Taking a Time Derivative of a Vector Equation

Question

I have an expression for the energy of a system in time, $E(\phi)$, which is a function of the concentration in space-time, $\phi(\vec{x},t)$. Now I want to take a time derivative to see how this involves over time, however, I'm not quite sure I'm doing this properly: $$ E(\phi) = \int_{\Omega} f(\phi) + \frac{\epsilon^2}{2}|\nabla\phi|^2 d\vec{x} \qquad \epsilon>0 $$ It's clear that this will just be the chain rule for the first term but how do I deal with the second?. To me, it seems that: $$ |\nabla\phi|^2=\nabla\phi\cdot\nabla\phi $$ $$ \implies \frac{\partial}{\partial t}|\nabla\phi|^2 = 2\nabla\phi\left[\frac{\partial}{\partial t}\left(\nabla \phi\right)\right] = 2\nabla\phi\cdot\nabla^2\phi\frac{\partial \phi}{\partial t} $$ where to do this I used the fact that: $$\frac{d}{dt}(\vec{a}\cdot\vec{a}) = 2\vec{a}\cdot\frac{d\vec{a}}{dt}$$ However, using this to find $\frac{dE}{dt}$ I get: $$ \frac{dE}{dt} = \int_{\Omega} f'(\phi)\frac{\partial \phi}{\partial t} + \epsilon^2\left(\nabla\phi\cdot\nabla^2\phi\frac{\partial\phi}{\partial t}\right) d\vec{x} $$ I'm following a long with a published paper and the author simply goes from the first equation defining $E$ to this line: $$ \frac{dE}{dt} = \int_\Omega (f'(\phi)-\epsilon^2\nabla^2\phi)\frac{\partial\phi}{\partial t} d\vec{x} + \int_{\partial\Omega}\epsilon^2\frac{\partial \phi}{\partial t}\nabla\phi\cdot\vec{n}\ dS $$ I'm not quite sure if what I did is wrong in comparison to this or if I simply just need to rewrite the terms and then apply the Divergence Theorem forone of the terms (to get that last surface integral term in their equation). Any help or comments would be much appreciated!

Answer 1

If you write $$ |\nabla\phi|^2=(\partial_x\phi)^2+(\partial_y\phi)^2+(\partial_z\phi)^2 $$ it becomes clear that $$ \partial_t|\nabla\phi|^2=2\,\partial_x\phi\,\partial_{xt}\phi+2\,\partial_y\phi\,\partial_{yt}\phi+2\,\partial_z\phi\,\partial_{zt}\phi $$ which differs from what you got: $2\nabla\phi\cdot\nabla^2\phi\frac{\partial \phi}{\partial t}$ looks wrong. Neglecting constants and the first term with $f$ we get $$ \frac{dE}{dt} =\int_\Omega\,\underbrace{\partial_x\phi\,\partial_{xt}\phi+\,\partial_y\phi\,\partial_{yt}\phi+\,\partial_z\phi\,\partial_{zt}\phi}_{(*)}\,d\vec{x}\,. $$ The author of the published paper has surely applied Gauss' divergence theorem which we will now reverse engineer: the divergence of $$ \frac{\partial \phi}{\partial t}\nabla\phi $$ is $$ \underbrace{ \partial_x\phi\,\partial_{xt}\phi+\,\partial_y\phi\,\partial_{yt}\phi+\,\partial_z\phi\,\partial_{zt}\phi}_{(*)\text{ very good !}}+\underbrace{\partial_t\phi\,\partial_{xx}\phi+\partial_t\phi\,\partial_{yy}\phi+\partial_t\phi\,\partial_{zz}\phi}_{\partial_t\phi\,\nabla^2\phi\text{ looks familiar ?}}\,. $$ I now see his formula on the horizon. Do you see it, too?