I'm taking a course of PDE. I have a problem with the choice of the constants in the fundamental solution of the Laplace equation. In particular, observing that the equation is rotation invariant, we have obtained the fundamental solution by lookin for a radial, non-constant solution, getting, for dimension $d\geq 3$: $$ v(r)=C_1r^{2-d}+C_2,\qquad C_1,\,C_2\in\mathbb{R}. $$ Now our lecture notes go on like this:
We select now the arbitrary constants $C_1,C_2\in\mathbb{R}$ by setting $C_2=0$ and by choosing $C_1$ in order to have $$ \int_{\partial B(0,R)}v(|x|)d\sigma(x)=1 $$ where $\sigma(x)$ is the surface element at $x$. In particular, since $v(|x|)=v(R)$ on $\partial B(0,R)$, for $d\geq 3$ we have $$ 1=\int_{\partial B(0,R)}v(|x|)d\sigma(x)=\int_{\partial B(0,R)}C_1(2-d)R^{1-d}d\sigma(x)=C_1(2-d)R^{1-d}\text{meas}(\partial B(0,R))=C_1(2-d)\omega_d, $$ where $\omega_d$ denotes the $(d-1)-$dimensional measure of $\mathbb{S}^{d-1}=\partial B(0,1)\in\mathbb{R}^d$, and so $C_1=\frac{1}{(2-d)\omega_d}$.
I have two problems with this part:
- I don't get where the equality $$ \int_{\partial B(0,R)}v(|x|)d\sigma(x)=\int_{\partial B(0,R)}C_1(2-d)R^{1-d}d\sigma(x) $$ comes from. For me, by definition of $v(R)$, it should be $$ \int_{\partial B(0,R)}v(|x|)d\sigma(x)=\int_{\partial B(0,R)}C_1R^{2-d}d\sigma(x) $$
- I don't get what is the motivation for choosing the constant $C_1$ in order to have such a strange integral equality to hold.