I need some help in this calculus of variation exercise:
Let $X$ to be the set of functions $\chi\in H^1_{loc}(\mathbb{R})$ such that $\chi(0)=1/2$ and such that the following limits do exists: \begin{equation*} \chi(-\infty)=\lim\limits_{x\to-\infty}\chi(x)=0 \end{equation*} \begin{equation*} \chi(\infty)=\lim\limits_{x\to\infty}\chi(x)=1 \end{equation*} Consider the functional $F:X\to[0,\infty]$ \begin{equation*} F(\chi)=\int_\mathbb{R}(\chi'(x)^2+\chi(x)^2(1-\chi(x)^2)dx \end{equation*}
Compute the following:
1) Consider a minimizing sequence $\chi_n$, prove that it can be chosen monotone and such that $0\leq\chi_n\leq 1$ and extract a subsequence that is weekly converging in $H^1_{loc}(\mathbb{R})$.
2) Study lower semicontinuity.
3) Deduce the Euler Lagrange equation in the weak form for the minimizer.
4) Does the candidate minimizer satisfy the condition at $\pm\infty$?
I know that this problem could be solved using $\Gamma$-convergence and the properties of Modica-Mortola (Ginzburg-Landau) functional but I think that the goal is to avoid that theory and only use the properties that ensure the existence of minimizers in Sobolev spaces.
UPDATE
Proof:
1) I've started choosing a minimizing sequence $\chi_n$ and observing that since the functional $F(\chi)\geq 0$ that sequence admits limit: \begin{equation*} \lim\limits_{n\to\infty}F(\chi_n)=M \end{equation*} then proved that it is not restricting to assume that $0\leq\chi_n\leq 1$ in this way:
If we have a sequence $\chi_k$ which is $>1$ in some set of non negative measure we can consider the new sequence defined by: \begin{equation*} \bar\chi_k= \begin{cases} 0 \ \ \text{if} \ \chi_k(x)< 0\\ \chi_k(x) \ \ \text{if} \ 0\leq\chi_k(x)\leq 1\\ 1 \ \ \text{if} \ \chi_k(x)> 1 \end{cases} \end{equation*} We now notice that for this new sequence the functional is identically zero when $\chi_k\in\{0,1\}$ and then we have that $F(\chi_k)\geq F(\bar{\chi}_k)$. Passing now to the limit on both sides we obtain that since $\chi_k$ is minimizing: \begin{equation*} \lim\limits_{k\to\infty}F(\chi_k)=\lim\limits_{k\to\infty}F(\bar\chi_k) \end{equation*} so in conclusion also $\bar\chi_k$ is minimizing.
I have also proved in a similar way that it is not restrictive to assume that the sequence is monotone.
Now I have to prove that exists a minimizing sequence that is bounded in order to extract a subsequence which converges weekly to some $u$.
Can you help me with this part of the proof also giving me some hints?