# Non-linear dynamics

1. Jan 24, 2008

### Bill Foster

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 $$f(x^{*})=x^{*}$$

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:

$$\tanh{ax}=\frac{e^{ax}-e^{-ax}}{e^{ax}+e^{-ax}}=x$$

This reduces to $$e^{2ax}=\frac{1+x}{1-x}$$

Then I can take the Taylor expansion...

$$e^{2ax}=1+2ax+\frac{(2ax)^2}{2!}+\frac{(2ax)^3}{3!}+\frac{(2ax)^4}{4!}+...=\frac{1+x}{1-x}$$

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

Then I tried this:

$$\tanh{(ax)}=x$$
$$ax=\tanh^{-1}{(x)}$$

Differentiate both sides:

$$a=\frac{1}{1-x^2}$$

Solve for x:

$$x=\pm \sqrt{1-\frac{1}{a}}$$

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

Suppose I let a=2. That means $$x=\frac{1}{\sqrt{2}}$$

So the following should hold true:

$$\tanh{(\frac{2}{\sqrt{2}})}=\frac{1}{\sqrt{2}}$$

But it doesn't. The correct value is

$$\tanh{(\frac{2}{\sqrt{2}})}=0.88839$$

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

Thanks.

Last edited: Jan 24, 2008
2. Jan 24, 2008

### nicksauce

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

3. Jan 24, 2008

### Bill Foster

So would using the Taylor series approximation be the best way?

Or do you know of a better way?

4. Jan 29, 2008

### salai

As far as I know, the definition of fixed point you gave is wrong. The fixed point is defined as $$\frac{dx}{dt} = f(x*) = 0$$ (look at Nonlinear Dynamics and Chaos, Steven H. Strogatz) then the problem is to find x* satisfying: $$tanh(x*)= 0$$.. Additionally, although, $$tanh(x*) = x*$$ seems not to be solved anaytically, it can be solved numerically, but there may be inappropriate results if you use nonliiinear fixed point anaysis.

5. Jan 29, 2008

### HallsofIvy

Staff Emeritus
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.

6. Mar 20, 2008

### Bill Foster

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