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I am currently trying to solve a problem, but am having a few difficulties. The question is to find the Green's Function of the Laplacian on the disk centered at the origin with radius $a$. Denote $\Omega$ as the open ball centered at the origin with radius $a$, and $\partial\Omega$ to be its boundary. Consider any $\textbf x_0\in\Omega$. Then define: $$\textbf x_0^* = \frac{a^2\textbf x_0}{|\textbf x_0|^2}$$ Now, I proved that if we take any $\textbf x\in\partial\Omega$, then we have the equality: $$\frac{|\textbf x_0|^2}{a^2}|\textbf x-\textbf x_0^*|^2 = |\textbf x-\textbf x_0|$$ Now because of this scaling factor, we have that the Green's Function is given by: $$G(\textbf x,\textbf x_0) = \frac{1}{2\pi}\log|\textbf x-\textbf x_0| - \frac{1}{2\pi}\log\left(\frac{a}{|\textbf x_0|}\cdot|\textbf x-\textbf x_0^*|\right)$$ Where the Fundamental Solution in $2D$ is given by: $$\Phi(\textbf x-\textbf x_0) = \frac{1}{2\pi}\log|\textbf x-\textbf x_0|$$ Is this correct? Am I missing something here?

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Almost correct, but it should be $$ G(\mathbf x,\mathbf x_0) = \frac{1}{2\pi}\log|\mathbf x-\mathbf x_0| - \frac{1}{2\pi}\log\left(\frac{|\mathbf x_0|}{a}\cdot|\mathbf x-\mathbf x_0^*|\right) $$ to make sure that $G=0$ when $|\mathbf x|=1$.

Also, the special case $\mathbf x_0=0$ deserves a mention:
$$ G(\mathbf x,\mathbf x_0) = \frac{1}{2\pi}\log|\mathbf x-\mathbf x_0| - \frac{1}{2\pi}\log a$$ which can be observed either directly, or by letting $\mathbf x_0\to 0$ and using $$ \frac{|\mathbf x_0|}{a} |\mathbf x-\mathbf x_0^*| = \frac{|\mathbf x_0|}{a} \left(|\mathbf x_0^*| + O(1)\right) = \frac{|\mathbf x_0|}{a} \left(\frac{a^2|\mathbf x_0|}{|\mathbf x_0|^2} + O(1)\right) \to a $$

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  • $\begingroup$ Is my logic sound leading up to the solution however? $\endgroup$ – Felicio Grande Nov 26 '17 at 22:19
  • $\begingroup$ Yes... "because of this scaling factor, we have" doesn't really present a lot of detail of the reasoning. You could have said that $G$ is zero on the boundary and that $G-\Phi$ is harmonic in $\Omega$; these are the requirements. $\endgroup$ – user357151 Nov 26 '17 at 22:47

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