Consider the Dirichlet problem in two dimensions on the domain $\Omega = \mathbb{R}\times \mathbb{R}_{+}$ where $\mathbb{R}_{+}=\{y\in \mathbb{R}:y>0\},$ \begin{align} \Delta u &= 0 \text{ at } \Omega\\ u_y(x,y)&=0 \text{ on } \partial\Omega=\mathbb{R}\times \{y=0\}. \end{align} I would like to find bounded solutions to this problem using the separation of variables. Here is my attempt.
We first write the above equation in polar coordinates as follows, \begin{align} u_{rr} + \frac{u_r}{r} + \frac{u_{\theta\theta}}{r^2} &= 0 \text{ for }(r,\theta)\in \Gamma= \mathbb{R}_{+}\times [0,\pi]\\ u_\theta &=0 \text{ when } \theta =0,\pi. \end{align} where $u_{r}$ means the partial derivative wrt to the variable $r$ and so on. Writing $u(r,\theta)=A(r)B(\theta)$ we get the following system, \begin{align} B_{\theta \theta} &=-\lambda B\\ A_{rr} + \frac{A_r}{r} -\lambda \frac{A}{r^2} &= 0 \end{align} The solution to the first equation is $B(\theta) = C_1\sin(\sqrt{\lambda} \theta) + C_2\cos(\sqrt{\lambda}\theta).$ With the constraint that $B_\theta =0$ at $\theta=0$ we get that $C_1=0$ and when $\theta=\pi$ we get that $C_2\cos(\sqrt{\lambda}\pi)=0$ which implies $\sqrt{\lambda} \pi= (2n+1)\frac{\pi}{2}$ which implies $\lambda = \frac{(2n+1)^2}{4}.$
Thus we need to solve $$A_{rr} + \frac{A_r}{r} -\frac{(2n+1)^2}{4} \frac{A}{r^2}$$ whose solutions are of the form, $$A(r) = c_1 r^{(2n+1)/2} + c_2 r^{-(2n+1)/2}.$$ For solutions to be bounded as $r\to +\infty$ we set $c_1=0$ and thus the general solution to the above problem should be $$u(r,\theta) =\sum_{n\in \mathbb{Z}}c_n r^{-(2n+1)/2}\cos\left(\frac{2n+1}{2}\theta\right).$$
Is this computation correct, or am I missing something?