34.5. EXERCISES 693

23. The solution to Laplace’s equation on the disk DR which has boundary values f (α)was derived and it is

u(r,θ) =∫ 2π

0

12π

R2− r2

R2−2(cos(θ −α))Rr+ r2 f (α)dα

Show thatlim

r→R−u(r,θ) = f (θ)

This shows how the boundary values are obtained.

24. Recall that u(r,θ) =

12π

∫ 2π

0f (θ)dθ +

∑n=1

rn

Rn

((∫ 2π

0cos(nα) f (α)dα

)cos(nθ)

+

(∫ 2π

0sin(nα) f (α)dα

)sin(nθ)

)Explain why if f is a 2π periodic continuous function, it follows that there is atrigonometric polynomial which is uniformly close to f (θ) for θ ∈ [0,2π]. Hint:From the above problem, convergence to f (θ) as r→ R− takes place. Note thatfrom the argument, this actually happens uniformly thanks to the uniform continuityof f . Now argue that the tail ∑

∞n=N of the above series is uniformly small if N is

large.

25. Let

f (x) =

{x if x ∈ [0,1]2− x if x ∈ [1,2]

Solve the following initial boundary value problems

(a) ut = a2uxx,u(x,0) = f (x) ,u(0, t) = 0 = u(2, t)

(b) ut = a2uxx,u(x,0) = f (x) ,ux (0, t) = 0 = u(2, t)

(c) ut = a2uxx,u(x,0) = f (x) ,ux (0, t) = 0 = ux (2, t)

34.5.23.24.25.EXERCISES 693The solution to Laplace’s equation on the disk Dg which has boundary values f (a)was derived and it is2m | R27?u(7,8) = 0 2m R ~F(cos(0—a))Rra ret(a)daShow thatlim u(r,0) = f (6)r—>R-—This shows how the boundary values are obtained.Recall that u(r, 0) =1 *mal F(0)40-+ = Fe (( | “cos na) # a) ae) cos (n8)n ( [ sin (na) fade) sin (nd)Explain why if f is a 27 periodic continuous function, it follows that there is atrigonometric polynomial which is uniformly close to f(@) for 6 € [0,27]. Hint:From the above problem, convergence to f(0@) as r + R— takes place. Note thatfrom the argument, this actually happens uniformly thanks to the uniform continuityof f. Now argue that the tail ))_,, of the above series is uniformly small if N islarge.Let_ | xifxe [0,1]fe) -{ 2—xifx€ [1,2]Solve the following initial boundary value problems(a) uy, = a°Uxx,u(x,0) = f (x) ,w(0,t) =0=u(2,2)(b) up = a°Uxx,u(x,0) = f (x) ux (0, be =u(2,t)(Cc) uy = A Uy, U(x,0) = f (x) ux (0,1) = 0 = uy (2,1)