674 CHAPTER 34. SOME PARTIAL DIFFERENTIAL EQUATIONS

After some tedious computations, this yields

ak =16

π3k3

(1− (−1)k

)Thus when k is even, this is 0 and when k is odd, it equals 32

π3k3 . Thus

u(x, t) =∞

∑k=1

32

π3 (2k−1)3 e−(2k−1)2π2

4 (.1)t sin((2k−1)πx

2

)The next example has to do with the same equation but with one end insulated and

the other held at a temperature of 0. The physical modeling of this equation shows that toconsider an insulated boundary, say at L, you let ux (L, t) = 0.

Example 34.2.3 Solve the problem

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

u(x,0) = 1− (1− x)2

To do this, first look for eigenfunctions. Find solutions to

y′′+λy = 0, y(0) = 0,y′ (2) = 0

Then the eigenfunctions are in Example 33.2.1. They are

sin((2n−1)πx

4

), n = 1,2, · · ·

It follows that the solution desired is of the form∞

∑n=1

bn (t)sin((2n−1)πx

4

)and one needs

b′n (t) =−1

10

((2n−1)π

4

)2

bn (t)

so

bn (t) = bn exp

(− 1

10

((2n−1)π

4

)2

t

)Then the Fourier series expansion of the solution is

∞

∑n=1

bn exp

(− 1

10

((2n−1)π

4

)2

t

)sin((2n−1)πx

4

)where bn is an appropriate Fourier coefficient chosen to satisfy the initial condition. Thus

bn =22

∫ 2

0sin((2n−1)πx

4

)(1− (1− x)2

)dx

=32

π3 (2n−1)3 (2(−1)nπn− (−1)n

π +4)