Higher Engineering Mathematics, Sixth Edition

526 Higher Engineering Mathematics

the potential at any point within the rectangleOPQR.

The boundary conditions are:

u=0whenx=0i.e.u( 0 ,y)=0for0≤y≤b

u=0whenx=a i.e.u(a,y)=0for0≤y≤b

u=0wheny=b i.e.u(x,b)=0for0≤x≤a

u=f(x)wheny=0i.e.u(x, 0 )=f(x)

for 0≤x≤a

As with previous partial differential equations, a solu-

tion of the formu(x,y)=X(x)Y(y)is assumed, where

X isa function ofx only, andY is a function of

y only. Simplifying tou=XY, determining partial

derivatives, and substituting into

∂^2 u

∂x^2

+

∂^2 u

∂y^2

=0gives:

X′′Y+XY′′= 0

Separating the variables gives:

X′′

X

=−

Y′′

Y

Lettingeach side equal a constant,−p^2 ,givesthetwo

equations:

X′′+p^2 X=0andY′′−p^2 Y= 0

from which,X=Acospx+Bsinpxand

Y=Cepy+De−pyorY=Ccoshpy+Dsinhpy(see

Problem 5, page 480 for this conversion).

This latter form can also be expressed as:

Y=Esinhp(y+φ)by using compound angles.

Hence u(x,y)=XY

={Acospx+Bsinpx}{Esinhp(y+φ)}

or u(x,y)

={Pcospx+Qsinpx}{sinhp(y+φ)}

whereP=AEandQ=BE.

The first boundary condition is:u( 0 ,y)=0, hence

0 =Psinhp(y+φ) from which, P=0. Hence,

u(x,y)=Qsinpxsinhp(y+φ).

The second boundary condition is: u(a,y)=0,

hence 0=Qsinpasinhp(y+φ) from which,

sinpa=0, hence,pa=nπorp=

nπ

a

for

n= 1 , 2 , 3 ,...

The third boundary condition is: u(x,b)=0,

hence, 0=Qsinpxsinhp(b+φ) from which,

sinhp(b+φ)=0andφ=−b.

Hence,u(x,y)=Qsinpxsinhp(y−b)=

Q 1 sinpxsinhp(b−y)whereQ 1 =−Q.

Since there are many solutions for integer values ofn,

u(x,y)=

∑∞

n= 1

Qnsinpxsinhp(b−y)

=

∑∞

n= 1

Qnsin

nπx

a

sinh

nπ

a

(b−y)

The fourth boundary condition is:u(x, 0 )=f(x),

hence, f(x)=

∑∞

n= 1

Qnsin

nπx

a

sinh

nπb

a

i.e. f(x)=

∑∞

n= 1

(

Qnsinh

nπb

a

)

sin

nπx

a

From Fourier series coefficients,

(

Qnsinh

nπb

a

)

= 2 ×the mean value of

f(x)sin

nπx

a

fromx=0tox=a

i.e. =

∫a

0

f(x)sin

nπx

a

dxfrom which,

Qnmay be determined.

This is demonstrated in the following worked

problem.

Problem 7. A square plate is bounded by the

linesx= 0 ,y= 0 ,x=1andy= 1 .Apply the

Laplace equation

∂^2 u

∂x^2

+

∂^2 u

∂y^2

=0todeterminethe

potential distributionu(x,y)over the plate, subject

to the following boundary conditions:

u=0whenx= 00 ≤y≤ 1 ,

u=0whenx= 10 ≤y≤ 1 ,

u=0wheny= 00 ≤x≤ 1 ,

u=4wheny= 10 ≤x≤1.

Initially a solution of the formu(x,y)=X(x)Y(y)is

assumed, whereXis a function ofxonly, andYis a

function ofyonly. Simplifyingtou=XY, determining

partial derivatives, and substituting into

∂^2 u

∂x^2

+

∂^2 u

∂y^2

= 0

gives: X′′Y+XY′′= 0

Separating the variables gives:

X′′

X

=−

Y′′

Y

Lettingeach side equal a constant,−p^2 ,givesthetwo

equations:

X′′+p^2 X=0andY′′−p^2 Y= 0