I am trying to follow along a review paper regarding the derivation of the Cahn-Hilliard paper. The question I have is related to the following section of the paper:
I have been trying to follow along with how the author went from line 2 to line 3. In the first line, the author uses the definition of $\mu = F'(c) – \epsilon^2\Delta c$ and then the vector identity that states:
$$ \nabla c \cdot \nabla c_t = \nabla\cdot c_t\nabla c-c_t \Delta c $$
Using this, I can arrive at the rightmost expression on the first line. To get to the second line, the author then uses the fact that:
$$ c_t = \nabla\cdot(M\nabla\mu) $$
So everything makes sense to me up until that second line. However, I don't quite see how to then arrive at the third line. Using vector identities again, for a scalar function $M(c)$ and a vector function $\nabla\mu$, we have that:
$$ \nabla \cdot (M\nabla\mu) = M\nabla\cdot\nabla\mu+\nabla\mu\cdot\nabla M $$
If I make this substitution in the second line, I then have that:
$$ =\int_\Omega \mu M\nabla\cdot\nabla\mu\ d\vec{x} + \int_\Omega \mu\nabla\mu\cdot\nabla M \ d\vec{x} $$
Applying the Divergence theorem to the first integral immediately above reduces this to the surface integral shown in the image at the top (which the author then argues vanishes because by an earlier assumption $M\nabla\mu\cdot\vec{n}=0$ on $\partial\Omega$). However, it is not clear to me how the second term in my expression can be rearranged to arrive at what the author is showing.
Am I doing something wrong or is there simply an extra step required that I am not seeing? For reference, the paper that I am referring to is "Physical, Mathematical, and Numerical Derivations of the Cahn-Hilliard Equation" by Lee, Huh, Jeong, Shin, Yun, and Kim.
Best Answer
I think that the step you miss is the formula :
$\int_{\Omega} b \cdot \nabla a dx = \int_{\partial \Omega} a b \cdot n ds - \int_{\Omega} a \nabla \cdot b dx$
which is the so-called Divergence theorem, that you use with the fact that $\nabla \cdot (a b) = a \nabla \cdot b + \nabla a \cdot b$ for vectorial $b$ and scalar $a$.
In your case, you apply it with $a = \mu$, $b = M \nabla \mu$.