General Solution to Laplace’s Equation in $\mathbb{R^3}$


I am trying to find the Green’s function for $ \ \nabla^2\phi=S(x)\ $ for $ \ x\in\mathbb{R^3}$ and express the general solution to Laplace’s equation in $ \mathbb{R^3}$ .

To find the Green’s function, I considered \begin{align} \nabla^2G&=\delta\left(\underline{x}-\underline{x’}\right) \ \nabla^2G&=0 \ \ \ \ \ \ \left(\text{when} \ \underline{x}\neq\underline{x’}\right)\ \nabla^2G&=\frac{1}{r^2}\frac{d}{dr}\left(r^2\frac{DG}{dr}\right) \ G(r)&=-\frac{A}{r}+B, \ \ A,B\in\mathbb{R}. \end{align} Next consider a ball centred at $ \underline{x’}$ with radius $ a$ , denoted $ B_a$ . Then, \begin{align} \iiint_{B_a} \nabla^2G \ dV&=\iiint_{B_a}\delta\left(\underline{x}-\underline{x’}\right) \ dV \ \iint_{\delta B_a}\frac{dG}{dr}\bigg|_{r=a} ds &=1 \ \ \ \text{(Divergence theorem)}\ 4\pi A&=1 \ \implies A&=\frac{1}{4\pi} \end{align} Hence (letting $ B=0$ without loss of generality), $ $ G(r)=-\frac{1}{4\pi r}.$ $ Now, $ r=|\underline{x}-\underline{x’}|$ , which means $ $ G(\underline{x},\underline{x’})=-\frac{1}{4\pi|\underline{x}-\underline{x’}|}.$ $

Why can the general solution be expressed as $ $ \phi(\underline(x))=-\frac{1}{4\pi}\iiint_{\mathbb{R^3}}\frac{1}{|\underline{x}-\underline{x’}|} S(\underline{x’}) \ dV’?$ $

Homotopy type of $\mathbb{R}^3 \setminus \{ \mathrm{wedge \hspace{3pt} sum \hspace{3pt} of \hspace{3pt} 2 \hspace{3pt} circles} \}$

Is there something “nice” that the space $ \mathbb{R}^3 \setminus \{ \mathrm{wedge \hspace{3pt} sum \hspace{3pt} of \hspace{3pt} 2 \hspace{3pt} circles} \}$ is homotopy equivalent to? I know that $ \mathbb{R}^3 \setminus \{\mathrm{circle}\} \simeq S^{2} \vee S^{1}$ , and that $ \mathbb{R}^3 \setminus \{ \mathrm{2 \hspace{3pt} linked \hspace{3pt} circles}\} \simeq S^{2} \vee T^{2}$ . However, I can’t seem to apply either of the “visual methods” used to prove these homotopy equivalences to my case.

My wild guess is that it might be homotopic to $ M_{2}$ , the orientable closed surface of genus $ 2$ , but I’m not sure if this is true at all.