I am trying to solve the following problem: Let $f : [a,b] \times \mathbb{R} \times \mathbb{R} \rightarrow \mathbb{R}$ be defined as $f(x,u(x),u'(x)) = u'(x)^2 + \arctan(u(x))$ or $f(x,z,p) = p^2 + \arctan(z)$ and \begin{equation} F(u) = \int^{b}_{a}{u'(x)^2 + \arctan(u(x)) dx}. \end{equation} Now I want to figure out if $F$ has a global or local Minimizer in $C^1[a,b]$.
Therefore I already proofed a proposition which I should use:
Let $F: [a,b] \times \mathbb{R} \times \mathbb{R} \rightarrow \mathbb{R}$ be defined as $F(u) = \int^{b}_{a}{f(x,u(x),u'(x)) dx}$ and f is $C^2$.
If $u \in C^1[a,b]$ solves die euler lagrange equation
\begin{equation}
\frac{d}{dx}{f_p}(x,z,p) = f_z(x,z,p) \mbox{ in }[a,b]
\end{equation}
and $f_{pp}(x_0, u(x_0), u'(x_0)) \neq 0$ for some $x_0 \in (a,b)$ then $u \in C^2(U(x_0))$ for a neighborhood of $x_0$.
Now if $u$ would solve the euler lagrange equation than
for $f(x,z,p) = p^2 + \arctan(z)$ we get
\begin{equation}
f_z = \frac{1}{1+u(x)^2} \stackrel{!}{=} 2 u''(x) = \frac{d}{dx}f_p.
\end{equation}
This expression doesn't seem to have a solution.
Furthermore
\begin{equation}
f_{pp} = 2.
\end{equation}
On the other hand I also calculated the first and second variation for $\xi \in C^{\infty}(a,b)$ for $\epsilon_0 > 0$, $u + \epsilon \xi \in C^1[a,b]$ for $|\epsilon| < \epsilon_0$
\begin{align}
\frac{d}{d \epsilon} \phi(\epsilon) &= \frac{d}{d \epsilon} \int^{b}_{a}{(u'(x) + \epsilon \xi'(x))^2 + \arctan(u(x) + \epsilon \xi(x)) dx} \\
&\stackrel{\mbox{Maj. derivative}}{=} \int^{b}_{a}{2\xi'(x)(u'(x) + \epsilon \xi'(x)) + \xi(x) \frac{1}{1+ (u(x) + \epsilon \xi(x))^2} dx} \\
&\stackrel{\mbox{Dom. convergence}, \epsilon \rightarrow 0}{\rightarrow} \int^{b}_{a}{2\xi'(x)u'(x) + \frac{\xi(x)}{1+u(x)^2} dx} \\
&=\int^{b}_{a}{\frac{\xi(x)}{1+u(x)^2} - 2\xi(x)u''(x) dx} \\
&= \int^{b}_{a}{\xi(x)\left( \frac{1}{1+u(x)^2} - 2u''(x)\right) dx} \\
&= \delta F(u,\xi).
\end{align}
Therefore our natural boundary conditions are $u'(a) = u'(b) = 0$ such that we can get the euler lagrange equation in the integrand and $\delta F(u,\xi) = 0$ for $\xi \in C^{\infty}[a,b]$.
The second variation is given by \begin{align*} \frac{d}{d \epsilon} \phi'(\epsilon) &= \frac{d}{d \epsilon} \int^{b}_{a}{2\xi'(x)(u'(x) + \epsilon\xi'(x) ) + \frac{\xi(x)}{1+(u(x) + \epsilon \xi(x))^2} dx} \\ &\stackrel{\mbox{Maj. derivative}}{=} \int^{b}_{a}{2 \xi'(x)^2 -\xi(x) \frac{2 \xi(x)(u(x) + \epsilon \xi(x))}{(1+ (u(x) + \epsilon \xi(x))^2 )^2}} \\ &\stackrel{\mbox{Dom. convergence}}{=} 2 \left( \int^{b}_{a}{\xi'(x)^2-\xi(x)^2 \frac{u(x)}{(1 + u(x)^2)^2}} \right) = \delta^2 F(u, \xi) \end{align*} I tried to proof that there is no global minima of $F$, by assuming there is one and try to find a contradiction:
The functional $F$ is bounded below by $-(b-a)\frac{\pi}{2}$ and if I assume that there exists a $u_0$ which is a global minima of $F$, than $u_0$ solves the euler lagrange equation $2u_0'' = \frac{1}{1+u_0^2}$ and the natural boundary condition $u_0'(a) = u_0'(b) = 0$. Also from the proposition it follows that $u_0 \in C^2([a,b])$ since $f_{pp} = 2$. As stated in the answer, there is no $C^2$ function which is a solution of the euler lagrange equation satisfing the natural boundary condition.
Also multipling the euler lagrange equation with $u_0'$ gives
\begin{equation}
(\arctan(u_0))' = (u_0'^2)'
\end{equation}
and therefore $\arctan(u_0(x)) = u_0'(x)^2 + C$ with the natural boundary condition we get $\arctan(u_0(a)) = \arctan(u_0(b)) = C$, so $u_0(a) = u_0(b) = \tan(C)$.
I also figured out that if $u_0$ would be a global minimizer in $C^1[a,b]$ which satisfies the natural boundary conditions we would have with $\arctan(u_0(x)) = u_0'(x)^2 + C$ since $u_0$ is a global minimizer and $\arctan$ is strictly monotone decreasing and in $C^1[a,b]$ \begin{equation} F(u_0) = \int^{b}_{a}{u'(x)^2 + \arctan(u(x))dx} \geq \int^{b}_{a}{\arctan(u(x))dx} \geq (b-a)\inf_{x \in [a,b]}\arctan(u(x)) = (b-a) \arctan(\min_{x \in [a,b]}u(x)) \geq (b-a) \arctan(C) \end{equation} for some $C \leq 0$ since $-\infty < \min_{x \in [a,b]}u(x) < \infty$ we can choose $u \equiv C \in C^1[a,b]$ with $\frac{\partial }{\partial p}F(x,z,p) = 2p = 2u'(x) = 0$. Therefore $F$ can't have a globale minimizer since we look at $u \in C^1[a,b]$ which are satisfying the natural boundary conditions. But $F$ can have local minimizers $u \equiv C \in C^1[a,b]$ such that for $\delta > 0$ we have $C < \min_{x \in [a,b]}u(x) $ or $|C - u(x)| < \delta$ for all $x \in [a,b]$ and $|u'(x)| < \delta$ for $x \in [a,b]$.
I would appreciate it if someone can give me a hint.
The differential equation $$ 2u_0'' = \frac1{1+u_0^2}, \qquad u_0'(a)=u_0'(b)=0, $$ does not have a solution:
Any solution $u_0$ has to satisfy that $u_0'$ is strictly monotone, because the second derivative is positive. But this is not possible if $u_0'$ also has to be equal at both endpoints. Thus, $u_0$ cannot exist.