Let $B_R:=B_R(0)\subseteq\mathbb{R}^n$ be the open ball with radius $R>0$ around $0$, $f\in\mathcal{L}^\infty(B_R)$ be radial and $u$ be a very weak solution² of $$\left\{\begin{matrix}-\Delta u&\equiv&f&\text{on }B_R\\u(R)&= &0\end{matrix}\right.\tag{1}$$
I want to calculate $$\int_{B_R}u\tag{2}$$
Ideas:
- $\varphi\in C^2\left(\overline{B_R}\right)$ with $$\varphi(r):=\frac{1}{2}r^2\left\{\frac{c_1r^{-n}}{n-2}+\frac{1}{n}\right\}+c_2\;\;\;\text{for }c_1,c_2\in\mathbb{R}\tag{3}$$ is radial and satisfies $\Delta\varphi=1$.
- If we choose $c_1=0$ and $$c_2=-\frac{R^2}{2n},$$ we additionally get $\varphi(R)=0$.
- Thus, we've got $$-\int_{B_R}u=-\int_{B_R}u\Delta\varphi\stackrel{u\text{ is a very weak solution of }(1)}{=}\int_{B_R}f\varphi=\frac{1}{2n}\int_{B_R}r^2f-\frac{R^2}{2n}\int_{B_R}f\tag{4}$$
Question: I'm not very familiar with such integrals. Is there anything we can further do (in order to get a more precise form of $(2)$)?
² $\;\;u$ is called a very weak solution of $(1)$ iff
- $u(R)=0$
- It holds $$-\int_{B_R}u\Delta\varphi =\int_{B_R}f\varphi\tag{5}$$ for all radial $\varphi\in C^2\left(\overline{B_R}\right)$ with $\varphi(R)=0$
Your approach is right, although it wasn't necessary to write out (3) with both $c_1$ and $c_2$ in there. When we need a function with constant Laplacian, the quadratic polynomials are to look for: so, $\varphi(x)=(R^2-|x|^2)/(2n)$ is the test function to use.
And you used it correctly: $$ -\int_{B_R} u =\frac{1}{2n} \int_{B_R} f (x) (R^2-|x|^2) $$ which for radial $f$ simplifies to $$ \int_{B_R} u =-\frac{\omega_{n-1}}{2n} \int_0^R f (r) (R^2-r^2)r^{n-1}\,dr $$ where $\omega_{n-1}$ is the surface of the unit sphere. This is as far as it goes without having an explicit $f$.