Here $w$ is the <harmonic replacement> of $u$ in $B_r=B(x_0,r)$, so $v=u-w\in H_0^1(B_r)$. Subtract the weak equations and test with $v$:
$$
\int_{B_r}|\nabla v|^2
=\frac12\int_{B_r}x_1^2u_{x_1}v_{x_1}+\int_{B_r}fv.
$$
Because $B_r\subset B(0,1)$, $|x_1|\leq1$. The <Sobolev inequality> $\|v\|_6\leq C\|\nabla v\|_2$ and the <Holder inequality> give
$$
\|\nabla v\|_2^2
\leq\frac12\|\nabla u\|_2\|\nabla v\|_2
+C\|f\|_{6/5}\|\nabla v\|_2.
$$
After division and squaring,
$$
\boxed{\int_{B_r}|\nabla v|^2
\leq C_1\int_{B_r}|\nabla u|^2
+C_2\left(\int_{B_r}|f|^{6/5}\right)^{5/3}.}
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 107 / 3 / i / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/iii/paper-107/3/i/solution