Linear PDE on a Disk

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Metronome

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I am trying to solve the PDE [math]\bigtriangleup u = 0,\ u(1,\ \theta) = \sin^2 \theta[/math]
I plugged [imath]u = R(r)T(\theta)[/imath] into the polar Laplacian formula and ended up with [imath]\frac{r^2}{R}\frac{\partial^2 R}{\partial r^2} + \frac{r}{R}\frac{\partial R}{\partial r} = -\frac{1}{T}\frac{\partial T}{\partial t} = \lambda[/imath]. My solutions to the Sturm-Louiville problems are [imath]R = Ar^{-\sqrt\lambda} + Br^{\sqrt\lambda}[/imath] and [imath]T = Ce^{-\sqrt{-\lambda}\theta} + De^{\sqrt{-\lambda}\theta}[/imath] (I have ruled out the [imath]\lambda = 0[/imath] case). Thus I have [math]u = (Ar^{-\sqrt\lambda} + Br^{\sqrt\lambda})(Ce^{-\sqrt{-\lambda}\theta} + De^{\sqrt{-\lambda}\theta})[/math] Plugging in thee boundary condition and absorbing [imath]A + B[/imath] yields [math]\sin^2 \theta = Ce^{-\sqrt{-\lambda}\theta} + De^{\sqrt{-\lambda}\theta}[/math] I tried forcing this into trig form as [imath]\sin^2 \theta = Ce^{\sqrt{-\lambda}\theta i^2} + De^{-\sqrt{-\lambda}\theta i^2}[/imath] becomes via Euler's Formula [math]\sin^2 \theta = (C + D)\cos(\sqrt{-\lambda}\theta i) + (Ci - Di)\sin(\sqrt{-\lambda}\theta i)[/math] How can I solve either of the blue equations for [imath]\lambda[/imath]?
 
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It's been ages since I looked at this stuff, but I do have a couple of questions. For your R(r) equation, what are its boundary conditions and how did you get them? Also, in your T equation, that is a first order DE so you wouldn't get two independent solutions. Do you have other conditions you haven't told us, like for example the solution is bounded? I don't see your eigenvalue problem anywhere.
 
Metronome said:
I am trying to solve the PDE [math]\bigtriangleup u = 0,\ u(1,\ \theta) = \sin^2 \theta[/math]
I plugged [imath]u = R(r)T(\theta)[/imath] into the polar Laplacian formula and ended up with [imath]\frac{r^2}{R}\frac{\partial^2 R}{\partial r^2} + \frac{r}{R}\frac{\partial R}{\partial r} = -\frac{1}{T}\frac{\partial T}{\partial t} = \lambda[/imath]. My solutions to the Sturm-Louiville problems are [imath]R = Ar^{-\sqrt\lambda} + Br^{\sqrt\lambda}[/imath] and [imath]T = Ce^{-\sqrt{-\lambda}\theta} + De^{\sqrt{-\lambda}\theta}[/imath] (I have ruled out the [imath]\lambda = 0[/imath] case). Thus I have [math]u = (Ar^{-\sqrt\lambda} + Br^{\sqrt\lambda})(Ce^{-\sqrt{-\lambda}\theta} + De^{\sqrt{-\lambda}\theta})[/math] Plugging in thee boundary condition and absorbing [imath]A + B[/imath] yields [math]\sin^2 \theta = Ce^{-\sqrt{-\lambda}\theta} + De^{\sqrt{-\lambda}\theta}[/math] I tried forcing this into trig form as [imath]\sin^2 \theta = Ce^{\sqrt{-\lambda}\theta i^2} + De^{-\sqrt{-\lambda}\theta i^2}[/imath] becomes via Euler's Formula [math]\sin^2 \theta = (C + D)\cos(\sqrt{-\lambda}\theta i) + (Ci - Di)\sin(\sqrt{-\lambda}\theta i)[/math] How can I solve either of the blue equations for [imath]\lambda[/imath]?
Click to expand...
This is a Dirichlet problem in the unitary disk. Did you try the Poisson Integral Formula?
 
Metronome said:
I am trying to solve the PDE [math]\bigtriangleup u = 0,\ u(1,\ \theta) = \sin^2 \theta[/math]
Click to expand...
This is the most basic and fun Dirichlet problem for a circle.

[imath]\displaystyle \Delta u = \nabla^2 u = \frac{\partial^2 u}{\partial r^2} + \frac{1}{r}\frac{\partial u}{\partial r} + \frac{1}{r^2}\frac{\partial^2 u}{\partial \theta^2} = 0[/imath]

[imath]u(1,\theta) = \sin^2\theta[/imath]

I will solve it, but I am not going to explain the solution in details unless the OP is interested.

[imath]\displaystyle u(r,\theta) = A_0 + \sum_{n=1}^{\infty} r^n(A_n\cos n\theta + B_n\sin n\theta)[/imath]

where,

[imath]\displaystyle A_0 = \frac{1}{2\pi}\int_{0}^{2\pi}\sin^2 \theta \ d\theta[/imath]


[imath]\displaystyle A_n = \frac{1}{\pi}\int_{0}^{2\pi}\sin^2 \theta \cos n\theta \ d\theta[/imath]


[imath]\displaystyle B_n = \frac{1}{\pi}\int_{0}^{2\pi}\sin^2 \theta \sin n\theta \ d\theta[/imath]


Note: This problem is sometimes called the steady temperature in a circular plate.
 
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