Problem statement

Let $u(t) \in L^{2}(0, 1)$ and $J(u) = \int_0^1 tu(t) \int_0^t u(s)dsdt$

Compute first and second Frechet derivatives.

Attempted solution

$$ \begin{split} J(u + h) - J(u) &= \int_0^1t(u(t) + h(t))\int_0^t u(s) + h(s) dsdt - \int_0^1tu(t)\int_0^t u(s) dsdt \\ &= \int_0^1th(t)\int_0^t u(s) + h(s) dsdt + \int_0^1 tu(t) \int_0^t h(s) dsdt \\ &= \int_0^1th(t)\int_0^t h(s) dsdt + \int_0^1th(t)\int_0^t u(s) dsdt + \int_0^1 tu(t)\int_0^t h(s)dsdt \end{split} $$

Now, in the previous question of mine I was taught a nice trick of "exchange order of integration, rename back the variables", which applies nicely to the last integral:

$$ \begin{split} \int_0^1 tu(t)\int_0^t h(s)dsdt &= \int_{t = 0}^1 \int_{s = 0}^t tu(t)h(s)dsdt\\ &= \int_{s = 0}^1 h(s) \int_{t = 0}^s tu(t)dtds \\ &= \int_{t = 0}^1 h(t) \int_{s = 0}^s su(s)dsdt \end{split} $$

After that the derivative computation looks like:

$$ J(u + h) - J(u) = \int_0^1th(t)\int_0^t h(s) dsdt + \int_0^1h(t)\int_0^t u(s)(t + s) dsdt $$

I believe the first one to be $o(\|h\|_{L_2})$ and the second one to be $DJ(u)(h)$ but I have no idea of how to prove that. How is that shown?


Since you already know how to deal with functionals of the form $J(u)=\left< Au,u\right>$, it makes sense to put this one in the same form (which is possible, because it's quadratic in $u$). Namely, let $Au(t) = t\int_0^t u(s)\,ds$. To find the adjoint, work as usual (this is similar to what you do above, but simpler): $$\int_0^1 v(t)\, t\,dt \int_0^t u(s)\,ds = \int_0^1 u(s) \,ds \int_s^1 v(t)\, t\,dt \tag{1}$$ from where $A^* v (t) = \int_t^1 v(s)\,s\,ds$. Hence, the first derivative is $$Au+A^*u = t\int_0^t u(s)\,ds + \int_t^1 v(s)\,s\,ds = \int_0^1 u(s)\max(s,t)\,ds \tag{2}$$

This is not what you got, because you exchanged the order of integration incorrectly. From $0\le s\le t$ you should get $s\le t\le 1$.

Sanity check: the kernel $\max(s,t)$ in (2) is symmetric, which means the operator in (2) is self-adjoint, as $A+A^*$ must be.

The second derivative for this sort of functional was already discussed.


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.