# How to find the Green's function

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 $\nabla ^2 G=\delta ( \textbf x - \xi )\pm \delta ( \textbf x - \xi_1 ) \pm \delta ( \textbf x - \xi_2 )\pm \delta ( \textbf x - \xi_3 )$

Can someone please clearly explain how you determine if the plus/minus signs are in fact plus or minus please.

I know that it is to do with the boundary conditions but don't understand how.

## 1 Answer

The correct extension is $$\nabla^2G = \delta(\mathbf{x}-\xi) + \delta(\mathbf{x}-\xi_1) - \delta(\mathbf{x}-\xi_2) - \delta(\mathbf{x}-\xi_3)$$ To see why, interpret $G$ as electric potential. Then, the delta functions correspond to point charges, and we have the conditions that (i) the potential is symmetric across $x=0$, and (ii) the potential is zero along $y=0$. The first condition implies that the sign on the delta function at $\xi_1$ should be the same as the sign on the delta function at $\xi$, while the second condition implies that the sign on delta function at $\xi_2$ should be opposite the sign at $\xi$. Then, since $\xi_3$ is the reflection of $\xi_2$ over $x=0$, the delta function there should have the same sign as $\xi_2$ (alternatively, since it is the reflection of $\xi_1$ over $y=0$, it should have the opposite sign from the delta function at $\xi_1$).

Solving for $G$ yields $$G = \ln\frac{1}{|\mathbf{x} - \xi|} + \ln\frac{1}{|\mathbf{x} - \xi_1|} - \ln\frac{1}{|\mathbf{x} - \xi_2|} - \ln\frac{1}{|\mathbf{x} - \xi_3|}$$ and we can verify that $G$ satisfies the required boundary conditions.

• Does $\xi$ basically mean an arbitrary point anywhere on the xy plane? And does $\xi_1$ mean an arbitrary point anywhere on left quadrants? And $\xi_2$ etc...? – snowman May 7 '16 at 10:58
• Is there an informal way to do this? I still don't get it... – snowman May 7 '16 at 10:59
• Remember that the Green's function is defined such that the solution to $$\Delta u = f$$ is $u(x) = (G(x,\cdots) * f(\cdot))(x)$. In other words, the Greens function tells you how the differential equation responds to an impulse of one unit at the point $\xi$. The points $\xi_1$, $\xi_2$, and $\xi_3$ are obtained by reflecting $\xi$ over the boundary lines $x=0$ and $y=0$. – Strants May 7 '16 at 21:03
• I'm also really not sure what you mean by an 'informal' way to find the Greens function. What do you mean by 'informal'? – Strants May 7 '16 at 21:06
• (1): $\xi_x$ and $\xi_y$ are positive real numbers, but they are not fixed. Remember that for $$\Delta u = f$$ with the given boundary conditions, we have $$u(\mathbf{x}) = \iint_{x>0,y>0} G(\mathbf{x}, \xi) f(\xi)\;d\xi_x d\xi_y$$ (2): Sorry, I meant to write $\xi_2$. (3): Well, you've already noticed that if you reflect over a line with Neumann boundary conditions, you don't change the sign of the delta function. Why would that change when you reflect a second time? – Strants May 11 '16 at 5:55