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

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.


2 Answers 2


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.

  • $\begingroup$ Oh, Jesus, I can't believe it , you just got it, I cut it short from a long equation. I would modify my question to give the primary condition. And, thanks ! $\endgroup$
    – Jimmy
    Commented Apr 15, 2015 at 13:50

I guess is not like:

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

  • $\begingroup$ This is not correct. $\phi$ is a function, so you can't just do the derivative as you usually would. That should also be clear from the fact that you have no way of explaining where the minus sign comes from. $\endgroup$ Commented Apr 15, 2015 at 11:22

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