I have a nonlinear dynamic model in discrete time. A simplified version of my dynamic system is:
\begin{equation} x_{t+1} = \frac{1}{1 + \exp(f(x_t))} \end{equation}
where $$f(x_t) = −\beta \left(2d \left(c+ \frac{1−c}{1+a(1−x_t)d} \right)−b−d \right)$$ with $\beta \geq 0$, $a > 0$, $d > b > 0$ and $0 \leq c \leq 1$.
I know that if $c$ is greater than a certain threshold, $\exp( f(x_t) )$ converges to 0 and $x_{t+1}$ converges to 1, thus:
\begin{equation} lim_{t \rightarrow \infty} (x_{t+1}) = 1 \end{equation}
If $c$ is smaller than a certain threshold, $\exp(f(x_t))$ goes to $\infty$ and $x_{t+1}$ converges to 0, thus:
\begin{equation} lim_{t \rightarrow \infty} (x_{t+1}) = 0. \end{equation}
(I proved the two statements above by imposing $\exp(f(x_t))<\epsilon$ and $\exp(f(x_t)) >\frac{1}{\epsilon}$, respectively, where $\epsilon$ is a sufficiently small positive number, and solving for $c$).
However, $x_{t+1} = 1$ and $x_{t+1} = 0$ do not exist. So I am wondering the following:
- can these two points ($x_{t+1} = 1$ and $x_{t+1} = 0$) be considered steady states of the system? Or they have to be defined in a different way (e.g. fixed points, or limit fixed points?)
- How can I prove their existence and stability?
Thanks!
An equilibrium point or a fixed-point is a state-value for which the dynamics does not change anymore. For a continuous-time system $\dot{x}=g(x)$, the equilibrium points are the state values $x^*$ for which we have $g(x^*)=0$. For a discrete-time system $x^+=g(x)$, the equilibrium points are the state values $x^*$ for which we have $g(x^*)=g^*$.
In your case, $x^*=0$ can not be equilibrium point because that would mean that $1/(1+\exp(f(0)))=0$, which is not possible.
Similarly, the point $x^*=1$ can not be an equilibrium point since we would require $1/(1+\exp(f(1)))=1$, which is not possible.
However, it is possible that other equilibro points exist and, for that, you will need to find the solution to $x^*=1/(1+\exp(f(x^*)))$. You will need to find them numerically.
Their local stability can be analyzed by linearizing the system about that equilibrium points.
For certain values of the parameters, it turns out that when we enter a neighborhood of 0, then $f(x)\approx \omega>>0$ there and $c$ is sufficiently large. This value will be amplified by the exponential function. So, in that case we have that
$$x^+=\dfrac{1}{1+\exp(f(x))}\approx \dfrac{1}{1+\exp(\omega)}\approx 0$$
but the trajectories do not converge to zero. They will stay in a neighborhood of zero.
The same reasoning applies to the point 1 with the difference that $\omega<<0$.
For a convergence to 0, a necessary condition would be that $f(x)\to\infty$ as $x\to0$ whereas for a convergence to 1, you will need to have that $f(x)\to-\infty$ as $x\to1$.