0
$\begingroup$

Let $u(t) \in L^{2}(0, 1)$. I need to calculate the first and second Frechet derivatives of $$J(u) = \int_0^1 \left(\int_0^{t^3}u(s)ds\right)^2dt$$

I am completely at a loss here: I know several tricks that facilitate computation of Frechet derivatives:

  1. Decompose operator to a composition of familiar operators and apply chain rule.
  2. Understand the action performed by the operator (like the shift operator), then apply matrix calculus.

However, my problem has an intergral with a changing limit. I don't even know if I can apply the common-calculus approach:

If $I(t) = \int_{x_1(t)}^{x_2(t)} f(x,t)dx$, then $I'(t) = f(x_2, t)\frac{dx_2}{dt} - f(x_1, t)\frac{dx_1}{dt} + \int_{x_1(t)}^{x_2(t)} \frac{\partial f}{\partial t}dx$.

And even if I could, how do I start?

Update:

@Stephen's comment got me thinking:

$$ \lim_{h\to0} \frac{|J(u + h) - J(u) - DJ(u)(h)|}{\|h\|_{L^2}} = \lim_{h\to0} \frac{|2 \int_0^1 \left( \int_0^{t^3} u(s) ds \int_0^{t^3}h(s)ds\right)dt + \int_0^{1} \left(\int_0^{t^3}h(s)ds\right)^2dt - DJ(u)(h)|}{\|h\|_{L^2}} $$

Now I am very tempted to introduce an operator $Au = \int_0^{t^3}u(s)ds$, rewrite the above as

$$ \frac{|2\left<Au, Ah\right> + \left<Ah, Ah\right> - DJ(u)(h)|}{\|h\|_{L^2}} $$

and find $DJ(u)$. However, that $t^3$ that is the upper limit of the integral when I introduce $A$ is confusig... Am I making a mistake?

$\endgroup$
  • 2
    $\begingroup$ $t$ is a dummy variable. You shouldn't be differentiating with respect to $t$. $J$ is a linear function, therefore its derivative is itself. $\endgroup$ – Stephen Montgomery-Smith Mar 23 '14 at 18:06
  • $\begingroup$ Also, nothing to do with answering your question, but you can use Fubini's Theorem and see that $J(u) = \int_0^1 \int_{s^{1/3}}^1 \,dt u(s) \,ds = \int_0^1 (1-s^{1/3}) u(s) \, ds$. $\endgroup$ – Stephen Montgomery-Smith Mar 23 '14 at 18:07
  • $\begingroup$ @Stephen sorry, my misprint. J(u) isn't linear, I am now trying to understand why $t$ is a dummy. $\endgroup$ – alisianoi Mar 23 '14 at 18:09
  • 1
    $\begingroup$ You are essentially correct, although your notation is messed up. No remember what the inner product is, and then apply Fubini's Theorem one or two times. In my answer, you can replace $\delta u$ by $h$ if you prefer. $\endgroup$ – Stephen Montgomery-Smith Mar 23 '14 at 18:30
2
$\begingroup$

Think of $J(u + h) - J(u) = F(u) \cdot h + o(h)$, where $u\cdot v$ in this context means $\int_0^1 u(s) v(s) \, ds$. So $$ \begin{split} J(u + h) -J(u) &= \int_0^1 \left(\int_0^{t^3} u(s)+h(s) \, ds \right)^2 \, dt - \int_0^1 \left(\int_0^{t^3} u(s)\, ds \right)^2 \, dt\\ &= \int_0^1 2 \left(\int_0^{t^3} u(s)\, ds \right) \left(\int_0^{t^3} h(s)\, ds \right) \, dt + o(h) \\ &= 2 \int_0^1 \int_0^{t^3} \int_0^{t^3} u(s) h(r) \, ds \, dr \, dt + o(h) . \end{split} $$ Now interchange the integrals between $r$ and $t$, and you get $$ 2 \int_0^1 \int_{r^{1/3}}^{1} \int_0^{t^3} u(s) h(r) \, ds \, dt \, dr + o(h) .$$ Hence the derivative of $J$ at $u$ is $$ F(u)(r) = 2 \int_{t=r^{1/3}}^{1} \int_{s=0}^{t^3} u(s) \, ds \, dt = 2 \int_{s=0}^1 \int_{t=\max\{r^{1/3},s^{1/3}\}}^1 u(s) \, ds \\ = \int_0^1 (1-\max\{r^{1/3},s^{1/3}\}) u(s) \, ds \\ = (1-r^{1/3}) \int_0^r u(s) \, ds + \int_r^1 (1-s^{1/3}) u(s) \, ds. $$

$\endgroup$
  • $\begingroup$ After the edit makes a bit more sense. But I certainly do not understand how you got a triple integral. $\endgroup$ – alisianoi Mar 23 '14 at 18:30
  • 1
    $\begingroup$ It is exactly the same as your $2 \langle Au, Ah \rangle$. $\endgroup$ – Stephen Montgomery-Smith Mar 23 '14 at 18:31
  • 1
    $\begingroup$ But also, I used that $s$ is a dummy variable, so I can change the name of one of them to $s'$. Then I can pull the integrals into each other and write it as one double integral (inside the $t$ integral, hence a triple integral). $\endgroup$ – Stephen Montgomery-Smith Mar 23 '14 at 18:33
  • $\begingroup$ I no longer believe $t$ to be dummy. It just can't be: I have $\int_0^1 I(u, t) dt$ where $I(u, t)$ is some integral. You are telling to exchange $s'$ and $t$ but that can't be done: the upper limit in $I(u, t)$ depends on $t$! $\endgroup$ – alisianoi Mar 23 '14 at 19:10
  • 1
    $\begingroup$ $\int_{t=0}^1 \int_{s=0}^{t^3} = \int_{s=0} ^1 \int_{t=s^{1/3}}^1 $. Or write it as $\int \int _ {0 \le s \le t^3 \le 1}$. $\endgroup$ – Stephen Montgomery-Smith Mar 23 '14 at 19:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.