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I am give the following ODE:

$x^{'}=\frac{1}{e^{x(t)}+t^2}$ and $x(t_0)=x_0$.

I proved that for all $t\geq t_0$, the solution $x(t)$ is bounded. I also proved the existence and uniquence of the solution $x(t)$ to the above ODE when $t\geq t_0$. I am trying to show that as $t\to \infty$, $x(t)$ tends to a finite limit. Here is my attempt: I will go by contradiction by assuming that the limit does not exist. Since the solution $x(t)$ is bounded, then there should be two sequences $\{\alpha_n\}$ and $\{\beta_n\}$ such that $\alpha_n \to \infty$ and $\beta_n \to \infty$ but $\lim_{n\to \infty}(x(\alpha_n))=x_1\neq\lim_{n \to \infty}(x(\beta_n))=x_2$. Then, I did the following: $|x(\alpha_n)-x(\beta_n)|\leq|\int_{\alpha_n}^{\beta_n}\frac{1}{e^x+u^2}du| \leq |\int_{\alpha_n}^{\beta_n}\frac{1}{e^x}du|\leq M|\alpha_n-\beta_n|$ where $M$ is an upper bounded for $\frac{1}{e^x(t)}$ since the solution $x(t)$ is bounded. Any help on how I can take it from here to reach a contradiction that $x_1=x_2$. Thanks!

User267845467
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2 Answers2

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I propose a different method.

Since we have $x'(t)\geq0,\forall t\geq t_0$, the solution $x(t)$ is a non decreasing function . Moreover, one has \begin{multline*} x(t)=\int_{t_0}^tx'(u)du+x_0=\int_{t_0}^t\frac{du}{e^{x(u)}+u^2}+x_0\\\leq \int_{t_0}^t\frac{du}{e^{x_0}+u^2}+x_0\leq\int_{t_0}^{+\infty}\frac{du}{e^{x_0}+u^2}+x_0 \end{multline*} So $x(t)$ is bounded. Then the solution has a finite limit as $t\to+\infty$.

Driss Alami
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Note that $x'(t)$ is positive, hence $x(t)$ is actually monotone. :)