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I'm solving Laplace's equation ($\triangledown^2 \phi = 0$) on a semi-infinite domain ($ - \infty < x < \infty$ and $ 0 < y < \infty$) by taking a Fourier Transform in x.

So far, I have found a general solution for the FT in x ($Ae^{\omega y} + Be^{-\omega y}$) and am now using information given to find the constants.

The question setter at this point says that since we require $\phi$ to be bounded as x → ±∞, the same is required of it's FT as ω → ±∞.

And an answerer of a similar question (found here) says that the solution must die away as $y \rightarrow \infty$.

My question(s): why is $\phi$ bounded? This was never mentioned in the question, is it a general Laplace's eqn condition?

and

does this mean the solution tends to a constant at the limit or just that it doesn't tend to $+$ or $-$ $\infty$? I thought the FT was for a fixed y, so are we really interested at all in what the solution is as y approaches its limit?

Thanks

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The key point here is uniqueness. Write $\Omega = \{ (x,y) \in \mathbb{R}^2 : y >0\}$. What we would like in this scenario is to specify the boundary data $f : \partial \Omega \to \mathbb{R}$ and then to solve the problem $$ \begin{cases} \Delta \varphi =0 & \text{in } \Omega \\ \varphi = f & \text{on } \partial \Omega. \end{cases} $$ For this problem, as in most of PDE, we want the problem to be well-posed, meaning that we want (1) existence of a solution, (2) uniqueness of a solution, and (3) we want the solution to depend continuously on the data.

So for the problem at hand how do we make these work? Existence is coming from the Fourier transform tricks. Dependence on the data will come once you've written down a formula for the solution via inverse Fourier transform. What about uniqueness? It turns out that for the stated problem we don't actually get uniqueness without imposing a further condition. Why? Consider the function $\psi : \Omega \to \mathbb{R}$ given by $\psi(x,y) = y$. Then we compute $$ \Delta \psi(x,y) =0 \text{ and } \psi(x,0) =0, $$ which means that if $\varphi$ is any solution to our original problem, then $\varphi + \alpha \psi$ is also a solution for every $\alpha \in \mathbb{R}$. Thus, uniqueness fails in general for this problem, as stated.

To get around this we impose another sort of "generalized boundary condition" by specifying the behavior of $\varphi$ "at infinity." Namely, we impose the condition that $\varphi$ is bounded, which immediately eliminates the solution $\psi$ as a possibility. It turns out that if we impose the boundedness condition then in fact there will exist at most one solution for a given $f$ that is, say, bounded and integrable.

This answers the question of why we would impose the boundedness condition, but it does not answer the question of how boundedness of $\varphi$ translates to the FT. The reasoning you quote is a bit faulty. The real issue that in order to apply the inverse FT you're going to have to integrate: $$ u_1(x,y) = \int_{\mathbb{R}} A(\omega) e^{\omega y} e^{2\pi i \omega x} d\omega \text{ and } u_2(x,y)= \int_{\mathbb{R}} B(\omega) e^{-\omega y} e^{2\pi i \omega x} d\omega $$ are the two general solutions you've found. The issue is that you care about $y>0$, so the $e^{\omega y}$ term will grow exponentially, and you can't guarantee in general that the integral even makes sense. You can, for instance, if you require that $A(\omega)$ is compactly supported in $\omega$, but this is not a condition that will hold for data $f$ that is just bounded and integrable. So you throw that term away and look at the second, which is well behaved since $e^{-\omega y}$ is bounded for $y >0$. You can then solve for $B(\omega)$ in order to make the boundary condition satisfied.

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