Equilibrium heat equation in 2D cylindrical coordinates

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SUMMARY

The discussion focuses on solving the equilibrium heat equation in 2D cylindrical coordinates for a circular halo plate with inner radius a and outer radius b, where the inner edge is maintained at a constant temperature T_0 and the outer edge follows the function f(φ) = T_0 cos(2φ). The solution involves separating variables and solving the Laplace equation, resulting in a general solution expressed as a superposition of two parts: u(r, φ) = C ln(r) + D + Σ (r^m[A_m cos(mφ) + B_m sin(mφ)]) + Σ (r^(-m)[C_m cos(mφ) + D_m sin(mφ)]). The challenge lies in determining the coefficients A_m, B_m, C_m, and D_m using Fourier summation techniques.

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dingo_d
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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 [tex]T_0[/tex], and the outer at the temperature given by the function [tex]f(\phi)=T_0\cos(2\phi)[/tex]. Find the equilibrium distribution of the heat everywhere inside the plate.

Homework Equations


Laplace equation:

[tex]\nabla^2 u(r,\phi)=0[/tex]

Boundary conditions:
[tex]u(a,\phi)=T_0[/tex]
[tex]u(b,\phi)=T_0\cos(2\phi)[/tex]

The Attempt at a Solution



By separation of variables I get:

[tex]r^2\frac{R''}{R}+r\frac{R'}{R}=-\frac{\Phi''}{\Phi}=\lambda^2[/tex]

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

[tex]R(r)=Ar^\lambda+Br^{-\lambda}[/tex]

[tex]\Phi(\phi)=C\cos(\lambda\phi)+D\sin(\lambda\phi)[/tex]

With the condition of consistency of the azimuthal part:

[tex]\Phi(\phi)=\Phi(\phi+2\pi)[/tex]

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

So my first solution is:

[tex]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)][/tex]

For [tex]\lambda=0[/tex] I have these solutions:

[tex]R(r)=C_0\ln(r)+D_0[/tex]

[tex]\Phi(\phi)=A_0\phi+B_0[/tex]

With the condition of consistency I get:

[tex]A_0\phi+B_0=A_0\phi+A_02\pi+B_0\Rightarrow A_0=0[/tex]

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

[tex]u_2(r\phi)=C\ln(r)+D[/tex]

The general solution is the superposition of the two solutions:

[tex]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)][/tex]

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:

[tex]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[/tex]

And the integral on the right would give me Kronecker delta which would 'kill' the sum so I would get the [tex]A_n[/tex] from that. But here I don't have this simple way. How should I get the [tex]A_m,\ B_m,\ C_m,\ D_m[/tex]?
 
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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:

[tex]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[/tex]

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

Use the same procedure for [tex]r=a[/tex] and [tex]r=b[/tex], multiplying your solution by [tex]sin(n\phi)[/tex] and [tex]cos(n\phi)[/tex]. You will get equations for the unknown coefficients.
 

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