Let $\Omega$ be an open set in $\mathbb{R}^n$ for $n \geq 2$. Consider the Schrodinger equation $iu_t+\Delta u = 0$ for $(x,t) \in \Omega \times (0,T)$. such that we have the Dirichlet boundary condition $u(x,t) = 0$ for $(x,t) \in \partial \Omega \times (0,T)$. We define the energy as:

$E(t) = \frac{1}{2}\int_{\Omega}|\nabla u(x,t)|^2\,dx$

How can we prove that this enery is conserved ? i.e: $E(t) = E(0)$ for all $t$. I try with multiplying method but I found not do that. Can anyone help me ?

  • $\begingroup$ Did you try calculating the derivative ${d \over dt} E(t)$ ? $\endgroup$ – lisyarus Nov 18 '17 at 9:00
  • $\begingroup$ I try it, but I could not prove E'(t) =0 $\endgroup$ – hoangimb Nov 18 '17 at 9:07
  • $\begingroup$ So, what is the derivative of E? $\endgroup$ – Mariano Suárez-Álvarez Nov 18 '17 at 9:20
  • $\begingroup$ Actually, I computed $E'(t) = \frac{-i}{2}\int_{\Omega}\Bigl(\nabla(\Delta u).\nabla\overline{u}-\nabla u.(\nabla(\Delta \overline{u}))\Bigr)\,dx$ where "." denote the scalar product in $\mathbb{R}^n$. How can I prove this one equal to 0 ? $\endgroup$ – hoangimb Nov 18 '17 at 9:25
  • $\begingroup$ @hoangimb Could you show the derivation of this in the post? By the way, use \cdot for the scalar product. $\endgroup$ – lisyarus Nov 18 '17 at 14:22

By the product rule, $$ \frac{\partial}{\partial t} \langle \nabla u, \nabla u \rangle = 2 \operatorname{Re}\langle \nabla u_t, \nabla u \rangle $$ Integrate this over $\Omega$ by parts, transferring $\nabla$ from first factor to the second. This gives $$ -2 \operatorname{Re} \int_\Omega u_t \overline{\Delta u} $$ (The inner product had complex conjugate over the second term, which now manifests itself.)

Since $u_t = -i\Delta u$, we end up with $$ 2 \operatorname{Re} \int_\Omega i |\Delta u|^2 = 0 $$ as claimed.

  • $\begingroup$ Probably worth noting that the last value is zero due to the integrand being pure imaginary, - this took me a minute or two to realize. $\endgroup$ – lisyarus Nov 20 '17 at 10:40

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