Find the solution of the problem $$u_{xx}(x,y)+u_{yy}(x,y)=0, x^2+y^2>1 \\u=1+3\sin^3 \theta , 0 \leq \theta <2\pi$$ $u$ is bounded.
I have done the following:
$$u(x,y)=v(\rho, \theta) \\ x=\rho \cos \theta \\ y=\rho \sin \theta$$
$$\Rightarrow v_{\rho \rho}(\rho , \theta)+\frac{1}{\rho}v_{\rho}(\rho, \theta)+\frac{1}{\rho^2}v_{\theta \theta}(\rho, \theta)=0, \rho>1 , 0\leq \theta <2\pi \\ v(1, \theta )=1+3 \sin^3 \theta$$
We are looking for the solution of $$v_{\rho \rho}+\frac{1}{\rho}v_{\rho}+\frac{1}{\rho^2}v_{\theta \theta}=0$$ of th form $$v(\rho, \theta )=R(\rho)\Theta(\theta)$$
We have the following two problems: $$\left.\begin{matrix} \Theta ''(\theta) +\lambda\Theta(\theta)=0, 0 \leq \theta <2 \pi\\ \Theta(0)=\Theta(2 \pi)\\ \Theta '(0)=\Theta '(2 \pi) \end{matrix}\right\}(1)$$ $$\left.\begin{matrix} \rho^2 R''(\rho)+\rho R'(\rho)-\lambda R(\rho)=0 \end{matrix}\right\}(2)$$
Solving the problem $(1)$ we get that the eigenvalues are $\lambda_0=0, \lambda_k=k^2, k \in \mathbb{N}$ and the corresponding eigenfunctions are $\Theta_0 (\theta)=1, \Theta_k (\theta)=\cos (k \theta), \Theta_k (\theta) =\sin (k \theta)$.
For the problem $(2)$ we have the following:
For $\lambda_0=0$ we have $R(\rho)=c_1 \ln \rho +c_2 , \rho>1$. So that the solution is bounded we choose $c_1=0$. So, we have $R_0(\rho)=1$.
For $\lambda_k=k^2$ we have $\rho^2 R''(\rho )+\rho R'(\rho )-k^2 R(\rho )=0$ (Euler).
$R(\rho )=\rho^m$
$\Rightarrow \rho^2 m (m-1) \rho^{m-2}+\rho m \rho^m-k^2 \rho^m=0 \Rightarrow \dots \Rightarrow m^2-k^2=0 \Rightarrow m=\pm k$
Genereal solution: $R_k(\rho )=c_1 \rho^k+c_2 \rho^{-k}, k \in \mathbb{N}, \rho >1$.
So that the soution is bounded we choose $c_1=0$. S, we have $R_k(\rho )=\rho^{-k}$.
So, we have the following $$v(\rho , \theta )=\frac{a_0}{2}+\sum_{k=1}^{\infty}\rho^{-k}(a_k \cos (k \theta)+b_n \sin (k \theta) \\ v(1, \theta )=1+3\sin^3 \theta$$
$$1+3\sin^3 \theta=\frac{a_0}{2}+\sum_{k=1}^{\infty}(a_k \cos (k \theta)+b_n \sin (k \theta))$$
$$a_k=\frac{1}{\pi} \int_0^{2 \pi}(1+3 \sin^3 \theta)\cos (k \theta)d \theta , k=0,1,2 \dots \\ b_k=\frac{1}{\pi}\int_0^{2\pi}(1+3 \sin^3 \theta)\sin (k \theta)d \theta , k=1,2, \dots $$
After calculations I found that $$v(\rho, \theta)=1+\frac{9}{4 \rho}\sin \theta-\frac{3}{4 \rho^3}\sin ( 3\theta)$$
Is this correct??
Do I have to transform it into $u(x, y)$ ?? How could I do that??
Yes, your solution is correct.
You do not have to (unless someone asks for it) transform it back to $u(x,y)$, but here is how it is done:
Since $x=\rho\cos\theta$ and $y=\rho\sin \theta$ (and $\rho=\sqrt{x^2+y^2}$), we find that $$ \sin \theta=y/\sqrt{x^2+y^2}. $$ Moreover, $\sin 3\theta=3\cos^2\theta\sin\theta-\sin^3\theta$, so $$ \sin 3\theta=\cdots=\frac{3x^2y-y^3}{(x^2+y^2)^{3/2}}. $$ I'm sure you can now put the pieces together, to write your solution $u(x,y)$.