Question

When
\[\mathcal{F}=\int \frac{1}{2}\alpha | \nabla C(\vec{x}) |^2 + \beta \Psi(C(\vec{x})) d\vec{x}\],
with alpha and beta some certain constants and psi a certain function of C, what is
\[\frac{\partial \mathcal{F}}{\partial C}\]

Answer 1

Ok as the equation interpreter has died:

When \[\mathcal{F}=\int \frac{1}{2}\alpha | \nabla C(\vec{x}) |^2 + \beta \Psi(C(\vec{x})) d\vec{x}\], with alpha and beta some certain constants and psi a certain function of C, what is \[\frac{\partial \mathcal{F}}{\partial C}\]

Answer 2

Do you think the chain rule could help?

Answer 3

\[
\frac{d\mathcal{F}}{d\vec{x}} =\frac{1}{2}\alpha | \nabla C(\vec{x}) |^2 + \beta \Psi(C(\vec{x}))
\]

Answer 4

I would suppose the chain rule is something like: \[
\frac{d\mathcal{F}}{d\vec{x}} = \frac{\partial \mathcal{F}}{\partial C} \cdot \huge ?

\]

Answer 5

\frac{dC}{d\vec{x}}

Answer 6

Ah yes
\[\frac{dC}{d\vec{x}}\]

Answer 7

So just divide the integrand by that and you got it? I dunno how else to simplify it.

Answer 8

Ah of course, as
\[\frac{\partial \mathcal{F}}{\partial \vec{x}}\]
is the integrand

Answer 9

Though I'm a bit concerned because I don't remember the chain rule for the multivariable case.

Answer 10

Well I have the answer at hand but the road to it is a bit unclear, would it help?

Answer 11

Sure.

Answer 12

\[\frac{\partial \mathcal{F}}{\partial C} = \beta \Psi'(C) - \alpha \nabla^2C\]

Answer 13

So I guess the chain rule on the part with alpha would:
\[\nabla \cdot \left(\frac{1}{2} \alpha |\nabla C|^2 \right) = \alpha \nabla \cdot \left(\nabla C\right)\]

Answer 14

I'm just not sure how to 'divide the integrand by'
\[\frac{dC}{d\vec{x}}\]

Answer 15

Oh, it's just division by a function, but in this case it is clearly not what they're looking for anyway.

Answer 16

Do you know Leibniz integral rule?

http://mathworld.wolfram.com/LeibnizIntegralRule.html

Answer 17

I do now :)

Answer 18

Ah, that seems promising.

Answer 19

Though I have an indefinite integral

Answer 20

It seems that using the variational derivative:
\[F(C(x)) = \int f(C,\nabla C)\, dx\]
Then the Euler-Lagrange equation:
\[\frac{\partial \mathcal{F}(C)}{\partial C} = \frac{\partial f(C,\nabla C)}{\partial C} - \nabla \cdot \frac{\partial f(C, \nabla C)}{\partial \nabla C}\]
Then:
\[\frac{\partial \mathcal{F}(C)}{\partial C} = \beta \Psi'(C) - \nabla \cdot \left(\alpha \nabla C\right)\]
where
\[\frac{\partial f(C, \nabla C)}{\partial \nabla C}=\frac{\partial}{\partial \nabla C} \left(\frac{1}{2}\alpha |\nabla C|^2\right)\]