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In this problem we consider one dimensional discrete time dynamical system $x(k+1)=f(x(k))$ with a fixed point $u$, at which $|f'(u)|=1$. For each of the following systems, find out the stability of the fixed point $u$:

  1. $f(x)= \sin x, u=0$.

  2. $f(x)=x^3+x, u=0$.

  3. $f(x)=1+ \log x, u=1$.

  4. $f(x)=x^{2}+\dfrac{1}{4}, u=\dfrac{1}{2}$.

  5. $f(x)=\dfrac{1}{x}, u=1$.

  6. $f(x)=\dfrac{1}{2x-2}+\dfrac{1}{2x+2}, u=0$.

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What have you tried? –  Alex Becker Dec 29 '11 at 10:39
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Also, it would be nice of you to start accepting answers to your previous questions. It's a good way to give feedback and reward answerers, and it makes people more willing to answer your questions. –  Alex Becker Dec 29 '11 at 10:40

2 Answers 2

up vote 3 down vote accepted

Even if $|f'(u)| = 1$, the following criteria still work:

  • The fixed point $u$ is locally stable if there exists a neighborhood $U$ of $u$ such that $|f(x)-u| < |x-u|$ for all $x \in U$, $x \ne u$.

  • The fixed point $u$ is locally unstable if there exists a neighborhood $U$ of $u$ such that $|f(x)-u| > |x-u|$ for all $x \in U$, $x \ne u$.

Often, it helps to look at the graph of the function $f$ (here $f(x)=1+\log x$) near the fixed point:

$\hspace{160px}$Graph of $f$ near its fixed point

If the graph of the function near the fixed point lies entirely within green area delimited by the line $y=x$ and its reflection through the fixed point, the fixed point is locally stable. If the graph lies entirely in the red area near the fixed point, it is locally unstable.

If, as in this example, the graph of $f$ lies in the green area on one side of the fixed point and in the red area on the other, things get more complicated. If, like here, $f$ is increasing near the fixed point, it will be locally stable on one side and unstable on the other, and is thus a saddle point. If $f$ is decreasing, you'll need to repeat the stability analysis for its second iterate $f^{(2)}$ (which will be increasing) and see what happens to it.

Finally, if the graph of $f$ lies on the line $y=x$, it has a continuum of neutrally stable fixed points, while if it lies on the reflected line (or, more generally, if the graph of $f^{(2)}$ lies on the $y=x$ line), it will have a single neutrally stable fixed point surrounded by neutral 2-cycles.

Of course, just looking at the graph is in general not enough: you have to prove that the graph doesn't cross over into another area even closer to the fixed point. But as long as $f$ is fairly well-behaved, looking at the graph at least gives you a good idea what the answer is going to be.

Since this is homework, I won't spoon-feed you the answers, but you should be able to figure them out using the criteria I gave above.

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Since this is homework, you ought to have seen in class that a sufficient condition for stability (or instability) is that $|f'(u)|<1$ (or $|f'(u)|>1$, respectively). (When $|f'(u)|=1$, no conclusion can be drawn.)

So what you have to do in each case is simply to compute $f'(x)$ and plug in $x=u$, and see what you get.

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No, I do not think so. This is not the homogenous case, therefore we have to check by hand calculation. I struggle with the $\delta, \epsilon$ language, that's why I could not solve this problem. –  Arsenaler Oct 30 '11 at 1:15
    
Sorry, I mean this is not the linear case. –  Arsenaler Oct 30 '11 at 1:22
    
@msnaber: This criterion is for the nonlinear case! (It's called "linear stability analysis" because you use Taylor series to approximate your nonlinear system by a linear one in the vicinity of the equilibrium point, and then determine the (in)stability of the original system from the (in)stability of the simpler linear system.) No need for epsilons or deltas. –  Hans Lundmark Oct 30 '11 at 7:14

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