34.2. HEAT AND WAVE EQUATIONS 673

PROCEDURE 34.2.1 To find the solution to an equation

ut = α2uxx, zero boundary conditions, Initial condition

you do the following.

1. First find eigenfunctions, nonzero solutions to

y′′+λ2y = 0, boundary conditions

There will typically be infinitely many of these {yn (x)}∞

n=1 corresponding to eigen-values λ n where limn→∞ λ n = ∞.

2. Your solution will then be of the form

u(x, t) =∞

∑n=1

bn (t)yn (x)

3. Choose bn (t) to satisfy the equation b′n (t) =−λ2nbn (t) in order that the terms of the

sum satisfy the partial differential equation. Thus

bn (t) = bn exp(−tλ 2

n

)Then the solution to the problem is

u(x, t) =∞

∑n=1

bn exp(−tλ 2

n

)yn (x)

where bn is chosen such that ∑∞n=1 bnyn (x) is the Fourier series expansion for the

initial condition.

Example 34.2.2 Find the solution to the initial boundary value problem

ut = .1uxx, u(0, t) = u(2, t) = 0

u(x,0) = 1− (1− x)2

where

f (x) =

{x if x ∈ [0,1]1− x if x ∈ [1,2]

From the above discussion,

u(x, t) =∞

∑k=1

ake−.1k2π2

22 t sin(

kπx2

)the eigenfunctions being sin

( kπx2

), and to satisfy the initial condition, you need

ak =22

∫ 2

0

(1− (1− x)2

)sin(

kπx2

)dx

34.2. HEAT AND WAVE EQUATIONS 673PROCEDURE 34.2.1 1% find the solution to an equationuy = Uy, zero boundary conditions, Initial conditionyou do the following.1. First find eigenfunctions, nonzero solutions toy+ Ay = 0, boundary conditionsThere will typically be infinitely many of these {yn (x)},_, corresponding to eigen-values A» where limy_yoo An = °°.2. Your solution will then be of the formMsu(x,t) = bn (t) Yn (x)n=13. Choose by (t) to satisfy the equation bi, (t) = —A?by (t) in order that the terms of thesum satisfy the partial differential equation. Thusby (t) = bnexp (-1a)Then the solution to the problem isu(x,t) = Dy bnexp (—t2s) yn (2)n=1where by, is chosen such that Y~_, bnYn (x) is the Fourier series expansion for theinitial condition.Example 34.2.2 Find the solution to the initial boundary value problemuw = .luyy, u(0,t) =u(2,t) =0u(x,0) = 1—(1—x)*where_ J xifxe [0,1]ro) 1—xifx€ [1,2]From the above discussion,co _ en? ko.u(x,t) = y ae © ‘sin (S)k=1 2the eigenfunctions being sin (43) , and to satisfy the initial condition, you needa = an (1- (1-2?) sin (S) dx