Decay estimate of energy minimizer and the linear ODE

Question

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?

Answer 1

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.