= Solution
Let
$$
I(t)=\int_{\mathbb R^4}|x|^2|u(t,x)|^2\,dx.
$$
The first <virial identity>, obtained from the equation by integration by parts, is
$$
I'(t)=4\operatorname{Im}\int_{\mathbb R^4}\overline u\,x\mathbin{\cdot}\nabla u\,dx.
$$
Differentiating once more gives
$$
I''(t)=8\int|\nabla u|^2-4\int x\mathbin{\cdot}\nabla\phi\,|u|^2.
$$
Write $\phi=W*|u|^2$, where $W(x)=-1/(C_4|x|^2)$ is homogeneous of degree $-2$. Symmetrizing the double integral and applying Euler's identity $z\cdot\nabla W(z)=-2W(z)$ yields
$$
\int x\mathbin{\cdot}\nabla\phi(x)|u(x)|^2\,dx
=-\int\phi|u|^2
=\int|\nabla\phi|^2.
$$
Therefore
$$
\boxed{I''(t)
=8\int|\nabla u|^2-4\int|\nabla\phi|^2
=16E(u)}.
$$
Not all solutions are global. Choose smooth finite-variance data of negative energy, which is possible by multiplying any nonzero test function by a sufficiently large constant: the kinetic term is quadratic in the amplitude and the attractive potential term is quartic. If such a solution were global, the <Virial identity for the four-dimensional gravitational Hartree equation> would make the nonnegative function $I(t)$ strictly concave with constant negative second derivative, forcing it below zero in finite time. The solution must therefore blow up in finite time.
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