Behavior of ODE with time : Does not cross boundary

Question

Let $a>0$, and for any time point $t\in [0,1]$, define $\sigma_t^2:= t^2 + (1-t)^2$. Next, we define the following ODE: $$ \frac{d X_t}{dt} = \frac{2t-1}{\sigma_t^2} X_t + \frac{a(1-t)}{\sigma_t^2} \tanh\left(\frac{a t X_t }{\sigma_t^2}\right) \quad \text{for $t \in [0,1]$}, $$ where $\tanh(u) = \frac{1 - e^{-2u}}{1+ e^{-2u}}$. Assume that initially $X_0$ is strictly positive, i.e., $X_0>\delta$ for some $\delta>0$. I conjecture that $X_t>0$ for all $t\in [0,1]$ and I also want prove this. For now, I simulated the ODE and got the following results: .

For now, this somewhat validates my claim. Note that the velocity at $t =0$ is $(-\delta)$ which means that initially, the particle would start going in the negative direction. Still, eventually it should it should start going the other way. Is there any way to formalize this? Thank you for any help.

Edit: I have posted a solution below.

Answer 1

I have found a solution to my question. I have posted the solution below. Thanks to everyone for insightful discussions.

Define a new variable $U_t := X_t/\sigma_t$. Then, using the ODE , we get the following new ODE:

\begin{equation} \label{eq: ODE U} \frac{dU_t}{dt} = \frac{y (1-t)}{\sigma_t^3} \tanh \left( \frac{a t U_t}{\sigma_t}\right), \quad U_0>0. \end{equation} First, note that to prove our claim, it is enough to show that $U_t>0$ for all $t>0$. We will now proceed via a contra-positive argument. Let there exists $t\in [0,1]$ such that $U_t \le 0$. To be more precise, let $$t_0 = \inf\{t \in [0,1] : U_t \le 0 \}.$$ Note that $t_0>0$ (as $U_0>0$) and $U_{t_0} \le 0$ as $t_0$ is infimum of a closed set by inverse map theorem. Now, we have $$ \int_0^{t_0} \dot{U}_t\; dt = U_{t_0} - U_0 <0. $$ Due to mean value theorem there exists $\tau \in (0,t_0)$ such that $\dot{U}_\tau <0$ , which implies $$ \frac{a (1-\tau)}{\sigma_\tau^3} \tanh \left( \frac{a \tau U_\tau}{\sigma_\tau}\right) <0 \Leftrightarrow U_\tau <0. $$ This is a direct contradiction to the definition of $t_0$. Therefore, we have $U_t>0$ for all $t>0$, which evidently yields that $X_t>0$ for all $t \in [0,1]$. This further shows that if $U_t$ is a non-decreasing (maybe strictly increasing) of $t$ if $U_0 = X_0>0$.