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Compute the Green's function for the Laplacian, for the region $$\Omega=\{ (x_1,x_2,x_3)\in \mathbb{R}^3:x_2,x_3>0 \}.$$

My approach is to use a reflection argument similar to the one used for computing the Green's function for the upper half-plane, where the last coordinate is taken to be negative. The problem is reduced to finding the corrector function $\phi^{x}(x,y),$ such that \begin{cases} \Delta_{y}\phi^{x}(x,y)=0&\text{for }y\in\Omega \\\phi^{x}(x,y)=\Phi(y-x)&\text{for }y\in\partial\Omega. \end{cases} Where $\Phi(y-x)$ is the fundamental solution of Laplace's equation, given in this case by $\Phi(y-x)=\frac{1}{3\alpha(3)}\frac{1}{|y-x|}$, and $\alpha(3)$ denotes the volume of the unit ball in $\mathbb{R}^3.$

If we define our corrector function to be $\phi^{x}(x,y)=\Phi(y-\tilde{x})+\Phi(y-x^{\star})-\Phi(y-x^{'})$, where $\tilde{x}=(x_1,-x_2,x_3), x^{\star}=(x_1,x_2,-x_3),x^{'}=(x_1,-x_2,-x_3).$ Then I think this function satisfies the conditions above.


Is any of the above correct? Could anyone point out what's the strategy to find Green's functions for domains such as the above.

Any feedback is much appreciated and thank you for your time.

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    $\begingroup$ you can place a sequence of image charges in the $x_2<0$ and $x_3<0$ half volumes $\endgroup$ – phdmba7of12 Feb 22 at 13:21
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The main issue here is to find the corrector function $e(x,y).$ Let's try the following, set $e(x,y)=\Phi(y-\tilde{x})-\Phi(y-x^{\star})+\Phi(y-x^{'})$, where $\tilde{x}=(x_1,-x_2,x_3), x^{\star}=(x_1,-x_2,-x_3),x^{'}=(x_1,x_2,-x_3).$ We need to check that $$\begin{cases}\Delta_{y}e(x,y)=0&\text{for }x,y\in\Omega \\ e(x,y)=\Phi(y-x)&\text{for }y\in\partial\Omega, x\in\Omega. \\ \end{cases}$$ Note that our corrector function is given by $$e(x,y)=\frac{1}{3\alpha(3)}\left[ \frac{1}{|(y_1,y_2,y_3)-(x_1,-x_2,x_3)|} -\frac{1}{|(y_1,y_2,y_3)-(x_1,-x_2,-x_3)|} +\frac{1}{|(y_1,y_2,y_3)-(x_1,x_2,-x_3)|} \right].$$ For $x \in \Omega,y \in \partial \Omega$, we have three cases, as computed below $$e(x,y)=\frac{1}{3\alpha(3)}\left[ \frac{1}{|(y_1,y_2,0)-(x_1,-x_2,x_3)|} -\frac{1}{|(y_1,y_2,0)-(x_1,-x_2,-x_3)|} +\frac{1}{|(y_1,y_2,0)-(x_1,x_2,-x_3)|} \right]$$ $$e(x,y)=\frac{1}{3\alpha(3)}\left[ \frac{1}{|(y_1,0,0)-(x_1,-x_2,x_3)|} -\frac{1}{|(y_1,0,0)-(x_1,-x_2,-x_3)|} +\frac{1}{|(y_1,0,0)-(x_1,x_2,-x_3)|} \right]$$ $$e(x,y)=\frac{1}{3\alpha(3)}\left[ \frac{1}{|(y_1,0,y_3)-(x_1,-x_2,x_3)|} -\frac{1}{|(y_1,0,y_3)-(x_1,-x_2,-x_3)|} +\frac{1}{|(y_1,0,y_3)-(x_1,x_2,-x_3)|} \right].$$ In each case, we obtain $e(x,y)=\Phi(x,y),$ so that the equality holds for all $x\in\Omega,y\in\partial\Omega.$

For $x,y\in\Omega,$ it is trivial that $\Delta_{y}e(x,y)=0,$ since we avoided the singularities by our reflection argument and $e(x,y)$ is a linear combination of modifications of the fundamental solution. It follows that $G(x,y)=\Phi(x,y)-e(x,y).$

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