Frechet differentiability of ODE functional 
Setup:
Suppose that $q(x)$ is a strictly positive $C^{2}$ function and consider the initial value problem:
\begin{equation}
\begin{aligned}
y''+q(x)y=&0\\
y(0)=&k\\
y'(0)=&c
\end{aligned}
\end{equation}
We know that such an initial value problem has a unique solution which we will denote by $f^q(x)$.  
Define the functional $F_{x}$ by
\begin{equation}
\begin{aligned}
F_{x}:C^{2}(\mathbb{R};\mathbb{R})&\rightarrow \mathbb{R}\\
%C^{1,2}(\mathbb{R}^2;\mathbb{R})\\
F_{x}(q)&\mapsto f^{q}(x).
\end{aligned}
\end{equation}
Question:
Is the functional $F_{x}$ Frechet differentiable?  How can we determine if it is?
 A: Fix $q_{0}\in C^{2}$. To find the Gateaux derivative let $p\in C^{2}$ and
consider $q_{0}+tp$ where $t\in\mathbb{R}$. Consider
\begin{gather*}
y^{\prime\prime}+(q_{0}(x)+tp(x))y(x)=0,\\
y(0)=k,\quad y^{\prime}(0)=c.
\end{gather*}
The $y$ depends on $x$, $t$, $q_{0}$, $p$, so we can write it as
$y=y(x,t;q_{0},p)$. In turn, we have
\begin{gather*}
\frac{\partial^{2}y}{\partial x^{2}}(x,t;q_{0},p)+(q_{0}(x)+tp(x))y(x,t;q_{0}%
,p)=0,\\
y(0,t;q_{0},p)=k,\quad\frac{\partial y}{\partial x}(0,t;q_{0},p)(0)=c.
\end{gather*}
Formally, if we take the derivative with respect to $t$ at $t=0$ we get
\begin{gather*}
\frac{\partial^{3}y}{\partial x^{2}\partial t}(x,0;q_{0},p)+(q_{0}%
(x)+0p(x))\frac{\partial y}{\partial t}(x,0;q_{0},p)+1p(x)y(x,0;q_{0},p)=0,\\
\frac{\partial y}{\partial t}(0,0;q_{0},p)=0,\quad\frac{\partial^{2}%
y}{\partial x\partial t}(0,0;q_{0},p)=0.
\end{gather*}
So if we write $y_{q_{0}}(x):=y(x,0;q_{0},p)$ and $z_{p}(x):=\frac{\partial
y}{\partial t}(x,0;q_{0},p)$, we have that $z$ solves the differential
equation
\begin{gather*}
z_{p}^{\prime\prime}+q_{0}z_{p}+py_{q_{0}}=0,\\
z_{p}(0)=0,\quad z_{p}^{\prime}(0)=0.
\end{gather*}
This can actually made rigorous.
So if we consider $F_{x}:C^{2}\rightarrow C$ and fix $q_{0}$ and a direction
$p$, the conjecture is that the directional derivative $\frac{\partial F_{x}%
}{\partial p}(q_{0})=z_{p}(x)$. If this is the case, then the Frechet
derivative at $q_{0}$ is given by $dF_{x}(q_{0})(p)=\frac{\partial F_{x}%
}{\partial p}(q_{0})=z_{p}(x)$ and so we need to prove that$$
\lim_{q\rightarrow q_{0}}\frac{F_{x}(q)-F_{x}(q_{0})-dF_{x}(q_{0})(q-q_{0}%
)}{\Vert q-q_{0}\Vert_{C^{2}}}=0.
$$
Let $y_{q}$ be the solution of the differential equation
\begin{gather*}
y_{q}^{\prime\prime}+q(x)y_{q}(x)=0,\\
y_{q}(0)=k,\quad y_{q}^{\prime}(0)=c.
\end{gather*}
Then we can rewrite the limit as$$
\lim_{q\rightarrow q_{0}}\frac{y_{q}(x)-y_{q_{0}}(x)-z_{q-q_{0}}(x)}{\Vert
q-q_{0}\Vert_{C^{2}}}=0,
$$
where $z_{q-q_{0}}$ solves
\begin{gather*}
z_{q-q_{0}}^{\prime\prime}+q_{0}z_{q-q_{0}}+(q-q_{0})y_{q_{0}}=0,\\
z_{q-q_{0}}(0)=0,\quad z_{q-q_{0}}^{\prime}(0)=0.
\end{gather*}
Since $y_{q}^{\prime\prime}+qy_{q}=0$, $y_{q_{0}}^{\prime\prime}+q_{0}%
y_{q_{0}}=0$, and $z_{q-q_{0}}^{\prime\prime}+q_{0}z_{q-q_{0}}+(q-q_{0}%
)y_{q_{0}}=0$, by subtracting the differential equations we get
\begin{align*}
y_{q}^{\prime\prime}-y_{q_{0}}^{\prime\prime}-z_{q-q_{0}}^{\prime\prime}  &
=-qy_{q}+q_{0}y_{q_{0}}+q_{0}z_{q-q_{0}}+(q-q_{0})y_{q_{0}}\\
& =-q(y_{q}-y_{q_{0}}-z_{q-q_{0}})+(q_{0}-q)z_{q-q_{0}}%
\end{align*}
with $(y_{q}-y_{q_{0}}-z_{q-q_{0}})(0)=0$ and $(y_{q}-y_{q_{0}}-z_{q-q_{0}%
})^{\prime}(0)=0$. By integration$$
(y_{q}-y_{q_{0}}-z_{q-q_{0}})(x)=\int_{0}^{x}\int_{0}^{s}-q(t)(y_{q}-y_{q_{0}%
}-z_{q-q_{0}})(t)+(q_{0}-q)(t)z_{q-q_{0}}(t)\,dtds
$$
and so$$
|(y_{q}-y_{q_{0}}-z_{q-q_{0}})(x)|\leq\int_{0}^{x}\int_{0}^{s}|q(t)||(y_{q}%
-y_{q_{0}}-z_{q-q_{0}})(t)|+|(q_{0}-q)(t)z_{q-q_{0}}(t)|\,dtds.
$$
It follows by Gronwall's inequality that%
\begin{align*}
|(y_{q}-y_{q_{0}}-z_{q-q_{0}})(x)|  & \leq\int_{0}^{x}\int_{0}^{s}%
|(q_{0}-q)(t)z_{q-q_{0}}(t)|\,dtds\exp\left(  \int_{0}^{x}\int_{0}%
^{s}|q(t)||(y_{q}-y_{q_{0}}-z_{q-q_{0}})(t)|\,dtds\right)  \\
& \leq\Vert q-q_{0}\Vert_{C^{2}}\int_{0}^{x}\int_{0}^{s}|z_{q-q_{0}%
}(t)|\,dtds\exp\left(  \int_{0}^{x}\int_{0}^{s}|q(t)||(y_{q}-y_{q_{0}%
}-z_{q-q_{0}})(t)|\,dtds\right)
\end{align*}
and so$$
\frac{|(y_{q}-y_{q_{0}}-z_{q-q_{0}})(x)|}{\Vert q-q_{0}\Vert_{C^{2}}}\leq
\int_{0}^{x}\int_{0}^{s}|z_{q-q_{0}}(t)|\,dtds\exp\left(  \int_{0}^{x}\int%
_{0}^{s}|q(t)||(y_{q}-y_{q_{0}}-z_{q-q_{0}})(t)|\,dtds\right)  .
$$
It remains to prove that the right hand-side goes to zero as $q\rightarrow
q_{0}$. By integration$$
z_{q-q_{0}}(x)=\int_{0}^{x}\int_{0}^{s}-q_{0}(t)z_{q-q_{0}}(t)+(q_{0}%
-q)(t)y_{q_{0}}(t)\,dtds
$$
and so by Gronwall's inequality
\begin{align*}
|z_{q-q_{0}}(x)|  & \leq\int_{0}^{x}\int_{0}^{s}|(q_{0}-q)(t)y_{q_{0}%
}(t)|\,dtds\exp\left(  \int_{0}^{x}\int_{0}^{s}|q_{0}(t)||\,dtds\right)  \\
& \leq\Vert q-q_{0}\Vert_{C^{2}}L.
\end{align*}
Hence, $\Vert z_{q-q_{0}}\Vert_{\infty}\leq\Vert q-q_{0}\Vert_{C^{2}%
}L\rightarrow0$ as $q\rightarrow q_{0}$. I am skipping the proof that $y_{q}$
is bounded.
