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I'm in a Masters level course on difference equations, and last week we discussed theorems which can be applied to show the stability characteristics of non-hyperbolic equilibria of first order difference equations. One of these theorems involved an application of the Schwarzian derivative:

$$ S(f) = \frac{f'''}{f'} - \frac{3}{2} (\frac{f''}{f'})^2 $$

The theorem states:

Given $ f: I \mapsto I$ and $ x_{n+1} = f(x_n), n = 0, 1, ...$ and $\bar{x}$ an equilibrium point of f.

We have for

1) ${f'}(\bar{x}) = -1$

2) ${f'}(\bar{x}) = 1$

that

i) $Sf(\bar{x}) < 0 \to \bar{x}$ is locally asymptotically stable.

ii) $Sf(\bar{x}) > 0 \to \bar{x}$ is unstable.

So, I asked myself, "What happens when $Sf(x)=0$?"

By setting the Schwarzian derivative equal to zero and simplifying I arrived at the following third order non-linear differential equation:

$$ \frac{3}{2} {f''}^2 = {f'''}{f'} $$

By guessing the form of the solution to be $f(x) = a x^m$, I was able to find a family of solutions:

$$ f(x) = \frac{a}{x} + c $$

and by intuition found a second

$$ f(x) = -x + c $$

where a and c are arbitrary constants.

Interestingly, both these solutions satisfy the condition that ${f'}(\bar{x}) = -1$. Also, both of these functions when taken as difference equations, (ie:

$$ x_{n+1} = \frac{a}{x_n} $$

exhibit solutions that are stable and (aside from the equilibrium points themselves) periodic with period 2, but not locally asymptotically stable.

My question is, have I found the only solutions to the ODE above? If not, what are some other methods of solution/approaches to attempt to find other solutions to the ODE?

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  • $\begingroup$ the solutions are $(ax+b)/(cx+d)$ $\endgroup$ – user8268 Sep 30 '13 at 17:22
  • $\begingroup$ Thank you! Do you know where I can find a discussion of how that answer is obtained? $\endgroup$ – tneff Sep 30 '13 at 17:33
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Let $g=f'$, so your equation becomes $$\frac{3}{2}g'^{2}=g''g \implies \frac{3}{2}\frac{g'}{g}=\frac{g''}{g'}$$ Integrating, $$\frac{3}{2}\ln(g)+C=\ln(g')$$ or $Ag^{3/2}=g'$, which is a first order separable O.D.E. solved as follows (separating variables with $A$ still an arbitrary constant): $$Ag^{-3/2}dg=dx \implies Ag^{-1/2}=x+B \implies g=\frac{1}{(Ax+B)^{2}}$$

Since $g=f'$, we then have that $$f(x)=\frac{-1}{A}\frac{1}{Ax+B}+C \text{ or }f(x)=\frac{1}{ax+b}+c$$ where $a$, $b$ and $c$ are arbitrary constants.
This solution is valid when neither $g$ or $g'$ is zero, which gives the trivial solution $f(x)=mx+k$

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