PDE: Annulus question, Steady State Temperature

[RJLiberator]

Homework Statement

Suppose the inner side of the annulus {(r,Φ): r_0 ≤ r ≤ 1} is insulated and the outer side is held at temperature u(1,0) = f(Φ).

a) Find the steady-state temperature
b) What is the solution if f(Φ) = 1+2sinΦ ?

Homework Equations

The Attempt at a Solution

a)
A = {(rcosΦ,rsinΦ): r_0 ≤ r ≤ 1}
We have the conditions

U_rr + r^(-1)*U_r+r^(-2)*U_ΦΦ = 0 in A
U(1,Φ) = f(Φ)
U(r_0,Φ) = g(Φ)

Φ"+v^2Φ = 0
ϑ^2R"+rR'-v^2R = 0

Φ => e^(ivΦ) , v= 0,+/-1, +/-2, ...

r = a)n*r^v+b_n*r^(-v)

Hence, u(r,Φ) = ∑ c_n * r^(|n|)*e^(inΦ)

We know that
C_n = 1/(2*pi) integral from -pi to pi f(Φ)e^(-inΦ)dΦ

u(r,Φ) = ∑C_n * ( (r^n+r_0^(2n)*r^(-n)) / (1+r_0^(2n)) * e^(inΦ))

That is the answer for part a. Finding the steady state temperature.

My question is how did we get to that messy r fraction part. We see what I stated r to equal, and we know cn is right based off Fourier coefficients.

WE went from u(r,Φ) = ∑ c_n * r^(|n|)*e^(inΦ)
to u(r,Φ) = ∑C_n * ( (r^n+r_0^(2n)*r^(-n)) / (1+r_0^(2n)) * e^(inΦ))

Must I use the Poisson Kernel?

b) b is a little tougher for me to understand.

answer is u(r,Φ) = 1+2*((r^2+r_0^2)/ (r(1+r_0^2))) sinΦ

I started off by analyzing the c_n by inputting f(Φ) = 1+2sinΦ
we get that the integral =

 

[pasmith]

Homework Statement

Suppose the inner side of the annulus {(r,Φ): r_0 ≤ r ≤ 1} is insulated and the outer side is held at temperature u(1,0) = f(Φ).

a) Find the steady-state temperature
b) What is the solution if f(Φ) = 1+2sinΦ ?

Homework Equations

The Attempt at a Solution

a)
A = {(rcosΦ,rsinΦ): r_0 ≤ r ≤ 1}
We have the conditions

U_rr + r^(-1)*U_r+r^(-2)*U_ΦΦ = 0 in A
U(1,Φ) = f(Φ)
U(r_0,Φ) = g(Φ)

Does not "the inner boundary is insulated" mean that $\frac{\partial u}{\partial r} = 0$ on $r = r_0$?

Φ"+v^2Φ = 0
ϑ^2R"+rR'-v^2R = 0

Φ => e^(ivΦ) , v= 0,+/-1, +/-2, ...

r = a)n*r^v+b_n*r^(-v)

Hence, u(r,Φ) = ∑ c_n * r^(|n|)*e^(inΦ)

Umm... no. I'm happy with assuming a solution of the form $$u(r,\phi) = \sum_{n=-\infty}^\infty c_n R_n(r)e^{in\phi}$$ where $$u(1,\phi) = f(\phi) = \sum_{n=-\infty}^\infty c_ne^{in\phi}$$ and the $R_n$ are an as yet unspecified sequence of functions, but $R_n = r^{|n|}$ is not the solution you are looking for as it doesn't satisfy the boundary condition on $r = r_0$.

We know that
C_n = 1/(2*pi) integral from -pi to pi f(Φ)e^(-inΦ)dΦ

u(r,Φ) = ∑C_n * ( (r^n+r_0^(2n)*r^(-n)) / (1+r_0^(2n)) * e^(inΦ))

That is the answer for part a. Finding the steady state temperature.

My question is how did we get to that messy r fraction part. We see what I stated r to equal, and we know cn is right based off Fourier coefficients

WE went from u(r,Φ) = ∑ c_n * r^(|n|)*e^(inΦ)
to u(r,Φ) = ∑C_n * ( (r^n+r_0^(2n)*r^(-n)) / (1+r_0^(2n)) * e^(inΦ))

If $u(r,\phi) = \sum_{n=-\infty}^\infty c_n R_n(r) e^{in\phi}$ and $f(\phi) = \sum_{n=-\infty}^{\infty} c_ne^{in\phi}$ then $R_n$ is the solution of the boundary value problem $$\frac{1}{r}\frac{d}{dr}\left(r\frac{dR_n}{dr}\right) - \frac{n^2 R_n}{r^2} = 0$$ subject to $R_n(1) = 1$ and $R_n'(r_0) = 0$, obtained by substituting the series definition of $u$ into Laplace's equation and considering the coefficients of $e^{in\phi}$.

Now there are two points to note:

(1) The BVP for $n > 0$ is the same as that for $-n < 0$ and therefore they have the same solution: $R_n = R_{|n|}$.

(2) The BVP for $n = 0$ is a special case.

Thus we have $$u(r,\phi) = c_0R_0(r) + \sum_{n=1}^\infty (c_n e^{in\phi} + c_{-n}e^{-in\phi}) R_n(r)$$ which is easily rearranged into $$u(r,\phi) = c_0R_0(r) + \sum_{n=1}^\infty (a_n \cos n\phi + b_n \sin n\phi) R_n(r)$$ if that's more convenient, and I'll let you go back and solve the BVPs for $R_n$ for $n \geq 0$.

Must I use the Poisson Kernel?

b) b is a little tougher for me to understand.

answer is u(r,Φ) = 1+2*((r^2+r_0^2)/ (r(1+r_0^2))) sinΦ

I started off by analyzing the c_n by inputting f(Φ) = 1+2sinΦ
we get that the integral =
View attachment 97271

The Fourier series of $1 + 2 \sin \phi$ is, astoundingly, $1 + 2 \sin \phi$. Substitute this into the general solution.