC : also gives

(

) ( Acos λθ + B sin λθ ) (70)

u ( r , 0 ) = 0,

)

u ( r,π ) = 0 we get A = 0; however, the BC :

B sin λπ cr λ + Dr λ − = 0

Differential Equations: Theory and Applications

226

Figure 4

λπ 0= . B 0 gives a trivial solution. For Implying either B = 0 or sin = a non-trivial solution. We λ = n . Hence the possible solution is

(

= u ( r ,θ ) B sin nθ Cr λ + Dr − λ

)



(71)

In eq. (71), we observe that as r → 0, the term r → ∞. But the solution = and so D 0 . Then after adjusting the constants, it should be finite at r 0= follows form the superposition principle that −λ



u ( r ,θ ) = ∑ B n a n sin nθ n =1

which is a half –range Fourier sine series . Therefore,

 4U 0 π , for = n 1 , 3 ,… 2  = Bn a = U 0 sin n θ d θ  nπ ∫ π 0  0 , for n = 2 , 4. , , , , , n

Hence,

= Bn

4U 0 = , n 1 , 3 ,… nπ a n With these values of

4U 0

Bn , the required solution is n

1 r u ( r ,θ ) = ∑ π π n  a  sin nθ ∞

Elliptic Differential Equation

227

4.10. INTERIOR NEUMANN PROBLEM FOR A CIRCLE The interior Neumann problem for a circle is described by

PDE : ∇ 2 u = 0, 0 ≤ r < a ; 0 ≤θ 2π (72) ∂u ∂u ( a , r ) BC = : = g= (θ ) ,r a ∂n ∂r Following the method of separation of variables, the general solution

2 0 in polar coordinates is given by (60) of equation ∇ u = ∞

(

u ( r ,θ ) = ∑ cn r n + d n r − n n=0

)( a

n

cosnθ + bn sin nθ )

At r = 0, the solution should be finite and , therefore, after adjusting the constant the general solution becomes

d n = 0.



u ( r , θ ) = ∑r n ( An cos nθ ) n=0

With no loss of generally , this equation can be written as ∞

= u ( r ,θ )

∑r

n

n=0 0

∂u ∂r

∑ nr

n −1

n =1

( An cosn θ + Bn sin nθ )



( An cos nθ + Bn sin nθ )

Using the BC:

∂u ( a ,θ ) = g ( θ ) ∂r we get

g (θ )



∑ na ( A cos nθ n –1

n

n =1

+ Bn sinθ )

which is a full-range. Fourier series in

na n −1 An =

1

π



∫ g ( θ ) cos nθ 0



g (θ )

, where

(73)

Hence ,

228

Differential Equations: Theory and Applications

na n − 1 Bn =

1

π



∫ g ( θ ) sin nθ



0

Here, we replace the dummy variables θ by φ to distinguish form the θ in Eq ( 74 ) . current variables now introducing Eq. (75) into Eq.(73), we obtain

u ( r , θ= )

A0 ∞ rn +∑ 2 n = 1 nπ a n − 1



∫ g ( φ ) ( cos nφ cos nθ sin nφ sin nθ )dφ 0

or 2π

n

∞ A0 r  u ( r , θ= + ∫ g (φ ) ∑   1/ nπ cos n (φ –θ ) dφ ) 2 0 n = 1 a 

(76)

or this solution can also be expressed in an alternative integral form as follows: Let n

r a C= ∑  cos n (φ − θ )  a  nπ n

r a S= ∑  sin n (φ − θ )  a  nπ Therefore, n

n

 r  a in (φ −θ ) a ∞  r i (φ −θ )  1 ∑  =∑ e C + iS = e π n =1  a  a  nπ  n   r i (φ −θ )   r i (φ −θ )   r i (φ −θ )    e   e   e    a a  + a  + a  +…  = 1 2 3 π    or a  r a  r r   − In 1 − cos (φ − θ ) − i sin (φ − θ )  C + iS = In 1 − ei (φ −θ )  = a π  a π  a   (77)

Elliptic Differential Equation

229

To get the real part of In z , we may note that

= W In = z or z e w i.e.,

x + iy= eu +iv= eu cos v + ieu sin v. Therefore, u = x e= cos v, y eu sin v

e 2u = x 2 + y 2 = z

2

u = In z .

i.e., Therefore ,

2

a  r  r  C= − In 1 − cos (φ − θ )  +  sin (φ − θ )  π  a  a 

= −

a

π

In

2

a 2 − 2ar cos (φ − θ ) + r 2 a2

Thus the required solution is

u ( r , θ= )

A0 a − 2 π



∫ In

a 2 − 2a r cos ( φ − θ ) + r 2 a2

0

g (φ ) d φ



(78)

which is again an integral equation.

4.11. SOLUTION OF LAPLACE EQUATION IN CYLINDRICAL COORDINATES The Laplace equation in cylindrical coordinates assumes the following form:

1 1 ∇ 2 u = urr + ur + 2 uθθ + u zz = 0 r r We now seek a separable solution of the form u ( r ,θ , z ) = F ( r ,θ ) Z ( z )



Subtracting Eq. (80) into Eq. (79), we get

(79)

(80)

230

Differential Equations: Theory and Applications

∂2 F 1 ∂F 1 d 2F d 2Z Z + Z + Z + F = 0 ∂r 2 r ∂r r 2 dθ dz 2

 ∂2 F 1 ∂ F 1 d 2F  1 d2 Z 1 k ( say ) + + = − =  2  r ∂ r r 2 dθ 2  F d z2 Z ∂r  or where k is a separation constant. Therefore , either d 2Z + kZ = 0 ( 81) d z2 or ∂2 F 1 1 ∂2 F + ∂ F \ ∂ r + − KF =0 ( 82 ) ∂r2 r r 2 ∂θ 2 If k is real and positive , the solution of Eq. (81) is

= Z c1 cos kz + c2 e −

kz

If k is equal to zero, the solution of Eq. (81) is

= Z c1 z + c2 From physical considerations, one would except a solution which decays with increasing z and, therefore, the solution corresponding to negative k is acceptable. Let k = −λ 2 . Then

= Z c1 e λ z + c2 e – λ z ( 83 ) Equation (82) now become

∂ 2 F 1 ∂F 1 ∂ 2 F + + 2 + λ2F = 0 2 2 ∂r r ∂r r ∂θ Let

f '' H +

F ( r , θ ) = f ( r ) H (θ )

.Substituting into the above equations we get

1 ' 1 f H + 2 f H '' + λ 2 f H = 0 r r

or

(

r 2 f '' + rf ' + λ 2 r 2

)

1 H '' = − = K '' ( say ) f H

Elliptic Differential Equation

231

Form physical consideration, we except the solution to be periodic in

θ , which can be obtained when k ′ is positive and k ' = n 2 . therefore , the

acceptable solution will be

( 84 )

= H c3 cos n θ + c4 sin nθ ' 2 When k = n , we will also have

d2 f d f +r + λ 2 r 2 − n2 f = 0 2 ( 85 ) dr dr which is a Bessel ‘s equation whose general solution is given by

(

r2

)

= f Ajn ( λ r ) + BY n ( λ r ) Jn ( λr )



( 86 )

Y n ( λr )

are the nth order Bessel function of first Y ( λr ) → − ∞ r → 0 ,Yn ( λ r ) and second kind, respectively. Since n as becomes unbounded at r = 0. Continuity of the solution demands B = 0 . Here ,

and

2 0 is Hence the most general and acceptable solution of ∇ u =

(

u ( r ,θ , r ) = J n ( λ r ) c1eλ z + c2 e – λ z

) ( c cos nθ + c 3

4

sin nθ )

( 87 )

Example. A homogeneous thermally conducting cylinder occupies

the region 0 ≤ r ≤ a, 0 ≤ θ ≤ 2π , 0 ≤ z ≤ h , where r ,θ , z are cylindrical o coordinates . The top z = h and the lateral surface r = a are held at 0 , 0 while the base z = 0 is held at 100 . Assuiming that there are no sources of heat generation within the cylinder, find the steady –temperature distribution within the cylinder. Solution. The temperature u must be a single valued continuous function. The steady state temperature satisfies the Laplace equation ‡E inside the cylinder. To compute the temperature distribution inside the cylinder , we have to solve the following BVP: 2 0 PDE: ∇ u =

o = on z h , BCs : u 0= o = u 0= on r a , o = u 100 = on z 0

Differential Equations: Theory and Applications

232

is

The general solution of the Laplace equation in cylindrical coordinates

(

r ( r , θ , z ) = J n ( λ r )( c1 cosnθ +c2 sin nθ ) c3 e λ z + c 4 e – λ z

)

0 Since the face z = 0 is maintained at 100 and since the other and lateral

0

surface of the cylinder are maintained at 0 , the temperature at any point inside the cylinder is obviously independent of θ . This is possible only when n = 0 in the general solution .Thus,

(

= u ( r , z ) j 0 ( λ r ) Ae λ z + Be − λ z

)

on z h, we get Using the BC= : u 0=

(

= 0 J 0 ( λ r ) Ae λ h + Be − λ h

)

λ −λh = 0 , from which Implying thereby Ae + Be

B= −

Aeλ h e −λh

Therefore, the solution is

= u( r ,z )

J0 ( λr ) A e

−λh

 λ ( z −h ) − e –λ ( z − h )  e 

or

= u= ( r , z ) J 0 ( λ r ) A1 sinh λ ( z − h )

A = 2 A e −λh where 1 / , Now using the BC : u = 0 on r = a, we have 0 A1 J 0 ( λ a ) sinh λ ( z − h ) Implying

j0 ( λ a ) = 0 ,

Denoting them by

λ=

which has infinitely many positive roots.

ξ n , we have ξ n = λ a , and therefore,

ξn a Thus the solution is

Elliptic Differential Equation

233

ξ r  ξ  u( r ,z ) = A1 j0  n  sinh  n ( z − h )  , n = 1 , 2 ,…  a  a  Using the principle of superposition , we have ∞

u( r ,z )

∑A n =1

n

ξr ξ  j0  n  sinh  n ( z − h )   a  a 

0 BC : u 100 = on z 0 gives The= ∞

∑A

100

n

n =1

ξ r  ξ  j0  n  sin h  n ( z − h )   a  a 

BC : u 100 = on z 0 gives The = ∞

∑A

100 =

n =1

n

 ξ h ξ r  j0  − n  j0  n   a   a 

which is a Fourier-Bessel series. Multiplying both sides with and integrating, we get

ξ r  j0  m  dr 100 ∫ r =  a  0 a

rJ 0 (ξ m r / a )

 ξh ξ r  ξ  An sin h  − n  ∫ rjo  n  j0  mr  dr ∑  a 0  a   a  n =1 a



Using the orthogonality property of Bessel’ s function , namely,

0, if i ≠ j   2 ∫0 xJ n ( ai x ) J n ( a j x ) dx =  a j 2 ( α ) , if i = j n +1 i   2 a

Where

ai , a j

1

are the zeros at

 ξn r  100 ∫ r j0=  a  dr   0

jn ( x ) = 0 ,

we have

2  ξn h  a 2 A − sinh ∑ n  a  2 j1 ( ξ n )   n =1 ∞

Therefore,

An =

200  ξh a 2 sin h  − n  j12 ( ξ n  a 

a

∫ rJ

)0

0

 ξn r  a 

  dr 

Differential Equations: Theory and Applications

234

Setting

ξn r a = x= , dr dx a ξn The relation for

An can also be written as

200 An =  ξ h ξ n2 sinh  − n  j12 ( ξ n )  a 

ξn

∫ x j ( x ) dx 0

0

Using the recurrence relation

x n jn –1 ( x ) =

d  x n jn ( x )  , dx 

For n =1 , we get

∫ x j0 ( x ) dx = x j1 ( x ) Now,

An can be written as ξ

An

    200 x j1 ( x ) 200 =   ξ 2 sinh  − ξ n h  j 2 ξ  ξ n sin h \ a ) j1 ( ξ )  a  1 ( n )  n   0 Hence, the required temperature distribution inside the cylinder is ∞

u ( r , z ) = 200 ∑

n =1

where

 ξ    r  j0  ξ n  sin h   n  ( z − h )   a  a   ξ n sin h ( − ξ n h / a ) j1 (ξ n )

xn are the positive zeros of j 0 ( ξ ) .

Example

.

Find

the

potential

u inside

the cylinder 0 ≤ r ≤ a, 0 ≤ θ ≤ 2π , 0 ≤ z ≤ h, if the potential on the top z = h and on the lateral surface r = a is held at zero, while on the base z = 0, the potential u ( r ,θ = , 0 ) V0 (1 − r 2 / a 2 ) , V is given by where 0 is a constant r , θ , z are cylindrical polar coordinates .

Elliptic Differential Equation

235

Solution. The potential u must be a single-valued continuous function and satisfy the Laplace equation inside the cylinder. To compute the potential inside the cylinder, we have to solve the following BVP: 2 0 PDE: ∇ u =

= u 0= on z h, BCs: = u 0= on r a,  r2  u= V0 1 − 2  on z = 0  a  In cylindrical coordinates , the general solution of the Laplace equation\  /\ is

(

u ( r , θ , z ) = J n ( λ r )( c1 cos nθ + c2 sin nθ ) c3eλ z + c4 e − λ z

)

 r2  V0 1 − 2   a  which is purely, a Since the face z = 0 has potential function of r and is independent of θ . This is possible only when n = 0 in the general solution. Thus ,

(

= u ( r , z ) j0 ( λ r ) Aeλ z + Be − λ z

)

on z h , we obtain Using the BC= : u 0=

(

= 0 j0 ( λ r ) Aeλ h + Be − λ h

)

λh −λh 0, which yields Implying Ae + Be =

Ae λ h B = − –λ h e Hence , the solution is

A

u( r ,z )

e

–λ h

j 0 ( λ r )  eλ ( z − h ) − e – λ ( z − h ) 

or

u ( r, z )

A e

−λh

j 0 ( λ r ) sinh λ ( z − h )

Differential Equations: Theory and Applications

236

A = A / e λ h , Now, using the BC : u = 0 on the lateral surface , Where 1 i.e., on r = a, we get

0

A1 J 0 ( λ a ) sinh λ ( z − h ) j0 ( λa ) = 0 .

Implying

ξ n we shall have = ξ n λ= a or λ ξ n / a

This has infinitely many positive roots; demoting

them by

The solution now takes the form

ξ r  ξ  A 1 j 0  n  sinh  n ( z − h )  , n 1 , 2 , ,….  a  a =

u( r ,z )

The principle of superposition gives

u( r ,z )



∑A n =1

n

ξ r ξ h  J 0  n  sin h  n ( z − h )   a   a 

 r2 u= V 0  1− 2  a The last BC : r2 V0 = (1 − 2 a



∑A n =1

n

 0  on z =  yeilds

 ξ h ξ r  Sinh  − n  J 0  n  a   a  

This is a Fourier-Bessel’s series . Multiplying both sides by and integrating, we get a  r2 V 0 ∫  1− 2 a 0

∞   ξm r   ξ h ξ r  rj dr An sinh  − n  j0  n  =  0 ∑   a   a  n =1   a 

Using the orthogonality property of Bessel functions a

∫ xj ( α x ) n

i

0

Where

0, if i = j   2 jn ( a j x ) dx =  a 2 jn + 1 ( α i ) , if i ≠ j   2

αi , α j

are the zeros of

jn ( x ) = 0 ,

we get

rj0 (ξ m r / a )

Elliptic Differential Equation a  r2 V0 ∫  1 − 2 a 0

  ξn r  r j0   a 

2 ∞   ξn h  a 2 dr A sinh = − ∑ n   a  2 j1 ( ξ n )    n =1

which gives a  r2 2V0 ∫0  1− a 2  ξn h  2 2 a sinh  −  j 1 ( ξn )  a 

An

  ξn r   r j0   dr  a  

Be letting ξ r a = x , this equation can be modified to

2V 0  ξ h ξ n4 sinh  − n  j12 ( ξ n ) a  

An

ξn

∫ (ξ

2 n

− x 2 ) x j0 ( x ) dx

0

Using the well-known recurrence relation

x α j α − 1 ( x )=

d  x α jα ( x )  for= α= 1 , 2 ,… dx 

We get

∫ x j 0 ( x ) = xj 1 ( x ) , ∫ x 2 j1 ( x ) = x 2 j 2 ( x ) From these relations, we obtain

An =

=

2V0  ξh ξ n4 sinh  − n  a

2V0  ξ h ξ n4 sinh  − n a  Thus,

An =

ξn

 2  J1 ( ξ n ) 



x 2 j1 ( x ) dx

0

ξn

 2  J1 ( ξ n ) 

 x 2 j2 ( x )  0

4V 0 j2 ( ξ n )  ξ h ξ n4 sinh  − n  J12 ( ξ n )  a 

237

238

Differential Equations: Theory and Applications

The recurrence relation

2n J n −1 ( x ) + jn +1 ( x ) = jn ( x ) x For n =1 gives

2 j0 ( ξ n ) + j2 ( ξ n ) = j1 ( ξ n )

ξn

Hence,

j2 ( ξ n ) = Since

An =

2

ξn

j1 ( ξ n )

j0 ( ξ n ) = 0 .

Therefore,

8V0 j1 ( ξ n ) h  ξ 3n sinh  − ξ n  j12 ( ξ n ) a 

Thus, the required potential inside the cylinder is

ξr   ξ  8V0 j0  n  sinh  n ( z − h )   a   a  u( r ,z )= ∑ ξ h   n =1 ξ n3 j1 ( ξ n ) sinh  − n   a  ∞

4.12. SOLUTION OF LAPLACE EQUATION IN SPHERICAL COORDINATES In example 3 the Laplace equation is expressed in spherical and has the following form:

1 ∂  ∂u  1 ∂2 u  2 ∂u  r + sin + = 0 θ     ∂θ  sin 2θ ∂φ 2  ∂r  sin θ ∂θ  Let us assume the separable solution in the form

= ∇ 2u

∂ ∂r

u ( r ,θ ,φ ) = R ( r ) F ( θ ,φ )



Subtracting of variables gives

( 88 ) ( 89 )

Elliptic Differential Equation

239

2 d  2 dR  − 1  ∂  sinθ ∂F  + 1 ∂ F      r  sin θ  ∂θ  ∂θ  sin θ ∂ φ 2  dr  dr  = = −µ R F where µ is separation constant. Therefore,

1 d  2 dR  r = −µ R dr  dr 

1  ∂  F sinθ  ∂θ Equation

r2

( 90 )

∂u  1 ∂2 F   sin + θ µ   = ∂θ  sin θ ∂φ 2  

( 90 )

( 91 )

gives

2

d R dR +2r + µR = 0 2 dr dr

z which is a Euler’s equation. Hence, using the transformation r = e , the auxiliary equation can be written as

D ( D −1) + 2 D + µ = D 2 + µ = 0 where

D=

D=

d . dz Its roots are given by

−1 ± 1 − 4 µ 2

Let

µ= − α (α + 1 ) ;

then we get

2  1     −1 ± 2  a +   2    1  1 = D= − ± a +  2 2  2

is

– ( a + 1) Hence, D = α and Therefore, the solution of Euler’s equation

= R c1 e a + c2 r – ( a + 1 )



( 92 )

Differential Equations: Theory and Applications

240

Taking

–= µ a ( a +1 ) , Eq.

becomes

2

∂  ∂F  1 ∂ F + a ( a +1 ) F sinθ = 0  sin θ + ∂θ  ∂θ  sin θ ∂ φ 2 Inserting Into the above equation and separating the variables, we obtain

2 where v is another separation constant. Then



 2 dH  sin θ  d  v  sin θ  + a ( a +1 ) sin ( θ ) H  =  H  dθ  dθ  

( 93 )

( 94 )

The general solution of Eq. (93) is

0 , the solution is independent of φ which Provided v ≠ 0 , if v = ( 94 ) becomes , for the ax corresponds to the ax symmetric case. Equation symmetric , case, d dθ

dH dx   sin θ dx dθ 

 0  + a ( a +1 ) sin ( θ ) H = 

i.e.,

d  dH  + a ( a +1 ) H = 1 − cos 2 θ 0  dx  dx 

(

)

or

d  dH  1− x 2 + a ( a +1 ) H = 0  dx  dx 

(

)

Elliptic Differential Equation

by

241

This is the well-known Legendre equation. Its general solution is given

= H c5 pa ( x ) + c6 Qa ( x ) , −1 ≤ x ≤1

p ,Q

where a a are Legendre function of the first and second kind respectively. For connivance let α be a positive integer, say a = n . Then

= H c5 pn ( cos θ ) + c6 Qn ( cosθ )



( 98 )

Continuity of implies the continuity of H ( x ) at x = ±1 . Q ( x) Since n has a singularity at x = 1, we choose

c6 = 0.

Therefore, in ax symmetric case the solution of Laplace equation in spherical coordinates is given by

u ( r ,θ = ,φ )

{c r 1

a

+ c2 r − ( a + 1 )

} ( c )  c p ( cosθ )  3

5

n

After renaming the constant s and using the principle of superposition , we find the solution to be

= u ( r ,θ )



∑  A

n=0

n

r n + B n r – ( n + 1 )  p n ( cos θ )

Example. in a solid sphere of radius ' a′, the surface is maintained at the temperature given by

π  k cos θ , ≤ θ 2 f (θ )=   0 , π 0 (1)

u= ( 0 , t ) u= ( L ,t ) 0

(2)

 πx  u ( x , 0 ) = u0 sin  , l   (3) ICs: ∂u ( x ,0 ) = 0 ∂t (4) We have shown the solution of the one-dimensional wave equation by canonical reduction as

u( x= , t ) φ ( x – ct ) +ψ ( x + ct )

(5)

One of the known method for solving this problem is based on trial function approach. Let us choose a trial function of the form

Hyperbolic Differential Equation

π π u= ( x , t ) A  sin ( x + ct ) + sin ( x − ct ) L L   where A is an arbitrary constant. Now, we rewrite Eq. (6) as

291

(6)

 πx   cπ t  u ( x , t ) = 2 A sin   cos    L   L  (7) A=

u0

,

Obviously, (7) satisfies the initial condition (3) with 2 while the second initial condition (4) is satisfied identically. In fact Eq. (7) also satisfies the boundary condition (2). Therefore, the final solution is found to be

 πx   cxπ u ( x , t ) = u0 sin   cos   L   L

   (8)

It may be noted that the trial function approach is easily adoptable if the initial condition is specified as a sin function . However, it is difficult if the initial conditions are specified as a general function such as f ( x ) . In such case, it is better to follow variable separable method as explained in previous sections .

Example. Solve the following initial value problem of the wave equation (Cauchy problem), described by the inhomogeneous wave equation PDE:

utt − c 2 u xx = f ( x ,t )

subject to the initial conditions

= u ( x , 0 ) η= ( x ) , ut ( x , 0 ) v ( x )

u= u + u

u

1 2 , so that 1 is Solution. To make the task easy, we shall set a solution of the homogenous wave equation subject to the general initial

conditions given above. Then

u2 will be a solution of

2 ∂ 2u 2 2 ∂ u2 − = c f ( x, t ) ∂t 2 ∂x 2 (20)

subject to the homogenous ICs

∂u 2 = u2 ( x , 0 ) 0= , ( x ,0 ) 0 ∂t (21)

292

Differential Equations: Theory and Applications

p ( x0 , t0 ) , To obtain the value of u at we integrate the partial differential equation ( 20 ) over the region IR as shown in , we obtain .

2  ∂ 2 u2 2 ∂ u2 ∫ − c ∫  ∂t 2 ∂x 2 IR 

.  dx dt =  ∫IR ∫ f ( x , t ) dx dt 

Using Green’s theorem in a plane to the left-hand side of the above equation to replace the surface integral over IR by a line integral around the boundary ∂IR of IR , the above equation reduces to .

 ∂  2 ∂ u2  ∂ −∫ ∫  c − ∂x  ∂x  ∂t IR 

 ∂u2      dx dt =  ∂t  

.

∫ ∫ f ( x , t ) dx dt

IR

And finally to .

.

 ∂u 2 ∂u  ∫∂IR  ∂t2 dx + c ∂x2  dt = ∫ ∫ f ( x , t ) dx dt IR (22) Now, the boundary ∂IR comprises three segments BP, PAand AB .

dx dx = −c; PB. = c. dt Along BP , dt along Using these results , Eq. (22) .

.

∂u ∂u  ∂u   ∂u  ∫BP c  ∂t2 dt + ∂x2 dx  − PA∫ c  ∂t2 dt + ∂ x2 dx  .

.

∂u  ∂u  − ∫  2 dx + c 2 2 dt  = ∫ f ( x , t ) dx dt ∫ ∂ t ∂ x   AB IR The integrands of the first two integrals are simply the total differentials, while in the third integral, the first term vanishes on AB in view of the second IC in Eq. (21), and the second term also vanishes because AB is dt = 0. directed along the x − axis on which dx Then we arrive at the result .

.

BP

IR

∫ c du2 – ∫ ∫ f ( x , t ) dx dt

which can be rewritten as

Hyperbolic Differential Equation

293

.

cu2 ( p ) − cu2 ( B ) + cu2 ( p ) − cu2 ( A ) = ∫ ∫ f ( x, t ) dxdt IR

Using the first IC of Eq. ( 21), we get Eq. (23) becomes

1 2c

u2= ( p)



u= u= 0, 2 ( A) 2 ( B)

(23) and hence

.

∫ ∫ f ( x , t ) dx dt

IR

with the help of , we deduce

1 2c

u2 ( p ) =

t0 x 0 + ct0 − c t

∫ ∫

f ( x , t ) dx dt (24)

0 x 0 − c t 0 + ct

u= u + u ,

1 2 Now, using the fact that as also using Eq. (24) and D’Alembert’s solution (18) the required solution of the inhomogeneous wave equation subject to the given ICs is given by

u ( x,= t)

1 1 η ( x + ct ) + η ( x – ct ) + { 2 c

x + ct



v (ξ ) d ξ +

x − ct

t x + ct − ct

10 0 0 f ( x, t ) dxdt 2 ∫0 x0 − ct∫0 + ct

(25)

The solution is known as the Riemann-Volterra solution. Following Tychonov and Samarski, it is known that transverse vibration of a string is normally generated in musical instruments. We distinguish the string instrument depending on whether the string is plucked as in the case of guitar or struck as in the case of harmonic or piano. In the case of string which are struck we give a fixed initial velocity but does not undergo any initial displacement. In the case of plucked instruments, the string vibrate form a fixed initial displacement without any initial velocity . Let a thin homogeneous string which is perfectly flexible under uniform tension lie in it equilibrium position along the x − axis . The ends of the string are fixed at x = 0 and x = L . The string is pulled aside a short distance and released. If no external forces are present which correspond to the case of free vibrations, the subsequent motion of the string is described by the u ( x, t ) solution of the following problem: PDE: BCs:

utt − c 2 u= 0, 0 ≤ x ≤ L, t > 0 tt

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

(26)

294

Differential Equations: Theory and Applications

u ( L,= t ) 0, t > 0

(27) = u ( x , 0 ) f ( x ) , ut ( x, 0 ) − g ( x ) ICs: (28) To obtain the variables separable solution, we assume

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

(29)

Substituting into Eq. (26), we obtain

X

d 2T d2X 2 = c T dt 2 dx 2

i.e.,

d 2 X / dx 2 d 2 / dt 2 = = k X c 2T (a separation constant) 2 Case I. When K > 0 , we have k = λ . Then

d2X −λ2X = 0 2 dx d 2T − c 2 λ 2T = 0 2 dt Their solution can be put in the form

= X c1eλ x +c2 e − λ x (30) = T c3ecλt + c4 e − cλt (31) Therefore,

(

)(

u ( x, t ) = c1eλ x + c2 e − λ x c3ecλt + c4 e − λt Now, use the BCs:

(

u ( 0, t ) = 0= ( c1 + c2 ) c3eλt + c4e−λt

)





(32)

)

(33)

0. Also, u ( L, t ) = 0 gives which imply that c1 + c2 = c1e − λ L + c2 e − λ L = 0 (34) Equation (33) and (34) possess a non- trivial solution if

Hyperbolic Differential Equation

1 e

λL

1 e−λ L

295

= e − λ L − eλ L = 0

or L L 1 − e 2 λ= 0 implying e 2 λ= 1 or λ = L 0

This implies that λ = 0, since L cannot be zero, which is against the assumption as in the Case I, Hence, the solution is not acceptable. Case II. Let K = 0 . Then we have

d2X d 2T = 0, = 0 dx 2 dt 2 Their solutions are found to be

X =+ Ax B, T = ct + D Therefore, the required solution of PDE (26) is

u ( x ,t ) = ( Ax + B )( ct + D ) Using the BCs, we have

u ( 0, t ) = 0= B ( ct + D ) , implying B = 0 u ( L ,t) = 0= AL ( ct + D ) , implying A = 0 Hence, only a trivial solution is possible. Since we are looking for a nontrivial solution, consider the following case. 2 Case III. When k = 0, say K = −λ , the differential equations are

d2X d 2T 2 + λ X = 0, + c 2 λ 2 T= 0 2 2 dx dt Their general solution gives

u ( x, t ) = ( c1 cos λ x + c2 sin λ x )( c3 cos cλt + c4 sin cλt ) Using the BC:

u ( L , t ) = 0,

u ( 0, t ) = 0

we obtain



(35)

c1 = 0. Also, using the BC:

λ we get sin λ L = 0 implying that= the eigenvalue. Hence the possible solution is

nπ = , n 1 , 2 ,…. which are L

Differential Equations: Theory and Applications

296

nπ x  nπ ct nπ ct  u n ( x, t ) = sin + B sin 1 , 2 ,…  A cos , n = L  L L 

(36)

Using the superposition principle, we have (37) The initial condition give

u ( x= , 0 ) f= ( x)



∑ A sin n

n =1

nπ x L

which is half – range Fourier sine series, where

2 nπ x f ( x ) sin dx L L (38) Also,

An =

ut ( x= , 0 ) g= ( x)



∑B

n

n =1

sin

nπ x  nπ  L  L

 c 

which is also a half range sine series , where

Bn =

2

L

g ( x ) sin nπ c ∫ 0

nπ x dx L

Hence the required physically meaningful solution is obtained from Eq. u ( x, t ) givenby Eq. A B (37), where n and n are given by Eq. (38) and (39). n nπ c = ωn= , n 1 , 2 ,… (36) are called normal modes vibrations and L are called frequencies . Example. Obtain the solution of the wave equation

utt = c 2u xx under the following conditions: (i) (ii)

u= ( 0, t ) u= ( 2, t ) 0

u ( x, 0 ) = sin 3

πx 2

Hyperbolic Differential Equation

(iii)

297

ut ( x, 0 ) = 0.

Solution. We have noted in Example 4 that the physically acceptable solution of the wave equation is given by Eq. (35), and is of the form

u ( x ,t)

( c1 cos λ x c2 sin λ x ) c3 cos ( cλt ) + c4 sin ( cλt )

Using the condition

u (0 , t ) ,

c = 0. Also, condition (iii)

we obtain 1 c = 0. The condition u ( 2, t ) = 0 gives implies 4

sin 2λ = 0, Implying that

= λ

nπ ,= n 1, 2 ,… 2 Finally, using condition (ii), we obtain



nπ x nπ ct cos 2 2 (40) Finally, using the condition (ii), we obtain

∑ A sin n =1

n

nπ x



∑A sin= 2 n =1

n

which gives

3 πx 1 3π x 3 πx sin sin = − sin 2 4 2 4 2

A1 =

3 4.

5.4. SUMMARY AND DISCUSSION One of the most important and typical homogenous hyperbolic differential equations is the wave equation. It is of the form

∂2u = c 2 ∇ 2u ∂r 2 where c is the wave speed. This differential equation is used in many branches of physics and longitudinal vibrations in a bar , propagation of sound waves, electromagnetic waves, sea waves, elastic waves in solids, and surface waves as in earthquakes. The solution of a waves equation is called a wave function.

298

Differential Equations: Theory and Applications

∂ 2u 2 2 −c ∇ u = F An example for inhomogeneous wave equation is ∂t 2 ,Where F is a given function of spatial variables and time. In physical problem F represents an external driving force such as gravity force. Another related equation is ,Where γ is a real positive constant . This equation is called a wave equation with damping term. The amplitude of which decreases exponentially as t increases . , we shall derive the partial differential equation describing the transverse vibration of a string.

u= ( x , t ) φ  k ( x – ct )  +ψ  k ( x + ct )  This is also a solution of the one-dimensional wave equation. Further,

let ω = kc. Then

= u ( x , t ) φ  ( kx –ω t )  +ψ  k + ( kx + ωt )  A function of the type given in above equation is a solution of onedimensional wave equation if ω = kc. Therefore, waves travelling with speeds which are not the same as c cannot be described by the solution of ( kx + ωt ) is called the phase for the left travelling the wave equation. Here, wave. We have already noted that x ± ct are the characteristics of the onedimensional wave equation. Solution of the wave equation is

  ω    ω   = u ( x , t )  c1 exp  i   x  + c2 exp  – i   x   e ± iωt   c     c    Since

k=±

ω

, c the time- dependent wave function are of the form

u ( x, t ) = Aei( kx ±ωt ) u ( x, t ) = Aei( kx ±ωt )

Hence, is a solution of the wave equation, and is called a wave function . it is also called a plane harmonic wave or monochromatic wave. Here, A is called the amplitude , ω the angular or circular frequency, and k is the wave number, defined as the number of waves per unit distance

Hyperbolic Differential Equation

299

. By Taking the real and imaginary parts of the solution. We find the linear combination of term of the form

A cos ( kx ± ωt ) , A sin ( kx ± ωt ) Representing periodic plane waves. For instance , consider the function u ( x, t ) = A sin ( kx –ωt ) . This is a sinusoidal wave profile moving towards x − axis the right along the with speed c. Defining the wave length λ as the 2π λ= k . Thereby length over which one full cycle is completed . we have 2π k= . λ implying that

1 1 η ( x + ct ) +η ( x − ct )  + ,t ) u( x = 2 2c

x + ct

∫ v ( ξ ) dξ

x − ct

Thus is known as the D’ALEMBERT’s Solution of the one-dimensional

wave equation. If v = 0 , i. e. , if the string is released form rest, the required solution is

u( x = ,t )

1 η ( x + ct ) +η ( x − ct )  2

The D’ALEMBERT’s solution has an interesting interpretation.

u ( x,= t)

1 1 η ( x + ct ) + η ( x – ct ) + { 2 c

x + ct



v (ξ ) d ξ +

x − ct

t x + ct − ct

1 0 0 0 f ( x, t ) dx dt 2 ∫0 x0 − ct∫0 + ct

The solution is known as the Riemann-Volterra solution

CHAPTER

6

PARABOLIC DIFFERENTIAL EQUATIONS

CONTENTS 6.1. Introduction..................................................................................... 302 6.2. Boundary Conditions....................................................................... 304 6.3. Elementary Solutions Of The Diffusion Equation ............................. 305 6.4. Dirac Delta Function....................................................................... 310 6.5. Separation Of Variables Method....................................................... 316 6.6. Maximum-Minimum Principle and Consequences........................... 340 6.7. Miscellaneous Example. ................................................................. 343 6.8. Boundary Conditions....................................................................... 352

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Differential Equations: Theory and Applications

6.1. INTRODUCTION The diffusion phenomena such as conduction of heat in solids and diffusion of vorticity in the case of viscous flow past a body are governed by a partial differential equation of parabolic type. For example , the flow of heat in a conducting medium is governed by the parabolic equation

∂T ∂t

ρ  = div ( K ∇T ) + H ( r.T , t )

(1)

where ρ is the density,  is the specific heat of the solid, T is the temperature

at a point with position vector r. K is the thermal conductivity , t is the time H ( r,T , t ) ,and is the amount of heat generated per unit time in the element dV situated at a point ( x, y, z ) whose position vector is r. This equation is known as diffusion equation or heat equation. We shall now derive the heat equation form the basic concepts. Let V be an arbitrary domain bounded by a closed surface

S and letV= V ∪ S . Let T ( x, y, z , t ) be the temperature at a point ( x, y, z ) at time t. if the temperature is not constant , heat flows from a region of high temperature to a region of low temperature and flows the Fourier law which q ( r ,t ) states that heat flux across the surface element dS with normal nˆ is proportional q ( r , t ) =− k ∇T ( r , t )

(2)

where k is the thermal conductivity of the body. The negative sign indicates that the heat flux vector points in the direction of decreasing temperature . Let nˆ be the outward unit normal vector and q be the heat flux at the surface element dS . Then the rate of heat flowing through the element surface dS in unit time as shown in

dQ = ( q . nˆ ) dS

(3) Heat can be generated due to nuclear reactions or movement of mechanical parts as in inertial measurement unit ( IMU ) , or due to chemical sources which may be a function of position, temperature and time and H ( r ,T , t ) . may be denoted by We also may be function of position , temperature and time may be substance as the amount of heat needed to

Parabolic Differential Equations

303

raise the temperature of a unit mass by a unit dm = ρ dV the value T is given by

therefore,

Q = ∫∫∫ C ρT dV V

dQ ∂T = ∫∫∫  ρ dV dt ∂t V These energy balance equation for a small control volume V is: The rate of energy storage in V equate to the sum of rate of heat entering V theorugh its bounding surface and the rate of heat generation in V . Thus,

∫∫∫  ρ

∂T ( r , t ) ∂t

V

.

dV = − −∬q . nˆ dS + ∫∫∫ H ( r ,T , t ) dV S

V



(4)

Using the divergence theorem, we get



V

ρ

∂T ( r , t )

 dV = − div q ( r , t ) + H ( r ,T , t )  dv = 0 ∂t  (5) Since the volume is arbitrary , we have

∫∫∫   ρ

∂T ( r , t ) ∂t

= − div q ( r , t ) + H ( r ,T , t )

(6)

Substituting Eq. (2) into Eq. (6), we obtain

∂T ( r , t )

ρ  = div q ( r , t ) + H ( r ,T , t ) ∂t

(7)

If we define thermal diffusivity of the medium as

α=

k ρ Then the differential equation of heat conduction with heat source is

H ( r,T , t ) 1 ∂T ( r , ) = ∇ 2T ( r , t ) + K α ∂t (8) In the absence of heat sources, Eq. (8) reduces to

304

Differential Equations: Theory and Applications

∂T ( r , t )

= α∇ 2T ( r , t ) ∂t (9) This is called Fourier heat conduction equation or diffusion equation. The fundamental problem of heat conduction is to obtain the solution of Eq. (8) subject to the initial and boundary conditions which are called initial boundary value problems, hereafter referred to as IBVPs.

6.2. BOUNDARY CONDITIONS The heat conduction equation may have numerous solution unless a set of initial and boundary conditions are specified .The boundary conditions are mainly of three types which we now briefly explain. Boundary Condition 1: The temperature is prescribed all over the T ( r ,t ) boundary surface. That is, the temperature is a function of both T = G ( r ,t ) position and time. In other words, which is some prescribed function on the boundary. This type of boundary condition is called the Dirichlet condition. Specification of boundary conditions depends on the problem under investigation. Sometimes the temperature on the boundary surface is a function of position only or is a function of time only or a constant. T ( r, t ) = 0 A special case includes on the surface of the boundary, which is called a homogeneous boundary condition Specification of boundary conditions depends on the problem under investigation. Sometimes the temperature on the boundary surface is a function of position only or is a T ( r ,t ) = 0 function of time only or a constant . A special case includes on the surface of the boundary, which is called a homogeneous boundary condition. Boundary Condition II: The flux of heat, i.e., the normal derivative of ∂T , the temperature ∂n is prescribed on the surface of the boundary. It may be a function of both position and time, i.e.,

∂T = f ( r ,t ) ∂n This is called the Neumann condition. Sometimes, the normal derivatives of temperature may be a function of position only or a function of time

Parabolic Differential Equations

305

only. A special case includes boundary. This homogenous boundary condition is also called insulated boundary condition which states the heat flow is zero. Boundary condition III. A linear combination of the temperature and its normal derivatives is prescribed on the boundary , i.e.,

K

∂T + hT = G ( r ,t ) ∂n

where K and h are constants. This type of boundary condition is called Robin’s condition. It means that the boundary surface dissipates heat by convection. Following Newton’s law of the cooling, which states that the rate at which heat is transferred from the body to the surrounding is Proportional to the difference in temperature between the body and the surrounding , we have

−K

∂T =h ( T − Ta ) ∂n

As a special case, we may also

K

∂T + hT = 0 ∂n

which is homogeneous boundary. This mean that heat is converted by dissipation from the boundary surface into a surrounding maintained at zero temperature. The other boundary conditions such as the heat transfer due to radiation obeying the fourth power temperature law and those associated with change of phase .like melting, ablation, etc, give rise to non-linear boundary conditions.

6.3. ELEMENTARY SOLUTIONS OF THE DIFFUSION EQUATION Consider the one-dimensional diffusion equation

∂ 2 T 1 ∂T = , − ∞ < x ∞ ,t 0 ∂ x 2 α ∂t

(10)

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Differential Equations: Theory and Applications

The function

= T ( x ,t )

1 2 exp  − ( x –ξ ) / ( 4at )    4 x at (11)

where ξ is an arbitrary real constant , is a solution of Eq. (10) . It can be verified easily as follows:

 ( x –ξ ) 2    exp[ − ( x –ξ ) 2 / ( 4at )]  4α t 

∂T ∂t

1 4 xat

∂T ∂t

1  −2 ( x –ξ )  2   exp [ − ( x –ξ ) / ( 4at )] 4 xat  4α t  Therefore

∂2T 1 = 2 ∂x 4 xat

2  x –ξ ) ( 1 + – 4α 2 t 2  2α t

 1 ∂T 2  exp [ − ( x –ξ ) / ( 4at ) ] = a ∂t 

which show that the function (11) is a solution of Eq. (10). The function (11), known as Kernel, is the elementary solution or the fundamental solution of t > 0 , the Kernel T ( x , t ) the heat equation for the infinite interval. For is T ( x ,t ) an analytic function of x and t and it can also be noted that is positive for every x . Therefore, the region of influence for the diffusion x →∞ , equation includes the entire x-axis. It can be observed that as the amount of heat transported decrease exponentially . In order to have an idea about the nature of the solution to the heat equation, consider a one-dimensional infinite region which is initially at f ( x) temperature . Thus, the problem is described by

∂T ∂2T PDE = : α 2 , − ∞ < x ∞ ,t 0 ∂t ∂x IC:

T= < ∞ ,t 0 ( x , 0 ) f ( x ) , − ∞ < x= Following the method of variables separable, we write

Parabolic Differential Equations

307

(14) Eq . ( 12 ) . Substituting into We arrive at

X '' 1 β ′ = =λ X α β (15) where λ is a separation constant. The separated solution for β gives

β = C e aλt (16) If λ > 0 , we have β and , therefore, T growing exponentially with

time. From realistic physical considerations, it is reasonable to assume that T ( x , t ) < M as x →∞ . f ( x ) → 0 as x →∞ , T ( x ,t ) while But for 2 to remain bounded , λ should be negative and thus we take λ = − µ . Eq . ( 15 ) . Now from we have

X '' + µ 2 X = 0 Its solution is found to be

= X c1 cos µ x +c 2 sin µ x Hence

T ( x , t , µ ) = A cos µ x B sin µ x ) e − aµ

2

t

(17)

Eq . ( 12 ) , is a solution of where Aand B are arbitrary constants . Since f ( x) is in general not periodic , it Is natural to use Fourier integral instead of Fourier series in the present case . Also, since A and B are arbitrary, we = A A= ( µ ) ,B B( µ ) . may consider them as functions of µ and take In this particular problem, since we do not have any boundary conditions which limit our choice of µ , we should consider all possible values. From the principles of superposition, this summation of all the product solution will give in the relation

T ( x, t ) =





0

0

T ( x, t , µ ) d µ ∫  A ( µ ) cos µ x + B ( µ ) sin µ x  e ∫=

−αµ 2 t

dµ (18)

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Differential Equations: Theory and Applications

which is the solution of

T (= x, 0 ) f= ( x)

Eq . ( 12 )

From the initial condition (13) , we have



∫  A ( µ ) cos µ x + B ( µ ) sin µ x  d µ 0

(19)

In addition, if we recall the Fourier integral theorem, we have

f ù( t )

 ∞  ùf ( x ) cos π ∫0  −∫∞ ∞

1



( t − x ) dx  d 



(20)

Thus, we may write

f ( x)=

 ∞   ∫ f ( y ) cos µ ( x – y ) dy  d µ ∫ π 0  −∞  1



 ∞  cos cos sin sin f y µ x µ y µ x µ y dy + ( )( )   dµ π ∫0  −∫∞  1



∞ ∞ ∞  1  cos cos sin µ x f y µ dy µ x f ( y ) sin µ ydy  d µ + ( )  ∫ ∫ ∫ π 0 −∞ −∞  Let

A( µ ) = B( µ ) =

1

π 1

π

(21)



∫ f ( y ) cos µ dy

−∞ ∞

∫ f ( y ) sin µ y dy

−∞

Then Eq. (21 ) can be written in the form

f ( x) =



∫  A ( µ ) cos µ x + B ( µ ) sin µ x  d µ 0

Comparing

Eq . ( 19 ) and ( 22 ) ,





we shall write relation (19) as

 ∞  cos – T (= x, 0 ) f= f y µ x y dy ( x) ( ) ( )   dµ π ∫0  −∫∞  1





(23)

(22)

Parabolic Differential Equations

309

Assuming that the conditions for the formal interchange of orders of integration are satisfied, we get

∞  2 exp cos – T ( x, t ) f y αµ t µ x d µ = − ( ) ( )   dy ∫ π ∫0 −∞  1



(

)

(25)

Using s = µ α t , and choosing

b=

x− y 2 αt Equation (26) becomes



∫e

−αµ 2t

cos µ ( x − y ) = dµ

0

Substituting

Eq . ( 27 )

1 4απ t

T ( x, t )

π 2 exp  − ( x – y ) / ( 4at )    4α t into

Eq . ( 25 ) ,



(27)

we obtain

∫ f ( y ) exp − ( x – y ) / ( 4at ) dy 2

−∞



(28)

f ( y) Hence, if is bounded for all real values of y , Eq. ( 28 ) is the solution of the problem described by Eq.( 12 ) and ( 13 ) . Example. In a one-dimensional infinite solid, − ∞ < x < ∞ , the surface

a < x < b is initially maintained at temperature T0 and at zero temperature everywhere outside the surface. Show that

T ( x ,t ) =

T0   b − x  erf 2   4at

  a−x   – erf     4at  

where erf is an error function. Solution. The problem is described as follows:

= Tt α T xx , − ∞ < x < ∞

PDE:

= T T0 ,

IC:

a< x

δ (t ) Obviously, this contradiction implies that cannot be a function in the ordinary sense. Some important properties of Dirac delta function are presented now:

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Differential Equations: Theory and Applications

Property 1: ∞

∫ δ ( t ) dt =1

−∞

Property II: For any continuous functions

f (t ) ,



∫ f ( t ) δ ( t ) dt = f ( 0 )

−∞

Proof (of property 2) Consider the equation ∞

LTε → 0

) δ ε ( t ) dt ∫ f ( t=

LT ε → 0 f ( ξ ) − ε < ξ < ε

−∞

As ε → 0 , we have ξ → 0 . Therefore, ∞

∫ f ( t ) δ ( t ) dt = f ( 0 )

−∞

Property III: Let

f (t )

be any continuous function. Then



∫ δ ( t – a ) f ( t ) dt = f ( a )

−∞

Proof Consider the function

1  . a 0 ∂ x2 PDE: ∂t θ ( 0= ,t ) 0 , t ≥ 0 BCs: ∂θ ,t ) 0 , t > 0 (L= ∂x θ ( x , 0= ) θ0 , 0 ≤ x ≤ L IC: 2 θ ( x , t ) = e − λ t ( A cos λ x B sin λ x ) Using the first boundary condition, we obtain A = 0. Thus the acceptable solution is 2

θ = Br a λ t sin λ x 2 ∂θ = λ Be − aλ t cosλ x ∂x

Using the second boundary condition, we have

0 = λ Be aλ

2

t

cos λ L

Implying cos λ L = 0. Therefore, the eigen value and the corresponding eigen functions are

= λn

1)π ( 2n += n 2I .

0 ,1 , 2 ,…

Thus the acceptable solution is of the form 2   2n + 1  2   2n + 1  = θ B exp  − α   π t  sin  π x  2L     2 L 

Using the principle of superposition, we obtain

= θ ( x ,t )

  2n + 1  2 2   2n + 1  B exp ∑ n  − a  2l  π t  sin  2 L π x  n=0   ∞

Parabolic Differential Equations

321

Finally, using the initial condition , we have

 2n + 1





θ 0 = ∑ B n sin  πx  2L  n=0 which is a half range Fourier-sine series and, this

Bn = 2 = L

2 L

L

∫θ

0

0

 2n + 1  sin  π x  dx  2L 

L    2n + 1    2L π x    –θ 0 cos ( 2n +1 ) π   2 L   0  

 ( 2n + 1 ) π  4θ 0 4θ 0 = − [ cos   − cos 0 = 2 ( 2n + 1 ) π ( 2n + 1 ) π   Thus, the required temperature distribution is 2    ( 2 n +1 )  4θ 0 2 n +1  2  sin  exp  –α  t θ ( x ,t ) = ∑ π πx     2L  n = 0 ( 2n + 1 ) π  2L  ∞

Example. A conducting bat of uniform cross-section lies among the and x L . it kept initially at temperature 00 x − axis with ends= at x 0= and its lateral surface is insulated. There are no heat sources in the bar. The 0 end x = 0 is kept at 0 , and heat is suddenly applied at the end x = L , so

that there is a constant flux in the bar for t > 0 .

q0

at x = L . find the temperature distribution

Solution. The given initial boundary value problem can be described as follows:

∂T ∂2T =α ∂ x2 PDE: ∂t BCs:

T ( 0= ,t ) 0 , t > 0

∂T L , t ) q0 , t > 0 (= ∂x IC:

T(x= ,0 ) 0 ,

0≤ x≤ L

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Differential Equations: Theory and Applications

Prior to applying heat suddenly to the end x = L , when t = 0 , the heat flow in the bar is independent of time (steady state condition ) . Let

T (= x , t ) T s ( x ) + T1 ( x , t )

T

where s is a steady part and Therefore,

T1 is the transient part of the solution .

∂ 2 Ts =0 ∂ x2 whose general solution is

T= Ax + B s = x 0= ,T s 0 , implying B = 0 . Therefore, T s = Ax

when

∂T s = q0 . A = q0 . hence , the steady state Using the other BC: ∂x w get solution is T s = q0 x For the transient part, the BCs and IC are redefined as

( i ) T1 ( 0 , t ) = T ( 0 , t ) – T s ( 0 ) = 0 − 0 = 0 ( ii ) ∂T1 ( L , t ) / ∂ x ( iii )

= ∂T / ∂x – ∂T s ( L t ) / ∂x = q 0 − q 0 = 0

T 1 ( x , 0 ) =T ( x , 0 ) − T s ( x ) =− q 0 x , 0 < x < L .

Thus, for the transient part, we have to solve the given PDE subject to Eq. ( 48 ) , i. e. these conditions. The acceptable solution is given by

= T1 ( x , t ) e − aλ

2t

( A cosλ x + B sin λ x )

Applying the BC

T 1 ( x , t ) = Be – a λ

2

t

Applying the BC

( i ) , we get

A = 0 . Therefore ,

sin λ x )

( i ) , we get

A = 0 . Therefore ,

Parabolic Differential Equations

323

T1 ( x , t ) = Be – a λ t sin λ x 2

And using the BC (ii), we obtain 2 ∂T 1 λ – a λ t cos λ L 0 = B= x=L ∂x

= λL

Implying principle , we have

π

,n ( 2n –1 ) = 2

1 , 2 , ,…

Using

the

superposition

2   2n −1  2   2n −1  T1 ( x , t ) = ∑B n exp  –α  πx  π t  sin   2L   2L  n =1   ∞

Now, applying the IC

( iii ) , we obtain

∞  2 n −1  T1 ( x , 0 ) = − q0 x = B n sin  πx ∑  2L  n =1

 2m −1 sin  π 2 L  Multiplying both sides by L to and noting that

 x  and integrating between 0

 0 , n≠ m   2n −1   2m −1  ∫0 B n sin  2L  π x sin  2L π x  dx =  Bm L , n = m  2 L

We get at once , after integrating by parts, the equation

− q0

4 L2

 L  2m −1   = sin B m     2 ( 2m −1 ) π 2   2 L   2

which gives

( −1 ) 8Lq 0 = 2 ( 2m −1 ) π 2 m

Bm

Hence, the required temperature distribution is

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Differential Equations: Theory and Applications

m   ( 2m –1) 2   2m − 1   8 Lq0 ∞  ( −1 ) 2   sin  T ( x, t ) = q0 x + 2 ∑ exp a t nπ   π −   π n =1  ( 2m − 1)2   L    2 L   

Example. The end A and B of a rod, 10 cm on length , are kept at

0 0 temperature 0 C and 100 C until the steady state condition prevails.

0

Suddenly the temperature at the end A is increased to 20 C , and the end 0 B is described to 60 C . find the temperature distribution in the rod at time t. Solution. The problem is described by

∂T ∂2 T = α , 0 < x 0 , the boundary surface is being subject is being subject to a periodic heat flux

g 0 cos ω t . investigate the

Differential Equations: Theory and Applications

326

penetration of these temperature variations into the earth’s surface and show that at a depth x, the temperature fluctuates and the amplitude of the steady temperature is given by

  ω   exp  −  x  ω  2a   



g0 2

Solution. The given IBVP is described by

∂T ∂2 T =α ∂x (49) PDE: ∂t ∂T − =g 0 cos ωt at x =0 , t > 0 BC: ∂x (50) IC:

T ( x ,0 ) = 0

(51)

We shall introduce an auxiliary function T satisfying and then define the complex function Z such that

Eq . ( 49 ) – ( 51 )

Z= T + i T We can easily verify that Z satisfies

∂Z ∂ 2Z =α ∂x 2 (52) PDE: ∂t ∂Z x 0 ,t > 0 = g 0 e ωt at= BC:- ∂x IC: Z = 0 in the region. t = 0 Let us assume the solution of

Eq. ( 52 )

in the form

Z = f ( x ) e iωt f ( x)

where

d2 f (x ) dx −

2

df ( x ) dx

−i

satisfies

ω f ( x)= 0 α (53)

= g 0 at x = 0

(54)

Parabolic Differential Equations

327

Also,

f ( x)

is finite for large x.

The solution of

Eq . ( 53) ,

  iω = f ( x ) A exp  −    α

satisfying the

BC ( 55 ) ,

( 55 )

is

  x   

BC ( 54 ) . The constant A can be determined by using the Therefore ,

1 g0 i

= f ( x)

 α  iω   exp  −   x ω  α   

Thus,

 a 1  iω   exp iωt –  x  ω i  α    (56)

Z = g0

It can be shown for convenience that

= i

1+ i 1 1− i = , 2 i 2 Thus,

q. ( 56 )

can be written in the form

  g 0 2a ω  ω  exp  − Z= x  (1 – i ) exp i (ωt − x 2 ω 2a   2a  

   2a   ω  ω  x  (1 − i ) cos  ωt – x  + i sin  ωt − x   − ω  2a  2a    ω exp   

2a

g = 0 2

Its real part gives the fluctuation in temperature is

T= ( x, t )

g0 2

g0 2

  ω  ω ω  x  cos(ωt – x + sin  ω − x  exp  − ω 2a 2a    2a   

2a

  ω  ω π  x  cos  ωt − x −   − 2a  2a 4 ω exp  

2a

Differential Equations: Theory and Applications

328

Hence, the amplitude of the steady temperature is given by the factor

 ω  exp  − x ω 2a  

2a

g0 2

Example. Find the solution of the non-dimensional diffusion equation satisfying the following BCs.

(i) T

( ii )

is bounded as t →∞

∂T ∂x

( iii )

∂T ∂x

=0 , x=0 ,

for all t

x=0

= 0 , for all t

Solution. This is an example with insulated boundary conditions. It can be seen that a physically acceptable general solution of the diffusion equation is

(

T ( x , t ) exp α λ 2 t

) ( A cos λ x + B sin λ x )

Thus,

∂T =exp −αλ 2t (− Aλ sin λ x + Bλ cos λ x) ∂x (57)

(

)

Using the BC (ii ) , Eq. ( 57 ) , gives B = 0 , Since we are looking for a non-trivial solution, the use of BC ( iii ) into Eq. (57) at once gives

λα n= π , n 0 ,1 , 2 ,… sin λα = 0 implying= Using the principle of superposition, we get ∞   nπ  2   nπ T ( x, t ) = A exp ∑ An exp −αλ 2t cos λ x =  −α  ∑ n  t  cos   a n=0   a  

(

)

The boundary condition (iv ) gives

  nπ  2   nπ  T ( x, 0 ) = x (a – x) = A0 + ∑ A n exp  − a   t  cos  x  a  n =1   a   ∞

where

  x. 

Parabolic Differential Equations

2 a

A0 = An = =

a

∫(

)

ax − x 2 dx =

0

2 a

a

∫(

)

ax − x 2 dx =

0

329

a2 6 a2 6

2a2 2a 2  n 1 + cos = n 1 + ( −1 )  π ( ) 2 2 2 2   n π n π Therefore,

 4a 2 , for n even − A n =  n 2π 2  0 , for n odd  Hence, the required solution is ∞ a 2 4a 2 1  nπ T ( x, t ) = cos  − 2 ∑ 2 6 π =n 2 , 4 ,… even n  a

  nπ  2    x exp  −α   t    a  

Example. The boundaries of the rectangle 0 ≤ x ≤ a, 0 ≤ y ≤ b are maintained at zero temperature. if at t = 0 the temperature T has the F(x ,y) , prescribed value show that for t > 0 , the temperature at a point within the rectangle is given

T ( x, y , t ) =

4 ab





∑∑ f ( m, n ) exp ( −aλ t ) sin

= 1= m n 1

2 mn

mπ x nπ y sin a b

where a b

f ( m , n ) = ∫ ∫ f ( x , y ) sin 0 0

mπ x nπ y sin dx dy a b

and

 m2 n2  = λ π  2+ 2 b   a 2 mn

2

Solution. The problem is to solve the diffusion equation described by

Differential Equations: Theory and Applications

330

 ∂ 2T ∂ 2 T ∂T PDE = : α 2 + ∂t ∂ y2  ∂x BCs :

  , 0 < x < a , 0 < y b ,t 0 

T ( 0 , y= ,t ) T ( a , y = , t ) 0 , 0 < y b ,t 0

T= ( x , 0 , t ) T= ( x ,b ,t ) 0 , 0 < x a ,t 0 IC:

T= ( x , y ,0 ) f ( x , y ) ,

0 < x < a ,0 < y < b

Let the separable solution be

T = X ( x )Y ( y ) β ( t ) Substituting into PDE , we get

X Y '' 1 β ' + = = −λ 2 X Y a β ''

Then

β +aλ 2 β = 0 '

X '' Y '' = − (λ 2 + = − p 2 ( say ) X Y Hence, ''

X + p2 X = 0 Y '' = −λ 2+ p 2 = − q 2 ( say ) Y Therefore , ''

Y +q2Y = 0 Thus, the general solution of the given PDE is

T ( x, y , t ) = ( A cos px + B sin px )( c cos qy + D sin qy ) e ^ −a

2

t

where

λ=2 p 2 + q 2 Using the BC: the form

T= , we get A 0 . ( 0 , y , t ) 0=

Then, the solution is of

Parabolic Differential Equations

= T ( x , y , t ) B sin px ( c cos qy + D sin qy ) e − aλ Applying the BC: given by

2

T= , we get c 0 ( x , 0 , t ) 0=

T ( x , y , t ) = BDsin px sin qye − aλ

2

331

Thus , the solution is

t

T ( a , y ,t ) =0 Applying the BC : gives sin pa = 0 , implying pa = nπ or

= p

nπ = , n 1 , 2 ,… a

Using the principle of superposition the solution can be written in the form 2  nπ  sin  x  sin qye − aλ t  a 



( x , y ,t ) = ∑ A n n =1

Using the solution is found to be ∞ ∞  nπ   mπ T ( x , y , t ) = ∑ ∑ A mn sin  x  sin   a   b = m 1= n 1

 − aλ 2 t e 

 m2 n2  + 2 2 a   b

λ 2 = p2 +q2 = π 2 

Finally, using the IC , we get

 nπ   mπ  T ( x , y , 0 ) = f ( x , y ) = ∑ Amn sin  x  sin  y  a   b  which is a double Fourier series , where

2 2  mπ   nπ  = . ∫ ∫ f ( x , y ) x  sin  y  dx dy a a00  a   b  a b

Amn

Hence, the required general solution is

T ( x, y , t ) =





∑ ∑ f ( m, n ) e

= 1= m n 1

− aλ 2 t

 mπ   nπ  sin  x  sin  y  dxdy  a   b 

332

Differential Equations: Theory and Applications

and

 m2 n2  = λ2 π 2 2 + 2  a   b

6.5.1. Solution of Diffusion Equation in Cylindrical Coordinates Consider a three dimensional equation

∂T = a∇ 2T ∂t In cylindrical coordinates

( r ,θ

, z ),

it becomes

1 ∂T ∂ 2 T 1 ∂T 1 ∂ 2 T ∂ 2 T = + + + a ∂t ∂ r 2 r ∂r r 2 ∂θ 2 ∂z 2 (58) where

T = T ( r ,θ , z , t )

.

Let us assume separation of variables in the form

T ( r ,θ , z , t ) = R ( r ) H ( θ ) Z ( z ) β ( t ) Substituting into Eq. (58) , it becomes

R′′ HZ β +

1 ' 1 β' R HZ β + 2 H '' RZ β + Z '' RH β = RHZ r r α

or

R '' 1 R ' 1 H ′′ Z ′′ 1 β ' + + 2 + = = −λ 2 R r R r H Z a β 2 where – λ is separation constant . Then

β ' + aλ 2 β = 0 (59) R '' 1 R′ 1 H ′′ Z ′′ + + 2 + λ2= − = − µ 2 ( say ) R R R r H Z Thus, the equation determining Z , R and H become

Z '' − µ 2 Z = 0 (60)

Parabolic Differential Equations

333

R′′ 1 R′ 1 H '' + + +λ 2 +µ 2= 0 R r R r2 H or

R '' R' H '' 2 +r + λ 2 + µ 2 r2 = − = v ( say ) R R H

(

r2

)

Therefore , ''

H +v2 H = 0 (61) R′′ +

v2  1 '  2 R +  λ + µ2 − 2  R = 0 r r   (62)

(

Equation

β = e − aλ

2

)

( 59 ) − ( 61 ) have particular solution of the form

t

= H c cos cθ + D sin vθ

= Z Ae µ z + Be – µ z The differential equation ( 62 ) is called Bessel’s equation of order v and its general solution is known as

R (= r ) c1 jv

(

)

λ 2 + µ 2 r + c 2 Yv

(

λ2 +µ2 r

)

jv ( r ) and Y v ( r )

are Bessel functions of order v of the first and Eq. ( 62 ) second kind, respectively. Of course, is singular when r = 0 . The physically meaningful solutions must be twice continuously differentiable Eq. ( 62 ) in 0 ≤ r ≤ a . Hence, has only one bounded solution, i.e., where

= R( r ) j v

(

λ2 +µ2 r

)

Finally, the general solution of

Eq. ( 58 )

is given by

T ( r ,θ , z , t ) = e − aλ t  Ae µ z + Be − µ z  [C cos vθ + D sin vθ ] jv 2

(

λ2 + µ2 r

)

334

Differential Equations: Theory and Applications

T ( r ,t ) Example. Determine the temperature in the infinite cylinder 0 ≤ r ≤ a when the initial temperature is T ( r , 0 ) = f ( r ) , and the surface r = a is maintained at 00 temperature. Solution. The governing PDE from the data of the problem is

∂T = a∇ 2T ∂t where T is a function of r and t only. Therefore,

∂ 2 T 1 ∂T 1 ∂T + = ∂r 2 r ∂r a ∂t (63) The corresponding boundary and initial coordination are given BC: IC:

T ( a ,t ) = 0 T ( r,0 ) = f ( r )

(64)

The general solution of

(

Eq. ( 63 )

)

is

T (= r , t ) A exp − a λ 2 t j 0 ( λ r ) Using the BC ( 64 ) , we obtain

j0 ( λa ) = 0

ξ a (= n 1 , 2 ,…. , ∞ ) . which has an infinite number of roots, n Thus , we get from the superposition principle the equation ∞

(

)

T ( r , t ) = ∑ An exp – a ξ n2 t j 0 ( ξ n r ) n =1

Now using the IC:

T ( r , 0 ) = f ( r ) , we get



f ( r ) = ∑ A n j0 ( ξ n r ) n =1

A ,

To compute n we multiply both side of the above equation by rj0 ( ξ m r ) and integrate with respect to r to get

Parabolic Differential Equations a



0

n =1

∫r f ( r ) j 0 ( ξm r ) dr = ∑ An

335

a

∫ r j ( ξ r ) J ( ξ r ) dr 0

0

m

n

0=

0 for n ≠ m   =  a2  2  Am  2  J 1 (ξ m a ) for n = m    which gives

Am =

2

a

a 2 j12 ( ξ m a )

∫uf ( u ) j ( ξ u ) du 0

m

0

Hence, the final solution of the problem is given by

T ( r ,t ) =

2 a2





m =1

j0 ( ξ m r )

a  2 exp αξ t −  ∫uf ( u ) j0 ( ξ u ) du  m 2 j 1 ( ξm a ) 0 

(

)

6.5.2. Solution of Diffusion Equation In spherical Coordinates In this section, we shall examine the solution of diffusion or heat conduction equation in the spherical coordinates system. Let us consider the three – T = T ( r ,θ ,φ , t ) . dimensional diffusion Eq. ( 9 ) and let In the spherical coordinates system Eq. ( 9 ) can be written as

∂ 2 T 2 ∂T 1 ∂  ∂T  1 ∂ 2T 1 ∂T + + =  sin θ + ∂ r 2 r ∂r r 2sin θ ∂θ  ∂θ  r 2 sin 2 θ ∂φ 2 α ∂t (65) This equation is separated by assuming the temperature function T in the form T = R ( r ) H ( θ )φ ( φ ) β ( t ) Substituting

(66)

Eq. ( 66 ) into Eq. ( 65 )

, we get

R′′ 2 R′ dH  d 2φ 1 β ′ 1 1 d  1 + + 2 + = = −λ 2 ( say ) sin θ   R r R r sin θ H dθ  dθ  φ r 2 sin 2 θ dφ 2 a β 2 where λ is a separation constant. Thus,

d β + λ 2 αβ = 0

Differential Equations: Theory and Applications

336

whose solution is

β = c1 e aλ Also,

2

t

(67) .

ich gives

whose solution is (68) Now, the other separated equation is

1  d 2 R 2 dR  1 d  dH  m2 2 +  +  sin θ  +λ = R  dr 2 r dr  Hr 2sin θ dθ  dθ  r 2 sin 2 θ or

r2  m2 d  dH  2  1 2 2 ′′ ′  R + R  + λ r =2 –  sin θ  R r  dθ  sin θ H sin θ dθ  On re-arrangement, this equation can be written as

R′′ +

 n ( n +1 )  2 R′ +  λ 2 – 0 R= 2 r r   (69)

and



1  d2H dH  m2 sin + cos + = n ( n +1 ) θ θ   H sin θ  dθ 2 dθ  sin 2 θ

or

d 2H dH  m2 + cot θ +  n ( n +1 ) − 2 sin θ dθ dθ  Let

 0 H = 

then Eq. (69 ) becomes

(70)

Parabolic Differential Equations

( λr )



1 2

  ψ ' ' ( r ) + 1 ψ  r  

Since

(λr ) ≠ 0 ,

2  1     n+    2  2  ' ( r ) + λ – ψ 2 r    

337

  = 0   

we have

 1   n+ Z2    1  2  0 ψ ′′ ( r ) + ψ ′ ( r ) +  λ 2 –  ψ ( r ) = r r2     1   n+  , 2  whose solution is which is Bessel’s differential equation of order 

ψ ( r ) Aj

n+

1 2

( λ r ) + BY n+ 1 ( λ r ) 2

Therefore

R( r ) where Now,

( λr )



   Ajn + 1 ( λ r ) + BY n + 1 ( λ r )  2 2  

1 2

(71)

jn and Yn are Bessel functions of first and second kind, respectively . Eq. ( 70 )

can be put in a more convenient from by introducing a new

independent variable , µ = cos θ . So that 2 cot θ = µ / 1 − µ dH dH = − 1− µ 2 dθ dµ 2 d 2H dH 2 d H = 1 − −µ µ 2 2 dθ dµ dµ

(

Thus,

)

Eq. ( 70 )

becomes

Differential Equations: Theory and Applications

338

 0 H=  (72) which is an associated Legendre differential equation whose solution is

(

1− µ 2

)

d 2H dH  m2 n n − 2 + + 1 − µ )  ( dµ 2 dµ  1− µ 2

= H (θ ) A′ pnm ( µ ) + B ' Qnm ( µ )





(73)

where p mn ( µ ) and Q mn ( µ ) are associated Legendre functions of degree n m , and of order of the and second kind, respectively. Hence the physically

meaningful general solution of the diffusion equation in spherical geometry is of the form

T ( r ,θ ,φ , t ) =

.

∑ λ

,m ,n

Aλ mn ( λ r )



1 2

j

1 n+ 2

( λ r ) pnm (cos θ )e± imφ − aλ t 2

In this general solution, the functions

Q mn ( µ )

and

(λr )

(74)

1 – 2

Y

n+

1 2

( λr )

±1 and r = 0 are exclude because these functions have poles at µ = respectively. Example. Find the temperature in a sphere of radius a. when its surface is kept at zero temperature and its initial temperature is

f ( r ,θ ) .

Solution. Here, the temperature is governed by three-dimensional heat

equation in spherical polar coordinates independent of . Therefore, the task is to find the solution of PDE 1 ∂T ∂ 2T 2 ∂T 1 ∂  ∂T  = + + sin θ  ∂θ  (75) α ∂t ∂r 2 r ∂r r 2 sin θ ∂θ 

Subject to BC: IC:

T ( a ,θ , t ) = 0

T ( r ,θ , 0 ) = f ( r ,θ )

The general solution of written as .

T ( r ,θ , t ) = ∑ Aλ n ( λ r ) λ ,n



1 2

j



Eq. ( 75 )

1 n+ 2



with the help of

( λ r ) pn ( cos θ ) e− a λ

2

(76)

Eq. ( 74 ) ,

t



(78)

can be

Parabolic Differential Equations

BC ( 76 ) ,

Applying the

j

n+

1 2

339

we get

( λa ) = 0

This equation has infinitely many positive roots. Denoting them by we have

T ( r ,θ , t )





∑∑Ani (ξi r )



1 2

J

n 0= n 1 =

1 2

n+

(ξi r ) Pn ( cos θ ) exp ( −αξt 2t )



ξi ,

(79)

Now applying the IC and denoting cos θ by µ , we get

(



)



f r , cos −1 ( µ ) = ∑∑ Ani (ξ1r )

−1/2

J

n 0=i 1 =

n+

pn ( µ ) d µ

Multiplying both side by

−1to1, we obtain 1

(



)

1 2



(ξi r ) pn ( µ ) . and integrating between the limits,

−1 ∫ f r , cos ( µ ) pm ( µ ) d µ = ∑∑Ani (ξi r )

−1



−1/2

n 0=i 1 =



= ∑∑Ani (ξi r )



1 2

n 0=i 1 =

j

1 n+ 2

1

jn +1/2 (ξi ) ∫ pm ( µ ) pn ( µ ) d µ −1

2    2n + 1 

( ξi r ) 

or 1

 2n + 1  −2 µ) dµ   ∫ pn ( µ ) f r , cos ( =  2  −1

(

)

Now, to evaluate the constants 3 2

r j

n+

1 2



∑A (ξ r ) i =1

ni

Ani ,

i

1 2

j

n+

1 2

(ξi r= ) for n

0 ,1, 2,3, …

we multiply both sides of the above

(ξ r ) j

equation by and integrating with respect to r between limits 0 to a and use the orthogonality property of Bessel functions to get a 3 1   2n + 1  2 −1 ξi   ∫r jn + 1 (ξ j r ) dr  ∫ pn ( µ ) f r , cos ( µ ) d µ   2 0 2 − 1  1 2

(

)

340 ∞

Differential Equations: Theory and Applications a

= ∑Ani ∫rj i =1

0

n+

1 2

( ξ i r ) j n + 1 ( ξ j r ) dr 2

2

  1 ∞ Ani  j ' 1 ( ξi= r )  , n 0 ,1 , 2 , 3 ,… = ∑ 2 i =1  n+ 2  (80) Thus, Eq. (79 ) and ( 80 ) together constitute the solution for the given problem.

6.6. MAXIMUM-MINIMUM PRINCIPLE AND CONSEQUENCES Theorem. (Maximum- minimum principle ) . Let

u ( x ,t )

be a continuous function and a solution of

ut = α u xx (81) for 0 ≤ x ≤ l , 0 ≤ t ≤ T , whereT > 0 is a fixed time. Then the maximum and minimum values u are attained either at time t = 0 or at the end points x = 0 and x = l at some time in the interval 0 ≤ t ≤ T .

Proof. To start with, let us assume that the assertion is false . Let the u ( x, t ) t 0 ( 0 ≤ x < l ) or for x= 0 or x= l ( 0 ≤ t ≤ T ) maximum value of for = u ( x ,t ) be denoted by M . we shall assume that the function attains ( x0 , t0 ) in the rectangle defined by its maximum at some interior point

0 ≤ x 0 ≤ l , 0 ≤ t0 ≤ T , and them arrive at a contradiction. This mean that

u ( x0 , t 0 = ) M +ε

(82)

Now, we shall compare the signs in Eq. (81) at point

( x0 , t0 ) ,

it is well u ( x, t ) known from calculus that the necessary condition for the function (x , t ) to posses maximum at 0 0 is.

∂u ∂2u x , t = 0 , ( 0 0) ( x0 , t0 ) ≤ 0 ∂x ∂ x2



(83)

Parabolic Differential Equations

In addition,

u ( x0 , t0

)

attains maximum for

t = t0 ,

341

implying

∂u ( x0 , t 0 ) ≥ 0 ∂t (84) Thus, with the help of Eq. (83) and (84) we observe that the signs on the left - and right - hand sides of Eq. (81) are different. However, we cannot claim that we have reached contradiction, since the left- and right hand sides can simultaneously by zero. To complete the proof, let us consider another ∂2 u ∂u ≤ 0 and > 0. 2 xl , tl ) ( ∂t point at which ∂x v= ( x , t ) u ( x , t ) + λ ( t0 – t )



Where λ is a constant. Obviously ,



(85)

v ( x 0 ,t 0 ) = M + ε and λ ( t 0 – t ) ≤ λT .

Suppose we choose λ > 0 , such that

λ
0 is given by

= T ( r ,t )

1 r



∑B n =1

n

cnπ  nπ  − λ 2n t , ) λn sin  r  exp ( = R  R 

Solution. The temperature distribution in a solid sphere is governed by the parabolic heat equation

T= c 2 ∇ 2T t From the data, T is a function of r and t alone. In view of the symmetry of the sphere the above equation with the help of Eq. reduce to

2   = Tt c 2  Trr + T r  r   Setting v = rT , the given BC gives v= ( R , t ) rt= ( R ,t ) 0 which the IC gives

= v ( r , 0 ) rT = ( r , 0 ) rf ( r ) Since T must be bounded at r = 0 , we require

v ( 0 ,t ) = 0 Now,

(86 )

Differential Equations: Theory and Applications

344

Similarly, finding

Trr and substituting into Eq. (86), we obtain

vt = c 2 v rr Using the variables separable method, we may write and get

v ( r , t ) = R ( r )τ ( t )

= R ( r ) A cos kr + B sin kr

τ= ( t ) exp ( − c 2 k 2 t ) Thus, using the principle of superposition , we get ∞

v ( r , t ) =∑( An cos kr + Bn sin kr ) n =1

Also, using

( An

v ( 0 ,t ) = 0 ,

cos kr + Bn sin kr )

r =0

r =0

=0

we have

= 0

An = 0. Also, v ( R , t ) = 0 gives Bn sin kr = 0 , implying sin kr = 0 , as Bn ≠ 0 . Therefore, nπ = kR nπ = , k = , n 1 , 2 ,… R Implying

Thus, the possible solution is ∞  c 2 n 2π 2 t   nπ  v ( r , t ) = ∑ Bn sin  r  exp  –  R2   R  n =1 

Finally, applying the IC:

v ( r , 0 ) = rf ( r ) ,

we get

∞  nπ  rf ( r ) = ∑ Bn sin  r  R  n =1

which is a half-range Fourier series . Therefore,

2 Bn = R

R

 nπ  r  dr R 

∫ rf ( r ) sin  0

Parabolic Differential Equations

by

But

v ( r , t ) = rT ( r , t ) .

345

Hence, the temperature in the sphere is given

 c2 n2 π 2 t   nπ  B sin r exp −   ∑ n   2  R  n =1  R  Example. A circular cylinder of radius a has its surface kept at a T. constant temperature 0 If the initial temperature is zero throughout the cylinder , prove that for t > 0 . T ( r ,t )

1 r



 2 T ( r , t ) T0 1 –  a



∑ξ n =1

j 0 (ξ n a ) n

 exp − ξ n2 kt  j1 ( ξ n a ) 

(

)

± ξ1 , ± ξ 2 ,…., ± ξ n are the roots of j0 ( ξ a ) = 0 , and k is the thermal where conductivity which is a constant. Solution. It is evident that T is a function of r and t alone and , therefore , the PDE to be solved is ∂ 2 T 1 ∂T 1 ∂T + = ∂ r 2 r ∂r k ∂t (87) Subject to

IC : T ( r , 0= ) 0 , 0≤ r < a BC : T ( a= , t ) T0 , t ≥ 0 Let

T ( r ,t = ) T0 + T1 ( r , t ) So that

T1 ( r , 0 ) = − T0 T1 ( a , t ) = 0 where have ,

(88)

(89)

T1 is the solution of Eq. (87). By the variables separable method we

(

= T1 ( r , t ) Aj0 ( λ r ) exp − λ 2 kt

)

Differential Equations: Theory and Applications

346

Using the BC:

T1 ( a , t ) = 0 ,

(

)

we get

Aj0 ( λ a ) exp − λ 2 kt = 0 j0 ( λ a ) = 0 as A ≠ 0 .let ξ1 ,ξ 2 ,…., ξ n ,

which gives j0 ( λ a ) = 0 .

Then the possible solution using the superposition principle is



∑A j (ξ r ) exp ( − ξ

= T1 ( r , t )

n =1

n

Using the ∞

0

n

n

0

n

2 n

kt

IC : T1 ( r , 0 ) = − T0

∑A j ( ξ r ) = n =1

be the roots of

)

(90)

into Eq. ( 90 ) , we obtain

− T0

Multiply both sides by a



a

0

n =1

0

rj0 ( ξ m r )

and integrating, we get

− T0 ∫ rj 0 ( ξ m r ) dr = ∑An ∫ r j0 ( ξm r ) j0 ( ξn r ) dr a

2 A= m ∫ rj0 ( ξ m r ) if m n ; otherwise 0 0

a2 2 j 1 ( ξm a ) 2

= Am

But, a

ξm a

0

0

x dx − T0 ∫rj0 ( ξ m r ) dr = − T0 ∫ j 0( x ) ξm r ) ( x= = −

T0

ξ

2 m

ξm a

∫ 0

ξm

d  xj1 ( x )  dx dx 

T aT = − 02  xj1 ( x )  ξ0m a = − 0 j1 ( ξ m a ) ξ ξm m

Therefore,

ξm

Parabolic Differential Equations

Am

347

aT 0 a2 2 J 1 ( ξm a ) = − j (ξ a ) 2 ξm 1 m

or

An = −

2T0 1 nξ n j1 (ξ n a )

Hence, Eq. (90) becomes

(

2 2 ∞ j0 ( ξ n r ) exp −ξ n kt T1 ( r , t ) = − T0 ∑ a n =1 j1 (ξ n a ) ξn

)

Finally, the complete solution is found to be

 2 T (r= , t ) T0 1 −  a





n =1

(

)

2 j0 ( ξ n r ) exp –ξ n kt   j1 ( ξ n a ) ξn 

Example. Determine the temperature in a sphere of radius a, when its surface is maintained at zero temperature while its initial temperature if f ( r ,θ ) . Solution. Here the temperature is governed by the three-dimensional

heat equation in polar coordinates independent of φ , which is given by

1 ∂u ∂ 2 u 2 ∂u 1 ∂  ∂u  = 2+ + 2  sinθ  k ∂t ∂r r ∂r r sin θ ∂θ  ∂θ  Let

(91)

u ( r ,θ , t ) = R ( r ) H ( θ ) T ( t ) By the variables separable method, the general solution of Eq. (91) is found to be

v ( r ,θ , t )

.

.

∑∑Aλn ( λ r ) λ

n



1 2

j

1 n+ 2

( λ r ) pn ( cos θ ) exp ( −k λ 2t )



In the present problem, the boundary and initial conditions are

BC : u ( a ,θ , t ) = 0

(93)

(92)

348

Differential Equations: Theory and Applications

IC : u ( r ,θ , 0 ) = f ( r ,θ )

(94)

Substituting the BC (93) into Eq. (92), we get

j

n+

1 2

( λa ) = 0

(95)

ξ a , ξ a,… . ,ξ a ,…

2 1 Let 1 be the roots of Eq. (95). Then the general solution can be put in the form

u ( r ,θ , t )





∑∑A (ξ r )

= n 1= i 1

ni

l



1 2

j

1 n+ 2

(ξl r ) pn ( cos θ ) exp ( − kξ t2t )

(96)

Now using the IC, we obtain ∞



f ( r ,θ ) = ∑∑ Ani ( ξ i r )



1 2

J

= n 1= i 1

Multiplying both sides by 1

n+

1 2

( ξi r ) pn ( cosθ )

pn ( cos θ ) d ( cos θ ) ∞



∫ pn ( cosθ ) f ( r ,θ ) d ( cos θ ) = ∑ ∑ Ani ( ξi r ) n 1= i 1 =

−1

1 − 2

and integrating, we have 1

j

1 n+ 2

( ξi r ) ∫ p n2 ( cosθ ) d ( cos θ ) −1

Using the orthogonality property of Legendre polynomials, we get 1



−1





pn ( cosθ ) f ( r ,θ ) d ( cosθ ) = ∑∑ A ni (ξi r ) n= 1= i 1



1 2

j

n+

1 2

2    2n + 1 

( ξi r ) 

Rearranging and multiplying both sides of the above equation by r J 1 (ξi r ) n+ 2 and integrating between the limits 0 to a with respect to r , we get 3/2

a 1 a 1 − 2n + 1 32 2 2 r j r dr p cos f r , d (cos ) = A rj ( dr ξ θ θ θ ξ ) ( ) n( ni ∫ i 1 ( i ) 1 ∫ n+ n+ 2 ∫0 2 2 −1 0

 a2  = Ani  j ' 1 (ξi a )  2  n+ 2  Therefore,

2

Parabolic Differential Equations

Ani =

( 2n + 1) ξi1/2

1

a

2

349

∫ J n+1/2 (ξi r ) dr ∫ pn ( cos θ ) f ( r ,θ ) d ( cos θ )

  0 −1 a  j ' 1 (ξi a )   n+ 2  (97) Hence, we obtain the solution to the given problem for Eq. (96), where 2

Ani is given by Eq. (97). Example. The heat conduction in a thin round insulated rod with heat sources present is described by the PDE

( x, t ) , 0 ut − au xx = F ρc

0 < x l,t 0

(98)

Subject to ‘

BCs : u= ( 0, t ) u= (l, t ) 0 IC : u ( = x, 0 ) f ( x ) , 0 ≤ x ≤ l

(99)

where ρ and c are constant and F is a continuous function of x and t. Find u ( x, t ) . Solution. It can be noted that the boundary conditions are of homogeneous type. Let us consider the homogeneous equation ut − au xx = 0 (100) u ( x, t ) = X ( x ) T ( t ) , we get Setting T ′ X ′′ = = −λ 2 ( say ) αT X (101) 2 0. The corresponding BCs are which gives X ′′ + λ X = X= ( 0 ) X= (l ) 0 The solution of Eq. (101) gives the desired eigen functions and eigen values, which are 2

 nπ  X n ( x ) sin λn x, λn2   , n ≥ 1 = =  l  (102) For the non-homogenous problem (98), let us propose a solution of the form

Differential Equations: Theory and Applications

350



u ( x, t ) = ∑ Tn ( t ) X n ( x )

(103) It is clear that Eq. (103) satisfies the BCs (99).From the orthogonality of eigen functions, it follows that n =l

Tm ( t ) =

2 u ( x, t ) X m ( x ) dx l ∫0 l

However, 1

2  mπ  Tm ( 0 ) = ∫ f ( x ) sin  x  dx l 0  l  (104) which is an IC for (98) . we get

Tm ( t ) .

introducing Eq. (103) into the governing equation

∞ F ( x, t ) ' T X − a Tn X n'' = ∑ ∑ n n ρ c (105) = n 1= n 1 Now, we shall expand F x, t ) / ρ c, so that it is represented by a 0 < x l,t 0 convergent series on in the form ∞

∞ F = ∑ qn ( t ) X n ( x ) ρ c n =1 (106) where

qn ( t ) =

2 F ( x, t )  nπ sin  l ∫0 ρ c  l 1

Thus, becomes ∞

∑X ( T n =1

n

' n

qn ( t )

 x  dx  (107)

is known. Now, Eq. (105), with the help of Eq.(101)

)

+ λn2 α Tn − qn = 0

Therefore, it follows that

Tn' ( t ) + λn2 α Tn ( t ) = qn ( t ) (108) Its solution with the help of IC (104) is

Parabolic Differential Equations

(

351

t

)

Tn = ( t ) Tn ( 0 ) exp −λn2α t + ∫ exp λn2 α (τ − t ) qn (τ ) dτ 0



(109)

From Eq. (103) and (109) , the complete solution is found to be

u ( x, t ) =

1   2 T 0 exp λ α t exp[ λn2α (τ − t ) qn (τ ) dτ  X n ( x ) − + ( )  n ∑ n ∫ n =1  0  ∞

(

)

In the expanded form, it becomes

u ( x, t )

  2 l  2   ∫ f (ξ ) X n (ξ ) d ξ  exp(−λn at ) ∑ l n =1     0 ∞

 F ξ ,τ } ∫ (ρ c ) X (ξ ) dξ dτ  X ( x )

2 + ∫ exp −λn2 a (τ − t ) l 0 l

{

l

n

0



n

(110)

It can be verified that the series in Eq. (110) converges uniformly for t > 0. By changing the order of integration and summation in Eq.(110) , we get

(

)

 ∞ exp −λn2 at X n ( x ) X n (ξ )   f (ξ ) d ξ u ( x, t ) = ∫  ∑ 1/ 2  n =1 0   l

{

}

 ∞ exp − λn2 a ( l − t ) X n ( x ) X n (ξ )  F (ξ , t )  + ∫∫  ∑ d ξ dt l/2 ρc  n =1 00    11

which can also be written in the form

= u ( x, t )

1

11

0

00

∫G ( x,ξ ; t ) f (ξ ) dξ + ∫∫G ( x, ξ ; t − τ )

where ∞

G ( x ,ξ ; t ) = ∑ n =1

(

F ( ξ ,τ )

ρc

d ξ dτ (111)

)

exp −λn2 at X n ( x ) X n (ξ ) l/2

is called Green’s function. More details on Green’s function are given .

352

Differential Equations: Theory and Applications

6.7.1. Summary and Discussion The diffusion phenomena such as conduction of heat in solids and diffusion of vorticity in the case of viscous flow past a body are governed by a partial differential equation of parabolic type. For example, the flow of heat in a conducting medium is governed by the parabolic equation

∂T ∂t

ρ  = div ( K ∇T ) + H ( r.T , t )

(1)

where ρ is the density,  is the specific heat of the solid, T is the temperature at a point with position vector r. K is the thermal conductivity , t is the time H ( r,T , t ) ,and

is the amount of heat generated per unit time in the element dV situated at a point ( x, y, z ) whose position vector is r. This equation is known as diffusion equation or heat equation. We shall now derive the heat equation form the basic concepts. Then the differential equation of heat conduction with heat source is

H ( r,T , t ) 1 ∂T ( r , ) = ∇ 2T ( r , t ) + K α ∂t (2) In the absence of heat sources, Eq. (8) reduces to ∂T ( r , t )

= α∇ 2T ( r , t )

∂t (3) This is called Fourier heat conduction equation or diffusion equation. The fundamental problem of heat conduction is to obtain the solution of Eq. (2) subject to the initial and boundary conditions which are called initial boundary value problems, hereafter referred to as IBVPs.

6.8. BOUNDARY CONDITIONS The heat conduction equation may have numerous solution unless a set of initial and boundary conditions are specified. The boundary conditions are mainly of three types which we now briefly explain. Boundary Condition 1: The temperature is prescribed all over the T ( r ,t ) boundary surface. That is, the temperature is a function of both T = G ( r ,t ) position and time. In other words, which is some prescribed function on the boundary .This type of boundary condition is called the

Parabolic Differential Equations

353

Dirichlet condition. Specification of boundary conditions depends on the problem under investigation.

T ( r ,t ) = 0 A special case includes on the surface of the boundary, which is called a homogeneous boundary condition. Boundary Condition II : The flux of heat , i.e., the normal derivative of ∂T , the temperature ∂n is prescribed on the surface of the boundary . it may be a function of both position and time, i.e.,

∂T = f ( r ,t ) ∂n This is called the Neumann condition .Sometimes, the normal derivatives of temperature may be a function of position only or a function of time only. A special case includes

∂T = 0 onthe ∂n boundary This homogenous boundary condition is also called insulated boundary condition which sates the heat flow is zero. Boundary condition III. A linear combination of the temperature and its normal derivatives is prescribed on the boundary , i.e.,

K

∂T + hT = G ( r ,t ) ∂n

Where K and h are constants. This type of boundary condition is called Robin’s condition. it means that the boundary surface dissipates heat by convection .Following Newton’s law of the cooling, which states that the rate at which heat is transferred from the body to the surrounding is Proportional to the difference in temperature between the body and the surrounding , we have

−K

∂T =h ( T − Ta ) ∂n

As a special case, we may also

K

∂T + hT = 0 ∂n

Differential Equations: Theory and Applications

354

which is homogeneous boundary . This mean that hear is converted by dissipation from the boundary surface into a surrounding maintained at zero temperature. The other boundary conditions such as the heat transfer due to radiation obeying the fourth power temperature law and those associated with change of phase .like melting, ablation, etc., give rise to non-linear boundary conditions.

δ (t ) Eq. ( 33 ) ( 34 ) is known as This limiting function defined by and Dirac delta function or the unit impulse function. Its profile is depicted in . Dirac originally called it an improper function as there is no proper function with these properties . In fact, we can observed that = 1





= δ ( t ) dt LTε → 0

.



δ= LT = ε ( t ) dt ε →0 0 0

t >ε

−∞

δ (t ) Obviously , this contradiction implies that cannot be a function in the ordinary sense. Some important properties of Dirac delta function are presented now: ∞

Property 1:

∫ δ ( t ) dt =1

−∞

Property II: For any continuous functions

. ∞

Property III: let

f (t )

∫ δ ( t – a ) f ( t ) dt = f ( a )

−∞

Property IV:

f (t ) ,



∫ f ( t ) δ ( t ) dt = f ( 0 )

−∞

be any continuous function . Then

.

δ ( −t ) = δ (t )

1 = δ ( at ) δ ( t ) , a >0 a Property V:

Parabolic Differential Equations

.



Property VI: Dirac delta

δ (t ) if is a continuously differentiable t, function Vanishing for large then ∞

f ( t ) δ ' ( t ) dt  f ( t ) δ ( t )  − ∞ − ∫=

−∞



∫ f ( t ) δ ( t ) dt '

−∞



Since δ t → 0 as t →∞ , − ∞ , we have ∞

operty VII:

355

∫ f ( t ) δ ( t ) dt = − f ( 0 ) '

'

−∞

− f (a) ∫ δ ( t − a ) f ( t ) dt = '

'

−∞

Having discussed the one-dimensional Dirac delta function , we

can extend the definition to two dimension, Thus, for every f which is ( ξ ,η ) , we define continuous over the region S containing the point δ ( x –ξ , y –η ) in such a way that .

∬δ ( x –ξ y –η ) f ( x , y ) dσ = f ( ξ ,η ) S

Note that functions, i.e.,

δ ( x − ξ , y −η )

. is a forma limit of a sequence of ordinary

δ ( x − ξ , y −η ) = LTε → 0 δ ε ( r )

r 2 = ( x − ξ ) + ( y −η ) . 2

,Where

2

CHAPTER

7

LAPLACE TRANSFORM METHODS

CONTENTS 7.1. Introduction..................................................................................... 358 7.2. Transform of Some Elementary Functions......................................... 362 7.3. Properties of Laplace Transform....................................................... 364 7.4. Transform of A Periodic Function..................................................... 370 7.5. Transform of Error Function.............................................................. 372 7.6. Transform of Bessel’s Function......................................................... 374 7.7. Transform of Dirac Delta Function................................................... 376 7.8. Convolution Theorem (Faltung Theorem).......................................... 382 7.9. Convolution Theorem ( Faltung Theorem)......................................... 387

358

Differential Equations: Theory and Applications

7.1. INTRODUCTION Laplace transform is essentially a mathematical tool which can be used to solve several problems in science and engineering. This transform was first introduced was first introduced by Laplace, a French mathematician, in the year 1790 in his work on probability theorem. This technique became popular when Heaviside applied to the solution of an ordinary differential equation referred hereafter as ODE. representing a problem in electrical engineering. To the basic question as to why one should learn Laplace transform technique when other technique available. The answer is very simple. Transform are used to accomplish the solution of certain problems with less effort and in a simple routine way. To illustrate, consider the problem of finding the value of x from the equation simple routine way. To illustrate, consider the problem of finding the value of x from the equation

x 1 . 85 = 3 (1) It is an extremely tedious task to solve this problem algebraically. However, taking logarithms on both side, we have the transformed equation as 1.85 In x = In 3 (2) In this transformed equation, the algebraic operation and exponentiation have been changed to multiplication which immediately gives

x= In

In 3 1.85

To get the required result, it is enough if we take the antilogarithm on both side of the above equation, which yields

 In 3  x = In −1    1.85  with the help of any ordinary calculator , we can now compute x. Following this simple example, the Laplace transform method reduce the solution of an ODE to the solution of an algebraic equation. In fact, this method has a particular advantage in finding the solution of an ODE with appropriate. ICs, without first finding the general solution and then using ICs for evaluating the arbitrary constants. Also, when the Laplace transform technique is applied to a PDE, it reduce the number of independent variables by one.

Laplace Transform Methods

Definition. Suppose

f (t )

359

is a piecewise continuous function and if it

has an additional property that there exists a real numbers positive number M such that

ã 0 and a finite

Ltt →∞ f ( t ) e −γ t ≤ M for γ > γ 0 And the limit does not exist when be of exponential order

(

f ( t ) = 0 e γ 0t

)

γ < γ 0 , then such a function is said to

γ 0 . Also written as

Variables such as velocity and current are always finite. Which means that f (t ) f ( t ) , | f ( ( t ) e −γ t → 0 is bounded . Thus for any bounded function

for all γ > 0 . The order of such a function in zero. However , variables such as electrical charge and mechanical displacement may increase without limit but of course proportional to t. Such functions are also of exponential order. For illustration, let us consider the following examples:

(i )

Ltt

→∞

te − γ t = 0

n The fact that t is of exponential order zero can be seen as follows:

 tn   nt n − 1  n −γt Lt t e Lt Lt = =   t →∞ t →∞ t →∞  γt γt   e   γ e  ( using L’ Hospital’s rule ) Applying the L’ Hospital rule repeatedly, we get LTt → ∞ t n= e – γ t 0 if γ > a at Thus the function e is of exponential order a.

( t ) ( n >1 ) is not of exponential order, since (iii) exp LT exp ( t ) e = LT exp  t – γ )  = ∞ n

t →∞

n

−γt

t →∞

n− 1

for any finite value of γ .

f (t ) Definition. Let be a continuous and single-valued function of the real variable t definition for all t , 0 < t < ∞, and is of exponential order.

360

Differential Equations: Theory and Applications

Then the Laplace transform of by the integral

L  f ( t )= ; s  F= (s)



∫e

− st

0

For any finite value of γ .

f (t )

is defined as a function

f ( t ) dt

F (s)

denoted

(3)

Over the range of values of s for which the integral exist. Here, s is L  f ( t ) : s  a parameter, real or complex . Obviously,  is a function of s Thus,

f ( t ) into F ( s ) Where L is the operator which transform called laplace −1 transform operator, and L is the inverse Laplace transform operator. The Laplace transform belongs to the family of “ integral transform”. F (s) f (t ) An integral transform pf the function is defined by an integral of the form b

∫k ( s, t ) f ( t ) dt = F ( s ) c

(4)

k ( s, t ) , Where a function of two variables s and t , is called the Kernel of the integral transform The kernel and limits of integration of various integral transform are given in Table 1 ( which is not exhaustive ).

Laplace Transform Methods

361

Table 1. Kernel and Limits for Various Integral Transforms

k ( s, t )

Name of the transform Laplace transform

e − st

Fourier transform

e ist /

0



2

π

Fourier sine transform

2

π

Fourier cosine transform

Mellin transform

sin st

t s −1

b

∝ −∞



0



cos st

tJ n ( st )

Hankel transform

a

0



0



0



The integral transform defined above are applicable , either for semiinfinite or infinite domains. Similarly , finite integral transform can be defined domains. Now, we are in a position to verify the following important result.

f (t )

is piecewise continuous in the range t ≥ 0 and is of F ( s ) of f ( t ) exponential order γ . then the Laplace transform exists for s > γ . all Theorem. If

Proof. From the definition of Laplace transform,

L  f ( t ) ; s =



− st ∫ e f ( t ) dt= 0

T

∫e

− st

f ( t ) dt= I 1 + I 2

0

f (t ) 0 < t < T , I1 Since is piecewise continuous on every finite interval exists, whereas ∞

I 2 ≤ ∫ | e − st f ( t ) | dt T

But

(t)

is a function of exponential order’ therefore,

Differential Equations: Theory and Applications

362

f ( t ) < Meλt for λ real Hence,

e

− st

f ( t ) < Me −( s – λ )t

Thus, ∞

I2 =

e − ( s − γ )t M dt ∫= T

Me – ( s – γ ) T , s >γ s −γ

l2 can be made as small as we like provided T is large L  f ( t ) ; s  l2 s >γ .

In other words,

enough and , therefore

exists. Hence

exist for

7.2. TRANSFORM OF SOME ELEMENTARY FUNCTIONS Following the definition of Laplace transform by the integral (3), we shall compute the Laplace transform of some elementary functions. Example. Find the Laplace transform of

( i )1 . ( ii ) 0 , ( iii ) t , ( iv ) eat , ( v ) e – at Solution. Using the definition of Laplace transform, we have ∞



l ;s ] ∫ ( i ) L [=

e

−st

0

(ii)

(iii)

= L [ 0 ; s )]

e ( 0 ) dt ∫=

0

− st

0

L[ t ; s ]=



∫e

− st

0

L [= t ;s ] (iv)



 e − st  1 1 ) dt  = if s > 0 (=   − s 0 s



∫e 0

– st





∞  e − st   e − st  e − st 1 . dt = ∫ td  t dt = 2 = −    ∫ s  − s   −s  0 0 − s 0



e dt = at

∫e 0



 e − ( s − a )t  1 dt  s>a = =   − ( s − a )  0 s – a

−( s − a ) t

Laplace Transform Methods ∞

363



1 e –( s +a ) d t ∫0= L  e − at ; s  0 s+a (v) = Example. Find the Laplace transform of − st e − at d t ∫e =

(i )

cos at ,

( ii ) sin at.

Solution Following the definition of Laplace transform, we have ∞

cos at dt ( i ) L= [cos at ; s ] ∫ e= − st

0



− st iat iat Re = ∫e e dt Re L e ; s  0

1 s + ia s = Re Re = 2 2 2 s – ia s +a s +a2 s + ia a iat L [ sin at ; s ] Im L = = = 2 2 2  e ; s  Im s +a s + a2 (ii) Example. Find the Laplace transform of

(i )

cosh at ,

( ii ) at .

Solution. Using the result established in , we have

 e at + e − at  1 i L cosh a t ; s L L  e at ; s  + L  e − at ; s  = = () [ ]   2   2

{

=

}

s 1  1 1  + =   2 2  s − a s + a  s − a2 c

 e at − e − at  1 L [sinh at ; s ] L  = ;s  L  e at ; s  − L  e − at ; s  = 2   2 (ii)

{

=

1 1 1 −  2  s –a s + a

}

a  = 2 2  s −a

n Example. Find the Laplace transform of t , where n is a positive integer.

Solution . Using the definition of Laplace transform , we have

364

Differential Equations: Theory and Applications ∞





 e − st   n e − s t  n L  t ; s  = e t dt t d =  t  + ∫0 ∫0  − s  = − s 0 s  n

− st

n

n





n n − 1 − st n −1 − st t e dt ∫0 t e dt = s ∫0

Hence ,

n L  t n ; s  = L  t n − 1 ; s  s Similarly, we can prove the following

n −1 L t n –1 ; s  = L t n − 2 ; s  s L t n − 2 ; s  =

n –2 L t n − 2 ; s  s



2 L t 2 ; s  = L [ t ; s ] s L[ t ; s ]=

1 s2

n n −1 n − 2 2 1 n! . . . 2 …= n +1 s s 2 s s s Therefore, expressed in Gamma functions as L  = t n ; s 

which can be

n +1 L  t n ; s  = n + 1 s

7.3. PROPERTIES OF LAPLACE TRANSFORM We present a few important properties of the Laplace transform in the following theorems which will be enable us to find the Laplace transform of a combination of functions whose transform are known.

c and c

2 are any two constants Theorem. (Linearity property ) . if 1 F ( s ) and F2 ( s ) and if 1 are the Laplace transforms, respectively of f1 ( t ) and f 2 ( t ) , then

Laplace Transform Methods

L  { c1 f 1 ( t ) + c2 f

2

365

c1 L  f1 ( t ) ; s  + c2 L  f 2 ( t ) ; s  = c1 F1 ( s ) + c2 F2 ( s ) ( t )} ; s  =

Proof. Following the definition of Laplace transform , we have ∞

L  { c1 f1 ( t ) + c2 f 2 ( t )} ; s  = ∫e − st { c1 f 1 ( t ) + c2 f 2 ( t )} dt 0



c1 ∫ e

− st

0



f1 ( t ) dt + c2 ∫e − st f 2 ( t ) dt 0

c1 L  f 1 ( t ) ; s  + c2 L  f 2 ( t ) ; s  = c1 F1 ( s ) + c2 F2 ( s ) Theorem. (shifting property ) . if a function is multiplied by e , the ( s − a ) in the transform of the resultant is obtained by replacing s by transform of the original function. That is , if at

L  f ( t ) ; s  = F ( s ) Then

L  e at f ( t ) = ; s  F ( s − a ) Proof . From the definition of Laplace transform,

L  e

at

f (t )= ; s 



∫e

− st

e

at

f (t = ) dt

0



∫e

–(s −a) t

f ( t= ) dt F ( s − a )

0

Similarly,

L  e – at f ( t )= ; s F ( s + a ) 



Theorem . (Multiplication by power of ) . if

L  f ( t ) ; s  = F ( s ) then n n d n   L  t f ( t ); s  = ( −1 ) n F ( s ) = ( −1) n F ( n ) ( s ) ds

= n 1 , 2 , 3 ,… where

Differential Equations: Theory and Applications

366

Proof. Form the definition of Laplace transform

 f ( t ) ; s  F ( s ) L= =



∫e

− st

f ( t ) dt

0

Hence, ∞  d d  − st  F ( s )  =  ∫e f ( t ) dt  dt ds  0 

Interchanging the operations of differentiation and integration for which we assume that the necessary conditions are satisfied and since there are two variables s and t , we use the notation of partial differentiation and obtain ∞



d ∂ F ( s )} = e – st f ( t ) dt = − ∫e − st tf ( t )}dt = − L tf ( t ) ; s  { ∫ dt s ∂ 0 0

{

}

Therefore ,

L  tf ( t ) ; s  = −

ds

F(s)

By repeated application of the above result, it can be shown that n n d n L  t n f ( t ) ; s  = ( −1 ) n F ( s ) = ( −1 ) F n ( s ) ds Theorem. ( Differentiate property) , if

L  f ( t ) ; s  = F ( s ) Then

L  = f n t ; s  s n F ( s ) − s n − 1 f ( 0 ) − s n − 2 f

'

( 0 ) −…− s f n − 2 ( 0 ) − f n − 1 ( 0 )

Proof Form the definition of Laplace transform , we have

L  f ( t ) ; s  = '



e f ( t ) dt ∫= – st

0

 e

− st



f ( t )  + s ∫ e − st f ( t ) dt ∞ 0

0

Similarly , it can be shown that L  f n ( t ) ; s = sL  f ' ( t ) ; s  – f

'

( 0 )= s { sF ( s ) − f ( 0 )} − f ' ( 0 )=

s 2 F ( s ) − sf ( 0 ) − f ' ( 0 )

Laplace Transform Methods

L  f

m

( t ) ; s  =

s 3 F ( s ) − s 2 f ( 0 ) − sf ' ( 0 ) – f

367

(0)

''

Thus, in general,

L  = f n t ; s  s n F ( s ) − s n − 1 ( 0 ) − s n − 2 f ( 0 ) −…− sf n – 2 ( 0 ) − f n − 1 ( 0 ) This property is very useful for solving differential equations. Example. Find the Laplace transform of

( i ) e at cos bt , ( ii ) e at sin bt , ( iii ) e at cosh bt , ( iv ) e at t n

, and ( v ) cos at cosh bt.

Solution. Using the shifting property

 e at sin bt ; s  ( i ) L=

s = 2 s + b 2 s → ( s –a )

 e at sin bt ; s  ( ii ) L=

b = 2 s −b 2 s → ( s –a )

at L e= cosh bt ; s 

( iii )

 e at t n ; s  ( iv ) L=

s−a ( s − a ) 2 +b 2

s –a

( s − a)



+ b2

s−a

s = 2 s − b2 s → ( s – a )

n! = s n + 1 s →( s − a )

2

(s − a)

2

−b 2

n!

( s − a)

n +1

 ebt − e − bt   ;s  2   

( v ) L [ cos at cosh bt ; s ] = L  cos at  

1 L  ebt cos at ; s  − L  e – bt cos at ; s  2

{

}

 1  s−a s +b −   2  ( s – b )2 + a 2 ( s + b ) 2 + a 2    Example. Find the Laplace transform of the following :

(i ) t 2

e at , ( ii ) t sin at , ( iii ) t 2 cos at , ( iv ) t n e − at

Differential Equations: Theory and Applications

368

Solution. Using the result established in the Theorem , we have

( i ) L  t 2

2 d 2  1  d  −1  2 2 d  = e at ; s  = ( −1 ) 2 L  e at ; s  = = 2 3 2    ds d s  s − a  ds ( s − a )   ( s –a )

Alternatively,

2! = s 3 s → ( s –a )

L = e at t 2 ; s 

( ii ) L [ t

2

( s−a )

3

(using the shifting property)

d  a  2as 1 d −  2 2 = sin at ; s ] = ( −1 ) L [ sin at ; s ] = ds ds  s + a  s 2 + a 2 2

(

2

)

2

d  s  2 d L t 2 cos at ; s  = ( −1 ) 2 L [ cos at ; s ] =  2  2 ds ds  s + a 2  (iii)

  d  a 2 − s 2  2 s 3 − 6 sa 2 = 3 ds  s 2 + a 2 2  s2 + a2  

(

)

 e − at t n ; s  ( iv ) L =

(

)

n! = s n +1 s → ( s + a )

n!

( s + a)

property )

n +1

(

using

Example. Verify the initial value theorem for the function

f (t ) = 1 + e − t ( sin t + cos t ) Solution . Given

f ( t ) = 1 + e − t (sin t + cos t ) , we have

(

)

F (s) = L  f ( t ) ; s  = L [ 1 ; s ] + L  e − t sin t + e − t cost ; s  1  1 s  =+  2 + 1 + 2  s s s +1  s → ( s + 1 ) =

1  2+s  +  s  s 2 + 2 s +2 

the

shifting

Laplace Transform Methods

369

Hence,

2s + s 2 sF ( s ) = 1 + 2 s + 2s + 2 Therefore,

Lts → ∞

2 s +1 sF ( s ) = Lt s → ∞ 1 + = 1+1= 2 2 2 1+ + 2 s s

But

f ( 0 ) =1 +1 = 2 .

Thus,

Lt s → ∞ sF ( s ) = f ( 0 ) Hence the result Theorem . ( Division by t ) . if

L  f ( t ) ; s = F ( s )  Then

 f (t )  ∞ L ; s  = ∫ F ( s ) ds  t  s Proof. From the definition of Laplace transform

L  f ( t )= ; s  F= (s)



∫e

− st

f ( t ) dt

0

Integrating the above with respect to s between the limits s to ∞, we get ∞  ∞ − st  F s ds e f ( t ) dt  ds = ( ) = ∫s ∫ s 

 ∞ − st  f t ( ) ∫0  ∫0e ds  dt



(by changing the order of integration ) ∞

∫ 0



 e − st  f= ( t )   dt  −t  s Hence the result .



f (t )

e dt ∫= t 0

– st

 f (t ) L  ;s  t 

Differential Equations: Theory and Applications

370

f (t ) Note: In applying this rule , one should be careful . Since

t

may

f (t )

have an infinite discontinuity at t = 0. it may not be integrable. If t is not integrable, then its Laplace transform does not exist. For example , at t = 0 , the function sin t t doses not have an infinite discontinuity , while t cos has an t the function infinite discontinuity .

7.4. TRANSFORM OF A PERIODIC FUNCTION f ( t ) is T , if ( t + T ) = f (t ) A function called periodic with period for all values of t and T > 0. For example, the trigonometric functions sin t and cos t are periodic functions of periods 2π . Periodic functions occur very often in a variety of engineering problems. Theorem.

L  f ( t ) ; s  =

if f ( t ) T

∫e

− st

is a periodic function with periods T , then

(

f ( t ) dt / 1 − e – st

0

)

Proof. From the definition of Laplace transform , we have

L  f ( t ) ; s  =



T

e f ( t ) dt ∫ e ∫= − st

0

− st

0



f ( t ) d t + ∫ e − st f ( t ) dt T

If we substitute t= u + T in the second integral on the right-hand side and write dt = du. we obtain T

L  f ( t ) ; s  =∫ e

– st

0

T

=

∫e

– st

0

T

=

∫e 0

f ( t ) dt + e



f ( t ) d t + ∫ e − s( u + T ) f ( u + T ) du 0

− st



∫e

– st

f ( u ) du

0

– st

f ( t ) dt + e – sT L  f ( t ) ; s 

Laplace Transform Methods

371

Rearranging , we get T

( 1− e ) L  f ( t ) ; s  = ∫ e f ( t ) dt − st

sT

0

Thus, T

L  f ( t ) ; s  = ∫ e – st f ( t ) dt 0

/

(1 – e ) −st

Hence the result. Example. Obtain the Laplace transform of the periodic saw-tooth wave function given by

t T of period T , 0 < t < T Solution . The graph of the periodic saw-tooth function is described in f (t ) . since is periodic with period T , we have f (t ) =

1 t 1 1 e − st dt e − st tdt = = − st ∫ − sT ∫ 1− e 0 T T 1− e T 1 − e − st 0 T

L f (t ) ; s ] =

(

1 T 1 − e − st

(

=

T

)

  te − st T 1 T  − st   + ∫ e dt    − s 0 s 0 

 Te − st 1 − st  1 e 1 − −   2 s T 1 − e − st  − s 

(

(

)

Therefore,

L  f ( t ) ;= s 

1 e − st − s 2T s 1 − e − st

(

)

)

Example. Find the Laplace transform of

1 , 0 ≤ t < 2 f (t ) =  −1, 2 ≤ t < 4 f ( t + 4) = f (t )

)

(

 e − st  td ∫0  −s 

T

)

372

Differential Equations: Theory and Applications

Solution. In this problem, therefore, have \

f (t )

is a periodic function of period 4; we,

4

1 e − st f ( t ) dt − 4s ∫ 1− e 0

L  f ( t ) ; s  =

2 4  1  − st − st 1 1 e dt e dt = + − ( ) ( )   ∫2 1 − e −4 s  ∫0 

=

1  −2 e − 2 s e −4 s 1  + +   1− e − 4 s  s s s

7.5. TRANSFORM OF ERROR FUNCTION The error function denoted by erf

erf ( t ) =

2

π

t

∫e

− u2

(t )

is defined as

du

0

This function occurs in many branches of science and engineering ; for example , in probability theory, the theory of heat condition, and so on . In terms of the power series , we have

erf ( t ) =

2

π

( −1 ) n t 2 n + 1 ∫0 n∑= 0 n !( 2n +1) t



We can easily verify that these series converge everywhere and, therefore erf ( t ) is , an entire function .From the definition , it can be verified at once that

erf ( 0 ) = 0

erf (= ∞)

2

π



∫e 0

−u2

= du

1/ 2 = 1

π

Laplace Transform Methods

373

The graph of error function is shown in In solving heat conduction equation, it has been found useful to introduce the complementary error Function defined as ∞ t  2  −u2 –u 2 du   ∫ e du − ∫ e π 0 0 



2 −u 2 = ∫ e du

erfc ( t ) =

π

0

Therefore,

erfc ( t ) = 1 – erf ( t ) Now we shall the Laplace transform of erf (t): From the definition Of Laplace Transform ∞

L  erf ( t ) ; s  = ∫ e

2

− st

t

∫e

π

0

−u2

du dt

0

Changing the order on integration , we obtain

=

=

2 s π 2 s π



∫e



2

L  erf ( t ) ; s  =

π

(

− u 2 + su

)

−u ∫e

2

0



∫e

– st

dt du

0

du

0

e

s2 8 −u + s    2   4

∫e

2

du

0

s x= u + , 2 we get Setting L  erf ( t ) ; s  =

2 s π

e

s2 ∞ 4

∫e

− x2

dx

s 2

1 2 s = e s /4 erfc   s 2 Example. Find the Laplace transform of erf

(t ). 1/2

Solution. From the definition of Laplace transform , we have

Differential Equations: Theory and Applications

374

1   ∞ − st 2 2 L erf (t ; s  = ∫ e π   0

r1/2

∫e

−u2

du dt

0

Changing the order of integration, we get

( ) ; s  =

L erf t =

s

2



e π ∫

− st

0

2

π

r1/2

∫e

− n2

du dt

0



−u − st ∫e du ∫ e dt

s π

2

u2

0



e π ∫

− u 2 − su 2

du

0



2 s



2 2

1/2

e π ∫

−(1+ s )u 2

du

0

Setting

u2 ( 1 + s )=

x 2 or 1 += su x,

we have

du = dx / 1+ s Then ∞ ∞   1   2 1  2 − x2 − x2 2 L erf  t  ; s  = e dx e dx =   ∫ s 1 + s  π ∫0      s π 1 + S 0

or

1 L erf t1/2 ; s  = s 1+ s

( )

7.6. TRANSFORM OF BESSEL’S FUNCTION Bessel function arise in several problem involving circular or cylindrical geometry, it is therefore useful to find the Laplace transform of Bessel functions if the first kind . Example. Find the Laplace transform of

( i ) j0 ( t ) , ( ii ) tj0 ( t ) , ( iii ) e−at j0 ( t ) .

Laplace Transform Methods

Solution.

( i ) From the definition of the Bessel function, we have

( −1)



r

t   r =0 r ! n + r + 1  2 

jn ( t ) = ∑

n+2r

For n = 0, we have 2r −1)  t  ( j0 ( t ) ∑ = ∑    2  r! r +1  2  = r 0= r 0 ( r !)  2  2r −1)  t  ( = r



= 1−



r

t2 t4 t6 + − +… 2 2 2 2 ×4 2 2 2 ×4 2 ×6 2

Thus,

L  j0 ( t= ) ; s  L [1; s ] − =

1 1 L t 2 ; s  + 2 2 L t 4 ; s  −… 2 2 2 ×4

1 1 2! 1 4! 1 6! − 2 3 + 2 2 5 − 2 2 2 7 +… s 2 s 2 ×4 s 2 ×4 ×6 s 2 3  1  1  1  1× 3  1  1×3 × 5  1  1 − + −   2  2  2  +… s  2  s  2 × 4  s  2 ×4 × 6  s  

1 1 =1 + 2  s s 

−1/2

1 = 1 + s2

Hence,

L  j0 ( t ) ; s  =

1 1 + s2

(ii) From the properties of Laplace transform, we have

L t n f ( t ) ; s  =

( −1)

n

dn F (s) ds n

Therefore,

d d  1  L tj0 ( t ) ; s  = −  ( −1)  L  j0 ( t ) ; s   =  ds ds  1 + s 2 

375

376

Differential Equations: Theory and Applications

Thus,

L tj0 ( t ) ; s  =

s

(

1 + s2

)

3 2

(iii) From the shifting property of Laplace transform , we have

L  e − at f ( t ) = ; s  F ( s + a ) Therefore,

1 = 1 + s 2 s →( s + a )

− at L= e j0 ( t ) ; s 

1 1+ (s + a)

2

7.7. TRANSFORM OF DIRAC DELTA FUNCTION The concept of impulse function or Dirac delta function has been introduced in itself. In certain applications involving a sudden excitation of a system or a large voltage over a short interval of time, the Laplace transform of Dirac Delta function is useful. Form the property of Dirac Delta function, we have ∞

f (a) ∫δ ( t − a ) f ( t ) dt = 0

In particular , if

L δ ( t − a ) ; s  =



∫e

f ( t ) = e − st , − st

then

δ ( t − a ) dt = e − as , a > 0

0

7.7.1. Inverse Transform So far we have discussed various properties of the Laplace transform and studies the Laplace transform of some simple functions. However, if the Laplace transform technique is to be useful in applications, we have to consider the reverse problem too, i.e., we have to find the original function f (t ) F ( s ). x when we know its Laplace transform Thus, if

L  f ( t ) ; s  = F ( s ) Then

Laplace Transform Methods

377

f ( t ) = L−1  F ( s ) ; t  In other words , the inverse Laplace transform of a given function F (s) f (t ) F ( s ). is that function whose Laplace transform is It can be −1 f (t ) established that is unique . Here, L is known as inverse Laplace transform operator. Form the elementary definition (24) and form the results obtained thus far in finding the Laplace transform of some elementary functions , we can immediately generate the following table of transforms: Table of Laplace Transform

f (t )

L  f ( t ) ; s 

0

0

1

1 s

e at

(s − a)

e

−at

t

tn

F= F (s) L−1  F ( s ) ; t  f ( t ) (s)

0 1 s

0 1

1 (s − a)

1

1

e at

1

(s + a)

e −at

(s + a)

1/ s 2 n! s n +1

1/ s 2

t

1/ s n +1

t n / n!

sin at

a s2 + a2

a s2 + a2

sin at

cos at

s s2 + a2

s s2 + a2

sinh at

sinh at

a s2 − a2

a 2 s − a2

sinh at

cosh at

s s − a2

s s − a2

t sin at t cos at

2

(s

(s

2as 2

+ a2

2

s a

2

2

+a

2

cosh at

2

)

)

2

(s

2

(s

2as 2

+ a2

s2 2

)

2

t sin at

2

+ a2

)

2

t cos at

Differential Equations: Theory and Applications

378

L  f ( t ) ; s  In most of the problems we have considered earlier,  is a simple rational function. The linearity property holds true even in the case F (s) F (s) of inverse transform. That is, if 1 and 2 Are the Laplace transform of two constants, then

f1 ( t ) and f 2 ( t )

, and if

c1 and c2 are any

 c1 L−1  F1 ( s ) ; t  ± c2 L−1  F2 ( s ) : t  L−1 { c1 F1 ( s ) ± c2 F2 ( s )} ; t =  L  f ( t ) ; s  By expressing  as partial fractions, we should be able to recognize them as the transform of some known function , with the helps of which we can write down the inverse transform . Similarly shifting property is also useful in constructing the inverse transform of some functions, which is stated in the following theorem: Theorem. If

L  f ( t ) ; s  = F ( s ) ,

then

L−1  F ( s + a ) ; t  = e − at L−1  F ( s ) ; t  Proof. Since

L  f ( t ) ; s  = F ( s ) ,

we have

L−1  F ( s ) ; t  = f ( t ) Recalling the shifting property of Laplace transform, we find that

L e − at ( t ) = ; s  F ( s + a ) L−1  F ( s + a ) ; t  = e − at f ( t ) Thus,

L−1  F ( s + a ) ; t  = e at L−1  F ( s ) ; t  Example. Obtain the inverse Laplace transform of

4 s 2 − 3s + 5 ( s + 1) s 2 − 2s + 2

(

)

Laplace Transform Methods

379

Solution Using partial fraction expansion , we can write

4 s 2 − 3s + 5 A Bs + C = + 2 2 ( s + 1) s − 2s + 2 s + 1 s − 2s + 2

(

)

Therefore,

4 s 2 − 3s + 5 A Bs + C = + 2 ( s + 1) s 2 − 2s + 2 s + 1 s − 2s + 2

(

)

Therefore,

(

)

4 s 2 − 3s += 5 A s 2 − 2 s +2 + ( Bs + C )( s + 1) Let s = −1 ; then we have

A=

12 . 5 Equating the coefficient of s on both sides,

9 B+C = 5 5 Equating the coefficient of constant on both sides, we get 2 A + C = 1 8 C= , B= . 5 and hence 5 The given expression can now be which gives written as 4 s 2 − 3s + 5 12 1 8 s 1 1 = + + 2 2 2 ( s + 1) s − 2s +2 5 s + 1 5 ( s − 1) +1 5 ( s − 1)2 + 12

(

)

Thus, we find that   12  1 −1   4 s 2 − 3s + 5 1  1  8 −1  ( s − 1) + 1 L−1  L−1  t + L t ;t = ;t + L  ; ;    2 2 2  s + 1  5  ( s − 1) +12  5  ( s − 1) + 12   ( s + 1) s − 2 s + 2  5

(

)

1  1  + et L−1  2 ; t  5  s + 1  ( by using the shifting property) 12 − t 8 t 1 = e + e ( cos t sin t ) + et sin t 5 5 5 12 8 9 = e − t + et cos t + et sin t 5 5 5

Differential Equations: Theory and Applications

380

Theorem. (change of scale property). If

L−1  F ( s ) , t  = f ( t ) then

L−1  F (α s ) ; t  =

1

t  f  α α 

Proof. Form the definition of Laplace transform, we have ∞

F ( s ) = ∫e − st f ( t ) dt 0

Therefore, ∞

f (α s ) = ∫ e −α st f ( t ) dt 0

Let α t = x, so that

= F (α s )

dt =

dx .Then a we get



1 − sx  x  1  x  1   t   = e f   dx = L f  ; s L f  ; ∫ a0 a   a   a   a   a

Thus,

L−1  F ( as ) ; t  =

1 t f  a a

Example. Find the inverse Laplace transform of

( i ) In

s2 + 1 , s ( s + 2)

( ii ) cot −1 

s  k

Solution. (i) From Theorem , we have

L t n f ( t ) ; s  =

( −1)

n

F n (s)

In particular n = 1 gives

L tf ( t ) ; s  = − F ′ ( s )

Laplace Transform Methods

i.e.,

d F ( s ) = − L  tf ( t ) ; s  ds Let

L  f ( t ) ; s  = In

s 2 +1 s ( s +1)

Then,

d d  = F(s) In s 2 +1 − I n s − In ( s +1 )  ds ds 

(

=

)

2s 1 1 − − = L tf ( t ) ; s  s +1 s s −1 2

We get

1 1 2s L  ft ( t ) ; s  =+ − 2 s s +1 s +1 Hence,

1   1  1   s  tf ( t ) = L−1  ; t  + L−1  ; t  ; t  − 2 L−1  2 ;t  s   s +1  s +1   s +1 

=+ 1 e − t − 2 cos t Therefore,

 s 2 +1  1 + e − t − 2 cost −1 f ( t ) L= In ;t =  t  s ( s +1)  −1  s  L = f ( t ) ; s  cot =   F (s) ( ii ) Let k Then ,

d d  k −1 s  F(s)=  cot = − 2 2 ds ds  k k +s Now we obtain d k = F(s) = L  tf ( t ) ; s  2 ds k + s2

381

Differential Equations: Theory and Applications

382

Therefore,

k  −1  = tf ( t ) L= 2  k + s 2 ; t  sin kt   Hence,

 −1  −1  s  L= cos  k  ; t     

f (t )

sin kt t

7.8. CONVOLUTION THEOREM (FALTUNG THEOREM) We often across function which are not the transform of some known function, but then, they can possibly be expressed as a product of two functions, each of which is the transform of a known function. Thus we may F ( s )G ( s ). be able to write the given function as where of a known to be f (t ) g (t ) , transform of the functions and respectively.

F ( s ) and G ( s ) Theorem. If are the Laplace transform of f ( t ) and g ( t ) F (s)G (s) respectively, then is the Laplace transform of t

∫ f ( t – u ) g ( u ) du 0

i.e.,

t

L  F ( s ) G ( s )  = ∫ f ( t – u ) g ( u ) du −1

0

This integral is called the convolution of f and g and is denoted by the

* symbol f g .

Proof. Form the definition of Laplace transform ,we have

∞  ∞  F ( s ) G ( s ) =  ∫ e − sv f ( v ) dv   ∫ e − su g ( u ) du  0  0  ∞∞

= ∫∫ e − s( v + u ) f ( v ) g ( u ) dv du 0 0

Laplace Transform Methods

383

 ∞ ( v + u )  = ∫ g ( u )  ∫e f ( v ) dv  du  0  0 ∞

t in the inner integral. Then, Let u + v =

F ( s )G ( s )

 ∞ – st  g u e f t – u dt ( ) ( )   du ∫0 ∫  0  ∞

Change the order of integration Then, we get

 t − st  = F ( s ) G ( s ) ∫  ∫ e f ( t − u ) g ( u ) du  dt  0 0 ∞

∞ t   = ∫ e − st  ∫ f ( t – u ) g ( u ) du  dt 0  0 

t  = L  ∫ f ( t – u ) g ( u ) du ; t  0  Hence the result. Definition. We define the Laplace convolution of integral

f ( t ) and g ( t )

by the

t

*

= f g

∫ f ( t − u ) g ( u ) du 0

It can be verified that f and g can be interchanged in the convolution, i.e., f and g are commutative Let t − u = v so that – du = dv . Then , Therefore,

f *g = g* f Example. Prove that t

sin t ∫ j ( t ) j ( t − u ) du = 0

0

0

Solution. Form table ,we not that

384

Differential Equations: Theory and Applications

 1  L−1  2 ; t  = sin t  s +1  We shall also write

1 = s +1 2

1

1

s 2 +1 s 2 +1

Now taking

= F(s)

1 = ,G ( s ) s 2 +1

1 2

s +1

Their inverse transform give

 1  L− 1  ; t  = j0 ( t ) 2 = f= (t ) g(t )  s +1  Hence, using the convolution theorem, we have

 1  ;t  L  2 =  s +1  −1

t

) du ∫ j ( t ) j ( t − u= 0

0

sin t

0

7.8.1. Summary and Discussion Laplace transform is essentially a mathematical tool which can be used to solve several problems in science and engineering . This transform was first introduced was first introduced by Laplace, a French mathematician, in the year 1790 in his work on probability theorem. This technique became popular when Heaviside applied to the solution of an ordinary differential equation referred hereafter as ODE. representing a problem in electrical engineering . To the basic question as to why one should learn Laplace transform technique when other technique available. The answer is very simple. Transform are used to accomplish the solution of certain problems with less effort and in a simple routine way.

f (t ) Definition. Let be a continuous and single-valued function of the real variable t definition for all t , 0 < t < ∞, and is of exponential order. Then the Laplace transform of by the integral

f (t )

is defined as a function

F (s)

denoted

Laplace Transform Methods

L  f ( t )= ; s  F= (s)



∫e

− st

f ( t ) dt

0

For any finite value of γ .

385

(3)

Over the range of values of s for which the integral exist. Here, s is L  f ( t ) : s  a parameter, real or complex . Obviously,  is a function of s Thus,

f ( t ) into F ( s ) Where L is the operator which transform called laplace −1 L transform operator, and is the inverse Laplace transform operator. The Laplace transform belongs to the family of “ integral transform”. F (s) f (t ) An integral transform pf the function is defined by an integral of the form b

∫k ( s, t ) f ( t ) dt = F ( s ) c

(4)

k ( s, t ) , where a function of two variables s and t , is called the Kernel of the integral transform The kernel and limits of integration of various integral transform are given in Table 1 ( which is not exhaustive ) . f (t ) Theorem. If is piecewise continuous in the range t ≥ 0 and is of F ( s ) of f ( t ) exponential order γ . then the Laplace transform exists for all s > γ . ∞



l ;s ] ∫ ( i ) L [=

e

−st

0

(ii)

(iii)

= L [ 0 ; s )]



 e − st  1 1 ) dt  = if s > 0 (=   − s 0 s

e ( 0 ) dt ∫= − st

0

0

L[ t ; s ]=



∫e 0

− st





∞  e − st   e − st  e − st 1 . dt = ∫ td  t dt = 2 = −    ∫ s  − s   −s  0 0 − s 0

Differential Equations: Theory and Applications

386

L [= t ;s ]



∫e

– st

e dt =

0

(iv)

L  e

− at

; s 

∫e



 e − ( s − a )t  1 dt  s>a = =   − ( s − a )  0 s – a

−( s − a ) t

0



(v)

∞ at



–( s +a )

e d t ∫= e dt ∫e =

=

− st

− at

0



0

cos at dt ( i ) L= [cos at ; s ] ∫ e= − st

0

1 s+a



− st iat iat Re = ∫e e dt Re L e ; s  0

1 s + ia s = Re Re = 2 2 2 s – ia s +a s +a2 s + ia a iat L [ sin at ; s ] Im L = = = 2 2 2  e ; s  Im s +a s + a2 (ii)  e at + e − at  1 i L cosh a t ; s L L  e at ; s  + L  e − at ; s  = = () [ ]   2   2

{

=

}

s 1  1 1  + =   2 2  s − a s + a  s − a2 c

 e at − e − at  1 L [sinh at ; s ] L  = ;s  L  e at ; s  − L  e − at ; s  = 2   2 (ii)

{

=

1 1 1 −  2  s –a s + a

a  = 2 2  s −a

}

Laplace Transform Methods

Table Of Laplace Transform f (t )

0

L  f ( t ) ; s 

F= F (s) L−1  F ( s ) ; t  f ( t ) (s)

0 1 s

1 s

e at

1 (s − a)

1 (s − a)

e −at

1 (s + a)

1 (s + a)

e −at

1/ s 2

t

1

t

tn sin at cos at

sinh at

cosh at

t sin at t cos at

7

0

1/ s 2 n! s n +1 a s + a2 s 2 s + a2 a 2 s − a2 s 2 s − a2 2as

(s

2

+ a2

2

s a

2

2

+a

2

)

)

2

1

e at

1/ s n +1 a 2 s + a2 s 2 s + a2 a 2 s − a2 s 2 s − a2 2as

2

(s

0

2

(s

(s

t n / n! sin at sinh at sinh at cosh at

2

+ a2

s

2 2

2

+ a2

)

)

2

2

t sin at t cos at

387

CHAPTER

8

GREEN’S FUNCTION

CONTENTS 8.1. Introduction..................................................................................... 390 8.2. The Eigenfunction Method............................................................... 403 8.3. Summary and Discussion................................................................. 406 References.............................................................................................. 408

390

Differential Equations: Theory and Applications

8.1. INTRODUCTION Consider the differential equation

Lu ( x ) = f ( x ) L−1 (1) Where L is an ordinary linear differential operator, f ( x ) is a known function, while is a unknown function. To solve the above differential equation, one method is to find the operator in the form of an integral operator with a kernel G ( x,ξ ) such that

u ( x ) = L−1 f ( x ) = ∫ G ( x, ξ ) f (ξ ) d ξ (2) The kernel of this integral operator is called Green’s function for the differential operator. Thus the solution of the non-homogenous differential equation can be written down, once the Green’s function for the problem is known. Applying the differential operator to both sides of , we get f ( x ) = LL−1 f ( x ) = ∫ LG ( x ,ξ ) f ( ξ ) d ξ (3) This equation is satisfied if we choose G ( x, ξ ) such that

LG ( x = , ξ ) δ ( x − ξ ) (4) Where δ ( x − ξ ) is a Dirac δ - function. The solution of Eq. is called a singularity solution of Eq. In previous section , we have already introduced the concept of Dirac δ − function and studied its various properties. Now, they become handy to understand more about Green’s function Definition . Let us consider an auxiliary function φ ( x ) of a real variable x which possess derivative of all orders and vanishes outside a finite interval. Such function are called test functions. Now we shall introduced the concept of the derivative of a δ − function in terms of the derivative of a test function . we say that δ ′ ( x ) is the derivative of δ ( x ) if

Green’s Function ∞

∫ δ ( x ) φ ( x ) dx = −φ ′ ( 0 )

391

'

−∞

(5)

For every test function φ ( x ) . similarly , we define φ ′′ ( x ) by ∞

∫ δ ( x )φ ( x ) dx = −φ ′′ ( 0 ) (6) ''

−∞

With this definition of a derivative, we can show that the δ − function is the derivative of a Heavside unit step function H ( x ) defined by

 1 for x ≥ 0 H ( x) =   0 for x < 0 (7) To see this result, we consider ∞





−∞

−∞

0

− ∫ H ( x )φ ′ ( x ) dx = − ∫ φ ′ ( x ) dx = φ(0) ∫ H ′ ( x )φ ( x ) dx =

By comparing the above result with property III of i.e., with ∞

∫ δ ( x ) φ ( x ) dx = φ ( 0 )

−∞

we obtain

H ′( x) = δ ( x)

(8)

Similarly, the notion of δ − function and its derivative enables us to give a meaning to the derivative of a function that has a jump discontinuity at x = ξ of magnitude unity. Let

 1, x ≥ ξ H (x −ξ ) =  0, x < ξ Then, for any test function ∞



−∞

H ′ ( x − ξ )φ ( x ) dx

φ ( x).

we have



=

∫ δ ( x –ξ ) φ ( x ) dx =φ ( ξ )

−∞



(9)

Differential Equations: Theory and Applications

392

It can also be noted that

d  ( x –ξ ) H ( x − ξ )  = ( x − ξ ) H ' ( x −ξ ) + H ( x − ξ ) dx Integrating both sides with respect to x between the limits − ∞ to ∞ ,we get ∞



−∞

−∞

( x − ξ ) H= ( x –ξ ) ∫ ( x –ξ ) H ' ( x –ξ ) dx + ∫ H ( x –ξ ) dx ∞

=

∫ H ( x –ξ ) dx

−∞

(10)

We now consider an example to illustrate the inversion of a differential operator by considering the BVP:

d2 u = f ( x ) , u= ( 0 ) u= (1) 0 dx 2 In this Eq. (4) becomes

d 2G = δ ( x –ξ ) d x2 (11) Noting that the δ − function is the derivative of the Heaviside unit step function and integrating d G= ( x , ξ ) H ( x –ξ ) + C1 ( ξ ) dx 

= LG

C (ξ ) where 1 is an arbitrary function. Integrating the above result once with respect to x. we get G ( x ,ξ ) = ∫ H ( x –ξ ) dx + C1 ( ξ ) x + C 2 (ξ ) = ( x − ξ ) H ( x − ξ ) + C1 ( ξ ) x + C 2 ( ξ ) C ( ξ ) and C2 ( ξ ) where 1 can be determined form the boundary conditions. Thus , from Eq. (2) we have = u ( x)

x



0

−∞



∫ ( x –ξ ) H ( x –ξ ) f ( ξ ) d ξ + x ∫ C ( ξ ) f ( ξ ) d ξ + ∫ C ( ξ ) f ( ξ ) d ξ 1

Now. Using the boundary condition

2

−∞

: u ( 0 ) = 0,

we get

Green’s Function

393



0 = 0 + 0 + ∫ C2 ( ξ ) f ( ξ ) d ξ −∞

C2 (ξ ) = 0.

Implying that u (1) = 0. we have

0

1



0

−∞

Using the second boundary condition :

∫ ( 1 –ξ ) f ( ξ ) dξ + ∫ C1 ( ξ ) f ( ξ ) dξ Implying that

C1 (ξ ) = − (1 − ξ ) , 0 ≤ ξ ≤ 1,

of ξ . Hence

u( x)

x

∫ ( x –ξ ) H ( x –ξ ) f ( ξ ) d ξ 0

and zero for all other values

1

– x ∫ ( 1 − ξ ) f (ξ ) d ξ 0

(12)

Comparing this result with Eq. (2) , we have the Kernel of the integral operator, which is known as Green’s function or source function given by

G ( x, ξ ) = ( x − ξ ) H ( x − ξ ) − x (1 − ξ ) , 0 ≤ ξ ≤ 1 Satisfying the boundary condition:



(13)

G= ( 0, ξ ) G= (1, ξ ) 0

This concept can be extended to partial differential equation also. To make the ideas clearer, let us consider

L u ( X )  = f ( X )

(14)

where L is some linear v partial differential operator in three independent variables x , y , z and X is a vector in three-dimensional space. Then the G( X; X ) Green’ which satisfies the equation.

′ )  δ ( X − X ′ ) L G ( X ; X=

(15)

On expansion , this equation becomes

L G ( x, y, z; x′ , y′, z ′ )  = δ ( x − x′ ) δ ( y − y ′ ) δ ( z − z ′ ) Here the expansion

δ ( X − X ')

(16)

is the generiliazation of the concept G( X, X ) of delta function in three-dimesnional space IR and represents

394

Differential Equations: Theory and Applications

the effect at that point X due to a source function or delta function input applied at ′ .Equation has the following interpretation in heat conduction or G( X; X ) electrostatic: can be viewed as the temperature ( the electrostatic potential) at any point X in IR due to a unit source ( due to a unit charge ) f (X ) located at X , we get Eq. (15) on both sides by and integrating over ′ V X . the volume with respect to we get

.  L  ∫ G ( X ; X ) f ( X ) dV= X  Vx 

.

= ∫ f ( X ) δ ( X − X ′) dV ' x

f (X )

VX

Comparing with Eq. (14) ,we arrive at .

u ( X ) = ∫ G ( X ; X ) f ( X ) dVx VX

(17)

Which is the solution of Eq. (14) . This leads to the simple definition that u ( x, y, z ; x′, y′, z ′ ) a function is a fundamental solution of the equation , for 2 0 if is a solution of the non-homogenous equation example, ∇ u =

∇ 2 u = δ ( x, y , z; x′ , y′, z ′ ) This idea can be easily extended to higher dimension . Thus the Green’s function technique can be applied, In principle , to find the solution of any linear non-homogenous partial differential equation . Although a neat formula (17) has been for solution of non-homogenous PDE, in practice , it is not any easy task to construct the Green’s function . We shall now present a few singularity solutions. Also called fundamental solutions to the wellknown operators. These will guide us to construct Green’s function for the solutions of partial differential equations which occurs most frequently in mathematical physics. To start with, the fundamental solution for a three-dimensional potential problem satisfies

∇ 2u = δ (X )

u (18)

or

div grad u = δ ( X )

Green’s Function

395

where u can be interpreted, for example, as the electrostatic potential. We r=X ; seek a solution which depends only on the source distance thus, for r > 0, u ( r ) satisfies 1 ∂  2 ∂u  = ∇2 u = r  0 2 r ∂r  ∂r  On integration, we get

A +B r

u=

Using the fact that the potential vanishes at infinity ,we have

u=

A r Integrating Eq. (18) the potential vanishes at infinity ,we have

.

∫ [ div grad u ] dV = 1



Using the divergence theorem, we obtain .

∂u

∫ σ ∂r ε

dS = 1 r =ε

Where

σ ò is the surface of Rε .Hence ,

Therefore,

A= −

1 4π

2 0 is Thus, the singularity solution or the fundamental solution of ∇ u =

u= −

1

4π r (19) The two-dimensional case of Eq. (18) is

Differential Equations: Theory and Applications

396

1 ∂  ∂u  (r  =  0, r > 0 r ∂r  ∂r  On integration, we get = u A In r + B Integrating Eq. (18) over a disc

σ ò , we get

Rò of small radius ò, whose surface is

.

∫ div grad u = 1



Hence, .

∂u

∫ σ ∂r ε

A A dS = ∫ dS = ×2π ò = 1 rò ò r =ò

1 . 2π The constant B remains arbitrary and can be set Therefore, equal to zero for convenience Thus, the fundamental solution is 1 u (r ) = In r 2π (20) A=

if r ( x, y , z ) and r ′ ( x′, y′, z ′ ) If are two distinct points in threedimensional space IR, then the singularity solution of Laplace equation is

u=

1 4π r − r ′

(21)

Similarly, the singularity solution for the diffusion equation

ut − k ∇ 2 u =0 In three-space variables assumes the form

 − r − r′ 2  exp   3/2  4k ( t − τ )    8 π k ( t − τ )  (22) For Helmholtz equation in three space variables, viz. 1

Green’s Function

397

∇2 u + k 2 u = 0 The singularity solution is ik r − r

e r − r′

(23)

Before we attempt to solve Eq.(17) we shall examine the form of three-

dimensional δ − function in genral curvilinear coordinates as a preparation to study the solution of partial differential equations in polar coordinates, sphereical polar coordinates etc. suppose we are looking for a transformation form Cartesian coordinates , x, y , to curvilinear coordinates ξ ,η through the relations x=

f ( ξ ,η ) , y = g (ξ ,η )

(24)

where f and g are single-valued, continuously differentiable functions of

= ξ β= β2 1 and η

their arguments. Suppose that under this transformation, correspond to

x = α1 and y = α 2 respectively Also,

 dx   fξ fη   d ξ  ∂ ( f , g )  dξ  = = j     g g =   dy   ξ η   dη  ∂ (ξ ,η )  dη  If we transform the coordinates following Eq. (24) , the relation

∫ ∫ φ ( x, y ) δ ( x − α1 ) δ ( y − α 2 ) dx dy =φ (α1 ,α 2 ) Becomes

∫ ∫ φ ( f , g ) δ  ( ξ ,η ) − a1  δ  g (ξ ,η ) – a 2  J= dη = φ ( α1 , α 2 ) (25) where j is the Jacobian of the transformation Eq. (25) states that the Dirac

δ -function δ  f ( ξ ,η ) − a1  ∂  g (ξ ,η ) − a2  j assign to any test function φ( f , g) = f a= a2 , i.e., 1 ,g the value of that test function at the point where = ξ β= β . Thus we may write 1 ,η

at the points

δ  f ( ξ ,η ) − a1  δ  g (ξ ,η ) − a2  j =δ ( x − a1 ) δ ( y − a2 ) j =δ ( ξ − β1 ) δ (η − β 2 )

Hence

Differential Equations: Theory and Applications

398

δ (ξ − β1 ) δ (η − β 2 ) δ ( x − a1 ) δ ( y − a2 ) = j

(26) In the next few sections, we shall discuss the Green’s function method for solving partial differential equations with particular emphasis on elliptic equations. The discussion on wave equation and heat equation is also included in respectively .

8.1.1. Green’s Function for Laplace Equation To find analytic solution of the boundary value problems , the Green’ function method of the convenient technique. In this section, we shall give the definition of Green’s function for the Laplace equation and study its basic properties . To begin with, we shall define Green’s function for the 2 0 is valid inside Dirichlet problem, i.e., we shall find u such that ∇ u = some finitely bounded region IR, enclosed by ∂IR, as sufficiently smooth surface , when u = f is prescribed on the boundary ∂IR . Suppose that u is known at every poibt of the boundary ∂IR and that it satisfies the relation

∇2 u = 0 in IR u ( p ) when P ∈ IR The task is to find . Let = r , and let c be a sphere with centre at p and radius . ò. Also , let ∑ be denoted by ∂ ∑ , and

1 r − r′

u′ =

(27)

where r ′ is another point Q either in ∑ or on the boundary ∂ ∑ . if u ∂ and ∂n are twice continuously differentiable functions in IR and have first

order derivatives on ∂IR, then by Green ‘s theorem the region IR we have , form Eq. (19) , the relation .

. .

 ∂u ' ′ ∂u ∫ ∫ ∫ u ∇ 2u ′ − u ′ ∇ 2 u dV = −u u ∫∫ ∂n ∂n IR ∂ IR 

(

)

  dS  (28)

Green’s Function

Here, n I the unit vector normal to dSIR

399

drawn outwards from and

∂ ∇ 2u ′ = 0 within ∂n denotes differentiation in that direction. Since ∇ 2 u = ∂ ∑ , we have , in the region ∑ , the relation .

. .

∂u   ∂u ′  ∂u ′ ′ ∂u −u u ∫C  u ∂n − u′ ∂n  dS ′ + ∫∫ ∂ n ∂n  ∂ IR

 0  dS = 

Inserting u ' form relation (27), the above equation reduces to

  ∂  1 1 ∂u  ′ ′ ∫ − u r r ( ) ( )   dS ′   ∫ ∂n  r − r ' r − r ' ∂n   C   . .

. .  ∂  1 + ∫∫  u ( r )  ∂n  r − r ∂ IR  

 1 ∂u  0 ( r ) dS =  −  r − r ∂n  .

(29)

When Q is on C , we have

2 Also, dS ′, the surface element on C , is ò sin θ dθ dφ . d on C ,

u (= r ′ ) u ( r ) + du or Therefore,

Now,

400

Differential Equations: Theory and Applications

Employing these results and taking the limit as . .  ∂  1 4π u ( r ) + ∫∫ u ( r )  ∂n  r − r ′ ∂ IR  

Eq. (29) becomes

  1 ∂ 0 u ( r ′ )  dS =  − − ∂ ' r r n   

Implying thereby

   dS   (30) Therefore , the value of at an point of the region IR is determined , if ∂u u and ∂n are known on the boundary , ∂IR . This lead to the conclusion ∂u that both the values of u and ∂n are required solution to obtain the solution of Dirichlet’ s problem. But this is not so, as can be seen from the concept H ( r, r ) of the Green’s function defined as follows : Let be a function h u armonic in IR. Then the Green’s function for the Dirichlet problem involving the Laplace operator is defined by are the G, the two point function for u (r )

1 4π

 1 ∂u ∂  1 − r u r ( ) ( )   ∫∫ ∂n  r − r '  r − r ' ∂n ∂ IR  . .

the Dirichlet problem involving the Laplace operator is defined by point function of position, as

G (= r, r′) where

1 + H ( r , r ') r −r'

H ( r, r′)

(31)

satisfies the following

Green’s Function

(i ) ( ii )

401

 ∂2 ∂2 ∂2  H ( r, r′) = + + 0  2 2 2  x y z ∂ ∂ ∂   . 1 G= + H ( r, r ) = 0 on ∂IR r −r'

.

(33)

Thus the Green’s function for the Dirichlet problem involving the Laplace operator is a function

G ( r, r )

(i ) ( ii )

which satisfies the following properties:

∇ G ( r, r ) = δ ( r − r ′ ) in IR 2

G ( r= , r ) 0 on ∂IR

(34)

(35)

( iii )

G is symmetric, i.e., (36) ( iv ) G is continuous , but ∂G / ∂n . has . dcontinuity at the point r. which is given by the equation . ∂G Ltò → 0 ∫ ∫ dS = 1 ∂n C (37) Following the procedure adopted in the derivation of Eq. (30), replacing u ′ by G ( r , r ) , we can show that . .

1 ∂u ∂G  u (r ) = − G (r , r ) ( r ) −u ( r ) ( r , r ) dS ∫∫  4π ∂ IR  ∂n ∂n 

. (38) G 0 on ∂IR. . Thus the solution From Eq. (31) and (33) we can see that= u . at an interior point is given by the relation

u (r ) = −

1 4π

. .

∂G

∫∫ u ( r ) ∂n ( r , r ′) dS ∂ IR

(39)

And therefore, the solution of the interior Dirichlet ‘ s problem is reduced G ( r, r′) . to the determination of Green’s function

402

Differential Equations: Theory and Applications

Green’s function can be interpreted physically as follows: Let ∂IR be a grounded electrical conductor ( boundary potential zero ) and if a unit charge is located athe source point r, then G . is the sum of the potential at the point r ' . due to the charge at the source point r in the free space and the potential due to the charges induced on ∂IR. Thus

G ( r, r ) =

1 + H ( r, r′) r, r '

(40)

( i ) , viz, . Eq. (34) , essentially mns that ∇ 2G = 0 Hence, property everywhere except at the source point (r) .. Example. Consider a sphere with centre at the origin and radius ‘a’. Apply the divergence theorem to the sphere and show that

1 ∇2   = −4πδ ( r ) r ..

δ (r )

is a Dirac delta function.

Solution . Applying the divergence theorem to

1 1 grad   = ∇   r r . t .

1 ∫ ∫ ∫ ∇.∇   dV = r V

.

1

∫ ∫ ∇  r  .nˆ dS S

u = u ( r ,θ , φ ) , where nˆ is an outward drawn normal . if then grad u = eˆr .

∂u 1 ∂u 1 ∂u + eˆθ + eθ sin θ ∂r r ∂θ r ∂φ .

1 1 1 − 2 ×4π a 2 = −4π ∫S ∫ grad  r  . eˆr dS = ∫S ∫ r 2 dS = a 1 ∇2    r  has the following properties : Thus, we observe that

Green’s Function

403

( i ) it is undefined at the origin . ( ii )

it vanishes if r ≠ 0.

( iii )

its integral over any sphere with centre at the origin is – 4π . Hence, we conclude that .

1 ∫ ∫ ∫ ∇ 2   dV = −4πδ ( r ) r V Now, we shall prove the symmetric property through the following theorem.

8.2. THE EIGENFUNCTION METHOD Let us consider the Dirichlet boundary value problem described by

∇2 u = f (41) Valid in certain region IR, subject to the boundary condition u = g (42) On ∂ IR, the boundary of IR. From the definition of Green’s function which has been introduced in The Green’s function must satisfy the relations

∇2 G = ∂ ( x − ξ , y − η ) in IR

(43)

= G 0 , on ∂IR (44) 2 Now, consider the eigenvalue problem associated with the operator ∇ in the domain IR,

∇ 2 φ + λφ =i 0 nIR (45) = φ 0 on ∂IR (46)

λ

φ

Let mn be the eigenvalue and mn be the corresponding eigen functions. Suppose we give Fourier series expansion to G and δ in terms of the eigen functions

φmn in the following form:

Differential Equations: Theory and Applications

404

.

.

m

n

G ( x, y ; ξ ,η ) = ∑∑ amn ( ξ ,η ) φmn ( x, y ) .

.

m

n

(47)

δ ( x − ξ , y −η ) = ∑∑ bmn ( ξ ,η ) φmn ( x, y )

(48)

where .

1

b= mn

2

φmn .

IR

∫ ∫φ

φmn 2=

dy ∫ ∫ δ ( x − ξ , y − η )φmn ( x, y ) dx =

2 mn

φmn ( ξ ,η ) φmn 2



(49)

dx dy

IR

Now, substituting Eq. (47) and ( (48) into Eq. (43) and (44) and noting that Eq. (45) has the form (50) We obtain .

.

.

.

m

n

m

n

∇ 2 ∑∑amn (ξ ,η ) φmn ( x, y ) = ∑∑ bmn ( ξ ,η )φmn ( x, y ) Using Eq. ( 49) and (50) , the above equation reduce to

. . . . φ ( ξ ,η )φmn ( x, y ) − ∑∑ λmn amn (ξ ,η ) φmn ( x, y ) = ∑∑ mn 2 m

n

m

n

φmn

.

From which

amn ( ξ ,η ) = −

φmn ( ξ ,η ) λmn φmn

2

(51)

Hence, Eq.(47) and (51) give the required Green’s function for the Dirichlet problem, in the form .

.

m

n mn

∑ ∑ φ (ξ ,η ) φ ( x, y ) G ( x, y; ξ ,η ) = − λmn φmn

mn

2

(52) 0 ≤ y ≤ b, ∇ 2 + λ u =0 in IR Example. find the Green’s function f or

(

)

Green’s Function

405

theirichlet problem on the rectangle IR : 0 ≤ x ≤ a, described by the PDE (53) u 0 on ∂ IR. And the BC= Solution. The eigen functios of the given PDE can be obtained easily by using the variables separable method. Let us assume the solution o the u ( x, y ) = X ( x ) Y ( y ) . given PDE in the form Substituting into the given PDE, we obtain

X ′′  Y ′′  = −  + λ  = −v X Y  ( a separation constant ) Since u is zero on the boundary ∂ IR , X satisfies X ′′ + = vX 0, X= ( 0 ) X= (a) 0

(54)

(55)

Its solution, in general , is

= X ( x ) A cos v x + B sin vx

sin va 0,= implying v = A 0.= X (a) 0 = implies gives The corresponding real valued eigen functions  nπ x  X n sin  = =  , n 1 , 2 ,… while  a  the eigen values X (0) = 0

= vn n

2 2 π

a

= , n 1 , 2 ,…

nπ . a are are

Now, the factor Y satisfies

Y ′′ + ( λ − vn ) Y = 0, Y ( 0 ) = Y ( b ) = 0 Following the above procedure, we can show at once that the eigen functions are given by

Y= sin m

mπ y , m = 1, 2 …, b

And the corresponding eigen values are

λ − vn =

m 2π 2 , m = 1, 2 ,… b2

Differential Equations: Theory and Applications

406

Thus, we obtain the eigen functions to the given problem in the form

mπ x nπ y sin , m = 1, 2,…; n = 1 , 2,.. a b while the eigen values are given by

φmn ( x, y ) = sin

(56)

2 m 2π 2 n 2π 2 n2  2m λmn − 2 + 2 + π  2 + 2  a b b  a (57)

Computation of

φmn

2

ab

mπ x

φmn

gives

nπ x

dx dy sin sin ∫∫= a b 2

2

00

ab 4

(58)

Hence , the Green’s function for the given Dirichlet problem can be obtained with the help of Eq. (56) and (57) as

nπ x mπξ nπη  mπ x  sin  sin sin sin  4ab b a b  a  G ( x , y ,ξ ,η ) = − 2 ∑∑ 2 2 2 2 π m= 1=n 1 m b +n a (59) ∞



8.3. SUMMARY AND DISCUSSION Consider the differential equation

Lu ( x ) = f ( x )

(1)

f ( x) Where L is an ordinary linear differential operator, is a known u ( x) function, while is a unknown function. To solve the above differential −1 equation, one method is to find the operator L in the form of an integral G ( x, ξ ) operator with a kernel such that u ( x ) = L−1 f ( x ) = ∫ G ( x, ξ ) f (ξ ) d ξ

(2)

The kernel of this integral operator is called Green’s function for the differential operator .

φ ( x ) of Definition . Let us consider an auxiliary function a real x variable which possess derivative of all orders and vanishes outside a finite interval. Such function are called test functions.

Green’s Function

407

The idea of Green’s function can be easily extended to higher dimension . Thus the Green’s function technique can be applied, In principle , to find the solution of any linear non-homogenous partial differential equation . Green’s function can be applied for the solutions of partial differential equations which occurs most frequently in mathematical physics. Green’s function for the Dirichlet problem involving the Laplace operator is a function

G ( r, r )

(i )

which satisfies the following properties:

∇ G ( r, r ) = δ ( r − r ′ ) in IR 2

( ii )

G ( r= , r ) 0 on ∂IR

( iii )

G is symmetric, i.e., G ( r , r ′ ) = G ( r ′, r )

( iv )

.

G is continuous , but ∂G / ∂n has a discontinuity at the point r. . ∂G Ltò → 0 ∫ ∫ dS = 1 ∂n C which is given by the equation .

408

Differential Equations: Theory and Applications

REFERENCES 1. 2. 3. 4. 5.

6.

7.

C.W. Gear Numerical Initial Value Problems in ordinary Differential Equation, Prentice-Hall, 1971. E Hairer , S. P. Norsett and G Wanner , Solving Ordinary Differential Equations I, Springer-Verlag, 1987. Differential equation with boundary value Problem, D . G Zill and M. R. Cullen, Brooks/Cole, USA Paul Blanchard, Robert L. Devaney, Glenn R. Hall,  Differential Equations (Preliminary Edition),PWS Publishing, Boston, 1996. William H. Boyce and Richard C. Diprima, Elementary Differential Equations and Boundary Value Problems (6th Edition), Wiley, New York, 1996. C.H. Edwards, Jr., David E. Penney, Elementary Dfifferential Equations with Applications (Third Edition), Prentice-Hall, Englewood Cliffs, NJ, 1996. R. Kent Nagle and Edward B. Saff,  Fundamentals of Differential Equations (Third Edition), Addison Wesley, Reading, MA, 1993.

INDEX

A Across function 382, 387 Arbitrary function 20 B Boundary condition 196, 197, 205, 213, 215, 260, 262, 263, 265, 268, 270 Boundary curve 131, 137 Boundary surface 304, 305, 325, 352, 353, 354 Boundary value problems (BVPs) 195 C Canonical equation 117, 119, 120, 121, 122, 123, 124, 125, 127, 128 Constant Acceleration 144 Constant horizontal velocity 161, 187

Constant temperature 345 D Dependent function 112, 141 Differential equation 2, 3, 4, 5, 6, 7, 10, 13, 16, 19, 23, 24, 29, 31, 41, 42, 44, 47, 50, 51, 52, 53, 57, 60, 65, 77, 91, 92, 94, 95, 97, 98, 99, 100, 101, 102, 103, 104, 108 Diffusion Equation 332, 335 Diffusion phenomena 302, 352 Distance Dependent Acceleration 153, 187 E Electrical engineering 358, 384 Electric field 283 Elliptic parabolic 113, 141 Exact equation 19, 20, 25, 27, 28

410

Differential Equations: Theory and Applications

F Flexible string 280 G Gravity 145, 161, 186, 187 H Harmonic function 199, 204, 273, 275, 276 Heat conduction equation 304, 317, 319, 335, 342, 352 Homogeneous 13, 14, 15, 16, 17, 24, 26, 51, 56, 61, 85, 87, 91, 92, 100, 101, 104, 106 Homogenous hyperbolic 280, 297 Horizontal range 166, 182, 184, 188 Horizontal velocity 166, 177, 188 Hyperbolic equation 117, 132, 138 Hyperbolic system 132 I Inclined plane 167, 168, 182 Independent variable 112, 113, 115, 122, 141, 142 Initial boundary value 304, 310, 320, 321, 352 Initial boundary value problem (IBVP) 310 Initial temperature 334, 338, 343, 345, 347 Integrate function 122 L Laplace equation 194, 196, 199, 205, 206, 207, 208, 212, 215, 225, 229, 231, 235, 241, 243, 249, 253, 256, 257, 264, 271, 276, 278

Laplace transform 358, 360, 361, 362, 363, 364, 365, 366, 367, 369, 370, 371, 373, 374, 375, 376, 377, 378, 380, 382, 384, 385, 388 Laplace transform technique 358, 376, 384 Legendre polynomial 244, 245, 258 Linear combination 305, 353 Linear homogeneous equation 210, 276 M Magnetic potential 192 Mathematical formulation 5, 99 Mathematical tool 197, 358, 384 Monochromatic wave 287, 298 N Normal direction 196 Notion mechanic 310 O One-dimensional Dirac delta function 313, 355 One-dimensional wave 284, 285, 286, 289, 290, 298, 299 Ordinary calculator 358 Ordinary differential equation (ODE) 27, 102 Original equation 114 P Parabolic equation 118 Parabolic of hyperbolic 113, 142 Partial derivative 112, 133, 141 Partial differential equation 112, 130, 132, 137, 141 Physical phenomena 192, 276

Index

Physical situation 192 Plane harmonic wave 287, 298 Potential 192, 193, 194, 196, 222, 234, 235, 238, 243, 245, 246, 247, 249, 252, 253, 255, 256, 264, 265, 276, 277 Prismatic beam 192 Probability theorem 358, 384 Problem under investigation 304, 353 R Riemann-Green function 136 S Several problem 358, 384 Spatial variable 280, 298 Specific heat 302, 352 Steady temperature distribution 222, 277 Subsequent analysis 113 Supplementary boundary 7, 100 Surrounding medium 196, 197

411

T Temperature distribution 225, 231, 234, 241, 250, 259, 261 Time Dependent Acceleration 148, 187 Transformation 113, 114, 115, 119, 120, 121, 122, 123, 126, 128, 142 Transformed equation 113, 114, 119 Transformed equation assume 113 Tthermal conductivity 302, 345, 352 U Uniform cross-section 321 V Velocity Dependent Acceleration 150, 187