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Suppose $u$ is a twice continuously differentiable function with linear growth, $$\lim_{x\rightarrow \infty} u'(x)-\frac{1}{g(x)} u(x) = 0 $$ and $g$ is a Lipschitz continuous function with Lipschitz constant $L<1$.

Consider the first order linear homogenous differential equation $$y'(x)- \frac{1}{g(x)} y(x) = 0 .$$ The general solution is $$y(x)=c \exp\left(\int \frac{1} {g(x)} dx\right)$$ for constant $c \in \mathbb R$. In any solution with linear growth, $\lim_{x\rightarrow \infty} y(x)=0$.

Is it possible to conclude that $\lim_{x\rightarrow \infty}u(x)=0$?

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  • $\begingroup$ What do you mean by "a solution with linear growth? Do you mean y(x)=O(x)? $\endgroup$ Aug 7, 2015 at 7:16
  • $\begingroup$ yes i mean y(x) is O(x) $\endgroup$
    – Jong
    Aug 7, 2015 at 7:24

1 Answer 1

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If we only consider the limit

$$ \lim_{x\rightarrow \infty} u'(x)-\frac{1}{g(x)}\, u(x) = 0 $$ and impose conditions $\,u \in \mathcal{C}^2(\mathbb R),\,$ $\,u(x) = O(x),\,$ and $g$ – Lipschitz continuous with $\,L<1$, then the answer is no, we cannot conclude that $$ \lim_{x\rightarrow \infty}u(x)=0. $$

Counterexample: set $u(x) = 2x$, and $g(x) = x$. Then all conditions are satisfied, $$ \lim_{x\rightarrow \infty} u'(x)-\frac{1}{g(x)} u(x) = \lim_{x\rightarrow \infty} \left(2-\frac{1}{x} 2x\right) = 0, $$ and yet $$ \lim_{x\rightarrow \infty} u(x) = \infty. $$

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