Show that a functional has no minimum in a given set

This is a problem from my calculus of variations class:

Let $$X=\{v:[0,1]\rightarrow\mathbb{R}$$ of class $$C^1$$, $$v(0)=1$$,$$\ v(1)=0\}$$, let $$F:X\rightarrow\mathbb{R}$$ be the functional $$F(v):=\int_0^1 (e^{v'(x)}+v^2(x))dx.$$ Integrate the Euler-Lagrange equation with conditions $$v(0)=1$$, $$\ v'(0)=a$$, $$a\in\mathbb{R}$$ and show that $$F$$ has no minimum on $$X$$.

The professor also said that we could restrict ourselves to the case $$a<0$$.
Ok, I did my homework: first we define $$X_a:=\{v:[0,1]\rightarrow\mathbb{R}$$ of class $$C^1$$, $$v(0)=1$$,$$\ v'(0)=a\}$$. The Euler-Lagrange equation for $$F$$ is $$v''e^{v'}=2v$$. Since the functional $$F$$ doesn't depend explicitely on $$x$$ the quantity $$v'e^{v'}-e^{v'}-v^2$$ is some constant, say $$k$$, thus imposing the boundary conditions at $$x=0$$ we get $$e^{v'}(v'-1)-v^2=e^{a}(a-1)-1$$ (which is the du Bois-Reymond equation and it is exactly what we would have got integrating $$v'(v''e^{v'}-2v)=0$$). Now, if a minumum exists in $$X$$ it must have bounded derivative in $$x=0$$, so it must be in $$X_a$$ for some $$a<0$$. Let's see if the boundary conditions of $$X$$ are compatible with the boundary conditions of $$X_a$$. In $$x=1$$ we have $$e^{v'(1)}(v'(1)-1)=e^{a}(a-1)-1$$. Since $$v'(1)$$ is something in $$\mathbb{R}$$, we reduced the problem to the study of the zeros of the function $$g(y):=e^{y}(y-1)-e^{a}(a-1)+1$$. It turns out that $$g$$ has no zeros for $$a<0$$, so we are done.
Let's get to the question: why can we restrict ourselves to $$a<0$$?
For $$a\geq0$$ g has one zero, so there could be a candidate minimum in $$X\cap X_a$$. Graphically, I can believe that the function $$v$$ has to be decreasing to minimize the functional, but this is not a proof, nor a satisfactory justification.
I thought that if we can bound from below the values that $$F$$ assumes on $$X\cap X_a$$, $$a\geq0$$ with $$F(w)$$ for some $$w$$ not in $$X\cap X_a$$, this will be enough to show that $$F$$ has no minimum in $$X$$. Nevertheless this seems not so simple, since we don't have an explicit formula for the candidate minimum.

$$\boldsymbol{Edit}$$: A colleague of mine told me that for the case $$a>0$$ I have to look at the sign of the second derivative. From the Euler-Lagrange equation we get $$sign(v''(x))=sign(v(x))$$ $$\forall x\in(0,1)$$, which implies that the first derivative is increasing in $$(0,1)$$. By continuity and positivity at 0 of the first derivative we get that $$v$$ is increasing in $$[0,1]$$, thus it cannot reach the point 0 at $$x=1$$.
My question is: does this argument cover also the case $$a=0$$?

1. Summarizing OP's analysis: Define function $$f(a)~:=~(a-1)e^a.\tag{1}$$ Now it is a fact that $$\exists ! a_0 >0:~~ f(a_0)~=~0. \tag{2}$$ $$\forall a< a_0: ~~ -1~\leq~f(a)~<~0.\tag{3}$$ $$\forall a> a_0: ~~ f(a)~>~0.\tag{4}$$ Here $$a_0=1$$. There is a first integral to OP's EL equation. Together with OP's boundary conditions, it becomes $$f(v^{\prime}(0))-1~=~ f(v^{\prime}) - v^2 ~=~f(v^{\prime}(1)) .\tag{5}$$ In light of eqs. (2)-(4) we see that eq. (5) is impossible to satisfy for $$v^{\prime}(0) because then $$f(v^{\prime}(0))<0$$ while $$f(v^{\prime}(1))\geq -1$$.
2. Next consider the case $$v^{\prime}(0)~\geq~ 0.\tag{6}$$ The EL equation reads $$\left(e^{v^{\prime}}\right)^{\prime}~=~2v.\tag{7}$$ If in the interval $$[0,x_0]$$ the function $$v > 0$$ is positive, then $$e^{v^{\prime}}$$ (and thereby $$v^{\prime}$$) is increasing, and hence $$v^{\prime}\geq 0$$ is non-negative by assumption (6). Therefore the function $$v > 0$$ is weakly increasing. Therefore we may assume $$x_0=1$$. Contradiction with the boundary condition $$v(1)=0$$. $$\Box$$
• Actually I think I made a mistake studying the zeros of the function $f$: $f$ is strictly larger that zero even for $a=0$. Thus the case $a=0$ was included in the first case. Anyway, thanks for your answer. Apr 3, 2017 at 17:52
• How did you get this expression $v'e^{v'}-e^{v'}-v^2$? How did you solve the E-L eqn? Feb 6, 2021 at 22:48