I am trying to prove that there exists a function $u$ satisfies the following conditions:
Here $\eta>0$ small is a given constant.
- $u$ is defined on $[0,T]$ where $T$ might depends on $\eta$.
- $0\leq u\leq 1$
- $u(0)=0$ and $u(T)=1$.
- $\int_0^T (|u'|^2+(1-u)^2)dx\leq 1+\eta$
I know a linear function might work, but I hope that $u'(0)=u'(1)=0$ and hence $u$ might be extend to $(-\infty,0]$ by $0$ and $[T,\infty)$ by $1$.
Tweaking the slope at the end comes at arbitrarily small cost to the functional $F(u)=\int_0^T (|u'|^2+(1-u)^2)\,dx$. After creating a function as in 1-4, extend it by $0$ and $1$, them convolve with mollifier $\eta_\epsilon$. Mollification changes $T$ slightly, and changes $F(u)$ slightly; but both effects go to zero as $\epsilon\to 0$. (Simply put, because $u\circ \eta_\epsilon\to u$ in the $W^{1,2}$ norm.)
So, it suffices to construct a function as in 1-4. Linear ones wouldn't work, because the linear function $u(x)=x/T$ on $[0,T]$ has $$ F(u) = \int_0^T (1/T^2+(1-x/T)^2)\,dx = \frac{1}{T}+\frac{T}{3} \ge\frac{2}{\sqrt{3}}>1 $$
Instead, solve the Euler-Lagrange equation $-u''-(1-u)=0$, the general solution is $u(x)=1+A\cosh x+B\sinh x$. With the boundary conditions, the solution is $$u(x)=1-\frac{\sinh(T-x)}{\sinh T}$$ for which
$$ F(u)=\frac{1-e^{-4T}}{(1-e^{-2T})^2} $$ This tends to $1$ as $T\to\infty$, so the existence of $T=T(\eta)$ is assured.