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:
- Decompose operator to a composition of familiar operators and apply chain rule.
- 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?
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 $$ F(u)(r) = 2 \int_{t=r^{1/3}}^{1} \int_{s=0}^{t^3} u(s) \, ds \, dt = 2 \int_{s=0}^1 \int_{t=\max\{r^{1/3},s^{1/3}\}}^1 u(s) \, ds \\ = \int_0^1 (1-\max\{r^{1/3},s^{1/3}\}) u(s) \, ds \\ = (1-r^{1/3}) \int_0^r u(s) \, ds + \int_r^1 (1-s^{1/3}) u(s) \, ds. $$