Explain this multivariate differential identity

Question

$$ \frac{\partial|\nabla\phi|^2}{\partial\phi}=-2\nabla\cdot\nabla\phi$$

I would very appreciate that you help me . Please do it in detail, I am quite not good at such problems.

There is something wrong with this equation, and it is the primary condition below :

$$F = \int\limits_\Omega (f_\text{gr} + f_\text{loc} + f_\text{el})\, dV$$

and then it gives $$\psi = \frac{\delta F}{\delta \phi} = -\nabla\cdot \lambda\nabla\phi + \frac{\lambda}{\epsilon^2}(\phi^2 - 1)\phi + \frac{\partial f_\text{el}}{\partial\phi}$$

so I cut it short (wrongly ) as given before.

Answer 1

The equation is not correct as stated, but needs to be interpreted in the variational sense when multiplied by a test function. To this end, consider the operator $$ L(\phi) = |\nabla \phi|^2. $$ Here, $\phi=\phi(x)$ is a function. Now, imagine what its derivative would be. Derivatives of operators are most commonly considered in certain directions $v(x)$. So we could denote $$ L'(\phi)(v) = \lim_{\varepsilon\rightarrow 0} \frac{L(\phi+\varepsilon v)-L(\phi)}{\varepsilon}. $$ If you go through the motions with the limit, you will find that $$ L'(\phi)(v) = 2 \nabla \phi \cdot \nabla v. $$ If you wanted to, you could then define the derivative of the operator, $L'(\phi)$ (or, in your notation, $\frac{\partial}{\partial\phi}L(\phi)$ as $$ L'(\phi) = 2 \nabla \phi \cdot \nabla $$ which is an operator that acts on whatever it is applied to.

You can extend this. Assume that the functions $\phi$ you consider live on a domain $\Omega\subset {\mathbb R}^d$ and that $\phi|_{\partial\Omega}=0$. (In other words, you assume that $\phi \in H^1_0(\Omega)$.) If you define a functional $$ J(\phi) = \int_\Omega L(\phi) = \int_\Omega |\nabla \phi|^2 $$ that associates a number to every function $\phi$, then you can again ask what the derivative of $J$ is in a direction $v$. You get $$ J'(\phi)(v) = \int_\Omega L'(\phi)(v) = \int_\Omega 2\nabla \phi \cdot \nabla v. $$ You can integrate the last formula by parts, using that both $\phi$ and $v$ are zero on the boundary, to get $$ J'(\phi)(v) = \int_\Omega\left[ -2\nabla \cdot \nabla \phi\right] v. $$ It is in this context that the term you cited in the question appears. But the formula you cite is, by itself, wrong.

Answer 2

I guess is not like:

$\frac{\partial}{\partial \phi}\left (\frac{\partial \phi}{\partial x_i}\frac{\partial \phi}{\partial x_i}\right) =-2 \frac{\partial }{\partial x_i}\frac{\partial \phi }{\partial x_i}$