I have this functional $J(u)=\frac12 \|u\|^2+\int_0^1 F(t,Ku(t))dt$ where $F(t,u)=\int_0^u f(t,\xi) d\xi$,$\displaystyle Ku(t)=\int_0^1 G(t,s)u(s) ds$ with $G(t,s)=\begin{cases} s(1-t),&0\leq s \leq t\leq 1\\t(1-s), &0\leq t\leq s \leq 1\end{cases}$
we define the Nemytskii operator $N_f : H\rightarrow H$ where $N_{f}u(t)=f(t,u(t)),\ t\in [0,1]$
and we have $J'(u)v=(u-KN_fK u,v)$ ,
For $\tau>0$
I'm trying to prove that $\frac{d}{d\tau} J(\tau u)=(J'(\tau u),u)$
it's written like that, and it gives the area that we applied the derivative of a composed function!
But whene i used the definition (Gateaux) i found: $$\displaystyle\lim_{s\rightarrow 0} \frac{J(\tau u+sh)-J(\tau u)}{s}=(J'(\tau u),h)$$ for all $h\in H$ so $\nabla J(\tau u)=J'(\tau u)$
how to find that $\frac{d}{d\tau}J(\tau u)=(J'(\tau u),u)$ ??
Thank you.
Assume that $J \colon H \to \mathbb{R}$ is G-differentiable. By definition, given $u \in H$, there is a linear functional $J'(u) \in H^*$, the G-derivative at $u$. By the Riesz representation theorem, this element is isometrically identified to a vector $\nabla J(u) \in H$, called the gradient of $J$ at $u$. Now, consider thae map $$ \phi \colon \tau \xrightarrow{\ell}\tau u \xrightarrow{J}J(\tau u), $$ and apply the chain rule ($\ell'(\tau)=u$ because $\ell$ is linear in $\tau$): $$ \phi'(\tau) = \langle \nabla J(\tau u), u \rangle, $$ which you can rewrite as $$ \frac{d}{d\tau}J(\tau u)=\langle \nabla J(\tau u),u \rangle. $$