Green's functions (Fourier Series)

  • #1
1
0
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(x-x0) \delta(y-y0)
\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(b-y_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(b-y))}{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 (y-y_0)
\end{equation}

And then solve this one-dimensional 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:

Answers and Replies

  • #2
175
39
Try properties of Green's function. First consider case ##y<y_0##. In this case R.H.S is 0. Ditto for ##y>y_0##. Now Apply boundary conditions,fact that they are continuous at y=##y_0## and the first partial derivative is discontinuous at y=##y_0## with discontinuity equal to ##e^{ik(x-x_0)}## if you use (6). Alternatively,you can use eigenfunction expansion of Green's Function (See Arfken and Weber,Mathematical methods for physicist; Morse and Feshback ,Methods of Theoretical physics(The later is detailed but notations are bit outdated))
 

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