Equilibrium heat equation in 2D cylindrical coordinates

dingo_d
Messages
199
Reaction score
0

Homework Statement


Plate in the shape of the circular halo (inner radius a, outer radius b>a), the inner edge is being kept at a constant temperature T_0, and the outer at the temperature given by the function f(\phi)=T_0\cos(2\phi). Find the equilibrium distribution of the heat everywhere inside the plate.

Homework Equations


Laplace equation:

\nabla^2 u(r,\phi)=0

Boundary conditions:
u(a,\phi)=T_0
u(b,\phi)=T_0\cos(2\phi)

The Attempt at a Solution



By separation of variables I get:

r^2\frac{R''}{R}+r\frac{R'}{R}=-\frac{\Phi''}{\Phi}=\lambda^2

For \lambda\neq 0 I have 2 DE, one is Euler's DE, and the other the equation of harmonic oscillator. Their solution is:

R(r)=Ar^\lambda+Br^{-\lambda}

\Phi(\phi)=C\cos(\lambda\phi)+D\sin(\lambda\phi)

With the condition of consistency of the azimuthal part:

\Phi(\phi)=\Phi(\phi+2\pi)

I get that \lambda=m\in\mathbb{Z}, the \lambda=0 gives different solutions.

So my first solution is:

u_1(r,\phi)=\sum_{m=1}^\infty r^m[A_m\cos(m\phi)+B_m\sin(m\phi)]+\sum_{m=1}^\infty r^{-m}[C_m\cos(m\phi)+D_m\sin(m\phi)]

For \lambda=0 I have these solutions:

R(r)=C_0\ln(r)+D_0

\Phi(\phi)=A_0\phi+B_0

With the condition of consistency I get:

A_0\phi+B_0=A_0\phi+A_02\pi+B_0\Rightarrow A_0=0

So the second solution is: (I have put the constants together)

u_2(r\phi)=C\ln(r)+D

The general solution is the superposition of the two solutions:

u(r,\phi)=C\ln(r)+D+\sum_{m=1}^\infty r^m[A_m\cos(m\phi)+B_m\sin(m\phi)]+\sum_{m=1}^\infty r^{-m}[C_m\cos(m\phi)+D_m\sin(m\phi)]

So here is where my problem starts. I know that I'm supposed to use formulas for Fourier summation but in the case of a string that was the more straightforward, for example:

f(x,0)=\sum_{m=0}^\infty A_m\sin\left(\frac{m\pi}{L}x\right)\Big/ \cdot \sin\left(\frac{n\pi}{L}x\right),\ \int_0^L dx

And the integral on the right would give me Kronecker delta which would 'kill' the sum so I would get the A_n from that. But here I don't have this simple way. How should I get the A_m,\ B_m,\ C_m,\ D_m?
 
Physics news on Phys.org
dingo_d said:
I know that I'm supposed to use formulas for Fourier summation but in the case of a string that was the more straightforward, for example:

f(x,0)=\sum_{m=0}^\infty A_m\sin\left(\frac{m\pi}{L}x\right)\Big/ \cdot \sin\left(\frac{n\pi}{L}x\right),\ \int_0^L dx

And the integral on the right would give me Kronecker delta which would 'kill' the sum so I would get the A_n from that.

Use the same procedure for r=a and r=b, multiplying your solution by sin(n\phi) and cos(n\phi). You will get equations for the unknown coefficients.
 
Thread 'Need help understanding this figure on energy levels'
This figure is from "Introduction to Quantum Mechanics" by Griffiths (3rd edition). It is available to download. It is from page 142. I am hoping the usual people on this site will give me a hand understanding what is going on in the figure. After the equation (4.50) it says "It is customary to introduce the principal quantum number, ##n##, which simply orders the allowed energies, starting with 1 for the ground state. (see the figure)" I still don't understand the figure :( Here is...
Thread 'Understanding how to "tack on" the time wiggle factor'
The last problem I posted on QM made it into advanced homework help, that is why I am putting it here. I am sorry for any hassle imposed on the moderators by myself. Part (a) is quite easy. We get $$\sigma_1 = 2\lambda, \mathbf{v}_1 = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} \sigma_2 = \lambda, \mathbf{v}_2 = \begin{pmatrix} 1/\sqrt{2} \\ 1/\sqrt{2} \\ 0 \end{pmatrix} \sigma_3 = -\lambda, \mathbf{v}_3 = \begin{pmatrix} 1/\sqrt{2} \\ -1/\sqrt{2} \\ 0 \end{pmatrix} $$ There are two ways...
Back
Top