A problem with the functional $F(\chi) = \int_{\mathbb{R}}(\chi'^2+\chi^2(1-\chi)^2)dx$.

Question

Let $X:=\{\chi \in H^1_{\text{loc}}(\mathbb{R})\}$ such that $\chi(0)=\frac {1}{2},\; \chi(-\infty)=(0)$ and $\chi(\infty)=1$.

Consider the functional $$F(\chi) = \int_{\mathbb{R}}(\chi'^2+\chi^2(1-\chi)^2)dx$$ Prove that $F$ has a minimizer on $X$ and compute it.

Hints: Let ($\chi_n)_{n\in \mathbb{N}}$ be a minimizing sequence. We may assume $0 \le \chi_{n} \le 1$, why?

Can we assume that each $\chi_n$ is monotone?

Extract a subsequence that is weakly converging in $H^1_{\text{loc}}(\mathbb{R})$, i.e., weakly on each bounded interval of $\mathbb{R}$. Study lower semicontinuity.

Deduce the Euler-Lagrange equation in the weak form for the minimizer.

Does the candidate minimizer satisfy the conditions at $\pm\infty$?

Show that the minimizer $\chi$ is more regular. Integrate the equation to compute the explicit solution.

This method of solving such a problem is new to me so I don't really understand the hints, can somebody help me in any way?

Answer 1

In this answer, we show how this is solved in physics. We think this will be useful to OP, even if OP ultimately is after another method.

1. In physics the model is known as a kink/soliton. The Lagrangian is $$F[\chi]~:=~\int_{\mathbb{R}}\!\mathrm{d}x~{\cal L}.\tag{1}$$ The Lagrangian density is$^1$ $${\cal L}~:=~\chi^{\prime 2} + V. \tag{2}$$
   The potential is $$V~:=~\chi^2(1-\chi)^2~\stackrel{(4)}{=}~ W^{\prime 2}. \tag{3}$$ The superpotential is $$W~:=~\frac{1}{2}\chi^2-\frac{1}{3}\chi^3. \tag{4}$$

2. The Beltrami identity leads to constant energy solutions: $$\chi^{\prime 2} - V~=~{\rm const}.\tag{5}$$ Zero-energy solutions are kink/antikink solutions $$ \pm\chi^{\prime}~\stackrel{(3)+(5)}{=}~ W^{\prime} ~\stackrel{(4)}{=}~\chi(1-\chi).\tag{6}$$ The explicit solution can be found by separation of variables and then integration: $$\chi~\stackrel{(6)}{=}~\frac{1}{1+e^{\mp (x-x_0)}}.\tag{7} $$ When we compare with OP's boundary conditions (BCs) $$ \chi(-\infty)~=~0, \qquad \chi(0)~=~\frac{1}{2}, \qquad \chi(\infty)~=~1, \tag{8}$$ we get the unique solution $$\chi~\stackrel{(7)+(8)}{=}~\frac{1}{1+e^{-x}}. \tag{9}$$

3. The Bogomol'nyi-Prasad-Sommerfield (BPS) bound $$ F[\chi]~\stackrel{(1)+(2)+(3)}{=}~\int_{\mathbb{R}}\!\mathrm{d}x~\underbrace{(\chi^{\prime}\mp W^{\prime})^2}_{\geq 0} \pm 2\underbrace{[W(\chi(x))]_{x=-\infty}^{x=\infty}}_{=1/6} \tag{10} $$ shows that the solution (6) minimizes the Lagrangian (1) with the given BCs (8).

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$^1$ The unusual plus sign in front of the potential $V$ is related to the fact that $x$ is a space coordinate rather than a time coordinate. This is strictly speaking a static model with no time. That being said, it is often helpful to think of $x$ as time and with the potential being $-V$. In this analogy $\chi^{\prime 2} - V$ is the energy, cf. eq. (5).

Answer 2

The first two parts are quite easy:

1. Let's suppose that $\chi_n$ is in a minimizing sequence and that $\chi_n(x)>1$ for some $x\in \mathbb{R}$ we can then consider $$ \chi_n^*(x)=\begin{cases} 1&\Leftarrow \chi_n(x)\ge 1 \\ \chi_n(x)&\Leftarrow \chi_n(x)\le 1 \end{cases}$$, we can suppose without loss of generality that $\chi_n(x)>1$ in an interval (a,b) and we can ultimately see what happens in here to the integrals inside $F(\chi_n)$ and $F(\chi_n^*)$, in particular for the latter $$F(\chi_n^*)=\int_{(a,b)}(\chi_n^{*'}+\chi^2(1-\chi)^2)dx $$ but this function is constant so the derivative is 0, moreover it's constantly equal to 1 which implies that also $\chi^2(1-\chi)^2$ goes to 0 therefore, since outside the interval the two functions are equal, we have $$ F(\chi_n)> F(\chi_n^*)\ge_{minimality} F(u).$$The same is true if $\chi_k(x)<0$ for some x since we can define $$ \chi_k^*(x)=\begin{cases} 1&\Leftarrow \chi_n(x)\le 0 \\ \chi_n(x)&\Leftarrow \chi_n(x)\le 1 \end{cases}$$ and again the derivative is 0 and $\chi^2(1-\chi)^2$ goes to 0 showing again that $$ F(\chi_k)> F(\chi_k^*)\ge_{minimality} F(u).$$

   We have shown that we can suppose that $0\le \chi_n(x)\le 1$ $\forall n\in \mathbb{N}, x\in \mathbb{R}$ and the same has to hold for the limit (the minimizer);

2. Regarding monotony we notice some facts

   1. If c is a constant $$ F(c)=\int_\mathbb{R} c^2(1-c)^2:+;$$
      2. $\int_{\mathbb{R}}\chi'^2dx\ge 0$;
      3. $\frac{\partial u(\chi)}{\partial \chi}\left(=\frac{ \chi^2(1-\chi)^2}{\partial \chi}\right)=(1-\chi)2\chi(1-2\chi)=0\iff \chi=\frac{1}{2}\wedge 0\wedge 1$ where $\frac{1}{2}$ is a maximum and 0,1 minimums, moreover the function is symmetric with respect to $\frac{1}{2}$ id est $u\left(\frac{1}{2}+\chi\right)=u\left(\frac{1}{2}-\chi\right)$ since we can explicitly compute $$ u\left(\frac{1}{2}+\chi\right)=\left(\frac{1}{2}+\chi-\frac{1}{4}-\chi-\chi^2\right)^2$$ $$ u\left(\frac{1}{2}-\chi\right)=\left(\frac{1}{2}-\chi-\frac{1}{4}+\chi-\chi^2\right)^2$$

   These imply that we can minimize any non-monotone function in the following way Given $(x_a,x_b)$ the interval where the function $\chi_n$ is decreasing, a the maximum value, b the minimum value, d (and the corresponding $x_d$) the point before a such that b=d) and c (and the corresponding $x_c$) the point after b such that a=c we first check the values $\left|\frac{1}{2}-a\right|$ and $\left|\frac{1}{2}-b\right|$ and have two cases

   1. $\left|\frac{1}{2}-a\right|\ge \left|\frac{1}{2}-b\right|$, in this case we consider the function $$ \chi_n^*=\begin{cases} a&\Leftarrow x\in (x_a,x_c)\\ \chi_n(x)&\Leftarrow x\in \mathbb{R}\setminus (x_a,x_c) \end{cases}$$ then thanks to symmetry and decreasingness of the function $u(\chi)$ if we move away from $\frac{1}{2}$ we have that $$u(a)<u(\chi_n(x))\; \forall x\in (x_a,x_c)$$ and so $$\int_{(x_a.x_c)}\chi_n'^{*2}+u(\chi_n^*)dx=\int_{(x_a.x_c)}u(a)<\int_{(x_a.x_c)}\chi_n'^{*2}+u(\chi_n)dx$$ and so $F(\chi_n)>F(\chi_n^*)\ge_{minimality} F(u)$;

   2. $\left|\frac{1}{2}-a\right|\le \left|\frac{1}{2}-b\right|$, in this case we consider the function $$ \chi_n^*=\begin{cases} b&\Leftarrow x\in (x_d,x_b)\\ \chi_n(x)&\Leftarrow x\in \mathbb{R}\setminus (x_d,x_b) \end{cases}$$ then thanks to symmetry and decreasingness of the function $u(\chi)$ if we move away from $\frac{1}{2}$ we have that $$u(b)<u(\chi_n(x))\; \forall x\in (x_d,x_b)$$ and so $$\int_{(x_d.x_b)}\chi_n'^{*2}+u(\chi_n^*)dx=\int_{(x_d.x_b)}u(b)<\int_{(x_d.x_b)}\chi_n'^{*2}+u(\chi_n)dx$$ and so $F(\chi_n)>F(\chi_n^*)\ge_{minimality} F(u)$

It's worth noticing that we know the points $x_c,x_d$ exists since out functions need to go from 0 to 1 and moreover it's bounded by the same values therefore b can't be lower than 0 (and for sure there is $x_d<x_b$ such that $\chi(x_d)=b$ since the function had to go from 0 to $a$ in a continuous way) and in the same way the function need to go from b to 1 continuously and so it will pass again through a proving the existence of $x_c$.

We have shown that we can suppose that $\chi_n$ is a monotone increasing function and the same has to hold for the limit (the minimizer).