Prove that the minimum of a functional doesn't exist Prove that there is no smooth solution ho the minimization problem:
$$\mathcal{L} (u)= \int_{0}^1 e^{-u'}+u^2 dx$$
Where the admissible space is $X =\{ u \in \mathcal{c}^2 [0,1] | u(0)=0,  u(1)=1
       \} $
UPDATE:
GOAL: I am trying to define a sequence of functions $u_n(x)$ in $X$ s.t their integrals $\mathcal{L}(u_n) \to 0$
What I have done so far:
Define $u_n(x)$ in the following manner:
$$u_n (x)= \begin{cases}
0, \quad 0 \leq x\leq 1-\frac{1}{n} \\[2ex]
ax^2 +bx+ c, \quad 1-\frac{1}{n}<x \leq 1 
\end{cases}$$
where $n=1,2,3,..$. Require that $u_n$ satisfy:
\begin{align}
 u_n(1-\frac{1}{n})=0\\
 u_n(1)=1\\
 u_n'(1-\frac{1}{n})=0\\
\end{align}
We then have:
\begin{align}
a+b+c=1\\
a(1-\frac{1}{n})^2+b(1-\frac{1}{n})+c=0\\
2a(1-\frac{1}{n})+b=0 
\end{align}
Find $a,b,c$ then:
$$u_n (x)= \begin{cases}
0, \quad 0 \leq x\leq 1-\frac{1}{n} \\[2ex]
n^2x^2 -2n(n-1)x+ (n-1)^2, \quad 1-\frac{1}{n}<x \leq 1 
\end{cases}$$
\begin{align}
\int_0^1 e^{-u_n(x)'} + u_n^2(x) dx &= \int_0^{1-\frac{1}{n}} e^{0} + 0 dx +  \int_{1-\frac{1}{n}}^1 e^{-(2n^2 x -2n(n-1))} + (n^2x^2 -2n(n-1)x+ (n-1)^2)^2 dx\\& = 1- \dfrac{1}{n}+ \dfrac{\mathrm{e}^{-2n}\left(\left(2n+5\right)\mathrm{e}^{2n}-5\right)}{10n^2}
\end{align}
$$\dfrac{\mathrm{e}^{-2n}\left(\left(2n+5\right)\mathrm{e}^{2n}-5\right)}{10n^2} \to 0$$ and $$1-\dfrac{1}{n} \to 1$$
 A: We can more or less use the same proof technique as in my Math.SE answer here.

*

*Define function $$f(a)~:=~(a+1)e^{-a}.\tag{1}$$
Now it is a fact that
$$ \exists ! a_0 <0:~~ f(a_0)~=~0. \tag{2}$$
$$ \forall a> a_0: ~~ 0~<~f(a)~\leq~1.\tag{3}$$
$$ \forall a< a_0: ~~ f(a)~<~0.\tag{4}$$
Here $a_0=-1$. There is a first integral to OP's EL equation. Together with OP's boundary conditions, it becomes $$  f(u^{\prime}(0))~=~ f(u^{\prime}) + u^2 
~=~f(u^{\prime}(1))+1 .\tag{5} $$
In light of eqs. (2)-(4) we see that eq. (5) is impossible to satisfy for $u^{\prime}(1)>a_0$ because then $f(u^{\prime}(1))>0$ while $f(u^{\prime}(0))\leq 1$.


*The EL equation reads
$$ \left(-e^{-u^{\prime}}\right)^{\prime}~=~2u.\tag{6} $$
If in the interval $[x_0,1]$ the function $u > 0$ is positive, then $-e^{-u^{\prime}}$ (and thereby $u^{\prime}$) is increasing, and hence $u^{\prime}\geq 0$. Then
$$u^{\prime}(1)~\geq~ 0,\tag{7}$$
which according to section 1 leads to a contradiction. $\Box$
A: Observe that you cannot have $\mathcal{L}(u)=0$, since that would imply $u=0$ a.e., which contradicts the fact that $u(1)=1$ (+continuity). Therefore, $\mathcal{L}(u)>0$ for all $u\in X$.
As someone already mentioned, you can consider the sequence $u_n(x)=x^n$, which satisfies $\mathcal{L}(u_n)\to 0$ (by the dominated convergence theorem).
This implies that $\mathcal{L}$ does not attain its minimum on $X$.
