Write $q=w_t+w_r$. The equation in part 5 says
$$
q_t-q_r=-\frac{|w|^{p-1}w}{r^{p-1}}.
$$
Differentiate $J_\psi$, replace $q_t$ by $q_r-|w|^{p-1}w/r^{p-1}$, and integrate the $qq_r$ and $|w|^{p-1}ww_r$ terms by parts. This gives the <outgoing-energy identity for a radial defocusing wave>
$$
\frac{dJ_\psi}{dt}
=-\frac12\int_0^\infty\psi' q^2dr
+\frac1{p+1}\int_0^\infty
\frac{|w|^{p+1}}{r^{p-1}}
\left(\psi'-(p-1)\frac\psi r\right)dr.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 154 / 2 / 6 / Solution

For $q=w_t+w_r$ and a radial weight $\psi$,
$$
\frac d{dt}\int_0^\infty\psi\left(\frac12q^2+\frac{|w|^{p+1}}{(p+1)r^{p-1}}\right)dr
=-\frac12\int_0^\infty\psi'q^2dr
+\frac1{p+1}\int_0^\infty\frac{|w|^{p+1}}{r^{p-1}}
\left(\psi'-(p-1)\frac\psi r\right)dr.
$$


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 Incoming links: Outgoing-energy identity for a radial defocusing wave