I am working on this problem
Consider the functional
$I=\int^a_b f (u'(x))dx$;
where $f \in C^1(\mathbb{R}; \mathbb{R})$ and $u \in \Phi =\{ u \in$ $D^1([a, b]; \mathbb{R})| u(a) = \alpha; u(b) = \beta \}$, for given
$\alpha; \beta \in \mathbb{R}$. ($D^1$: piecewise differentiable functions) Show that $\bar u(x) = \frac{\beta -\alpha}{b-a} (x-a)+ \alpha$ is a solution of the corresponding Euler-Lagrange equation and $u \in \Phi$.Is the function $ \bar u$ a minimizer in $\Phi_2 = D^0_1([0; 1];$ $\mathbb{R})$ (i.e piecewise differentiable functions that vanish at the endpoints) of $I(u) = \int_0^1 e^{−u'^2(x)}dx$ ?
I have already shown that $\bar u(x) $ is a solution of the corresponding Euler-Lagrange equation and $\bar u \in \Phi$. I am stuck in the second part. Any advice on how to proceed would be much appreciated!
My idea is that there is no minimizer because that exponential $e^{-x^2}$ in the xy plane has asymptote in y=0, so I need a sequence to prove the inf of the integrand is 0. On the other hand $ \bar u$ does not seem to belong to $\Phi_2$, because at the endpoints of $[0, 1]$ I don't get 0 but instead $ \bar u(0) = \alpha; \bar u(1) = \beta$ because of the first part
We can transcribe OP's variational problem into a related problem:
It is not hard to see that the constant function $v(x)=\frac{\beta -\alpha}{b-a}$ is a minimum (maximum) for the functional $J$ if and only if $\frac{\beta -\alpha}{b-a}$ is a minimum (maximum) point for the function $f$, respectively.
There is a bijective map $$\Phi^{\prime}~\ni~v\quad \mapsto\quad \left[ x\mapsto\alpha+\int_a^x\! dx^{\prime}~v(x^{\prime}) \right] ~\in~\Phi.$$
It follows that the affine function $\bar{u}$ is a minimum (maximum) for the functional $I$ if and only if $\frac{\beta -\alpha}{b-a}$ is a minimum (maximum) point for the function $f$, respectively.
In particular, the affine function $\bar{u}$ happens to be a maximum for the functional $I$ in OP's title example.