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Non-linear dynamics

  1. Jan 24, 2008 #1
    1. The problem statement, all variables and given/known data

    Obtain the fixed points for the following as a function of a, where 0<a<infinity:

    f(x)=tanh(ax)

    2. Relevant equations

    The fixed point is given by [tex]f(x^{*})=x^{*}[/tex]

    3. The attempt at a solution

    It's basically a math problem: tanh(ax)=x Find x in terms of a.

    I tried it two ways:

    [tex]\tanh{ax}=\frac{e^{ax}-e^{-ax}}{e^{ax}+e^{-ax}}=x[/tex]

    This reduces to [tex]e^{2ax}=\frac{1+x}{1-x}[/tex]

    Then I can take the Taylor expansion...

    [tex]e^{2ax}=1+2ax+\frac{(2ax)^2}{2!}+\frac{(2ax)^3}{3!}+\frac{(2ax)^4}{4!}+...=\frac{1+x}{1-x}[/tex]

    But as you can see, that doesn't really help.

    Then I tried this:

    [tex]\tanh{(ax)}=x[/tex]
    [tex]ax=\tanh^{-1}{(x)}[/tex]

    Differentiate both sides:

    [tex]a=\frac{1}{1-x^2}[/tex]

    Solve for x:

    [tex]x=\pm \sqrt{1-\frac{1}{a}}[/tex]

    Problem is, it doesn't work out in the calculator.

    Suppose I let a=2. That means [tex]x=\frac{1}{\sqrt{2}}[/tex]

    So the following should hold true:

    [tex]\tanh{(\frac{2}{\sqrt{2}})}=\frac{1}{\sqrt{2}}[/tex]

    But it doesn't. The correct value is

    [tex]\tanh{(\frac{2}{\sqrt{2}})}=0.88839[/tex]

    Is there an analytical solution to tanh(ax)=x ?

    Thanks.
     
    Last edited: Jan 24, 2008
  2. jcsd
  3. Jan 24, 2008 #2

    nicksauce

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    Is there an analytical solution to tanh(ax)=x ?

    Answer: no.
     
  4. Jan 24, 2008 #3
    So would using the Taylor series approximation be the best way?

    Or do you know of a better way?
     
  5. Jan 29, 2008 #4
    As far as I know, the definition of fixed point you gave is wrong. The fixed point is defined as [tex]\frac{dx}{dt} = f(x*) = 0 [/tex] (look at Nonlinear Dynamics and Chaos, Steven H. Strogatz) then the problem is to find x* satisfying: [tex] tanh(x*)= 0 [/tex].. Additionally, although, [tex] tanh(x*) = x* [/tex] seems not to be solved anaytically, it can be solved numerically, but there may be inappropriate results if you use nonliiinear fixed point anaysis.
     
  6. Jan 29, 2008 #5

    HallsofIvy

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    No, his definition of "fixed point" is correct. What you are referring to is an "equilibrium" point of a differential equation.

    As nicksause said, there is no "analytic" solution to that equation. I would use Newton's method to numeraically solve the equation.
     
  7. Mar 20, 2008 #6
    That is incorrect. A fixed point is where x=f(x).

    If f(x)=ax(1-x) where a<1, then the only fixed point will be at x=0, which is a trivial fixed point. When a>1, then fixed point will be given by x*=1-1/a
     
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