Given nonlinear dynamical system, described by the following equation:

$\frac{dx}{dt}=−x(x − 1)(x − 2)(x − 3)(x − 4)$,

with initial condition $x(t_0)=x_0$;

$F$ function denotes the time-development function of the system How to find time-asymptotic state:

$\lim_{t\rightarrow\infty}(F(t,x_0))$ as a function of the initial condition $x_0$.

How to perform it without solving this equation.

Thanks much in advance!!!

  • $\begingroup$ Please do not deface the question. It orphans the answers that people have given. $\endgroup$ – robjohn Feb 11 '13 at 18:40

If $x_0\leq 0$ then the derivative with respect to time will be positive (plug it into the formula) and so $\lim_{t\rightarrow\infty}(F(t,x_0))=0$ because at $x=0$ the derivative is exactly zero so it can never go past zero.

if $0<x_0\leq 1$ then the derivative negative and so $\lim_{t\rightarrow\infty}(F(t,x_0))=0$ because again $F(t,1)$ will keep increasing but wont go past $x=1$ because here the derivative is zero.

If $1<x_0\leq 2$ the derivative is again increasing and so $\lim_{t\rightarrow\infty}(F(t,x_0))=2$ for similar reasons as above.

If $2<x_0\leq 3$ the derivative is decreasing and so $\lim_{t\rightarrow\infty}(F(t,x_0)) =2$.

If $3<x_0\leq 4$ the derivative is increasing and so $\lim_{t\rightarrow\infty}(F(t,x_0))=4$.

Finally if $x_0>4$ the derivative is decreasing to $\lim_{t\rightarrow\infty}(F(t,x_0))=4$.


Let $f(x)$ be the right hand side of the equation. Then $f(x)=0$ for $x=0,1,2,3,4$. Those are stationary solutions, that is, if $x_0$ is one on those values, the solution is constant and equal to $x_0$ for all $t>0$.

Suppose now that $0<x_0<1$. We have $f(x)>0$ for all $x\in(0,1)$. Then $F(t,x_0)$ is increasing while $F(t,x_0)\in(0,1)$. Is it possible to have $F(t,x_0)=1$ for some $t>0$? No, because of uniqueness of solution. This means that as long as $F(t,x_0)$ is defined, $0\le F(t,x_0)<1$ (which in turn implies hat $F$ is defined for all $t>0$.) Now $F(t,x_0)$ is increasing and bounded by $1$. It follows that $\lim_{t\to\infty} F(t,x_0)$ exists and is $\le1$. I leave to you the proof that the limit has to be $1$.

A similar analysis gives the behaviour for other values of $x_0$.


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