Part 4: Linearity and Fourier Series

We say that a partial differential equation is linear if the linear combination of any two solutions is also a solution. For example, suppose that p( x,t) and q( x,t) are both solutions to the heat equation-i.e., suppose that

¶p ¶t = k  ¶2p ¶x2        and         ¶q ¶t = k  ¶2q ¶x2

(8)

A linear combination of p and q is of the form u( x,t) = Ap( x,t) +Bq( x,t) where A,B are both constants. Moreover,

¶u ¶t =  ¶ ¶t ( Ap(x,t) +Bq( x,t) ) = A  ¶p ¶t +B  ¶q ¶t

so that (8) implies that

¶u ¶t = A  ¶p ¶t +B  ¶q ¶t = Ak  ¶2p ¶x2 +Bk  ¶2q ¶x2 = k  ¶2 ¶x2 ( Ap(x,t) +Bq( x,t) )

That is, the linear combination u( x,t) = Ap( x,t)+Bq( x,t) is also a solution to the heat equation, and consequently, we say that the heat equation is a linear partial differential equation.

Suppose now that a linear partial differential equation has both boundary conditions and initial conditions, where initial conditions are constraints on the solution and its derivatives at a fixed point in time. Then a complete solution to the partial differential equation can often be obtained from the Fourier series decompositions of the initial conditions.

For example, let us suppose that the vibrating string in example 5 is plucked at time t = 0, which is to say that it is released from rest at time t = 0 with an initial shape given by the graph of the function y = f( x) : Then the initial conditions for the vibrating string are 

u( x,0) = f( x)         and         ¶u ¶t ( x,0) = 0

Let's apply the initial conditions to the separated solution (7). The initial condition u_t( x,0) = 0 implies that f( x) T^¢( 0) = 0, so that to avoid the trivial solution we suppose that T^¢( 0) = 0. Thus,

0 = T¢( 0) = -awA1sin( 0)+B1awcos( 0) = B1

As a result, we must have T( t) = A₁cos( anpt/l) , and if we define b_n = A₁B₂, then (7) reduces to

un( x,t) = bncos æ è  anp l t ö ø sin æ è  np l x ö ø

As will be shown in the exercises, the 1 dimensional wave equation is linear. Consequently, the general solution to the 1 dimensional wave equation with the given boundary and initial conditions is

u( x,t) = ¥ å n = 1  bncos æ è  anp l t ö ø sin æ è  np l x ö ø

(9)

Hence, the only task left is that of determining the values of the constants b_n. However, (9) implies that

u( x,0) = ¥ å n = 1  bncos( 0) sin æ è  np l x ö ø

and since u( x,0) = f( x) , this reduces to

f( x) = ¥ å n = 1  bnsin æ è  np l x ö ø

As a result, if f( x) is continuous and if f( 0) = f( l) , then the constants b_n are the Fourier Sine coefficients of f( x) on [ 0,l] , which are given by

bn =  2 l ó õ l 0  f( x) sin æ è  np l x ö ø dx

(10)

For more on Fourier series and their relationship to partial differential equations, see the Maple worksheet associated with this section.       

EXAMPLE 6    What is the solution to the vibrating string problem for a 2 foot long string which is initially at rest and which has an initial shape that is the same as the graph of the function

u( x,0) =  1 12  -   | x-1| 12

Solution: We begin by finding the Fourier coefficients b_n, which according to (10) are for an l = 2 foot long string given by

bn =  2 2 ó õ 2 0  æ è  1 12 -  | x-1| 12 ö ø sin æ è  np 2 x ö ø dx

:Evaluating using the computer algebra system Maple then yields

bn = 2sin æ è  np 2 ö ø -sin( np) 3n2p2

However, since n is an integer, sin( np) = 0 for all n. Thus, b_n reduces to

bn = 2sin æ è  np 2 ö ø 3n2p2

But sin( [(np)/2]) = 0 when n is even, so that b₀ = b₂ = ¼ = b_2n = 0. Thus, we only have odd coefficients of the form

b1 = 2sin æ è  p 2 ö ø 3·12·p2 ,        b3 = 2sin æ è  3p 2 ö ø 3·32·p2 ,    b5 = 2sin æ è  5p 2 ö ø 3·52·p2 ,  ¼

which simplify to

b1 =  2( 1) 3·12·p2 ,        b3 =  2( -1) 3·32·p2 ,    b5 =  2( 1) 3·52·p2 ,  ¼

Odd numbers are of the form 2n+1 for n = 0,1,¼. Thus, we have

b2n+1 =  2( -1) n 3p2( 2n+1) 2

and the solution (9) is of the form

u( x,t) = ¥ å n = 0   2( -1) n 3p2( 2n+1) 2 cos æ è  a( 2n+1)p l t ö ø sin æ è  ( 2n+1)p l x ö ø