Solving the Dirichlet problem$\nabla^2u=0$, $u|_{\partial \Omega}=x$ where $\Omega=\{(x,y)|x^2+y^2<2\}$

Question

Attempt: (Seperation of variables in polar coordinates)

Assume $$u=\Theta(\theta)R(r)$$

$$\nabla^2=u_{rr}+1/r .u_r+1/r^2.u_{\theta\theta}$$

so

$$0=\nabla^2u=\Theta(\theta)\dfrac{\partial^2R}{\partial r^2}+\dfrac1r\Theta(\theta)\dfrac{\partial R}{\partial r}+\dfrac{R(r)}{r^2}\dfrac{\partial^2 \Theta}{\partial\theta^2}$$

so $$0=\dfrac{r^2}{R}\dfrac{\partial^2R}{\partial r^2}+\dfrac r R\dfrac{\partial R}{\partial r}+\dfrac{1}{\Theta}\dfrac{\partial^2 \Theta}{\partial\theta^2}$$

$Q_1$:Assuming we want $\Theta$ to have sinusoidal solution so $$\dfrac{1}{\Theta}\dfrac{\partial^2 \Theta}{\partial\theta^2}=-l^2,l\in \mathbb R$$ i.e.$$\Theta''+l^2\Theta=0\Rightarrow \Theta(\theta)=A\cos(l\theta)+B\sin(l\theta)$$

$$\dfrac{r^2}{R}\dfrac{\partial^2R}{\partial r^2}+\dfrac r R\dfrac{\partial R}{\partial r}=l^2$$

so $$r^2R''+rR'-l^2R=0\Rightarrow R(r)=Cr^{-l}+Dr^l$$

Now $u|_{\partial \Omega}=x$ where $\Omega=\{(x,y)|x^2+y^2<2\}$ means $$u|_{\partial \Omega}=r\cos\theta=\left(A\cos(l\theta)+B\sin(l\theta)\right)\left(Cr^{-l}+Dr^l\right)$$ means $B=0$ and $l=1$ so $C=0$

so we have:

$$u=K\cos(l\theta)r^l$$

But I am pretty sure I have wrong answer 1st because of assumption that I made in $Q_1$ point and why we do it in general and I cannot play with boundary conditions very well.

Answer 1

Your solution is correct (correctly written down as $u=r\cos\theta$). But for this specific example, you don't need to do a lot of work. Any linear polynomial is harmonic, so since we require $u|_{\partial \Omega} = x$, a good guess is $u(x)=x$ inside $\Omega$ as well. If $v$ was any other solution, then $w:=u-v$ solves $$ \Delta w = 0 \text{ in } \Omega,\quad w|_{\partial \Omega} =0. $$ By the maximum principle for harmonic functions, $w=0$. Thus, this is the unique solution.

Answer 2

When you solve the Laplace equation over some disc of radius $L$, the solution is of the form (this is a theorem)

$$u(r,\theta) = a_0/2 + \sum_{n=1}^\infty (a_n cos(n\theta) + b_n sin(theta))\mathbb{R}^n,$$ where $a_n = \frac{ 1}{L^n \pi } \int_{-\pi}^\pi h(\theta) cos(n\theta) d\theta$, and $b_n = \frac{ 1}{L^n \pi } \int_{-\pi}^\pi h(\theta) sin(n\theta) d\theta$, and $h(\theta)$ is the term in the boundary condition, in particular $h(\theta) = 2 cos(\theta)$ in this case.

If you do the computation, you can see that $$u(r,\theta) = r cos(\theta).$$

This is basically because $a_n, b_n$ are the Fourier coefficients of $h(\theta)$, and it is known that those coefficients are unique.