Show that a functional has no minimum in a given set

Question

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 $a=0$?

Answer 1

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 $v^{\prime}(0)<a_0$ 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 $v(1)=0$. $\Box$