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?