# Find Green's function of quarter-plane with method of images

Consider a domain $$D= \{(x,y) : x>0, y>0 \}$$ Let $\textbf x = (x,y)$ and $\xi =(\xi_x , \xi_y)$. Find the Green's function, $G(\textbf x , \xi)$ such that $$\nabla ^2 G=\delta (\textbf x - \xi), \, \, \, \, \, \, \textbf x \in D$$ subject to $$\frac{\partial G}{\partial x}(0,y,\xi)=0, \, \, \, \, \, \text{for} \, \, \, \, y>0$$ and $$G(x,0,\xi)=0, \, \, \, \, \, \text{for} \, \, \, \, x>0$$

Using method of images:

We have the source $\xi=(\xi_{x},\xi_{y})$ and images sources:

$\xi_1=(-\xi_{x},\xi_{y})$, $\xi_2=(\xi_{x},-\xi_{y})$, $\xi_3=(-\xi_{x},-\xi_{y})$

So we have $\nabla ^2 G=\delta ( \textbf x - \xi )-\delta ( \textbf x - \xi_1 )-\delta ( \textbf x - \xi_2 )+\delta ( \textbf x - \xi_3 )$

So $G(\textbf x , \xi)=\frac1{2 \pi} \ln |\textbf x - \xi |-\frac1{2 \pi} \ln |\textbf x - \xi_1 |-\frac1{2 \pi} \ln |\textbf x - \xi_2 |+\frac1{2 \pi} \ln |\textbf x - \xi_3 |$

The differentiation of that w.r.t. $x$, I think is (checked twice): $$\frac{\partial G}{\partial x}=\frac{1}{2 \pi} \frac{x - \xi_x - x \xi_{xx} +\xi_{xx} - y \xi_{yx} + \xi_{yx}}{(x-\xi_x)^2+(y-\xi_y)^2} -\frac{1}{2 \pi} \frac{x + \xi_x + x \xi_{xx} +\xi_{xx} - y \xi_{yx} + \xi_{yx}}{(x+\xi_x)^2+(y-\xi_y)^2} -\frac{1}{2 \pi} \frac{x - \xi_x - x \xi_{xx} +\xi_{xx} + y \xi_{yx} + \xi_{yx}}{(x-\xi_x)^2+(y+\xi_y)^2} +\frac{1}{2 \pi} \frac{x + \xi_x + x \xi_{xx} +\xi_{xx} + y \xi_{yx} + \xi_{yx}}{(x+\xi_x)^2+(y+\xi_y)^2}$$

I find that the second boundary condition holds but the first one doesn't (after differentiating the above).

If we sub in $x=0$ in the above expression, I end up with two terms with different denominators and positive $\xi_x$ on both the numerators.

When $x=0$, does that mean $\xi_x=0$ too? Because if it doesn't, I cant see how this boundary condition can hold.

The point $\xi$ should be considered fixed in this context: we plant it somewhere and then look for Green's function, considered as a function of $x$ only. So, no such things as $\xi_x$ are needed in the calculations.

Also, the signs you chose are incorrect. You need the symmetry properties $$G((-a,b),(c,d)) = G((a,b),(c,d))\tag1$$ $$G((a,b),(-c,d)) = -G((a,b),(c,d))\tag2$$ Even reflection gives zero normal derivative, odd reflection gives zero value.

So, the formula is

$$G(\textbf x , \xi)=\frac1{2 \pi} \ln |\textbf x - \xi | + \frac1{2 \pi} \ln |\textbf x - \xi_1 |-\frac1{2 \pi} \ln |\textbf x - \xi_2 | -\frac1{2 \pi} \ln |\textbf x - \xi_3 |$$

which can be expanded in coordinates as

$$G((a,b),(c,d))=\frac1{ 4\pi}\left( \ln ((a-c)^2+(b-d)^2) + \ln ((a+c)^2+(b-d)^2) \\ - \ln((a-c)^2+(b+d)^2) - \ln ((a+c)^2+(b+d)^2)\right)$$ This function satisfies (1) and (2), and therefore satisfies the required boundary conditions.

and $$G((a,b),(c,d)) = -G((a,b),(c,d))$$

• @4o4 you said no such things as $\xi_x$ are needed in calculations but our lecturer seemed to have used them as how you have used $c$ and $d$. Mar 13, 2016 at 19:07
• If $\xi$ is a fixed point like you say, does that mean it is a constant. So when it is differentiated wrt $x$ or $y$, it will be zero? Mar 13, 2016 at 19:13
• Yes, yes, and yes.
– user147263
Mar 13, 2016 at 19:18
• Hi, how do you know which image sources to plus or minus when making the green formula? I don't understand. Please help. May 6, 2016 at 13:55
• @sandwich Are you available to response the question : math.stackexchange.com/questions/1785466/…? You seem clever to solve this sort of question.
– user332681
May 14, 2016 at 22:11