Inner product of integral with partial derivative?

Question

I have the norm and the inner product \begin{align} \|u(x)\|^2&=\int_a^b |u(x)|^2 \, \mathrm{d}x,\\ \left<u(x),v(x)\right>&=\int_a^b u(x)v(x) \, \mathrm{d}x \end{align} Why can this integral be written as \begin{align} \int_\Omega u(x,t)\frac{\partial u(x,t)}{\partial t} \, \mathrm{d}x = \frac{\mathrm{d} \|u(x,t)\|^2}{\mathrm{d}t} \end{align} Also, why no partial derivative in the RHS?

Answer 1

The expression as written in the OP is not quite correct. If $u\in \mathbb{C}$, then we have

$$\begin{align} \frac{d||u||^2(t)}{dt}&=\frac{d}{dt}\int_a^b |u(x,t)|^2\,dx\\\\ &=\int_a^b \frac{\partial}{\partial t}(u(x,t)\,\overline{u}(x,t))\,dx\\\\ &=\int_a^b \left(u(x,t)\frac{\partial \overline{u}(x,t)}{\partial t}+\overline{u}(x,t)\frac{\partial u(x,t)}{\partial t}\right)\,dx \tag 1 \end{align}$$

If $u\in \mathbb{R}$, then $(1)$ becomes

$$\begin{align} \frac{d||u||^2(t)}{dt}&=2\int_a^b u(x,t)\frac{\partial u(x,t)}{\partial t}\,dx \tag 1 \end{align}$$

The reason that the total derivative appears outside the integral while the partial derivative appears inside the integral is that the integration of $|u(x,t)|^2$ is over the dummy variable $x$. Hence, $||u||^2(t)$ is only a function of $t$.