Since $w=ru$,
$$
w_t+w_r=r\left(u_t+u_r+\frac ur\right),
$$
and radial integration satisfies $dx=4\pi r^2dr$. Multiplying the identity from part 6 by $4\pi$ therefore gives
$$
\frac d{dt}\int_{\mathbb R^3}\psi\left[
\frac12\left(u_t+u_r+\frac ur\right)^2
+\frac{|u|^{p+1}}{p+1}
\right]dx
$$
$$
=-\frac12\int_{\mathbb R^3}\psi'
\left(u_t+u_r+\frac ur\right)^2dx
+\frac1{p+1}\int_{\mathbb R^3}|u|^{p+1}
\left(\psi'-(p-1)\frac\psi r\right)dx,
$$
which is the modified Morawetz identity.

Choose the constant weight $\psi=1$. The first term on the right vanishes and the second is
$$
-\frac{p-1}{p+1}\int_{\mathbb R^3}\frac{|u|^{p+1}}r,dx.
$$
The functional on the left is bounded by the conserved energy using the <Hardy inequality in Euclidean space>. Integrating in time gives another proof of the <Morawetz estimate for the defocusing wave equation>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 154 / 2 / 7 / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/iii/paper-154/2/7/solution