 Solution

The displayed estimate requires the standard choice $\psi (r)=r$; read literally, the printed ${\psi }^{\prime }(r)=r$ 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 isThe 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 givesuniformly 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 $u=0$.