# Method of Images - Green's Function

1. Oct 19, 2010

### OliviaB

Use the method of images to find a Green's function for the problem in the attached image.

Demonstrate the functions satisfies the homogenous boundary condition.

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• ###### Chapter Review Q17.bmp
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2. Oct 27, 2010

### OliviaB

It's taken me ages, and I really am struggling to understand this, so Im not sure if this is correct but here goes....

$$\bigtriangledown^2 G = \delta(\underline{x} - \underline{x}_0)$$ $$\frac{\partial G(x,0)}{\partial y} = 0$$ for $$y \geq 0 \ - \infty < x < \infty$$

Let $$r = |\underline{x} - \underline{x}_0 | = \sqrt{(x - x_0)^2 + (y - y_0)^2}$$

be the distance between $$\underline{x}$$ and $$\underline{x}_0$$

Then $$\bigtriangledown^2 G = \delta(\underline{x} - \underline{x}_0)$$ becomes

$$\bigtriangledown^2 G = \frac{1}{r} \frac{\partial}{\partial r} \Big( r \frac{\partial}{\partial r} \Big) = 0$$

everywhere (although not at $$r = 0$$) and subject to

$$\iint\limits_{\infty} \bigtriangledown G dV = \iint\limits_{\infty} \delta (0) dV = 1$$

which gives

$$G(r) = A \ln r + B$$

$$A = \frac{1}{2 \pi}$$

Choosing $$B = 0$$

$$G = \frac{1}{2 \pi} \ln r = \frac{1}{2 \pi} \ln |\underline{x} - \underline{x}_0|$$

I don't think this is the method of images though......(Headbang)