1540470959-Boundary_Value_Problems_and_Partial_Differential_Equations__Powers

1.11 Applications of Fourier Series and Integrals 119

Next, suppose thatr(t)is a square-wave function with Fourier series

r(t)=

∑∞

n= 1

2 ( 1 −cos(nπ))

nπ sin(nt).

The corresponding response is

y(t)=

∑∞

n= 1

2 ( 1 −cos(nπ))

nπ ·

− 0. 4 ncos(nt)+( 1. 04 −n^2 )sin(nt)

( 1. 04 −n^2 )^2 +( 0. 4 n)^2.

Note that the term forn=1 has a small denominator, causing a large
response.

B. Boundary Value Problems

By way of introduction to the next chapter, we apply the idea of Fourier series
to the solution of the boundary value problem

d^2 u

dx^2 +pu=f(x),^0 <x<a,

u( 0 )= 0 , u(a)= 0.

First, we will assume thatf(x)is equal to its Fourier sine series,

f(x)=

∑∞

n= 1

bnsin

(nπx

a

)

, 0 <x<a.

And second, we will assume that the solutionu(x), which we are seeking,
equals its Fourier sine series,

u(x)=

∑∞

n= 1

Bnsin

(

nπx

a

)

, 0 <x<a,

and that this series may be differentiated twice to give

d^2 u

dx^2 =

∑∞

n= 1

−

(n (^2) π 2
a^2 Bn

)

sin

(nπx

a

)

, 0 <x<a.

When we insert the series forms foru,u′′,andf(x)into the differential
equation, we find that

∑∞

n= 1

(

−n

(^2) π 2
a^2
Bn+pBn

)

sin

(

nπx

a

)

=

∑∞

n= 1

bnsin

(

nπx

a

)

, 0 <x<a.