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 Homework Statement:
 Prove that the Green's function satisfying (1) subject to the boundary conditions (2) can be expressed in the form of (3).
 Relevant Equations:

\begin{equation}
\partial^2_x G + \partial^2_y G = \delta(xx0) \delta(yy0)
\end{equation}
\begin{equation}
G(0,y) = G(x,0) = G(x,b), G(x,y)< \infty
\end{equation}
\begin{gather}
y<y_0: G = \frac{2}{\pi} \int_0^{\infty}
\frac{sin (k x_0) sin (kx) sinh(ky) sinh (k(by_0))}{k sinh (kb)} \\
y>y_0: G = \frac{2}{\pi} \int_0^{\infty}
\frac{sin (k x_0) sin (kx) sinh(ky_0) sinh (k(by))}{k sinh (kb)}
\end{gather}
In order to obtain equation (3), I think I have to do the Fourier transform in the x direction:
\begin{equation}
\tilde{G}(k,y,x_0,y_0) = \int_{ \infty}^{\infty} G(x,y,x_0,y_0) e^{i k x} dx
\end{equation}
So I have:
\begin{equation}
k ^2 \tilde{G}(k,y,x_0,y_0) + \frac{\partial^2 \tilde{G}(k,y,x_0,y_0}{\partial y^2} = e^{i k x_0} \delta (yy_0)
\end{equation}
And then solve this onedimensional equation for ##\tilde{G}##. I'm not sure doing this is correct, because the variable ##y## is delimited and ##x## is not.
In case this is right, do I have to demonstrate first any property? And finally, would it be correct to solve the 1D equation with the Wronskian method?
Thanks in advanced
\begin{equation}
\tilde{G}(k,y,x_0,y_0) = \int_{ \infty}^{\infty} G(x,y,x_0,y_0) e^{i k x} dx
\end{equation}
So I have:
\begin{equation}
k ^2 \tilde{G}(k,y,x_0,y_0) + \frac{\partial^2 \tilde{G}(k,y,x_0,y_0}{\partial y^2} = e^{i k x_0} \delta (yy_0)
\end{equation}
And then solve this onedimensional equation for ##\tilde{G}##. I'm not sure doing this is correct, because the variable ##y## is delimited and ##x## is not.
In case this is right, do I have to demonstrate first any property? And finally, would it be correct to solve the 1D equation with the Wronskian method?
Thanks in advanced
Last edited: