- #1
alicexigao
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For my current research, I need to prove the following:
[tex]\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[/tex]
where [tex]C(q(x)) = \int_0^1 \int_L^U p(kq(x)) q(x)\,dx\,dk [/tex]
Here's what I've tried using the definition of functional derivative:
[tex]
\frac{\partial C(q(x))}{\partial q(x)}
[/tex]
[tex]
= \lim_{\delta q(x) \rightarrow 0} \frac{C[q(x) + \delta q(x)] - C[q(x)]}{\delta q(x)}
[/tex]
[tex]
= \int_L^U \int_0^1 \frac{\partial p(kq(x))}{\partial q(x)}q(x) + p(kq(x)) dk dx
[/tex]
My guess is that
[tex]
\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'}
[/tex]
but I'm not sure what to do next. Any help will be greatly appreciated!
Thank you very much!
Alice
[tex]\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[/tex]
where [tex]C(q(x)) = \int_0^1 \int_L^U p(kq(x)) q(x)\,dx\,dk [/tex]
Here's what I've tried using the definition of functional derivative:
[tex]
\frac{\partial C(q(x))}{\partial q(x)}
[/tex]
[tex]
= \lim_{\delta q(x) \rightarrow 0} \frac{C[q(x) + \delta q(x)] - C[q(x)]}{\delta q(x)}
[/tex]
[tex]
= \int_L^U \int_0^1 \frac{\partial p(kq(x))}{\partial q(x)}q(x) + p(kq(x)) dk dx
[/tex]
My guess is that
[tex]
\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'}
[/tex]
but I'm not sure what to do next. Any help will be greatly appreciated!
Thank you very much!
Alice
Last edited: