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 978-93-5274-099-4

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LAPLACE AND FOURIER TRANSFORMS

ISBN 978-93-5274-099-4

9 789352 740994

LAPLACE AND FOURIER TRANSFORMS

TOPICS IN

LAPLACE AND FOURIER TRANSFORMS

Salient features of the present edition : H It has detailed theory supplemented with well explained examples. H It has adequate number of unsolved problems of all types in exercises. H It has working rules for solving problems before exercises. H It has hints of tricky problems after relevant exercises.

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TOPICS IN

LAPLACE AND FOURIER TRANSFORMS

By PARMANAND GUPTA B.Sc.(Hons.), M.Sc. (Delhi) M.Phil (KU), Pre. Ph.D. (IIT Delhi) Associate Professor of Mathematics Former Head of Department of Mathematics Indira Gandhi National College, Ladwa Kurukshetra University, Haryana

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CONTENTS Chapter

Pages

1. Laplace Transforms ........................................................................................................ 1–38 1.1. 1.2. 1.3. 1.4. 1.5. 1.6. 1.7. 1.8. 1.9. 1.10. 1.11. 1.12. 1.13. 1.14. 1.15.

Introduction ................................................................................................................................. 1 Laplace Transform of a Function .............................................................................................. 1 Laplace Transforms of Elementary Functions ......................................................................... 2 Linearity of the Laplace Transform .......................................................................................... 7 Shifting Theorems .................................................................................................................... 11 First Shifting Theorem ............................................................................................................. 11 Unit Step Function ................................................................................................................... 15 Second Shifting Theorem ......................................................................................................... 15 Change of Scale Property ......................................................................................................... 17 Piecewise Continuous Function ............................................................................................... 21 Existence Theorem for Laplace Transforms ........................................................................... 21 Laplace Transforms of Derivatives ......................................................................................... 22 Laplace Transforms of Integrals ............................................................................................. 27 Differentiation of Laplace Transforms .................................................................................... 30 Integration of Laplace Transforms .......................................................................................... 34

2. Inverse Laplace Transforms ........................................................................................ 39–81 2.1. 2.2. 2.3. 2.4. 2.5. 2.6. 2.7. 2.8. 2.9. 2.10.

Introduction ............................................................................................................................... Inverse Laplace Transform of a Function ............................................................................... Existence and Uniqueness of Inverse Laplace Transform .................................................... Elementary Inverse Laplace Transform Formulae ............................................................... Linearity of the Inverse Laplace Transform .......................................................................... Value of L–1(F(s – a)) in Terms of L–1 (F(s)) ............................................................................ Value of L–1 (e–as F(s)) in Terms of L–1 (F(s)) .......................................................................... Value of L–1 F(s/a)) in Terms of L–1 (F(s)) .............................................................................. Value of L–1 (F(s)/s) in Terms of L–1 (F(s)) .............................................................................. Value of L–1 (F′(s)) in Terms of L–1(F(s)) .................................................................................

2.11. Value of L−1

F GH

z



s

I JK

39 39 39 40 42 44 48 49 52 53

F (s) dS in Terms of L–1 (F(s)) ..................................................................... 55

2.12. Convolution Theorem ............................................................................................................... 58 2.13. Inverse Laplace Transforms by the Method of Partial Fractions ................................. 65 2.14. Solution of Differential Equations by Using Laplace Transformation ................................ 76

3. Solution of Integral Equations Using Laplace Transformation ............................... 82–88 3.1. Introduction ............................................................................................................................... 82 3.2. Definition of Integral Equation ............................................................................................... 82

(v)

Chapter

Pages

3.3. Method of Solving Integral Equation of Convolution Type ................................................... 82 3.4. Integro-differential Equation ................................................................................................... 85

4. Solution of Systems of Differential Equations Using the Laplace Transformation .............................................................................................................. 89–98 4.1. Introduction ............................................................................................................................... 89 4.2. Method of Solving System of Differential Equations ............................................................. 89

5. Fourier Transforms ..................................................................................................... 99–137 5.1. 5.2. 5.3. 5.4. 5.5. 5.6. 5.7. 5.8. 5.9. 5.10. 5.11.

Introduction ............................................................................................................................... 99 Fourier’s Integral Theorem ...................................................................................................... 99 Fourier Transform and Its Inverse ....................................................................................... 100 Shifting Property of Fourier Transforms .............................................................................. 101 Modulation Property of Fourier Transforms ........................................................................ 102 Convolution Theorem ............................................................................................................. 102 Fourier Sine and Cosine Transforms .................................................................................... 110 Linearity of Transforms ......................................................................................................... 129 Change of Scale Property of Transforms .............................................................................. 130 Transforms of Derivatives ...................................................................................................... 131 Parseval’s Identities ............................................................................................................... 132

6. Solution of Differential Equations Using Fourier Transforms ............................. 138–149 6.1. Introduction ............................................................................................................................. 138 6.2. Partial Differential Equation ................................................................................................. 138 6.3. Method of Solving Partial Differential Equation by Using Fourier Transforms .............. 139

(vi)

PREFACE The present book on ‘‘Laplace and Fourier Transforms’’ has been written as a textbook according to the latest guidelines and syllabus in Mathematics issued by the U.G.C. for various universities. The text of the book has been prepared with the following salient features: (i) The language of the book is simple and easy to understand. (ii) Each topic has been presented in a systematic, simple, lucid and exhaustive manner. (iii) A large number of important solved examples properly selected from the previous university question papers have been provided to enable the students to have a clear grasp of the subject and to equip them for attempting problems in the university examination without any difficulty. (iv) Apart from providing a large number of examples, different type of questions in ample quantity have been provided for a thorough practice to the students. (v) A large number of ‘notes’ and ‘remarks’ have been added for better understanding of the subject. A serious effort has been made to keep the book free from mistakes and errors. In fact no pains have been spared to make the book interesting and useful. Suggestions and comments for further improvement of the book will be welcomed. —AUTHOR

(vii)

SYMBOLS Greek Alphabets A B Γ D E Z H Θ

α β γ δ ε ζ η θ ∃

Alpha Beta Gamma Delta Epsilon Zeta Eta Theta there exists

I K Λ M N Ξ O Π

ι κ λ μ ν ξ ο π V

Iota Kappa Lambda Mu Nu Xi Omicron Pi for all

P Σ T Y Φ X Ψ Ω

ρ σ τ υ ϕ χ ψ ω

Rho Sigma Tau Upsilon Phi Chi Psi Omega

Metric Weights and Measures LENGTH 10 millimetres 10 centimetres 10 decimetres 10 metres 10 decametres 10 hectometres

CAPACITY 10 millilitres 10 centilitres 10 decilitres 10 litres 10 decalitres 10 hectolitres

= 1 centilitre = 1 decilitre = 1 litre = 1 decalitre = 1 hectolitre = 1 kilolitre

VOLUME 1000 cubic centimetres = 1 centigram 1000 cubic decimetres = 1 cubic metre

AREA 100 square metres 100 ares 100 hectares

= 1 are = 1 hectare = 1 square kilometre

WEIGHT 10 milligrams 10 centigrams 10 decigrams 10 grams 10 decagrams 10 hectograms 100 kilograms 10 quintals

ABBREVIATIONS kilometre km metre m centimetre cm millimetre mm kilolitre kl litre l millilitre ml

tonne quintal kilogram gram are hectare centiare

= = = = = =

= = = = = = = =

1 centimetre 1 decimetre 1 metre 1 decametre 1 hectometre 1 kilometre

1 centigram 1 decigram 1 gram 1 decagram 1 hectogram 1 kilogram 1 quintal 1 metric ton (tonne)

( ix )

t q kg g a ha ca

1

Laplace Transforms

1.1. INTRODUCTION The Laplace transform* of a suitably defined function f of a real variable t is a related function F of a real variable s. The use of Laplace transforms provide a powerful method of solving differential and integral equations. The Laplace transform method also has the advantage that it solves initial value problems directly without first finding a general solution. The ready tables of Laplace transforms has reduced the problem of solving differential equations to merely algebraic manipulation. 1.2. LAPLACE TRANSFORM OF A FUNCTION Let f be a real valued function of the real variable t, defined for t ≥ 0. Let s be a real variable and consider the function F of s defined by F( s ) =

z



0

e − st f (t ) dt

for all values of s for which this integral exists (that has some finite value). The function F(s) defined by the integral is denoted by L(f).

z



0

e − st f (t ) dt is called the Laplace transform of the function f and F(s) = L(f) =



z



0

e − st f(t) dt .

Remarks 1. The original function f depends on t and its Laplace transform (F) depends on s. 2. Original functions are denoted by lower case letters and their Laplace transforms by the same letters in capitals.

WORKING STEPS TO FIND LAPLACE TRANSFORMS OF f(t), t ≥ 0. Step I.

Write L( f ) =

z



0

e − st f (t ) dt .

Step II. Write this improper integral as lim Step III. Evaluate

z

T→∞

T

0

Step IV. Find limit of

e − st f (t ) dt .

z

T

0

z

T

0

e − st f ( t ) dt and simplify e − st f (t ) .

e − st f (t ) dt as T → ∞ . This gives the value of L(f).

*Pierre Simon Marquis De Laplace (1749––1827) French mathematician, made contribution to special functions, probability theory, potential theory and astronomy.

1

2

LAPLACE AND FOURIER TRANSFORMS

1.3. LAPLACE TRANSFORMS OF ELEMENTARY FUNCTIONS In this section, we shall find the Laplace transforms of some elementary functions. Theorem. Prove that for t ≥ 0, 1. L(1) =

1 ,s>0 s

n 3. L(t ) =

n! sn + 1

5. L(sinh at) = 7. L(sin at) =

2. L(t a ) =

, n = 0, 1, 2, . ... .. ; s > 0 a 2

s −a

2

, s > |a|

4. L(e at ) =

a , s>0 s + a2

8. L(cos at) =

Proof. By definition, the improper integeral



z



1 , s>a s−a

6. L(cosh at) =

2

1. By definition, L(1) =

Γ(a + 1) , a > − 1, s > 0 sa + 1

z



s 2

s − a2

, s >|a|

s , s > 0. s2 + a 2

g(t) dt is equal to lim

T→∞

k

z

T

k

g(t) dt .

e − st . 1 dt

0

z

L e OP L(1) = lim e dt = lim M N−sQ 1 L 1 OP = 1 lim L− 1 = lim M− s N e Q s MN e T→∞

T

− st

− st

T→∞

0

T

0

T

st

T→∞

0

sT

T→∞

OP Q

+1

1 1 [ 0 + 1] = for s > 0 . s s 1 L(1) = , s > 0 . s =



z

2. By definition, L(t a ) =

0

Let st = u. ∴ s dt = du. ∴

L(t a ) =

=



z



e − st t a dt .

e−u

0

FG u IJ H sK

a

du = 1 sa +1 s

z



0

e − u u a du

Γ ( a + 1)

, provided a > – 1 sa +1 Γ (a + 1) L(t a ) = , a > − 1, s > 0 . sa + 1

3. By definition, L(tn ) = Let



z



0

e − st tn dt .

st = u. ∴ s dt = du

(∵

t ≥ 0, u ≥ 0, u = st ⇒ s ≥ 0)

3

LAPLACE TRANSFORMS



z

L(t n ) =

0

=

Let ∴



e− u

FG u IJ H sK

n



0

Γ ( n + 1)

e − u u n du

, provided n + 1 > 0. sn + 1 n = 0, 1, 2, ...... ∴ Γ (n + 1) = n ! n! L(t n ) = n + 1 , n = 0, 1, 2, ...... ; s > 0. s

z

4. By definition, L( e at ) =



0

T→∞

=

L(e at ) =



(∵

t ≥ 0, u ≥ 0, u = st ⇒ s ≥ 0)

e − st e at dt

= lim

5.

z

du 1 = n +1 s s

z

T

0

T→∞

1 lim a − s T→∞

LM Ne

OP Q

1

−1 =

( s − a) T

1 , s > a. s−a

z

LM e OP MN a − s PQ ( a − s )t

e ( a − s )t dt = lim

T

0

1 1 [ 0 − 1] for s > a = a−s s−a

F e − e I dt GH 2 JK 1 = −e −e ee j dt = 12 lim ee 2 Le OP 1 e + = lim M 2 N a−s s+a Q L 1 FG 1 − 1IJ + 1 FG 1 − 1IJ OP 1 = lim M K s + a He KQ 2 Na − s H e O 1L 1 1 = M ( 0 − 1) + ( 0 − 1)P provided s > a, s > – a 2 Na − s s+a Q 1 L 2a O = M P = a , s > | a |. 2 MN s − a PQ s − a

L(sinh at ) = L

Fe GH

z

at

− e − at 2



I JK

=

− ( s − a )t



0

at

e − st

− at

− ( s + a )t

T→∞

0

− ( s − a) t

− ( s + a) t

T→∞

(s − a) T

2

2

( s + a )T

2

a , s > | a |. s − a2



L(sinh at) =

7.

L(sin at ) =



L(sin at) = lim

2

z



0

z

e − st sin at dt

T→∞

z

T

0

e − st sin at dt

e − st sin at dt

Let

I=



− st I= e .

0

0

T→∞

2

T

z

T

− cos at − a

z

− se − st .

− cos at dt a

− ( s − a )t

− ( s + a )t

j dt

4

LAPLACE AND FOURIER TRANSFORMS

z

e − st cos at s e − st cos at dt − a a e − st cos at s − st sin at sin at =− − − − se − st . e dt a a a a =−



I=−



I=−



s2 + a 2

T→∞

= =

L(sin at) =

[ a cos at + s sin at ] −

a2 e − st

L(sin at) = lim =



e − st

LM N

z

s2

I

a2

[ a cos at + s sin at ]

LM OP N Q LM a cos aT + s sin aT − a + 0 OP 1 Q e N

− e − st a cos at + s sin at s2 + a2

−1 2

s + a2 −1

s2 + a2 a s2 + a 2 a

s2 + a 2

OP Q

lim

T

0

sT

T→∞

[0 − a] , provided s > 0

, s > 0. , s > 0.

Remark. The proofs of the parts 6 and 8 are left for readers as exercises.

ILLUSTRATIVE EXAMPLES Example 1. Find the Laplace transform of the following function :

Sol.

L f (t ) = = =

z z zL

f(t) =



0 2

0 2

0

0≤t≤2 t > 2.

e − st f (t ) dt

z

e − st f (t ) dt +



2

e − st (t + 1) dt +

e = M(t + 1) . MN −s

=−

RSt + 1, T 3,

− st

2

− 0

z

3 − 2 s 1 e − st e + − 2 s s s

z

e − st f (t ) dt ∞

2

2

0

OP PQ

e − st e − st 1. dt + 3 . −s −s

2

− 0

e − st . 3 dt

3 s

LM lim N

T→∞



2

e − sT − e − 2 s

OP Q

3e − 2 s 1 e − 2 s 1 3 =− + − 2 + 2 − (0 − e − 2 s ) s s s s s − 2s − 2s 3e 1 e 1 3e − 2 s 1 =− + − 2 + 2 + = 2 [s − e − 2s + 1] . s s s s s s

5

LAPLACE TRANSFORMS

Example 2. Find the Laplace transform of the following function :

R| S| T

0, 0 ≤ t < 1 f(t) = t, 1 < t < 2 1, t > 2.

Sol.

L( f (t )) = = = =

z z z z



0 1

=−

e − st f (t ) dt +

0 1

z

e − st . 0 dt +

0

z

2

2 se 2 s 2

se

2s

2 se

2s

1

+

+ +

+

se s 1

se

s



s



1 se

s

2

z

2



z

2

1

2s

1 2 2s

s e

+



2

2

e − st f (t ) dt

e − st . 1 dt

OP PQ

1 lim s T→∞



LM 1 Ne





LM e OP MN − s PQ LM 1 − 1 OP Ne e Q

e − st 1. dt + lim T→∞ −s

2

1

1

e − st − s2

1

z

e − st f (t ) dt +

e − st t dt +

1

− st

0

2

1

L e 0 dt + Mt . MN − s

1

=0− =−

e − st f (t ) dt

e

s

1

1

+

2 s

s e

se 2s

T

2

2s

sT

OP − 1 [ 0 − 1] , Q s

1

− st

s>0

, s>0

1 [ − se − 2 s + se − s − e − 2 s + e − s ] , s > 0 s2 1 = 2 (e − s − e − 2s ) (s + 1) , s > 0. s Example 3. Find the Laplace transform of the function shown in the graph. Sol. From the graph, we have t tan 135° + 1 0 ≤ t ≤ 1 f (t) = 0 t≥1 =

or

RS T R− t + 1 f ( t) = S T 0

∴ L( f (t )) = = =

z z z



0 1

0 1

0

=0+

f(t) 1

0≤t≤1 t ≥ 1. O

e − st f (t ) dt e − st f (t ) dt + (1 − t ) e

− st

e0 e − st − s − s2

z

1

dt + 1

= 0



e − st f (t ) dt =

z



0

z

1

0

e − st ( − t + 1) dt +

e − st 0 dt = (1 − t ) . −s

LM N

OP Q

1

1

− 0

LM N

z

1

0

− 1.

OP Q

z



1

e − st . 0 dt

e − st dt + 0 −s

1 1 1 1 1 1 1 + 2 − 0 = + 2 − 1 , s ≠ 0. s s s es s s e e

t

6

LAPLACE AND FOURIER TRANSFORMS

Example 4. Show that Sol. Let

I=

z



z



cos x 2 dx =

0

1 2

π . 2

cos x 2 dx .

0

t = x2 ⇒ dt = 2x dx ⇒ dx = x → ∞ ⇒ t → ∞, x = 0 ⇒ t = (0)2 = 0 ∴

I=

z

cos t dt



0

LM N

z

2 t

=

1 2

z



0

e it + e − it

z

dt 2 t

dt

2 t

∞ 1 ∞ it −1/ 2 e .t dt + e − it . t −1/2 dt 0 4 0 1 = [L(t–1/2) |s = –i + L(t–1/2) |s = i] 4

=

LM MN 1L = M 4 MN =

=

1 Γ (1 /2) 4 s 1/ 2

π −i

+

LM N

s =− i

+

Γ (1 /2) s 1/2

OP = i PQ

π

π 4

OP Q

2π 1 1 + = 4 1− i 1+ i

LM MM MN

s= i

OP Q

OP PQ

1 1 1 i − 2 2

LM OP NQ

+

1 2π 2 = 2 4 2

1 1 + 2

π . 2

OP P 1 P i 2 PQ

(Note this step)

TEST YOUR KNOWLEDGE 1. By using standard formulae, write the Laplace transform of the following functions : (i) t3/2 (ii) t5/3 (iii) t7 (iv) e3t (v) e–10t (vi) sinh 5t (vii) cosh 7t (viii) sin 4t (ix) cos 6t. 2. Find the Laplace transform of the function f(t) = t, t ≥ 0. 3. Find the Laplace transform of the function f(t) = 4t + 3, t ≥ 0. 4. Find the Laplace transform of the function f(t) = cosh at, t ≥ 0. 5. Find the Laplace transform of the function f(t) = cos at, t ≥ 0. 6. Find the Laplace transform of the following functions : (i) f (t ) =

RS4, 0 ≤ t < 3 T 2, t > 3

(ii) f (t ) =

7. Find the Laplace transform of the following function :

R| S| T

t f (t) = t − 1 0

0 ≤ t ≤ 1/2 1/2 < t ≤ 1 . t>1

RSt, 0 ≤ t < 2 . T3, t > 2

7

LAPLACE TRANSFORMS

8. Find the Laplace transform of the functions whose graphs are given : f(t)

f(t) k

(i)

(ii)

k

O

t

a

O

9. Show that : (i)

z



0

2

e− x dx =

π 2

(ii)

z



sin x2 dx =

0

t

k

1 2

π . 2

Answers 1.

(i)

3 π

5/ 2

,s>0

(ii)

,s>0 9 s8 / 3 1 (v) , s > – 10 s + 10 4 (viii) 2 ,s>0 s + 16 3s + 4 , s>0 3. s2

4s 1 (iv) ,s>3 s−3 s ,s>7 (vii) 2 s − 49 1 , s>0 2. s2

5. 7.

s s2 + a 2

1 s

2

(ii)

, s>0

1 − se− s / 2 − e− s , s > 0

1 s2

(1 − e − ks ) −

10 Γ(2/3)

(iii) (vi) (ix) 4.

6. (i)

2 ( 2 − e − 3s ) , s > 0 s

8. (i)

k (1 − e − as ) , s > 0 s

(ii)

5040 s8 s

,s>0

s2 − 25 s s2 + 36 s

s2 − a 2

1 s2

+

,s>5 ,s>0 , s > |a |

e − 2s e − 2 s − 2 , s>0 s s

k − ks , s>0. e s

1.4. LINEARITY OF THE LAPLACE TRANSFORM Theorem. If f(t) and g(t) be any functions of t for t ≥ 0 whose Laplace transforms exists and a and b be any constants, then L{af(t) + bg(t)} = aL{f(t)} + bL{g(t)}. Proof. We have L{ f (t)} = Now

L{ af (t) + bg (t)} =

z z z

=a



0



0

e − st f (t) dt

e − st f (t) dt + b

z



0

= aL { f ( t )} + bL { g(t )}



z

e − st [ af (t) + bg (t)] dt



a

L{g(t)} =

and

L{af(t) + bg(t)} = aL{f(t)} + bL{g(t)}.

e − st g (t) dt



0

e − st g (t) dt .

8

LAPLACE AND FOURIER TRANSFORMS

ILLUSTRATIVE EXAMPLES Example 1. Find Laplace transform of the function cosh at, t ≥ 0 using linearity property.

Fe +e I GH 2 JK F1 1 I =LG e + e J H2 2 K at

Sol. We have L(cosh at) = L

− at

at

=

− at

=

1 1 L(e at ) + L(e − at ) 2 2

1 1 1 1 . + . , s > a, s > − a 2 s − a 2 s − (− a)

s , s >|a| s − a2 s L(cosh at) = 2 , s > |a|. s − a2 =



2

FG∵ H

L ( e at ) =

1 ,s > a s−a

IJ K

Example 2. Find the Laplace transform of the following functions of t, t ≥ 0 : (i) 6t + 9 (iv) sin 2t sin 3t

(ii) 2t2 – t + 5

(iii) sin2 t

(v) cos3 t

(vi) e4t + e2t + t3 + sin2 t.

Sol. (i) L(6t + 9) = 6L(t) + L(9) = 6 =

(By linearity)

FG 1 ! IJ + 1 , s > 0 Hs K s 1+1

6 + 9s s2

,s>0

(ii) L(2t2 – t + 5) = 2L(t2) – L(t) + L(5) = 2. =

(iii)

=

3

4 − s + 5s2 s3

L(sin 2 t) = L =

FG 2 ! IJ − 1 ! + 5 , s > 0 Hs K s s 2

, s > 0.

FG 1 − cos 2t IJ H 2 K

=L

FG 1 − 1 cos 2tIJ H2 2 K FG IJ HK

F GH

1 1 s L(1) − L(cos 2t) = 1 1 − 1 2 2 2 s 2 s2 + 4 2 2

s (s + 4)

I JK

,s>0

, s > 0.

FG 1 (2 sin 2t sin 3t)IJ H2 K 1 1 F1 I 1 = L G cos t − cos 5tJ = L (cos t) − L (cos 5t) H2 K 2 2 2

(iv) L(sin 2t sin 3t) = L

9

LAPLACE TRANSFORMS

=

I JK

F GH

I JK

F GH

1 s 1 s . 2 − . 2 s >|1|, s >|5| 2 2 s +1 2 s + 52

=

12s 2

(s + 1) (s 2 + 25)

,s>5.

(v) We have cos 3θ = 4 cos3 θ – 3 cos θ. ⇒

cos3 θ =

1 3 cos 3θ + cos θ 4 4



cos3 t =

1 3 cos 3t + cos t . 4 4



L(cos3 t) =

1 3 L(cos 3t) + L(cos t) 4 4

(By linearity)

I F I F JK GH JK GH s L 1 3 O = M + P,s>0 4 MN s + 9 s + 1 PQ 1 3 s s + 4 s2 + 32 4 s 2 + 12

=

2

=

,s>0

2

(s 2 + 7) s (s2 + 9) (s 2 + 1)

, s > 0.

(vi) L(e 4 t + e 2t + t 3 + sin 2 t) = L(e 4 t ) + L(e 2t ) + L(t 3 ) + L = L(e 4 t ) + L(e 2 t ) + L(t 3 ) +

FG 1 − cos 2t IJ H 2 K

(By linearity)

1 1 L(1) − L(cos 2t) 2 2

FG IJ HK

F GH

I JK

=

1 s 1 3! 1 1 1 + + + − , s > 4, s > 2, s > 0 s − 4 s − 2 s3 + 1 2 s 2 s 2 + 22

=

1 1 6 1 s , s > 4. + + 4 + − 2 s−4 s−2 s 2s 2(s + 4)

Example 3. Find the Laplace transform of the function sin (ωt + δ), t ≥ 0. Sol. L(sin (ωt + δ)) = L(sin ωt cos δ + cos ωt sin δ) = cos δ L(sin ωt) + sin δ L(cos ωt) = cos δ

=

F GH s

1 2

s + ω2

2

I JK

F GH

I JK

ω s + sin δ 2 , s>0 + ω2 s + ω2

[ω cos δ + s sin δ ] , s > 0.

(By linearity)

10

LAPLACE AND FOURIER TRANSFORMS

WORKING STEPS FOR SOLVING PROBLEMS Step I.

Simplify the given functions and write it as a linear combination of functions with known Laplace transforms. Step II. Apply linearity theorem. Step III. Simplify the terms as far as possible.

TEST YOUR KNOWLEDGE 1. Find the Laplace transform of the following functions of t, t ≥ 0 : (i) 5t + 9

(ii) t2 + 8t – 15

t3

(iv) t10 + t4 + 2.

(iii)

+4

2. Find the Laplace transform of the following functions of t, t ≥ 0 : (i) e5t + t2

(ii) e– 4t + e2t

(iii) e3t + t3

(iv) e–5t + t2 + 6t.

3. Find the Laplace transform of the following functions of t, t ≥ 0 : (i) sinh 4t + cosh 4t (iii) sin πt +

e6t

(ii) t – sinh 2t (iv) cos (at + b).

4. Find the Laplace transform of the following functions of t, t ≥ 0 : (i) sin2 2t

(ii) sin2 3t

cos2

(iv) 7 cos2 t.

(iii)

4t

5. Find the Laplace transform of the following functions of t, t ≥ 0 : (i) sin3 t

(ii) sin3 2t

(iii) cos3 2t + t2

(iv) cos3 4t + t.

6. Find the Laplace transform of the following functions of t, t ≥ 0 : (i) sin 3t cos 2t (iii) cos 4t cos t

(ii) cos 5t sin 2t (iv) sin 3t sin 7t.

7. Find the Laplace transform of the following functions of t : (i) (sin t – cos t)2, t ≥ 0 (iii) sin at sin bt, t ≥ 0

3

t+

,t > 0 t (iv) cosh at – cos at, t ≥ 0. (ii) 1 +

Answers 1. (i) (iii)

5 s

2

6 s

4

+

9 ,s > 0 s

(ii)

+

4 ,s > 0 s

(iv)

2 3

s

+

10 ! 11

s

8 s

+



2

24 5

s

15 ,s > 0 s +

2 ,s > 0 s

2. (i)

1 2 + ,s > 5 s − 5 s3

(ii)

2 ( s + 1) ,s > 2 ( s + 4 ) ( s − 2)

(iii)

1 6 + ,s > 3 s − 3 s4

(iv)

1 2 6 + + ,s > 0 s + 5 s3 s2

11

LAPLACE TRANSFORMS

3. (i) (iii) 4. (i)

(iii)

5. (i)

(iii) 6. (i)

(iii) 7. (i)

(iii)

1 ,s > 4 s−4 π

s2 + π 2

(ii)

1 ,s > 6 s−6

+

8 s ( s2 + 16)

s2 + 32

s ( s2 + 64 )

,s > 0

(ii)

,s > 0

(iv)

6 ( s2 + 1) ( s2 + 9)

2

,s > 0

s (s2 + 28)

+

2

(s + 4) (s + 36) 3s2 + 15

s (s2 + 17)

(s2 + 9) (s2 + 25) 2

s (s + 4)

(ii)

2

,s>0 s3

(ii)

,s>0

(iv)

,s > 0

(ii)

2abs 2

(iv)

,s > 0

( s2 + 1) ( s2 + 25)

s 2 − 2s + 4

(iv)

2

2

2

(s + (a − b) ) ( s + (a + b) )

,s>0

(iv)

4 + s2

2

s (4 − s 2 )

,s>2

s cos b − a sin b s2 + a 2

18 s ( s2 + 36 )

7 ( s2 + 2) s ( s2 + 4 )

,s > 0

,s > 0

,s > 0

48 2

( s + 4) ( s2 + 36) 2

,s>0

s (s2 + 112)

(s + 16) (s2 + 144) 2

2s2 − 42

( s + 9) ( s2 + 49)

+

1 s2

,s>0

,s > 0

42 s ( s2 + 16) (s2 + 100)

,s>0

π 1 3 π + + 1/2 , s > 0 s 2s3 / 2 s 2a 2 s 4

s − a4

, s > |a |.

1.5. SHIFTING THEOREMS There are two shifting theorems. In the first theorem the variable s of the function L(f(t)) is replaced by s – a and in the second theorem the variable t of the function f(t) is replaced by t – a. 1.6. FIRST SHIFTING THEOREM If f(t) be a function of t for t ≥ 0 whose Laplace transform F(s) exists for s > k then for any constant ‘a’ the function eat f(t) has the Laplace transform F(s – a) for s > k + a. Proof. We have F( s ) = L( f ) =

z



0

e − st f (t ) dt, s > k .

Replacing s by s – a, we get F(s − a) = ⇒

F( s − a) =

z



0

z



0

e − ( s − a) t f (t) dt, s − a > k .

e − st (e at f (t)) dt, s > k + a

∴ Laplace transform of the function eat f(t) is F(s – a) for s > k + a. Remark. The result of above theorem can be easily remembered as follows : If L(f(t)) = F(s), s > k then for any a, L(eat f(t)) = F(s – a), s > k + a.

12

LAPLACE AND FOURIER TRANSFORMS

Theorem. Using first shifting theorem. Show that for t ≥ 0 : 1. L(eat ) =

1 , s>a s−a

2. L(eat t b ) =

Γ ( b + 1)

(s − a) b + 1

3. L(e at t n ) =

, b > – 1, s > a

n! , n = 0, 1, 2, ...... , s > a (s − a) n + 1

4. L(e at sinh bt) =

b , s > |b| + a (s − a)2 − b 2

5. L(e at cosh bt) =

s−a , s > |b| + a (s − a) 2 − b 2

at 6. L(e sin bt) =

b , s>a (s − a) 2 + b 2

7. L(e at cos bt) =

s−a , s > a. (s − a)2 + b 2

Proof. 1. We have

L(1) =

1 , s>0. s

∴ By first shifting theorem,

1 , s>a s−a

L( e at . 1) =

L(eat ) =

∴ 2. We have L(t b ) =

1 , s>a . s−a

Γ ( b + 1)

, b > − 1, s > 0 . sb + 1 ∴ By first shifting theorem,

L(e at t b ) = n 3. We have L(t ) =

Γ (b + 1) (s − a) b + 1

, b > – 1, s > a.

n!

, n = 0, 1, 2, ...... ; s > 0 . sn + 1 ∴ By first shifting theorem,

L(e at t n ) =

4. We have L(sinh bt) =

n! , n = 0, 1, 2, ...... ; s > a . (s − a) n + 1

b

, s >|b|. s − b2 ∴ By first shifting theorem, 2

L(e at sinh bt) =

b , s > |b| + a . (s − a)2 − b 2

13

LAPLACE TRANSFORMS

s

5. We have L(cosh bt ) =

, s > |b|. s − b2 ∴ By first shifting theorem, 2

L(e at cosh bt) =

s−a , s > |b| + a . (s − a) 2 − b 2

b

6. We have L(sin bt ) =

, s>0. s + b2 ∴ By first shifting theorem, 2

L(e at sin bt) =

b , s > a. (s − a)2 + b 2

s

, s > 0. s + b2 ∴ By first shifting theorem,

7. We have L(cos bt ) =

2

L(e at cos bt) =

s−a , s > a. (s − a)2 + b 2

ILLUSTRATIVE EXAMPLES Example 1. Find the Laplace transform of the following functions : (i) t3 e–3t, t ≥ 0

(ii) e–3t (2 cos 5t – 3 sin 5t), t ≥ 0

(iii) e4t sin 2t cos t, t ≥ 0

(iv) sinh t cos t, t ≥ 0

(v) e–t (3 sinh 2t – 5 cosh 2t). Sol. (i) We have L(t 3 ) =

3!

=

6

s s4 ∴ By first shifting theorem,

L(t 3 e − 3t ) =

3+1

6 (s − (− 3))

4

, s > 0.

, s > 0 + (− 3) =

6 , s > − 3. (s + 3)4

(ii) We have L(2 cos 5t – 3 sin 5t) = 2L(cos 5t) – 3L(sin 5t) =2

F GH s

s 2

+5

2

I − 3F JK GH s

5 2

+5

∴ By first shifting theorem,

2

I , s>0 JK

(By linearity) =

2s − 15 s 2 + 25

, s > 0.

L(e–3t (2 cos 5t – 3 sin 5t)) =

(iii)

2s − 9 2 (s − (− 3)) − 15 , s > − 3. , s > 0 + (− 3) = 2 2 s + 6s + 34 (s − (− 3)) + 25

L(sin 2t cos t) = L

FG 1 (sin 3t + sin t)IJ H2 K

14

LAPLACE AND FOURIER TRANSFORMS

=

LM N

OP Q

1 3 1 + 2 , s > 0. 2 2 s +9 s +1 ∴ By first shifting theorem, =

LM MN 1L = M 2 MN s

L( e 4t sin 2t cos t ) =

I JK

F GH

I JK

F GH

1 1 1 3 1 1 L(sin 3t) + L(sin t) = + ,s>0 2 2 2 2 2 2 s +3 2 s + 12

OP PQ

1 3 1 + ,s > 0 + 4 2 2 ( s − 4 ) + 9 ( s − 4)2 + 1

OP , s > 4 . − 8s + 25 s − 8s + 17 PQ LM∵ e −e F cos t IJ − e FG cos t IJ . cos t = e G (iv) sinh t cos t = H 2K H 2K 2 N F cos t IJ = 1 L (cos t) = 1 FG s IJ = s , s > 0 . Now LG H 2K 2 2 H s + 1 K 2 (s + 1) F cos t − e . cos t IJ ∴ L(sinh t cos t) = L G e . H 2 2 K F cos t IJ − L FG e . cos t IJ = L Ge . H 2 K H 2K t

3

1

+

2

−t

2

t

−t

2

t

2

sinh at =

OP Q

2

−t

t

−t

(By linearity)

=

s−1 s − (− 1) , s > 0 + 1, s > 0 + (− 1) − 2 2((s − 1) + 1) 2((s − (− 1)) 2 + 1)

=

s−1 s+1 − , s>1 2 2(s − 2s + 2) 2(s + 2s + 2)

=

e at − e − at 2

2

s2 − 2 s4 + 4

, s > 1.

(v) We have L(3 sinh 2t – 5 cosh 2t) = 3L(sinh 2t) – 5L(cosh 2t) = 3 . =

6 − 5s s2 − 4

2 2

s −4

− 5.

s 2

s −4

, s > |2|

, s > 2.

By first shifting theorem, L(e − t (3 sinh 2t − 5 cosh 2t)) =

that

1 − 5s 6 − 5 (s + 1) , s > 2 + (− 1) = 2 , s > 1. 2 (s + 1) − 4 s + 2s − 3

Example 2. If the Laplace transform of the function f(t) of t for t ≥ 0 is F(s), then show L[(sinh at) f(t)] =

Hence evaluate L(sinh 2t sin 3t).

1 [F(s − a) − F(s + a)]. 2

15

LAPLACE TRANSFORMS

LM e − e f (t)OP Q N 2 1 L1 = L M e f (t ) − e 2 2 N at

Sol. L[(sinh at) f (t)] = L

at

Also,



− at

f (t )

OP Q

=

1 1 L(e at f (t)) − L(e − at f (t)) 2 2

=

1 1 F(s − a) − F(s − (− a)) 2 2

=

1 [F(s − a) − F(s + a)]. 2

L(sin 3t ) =

3 2

s +3

LM N 3L = M 2 MN s

L (sinh 2t sin 3t) =

=

F∵ sinh at = e GH

− at

2

=

3

at

− e − at 2

I JK

(By linearity) (By first shifting theorem)

, s>0.

2

s +9

OP Q

1 3 3 − , s > 0 + 2, s > 0 + (− 2) 2 ( s − 2) 2 + 9 (s + 2) 2 + 9

1 2

− 4s + 13



12s 4

s + 10s 2 + 169

OP , s > 2 + 4s + 13 PQ 1

s

2

, s > 2.

1.7. UNIT STEP FUNCTION Let a ≥ 0. The function of t taking value 0 if t < a and 1 if t > a, is called a unit step function and is denoted by ua(t). ua (t ) =



RS0 T1

if t < a if t > a.

ua(t) 1

O

a

t

1.8. SECOND SHIFTING THEOREM If f(t) be a function of t for t ≥ 0 whose Laplace transform F(s) exists then for any constant a (≥ 0), the function f(t – a) ua(t) has the Laplace transform e–as F(s). Proof. t < a ⇒ f(t – a) ua(t) = f(t – a) . 0 = 0 and t > a ⇒ f(t – a) ua(t) = f(t – a) . 1 = f(t – a) ∴

f(t – a) ua(t) =

RS 0 Tf (t − a )

if t < a if t > a

16

LAPLACE AND FOURIER TRANSFORMS

Now

L(f (t − a) ua (t)) =

= =

z z z



0



0

e − st f ( t ) dt

...(1)

e − st f (t − a) ua (t) dt

a

e − st f (t − a ) u a (t ) dt +

0 a

0

e − st . 0 dt +

=0+ = e − sa



z

F( s ) = L( f (t )) =

Also,

=

e– as

L(f(t – a) ua(t)) =

e–as

z z



0



a

e − st f ( t − a ) u a ( t ) dt

a

e − st f (t − a) dt

e − s( a + u ) f ( u ) du , where u = t − a



0

z

z



e − su f ( u ) du = e − sa

z



0

e − st f ( t ) dt

F(s) F(s), where F(s) = L(f(t)).

(By replacing u by t) (By using (1))

Remark. The result of above theorem can be easily remembered as follows : L(f(t)) = F(s), then for any a ≥ 0, L(f(t – a) ua(t)) = e–as F(s).

If

Example 3. Find the Laplace transform of the function t2 u1(t). Sol. Let

g(t) = t2 u1(t).



g(t) = ((t – 1) + 1)2 u1(t) = f(t – 1) u1(t), where f(t) = (t + 1)2.

Now

L(f(t)) = L((t + 1)2) = L(t2 + 2t + 1) = L(t2) + 2L(t) + L(1) =

2! s3

+2

FG 1 ! IJ + 1 , ( s > 0) = 2 + 2s + s Hs K s s 2

∴ By second shifting theorem, L(g(t)) = L(f(t – 1)) u1(t)) = e–1.s L(f(t)) = e− s .

2 + 2s + s2

, s > 0. s3 Example 4. Find the Laplace transform of the following function of t :

RS T

0, 0 < t < π/2 g(t) = sin t, t > π/2.

Sol. Given function g(t) can be written as

0 < t < π/2 R| 0, S|cos FG t − π IJ , t > π . T H 2K 2 F πI g(t ) = cos G t − J u (t ) H 2K F πI g(t) = f G t − J u (t ) , where f(t) = cos t H 2K g(t) =

or

π/ 2



π/2

3

2

, s > 0.

17

LAPLACE TRANSFORMS

s s = ,s>0 s 2 + 12 s 2 + 1 ∴ By second shifting theorem,

Now

L( f (t)) = L(cos t) =

FG FG H H

L( g( t )) = L f t −

IJ K

IJ K

π u π / 2 (t ) 2

e − πs/ 2 s , s > 0. s2 + 1

= e − ( π / 2) s L( f (t)) =

Example 5. Find the Laplace transform of the function g(t) =

RS0, Tt,

0 0 1+ 1 s s2 s By second shifting theorem,

=

L(g(t)) = L(f(t – 2) u2(t)) = e–2s L(f(t)) 2s + 1 1 + 2s = e–2s . , (s > 0) = 2 2s , s > 0. 2 s e s 1.9. CHANGE OF SCALE PROPERTY Theorem. If f(t) be a function of t for t ≥ 0 whose Laplace transform F(s) exists, then for any positive constant ‘a’, the function f(at) has the Laplace transform

FG IJ H K

1 s F . a a

18

LAPLACE AND FOURIER TRANSFORMS

Proof. We have Now

F( s ) = L( f (t )) = L( f (at)) =

=

z



0

1 a

z



0

e − st f (t ) dt .

z

e − st f (at) dt =

z



0



0

e − sz / a f ( z ) 1 a

e − ( s/ a ) z f ( z ) dz =

FG IJ H K 1 F sI L(f(at)) = F G J . a H aK =



z



0

dz , where z = at a

e − ( s/ a )t f (t ) dt

(By replacing variable z by t)

1 s F . a a

Remark. The above theorem can be easily remembered as follows : L(f(t)) = F(s), then for any a (> 0), L(f(at)) =

If

FG IJ H K

1 s F . a a

Example 6. Find L(f(λt)) where λ is any positive constant and f(t) is a function of t for t ≥ 0 : (i) f(t) = tn , n ∈ N (ii) f(t) = et (iii) f(t) = sinh t (iv) f(t) = cosh t (v) f(t) = sin t (vi) f(t) = cos t. n Sol. (i) f(t) = t , n ∈ N ∴

L ( f (t)) = L (t n ) =

n! sn + 1

,s>0

1 n! s . , >0 λ (s/λ) n + 1 λ

By change of scale property, L( f (λt)) =

λn n ! , s > 0. sn + 1

=

(∵ λ > 0)

(ii) f(t) = et

L( f (t)) = L(e t ) =



By change of scale property, L( f (λt)) =

=

1 , s>1 s−1

1 1 . , λ (s/λ) − 1

s >1 λ

1 , s > λ. s−λ

(iii) f(t) = sinh t ∴ L( f ( t )) = L(sinh t ) =

1 2

s − (1)

2

, ( s > |1|) =

By change of scale property, L( f (λt)) = =

1 2

s −1

, s >1

1 1 s . , >1 2 λ (s/λ) − 1 λ λ , s > λ. s − λ2 2

19

LAPLACE TRANSFORMS

(iv) f(t) = cosh t ∴

L( f (t)) = L(cosh t) =

s ,s>1 s −1 2

1 s/λ s . , >1 2 λ (s/λ) − 1 λ

By change of scale property, L( f (λt)) = =

s s 2 − λ2

, s > λ.

(v) f(t) = sin t ∴

L( f (t)) = L(sin t) =

1 ,s>0 s +1 2

By change of scalar property, L( f (λt)) = =

1 1 s . , >0 λ (s/λ) 2 + 1 λ λ , s > 0. s + λ2 2

(vi) f(t) = cos t ∴

L( f (t)) = L(cos t) =

s ,s>0 s +1 2

By change of scalar property, L( f (λt)) = =

1 s/λ s . >0 , λ (s/λ) 2 + 1 λ

s , s > 0. s + λ2 2

WORKING RULES FOR SOLVING PROBLEMS Rule I.

If L(f(t)) = F(s), s > k then for any a, L(eat f(t)) = F(s – a), s > k + a.

Rule II. For a ≥ 0, the unit step function u(t – a) is defined as u(t − a ) =

RS0 if t < a T1 if t > a.

Rule III. If L(f(t)) = F(s), then for any a ≥ 0, L(f(t – a) u(t – a)) = e–as L(f(t)). Rule IV. If L(f(t)) = F(s), then for any a(> 0), L( f (at)) =

FG IJ H K

1 s F . a a

TEST YOUR KNOWLEDGE 1. Find the Laplace transform of the following functions of t for t ≥ 0 : (i) et tk, k > – 1 (iv) e3t cosh 3t

(ii) e2t t4

(iii) et sinh t

(v) e2t sin t

(vi) e2t cos 2t.

20

LAPLACE AND FOURIER TRANSFORMS

2. Find the Laplace transform of the following functions of t for t ≥ 0 : (ii) e– 6t t7 (iii) e–2t sinh 6t (i) e– 4t t3/2 – 5t –5t cosh 3t (v) e sin 8t (vi) e–3t cos 4t. (iv) e 3. Find the Laplace transform of the following functions of t for t ≥ 0 : (ii) (t + 2)2 et (i) e3t sin2 t 2 (iii) sinh 3t cos t (iv) cosh at sin at. (v) e–2t sin t cos 3t. 4. Find the Laplace transform of the following functions of t for t ≥ 0 : (ii) 5e2t sinh 2t (i) e–t sin2 t (iii) e– at sinh bt (iv) cosh 4t sin 6t. 5. If the Laplace transform of the function f(t) of t for t ≥ 0 is F(s), then show that

L[(cosh at) f (t)] =

1 [F(s − a) + F(s + a)]. 2

Hence evaluate L(cosh 3t cos 2t). 1 3 6 6 , s > 0. + + + s (s + 1)2 (s + 2)3 (s + 3)4 Find the Laplace transform of the following functions : (i) (t – 1) u1(t) (ii) (5 cos t) uπ(t).

6. Show that L {(1 + te− t )3 } = 7.

RS0, 0 < t < π/2 Tcos t, t > π/2 . Rcos (t − 2π/3), t > 2π/3 Find the Laplace transform of the function : g(t) = S 0 < t < 2π/3 . T0, Find the Laplace transform of the function g(t) = RS0, 0 < t < 5 by using (i) definition (ii) second T t, t > 5

8. Find the Laplace transform of the function : g(t) = 9. 10.

shifting theorem. Verify that the results are same.

Answers 1. (i) (iv) 2. (i) (iv) 3. (i)

(iii) (v)

Γ ( k + 1) ( s − 1)

k +1

s 2

s − 6s

3 π 4 (s + 4)

, s >1

, s>6

5/ 2

(v)

,s>–4

s+5

2

s + 10s + 16

(ii)

, s>−2

2 2

(s − 3) ( s − 6 s + 13)

LM MN

(ii)

(v)

, s > 3 (ii)

24 5

( s − 2)

, s>2

1 2

, s>2

(vi)

, s>−6

(iii)

s − 4s + 5 5040 ( s + 6 )8

8 2

s + 10s + 89 2 ( s − 1)3

, s>−5

2



1 s2 + 4 s + s

,s>−2

(vi)

1 2

s − 2s 2

, s>2

s−2

s − 4s + 8

, s>2

6 s2 + 4s − 32 2

s+3

s + 6s + 25

, s>4 , s>−3

[(2 s2 − 2 s + 1)] , s > 1

OP PQ

1 s2 − 6s + 11 s2 + 6s + 11 , s>3 + 2 ( s − 3) ( s2 − 6s + 13) ( s + 3) ( s2 + 6s + 13) s2 + 4 s + 20

(iii)

(iv)

a(s2 + 2 a2 ) s4 + 4 a 4

, s > | a|

21

LAPLACE TRANSFORMS

4. (i)

(iii) 5.

8.

2

b s2 + 2as + a 2 − b2

s3 − 5s

s4 − 10s2 + 169



, s > −1

( s + 1) ( s2 + 2s + 5 )

e− πs /2 2

s +1

(ii)

, s > |b | − a

(iv)

,s>0

9.

FG H

8. Use cos t = – sin t −

π 2

IJ K

s − 4s 4

s2

6 (s2 + 52)

,s>0

e− 2 πs / 3 s 2

, s>4

, s>4

s + 40 s2 + 2704

e− s

7. (i)

, s>3

10 2

s +1

(ii)

,s>0

− 5se − πs s2 + 1

10.

,s>0

(5s + 1) e − 5s s2

,s>0

Hint .

1.10. PIECEWISE CONTINUOUS FUNCTION A function f(t) is called piecewise continuous (or sectionally continuous) on a finite interval [a, b] if this interval can be divided into a finite number of subintervals such that (i) f(t) is continuous in the interior of each of these sub-intervals and (ii) f(t) approaches a finite limit as t approaches either endpoint of each of the sub-interval from its interior. Thus a piecewise continuous function has finitely many jumps as discontinuities. Clearly every continuous function is a piecewise continuous function. Also, if f(t) is a piecewise continuous function on [a, b], then f(x) is integrable on [a, b]. For example,

R2, let f (t ) = S T5,

f(t)

O

0 k.

22

LAPLACE AND FOURIER TRANSFORMS

Proof. Since f(t) is piecewise continuous on every finite interval in the range t ≥ 0, f(t) is piece wise continuous on the finite interval [0, T] for T > 0. Also e–st is continuous on the finite interval [0, T] for T > 0 ∴ e–st f(t) is piecewise continuous on [0, T] for T > 0. ∴ e–st f(t) is integrable on [0, T] for T > 0.

z z z



T

0

Now,



0



and

0

e − st f (t ) dt exists for all T > 0.

T→∞

e − st f ( t ) dt ≤





z



0

z



0

e

− st

z



0

T

e − st f (t ) dt

0

. Me

kt

LM Ne

dt = M

M lim k − s T→∞

=

M for s > k . s−k

L( f ) =

z

e − st f (t ) dt =

=

e − st f (t ) dt ≤



z

e − st f ( t ) dt = lim

1 ( s − k )T

z



0



0

e

e − st |f ( t )|dt

( k − s) t

OP Q

−1 =

L e OP dt = M lim M N k−s Q ( k − s) t

T→∞

T

0

M [ 0 − 1] for s > k k−s

M for s > k s−k

z



0

e − st f (t ) dt exists for s > k.

1.12. LAPLACE TRANSFORMS OF DERIVATIVES Theorem 1. (Laplace transform of the first derivative). Let f(t) be a real function such that (i) f(t) is continuous for all t ≥ 0 (ii) there exists constants k and M such that |f(t)| ≤ Mekt for t ≥ 0 (iii) f ′(t) is piecewise continuous on every finite interval in the range t ≥ 0, then the Laplace transform of the derivative f ′(t) exists and L(f ′) = sL(f) – f(0) for s > k. Proof. By definition

L( f ′) = lim

T→∞

z

T

0

e − st f ′(t) dt , provided this limit has some finite value.

Let T be any positive number. Using (iii), f ′(t) is piecewise continuous on [0, T]. ∴ f ′(t) has at most a finite number of discontinuities, say t1, t2, ......, tn, where 0 < t1 < t2 < ...... < tn ≤ T ∴

z

T

0

e − st f ′(t ) dt =

z

t1

0

e − st f ′(t ) dt + ...... +

z

T

tn

e − st f ′(t ) dt

23

LAPLACE TRANSFORMS

The integrand of each of the integrals on the right is continuous. ∴ By using integration by parts, we have

z

T

0

e − st f ′(t) dt = e − st f (t)

t1 0

z



T

0

z

T

z

t1

T→∞ 0

tn

+s

z

T

tn

e − st f (t) dt

+ e − sT f (T) − e − stn f (tn ) + s e − st f ( t ) dt + ...... +

0

T

e − st f (t) dt + ......

0

z

T

z

T

tn

e − st f (t ) dt

OP Q

z

T

tn

(∵ T→∞

T→∞

e − sT f (T) ≤ e − sT | f (T)|≤ e − sT . Me kT =

z

f is continuous for t ≥ 0) T

0

e − st f (t) dt

M e (s – k)T

lim e − sT f (T) = 0 for s > k



T→∞

lim s

T→∞



z

lim

T

0

T→∞

e − st f (t) dt = sL( f )

z

T

0

e − st f ′( t ) dt = 0 − f ( 0) + sL( f ) .

∴ L(f ′) = sL(f) – f(0) for s > k.

ILLUSTRATIVE EXAMPLES Example 1. Using L( t n ) = Sol. Let ∴ We have ∴ ∴

n!

s n +1 f(t) = tn + 1.

, find the value of L(tn + 1).

f ′(t) = (n + 1) tn. L(f ′) = sL(f) – f(0). L((n + 1) tn) = sL(tn+1) – (0)n+1 ⇒ (n + 1) L(tn) = sL(tn+1) L( tn + 1 ) =

n +1 n +1 n! L( t n ) = . n +1 s s s

=

(n + 1) ! sn + 2

.

Example 2. Find the Laplace transform of the function sin2 at cos at, t ≥ 0. Sol. Let ∴

e − st f (t) dt

e − st f (t ) dt

0

e − st f ′ (t) dt = lim e − sT f (T) − f (0) + lim s

Now using (ii),

∴ (1)

0

− st f (t) e − st f (t) dt + ...... + e

e − st f ′ (t) dt = e − sT f ( T ) − f ( 0) + s

lim



LM N

t1

t1

= e − st1 f (t1 ) − f (0) + s

= e − sT f ( T ) − f ( 0) + s

Also

z z

+s

f(t) = sin3 at. f ′(t) = 3 sin2 at . a cos at = 3a cos at sin2 at

...(1)

24

LAPLACE AND FOURIER TRANSFORMS

We have L(f ′) = sL(f) – f(0). ∴ L(3a cos at sin2 at) = sL(sin3 at) – sin3 0 ⇒ 3a L(cos at sin2 at) = sL(sin3 at) Now, ∴ ∴

F GH

=

F GH

I JK

L(cos at sin2 at) =

F GH

I JK

I JK

3a 1 1 6a3 = − 4 s2 + a2 s2 + 9a2 ( s 2 + a 2 )(s 2 + 9 a 2 )

∴ (1) ⇒ 3a L(cos at sin2 at) = s . ∴

...(1)

3 1 sin 3x = 3 sin x – 4 sin3 x ⇒ sin3 x = sin x − sin 3x 4 4 3 1 sin3 at = sin at – sin 3at 4 4 3 1 L(sin3 at) = L(sin at) – L(sin 3at) 4 4 3 a 1 3a − = 2 2 2 4 s +a 4 s + (3a) 2

6a3 –0 (s + a )(s 2 + 9 a 2 ) 2

2

1 6a3 s 2a 2 s . 2 = 2 . 2 2 2 2 3a (s + a )(s + 9 a ) (s + a )(s 2 + 9a 2 )

Alternative method 1 . sin at . 2 sin at cos at 2 1 1 = sin at sin 2at = . 2 sin 2at sin at 2 4 1 = (cos at − cos 3at) . 4 1 2 L(sin at cos at) = L (cos at − cos 3at) 4 sin 2 at cos at =



FG H

=

IJ K

LM MN

1 s s 1 − [L(cos at) − L(cos 3at)] = 4 s 2 + a 2 s 2 + 9a 2 4

LM N

OP Q

OP PQ

2a 2s s 8a2 = . 4 (s 2 + a 2 ) ( s 2 + 9 a 2 ) (s2 + a 2 ) (s2 + 9a 2 ) Theorem 2. (Laplace transform of the derivative of any order n). Let f(t) be a real function such that

=

(i) f(t), f ′(t), f ″(t), ......, f(n–1)(t) are continuous for all t ≥ 0 (ii) there exists constants k and M such that | f(t) | ≤ Mekt for t ≥ 0 and | f(i) (t)|≤ Mekt for t ≥ 0 and i = 1, 2, ......, n – 1 (iii) f(n)(t) is piece wise continuous on every finite interval in the range t ≥ 0 then the Laplace transform of f(n)(t) exists and

L(f (n) ) = sn L(f) − sn − 1 f(0) − sn − 2 f ′ (0) − ...... − f (n − 1) (0) for s > k.

25

LAPLACE TRANSFORMS

or

Proof. Using (i), f ′(t) is piece wise continuous on every finite interval in the range t ≥ 0. ∴ By theorem 1, L(f ′) = s L(f) – f(0), s > k By applying theorem 1 to the function f ′(t), we get L(f ″) = s L(f ′) – f ′(0). ∴ L(f ″) = s [s L(f) – f(0)] – f ′(0) L(f ″) = s2 L(f) – s f(0) – f ′(0)

Let the result be true for f (m)(t), the mth derivative of f(t). ∴ L(f (m)) = sm L(f) – sm–1 f(0) – ...... f(m–1)(0) ...(1) Now (i) f (m)(t) is continuous for all t ≥ 0. (ii) | f (m)(t) | ≤ M ekt for t ≥ 0. (iii) By given condition (i), f (m+1)(t) is piece wise continuous on every finite interval in the range t ≥ 0. ∴ By theorem 1, the Laplace transform of the derivative f (m+1)(t) exists and L(f (m+1)) = s L(f (m)) – f (m)(0). ...(2) Using (1), we get L(f (m+1)) = s [sm L(f) – sm–1 f(0) – ...... – f (m–1)(0)] – f (m)(0) or L(f (m+1)) = sm+1 L(f) – sm f(0) – sm–1 f(0) – ...... – f m(0) ∴ By P.M.I., we have L(f(n)) = sn L(f) – sn–1 f(0) – sn–2 f ′(0) – ...... – f(n–1)(0) for s > k. In particular n=1 ⇒ L(f ′) = sL(f) – f(0) n=2 ⇒ L(f ″) = s2L(f) – sf(0) – f ′(0) n=3 ⇒ L(f ″′) = s3L(f) – s2f(0) – sf ′(0) – f ″(0). 6

Example 3. Given L(t3) =

s4 f(t) = t6.

Sol. Let

, find the value of L(t 6).

f ′(t) = 6t5, f ″(t) = 30t4 and f ″′(t) = 120t3.



L(f ″′) = s3L(f) – s2 f(0) – sf ′(0) – f ″(0).

We have ∴

L(120t3) = s3L(t6) – s2 . (0)6 – s . 6(0)5 – 30(0)4



120L(t3) = s3L(t6) – 0 – 0 – 0

⇒ ⇒

120

FG 6 IJ = s L(t ) Hs K 1 F 720 I 720 L(t ) = G J= s . s H s K 3

4

6

3

6

4

7

Example 4. Find the Laplace transform of the function cos at, t ≥ 0. Sol. Let ∴ We have

f(t) = cos at, t ≥ 0. f ′(t) = – a sin at and f ″(t) = – a2 cos at. L(f ″) = s2 L(f) – sf(0) – f ′(0).

(Given)

26

LAPLACE AND FOURIER TRANSFORMS



L(– a2 cos at) = s2 L(cos at) – s cos 0 – (– a sin 0)



– a2 L(cos at) = s2 L(cos at) – s – 0 (s2 + a2) L(cos at) = s



L(cos at) =



s . s2 + a 2

Example 5. Find the Laplace transform of the function t sin bt. Sol. Let

f(t) = t sin bt. f ′(t) = t(b cos bt) + 1 . sin bt = bt cos bt + sin bt



f ″(t) = b[t(– b sin bt) + 1 . cos bt] + b cos bt = – b2t sin bt + 2b cos bt

and

L(f ″) = s2L(f) – sf (0) – f ′(0).

We have ∴

L(– b2t sin bt + 2b cos bt) = s2L(t sin bt) – s(0 . sin 0) – (b . 0 . cos 0 + sin 0)



– b2L(t sin bt) + 2bL(cos bt) = s2L(t sin bt) – 0 – (0 + 0)



(s2 + b2) L(t sin bt) = 2bL(cos bt) = 2b L(t sin bt) =



1 2

s +b

2

F 2bs GH s + b 2

F s I GH s + b JK I = 2bs JK (s + b ) 2

2

2

2

2 2

WORKING RULES FOR SOLVING PROBLEMS Rule I. L(f ′) = sL(f) – f(0). Rule II. L(f (n)) = sn L(f) – sn–1 f(0) – sn–2 f ′(0) – ...... – f Rule III. (i) L(f ″) = s2L(f) – sf(0) – f ′(0) (ii) L(f ″′) = s3L(f) – s2f(0) – sf ′(0) – f ″(0).

(n–1)(0).

TEST YOUR KNOWLEDGE 1 , find L(t). s a (i) Using L (sinh at ) = , find L(cosh at). s2 − a 2

1. Using L (1) = 2.

s , find L(sinh at). s2 − a 2 b , find L(cos bt). 3. (i) Using L (sin bt) = 2 s + b2 s , find L(sin bt). (ii) Using L (cos bt) = s2 + b2 4. Find the Laplace transform of the function sin bt cos2 bt, t ≥ 0.

(ii) Using L (cosh at ) =

1 , find L(t2). s 2 Given L (t2 ) = 3 , find the value of L(t5). s

5. Using L (1) = 6.

.

27

LAPLACE TRANSFORMS

7. Find the Laplace transform of the following functions by using the result of Laplace transform of second derivative of a function : (ii) sinh at (iii) cosh at (iv) sin at. (i) eat 8. Find the Laplace transform of the function t cos at.

Answers 1. 4.

1 s

2. (i)

2

2

s 2

s −a

(ii)

2

b (s2 + 3b2 )

5.

(s + b2 ) (s2 + 9b2 )

(ii)

a

(iii)

s2 − a 2

s

(iv)

s2 − a 2

a 2

s −a

3. (i)

2

2

6.

3

s

a 2

s +a

8.

2

s

(ii)

2

s + b2

120

b s 2 + b2

7. (i)

s6 s2 − a 2

( s 2 + a 2 )2

1 s−a

.

Hint 8. Let f(t) = t cos at. Use L(f ″) = s2L(f) – sf(0) – f ′(0).

1.13. LAPLACE TRANSFORMS OF INTEGRALS Theorem. Let f(t) be a real function such that (i) f(t) is piece wise continuous on every finite interval in the range t ≥ 0 (ii) there exists constants k and M such that |f(t)| ≤ Mekt for t ≥ 0, then the Laplace transform of the integral L

FG H

z

t

0

z

t

0

f(T) dT exists and

f(T) dT

IJ K

1 L(f(t)) for s > k. s

=

Proof. If k in (ii) is negative, then – k > 0 and M ≤ Me − kt . e − kt ∴ We can assume that k is positive.

(ii) ⇒ |f (t )|≤ Me kt =

g (t) =

Let

∴ g(t) is continuous for all t ≥ 0. Now | g(t )|=

z

t

0

f ( T ) dT ≤

z

t

0

z

t

0

f (T) dT,

|f ( T )|dT ≤

(∵

e–kt > 1)

t≥0

z

t

0

|Me kT | dT = M

z

t

0

e kT dT

M kt M kt ( e − 1) ≤ e . ( ∵ k > 0 ⇒ ekt > 1) k k M kt ∴ | g(t)|≤ e for t ≥ 0 k Also g′(t) = f(t) except for points at which f(t) is discontinuous. ∴ g′(t) is piecewise continuous on every finite interval in the range t ≥ 0. =

28

LAPLACE AND FOURIER TRANSFORMS

∴ The Laplace transform of g′(t) exists and L(g′) = sL(g) – g(0) for s > k ⇒

L(g′) = sL(g) for s > k



L( g ) =



Rule

L

FG H

L

FG H

z

t

0

z

IJ K

FG∵ H

g (0) =

z

0

0

IJ K

f (T ) dT = 0

1 L( g ′ ) s

1 L(f(t)) for s > k. s

f(T) dT =

WORKING RULE FOR SOLVING PROBLEMS t

0

IJ K

f ( T ) dT =

1 L( f (t)) . s

ILLUSTRATIVE EXAMPLES 1 , find L(t 2 ). s 1 L(f) = L(1) = s

Example 1. Using L(1) = Sol. Let f(t) = 1. We have

L

FG H



z



IJ 1 L( f ) . K s F 1 dTIJ = 1 FG 1IJ LG K s H sK H F I 1 ⇒ L(t) = 1 . L GT J = s H K s t

0

f ( T ) dT =

z

t

0



t

0

Let

L

FG H

z



2

1 s2

IJ 1 L( g ) K s F T dTIJ = 1 . 1 LG K s s H FT I 1 Ft I 1 LG GH 2 JJK = s ⇒ L GH 2 JK = s t

0

g ( T ) dT =

z

t

2

0

2



2

L(g) = L (t) =

g(t) = t. ∴

We have ∴

(Given)

t

2

3

3

0

1 1 L (t 2 ) = 3 2 s

⇒ L (t 2 ) =

2 s3

.

29

LAPLACE TRANSFORMS

Example 2. Using L (sin bt) = Sol. Let f(t) = sin bt. ∴ We have





z

b 2

s + b2

, find L (cos bt).

L( f ) = L (sin bt) =

IJ 1 L( f ) . K s F sin bT dTIJ = 1 . b LG H K s s +b F 1 cos bT IJ = b LG − H b b K s (s + b ) L

FG H

z

t

0

b . s + b2

(Given)

2

f ( T ) dT =

t

0

2

2



2

2



F GH

L −

cos bT b

t 0

I= b JK s (s + b ) 2

2

b 1 1 L(1) − L (cos bt ) = 2 b b s ( s + b2 )

1 1 1 b s L(cos bt) = . − = b b s s ( s 2 + b2 ) b (s 2 + b 2 )



L(cos bt) =



Example 3. Using L(1) =

s . s + b2 2

1 , find the Laplace transform of the function cos bt. s

Sol. Let f(t) = cos bt. We have

⇒ ⇒

z

IJ 1 L( f ) . K s F cos bT dTIJ = 1 L(cos bt) LG K s H F sin bt − 0IJ = 1 L(cos bt) LG H b K s L

FG H

z

0



f ( T ) dT =

t

0

Let We have

t

z

F sin bT I = 1 L(cos bt) GH b JK s b Lbsin btg = L(cos bt) s t

L

0



g(t) = sin bt.

z

t

0

g( T ) dT =

t

0

t

⇒ ⇒ ⇒

...(1)

IJ 1 L( g ) . K s F sin bT dTIJ = 1 L(sin bt) = 1 . FG b L(cos bt)IJ (By using (1)) LG K H K s s Hs F cos bT I = b L(cos bt) F cos bt + 1IJ = b L(cos bt) L G− ⇒ L G− J H b bK s H b K s 1 1 F b 1I b 1 1 ⇒ . =G + J L(cos bt) L(cos bt) − L (cos bt) + L(1) = b s Hs bK b b s L

FG H



0

2

2

2

2

1 b2 + s 2 = L(cos bt) bs bs 2



L(cos bt) =

s s2 + b2

.

30

LAPLACE AND FOURIER TRANSFORMS

TEST YOUR KNOWLEDGE 1. Using L(t ) =

1 s2

, find the values of L(t2) and L(t3).

2. Using L(cos at ) = 3. Using L(1) =

s 2

s + a2

, find L(3 sin 8t).

1 a and L(sinh at ) = , find L(4 cosh 6t). s s2 − a 2

4. Using L(cosh at ) =

s 2

s − a2

, find L(2 sinh pt).

5. Using L(1) =

1 1 , find the Laplace transform of the function sin πt . 5 s

6. Using L(1) =

1 , find the Laplace transform of the function b cosh pt. s

7. Using L(1) =

1 , find the Laplace transform of the function b sinh at. s

8.

Evaluate L

F GH

z

t

0

I JK

e− x cos x dx .

Answers 1.

5.

2 3

s

,

6 s

4

π

5 ( s2 + π 2 )

24

2.

s2 + 64

6.

s 2 − p2

bs

4s

3.

s2 − 36

7.

s2 − a2

ab

2p

4.

s 2 − p2

8.

s( s2 + 2 s + 2)

s+1

.

1.14. DIFFERENTIATION OF LAPLACE TRANSFORMS Theorem. Let f(t) be a real function such that (i) f(t) is piecewise continuous on every finite interval in the range t ≥ 0 (ii) there exists constants k and M such that | f(t) | ≤ Mekt for t ≥ 0, then F′(s) = – L(t f(t)), where F(s) = L(f(t)). Note. The proof of this theorem is beyond the scope of this book. Remark. The result of the above theorem can be easily remembered as follows : If L(f(t)) = F(s), then L(t f(t)) = – F′′(s).

Corollary. We have

L(t f(t)) = – F′(s).

Applying this result successively, we get L(t2 f(t)) = (– 1)2 F″(s) L(t3 f(t)) = (– 1)3 F″′(s) ................................. ∴

L(tn f(t)) = (– 1)n F(n)(s).

31

LAPLACE TRANSFORMS

ILLUSTRATIVE EXAMPLES Example 1. Find the Laplace transform of the following functions of t for t ≥ 0 : (i) t eat (ii) t2 cosh πt (iii) t2 sin 3t. 1 Sol. (i) We have L(eat) = s−a ∴ By differentiability of Laplace transform,

(ii) We have

s

L(cosh πt) =

IJ K

FG H

1 d 1 = – (– (s – a)–2) = . (s − a) 2 ds s − a

L(t . eat) = –

2

s − π2 ∴ By differentiability of Laplace transform,

L(t . cosh πt) = –

F GH

I = – LM (s JK N

F GH

I JK

d s 2 ds s − π 2

Again by using the same formula, L(t . t cosh πt) = – =–

d s2 + π 2 ds (s 2 − π 2 ) 2

LM (s N

2

L(t2 cosh πt) =

(s 2 − π 2 ) 2

2

2 4

(s − π )

F GH

L(t2 cosh πt) = L t 2 .

2

e πt + e − πt 2

I = 1 L(e JK 2

πt t2

LM N

=

2

We have

L(sin 3t) =

2 2

OP = 2s(s + 3π ) . Q (s − π )

1 ( s − π) 3

+

1 ( s + π) 3

2s 3 + 6 sπ 2 2

2 3

(s − π ) 3

=

=

2! s3

(s 2 − π 2 ) 3 3

L(t · sin 3t) = –

F GH

I JK

2

2 3

OP Q

and first shifting theorem)

( s + π) 3 + ( s − π) 3 (s 2 − π 2 ) 3

2s(s 2 + 3π 2 )

2

2

+ e–πt t2)

.

= s 2 + (3) 2 s 2 + 9 By differentiability of Laplace transform,

(iii)

2

1 1 2! 2! + [L(eπt t2) + L(e–πt t2)] = 3 2 2 ( s − π) (s − (− π)) 3

(Using L(t2) = =

OP = s + π Q (s − π ) 2

2s(s + 3π ) . (s 2 − π 2 )3

Alternative method

=

− π 2 ) . 1 − s . 2s

− π 2 ) 2 . 2s − (s 2 + π 2 ) . 2(s 2 − π 2 ) . 2s

2



2

6s d 3 = – 3(– (s2 + 9)–2 · 2s) = 2 2 ds s + 9 (s + 9) 2

32

LAPLACE AND FOURIER TRANSFORMS

Again by using the same formula,

F GH

d 6s ds (s 2 + 9) 2

L(t . t sin 3t) = –

LM (s N

=–6

2

I JK

+ 9) 2 . 1 − s . 2 (s 2 + 9) . 2s (s 2 + 9) 4

OP = 18(s − 3) Q (s + 9) 2

2

3

18(s 2 − 3) . (s 2 + 9)3 Example 2. Find the Laplace transform of the function 3te5t sin 7t for t ≥ 0. 7 Sol. We have L(sin 7t) = 2 . s + 49 By first shifting theorem,

L(t2 sin 3t) =



L(e 5t sin 7t) =

7 7 = 2 . (s − 5) 2 + 49 s − 10s + 74

By differentiability of Laplace transform,

I JK

F GH

7 . ( 2s − 10 ) 14 ( s − 5 ) d 7 = 2 = 2 2 2 ds s − 10 s + 74 ( s − 10s + 74 ) ( s − 10s + 74 ) 2 By linearity of Laplace transform, 42 (s − 5) . L(3te5t sin 7t) = 3L(te 5t sin 7t) = 2 (s − 10s + 74) 2 L(te5t sin 7t) = −

Example 3. Prove that L

LM 1 N 2a

OP Q

(sin at − at cos at) =

3

a . s + a2 s Also L(cos at) = 2 s + a2 By differentiability of Laplace transform,

Sol. We have L(sin at) =

=– L

LM 1 N 2a

3

F GH

I JK

d s . 2 ds s + a 2

LM (s N

2

= =

OP Q

+ a 2 ) . 1 − s . 2s s2 − a2 = (s 2 + a 2 ) 2 (s 2 + a 2 ) 2

(sin at − at cos at)

=

...(1)

2

L (t cos at ) = −

Now

1 . (s + a 2 ) 2 2

OP Q

1 [L(sin at) – aL(t cos at)] 2a3

1 2a3

1 2a 3

LM a − a F s − a I OP MN s + a GH (s + a ) JK PQ LM as + a − as + a OP = N (s + a ) Q (s 2

2

2

2

2

2

3

2

2

2 2

2 2

3

2

1 + a 2 )2

.

33

LAPLACE TRANSFORMS

Example 4. Evaluate Sol. We have

z



0

te −2t sin t dt .

1

L(sin t) =

2

s +1

L(t sin t) = –



z







z



0

1

=

2

s +1

.

I JK

F GH

2s d 1 = – (– (s2 + 1)2 . 2s) = 2 2 (s + 1) 2 ds s + 1

2s (s + 1) 2

e − st . t sin t dt =

0

2

2

e −2t t sin t dt =

2s (s + 1) 2

=

2

s= 2

2(2) 4 = . ((2) 2 + 1) 2 25

WORKING RULES FOR SOLVING PROBLEMS Rule I. If L(f(t)) = F(s), then L(t f(t)) = – F′(s). Rule II. If L(f(t)) = F(s), then L(t2 f(t)) = (– 1)2 F″(s) L(t3 f(t)) = (– 1)3 F″′(s) ................................. L(tn f(t)) = (– 1)n F(n)(s).

TEST YOUR KNOWLEDGE 1. Find the Laplace transform of the following functions of t for t ≥ 0 : (i) t e5t

(ii) 4t cos 3t

(iv) 2t sinh 2t

(iii) 6t sin 5t

(v) 3t cosh 4t.

2 , find the Laplace transform of t3 and t4. s3 3. Find the Laplace transform of the following functions of t for t ≥ 0 :

2. Using L (t2 ) =

(i) 2t2 e3t

(ii) t2 cos 4t

(iv) 6t2 sinh 3t

(iii) t2 sin at

(v) 5t2 cosh 7t.

4. Find the Laplace transform of the following functions of t for t ≥ 0 : (i) t3 e2t (iv)

5t3

(ii) 3t3 cos πt

sinh 4t

(v)

6t3

(iii) 4t3 sin 4t

cosh (t/2).

5. Find the Laplace transform of the following functions of t for t ≥ 0 : (i) t e2t cos 4t (iii) t

e3t

(ii) t e–t sin 3t

sinh 3t

6. Prove that L 7. Evaluate

z



0

(iv) 3t e2t cosh 7t

FG 1 (sin at + at cos at)IJ = H 2a K (s t3 e− t sin t dt .

s2 2

+ a2 ) 2

(v) te–2t sin 3t .

34

LAPLACE AND FOURIER TRANSFORMS

Answers 1. (i)

(iv)

3. (i)

(iv)

(ii)

(v)

(iii)

1 ( s − 5)

(ii)

2

8s 2

(s − 4)

(v)

2

4

(ii)

3

( s − 3)

108 ( s2 + 3) 2

(v)

3

( s − 9)

18(π 4 − 6 π 2 s2 − s4 ) 2

(iii)

2 4

(s + π )

576 (16 s4 + 24 s2 + 1) 2

(4 s − 1) ( s − 6s )

2

(s + 9)

4

(iv)

2

(iii)

2

3 ( s2 + 16 )

2.

( s2 − 16 )2

2s ( s2 − 48)

(iii)

( s2 + 16)3

10 s(s2 + 147) 2

(s − 49) 2

( s + 16)

60s ( s2 + 25)2

6 s

4

(iv)

4

s2 − 4s − 12

(ii)

( s2 − 4 s + 20)2

3 ( s2 − 4s + 53)

(v)

( s2 − 4s − 45)2

,

24 s5

2 a (3s2 − a2 ) ( s 2 + a2 ) 3 6

4. (i)

3

384s ( s2 − 16)

5. (i)

6 ( s − 3) 2

4(s2 − 9)

(s − 2)4

480s ( s2 + 16) ( s2 − 16)4 6 ( s + 1)

( s2 + 2 s + 10)2

2

6 ( s + 2)

( s + 4 s + 13)2

7. 0.

1.15. INTEGRATION OF LAPLACE TRANSFORMS Theorem. Let f(t) be a real function such that (i) f(t) is piecewise continuous on every finite interval in the range t ≥ 0 (ii) there exists constants k and M such that |f(t)| ≤ Mekt for t ≥ 0 (iii) lim+ t→0

f ( t) exists finitely, then t

z



s

F (S) dS = L

FG f (t) IJ , s > k where F (s) = L( f (t)). H tK

Note. The proof of this theorem is beyond the scope of this book. Remark. The result of the above theorem can be easily remembered as follows : If

L(f(t)) = F(s), then

L

FG f (t) IJ = H tK

z



s

F(S) dS .

ILLUSTRATIVE EXAMPLES 2 (1 − cos at), for t ≥ 0. t Sol. We have L(2(1 – cos at)) = 2L(1 – cos at) = 2[L(1) – L(cos at)]

Example 1. Find the Laplace transform of the function

35

LAPLACE TRANSFORMS

LM 1 − Ns s

=2

2

OP F Q GH

2 2s s = − 2 2 s s + a2 +a

I*. JK

By integration of Laplace transform, L

z

FG 2(1 − cos at) IJ = LM 2 − 2S OP dS H t K NS S + a Q L O S = M2 log S − log (S + a ) P = log S +a MN PQ F S I − log s = lim G log H S + a JK s + a s 1 I – log = log F lim GG 1 + a JJ s +a H S K F 1 IJ − log s = 0 + log s = log G H 1 + 0K s + a s +a s +a 2 L FG (1 − cos at)IJ = log . Ht K s ∞

2

s

2



2



2

2

2

2

s

s

2

2

2

S→∞

2

2

2

2

2

2

S→∞

2

2

2

2

2

2



2

2

2

= log

s2 + a2 . s2

2

2

Example 2. Find the Laplace transformation of the function

2 ( e t − cos t ) . t

Sol. We have L(2(et – cos t)) = 2L(et – cos t)

LM 1 − Ns − 1 s

=2

OP = 2 − 2s . + 1Q s − 1 s + 1

s 2

By integration of Laplace transform,

L

F 2 (e GH

t

I JK

− cos t) = t

z FGH ∞

s

I JK

2 2S dS − S − 1 S2 + 1

L = M2 log (S − 1) – log (S MN

2

O + 1)P PQ



s

(S − 1) 2 = log 2 S +1

= lim log

(S − 1) 2 (s − 1) 2 − log S2 + 1 s2 + 1

= log lim

(1 − 1/S) 2 (s − 1) 2 − log 1 + 1/S 2 s2 + 1

S→∞

S→∞

*Why this step. We have not simplified step.

2



s

2 2s because we shall be integrating it in the next − s s2 + a2

36

LAPLACE AND FOURIER TRANSFORMS

s2 + 1 (1 − 0) 2 ( s − 1) 2 − log 2 = 0 + log (s − 1)2 1+0 s +1

= log



L

F 2 (e GH

t

I JK

s2 + 1 − cos t) = log . t (s − 1)2

Example 3. Find the Laplace transform of the function Sol. We have L (sin at) =

sin at . t

a = F(s) (say) s + a2 2

By integration of Laplace transform,

FG sin at IJ = H t K F sin at IJ = LG H t K L



=



L

z z



s

a



S2 + a 2

s

dS = a .

1 S tan − 1 a a

∞ s

π s s − tan − 1 = cot − 1 . 2 a a

FG sin at IJ = cot H t K

Example 4. Show that :

F( S) dS .

z

−1

s . a

e − t − e −3t dt = log 3. 0 t 1 1 − L(e–t – e–3t) = L(e–t) – L(e–3t) = s+1 s+3

Sol.



∴ By integration of Laplace transform, L

Fe GH

−t

− e −3t t

z

I = F 1 − 1 I dS JK GH S + 1 S + 3 JK L O S+1 = Mlog (S + 1) − log (S + 3)P = log S+3 N Q F 1+ 1I F s+1 S + 1I G S J − log s + 1 = lim G log = log G lim − log J H S + 3K s + 3 GH 1 + S3 JJK s + 3 F 1 + 0 IJ + log s + 3 = log s + 3 = log G s+1 H 1 + 0K s + 1 I = log s + 3 JK s + 1 ∞

s

S→∞



L

Fe GH

−t

− e −3t t





s

s

S→∞

37

LAPLACE TRANSFORMS

z

s+3 e − t − e −3 t dt = log s+1 0 t In particular, let s = 0, ∴



e − st .

z





0



e0 .

z



0

0+3 e − t − e −3t dt = log 0+1 t

e − t − e −3t dt = log 3. t

WORKING RULE FOR SOLVING PROBLEMS Rule

If L(f(t)) = F(s), then L

FG f (t) IJ = H tK

z



s

F( S) dS .

TEST YOUR KNOWLEDGE 1. Find the Laplace transform of the following functions of t for t > 0 : (i)

sin t t

(ii)

1 − cos 2t . t

2. Find the Laplace transform of the following functions of t for t > 0 : 1 − et t

− at − e − bt . (ii) e t 3. Find the Laplace transform of the following functions of t for t > 0 :

(i)

(i)

e at − cos bt t

(i)

cos αt − cos βt t

(ii) cos 2t − cos 3t . t 4. Find the Laplace transform of the following functions of t for t > 0 : (ii)

e− t sin t . t

cos at does not exist. t

5. Show that the Laplace transform of the function 6. Show that : (i)

z z



0

π sin t dt = t 2

(ii)



2 cos 6t − cos 4t dt = log . 3 t 0 7. Show that : (iii)

F GH F (iii) L G H (i) L

z z

t

0

t

0

I JK

sin x 1 dx = cot − 1 s x s

I JK

e x sin x cot −1 (s − 1) dx = . x s

z



0

(ii) L

F GH

e− t sin t π dt = . 4 t

z

t

0

I JK

FG H

IJ K

1 − e − 2x 1 2 . dx = log 1 + x s s

38

LAPLACE AND FOURIER TRANSFORMS

Answers 1. (i) cot–1 s

(ii)

1 s2 + 4 log 2 s2

s 2 + b2 s2 + 9 1 1 log 2 3. (i) 2 log 2 (ii) 2 (s − a) s +4

2. (i) log

4. (i)

s −1 s

s2 + β 2 1 log 2 (ii) cot–1 (s + 1) 2 s + α2

Hints 6. We have

L

z





0

FG sin t IJ = cot H t K

−1

s

e− st sin t dt = cot − 1 s t

For part (i), take s = 0. For part (ii), take s = 1. 7. (i) L(sin t) =

1

⇒ L

s2 + 1

FG sin t IJ = cot H t K

–1

s.

Now using Laplace transform of integral of L

F GH

z

t

0

I JK

sin T 1 dT = . cot–1 s. T s

s+b (ii) log s + a

sin t , we have t

2

Inverse Laplace Transforms

2.1. INTRODUCTION We have already noted that the main purpose of studying Laplace transforms is for solving various types of differential equations. During the process of solving a differential equation, we shall also require to find a function when its Laplace transform is known. This is the reverse process of finding the Laplace transform of a function. In the present chapter, we shall learn to find the function whose Laplace transform is known. 2.2. INVERSE LAPLACE TRANSFORM OF A FUNCTION Let f be a real valued function of the real variable t, defined for t ≥ 0. Let the Laplace transform F(s) of f(t) exists. Therefore the infinite integral

z



0

e − st f (t ) dt exists and equals F(s).

The function f(t) is called the inverse Laplace transform of the function F(s) and we write L–1(F(s)) = f(t). In other words, L–1(F(s)) is that function whose Laplace transform is the function F(s). For example, 1 1 (i) L– 1 because L(eat) = = e at , s−a s−a (ii) L–1

FG IJ H K F s I = cos 3t , GH s + 9 JK 2

because L(cos 3t) =

s s2 + 9

2.3. EXISTENCE AND UNIQUENESS OF INVERSE LAPLACE TRANSFORM A given function F(s) of s may or may not have its inverse Laplace transform. So far as the uniqueness of inverse Laplace transforms, we have the following result : If f1(t) and f2(t) be two continuous functions for t ≥ 0 having the same Laplace transform F(s) i.e. L–1(F(s)) = f1(t) and L–1(F(s)) = f2(t), then f1(t) = f2(t) ∀ t ≥ 0. We accept this result without proof. Illustration : The functions f1(t) = 1 1, 0 ≤ t < 4 and f2(t) = 5, t = 4 1, t > 4

R| S| T

39

40

LAPLACE AND FOURIER TRANSFORMS

1 . Here the above result is not applicable because the function s f1(t) is continuous for t ≥ 0 but the function f2(t) is not continuous for t ≥ 0.

have the same Laplace transform

2.4. ELEMENTARY INVERSE LAPLACE TRANSFORM FORMULAE In this section, we shall find some elementary inverse Laplace transform formulae.

FG 1IJ = 1 , s > 0 . H sK Γ ba + 1g L (t ) = , a > −1, s > 0 .

L– 1

∴ By definition,

a

2. We have

sa +1

L–1

∴ By definition,

L– 1

∴ By definition,

L– 1

F GH s

a 2

−a

2

L– 1

F GH s

s 2

−a

2

8. We have

F GH s

2

F GH s

a 2

s − a2

at

s s2 − a 2

, s > a.

, s >|a | s > |a|.

, s >|a |

I = L (cosh at) , JK

I JK

a 2

s + a2

s > |a|.

, s>0

a = sin at , s > 0 . + a2

s 2

, n = 0, 1, 2, . . .. .. ; s > 0 .

1 , s>a. s−a

I = sinh at , JK

L (cos at ) = L– 1

n

FG 1 IJ = e H s − aK

L (sin at ) =

–1 ∴ By definition, L

, n = 0, 1, 2, ...... ; s > 0

n+1

L (cosh at ) =

7. We have

∴ By definition,

sn + 1 n!

, a > – 1, s > 0.

FG IJ = t Hs K

L (sinh at ) =

5. We have

∴ By definition,

n!

L ( e at ) =

4. We have

a

a+1

L– 1

∴ By definition,

6. We have

FG Γ(a + 1)IJ = t H s K

L (t n ) =

3. We have

∴ By definition,

1 , s>0 s

L(1) =

1. We have

+a

2

s 2

s + a2

, s > 0.

I = cos at , JK

s > 0.

41

INVERSE LAPLACE TRANSFORMS

F Γ ba + 1g I = t ⇒ L FG 1 IJ = t ; a > − 1 , s > 0 . GH s JK H s K Γ (a + 1) F 1 I t , a > 0 , s > 0. L G J= H s K Γ (a) F n ! IJ = t ⇒ L FG 1 IJ = t , n = 0, 1, 2, ...... ; s > 0 L G Hs K Hs K n! F 1 I t , n ∈ N, s > 0 . L G J= H s K (n − 1) !

Remarks 1. L– 1 ∴ 2. ∴

a

a+1

a

n

−1

n

n+1

a+ 1

a−1

−1

−1

a

–1

n+1

n−1

−1

n

WORKING RULES FOR SOLVING PROBLEMS If L(f(t)) = F(s), then L–1 (F(s)) = f(t).

Rule I.

Rule II. (i) L− 1 (iii) L− 1 (v) L− 1 (vii) L− 1

FG 1IJ = 1 H sK FG n ! IJ = t Hs K F a I = sinh at GH s − a JK F a I = sin at GH s + a JK

(ii) L− 1 (iv) L− 1

n

n+1

2

2

2

2

(vi) L− 1 (viii) L− 1

FG Γ ( a + 1) IJ = t , a > − 1 H s K FG 1 IJ = e H s − aK F s I = cosh at GH s − a JK F s I = cos at . GH s + a JK a

a+1

at

2

2

2

2

ILLUSTRATIVE EXAMPLES Example 1. Find the values of : (i) L− 1 (iii) L− 1 (v) L− 1

FG Γ (5/2)IJ H s K FG 1 IJ Hs + 5K F s I GH s + 16 JK

(iv) L− 1 (vi) L− 1

2

Sol. (i) We have ∴ (ii) We have ∴

(ii) L− 1

5/ 2

FG 5040 IJ H s K F 3 I GH s − 9 JK F 6 I. GH s + 36 JK 8

2

2

FG Γ ( a + 1) IJ = t . H s K F Γ (5/2) IJ = t = t . L G H s K F n ! IJ = t . L G Hs K F 5040 IJ = L FG 7 ! IJ = t . L G H s K Hs K a

L− 1

a +1

−1

−1

−1

5/ 2

5 −1 2

3/ 2

n

n +1

8

−1

7

7+1

42

LAPLACE AND FOURIER TRANSFORMS

(iii) We have

FG 1 IJ = e . H s − aK F 1 IJ = L FG 1 IJ = e . L G H s + 5 K H s − ( − 5) K F a I = sinh at. L G H s − a JK F 3 I = L F 3 I = sinh 3t. L G H s − 9 JK GH s − 3 JK F s I = cos at L G H s + a JK F s I = L F s I = cos 4t. L G H s + 16 JK GH s + 4 JK F a I = sin at L G H s + a JK F 6 I = L F 6 I = sin 6t. L G H s + 36 JK GH s + 6 JK L− 1

at

−1

∴ (iv) We have

−1

−1

2

−1

∴ (v) We have

−1

−1

∴ (vi) We have

−1

−1



− 5t

2

−1

2

2

2

2

2

2

2

2

2

−1

2

2

2

2

−1

2.5. LINEARITY OF THE INVERSE LAPLACE TRANSFORM Theorem. If f(t) and g(t) be any functions of t for t ≥ 0 such that L(f(t)) = F(s) and L(g(t)) = G(s) and a and b be any constants, then L–1(aF(s) + bG(s)) = aL–1(F(s)) + bL–1(G(s)). Proof. We have L(f(t)) = F(s) and L(g(t)) = G(s). aF(s) + bG(s) = aL(f(t)) + bL(g(t))



= L(af(t) + bg(t)) L–1(aF(s)

∴ or

L–1(aF(s)

+ bG(s)) = af(t) + bg(t)

+ bG(s)) = aL–1(F(s)) + bL–1(G(s)).

Example 2. Find the value of L− 1 Sol.

L− 1

FG 1 + 2 + 6 IJ . Hs +3 s +5 s K 4

FG 1 + 2 + 6 IJ Hs+3 s +5 s K F 1 IJ + 2L FG 1 IJ + L FG 6 IJ =L G H s + 3K H s + 5K H s K 4

−1

−1

−1

4

(Using linearity)

43

INVERSE LAPLACE TRANSFORMS

= L− 1

FG 1 IJ + 2L FG 1 IJ + L FG 3 ! IJ H s − (− 3) K H s − (− 5) K H s K −1

= e–3t + 2e–5t + t3. Example 3. Find the value of L− 1 −1 Sol. L

F GH 4s

s 2

− 16

= L− 1

F GH 4s

+

9 2

s + 25

+

4s 2

−1

F GH

I JK

s 2

− 16

I JK

2

s + 25

+

3+ 1

4s 2

9s + 4

+

I JK

1 . 4s − 1

I + L F 4s I + L FG 1 IJ J GH 9s + 4 JK H 4s − 1K + 25 K I F F 5 I + 4 L GG s JJ + 1 L GH s + 5 JK 9 G F 2 I J 4 GH s + GH 3 JK JK 9

−1

2

s 1 9 = L− 1 2 + L− 1 4 5 s − 22

9

+

1 4s − 1

+

9s + 4

I +L F J GH s − 16 K

s 2

F GH 4s

−1

2

−1

2

−1

2

−1

2

2

1

F I GG 1 JJ GH s − 41 JK

t 1 9 4 2 1 = cosh 2t + sin 5t + cos t + e 4 . 4 5 9 3 4 Example 4. Find the inverse Laplace transform of the following functions :

(i)

4s − 8 9 − s2

F s−2 I = −4. s + 8 . 3 GH s − 9 JK s −9 3 s −9 9−s F 4s − 8 I = L F − 4 . s + 8 . 3 I GH 9 − s JK GH s − 9 3 s − 9 JK F s I+8L F 3 I = − 4L G H s − 3 JK 3 GH s − 3 JK 4s − 8

Sol. (i)



2

L−1

=−4

2

2

−1

2

2

2

F GG GH

5 s− 2 2 = 4s 2 + 25 4 s 2 + 25 4

(ii)

L− 1

−1

2

= − 4 cosh 3t +

2s − 5

2

2

−1



2s − 5 4s 2 + 25

(ii)

I JJ JK

2

2

8 sinh 3t. 3

=

1 . 2

s 25 s + 4 2



5 . 4

1 2

s +

25 4

F 2s − 5 I = L F 1 . s − 1 . 5/2 I GH 4s + 25 JK GH 2 s + (5/2) 2 s + (5/2) JK F s I − 1 L F 5/2 I 1 = L G 2 H s + (5/2) JK 2 GH s + (5/2) JK −1

2

2

−1

=

2

2

2

1 5 1 5 cos t − sin t. 2 2 2 2

2

−1

2

2

2

44

LAPLACE AND FOURIER TRANSFORMS

TEST YOUR KNOWLEDGE Find the inverse Laplace transform of the following functions : 1. (i) 2. (i) 3. (i) 4. (i) 5. (i) 6. (i)

720

(ii)

s7 s

(ii)

s2 + 25

60 + 6 s2 + s4 s7

(ii)

40320

(iii)

s9 s

(iii)

s2 − 49

s−4

(iii)

s2 − 16

6 3 9 + + s−4 s+5 s−7 4 s2 + 1

3 2

s −1





9s s2 + 5

4 2

s +9

+

+

6 4 s2 + 1

s 2

s + 16

+



8s s2 + 25

5 s − 16

(iv)

s3 / 2 8

(iv)

s2 − 64

01 . s + 0.9

(iv)

s2 + 3.24

1 s−7 4 s2 + 16

s a 2 s 2 + b2 c 2

(ii)

2 4 11 − + 2s + 1 5s + 7 1 + 5s

(ii)

2 6s 4s 1 + − + ( s − 1) ( s + 1) s2 − 25 s2 − 9 6s2 − 1

(ii)

2

Γ (3/2)

5 2

3 s + 27

6

+

2

5s − 9



5 2

36 s + 1

+

9 5 + s2

.

Answers 1. (i) t6

(ii) t8

(iii)

2. (i) cos 5t

(ii) cosh 7t

(iii) sinh 8t

3. (i)

1 6 1 4 1 2 t + t + t 12 4 2

(iii)

(iv) sin 4t

(ii) e– 4t

1 1 cos 1.8 t + sin 1.8 t 10 2

(iv)

4. (i) 6e4t + 3e–5t + 9e7t

1 a

2

cos

(ii) e

5. (i) 4 sin t − 9 cos 5 t + 3 sin 6. (i) 3 sinh t −

(iv) e7t

t

t + 8 cos 5t 2

4 5 sin 3t + cos 4t − sinh 4t 4 3



bc t a 1 t 2

7



1

4 − 5 t 11 − 5 t e + e 5 5

(ii) 2 sinh t + 6 cosh 5t − 4 cosh 3t + (ii)

1 6

sinh

t 6

t 2 3 5 9 5 sin 3t + sinh t − sin + sin 5t. 6 6 9 5 5 5

2.6. VALUE OF L–1(F(s – a)) IN TERMS OF L–1 (F(s)) By the first shifting theorem of Laplace transforms, if L(f(t)) = F(s), then L(eat f(t)) = F(s – a). –1 ∴ If L (F(s)) = f(t) then L–1 (F(s – a)) = eat f(t). Equivalently, L–1 (F(s – a)) = eat L–1 (F(s)). form,

Remark. In the formula, L–1 (F(s – a)) = eat L–1 (F(s)), if we replace a by – a, the formula takes the L–1 (F(s + a)) = e–at L–1 (F(s)). For example,

L− 1

F s−2 I GH (s − 2) + 5 JK 2

2

= e2t L− 1

F GH s

s 2

+5

I =e JK

2t

2

cos 5t.

45

INVERSE LAPLACE TRANSFORMS

Theorem. Using the formula L–1 (F(s – a)) = eat L–1 (F(s)) prove that :

FG 1 IJ = e H s − aK F Γ (b + 1) I = e t , b > − 1 2. L G H (s − a) JK F n ! I = e t , n = 0, 1, 2, . ... .. 3. L G H (s − a) JK F b IJ = e sinh bt 4. L G H (s − a) − b K F s − a IJ = e cosh bt 5. L G H (s − a) − b K F b IJ = e sin bt 6. L G H (s − a) + b K F s − a IJ = e cos bt. 7. L G H (s − a) + b K F 1I Proof. 1. We have L G J = 1 . H sK F 1 IJ = e . L FG 1IJ = e . 1 ∴ L G H sK H s − aK F 1 IJ = e . ∴ L G H s − aK F Γ (b + 1) IJ = t , b > − 1 . 2. We have L G H s K F Γ (b + 1) I = e L FG Γ (b + 1) IJ = e t ∴ L G H s K H (s − a ) JK F Γ (b + 1) IJ = e t , b > − 1 . L G ∴ H (s − a) K F n ! IJ = t , n = 0, 1, 2, ...... 3. We have L G Hs K F n ! I = e L FG n ! IJ = e t ∴ L G Hs K H (s − a) JK F n ! I = e t , n = 0, 1, 2, .. . .. . ∴ L G H (s − a) JK 1. L− 1 −1

−1

−1

−1

−1

−1

at

at

b

at

n

b+ 1

n+ 1

2

2

2

2

2

at

at

at

2

2

at

2

−1

−1

at

−1

−1

at

b

at

b+1

−1

b+ 1

−1

−1

at

b+1

−1

−1

−1

at

−1

at b

b+ 1

b

n

n+ 1

at

n+1

at

n+ 1

−1

n

at n

n+1

46

LAPLACE AND FOURIER TRANSFORMS

I = sinh bt . J −b K F b I =e L F b I =e ∴ L G GH s − b JK H (s − a ) − b JK F b IJ = e sinh bt . L G ∴ H (s − a) − b K F s I = cosh bt . 5. We have L G H s − b JK F s−a I =e L F s I =e ∴ L G GH s − b JK H ( s − a) − b JK F s − a IJ = e cosh bt . L G ∴ H (s − a) − b K F b I = sin bt . 6. We have L G H s + b JK F b I =e L F b I =e L G ∴ GH s + b JK H (s − a ) + b JK F b I = e sin bt . ∴ L G H (s − a) + b JK F s I = cos bt. 7. We have L G H s + b JK F s−a I =e L F s I =e ∴ L G GH s + b JK H ( s − a) + b JK F s − a IJ = e cos bt. L G ∴ H (s − a) + b K 4. We have

L− 1

F GH s

b

2

2

2

2

−1

at

−1

2

−1

2

2

2

2

at

−1

2

−1

2

2

2

at

−1

2

2

2

2

2

at

sinh bt

−1

at

cosh bt

at

sin bt

at

2

2

−1

2

at

2

−1

−1

−1

at

2

−1

−1

2

2

2

2

2

at

−1

2

−1

at

2

2

cos bt

at

2

ILLUSTRATIVE EXAMPLES Example 1. Find the inverse Laplace transform of the function Sol.

L− 1

F 12 I = 6L F 2 I GH ( s − 4) JK GH (s − 4) JK F 2I = 6e L G J = 6e Hs K −1

3

4t

=

6e4t

12 (s − 4) 3

.

3

−1

t 2.

3

4t

L− 1

FG 2 ! IJ Hs K 2+1

(∵ L–1 (F(s – a)) = eat L–1 (F(s)))

47

INVERSE LAPLACE TRANSFORMS

Example 2. Find the inverse Laplace transform of the following functions : (i)

5

3s + 1 (s + 1)

(ii)

4

F 3s + 1 I = L F 3(s + 1) − 2 I GH (s + 1) JK GH ( s + 1) JK F 3s − 2 IJ = e =e L G H s K

Sol. (i) L− 1

−1

4

−1

F GG ∑ H 5

λk

k=1

s+k

−1

5

2

4

−t

.

FG 3 − 2 IJ Hs s K (∵ L (F(s – a)) = e FG 3 ! IJ OP = e LM 3 t − 1 t OP . H s KQ N2 3 Q

L− 1

3

LM 3 L FG 2 ! IJ − 2 L N2 H s K 6 I JJ = ∑ λ L FGH s +1k IJK = ∑ λ K = e− t

L− 1

s + k2

k=1

4

−t

(ii)

λk



−1

3

–1

−t

4

5

−1

k

2

k=1

4

k

e− k

2

t

at

2

L–1 (F(s)))

3

.

k=1

Example 3. Find the inverse Laplace transform of the following functions : (i)

s+3

(ii)

(s + 3) 2 + 4

Sol. (i) L

2

3s

(ii)



F s+3 I =e GH (s + 3) + 4 JK s 2 + 2s − 8

L− 1

F GH s

=

L–1

F GH s

3s ( s + 1) 2 − 9

s + 22

2

I =e JK

–3t

cos 2t. (∵ L–1 (F(s – a)) = eat L–1 (F(s)))

3s + 3 − 3

=

.

3( s + 1)

=

( s + 1) 2 − 9

2

( s + 1) − 9



I = L F 3(s + 1) − 3 I J GH (s + 1) − 9 (s + 1) − 9 JK + 2s − 8 K F s +1 I − L F 3 I = 3L G H (s + 1) − 9 JK GH (s + 1) − 9 JK F s I−e L F 3 I = 3e L G GH s − 3 JK H s − 3 JK 3s

2

− 3t

3s s 2 + 2s − 8

−1

2

−1

−1

( s + 1) 2 − 9

2

−1

2

− 1. t

3

2

2

− 1. t

2

−1

2

2

= 3e–t cosh 3t – e–t sinh 3t = e–t (3 cosh 3t – sinh 3t). Example 4. Find the inverse Laplace transform of the function 1 2

s + 4s + 13

Sol.

1 2

s + 4s + 13

− −

=

s+4 2

s + 8s + 97 s+4 2

s + 8s + 97

1 2

( s + 2) + 9

+ +



s+2 2

s − 4s + 29

.

s+2 2

s − 4 s + 29

s+4 2

( s + 4 ) + 81

+

s+2 ( s − 2 ) 2 + 25

48

LAPLACE AND FOURIER TRANSFORMS

=



1 4 s+4 s−2 − + + (s + 2) 2 + 9 (s + 4) 2 + 81 (s − 2) 2 + 25 (s − 2) 2 + 25

I J + 4s + 13 s + 8s + 97 s − 4s + 29 K F 1 I −L F s+4 I +L F s−2 I +L F 4 I =L G H (s + 2) + 9 JK GH (s + 4) + 81JK GH (s − 2) + 25 JK GH (s − 2) + 25 JK F 1 I −e L F s I +e L F s I +e L F 4 I =e L G GH s + 81JK GH s + 25 JK GH s + 25 JK H s + 9 JK F 3 I −e L F s I +e L F s I + 4 e L F 5 I 1 = e L G GH s + 9 JK GH s + 5 JK 5 GH s + 5 JK 3 H s + 3 JK

L− 1

F GH s

1



2

−1

−1

− 2t

− 4t

2

−1

s+2

+

2

−1

2

− 2t

=

s+4

2

2

−1

2t

−1

2t

2

−1

2

−1

2

− 4t

2

−1

2

2

2t

2

−1

2

−1

2

2t

2

2

−1

1 − 2t 4 e sin 3t − e − 4t cos 9t + e2t cos 5t + e2t sin 5t. 5 3

2.7. VALUE OF L–1 (e–as F(s)) IN TERMS OF L–1 (F(s)) By the second shifting theorem of Laplace transforms, if L(f(t)) = F(s) then for any a ≥ 0, L(f(t – a) ua(t)) = e– as F(s). ∴ If L–1(F(s)) = f(t) then L–1(e–as F(s)) = f(t – a) ua(t) for any a ≥ 0. Remark. It may be recalled that for a ≥ 0, ua(t) is the unit step function defined as : ua ( t ) =

RS0 T1

if t < a if t > a.

∴ The function f(t – a) ua(t) can also be written as f (t − a ) ua (t ) =

RS0 T f (t − a )

if t < a if t > a.

Example 5. Find the inverse Laplace transform of the function

s 2 e 2s 2s + 1 2s + 1 2s + 1 = e −2 s . = e–2s F(s), where F(s) = . s2 e2s s2 s2 2s + 1 2 1 1 1 = L−1 + 2 = 2L−1 + L−1 2 L–1 (F(s)) = L–1 2 s s s s s = 2(1) + t = t + 2 = f(t), say.

Sol.

FG H



2s + 1

L–1

F 2s + 1I GH s e JK 2

2s

IJ K

FG H

IJ K

FG IJ HK

L–1

F 2s + 1I = g(t), where g(t) = RS0 GH s e JK Tt 2

2s

FG IJ H K

= L–1(e–2sF(s)) = f(t – 2) u2(t) = ((t – 2) + 2) u2(t) = t . u2(t) =



.

if t < 2 . if t > 2

RS0 Tt

if t < 2 . if t > 2

2

2

49

INVERSE LAPLACE TRANSFORMS

Example 6. Find the inverse Laplace transform of the function

s e − πs / 2 s s = e–(π/2)s F(s), where F(s) = 2 . = e − ( π / 2) s . 2 2 s +1 s +1 s +1

Sol.

L–1(F(s) = L–1 ∴

e − π s/2 s . s2 + 1

L–1

Fe GH s

I JK

− πs /2 2

F GH s

I = cos t = f(t), say. J + 1K

s 2

s = L–1 (e–(π/2)s F(s)) +1

FG H

=f t−

IJ K

IJ K

FG H

π π u π / 2 (t ) = cos t − u π / 2 (t ) 2 2

= sin t . u π / 2 (t ) =

RS 0 Tsin t

if t < π/2 . if t > π/2

Example 7. Find the inverse Laplace transform of the function Sol.

L–1



L–1

3 (1 − e − πs ) . s2 + 9

F 3(1 − e ) I = L F 3 − 3e I = L F 3 I − L F 3e I GH s + 9 JK GH s + 9 s + 9 JK GH s + 9 JK GH s + 9 JK F 3(1 − e ) I = sin 3t – L F 3e I GH s + 9 JK GH s + 9 JK − πs

−1

2

− πs

2

−1

2

− πs

–1

2

−1

2

− πs

2

− πs

2

...(1)

3 3e − πs 3 = e–πs F(s), where F(s) = 2 . = e − πs . 2 2 s +9 s +9 s +9

L–1 (F(s)) = L–1 ∴

L–1

F 3e I = L GH s + 9 JK − πs

2

F 3 I = sin 3t = f(t), say. GH s + 9 JK 2

–1(e–πs

F(s)) = f(t – π) uπ(t) = sin 3(t – π) uπ(t)

= – sin (3π – 3t) uπ(t) = – sin 3t . uπ(t).

∴ (1) ⇒ L–1

F 3(1 − e ) I = sin 3t – (– sin 3t . u (t)) GH s + 9 JK − πs

2

π

= (1 + uπ(t)) sin 3t.

2.8. VALUE OF L–1 F(s/a)) IN TERMS OF L–1 (F(s)) By the change of scale property, if

L(f(t)) = F(s) then for any a > 0, L ( f (at)) = ∴

FG IJ H K

1 s F a a

i.e. L(af(at)) = F

FG FG s IJ IJ = a f(at) for any a > 0. H H aKK

If L–1 (F(s)) = f(t) then L–1 F

FG s IJ . H aK

50

LAPLACE AND FOURIER TRANSFORMS

Example 8. Find the inverse Laplace transform of the following functions : (i)

FG s IJ H λK

a 2

(ii)

+ a2 F( s ) =

Sol. (i) Let

L− 1

F GG a GG FG s IJ + a H H λK 2

+9

.

2

s + a2

L− 1

F GG a GG FG s IJ + a H H λK 2

(ii) Let

I JJ F F s I I JJ = L GH F GH λ JK JK = λ f(λt) = λ sin λat. K −1

2

Alternative method

I JJ F JJ = L GH s K −1

2

F( s ) =

2

s 2

s +9

I JK

F GH

λ2 a λa = λL− 1 2 2 2 s + (λa) 2 +a λ

I = λ sin λat. JK

.

L–1 (F(s)) = cos 3t = f(t), say





2

L–1 (F(s)) = sin at = f(t), say





a

FG s IJ H7K

s

L− 1

I F GG s JJ F s/7 I = 7(7 f(7t)) = 7L G GG FG s IJ + 9 JJ H (s/7) + 9 JK H H 7K K 2

−1

2

= 49 cos (3(7t)) = 49 cos 21t.

WORKING RULES FOR SOLVING PROBLEMS Rule I.

If

L–1(F(s))

= f(t), then L–1(F(s – a)) = eatf(t).

Rule II. If L–1 (F(s)) = f(t), then L–1 (e– as F(s)) = f(t – a) ua(t), a ≥ 0. Rule III. If L–1 (F(s)) = f(t), then

FG FG s IJ IJ = a f(at), a > 0. H H aKK

L–1 F

51

INVERSE LAPLACE TRANSFORMS

TEST YOUR KNOWLEDGE Find the inverse Laplace transform of the following functions : 1. (i) 2. (i) 3. (i)

4. (i)

5. (i) 6. (i)

1

(ii)

( s − 4 )5 1

(ii)

s2 + 6s + 13

s

FG s + 1 IJ H 2K

(ii)

2

+1

5

(ii)

2

s − 4s − 3 s+2

2

5

(s − 4)

+

s+3

s2 − 4 s + 13 4 2

s + 6 s + 18

9 2

5 2

s − 4 s + 29

1 2 s + 10

s +s+

(ii)

s + 4s + 7 1

4 ( s + 2)7

s2 + 8 s + 20 1 s+3 + 2 (ii) 4 ( s − 3) s + 6 s + 45

s +1 1 5s − 2 + + 7. (i) 2s − 5 3s2 + s + 8 s2 + 2s + 2 1 1 3s − 2 3s + 7 + + + (ii) 2 s − 6 s + 10 s2 + 8 s + 16 s2 − 4 s + 20 s2 − 2 s − 3

8. (i) 9. (i) 10.

e− s s2 (5s + 1) e − 5s s

(i) −

11.

(i)

12.

(i)

(ii) (ii)

2

e − πs /2 s +1 − 2s

2

s +π

s3 2 + 2 s + s2 s3 es

5se− πs

(ii) −

2

se

e− 3 s

(ii)

2

e 4 − 3s

(ii)

( s + 4 )5/ 2

s2 + 1

e− 2 πs / 3 s2 + 1 ( s + 1) e − πs s2 + s + 1

.

Answers 1. 2.

1 4 4t t e (i) 24 1 − 3t e sin 2t (i) 2

FG H

− t/2 cos t − 3. (i) e

4. (i)

5 7

(ii)

FG H

2t cos 3t + (ii) e

1 sin t 2

e 2t sinh 7t

5. (i) e − 2t cos 3t

1 6 − 2t t e 180

IJ K

IJ K

5 sin 3t 3

4 − 3t e sin 3t 3 t (ii) 18 e− t / 2 sin 2

(ii)

(ii) e− 4 t [cos 2t + 3 sin 2t]

52

LAPLACE AND FOURIER TRANSFORMS

6. (i)

1 4t 4 e t + e 2t sin 5t 24

7. (i)

− 1 5t / 2 5 e +e 6 cos 2 3

t

LM MN

(ii)

95 25 t− sin 6 3 95

OP PQ

1 3t 3 e t + e − 3t cos 6t 6

95 t + e− t cos t 6

(ii) e3t sin t + te–4t +3e2t cos 4t + e2t sin 4t + 3et cosh 2t + 5et sinh 2t

1 (t − 3)2 u3 (t) 2 (ii) t2u1(t)

8. (i) (t – 1) u1(t)

(ii)

9. (i) t . u5(t) 10.

(i) cos t . uπ / 2 ( t )

(ii) 5 cos t . uπ(t)

11.

(i) cos πt . u2(t)

(ii) − cos

12.

(i)

4 3 π

(ii) e



FG π − tIJ . u H6 K

2π / 3

(t )

e16 − 4t ( t − 3 )3/ 2 u3 ( t )

1 (t − π ) 2

LMcos MN

OP PQ

3 1 3 (t − π ) + sin (t − π ) uπ (t ). 2 2 3

2.9. VALUE OF L–1 (F(s)/s) IN TERMS OF L–1 (F(s)) By the property of Laplace transforms of integrals,

FG H

L( f (t)) = F( s), then L

if ∴ If

z

L–1 (F(s)) = f(t), then L−1

t

0

IJ K

f ( T ) dT =

FG 1 F(s)IJ = Hs K

z

t

0

1 F(s). s

f(T) dT .

ILLUSTRATIVE EXAMPLES Example 1. Find the inverse Laplace transform of the function

F GH

I JK

1 1 1 1 1 = F(s), where F(s) = 2 = . 2 2 2 s s + a2 s(s + a ) s s + a

Sol.

2

L–1 (F(s)) = L–1

F GH s

1 2

2

I= 1L F JK a GH s −1

a 2

2

I = 1 sin at = f(t), say. JK a

+a +a ∴ By the property of Laplace transform of integrals,

L–1

FG 1 F(s)IJ = Hs K

z

t

0

=– ∴

1 s (s 2 + a 2 )

L–1

F GH s(s

2

I JK

f ( T ) dT =

z

t

0

1 1 − cos at (cos at − cos 0) = a2 a2

1 − cos at 1 = . a2 + a2 )

FG H

1 1 cos aT − sin aT dT = a a a

IJ K

t 0

.

53

INVERSE LAPLACE TRANSFORMS

Example 2. Find the inverse Laplace transform of the function 1 1 1 = 3 2 = 3 3 s +s s ( s + 1) s

Sol.

5

F 1 I= 1 GH s + 1JK s 2

3

1 5

s + s3

F(s), where F(s) =

F 1 I = sin t = f(t), say GH s + 1JK

L–1(F(s)) = L–1

.

1 s2 + 1

.

2

∴ By the property of Laplace transform of integrals, L–1

FG 1 F(s)IJ = Hs K

z z

t

0

F (T ) dT =

Again by using the same formula,

z

t

0

t 0

FG 1 . 1 F(s)IJ = (1 − cos T) dT = (T − sin T) Hs s K F 1 F(s)IJ = t – sin t. L G Hs K

L–1

t

0

–1



z

FG 1 . 1 F(s)IJ = (T − sin T) dT = FG T Hs s K H2 F 1 F(s)IJ = t + cos t – 1 L G Hs K 2 F 1 I = t + cos t – 1. L G H s + s JK 2 2

2

t

0

–1



t

= t – sin t. 0

2

Again by using the same formula, L–1

= – cos t + cos 0 = 1 – cos t.

sin T dT = − cos T

I + cos TJ K

t

= 0

t2 + cos t – 1 2

2

3

2

–1

or

5

3

Remark. The above problem can also be solved by the ‘‘method of partial fractions’’.

2.10. VALUE OF L–1 (F′′(s)) IN TERMS OF L–1(F(s)) By the property of differentiation of Laplace transforms if

L(f(t)) = F(s), then L(t f(t)) = – F′(s). L–1 (F(s)) = f(t), then F′(s) = L (– t f(t)).

∴ If

L–1 (F(s)) = f(t), then L–1 (F′(s)) = – t f(t).

∴ If

Example 3. Find the inverse Laplace transform of the function

Sol. ∴ Also

z

s ( s2 + 4 )2

ds =

1 2

F′(s) =

z

2s ( s2 + 4 )2

ds =

s 2

(s + 4) 2

.

1 (s 2 + 4) − 1 1 . =− = F(s), say. 2 2 −1 2(s + 4)

s 2

(s + 4) 2

F GH

L– 1 (F( s)) = L– 1 −

I JK

F GH

1 1 2 = − L– 1 2 4 s + 22 2(s + 4) 2

I = − 1 sin 2t = f (t), JK 4

say.

54

LAPLACE AND FOURIER TRANSFORMS

∴ By the property of differentiation of Laplace transforms, we have L–1 (F′(s)) = – t f(t). ∴

L– 1



L– 1

F s I = − t FG − 1 sin 2tIJ = 1 t sin 2t GH (s + 4) JK H 4 K 4 F s I = 1 t sin 2t. GH (s + 4) JK 4 2

2

2

2

Example 4. Find the inverse Laplace transform of the function Sol.

L− 1

LM N (s

2

OP Q

2s 2

OP Q

LM N

2s + 6 2(s + 3) 2s = L− 1 = e − 3t L− 1 2 2 2 2 ((s + 3) + 1) (s + 1) 2 + 6 s + 10)

Now, we require the value of L− 1

z

LM N LM 2s MN (s + 1)

2s + 6 . (s + 6s + 10) 2 2

(s + 1)

2

ds =

2

2

OP . PQ

OP Q

...(1)

(s 2 + 1) − 1 1 =− 2 = F(s), say. −1 s +1

F′(s) =



2s (s 2 + 1) 2

F GH

L− 1 ( F( s )) = L− 1 −

Also

I = − sin t = f (t), J + 1K

1 s

2

say.

∴ By the property of differentiation of Laplace transforms, we have L–1 (F′(s)) = – t f(t).

∴ (1)

F 2s I = − t (− sin t) ⇒ GH (s + 1) JK F 2s + 6 I = e t sin t . GH (s + 6s + 10) JK

L− 1

⇒ ⇒

L− 1

2

2

2

2

L− 1

F 2s I = t sin t. GH (s + 1) JK 2

− 3t

Example 5. Find the inverse Laplace transform of the function F( s ) =

Sol. Let ∴ Using,

1 2

s + a2

L− 1 (F(s)) = L− 1 −1

L

F GH s

+a

2

I= 1L F JK a GH s −1

a 2

+a

bF′ (s)g = − t f (t), we get FF 1 I ′I 1 J L GG = − t . sin at . J GH H s + a K JK a −1

2

2

1 2

(s + a 2 ) 2

. 1

2

2

2

I = 1 sin at = f (t), say. JK a

.

55

INVERSE LAPLACE TRANSFORMS



L− 1



L− 1

F − 2s GH ( s + a ) F s GH (s + a ) 2

2 2

2

2 2

I = − 1 t sin at JK a I = 1 t sin at = g(t), JK 2a

say .

Using formula for Laplace transform of integrals, we get

F1 . GH s (s F L G H (s

L− 1

−1



s

2

2

2 2

+a )

1 + a2 )2

I= JK I= JK

z z

L− 1

2.11. VALUE OF L−1

FG H

z

F GH (s

2

LM N

1 1 T cos aT sin aT T sin aT dT = − + a 2a 2a a2

t

0

1 2a 3

I JK

[ − at cos at + sin at ]

1 1 = [sin at − at cos at] . 2a 3 + a2 )2

IJ K



s

g( T ) dT .

0

=



t

F(s) dS IN TERMS OF L–1 (F(s))

By the property of integration of Laplace transforms

z

FG f (t) IJ = F (S) dS . H tK F I f(t) . (F(s)) = f(t), then L G F(S) dSJ = H K t

L( f ( t )) = F( s ) , then L

if ∴ If

L– 1

–1

z



s



s

F GH

Example 6. Find the inverse Laplace transform of the function log 1 +

Sol.

∴ Also,

F F GH GH

I I = d log ( s + a ) − 2 log s = 2s − 2 = F( s), j s +a s JK JK ds e F aI F(s) ds = log G 1 + H s JK F 2s − 2 I = 2 L F s I − 2 L F 1 I L (F( s)) = L G GH s JK GH s + a JK H s + a s JK

d a2 log 1 + 2 ds s

2

z

2

2

2

a2

2

−1

2

2

−1

2

2

−1

= 2 cos at – 2(1) = 2 cos at – 2 = f(t), say. ∴ By the property of integration of Laplace transforms, we have L– 1

FG H

z



s

IJ K

F (S) dS =

f (t ) . t

I. JK

say .

2

−1

s

2

OP Q

t 0

56

LAPLACE AND FOURIER TRANSFORMS

F F a I I 2 cos at − 2 L G log G 1 + GH H S JK JJK = t F F a I I = 2 cos at − 2 L G 0 − log G 1 + t H s JK JK H F F a I I = 2 − 2 cos at . L G log G 1 + t H H s JK JK





2

–1

2

s

2

–1



2

2

–1



2

Example 7. Find the inverse Laplace transform of the function log

Sol.



Also

F GH

I JK

s2 + 1 (s − 1) 2

.

2s 2 d s2 + 1 d − = F( s ) , say . log (log (s 2 + 1) − 2 log (s − 1)) = 2 = 2 −1 s s +1 ds ds (s − 1)

z

F(s) ds = log

L− 1 (F(s)) = L− 1

s2 + 1 (s − 1) 2

F 2s − 2 I GH s + 1 s − 1JK

= 2L− 1

2

F GH s

2

I JK

FG H

s 1 − 2L− 1 1 s − +1

IJ K

= 2 cos t – 2et = f(t), say. By the property of integration of Laplace transforms, we have ∞

s

F(S) dS = ∞

2

−1



FG H

z

IJ f (t) . K t F F S + 1 I I 2 cos t − 2e L G G log GH H (S − 1) JK JJK = t F s +1I 2e − 2 cos t L G 0 − log =− J t (s − 1) K H F s + 1 I = 2e − 2 cos t . L G log t H (s − 1) JK L− 1

t

2

s

2

−1



t

2

−1



t

2

2

Example 8. Find the inverse Laplace transform of the function cot –1 (s + 1). Sol. ∴

d −1 1 (cot − 1 (s + 1)) = .1= − = F(s) , say. 2 ds (s + 1) + 1 (s + 1) 2 + 1

z

F(s) ds = cot −1 (s + 1)

57

INVERSE LAPLACE TRANSFORMS

F GH

L− 1 (F(s)) = L− 1 −

Also,

1 ( s + 1)

= − e − t L− 1

2

I =−L F 1 J GH (s + 1) + 1K −1

1 2

s +1

2

I J + 1K

= – e–t sin t = f(t), say.

∴ By the property of integration of Laplace transforms, we have L− 1

FG H

FG H

z



s

IJ f (t) . K t IJ = − e sin t K t

F(S) dS = ∞

−t



L− 1 cot − 1 (S + 1)



L− 1 (0 − cot − 1 (s + 1)) = −



L− 1 (cot − 1 (s + 1)) =

s

e − t sin t t

e − t sin t . t

WORKING RULES FOR SOLVING PROBLEMS If L− 1 (F(s)) = f (t) , then L− 1

Rule I.

FG 1 F(s)IJ = Hs K

z

t

0

f (T) dT.

−1 −1 Rule II. If L (F(s)) = f (t) , then L (F ′(s)) = − t f (t).

Rule III. If L− 1 (F(s)) = f (t) , then L− 1

FG H

z



s

IJ K

F(S) dS =

f (t) . t

TEST YOUR KNOWLEDGE Find the inverse Laplace transform of the following functions : 1. (i) 3. (i) 5. (i)

1

1 s

(ii)

( s 2 + a 2 )2

s+a s+b 2

11.

(ii)

s( s + a )3

7. (i) log 9. (i)

(ii)

s2 + 4s

1 s( s2 + 16) 1 2

s( s + 2s + 2 )

s ( s 2 − 9 )2

(ii) log 2

s +1 s −1 2

s +b 1 s +1 log 2 (ii) log 2 2 s( s + 1) s +a

−1 (i) cot

s π

−1 (ii) tan

2 . s

2. (i) 4. (i) 6. (i)

1

(ii)

s3 − s 7

(ii)

s4 − s2

s2 − π 2

( s 2 + π 2 )2

F GH

8. (i) log 1 − 10. (i) log

a2 s

2

(ii)

I JK

(s + 1)2 (s + 2)(s + 3)

4 s3 − 2 s 2 4

π5

2

s (s + π 2 ) s +1

( s2 + 2s + 2)2

(ii) log (ii) log

1+s s

s2 + 1

( s − 1) 2

58

LAPLACE AND FOURIER TRANSFORMS

Answers 1. (i) 3. (i)

1 e − 4t − 4 4

1 a

3



(ii)

Fa GH 2

e − at a

2

3

4. (i) 7(sinh t – t)

1 cos 4t − 16 16

I JK

t2 + at + 1

1 1 − (sin t + cos t ) e − t 2

(ii)

(ii) sin π t +

π3 3 t − πt 6

5. (i)

1 t sin at 2a

(ii)

1 t sinh 3t 6

6. (i) t cos πt

7. (i)

e − bt − e − at t

(ii)

et − e − t t

8. (i)

9. (i)

cos at − cos bt t

10. 11.

(ii) e2t – 2t – 1

2. (i) cosh t – 1

(ii)

1 −t e t sin t 2

1 − e− t 2(1 − cos at ) (ii) t t

−t (ii) 1 + e − 2 cos t t

2( et − cos t) t sin 2t . (ii) t

e − 2t + e − 3t − 2e − t t sin πt (i) t

(i)

(ii)

Hint 6. (i) Use

F GH s

2

I′ = − s − π JK (s + π ) 2

s + π2

2

2

2 2

.

2.12. CONVOLUTION THEOREM The convolution theorem is used to find the inverse Laplace transform of the product of two functions with known inverse Laplace transforms of the factors of the product. Let F(s) and G(s) be two functions with known inverse Laplace transforms f(t) and g(t) respectively. The convolution theorem would help us to find the inverse Laplace transform of the product F(s) G(s). In this theorem, we shall prove that the inverse Laplace transformation of the product

z

F(s) G(s) is given by the function

t

0

f ( T ) g(t − T ) dT of t. This integral is called the convolution

of the functions f and g and it is denoted by f ∗ g. ∴

( f ∗ g)(t) =

z z z

t

0

f ( T ) g ( t − T ) dT .

Let z = t – T. ∴ dz = – dT ∴

( f ∗ g)(t) = =

0

t t

0

f (t − z) g ( z) . − dz = − g ( z) f (t − z) dz =

= (g ∗ f)(t) ∴

f∗g=g∗f

z

t

0

z

0

t

g ( z) f (t − z) dz

g ( T ) f (t − T ) dT

(Replacing z by T)

59

INVERSE LAPLACE TRANSFORMS

that

Statement. Let f(t) and g(t) be functions such that their Laplace transforms exist. Prove

L(f ∗ g) = L(f) L(g). Proof. It can be proved mathematically that the Laplace transform of the convolution of f and g (i.e., f ∗ g) exists.

zL z MNz zz zz

By definition, L( f ∗ g) = ∞

L ( f ∗ g) =



= =

0

0



t

0

0

e

0

t

− st

e



e − st ( f ∗ g) (t) dt

O f (T) g (t − T ) dT P dt Q

R T=0

T

e − st f ( T ) g(t − T ) dT dt

− st

J=



∂T ∂v

∂t ∂u

∂t ∂v

v

R′

=

1 0 1 1

=1

New variables transform the region R into the region R′ in the uv-plane defined by u ≥ 0, v ≥ 0.

zz zz FG Hz

T

O

∂T ∂u

t

45°

where R is the 45° wedge bounded by the lines T = 0 and T = t in the Tt-plane. Let u = T and v = t – T. ∴ T = u and t = u + v

L ( f * g) =

=

f ( T ) g(t − T ) dT dt ,

R



t

O

u

e − s(u + v) f (u) g (v)| J | du dv

R′

= =





0

0



0

e − su e − sv f (u ) g(v) du dv

e − su f (u) du

IJ FG KH

z



0

(∵ |J| = |1| = 1)

IJ K

e − sv g (v) dv

= L(f) L(g). ∴ L(f ∗ g) = L(f) L(g). Corollary 1. Let L(f) = F(s) and L(g) = G(s). ∴ The result of the convolution theorem can be written as L(f ∗ g) = F(s) G(s). ⇒ L–1 (F(s) G(s)) = f ∗ g L− 1 ( F( s ) G( s )) =

⇒ ∴ then

If

z z

t

f ( T ) g(t − T ) dT.

0

L− 1 (F(s)) = f(t) and L− 1 (G(s)) = g(t), L− 1 (F(s) G(s)) =

t

0

f(T) g(t − T) dT.

60

LAPLACE AND FOURIER TRANSFORMS

This formula is used to find the inverse Laplace transform of the product of two functions. −1 Corollary 2. Since f ∗ g = g ∗ f, we can also write L ( F( s ) G( s )) =

z

t

0

g( T ) f (t − T ) dT.

ILLUSTRATIVE EXAMPLES Example 1. Find the inverse Laplace transform of the function

FG H

IJ FG KH

IJ K

1 . (s + 3)(s − 2)

1 1 1 = = F(s) G(s), (s + 3)(s − 2) s+3 s−2 1 1 F(s) = and G(s) = s+3 s−2

Sol. where

L– 1 (F(s)) = L– 1 L– 1 (G (s)) = L– 1

and

FG 1 IJ = e H s + 3K FG 1 IJ = e H s − 2K

− 3t

= f (t), say

2t

= g(t), say .

By convolution theorem, we have L− 1 (F( s) G ( s)) = f ∗ g =



L− 1

FG 1 IJ = H (s + 3) (s − 2) K

=

z z

t

z

t

f (T) g (t − T) dT.

0

e − 3T e 2(t − T) dT

0 t

e

0

2t − 5T

e 2t − 5T dT = −5

t

=− 0

Example 2. Use convolution theorem to evaluate L–1 Sol.

1 2

2

2

s (s + a )

=

F(s) =

FG 1 IJ FG Hs K Hs 2

1 s

2

L–1(F(s)) = L–1 and

L–1(G(s)) = L–1

1 2

+a

2

1 − 3t e 2t e − 3t . [e − e 2t ] = − 5 5 5

F 1 I. GH s (s + a )JK 2

2

2

I = F(s) G(s), where JK

and G(s) =

1 2

s + a2

FG 1 IJ = t = f(t), say Hs K F 1 I = 1L F GH s + a JK a GH s

.

2

2

−1

2

By convolution theorem, we have

L–1(F(s) G(s)) = f ∗ g = g ∗ f =

z

t

0

a 2

+a

2

I = 1 sin at = g(t), say JK a

g (T) f(t – T) dT

61

INVERSE LAPLACE TRANSFORMS

L–1



z

F 1 I = FG 1 sin aTIJ (t – T) dT GH s (s + a ) JK H a K 2

2

2

t

0

t a

=

z

t

sin aT dT −

0

FG H

z

t

0

T sin aT dT

z

LM F IJ − 1 . FG − cos aT IJ dTOP G K H a K PQ MN H 1 L t cos at 1 sin aT O +0+ . (cos at – 1) – M− PP a MN a a a Q

t cos aT − = a a t =– 2 a

1 a

IJ K

t

0

t

1 cos aT T − – a a

0

t

0

t

0

t t 1 (1 – cos at) + 2 cos at – 3 sin at – 0 2 a a a 1 t 1 = 2 − 3 sin at = 3 (at – sin at). a a a

=

Example 3. Find the inverse Laplace transform of the function s

Sol.

2

(s + π )

where

F GH s

=

2 2

s 2

IF JK GH s

s

F(s) =

2

s +π

L– 1 (F(s)) = L– 1 L– 1 (G (s)) = L– 1

and



2

2

2



2

I = F(s) G(s), JK

and G(s) =

2

F GH s F GH s

1 2

1

s + π2

I JK I= 1L F JK π GH s

2

s 2

(s + π 2 ) 2

.

s = cos πt = f (t) , + π2

say

1 + π2

I = 1 sin πt = g(t) , JK π

−1

2

π + π2

.

say.

By convolution theorem, we have L− 1 ( F( s ) G( s )) = f * g =

∴ L– 1

F GH (s

2

I JK

s = + π 2 )2 =

z

t

0

1 2π

1 = 2π

=

z

t

0

cos πT .

z z

t

0 t

0

LM N

f ( T ) g(t − T ) dT. 1 sin π (t − T) dT π

[2 sin π (t − T) cos πT dT [sin πt + sin π (t − 2T)] dT =

LM N

1 cos π (t − 2T ) T sin πt − − 2π 2π

1 1 1 t sin πt + cos πt − 0 − cos πt 2π 2π 2π

OP Q

=

1 t sin πt. 2π

OP Q

t 0

62

LAPLACE AND FOURIER TRANSFORMS

Example 4. Find the inverse Laplace transform of the function s+3

Sol.

2

(s + 6 s + 13)

where

2

=

F(s) =

F GH s

s+3 s + 6 s + 13

= e–3t L–1 (G(s)) = L–1

(s + 6s + 13) 2

.

I = F(s) G(s), J + 6 s + 13 K 1

2

and

2

L–1 (F(s)) = L–1

and

IF JG + 6 s + 13 K H s s+3

2

s+3 2

G(s) =

F s+3 I =L GH s + 6s + 13 JK cos 2t = f(t), say F 1 I =L GH s + 6s + 13 JK

−1

2

−1

2

sin 2t = g(t), say. 2 By convolution theorem, we have

1 . s + 6 s + 13 2

F s+3 I =e GH (s + 3) + 4 JK F 1 I GH (s + 3) + 4 JK = e

–3t

2

L–1

–3t L–1

2

F GH s F GH s

I J + 4K 1 I J + 4K s

2

2

= e −3t

L–1 (F(s) G(s)) = f ∗ g =

∴ L–1

F GH (s

s+3 2

+ 6 s + 13)

2

I= JK

z

t

0

z

t

0

f (T) g (t − T) dt .

1 –3(t – T) e sin 2(t – T) dT. 2

e −3T cos 2T .

z

1 t −3 t e cos 2T sin 2(t − T) dT 2 0 e −3t t 2 sin 2(t − T) cos 2T dT = 4 0 e −3t t = (sin 2t + sin (2t − 4T)) dT 4 0

=

z z

LM OP N Q LMt sin 2t + 1 cos 2t − 0 − 1 cos 2tOP = 1 te 4 4 N Q 4

cos (2t − 4T) e −3t (sin 2t) T − = 4 −4 =

e

−3t

4

t

0

Example 5. Find the inverse Laplace transform of the function

where F(s) =

F GH

IF JK GH s

G(s) =

2

s 1 = 2 2 3 (s + a ) s + a2

Sol.

2

1 2

s +a

2

and

L–1 (F(s)) = L–1 and

L–1 (G(s)) = L–1

F GH s F GH s

2

1 + a2

s

s + a2 2

2

s + a2

sin 2t.

s

(s 2 + a 2 ) 3

.

I = F(s) F(s) G(s), JK

.

I F I JK GH JK I = cos at = g(t), say. JK

1 1 1 a = L−1 2 = sin at = f(t), say 2 2 a a +a s +a s

2

IF JK GH s

–3t

+ a2

63

INVERSE LAPLACE TRANSFORMS

By convolution theorem, we have

z

L–1 (F(s) F(s)) = f * f = L–1(F(s) F(s)) =



z

t

0

1 2a 2

=

1 2a 2

= =

f ( T ) f (t − T ) d T .

0

1 1 sin aT . sin a(t – T) dT a a

=

=

t

1 2a 2 1 2a 2 1 2a 2

z z

t

0

t

0

2 sin aT sin (at − aT) dT

[cos (2aT − at) − cos at] dT

LM sin (2aT − at) − T(cos at)OP N 2a Q LM sin at − t cos at − sin (− at) + 0OP 2a N 2a Q LM sin at − t cos atOP = 1 [sin at – at cos at] = h(t), say. N a Q 2a t

0

3

Again, by convolution theorem, we have L–1 (F(s) F(s) . G(s)) = h ∗ g = ∴ L–1

F GH (s

s 2

2 3

+a )

I= JK

z

t

1

0

2a3

z z z

1 2a 3

=

1 4 a3

=

1 4 a3

=

1 sin at 4 a3

0 t

0

LM N

t

0

t

0

h(T) g (t − T) dT .

(sin aT – aT cos aT) cos a(t – T) dT

=

t

z

(sin aT cos (at − aT) – aT cos aT cos (at – aT)) dT (2 sin aT cos (at − aT) – aT . 2 cos aT cos (at – aT)) dT (sin at + sin (2aT − at) – aT(cos at + cos (2aT – at))) dT

z z t

0

dT +

t

0

−a

=

1 4a 3

LM MN(sin at)T

t

0

z

sin (2aT − at) dT − a cos at

cos (2aT − at) − 2a

F GH

t 0

z

t

0

0

t

− 0

z

t

0

Tdt

T cos (2 aT − at) dT

T2 − (a cos at) 2

– a T . sin (2aT − at) 2a

t

1.

OP Q

t

0

sin (2aT − at) dT 2a

I OP JK P Q

64

LAPLACE AND FOURIER TRANSFORMS

=

1 4a

3

LMt sin at − 0 − 1 (cos at − cos at) − at 2a N

2

cos at 1 − (t sin at – 0) 2 2 +

=

1

4 a3 1 = 4 a3

FG H

1 − cos (2aT − at) 2 2a

LMt sin at − 1 at cos at − 1 t sin at − 1 (cos at − cos at)OP 2 2 4a N Q LM 1 t sin at − 1 at cos atOP . = t [sin at − at cos at]. 2 N2 Q 8a

IJ K

2

2

t 0

OP PQ

3

WORKING STEPS FOR SOLVING PROBLEMS Step I. Write the given function as the product of two functions say F(s) and G(s). Step II. Find the inverse Laplace transforms of the functions F(s) and G(s) and call these functions as f(t) and g(t) respectively. Step III. Write the result of convolution theorem as L− 1 ( F( s ) G( s )) =

z

t

0

f ( T ) g(t − T ) dT.

Step IV. Substitute the values of f(T) and g(t – T) and simplify the integral on the right side.

TEST YOUR KNOWLEDGE Find the inverse Laplace transform of the following functions by using convolution theorem : 1 4 1. (i) ( s − 2) ( s + 4 ) (ii) ( s − 3) ( s + 7) 2. (i) 3. (i) 4. (i) 5. (i) 6. (i) 7. (i) 8. (i) 9. (i)

1

(ii)

2

s ( s − 1)

1 2

(ii)

2

(ii)

2

s ( s + 1) 1

( s + 1) ( s + 1) 1 2

2

2

s (s − a ) 1

(ii)

s( s + 1)3

s2 2

(s + 4)

2

1 3

s − a3

s 2

( s + 2s )

(ii)

2

1 2

s ( s + 1)

1 3

s ( s2 + 1) 1 2

( s + 1)2 a2 s( s + a )2 4 s3 + s2 + s + 1

s2

(ii)

( s − 9 )2

(ii)

( s 2 + 4 s + 5 )2

(ii)

2

s+2

1 s( s + 2)3

.

65

INVERSE LAPLACE TRANSFORMS

Answers 1. (i)

1 2t [e – e–4t] 6

(ii)

2 3t [ e − e − 7t ] 5

2. (i) et – t – 1

(ii) 1 – cos t

3. (i) (t + 2) e–t + t – 2

(ii) cos t +

4. (i) 5. (i)

1 −t [ e + sin t − cos t ] 2

1

[sinh at − at]

a3

6. (i) −

1 et

LM N

LM t + t + 1OP MN 2 PQ + 1

1 3a

9. (i) −

2

1 (sin t − t cos t) 2

(ii) 1 − e − at (1 + at )

2

1 1 7. (i) 2 t cos 2t + 2 sin 2t

8. (i)

(ii)

t2 −1 2

LM MNe

at

−e



at 2

F cos GH

(ii) 2 [ e− t + sin t − cos t]

OP Q

LM N

1 1 (ii) 2 t cosh 3t + 3 sinh 3t

I OP JK P Q

3 3 at + 3 sin at 2 2

1 [ 2te −2t + e −2t − 1] 4

(ii)

OP Q

1 − 2t te sin t 2

(ii) −

1 8e

2t

( 2t2 + 2t + 1) +

1 . 8

2.13. INVERSE LAPLACE TRANSFORMS BY THE METHOD OF PARTIAL FRACTIONS Let

a 0 s m + a1s m − 1 + ...... + am

be a proper rational algebraic function of s with m < n. The b0 s n + b1s n − 1 + ...... + bn denominator of this quotient can be factorised into linear and quadratic factors. The given rational function can be expressed as the sum of partial fractions as per the rules given below : (i) If as + b is any linear non-repeated factor in the denominator, then there corresponds

A . as + b (ii) If as + b is any linear factor repeated r (∈ N) times in the denominator, then there

a partial fraction of the form

A B C , , ,...... r terms. 2 as + b ( as + b) ( as + b)3 (iii) If as2 + bs + c is any irreducible quadratic factor in the denominator, then there

corresponds partial fractions of the form

corresponds partial fraction of the form

As + B . as 2 + bs + c

(iv) If as2 + bs + c is any irreducible factor repeated r (∈ N) times in the denominator, then there corresponds partial fractions of the form

As + B Cs + D , , ...... r terms. as 2 + bs + c ( as 2 + bs + c ) 2

66

LAPLACE AND FOURIER TRANSFORMS

The quantities A, B, C, D, ...... are all constants independent of s. The constants A, B, C, D, ...... occurring in the numerators of the partial fractions are determined by simplifying the sum of partial fractions and then giving various values to s, to obtain equations involving unknown constants or by comparing the coefficients of like powers of s. ∴ The given proper fraction can be expressed as the sum of its partial fractions. Thus, by using linearity of inverse Laplace transform and elementary inverse Laplace transform formulae, the inverse Laplace transform of the given proper fraction is found. first.

If the given fraction is not proper, then division of numerator by denominator is carried

ILLUSTRATIVE EXAMPLES Example 1. Find the inverse Laplace transform of the following functions : (i)

s2 + s − 2 s(s + 3)(s − 2)

Sol. (i)

(ii)

s 2 − 10s + 13 . (s − 7)(s 2 − 5s + 6)

s2 + s − 2 is a proper fraction. s( s + 3) ( s − 2) s2 + s − 2 A B C = + + s( s + 3) ( s − 2) s s + 3 s − 2

Let

Multiplying both sides by s(s + 3)(s – 2), we get s2 + s – 2 = A(s + 3)(s – 2) + Bs(s – 2) + Cs(s + 3) s = 0 in (1)



– 2 = A(3)(– 2) + 0 + 0

...(1)

⇒ A = –2/– 6 = 1/3

s = – 3 in (1) ⇒ 9 – 3 – 2 = 0 + B(– 3)(– 5) + 0 ⇒ B = 4/15 s = 2 in (1)

⇒ C = 4/10 = 2/5

FG IJ FG IJ FG HK H K H FG 1 IJ + 2 L FG 1 IJ H s + 3K 5 H s − 2K

1 1 4 1 2 1 s2 + s − 2 1/3 4/15 2/5 + + = = + + 3 s 15 s + 3 5 s−2 s(s + 3) (s − 2) s s+3 s−2





⇒ 4 + 2 – 2 = 0 + 0 + C(2)(5)

L− 1

F s + s − 2 I = 1 L FG 1IJ + 4 L GH s(s + 3)(s − 2) JK 3 H s K 15 2

−1

=

−1

IJ K

−1

1 4 − 3t 2 2 t 1 4 − 3t 2 2t e (1) + e + e = + + e . 3 15 5 3 15 5

Note. Since the factors in the denominator are linear and non-repeated, the short-cut method can also be used.

67

INVERSE LAPLACE TRANSFORMS

(ii)

s 2 − 10 s + 13 s 2 − 10 s + 13 = is a proper fraction. 2 (s − 7)(s − 5s + 6) (s − 7)(s − 2)(s − 3)

The factors in the denominator are linear and non-repeated. ∴ By short-cut method, s 2 − 10 s + 13 7 2 + 10(7) + 13 2 2 − 10(2) + 13 3 2 − 10(3) + 13 = + + (s − 7) (5) (4) (− 5) (s − 2) (− 1) (− 4) (1) (s − 3) (s − 7) (s 2 − 5s + 6)

=

−8 −3 −8 + + 20 ( s − 7) 5 ( s − 2) − 4 ( s − 3) 2 3 1 − + 2. 5 ( s − 7) 5 ( s − 2 ) s−3

=−



L− 1

F s − 10s + 13 I = L F − 2 − 3 + 2 . 1 I GH (s − 7) (s − 5s + 6) JK GH 5 (s − 7) 5 (s − 2) s − 3 JK F 1 IJ − 3 L FG 1 IJ + 2L FG 1 IJ 2 =− L G 5 H s − 7K 5 H s − 2K H s − 3K 2

−1

2

−1

=−

−1

−1

2 7t 3 2t e − e + 2e3t . 5 5

Example 2. Find the inverse Laplace transform of the function

s2 by (s 2 + a 2 ) (s 2 + b 2 )

using (i) method of partial fractions (ii) convolution theorem. Sol. (i)

s2 2

2

2

2

(s + a ) (s + b )

=

=

=

=



L− 1

F GH (s

s2 2

2

+ a ) (s

2

z 2

( z + a ) ( z + b2 ) − a2

( z + a 2 ) ( − a 2 + b2 ) a2 2

a −b

2

a 2

a −b

I= J + b )K a 2

, where z = s 2

2

a 2

−b

2

.

.

1 z+a

2



a 2

s +a

L− 1

F GH s

2

+

( − b2 + a 2 ) ( z + b2 )

b2 2

a −b



+a

.

2

b 2

a −b

a 2

− b2

2

2

I− JK a

1 z + b2 .

b 2

s + b2

b 2

−b

2

L− 1

.

F GH s

b 2

+b

2

I JK

68

LAPLACE AND FOURIER TRANSFORMS

a

=

a 2 − b2

F GH

a 2 − b2

s2 s = 2 2 2 2 2 (s + a )(s + b ) s + a2

where F(s) =

s 2

s +a

2

and G(s) =

L− 1

and

IF JK GH s

s

2

s + b2

sin bt

.

I = F(s) G(s), JK

.

2

s + b2

F s I = cos at = f (t), say GH s + a JK F s I = cos bt = g(t), say. (G( s )) = L G H s + b JK

L− 1 (F(s)) = L− 1

Also

b a 2 − b2

a sin at − b sin bt

=

(ii)

sin at −

2

−1

2

2

2

By convolution theorem, we have L− 1 (F( s) G( s)) = f ∗ g =



L− 1

F GH (s

s2 2

2

+ a ) (s

2

z

t

0

I= J + b )K 2

=

f (T) g (t − T) dT.

z

t

cos aT cos b(t − T ) dT

0

1 2

z

t

0

cos ( aT + bt − bT ) + cos ( aT − bt + bT ) dT

LM N 1 L sin at sin at sin bt sin bt O = M + − + P 2 Na−b a+b a−b a+bQ O 1 L 2a 2b = M sin at − sin bt P 2 Na − b a −b Q =

1 sin (( a − b) T + bt ) sin (( a + b) T − bt ) + 2 a−b a+b

2

=

2

2

1 (s − 1) (s + 2)

=

1 5

0

a sin at − b sin bt . a 2 − b2

(i) method of partial fractions (ii) convolution theorem. 5

t

2

Example 3. Find the inverse Laplace transform of the function

Sol. (i)

OP Q

z ( z + 3)

, where

z = s – 1.

1 by using (s − 1) 5 (s + 2)

69

INVERSE LAPLACE TRANSFORMS

We divide 1 by z + 3 as follows : 1 z z2 z3 z4 − + − + 3 9 27 81 243 1

3+ z

1+

z 3



z 3



z z2 − 3 9 z2 9 z 2 z3 + 9 27 −

z3 27



z3 z4 − 27 81 z4 81 z4 z5 + 81 243 −

z4 z 5 / 243 1 1 z z2 z3 = − + − + − 3 + z 3 9 27 81 243 3+z

∴ ∴ ∴

1 5

z ( z + 3) 1 5

(s − 1) (s + 2)

1

=

z

=

5

FG 1 IJ = 1 H 3 + z K 3z

5

1 3(s − 1)



∴ L− 1

z5 243

5





1 9z

1 9(s − 1)

4

4

+

+

1 27 z

3



1 81z

2

+

1 1 − 243 z 243(3 + z)

1 27(s − 1) 3

1 1 1 + − 2 243(s − 1) 243(s + 2) 81(s − 1)

F 1 I = 1L F 1 I − 1L F 1 I GH (s − 1) (s + 2) JK 3 GH (s − 1) JK 9 GH (s − 1) JK F 1 I− 1 L F 1 I + 1 L F 1 I− 1 L F 1 I 1 + L G 27 H (s − 1) JK 81 GH (s − 1) JK 243 GH s − 1JK 243 GH s + 2 JK −1

5

−1

3

−1

5

−1

4

2

−1

−1

70

LAPLACE AND FOURIER TRANSFORMS

FG IJ + 1 e L FG 1 IJ H K 27 Hs K 1 F 1 I 1 e L FG 1IJ − 1 e L FG 1IJ − e L G J+ H s K 243 H s K 243 H sK 81 e 1 F 4 !I e 1 F 3 ! I e . 1 L FG 2 !IJ − e . t + e . 1 − e . 1 = . L G J− . L G J− H s K 9 6 H s K 27 2 H s K 81 243 243 3 24 =

FG IJ H K

1 t −1 1 1 1 e L − e t L− 1 4 5 3 9 s s t

t

= (ii)

−1

−1

t

F GH

t

−1

4

−1

t

− 2t

3

I FG 1 IJ = F(s) G(s), JK H s + 2 K

1 1 = (s − 1) (s + 2) (s − 1) 5

where

F(s) =

1 (s − 1) 5

F 1 I = e L FG 1 IJ GH (s − 1) JK Hs K 1 F 4 !I e t = f (t), say =e . L G J= H s K 24 24 F 1 IJ = e = g(t), say . (G (s)) = L G H s + 2K 1.t

–1

5

t

L− 1

1 . s+2

and G(s) =

L− 1 (F(s)) = L− 1

and

– 2t

t

−1

5

3

−1

2

t

−1

et 4 et 3 et 2 et et e− 2t t − t − t − t+ − . 72 54 54 81 243 243

5

Also

t

5

t

−1

4

5

−1

– 2t

By convolution theorem, we have L− 1 (F( s) G( s)) = f ∗ g =



Now

L− 1

F 1 I= GH (s − 1) (s + 2) JK 5

z

t

0

4

T e

3T

z

t

0

z

t

0

f (T) g (t − T) dT.

e − 2t e T T 4 − 2( t − T ) .e dT = 24 24

T 4 e 3T dT = 3

t

− 0

LM MN

z

t

0

4T 3 .

t 4 e 3t 4 T 3 e 3 T = − 3 3 3



LM MN

t

0

T 4 e 3T dT.

e 3T dT 3

t

0

z

z

t

0

3T 2 .

t 4 e 3t 4 3 3t 4 T 2 e 3 T = − t e + 3 9 3 3

t

− 0

LM MN

e3T dT 3

z

t

0

t 4 e3t 4 3 3t 4 2 3t 8 Te3T = − t e + t e − 3 9 9 9 3

OP PQ

2T . t

− 0

e3T dT 3

z

t

0

1.

OP PQ

e3T dT 3

OP PQ

71

INVERSE LAPLACE TRANSFORMS

= L− 1



t 4 e 3t 4 3 3t 4 2 3t 8 3t 8 3t − t e + t e − te + ( e − 1) 3 9 9 27 81

F 1 I = e LM t e GH (s − 1) (s + 2) JK 24 MN 3 − 2t

5

4 3t



4 3 3t 4 2 3t 8 3t 8 3t 8 t e + t e − te + e − 9 9 27 81 81

OP PQ

t 4e t t 3e t t 2et te t et e− 2t − + − + − . 72 54 54 81 243 243 Example 4. Find the inverse Laplace transform of the following functions : =

(i)

4s + 5

(ii)

(s − 1) 2 (s + 2)

Sol. (i) Let

4s + 5 ( s − 1) 2 ( s + 2 )

1 + 2s (s + 2) 2 (s − 1) 2

.

is a proper fraction.

4s + 5 A B C = + + 2 2 s+2 (s − 1) (s + 2) s − 1 (s − 1)

Multiplying both sides by (s – 1)2 (s + 2), we get 4s + 5 = A(s – 1) (s + 2) + B(s + 2) + C(s – 1)2. s = 1 in (1) ⇒ 9 = 0 + 3B + 0 ⇒ B=3 s = – 2 in (1) ⇒ – 3 = 0 + 0 + 9C ⇒ C = – 1/3 Let s = 0. ∴ 5 = – 2A + 2B + C ⇒ 2A = 2(3) – (1/3) – 5 = 2/3 ⇒ A = 1/3.

FG H

L− 1



F 4s + 5 I = 1 L F 1 I + 3L F 1 GH ( s − 1) (s + 2) JK 3 GH s − 1JK GH (s − 1) 1 F 1I 1 = e + 3e L G J − e Hs K 3 3 −1

2

1. t

=

(ii)

I FG IJ FG JK H K H I−1L F 1 I JK 3 GH s + 2 JK

1 1 1 1 1 4s + 5 1/3 3 − 1/3 +3 − = + + = 2 2 2 + s − s 3 1 3 2 s+2 (s − 1) (s − 1) (s + 2) s − 1 (s − 1)



1 + 2s

−1

1. t

−1

2

2

IJ K

−1

− 2t

1 t 1 1 e + 3e t t1 − e − 2t = ((9t + 1) e t − e − 2t ) . 3 3 3

is a proper fraction.

( s + 2) 2 ( s − 1) 2

A B C D + + + 2 s + 2 s − 1 (s + 2) (s − 1) (s + 2) (s − 1) 2 Multiplying both sides by (s + 2)2 (s – 1)2, we get 1 + 2s = A(s + 2) (s – 1)2 + B(s – 1)2 + C(s + 2)2 (s – 1) + D(s + 2)2. s = 1 in (1) ⇒ 3 = 0 + 0 + 0 + 9D ⇒ D = 1/3 s = – 2 in (1) ⇒ – 3 = 0 + 9B + 0 + 0 ⇒ B = – 1/3

Let

...(1)

1 + 2s 2

2

=

...(1)

72

LAPLACE AND FOURIER TRANSFORMS

Comparing coefficients of s3 and s2 in (1), we get 0=A+C

...(2)

2

(s + 2) ( s − 1) L− 1



...(3)

A = – C = 0.

1 + 2s



0 = B + 3C + D

1 1 − =0 ⇒ C=0 3 3

(3) ⇒ 3C = – B – D = ∴ (2) ⇒

and

=

2

I JK

F GH

F 1 + 2s I = − 1 L F 1 I + 1 L F 1 I GH (s + 2) (s − 1) JK 3 GH (s + 2) JK 3 GH (s − 1) JK 1 F1I 1 F1I =− e L G J+ e L G J Hs K 3 Hs K 3 2

F GH

0 − 1/3 0 1/3 1 1 1 1 + + + + =− s + 2 (s + 2) 2 s − 1 (s − 1) 2 3 (s + 2) 2 3 (s − 1) 2 −1

2

− 2t

=−

−1

2

−1

1.t

2

I JK

2

−1

2

1 − 2t 1 t e . t + e t . t = (e t − e − 2t ). 3 3 3

Example 5. Find the inverse Laplace transform of the following functions : (i)

3s + 1

(ii)

(s + 1) (s 2 + 1)

Sol. (i)

3s + 1 ( s + 1) ( s 2 + 1)

(s 2 + 1) (s 2 + s + 1)

.

is a proper fraction.

3s + 1

Let

2s 3 + 2s 2 + 4s + 1

2

( s + 1) ( s + 1)

=

A Bs + C + 2 . s +1 s +1

Multiplying both sides by (s + 1)(s2 + 1), we get 3s + 1 = A(s2 + 1) + (Bs + C) (s + 1). s = – 1 in (1)



...(1)

– 2 = 2A + 0 ⇒ A = – 1

Comparing coefficients of s2 and s in (1), we get 0=A+B

...(2)

and

3=B+C

...(3)

(2) ⇒ B = – A = – (– 1) = 1 (3) ⇒ C = 3 – B = 3 – 1 = 2. 3s + 1





( s + 1) ( s 2 + 1) L− 1

=

1 2 s − 1 1. s + 2 =– + 2 + 2 + 2 1 s + s +1 1 s + s +1 + s 1

F 3s + 1 I = − L FG 1 IJ + L F GH (s + 1) (s + 1) JK H s + 1K GH s 2

−1

−1

I + 2L F 1 I J GH s + 1JK + 1K

s 2

= – e t + cos t + 2 sin t.

–1

2

73

INVERSE LAPLACE TRANSFORMS

(ii) Let

2s 3 + 2s 2 + 4s + 1

F( s ) =

.

( s 2 + 1) ( s 2 + s + 1)

∴ F(s) is a proper fraction. Let

2s 3 + 2s 2 + 4s + 1

=

( s 2 + 1) ( s 2 + s + 1)

As + B s2 + 1

+

Cs + D s2 + s + 1

.

Multiplying both sides by (s2 + 1) (s2 + s + 1), we get 2s3 + 2s2 + 4s + 1 = (As + B) (s2 + s + 1) + (Cs + D) (s2 + 1). Comparing coefficients of s3, s2, s and constant terms, we get 2=A+C

...(1)

2=A+B+D

...(2)

4=A+B+C

...(3)

1=B+D

...(4)

(2) – (4)

⇒ A=1

(1)

⇒ C=2–A=2–1=1

(3)

⇒ B=4–A–C=4–1–1=2

(4)

⇒ D = 1 – B = 1 – 2 = – 1.

∴ ∴

F( s ) =

1. s + 2 2

s +1

+

1. s −1

=

2

s + s +1

F GH

I JK

s s−1 1 +2 2 + 2 s +1 s +1 s +s+1 2

I + 2L F 1 I + L F s − 1 I J GH s + 1JK GH s + s + 1JK + 1K F FG s + 1IJ − 3 I G H 2 K 2 JJ = cos t + 2 sin t + L G GG FG s + 1IJ + 3 JJ H H 2K 4 K F s− 3 I G 2 JJ = cos t + 2 sin t + e L G GH s + 43 JK F I s 3 /2 J G = cos t + 2 sin t + e L G − 3 GH s + 43 s + 43 JJK F cos 3 t − 3 sin 3 tI = cos t + 2 sin t + e . GH 2 2 JK

L− 1 (F( s)) = L− 1

F GH s

s

2

−1

−1

2

–1



1 t 2

2

2

−1

2



1 t 2

−1

2



1 t 2

2

74

LAPLACE AND FOURIER TRANSFORMS

Example 6. Find the inverse Laplace transform of the function s

Sol.

4

s + 4a

4

=

4

=

s

∴ Let

4

s + 4a

s 2

2 2

2 2

( s + 2a ) − 4a s

=

s . s + 4a 4 4

s 2

2

( s + 2as + 2a ) ( s 2 − 2as + 2a 2 )

As + B Cs + D . + 2 2 s + 2as + 2a s − 2as + 2a 2 2

Multiplying both sides by s4 + 4a4, we get s = (As + B)(s2 – 2as + 2a2) + (Cs + D) (s2 + 2as + 2a2). Comparing coefficients of s3, s2, s and constant terms, we get 0=A+C

...(1)

0 = – 2aA + B + 2aC + D

...(2)

1 = 2a2A – 2aB + 2a2C + 2aD

...(3)

(2) – (5)

+

2a2

D

...(4)

⇒ B+D=0

...(5)

⇒ – 2a(A – C) = 0 ⇒ A = C

...(6)

0= (4)

2a2B

∴ (1) and (6) ⇒ A = 0, C = 0 (3)

⇒ 1 = – 2a (B – D) ⇒ B – D = – 1/2a

...(7)

Solving (5) and (7), we get B = – 1/4a, D = 1/4a. ∴

s 4

s + 4a

4

=

0. s − 1/4 a 0. s + 1/4 a + 2 2 s + 2as + 2a s − 2as + 2a 2 2

F I F I GH JK GH JK I+ 1 F 1 I 1 F 1 =− G 4 a H ( s + a) + a JK 4 a GH ( s − a) + a JK I =− 1 L F 1 I+ 1 L F 1 JK 4a GH (s + a) + a JK 4a GH (s − a) + a F 1 I+ 1 e L F 1 I 1 e =− L G GH s + a JK 4a H s + a JK 4a F a I = 1 Fe −e 1 1 = (e − e ). L G a 4a H s + a JK 2a GH 2 =–

1 1 1 1 + 2 2 2 4 a s + 2as + a 4 a s − 2as + 2a 2 2

∴ L− 1

F GH s

s 4

+ 4a

4

2

−1

− at

at

=

1 2a 2

2

2

−1

− at

−1

2

at

2

2

−1

sinh at sin at.

2

2

−1

2

2

2

I JK

2

at

2

2

− at

I . sin at JK

75

INVERSE LAPLACE TRANSFORMS

TEST YOUR KNOWLEDGE Find the inverse Laplace transform of the following functions by the method of partial fractions (Q No 1–12) : 1. (i) 2. (i) 3. (i) 4. (i)

5. (i) 6. (i) 7. (i) 8. (i) 9. (i) 10.

(i)

11.

(i)

12.

(i)

1 ( s − 1) ( s − 2) 2

(ii)

3s + 7

(ii)

s − 2s − 3 2s2 + 5s − 4

(ii)

s3 + s2 − 2s

1

(ii)

s2 (s2 + 1) s2 − 2s + 3 2

( s − 1) ( s + 1) 2

5s − 2

s ( s + 2 ) ( s − 1)

5s2 − 15s − 11 3

( s + 1) ( s − 2 ) 1 2

( s + 1) ( s + 1) 3s + 1

( s − 1) ( s2 + 1) s s 4 + s2 + 1

s2 + s

( s2 + 1) ( s2 + 2s + 2) s 2

2

( s + 1) ( s + 1)

(ii) (ii) (ii) (ii) (ii) (ii) (ii) (ii)

1 ( s + a ) ( s + b) 3s − 11

s2 − 7s + 12

s2 − 6

s3 + 4 s 2 + 3 s

1 s2 ( s2 + 1) ( s2 + 9) 1 2

( s + 2 ) ( s − 2) 1 2

s ( s − a )2

3s3 − 3s2 − 40s + 36 ( s 2 − 4 )2 5s + 3

( s − 1) ( s2 + 2s + 5) 1 s3 ( s2 + 1)

1 s3 − a3

s3 s4 − a 4 a ( s2 − 2a 2 ) s4 + 4a 4

.

Answers

2. (i) 4e3t – e–t

e − at − e − bt b−a (ii) e4t + 2e3t

3. (i) 2 + et – e–2t

(ii) − 2 +

4. (i) t – sin t

(ii)

1. (i) e2t – et

FG H

5. (i) t −

IJ K

1 t 3 −t e + e 2 2

(ii)

5 − t 1 − 3t e + e 2 2

1 1 1 t − sin t + sin 3t 9 8 216 1 2t (e − (4t + 1) e− 2t ) (ii) 16

76

LAPLACE AND FOURIER TRANSFORMS

6. (i) t + et + e–2t – 2

(ii)

OP Q

LM N

7 1 2t 1 + 4t − t2 − e − t 7. (i) e 3 2 3

8. (i)

(i)

FG H

1 −t [ e − cos t + sin t] 2

(ii)

2

sin

3 1 3 a2

LM MNe

t −t cos 2t − (ii) e − e

(ii)

F F GH GH

− e− at / 2 cos

I JK

3 at + 3 sin 2

1 3 (1 + e − t ) sin t + (1 − e − t ) cos t 5 5 1 −t (i) [sin t − te ] 2

11.

(i)

12.

F GH

3 at 2

(ii)

3 sin 2t 2

IJ K

1 2 t + cos t − 1 2

t 3 t sinh 2 2

at

[(2 + at) − (2 − at) eat ]

(ii) – 10t sinh 2t + 3e–2t + 3te2t

9. (i) 2et – 2 cos t + sin t 10.

1 a3

I I OP JK JK P Q

1 cos at + cosh at 2

(ii) cos at sinh at.

Hints 4. (i)

1 s2 (s2 + 1)

=

=

1 2 z ( z + 1) , where z = s

1 1 1 1 1 1 + = − . = 2 − 2 z(1) (− 1)( z + 1) z z + 1 s s +1

(i) s4 + s2 + 1 = (s4 + 2s2 + 1) – s2.

10.

2.14.

SOLUTION OF DIFFERENTIAL EQUATIONS BY USING LAPLACE TRANSFORMATION

We know that a differential equation with initial conditions is solved by first finding the general solution of the given equation and then finding the values of the arbitrary constants by using the given initial conditions. By using Laplace transformations, we find the required solution without first finding the general solution of the given equation.

ILLUSTRATIVE EXAMPLES Example 1. Solve the following equations : (i)

d2 y dt 2 d2 y

+ y = 6 cos 2t, where y′(0) = 1, y(0) = 3.

dy dy – 3y = sin t, where y = = 0, when t = 0. dt dt dt dy (iii) (D2 + m2) y = a cos nt, where y = 0 = when t = 0. dt

(ii)

(iv)

2

d3 y dt 3

+2

+2

dy d 2 y dy d2 y = 0, where y = 1, = 2, = 2 at t = 0. 2y − − dt dt dt 2 dt 2

77

INVERSE LAPLACE TRANSFORMS

Sol. (i) We have d2 y + y = 6 cos 2t. dt 2 y″ + y = 6 cos 2t

⇒ Taking Laplace Transform, we have L(y″) + L(y) = 6L(cos 2t)

s

[s2L(y) – s . y(0) – y′(0)] + L(y) = 6



s2 y – s . 3 – 1 + y =



(s2 + 1) y =



2

s + ( 2) 2

6s s +4

(Putting y = L(y))

2

6s + 3s + 1 s2 + 4

6s 3s 1 + 2 + 2 2 (s + 1) (s + 4) s + 1 s + 1 Taking inverse Laplace transform, we have y =



y = 6L–1

F GH (s

2

s 2

+ 1) (s

2

I + 3L F J GH s + 4) K −1

I +L F 1 I J GH s + 1JK + 1K

s 2

s s 1 . 2 = 2 = F(s) G(s), 2 (s + 1) (s + 4) s +1 s +4

−1

2

2

where

s

F(s) =

s +1

L–1(F(s)) = L–1 and

and G(s) =

2

L–1(G(s)) = L–1

1 2

s +4

.

F s I = cos t = f(t), say GH s + 1JK F 1 I = 1 sin 2t = g(t), say GH s + 4 JK 2 2

2

By convolution theorem, L–1(F(s) G(s)) = f ∗ g = = =

z

t

0

1 4

z

0

(cos T)

z

t

0

t

f ( T ) g (t − T ) d T

FG 1 sin 2(t − T)IJ dT = 1 H2 K 4

z

t

0

2 sin (2t – 2T) cos T dT

(sin (2t − T) + sin (2t − 3T)) dT

LM OP N Q 1L cos t cos 2t O − = Mcos t − cos 2t + 4N 3 3 PQ 1 L4 4 O 1 = M cos t − cos 2t P = [cos t – cos 2t] 4 N3 3 Q 3 =

1 cos (2t − T) cos (2t − 3T) − − 4 −1 −3

t

0

...(1)

78

LAPLACE AND FOURIER TRANSFORMS

L–1

Also,

F GH s

I J = cos t + 1K

s 2

∴ (1) ⇒

and L–1

F 1 I GH s + 1JK = sin t 2

1 (cos t – cos 2t) + 3 cos t + sin t 3 y = 5 cos t + sin t – 2 cos 2t.

y=6.

or (ii) We have

d2 y

dy – 3y = sin t. dt dt y″ + 2y′ – 3y = sin t 2

+2

⇒ Taking Laplace transform, we have L(y″) + 2L(y′) – 3L(y) = L(sin t) ⇒

[s2L(y) – s . y(0) – y′(0)] + 2[sL(y) – y(0)] – 3L(y) =



s2 y – s.0 – 0 + 2[s y – 0] – 3 y = (s2 + 2s – 3) y =



1 2

s +1 1 s2 + 1

1 2

s +1

1 1 = 2 (s 2 + 1) (s 2 + 2s − 3) (s + 1) (s + 3) (s − 1) Taking inverse Laplace transform, we have y =



y = L–1 1

F GH (s

I J + 1) (s + 3) (s − 1) K 1

2

As + B C D + + 2 s +1 s+3 s−1 (s + 1) (s + 3) (s − 1) 2 Multiplying by (s + 1) (s + 3) (s – 1), we get

Let

2

=

1 = (As + B) (s + 3) (s – 1) + C(s2 + 1) (s – 1) + D(s2 + 1) (s + 3) s = – 3 ⇒ 1 = 0 + c(10) (– 4) + 0

⇒ C=–

...(1)

1 40

1 8 Comparing the coefficients of s3 and constant terms in (1), we get

s=1

⇒ 1 = 0 + 0 + D(2) (4)

⇒ D=

0=A+C+D 1 = – 3B – C + 3D 1 1 1 − =– 40 8 10 1 3 3 + =– (3) ⇒ 3B = – 1 – C + 3D = – 1 + 40 8 5 1 1 1 1 − s− − 1 10 5 + 40 = + 8 s+3 s2 + 1 (s 2 + 1) (s + 3) (s − 1) s −1

(2)



...(2) ...(3)

A=–C–D=

∴ B=–

1 5

79

INVERSE LAPLACE TRANSFORMS

=– y=–



1 1 1 1 1 1 1 s . − . − . + . 10 s 2 + 1 5 s 2 + 1 40 s + 3 8 s − 1

I JK

F GH

FG H

I JK

F GH

IJ K

FG H

IJ K

1 −1 1 1 1 1 −1 s 1 1 L + L−1 – L − L−1 2 2 40 8 s+3 s−1 10 5 s +1 s +1

1 1 1 −3t 1 t cos t − sin t − e + e . 10 5 40 8 (D2 + m2) y = a cos nt.

=–

(iii) We have

y″ + m2y = a cos nt



Taking Laplace transform, we have L(y″) + m2L(y) = a L(cos nt) ⇒

[s2L(y) – s . y(0) – y′(0)] + m2L(y) = a . s2 y – s.0 – 0 + m2 y =



s 2

s + n2 as s + n2 2

as s2 + n2



(s2 + m2) y =



y =



y = aL–1

as (s + m ) (s 2 + n 2 ) 2

2

F GH (s

2

s + m ) (s 2 + n 2 ) 2

I JK

...(1)

s s 1 s . 2 2 2 = 2 2 2 = F(s) G(s), where F(s) = 2 (s + m ) (s + n ) s +m s +n s + m2 2

and

2

1

G(s) =

2

s + n2

L–1(F(s)) = L–1 and

L–1(G(s)) = L–1 By convolution theorem, L–1(F(s) G(s)) = f ∗ g = =

z

t

0

F GH s F GH s

z

t

0

1 2n

=

1 2n

z z

t

0 t

0

2

I JK = cos mt = f(t), say I 1 JK = n sin nt = g(t), say

s + m2 1 + n2

f (T) g (t − T) dT

cos mT .

=

2

.

1 sin n(t – T) dT n

2 sin (nt − nT) cos mT dT [sin (nt + (m − n) T) + sin (nt − (m + n) T)] dT

80

LAPLACE AND FOURIER TRANSFORMS

LM OP N Q 1 L cos mt cos nt cos mt cos nt O − + + − = 2n MN m − n m − n m + n m + n PQ 1 LF 1 1 I = − MG J cos mt + FGH m 1− n − m 1+ n IJK cos ntOPQ 2n NH m + n m − n K 1 1 L 2n cos mt 2n cos nt O − + = = (cos nt – cos mt) M P 2n N m − n m −n m −n Q =

1 cos (nt + (m − n) T) cos (nt − (m + n) T) − − 2n − (m + n) m−n

2



(1)

2

2

2

0

2

a (cos nt – cos mt). m2 − n2

y=



2

t

(iv) We have d3 y dt

3

+2

d2 y dt

2



dy – 2y = 0. dt

y′″ + 2y″ – y′ – 2y = 0



Taking Laplace transform, we have L(y′″) + 2L(y″) – L(y′) – 2L(y) = 0 ⇒

[s3L(y) – s2 . y(0) – s . y′(0) – y″(0)] + 2[s2L(y) – s . y(0) – y′(0)] – [sL(y) – y(0)] – 2L(y) = 0



(s3 y – s2.1 – s.2 – 2) + 2(s2 y – s.1 – 2) – (s y – 1) – 2 y = 0



s3 y – s2 – 2s – 2 + 2s2 y – 2s – 4 – s y + 1 – 2 y = 0



(s3 + 2s2 – s – 2) y = s2 + 4s + 5





y =

s2 + 4 s + 5 s2 + 4s + 5 = s 3 + 2s 2 − s − 2 s 2 (s + 2) − (s + 2)

y =

s2 + 4s + 5 (s − 1) (s + 1) (s + 2)

=

1+ 4 + 5 1− 4 + 5 4−8+5 + + (s − 1) (2) (3) (− 2) (s + 1) (1) (− 3) (− 1) (s + 2)

=

5 1 1 1 1 . − + . 3 s−1 s+1 4 s+2

y= =

FG H

IJ K

FG H

IJ K

FG H

5 −1 1 1 1 1 L − L−1 + L−1 3 3 s−1 s+1 s+2 5 t 1 e − e − t + e −2t . 3 3

IJ K

81

INVERSE LAPLACE TRANSFORMS

TEST YOUR KNOWLEDGE Solve the following differential equations : 1. 2.

d2 y

+ y = t, where y(0) = 1, y′(0) = – 2.

dt 2

d2 y dt

2

+4

dy + 3y = e–t, where y(0) = y′(0) = 1. dt

d 2x

dx + = 2, where x(0) = 3, x′(0) = 1. dt dt2 4. y″ – 3y′ + 2y = 4t + e3t, where y(0) = 1, y′(0) = – 1.

3.

d3 y

d2 y

dy +3 – y = t2et, where y(0) = 1, y′(0) = 0, y″(0) = – 2. dt dt dt 2 6. (D4 + 2D2 + 1) y = 0, where y(0) = 0, y′(0) = 1, y″(0) = 2, y′″(0) = – 3.

5.

7.

3

d2y dt

2

−3

+

dy = t2 + 2t, where y(0) = 4, y′(0) = – 2. dt

Answers 3 −3t 7 −t 1 −t e + e + te 4 4 2

1. y = t – 3 sin t + cos t

2. y = –

3. x = 2t + e–t + 2

4. y = 2t + 3 +

LM t − t − t + 1OP MN 60 2 PQ 5

5. y = et 7.

y=

1 3t 1 t e − e – 2e2t 2 2

2

1 3 t + 2e −t + 2 . 3

6. y = t(sin t + cos t)

3

Solution of Integral Equations Using Laplace Transformation 3.1. INTRODUCTION

Integral equations occur very frequently in the fields of mechanics and mathematical physics. The development of the theory of integral equations began with the works of Italian mathematician V. Volterra (1896) and the Swedish mathematician I. Fredholm (1900). 3.2. DEFINITION OF INTEGRAL EQUATION An equation involving an unknown function under the integral and perhaps also outside it is called an integral equation. For example the equation y(t ) = f (t ) +

z

b

a

y(u ) k(t, u ) du , where f(t), k(t, u) are known

functions and the function y(t) is to be determined, is an integral equation. If the integral involved in an integral equation is of the form

z

t

0

y(u) k(t − u) du, then

the integral equation is said to be an integral equation of convolution type. 3.3. METHOD OF SOLVING INTEGRAL EQUATION OF CONVOLUTION TYPE y(t) = f (t) +

Let

z

t

0

y(u) k(t − u) du

...(1)

be an integral equation of convolution type. The integral on the right side is the convolution of the functions y(t) and k(t). ∴ (1) can be written as y(t) = f(t) + (y(t)∗k(t)) Taking the Laplace transform on both sides, we get L(y(t)) = L(f(t)) + L(y(t)∗k(t)). ⇒ ⇒ where

L(y(t)) = L(f(t)) + L(y(t)).L(k(t))

(By convolution theorem)

Y(s) = F(s) + Y(s) K(s), L(y(t)) = Y(s), L(f(t)) = F(s) and L(k(t)) = K(s).



(1 – K(s)) Y(s) = F(s) ⇒

Y (s) = 82

F(s) 1 − K (s)

83

SOLUTION OF INTEGRAL EQUATIONS USING LAPLACE TRANSFORMATION

FG F(s) IJ H 1 − K(s) K

L– 1 (Y (s)) = L– 1





y(t) = L– 1

This gives the solution of the given integral equation.

FG L(f(t)) IJ . H 1 − L(k(t))K

ILLUSTRATIVE EXAMPLES Example 1. Solve the following integral equation by using Laplace transforms :

z z

y(t) = 1 +

0

y (t ) = 1 +

Sol. We have

t

t

0

y(u) sin (t − u) du. y(u) sin (t − u) du .

⇒ y(t) = 1 + (y(t) ∗ sin t) Taking the Laplace transform, we get L(y(t)) = L(1) + L(y(t) ∗ sin t) 1 ⇒ (By convolution theorem) L( y(t)) = + L( y(t)) . L(sin t) s 1 1 Y ( s ) = + Y( s ) . 2 , where Y( s ) = L( y( t )) ⇒ s s +1 ⇒

F1 − 1 I Y(s) = 1 GH s + 1JK s

Y( s ) =



2

y(t) = L– 1 (Y (s)) = L– 1



s2 + 1 s

3

=

1 1 + s s3

FG 1IJ + L FG 1 IJ = 1 + 1 L FG 2 ! IJ = 1 + 1 t H sK H s K 2 H s K 2 –1

–1

3

2+1

2

.

Example 2. Solve the following integral equation by using Laplace transforms : y(t) = te t − 2e t

Sol. We have

y(t ) = te t − 2e t



y(t ) = te t − 2

z z

t

0

z

t

0

t

0

y(u) e − u du.

y(u ) e − u du .

y(u ) e t − u du

⇒ y(t) = te t − 2 ( y(t) ∗ e t ) Taking the Laplace transform, we get L(y(t)) = L(tet) – 2L(y(t) ∗ et). ⇒ L(y(t)) = L(tet) – 2(L(y(t)) . L(et)) ⇒ ⇒ ∴

Y (s) = −

FG H

IJ K

FG H

IJ K

d 1 1 − 2 Y (s) ds s − 1 s−1

FG 1 + 2 IJ Y(s) = 1 H s − 1K (s − 1)

2

y(t ) = L− 1 (Y( s )) = L− 1

(By convolution theorem) ⇒

Y (s) =



Y( s ) =

F 1 I = sinh t. GH s − 1JK 2

1 (s − 1)

2

− 2 Y (s)

FG 1 IJ H s − 1K

1 1 = ( s − 1) ( s + 1) s 2 − 1

84

LAPLACE AND FOURIER TRANSFORMS

Example 3. Solve the following integral equation by using Laplace transforms : y(t) = e − t − 2 y(t) = e − t − 2

Sol. We have

z z

t

0

y(u) cos (t − u) du.

t

0

y(u) cos (t − u) du.

⇒ y(t) = e–t – 2 (y(t) ∗ cos t) Taking the Laplace transform, we get L(y(t)) = L(e–t) – 2L(y(t) ∗ cos t) ⇒ L(y(t)) = L(e–t) – 2(L(y(t)) . L(cos t)) (By convolution theorem) 1 s − 2 Y( s ) . 2 s +1 s + 12



Y( s ) =



Y( s ) =

F1 + 2s I Y(s) = 1 GH s + 1JK s +1



2

s2 + 1 ( s + 1)3

F s + 1 I = e L FG (s − 1) + 1IJ y(t ) = L (Y( s )) = L G H s K H (s + 1) JK F s − 2s + 2 I = e LL FG 1IJ − 2L FG 1 IJ + L FG 2 !IJ O =e L G H s K H s K PQ H s JK MN H s K −1



2

−1

−t −1

2

−t −1

3

3

2

−t

3

−1

−1

2

−1

3

= e − t [1 − 2t + t 2 ] = e − t (1 − t)2 .

Example 4. Solve the following integral equation by using Laplace transforms : y(t) = 1 − sinh t + y(t ) = 1 − sinh t +

Sol. We have

z z

t

(1 + u) y(t − u) du.

0

t

(1 + u ) y(t − u ) du.

0

y(t) = 1 − sinh t + ((1 + t) ∗ y(t))



Taking the Laplace transform, we get L(y(t)) = L(1) – L(sinh t) + L((1 + t) ∗ y(t)). ⇒

L( y(t)) =



Y (s) =

1 1 − 2 + L(1 + t) . L( y(t)) s s −1

FG H

IJ K

1 1 1 1 − + + Y( s), where Y ( s) = L( y(t)) s s2 − 1 s s2

FG1 − 1 − 1 IJ Y( s) = s − 1 − s H s sK s( s − 1) 2

⇒ ⇒ ∴

2

2

Y( s ) =

(By convolution theorem)

s 2 ( s 2 − 1 − s) 2

2

( s − s − 1) s( s − 1)

y(t) = L− 1 ( Y (s)) = L− 1

=

s 2

s −1

F s I = cosh t. GH s − 1JK 2

85

SOLUTION OF INTEGRAL EQUATIONS USING LAPLACE TRANSFORMATION

Example 5. Solve the following integral equation using Laplace transforms :

Sol. We have

z

z

t

t

y(u)

0

t−u

du = 1 + 2t − t 2 .

y(u ) (t − u ) − 1/ 2 du = 1 + 2t − t 2 .

0

⇒ y(t) ∗ t − 1/2 = 1 + 2t − t 2 Taking the Laplace transform, we get L ( y(t) ∗ t − 1/2 ) = L (1 + 2t − t 2 ) L( y(t)) . L(t − 1/2 ) = L(1) + 2L(t) − L(t 2 )



Y (s) .



Γ (1/2) s

Y( s )

π s

= =

(By convolution theorem)

1 2 2! , where Y (s) = L( y(t)) + − s s2 s3 s 2 + 2s − 2 s

3

Y( s ) =



1

LM 1 N

π s1/ 2

+

2 s 3/ 2



2 s 5/ 2

OP Q

LML FG 1 IJ + 2L FG 1 IJ − 2L FG 1 IJ OP Hs K Hs K H s KQ π N OP 1 L t t t = + 2. − 2. M 1 2 3 2 5 2 Γ ( / ) Γ ( / ) Γ ( / ) π N Q L OP 1 Mt 2t t = M + 1 − 2 . 3 1 PP π M π . π . . π 2 2 2 N Q 1L 8 O y( t ) = M t + 4t − t P . 3 πN Q

y(t) = L– 1 (Y (s)) =



1

–1

− 1/2

− 1/2

1/2

1/2

–1

3/ 2

1/2

− 1/2



–1

1/2

5/ 2

3/ 2

3/ 2

3/2

3.4. INTEGRO-DIFFERENTIAL EQUATION An equation involving an unknown function under the integral and perhaps also outside it and the derivatives of the unknown function is called an integro-differential equation. The method of solving an integro-differential equation is same as that of solving an integral equation. Example 6. Solve the following integro-differential equation using Laplace transforms : y ′(t) = t +

Sol. We have ⇒

y ′ (t ) = t +

z

t

0

z

(cos u) y(t − u) du, y(0) = 4. t

0

(cos u) y(t − u) du, y(0) = 4 .

y ′(t) = t + (cos t ∗ y(t))

Taking the Laplace transform, we get L ( y ′(t)) = L (t) + L(cos t ∗ y(t)) .

86

LAPLACE AND FOURIER TRANSFORMS

1 + L(cos t) . L( y(t)) s2 1 s sY (s) − 4 = 2 + 2 . Y (s) , where Y (s) = L( y(t)) s s +1

sL ( y(t)) − y(0) * =

⇒ ⇒

F s − s I Y(s) = 1 + 4 GH s + 1JK s 2

2

Y( s ) =





1 s

3

+

1 s

5

Y( s ) =

=5

s3

2

FG 1 IJ + L FG 1 IJ + 4L FG 1IJ Hs K Hs K H sK –1

3

F t I + t + 4(1) = 5 t GH 2 !JK 4 ! 2 2

FG 1 + 4IJ Hs K

4 4 5 1 4 + 3 = 3 + 5 + s s s s s

+

y(t) = L– 1 (Y (s)) = 5L– 1



s2 + 1

(By convolution theorem)

4

–1

5

2

+

1 4 t +4. 24

Example 7. Solve the following integro-differential equation by using Laplace transforms : y ′(t) + 4y(t) + 5

Sol. We have

z z

t

0

y ′(t) + 4 y(t) + 5

t

0

y(u) du = e − t , y(0) = 0. y(u) du = e − t , y(0) = 0 .

y′(t) + 4y(t) + 5(y(t) ∗ 1) = e–t



Taking the Laplace transform, we get L(y′(t) + 4L(y(t)) + 5L(y(t) ∗ 1) = L(e–t). ⇒ (sL(y(t)) – y(0)) + 4L(y(t)) + 5(L(y(t)) . L(1)) = ⇒ ⇒

sY(s) – 0 + 4Y(s) + 5Y(s) .

FG s + 4 + 5 IJ Y( s) = 1 H sK s +1



Y(s) =

1 s − (− 1)

(By convolution theorem)

1 1 = . s s+1 s 2

(s + 1) (s + 4 s + 5)

Using partial fractions, we have Y( s ) = −



1 s+5 1 ( s + 2) + 3 + =− + 2 2( s + 1) 2( s + 1) 2( s + 4s + 5 ) 2(( s + 2) 2 + 1)

y(t) = L– 1 (Y (s)) = − =−

FG IJ H K F s+3I GH s + 1JK

F GH

I JK

(s + 2) + 3 1 –1 1 1 L + L– 1 s+1 2 2 (s + 2) 2 + 1

1 − t 1 − 2t – 1 e + e L 2 2

*Why this step. Using L(f ′ ) = sL(f ′ ) – f(0).

2

87

SOLUTION OF INTEGRAL EQUATIONS USING LAPLACE TRANSFORMATION

=− y( t ) = −



F F I F GH GH GH JK bcos t + 3 sin tg.

II JK JK

1 − t 1 − 2t – 1 1 s e + e L + 3L– 1 2 2 2 2 s +1 s +1 1 − t 1 − 2t e + e 2 2

WORKING STEPS FOR SOLVING PROBLEMS Step I. Take the Laplace transform of the given equation. Step II. Substitute the values of Laplace transforms. Step III. Simplify the equation to get the value of Laplace transform of the unknown function. Step IV. Find the unknown functions by taking the inverse Laplace transform.

TEST YOUR KNOWLEDGE Solve the following integral equations by using Laplace transforms :

z

t

1.

y( t ) = t +

3.

y( t ) = t + 2

5.

y(t ) = sin t + 2

7.

y( t ) = t2 +

9.

11.

z z

t

y(u )

0

t−u

t

0

0

z

t

y(u ) cos (t − u ) du

0

z

t

0

z

t

6. y(t ) = 3t2 +

y( u ) sin (t − u ) du

10.

y( u ) y(t − u ) du = 2 y(t ) + t − 2

13.

y′(t) + 3 y(t) + 2

15.

y ′(t ) = t +

z

t

0

z

t

0

1 6

8. y(t ) = t +

du = 1

z

t

y( u )

0

t−u

12. y′(t) = 3

y(u) du = t, y(0) = 1

14. y′(t) =

z

t

0

4. y(t ) = 1 + 2

y(u ) cos ( t − u ) du

0

z

2. y(t ) = 1 −

y( u ) sin (t − u ) du

t

0

0

z z z

t

0

y(u ) cos (t − u ) du

t

y(u ) sin ( t − u ) du

0 t

0

y(u ) (t − u )3 du

du = 1 + t + t2

z

t

y(u ) (t − u ) du

y(u) cos 2(t − u) du + 2, y(0) = 1

y(u) cos (t − u) du, y(0) = 1

(cos u ) y( t − u ) du, y( 0) = 1.

Answers 1.

y(t ) = t +

5. y = tet

t3 6

2. y = cos t 6. y( t ) = 3t2 +

3. y(t) = 2et (t – 1) + 2 + t t4 4

7. y(t ) = t2 +

t4 12

4. y(t) = 1 + 2tet 8. y( t ) =

1 (sin t + sinh t ) 2

88

LAPLACE AND FOURIER TRANSFORMS

9.

y(t ) =

1 π t

10. y(t ) =

1 π

12.

y(t) = 4 + 8t – 3 cos t – 6 sin t

15.

y(t ) = 1 + t2 +

LM 1 + 2 Nt

t+

8 3/ 2 t 3

13. y( t ) =

OP Q

11. y(t) = 1

1 5 − 2e − t + e − 2t 2 2

14. y(t ) = 1 +

t4 . 24

Hint 11.

If L(y(t)) = Y(s), then ( Y( s ))2 − 2Y( s ) − ∴

Y( s ) = 2 −

Rejecting Y( s ) = 2 −

1 s

or

1 − 2s s2

1 s

1 1 , we have Y( s ) = . s s

= 0.

t2 2

Solution of Systems of Differential Equations Using the Laplace Transformation

4

4.1. INTRODUCTION In the present chapter, we shall study the method of solving a system of differential equations using Laplace transformations of functions. Our knowledge of finding Laplace transforms and inverse Laplace transforms would help us to solve the systems of differential equations. In the method of Laplace transforms, we would be able to find the required particular solution of the given system of differential equations with known initial conditions, without the necessity of first finding the general solution and then evaluating the arbitrary constants by using the given conditions. 4.2. METHOD OF SOLVING SYSTEM OF DIFFERENTIAL EQUATIONS Let t be the independent variable and x, y the dependent variables of a given system of differential equations. We shall denote the Laplace transforms of the functions x(t) and y(t) by x ( s ) and y ( s ) respectively. Thus, L ( x(t)) = x ( s) and L ( y(t)) = y (s).

We take the Laplace transform of each differential equation of the given system. These equations are simplified to get the values of the functions x ( s ) and y ( s ) . The values of the functions x(t) and y(t) are found by taking the inverse Laplace transforms of the functions x ( s ) and y ( s ) respectively.

ILLUSTRATIVE EXAMPLES Example 1. Solve the following system of differential equations by the method of Laplace transforms : x′ + y = 2 cos t, x + y′ = 0, x(0) = 0, y(0) = 1. Sol. Given system is x′ + y = 2 cos t ...(1) x + y′ = 0. ...(2) Taking the Laplace transform, we get L(x′ ) + L(y) = L(2 cos t) ...(3) and L(x) + L(y′ ) = L(0). ...(4) 89

90

LAPLACE AND FOURIER TRANSFORMS

(3) ⇒

[ sL( x ) − x ( 0)] + y =

sx + y =

⇒ (4) ⇒

s +1

2s s +1

(∵ x(0) = 0) ...(5)

2

(∵ y(0) = 1)

x + [ sL( y ) − y( 0)] = 0 ⇒ x + sy − 1 = 0



...(6)

x + sy = 1

(5) – s × (6) ⇒

y − s2 y =

⇒ (6)

2s 2

y= ⇒

2s s2 + 1

s

−s=

s +1

x = L− 1

s2 + 1

⇒ y(t ) = L− 1

2

x = 1 − sy = 1 − s



s − s3

F s I = cos t GH s + 1JK 2

F s I= 1 GH s + 1JK s + 1 2

2

F 1 I = sin t. GH s + 1JK 2

∴ x = sin t, y = cos t. Example 2. Solve the following system of differential equations by the method of Laplace transforms : x′ + 5x – 2y = t, y′ + 2x + y = 0 ; x(0) = 0, y(0) = 0. Sol. Given system is x′ + 5x – 2y = t ...(1) y′ + 2x + y = 0. ...(2) Taking the Laplace transform, we get L(x′) + 5L(x) – 2L(y) = L(t) ...(3) and L(y′) + 2L(x) + L(y) = L(0). ...(4) (3)



sx − 0 + 5x − 2 y =

⇒ (4) ⇒

[ sL( x ) − x ( 0)] + 5 x − 2 y =



1 s2 1 s



2

1

=0

...(5)

2x + ( s + 1) y + 0 = 0

...(6)

( s + 5) x − 2 y −

s2

[ sL( y ) − y( 0)] + 2x + y = 0



sy − 0 + 2 x + y = 0

Solving (5) and (6), we get

∴ Also

1 x y = = 2 s +1 ( s + 5) ( s + 1) + 4 0+ 2 − 2 −0 s s s +1 s +1 x= 2 2 = 2 s ( s + 6s + 9) s ( s + 3 ) 2 y=−

2 2 =− 2 . s (s + 6 s + 9) s (s + 3) 2 2

2

91

SOLUTION OF SYSTEMS OF DIFFERENTIAL EQUATIONS USING THE LAPLACE TRANSFORMATION

Using partial fractions, x=

1 1 1 2 . + − − 27s 9s 2 27( s + 3) 9( s + 3) 2

x = L− 1 ( x ) =



=

FG IJ HK

FG IJ H K

FG H

F GH

IJ K

1 −1 1 1 1 1 −1 1 2 1 L + L− 1 2 − L − L− 1 s s+3 27 9 27 9 s ( s + 3) 2

FG IJ H K

I JK

1 1 1 − 3t 2 1 (1) + . (t) − (e ) − e − 3t L− 1 2 27 9 27 9 s

1 t 1 − 3t 2 − 3t e t. + − − e 27 9 27 9 Using partial fractions, x=



y=

4 2 4 2 − − − . 27s 9s 2 27( s + 3) 9( s + 3) 2

y = L− 1 ( y ) = −



=−

FG IJ HK

FG H

IJ K

F GH

4 −1 1 2 4 −1 1 2 1 L − L− 1 (s − 2 ) − L − L− 1 s s+3 27 9 27 9 ( s + 3) 2

FG IJ H K

I JK

4 2 4 − 3t 2 1 (1) − (t) − (e ) − e − 3t L− 1 2 27 9 27 9 s

4 2 4 − 3t 2 − 3t e t. − t− − e 27 9 27 9 Example 3. Solve the following system of differential equations by the method of Laplace transforms :



and

y=−

x′ + y = 3e2t, y′ + x = 0, x(0) = 2, y(0) = 0. Sol. Given system is x′ + y = 3e2t y′ + x = 0. Taking the Laplace transform, we get L(x′) + L(y) = 3L(e2t) L(y′) + L(x) = 0. (3)



[ sL( x ) − x ( 0)] + y = 3 .

1 s−2

sx + y = 2 +

⇒ (4) ⇒

x + sy = 0

(5) – (6) × s ⇒

y − s2 y = 2 +



...(3) ...(4) ⇒

sx − 2 + y =

3 s−2

3 s−2

...(5)

[ sL( y ) − y( 0)] + x = 0 ⇒ sy − 0 + x = 0



(6)

...(1) ...(2)

...(6)

3 2s − 1 = s−2 s−2

x = − sy =



y=

2s − 1 (s − 2) (1 − s)(1 + s)

– s(2s − 1) − s(2s − 1) = (s − 2) (1 − s)(1 + s) (s − 2) (1 − s)(1 + s)

92

LAPLACE AND FOURIER TRANSFORMS

Using partial fractions,

x=

= ∴

(− 2) (3) (− 1) (1) (1) (− 3) + + (s − 2) (− 1) (3) (− 1) (1 − s) (2) (− 3) (2) (1 + s)

2 1 1 − + . s − 2 2( s − 1) 2( s + 1)

x = L− 1 ( x ) = 2L− 1

FG 1 IJ − 1 L FG 1 IJ + 1 L FG 1 IJ H s − 2 K 2 H s − 1K 2 H s + 1K −1

−1

1 t 1 −t e + e . 2 2 Using partial fractions, = 2e 2t −

y=

3 1 −3 + + ( s − 2) ( − 3) ( − 1) (1 − s ) ( 2) ( − 3) ( 2) (1 + s )

=− ∴

1 1 1 + + . s − 2 2( s − 1) 2( s + 1)

y = L− 1 ( y) = − L− 1

FG 1 IJ + 1 L FG 1 IJ + 1 L FG 1 IJ H s − 2 K 2 H s − 1K 2 H s + 1K −1

−1

1 t 1 −t e + e . 2 2 1 1 1 1 x = 2e 2t − e t + e − t and y = − e 2t + e t + e − t . ∴ 2 2 2 2 Example 4. Solve the following system of differential equations by the method of Laplace transforms : = − e 2t +

(D – 2)x – (D + 1)y = 6e3t (2D – 3)x + (D – 3)y = 6e3t, x(0) = 3, y(0) = 0. Sol. Given system is Dx – Dy – 2x – y = 6e3t

...(1)

6e3t.

2Dx + Dy – 3x – 3y =

...(2)

Taking the Laplace transform, we get and

L(Dx) – L(Dy) – 2L(x) – L(y) = 6L(e3t)

...(3)

6L(e3t).

...(4)

2L(Dx) + L(Dy) – 3L(x) – 3L(y) = (3) ⇒ ⇒ (4) ⇒ ⇒

[sL(x) – x(0)] – [sL(y) – y(0)] – 2x − y =

sx − 3 − sy + 0 − 2 x − y =

6 s−3



6 s−3 ( s − 2)x − ( s + 1) y = 3 +

2 [ sL( x ) − x ( 0)] + [ sL( y ) − y( 0)] − 3x − 3 y = 2(sx − 3) + sy − 0 − 3 x − 3 y =

6 s−3



6 s−3

...(5)

6 s−3

( 2s − 3)x + ( s − 3) y = 6 +

6 s−3

...(6)

SOLUTION OF SYSTEMS OF DIFFERENTIAL EQUATIONS USING THE LAPLACE TRANSFORMATION

93

(5) × (s – 3) + (6) × (s + 1) ⇒ ⇒

[( s − 3) ( s − 2) + ( 2s − 3) ( s + 1)] x = 3( s − 3) + 6 + 6( s + 1) + 3( s − 1) 2 x =

3( 3s 2 − 6s − 1) s−3



x=

3s 2 − 6s − 1 ( s − 1) 2 ( s − 3)

6( s + 1) s−3

.

Using partial fractions, x=



1 2 2 . + + 2 s − 1 ( s − 1) s−3

= et

(5) × (2s – 3) – (6) × (s – 2) ⇒ ⇒ ⇒

−1

t

− 3( s − 1) 2 y = 3 + y=−

=−

3t

t

2

I I + 2L FG 1 IJ JK JK H s − 3K −1

+ 2e t t + 2e 3t .

6( s − 2) 6( 2s − 3) − 6( s − 2) − s−3 s−3

6s − 6 s−3 1

( s − 1)

2 ( s − 1) ( s − 3)

2



2

−2

2

+

1 (s − 1) 1 ( s − 1)

LM 1 + 1 OP N (s − 1) (− 2) 2(s − 3) Q

1 1 − . s −1 s − 3

F 1 I +L F 1 I −L F 1 I GH (s − 1) JK GH s − 1JK GH s − 3 JK FG 1 IJ + e − e = − e t + e − e . Hs K

y = − L− 1 = − e t L− 1



−1

2

[ − ( 2s − 3) ( s + 1) − ( s − 2) ( s − 3)] y = 3( 2s − 3) +

=−



FG 1 IJ + 2 FG L FG 1 H s − 1K H H (s − 1) F1I + 2e L G J + 2e = e Hs K

x = L− 1

−1

2

t

3t

−1

t

t

3t

2

x = e t + 2te t + 2e 3t and y = − te t + e t − e 3t .

Example 5. Solve the following system of differential equations by the method of Laplace transforms : x″ = x + 3y, y″ = 4x – 4et, x(0) = 2, x′(0) = 3, y(0) = 1, y′(0) = 2. Sol. Given system is x″ = x + 3y ...(1) t y″ = 4x – 4e . ...(2) Taking the Laplace transform, we get L(x″) = L(x) + 3L(y) ...(3)

94

LAPLACE AND FOURIER TRANSFORMS

L(y″) = 4L(x) – 4L(et).

and (3)



[ s 2 L( x) − sx(0) − x ′(0)] = x + 3 y



s 2 x − s.2 − 3 = x + 3 y



(s 2 − 1) x − 3 y = 2s + 3

(4) ⇒

(5) × 4 + (6) × (s2 – 1) ⇒

4 s −1

...(6)

− 12 y + s 2 (s 2 − 1) y = 8 s + 12 + (s 2 − 1) (s + 2) − 4(s + 1)

(s 2 − 4) (s3 + 3) y = 4(s + 2) + (s 2 − 1) (s + 2)



y=



( s + 2) ( s 2 + 3) 2



(s 2

2

( s − 4 ) ( s + 3)

=

1 . s−2

FG 1 IJ = e H s − 2K F 1 IJ + 2s + 3 = (s + 1) (2s − 3) − 1) x = 3 G s−2 H s − 2K y = L− 1

∴ (5)

1 s −1

4 s −1

− 4x + s2 y = s + 2 −



...(5)

[ s 2 L( y ) − sy( 0) − y ′( 0)] = 4 x − 4 . s2 y − s . 1 − 2 = 4x −



...(4)

2t

2s − 3 ( − 1) 1 1 1 = + = + . ( s − 1) ( s − 2) ( s − 1) ( − 1) 1 . ( s − 2) s −1 s − 2



x=



x = L− 1

FG 1 IJ + L FG 1 IJ = e H s − 1K H s − 2 K −1

t

+ e 2t .

x = e t + e 2t and y = e 2t .



Example 6. Solve the following system of differential equations by the method of Laplace transforms : (D2 + 2)x – Dy = 1, Dx + (D2 + 2)y = 0, x(0) = 0, x′(0) = 0, y(0) = 0, y′(0) = 0. Sol. Given system is D2x + 2x – Dy = 1 ...(1) Dx + D2y + 2y = 0. ...(2) Taking the Laplace transform, we get L(x″) + 2L(x) – L(y′) = L(1) ...(3) and L(x′) + L(y″) + 2L(y) = L(0). ...(4) (3) ⇒ ⇒



[ s 2 L( x) − sx(0) − x ′(0)] + 2 x − [ sL( y) − y(0)] = L(1) s 2 x − s . 0 − 0 + 2x − sy + 0 = (s 2 + 2) x − sy −

1 s

1 =0 s

...(5)

SOLUTION OF SYSTEMS OF DIFFERENTIAL EQUATIONS USING THE LAPLACE TRANSFORMATION

(4)



95

[ sL( x) − x(0)] + [ s 2 L( y) − sy(0) − y ′(0)] + 2 y = 0



sx − 0 + s 2 y − s . 0 − 0 + 2 y = 0



sx + (s 2 + 2) y + 0 = 0

...(6)

Solving (5) and (6), we get x y 1 = = 2 2 s + 2 − 1 − 0 ( s + 2) 2 + s 2 0+ s

x=

∴ and

y=

F GH

I JK

s2 + 2 s2 + 2 1 = 4 2 2 s s + 5s + 4 s( s + 1) ( s 2 + 4 ) −1 2

( s + 1) ( s 2 + 4 )

.

Using partial fractions, we get x=

and

1 s s − − 2s 3( s 2 + 1) 6( s 2 + 4 )

y=−

x=



=

3( s + 1)

+

1 2

3( s + 4 )

.

F GH

FG IJ HK

I JK

F GH

s s 1 −1 1 1 1 L − L− 1 2 − L− 1 2 s 2 3 6 s +1 s +4 1 1 1 . 1 − . cos t − cos 2t 2 3 6

y=−

and

1 2

F GH

I JK

F GH

1 −1 1 1 1 L + L− 1 2 2 3 3 s +1 s +4

I JK

=−

I JK

1 1 1 sin t + . sin 2t. 3 3 2

1 1 1 1 1 − cos t − cos 2t and y = − sin t + sin 2t. 2 3 6 3 6 Example 7. Solve the following system of differential equations by the method of Laplace transforms : x=



dx dy d2x + = t, − y = e − t with x(0) = 3, x′(0) = – 2, y(0) = 0. dt dt dt 2 dx dy + =t Sol. Given system is dt dt

d 2x

− y = e −t . dt 2 Taking the Laplace transform, we get

FG dx IJ + L FG dy IJ = L(t) H dt K H dt K F d x I − L( y ) = L(e LG H dt JK

...(2)

L

2

and

2

−t

...(1)

...(3)

).

...(4)

96

LAPLACE AND FOURIER TRANSFORMS

(3)



[ sL( x ) − x ( 0 )] + [ sL( y ) − y( 0 )] = [ sx − 3] + [ sy − 0 ] =







[ s 2 x − s(3) − ( − 2)] − y =



s2 x − y =

(1 + s 2 ) x =

(5) + (6) ⇒ x=



3 2

s( s + 1)

1

L− 1

We have

s2 + 1

+

−1

2

2

2

2

2

1 2

( s + 1) ( s + 1)

L− 1

F 1 GH ( s + 1) (s

...(5)

1 s +1 1 + 3s − 2 s +1

...(6)

3 1 1 + + + 3s − 2 s s3 s + 1

1 3

s ( s + 1)

z z z

t

+

1 2

( s + 1) ( s + 1)

+

3s 2

s +1



2 2

s +1

.

0 t

0

t

0

t

sin T dT = − cos T

= 1 − cos t 0 t

(1 − cos T) dT = (T − sin T)

= t − sin t 0

FT ( T − sin T) dT = G H2

2

Using partial fractions,

F GH

I + cos TJ K

t

= 0

t2 + cos t − 1 . 2

I JK

=

1 1 1−s + 2( s + 1) 2 s 2 + 1

=

1 1 s + − 2 2 2( s + 1) 2( s + 1) 2( s + 1)

I = 1 e + 1 sin t − 1 cos t . J 2 2 + 1) K 2 Ft I 1 x = 3(1 − cos t) + G + cos t − 1J + (e + sin t − cos t) + 3 cos t − 2 sin t 2 H K 2 −t

2

2



1

= sin t.

F1 1 I = L G . H s s + 1JK F1 1 I = L G . H s s(s + 1) JK F1 1 I = L G . H s s (s + 1) JK −1

−1



s2

s2 3 1 x+y= + 3 s s 1 [ s 2 L( x ) − sx ( 0) − x ′( 0)] − y = s +1





s2 1

sx + sy = 3 +



(4)

1

−t

SOLUTION OF SYSTEMS OF DIFFERENTIAL EQUATIONS USING THE LAPLACE TRANSFORMATION

97

1 1 1 3 cos t + t 2 + e − t − sin t 2 2 2 2 3 1 y= + 3 −x s s x=2+

or (5) ⇒

FG 1 IJ + L FG 1 IJ − L H sK H s K

y = 3L− 1



−1

3

−1

(x )

1 2 1 1 1 3 t − 2 − cos t − t 2 − e − t + sin t 2 2 2 2 2 1 1 3 y = 1 − cos t − e − t + sin t. 2 2 2 = 3(1) +

or

WORKING STEPS FOR SOLVING PROBLEMS Step I. Take the Laplace transform of the given equations. Step II. Substitute the values of Laplace transforms. Step III. Simplify the equations to get the values of x and y . Step IV. Find the inverse Laplace transforms of x and y to get the values of x and y.

TEST YOUR KNOWLEDGE Solve the following systems of differential equations by the method of Laplace transforms : 1. x′ + x – 4y = 0, y′ = 3x – 2y, x(0) = 3, y(0) = 4. 3. 3x′ + y′ + 2x = 1, x′ + 4y′ + 3y = 0, x(0) = 0, y(0) = 0. 5. x′ = 6x + 9y, y′ = x + 6y, x(0) = 3, y(0) = 3. 7. x′ = – y, y′ = x, x(0) = 1, y(0) = 0. 9. x′ = 2x + 4y, y′ = x + 2y, x(0) = – 4, y(0) = – 4. 11.

x′ + 2x – 4y = 0, y′ – 2x = t, x(0) = 0, y(0) = 3.

13.

x″ – 3x′ +y′ + 2x – y = 0, x′ + y′ – 2x + y = 0,

2. x′ + x – y = 0, y′ + x + y = 0, x(0) = 1, y(0) = 0. 4. (D – 2)x + 3y = 0, 2x + (D – 1)y = 0, x(0) = 8, y(0) = 3. 6. x′ = 5x + y, y′ = x + 5y, x(0) = – 3, y(0) = 7. 8. x′ = 2x – 4y, y′ = x – 3y, x(0) = 3, y(0) = 0. 10. x′ = – 2x + 3y, y′ = 4x – y, x(0) = 4, y(0) = 3. 12. 2x′ + y′ – x – y = e–t, x′ +y′ + 2x + y = et, x(0) = 2, y(0) = 1. 14. x″ = – 5x + 2y, y″ = 2x – 2y, x(0) = 3, x′(0) = 0, y(0) = 1, y′(0) = 0.

x(0) = 0, y(0) = – 1, x′ (0) = 0. 15.

x″ + 3x – 2y = 0, x″ + y″ – 3x + 5y = 0, x(0) = 0, x′(0) = 3, y(0) = 0, y′(0) = 2.

16. x″ – x + 5y′ = t, – 2x′ + y″ – 4y = – 2 x(0) = 0, x′(0) = 0, y(0) = 0, y′(0) = 0.

98

LAPLACE AND FOURIER TRANSFORMS

Answers 1. x = 4e2t – e–5t, y = 3e2t + e–5t

2. x = e–t cos t, y = – e–t sin t

4. 6. 8. 10.

1 1 ( 5 − 2e −t − 3e −6t /11 ), y = ( e −t − e −6t/11 ) 10 5 x = 5e–t + 3e4t, y = 5e–t – 2e4t 5. x = – 3e3t + 6e9t, y = e3t + 2e9t 4t 6t 4t 6t x = – 5e + 2e , y = 5e + 2e 7. x = cos t, y = sin t 9. x = 2 – 6e4t , y = – 1 – 3e4t x = 4et – e–2t, y = et – e–2t 2t –5t 2t –5t x = 3e + e , y = 4e – e

11.

x=

12.

x = 8 sin t + 2 cos t,

3.

x=

13 2t 49 − 4t 1 1 13 2t 49 − 4t 1 3 e − e − t− , y= e + e − t− 6 24 2 8 6 48 4 16

y = − 13 sin t + cos t +

13. x = 2et – e2t – 1, et e − t − 2 2

y = et – 2 15. x =

14.

x = cos t + 2 cos 6 t , y = 2 cos t − cos 6 t

16.

x = – t + 5 sin t – 2 sin 2t, y = 1 – 2 cos t + cos 2t.

11 1 11 1 sin t + sin 3t, y = sin t − sin 3t 4 4 4 12

5

Fourier Transforms

5.1. INTRODUCTION Fourier* transforms play an important part in the theory of many branches of science. The use of Fourier transforms is indispensible in solving the problems in the field of mathematics, physics and engineering. Many types of integrals and differential equations can be solved by using Fourier transforms. 5.2. FOURIER’S INTEGRAL THEOREM Statement. Let f be a real valued function of the real variable x such that f(x) and f ′(x) are piecewise continuous in every finite interval and the integral of |f(x)| from – ∞ to ∞ exists. Then f(x) can be expressed as f(x) =

z



( A(s) cos sx + B( s) sin sx) ds , where

z

0

z

1 ∞ 1 ∞ f (v) cos sv dv and B(s) = f (v) sin sv dv . π −∞ π −∞ At a point where f(x) is discontinuous the value of the integral on the right side equals the average of left hand limit and right hand limit of f(x) at that point.

A(s) =

Note. The proof of this theorem is beyond the scope of this book. The integral function f(x).

z



0

(A(s) cos sx + B(s) sin sx) ds is called the Fourier integral expression of the

Theorem. Prove that the Fourier integral expression of a function f(x) is same as the integral

1 2π

z FGH z ∞



−∞

−∞

IJ K

f (v) e is( x − v) dv ds .

Proof. The Fourier integral expression of f(x) is : f(x) = A(s) =

z



0

1 π

(Α(s) cos sx + Β(s) sin sx) ds, where

z



−∞

f (v) cos sv dv and

B(s) =

1 π

z



−∞

f (v) sin sv dv

* Baron Baptiste Joseph Fourier (1768–1830) French physicist and mathematician lived and taught in Paris. Fourier was appointed as Perfect of Isère by Nepoleon in 1802. He never married.

99

100

LAPLACE AND FOURIER TRANSFORMS





z FGHFGH z z FGH z z FGH z z FGH z 1 π



f(x) =

0

=

1 π

=

1 π

f(x) =

1 π



−∞





0

−∞





0

−∞





0

−∞

IJ K

f (v) cos sv dv cos sx +

FG 1 Hπ

z



−∞

IJ K

IJ K

f (v) sin sv dv sin sx ds

IJ K

f (v) [cos sv cos sx + sin sv sin sx ] dv ds

IJ K

f (v) cos (sv − sx) dv ds

IJ K

f (v) cos s( x − v) dv ds

...(1)

The function f(v) cos s(x – v) is an even function of s. ∴ The function

z



−∞

f (v) cos s( x − v) dv is also an even function of s.

z z

1 2π

∴ (1) ⇒ f(x) =



−∞

FG∵ H

FG H



−∞

IJ K

(∵ The integration is not w.r.t. s)

f (v) cos s( x − v) dv ds

...(2)

If φ( x ) is an even function of x then

The function f(v) sin s(x – v) is an odd function of s. ∴ The function ∴

z z FGH z ∞

−∞



−∞

−∞

z



0

φ( x ) dx

IJ K

IJ K

f (v) sin s( x − v) dv ds = 0

z FGH z z FGH z z FGH z

IJ K

∞ 1 ∞ f (v) sin s( x − v) dv ds 2π − ∞ −∞ Adding (2) and ‘i’ times (3), we get

0=

f(x) + i . 0 =

1 2π

f(x) =

1 2π



−∞

φ( x ) dx = 2

f (v) sin s( x − v) dv is also an odd function of s.





z



∴ The result holds.





−∞

−∞





−∞

−∞

...(3)

IJ K

f (v) (cos s( x − v) + i sin s( x − v)) dv ds

IJ K

f (v) e is( x − v) dv ds *.

5.3. FOURIER TRANSFORM AND ITS INVERSE Let f be a real valued function of the real variable x such that f(x) and f ′(x) are piecewise continuous in every finite interval and the integral of |f(x)| exists from – ∞ to ∞. The integral

z



−∞

f ( x) e − isx dx is called the Fourier transform of the function f(x) and

we write this as F(f(x)) or as f (s) . *Why this step. We have used the Euler formula : eix = cos x + i sin x.

101

FOURIER TRANSFORMS

z

f(s) = F(f(x)) =

Thus,



−∞

f(x) e − isx dx .

By Fourier integral expression of f(x), we have f(x) =

1 2π



f(x) =

1 2π



f(x) =

1 2π

The function

1 2π

z



−∞

zz zz z ∞

−∞ ∞

−∞ ∞

−∞

FG H FG H



IJ K

f (v) e is( x − v) dv ds .

−∞

IJ K



f (v) e − isv dv e isx ds

−∞

f (s) e isx ds .

f (s) e isx ds

i.e., f(x) is called the inverse Fourier transform of

–1

–1

the function f (s) and we write F ( f (s) ) = F (F(f(x)) = f(x). The constants 1 and

1 preceding the integrals 2π

could be replaced by any constants whose product is the above choice.

z



−∞

f ( x) e − isx dx and

1 2π

z



−∞

f (s) e isx ds

1 . In our discussion, we shall stick to 2π

Remark 1. The Fourier transform of a function will in general be a complex-valued function. Remark 2. In finding the Fourier transform of a function, we may require the following ‘general rule of integration by parts’ :

z

uv dx = uv1 – u′′v2 + u″″v3 – u′″ ′″ v4 + ...... ,

where dashes denote the successive differentiation and suffixes denote the successive integration.

5.4. SHIFTING PROPERTY OF FOURIER TRANSFORMS Theorem I. If the Fourier transform of the function f(x) is f (s) then prove that the Fourier transform of the function f(x – a) is e–ias f (s) . Proof. We have

z

F(f(x)) = f (s) = ∴ Let ∴

F(f(x – a)) =

z z



−∞



−∞

f ( x) e − isx dx .

f ( x − a) e − isx dx

t = x – a. ∴ dt = dx F(f(x – a)) =



−∞

f (t) e − is(t + a) dt =

= e–ias

z



−∞

z



−∞

f (t) (e − ist . e − ias ) dt

f (t) e − ist dt = e–ias f (s) .

Theorem II. If the Fourier transform of the function f(x) is f (s) then prove that the Fourier transform of the function eiax f(x) is f (s − a) .

102

LAPLACE AND FOURIER TRANSFORMS

Proof. We have F(f(x)) = f (s) = F(eiax f(x)) =



z



z



−∞

f ( x) e − isx dx .

( e iax f ( x)) e − isx dx =

−∞

= f (s − a) .

z



−∞

f ( x) e − i( s − a) x dx

5.5. MODULATION PROPERTY OF FOURIER TRANSFORMS Theorem. If the Fourier transform of the function f(x) is f (s) then prove that the Fourier 1 1 f (s − a) + f (s + a) . 2 2

transform of the function f(x) cos ax is Proof. We have F(f(x)) = f (s) = ∴

F(f(x) cos ax) = =

z z



−∞

f ( x)

−∞



f ( x) e − isx dx .

( f ( x ) cos ax ) e − isx dx

−∞



z

z z

Fe GH

iax

I JK

+ e − iax − isx e dx 2

z

1 ∞ 1 ∞ f ( x ) e iax e − isx dx + f ( x ) e − iax e − isx dx 2 −∞ 2 −∞ 1 ∞ 1 ∞ = f ( x) e − i( s − a) x dx + f ( x) e − i( s + a) x dx 2 −∞ 2 −∞ 1 1 = f (s − a) + f (s + a) . 2 2

=

z

5.6. CONVOLUTION THEOREM The convolution theorem is used to find the inverse Fourier transform of the product of two functions with known inverse Fourier transforms of the factors of the product. This theorem is also used in solving partial differential equations. The convolution of the functions f(x) and g(x) is defined by the integral and denoted by f ∗ g. ∴

(f ∗ g)(x) =

Let v = x – u. ∴ ∴ (1) ⇒

z z



−∞

f(u) g(x – u) du

...(1)

f (u) g ( x − u) du

−∞

z



dv = – du

(f ∗ g)(x) =

−∞



=– =

z

z

f ( x − v) g (v) (− dv) −∞





−∞

g (v) f ( x − v) dv

g (v) f ( x − v) dv =

z



−∞

g (u) f ( x − u) du

(Replacing v by u)

103

FOURIER TRANSFORMS

= (g ∗ f )(x) ∴ f ∗ g = g ∗ f. Convolution Theorem. Let f(x) and g(x) be functions such that their Fourier transforms exist. Prove that F(f ∗ g) = F(f ) F(g). Proof. It can be proved mathematically that the Fourier transform of the convolution of f and g (i.e., f ∗ g) exists. By definition,

F(f ∗ g) = = =

Let v = x – u. ∴ ∴

z



−∞



( f ∗ g )( x ) e − isx dx

−∞





−∞

−∞



−∞

IJ K

f (u) g ( x − u) du e − isx dx

f (u)



−∞

IJ K

g ( x − u) e − isx dx du

(By changing the order of integration)

dv = dx

g ( x − u) e − isx dx =

F(f ∗ g) =



z z FGH z z FGH z

= = =

z F z GH z z FGH z z FGH z FG IJ FG Hz K Hz ∞

−∞



−∞ ∞

−∞



−∞

g (v) e − is(u + v) dv

f (u) f (u)



−∞ ∞

−∞

−∞

IJ K I dvJ du K

g (v) e − isu e − isv dv du ∞

f (u) e − isu



IJ K

g (v) e − is(u + v) dv du

−∞

f (u) e − isu du

g (v) e − isv ∞

−∞

= F(f ) F(g) F(f ∗ g) = F(f) F(g).



IJ K

g (v) e − isv dv

WORKING RULES FOR SOLVING PROBLEMS Rule I.

If A(s) = f(x) =

z



0

1 π

z



−∞

f(v) cos sv dv

and B(s) =

1 π

z



−∞

f(v) sin sv dv then

(A(s) cos sx + B(s) sin sx) ds .

z FGH z z z

IJ K

∞ 1 ∞ f(v) e is(x − v) dv ds . −∞ 2 π −∞ Rule III. The Fourier transform of the function f(x) is given by :

Rule II. f(x) =

f(s) = F(f(x)) =



−∞

f(x) e − isx dx .

Rule IV. The inverse Fourier transform of the function g(s) is given by F–1(g(s)) =

1 2π



−∞

g(s) e isx ds .

104

LAPLACE AND FOURIER TRANSFORMS

Rule V.

F–1 (f(s)) = f(x). F(f(x – a)) = e–ias f(s)

Rule VI. (i) F(f(x)) = f(s) ⇒

F(eiax f(x)) = f (s − a) . 1 1 ⇒ F(f(x) cos ax) = f(s − a) + f(s + a) . 2 2

(ii) F(f(x)) = f(s) ⇒ Rule VII.

F(f(x)) = f(s)

Rule VIII. (i) (f ∗ g)(x) =

z



−∞

f(u) g(x − u) du

(ii) F(f ∗ g) = F(f ) F(g).

ILLUSTRATIVE EXAMPLES Example 1. Find the Fourier transform of the function f(x) = Sol.

F(f(x)) = = =

RSλ, T0,

z z z



−∞ 0 −∞ 0 −∞

if 0 < x < a . otherwise f ( x) e − isx dx

z

f ( x) e − isx dx + 0 . e − isx dx +

=0+λ

z

a

0

z

a

0

f ( x) e − isx dx +

a

0

λe − isx dx +

e − isx dx + 0 = λ

zI λ

a a

z



a

f ( x) e − isx dx

0 . e − isx dx

Fe GG −is JJ = isλ (e H K − isx

–isa

– 1).

0

Example 2. Find the Fourier transform of the function f(x) = Sol.

F(f(x)) = = =

RSe , T0, x

z z z



−∞ −a

−∞ −a −∞

=0+

if − a < x < a . otherwise

f ( x) e − isx dx f ( x) e − isx dx + 0 . e − isx dx +

z

a

−a

e (1 − is) x = 1 − is

z

z

a

−a

a

−a

e x e − isx dx +

e (1 − is) x dx + 0 a

f ( x) e − isx dx +

z



a

z



a

f ( x) e − isx dx

0 . e − isx dx

1 [e(1 – is)a – e–(1 – is)a]. 1 − is −a Example 3. Find the Fourier transform of the function f(x) = Sol.

F(f(x)) =

RS5x , T0,

z



−∞

=

if 0 < x < 3 otherwise .

f ( x) e − isx dx

105

FOURIER TRANSFORMS

= =

z z

0

−∞ 0

−∞

f ( x) e − isx dx + 0 . e − isx dx +

z

=0+5

3

3

0

0

f ( x) e − isx dx +

5 xe − isx dx +

z



z



3

f ( x) e − isx dx

0 . e − isx dx

3

x . e − isx dx + 0

0 I

II

L e = 5 Mx MN − is

3

− isx

=

z

z

3

z

OP PQ

3

5 e − isx 5i e − isx . dx = [3e–3is – 0] + − 1. is − is s 0 − is

0

3

15ie −3is 5 + 2 [e–3is – 1] s s

0

5 –3is (3is + 1) – 1]. 2 [e s s Example 4. Find the Fourier transform of the function 5

=

f(x) = Sol.

F(f(x)) = = =

z z z



−∞ 0

−∞ 0

−∞

2

[3is e–3is + e–3is – 1] =

RSxe , T 0, −x

if x > 0 . if x < 0

f ( x) e − isx dx f ( x) e − isx dx + 0 . e − isx dx +

z

z



f ( x) e − isx dx

0



0

L e OP e = M x . − (1 + is ) − 1 . ( − (1 + is )) PQ MN − (1 + is ) x

z

xe − x . e − isx dx = 0 +

x . e − (1 + is) x dx

0 I



− (1 + is ) x



II

2

1 x =− . x 1 + is e (cos sx + i sin sx ) =–

LM N

=

0

1 (1 + is)

2

LM lim N

x→∞

1 x

e (cos sx + i sin sx)

1 1 [0 – 0] – [0 – 1] 1 + is (1 + is) 2

1 . (1 + is) 2



1 1 . x – 2 (1 + is) e (cos sx + i sin sx)

x 1 1 0 lim x . lim − 0 1 + is x → ∞ e x → ∞ cos sx + i sin sx e (cos 0 + i sin 0)

– =–

0



FG∵ H

lim

x→∞

x e

x

= lim

x→∞



0

OP Q

OP e (cos 0 + i sin 0) Q 1

0

( x) ′ x

(e ) ′

= lim

x→∞

1 e

x

IJ K

=0

106

LAPLACE AND FOURIER TRANSFORMS

Example 5. Find the Fourier transform of the function f(x) = Sol.

F(f(x)) = = =

z z z



−∞ −a

−∞ −a

−∞

R|− 1 , S| 1 , T0,

if if

− a < x 1

x cos x − sin x x cos dx . 3 2 x f(x) = F(f(x)) = =

RS1 − x T 0,

z z



−∞ −1

−∞

2

, if − 1 < x < 1 . if x < − 1 or x > 1

f ( x) e − isx dx f ( x) e − isx dx +

z

1

−1

f ( x) e − isx dx +

z



1

f ( x) e − isx dx

a = 1)

109

FOURIER TRANSFORMS

=

z

−1

−∞

0 . e − isx dx +

LM MN

= 0 + (1 − x 2 ) .

z

1

−1

(1 − x 2 ) e − isx dx +

z



1

0 . e − isx dx

e − isx e − isx e − isx − ( − 2x ) . + ( − 2 ) . − is ( − is ) 2 ( − is )3

OP PQ

1

+0 −1

(By using general rule of integration by parts)

LM N

=0–0+ − =– =– =–

2x s2

e − isx +

2i

e − isx

s3

OP Q

1 −1

2i 2 (e–is + eis) + 3 (e–is – eis) 2 s s

Fe GH

4 s

2

is

F GH

I JK

+ e − is 4i 2 e is − e − is − 3 2 2i s

I JK

4 4 4 cos s + 3 sin s = 3 (sin s – s cos s) 2 s s s

∴ Fourier transform of f(x) = f (s) = F(f(x)) =

4 s3

(sin s – s cos s).

By Inverse Fourier transform, we have f(x) =

1 2π





z



−∞

z

−∞

z



−∞

f ( s) e isx ds

4 (sin s − s cos s) e isx ds = f(x) s3

RS T FG (sin s − s cos s) cos sx + i (sin s − s cos s) sin sx IJ ds = R|S π2 (1 − x ) , H K |T 0 , s s 2 π



z



1 2π

1

2 (sin s − s cos s ) (cos sx + i sin sx ) ds = 1 − x , if |x |< 1 0, if |x |> 1 s



3

−∞

2

3

3

if | x|< 1 if | x|> 1

Equating the real parts, we have

z z



−∞



2



0

R| S| T R| π (1 − x ) , (sin s − s cos s) cos sx ds = S 2 s |T 0 ,

π (sin s − s cos s) cos sx (1 − x 2 ) , if | x|< 1 ds = 2 3 s 0, if | x|> 1 2

3



z

Taking x =

1 , we get 2



0

R| S| T

if | x|< 1 if | x|> 1

(∵ The integrand is an even function of s)

π 2 (s cos s − sin s) cos sx ( x − 1) , if | x|< 1 ds = 4 s3 if | x|> 1 0,

110

LAPLACE AND FOURIER TRANSFORMS

z



( s cos s − sin s ) cos s / 2 s

0

3

ds =

π 4

FF 1I GH GH 2 JK

I JK

2

−1 = −

FG∵ H

3π 16

∴ By changing the variable of integration, we get

z



x cos x − sin x x

0

3

cos

IJ K

1 l

1 , if 3 < x < 9 2. f(x) = 0 , otherwise

4 , if 0 < x < 2 1. f(x) = 0 , otherwise 3.

5.

7.

9. f(x) = 11.

0 ε a > 0.

e–a| x | ,

1 , if | x | < 1 Find the Fourier transform of f(x) = 0 , if | x | > 1 . Hence evaluate

z



0

sin x dx . x

Answers 1. 4. 7. 10.

4i –2is (e – 1) s e− ias (1 + ias) − 1 s2

2π sin ls ls 2a

a2 + s2

2.

i – 9is (e – e–3is) s

3.

5.

2i (cos s – 1) s

6.

sin εs εs 2 sin s π 11. , . s 2 8.

9.

2(e−7is (1 + 7is) − 1) s2

2 sin as s 2 1 + s2

5.7. FOURIER SINE AND COSINE TRANSFORMS Let f be a real valued function of the real variable x such that f(x) and f ′(x) are piecewise continuous in every finite interval and the integral of | f(x) | exists from – ∞ to ∞. The integral

111

FOURIER TRANSFORMS

z



0

f ( x ) sin sx dx is called the Fourier sine transform of the function f(x) and we write this

as FS(f(x)) or as fS (s) . Thus, fS (s) = FS(f(x)) = Similarly, the integral

z



0

z



0

f(x) sin sx dx .

f ( x) cos sx dx is called the Fourier cosine transform of the

function f(x) and we write this as FC(f(x)) or as fC ( s ) . Thus, fC (s) = FC(f(x)) =

z z



f(x) cos sx dx .

0

By Fourier integral expression of f(x), we have f(x) = where



0

( A ( s) cos sx + B(s) sin sx) ds ,

z

1 ∞ f (v) cos sv dv π −∞ Case I. f(x) is an odd function. ∴ f(v) cos sv is an odd function of v.

A(s) =

z





−∞

and B(s) =

1 π

z

...(1) ∞

−∞

f (v) sin sv dv .

f (v) cos sv dv = 0

1 .0=0 π Also, f(v) sin sv is an even function of v.

A(s) =



z





−∞

B(s) =

∴ ∴

(1)

z zF



f (v) sin sv dv = 2

f(x) =



f(x) =

⇒ ⇒ The function

2 π

z

f(x) = ∞

0

2 π

0 ∞

f (v) sin sv dv

f (v) sin sv dv

z GH z FGH z z 0



0

2 π 2 π

0 . cos sx +





0

0



0

FG 2 Hπ

z



0

IJ K

IJ K

f (v) sin sv dv sin sx ds

fS (s) sin sx ds .

fS (s) sin sx ds i.e., f(x) is called the inverse Fourier sine transform

of the function fS (s) and we write FS–1 ( fS (s) ) = FS–1(FS(f(x)) = f(x). Case II. f(x) is an even function. ∴ f(v) cos sv is an even function of v. ∴ ∴

z



−∞

IJ K

f (v) sin sv dv sin sx ds

f (v) cos sv dv = 2

A(s) =

z

2 π



z

0

f (v) cos sv dv ∞

0

f (v) cos sv dv

112

LAPLACE AND FOURIER TRANSFORMS

Also, f(v) sin sv is an odd function of v.

z



−∞

∴ ∴



(1)

f (v) sin sv dv = 0

1 .0=0 π ∞ 2 ∞ f(x) = f (v) cos sv dv cos sx + 0 . sin sx ds 0 π 0 2 ∞ ∞ f(x) = f (v) cos sv dv cos sx ds π 0 0 2 ∞ fC (s) cos sx ds . f(x) = π 0

B(s) = ⇒

⇒ ⇒ The function

2 π

z



0

z FGHFGH z z FGH z z

IJ K

IJ K IJ K

fC (s) cos sx ds i.e., f(x) is called the inverse Fourier cosine trans-

form of the function fC (s) and we write FC–1( fC (s) ) = FC–1 (FC(f(x)) = f(x).

Y

Important Note. If a function is defined on the interval (0, ∞) then it can be defined on the interval (– ∞, 0) so that the extension of the function under consideration is an even (resp. odd) function. Let f(x) be a function defined on (0, ∞) with graph as shown in the figure. The function shown in the figure I and figure II are respectively the even extension of f(x) and the odd extension of f(x). ∴ Any function defined on (0, ∞) can be considered to be an even function as well as an odd function.

f(x)

O

X

Y

Y

O

X

Figure I. Even extension of f(x)

O

X

Figure II. Odd extension of f(x)

∴ If a function is defined on (0, ∞) then the formulae regarding inverse Fourier sine transform and inverse Fourier cosine transform are applicable.

z z

Remark. We know that

and

eax sin (bx + c) dx = eax cos (bx + c) dx =

eax a2 + b2 eax a2 + b2

(a sin (bx + c)) – (b cos (bx + c)) + C (a cos (bx + c) + b sin (bx + c)) + C.

113

FOURIER TRANSFORMS

z





0

e− ax sin bx dx =

a2 + b2

= lim

z



0

e − ax

(− a sin bx − b cos bx)

1 e

ax

0

.

− ( a sin bx + b cos bx ) 2

2

a +b b b = =0+ 2 . a + b2 a 2 + b 2 a . cos bx dx = 2 a + b2 x→∞

Similarly,



e− ax



1 2

a + b2

(– a . 0 – b . 1)

WORKING RULES FOR SOLVING PROBLEMS Rule I.

(i) The Fourier sine transform of the function f(x) is given by fS (s) = FS(f(x)) =

z



0

f(x) sin sx dx .

(ii) The Fourier cosine transform of the function f(x) is given by fC (s) = FC(f(x)) =

z



0

f(x) cos sx dx .

Rule II. (i) The inverse Fourier sine transform of the function g(s) is given by 2 ∞ g(s) sin sx ds . FS–1(g(s)) = π 0 (ii) The inverse Fourier cosine transform of the function g(s) is given by 2 ∞ g(s) cos sx ds . FC–1(g(s)) = π 0 Rule III. (i) If f(x) is either an odd function of x or defined on (0, ∞), then

z z

FS–1( fS (s) ) = f(x). (ii) If f(x) is either an even function of x or defined on (0, ∞), then FC–1( fC (s) ) = f(x). TYPE I. Problems Based on Fourier Sine and Cosine Transforms

ILLUSTRATIVE EXAMPLES Example 1. Find the Fourier sine and cosine transforms of the function e–ax, a > 0. Sol. Let ∴

f(x) = e–ax , a > 0.

FS(f(x)) = =

Also,

FC(f(x)) = =

z z z z



0 ∞

0



0 ∞

0

f ( x) sin sx dx e − ax sin sx dx =

s 2

a +s

2

.

F Using GH

.

F Using GH

z



0

e − ax sin bx dx =

b a + b2

I JK

a

I JK

2

f ( x) cos sx dx e − ax cos sx dx =

a a2 + s2

z



0

e − ax cos bx dx =

a 2 + b2

114

LAPLACE AND FOURIER TRANSFORMS

Example 2. Find the Fourier sine transform and Fourier cosine transform of the function 2e–5x + 5e–2x. Sol. Let f(x) = 2e–5x + 5e–2x. ∴

fS ( s ) =

z



z

0

=2



0

fC ( s ) =

2

(5) + s

2

+ 5.



=2

z



0

(5) 2 + s 2

F 1 GH 25 + s

= 10

2

+ 5.

+

0

e −2 x sin sx dx

(2) + s

z



0

e −5 x cos sx dx + 5

5

=2.



F Using GH

s

f ( x ) cos sx dx =

0

(2 e −5 x + 5 e −2 x ) sin sx dx

z

0

2

2s 5s . + 25 + s 2 4 + s 2

z



e −5 x sin sx dx + 5

s

=2. =

z

f ( x ) sin sx dx =

2

4+s

0

e − ax sin bx dx =

b 2

a +b

2

I JK

z



0

e −2 x cos sx dx

F Using GH

(2) 2 + s 2

I. JK

2



( 2e −5 x + 5e −2 x ) cos sx dx

2

1

z

z



0

e − ax cos bx dx =

a a 2 + b2

I JK

Example 3. Find the Fourier sine transform and Fourier cosine transform of the function

RS T

sin x , if 0 < x < a f(x) = 0 , if x > a. Sol. Fourier sine transform of f(x)

= = =

z z



f ( x) sin sx dx =

0 a

0

1 2

z z

a

0

sin x sin sx dx +

z

a

0

f ( x) sin sx dx + ∞

a

0 . sin sx dx

2 sin sx sin x dx + 0 =

1 2

z

a

0

a

0

Fourier cosine transform of f(x) = =

z z



0

a

0

f ( x) cos sx dx

f ( x) cos sx dx +

z



a



a

f ( x) sin sx dx

(cos (s − 1) x − cos (s + 1) x dx

LM OP N Q 1 L sin (s − 1)a sin (s + 1)a O − = M PQ . 2N s−1 s+1 1 sin (s − 1) x sin (s + 1) x − = 2 s−1 s+1

z

f ( x) cos sx dx

115

FOURIER TRANSFORMS

z

=

a

0

1 2

=

z

sin x cos sx dx +

z

a

0



a

0 . cos sx dx 1 2

2 cos sx sin x dx + 0 =

z

a

0

(sin (s + 1) x − sin (s − 1) x) dx

LM OP N Q 1 L cos (s + 1) a cos (s − 1) a 1 1 O = M− + + − 2N s+1 s−1 s + 1 s − 1PQ 1 L cos (s − 1)a cos (s + 1)a O − = M PQ − s 1− 1 . 2N s−1 s+1 a

1 cos (s + 1) x cos (s − 1) x − + = 2 s+1 s−1

0

2

Example 4. Find the Fourier sine and cosine transforms of the function xm – 1 , m > 0. f(x) = xm – 1, m > 0.

Sol. Let

fS (s) =

∴ and

fC ( s) =

Let

g(s) =

z z z



f ( x) sin sx dx =

0 ∞

0

f ( x) cos sx dx =



0

x m − 1 e − isx dx .

z z



0

x m − 1 sin sx dx



0

x m − 1 cos sx dx

y = isx ⇒ dy = is dx g(s) =



=

z





0

⇒ ⇒

z z

0

0

FG y IJ H is K

z



m− 1

1 m

s (e

e− y

1 dy = (is) m is

Γ(m) =

iπ / 2 m

x m − 1e − isx dx =

0



z



)

z

x m − 1 (cos (− sx) + i sin (− sx) dx =

0

∴ (1)

x m − 1 cos sx dx =



fS (s) =

0

z



0

x m − 1 sin sx dx =

Γ (m) mπ cos m 2 s Γ(m) s

m

sin

....(2)

e − y y m − 1 dy =

FG∵ H

Γ(m) –imπ/2 e sm

Comparing real and imaginary parts, we get ∞



Γ(m) − imπ / 2 e sm

x m − 1 cos sx dx − i

...(1)

mπ 2

and and

Γ (m) s

m

1 Γ (m) (is) m

e iπ / 2 = cos

IJ K

π π + i sin = i 2 2

FG cos FG − mπ IJ + i sin FG − mπ IJ IJ H 2 KK H H 2 K

Γ (m) Γ (m) mπ mπ cos − i m sin 2 2 sm s

z



0

x m − 1 sin sx dx =

(2)



fC (s) =

Γ (m) mπ . sin 2 sm

Γ(m) s

m

cos

mπ . 2

116

LAPLACE AND FOURIER TRANSFORMS 2

Example 5. Find the Fourier cosine transform of the function e − x .

e j=

FC e − x

Sol.

2

I=



z z



0



2

e − x cos sx dx = I, say 2

e − x cos sx dx

0

...(1)

Differentiating w.r.t. s, using Leibnitz’s rule of differentiation under integral sign, we have

z

∞ 2 dI = e − x ( − x sin sx ) dx 0 ds 2 dI 1 ∞ = (sin sx ) ( − 2x e − x ) dx ds 2 0 I II



z

LM N

=

2 1 sin sx . e − x − 2

=

1 s (0 – 0) – 2 2

dI s + I=0 ds 2



log



I s2 =− c 4 /4

z

Putting s = 0 in (1) and (2), we get

z

⇒ ∴



0

(2)

π − s2 / 4 e 2

I=

Sol. ∴

FC

F 1 I= GH 1 + x JK 2

I =

z z



0

0

s I 2

⇒ log I = –

s2 + log c. 4

...(2) ∞

0

2

e − x cos 0 dx = ce 0

F∵ GH

i.e., FC ( e

− x2

)=

x 1 + x2 1

1 + x2



cos sx

0

1 + x2

z



0

2

e − x dx =

π 2

I JK

π −s2 / 4 e . 2

Example 6. Find the Fourier cosine transform of the function Fourier sine transform of the function



π =c 2

2

e − x dx = c ⇒



2

e − x cos sx dx = −

OP Q

2 I = e −s / 4 c

⇒ 2

0

2

s cos sx . e − x dx

dI s = − ds I 2



−s I = ce



z



z

1 . Hence deduce the 1 + x2

.

cos sx dx = I, say dx

...(1)

Differentiating w.r.t. s, using the Leibnitz’s rule of differentiation under integral sign, we have

117

FOURIER TRANSFORMS

dI = ds

We have ∴

(2)

z

− x sin sx



1 + x2

0

...(2)

dx

x x2 1 1 (1 + x 2 ) − 1 =− =− =− + 2 2 2 x 1+ x x(1 + x ) x(1 + x ) x (1 + x 2 )



dI = ds



z



0

=–

z

F − 1 + 1 I sin sx dx GH x x(1 + x ) JK 2

z

sin sx dx + x



0

dI π∗ =− + ds 2

z

sin sx



0

x(1 + x 2 )

dx

sin sx dx x(1 + x 2 ) Differentiating again w.r.t. s, we get



d 2I ds



2

z

=0+

...(3)

0

x cos sx



0



z

dx =

2

x(1 + x )

(D2 – 1)I = 0, where D ≡

d ds



cos sx

0

1 + x2

dx = I

D2 – 1 = 0 ⇒ D = ± 1 I = c1 e1.s + c2 e–1.s = c1es + c2 e–s



dI = c1es – c2e–s ds Putting s = 0 in (1), we get



c1

e0

+ c2

e0

=

z



0

cos 0 dx = 1 + x2

π ∴ c1 + c2 = 2 Putting s = 0 in (3), we get

c1e0 – c2e0 = – ∴

c1 – c2 = –

Solving (4) and (5), we get ∴ *Why this step. We know that Let ∴

z



0

z



0

1 dx = tan −1 x 2 1+ x



= 0

π π −0= 2 2

...(4)

π + 2

π 2

z



0

sin 0 π dx = − + 2 2 x(1 + x )

z



0

0 dx = −

...(5)

π . 2 π –s π –s I = 0. es + e = e 2 2

c1 = 0 and c2 =

z

π 2



π sin x dx = . x 2 0 y = sx, s > 0.

sin sx dx = x

z



0

sin y dy = y/ s s

z



0

sin y π dy = . 2 y

118

LAPLACE AND FOURIER TRANSFORMS

FC

∴ Also

FS

F x GH 1 + x

2

F 1 I = πe . GH 1 + x JK 2 I = x sin sx dx = − dI JK 1 + x ds π F I = – (c e – c e ) = – G 0. e − e J H K 2 −s

2

z



(Using (2))

2

0

π −s e . 2 1 Example 7. Find the Fourier sine transform of the function . 2 x(x + a 2 ) 1

Sol.

FS

F 1 I= GH x(x + a ) JK 2

2

z z

s

2

s

–s

−s

=

1 sin sx dx = I, say x( x 2 + a 2 )



0

sin sx



dx ...(1) x( x 2 + a 2 ) Differentiating w.r.t. s, using Leibnitz’s rule of differentiation under integral sign, we have

I=



0

z z

∞ x cos sx dI = dx = 0 x( x 2 + a 2 ) ds Differentiating again w.r.t. s, we get

d 2I = ds 2

We have



x x 2 + a2

0

=−

d 2I = ds 2





z



0

=– d 2I



ds

2

– a2I = –

z



0

cos sx x 2 + a2

dx

...(2)

− x sin sx dx x 2 + a2 x2 x( x 2 + a 2 )

=−

1 a2 ( x 2 + a2 ) − a2 = – + x x( x 2 + a 2 ) x( x 2 + a 2 )

F − 1 + a I sin sx dx GH x x(x + a ) JK 2

z

2



0

2

sin sx dx + a 2 x

π 2

z



0

sin sx π dx = − + a2I 2 2 2 x( x + a )

FG Using H

z



0

π 2 D2 – a2 = 0 ⇒ D = ± a ∴ C.F. = c1eas + c2 e–as

(D2 – a2)I = –



FG IJ H K

π 1 π 1 π π 1 e0 s = − =− . 2 e0 s = – . 2 2 2 2 2 0 −a 2 2 D −a 2a 2 D −a π ∴ I = C.F. + P.I. = c1eas + c2e–as + 2a 2 dI ⇒ = ac1eas – ac2e–as ds Putting s = 0 in (1), we get

P.I. =

z



0

2

sin 0 π dx = c1e0 + c2e0 + . 2a 2 x( x 2 + a 2 )

IJ K

sin λx π dx = , if λ > 0 2 x

119

FOURIER TRANSFORMS

0 = c1 + c2 +



π 2a 2

⇒ c1 + c2 = –

π

...(3)

2a 2

Putting s = 0 in (2), we get

z



0

cos 0 dx = ac1e0 – ac2e0. x 2 + a2 ∞

x 1 tan −1 a a



0

c1 – c2 =



FG H

IJ K

1 π − 0 = a(c1 – c2) a 2

= ac1 – ac2 ⇒ π 2a 2

...(4)

Solving (3) and (4), we get c1 = 0 and c2 = –

FG H

I = 0.eas + −

∴ FS



F 1 I= π GH x(x + a ) JK 2a 2

2

2

π 2a

2

π

2a 2

IJ e K

–as

.

+

π 2a

2

=

π

2a 2

(1 – e–as)

(1 – e–as).

TEST YOUR KNOWLEDGE Find the Fourier sine and cosine transforms of the following functions (Q. 1– 6) : 1. f(x) = e–7x

2. f(x) = 3e–2x + 8e–11x

R|0 , if S|x , if T0 , if Rcos x , 6. f(x) = S T 0,

RS1 , if 0 ≤ x < 1 T0 , if x > 1 R| x , if 0 < x < 1 f(x) = S2 − x , if 1 < x < 2 |T 0 , if x > 2

3. f(x) =

5.

4. f(x) =

7. Find the Fourier sine transform of the function

1 . x

8. Find the Fourier sine transform of the function

e− ax , a > 0. x

0 0. s

f(x) = FC–1 =

1 π

=

1 π

z z

FG sin as IJ = 2 H s K π



0 ∞

0

z



0

sin as cos sx ds s

2 sin as cos sx 1 ds = s π sin s(a + x) 1 ds + s π

z



0

z



0

sin s(a + x) + sin s( a − x) ds s

sin s(a − x) ds s

121

FOURIER TRANSFORMS

R| 1 . π + 1 . π π 2 π 2 = S1 π 1 F π I || π . 2 + π . GH − 2 JK T R1 , if x < a =S T0 , if x > a R1 , if 0 < x < a . f(x) = S T0 , if x > a



if a − x > 0

(∵ x > 0)

Remark. If λ < 0, then

z



0

sin λx dx = x

z



0

− sin (− λ ) x dx = – x

z



sin (− λ) x π dx = − . x 2

0

R| 1 FG a − s IJ , S| 2 π H 2 K T 0,

Example 2. Find the function f(x) if FC(f(x)) =

Sol. We have



R| 1 FG a − s IJ , F (f(x)) = S 2π H |T 0, 2 K C

if 0 < s < 2a if

f(x) = FC–1(FC (f(x)) =

LM N 2L = M πN =

2 π

z z z

2a

0

2a

0

(∵ x > 0, a > 0 ⇒ x + a > 0)

if a − x < 0

2 π

z

if

.

s ≥ 2a

.

s ≥ 2a ∞

0

FC ( f ( x )) cos sx ds

FC ( f ( x )) cos sx ds +

FG H

if 0 < s < 2a

IJ K

z



FC ( f ( x )) cos sx ds

2a

s 1 a− cos sx ds + 2π 2

z



2a

OP Q

OP Q

0 . cos sx ds

FG a − s IJ cos sx ds + 0 H 2K π s I sin sx 1 LF F 1 I sin sx dsOP a− J − G− J = G M H K H 2K x Q x 2 π N 1 L cos 2ax cos 0 O cos sx O 1 L 2a − s = = 0− −0+ sin sx − M M P x 2 2x 2 x PQ π N π N 2x Q =

1

2

2

2a

0

z

2a

0

2a

2

=

2

0

1 sin 2 ax . 2 2 (1 – cos 2ax) = 2π x π2x2

2

2

2

122

LAPLACE AND FOURIER TRANSFORMS

Example 3. Find the inverse Fourier sine transform of the function e–as/s, a > 0, s > 0. Hence deduce the value of FS–1(1/s). FS–1

Sol.

Fe I = 2 GH s JK π − as

f(x) =



z z z

e − as sin sx ds = f(x), say s



0

2 π



0

e − as sin sx ds s

...(1)

Differentiating w.r.t. x, we get df 2 = dx π

=

e − as 2 . s cos sx ds = s π



0

2 a . π a2 + x 2

z



0

e − as cos sx ds

F∵ GH

z



0

e − ax cos bx dx =

a 2

a +b

2

I JK

Integrating w.r.t. x, we get f(x) =

2 x 2a 1 x . tan −1 + c = tan −1 +c π a π a a

...(2)

Putting x = 0 in (1) and (2) and equating, we get 2 π

z



0

2 0 e − as sin 0 ds = tan −1 + c s a π 2 .0=0+c ⇒ c=0 π

⇒ ∴ (2)

f(x) =



2 x tan–1 π a

F e I 2 tan GH s JK = π − as



FS–1

x . a

−1

Let a = 0. ∴

FS–1

F e I = 2 tan GH s JK π −0

−1

x 0

or FS–1

FG 1IJ = 2 . π = 1. H sK π 2

Example 4. Find the inverse Fourier cosine transform of the function

Sol.



FC–1

F 1 I=2 GH 1 + s JK π 2

f(x) =

z z

2 π



1 1 + s2

0



0

1 1 + s2

.

cos sx ds = f(x), say

cos sx ds 1 + s2

...(1)

123

FOURIER TRANSFORMS

Differentiating w.r.t. x, we get df 2 = dx π

We have



s 1 + s2

=−

df 2 = dx π



=–

=–

z

− s sin sx ds 1 + s2



0

s2 s(1 + s 2 )

z FGH z ∞

0

2 π

1 1 (1 + s 2 ) − 1 =− + 2 s s(1 + s 2 ) s(1 + s )

I JK

1 1 sin sx ds + s s(1 + s 2 )



sin sx 2 ds + s π



0

2 π 2 . + π 2 π

z

df 2 = − 1+ dx π



=−



0

z



0

z

s(1 + s 2 )

0

sin sx s(1 + s 2 )

d2 f



dx 2

z



0

ds

(Assuming x > 0)

ds

sin sx ds s(1 + s 2 )

Differentiating again w.r.t. x, we get d2 f 2 =0+ 2 π dx

sin sx



s cos sx 2 ds = 2 π s(1 + s )

...(2)

z



0

cos sx ds = f(x) 1 + s2

(By using (1))

– f(x) = 0 ⇒ (D2 – 1)f = 0



f(x) = c1e1.x + c2e–1.x = c1ex + c2e–x

...(3)



df = c1ex – c2e–x dx

...(4)

Putting x = 0 in (1) and (3) and equating, we get 2 π

z



0

cos 0 ds = c1e0 + c2e0 1 + s2

2 tan −1 s π



∞ 0

= c1 + c2 ⇒

FG H

IJ K

2 π − 0 = c1 + c2 π 2

c1 + c2 = 1



...(5)

Putting x = 0 in (2) and (4) and equating, we get –1+ ⇒

2 π

z



0

sin 0 s(1 + s 2 )

–1+

ds = c1e0 – c2e0

2 . 0 = c1 – c2 ⇒ c1 – c2 = – 1 π

...(6)

124

LAPLACE AND FOURIER TRANSFORMS

Solving (5) and (6), we get c1 = 0, c2 = 1 ∴ f(x) = 0.ex + 1 . e–x = e–x

F 1 I GH 1 + s JK = e

FC–1



–x,

2

x > 0.

TEST YOUR KNOWLEDGE 1. Find the inverse Fourier sine and cosine transforms of the function e–πs.

FG H

IJ K

2. Find the function FC–1 sin 7s . s

R| 4 − s , S| 4π T 0,

3. Find the function f(x) if FC(f(x)) =

if 0 < s < 4

if s ≥ 4. 4. Find the inverse Fourier sine transform of the function e–7s/s, s > 0. 5. Find the inverse Fourier sine transform of the function

s 1 + s2

.

Answers 1. FS–1(e–πs) =

2x π(π 2 + x 2 )

2. 1 if 0 < x < 7 and

sin 2x

4.

π2 x2

2 π 2 + x2

0 if x > 7

2

3.

, FC–1(e–πs) =

2 x tan −1 7 π

5. e–x.

TYPE III. Problems Based on Integrals and Integral Equations

ILLUSTRATIVE EXAMPLES Example 1. Show that the Laplace integral Sol. Let

f(x) = FS(f(x)) =



=

e–ax,

z



0

a +s

f(x) = FS–1



e–ax =

z



s sin sx

0

a2 + s2

z

2

a +s

0

2

ds , x > 0, a > 0 is equal to

F GH a

s 2

+s

2

s 2

a + s2

I JK

0

e − ax sin sx dx

F Using GH

z



0

e − ax sin bx dx =

b 2

a +b

2

I JK

(Note that f(x) is defined only for x > 0)

sin sx ds

ds = π e–ax, x > 0, a > 0. 2

π –ax e . 2

(Note this step) ∞

2



0

s sin sx

x > 0, a > 0.

s







f ( x) sin sx dx =

2

2 π

z z

125

FOURIER TRANSFORMS

Example 2. Show that :

z



0

RS T

π / 2 , if 0 < x < π 1 − cos πλ sin xλ dλ = . 0 , if x>π λ

Sol. Let

f(x) = FS(f(x)) =



= =

RSπ/ 2 , T0 ,

z z z



0

π

0 π

0

if 0 < x < π . if x>π

f ( x) sin sx dx

f ( x) sin sx dx + π sin sx dx + 2

FG H

π cos sx − = s 2

=

=

z



0

π



π

f ( x) sin sx dx

0 . sin sx dx

π

+0 0

π π(1 − cos πs) (– cos πs + cos 0) = 2s 2s

f(x) = FS–1(FS(f(x)) = FS–1





IJ K

z

z



z



0

FG π(1 − cos πs) IJ = 2 H 2s K π

1 − cos πs sin sx ds = s

RS T

π /2 , if 1 − cos πλ sin λx dλ = f ( x ) = 0 , if λ

z



0

z



0

π (1 − cos πs) sin sx ds 2s

1 − cos πλ sin λx dλ λ

(Changing s by λ)



Example 3. Using the Fourier sine transform of the function e–ax – e–bx, a, b > 0, x > 0, show that :

z



0

Sol. Let ∴

u sin ux π(e − ax − e − bx ) , x > 0. du = 2 2 2 (u + a ) (u + b ) 2(b 2 − a 2 ) 2

f(x) = e–ax – e–bx, x > 0. fS (s) =

=

=

z z



0 ∞

0

f ( x) sin sx dx = e − ax sin sx dx −

s 2

a +s

2



s 2

b +s

2

z z



0 ∞

0

(e − ax − e − bx ) sin sx dx e − bx sin sx dx

F Using GH

z



0

e − ax sin bx dx =

b a + b2 2

I JK

126

LAPLACE AND FOURIER TRANSFORMS

=



fS (s) =

s(b2 + s 2 − a 2 − s 2 ) s(b2 − a 2 ) = (a 2 + s 2 ) (b2 + s 2 ) (a 2 + s 2 ) (b2 + s 2 ) s(b2 − a 2 ) (a 2 + s 2 ) (b2 + s 2 )

Since the function f(x) is defined only for (0, ∞), we have FS–1 ( fS (s)) = f(x). 2 π







2 π

z



0



z



0

fS (s) sin sx ds = f(x)

s(b2 − a 2 ) sin sx ds = f(x) (a 2 + s 2 ) (b2 + s 2 )

2(b 2 − a 2 ) π



z

z z

2

(a + s 2 ) (b2 + s 2 )

0



0

z

s sin sx





0

ds = e–ax – e–bx

π(e − ax − e − bx ) s sin sx ds = (a 2 + s 2 ) (b2 + s 2 ) 2(b2 − a 2 ) u sin ux

( a 2 + u 2 ) (b2 + u 2 )

du =

π(e −ax − e − bx ) 2(b 2 − a 2 )

, x > 0.

(Replacing s by u)

Example 4. Using the Fourier cosine transform of e –x cos x, x > 0, show that ∞

0

(u 2 + 2) cos ux π du = e–x cos x, x > 0. 4 2 u +4

Sol. Let ∴

f(x) = e– x cos x, x > 0. fC ( s) =

z



0

f ( x) cos sx dx =

z z z

z



0

e − x cos x cos sx dx

=

1 2

=

1 2

=

1 2

=

1 1 1 1 . + . 2 2 1 + (s + 1) 2 1 + (s − 1) 2



0 ∞

0 ∞

0

e − x (2 cos sx cos x) dx e − x (cos (s + 1) x + cos (s − 1) x) dx e − x cos (s + 1) x dx +

1 2

z



0

e − x cos (s − 1) x dx

F Using GH

z



0

e − ax cos bx dx =

a 2

a +b

2

I JK

127

FOURIER TRANSFORMS

LM MN 1 Ls = M 2N =

2

2

2

2

2

4

2

2 4

+2 +4

s2 + 2 s4 + 4

fC (s) =



OP PQ 1 L 2s + 4 O s − 2s + 2 + s + 2s + 2 O = M P= P 2 (s + 2) − 4 s Q N s +4 Q s

1 1 1 + 2 2 2 s + 2s + 2 s − 2s + 2

Since the function f(x) is defined only for (0, ∞), we have FC–1( fC (s) ) = f(x). 2 π

∴ ⇒ ⇒



z

2 π

z

0



0

z z





s2 + 2 cos sx ds = f(x) s4 + 4

( s 2 + 2) cos sx

ds =

π . e–x cos x 2

du =

π –x e cos x, x > 0. 2

4

s +4

0



fC (s) cos sx ds = f(x)

(u 2 + 2) cos ux 4

u +4

0

(Replacing s by u)

Example 5. Solve the integral equation :

z z z



0



Sol. We have

0





0



f(x) cos λx dx = e–λ.

f ( x) cos λx dx = e–λ. f ( x) cos sx dx = e–s

FC(f(x)) = e–s

f(x) = FC–1 (e–s) =



=

f(x) =



2 π

z

π

0

e − s cos sx ds

F∵ GH

2 1 π 1 + x2 2 π(1 + x 2 )

.

Example 6. Solve the integral equation :

z



0

R| S| T

1 , if f(x) sin sx dx = 2 , if 0 , if

0 ≤ s 2

1 , if 0 ≤ s < 1 f ( x ) sin sx dx = 2 , if 1 ≤ s < 2 . s>2 0 , if



0

R| S| T

f(x) = FS–1(FS(f(x))) =



z

2 π

z



FS ( f ( x )) sin sx ds

0

z

z

LM 1 . sin sx ds + 2 . sin sx ds + 0.sin sx dsOP N Q O 2 L − cos sx 2(− cos sx) + + 0P = M π MN x x PQ 2 L − cos x 1 2 cos 2 x 2 cos x O + − + = M πN x x x x PQ 2 L 1 cos x 2 cos 2 x O 2 = M + = − P π Nx x x Q πx (1 + cos x – 2 cos 2x). =

2 π

1

0

2

1

1



2

2

0

1

TEST YOUR KNOWLEDGE 1. Show that the Laplace integral 2. Show that

z z



0

cos λx

dλ =

2

λ +1

z



cos sx 2

a +s

0

2

ds , x > 0, a > 0 is equal to

π –ax e . 2a

π e–x, x ≥ 0. 2

3. Find the Fourier sine transform of the function e–| x |, x > 0. Hence show that : ∞

0

x sin mx 1+ x

2

dx =

π –m e , m > 0. 2

4. Using the Fourier cosine transform of the function f(x) =

z



0

sin λ cos λx dλ . λ

5. Solve the integral equation : 6. Solve the integral equation :

Hence deduce that

z

z z





0



0

sin2 t t2

dt =

RS1 − α , T 0, R1 − λ , f ( x ) cos λx dx = S T 0,

f ( x ) sin αx dx =

0

RS1 , T0 ,

π. 2

if if

0 < x 1

if if

0 ≤ α ≤1 . α >1

if if

0 ≤ λ ≤ 1. λ >1

129

FOURIER TRANSFORMS

Answers 3.

s 1+ s

6. f(x) =

π if 0 < x < 1 and 0 if x > 1 2

4.

2

2(1 − cos x)

z



πx 2

Hint ∞

2(1 − cos x ) πx

0

2

cos λx dx =

4 π

z



sin 2

z

2 π

Taking limits as λ → 0, we get



2( x − sin x)

.

πx2

6. We have

5. f(x) =



0

x 2 dx = 1

RS1 − λ , T 0,

if if

1 − cos x x2

0 ≤ λ ≤1 λ >1.

. 1 dx = 1

x2 x/2 = t ⇒ dx = 2 dt

4 π

0

z



0

sin 2 t 4t 2

2 dt = 1.

5.8. LINEARITY OF TRANSFORMS In this section, we shall prove that the Fourier transform, the Fourier sine and cosine transforms are linear operations. Theorem I. If f(x) and g(x) be any functions whose Fourier transforms exist then for any constants a and b then prove that F(af(x) + bg(x)) = aF(f(x)) + bF(g(x)). Proof. By definition, F(af(x) + bg(x)) =

z



z

−∞

=a

(af ( x) + bg ( x)) e − isx dx ∞

−∞

f ( x) e − isx dx + b

z



−∞

g ( x) e − isx dx

= aF(f(x)) + bF(g(x)). Theorem II. If f(x) and g(x) be any functions whose Fourier sine transforms exist then for any constants a and b prove that FS(af(x) + bg(x)) = aFS(f(x)) + bFS(g(x)). Proof. By definition, FS (af(x) + bg(x)) =

z



z

( af ( x ) + bg( x )) sin sx dx

0

=a



0

f ( x) sin sx dx + b

z



0

g ( x) sin sx dx

= aFS(f(x)) + bFS(g(x)). Theorem III. If f(x) and g(x) be any functions whose Fourier cosine transforms exist then for any constants a and b then prove that FC (af(x) + bg(x)) = aFC(f(x)) + bFC(g(x)).

130

LAPLACE AND FOURIER TRANSFORMS

Proof. By definition, FC(af(x) + bg(x)) =

z



( af ( x) + bg ( x)) cos sx dx

z

0

=a



f ( x) cos sx dx + b

0

z



0

g ( x) cos sx dx

= aFC (f(x)) + bFC(g(x)).

Remark. On the same lines, we can also prove that the inverse Fourier transform, the inverse Fourier sine and cosine transforms are also linear operations.

5.9. CHANGE OF SCALE PROPERTY OF TRANSFORMS Theorem I. If the Fourier transform of the function f(x) is f (s) then prove that the Fourier transform of f(ax) is

FG IJ H K

1 s f , where a > 0. a a

Proof. We have

bg

f s = F( f ( x )) =

F(f(ax)) =

∴ Let t = ax. ∴

z z



z



f ( x ) e − isx dx .

−∞

f (ax) e − isx dx

−∞

dt = adx F(f(ax)) =



=



f (t) e − is(t / a)

z

−∞

1 a



−∞

dt a

(∵ a > 0)

f (t) e − i( s / a) t dt =

1 f a

FG s IJ . H aK

Theorem II. If the Fourier sine transform of the function f(x) is fS (s) then prove that the Fourier sine transform of f(ax) is Proof. We have

Let t = ax. ∴ ∴

fS (s) = FS(f(x)) =

FS(f(ax)) =



FG IJ H K

1 s fS , where a > 0. a a

z z



z



0

f ( x) sin sx dx .

f (ax) sin sx dx

0

dt = adx

FS (f(ax)) = =



0

1 a

z

f (t ) sin ( s(t / a )) ∞

0

dt a

f ( t ) sin (( s/a )t ) dt =

(∵ a > 0)

FG IJ H K

s 1 . fS a a

Theorem III. If the Fourier cosine transform of the function f(x) is fC (s) then prove that the Fourier cosine transform of f(ax) is Proof. We have

FG IJ H K

1 s , where a > 0. fC a a

fC ( s ) = FC ( f ( x )) =

z



0

f ( x ) cos sx dx .

131

FOURIER TRANSFORMS

FC(f(ax)) =



Let t = ax. ∴

f (ax) cos sx dx

0

dt = adx

FC(f(ax)) =



z z



=



0

1 a

z

f (t) cos (s(t / a) ∞

0

dt a

f (t) cos ((s / a)t) dt =

(∵ a > 0)

FG IJ H K

1 s . fC a a

5.10. TRANSFORMS OF DERIVATIVES In this section, we shall derive the formulae to find the Fourier transform and the Fourier sine and cosine transforms of the derivatives of a function. Theorem I. If f(x) is any function for which the Fourier transforms of f(x) and f ′(x) exist and f(x) → 0 as |x| → ∞ then prove that F(f ′(x)) = is F(f(x)). Proof. By definition, F(f(x)) = and

F(f ′(x)) = (2)



F(f ′(x)) =

z z z



−∞ ∞

−∞



−∞

f ( x) e − isx dx

...(1)

f ′( x) e − isx dx

...(2)

e − isx f ′( x) dx I

II

Integrating by parts, we get ∞

− isx f ( x) F(f ′(x)) = e



z

−∞

= 0 – 0 + is



−∞

z



−∞

(− is) e − isx f ( x) dx

f ( x) e − isx dx = is F(f(x)).

(∵ f(x) → 0 as | x | → ∞)

Remark. By repeated application of the above theorem, we have F(f ″(x)) = is F(f ′(x)) = (is). (is)F(f (x)) = – s2F(f(x)) ∴

F(f ″(x)) = – s2F(f(x)).

By using P.M.I., we can prove that F(f(n)(x)) = (is)nF(f(x)), n ∈ N.

Theorem II. If f(x) is any function for which the Fourier sine and cosine transforms of f(x) and f ′(x) exist and f(x) → 0 as x → ∞ then prove that (i) FS(f ′(x)) = – sFC(f(x)) Proof. By definition and

FS(f(x)) = FC(f(x)) =

z z

(ii) FC(f ′(x)) = sFS(f(x)) – f(0). ∞

0 ∞

0

f ( x) sin sx dx f ( x) cos sx dx .

132

LAPLACE AND FOURIER TRANSFORMS

(i)

FS(f ′(x)) =

z



0



= sin sx f ( x) =0–0– s

(ii)

FC(f ′(x)) =

z



0

z

f ′( x) sin sx dx =

z

− 0



0

z



0



0

sin sx . f ′( x) dx I

II

s cos sx f ( x) dx

f ( x) cos sx dx = – sFC(f(x)).

z

f ′( x) cos sx dx = ∞

= cos sx f ( x)

− 0

z



0

= 0 – (cos 0) f(0) + s

(∵ f(x) → 0 as x → ∞) ∞

0

cos sx . f ′( x) dx I

II

− s sin sx f ( x) dx

z



0

(∵ f(x) → 0 as x → ∞)

f ( x) sin sx dx

= – f(0) + sFS (f(x)) = sFS (f(x)) – f(0). Remarks. (i) FS(f ″(x)) = – sFC(f ′(x)) = – s[sFS(f(x)) – f(0)] = – s2FS(f(x)) + sf(0). (ii) FC(f ″(x)) = sFS(f ′(x)) – f ′(0) = s[– sFC (f(x))] – f ′(0) = – s2 FC (f(x)) – f ′(0). ∴ FS(f ″(x)) = – s2 FS(f(x)) + sf(0) and FC(f ″(x)) = – s2 FC(f(x)) – f ′(0).

ILLUSTRATIVE EXAMPLES Example 1. Find the Fourier sine and cosine transforms of the function e–mx, m > 0 by using its second derivative. Sol. Let f(x) = e–mx , m > 0. ∴ f ′(x) = – me–mx and f ″(x) = (– m) (– m) e–mx = m2 e–mx We have FS(f ″(x)) = – s2 FS (f(x)) + sf(0). ∴ FS(m2 e–mx) = – s2 FS(e–mx) + se0 ⇒ m2FS(e–mx) + s2FS(e–mx) = s ⇒ Also, ⇒ ⇒ ⇒

s

(m2 + s2) FS (e–mx) = s ⇒ FS(e–mx) =

m2 + s2 FC(f ″(x)) = – s2 FC(f(x)) – f ′(0).

.

FC (m2e–mx) = – s2 FC(e–mx) – (– me0) m2FC(e–mx) + s2FC(e–mx) = m (m2 + s2) FC(e–mx) = m ⇒ FC(e–mx) =

m 2

m + s2

.

5.11. PARSEVAL’S IDENTITIES Parseval’s identities are for Fourier transforms and also for Fourier sine and cosine transforms. These identities are proved in the following theorems. Theorem I. If f (s) and g ( s) are the Fourier transforms of the functions f(x) and g(x) respectively then prove that

133

FOURIER TRANSFORMS

z



−∞

f ( x) g( x) dx =

z

1 2π

bg



f (s) g s ds

−∞

z

and



−∞

|f ( x)|2 dx =

where g (s) represents the complex conjugate of the function g ( s) .

z

z FGH z z FGH z z FGH z z FGH z z FGH z z z z z z z z z

Proof. 1 2π



−∞

f ( s ) g ( s ) ds = =

1 2π

=

1 2π

=

1 2π



−∞

−∞ ∞

−∞

−∞



f ( s)

−∞ ∞

f ( s)

−∞

f ( s)

|f (s)|2 ds ,

I JK

g( x ) e − isx dx ds

−∞

IJ K I dx J ds (∵ g(x) is a real valued function) K I dsJ dx K (By changing the order of integration) I dsJ dx K

g( x ) e isx

−∞

1 2π



−∞

g( x ) e − isx dx ds



g( x )

g( x )



z





−∞

g( x ) f ( x ) dx =

f ( s ) e isx

f ( s ) e isx ∞

−∞

f ( x ) g( x ) dx

1 ∞ f(s) g(s) ds . 2π −∞ g(x) = f(x).

f(x) g(x) dx =

−∞

In particular, let



∴ (1) ⇒ ∴



−∞

=



−∞



=





1 2π

1 2π

1 2π 1 |f(x)|2 dx = 2π

f ( x ) f ( x ) dx =

−∞



−∞



−∞ ∞

−∞

...(1)

f ( s ) f ( s ) ds .

|f(s)|2 ds .

(∵ (f(x))2 = |f(x)|2)

Theorem II. If fS (s) and g S ( s) are the Fourier sine transforms of the functions f(x) and g(x) respectively then prove that ∞ ∞ 2 ∞ 2 ∞ (f ( x)) 2 dx = ( f S ( s)) 2 ds . f ( x) g ( x) dx = fS (s) g S s ds and 0 π 0 0 π 0 ∞ 2 ∞ 2 ∞ Proof. fS (s) gS (s) ds = fS (s) g ( x) sin sx dx ds 0 π 0 π 0 ∞ 2 ∞ = g ( x) fS (s) sin sx ds dx 0 π 0 (By changing the order of integration) ∞ 2 ∞ = g ( x) fS (s) sin sx ds dx 0 π 0

z z



z

z

= ∞

In particular, let ∴

z



0

z FGH z z FGH z z FGH z z z z z ∞

g ( x) f ( x) dx =

0

JK IJ K



0

2 ∞ fS (s) g S (s) ds . π 0 g(x) = f(x).

f(x) g(x) dx =

0

zI

bg

(f(x))2 dx =

2 π



0

(fS (s))2 ds .

z

IJ K

f ( x) g( x) dx

...(1)

134

LAPLACE AND FOURIER TRANSFORMS

Theorem III. If fC (s) and gC ( s) are the Fourier cosine transforms of the functions f(x) and g(x) respectively then prove that ∞ ∞ 2 ∞ 2 ∞ ( f ( x)) 2 dx = ( fC (s)) 2 ds . f ( x) g ( x) dx = fC (s) gC (s) ds and 0 π 0 0 π 0 ∞ 2 ∞ 2 ∞ Proof. fC ( s ) gC ( s ) ds = fC ( s ) g( x ) cos sx dx ds 0 π 0 π 0

z z

z F z GH z z FGH z z FGH z z z z z z z z z =

2 π

= =



0



0



0

g ( x)

g ( x)

2 π



0



0

z

IJ K

IJ K (By changing the order of integration) I (s) cos sx dsJ dx K

fC (s) cos sx ds dx

fC

g ( x) f ( x) dx =



0

f ( x) g ( x) dx

2 ∞ fC (s) g C (s) ds . 0 π 0 In particular, let g(x) = f(x). ∞ 2 ∞ ∴ (1) ⇒ f ( x) f ( x) dx = fC (s) fC ( s) ds 0 π 0 ∞ 2 ∞ (f(x))2 dx = (fC (s))2 ds . ∴ 0 π 0





f(x) g(x) dx =

WORKING RULES FOR SOLVING PROBLEMS

Rule I.

(i) F(f ′(x)) = isF(f(x)) (ii) F(f(n)(x)) = (is)n F(f(x)), n ∈ N Rule II. (i) FS(f ′(x)) = – sFC(f(x)) (ii) FS(f ″(x)) = – s2FS(f(x)) + sF(0) Rule III. (i) FC(f ′(x)) = sFS(f(x)) – f(0) (ii) FC(f ″(x)) = – s2FC(f(x)) – f ′(0) Rule IV.

(i) (ii)

Rule V.

(i) (ii)

Rule VI.

(i) (ii)

z z z z z z

z

1 ∞ F( f ( x)) F( g ( x)) ds −∞ 2π −∞ ∞ 1 ∞ | f ( x)|2 dx = |F( f ( x))|2 ds −∞ 2π −∞ ∞ 2 ∞ f ( x) g ( x) dx = FS ( f ( x)) FS ( g ( x)) ds 0 π 0 ∞ 2 ∞ ( f ( x)) 2 dx = (FS ( f ( x))) 2 ds 0 π 0 ∞ 2 ∞ f ( x ) g( x ) dx = FC ( f ( x )) FC ( g( x )) ds 0 π 0 ∞ 2 ∞ ( f ( x )) 2 dx = ( FC ( f ( x ))) 2 ds . 0 π 0 ∞

f ( x) g ( x) dx =

z z z z z

z

...(1)

135

FOURIER TRANSFORMS

ILLUSTRATIVE EXAMPLES Example 1. If f(x) = f(x) to show that

z

RS1 , T0 ,

sin 2 ax



x

0

2

Sol. We have

if |x|< a , use the Parseval’s identity for Fourier transform of if |x|> a

dx =

f(x) = F(f(x)) =



=

πa . 2

RS1 , T0 ,

z z



−∞ −a

−∞

if | x|< a . if | x|> a f ( x) e − isx dx 0 . e − isx dx +

e − isx =0+ − is

a

z

a

1 . e − isx dx +

−a

z



a

0 . e − isx dx

+0 −a

F GH

I JK

2 e isa − e − isa 1 –isa 2 (e – eisa) = = sin sa s 2i is s By Parseval’s identity for Fourier transforms, we have ∞ 1 ∞ | f ( x)|2 dx = |F( f ( x))|2 ds . −∞ 2π −∞ 2 a −a ∞ 1 ∞ 2 ⇒ (0) 2 dx + (1) 2 dx + (0) 2 dx = sin sa ds a −∞ −a 2π −∞ s =–

z

z



z z

z

0 + (a – (– a)) + 0 = 2 π

⇒ ⇒

z z



−∞ ∞

−∞

z

2

1 2π

sin ax dx = 2a x2

z



−∞

z

4 sin 2 as ds s2

sin 2 ax dx = πa ⇒ 2 x2

z

(Replacing s by x) ∞

0

sin 2 ax x2

dx = πa

F∵ GH

sin 2 ax

πa dx = . 2 x2 Example 2. Using a Parseval’s identity, show that : ∞ π dx , a, b > 0. = 0 (a 2 + x 2 ) (b 2 + x 2 ) 2ab(a + b) Sol. Let f(x) = e–ax and g(x) = e–bx. ∞ a ∴ fC ( s) = e − ax cos sx dx = 2 0 a + s2



0

z

and



I JK

sin 2 ax is an even function x2

gC (s) =

z z



0

e − bx cos sx dx =

b 2

b + s2

(Note this step)

136

LAPLACE AND FOURIER TRANSFORMS

∴ By Parseval’s identity for Fourier cosine transforms, we have

z z z



0







0

z

e − ( a + b) x dx =

π e − ( a + b) x . 2ab −(a + b)

⇒ ∴

e − ax e − bx

0





= 0

dx

∞ 2

2

2



fC (s) gC (s) ds .

0

2

a +s

z

2ab π

z

a



0

.

2

b

ds

2

b + s2

ds (a 2 + s 2 ) (b2 + s 2 )



0

dx (a + x ) (b2 + x 2 )



2

0

(Changing s by x)

2

π π (0 – 1) = . 2ab( a + b) 2ab(a + b)

=−

2

( a + x )( b + x )

0

z z

2 π 2 dx = π

f ( x) g ( x) dx =

Example 3. Using a Parseval’s identity, show that :

z

0

Sol. Let

F GH

π 1 − e−a = dx 2 x(a 2 + x 2 ) a2 sin ax



2

I , a > 0. JK

f(x) = e–ax, x > 0, a > 0

and

g(x) = fC (s) =

∴ and

gC (s) =

=

RS1 , T0 ,

z z z



if if

0a a a + s2

e − ax cos sx dx =

0 ∞

g ( x) cos sx dx

0

a

z

1 . cos sx dx +

0

2



a

a

0 . cos sx dx

1 sin as sin sx + 0 = (sin sa – sin 0) = s s s 0 ∴ By Parseval’s identity for Fourier cosine transforms, we have

=

⇒ ⇒

z



0

a

0

f ( x) g ( x) dx +

z

a

0

a

a

z

2 π 2 f ( x) g ( x) dx = π f ( x) g ( x) dx =



a

+0= 0

e − ax . 0 dx =



0

z z

2a π

1 − a2 2a − 1] = [e −a π



z



e − ax . 1 dx +

e − ax −a





z z



0



0

F GH

z z z



2a π

sin ax x( a 2 + x 2 )

I. JK

2

a +s

0



0

x(a 2 + x 2 )

2

a



sin ax

sin ax π 1 − e −a dx = 2 2 2 a2 x( a + x )

fC (s) gC (s) ds .

0

2

.

sin as ds s

sin as ds s(a 2 + s 2 )

dx dx

(Replacing s by x)

137

FOURIER TRANSFORMS

TEST YOUR KNOWLEDGE 1. Find the Fourier sine and cosine transforms of the function e–3x/2 by using its second derivative. 2. Using a Parseval’s identity, show that :

3.

z z



dx

π 60 0 ( x + 4 )( x + 9) Using a Parseval’s identity, show that :

(i)

(i)

2



0

=

2

dx 2

( x + 1)

=

2

π 4

(ii)

(ii)

z z



dx

−∞

( x 2 + 1)( x 2 + 16)



0

x2 dx ( x2 + 1)2

=

=

π . 80

π . 4

4. Use Parseval’s identity for Fourier sine transform of the function f(x) = show that

z



0

FG 1 − cos x IJ H x K

2

dx =

RS1 , T0 ,

z



sin2 x x2

0

dx =

RS1 , T0 ,

π . 2

Answer 1.

4s 9 + 4s

2

,

6 9 + 4 s2

0 < x ≤1 to x >1

π . 2

5. Use Parseval’s identity for Fourier cosine transform of the function f(x) = to show that

if if

.

Hint 3. (ii) Use Parseval’s identity for the Fourier sine transform of the function e–x, x > 0.

if if

0 < x ≤1 x >1

Solution of Differential Equations Using Fourier Transforms

6

6.1. INTRODUCTION When the function involved in a physical (or geometrical) problem involves more than one independent variable, a partial differential equation is obtained. The solution of this partial differential equation gives the required function. Fourier transforms are a very important tool in solving such partial differential equations. 6.2. PARTIAL DIFFERENTIAL EQUATION An equation containing one or more partial derivatives of an unknown function of more than one independent variable is called a partial differential equation. 2 2 For example, ∂ u = c 2 ∂ u is a partial differential equation. The order of a partial 2 ∂t ∂x 2 differential equation is defined as the order of the highest partial derivative occurring in the partial differential equation. The order of the above partial differential equation is two.

A solution of a partial differential equation is a relation between the variables by means of which and the partial derivatives derived there from the given partial differential equation is satisfied. In the present chapter, we shall confine only to the application of Fourier transforms in solving partial differential equations of the following types : (i) (ii) (iii)

∂u ∂ 2u = c2 2 ∂t ∂x ∂ 2u ∂t 2

∂ 2u ∂x 2

= c2

+

(One dimensional heat equation)

∂ 2u ∂x 2

∂ 2u ∂y 2

(One dimensional wave equation)

=0

(Two dimensional Laplace equation)

Here c is a constant, t is time, x, y are cartesian coordinates and dimension is the number of the coordinates in the equation. on.

The general methods of solving partial differential equations would be taken up later 138

139

SOLUTION OF DIFFERENTIAL EQUATIONS USING FOURIER TRANSFORMS

6.3. METHOD OF SOLVING PARTIAL DIFFERENTIAL EQUATION BY USING FOURIER TRANSFORMS Let the unknown function in the given partial differential equation be u(x, t). Fourier transform or Fourier sine (or cosine) transform is taken w.r.t. x on both side of the given equation. This equation is simplied by using the following formulae : 1. (i) F(f ′(x)) = isF(f(x)) (Assuming f(x) → 0 as | x | → ∞) 2 (ii) F(f ″(x)) = – s F(f(x)) (Assuming f(x), f ′(x) → 0 as | x | → ∞) 2. (i) FS(f ′(x)) = – sFC(f(x)) (Assuming f(x) → 0 as x → ∞) 2 (ii)FS(f ″(x)) = – s FS(f(x)) + sf(0) (Assuming f(x), f ′(x) → 0 as x → ∞) 3. (i) FC(f ′(x)) = sFS(f(x)) – f(0) (Assuming f(x) → 0 as x → ∞) 2 (ii)FC(f ″(x)) = – s FC(f(x)) – f ′(0) (Assuming f(x), f ′(x) → 0 as x → ∞) Formula 2(ii) is used if f(0) is given and formula 3(ii) is used if f ′(0) is given. In applying the above formulae to the function u(x, t), the requirement of u(x, t) → 0 or u(x, t) → 0, ux(x, t) → 0 both as x → ∞ or | x | → ∞ are generally given with the problem. If such requirements are not given then these are assumed to have been given with the problem. Remarks. 1. If u(0, t) is given then generally Fourier sine transform w.r.t. x of the given equation is taken. 2. If ux(0, t) is given then generally Fourier cosine transform w.r.t. x of the given equation is taken. 3. The solution of the equation

dy = ky is y = cekx. dx

4. The solution of the equation

dy + Py = Q is y(I.F.) = dx

z

(∵ D – k = 0 ⇒ D = k) Q (I. F.) dx + c, where I.F. = e

z

P dx

.

ILLUSTRATIVE EXAMPLES Example 1. If the flow of heat is linear so that the variation of θ(temperature) w.r.t. y and z may be neglected and if it is assumed that no heat is generated in the medium, then solve the one dimensional heat equation being a given function of x. Sol. The given equation is

∂θ ∂ 2θ = k 2 , where – ∞ < x < ∞ and θ = f(x) when t = 0, f(x) ∂t ∂x

∂θ ∂ 2θ =k 2 ∂t ∂x The given condition is : θ(x, 0) = f(x), – ∞ < x < ∞. Taking Fourier transform of (1) w.r.t. x, we get

F ⇒ ⇒

z



−∞

∂ ∂t

z

FG ∂θ IJ = FFG k ∂ θ IJ H ∂t K H ∂x K F ∂ θI dx = kFG H ∂x JK

∂θ − isx e ∂t ∞

−∞

...(1)

2

2

2

2

θ e − isx dx = k [– s2F(θ)]

(Using F(f ″(x)) = – s2 F(f(x))

140

LAPLACE AND FOURIER TRANSFORMS

∂ (θ) + ks 2 θ = 0 ∂t



The solution of (2) is

...(2)

θ = ce − ks

∴ ∴

z



−∞

(3)

t

...(3)

θ (s, 0) = ce0 = c.

Putting t = 0 in (3), we get ⇒

2

z

θ( x, 0) e − isx dx = c ⇒

−∞

c = f (s)

θ = f (s) e − ks



2

−1 θ(x, t) = F (θ ) =



=

z

θ(x) =

RSθ , T0 ,



f ( x) e − isx dx = c ⇒

t

1 2π

z



θ e isx ds

−∞

f (s) = c

z

2 2 1 ∞ 1 ∞ f(s) e −(ks t − isx) ds . f ( s ) e − ks t e isx ds = 2π −∞ 2π −∞ Example 2. If the initial temperature of an infinite bar is given by

0

for |x| < a for |x| > a ,

determine the temperature at any point x and at any instant t. Sol. To find the temperature θ(x, t), the required heat-flow equation is ∂θ ∂ 2θ = c2 ,t>0 ∂t ∂x 2 The given condition is

...(1)

RSθ , for |x |< a T 0 , for |x |> a. Taking Fourier transform of (1) w.r.t. x, we get F ∂ θI ∂θ e dx = c FG ∂t H ∂x JK 0

θ(x, 0) =

z



− isx

2

2

2

−∞



∂ ∂t

z





−∞

2 2 θ e − isx dx = c (− s θ)

∂θ + c2s2 θ = 0 ∂t

...(2)

The solution of (2) is θ = Ke − c Putting t = 0 in (3), we get ⇒ ⇒

z

z



−∞

−a

−∞



2 2

s t

...(3)

θ(s, 0) =

Ke0

θ ( x, 0) e − isx dx = K

0 . e − isx dx +

⇒ ⇒

(Using F(f ″(x)) = – s2F(f(x)))

z

a

−a

θ 0 e − isx dx +

z



a

or

0. e − isx dx = K

e − isx 0 + θ0 − is

θ 0 –isa (e – eisa) = K is



θ(s, 0) = K

a

+0=K −a

2θ 0 2θ 0 e ias − e − ias =K ⇒ K= sin as s s 2i

141

SOLUTION OF DIFFERENTIAL EQUATIONS USING FOURIER TRANSFORMS

2θ 0 sin as − c 2 s2 t e s

θ=

∴ (3) ⇒

θ(x, t) = F–1 ( θ ) =



1 2π

=

θ0 π

=

θ0 π

=

2θ 0 π

=

θ0 π

−∞

−∞

z z



0

0



0

–∞

θ e isx ds

iθ sin as − c 2s 2t e cos sx ds + 0 π s







sin as − c2s2t (cos sx + i sin sx ) ds e s



z

z

θ 0 sin as − c 2s 2t isx e e ds s



−∞

θ θ(x, t) = 0 π



z z z

=

1 2π

z



–∞

sin as − c2s 2t sin sx ds e s

sin as − c 2s 2t e cos sx ds + 0 s 2 2

e −c s t ( 2 sin as cos sx ) ds s 2 2

e− c s t (sin (a + x)s + sin (a − x)s) ds . s

Example 3. An infinite string is initially at rest and that the initial displacement is f(x), (– ∞ < x < ∞). Determine the displacement y(x, t) of the string. Sol. The equation for the vibration of string is 2 ∂2 y 2 ∂ y c = ∂t 2 ∂x 2 The given conditions are :

(i)

FG ∂y IJ H ∂t K

...(1)

=0

(∵ String is initially at rest)

t=0

(ii) y(x, 0) = f(x), – ∞ < x < ∞

(∵ Initial displacement is f(x))

Taking Fourier transform of (1) w.r.t. x, we get F ⇒ ⇒

∂2 ∂t

2

∂2 ∂t 2

F ∂ y I = FF c ∂ y I GH ∂t JK GH ∂x JK F ∂ yI (F(y)) = c F G H ∂x JK 2

2

2

2

2

2

2

(F(y)) = c2[– s2 F(y)]

∂2 y + c2s2 y = 0 ∂t 2 The roots of A.E. of (2) are ± csi.



2

(Using F(f ″(x)) = – s2 F(f(x))) ...(2)

142

LAPLACE AND FOURIER TRANSFORMS

∴ The solution of (1) is

We have

z





−∞

FG ∂ H ∂t



∴ (3)



FG ∂y IJ H ∂t K FG ∂y IJ H ∂t K

z

y = c1 cos cst + c2 sin cst

= 0. t=0

t=0



−∞

...(3)

FG H

e − isx dx = 0 ⇒

IJ K

y e − isx dx

z



−∞

IJ K

∂y − isx e dx ∂t

∂ ( y) ∂t

=0 ⇒ t=0

( − c1 cs sin cst + c2 cs cos cst )

=0 t=0

=0 t=0

=0 t=0

– c1 cs sin 0 + c2 cs cos 0 = 0 ⇒ 0 + c2 cs = 0 ⇒ c2 = 0

⇒ ∴

(3)



y = c1 cos cst

...(4)

Putting t = 0 in (4), we get

⇒ ∴ ∴

z

y ( s, 0) = c1 cos ( cs . 0) = c1 ∞

−∞

(4)

z

y( x, 0) e − isx dx = c1 ⇒

−∞

c1 = f (s) ⇒





y = f (s) cos cst

y = F–1 ( y ) = =

1 2π

=

1 2π

z z z



−∞



−∞

1 2π

f ( x) e − isx dx = c1 ⇒

z



−∞

f (s) = c1

y e isx ds

( f ( s ) cos cst ) e isx ds

f ( s)

Fe GH

icst

I JK

+ e − icst isx e ds 2

1 ∞ ( f (s) e is( x + ct) + f (s) e is( x − ct) ) ds 4π −∞ 1 1 ∞ 1 ∞ = f ( s) e is( x + ct) ds + f ( s) e is( x − ct) ds 2 2π −∞ 2π −∞

=

LM N

z

z

OP Q

1 [f(x + ct) + f(x – ct)]. 2 Example 4. Let u(x, t) denote the temperature at distance x and time t. Solve the heat-



y(x, t) =

∂u ∂ 2u = k 2 , x ≥ 0, t ≥ 0 under the boundary condition u = u0 when x = 0, t > 0 and ∂t ∂x the initial condition u = 0 when x > 0, t = 0.

flow equation

Sol. We have

∂u ∂ 2u =k 2 ∂t ∂x

...(1)

SOLUTION OF DIFFERENTIAL EQUATIONS USING FOURIER TRANSFORMS

143

Taking Fourier sine transform of (1) w.r.t. x, we get

z





0

∂ ∂t



F GH

I JK F ∂ uI ∂u sin sx dx = k F G ∂t H ∂x JK FG IJ H K

2 F S ∂u = F k ∂ u S ∂t ∂x 2

2

z

S



0

2

u sin sx dx = k (− s 2 FS (u) + s. u(0, t))

(Using FS(f ″(x)) = – s2FS(f(x)) + sf(0)) ∂ (uS ) = − ks 2 uS + ksu0 ∂t



∂ uS + ks 2 uS = ksu0 ∂t This is a linear differential equation.

...(2)



I.F. = e

Here

∴ The solution of (2) is 2

uS e ks t = ks2 t



uS e



uS e ks t =

z

z

ks2 dt

2

t

2

ksu0 e ks t dt + c

= ksu0

2

= e ks

e ks

2

t

ks 2

+c

u0 ks 2t e +c s

2 u0 + ce − ks t s Putting t = 0 in (3), we get

uS =



(∵ u(0, t) = u0)

...(3)

u0 + ce0 s u uS (s, 0) = 0 + c s ∞ u0 u( x, 0) sin sx dx = +c 0 s π u 0 . sin sx dx = 0 + c 0 s u0 u 0= +c ⇒ c=– 0 s s 2 u0 u0 − ks2 t u0 uS = − e = (1 − e − ks t ) (3) ⇒ s s s ∞ 2 u = FS–1 (uS ) = uS sin sx ds π 0 2 2 ∞ u0 (1 − e − ks t ) sin sx ds u(x, t) = π 0 s uS (s, 0) =

⇒ ⇒ ⇒ ⇒ ∴ ∴ ∴

z

z

z

z

(∵ u(x, 0) = 0)

144

LAPLACE AND FOURIER TRANSFORMS

2u0 = π =

2u0 π

LM MN

z

LM sin sx ds − MN s LM π − 1 − e s MN 2 ∞

0

0

− ks2 t



0

u(x, t) = u0 1 − 2 π



z z

z





0

1 − e − ks s

2

1 − e − ks s

2

sin sx ds

t

t

OP PQ

sin sx ds

OP PQ

OP PQ

FG∵ H

z



0

IJ K

sin sx π ds = if x > 0 2 s

sin sx ds .

Example 5. The temperature u in a semi-infinite rod is given by the heat-flow equation 2 ∂u = c2 ∂ u , 0 ≤ x < ∞ ∂t ∂x 2

subject to the conditions :

(i) u = 0 when t = 0, x ≥ 0 ∂u = – μ (a constant) when x = 0 ∂x ∂u (iii) → 0 as x → ∞. ∂x Making use of the Fourier cosine transform, show that the temperature u(x, t) is given by

(ii)

2μ π

z



1 − e− s s2

0

2 2

c t

cos sx ds .

Sol. We have

∂u ∂ 2u = c2 ∂t ∂x 2

...(1)

Taking Fourier cosine transform of (1) w.r.t. x, we get



z



∂ ∂t



F GH

I JK F ∂ uI ∂u cos sx dx = c F G ∂t H ∂x JK F F ∂u at x = 0IJ IJ u cos sx dx = c G − s F (u) − G KK H ∂x H FG IJ H K

2 FC ∂u = FC c 2 ∂ u ∂t ∂x 2



0

2

2

z



C

2

2

2

0

C

(Using FC(f ″(x)) = – s2 FC(f(x)) – (f ′(0))

∂ (uC ) = − c 2 s 2 uC – c2(– μ) ∂t

∂ (uC ) + s 2 c 2 uC = μc 2 ∂t This is a linear differential equation.

...(2)



Here

I.F. = e

z

s 2 c 2 dt

= es

2 2

c t

145

SOLUTION OF DIFFERENTIAL EQUATIONS USING FOURIER TRANSFORMS

∴ The solution of (2) is

uC e s uC e s



2 2

c t

2 2

c t

=

z

μc 2 e s

= μc 2

es

2 2

c t

dt + K

2 2

c t

s2c2

+K

2 2 μ + Ke − s c t s2 μ Putting t = 0 in (3), we get uC (s, 0) = 2 + Ke0. s μ uC (s, 0) = 2 + K ⇒ s

uC =



z





0

u( x , 0) cos sx dx =

(3)



∴ ∴

2

+K ⇒

z



0 . cos sx dx =

μ

s s2 μ μ 0= 2 +K ⇒ K=– 2 s s 2 2 2 2 μ μ μ uC = 2 − 2 e − s c t = 2 (1 − e − s c t ) s s s ∞ 2 uC cos sx ds u = FC–1 (uC ) = π 0 2 2 2 ∞ μ (1 − e − s c t ) cos sx ds u(s, t) = 2 π 0 s

⇒ ∴

μ

...(3)

0

+K

z

z z

2 2

2μ ∞ 1 − e− s c t cos sx ds . π 0 s2 Example 6. Use Fourier cosine transform to show that the steady temperature in the semi-infinite solid y > 0 when the temperature on the surface y = 0 is kept at unity over the strip | x | < a and at zero outside the strip is =

z

LM N

OP Q

1 a+ x a− x tan −1 + tan −1 . y y π

e − bx sin cx c dx = tan −1 , b > 0, c > 0 may be used. 0 b x Sol. Let u(x, y) denote the steady temperature at the point (x, y), on the solid under consideration.

The result



The steady temperature u(x, y) satisfies the two dimensional Laplace equation ∂ 2u ∂ 2u + = 0, – ∞ < x < ∞, 0 ≤ y < ∞ ∂x 2 ∂y 2

subject to the boundary conditions : (i) u = 1 when y = 0, – a < x < a (ii) u = 0 when y = 0, x ≤ – a or x ≥ a.

...(1)

146

LAPLACE AND FOURIER TRANSFORMS

Taking Fourier cosine transform of (1) w.r.t. x, we get

F ∂ u + ∂ u I = 0. GH ∂x ∂y JK F ∂ uI + F F ∂ uI = 0 F G H ∂x JK GH ∂y JK F ∂u at x = 0IJ OP + ∂ F (u) − G H ∂x K Q ∂y 2

FC

2

2

2

2

⇒ ⇒

C

LM− s N

2

2

C

2

2

2

C

2

– s2 uC – 0 +



FC (u) = 0 (Using FC(f ″(x)) = – s2FC(f(x)) – f ′(0))

FG Assuming ∂u = 0 at x = 0IJ H K ∂x

∂2 (uC ) = 0 ∂y 2

∂2 (uC ) – s2 uC = 0 ∂y 2



uC = c1esy + c2e–sy.

The solution of this equation is

uC being the value of a definite integral is a finite quantity.

∴ c1 = 0 for otherwise uC becomes infinite as y → ∞. ∴

uC = c2 e–sy

...(3)

Putting y = 0 in (3), we get

⇒ ⇒

z

z

uC (x, 0) = c2e0 = c2 ∞

0

a

0

u( x, 0) cos sx dx = c2

1 . cos sx dx + sin sx s

⇒ ∴ (3) ∴



a 0

z



a

0 . cos sx dx = c2

+ 0 = c2 ⇒ c2 = uC =

sin sa − sy e s

u = FC–1 (uC ) = =

2 π

=

1 π

=

1 π

=

1 π

z z z LM Nz



0



0



0

z

...(4) ∞

0

uC cos sx ds

sin sa − sy e cos sx ds s

e − sy (2 sin sa cos sx) ds s e − sy (sin (a + x) s + sin (a − x) s) ds s



0

2 π

sin sa s

e − sy sin (a + x) s ds + s

z



0

e − sy sin (a − x) s ds s

OP Q

147

SOLUTION OF DIFFERENTIAL EQUATIONS USING FOURIER TRANSFORMS

LM N

OP Q

1 a+x a−x tan −1 + tan −1 . π y y

u(x, y) =



F Using GH

z



0

e − bx sin cx c dx = tan −1 x b

I JK

∂ 4V ∂ 2V + = 0, – ∞ < x < ∞, y ≥ 0 subject to the following ∂x 4 ∂y 2

Example 7. Solve the equation conditions :

(i) V and its partial derivatives tend to zero as | x | → ∞ (ii) V = f(x),

∂V = 0 when y = 0. ∂y

Sol. We have

∂4 V ∂x 4

∂2V

+

= 0.

∂y 2

...(1)

The given conditions are : (i) V and its partial derivatives tend to zero as | x | → ∞ (ii) V(x, 0) = f(x), Vy(x, 0) = 0. Taking Fourier transform of (1) w.r.t. x, we get F

F ∂ V I + F F ∂ V I = 0. GH ∂x JK GH ∂y JK 4

2

4



(is)4 F(V) +



s4F(V) +

z

2



−∞

∂2 ∂y 2

z

∂ 2 V − isx e dx = 0 ∂y 2 ∞

−∞

s4F(V) +



V e − isx dx = 0 ∂2 F(V) = 0 ∂y 2

∂2 ( V ) + s4 V = 0 ∂y 2



(Putting F(V) = V )

This is a linear differential equation. The A.E. is D2 + s4 = 0. ∴ D = ± is2 ∴ The solution is

V = c1 cos s2y + c2 sin s2y

Putting y = 0 in (2), we get



z

V (s, 0) = c1 cos 0 + c2 sin 0 = c1 ∞

−∞

V( x, 0) e − isx dx = c1

...(2)

148

LAPLACE AND FOURIER TRANSFORMS

⇒ ∴

z



−∞

(2)

f ( x) e − isx dx = c1 ⇒

f (s) = c1

V = f (s) cos s2y + c2 sin s2y





∂V = – s2 f (s) sin s2y + c2 s2 cos s2y ∂y

Now,

∂V ∂ = ∂ y ∂y



∂V ∂y

= y=0

= ∴

∂V ∂y

z z

FG H

z



−∞

∂V − isx e dx ∂y



−∞



−∞

V e − isx dx =

IJ K

z



−∞

...(3)

∂V − isx e dx ∂y

= y=0

z



−∞

F ∂V GH ∂y

y=0

Ie JK

− isx

dx

0 . e − isx dx = 0

=0 y=0

∴ (3) ⇒ – s2 f (s) sin 0 + c2s2 cos 0 = 0 ⇒ c2s2 = 0 ⇒ c2 = 0 V = f (s) cos s2y

∴ ∴

V(x, y) = F–1 ( V ) = =

1 2π

z



−∞

1 2π

z

(∵ c2 = 0) ∞

−∞

V e isx ds

f(s) cos s 2 y e isx ds .

TEST YOUR KNOWLEDGE 1. Determine the distribution of temperature in the semi-infinite medium x ≥ 0, when the end x = 0 is maintained at zero temperature and the initial distribution of temperature is f(x). 2. Solve the equation

∂u ∂ 2u = 2 2 if : ∂t ∂x

(i) u(0, t) = 0

(ii) u(x, 0) = e–x, x > 0

(iii) u(x, t) is bounded when x > 0, t > 0. 3. Solve the equation

∂u ∂2u , x > 0, t > 0 subject to the conditions : = ∂t ∂x2

(i) u(0, t) = 0 when t > 0

(ii) u(x, 0) =

RS1 , T0 ,

if if

0 < x 1

(iii) u(x, t) is bounded. 2 4. Solve the equation ∂u = ∂ u subject to the conditions : ∂t ∂x2

(i) ux(0, t) = 0 (iii) u(x, t) is bounded when x > 0, t > 0.

(ii) u(x, 0) =

SOLUTION OF DIFFERENTIAL EQUATIONS USING FOURIER TRANSFORMS

2 π 2 2. u(x, t) = π

1. u(x, t) =

2 π 2 4. u(x, t) = π 3. u(x, t) =

z z zF z GH ∞

0 ∞

Answers

fS ( s )e − c

s

2 2

s t

sin sx ds , where u(x, t) represents the temperature 2

2

e−2 s t sin sx ds

0

1+ s



1 − cos s − s2 t e sin sx ds s sin s cos s − 1 − s2 t e cos sx ds . + s s2

0 ∞

0

149

IJ K

Hint

1. If u(x, t) represents the temperature then we are to solve the one dimensional heat-flow equation

∂u ∂ 2u = c2 2 , x > 0, t > 0 ∂t ∂x subject to the conditions : (i) u(0, t) = 0

(ii) u(x, 0) = f(x).

LAPLACE AND FOURIER TRANSFORMS

ISBN 978-93-5274-099-4

9 789352 740994

LAPLACE AND FOURIER TRANSFORMS