# Why Method of Images solves Laplace's Equation

We have been studying Laplace's and Poisson's equation in maths, and have derived fundamental Green's functions, eg. in 3D $$G(\vec{r},\vec{r_0}) = \frac{-1}{4\pi |\vec{r} -\vec{r_0}|}$$ When we have Poisson's equation for a point source, eg $\nabla^2\Phi = \delta(\vec{r}-\vec{r_0})$ we can swiftly add images charges on the form above, outside the region of concern to match boundary conditions. This far everything is okay,

However, it confuses me strongly when we start to use the same method and Green's functions for solutions of Laplace's equation $\nabla^2\Phi = 0$. In problems I have run into, the Green's function is found for these regions by supposing there is a point source $\delta(\vec{r} -\vec{r_0})$ inside the regios and then building up image charges in the same way.

How is this okay? Doens't that mean that we are actually solving Possion's equation and not Laplace's as required?

Edit: Here is an "example" from my notes

Example 3D half-space. What is the Green function for a domain D with Dirichlet boundary conditions, where D is the half-space of R3 with z > 0?$$.$$ Solution: Imagine there is a charge at $\vec{r_1} = (x,y,z)$ in D. Uniqueness of solutions allows us to solve this using a trick: remove the boundary at z = 0, consider all of space and add a point source of opposite sign, an `image source', at the image point $\vec{r_2} = (x,y,-z)$. The fundamental solution then gives $$G(\vec{r},\vec{r_1}) = \frac{-1}{4\pi |\vec{r} -\vec{r_1}|} +\frac{1}{4\pi |\vec{r} -\vec{r_2}|}$$

Note that this "example" doesn't mention anything about whether it is Poisson's or Laplace's equation we are solving. It just seems to associate a Green's function with the space, regardless of the charge distribution. 10 pages ahead in my notes there is a second example:

Example Laplace What is the solution of Laplace's equation in the 3D half-space with z > 0 subject to $\Phi$ = f(x; y) on z = 0? $$.$$ Solution: We derived the required Green function earlier using the method of images: $$G(\vec{r},\vec{r_1}) = \frac{-1}{4\pi |\vec{r} -\vec{r_1}|} +\frac{1}{4\pi |\vec{r} -\vec{r_2}|}$$ Hence [...]

The rest of the question is okay, but as you can see the Green's function used for the Laplace equation is the same as derived for when there is a charge inside the relevant domain D. How is this possible?

• Can you give a specific example of such a problem and this technique being used on it? – Chappers Apr 19 '18 at 12:14
• @Chappers I have now included an example from my notes to clarify what my confusion is about. – Jhonny Apr 21 '18 at 10:30