I am reading Methods in Nonlinear Analysis by Kung-Ching Chang and having trouble in obataining a Fréchet derivative in the text.
For those who has the book, it is on page 37, which concern Euler elastic rod.
Let \begin{aligned} &\ddot\varphi+\lambda\sin\varphi=0 \ \ \text{in} \ \ (0, \pi) \\ &\dot \varphi(0)=\dot \varphi(\pi). \end{aligned}
Also let \begin{aligned} &X=\{u\in C^2[0,\pi]: \dot u(0)=\dot u(\pi)=0\}\\ &Y=C[0,\pi], \end{aligned}
and let $F: X \times \mathbb{R} \rightarrow Y$ be the map
$$(u,\lambda) \rightarrow u''+\lambda \sin u.$$
The book claimed that $F''_{u\lambda}(0,n^2)=\cos (nu)|_{u=0}=I$, where $I$ is the identity map.
First, I am not quite sure about what the notation $F''_{u\lambda}(0,n^2)$ mean, I guess it mean double partial Fréchet derivative evaluated at $(0,n^2)$, but I am not able to arrive that result. So far, I can only obtain the partial Fréchet derivative
$$ F_u(u,\lambda)h \rightarrow h''+\lambda h \cos u. $$
Help and comments greatly appreciated. Thanks.
$$ F: \big(u(t),\lambda\big) \to \frac{d^2}{dt^2}u + \lambda \sin u $$
Expand $\sin u$ into series around $u=0$, get linearization of $F$ operator: $$ F\big(u,\lambda\big) = \frac{d^2}{dt^2}u + \lambda \left( u - \frac{u^3}{3!} + \cdots\right) \approx \bigg[ \frac{d^2}{dt^2} + \lambda I\bigg] u, $$ where $I$ is an identity mapping. Denote $L:X\times R\to X$ the linearization of $F$: $$ L u:= \bigg[ \frac{d^2}{dt^2} + \lambda I\bigg]u, $$ then the kernel $\ker L$ is the set of eigenvalues of $\frac{d^2}{dt^2}$ operator with corresponding eigenvalues $\lambda $, i.e. $$ \ker L = \big\{ C\cos\left(nt \right)\, \big| \ C - \operatorname{const}., \ \lambda = n^2 \big\} $$ Since $$ Fu = L u + r = \bigg[ \frac{d^2}{dt^2} + \lambda I\bigg] u + O(u^3) \implies F_u \approx L, $$ and $$ F_u(u, \lambda) = F_uu \approx Lu$$ the kernel of $F_u$ "coincides" with the kernel of $L$.
Now, $$ F_{u\lambda} = \frac{d}{d\lambda} F_u(u, \lambda) \approx \frac{d}{d\lambda} Lu = u \implies F_{u\lambda} \approx I - \text{ the identity map.} $$ On the space of eigenfunction of $L$ we have $$ F_{u\lambda}(u,\lambda)\Big|_{\lambda=n^2} = \frac{d}{d\lambda} Lu \Big|_{\lambda=n^2}= u \Big|_{\lambda=n^2} = C \cos (nu) $$ Finally, $$ F_{u\lambda}\left(u,n^2\right)\Big|_{u=0}= C \cos(nu) \Big|_{\lambda=n^2} = I, $$ which matches approximate (near zero) calculations for $F_{u\lambda}$.