1540470959-Boundary_Value_Problems_and_Partial_Differential_Equations__Powers

4.2 Potential in a Rectangle 261

must be thenth Fourier sine coefficient off 2 .Sinceanis known,bncan be
determined from the following computations:

ancosh(λnb)+bnsinh(λnb)=^2 a

∫a

0

f 2 (x)sin(λnx)dx=cn,

bn=cn−ancosh(λnb)

sinh(λnb)

.

If we use this last expression forbnand substitute into Eq. (9), we find the
solution

u(x,y)=

∑∞

n= 1

{

cnsinh(λny)

sinh(λnb)

+an

[

cosh(λny)−coshsinh(λ(λnb)

nb)

sinh(λny)

]}

sin(λnx). (11)

Notice that the function multiplyingcnis 0 aty=0andis1aty=b. Similarly,
the function multiplyinganis 1 aty=0and0aty=b. An easier way to write
this latter function is
sinh(λn(b−y))
sinh(λnb)

,

as can readily be found from hyperbolic identities.

Example.
Supposef 1 andf 2 are both given by

f 1 (x)=f 2 (x)=







2 x

a,^0 <x<

a

2 ,

2

(a−x

a

)

,

a

2 <x<a.

Then

cn=an=

8

π^2

sin(nπ/ 2 )
n^2.
The solution of the potential equation for these boundary conditions is

u(x,y)=

8

π^2

∑∞

n= 1

sin

(nπ

2

)

n^2

sinh

(nπ

ay

)

+sinh

(nπ

a(b−y)

)

sinh

(nπb

a

) sin

(nπx

a

)

. (12)

In Fig. 2 is a graph of some level curves,u(x,y)=constant, for the case where
a=b, and also a view of the surfacez=u(x,y). Also see color figures on the
CD.