I am interested in a free-energy functional given by: $$ E[\phi] = \int f(\phi) - \frac{\epsilon^2}{2}|\nabla\phi|^2 \ d\vec{x} = \int F(\phi)$$ My understanding of variational derivatives is quite weak, and so, I am wondering what to make of: $$ \frac{\delta E}{\delta (1-\phi)}=\ ???? $$ From what I understand, the variational derivative with respect to $\phi$ can be calculated as: $$ \frac{\delta E}{\delta \phi} = \frac{\partial F}{\partial \phi}-\nabla\cdot\frac{\partial F}{\partial \nabla\phi} = f'(\phi)-\epsilon^2\nabla^2\phi $$ My question then: is it true that: \begin{equation} \pmb{\frac{\delta E}{\delta (1-\phi)}=-\frac{\delta E}{\delta \phi}}\ ???? \end{equation} If it matters, in this context $\phi=\phi(\vec{x},t)$ represents the composition of component A in a binary mixture. (Hence, $1-\phi$ is the composition of component B).


1 Answer 1


Yes, that's right.

As a simple explanation, changing $1-\phi$ by $\epsilon \psi$ amounts to changing $\phi$ by $-\epsilon \psi$, so $E$ responds as if you had done that change to $\phi$.

  • $\begingroup$ so in that regard, it's the same as a partial derivative? $\endgroup$
    – Mjoseph
    Commented Oct 10, 2022 at 15:44
  • $\begingroup$ @Mjoseph More or less. The subtle thing is that in functional calculus you cannot lose track of the fact that the variational derivative of a scalar functional is linear functional-valued, not scalar-valued. That is, it takes in a starting point and a direction of perturbation and gives you the directional derivative in that direction from that starting point. (Technically this is how derivatives always work, but in finite dimensions you can hide it, and the normal pedagogy does so.) $\endgroup$
    – Ian
    Commented Oct 10, 2022 at 15:58

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