3.Applications of Fourier Transform

8/22/2019 3.Applications of Fourier Transform

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Application of transforms to Initial Boundary Value Problems

(IBVP):

The solution of a IBVP consisting of a partial differential equation together with

boundary and initial conditions can be solved by the Fourier Transform method. If the

boundary conditions are of the Dirichlet type where the function value is prescribes on

the boundary, then the Fourier sine transform is used. If the boundary conditions are of

the Neumann type where the derivative of function is prescribed on boundary, then

Fourier cosine transform is applied. In either case, the PDE reduces to an ODE in Fourier

transform, which is solved. Then the inverse Fourier sine (or cosine) transform will give

the solution to the problem.

In one dimensional boundary value problems, the partial differential equation can

easily be transformed into an ordinary differential equation by applying a suitable

transform. The required solution is then obtained by solving this equation and inverting

by means of the complex inversion formula or by any other method. In two dimensional

problems, it is sometimes required to apply the transforms twice and the desired solution

is obtained by double inversion.

3333rdrdrdrd TopicTopicTopicTopic

Fourier TransformFourier TransformFourier TransformFourier Transform

Applications of Fourier transforms to Initial boundaryvalue problems:Heat conductionVibrations of a stringTransmission lines

Prepared by:

Dr. SunilNIT Hamirpur (HP)

(Last updated on 24-10-2007)


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Fourier Transform: Prepared by: Dr. Sunil, NIT Hamirpur (HP) 2

(i). If in a problem u(x, t)x=0 is given then we use infinite sine transform to remove 2

2

x

u

from the differential equation.

In case ( )[ ] 0xx/t,xu = is given then we employ infinite cosine transform to

remove2

2

x

u

.

(ii). If in a problem u(0, t) and )t,(u l are given then we use finite sine transform to

remove2

2

x

u

from the differential equation.

In case

0x

2

2

x

u

=

and

l

=

x

2

2

x

uare given then we employ finite cosine transform

to remove2

2

x

u

.

The method of solution is best explained through the following examples.

Laplace equation

Q.No.1.: Solve the Laplaces equation in the semi-infinite strip shown in Fig.

Sol.: To solve the Laplaces equation 0y

u

x

u2

2

2

2

=

+

, (i)

with the boundary conditions

u(0, y), for 0 < y < b (ii)

u(x, b) = 0 for (iv)

Region is :


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Applying Fourier sin transform of (i) on both sides, we get

0sxdxsiny

usxdxsin

x

u2

2

02

2

0

=

+

. (v)

The first integral is simplified by successive integration by parts and gives

sxdxcos.s.x

usxsin.

dx

dusxdxsin

x

u

002

2

0

=

.

Assuming 0x

u

as x

( )

=

dxsxsin.s.usxcos.ussxdxsinx

u

002

2

0

using the boundary condition (ii), u(0, y) = 0 and assuming that u is bounded, i.e. 0u

as x

sxdxsinussxdxsinx

u

0

2

2

2

0

=

.

Put ( ) ( ){ } ( ) sxdxsiny,xuy,xuFy,sU0

s

==

Thus Ussxdxsinx

u 22

2

0

=

. (vi)

Substituting (vi) in (v) and rewriting

( ) 0sxdxsiny,xudy

dUs

02

22 =

+

0dy

UdUs

2

22 =+ . (vii)

Thus, Laplaces equation (i) has been reduced to an ordinary differential equation in y

(with s as a parameter).

Solution of (vii) is

( ) sysinhBsycoshAy,sU += , (viii)


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when y = b, ( ){ } 0}0{Fb,xuF)b,s(U ss === , since from (iii), u(x, b) = 0. From (viii), at

y = b,

sbsinhBsbcoshA)b,s(U0 +== , (ix)

when y = 0, ( ) ( ) sxdxsin0,xu0,sU0

= .

From (iv), ase)0,x(u = so ( )22

ax

0 sa

ssxdxsine0,sU

+==

. (x)

Putting y = 0 in (viii), we get ( ) 0.BA0,sUsa

s22

+== 22 sa

sA

+= . (xi)

Substituting (xi) in (ix), we get sbcoth

sa

sB

22

+

= . (xii)

Thus, the solution (viii) becomes ( ) sysinhsbcothsa

ssycosh

sa

sy,sU

2222 +

+= . (xiii)

Taking the inverse Fourier sine transform of (xii), we get

( ) ( ){ }y,sUFy,xu 1s= ( ) sxdssinsysinh.sbcothsycosh

sa

s222

0

+

=

,

which is the required solution of this problem.

Q.No.2.: Solve the Laplaces equation 0y

u

x

u2

2

2

2=

+

in square plate of length L with

the following conditions u(0, y) = u(L, y) = 0, u(x, 0) = 0, u(x, L) = f(x). Deduce

when 2x)x(f = and =L .

Sol.: Since the boundary conditions are Dirichlet type (u prescribes), take finite Fourier

sine transform of Laplace equation on both sides

0dx

L

xnsin

y

udx

L

xnsin

x

u2

2L

02

2L

0

=

+

Integrating by parts, we get

( ) 0L

xnsiny,xu

ydx

L

xncos.

x

u

L

n

L

xnnsi.

x

uL

02

2L

00

=

+


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Put ( ) dxL

xnsiny,xu)}y,x(u{FU

L

0

s

==

= U(x, y)

Then 0L

xnsin).y,x(uLn

Lxncos).y,x(u

Ln0

dyUd

L

0

L

02

2

=

++

[ ]0UL

n)y,0(u)1)(y,L(u

L

n

dy

Ud2

22n

2

2

=

Using the boundary condition u(L, y) = u(0, y) = 0.

0Udy

Ud 22

2

= whereL

n= (i)

The solution of 2nd

order ordinary differential equation is

y.sinhby.coshA)y,n(U += (ii)

Using the boundary condition u(x, 0) = 0

0 = U(n, 0) = A . 1 + B . 0 0A = (iii)

Using u(x, L) = f(x).

)}x(f{F)}L,x(u{F)L,n(U ss ==

dx

L

xnsin)x(f

L

0

= (iv)

Substituting (iii), (iv) in (ii)

( ) Lsinh.BL,nU =

( )Lsinh

L,nUB

= (v)

Thus (ii) reduces

ysinh.B)y,n(U = (vi)

Where B and are given by(v) and (i).Taking the inverse finite Fourier sine transform of (vi), we get

( ) { } ( )L

xnsin.y,nU

L

2UFy,xu

1n

1s

==

=

.


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( )

=

= L

ynsin.

L

ynsin.B

L

2y,xu

1n

(vii)

When 2x)x(f = and =L from (iv), we get

dxL

xnsin)x(f)L,n(U

L

0

= nxdxsinx

2

0

=

++

=

032

2nxcos.

n

12nxsin

n

1.x2

n

nxcos.x

( )( )

3

n

3

n2

n

21

n

2

n

1)L,n(U +

= (viii)

Substituting (v), (viii) in (vii), we get

( )

=

=3

2

3

n

1n n

2

n

2)1(nsinh

12y,xu

One-dimensional Heat equation

Q.No.1.: Determine the distribution oftemperature in the semi-infinite medium 0x ,

when the end x = 0 is maintained at zero temperature and the initial distribution

of temperature is f(x).

Sol.: Let U(x, t) be the temperature at any point x and at any time t. We have to solve the

heat-flow equation

2

22

x

uc

t

u

=

(x > 0, t > 0). (i)

Subject to the initial condition u(x 0) = f(x) (ii)

And the boundary condition u(0, t) = 0 (iii)

Taking Fourier sine transform of (i) and denoting ( )[ ]t,xuFs by su , we get

( )[ ]s22s ust,0sucdt

ud = 0uscdt

uds

22s == . (iv)

Also the Fourier sine transform of (ii) is )s(fu ss = at t = 0. (v)

Solving (iv) and using (v), we get tscss22

e)s(fu = .

Hence taking its inverse Fourier sine transform, we obtain


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( ) ( ) xsdssinesf2

t,xu tscs0

22

= .

Q.No.2.: Solve2

2

x

u2

t

u

=

, if u(0, t) = 0 , ( ) ( )0xe0,xu x >= , u(x, t) is bounded where

0x > , t > 0.

Sol.: Given2

2

x

u2

t

u

=

, x > 0, t > 0 (i)

with boundary conditions : u(0, t) = 0, u(x, t) is bounded (ii)

and initial boundary condition ( ) ( )0xe0,xu x >= (iii)

Since u(0, t) is given, we take Fourier sine transform of both sides of (i), so that

pxdxsinx

u2pxdxsintu

2

2

00 =

Integrating by parts, we get

( )

=

pxdxcosp.x

upxsin

x

u2pxdxsint,xu

dt

d

000

pxcosx

up2

dt

ud s

= , if 0

x

u

as x (where ( ) pxdxsint,xu)t,p(u

0

s

= )

Again integrating by parts, we get

( ) ( ) ( )

=

dxpxsinpt,xupxcost,xup2dt

ud

00

s .

( ) ( )

+=

pxdxsint,xupt,0u0p2

0

( )[ ](ii)byxas0t,xu Q

( ) s2 up2t,0pu2 = s

2s up2

dt

ud= .

Integrating, we get dtp2clogu

ud 2

s

s = tp2clogulog

2s =

( ) tp2s2

cet,pu= . (iv)

Taking Fourier sine transform of both sides of (iii), we get


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( ) pxdxsinepxdxsin0,xu x

00

=

( ) ( )2

02

x

sp1

ppxcosppxsin

p1

e0,pu

+=

+=

. (v)

Putting t = 0 in (iv) and using (v), we obtain( )

cp1

p2=

+.

Thus (iv) becomes ( ) tp22s

2

ep1

pt,pu

+= .

Now taking inverse Fourier sine transform, we get

( ) pxdpsinp1

pe2t,xu

2

tp2

0

2

+=

. Ans.

Q.No.3.: Find the temperature distribution in semi-infinite bar with its end point and

lateral surface insulated and with initial temperature distribution in the bar is

prescribed by f(x). Deduce the solution when axe)x(f = .

Sol.: The problem is represented by the one-dimensional heat equation2

22

t

uc

t

u

=

,

for 0

with boundary condition ( ) 0t,0u x = (insulated)

and with initial condition u(x, 0) = f(x) (given) for


9/24


Assuming 0x

u

as x and using the boundary condition 0

x

u

0x

=

=

, the first

term in RHS becomes zero.

Integrating by parts again, we obtain

sxdxcos.ut

0

=

sxdxcos.ussxsin..usc

00

2 .

Assuming that u is bounded i. e. 0u as x the first term in the RHS is zero.

Put ( ) ( ){ } ( ) sxdxcos.t,xut,xuFs,tU0

c

== .

Then UcsU

dt

d 22= .

Note that the result of Fourier transforming on x is to eliminate derivatives of x from the

heat equation, thus leaving an ordinary differential equation w.r.t. t.

Integrating by separation of variables, we get

( ) tsc22

Aet,sU = .

At t = 0, U(s, 0) = A

Thus ( ) ( ) ( )dxsxcos.0,xu0,sUA0

== .

From the initial condition u(x, 0) = f(x).

Then ( ) ( )dxsxcosxfA0

= .

Taking inverse Fourier cosine transform, we get

( ) { } tsc1c1

c

22

AeFUFt,xu == ( )dssxcos.Ae

2 tsc

0

22

= .

Special case: When ( ) axexf = then ( )22

ax

0 ba

adxsxcoseA

+==

.

Thus, the solution in this case is ( )

+= tsc

22

1c

22

eba

aFt,xu


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( ) ( )dssxcos.e.ba

a2t,xu tsc

220

22

+= .

Q.No.4.: Find the temperature distribution in a bar of length L, with its both ends and

lateral surface insulated when the initial temperature in the bar is f(x). Deduce

when 2x)x(f = and L = 10.

Sol.: This problem is to solve the one-dimensional heat equation2

22

x

uc

t

u

=

for , with

boundary conditions ( ) 0)t,L(ut,0u xx == and with initial condition u(x, 0) = f(x). Since

the boundary conditions are of the Neumann type (derivative) apply finite Fourier cosine

transform to both sides of the equation.

dxL

xncosx

ucdxL

xncos.t

u2

2L

0

2L

0

=

(i)

Put ( )[ ]t,xuFU c= and use (ii)

( ) ( ) ( )[ ] )u(FL

nt,0ut,Lu1

x

uF c2

22

xxn

2

2

c

=

The equation (i) reduces to

( ) ( )[ ]

=

)u(FLn

t,0u1cdsL

xn

cos.udt

d

c2

22

s

n2L

0

UL

nc

dt

dU2

222

= (iii)

Since the first term in the RHS is zero because of zero boundary conditions

( ) ( )( )0t,0ut,Lu xx == . Here U = U(n, t).

Solving (iii), we get

( )

=

2

222

L

tcn

Aet,nU (iv)

To determine the arbitrary constant A in (iv) use the initial condition.

Put t = 0 in (iv) this yields

( ) ( )[ ] ( ){ }xfF0,xuF0,nUA cc ===


11/24


dxL

xncos).x(f

L

0

= (v)

2222 L/tcnL

0

edx

L

xncos).x(f)t,n(U

= (vi)

Taking inverse finite Fourier cosine transform

)}t,n(U{F)t,x(u 1c=

+=

= L

xncos)n(F

L

2)0(F

L

1c

1n

c

where dx)x(f)0(F

L

0

c =

and )t,n(U)}t,x(u{F)n(F cc == given by (vi)

Thus ( )2222 L/tcn

L

01n

L

0

eL

xncosdx

L

xncosxf

L

2dx)x(f

L

1)t,x(u

=

+= (vii)

when 2x)x(f = and L = 10.

( )3

1000

3

xdxx0F

10

0

32

10

0

c ===

( ){ } dx10

xncos.x0,xuF)0,n(U

210

0

c

==

10

033

3

22

22

10

xnsin

n

102

10

xncos

n

10.x2

10

xnsin.

n

10.x

+

=

( )n22

1n

2000

= .

Thus( ) ( )

+=

=

10

xncos.e

n

12000.

10

2

3

100)t,x(u 100/tcn

2

n

1n2

222

.

Q.No.5.: If the initial temperature of an infinite bar is given by

( )

>


12/24


2

22

xc

t

=

(t > 0) . (i)

subject to the initial condition ( )

>

0, assuming that the surface of the bar is

insulated.

Sol.: Here we have to solve the differential equation


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2

22

x

uc

t

u

=

. (0 < x < a, t > 0) (i)

Subject to the conditions u(x, 0) = 0 (ii)

ux(0, t) = 0 (iii)

and u(a, t) = u0 . (iv)

The Laplace transform of (i), if ( )[ ] )s,x(ut,xuL = , is

2

22

dx

udc)0,x(uus = .

Using (ii), we get 0uc

s

dx

ud22

2

= . (v)

Similarly, the Laplace transforms of (iii) and (iv) are

( ) 0s,0us = , (vi)

( )s

us,au 0= . (vii)

Solving (v), we get c/sx2c/sx

1 eCeCu+= .

Using (vi), we find 21 CC = so that

( )c/sxcoshC2eeCu 1c/sxc/sx1

+= .

Now using (iii), we get( )c/sxcoshs

c/sxcoshuu 0= .

By the inversion formula (iii), we get

)t,x(u = sum of the residues of( )

( )c/sacoshsc/axcosu.e 0

st

at all poles which occur at s = 0

And ( ) 0c/sacosh = i.e. at s = 0, i2

1nc/sa

= , ,......2,1,0n =

at s = 0, ( )( )

,......2,1,0a4

c1n2ss

2

222

n =

==

Now( )

( ) 0st

0

0s0s u

c/sacoshs

c/axcosheu.sLt)Res( =

=

=


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( )( )

( )

=

=c/sacoshs

c/axcosheussLtu)Res(

st0

nss

00sn

=

s

c/sxcosheLt.

c/sacosh

ssLtu

st

ss

n

ss0

nn

form

0

0

( )( )( )

= s

c/sxcosheLt

c/s2/a.c/sasinh

1Ltu

st

ssss0

nn

( )

( )( ) ( )

a2

x1n2cose

1n2

1u4 2222 a4/tc1n2n

0

=

Thus we get( ) ( ) ( )

a2

x1n2cose

1n2

1u4u)t,x(u

2222a4/tc1n2

n

1n

00

+=

= .

Vibrations of a string

Q.No.1.: An infinite string is initially at and that the initial displacement is f(x),

( )


15/24


dxe.cstcos)s(F2

1)t,x(y ist

= dxe.2

ee)s(F

2

1 isxicsticst

+

=

( ) ( )[ ]dse)s(Fe)s(F4

1 ctxisctxis +

+

= [ ])ctx(f)ctx(f21

++= .

Q.No.2.: An infinitely long string having one end at x = 0, is initially at rest along the x-

axis. The end x = 0 is given a transverse displacement f(t), t > 0. Find the

displacement of any point of the string at any time.

Sol.: Let y(x, t) be the transverse displacement of any point x of the string at any time t.

Then we have to solve the wave equation

2

22

2

2

x

yc

t

y

=

(x > 0, t > 0). (i)

Subject to the conditions y(x, 0) = 0, yt(x, 0) = 0, y(0, t) = f(t) and the displacement

y(x, t) is bounded

The Laplace transform of (i), writing [ ] )s,x(y)t,x(yL = is

( )2

222

dx

ydc

t

0,xy)0,x(syys =

.

Using the first two conditions, we get

yc

s

dx

yd2

2

2

= . (ii)

Similarly the Laplace transform of the 3rd and 4th conditions are

( ) )s(fs,0y = at x = 0 (iii)

and )s,x(y is bounded (iv)

Solving (ii), we get ( ) c/sx2c/sx

1 eCeCs,xy+= .

To satisfy condition (iv), we must have C1 = 0.

Using the condition (iii), we get )s(fC2 = .

c/sxe)s(f)s,x(y = .

Using the complex inversion formula, we get

( ))c/xt(fds)s(fe

i2

1y sc/ct

ia

ia

=

= +

.


16/24


Q.No.3.: A tightly stretched flexible string has its ends fixed at x = 0 and l=x . At time

t = 0, the string is given a shape defined by ( ) ( )xxxF = l , where is a

constant and then released. Find the displacement of any point of the string at

any time t > 0.

Sol.: Then we have to solve the wave equation

2

22

2

2

x

yc

t

y

=

(x > 0, t > 0)

Subject to the conditions y(0, t) = 0, ( ) 0t,y =l

And ( ) ( ),xx0,xy = l ( ) 00,xy t =

Now taking Laplace transform, writing [ ] )s,x(y)t,x(yL = is

( )2

222

dx

ydct

0,xy)0,x(syys =

, (i)

where 0)s,0(y = , ( ) 0s,y =l (ii)

(i) reduces to( )

2

2

2

2

c

xsxy

c

s

dx

yd =

=

l.

Its solution is( )

3

2

21s

c2

s

xx)c/sxsinh(c)c/sxcosh(c)s,x)y

++=

l.

Applying the conditions (ii), we get

221 s/c2c = and

( )

( )( )c2/stanh

s

c2

c/ssinh

c/scosh1

s

c2c

3

2

3

2

2

=l

l

Thus( ){ }

( )

( )3

2

3

2

s

c2

s

xx

c2/scosh

c2/x2scosh

s

c2)s,x(y

+

=

l

l

l

Now using the inverse formula

)t,x(u = sum of the residues of{ }

( )

c2/scoshs

c2/)x2(scoshec2

3

st2

l

lat all poles which occur at

( ) 22 tcxx + l

sum of the residues of{ }

( )

c2/scoshs

c2/)x2(scoshec2

3

st2

l

lat all poles


17/24


+=

22

22

c2c2

x2tc

ll

( )

( )

( ) ( )

= ll

ll ct)1n2(acos

2

x21n2cos

1n2

1

c2

c323

n

1n

2

3

2

( )( )

( ) ( )

+=

= ll

ll

ct1n2cos

x1n2sin

1n2

18xxtc

31n

3

222

Hence ( )( )

( ) ( )

=

= ll

l ct1n2cos

x1n2sin

1n2

18t,xy

31n

3

2

Transmission lines

Q.No.1.: A semi-infinite transmission line of negligible inductance and leakance per

unit length has its voltage and current equal to zero. A constant voltage v0is

applied at the sending end (x = 0). Find the voltage and current at any point(x

> 0) and at any instant.

Sol.: Let v(x, t) and i(x, t) be the voltage and current at any point x and at any time t. If L

= 0 and G = 0, then the transmission line equation becomes

Rix

v=

,

t

vC

x

i

=

i.e.

t

vRC

x

v2

2

=

. (i)

The boundary conditions are v(0, t) = v0 and i(x, t) is finite for all x and t.

The initial conditions are v(x, 0) = 0, i(x, 0) = 0. (ii)

Laplace transform of (i) are

( )0vsRCdx

vd2

2

= 0vRCsdx

vd2

2

== . (iii)

Laplace transforms of the conditions in (ii) are

( )s

vs,0v 0= at x = 0 (iv)

and ( )s,xv remains finite as x . (v)

The solution of (iii) is

( ) RCsx2RCsx

1 eCeCs,xv+= .

To satisfy condition (v), we must have C1 = 0.


18/24


Using the condition (iv), we gets

vC 02 = .

Thus xRCs0 es

v)s,x(v = .

Using the inversion formula, we get

=

=

t2

RCxerfev

s

eLv)t,x(v 0

s.xRC1

0( )

dueu2

RCxv u4/RCx2/3

t

0

0

2

=

sincex

v

R

1i

= , we obtain by differentiation,

( )t4/RCx2/30 2etR

C

2

xv)t,x(i

= .

Q.No.2.: A transmission line of length l has negligible inductance and leakance. A

constant voltage v0 is applied at the sending end (x = 0) and is open circuited at

the far end. Assuming the initial voltage and current to be zero, determine the

voltage and current.

Sol.: For transmission line with L = G = 0, the voltage v and current i are given by the

equations

t

vRC

x

v2

2

=

and 0Ri

x

v=+

. (i)

The boundary conditions are (for t > 0)

0vv = at x = 0 and 0x

vi =

= at l=x . (ii)

The initial condition is v = 0 at t = 0 (x > 0)

Laplace transforms of (i) and (ii) are

( )0vsRCx

v2

2

=

. (iii)

And s/vv 0= at x = 0, 0x

v=

at l=x . (iv)

The solution of (iii) is ( ) ( )xRCssinhcxRCscoshcv 21 +=

Applying conditions (iv), it gives

10 cs/v = , ( ) ( )ll RCscoshcRCssinhc0 21 +=


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( )( )

( )( )

= xRCssinh

RCscosh

RCssinhsRCscosh

s

vv 0

l

l

spcoshs

spqcoshv0= where ( )lRCp = and ( ) ll /xq =

By the inversion formula, we get

V(x, t) = sum of the residues of ( )vest at all poles of vest . (v)

These poles are at s = 0 and ( ) ipk2/1n2isp == .

Now ( ) 00st

0s0s

stv

spcoshs

spqcoshvseLtveRes =

=

=

and

( ) ( )

+

=

=spcoshs

spqcoshveksLtveRes 0

st2

0sks

st2

( )2/1

2st0

ksps

2

1.spsinhsspcosh

(....)ksspqcoshe.vLt

2 +

++=

( )

( ) ( )

( )

pksinpk

pqkcosev2

ipksinhipk2/10

0ipqkcoshev tk0tk

0

22

=

+

+=

.

Adding up all the residues, (iv) gives

( )( ) ( )[ ] ( ) ( )[ ]ll 2/x1n2cose

1n2

1v4vt,xv

222 RCt4/t1n2n

1n

00

+=

=

( ) ( ) ( ) ( )

( )

=

===

2222

n

RC4/1n2k

,/x1n22

1pqk,1pksin,2/1n2pk

l

llQ

Alsox

v

R

1i

= .

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Laplace equation

Q.No.1.: Find the steady state temperature distribution u(x, y) in an infinite metal plate

covering the first quadrant with the edge along the y-axis held at 0o and the

edge along the axis held at

>


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Q.No.3.: Solve the Laplaces equation 0y

u

x

u2

2

2

2

=

+

in square metal plate of side

with ( ) 0u,xu = , u(0, y) = 0, ( ) 0y,u = , u(x, 0) = 0. Assume u(x, y) is

bounded.

Ans.:( )( ) ( )

( ) ( )( )++

++

=

= s1n2sinh.1n2

x1n2siny1n2sinhu4)y,x(u

0n

0 .

One dimensional heat equation

Q.No.1.: Find the temperature distribution u(x, t) in a semi-infinite metal bar with the

end x = 0 at zero temperature and initial temperature distribution f(x).

Deduce when xe)x(f = .

Ans.: ( ) ( ) sxdssine0,sU2

t,xu tsc

0

22

= , where

( ) ( ) ( ) sxdxsin.xfsxdxsin.0,xu0,sU00

==

When xe)x(f = : ( )2s1

s0,sU

+= ; sxdssin

s1

se2)t,x(u

2

tpc

0

22

+=

Q.No.2.:Find the temperature u(x, t) in a semi-finite bar with =

x

uwhen x = 0 and

with initial temperature 0. Assume that 0x

u

as x and is bounded.

Ans.: ( ) ( ) sxdscose1s

2t,xu tcs

20

22

= .

Q.No.3.: Solve2

22

x

uc

t

u

=

, x > 0, t > 0 with 0

x

u

0x

=

=

and

( )

>

=

1x0,

1x0,x0,xu .

Ans.: ( ) sxdscose1scoss

1

s

ssin2)t,x(u tcs

20

22

+

=

.


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Q.No.4.: Solve2

22

x

uc

t

u

=

, x > 0, t > 0 with 0)0,x(u = , x > 0 and ( ) 0ut,0u =

for t > 0.

Ans.:

=

sxdssin.se21u)t,x(utcs

0

0

22

.

Q.No.5.: Solve the one-dimensional heat equation2

22

x

uc

t

u

=

with:

( ) ( ) x20,xu,0t,4u)t,0(u ===

Ans.:( )

4

xnsin.e

n

132

4

2)t,0(u 16/tcn

1n

1n

x

222

=

+

= .

Q.No.6.: Solve the one-dimensional heat equation2

22

xuc

tu

=

with:

( ) ( ) x20,xu,0t,6u)t,0(u x ===

Ans.:( )

6

xncos.e

1126)t,x(u 36/tcn

2

n

1n2

222

+=

= .

Q.No.7.: Using the suitable transform, solve the differential equationt

V

x

V2

2

=

0t,x0 , where V(0, t) = 0 ( )t,V = and V(s, t) = V0 constant.

Ans.: ( ) 02 nxsintn0

1n

encos1n

V

=

.

Q.No.8.: Use the complex form of the Fourier transform to show that

( )( ) { }duut/)ux(euf

xt2

1V =

is the boundary value problem

2

2

x

V

t

V

=

, 0) is initially at temperature zero. At time t = 0, a

constant temperature 00 > is applied and maintained at the face x = 0. Show


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that the temperature at any point x and at any time t, is given by

( ) ( )tc2/xerfct,x 0= .

Ans.:

Q.No.10.: A solid is initially at constant temperature 0 , while the ends x = 0 and x = a

are maintained at temperature zero. Determine the temperature at any point of

the solid at any later time t > 0.

Ans.:

Vibrating string

Q.No.1.: An infinite string is initially at rest and has an initial transverse displacement

y(x, 0) = f(x), =

x/ct0,

x/ct,c/xtsina)t,x(y

0

Where c is the wave velocity.

Ans.:( )

( )( ) ( )

+

+

=

= kt2

xnerf

kt2

xnerf1

kt2

xerf)t,x(

n

1n

00

ll.

Q.No.6.: A. infinite string has an initial transverse displacement y(x, 0) = f(x),


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Transmission lines

Q.No.7.: A semi-infinite transmission line has negligible inductance and leakance per

unit length. A voltage v applied at the sending end (x = 0) which is given by

( )

>

0 at any time t > 0 is given by

( )

= t/RC

2

xerfcv)t,x(v 0 .

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3.Applications of Fourier Transform

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Transcript of 3.Applications of Fourier Transform