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I am stuck on a problem with the method of images.

The formulation is rather simple; Solve the for green's function given by $\nabla^2 G = \delta( \underline{x} - \underline{x}_0)$ in the wedge enclosed by the lines $\theta = 0$ and $\theta = \frac{\pi}{3}$ in the first quadrant. (So, $\underline{x} \in W, W \equiv \{ (r, \theta): 0 < \theta < \frac{\pi}{3}\}$).

Initially I thought the solution would be to place an image point $\underline{x}_1$ to cancel for the upper boundary, and $\underline{x}_2$ to cancel for the lower boundary. Of course, $\underline{x}_1$ would then get an image point $\underline{x}_4$ in the lower boundary, likewise $\underline{x}_2$ will then get an image point $\underline{x}_3$ in the upper boundary, and so forth.

Since a point $(r, \theta)$ has image point $(r, \frac{2\pi}{3})$ in the upper boundary, and $(r, -\theta)$ in the lower boundary, all the image points lie on a sphere of radius $r$. Sadly, this causes the image points in the lower boundary to be below $\theta=0$ at some point, and visa versa; I got extremely confused there. At some point, the image points become $\in W$, at which point the solution no longer satisfies the DE for G.

I have tried looking at singular points, but did not find a single $x^\star$ that satisfied the DE and the BC. Is the solution different? Is it perhaps a square?

A hint or solution would be marvellous.

Thanks in advance, Daimonie

Daimonie
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    When one says "solve for the Green's function", they mean a Green's function for a differential operator, meaning not only a differential equation and domain, but also boundary conditions. What are the boundary conditions for some function $u(x)$ such that $\nabla^2 u = f(x)\quad \forall x \in W$? – Ron Gordon Nov 23 '14 at 13:02
  • My bad, I thought I had typed it. $u(\underline{x}) = 0$ on the boundary of $W$, so on both $\theta=0$ and $\theta = \frac{pi}{3}$. – Daimonie Nov 23 '14 at 22:31

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I'm just going through my previous questions and resolving them.

The answer to this one turned out to be a grand total of 360/60-1 points. You just put +- charges to settle the $u=0$ boundary, and at some point they either cancel (one b.c. wants a +, the other a -; total = 0) or they are the same (both require a + at the same place). This is one of the latter; they require the same image point. Tricky to see, as my bookkeeping-method should've included a $0 < \theta < 2 \pi$; I hadn't realised that :)

Daimonie
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