Computing Fréchet derivative 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.
 A: $$
F: \big(u(t),\lambda\big) \to \frac{d^2}{dt^2}u + \lambda \sin u
$$
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 $F_{u\lambda}$.
