For an exercise (3.2.27 in P. Drábek, J. Milota, "Methods of Nonlinear Analysis"), I'm trying to differentiate (in the Fréchet sense) this operator
$$ F(\varphi) = \int_0^1 \left[ \int_0^t |\varphi(s)|^2 \text{d}s \right]^3 \text{d}t $$
both for $\varphi \in C[0,1]$ and $\varphi \in L^2(0,1)$.
I'm trying to informally guess an expression for the Fréchet derivative of $F$ using the chain rule and the fact that the derivative of the Volterra operator $V(\psi)(t) = \int_0^t \psi(s) \text{d}s$ (which is well defined both on $C[0,1]$ and $L^2(0,1)$) is given by $V'(\psi)(h) = \int_0^t h(s) \text{d} s$. But I don't know how to deal with the outer integral (can I just bring the differentiation operator inside?) and with the composition of the Volterra operator with the absolute value (or complex modulus) squared.
Can someone give me an idea on how to compute this kind of derivatives? An informal derivation would also suffice, then I would need to verify that this is indeed the Fréchet derivative of $F$ at $\varphi$ by showing that it is linear, bounded, and that it satisfies
$$ \lim_{h \to 0} \frac{\lVert F(\varphi + h) – F(\varphi) – F'(\varphi)h\rVert}{\Vert h \rVert } = 0 $$
(the norms are either the $L^2$-norm or the $C^0$-norm, depending on what we're considering to be the domain of the operator).
Edit: I think I can introduce the following operators (let's consider the case $C[0,1]$ for the moment):
$$ A : C[0,1] \to \mathbb{R} : \omega \mapsto \int_0^1 (\omega(t))^3 \text{d}t \\
B : C[0,1] \to C[0,1] : \psi \mapsto \int_0^t \psi(s) \text{d} s \\
C : C[0,1] \to C[0,1] : \varphi \mapsto |\varphi|^2,$$
and rewrite $F$ as the composition $A \circ B \circ C$, so that I can apply the chain rule:
$$ F'(\varphi) = A'(B \circ C(\varphi)) B'(C(\varphi)) C'(\varphi). $$
Would this approach work? Can someone help me with these computations?
Best Answer
I will handle the case of $C[0,1]$ on the reals. Firstly split the functional into four parts:
Define \begin{align} H : C[0,1] &\longrightarrow C[0,1] \\ \varphi &\longmapsto \big( s \mapsto |\varphi(s)|^2 \big), \end{align} \begin{align} G : C[0,1] &\longrightarrow C[0,1] \\ \psi &\longmapsto \big( t \mapsto \psi(t)^3 \big) \end{align} and $J : C[0,1] \to \mathbb{K}$ by $J(\psi) = \int_0^1 \psi (t) dt$. Then clearly $F = J \circ G \circ V \circ H$. Here $V$ is the Voltera operator. Since $J$ and $V$ are continuous and linear their derivatives are just themselves.
To find the derivative of $G$ let $\psi , \eta \in C[0,1]$. Then $$ G(\psi + \eta ) - G(\psi) = (\psi + \eta )^3 -\psi^3 = 3 \psi \eta^2 + 3 \psi^2 \eta + \eta^3 $$ from which it follows that $DG(\psi) = M_{3 \psi^2} $, where $M_{3 \psi^2} $ is the multiplication with $3 \psi^2$ operator and $DG(\psi)$ is the derivative of $G$ at $\psi$.
If the underlying field is the real numbers we can compute the derivative of $H$ in the same way to receive $DH(\psi) = M_{2 \psi}$.
The chain rule gives: $$DF (\psi)= J \circ DG (V H(\psi)) \circ V \circ DH(\psi) = J M_{3 (V \psi^2)^2} V M_{2\psi}.$$
Inserting all the definitions: $$ DF(\psi) (\eta) = \int_0^1 \bigg[3\bigg( \int_0^t \psi^2(s) ds \bigg)^2 \int_0^t 2 \psi(s) \eta(s) ds \bigg]dt .$$