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There is an equation for Green function of a half-plane: $$ \Delta G = \delta (x - x_{0})\delta(y - y_{0}), \quad G(x, 0) = 0.$$ Of course, the Green function which satisfies the boundary condition is represented as $$ G(x, y, x_{0}, y_{0}) = G_{0}(x, y, x_{0}, y_{0}) - G_{0}(x, y, x_{0}, -y_{0}), $$ where $G_{0}(x, y, x_{0}, y_{0})$ is the Green function for a plane, so I need to find a Green function for a plane.

After using Fourier transform (and setting $x_{0}, y_{0}$ to zero) I got $$-(k_{x}^{2} + k_{y}^2)\tilde {G_{0}} = \frac{1}{2 \pi} \Rightarrow \tilde G_{0} = -\frac{1}{2 \pi}\frac{1}{k_{x}^{2} + k_{y}^{2}}.$$

But after that I got divergent integral. Did I make a mistake?

Kirill
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John Taylor
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1 Answers1

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You didn't make a mistake, but it does mean that the Green function doesn't have a nice Fourier transform (else the integral would converge).

You can also try, for instance, to look for spherically symmetric Green functions $G=G(r)$, which will give you an easily-solved ODE for $G(r)$ (but the point $r=0$ will need some special attention).

Another thing is that if you get an integral somewhere of the form $$ I(r)=\int_{-\infty}^{\infty} \frac{e^{-|k|r}}{|k|}\,dk, $$ you can (not rigorously at all) differentiate it w.r.t. $r$, integrate the result, then integrate back. It depends on your tolerance towards such cheap tricks. Even if you get your answer using a cheap trick, you can in the end just verify that it does, in fact, solve the equation.

Kirill
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  • Maybe there will be better if I solve the equation in polar coordinates. So, for for an area that does not contain a singular point, $$\frac{1}{r}\partial_{r}(r\partial_{r}G) = 0 \Rightarrow G = Cln(r) + C_{1}.$$ But I can't determine the constants. Constants that can be virtually any. – John Taylor May 25 '13 at 20:57
  • Well, it's true that $C_1$ can be anything (because $\Delta C_1=0$), but $C$ can't be just anything because the inhomogeneous term is $\delta(x-x_0)\delta(y-y_0)$, and $\Delta \log r\not\equiv0$. Try integrating the lhs and the rhs over a small disk of infinitesimal radius around the origin and use Green's theorem. – Kirill May 25 '13 at 21:16
  • Maybe, $$ int \delta (r)dS = \int \Delta u dS = \int (\nabla u , \mathbf n )dl = \int \partial_{r}u dl = \int \limits_{0}^{2 \pi} \frac{A}{R}Rd\varphi = 2\pi A = 1 \Rightarrow A = \frac{1}{2 \pi}, $$ where I integrated over a disc with radius R. But A must be $-\frac{1}{2 \pi}$, if I didn't make the mistake. – John Taylor May 25 '13 at 22:04
  • Um, $\frac{1}{2\pi}$ seems correct. Why $-\frac{1}{2\pi}$? – Kirill May 25 '13 at 22:24
  • I read about that in a text by the some reference. – John Taylor May 25 '13 at 22:28
  • Wikipedia says $1/2\pi$, for example: http://en.wikipedia.org/wiki/Green%27s_function Are you sure? Could be a typo, for instance. – Kirill May 25 '13 at 22:35