Proof of Hartree Energy conservation

I recently encountered the problem of showing that given a solution $u\in H^2(\mathbb{R}^3)$ to the Hartree equation $$iu’(t)+\Delta u -(V*|u(t)|^2)u(t)=0$$ conserves Hartree energy $$\mathcal{E}(u(t)):= ||\nabla u||^2 +\frac{1}{2}\int (V*|u(t)|^2)(x)|u(t,x)|^2 dx.$$ We can assume bounded even real-valued $V$. How can this be done (ideally without invoking sledgehammers)? I tried differentiating the energy directly and modifying Duhamel’s principle to no avail. I would also appreciate any references.

Solution 1:

Observe that \begin{align} \frac{d}{dt}\int |\nabla u|^2 dx =&\ \int \partial_t \nabla u \cdot\overline{\nabla u} + \nabla u \cdot\overline{\partial_t\nabla u}\ dx = 2 \Re\left(\int \nabla\partial_t u \cdot \overline{\nabla u}\ dx\right)\\ =&\ 2 \Re\left(i\int \nabla (\Delta u- (V\ast|u|^2)u) \cdot \overline{\nabla u}\ dx\right)\\ =&\ 2 \Re\left( i \int \operatorname{div}((\Delta u-(V\ast|u|^2)u) \overline{\nabla u})-(\Delta u-(V\ast |u|^2)u)\overline{\Delta u}\ dx \right)\\ =&\ 2\Re\left(i \int -|\Delta u|^2+(V\ast |u|^2)u\overline{\Delta u}\ dx \right)= 2\Re\left(i \int (V\ast |u|^2)u\overline{\Delta u}\ dx \right) \end{align} and \begin{align} \frac{d}{dt}\frac{1}{2}\iint |u(t, x)|^2V(x-y)|u(t, y)|^2\ dxdy =&\ 2\Re\left(\iint u(t, x) \overline{\partial_t u(t, x)}V(x-y)|u(t, y)|^2\ dxdy\right)\\ =&\ 2\Re\left(i\iint u(t, x) \overline{i\partial_t u(t, x)}V(x-y)|u(t, y)|^2\ dxdy\right)\\ =&\ 2\Re\left(i \int (V\ast|u|^2)u\ \overline{i \partial_t u}\ dx \right). \end{align} Hence it follows that \begin{align} \frac{d}{dt}\mathcal{E}(u) = 2\Re\left(i \int (V\ast |u|^2)u\ \overline{(i \partial_t u+\Delta u)}\ dx\right) = 2\Re\left(i \int (V\ast |u|^2)^2|u|^2\ dx \right)=0. \end{align}

Notice: I made the implicit assumption that $V(x-y) = V(y-x)$.