Proving that an ODE satisfies a minimum

Question

The problem

Given that $\dot{x} = f(t,x) = \frac{x}{x^2 + t^2 + 1}$, $t, x \in \mathbb{R} \times \mathbb{R}$. Prove that $\exists \epsilon > 0$ and if $|x_0| < \epsilon $ and $\phi : I \subset \mathbb{R} \to \mathbb{R}$ is a solution to

$$\dot{x} = f(t, x) = \frac{x}{x^2 + t^2 + 1}$$

When $x(t_0) = x_0$,

Then $| \phi(t) | < 10^{-4}, \forall t \in [0, 1000]$

My attempt

I know that:

$$x^{2}(t) + t^{2} + 1 > t^{2}$$

So

$$\frac{x(t)}{x^{2}(t) + t^{2} + 1} < \frac{x(t)}{t^{2}}$$

If i find the solution to $g(t, x) =\frac{x(t)}{t^{2}}, g(t_0) = x_0$, then will find the boundaries of $x(t)$. Luckily, this is a a well known homogenous equation and it's solution is:

$$\phi_{2}(t) = x_{0}e^{-\frac{1}{t}}$$

And with that we can say that:

$$ \phi(t) < |x_{0}|e^{-\frac{1}{t}} < 10^{-4}$$

I don't feel that my proof is correct and I think that there is a better way to solve this, can someone please show some possible flaws in my proof and the correct way to solve this answer?

Answer 1

We notice that the $x(t)$ satisfies the Lipschitz Conditions as for any $z,y$ $$\begin{align} \left|f(t,z)-f(t,y) \right| &= \left|\frac{z}{z^2+t^2+1}-\frac{y}{y^2+t^2+1} \right|\\ &= |z-y|\cdot\frac{|zy+t^2+1|}{(z^2+t^2+1) (y^2+t^2+1)}\le|z-y| \end{align}$$ Then the solution must be unique. Therefore, if there exists any $t_1$ such that $x(t_1) = 0$ then $x(t) =0$ for all $t \in \mathbb{R}$.

Case 1: $x_0 = 0$

Then according to what we proved above, $x(t) =0$ is the unique solution. $$\Longrightarrow|x(t)| = 0 < 10^{-4} \tag{1}$$

Case 2: $x_0 >0$

If there are any value $t_1$ such that $x(t_1)\le0$, we return to the trivial result $(1)$.

Let us suppose that $x(t) > 0$ for all $t\in \mathbb{R}$, use Grönwall's inequality: $$\begin{align} &\Longrightarrow 0<\dot{x} \le \frac{x}{ t^2 + 1} \\ &\Longrightarrow x_0<x(t) \le x_0\cdot \exp{\left(\int_{t_0}^t \frac{1}{ s^2 + 1}ds \right)} = x_0 \cdot e^{\arctan(t)-\arctan(t_0)} \\ \end{align} $$

As $e^{\arctan(t)}$ is bounded by $e^{\pi/2}$, then for all $M>0$, there exists $\epsilon>0$ such that if $|x_0| < \epsilon $, $|x(t)| < M$.

Case 3 $x_0 <0$

By the same argument as the case 2 $x_0>0$, let us suppose that $x(t) <0$ for all $t\in \mathbb{R}$. Make a change of variable $y(t) = -x(t)$ then

$$\begin{align} y(t_0) &=-x_0>0 \\ y'(t) &= \frac{y}{y^2+t^2+1} \end{align} $$

Applying the result of the case 2, we deduce that for all $M>0$, there exists $\epsilon>0$ such that if $|x_0|=|y_0| < \epsilon $, $|y(t)| < M$. Then $$|x(t)| = |-y(t)| < M$$

Q.E.D

Answer 2

Here is a quick and dirty solution. I am assuming that $t_0=0$ and $x(0) =x_0$.

Note that $|f(x,t)| \le u(x)={|x| \over 1+x^2}$ for $t$. It is straightforward to check that $u$ is bounded, say $|u(x)| \le B$ for all $x$.

In particular, $|\dot{x}(t)| \le B$ for all $t$ and so $|x(t)| \le |x_0|+ tB$ for all $t \ge 0$.

Choose some $\epsilon>0$ such that $10^{-4} > 1000 B $ and choose $|x_0| < 10^{-4} - 1000 B $.