Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 154 / 2

The Gravitational Hartree equation can be writtenwhere $\varphi$ is real. Consequentlybecause both integrals multiplied by $i$ are purely imaginary after integration by parts. Hence Hartree mass conservation holds.

Since $\Delta \varphi =|u{|}^2$, integration by parts givesThe Newtonian convolution operator is self-adjoint, so differentiating this identity symmetrically givesIt follows thatThe equation says $-\Delta u+\varphi u=i{u}_t$, and therefore the final real part is $\mathrm{Re}\int i|u_t{|}^2=0$. This proves Hartree energy conservation.

In three dimensions, the Hardy–Littlewood–Sobolev inequality applied to the Newtonian kernel givesThe Gagliardo-Nirenberg interpolation inequality then givesWriting $X=\Vert \nabla u{\Vert }_2$ and using the conserved mass, conservation of energy impliesThus $X$ remains bounded. Together with the conserved $L^2$ norm this bounds $\Vert u(t){\Vert }_{H^1}$, and the supplied blowup criterion proves Global H1 solutions of the three-dimensional gravitational Hartree equation.

LetThe first virial identity, obtained from the equation by integration by parts, isDifferentiating once more givesWrite $\varphi =W\ast |u{|}^2$, where $W(x)=-1/(C_4|x{|}^2)$ is homogeneous of degree $-2$. Symmetrizing the double integral and applying Euler's identity $z\cdot \nabla W(z)=-2W(z)$ yieldsTherefore

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.