1540470959-Boundary_Value_Problems_and_Partial_Differential_Equations__Powers

120 Chapter 1 Fourier Series and Integrals

Since the coefficients of like terms in the two series must match, we may con-
clude that
(
p−n

(^2) π 2
a^2

)

Bn=bn, n= 1 , 2 , 3 ,....

If it should happen thatp=m^2 π^2 /a^2 for some positive integerm,thereis
no value ofBmthat satisfies
(
p−m

(^2) π 2
a^2

)

Bm=bm

unlessbm=0 also, in which case any value ofBmis satisfactory. In summary,
we may say that

Bn= bn

p−n^2 π^2 /a^2

and

u(x)=

∑∞

n= 1

a^2 bn

a^2 p−n^2 π^2

sin

(nπx

a

)

,

with the agreement that a zero denominator must be handled separately.

Example.
Consider the boundary value problem

d^2 u

dx^2 −u=−x,^0 <x<^1 ,

u( 0 )= 0 , u( 1 )= 0.

We have found previously that

−x=

∑∞

n= 1

2 (− 1 )n

πn sin(nπx),^0 <x<^1.

Thus, by the preceding development, the solution must be

u(x)=

∑∞

n= 1

2

π

(− 1 )n+^1

n(n^2 π^2 + 1 )sin(nπx),^0 <x<^1.

Although this particular series belongs to a known function, one would not,
in general, know any formula for the solutionu(x)other than its Fourier sine
series.