34.5. EXERCISES 689

Thus

u(r,θ) =1

2π

∫ 2π

0

(R2− r2

R2−2(cos(θ −α))Rr+ r2

)f (α)dα

Note that this shows that if r = 0 so you are at the center, then

u(r,θ) =1

2π

∫ 2π

0f (α)dα

so the value at the center is the average of the boundary values. This proves the followingfundamental result.

Theorem 34.4.3 The solution to the problem

∆u = urr +1r

ur +1r2 uθθ = 0

on the disc of radius R where on the boundary of this disk,

u(R,θ) = f (θ) , f (0) = f (2π)

is given by the formula

u(r,θ) =1

2π

∫ 2π

0

(R2− r2

R2−2(cos(θ −α))Rr+ r2

)f (α)dα

34.5 Exercises1. Solve the following initial boundary value problems.

(a) ut = uxx,u(x,0) = 1,u(0, t) = 0,u(4, t) = 0

(b) ut = 2uxx,u(x,0) = 1,ux (0, t) = 0,u(3, t) = 0

(c) ut = 3uxx,u(x,0) = 1− x,u(0, t) = 0,u(2, t) = 0

(d) ut = 4uxx,u(x,0) = 1− x,u(0, t) = 0,u(2, t) = 0

(e) ut = 5uxx,u(x,0) = 1− x,u(0, t) = 0,ux (1, t) = 0

(f) ut = uxx,u(x,0) = x+1,ux (0, t) = 0,u(2, t) = 0

(g) ut = 3uxx,u(x,0) = x,ux (0, t) = 0,u(1, t) = 0

(h) ut = 3uxx,u(x,0) = x2,u(0, t) = 0,u(5, t) = 0

(i) ut = 4uxx,u(x,0) = 1,u(0, t) = 0,ux (1, t) = 0

(j) ut = uxx,u(x,0) = x,u(0, t) = 0,ux (4, t) = 0

(k) ut = 2uxx,u(x,0) = 1,ux (0, t) = 0,ux (5, t) = 0

(l) ut = 2uxx,u(x,0) = x,ux (0, t) = 0,ux (4, t) = 0

(m) ut = 2uxx,u(x,0) = 1− x,ux (0, t) = 0,ux (3, t) = 0

2. Find the solution to the initial boundary value problem

utt = 3uxx,u(0, t) = 0,u(5, t) = 0,u(x,0) = 3x(x−5) ,ut (x,0) = x2