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I'm attempting to find a Green's function for the Laplacian. I was looking at examples for finding the Green's function for the 2D Laplacian on the unit disk and the upper half plane. In both examples when finding the Green's function they find an image point $x_0^*$ that is outside the domain of definition for the Laplacian and use it to find the Green's function.

My question is why must this image point be found for finding the Laplacian and why can't it be taken as a point inside the domain of definition? Thanks.

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  • $\begingroup$ Can i answer using its application to electrostatics? $\endgroup$ – Lelouch Apr 6 '17 at 15:44
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The laplace equation is a special form of the poisson equation:

$\nabla ^2 V = -\frac{\rho}{\epsilon_0}$

for charge free space($\rho =0$) we have the laplace equation $\nabla^2 V = 0$. Now suppose the laplace equation holds for some space having boundary $S$ (this may also be at infinity). We can find unique solutions of $V$ satisfying both the laplace equation and the boundary conditions, and it is guranteed by the uniqueness theorem that there exists no other solution. Using this property, we search for other configurations (image charge distibutions) which may generate the same boundary conditions, and then find the solution of this problem. However, the image charge cannot be inside $S$, as then the space bounded by $S$ is not charge free anymore. Thus laplace equation will not hold anymore is image charges are consiered inside the boundary(it will revert back to the poisson version).

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