The displayed estimate requires the standard choice $\psi(r)=r$; read literally, the printed $\psi'(r)=r$ would give $\psi=r^2/2$, $-\Delta^2\psi=0$, and would not imply the claimed estimate. For $\psi=|x|$, the <distributional bilaplacian of the radial coordinate in three dimensions> is
$$
-\Delta^2\psi=8\pi\delta_0,
\qquad
\Delta\psi=\frac2r,
\qquad
\psi''=0.
$$
The delta term is nonnegative. The Morawetz action is bounded by $CE$ using Cauchy-Schwarz and the <Hardy inequality in Euclidean space>. Integrating the identity from $0$ to $T$ and discarding the delta term gives
$$
\int_0^T\int_{\mathbb R^3}\frac{|u|^{p+1}}{|x|}\,dx\,dt
\leq CE(u(0)),
$$
uniformly in $T$, proving the <Morawetz estimate for the defocusing wave equation>.

A finite-energy stationary solution would make the nonnegative spatial integral on the left constant in time. Its integral over $[0,\infty)$ can be finite only when that spatial integral is zero, so the stationary solution is $u=0$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 154 / 2 / 4 / Solution

{title2=$-\Delta^2|x|=8\pi\delta_0$}

In three dimensions, $\Delta|x|=2/|x|$ away from the origin and $-\Delta(1/|x|)=4\pi\delta_0$. Hence $-\Delta^2|x|=8\pi\delta_0$ as a distribution.


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 Incoming links: Distributional bilaplacian of the radial coordinate in three dimensions