Assume that I have the following functional: $$ F[u_1,u_2,...u_N,\nabla u_1,...,\nabla u_N,...;t]=\int_{\Omega} f( u_1(x,t),u_2(x,t),...,u_N(x,t),\nabla u_1 , \nabla u_2,... ) dV $$ where $\Omega \subseteq \mathbb{R}^k$ . Is it true that: $$ \frac{dF}{dt}= \int_{\Omega} \Sigma_{i=1}^N \frac{\delta F}{\delta u_i } \frac{\partial u_i}{\partial t} $$ ? If so, will you please explain why?
Thanks in advance
On
Yes, with appropriate hypotheses (on $f$ and/or on $\Omega$), one can pass the derivative inside the integral, and so the second identity follows from $$ \frac{d}{dt}f( u_1(x,t),u_2(x,t),...,u_N(x,t) )=\sum_{i=1}^N\frac{\partial F}{\partial u_i}( u_1(x,t),u_2(x,t),...,u_N(x,t) )\frac{\partial u_i}{\partial t}(x,t). $$ This property is usually called Leibniz's rule .
This is basically the chain rule. The problem is how you move the derivative into the integration, i.e., why $$ \frac{dF}{dt}=\int\frac{df}{dt} $$ hold.
There are different ways to justify the above equation. You may put $f\in L^\infty$ or $\Omega$ is bounded and $f$ is uniformly conts. I would suggest you to read page 454 about this book to get more ideas.