Set
$$
M_\psi(t)=\int_{\mathbb R^3}u_t
\left(\nabla\psi\cdot\nabla u+\frac{\Delta\psi}{2}u\right)dx.
$$
Differentiate, substitute $u_{tt}=\Delta u-u|u|^{p-1}$, and integrate every second derivative off $u$. The mixed first-derivative terms cancel because of the correction $(\Delta\psi)u/2$. For radial $u$ and radial $\psi$, the Hessian term is $\psi''|\nabla u|^2$. The potential term uses $u|u|^{p-1}\nabla u=\nabla(|u|^{p+1})/(p+1)$. One obtains the <Morawetz identity for the defocusing wave equation>
$$
-\frac{dM_\psi}{dt}
=\int_{\mathbb R^3}\left[
\psi''|\nabla u|^2-\frac14(\Delta^2\psi)u^2
+\frac{p-1}{2(p+1)}(\Delta\psi)|u|^{p+1}
\right]dx.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 154 / 2 / 3 / Solution

{c}

Multiplying a defocusing wave equation by $\nabla\psi\cdot\nabla u+(\Delta\psi)u/2$ and integrating by parts gives a weighted spacetime identity. For a radial solution and radial weight, its coercive terms are
$$
\psi''|\nabla u|^2-\frac14(\Delta^2\psi)u^2
+\frac{p-1}{2(p+1)}(\Delta\psi)|u|^{p+1}.
$$


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 Incoming links: Morawetz identity for the defocusing wave equation