1540470959-Boundary_Value_Problems_and_Partial_Differential_Equations__Powers

260 Chapter 4 The Potential Equation

Under the new assumption, Eq. (6) separates into

X′′+λ^2 X= 0 , Y′′−λ^2 Y= 0. (8)

The first of these equations, along with the boundary conditions, is recogniz-
able as an eigenvalue problem, whose solutions are

Xn(x)=sin(λnx), λ^2 n=

(nπ

a

) 2

.

The functionsYthat accompany theX’s are

Yn(y)=ancosh(λny)+bnsinh(λny).

Thea’s andb’s are for the moment unknown.
We s e e t h a tXn(x)Yn(y)is a solution of the (homogeneous) potential Eq. (1),
which satisfies the homogeneous conditions Eqs. (4) and (5). A sum of these
functions should satisfy the same conditions and equation, soumay have the
form

u(x,y)=

∑∞

n= 1

(

ancosh(λny)+bnsinh(λny)

)

sin(λnx). (9)

The nonhomogeneous boundary conditions Eqs. (2) and (3) are yet to be
satisfied. Ifuis to be of the form of Eq. (9), the boundary condition Eq. (2)
becomes

u(x, 0 )=

∑∞

n= 1

ansin

(nπx

a

)

=f 1 (x), 0 <x<a. (10)

We recognize a problem in Fourier series immediately. Theanmust be the
Fourier sine coefficients off 1 (x),

an=^2 a

∫a

0

f 1 (x)sin

(

nπx

a

)

dx.

The second boundary condition reads

u(x,b)=

∑∞

n= 1

(

ancosh(λnb)+bnsinh(λnb)

)

sin

(nπx

a

)

=f 2 (x), 0 <x<a.

This also is a problem in Fourier series, but it is not as neat. The constant

ancosh(λnb)+bnsinh(λnb)