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I am working with integrals of complex functions. I assume all terms are well-defined.

If $u=u(x,t):\mathbb{R}^n\times\mathbb{R}_+ \to \mathbb{R}$ (a real function), I have \begin{equation} i\int_\Omega u u_t \,dx = \frac{i}{2} \frac{d}{dt} \int_\Omega u^2 \,dx. \end{equation} Assuming $u$ is $0$ on the boundary of $\Omega$, I have \begin{equation} \int_\Omega u\Delta u \,dx = -\int_\Omega Du \cdot Du \,dx. \end{equation} Now if $u=u(x,t):\mathbb{R}^n\times\mathbb{R}_+ \to \mathbb{C}$ (a complex function), can we have anything from these: \begin{equation} i\int_\Omega \overline{u} u_t \,dx = \,? \end{equation} Assuming $u$ is $0$ on the boundary of $\Omega$, I have \begin{equation} \int_\Omega \overline{u}\Delta u \,dx = \,? \end{equation} $\overline{u}$ is the complex conjugate of $u$.

Thank you.

PS: I am working with Schrodinger equation, and I would like to use energy method to show the equality \begin{equation} \int_\Omega |u(x,t)|^2 dx = \int_\Omega |u(x,0)|^2 dx \end{equation}

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    $\begingroup$ $\frac{d}{dt} |u|^2 = \frac{d}{dt} \bar{u}u = \bar{u} u_t + \bar{u}_t u = 2\Re(\bar{u}u_t)$ may be of use to you. $\endgroup$
    – Chappers
    Commented Apr 29, 2015 at 2:40
  • $\begingroup$ @Chappers Thank you. It helps a lot. $\endgroup$
    – dh16
    Commented Apr 29, 2015 at 4:13

1 Answer 1

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Start with $\bar uu_t=(\bar u u )_t-u\bar u_t$. Then,

$$\int_{\Gamma} \bar uu_t\, dx = \frac{d}{dt}\int_{\Gamma} \bar u u \,dx-\int_{\Gamma} u \bar u_t\, dx=\frac{d}{dt}\int_{\Gamma} |u |^2 \,dx-\int_{\Gamma} u \bar u_t\, dx$$

whereupon rearranging gives

$$i( \int_{\Gamma} (\bar uu_t+u \bar u_t)\, dx =i\frac{d}{dt}\int_{\Gamma} |u |^2\,dx$$

or

$$i\text{Re}\left(\int_{\Gamma} \bar uu_t\, dx \right)=\frac{i}{2}\frac{d}{dt}\int_{\Gamma} |u |^2\,dx$$

Similarly, start with $\bar u \Delta u=D \cdot (\bar u D u)-D \bar u \cdot Du$. Integration reveals

$$\int_{\Gamma} \bar u \Delta u \, dx=-\int_{\Gamma} D \bar u \cdot Du\, dx$$

where we tacitly used $\bar u =0$ on $\partial \Gamma$.

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