I`m reading a paper and it came in it the following:
$$J(t) := \int_{R^N} \vert x \vert^2 \vert u (t,x)\vert^2 dx $$ $$ J'(t)=4 Re \int x. \nabla u \overline u dx$$
and $$u$$ satisfy the Cauchy equation : $$ iu_t + \Delta u + \vert x\vert^b \vert u \vert^{p-2} u =0 , t\geq0, x \in R^N $$ $$ u(0,x)=u_\circ $$
Can anyone please explain to me how to get the derivative this way?
Hint: Let us just consider $i\partial_t u+\Delta u = 0$, then observe \begin{align} J'(t) =&\ \frac{d}{dt}\int |x|^2|u(t, x)|^2\ dx\\ =&\ \int |x|^2 \left\{\partial_t u(t, x)\overline{u(t, x)}+u(t, x)\overline{\partial_t u(t, x)}\right\}\ dx\\ =&\ \int |x|^2 \left\{i\Delta u(t, x) \overline{u(t, x)}+u(t, x)\overline{i\Delta u(t, x)}\right\}\ dx\\ =&\ i\int |x|^2 \left\{\Delta u(t, x) \overline{u(t, x)}-u(t, x)\overline{\Delta u(t, x)}\right\}\ dx. \end{align}
Next, note that \begin{align} \int |x|^2\overline{u(t, x)} \Delta u(t,x )\ dx =&\ -\int \nabla(|x|^2\overline{u(t, x)})\cdot \nabla u(t, x)\ dx\\ =&\ -\int \{2x\overline{u(t, x)}+|x|^2\overline{\nabla u(t, x)}\}\cdot \nabla u(t, x)\ dx \\ =&\ -2 \int [x\cdot \nabla u(t, x)]\overline{u(t, x)}\ dx -\int |x|^2|\nabla u(t, x)|^2\ dx \end{align} and \begin{align} \int |x|^2u(t, x)\overline{\Delta u(t, x)}\ dx = -2 \int [x\cdot \overline{\nabla u(t, x)}]u(t, x)\ dx -\int |x|^2|\nabla u(t, x)|^2\ dx \end{align} then it follows \begin{align} J'(t) =&\ -2i\int [x\cdot\nabla u(t, x)]\overline{u(t, x)}-[x\cdot\overline{\nabla u(t, x)}] u(t, x)\ dx \\ =&\ 4\operatorname{Im}\left(\int[x\cdot \nabla u(t, x)] \overline{u(t, x)}\ dx\right). \end{align}
Note: The paper made a typo. It should take the imaginary part instead of the real part.