This is an equation in a well known paper, which I cannot figure out: specifically, where does the 2nd power come from in the second integral?
Here $k_1$ is a scalar function of $c$, and $c$ is a scalar function of its location in space: $c(\mathbf{x})$.
This appears to be done using integration by parts:
$$ \int u dv = uv - \int v du $$
where in this case:
$$ u = k_1 $$ $$ du = dk_1/dc $$ $$ dv = \nabla^2 c $$ $$ v = \nabla c $$
which results in:
$$ \int_V (k_1 \nabla^2 c) dV = \int_S (k_1 \nabla c \cdot n) dS - \int_V \frac{dk_1}{dc} \nabla c\ dV $$
What am I missing? A similar result can be achieved using the divergence theorem, but again, I end up missing the: $(\nabla c)^2$.
I'll note that this cannot be a typo in the paper, and given how well established this paper and it's results are, I think it's overwhelmingly likely that I'm misunderstanding something basic about the math here.
Any help much appreciated!
Apparently $\kappa_1$ is a function of $c$. Then the divergence of $\kappa_1 \nabla c$ is $$ \nabla \kappa_1 \cdot \nabla c + \kappa_1 \nabla^2 c, $$ where the chain rule says that $$ \nabla \kappa_1 = \frac{d\kappa_1}{dc} \nabla c , $$ so that $$ \nabla \cdot (\kappa_1 \nabla c) = \frac{d\kappa_1}{dc} \nabla c \cdot \nabla c + \kappa_1 \nabla^2 c , $$ and the result follows from the divergence theorem.