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Let $u$ be a global sufficiently regular finite-energy solution of $u_{tt}-\Delta u+|u{|}^{p-1}u=0$ on ${\mathbb{R}}^3$, with $p>1$, and let $E$ be its conserved nonnegative wave energy. Then the Morawetz estimate for the defocusing wave equation isThe constant depends on $p$ and the spatial dimension, not the upper endpoint of the time interval. A finite-time version holds on every interval of smooth existence. A global spacetime bound does not itself assert existence of a global solution.^[ref]

Use the radial multiplier $u_r+u/r$ in the Morawetz identity for the defocusing wave equation. Its action is bounded by $\Vert u_t{\Vert }_2(\Vert \nabla u{\Vert }_2+\Vert u/r{\Vert }_2)\le CE$ using the Hardy inequality in Euclidean space. The radial weight has nonnegative angular terms and $-{\Delta }^2|x|=8\pi {\delta }_0$; the remaining nonlinear term has positive coefficient proportional to $(p-1)/(p+1)$. Integration in time bounds its spacetime integral by endpoint actions, and monotone convergence gives the infinite-time statement. The estimate limits how long nonlinear wave energy can remain concentrated near a fixed location.

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.