# Functional derivative

1. Jun 25, 2010

### alicexigao

For my current research, I need to prove the following:

$$\int_0^1 \frac{dC(q(x) + k'(q'(x) - q(x)))}{dk'}\,dk' = \int_0^1 \int_L^U p(q(x) + k(q'(x) - q(x)))(q'(x)-q(x)) dx dk$$

where $$C(q(x)) = \int_0^1 \int_L^U p(kq(x)) q(x)\,dx\,dk$$

Here's what I've tried using the definition of functional derivative:

$$\frac{\partial C(q(x))}{\partial q(x)}$$

$$= \lim_{\delta q(x) \rightarrow 0} \frac{C[q(x) + \delta q(x)] - C[q(x)]}{\delta q(x)}$$

$$= \int_L^U \int_0^1 \frac{\partial p(kq(x))}{\partial q(x)}q(x) + p(kq(x)) dk dx$$

My guess is that

$$\frac{dC(q(x) + k'(q'(x) - q(x)))}{dk'} = \frac{\partial C(q(x) + k'(q'(x) - q(x)))}{\partial (q(x) + k'(q'(x) - q(x)))} \frac{d(q(x) + k'(q'(x) - q(x)))}{dk'}$$

but I'm not sure what to do next. Any help will be greatly appreciated!

Thank you very much!
Alice

Last edited: Jun 25, 2010
2. Aug 17, 2010

### ross_tang

Hello, alicexigao. I can't really help you prove the identity. But I think it maybe able to evaluate to something more simple.

$$\int _0^1\frac{d C(q(x)+k'(q'(x)-q(x)))}{d k'}d k'$$

$$=\int _0^1d C(q(x)+k'(q'(x)-q(x)))$$

$$=[ C(q(x)+k'(q'(x)-q(x)))]_0^1$$

$$=C(q(x)+(q'(x)-q(x)))-C(q(x))$$

$$=C(q'(x))-C(q(x))$$

Now, you can do the substitution and continue.