What is the Steady State Solution for the Heat Equation in an Annulus?

[joelcponte]

Homework Statement

Heat equation in a annulus, steady state solution.

u(a,θ,t) = T_a
u(b,θ,t) = T_bcos(θ)

Homework Equations

Using separation of Variables

$\frac{}{}\frac{1}{r}\frac{d}{d r}(r\frac{d R}{d r}) + \frac{1}{r^2}\frac{d^2\Theta}{d \theta} = 0$

The Attempt at a Solution

I found

u(r,θ,t) = $\alpha_0 + \beta_0 ln(r) + \sum (\alpha_n r^n + \beta_n r^{-n})(\gamma_n cos(n\theta) + \sigma_n sin(n\theta))$

but the answer is


u(r,θ,t) = $\alpha_0 + \beta_0 ln(r) + \sum (\alpha_n r^n + \beta_n r^{-n})cos(n\theta)$

(this is before applying the boundary conditions)

 

[vela]

Your answer is correct. The second solution is what you get after you apply the boundary condition at r=b.

 

[joelcponte]

Hi, I can't see how does the sin term disappear when I use the boundary condition at b =\

 

[vela]

Actually, I was a little careless. What they did was to say that both boundary conditions are even functions of θ, so the solution must also be an even function of θ, which implies you can throw out the sine terms. So you're not exactly applying the boundary conditions yet, but you are using some information gleaned from them.

 

[joelcponte]

Oh, really? I didn't know that! Do you know anywhere I can read about it?

Thank you for your help!