PDE: find the steady-state temperature u(r,θ) in the upper half plate removing the upper half unit disk:

To get an eigenvalue problem in Θ, put

u(r, θ) = v(r, θ) + f(θ)

where f(0) = 0, and f(π) = 1. The simple choice for f is f(θ) = θ/π. The problem for v is then

v_rr + (1/r)*v_r + (1/r^2)*v_θθ = 0 , 1 v_r(1,θ) = -θ/π, 0 < θ < π
v(r,0) = 0, v(r,π) = 0, 1
Letting v(r,θ) = R(r)Θ(θ) give the problem for R and Θ

Θ'' + λΘ = 0, Θ(0) = 0 and Θ(π) = 0

r² R'' + rR' - λR = 0.

The eigenvalues and eigenfunctions for the Θ problem are

λ_n = n², n = 1, 2, 3, ....

Θ_n = sin(nθ).

The solutions for R are

R_n(r) = a_n r^(-n) + b_n r^n.

Since we expect R to be bounded as r-> ∞, b_n = 0 for all n. This gives the solution v(r,θ) =

∞
Σ a_n r^(-n) sin(nθ)
n=1

When r = 1, set this equal to -θ/π and find that a_n are the Fourier sine coefficients

a_n = 2(-1)ⁿ/(nπ).

Finally, the solution u(r,θ) =

∞
Σ {2(-1)ⁿ/(nπ) r^(-n) sin(nθ) }+ θ/π.
n=1

Everything looks realy good, But I am confused at one point
v_r(1,θ) = -θ/pi, isn't it supposed to be zero since d/dr(f(θ)) = 0 rather than θ/pi
and how did you do the integrals for coefficient a_n ?

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You are right! I totally didn't notice it is was v_r, I didn't read carefully enough. My solution was for v(1,θ) = -f(θ). So in this case, the answer is just u = θ/π. This would mean that a_n would be the cosine coefficients for 0 giving a_n = 0 for all n. That's actually a simpler problem!

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P.S. I hope you'll see this. I checked, but you don't allow email through Y!A.

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