Higher Engineering Mathematics

AN INTRODUCTION TO PARTIAL DIFFERENTIAL EQUATIONS 523

I

the potential at any point within the rectangleOPQR.
The boundary conditions are:

u=0 whenx= 0 i.e.u(0,y)= 0 for 0≤y≤b

u=0 whenx=a i.e.u(a,y)= 0 for 0≤y≤b

u=0 wheny=b i.e.u(x,b)= 0 for 0≤x≤a

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

As with previous partial differential equations, a
solution of the formu(x,y)=X(x)Y(y) is assumed,
whereXis a function ofxonly, andYis a function
ofyonly. Simplifying tou=XY, determining par-

tial 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
Letting each side equal a constant,−p^2 , gives the two
equations:

X′′+p^2 X= 0 and Y′′−p^2 Y= 0

from which,X=Acospx+Bsinpxand
Y=Cepy+De−py or Y=Ccoshpy+Dsinhpy
(see Problem 5, page 478 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)=0 andφ=−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=1 andy= 1 .Apply the

Laplace equation

∂^2 u

∂x^2

+

∂^2 u

∂y^2

=0 to determine

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

subject to the following boundary conditions:

u=0 whenx= 00 ≤y≤1,

u=0 whenx= 10 ≤y≤1,

u=0 wheny= 00 ≤x≤1,

u=4 wheny= 10 ≤x≤1.

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

is assumed, whereX is a function ofxonly, and

Yis a function ofyonly. Simplifying tou=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

Letting each side equal a constant,−p^2 , gives the

two equations:

X′′+p^2 X= 0 and Y′′−p^2 Y= 0