How can I know the behavior of this non-linear equation as t→∞?

Question

$ {dx\over dt} = ln(x^2+1) -1 $ as $ t→∞ $

$ x(0)=x_0 $

Answer 1

The function $\log(x^2+1)-1$ is Lipschitz in $x$ so we have existence and uniqueness for initial conditions (the derivative $\frac{2x}{x^2+1}$ is bounded by $\pm 1$).

All in all, the differential equation has two equilibrium points (constant solutions) at $x(t)=\pm\sqrt{e-1}$ which also act as asymptotes due to uniqueness, so we have 5 cases:

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Case $1$: $\Bigr(x_0 > \sqrt{e-1}\Bigr)$

Here $x'(t)>0$, so the solution will keep increasing forever hence

$$\lim_{t\to\infty}x(t)=\infty$$

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Case $2$: $\Bigr(x_0 = \sqrt{e-1}\Bigr)$

This is a constant solution with $x'(t)=0$ for all $t$, so

$$\lim_{t\to\infty}x(t)=\sqrt{e-1}$$

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Case $3$: $\Bigr(-\sqrt{e-1} < x_0 < \sqrt{e-1}\Bigr)$

Plugging into our function we get that $x'(t)<0$ so the solution decreases forever. It is not unbounded, though, as uniqueness prevents it from ever crossing the $x=-\sqrt{e-1}$ line. Therefore we have

$$\lim_{t\to\infty}x(t) = -\sqrt{e-1}$$

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Case $4$: $\Bigr(x_0=-\sqrt{e-1}\Bigr)$

Same as the constant solution from before, we get

$$\lim_{t\to\infty}x(t) = -\sqrt{e-1}$$

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Case $5$: $\Bigr(x_0 < -\sqrt{e-1}\Bigr)$

Similar to case $1$, since $x'(t)>0$, it will keep increasing forever, but this time it has an upper bound leading us to conclude that

$$\lim_{t\to\infty} x(t) = -\sqrt{e-1}$$

$$ $$

So really we could consolidate the last three cases into one big case for our final answer.