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For a smooth compactly supported $u$, expand the nonnegative squareSince $\nabla \cdot (x/|x{|}^2)=|x{|}^{-2}$ in three dimensions, integration by parts turns the cross term into $-\frac{1}{2}\int |u{|}^2/|x{|}^2$. HenceDensity extends this Hardy inequality in Euclidean space to $H^1({\mathbb{R}}^3)$.

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 $u=0$.

Since $w=ru$,and radial integration satisfies $dx=4\pi r^{2}dr$. Multiplying the identity from part 6 by $4\pi$ therefore giveswhich is the modified Morawetz identity.

Choose the constant weight $\psi =1$. The first term on the right vanishes and the second isThe 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.