Any idea to start this problem I got stuck with: $$-\Delta \phi +\beta^2\phi=0,\quad \beta>0 \quad (\beta \,constant)$$ in $\quad D=\{(x,y)\in \Bbb R^2:x\in \Bbb R, \quad y\in [-h,0]\}\quad$ with boundary condition: $$\phi(x,0)=f(x),\quad \phi_y(x,-h)=0\quad \forall x\in \Bbb R$$ where f(x) it's a given function such that $f:\Bbb R\rightarrow \Bbb R$
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This is Helmholtz equation. Look here http://eqworld.ipmnet.ru/en/solutions/lpde/lpde303.pdf – Nikita Evseev May 29 '15 at 03:31
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1Maybe using Fourier transform in $x$ does the trick? – Dmoreno May 29 '15 at 03:32
1 Answers
If $\phi_1$ and $\phi_2$ are solutions of the problem, then the difference is a solution of $-\Delta \phi +\beta^{2}\phi = 0$ and $\phi(x,0)=0 = \phi(x,-h)$. There are non-trivial solutions of this problem that can be obtained using separation of variables. $$ -\frac{X''}{X} +\beta^{2} = \mu = \frac{Y''}{Y},\\ Y(-h) = Y(0) = 0, $$ where $\mu$ is a separation parameter. In this case $\mu = -n^{2}\pi^{2}/h^{2}$ for $n=1,2,3,\cdots$ and $$ Y_{n}(y) = \sin(n\pi y/h). $$ The equation in $X$ becomes $$ X''-(n^{2}\pi^{2}/h^{2}+\beta^{2})X=0, $$ for which the general solution involves arbitrary constants $A_n$, $B_n$: $$ X_{n}(x) = A_{n}\exp\left(\sqrt{n^{2}\pi^{2}/h^{2}+\beta^{2}}x\right)+B_{n}\exp\left(-\sqrt{n^{2}\pi^{2}/h^{2}+\beta^{2}}x\right) $$ So your problem does not have unique solutions because any solution of your problem remains a solution if you add linear combinations of $X_{n}(x)Y_{n}(x)$ as described above.
As someone mentioned in a comment, you can Fourier transform in $x$ assuming, for example, that $\phi(x,y)$ is square integrable in $x$ for fixed $y \in [-h,0]$. Let $\psi(s,y)$ denote this Fourier transform. Then $$ s^{2}\psi(s,y)-\psi_{yy}(s,y)+\beta^{2}\psi(s,y) = 0,\\ \psi(s,y) = A(s)\sinh(\sqrt{s^{2}+\beta^{2}}(y+h)). $$ The form shown is convenient in order to force $\phi(x,-h)=0$. Then $$ \phi(x,y) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}A(s)\sinh(\sqrt{s^{2}+\beta^{2}}(y+h))e^{isx}ds. $$ And $\phi(x,0)=f(x)$ determines $A(s)$ through $$ f(x) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}A(s)\sinh(\sqrt{s^{2}+\beta^{2}}h)e^{isx}ds \\ \implies A(s) = \frac{\hat{f}(s)}{\sinh(\sqrt{s^{2}+\beta^{2}}h)} $$ So a particular solution is $$ \phi(x,y) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} \frac{\sinh(\sqrt{s^{2}+\beta^{2}}(y+h))}{\sinh(\sqrt{s^{2}+\beta^{2}}h)}\hat{f}(s)e^{isx}ds. $$ However, as mentioned before, this solution is not unique without further conditions.
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