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Contents

1 One Dimensional Wave and Heat Equations 51.1 One Dimensional Wave Equation . . . . . . . . . . . . . . . . . 5

1.1.1 Laplace transform formulae . . . . . . . . . . . . . . . . 51.1.2 Solution of Wave Equation by variable separable method 51.1.3 D Alemberts solution of the wave equation . . . . . . . 71.1.4 Fourier Transform Formulae . . . . . . . . . . . . . . . . 67

2 Elliptic Equations 932.1 Laplace Equation . . . . . . . . . . . . . . . . . . . . . . . . . . 93

3 Calculus of Variations 1073.1 Calculus of Variations . . . . . . . . . . . . . . . . . . . . . . . 1073.2 The variational problem: . . . . . . . . . . . . . . . . . . . . . . 107

3.2.1 Euler-Lagrange Equation . . . . . . . . . . . . . . . . . 1073.2.2 Other forms of Euler Equation . . . . . . . . . . . . . . 1103.3 Variational Problems involving several unknown functions . . . 1383.4 Functionals involving Higher Order Derivatives . . . . . . . . . 141

3.4.1 Euler Equation . . . . . . . . . . . . . . . . . . . . . . . 1413.5 The Raleigh Ritz method . . . . . . . . . . . . . . . . . . . . . 1473.6 Variational Problems involving several independent variables . 1543.7 Constraints and Lagrange multipliers . . . . . . . . . . . . . . . 155

3.7.1 Working Rule . . . . . . . . . . . . . . . . . . . . . . . . 1553.8 Variational problems with moving boundaries . . . . . . . . . . 159

4 Eigen Value Problems 171

4.1 Faddeev-Leverrier Method for Eigenvalues . . . . . . . . . . . . 1714.1.1 Faddeev-Leverrier Method . . . . . . . . . . . . . . . . . 1714.2 Power Method with deflation . . . . . . . . . . . . . . . . . . . 1804.3 Rayleigh-Ritz method . . . . . . . . . . . . . . . . . . . . . . . 187

5 Numerical Integration 1895.1 Introduc tion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1895.2 Gaussian Quadrature . . . . . . . . . . . . . . . . . . . . . . . . 189

5.2.1 Two Point Gaussian Quadrature Formulae . . . . . . . . 1895.2.2 Higher Point Gaussian Quadrature Formulae . . . . . . 1955.2.3 Gauss-Hermite Quadrature Formulae . . . . . . . . . . . 200

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MA 9203 APPLIED MATHEMATICS(ME STRUCTURAL ENGINEERING)


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MA 9203 APPLIED MATHEMATICS L T P C3 1 0 4

OBJECTIVE:

. To familiarize the students in the field of differential and elliptic equa-tions to solve boundary value problems associated with engineering ap-plications.

. To expose the students to variational formulation and numerical integra-tion techniques and their applications to obtain solutions for buckling,dynamic response, heat and flow problems of one and two dimensionalconditions.

UNIT I ONE DIMENSIONAL WAVE AND HEAT EQUATIONS10+3

Laplace transform methods for one-dimensional wave equation Displace-ments in a long string longitudinal vibration of an elastic bar Fourier trans-form methods for one-dimensional heat conduction problems in infinite andsemi-infinite rods.UNIT - II ELLIPTIC EQUATION 9+3

Laplace equation Properties of harmonic functions Solution of Laplacesequation by means of Fourier transforms in a half plane, in an infinite stripand in a semiinfinite strip Solution of Poisson equation by Fourier transformmethod.UNIT - III CALCULUS OF VARIATIONS 9+3

Concept of variation and its properties Eulers equation Functional depen-dant on first and higher order derivatives Functionals dependant on functionsof several independent variables Variational problems with moving boundariesDirect methods Ritz and Kantorovich methods.

UNIT - IV EIGEN VALUE PROBLEMS 9+3Methods of solutions: Faddeev Leverrier Method, Power Method with

deflation Approximate Methods: Rayleigh Ritz Method

UNIT - V NUMERICAL INTEGRATION 8+3

Gaussian Quadrature One and Two Dimensions Gauss Hermite Quadra-ture Monte Carlo Method Multiple Integration by using mapping function

TOTAL (L:30+T:15) : 45 PERIODS

REFERENCES:


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1. Sankara Rao, K., Introduction to Partial Differential Equations, PrenticeHall of India Pvt. Ltd., New Delhi, 1997.

2. Rajasekaran.S, Numerical Methods in Science and Engineering A Prac-tical Approach, A.H.Wheeler and Company Private Limited, 1986.

3. Gupta, A.S., Calculus of Variations with Applications, Prentice Hall ofIndia Pvt. Ltd., New Delhi, 1997.

4. Andrews, L.C. and Shivamoggi, B.K., Integral Transforms for Engineers,Prentice Hall of India Pvt. Ltd., New Delhi, 2003.


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Chapter 1One Dimensional Wave and Heat

Equations1.1 One Dimensional Wave Equation

1.1.1 Laplace transform formulae

(1) L [u(x, t)] = U(x, s)

(2) Lx u(x, t)

= d

dxU(x, s)

(3) L

2

x2 u(x, t)

= d2

dx2 U(x, s)

(4) L t u(x, t)

= sL[u(x, t)] su(x, 0) = sU(x, s) su(x, 0)

(5) L2

t2 u(x, t)

= s2L[u(x, t)] su(x, 0) ut(x, 0) = s2U(x, s) su(x, 0) ut(x, 0)

(6) L1 [F(s)] = 12i

+ i i

estF(s)ds = sum of residues of estF(s)

1.1.2 Solution of Wave Equation by variable separablemethod

Let the one dimensional wave equation be

2y

t2= a2

2y

x2or ytt = a

2yxx (1.1.1)

Clearly the solution y is in terms of x and t.Assume that the solution of equation (1.1.1) is y(x, t) = X(x)T(t) or simply y= XT.Then

y

x= yx = X

T,y

t= yt = XT

2y

x2

= yxx = XT, and

2y

t2

= ytt = XT.

5


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Therefore, equation (1.1.1) becomes

XT = a2XT

X

X=

T

a2T= k say

X

X= k and

T

a2T= k

X = kX and T = ka2T

2X

x2 kX = 0 and

2T

t2 ka2T = 0

d2X

dx2 kX = 0 and d

2T

dt2 ka2T = 0

D2X kX = 0 and D2T ka2T = 0 (D2 k)X = 0 and (D2 ka2)T = 0

Case i: Suppose that k is positiveThen k = 2 for some .Therefore, (D2 2)X = 0 and (D2 2a2)T = 0Hence the auxiliary equation is

m2 2 = 0 and m2 2a2 = 0 m2 = 2 and m2 = 2a2 m2 = and m = a

Consequently, X = Aex + Bex and T = Ceat + Beat

Therefore, the solution isy(x, t) =

Aex + Bex

Ceat + Beat

Case ii: Suppose that k is negativeThen k = 2 for some .Therefore, (D2 + 2)X = 0 and (D2 + 2a2)T = 0Hence the auxiliary equation is

m2 + 2 = 0 and m2 + 2a2 = 0

m2 = 2 and m2 = 2a2

m2 =

i and m =

ai


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Consequently, X = Acos(x) + Bsin(x) and T = Ccos(at) + Bsin(at)Therefore, the solution isy(x, t) = [Acos(x) + Bsin(x)] [Ccos(at) + Bsin(at)]Case iii: Suppose that k = 0Then,

D2X = 0 and D2T = 0

d2X

dx2= 0 and

d2T

dt2= 0

ddx

dX

dx

= 0 and

d

dt

dT

dt

= 0

dXdx

= A anddT

dt= C

X = Ax + B and T = Ct + D

Therefore, the solution isy(x, t) = (Ax + B)(Ct + D)

1.1.3 D Alemberts solution of the wave equation

Let the one dimensional wave equation be

2y

t2= a2

2y

x2or ytt = a

2yxx (1.1.2)

Let yt = D and yx = D. Then equation (1.1.2) becomes

D2y = a2D2y

D2 a2D2

y = 0

Therefore, the auxiliary equation is

m2 a2 = 0 m2 = a2 m = a

Therefore, the solution is

y(x, t) = f(x + at) + g(x at) (1.1.3)


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where f and g are arbitrary functions.Assume that at time t = 0, initial displacement is y(x, 0) = (x) and initialvelocity is y

t(x, 0) = v(x).

At time t = 0, equation (1.1.3) becomes y(x, 0) = f(x) + g(x)

(x) = f(x) + g(x) (1.1.4)Also differentiating equation partially with r. to t, we get yt(x, t) = af

(x +at) ag(x at)When t = 0, we have yt(x, 0) = af(x) ag(x)

v(x) = af(x) ag(x) (1.1.5)Integrating (1.1.5) between ( c, x ) w.r.t. x, we have

x

c

v(x)dx =x

c

[af(x)

ag(x)] dx

xc

v()d = axc

f(x)dx axc

g(x)dx

xc

v()d = af(x) ag(x)

Hence

f(x) g(x) = 1a

x

c

v()d (1.1.6)

Adding (1.1.4) and (1.1.6), we have

2f(x) = (x) +1

a

xc

v()d

f(x) = (x)2

+1

2a

xc

v()d

f(x + at) = (x + at)2

+1

2a

x + at

cv()d


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Hence

f(x + at) =

(x + at)

2 +

1

2a

x + atc v()d (1.1.7)

Subtracting (1.1.4) and (1.1.6), we have

2g(x) = (x) 1a

xc

v()d

g(x) = (x)2

12a

xc

v()d

g(x at) = (x at)2

12a

x atc

v()d

Hence

g(x at) = (x at)2

12a

x atc

v()d (1.1.8)

Substituting the values of f(x + at) and g(x at) in (1.1.3), we have

y(x, t) =(x + at)

2 +1

2a

x + atc

v()d +(x

at)

2 1

2a

x

at

cv()d

=1

2[(x + at) + (x at)] + 1

2a

x + atc

v()d x at

c

v()d

=1

2[(x + at) + (x at)] + 1

2a

x + atc

v()d +c

x atv()d

=1

2[(x + at) + (x at)] + 1

2a

x + at

x atv()d


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Hence the D Alemberts solution of the one dimensional wave equation is givenby

y(x, t) = 12

[(x + at) + (x at)] + 12a

x + atx at

v()d (1.1.9)

Remark 1.1.1 If the string is at rest at t = 0, then initial velocity = 0.Therefore, v(x) = 0. Hence y(x, t) = 12 [(x + at) + (x at)]Example 1.1.2 A tightly stretched string of length with its fixed ends at x= 0, x = executes transverse vibrations. Motion is started with zero initialvelocity by displacing the string into the form f(x) = k(x2 x3). Find thedeflection at any time t.

SolutionGiven that the ends x = 0, x = of a given tightly stretched string of length

are fixed. Therefore, there is no displacement at its end at any time. Hence,y(0, t) = 0, y(, t) = 0. Also the initial displacement is f(x) = k(x2 x3).Hence y(x, 0) = (x) = f(x) = k(x2 x3). Initial velocity is zero. Therefore,yt(x, 0) = v(x) = 0.

Therefore, the D Alemberts solution of the one dimensional wave equationis given by

y(x, t) =1

2[(x + at) + (x at)]

=1

2

k

(x + at)2 (x + at)3 + k (x at)2 (x at)3=

k

2 x2 + 2xat + (at)2 (x3 + 3x2(at) + 3x(at)2 + (at)3

+

x2 2xat + (at)2 (x3 3x2(at) + 3x(at)2 (at)3

=k

2

x2 + 2xat + (at)2 x3 3x2(at) 3x(at)2 (at)3

+x2 2xat + (at)2 x3 + 3x2(at) 3x(at)2 + (at)3

=k

2

2x2 + 2(at)2 2x3 6x(at)2

=2k

2

x2 + (at)2 x3 3x(at)2

= k

x2 + (at)2 x3 3x(at)2


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Example 1.1.3 Using laplace transform method, solve the initial value prob-lem ytt = yxx, 0 < x < 1 and t > 0 with the boundary conditions y(0, t) = 0,y(1, t) = 0, y(x, 0) = sin x and yt(x, 0) =

sin x.

SolutionLet the given equation be

ytt = yxx (1.1.10)

Taking laplace transform in equation (1.1.10), we have

L [ytt] = L [yxx]

s2y(x, s) sy(x, 0) yt(x, 0) = d2 [y(x, s)]

dx2where y(x,s) = L[y(x,t)]

s2y(x, s)

(s)sinx

{sinx

}=

d2 [y(x, s)]

dx2

d2 [y(x, s)]

dx2 s2y(x, s) = sinx (s)sinx

D2 [y(x, s)] s2y(x, s) = (1 s)sinx (D2 s2) [y(x, s)] = (1 s)sinx

Now, the auxiliary equation is

m2 s2 = 0

m2 = s2

m = s

Therefore, the complementary function isC.F = Aesx + Besx

Now, Particular integral

P.I =1

D2 s2 (1 s)sinx

=1

2 s2 (1 s)sinx

=1

2

+ s2

(s

1)sinx


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Therefore, the solution is Hence the solution isy(x, s) = Aesx + Besx + 12+s2 (s 1)sinx.Now, taking laplace transforms to the first two conditions, we have

L [y(0, t)] = L[0] and L [y(1, t)] = L[0]

y(0, s) = 0 and y(1, s) = 0

y(0, s) = 0 Ae0 + Be0 + 12 + s2

(s 1)sin(0) = 0 A + B = 0 A = B

y(1, s) = 0 Aes + Bes + 12 + s2

(s 1)sin() = 0

Aes + Bes = 0 Aes Aes = 0 A(es es) = 0 A = 0 B = 0

Hence the solution is

y(x, s) =1

2 + s2(s 1)sinx

L [y(x, t)] =1

2 + s2 (s 1)sinx

y(x, t) = L1

1

2 + s2(s 1)sinx

y(x, t) = sinx

L1

s

2 + s2

L1

1

2 + s2

y(x, t) = sinx

L1

s

2 + s2

1

L1

2 + s2

y(x, t) = sinx

cost

1

sint


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Example 1.1.4 A string is stretched and fixed between two points (0,0) and(, 0). Motion is started by displacing the string in the form ksin

x

andreleased from rest at time t = 0. Find the displacement of any point of thestring at any time t using laplace transform.

SolutionLet the one dimensional wave equation be

ytt = a2yxx (1.1.11)

Given that the ends x = 0, x = of a given tightly stretched string of length are fixed. Therefore, there is no displacement at its end at any time. Hence,y(0, t) = 0, y(, t) = 0. Also the initial displacement is y(x, 0) = ksin

x

.

Also the string is released from rest at time t = 0. Therefore, Initial velocityis zero. Therefore, yt(x, 0) = 0.


L [ytt] = L

a2yxx

s2y(x, s) sy(x, 0) yt(x, 0) = a2 d2 [y(x, s)]

dx2where y(x,s) = L[y(x,t)]

s2y(x, s) (s)ksinx

= a2

d2 [y(x, s)]

dx2

a2 d2 [y(x, s)]

dx2 s2y(x, s) = (s)ksin

x

a2D2y(x, s) s2y(x, s) = (s)ksin

x

D2y(x, s)

s2

a2y(x, s) =

ks

a2sinx

D2 s2

a2

y(x, s) = ks

a2sin

x


m2 s2

a2= 0

m2 = s2

a2

m =

s

a


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Therefore, the complementary function isC.F = Ae

s

ax + Be

s

ax


P.I =1

D2 s2a2

ksa2

sinx

=ksa2

2 s2a2

sin

x

=ksa2

2+ s

2

a2

sin

x


y(x, s) = Aes

ax + Be

s

ax +

ks

a2

( )2+ s

2

a2 sinx .

Now, taking laplace transforms to the first two conditions, we have

L [y(0, t)] = L[0] and L [y(, t)] = L[0]

y(0, s) = 0 and y(, s) = 0

y(0, s) = 0 Ae0 + Be0 +ksa2

2+ s

2

a2

sin(0) = 0

A + B = 0 A = B

y(, s) = 0 Ae sa + Be sa +ksa2

2+ s

2

a2

sin

= 0

Ae sa + Be sa +ksa2

2+ s

2

a2

sin() = 0

Ae sa + Be sa = 0 Ae sa Ae sa = 0 A(e sa e sa ) = 0 A = 0

B = 0


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y(x, s) =ksa2

2+ s

2

a2

sin

x

L [y(x, t)] =ksa2

1a2

a

2+ s2

sinx

y(x, t) = L1

ksa2

a2

a

2+ s2

sin

x

y(x, t) = L1

ks

a

2+ s2

sin

x

y(x, t) = k sin

x

L1

s

a

2+ s2

y(x, t) = k sinx

cos

at

Example 1.1.5 Solve using laplace transform method uxx =1c2 utt cost,

0 < x < , t > 0 u(x, 0) = ut(x, 0) = u(0, t) = 0 and u is bounded as x ..

SolutionThe given equation is

uxx =1

c2utt cost uxx = 1

c2

utt c2cost

c2uxx = utt c2cost utt = c2uxx + c2cost

Therefore,utt = c

2uxx + c2cost (1.1.12)

Also the initial and boundary conditions are given by

(i) u(x, 0) = 0


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(ii) ut(x, 0) = 0

(iii) u(0, t) = 0

(iv) u is bounded as x

Taking laplace transform in equation (??), we have

L [utt] = L

c2uxx + c2cost

L [utt] = L

c2uxx+ L

c2cost

s2U(x, s) su(x, 0) ut(x, 0) = c2 d2 [U(x, s)]dx2

+ c2 ss2 + 2

where U(x,s) = L[u(x,t)]

s2U(x, s) = c2 d2 [U(x, s)]

dx2+ c2

s

s2 + 2{by using the initial conditions (i) and (ii)}

c2 d2 [U(x, s)]

dx2 s2U(x, s) = c2 s

s2 + 2

d2 [U(x, s)]

dx2 s

2

c2U(x, s) =

ss2 + 2

D2 s2

c2

U(x, s) =

ss2 + 2


m2 s2

c2= 0

m2 = s2

c2

m = sc

Therefore, the complementary function is

C.F = Ae

s

cx

+ Bes

cx


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P.I = 1D2 s2c2

ss2 + 2

=

1

s2c2

1 c2

s2(D2)

ss2 + 2

=

c2

s2

1 c

2

s2(D2)

1s

s2 + 2

=

c2

s(s2 + 2)

1 +

c2

s2(D2) + ...

(1)

=c2

s(s2 + 2)(1)

=c2

s(s2 + 2)

Hence the solution isU(x, s) = Ae

s

cx + Be

s

cx + c

2

s(s2+2) .

Now,taking laplace transforms to the remaining two conditions, we have

L [u(0, t)] = L[0] and Lu(x, t) is bounded as x u(0, s) = 0 and u(x, s) is bounded as x

u(0, s) = 0 Ae0 + Be0 + c2

s(s2 + 2)= 0

A + B + c2

s(s2 + 2)= 0

A + B = c2

s(s2 + 2)

Also u(x, s) is bounded as x A = 0

B = c2

s(s2+2) .


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U(x, s) =

c2s(s2 + 2)

e

s

cx +

c2

s(s2 + 2)

L [U(x, t)] = c2

s(s2 + 2)

e

s

cx +

c2

s(s2 + 2)

U(x, t) = L1 c2

s(s2 + 2)

e

s

cx +

c2

s(s2 + 2)

U(x, t) =

c2L1

1

s(s2

+ 2

) e

s

cx + c2L1

1

s(s2

+ 2

)

Now,

L1

1

s(s2 + 2)

=

1

2L1

2

s(s2 + 2)

=1

2L1

s2 + 2 s2

s(s2 + 2)

=1

2

L1

s2 + 2

s(s2 + 2)

L1

s2

s(s2 + 2)

=1

2

L1

1

s

L1

s

s2 + 2

=1

2

{1

cost

}


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Also we know that

L1

eas

F(s)

= L1

[F(s)]tta H(t a) where H(t a)= 1 if t > a

= 0 if t < a

L1

es

cx

1

s(s2 + 2)

= L1

1

s(s2 + 2)

tt s

c

H

t sc

L1

es

cx

1

s(s2 + 2)

=

1

2{1 cost}tt s

c

H

t sc

L1

es

cx

1

s(s2 + 2)

=

1

2

1 cos

t s

c

H(t a)

L1

es

cx

1

s(s2 + 2)

= 1

2

1 cos t sc if t > a

= 0 if t < a

Therefore, the solution isU(x, t) = c2

12

1 cos t sc if t > a

0 if t < a

+ c2 12 {1 cost}

Example 1.1.6 Using laplace transform method, solve the initial value prob-lem utt = uxx, 0 < x < and t > 0 with the boundary conditions ux(0, t) =0, u(x, 0) = ex, ut(x, 0) = 0 and u(x, t) 0 as x .


utt = uxx (1.1.13)


L [utt] = L [uxx]

s2U(x, s) su(x, 0) ut(x, 0) = d2 [U(x, s)]

dx2where U(x,s) = L[u(x,t)]

s2U(x, s) (s)ex = d2 [U(x, s)]

dx2

d2 [U(x, s)]

dx2 s2U(x, s) = sex

D2 [U(x, s)] s2U(x, s) = sex

(D2

s2) [U(x, s)] =

sex


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m2

s2 = 0

m2 = s2 m = s

Therefore, the complementary function isC.F = Aesx + Besx


P.I =1

D2 s2 (s)ex

=1

(1)2 s2 (s)ex

= 11 s2 (s)e

x

=sex

s2 1

Therefore, the solution is U(x, s) = Aesx + Besx + sex

s21 .Now,taking laplace transforms to the remaining two conditions, we have

L [ux(0, t)] = L [0] and L [u(x, t)] 0 as x Ux(0, s) = 0 and U(x, s) 0 as x

Now, Ux(x, s) = Asesx + B(

s)esx + se

x

s2

1

Ux(0, s) = 0 Ase0 Bse0 + se0

s2 1 = 0

As Bs ss2 1 = 0

A B 1s2 1 = 0

A B = 1s2 1

Also U(x, s) 0 as x A = 0

B

= 1

s21 .


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U(x, s) =1

s2 1(

s)esx +sexs2 1

U(x, s) =s

s2 1 esx +

sexs2 1

L [u(x, t)] = ss2 1 e

sx +sexs2 1

u(x, t) = L1

s

s2 1 esx +

sexs2 1

u(x, t) = L1

s

s2 1 esx

exL1

s

s2 1

u(x, t) = L1 s

s2

1esx

excosht

Also we know that

L1

easF(s)

= L1 [F(s)]tta H(t a) where H(t a)= 1 if t > a= 0 if t < a

L1

esxs

s2 1

= L1

s

s2 1ttx

H(t x)

= [cosht]ttx H(t x)

= cosh(t x) 1 if t > x0 if t < x= cosh(t x) if t > x

0 if t < x


u(x, t) =cosh(t x) if t > x

0 if t < x excosht

u(x, t) = cosh(t x) excosht if t > x

excosht if t < xExample 1.1.7 Solve using laplace transform method uxx =

1c2 utt, 0 < x 0 u(x, 0) = u

t(x, 0) = u(0, t) = f(t) and u(x, t)

0 as x

. .


24/203

22


uxx = 1c2 utt

c2uxx = uttTherefore,

utt = c2uxx (1.1.14)


(i) u(x, 0) = 0

(ii) ut(x, 0) = 0

(iii) u(0, t) = f(t)

(iv) u(x, t) 0 as x Taking laplace transform in equation (??), we have

L [utt] = L

c2uxx

L [utt] = c2L [uxx]

s2U(x, s) su(x, 0) ut(x, 0) = c2 d2 [U(x, s)]


s2U(x, s) = c2 d2 [U(x, s)]

dx2{by using the initial conditions (i) and (ii)}

c2 d2 [U(x, s)]

dx2 s2U(x, s) = 0

d2 [U(x, s)]dx2

s2c2

U(x, s) = 0

D2 s2

c2

U(x, s) = 0


m2 s2

c2= 0

m2 = s2

c2

m = sc


25/203

23

Therefore, the solution isU(x, s) = Ae

s

cx + Be

s

cx


L [u(0, t)] = L [f(t)] and L [u(x, t)] 0 as x u(0, s) = F(s) and u(x, s) 0 as x

u(0, s) = F(s) Ae0 + Be0 = F(s) A + B = F(s)

Also u(x, s) 0 as x A = 0 B = F(s).Therefore, the solution is

U(x, s) = F(s)es

cx

L [u(x, t)] = F(s)e scx u(x, t) = L1 F(s)e scx

Also we know that

L1

easF(s)

= L1 [F(s)]tta H(t a) where H(t a)= 1 if t > a= 0 if t < a

L1 e scxF(s) = L1 [F(s)]tt sc

H

t sc


U(x, t) = L1 [F(s)]tt sc

H

t sc

= [f(t)]tt s

c

H

t sc

= f

t s

c

H

t sc

Example 1.1.8 Using laplace transform, solve a2yxx = ytt , 0 < x < ,t > 0subject to y = 0 at x = 0, t0 and Eyx = p at x = , y = yt(x, 0) = 0, 0 < x < .


26/203

24


ytt = a2yxx (1.1.15)


(i) y(x, 0) = 0 , 0 < x <

(ii) yt(x, 0) = 0 , 0 < x <

(iii) y(0, t) = 0, t > 0

(iv) yx(, t) =pE


L [ytt] = L

a2yxx

s2Y(x, s) sy(x, 0) yt(x, 0) = a2 d2 [Y(x, s)]

dx2where Y(x,s) = L[y(x,t)]

s2Y(x, s) = a2 d2 [Y(x, s)]

dx2

a2 d2 [Y(x, s)]

dx2 s2Y(x, s) = 0

a2D2Y(x, s) s2Y(x, s) = 0

D2Y(x, s) s2

a2Y(x, s) = 0

D2 s2a2

Y(x, s) = 0


m2 s2

a2= 0

m2 = s2

a2

m = sa


27/203

25

Therefore, the solution isY(x, s) = Ae

s

ax + Be

s

ax

Now, taking laplace transforms to the remaining two conditions, we have

L [y(0, t)] = L[0] and L [yx(, t)] = L p

E

Y(0, s) = 0 and Yx(, s) = p

Es

y(0, s) = 0 Ae0 + Be0 = 0 A + B = 0 A = B

Now, Yx(x, s) = Asa

es

ax + B

sa

e

s

ax

yx(, s) =p

Es A

sa

es

a + B

sa

e

s

a =

p

Es

sa

Ae

s

a Be sa = p

Es

Ae sa Be sa = paEs2

Ae sa + Ae sa = paEs2

A e sa + e sa = paEs2

A2cosh s

a

=pa

Es2

A =paEs2

2coshsa

A = pa2Es2cosh

sa

B = pa2Es2cosh

sa


28/203

26


Y(x, s) =pa

2Es2coshsae sax + pa

2Es2coshsa e sax

pa2Es2cosh

sa e sax e sax

pa2Es2cosh

sa 2sinh s

ax

y(x, t) = L1

pa

2Es2coshsa 2sinh s

ax

y(x, t) =pa

EL1 sinh sax

s2coshsa

Let F(s) =sinh( sax)s2cosh( sa )

Then the poles of F(s) are given by

s2cosh s

a

= 0

s = 0 and cosh sa = 0 s = 0 and cos

is

a

= 0

s = 0 and isa

=(2n + 1)

2, n is any integer

s = 0 and s = (2n + 1)a2i

s = 0 and s = (2n + 1)ia2i2

s = 0 and s = (2n + 1)ia2


29/203

27

Clearly s = 0 is a pole of order 2 and s = (2n+1)ic2 are the poles of orderone. Now,

Res

estF(s)

s=0

=Lim

s 01

1!

d

ds

(s 0)2estF(s)

=Lim

s 0d

ds

s2

est

sinhsax

s2coshsa

=Lim

s 0d

ds

est

sinhsa

x

coshsa

=Lim

s 0cosh

sa

estxa

cosh

sa

x

+ testsinhsa

x estsinh s

ax

a

sinh

sa

cosh

sa2

=cosh(0)

e0xa

cosh(0) + te0sinh(0)

e0sinh(0) a

sinh(0)

[cosh(0)]2

=(1)

(1)

xa

(1) + t(1)(0)

[(1)(0)] a

(0)

12

=x

a

Now, take estF(s) =estsinh( sax)s2[cosh( sa )]

= G(s)H(s) where G(s) = e

stsinhsax

and H(s)

= s2coshsa

Res

estF(s)

(2n+1)ia

2

=Lim

s (2n+1)ia2G(s)

H(s)


30/203

28

=Lim

s (2n+1)ia2estsinh

sax

(s2)

a

sinh

sa

+ (2s)cosh

sa

=e

(2n+1)iat

2 sinh(2n+1)ix2

(2n+1)ia

2

2 a

sinh

(2n+1)i

2

+2

(2n+1)ia

2

cosh

(2n+1)i

2

=e(2n+1)iat

2

1i

sin

(2n+1)i2x

2

1i

(2n+1)i2a22

42

a

sin

(2n+1)i2

2

+2

(2n+1)ia

2

cos

(2n+1)i2

2

=e(2n+1)iat

2 sin(2n+1)x

2 (2n+1)a24

sin

(2n+1)

2

+ 2

(2n+1)ia2

cos

(2n+1)

2

=e(2n+1)iat

2 sin(2n+1)x

2

(2n+1)a2

4

(1)n + 2

(2n+1)ia

2

(0)

=e(2n+1)iat

2 sin(2n+1)x

2

(2n+1)a2

4

(1)n

=e(

(2n+1)iat2 )sin

(2n+1)x

2

(1)n

(2n+1)2a24

(1)n(1)n=

4e((2n+1)iat2 )sin(2n+1)x

2

(1)n

(2n + 1)2a2(1)2n

=4e((2n+1)iat2 )sin

(2n+1)x

2

(1)n

(2n + 1)2a2

Now,

Res

estF(s)

(2n+1)ia

2

=Lim

s (2n+1)ia2G(s)

H(s)


31/203

29

=Lim

s (2n+1)ia2estsinh

sax

(s2)

a

sinh

sa

+ (2s)cosh

sa

=e

(2n+1)iat

2 sinh (2n+1)ix

2

(2n+1)ia

2

2 a

sinh

(2n+1)i

2

+2

(2n+1)ia

2

cosh

(2n+1)i

2

=e(2n+1)iat

2

1i

sin

(2n+1)i2x

2

1i

(2n+1)i2a22

42

a

sin

(2n+1)i2

2

+2

(2n+1)ia

2

cos

(2n+1)i2

2

=e(2n+1)iat

2 sin(2n+1)x

2 (2n+1)a24

sin

(2n+1)2

+ 2

(2n+1)ia

2

cos

(2n+1)2

=e (2n+1)iat2 sin

(2n+1)x

2

(2n+1)a2

4

sin

(2n+1)

2

+ 2

(2n+1)ia

2

cos

(2n+1)

2

=e (2n+1)iat2 sin

(2n+1)x

2

(2n+1)a2

4

(1)n + 2

(2n+1)ia

2

(0)

=e (2n+1)iat2 sin

(2n+1)x

2

(2n+1)a2

4 (1)n

=e( (2n+1)iat2 )sin

(2n+1)x

2

(1)n

(2n+1)2a2

4

(1)n(1)n

=4e( (2n+1)iat2 )sin

(2n+1)x

2

(1)n

(2n + 1)2a2(1)2n

=4e( (2n+1)iat2 )sin

(2n+1)x

2

(1)n

(2n + 1)2a2


32/203

30

Now, L1 [F(s)] = Sum of the residues of estF(s) at its poles

L1

sinh sax

s2

coshsa

= n = 0

4e((2n+1)iat

2 )sin[ (2n+1)x2 ](1)n

(2n+1)2a2

+4e(

(2n+1)iat2 )sin[ (2n+1)x2 ](1)n

(2n+1)2a2

=4

a2

n = 0

(1)nsin(2n+1)x

2

(2n + 1)2

e(

(2n+1)iat2 ) + e(

(2n+1)iat2 )

=8

a2

n = 0

(1)ncosh(2n+1)iat

2

sin

(2n+1)x

2

(2n + 1)2


u(x, t) =pa

EL1

sinh

sax

s2

coshsa

=pa

E

8

a2

n = 0

(1)ncosh(2n+1)iat

2

sin

(2n+1)x

2

(2n + 1)2

=

8p

E2

n = 0

(1)ncosh (2n+1)iat

2 sin(2n+1)x

2 (2n + 1)2

=8a

2

n = 0

(1)ncosh(2n+1)ict

2

sin

(2n+1)x

2

(2n + 1)2

Example 1.1.9 Prove that the solution of the wave equation uxx =1c2 utt

in the strip |x| ,t > 0 satisfying the boundary conditions u(,t)x = 0 ,t > 0 and the initial condition u(x, 0) = ax,

u(x,0)t = 0 , |x| < is ax

acL1 s2sinh xsc sech s

c.


33/203

31


uxx =1

c2utt

c2uxx = utt

Therefore,

utt = c2uxx (1.1.16)


(i) u(x, 0) = ax

(ii) u(x,0)t

= 0 , |x| <

(iii)u(,t)x = 0 , t > 0

(iv) u(,t)x

= 0 , t > 0


L [utt] = L c2uxx L [utt] = c2L [uxx]



s2U(x, s) sax = c2 d2 [U(x, s)]


c2 d2 [U(x, s)]

dx2 s2U(x, s) = sax

d2 [U(x, s)]

dx2 s

2

c2U(x, s) =

saxc2

D2 s2

c2 U(x, s) =sax

c2


34/203

32


m2

s2

c2 = 0

m2 = s2

c2

m = sc


s

cx + Be

s

cx


P.I =1

D2 s2

c2

saxc2

=1

s2c2

1 c2D2

s2

saxc2

=

axc2

c2s2

1 c2D2s2

(s)=

ax

s2

1 c

2D2

s2

1(s)

=ax

s2

1 +

c2D2

s2+ ...

(s)

=

ax

s2 (s)

=ax

s


s

cx + Be

s

cx + ax

s.


L [ux(, t)] = L [0] and L [ux(, t)] = L [0]

Ux(, s) = 0 and Ux(

, s) = 0


35/203

33

Now, Ux(x, s) = Asc

e(

s

c)x + Bs

c

e(

s

c)x + as

Ux(, s) = 0 A

sc

es + B

sc

es + a

s= 0

As

c

e(

s

c ) Bs

c

e(

s

c) =a

s

s

c

Ae(

s

c) Be( sc)

=a

s

Ae( sc) Be( sc ) = acs2

Ae( sc)e( sc) Be(sc)e( sc ) = esac

s2

Ae2( sc ) B = e( sc)ac

s2

(1.1.17)

Ux(, s) = 0 As

c

e(

s

c ) + B

sc

e(

s

c ) +a

s= 0

As

c e( sc ) B

s

c e( sc ) = a

s

sc

Ae(

s

c) Be(sc)

=a

s

Ae( sc ) Be( sc ) = acs2

Ae( sc )e( sc) Be( sc )e( sc ) = e( sc )ac

s2

Ae2(

s

c )

B = e(

s

c )ac

s2 (1.1.18)


36/203

34

Subtracting equations (5.2.16) and (5.2.12), we get,

Ae2(s

c )

Ae2(

s

c ) = e(s

c )acs2 e

( sc )acs2

A

e2(s

c) e2( sc )

=

acs2

e(

s

c) e( sc)

A

2sinh

2s

c

=

acs2

2sinh

s

c

A =ac

s2

2sinh

sc

2sinh

2sc

A =

acs2

sinh

sc

2sinh

sc

cosh

sc

A = acs

2 2cosh

sc

A =

ac2s2

sech

s

c

From equation (5.2.16), we have

Ae2(s

c ) B = e( sc )ac

s2

B = Ae2( sc) e( sc )ac

s2

B = ac2s2 sechsc e2(s

c )

e( sc )acs2

B =ac

s2

e(

s

c )

1

2

sech

s

c

e(

s

c) 1

B =ac

s2

e(

s

c )

1

2

2

e(s

c ) + e(s

c )e(

s

c) 1

B =ac

s2

e(

s

c )

e(

s

c)

e(s

c ) + e(s

c ) 1

B =ac

s2

e(

s

c )

e(

s

c ) e(sc ) e(sc )e(

s

c ) + e(s

c )


37/203

35

B =ac

s2

e(

s

c)

e(sc )

e(s

c ) + e(s

c )

B = acs2

1e(

s

c ) + e(s

c )

B =

ac2s2

2e(

s

c ) + e(s

c )

B = ac

2s2

sech

s

c


U(x, s) =ac

2s2

sech

sc

escx +

ac

2s2

sech

sc

e

scx + ax

s

=

ac2s2

sech

s

c

es

cx e scx + ax

s

=

ac2s2

sech

s

c

2sinh

sc

x

+ax

s

=ax

sac

s2

sech

s

c

sinh

sc

x

L [u(x, t)] = axs

ac

s2

sech

s

c

sinh

sc

x

u(x, t) = L1

ax

sac

s2

sech

s

c

sinh

sc

x

u(x, t) = (ax)L1

1

s

(ac)L1

sinh

sc

x

s2

coshsc

u(x, t) = (ax) (ac)L1

sinhscx

s2 cosh sc


38/203

36

Let F(s) =sinh( scx)

s2[cosh( sc )]Then the poles of estF(s) are given by

s2coshs

c

= 0

s = 0 and coshs

c

= 0

s = 0 and cos

is

c

= 0

s = 0 and isc

= (2n + 1)2

s = 0 and s = (2n + 1)c2i

s = 0 and s = (2n + 1)ic2i2

s = 0 and s = (2n + 1)ic2

s = 0 and s = (2n + 1)ic2


Res

estF(s)

s=0

=Lim

s 01

1!

d

ds

(s 0)2estF(s)

=Lim

s 0d

ds

s2

est

sinhsa

x

s2coshsc

=Lim

s 0d

ds

est

sinhsax

cosh sc


39/203

37

=Lim

s 0cosh

sc

estxc

cosh

scx

+ testsinhscx estsinh scx c sinh sc

cosh

sc

2

= cosh(0)

e0 x

c

cosh(0) + te0

sinh(0) e0sinh(0) c sinh(0)

[cosh(0)]2

=(1)

(1)

xc

(1) + t(1)(0)

[(1)(0)] c (0)12

=x

c

Now, take estF(s) =estsinh( scx)s2[cosh( sc )]

=G(s)H(s) where G(s) = e

stsinhsc

x

and H(s)

= s2coshsc

Res

estF(s)

(2n+1)ic

2

=Lim

s (2n+1)ic2G(s)

H(s)


40/203

38

=Lim

s (2n+1)ic2estsinh

scx

(s2)

c

sinh

sc

+ (2s)cosh

sc

= e

(2n+1)ict

2

sinh(2n+1)ix2

(2n+1)ic

2

2 c

sinh

(2n+1)i

2

+2

(2n+1)ic

2

cosh

(2n+1)i

2

=e(2n+1)ict

2

1i

sin

(2n+1)i2x

2

1i

(2n+1)i2c22

42

c

sin

(2n+1)i2

2

+2

(2n+1)ic

2

cos

(2n+1)i2

2

=e(2n+1)ict

2 sin

(2n+1)x

2 (2n+1)c24

sin

(2n+1)

2

+ 2(2n+1)ic

2

cos(2n+1)

2

=e(2n+1)ict

2 sin(2n+1)x

2

(2n+1)c2

4

(1)n + 2

(2n+1)ic

2

(0)

=e(2n+1)ict

2 sin(2n+1)x

2

(2n+1)c2

4

(1)n

=e(

(2n+1)ict2 )sin

(2n+1)x

2

(1)n

(2n+1)2c24 (1)n(

1)n

=4e((2n+1)ict2 )sin

(2n+1)x

2

(1)n

(2n + 1)2c2(1)2n

=4e((2n+1)ict2 )sin

(2n+1)x

2

(1)n

(2n + 1)2c2

Now,

Res

estF(s)

(2n+1)ic

2

=Lim

s (2n+1)ic2G(s)

H(s)


41/203

39

=Lim

s (2n+1)ic2estsinh

scx

(s2)

c

sinh

sc

+ (2s)cosh

sc

= e

(2n+1)ict

2

sinh (2n+1)ix

2

(2n+1)ic

2

2 c

sinh

(2n+1)i

2

+2

(2n+1)ic

2

cosh

(2n+1)i

2

=e(2n+1)ict

2

1i

sin

(2n+1)i2x

2

1i

(2n+1)i2c22

42

c

sin

(2n+1)i2

2

+2

(2n+1)ic

2

cos

(2n+1)i2

2

=e(2n+1)ict

2 sin

(2n+1)x

2 (2n+1)c24 sin(2n+1)2 + 2 (2n+1)ic2 cos(2n+1)2 =

e (2n+1)ict2 sin(2n+1)x

2

(2n+1)c2

4

sin

(2n+1)

2

+ 2

(2n+1)ic

2

cos

(2n+1)

2

=e (2n+1)ict2 sin

(2n+1)x

2

(2n+1)c2

4

(1)n + 2

(2n+1)ic

2

(0)

=e (2n+1)ict2 sin

(2n+1)x

2

(2n+1)c24 (1)n

=e( (2n+1)ict2 )sin

(2n+1)x

2

(1)n

(2n+1)2c2

4

(1)n(1)n

=4e( (2n+1)ict2 )sin

(2n+1)x

2

(1)n

(2n + 1)2c2(1)2n

=4e( (2n+1)ict2 )sin

(2n+1)x

2

(1)n

(2n + 1)2c2


42/203

40


L1 sinh scxs2 cosh sc =x

c +

n = 0

4e(

(2n+1)ict2 )sin[ (2n+1)x2 ](1)n

(2n+1)2

c2

+4e(

(2n+1)ict2 )sin[ (2n+1)x2 ](1)n

(2n+1)2c2

=x

c 4

c2

n = 0

(1)nsin(2n+1)x

2

(2n + 1)2

e(

(2n+1)ict2 ) + e(

(2n+1)ict2 )

=x

c 8

c2

n = 0

(1)ncosh(2n+1)ict

2

sin

(2n+1)x

2

(2n + 1)2


u(x, t) = (ax) (ac)L1

sinhscx

s2

coshsc

= ax (ac)xc 8c2

n = 0

(1)ncosh(2n+1)ict

2

sin

(2n+1)x

2

(2n + 1)2

= ax ax + 8a2

n = 0(1)ncosh

(2n+1)ict

2

sin

(2n+1)x

2

(2n + 1)2

=8a

2

n = 0

(1)ncosh(2n+1)ict

2

sin

(2n+1)x

2

(2n + 1)2

Example 1.1.10 An infinitely long string having one end at x = 0 is initiallyat rest on the x-axis and the end x = 0 undergoes a periodic transverse dis-placement described by A0sint, t > 0. Find the displacement of any point onthe string at any time t.

SolutionThe transverse displacement of the string is described by the partial differential


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41

equation

uxx =1

c2utt

c2uxx = uttTherefore,

utt = c2uxx (1.1.19)


(i) u(x, 0) = 0

(ii) ut(x, 0) = 0

(iii) u(0, t) = A0sint, t > 0

(iv) u is bounded as x

Taking laplace transform in equation (??), we haveL [utt] = L

c2uxx

L [utt] = c2L [uxx]



s2U(x, s) = c2 d2 [U(x, s)]


c2 d2 [U(x, s)]

dx2 s2U(x, s) = 0

d2 [U(x, s)]

dx2 s2

c2 U(x, s) = 0

D2 s2

c2

U(x, s) = 0


m2 s2

c2= 0

m2 = s2

c2

m = sc


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42


s

cx + Be

s

cx.


L [u(x, t)] = bounded and L [u(0, t)] = L [A0sint]

U(x, s) = bounded and U(0, s) = A0 s2 + 2

Since U(x, s) is bounded as x , we have A = 0Therefore, U(x, s) = Be

s

cx

U(0, s) = A0

s2 + 2 Be0 = A0

s2 + 2

B = A0 s2 + 2


U(x, s) = A0

s2 + 2e

s

cx

L [u(x, t)] = A0 s2 + 2

es

cx

u(x, t) = L1

A0

s2 + 2e

s

cx

u(x, t) = A0L1

s2 + 2e

s

cx

Also we know that

L1

eas

F(s)

= L1

[F(s)]tta H(t a) where H(t a)= 1 if t > a= 0 if t < a

L1

s2 + 2e

s

cx

= L1

s2 + 2

tt x

c

H

t xc

= [sint]tt xc

H

t xc

= sin

t x

c

1 if t > xc0 if t < x

c

= sin

t xc

if t > xc0 if t < xc


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u(x, t) =sin

t xc

if t > xc

0 if t < xc

Example 1.1.11 If the function u(x,t) satisfies the following PDE uxx =1c2 utt + k, 0 < x < , t > 0 subject to the boundary conditions u(0, t) =0 and ux(, t) = 0, t > 0 and the initial conditions u(x, 0) = ut(x, 0) = 0,0 < x < . .


uxx =

1

c2 utt + k uxx =1

c2

utt + c2

k

c2uxx = utt + c2k utt = c2uxx c2k

Therefore,

utt = c2uxx c2k (1.1.20)


(i) u(x, 0) = 0, 0 < x <

(ii) ut(x, 0) = 0, 0 < x <

(iii) u(0, t) = 0

(iv) ux(, t) = 0, t > 0


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L [utt] = L

c2uxx c2k L [utt] = L

c2uxx

c2kL[1] s2U(x, s) su(x, 0) ut(x, 0) = c2 d

2 [U(x, s)]

dx2 c2k

1

s


s2U(x, s) = c2 d2 [U(x, s)]

dx2 c

2k

s{by using the initial conditions (i) and (ii)}

c2 d2 [U(x, s)]

dx2 s2U(x, s) = c

2k

s

d2 [U(x, s)]

dx2 s

2

c2U(x, s) =

k

s

D2 s2c2

U(x, s) = ks


m2 s2

c2= 0

m2 = s2

c2

m = s

c


s

cx + Be

s

cx


P.I =1

D2 s2c2 k

s


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45

P.I =1

D2 s2c2

k

s

=

1s2c2

1 c2D2s2

ks

=

c2s2

1 c2D2s2

ks

=kc2

s3

1 c

2D2

s2

1(1)

=kc2

s3

1 +

c2D2

s2+ ...

(1)

=kc2

s3


s

cx + Be

s

cx kc2

s3.


L [ux(, t)] = L [0] and L [u(0, t)] = L [0]

Ux(, s) = 0 and U(0, s) = 0

Now, Ux(x, s) = A sc

e

( sc)x

+ Bsc e(

s

c)x

U(0, s) = 0 A + B kc2

s3= 0

A + B = kc2

s3

A + B =

kc2

s

3(1.1.21)


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46

Ux(, s) = 0 As

c

e(

s

c) + B

sc

e(

s

c ) = 0

As

c e( sc) B

s

c e( sc ) = 0

s

c

Ae(

s

c ) Be( sc)

= 0

Ae( sc) Be( sc) = 0 Ae( sc)e( sc) Be( sc)e( sc) = 0 Ae( 2sc ) B = 0

Ae(2sc ) B = 0 (1.1.22)

adding equations (1.1.21) and (1.1.22), we get,

Ae2(s

c ) + A =kc2

s3

Ae(sc)

e(s

c) + e2(s

c )

=kc2

s3

A = kc2

s3e(s

c )

e(s

c ) + e2(s

c)

A = kc2e(

s

c )

2s3coshsc

From equation (1.1.22), we have

Ae2(s

c) B = 0 B = Ae2( sc )

B =

kc2e(s

c )

2s3coshsc

e2(

s

c )

B =

kc2e(s

c)

2s3coshsc


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47


U(x, s) = Aes

cx + Be

s

cx

kc2

s3

=kc2e(

s

c )

2s3coshsc

es

cx +

kc2e(s

c)

2s3coshsc

es

cx kc

2

s3

=kc2

2s3coshsc

e(

(x)sc ) + e(

(x)sc )

kc

2

s3

=kc2

2s3coshsc

2cosh

( x)s

c

kc

2

s3

=kc2

s3cosh

sc

cosh

( x)s

c

kc

2

s3

L [u(x, t)] = kc2

s3coshsc

cosh

( x)s

c

kc

2

s3

u(x, t) = L1

kc2

s3coshsc

cosh

( x)s

c

kc

2

s3

u(x, t) = kc2

L1

cosh

(x)s

c

s3cosh sc

L1

1

s3

u(x, t) = kc2L1

cosh

(x)s

c

s3coshsc

1

2L1

2!

s3

u(x, t) = kc2L1

cosh

(x)s

c

s3coshsc

1

2t2

Let F(s) =cosh( (x)sc )s3cosh(sc)

Then estF(s) =estcosh( (x)sc )

s3cosh(sc)


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Then the poles of estF(s) are given by

s3cosh

sc

= 0

s = 0 and coshs

c

= 0

s = 0 and cos

is

c

= 0

s = 0 and isc

= (2n + 1)2

s = 0 and s = (2n + 1)c2i

s = 0 and s = (2n + 1)ic2i2

s = 0 and s = (2n + 1)ic2

s = 0 and s = (2n + 1)ic2


Res

estF(s)

s=0

=Lim

s 01

2!

d2

ds2

(s 0)3estF(s)

=1

2

Lim

s 0d2

ds2

s3

est cosh

(x)s

c

s3coshsc

=1

2

Lim

s 0d2

ds2

estcosh

(x)s

c

cosh

sc


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=1

2

Lim

s 0d

ds

coshsc

est

(x)c sinh

(x)s

c

+ testcosh

(x)s

c

estcosh

(x)s

c

c

sinh

sc

cosh sc

2

=1

2

Lim

s 0

cosh

sc

2

coshsc

test (x)c sinh

(x)s

c

+ est

xc

2 cos

(x)s

c

+ t2estcosh

(x)s

c

+ t

xc

estsinh

(x)s

c

+c

sinh

sc

est

(x)c sinh

(x)s

c

+ testcosh

(x)s

c

testcosh

(x)s

c

c

sinh

sc

est xc sinh (x)sc c sinh sc estcosh (x)sc c2 cosh sc

coshsc

est

(x)c sinh

(x)s

c

+ testcosh

(x)s

c

estcosh

(x)s

c

c

sinh

sc

2c

cosh

sc

sinh

sc

coshsc

4

=1

2

[cosh(0)]2

cosh(0)

te0(x)

c sinh(0) + e0xc

2cos(0) + t2e0cosh(0) + t

xc

e0sinh(0)

+c

sinh(0)

e0

(x)c sinh(0) + te

0cosh(0)

te0cosh(0) c sinh(0) e0 xc sinh(0) c sinh(0) e0cosh(0) c2 cosh(0)

cosh(0)

e0(x)

csinh(0) + te0cosh(0)

e0cosh(0)

c

sinh(0)

2c

cosh(0)sinh(0)

[cosh(0)]4

=1

2

0 +

x

c

2+ t2 + 0 + 0 0 0

c

2 {0 + t 0} 2(0)

=1

2

2 + x2 2x + c2t2 2

c2

=x2 2x + c2t2

2c2


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Now, take estF(s) =estcosh( (x)sc )

s3cosh( sc )= G(s)

H(s) where G(s) = estcosh

(x)s

c

and H(s) = s3cosh sc

Res

estF(s)

(2n+1)ic

2

=Lim

s (2n+1)ic2G(s)

H(s)

=Lim

s (2n+1)ic

2

estcosh

(x)s

c

3s2

cosh sc

+ s3

c

sinh sc

=

e[(2n+1)ic

2 ]tcosh

(x)[(2n+1)ic2 ]

c

3(2n+1)ic

2

2cosh

[(2n+1)ic2 ]

c

+

(2n+1)ic

2

3 c

sinh

[(2n+1)ic2 ]

c

=e[

(2n+1)ic2 ]tcos

(x)(2n+1)i2

2

3(2n+1)ic

2

2cos

(2n+1)i2

2

+(2n+1)ic

2

3 ic

sin

(2n+1)i2

2

=e[

(2n+1)ict2 ]cos

(x)(2n+1)

2

3(2n+1)2c2242 cos (2n+1)2 + (2n+1)3ic3383 ic sin (2n+1)2

=e[

(2n+1)ict2 ]cos

(x)(2n+1)

2

(2n+1)3ic23

8i2

sin

(2n+1)

2

=82

cos

(2n+1)ct

2

isin

(2n+1)ct

2

cos

(x)(2n+1)

2

[(2n + 1)3c23] (1)n

=82(1)n

cos

(2n+1)ct

2

isin

(2n+1)ct

2

cos

(x)(2n+1)

2

[(2n + 1)3c23]


53/203

51Res

estF(s)

(2n+1)ic

2

=Lim

s (2n+1)ic2G(s)

H(s)

= Lims (2n+1)ic2

estcosh (x)sc

3s2cosh

sc

+ s3

c

sinh

sc

=

e[(2n+1)ic

2 ]tcosh

(x)[ (2n+1)ic2 ]

c

3(2n+1)ic

2

2cosh

[ (2n+1)ic2 ]

c

+

(2n+1)ic

2

3 c

sinh

[ (2n+1)ic2 ]

c

=e[

(2n+1)ic2 ]tcos

(x)(2n+1)i2

2

3(2n+1)ic

2

2cos

(2n+1)i2

2

+(2n+1)ic

2

3 ic

sin

(2n+1)i2

2

=

e[(2n+1)ict

2 ]cos(x)(2n+1)2 3(2n+1)2c22

42

cos

(2n+1)

2

+(2n+1)3ic33

83

ic

sin

(2n+1)

2

=e[

(2n+1)ict2 ]cos

(x)(2n+1)

2

(2n+1)3ic23

8i2

sin

(2n+1)

2

=82

cos

(2n+1)ct

2

+ isin

(2n+1)ct

2

cos

(x)(2n+1)

2

[(2n + 1)3c23] (1)n

=82(1)n

cos

(2n+1)ct

2

+ isin

(2n+1)ct

2

cos

(x)(2n+1)

2

[(2n + 1)3c23]


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Now,L1 [F(s)] = sum of residues of estF(s) at its polesHence,

L1cosh (x)sc

s3coshsc

= sum of residues of est cosh(

x)s

c

s3cosh

sc

=

x22x+c2t22c2

+

n = 0

82(1)n{cos( (2n+1)ct2 )+isin( (2n+1)ct2 )}cos( (x)(2n+1)2 )

[(2n+1)3c23]

+

n = 0

82(1)n{cos( (2n+1)ct2 )isin( (2n+1)ct2 )}cos( (x)(2n+1)2 )

[(2n+1)3c23]

=x2

2x + c2t2

2c2 +

n = 0

162(

1)ncos (2n+1)ct2 cos (x)(2n+1)2

[(2n + 1)3c23]

The solution is

u(x, t) = kc2

L1

cosh

(x)s

c

s3coshsc

1

2t2

= kc2

x2 2x + c2t22c2

+

n = 0

162(1)ncos(2n+1)ct

2

cos

(x)(2n+1)

2

[(2n + 1)3c23]

t2

2

= kc2

x

2 2x2c2

+

n = 0

162(1)ncos(2n+1)ct

2

cos

(x)(2n+1)

2

[(2n + 1)3c23]

=kx(x 2)

2+

8k2

3

n = 0

(1)ncos(2n+1)ct

2

cos

(x)(2n+1)

2

(2n + 1)3

Example 1.1.12 The end x = 0 of an elastic bar is free while a constantlongitudinal force F0 per unit area acts longitudinally at length x = . Thebar is initially at rest and it is unstrained. Find the displacement u(x, t) in thebar.


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53

SolutionThe initial and boundary conditions are given by

(i) u(x, 0) = 0

(ii) ut(x, 0) = 0

(iii) ux(0, t) = 0

(iv) Eux(, t) = F0.

The transverse displacement of the string is described by the partial differentialequation

uxx = 1c2

utt

c2uxx = utt

Therefore,

utt = c2uxx (1.1.23)


L [utt] = L

c2uxx

L [utt] = c2L [uxx]



s2U(x, s) = c2 d2 [U(x, s)]


c2 d2 [U(x, s)]

dx2 s2U(x, s) = 0

d2 [U(x, s)]

dx2 s

2

c2U(x, s) = 0

D2 s2

c2

U(x, s) = 0


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54


m2 s2

c2= 0

m2 = s2

c2

m = sc


s

cx + Be

s

cx.


L [ux(0, t)] = L[0] and L [ux(, t)] = L

F0

E

Ux(0, s) = 0 and Ux(, s) = F0Es

Now, Ux(x, s) = A sc

es

cx + B

sc

es

cx

Ux(0, s) = 0 As

c

e0 + B

sc

e0 = 0

As

c

B

sc

= 0

s

c

(A B) = 0

A B = 0 A = B.


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55

Ux(, s) =F0

Es A

sc

es

c + B

sc

e

s

c =

F0

Es

s

cAesc Be

sc = F0

Es

Ae sc Be sc = F0cEs2

Ae sc Ae sc = F0cEs2

A e sc e sc = F0cEs2

A2sinhs

c

=F0c

Es2

A =F0cEs2

2sinh sc

A = F0c2Es2sinh

sc

B = F0c2Es2sinh

sc


U(x, s) =F0c

2Es2sinhsc e scx + F0c

2Es2sinhsc e scx

F0c2Es2sinh

sc e scx + e scx

pa2Es2sinh

sa 2cosh s

ax

u(x, t) = L1

F0c

2Es2sinhsc

2coshs

cx

u(x, t) = F0cE

L1

coshscx

s2sinh sc


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56

Let F(s) =cosh( scx)s2sinh( sc )

Then the poles of F(s) are given by

s2coshs

c

= 0

s = 0 and sinhs

c

= 0

s = 0 and 1i

sin

is

c

= 0

s = 0 and sin isc = 0 s = 0 and is

c= n, n is any integer

s = 0 and s = nci

s = 0 and s = nici2

s = 0 and s = nic

s = 0 and s = nic

, n is any integer


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Clearly s = 0 is a pole of order 2 and s = nic are poles of order 1 where n isany integer.

estcosh

scx

s2sinhsc =

1 + st1! +

(st)2

2! + ...

1 +( scx)

2

2! +( scx)

4

4! + ...

s2

( sc )1! +

( sc )3

3! +( scx)

5

5! + ...

=

1 + st + s

2t2

2 +s3t3

6 ...

1 + s2x2

2c2 + ...

s3c

1 + s

22

6c2 + ...

=c

s3

1 +

s2x2

2c2+ st +

s3x2t

2c2+

s2t2

2+

s3t3

6+ ...

1 +

s22

6c2+ ...

1

= c

s3 +x2

2cs +ct

s2 +x2t

2c +ct2

2s +ct3

6 + ...

1 s22

6c2 + ...

=c

s3+

x2

2cs+

ct

s2+

x2t

2c+

ct2

2s+

ct3

6

6cs sx

2

12c3 t

6c s

2x2t

6c3 st

2

12c s

2t3

36c ...

Res

estF(s)

s=0

= coefficient of1

sin the expansion of Laurent series of estF(s)

=

x2

2c +

ct2

2

6c

Now, take estF(s) =estcosh( scx)s2[sinh( sc )]

=G(s)H(s) where G(s) = e

stcoshscx

and H(s)

= s2sinhsc

Res

estF(s)

nic

=Lim

s nicG(s)

H(s)


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=Lim

s nicestcosh

scx

(s2)c

cosh

sc

+ (2s)sinhsc

=

enict

cosh nix

nic

2 c

cosh (ni)

+2nic

sinh (ni)

=enict

cosni2x

n2i2c22

2

c

cos

ni2+2

nic

1i sin

ni2

=enict

cosnx

n2c22

2

c

cos (n)

+2 nc sin (n)

=enict

cosnx

n2c2

(1)n + 2 nc

(0)

=enict

cos(nx

n2c2

(1)n

=e(

nict

)cosnx

(1)nn2c2

(1)n(1)n

=e(nict )cos nx (1)n

n2c2(1)2n

=e(

nict

)cosnx

(1)n

n2c2


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Now,

=Lim

s nicestcosh

scx

(s2)c

cosh

sc

+ (2s)sinhsc

=

enict

coshnix

nic

2 c

cosh (ni)

+2nic

sinh (ni)

=enict

cosni2x

n2i2c22

2 c cos ni

2+2

nic 1

i sin

ni2

=enict

cosnx

n2c22

2

c

cos (n)

+2nc

sin (n)

=enict

cosnx

n2c22

2

c

cos (n)

2 nc sin (n)

=enict

cos

nx

n2c2 (1)n

2 nc (0)

=enict

cos(nx

n2c2

(1)n

=e(

nict

)cosnx

(1)nn2c2

(1)n(1)n


n2c2(1)2n


n2c2


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L1 cosh scx

s2

sinhsc

=

x2

2c +

ct2

2

6c +

n = 0

e(

nict )cos[nx ](1)n

n2c2

+e(

nict )cos[nx ](1)n

n2c2

=x2

2c+

ct2

2

6c

c2

n = 0

(1)ncos nx

n2

e(

nict

) + e(nict

)

=x2

2c+

ct2

2

6c 2

c2

n = 0

(1)ncos nct cos nx n2


u(x, t) =F0c

EL1

cosh

scx

s2

sinhsc

=F0c

E

x

2

2c+

ct2

2

6c 2

c2

n = 0

(1)ncos nct cos nx n2

Example 1.1.13 A semi infinite string is stretched along the positive half ofa horizontal x-axis with its end x = 0 ties at the origin and with its end x = 0ties at the origin and with its distant end looped around a vertical support that

exerts no vertical force on the loop. The string is initially supported at restalong the x-axis. At time t = 0, the support is removed and the string movesunder the action of gravity. Obtain the displacement u(x, t) at any position x,at any time t.

SolutionWe know that the wave equation for a long string under its weight is

utt = c2uxx g (1.1.24)

where g is the gravitational force.Also the initial and boundary conditions are given by


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(i) u(x, 0) = 0

(ii) ut(x, 0) = 0

(iii) u(0, t) = 0

(iv) ux 0 as x


L [utt] = L

c2uxx g

L [utt] = L c2uxx L[g]

s2U(x, s) su(x, 0) ut(x, 0) = c2 d2 [U(x, s)]dx2

gs


s2U(x, s) = c2 d2 [U(x, s)]

dx2 g

s{by using the initial conditions (i) and (ii)}

c2 d2 [U(x, s)]

dx2 s2U(x, s) = g

s

d2 [U(x, s)]

dx2 s

2

c2U(x, s) =

g

sc2

D2 s2

c2

U(x, s) =

g

sc2


m2 s2

c2= 0

m2 = s2

c2

m = sc


s

cx + Be

s

cx


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P.I =1

D2 s2

c2

g

sc2

=1

D2 s2c2 g

sc2e0x

=1 s2c2

gsc2

e0x

=c2s2

g

sc2

=gs3


s

cx + Be

s

cx

g

s3 .

Now,taking laplace transforms to the remaining condition u(0, t) = 0, we have

L [u(0, t)] = L[0]

u(0, s) = 0

u(0, s) = 0 Ae0 + Be0 gs3

= 0

A + B gs3

= 0

A + B = gs3

Also, Ux(x, s) = Asc

e scx + B

sc

e scx

Now, by (iv), ux 0 as x Hence A = 0. B = g

s3.


U(x, s) = g

s3

e

s

cx g

s3

U(x, s) = g

s3

e

s

cx 1

L [U(x, t)] =

g

s3 e

s

cx 1


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U(x, t) = L1 g

s3

e

s

cx 1

= L1 ge

s

cx

s

3

g

s

3 = gL1

e

s

cx

s3

g

2!L1

2!

s3

= gu xc

(t)L1

1

s3

g

2!L1

2!

s3

= gu xc

(t)1

2!L1

2!

s3

ttx

c

g2!

t2

= g

0 if t < xc1 if t > xc

t2

2 ttxc

gt2

2

= g 0 if t < xc

1 if t > xc t

xc 2

2 gt2

2

= g

0 if t < xc

(txc )2

2 if t >xc

gt

2

2

Example 1.1.14 A tightly stretched string has ends fixed at x = 0 and x =. At time t = 0, the string is given a shape defined by y = x( x) and thenreleased. Find the displacement at any time t.

SolutionLet the one dimensional wave equation be

ytt = a2yxx (1.1.25)


(i) y(0, t) = 0, t > 0

(ii) y(, t) = 0

(iii) yt(x, 0) = 0 , 0 < x <

(iv) y(x, 0) = x( x) , 0 < x <


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L [ytt] = L

a2yxx

s2Y(x, s) sy(x, 0) yt(x, 0) = a2 d2 [Y(x, s)]

dx2where Y(x,s) = L[y(x,t)]

s2Y(x, s) sx( x) = a2 d2 [Y(x, s)]

dx2

a2 d2 [Y(x, s)]

dx2 s2Y(x, s) = sx( x)

a2D2Y(x, s) s2Y(x, s) = sx( x)

D2Y(x, s)

s2

a2

Y(x, s) =

sx(

x)

D2 s2

a2

Y(x, s) =

sxa2

( x)


m2 s2a2

= 0

m2 = s2

a2

m = sa


s

ax + Be

s

ax


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P.I = 1D2 s2a2

sxa2

( x)

=1

s2a2

1 D2a2s2

sxa2

( x)

=a2

s2

1 D

2a2

s2

1 sa2

(x x2)

=a2

s2

1 +

D2a2

s2+ ...

sa2

(x x2)

=a2

s2 s

a2(x x2) + a

2

s2s

a2(2)

=

s(x x2) 2a

2

s3


Y(x, s) = Aes

ax + Be

s

ax + s (x x2) 2a

2

s3


L [y(0, t)] = L[0] and L [y(, t)] = 0 Y(0, s) = 0 and Y(, s) = 0

y(0, s) = 0 Ae0 + Be0 2a2

s3= 0

A + B = 2a2

s3


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y(, s) = Aes

a + Be

s

a +

s(2 2) 2a

2

s3

Ae

s

a + Be

s

a

2a2

s3

= 0

Ae sa + Be sa = 2a2

s3

Ae 2sa + B = 2a2e

s

a

s3

Ae 2sa A = 2a2e

s

a

s3 2a

2

s3

Ae sa Aesa = 2a2

s3 2a

2es

a

s3

A es

a esa =

2a2

s3 1 es

a

2Asinh

sa

= 2a2

s3

1 esa

Asinh

sa

=a2

s3

1 esa

A = a

2

s3sinhsa

1 esa

a2

s3sinhsa 1 esa + B = 2a2

s3

B = 2a2

s3 a

2

s3sinh sa 1 esa


Y(x, s) =a2

s3sinhsa

1 esa e sax + 2a2

s3 a

2

s3sinhsa

1 esa e sax +

s(x x2) 2a

2

s3

L [y(x, t)] = a2

s3sinhsa 1 esa e sax + 2a2

s3 a

2

s3sinhsa 1 esa e sax +

s(x x2) 2a

2

s3

y(x, t) = L1

a2

s3sinh

sa

1 esa

es

ax +

2a2

s3 a

2

s3sinh

sa

1 esa

e

s

ax +

s(x x2) 2a

2

s3


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1.1.4 Fourier Transform Formulae

(1) Fourier Transform of f(x) is F[f(x)] = 1

2

f(x)e

isx d x = F(s).

(2) Inverse Fourier Transform of F[f(x)] is f(x) = 12

F[f(x)]eisx d

s.

(3) Fourier Cosine Transform of f(x) is FC[f(x)] =

2

0

f(x)cossx d x.

(4) Inverse Fourier Cosine Transform ofFC[f(x)] is f(x) = 20

FC[f(x)]cossx

d s.

(5) Fourier Sine Transform of f(x) is FS[f(x)] =

2

0

f(x)sinsx d x.

(6) Inverse Fourier Sine Transform ofFS[f(x)] is f(x) =

2

0

FS[f(x)]sinsx

d s.

(7) FS[f(x)] = sFC[f(x)]

(8) FC[f(x)] = sFS[f(x)] f(0)(9) FS[f

(x)] = s2FS[f(x)] + sf(0)

(10) FC[f(x)] = s2FC[f(x)] f(0)(11) F [u(x, t)] = U(s, t)

(12) Fdnfdxn

= (is)nF(s) if f, f,...,fn1 0 as x where F(s) =

F[f(x)].

(13) Fx u(x, t) = (is)U(s, t)


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(14) F

2

x2 u(x, t)

= (is)2U(s, t)

(15) F t u(x, t)

= Ut(s, t)

(16) FC

x u(x, t)

=

(17) FC

2

x2 u(x, t)

= s2FC[u(x, t)], if ux(0, t) = 0.

(18) FS

x u(x, t)

=

(19) FS 2x2 u(x, t)

= s

2

FS[u(x, t)], if u(0, t) = 0.

Remark 1.1.15 If u at x = 0 is given, then take Fourier sine transform andif u

xat x = 0 is given, then take Fourier cosine transform.

Example 1.1.16 Solve the diffusion equation ut

= K2u

x2, < x < ,

t > 0 with the conditions u(x, 0) = f(x) and ux , u tend to zero as x tend to.


u

t= K

2u

x2, < x < , t > 0 (1.1.26)

Here u = u(x, t). We know that the Fourier transform of u(x, t) = u(s, t) =

12

u(x, t)eisx d x. Hence taking Fourier transform of the given differ-

ential equation, we get KF[uxx] = F[ut]


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K[(is)2u(s, t)] = 12

u

teisxdx

=d

dt

12

u(x, t)eisxdx

=d

dtF[u(x, t)]

=du(s, t)

dt

Ks2u(s, t) = du(s, t)dt

du(s, t)u(s, t)

= Ks2dt

log[u(s, t)] = Ks2t + logA log[u(s, t)] logA = Ks2t

log

u(s, t)

A

= Ks2t

u(s, t)A

= eKs2t

u(s, t) = AeKs2t

u(s, 0) = Ae0 = ANow,

u(x, 0) = f(x) F[u(x, 0)] = F[f(x)] u(s, 0) = F(s) A = F(s)

Hence

u(s, t) = F(s)eKs2t

F[u(x, t) = F(s)eKs2t


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u(x, t) = F1

F(s)eKs2t

= F1 [F(s)] F1 eKs2t

= f(x)

ex24Kt2Kt

=12

f()

e(x)2

4Kt2Kt

d

Take = x2Kt

. Then

(2

Kt) = x

= x 2Kt

Also,

d

d=

12

Kt

d = 2

Ktd

Hence

u(x, t) = 12

f(x 2Kt)e

2

(2Kt)d

=2Kt

f(x 2

Kt)e2

d

This is the required solution.

Example 1.1.17 Solve ut = K2ux2 , t > 0 with the conditions u(x, 0) =

{ 1if |x| < a0if |x| > a .


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u

t = K

2u

x2 , t > 0 (1.1.27)

Here u = u(x, t). We know that the Fourier transform of u(x, t) = u(s, t) =

12

u(x, t)eisx d x. Hence taking Fourier transform of the given differ-

ential equation, we get KF[uxx] = F[ut]

K[(is)2u(s, t)] = 12

u

teisxdx

=d

dt

12

u(x, t)eisxdx

=d

dtF[u(x, t)]

=du(s, t)

dt

Ks2u(s, t) = du(s, t)dt

du(s, t)

u(s, t) = Ks2

dt

log[u(s, t)] = Ks2t + logA log[u(s, t)] logA = Ks2t

log

u(s, t)

A

= Ks2t

u(s, t)A

= eKs2t

u(s, t) = AeKs2t


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u(s, 0) = Ae0 = ANow,

u(x, 0) = { 1if |x| < a0if |x| > a F[u(x, 0)] =

12

u(x, 0)eisxdx

F[u(x, 0)] = 12

aa

eisxdx

u(s, 0) = 12

eisx

is

aa

u(s, 0) = 12

eisa eisa

is

u(s, 0) = 12

2isin(as)

is

u(s, 0) =

2

sin(as)

s

A =

2

sin(as)

s

Hence

u(s, t) =

2

sin(as)

s

eKs

2t

F[u(x, t)] =

2

sin(as)

s

eKs

2t


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u(x, t) = F1

2

sin(as)

s

eKs

2t

= 12

2

sin(as)

s

eKs2teisxds

=1

sin(as)

s

eKs

2t[cos(sx) isin(sx)]ds

=1

sin(as)

s

eKs

2tcos(sx)ds i

sin(as)

s

eKs

2tsin(sx)ds

=2

0

sin(as)

s eKs2tcos(sx)ds

Example 1.1.18 Solve ut

= K2u

x2, x > 0 with the conditions u(x, 0) = ex,

u(0, t) = 0.


u

t= K

2u

x2, x > 0 (1.1.28)

Here u = u(x, t). We know that the Fourier sine transform of u(x, t) = uS(s, t)

=2

0

u(x, t)sin(sx) d x. Hence taking Fourier sine transform of the given

differential equation, we get KFS[uxx] = FS[ut]


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74

K[(is)2uS(s, t)] = 12

0

u

tsin(sx)dx

=d

dt

12

0

u(x, t)sin(sx)dx

=d

dtFS[u(x, t)]

=duS(s, t)

dt

Ks2uS(s, t) = duS(s, t)dt

duS(s, t)

uS(s, t) = Ks2

dt

log[uS(s, t)] = Ks2t + logA log[uS(s, t)] logA = Ks2t

log

uS(s, t)

A

= Ks2t

uS(s, t)A

= eKs2t

uS(s, t) = AeKs2t


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uS(s, 0) = Ae0 = ANow,

u(x, 0) = ex

FS[u(x, 0)] = FS[e

x]

FS[u(x, 0)] =

2

0

u(x, 0)sin(sx)dx

FS[u(x, 0)] =

2

0

exsin(sx)dx

FS[u(x, 0)] =

2

ex

1 + s2[(1)sin(sx) scos(sx)]

0

FS[u(x, 0)] =

2

0 e0

1 + s2[(1)sin(0) scos(0)]

FS[u(x, 0)] = 2

0 1

1 + s2(s)

uS(s, 0) =

2

s

1 + s2

A =

2

s

1 + s2

Hence

uS(s, t) = AeKs2t

F[u(x, t)] =

2

s

1 + s2 eKs2t

u(x, t) = F1S

2

s

1 + s2

eKs

2t

=

2

0

2

s

1 + s2

eKs

2tsin(sx)ds

=2

0

s

1 + s2

eKs

2tsin(sx)ds


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Example 1.1.19 Solve ut =2ux2 , for x > 0, t > 0 with the conditions u(0, t)

= 0 for t > 0, u(x, 0) = { 1if 0 < x < 10if x 1

and u(x, t) is bounded.


u

t=

2u

x2, t > 0 (1.1.29)


=2

0

u(x, t)sin(sx) d x. Hence taking Fourier sine transform of the given

differential equation, we get FS[uxx] = FS[ut]

(s2)uS(s, t) =

2

0

u

tsin(sx)dx

=d

dt

2

0

u(x, t)sin(sx)dx

=d

dtFS[u(x, t)]

=duS(s, t)

dt


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s2uS(s, t) = duS(s, t)dt

duS(s, t)

uS

(s, t)=

s2dt

log[uS(s, t)] = s2t + logA log[uS(s, t)] logA = s2t

log

uS(s, t)

A

= s2t

uS(s, t)A

= es2t

uS(s, t) = Aes2t

uS(s, 0) = Ae0 = A

Now,

u(x, 0) = { 1if 0 < x < 10if x 1

FS[u(x, 0)] =

2

0

u(x, 0)sin(sx)dx

FS[u(x, 0)] =

2

10

sin(sx)dx

uS(s, 0) =

2

cos(sx)s

10

uS(s, 0) = 2

cos(s) + 1s

A =

2

1 cos(s)

s

Hence

uS(s, t) =

2

1 cos(s)

s

es

2t

FS[u(x, t)] =

2

1 cos(s)

s

es

2t


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u(x, t) = F1S

2

1 cos(s)

s

es

2t

= 2

0

2

1 cos(s)

s

es

2

tsin(sx)ds

=2

0

1 cos(s)

s

es

2tsin(sx)ds


= K2u

x2, 0 < x < , t > 0 with the conditions

u(x, 0) = 0, x 0, ux (0, t) = a, is a constant and u(x, t) is bounded.


u

t= K

2u

x2, 0 < x < , t > 0. (1.1.30)

Here u = u(x, t). We know that the Fourier cosine transform of u(x, t) =

uC(s, t) =2

0

u(x, t)cos(sx) d x. Hence taking Fourier cosine transform of

the given differential equation, we get KFC[uxx] = FC[ut]

Ks2uC(s, t)

2

xu(0, t)

=

2

0

u

tcos(sx)dx

=d

dt

2

0

u(x, t)cos(sx)dx

=d

dtFC[u(x, t)]

=duC(s, t)

dt


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Ks2uC(s, t)

2

(a)

=

duC(s, t)

dt

Ks2uC(s, t) + 2 Ka = duC(s, t)dt duC(s, t)

dt+ Ks2uC(s, t) =

2

Ka

This is linear equation in uC(s, t). Here P = Ks2 and Q =

2Ka

.

Hence the integrating factor

I.F = ePdt

= eKs2dt

= eKs2t


uC(s, t)ePdt =

Qe

Pdt

dt + C

uC(s, t)eKs2t =

2

(Ka)eKs

2t

dt + C

uC(s, t)eKs2t =

2

(Ka)

eKs

2t

Ks2

+ C

uC(s, t) =

2

as2

+ CeKs

2t


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uC(s, 0) =

2

as2

+ Ce0 =

2

as2

+ C

Now,

u(x, 0) = 0

FC[u(x, 0)] = 0

uC(s, 0) = 0

2

as2

+ C = 0

C =

2

as2

Hence

uC(s, t) =

2

as2

+ CeKs

2t

FC[u(x, t)] = 2 as2 + 2 as2 eKs2t

FC[u(x, t)] =

2

as2

2

as2

eKs

2t

FC[u(x, t)] =

2

as2

1 eKs2t

u(x, t) = F1C

2

a

s2 1 eKs2t

=

2

0

2

as2

1 eKs2t

cos(sx)ds

=2a

0

1

s2

1 eKs2t

cos(sx)ds

Example 1.1.21 Find the temperature distribution in semi-infinite bar withits end point and the lateral surface insulated and with initial temperaturedistribution in the bar is prescribed by f(x). Deduce the solution when f(x)= eax.


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SolutionThis problem is represented by one dimensional heat equation

u

t= c2

2u

x2, 0 < x < , t > 0. (1.1.31)

with boundary condition ux(0, t) = 0 (insulated end) and with the initial con-

dition u(x, 0) = f(x) (given) for 0 < x < . Here u = u(x, t). We know thatthe Fourier cosine transform of u(x, t) = uC(s, t) =

2

0

u(x, t)cos(sx) d x.

Hence taking Fourier cosine transform of the given differential equation, we getc2FC[uxx] = FC[ut]

c2s2uC(s, t)

2

xu(0, t)

=

2

0

u

tcos(sx)dx

=d

dt

2

0

u(x, t)cos(sx)dx

=d

dtFC[u(x, t)]

=duC(s, t)

dt


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c2s2uC(s, t)

2

(0)

=

duC(s, t)

dt

duC(s, t)

uC(s, t) = c2

s

2

dt

log[uC(s, t)] = c2s2t + logA log[uC(s, t)] logA = c2s2t

log

uC(s, t)

A

= c2s2t

uC(s, t)A

= ec2s2t

uC(s, t) = Aec2s2t

uC(s, 0) = Ae0 = ANow,

u(x, 0) = f(x) FC[u(x, 0)] = FC[f(x)]

FC[u(x, 0)] =

2

0

f(x)cos(sx)dx

uC(s, 0) =

2

0f(x)cos(sx)dx

A =

2

0

f(x)cos(sx)dx

uC(s, t) = Aec2s2t F[u(x, t)] =

2

0

f(x)cos(sx)dx

ec2s2t


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u(x, t) = F1C

2

0

f(x)cos(sx)dx

ec2s2t

= 2

0

2

0

f(x)cos(sx)dx ec2s2tcos(sx)ds

=2

0

0f(x)cos(sx)dx

ec2s2tcos(sx)ds

Suppose f(x) = eax. Then,

u(x, t) =2

0

0eaxcos(sx)dx

ec

2s2tcos(sx)ds

=2

0

eax

a2 + s2[(a)cos(sx) + ssin(sx)]

0

ec

2s2tcos(sx)ds

=2

0

0

e0

a2 + s2[(a)cos(0) + ssin(0)]

ec

2s2tcos(sx)ds

=2

0

1a2 + s2

(a)

ec2s2tcos(sx)ds

=

2

0

aa2 + s2

e

c2s2t

cos(sx)ds

Example 1.1.22 Solve ut = K2ux2 , 0 < x < , t > 0 under the conditions u

= u0 at x = 0, t > 0 with the initial condition u(x, 0) = 0, for x 0.SolutionLet the given equation be

u

t= K

2u

x2, x > 0 (1.1.32)


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84


=2

0u(x, t)sin(sx) d x. Hence taking Fourier sine transform of the given

differential equation, we get KFS[uxx] = FS[ut]

Ks2uS(s, t) +

2

su(0, t)

=

2

0

u

tsin(sx)dx

=d

dt

2

0

u(x, t)sin(sx)dx

=d

dtFS[u(x, t)]

=duS(s, t)

dt

Ks2uS(s, t) +

2

su0

=

duC(s, t)

dt

Ks2uS(s, t) +

2

Ksu0

=

duS(s, t)

dt

duS(s, t)

dt

+ Ks2uS(s, t) = 2

Ksu0

This is linear equation in uS(s, t). Here P = Ks2 and Q =

2 Ksu0.

Hence the integrating factor

I.F = ePdt

= eKs2dt

= eKs2t


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85


uS(s, t)ePdt =

Qe

Pdt

dt + C

uS(s, t)eKs2t =

2

(Ksu0)e

Ks2t

dt + C

uS(s, t)eKs2t =

2

(Ksu0)

eKs

2t

Ks2

+ C

uS(s, t)eKs2t =

2

(u0)

eKs

2t

s

+ C

uS(s, t) =

2

u0s

+ CeKs2t

uS(s, 0) =

2

u0s

+ Ce0

uS(s, 0) =

2

u0s

+ C

Now,

u(x, 0) = 0 FS[u(x, 0)] = 0 uS(s, 0) = 0

2

u0s

+ C = 0

C =

2

u0s


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Hence

uS(s, t) =

2

u0

s + CeKs2t

FS[u(x, t)] =

2

u0s

+

2

u0s

eKs

2t

FS[u(x, t)] =

2

u0s

2

u0s

eKs

2t

FS[u(x, t)] =

2

u0s

1 eKs2t

u(x, t) = F1

S2

u0

s

1 eKs2t

=

2

0

2

u0s

1 eKs2t

sin(sx)ds

=2u0

0

1

s

1 eKs2t

sin(sx)ds


= K2u

x2, 0 < x < , t > 0 under the conditions

u(x, 0) = eax, a > 0, ux(0, t) = 0, ux(x, t) = 0, t 0.


u

t= K

2u

x2, 0 < x < , t > 0. (1.1.33)

under the conditions u(x, 0) = eax, a > 0, ux(0, t) = 0, Ux(x, t) = 0, t 0.Here u = u(x, t). In this problem, the ends of the bar have been insulated andkept at zero temperature. We know that the Fourier cosine transform ofu(x, t)

= uC(s, t) =2

0

u(x, t)cos(sx) d x. Hence taking Fourier cosine transform


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of the given differential equation, we get KFC[uxx] = FC[ut]

Ks2uC(s, t)

2

xu(0, t)

=

2

0

u

tcos(sx)dx

=d

dt

2

0

u(x, t)cos(sx)dx

=d

dtFC[u(x, t)]

=duC(s, t)

dt

Ks2uC(s, t)

2

(0)

=

duC(s, t)

dt

duC(s, t)

uC(s, t)=

Ks2dt

log[uC(s, t)] = Ks2t + logA log[uC(s, t)] logA = Ks2t

log

uC(s, t)

A

= Ks2t

uC(s, t)A

= eKs2t

uC(s, t) = AeKs2t


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uC(s, 0) = Ae0 = ANow,

u(x, 0) = eax

FC[u(x, 0)] = FC[e

ax]

FC[u(x, 0)] =

2

0

eaxcos(sx)dx

uC(s, 0) =

2

eax

a2 + s2[(a)cos(sx) + ssin(sx)]

0

A =

2

0 e

0

a2 + s2[(a)cos(0) + ssin(0)]

A =

2

1

a2 + s2(a)

A =

2

aa2 + s2

uC(s, t) = AeKs2t F[u(x, t)] =

2

a

a2 + s2

eKs

2t

u(x, t) = F1C

2

a

a2 + s2

eKs

2t

= 2

0

2

a

a2 + s2

eKs

2tcos(sx)ds

=2a

0

1

a2 + s2

eKs

2tcos(sx)ds

Example 1.1.24 Determine the temperature distribution in semi-infinite mediumx 0 when the end x = 0 is maintained at zero temperature and the initialtemperature distribution is f(x).


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SolutionThis problem is represented by one dimensional heat equation

u

t= K

2u

x2, 0 < x < , t > 0. (1.1.34)

under the conditions u(x, 0) = f(x) (given) for 0 < x 0, described by the partial differential equation uxx+uyy = 0, < x < , y > 0 with the boundary conditions u(x, 0) = f(x), < x < , u isbounded as y , u and ux both vanishes as |x| .

SolutionLet the given the partial differential equation be

uxx + uyy = 0, < x < , y > 0

Since x has an infinite range of values, we take Fourier transform on both sidesto the given partial differential equation in the variable x. Therefore, we have

F[uxx] + [uyy ] = 0

1

2

uxxe

isxdx +1

2

uyye

isxdx = 0

12

eisxd[ux] +12

2u

y2

eisxdx = 0

12

eisxux

ux

eisx(is)

dx

+

d2

dy2

12

ueisxdx

= 0

93


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12

0 (is)

ux

eisx

dx

+ d2dy

2F[u(x, y)] = 0 1

2(is)

eisx

d(u)

+

d2

dy2

F[u(x, y)] = 0

12

(is)eisxu

u

eisx(is)

dx

+

d2

dy2

F[u(x, y)] = 0

12 (is)[0 0] (is) ueisxdx+

d2

dy2

F[u(x, y)] = 0

(is)2 12

ueisxdx +

d2

dy2

F[u(x, y)] = 0

s2F[u(x, y)] +

d2

dy2

F[u(x, y)] = 0

d2

dy2

U(s, y) s2

U(s, y) = 0

D2 s2U(s, y) = 0where U(s, y) = F[u(x, y)].Therefore, the auxiliary equation is

m2 s2 = 0 m2 = s2 m = s


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Hence the solution isU(s, y) = A(s)esy + B(s)esy .Since u must be bounded as y , its Fourier transform U(s, y) also shouldbe bounded. Hence we have A(s) = 0 for s > 0 and B(s) = 0 for s < 0.

Therefore, U(s, y) = B(s)esy for s > 0 and U(s, y) = A(s)esy for s < 0.Consequently, U(s, y) = Ke|s|y. U(s, 0) = Ke0 = KNow,

u(x, 0) = f(x) F[u(x, 0)] = F[f(x)] U(s, 0)] = F[F(x)] K = F[f(x)]

U(s, y) = F[f(x)]e|s|y

Hence,

U(s, y) =

1

2

f(x)eisxdx

e|s|y

F[u(x, y)] = 12

f(x)e|s|y e

isxdx

u(x, y) = F1 12

f(x)e|s|y

eisxdx

u(x, y) = 12

12

f()e|s|y

eisd

eisxds

u(x, y) = 12

f()d

eis(x)|s|yds


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Now,

eis(x)|s|yds =

0

eis(x)|s|yds +

0

eis(x)|s|yds

=0

eis(x)+syds +

0

eis(x)syds

=0

es[i(x)+y]ds +

0

es[i(x)+y]ds

=

es[i(x)+y]

i( x) + y0

+

es[i(x)+y]

[i(x ) + y]0

=1

i( x) + y 0 + 0

1

[i(x ) + y]=

1

i( x) + y +1

i(x ) + y=

i(x ) + y + i( x) + y[i( x) + y][i(x ) + y]

=ix i+ y + i ix + y

[y + i( x)][y i( x)}]=

2y

y2 + ( x)2

Therefore,

u(x, y) =1

2

f()d

2y

y2 + ( x)2

=y

f()

y2 + ( x)2

d

Example 2.1.2 Solve uxx + uyy = 0, y > 0 subject to the conditions u and

ux 0 as x2 + y2 , u(x, 0) = { 1if|x|

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