Without solving explicitly show that for the IVP $x'=x^{3}-x,x(0)=0.5$ the solution converges to $0$ for $t\rightarrow\infty$ I am trying to determine whether the solution to the following IVP converges to $0$ for $t\rightarrow\infty$. The IVP is $x'(t)=x^{3}-x=f(x)$,$x(0)=\frac{1}{2}$.
By looking at the "extended phase portrait" I would infer  that this is indeed the case. However I would like to have solid argument.
I would argue as following: Let $x(t)$ be the solution to the above IVP.
For $x\in(0,1)$ $x^{3}-x<0$. Therefore $x'(t)<0$ and $x(t)$ is decreasing. Since $\tilde{x}(t)=0$ is a (equilibrium) solution $x(t)>0$, which follows from the uniqueness theorem. Hence, $0$ is a lower bound. Because $x(t)$ is bounded below and monotonically decreasing $lim_{t\rightarrow\infty}x(t)$ exists and $lim_{t\rightarrow\infty}x(t)\geq 0$.
Here is the part were I am stuck. I have two approaches. One works and the other doesn't I think. What I want to show is that $\lim_{t\rightarrow\infty}x'(t)=0$. Because then
$0=\lim_{t\rightarrow\infty}x'(t)=lim_{t\rightarrow\infty}f(x(t))=f(lim_{t\rightarrow\infty}x(t))=f(c)=c^{3}-c\iff c=0,c=1,c=-1$.
$c=-1$ is not possible, because the $x(t)$ would intersect $\tilde{x}(t)=0$, hence by the uniqueness it follows that $x(t)=0$ and thus $0>x'{t}=0$.
If $c=1$, then we can find $t_{1}\in\mathbb{R}_{>0}$ such that
$|x(t_{1})-1|<\frac{1}{2}\iff x(0)=\frac{1}{2}<x(t_{1})$.
Hence, by the mean value theorem it follow that there is
$\eta\in(0,t_{1})$ such that $f'(\eta)=\frac{x(t_{1})-x(0)}{t_{0}-0}>0$, but $f'(\eta)<0$. Therefore $c=0$.
Approach 1. Here I am stuck:
From the inequality $x'(t)<0$ it follows that $0\geq lim_{t\rightarrow\infty}x'(t)$.
Since $x(t)$ is a solution and because of the continuity of the r.h.s. of the IVP it follows that
$0\geq lim_{t\rightarrow\infty}x'(t)=lim_{t\rightarrow\infty}f(x(t))=f(lim_{t\rightarrow\infty}x(t))=f(c)=c^{3}-c$. $(*)$
Therefore $c\in[0,1]$ or $c\in(-\infty,-1]$. If the latter is the case then $x(t)$ would intersect with at least one equilibrium solution. Hence, this is not possible. Therefore $c\in[0,1]$. I think I can rule out any $c\in[\frac{1}{2},1]$, since then by the mean value theorem. $f'(\eta)>0$, which is not possible.
Can I somehow show that $f(c)=c^{3}-c\geq 0$. Then it would follow that $\lim_{\rightarrow\infty}x'(t)=0$, wouldn't it?
But isn't it the case that $f(c)\leq 0$, since $c\in[0,1]$ in equation $(*)$? Therefore we would have $0\leq f(c)\leq 0\iff f(c)=0$. Hence, $\lim_{t\rightarrow\infty}x'(t)=f(c)=0$
Is there a shorter way?
Approach 2:
We know that $\lim_{t\rightarrow\infty}x(t)=c$. Then $\lim_{t\rightarrow\infty}x'(t)=\lim_{t\rightarrow\infty}\lim_{h\rightarrow 0}\frac{x(t+h)-x(t)}{h}=\lim_{h\rightarrow 0}\lim_{t\rightarrow\infty}\frac{x(t+h)-x(t)}{h}=\lim_{h\rightarrow 0}\frac{c-c}{h}=0$.
I think all requirements for changing the order of the limits to be legal are met...or at least I hope so.
I would really appreciate any help. Thank you very much in advance!
 A: The IVP
$$
x'=x^3-x, \quad x(0)=\frac{1}{2}, \tag{1}
$$
enjoys uniqueness, since $f(x)=x^3-x$ is continuously differentiable.
Claim. If  $\varphi$ is the solution of $(1)$, then
$$
0<\varphi(t)<1
$$
Proof. If not then $\varphi(t_1)\ge 1$, for some $t_1>0$. Hence, due to Intermediate Value Theorem, there would be a $t_1\in (0,t_1]$ such that
$\varphi(t_2)=1$. In that case, $\varphi$ would satisfy the IVP
$$
x'=x^3-x, \quad x(t_1)=1, \tag{2}
$$
But $(2)$ is satisfied by $\psi(t)\equiv1$, and since $(2)$ enjoys uniqueness, then $\varphi\equiv \psi$. Contradiction. Similarly we can arrive to a contradiction if we assume that $\varphi(t_1)\le 0$, for some $t>0$. $\Box$
So, since $\varphi(t)\in (0,1)$, for all $t$, then (1) possesses a global solution, and
$$
\varphi'(t)=\varphi^3(t)-\varphi(t)<0, \quad \text{for all $t\in\mathbb R$.}
$$
Thus $\varphi$ is strictly decreasing, and therefore the $\lim_{t\to\infty}\varphi(t)$
exists and
$$
\lim_{t\to\infty}\varphi(t)\ge 0.
$$
We shall now show that $\lim_{t\to\infty}\varphi(t)=0$. If not and
$\lim_{t\to\infty}\varphi(t)=a>0,\,$ since $\varphi$ is decreasing, then $a \le 1/2=\varphi(0)$. We would then have that for all $t>0$,
$$
\varphi(t)\ge a\quad\Longrightarrow\quad \varphi'(t)=\varphi^3(t)-\varphi(t)\le a^3-a=-b<0
$$
since $g(t)=t^3-t$ is decreasing in $(0,1/\sqrt{3})$. Thus
$$
\varphi(t)-\varphi(0)=\int_0^t \varphi'(s)\,ds\le -bt\to -\infty
$$
as $t\to\infty$. Contradiction. Thus $\lim_{t\to\infty}\varphi(t)=0$.
Similarly, we can show that $\lim_{t\to-\infty}\varphi(t)=1$.
A: This problem can be easily solved using Lyapunov functions. Consider the Lyapunov function
$$
V=x^2.
$$
Its derivative along the trajectories,
$$
\dot V= 2x\dot x= 2x^4-2x^2=-2x^2(1-x^2),
$$
is negative definite, so the origin is asymptotically stable. Furthermore,
$
\forall x\in\mathbb (-1,1)\setminus \{0\}$
$\dot V(x)<0. 
$
This means that the set
$$
S=\{x:\; V(x)<1\}=(-1,1)
$$
is a basin of attraction of the origin, thus, any solution $x(t)$ satisfying the initial condition $x(0)=C$, $C\in (-1,1)$, tends to zero.
