Radiality gives $\phi_\varepsilon(x-y)=\phi_\varepsilon(y-x)$. Since $\phi_\varepsilon$ is supported in $B(0,\varepsilon)$, the changes of variables $y=x+z$ and then $z=\varepsilon r\omega$ in <spherical coordinates> give
$$
\begin{aligned}
(u*\phi_\varepsilon)(x)
&=\int_{B(0,\varepsilon)}u(x+z)\varepsilon^{-d}\phi(z/\varepsilon)\,dz\\
&=\boxed{\int_0^1\int_{S^{d-1}}u(x+\varepsilon r\omega)\phi(r\omega)r^{d-1}\,d\omega\,dr.}
\end{aligned}
$$
The restriction $\varepsilon<\operatorname{dist}(x,\partial\Omega)$ ensures that every sampled point remains in $\Omega$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 107 / 1 / iii / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/iii/paper-107/1/iii/solution