Consider the autonomous Initial-Value Problem (IVP) $$\dfrac {dx}{dt}=\dfrac 12(x^3−x)$$ $$ x(0) =x_0 \in \mathbb{R}$$
where x is a function of time. I am supposed to indicate the time domain over which the solution is defined. I guess it is related to the equilibrium point and orbit of the differential equation, but I am not sure how to find a restriction in the intervals.
Note: all logarithm functions below have an absolute value function inside them as we work on the reals, unless specified otherwise.
If you were looking to describe how $\dot{x}$ changes (phase line):
To find the equilibrium points of the Autonomous DE we need to find its roots in terms of $x$.
\begin{align} \dfrac {dx}{dt}&=\dfrac 12(x^3−x):=f(x)\\ \dfrac {dx}{dt}&=\dfrac 12x(x^2−1)=\frac{1}{2}x(x-1)(x+1) \end{align}
The roots of the equation (when $dx/dt=0$) are $x\in\{0, -1, 1\}$ .
Let's find the slope of the function at its roots(zeros):
\begin{align*} \frac{d}{dx}f(x)&=\frac{1}{2}(3x^2-1)\\ \frac{d}{dx}f(-1)&=1\quad&\text{source}\\ \frac{d}{dx}f(0)&=-\frac{1}{2}\quad&\text{sink}\\ \frac{d}{dx}f(1)&=1\quad&\text{source}\\ \end{align*} Phase line of the DE:
If you actually want to find the domain of $t$
The solution of the equation is:
\begin{align} \dfrac {dx}{dt}&=\dfrac 12(x^3−x)\\ \int\dfrac{dx}{2(x^3-x)}&=\int dt\\ \frac{1}{2}\int\left(-\frac{1}{x}+\frac{1}{2(x-1)}+\frac{1}{2(x+1)}\right)\;dx&=\int dt\\ \frac{1}{2}\left(-\ln(x)+\frac{1}{2}\ln(x-1)+\frac{1}{2}\ln(x+1)\right)&=t+C\quad C\in\mathbb{R}\\ \frac{1}{4}\left(-\ln(\sqrt{x})+\ln(x-1)+\ln(x+1)\right)&=t+C\\ \frac{1}{4}\ln\left(\frac{1}{x^2}-1\right)&=t+C\\ \end{align}
For the IVP $x_0=x(t=0)$
$$\frac{1}{4}\ln\left(\frac{1}{{x_0}^2}-1\right)=C$$
IVP:
$$\frac{1}{4}\ln\left(\frac{1}{x^2}-1\right)=t+\frac{1}{4}\ln\left(\frac{1}{{x_0}^2}-1\right)$$
$$\frac{1}{x^2}=\exp\left(4t+\ln\left(\frac{1}{{x_0}^2}-1\right)\right)+1$$
$$x(t)=\pm\frac{1}{\sqrt{\exp\left(4t+\ln\left(\frac{1}{{x_0}^2}-1\right)\right)+1}}$$
We can have only positive numbers under the squareroot, thus:
$$\exp\left(4t+\ln\left(\frac{1}{{x_0}^2}-1\right)\right)+1<0$$
Let's assume we work on $\mathbb{C}$ to get a real solution:
$$\exp\left(4t+\ln\left(\frac{1}{{x_0}^2}-1\right)\right)<-1$$
By Euler's identity:
$$e^{i\pi}=-1$$ $$i\pi=\mathrm{Log}(-1)$$
where $\mathrm{Log(z)}$ is the complex logarithm.
$$4t+\ln\left(\frac{1}{{x_0}^2}-1\right)<i\pi$$
$t$ depends on $x_0$ but $x_0$'s domain is not the every single real.
The domain of $\dfrac{1}{{x_0}^2}-1$ is $x_0\in\mathbb{R}\setminus\{0\}$
If $x_0=\pm1$ then there will be a zero inside the logarithm which tends to $-\infty$. Because there is an absolute value function inside the logarithm, every other positive and negative number will be defined in the function. Thus
$$x_0\in\mathbb{R}\setminus\{-1,0,1\}$$
Let's cut the domain of $x_0$ into 4 parts and analyze $t$ with respect to them:
$$g(x_0)=\ln\left(\left|\frac{1}{{x_0}^2}-1\right|\right)$$
Let $a+\epsilon:=a^+$ and $a-\epsilon:=a^-$
a. $\mathcal{D}_{x_0}:(-\infty,-1)$ $$\begin{align*} \lim_{x_0\to-\infty}g(x_0)=\ln\left(\left|\frac{1}{(-\infty)^2}-1\right|\right)=\ln\left(|0^+-1|\right)=\ln(|-1^+|)=\ln(1^-)=0^- \end{align*}$$
$$\begin{align*} \lim_{x_0\to-1^-}g(x_0)=\ln\left(\left|\frac{1}{(-1^-)^2}-1\right|\right)=\ln(|1-1|)=\ln(0)=-\infty \end{align*}$$
$$x_0:(-\infty,-1)$$ $$g(x_0):(0^-,-\infty)$$
b. $\mathcal{D}_{x_0}:(-1,0)$ \begin{align*} \lim_{x_0\to-1^+}g(x_0)=-\infty \text{, as } x^2 \text{ is an even function.} \end{align*}
$$\begin{align*} \lim_{x_0\to0^-}g(x_0)=\ln\left(\left|\frac{1}{(0^-)^2}-1\right|\right)=\ln\left(\left|\frac{1}{0^-}-1\right|\right)=\ln(|\infty-1|)=\ln(|\infty|)=\infty \end{align*}$$
$$x_0:(-1,0)$$ $$g(x_0):(-\infty,\infty)$$
Between $0$ and $1$ and $1$ and $\infty$ the procedure is the same. to result in:
c. $\mathcal{D}_{x_0}:(0,1)$ $$x_0:(-1,0)$$ $$g(x_0):(\infty,-\infty)$$
d. $\mathcal{D}_{x_0}:(1,\infty)$ $$x_0:(-1,0)$$ $$g(x_0):(-\infty,0^-)$$
$$t<\dfrac{i\pi-g(x_0)}{4}$$
I thought i could do something with complex numbers here as the relation for $t$ is implicit, but i can't continue with this part. So i will just substitute back the 'domain' of $g(x_0)$ into the original problem.
Finding the proper numbers for the limits is kind of a guess-game for $t$. I hope the function between the two endpoints of the interval is at least $C^0$.
For $t\to-\infty$
$$\begin{align*} \lim_{\substack{g(x_0)\to0^- \\ t\to-\infty}}x(t)_+=\frac{1}{\sqrt{\exp\left(-\infty+0^-\right)+1}}=\frac{1}{\sqrt{e^{-\infty}+1}}=\frac{1}{\sqrt{0^++1}}=1 \end{align*}$$
$$\begin{align*} \lim_{\substack{g(x_0)\to-\infty \\ t\to-\infty}}x(t)_+=\frac{1}{\sqrt{\exp\left(-\infty+-\infty\right)+1}}=\frac{1}{\sqrt{e^{-\infty}+1}}=\frac{1}{\sqrt{0^++1}}=1 \end{align*}$$
$$\begin{align*} \lim_{\substack{g(x_0)\to+\infty \\ t\to-\infty}}x(t)_+=\frac{1}{\sqrt{\exp\left(-\infty++\infty\right)+1}}\stackrel{\text{the } 4t \text{ tends to }\infty\text{ much faster than }g(x_0)}{=}\frac{1}{\sqrt{e^{-\infty}+1}}=\frac{1}{\sqrt{0^++1}}=1 \end{align*}$$
For $t\to+\infty$
$$\begin{align*} \lim_{\substack{g(x_0)\to0^- \\ t\to\infty}}x(t)_+=\frac{1}{\sqrt{\exp\left(\infty+0^-\right)+1}}=\frac{1}{\sqrt{e^{\infty}+1}}=\frac{1}{\sqrt{\infty}}=0 \end{align*}$$
$$\begin{align*} \lim_{\substack{g(x_0)\to-\infty \\ t\to+\infty}}x(t)_+=\frac{1}{\sqrt{\exp\left(-\infty++\infty\right)+1}}\stackrel{\text{the } 4t \text{ tends to }\infty\text{ much faster than }g(x_0)}{=}\frac{1}{\sqrt{e^{\infty}+1}}=\frac{1}{\sqrt{\infty}}=0 \end{align*}$$
$$\begin{align*} \lim_{\substack{g(x_0)\to+\infty \\ t\to+\infty}}x(t)_+=\frac{1}{\sqrt{\exp\left(+\infty++\infty\right)+1}}=\frac{1}{\sqrt{e^{\infty}+1}}=\frac{1}{\sqrt{\infty}}=0 \end{align*}$$
To conclude: \begin{align*} \lim_{t\to-\infty}x(t)_+&=1\\ \lim_{t\to+\infty}x(t)_+&=0\\ \lim_{t\to-\infty}x(t)_-&=-1\\ \lim_{t\to+\infty}x(t)_-&=0 \end{align*}
To find the assymptotes of the function we need to find its poles:
$$\sqrt{\exp\left(4t+\ln\left(\frac{1}{{x_0}^2}-1\right)\right)+1}=0$$
$$\exp\left(4t+\ln\left(\frac{1}{{x_0}^2-1}\right)\right)=-1$$
But we know that the exponential function can never be negative on the reals.
Thus the domain of $t$ is $\mathbb{R}$.
Which checks out numerically - https://www.desmos.com/calculator/seubjf5tem