1540470959-Boundary_Value_Problems_and_Partial_Differential_Equations__Powers

4.3 Further Examples for a Rectangle 265

pected) that
X′′(x)
X(x)=−

Y′′(y)
Y(y)=constant.
The conditions atx=0andx=abecome

X′( 0 )= 0 , X′(a)= 0.

If we make the separation constant−λ^2 , we find a familiar eigenvalue problem
forXwhose solution is

X 0 (x)= 1 ,λ 0 = 0 ,

Xn(x)=cos(λnx), λn=nπ/a, n= 1 , 2 ,....

For the factorY(y), the differential equation is

Y 0 ′′= 0 , or Yn′′−λ^2 nYn= 0

with solution

Y 0 (y)=a 0 +b 0 y or Yn(y)=ancosh(λny)+bnsinh(λny).

Thus, the principle of superposition leads to the series solution

u(x,y)=a 0 +b 0 y+

∑∞

n= 1

(

ancosh(λny)+bnsinh(λny)

)

cos(λnx).

The boundary condition aty=0becomes

a 0 +

∑∞

1

ancos(λnx)= 0 , 0 <x<a,

from which we see that all thea’s are 0. Then aty=bwe have

b 0 b+

∑∞

n= 1

(

bnsinh(λnb)

)

cos(λnx)=

V 0 x

a ,^0 <x<a.

This is a slightly disguised cosine series. The coefficients are

b 0 b=^1

a

∫a

0

V 0

(

x

a

)

dx,

bnsinh(λnb)=^2

a

∫a

0

V 0

(

x

a

)

cos(λnx)dx.

See a color graphic of the solution on the CD.