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Suppose $a$ is constant, $b(x)=C_b(1-x)$ and $c(x)=C_c(1+x^2)$ for some positive constants, we consider the minimizer of the energy $$E(u)=\int_{\mathbb{R}}\left[a^2(u'(x))^2/2+c(x)\left(\dfrac{1}{c(x)}-u(x)\right)^2-\dfrac{1}{c(x)}\right]\dfrac{e^{2\int^x_0\frac{b(y)}{a^2}}dy}{a^2}dx$$ in the weighted Sobolev space $$H:=\left\{u\,\middle|\,\int_{\mathbb{R}}\left[a^2(u'(x))^2+u^2\right]\dfrac{e^{2\int^x_0\frac{b(y)}{a^2}}dy}{a^2}dx<\infty\right\}.$$ The energy minimizer solves the ODE $$a^2u''/2+bu'-cu=-1$$ on $\mathbb{R}$. We can truncate $u$ to show that $0 \leq u \lesssim 1$ as $u$ is the unique minimizer. More precisely, we define $u^*(x):=u(x)$ if $\inf_{\mathbb{R}}(1/c(x))\leq u(x)\leq \sup_{\mathbb{R}}(1/c(x))$; $u^*(x):=\sup_{\mathbb{R}}(1/c(x))$ if $u(x)\geq \sup_{\mathbb{R}}(1/c(x))$; $u^*(x):=\inf_{\mathbb{R}}(1/c(x))$ if $\inf_{\mathbb{R}}(1/c(x))\geq u(x)$. We see that $E(u^*)\leq E(u)$.

Suppose that $u$ is $C^2$. My problem is: is there any decay estimate of $u$ as $|x| \to \infty$? I can construct some comparsion function $u_*(x)= \dfrac{C_*}{1+x^2}$ to show that $$a^2(u-u_*)''/2+b(u-u_*)'-c(u-u_*)\leq 0.$$ By the maximum principle, if the minimum is attained in the interior of $\mathbb{R}$, then either $(u-u_*)\geq\min(u-u_*)>0$ or $(u-u_*)$ is constant. In both case, I can obtain that $u\geq u_*$. If the minimum is not attained in the interior, then $(u-u_*)\geq\liminf_{|x|\to \infty}(u-u_*)\geq 0$, we see that $u\geq u_*$ again.

I could like to do the same for the upper bound, but I can at most have the result $u\lesssim \dfrac{1}{1+x^2}+\limsup_{|x|\to \infty} u$. What I expect is that $\limsup_{|x|\to \infty} u=0$, do we can any method to show this? or can we show that $u\lesssim \dfrac{1}{1+x^2}$ by another method?

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    $\begingroup$ (i) Minimizer under what conditions on $u$? (ii) "truncate" in what sense? (iii) "We can truncate $u$ to show that ..." -- How do you do that? (iv) "I can construct some comparsion function" -- How do you do that? $\endgroup$ Apr 2 at 21:37
  • $\begingroup$ Thank you for the comments I have made it more detailed, (i) in the weighted Sobolev space (ii) & (iii) truncate by the sup and inf of $1/c(x)$ (iv) assuming the solution is smooth enough, we use the maximum principle to show the inequality by choosing $C_*$ small enough. The calculation is long, I omit here. $\endgroup$
    – mnmn1993
    Apr 2 at 22:04

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You can just play with the same truncation techniques you used already.

  1. $u\ge 0$ (replace by $\max(u,0)$ on $\mathbb R$)

  2. If $x\ge 0$ and $u(x)\ge \frac{1}{c(x)}$, then $u(y)\le u(x)$ for $y\ge x$ (replace with $u=\min(u(x),u)$ on $[x,+\infty)$).

This means that the only alternative to tending to $0$ at $+\infty$ is decreasing to a positive value $u_*$ from some point on.

Now assume it is the case. Take a smooth flat bump $\psi$ supported on $[T,2T]$ such that $0\le\psi\le 1$, $\psi\equiv 1$ on $[1.3 T, 1.6 T]$, $|\psi'|\le \frac CT$, $\psi''\le \frac C{T^2}$ and integrate the differential equation with respect to it using integration by parts wherever applicable (so you will need the equation to hold just in the sense of distributions, which is certainly the case).

$\int u''\psi=\int u\psi''=O(1/T)$

$\int bu'\psi=-\int u(b'\psi+b\psi')=O(T)$

$\int cu\psi\approx u_*\int c\psi\approx u_* T^3$

$\int (-1)\psi=O(T)$

Thus, not to have a severe mismatch, one must have $u_*=0$. In fact, this proves that $u(x)=O(x^{-2})$ as $x\to+\infty$.

The other half ($x\to-\infty$) can be treated in the same way.

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  • $\begingroup$ I am also confused why the first part (the truncation) leads to the result that $u(x)$ is decreasing to a positive number from some point. $\endgroup$
    – mnmn1993
    Apr 12 at 12:05
  • $\begingroup$ Your arguments are perfect but I cannot ensure that $u(x)>u^*$ for any sufficiently large $x$. $\endgroup$
    – mnmn1993
    Apr 12 at 15:43
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    $\begingroup$ @mnmn1993 What other possible scenario may be there? If $u$ crosses $1/c(x)$ infinitely far to the right, then, after each crossing, it cannot get higher any more, so it must tend to $0$. Otherwise it stays above $1/c(x)$ beyond a certain point and (2) means that it is decreasing from that point on. $\endgroup$
    – fedja
    Apr 12 at 23:09

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