Is there a piecewise differentiable minimizer that vanishes at the endpoints of [0,1] of $I(u) = \int_0^1 e^{−u'^2(x)}dx$? 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
 A: We can transcribe OP's variational problem into a related problem:

Consider the functional
$$J[v]~:=~\int^a_b\! dx~f (v(x));$$
where $f \in C^1(\mathbb{R}; \mathbb{R})$ and
$$v ~\in~ \Phi^{\prime} ~:=~\left\{ v \in D^0([a, b]; \mathbb{R})\left| \int^a_b\! dx~v(x)=\beta-\alpha\right.\right\},$$
for given $\alpha; \beta \in \mathbb{R}$. ($D^0$: piecewise continuous functions.)

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.
