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The problem is $$\Delta u = 0$$ in the semi infinite strip $S = \{(x,y)|0<x<1,y>0\}$ subject to the boundary conditions $$ \begin{cases} u(0,y) = 0,\ y\ge0\\ u(1,y) = 0,\ y\ge0\\ u(x,0) = f(x),\ 0\le x \le 1 \end{cases} $$ where $f$ is a given function with $f(0) = f(1) = 0$. Write $$f(x) = \sum_{n=1}^{\infty} a_nsin(n\pi x)$$ and expand the general solution in terms of the special solutions given by $u_n(x,y) = e^{-n \pi y}sin(n \pi x)$.

The above is the problem.

So by using separation of variables and analyzing the first two boundary conditions, I got $u(x,y) = \sum_{n=1}^{\infty} sin(n\pi x)(C_n e^{-n \pi y}+D_n e^{n \pi y})$, where $C_n$ and $D_n$ are constants. My question is if I could get rid of the $e^{n \pi y}$ part from the general solution. I've seen someone did so but usually the boundary conditions would include for example $\lim_{y\to\infty} u(x,y)$ is finite. But here the boundary conditions do not say something like this. Can I still get rid of that part based on the give information? I'm a bit confused since the later instructions of the problem seem to suggest only $e^{-n \pi y}$ part remains.

Thank you so much for any insights.

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    $\begingroup$ Unfortunately you need two conditions to determine two constants. The solution may be implied to be bounded at $y \to \infty$ which would give $D_n=0$ $\endgroup$ – Dylan Feb 3 at 9:00
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Yes, there is one condition missing, as your semi-infinite strip has, very simply put, four boundary segments, at $x \in \{0,1\}$ and at $y \in \{0,\infty\}$. But you have prescribed conditions on only three of the boundary segments so far, which leaves you with arbitrary constants in the solution (you obtain $C_n + D_n = a_n$ where only $a_n$ is known).

The boundedness of the solution as $y \rightarrow \infty$ usually follows from some physical considerations.

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