April 3, 2014
Let H(u)=\int \frac{1}{2}|\nabla u|^2+\frac{2\mu}{p+1}|u|^{p+1}dx. We have H(u+\epsilon v)= \int \frac{1}{2} \nabla (u+\epsilon v) \nabla (\overline{u}+\epsilon \overline{v})+ \frac{2\mu}{p+1} (u+\epsilon v)^{\frac{p+1}{2}} (\overline{u}+\epsilon \overline{v})^{\frac{p+1}{2}}. We have, for v \in T_u, \begin{aligned} dH(u)v&=&\frac{d}{d\epsilon}|_{\epsilon=0} H(u+\epsilon v)\\ &=&\int \frac{1}{2}\left( \nabla v \nabla \overline{u}+\nabla u \nabla \overline{v} \right)\\ &&+\frac{2\mu}{p+1}\left( \frac{p+1}{2} u^{\frac{p-1}{2}}v \overline{u}^{\frac{p+1}{2}} +\frac{p+1}{2}u^{\frac{p+1}{2}} \overline{u}^{\frac{p-1}{2}}\overline{v} \right) dx\\ &=&\Re\int \nabla \overline{v} \nabla u+2\mu|u|^{p-1} u \overline{v} dx\\ &=&\Re \int \left(-\Delta u+2\mu|u|^{p-1} u\right) \overline{v} dx.\end{aligned}
Define \omega(v_1,v_2)=-2 \Im \int v_1 \overline{v_2} dx. The Hamiltonian vector field X_H of H is defined, for v \in T_u, by \omega(X_H(u),v)=dH(u)v. Therefore we have for v \in T_u that \begin{aligned} -2\Im \int X_H(u) \overline{v} dx&=&\Re \int \left(-\Delta u+2\mu|u|^{p-1} u\right) \overline{v} dx\\ &=&\Im \int \left(-i\Delta u+2i\mu |u|^{p-1} u\right) \overline{v} dx.\end{aligned} It follows that -2X_H(u)=-i\Delta u+2i\mu |u|^{p-1} u, i.e., X_H(u)=\frac{i}{2}\Delta u-i\mu |u|^{p-1} u.
The flow S(t) of the Hamiltonian vector field X_H satisfies, for any u_0 and for u(t)=S(t)u_0, u_t(t) = X_H(u(t))=\frac{i}{2}\Delta u(t)-i\mu |u(t)|^{p-1} u(t). This equation can be written as iu_t(t)+\frac{1}{2} \Delta u(t)=\mu |u(t)|^{p-1} u(t).