Let $$G(r)=\begin{cases}-\frac{1}{2\pi}\ln(r)&d=2\\ \frac{1}{d(d-2)|B_1|r^{d-2}}&d\geq 3\end{cases}$$ the fundamental solution of Laplace equation where $|B_1|$ is the volume of the unit sphere in $\mathbb R^d$. Formally, $$-\nabla G(x,y)=\delta_{x=y}$$ where $G(x,y)= G(|x-y|)$.
Quest. 1 : Isn't it strange the notation $G(x,y)= G(|x-y|)$ ? I don't really see what it mean (since in one way $G:\mathbb R^{2d}\longrightarrow \mathbb R$ and and the other $G:\mathbb R\longrightarrow \mathbb R$...
Let $f\in \mathcal C^2(\mathbb R^d)$ compact supported and define $$u(x)=\int_{\mathbb R^d}G(x,y)f(y)\mathrm d y.$$ I'm trying to show that $\Delta u=-f$ is $\mathbb R^d$. In the correction it's written $$-\Delta u(x)=-\int_{\mathbb R^d}\nabla _xG(x,y)\cdot \nabla f(y)\mathrm d y=\int_{\mathbb R^d}\nabla _y G(x,y)\cdot \nabla f(y)\mathrm d y,$$ but I don't understand those to equality. To me, we only have $$-\Delta u(x)=-\int_{\mathbb R^d} \Delta _x G(x,y)f(y)\mathrm d y.$$
Any idea ?
Hint
$$\nabla _x G(x,y)=G'(|x-y|)\nabla _x(|x-y|)=-G'(|x-y|)\nabla _y(|x-y|).$$