# Method of Images - Green's Function

• October 19th 2010, 04:32 PM
OliviaB
Method of Images - Green's Function
Use the method of images to find a Green's function for the problem

Attachment 19388

Demonstrate the functions satisfies the homogenous boundary condition.
• October 27th 2010, 06:24 AM
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)
• October 27th 2010, 04:17 PM
zzzoak
$
\displaystyle { G=\frac{1}{2 \pi} \; ( \; ln(\sqrt{(x-x_0)^2+(y-y_0)^2}+ln(\sqrt{(x-x_0)^2+(y+y_0)^2} \; )
}
$

$
\displaystyle { \frac{ \partial G}{\partial y}=\frac{1}{4 \pi} \; ( \; \frac{ 2(y-y_0)}{(x-x_0)^2+(y-y_0)^2}+\frac{ 2(y+y_0)}{(x-x_0)^2+(y+y_0)^2 } \; )
}
$

$
\displaystyle { \frac{ \partial G}{\partial y}|_{y=0}=0
}
$
• October 27th 2010, 06:15 PM
OliviaB
Is this the method of images though?
• October 28th 2010, 09:54 AM
zzzoak