Proving Functional Derivative for Current Research - Alice

[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:

<br /> \frac{\partial C(q(x))}{\partial q(x)} <br />

<br /> = \lim_{\delta q(x) \rightarrow 0} \frac{C[q(x) + \delta q(x)] - C[q(x)]}{\delta q(x)} <br />

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

My guess is that

<br /> \frac{dC(q(x) + k&#039;(q&#039;(x) - q(x)))}{dk&#039;} = \frac{\partial C(q(x) + k&#039;(q&#039;(x) - q(x)))}{\partial (q(x) + k&#039;(q&#039;(x) - q(x)))} \frac{d(q(x) + k&#039;(q&#039;(x) - q(x)))}{dk&#039;}<br />

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

Thank you very much!
Alice

 

[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&#039;(q&#039;(x)-q(x)))}{d k&#039;}d k&#039;

=\int _0^1d C(q(x)+k&#039;(q&#039;(x)-q(x)))

=[ C(q(x)+k&#039;(q&#039;(x)-q(x)))]_0^1

=C(q(x)+(q&#039;(x)-q(x)))-C(q(x))

=C(q&#039;(x))-C(q(x))

Now, you can do the substitution and continue.