Linear PDE on a Disk

[Metronome]

I am trying to solve the PDE $$\bigtriangleup u = 0,\ u(1,\ \theta) = \sin^2 \theta$$
I plugged $u = R(r)T(\theta)$ into the polar Laplacian formula and ended up with $\frac{r^2}{R}\frac{\partial^2 R}{\partial r^2} + \frac{r}{R}\frac{\partial R}{\partial r} = -\frac{1}{T}\frac{\partial T}{\partial t} = \lambda$. My solutions to the Sturm-Louiville problems are $R = Ar^{-\sqrt\lambda} + Br^{\sqrt\lambda}$ and $T = Ce^{-\sqrt{-\lambda}\theta} + De^{\sqrt{-\lambda}\theta}$ (I have ruled out the $\lambda = 0$ case). Thus I have $$u = (Ar^{-\sqrt\lambda} + Br^{\sqrt\lambda})(Ce^{-\sqrt{-\lambda}\theta} + De^{\sqrt{-\lambda}\theta})$$ Plugging in thee boundary condition and absorbing $A + B$ yields $$\sin^2 \theta = Ce^{-\sqrt{-\lambda}\theta} + De^{\sqrt{-\lambda}\theta}$$ I tried forcing this into trig form as $\sin^2 \theta = Ce^{\sqrt{-\lambda}\theta i^2} + De^{-\sqrt{-\lambda}\theta i^2}$ becomes via Euler's Formula $$\sin^2 \theta = (C + D)\cos(\sqrt{-\lambda}\theta i) + (Ci - Di)\sin(\sqrt{-\lambda}\theta i)$$ How can I solve either of the blue equations for $\lambda$?

 

[LCKurtz]

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.

 

[Eugenio]

I am trying to solve the PDE $$\bigtriangleup u = 0,\ u(1,\ \theta) = \sin^2 \theta$$
I plugged $u = R(r)T(\theta)$ into the polar Laplacian formula and ended up with $\frac{r^2}{R}\frac{\partial^2 R}{\partial r^2} + \frac{r}{R}\frac{\partial R}{\partial r} = -\frac{1}{T}\frac{\partial T}{\partial t} = \lambda$. My solutions to the Sturm-Louiville problems are $R = Ar^{-\sqrt\lambda} + Br^{\sqrt\lambda}$ and $T = Ce^{-\sqrt{-\lambda}\theta} + De^{\sqrt{-\lambda}\theta}$ (I have ruled out the $\lambda = 0$ case). Thus I have $$u = (Ar^{-\sqrt\lambda} + Br^{\sqrt\lambda})(Ce^{-\sqrt{-\lambda}\theta} + De^{\sqrt{-\lambda}\theta})$$ Plugging in thee boundary condition and absorbing $A + B$ yields $$\sin^2 \theta = Ce^{-\sqrt{-\lambda}\theta} + De^{\sqrt{-\lambda}\theta}$$ I tried forcing this into trig form as $\sin^2 \theta = Ce^{\sqrt{-\lambda}\theta i^2} + De^{-\sqrt{-\lambda}\theta i^2}$ becomes via Euler's Formula $$\sin^2 \theta = (C + D)\cos(\sqrt{-\lambda}\theta i) + (Ci - Di)\sin(\sqrt{-\lambda}\theta i)$$ How can I solve either of the blue equations for $\lambda$?

This is a Dirichlet problem in the unitary disk. Did you try the Poisson Integral Formula?

 

[LCKurtz]

After looking at this a bit more, I believe your separation of variables should give a second order T equation. I also think you would do well to look at this link:

https://math.okstate.edu/people/binegar/4263/4263-l15.pdf

 

[mario99]

I am trying to solve the PDE $$\bigtriangleup u = 0,\ u(1,\ \theta) = \sin^2 \theta$$

This is the most basic and fun Dirichlet problem for a circle.

$\displaystyle \Delta u = \nabla^2 u = \frac{\partial^2 u}{\partial r^2} + \frac{1}{r}\frac{\partial u}{\partial r} + \frac{1}{r^2}\frac{\partial^2 u}{\partial \theta^2} = 0$

$u(1,\theta) = \sin^2\theta$

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

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

where,

$\displaystyle A_0 = \frac{1}{2\pi}\int_{0}^{2\pi}\sin^2 \theta \ d\theta$


$\displaystyle A_n = \frac{1}{\pi}\int_{0}^{2\pi}\sin^2 \theta \cos n\theta \ d\theta$


$\displaystyle B_n = \frac{1}{\pi}\int_{0}^{2\pi}\sin^2 \theta \sin n\theta \ d\theta$


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