# Limit of a Discrete Dynamical System

For the system defined below, the point by point evolution remains bounded for all $t$ so I could see that some sort of limit exists. However, the question is what sort of limit is it -- a single point limit or a limit cycle? (Refer to the attached graphic.) Can someone please prove the exclusion of one over the other? If the limit is a point limit, then how could I derive the actual numeric limit? $$\left[\begin{array}{c}x_{t+1}\\y_{t+1}\end{array}\right]=\left[\begin{array}{c}y_{t}/b\\x_{t}-1+\left(\frac{a}{b^2}\right)y_t^2\end{array}\right]\;\;\;\text{with}\;\left[\begin{array}{c}x_{0}\\y_{0}\end{array}\right]=\left[\begin{array}{c}0\\0\end{array}\right]$$ and where $a=-0.5$ and $b=-1.0003$

I have never studied discrete dynamical systems so I am not familiar with the techniques that would be used to answer such questions. It would be interesting to see how these questions are answered.

Below is the graphic of the system specified above (I hope this works, I never linked in a graphic before).

Define $T(x,y) = (y/b, x-1+(a/b^2)y^2)$ with $a=-0.5$ and $b=-1.0003$. Then, the equation $T(x,y)=(x,y)$ has two solutions at approximately $$(3.41494, -3.41596) \; \mbox{ and } \; (0.585662, -0.585838).$$ If the sequence converges to a single fixed point, then it must be one of these two. Your image suggests that the sequence might converge to the second point. I computed the first $25000$ points of the orbit to high precision and generated a similar image: We can verify that the second point is attractive by computing the Jacobian of $T$ and evaluating at the fixed point:
$$\left( \begin{array}{cc} 0 & 1 \\ -0.9997 & -0.9994 y \\ \end{array} \right) = \left( \begin{array}{cc} 0 & 1 \\ -0.9997 & 0.585486 \\ \end{array} \right)$$
Now, near the fixed point, your nonlinear system behaves like the linear system defined by the second matrix above. The fixed point is attractive if the eigenvalues of that matrix are both less than 1 in absolute value. In fact, the absolute values of both eigenvalues are $0.99985$, thus the point is indeed attractive.
• @user54738 Many software packages provide the ability to do numerical computations to much higher precision that just the 16 decimal digits provided by the CPU. I used Mathematica which provides significance tracking so that you can actually measure how precise your computations are. As a result, you can have confidence of your results. To see the need for this sort of thing, try iterating $f(x)=4x(1-x)$ from $x_0=(2+\sqrt{3})/4$ at machine precision. It will appear to converge to $0.75$ (correctly) but, after around 40 iterates, it moves away from $0.75$, even though $3/4$ is fixed. – Mark McClure Aug 1 '14 at 15:11