How to solve Laplace's equation in an infinite domain? Consider Laplace's equation,
$$\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2}=0,$$
with boundary conditions
$$u(0,y)=u(L,y)=0,$$
$$u(x,0)=u_0(x),$$
and $u$ must remain finite $\forall y \ge 0$. We can solve this by assuming $u$ is of the form
$$u(x,y) = \sum_{n=1}^\infty A_n\sin\left(k_n x\right)\exp(-k_ny),$$
where
$$k_n = \frac{n\pi}{L}.$$
My question is how do we solve for the case where the boundary conditions are
$$u(0,y)=f(y),$$
and
$$u(L,y)=g(y).$$
Is the solution unique, if so, how do you calculate it?
 A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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*

*Lets $\ds{\Phi\pars{x,y} \equiv \on{f}\pars{y} + \bracks{\on{g}\pars{y} - \on{f}\pars{y}}{x \over L} - \on{u}\pars{x,y}}$.

*$\ds{\Phi\pars{x,y}}$ satisfies $\ds{\Phi\pars{0,y} = \Phi\pars{L,y} = 0}$.

*In addition,
\begin{align}
&\Phi_{xx}\pars{x,y} + \Phi_{yy}\pars{x,y} =
-\on{u}_{xx}\pars{x,y}
\\[2mm] & +\ 
\underbrace{\on{f}''\pars{y} + \bracks{\on{g}''\pars{y} - \on{f}''\pars{y}}{x \over L}}
_{\ds{\equiv \varphi\pars{x,y}}}\ -\ \on{u}_{yy}\pars{x,y}
\\[5mm] &\ \implies
\bbx{\Phi_{xx}\pars{x,y} + \Phi_{yy}\pars{x,y} =
\varphi\pars{x,y}}\label{1}\tag{1} \\ &
\end{align}

*Lets $\ds{\Phi\pars{x,y} \equiv
\sum_{n = 1}^{\infty}a_{n}\pars{y}\sin\pars{k_{n}x}\quad}$ with
$\ds{\quad k_{n} \equiv n\,{\pi \over L}}$.

Then, with (\ref{1}),
\begin{align}
&\sum_{n = 1}^{\infty}\bracks{a_{n}''\pars{y} - k_{n}^{2}a_{n}\pars{y}}\sin\pars{k_{n}x} = \varphi\pars{x,y}
\end{align}
Multiply both members by $\ds{2\sin\pars{k_{n}x}/L}$ and integrate over $\ds{x \in \pars{0,L}}$:
\begin{align}
&a_{n}''\pars{y} - k_{n}^{2}\,a_{n}\pars{y} =
{4\bracks{n\ odd} \over n\pi}\on{f}''\pars{y} -
{2\pars{-1}^{n} \over n\pi}
\bracks{\on{g}''\pars{y} - \on{f}''\pars{y}}
\end{align}
$$
\bbx{\mbox{Now, you have a simple equation to solve}} \\
$$
