How to solve Laplace's equation in an infinite domain?

Question

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?

Answer 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} $$ \bbx{\mbox{Now, you have a simple equation to solve}} \\ $$