Discrete equation

aidan

Investigate the stability of the constant solutions $(u_{n+1} = u_n)$ of the discrete equation
$u_{n+1} = 4u_n (1 - u_n)$ In the case $0 \le u_0 \le 1$ , use the substitution $u_0 = \sin^2 \theta$ to find the general solution and verify your stability results. Can you find an explicit form of the general solution in the case $u_0 > 1$ ?

**Posted:** Sun Nov 01, 2009 5:26 am

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J.Y.Choi

I want to type for you but I don't know how...(I'm a computer illiteracy, sorry.) It's about the logistic map. See the book [Mathematical Methods for Physicists] by Arfken.

Thinking graph of the function $y = 4x(1 - x)$ and $y=x$ will help.

**Posted:** Sun Nov 01, 2009 7:11 am

Kent Merryfield

The interation $u_{n+1}=f(u_n)$ can only possibly converge to a fixed point of the function, namely an $x$ such that $x=f(x).$

Such a fixed point $x$ is stable (all sequences starting sufficiently close to $x$ converge to $x$ ) if an only if $|f'(x)|<1.$ If $|f'(x)|>1,$ then the fixed point $x$ is unstable, and sequences that start very close to (but not exactly at) $x$ do not remain very close to $x.$

(Point for which $|f'(x)|=1$ require more delicate analysis; I will not attempt such analysis below.)

The family of logistic mappings $f(x)=rx(1-x)$ maps $[0,1]$ into $[0,1]$ as long as $0\le r\le 4.$

Solving $x=rx(1-x)$ gives us two roots: $x=0$ and $x=1-\frac1r.$

$f'(x)=r$ for $x=0$ and $f'(x)=2-r$ for $x=1-\frac1r.$

For $0\le x<1,$ the point $x=0$ is a stable fixed point and the point $x=1-\frac1r$ does not lie in $[0,1]$ and cannot be accessed by a sequence starting inside $[0,1]$ . Hence, $x=0$ is the unique stable solution, and all sequences starting in $[0,1]$ tend to zero.

For $1<x<3,$ the fixed point $x=0$ is unstable but the fixed point $x=1-\frac1r$ is stable. All sequences starting in $[0,1]$ tend to $1-\frac1r.$

But for $3<x\le 4$ (which includes the $r=4$ case that aidan is asking about), both the $x=0$ and $x=1-\frac1r$ fixed points are unstable. Unless the sequence starts precisely at one of those fixed points, it will not converge.

The output of the following program (written in pseudo-code) is a famous picture.

Set up a $[0,4]\times[0,1]$ window for plotting. (Or perhaps $[3,4]\tiimes[0,1].$ )

Set up a loop, incrementing $r$ by many very small steps from $0$ to $4$ (or perhaps from $3$ to $4$ ).

Let $x: =\frac12.$

Loop 250 times:

Let $x: =rx(1-x).$

For the first 50 iterations, plot nothing; for the next 200 iterations, plot a dot at $(r,x).$

End of loop.

End of loop on $r.$

**Posted:** Sun Nov 01, 2009 10:53 am

aidan

Ah this is interesting - I've been reading a wikipedia article about the logistic map too. Thanks!

**Posted:** Mon Nov 02, 2009 5:19 am

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