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I have been trying to solve the following difference equation for some time now : $$u^3(n+1) = a - b\cdot u^2(n) + u^3(n), \qquad a \ne 0 \ne b$$ I have tried various substitutions, simplifications but nothing did.

My last attempt was to transform it into a continuous differential equation by letting $u(n) = u(t_n)$, with $t_n = h + t_{n-1}$ and $t_0 = 0$ ie $t_n = h\cdot n$. From there I transformed the above equation in : $$\frac{u^3(t_{n+1}) - u^3(t_n)}{h} = \frac{a - b\cdot u^2(t_n)}{h}$$ $$\frac{u(t_{n+1}) - u(t_n)}{h} \cdot (u^2(t_{n+1}) + u(t_{n+1})\cdot u(t_n) + u^2(t_n)) = \frac{a - b\cdot u^2(t_n)}{h}$$ Taking the limit as $h\longrightarrow 0$, I get $$3u^2\cdot\frac{du}{dt_n} = \frac{a - b\cdot u^2}{h}$$

From here I can solve $\int\frac{3h\cdot u^2}{a-b\cdot u^2} du = \int dt_n$,

integrating gives me $$\frac {t_n} h = \frac{3\sqrt a \cdot tanh^{-1}(u\cdot \sqrt{\frac ba})}{b^{3/2}} - \frac{3u}{b} + C$$ but I'm interested in the inverse of this function, as in $u(t_n) = ...$

Does anybody know how to either solve anatically the initial difference equation or find an approximation?

Also is my continuous approach any good and is it possible to invert it? I tried using a polynomial expansion but it wouldn't converge quickly enough.

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    $\begingroup$ I guess * means multiply? This is a discrete dynamical system - probably no explicit solution, and the tag "dynamical-systems" should be added. $\endgroup$
    – user25199
    Commented Jul 31, 2013 at 12:32
  • $\begingroup$ The word "solve" has different meanings for different people. For instance, what is wrong with selecting your favorite initial condition and obtaining next, say, million steps on a computer? If you want certain limiting or qualitative information about the solutions, you should be more specific. As Carl already mentioned, a simple formula for the general solution is unlikely to exist. $\endgroup$
    – Igor Khavkine
    Commented Jul 31, 2013 at 19:04
  • $\begingroup$ @Igor A million steps on a computer can be quite misleading for 1D maps and ordinary precision (say, 16 digits). There is a significant probability of finding a "numerical" periodic orbit, even if all periodic orbits in the real system are unstable; the classic example is the Ulam map $x\to4x(1-x)$ for which most initial conditions reach $x=0$ at remain there after a few million iterations. $\endgroup$
    – user25199
    Commented Aug 1, 2013 at 10:18
  • $\begingroup$ perhaps the Carleman linearization method can be usefull but the linearized form may become an infinte system of equations. $\endgroup$
    – user36539
    Commented Sep 19, 2013 at 16:14

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Some ideas: You can find approximate solutions near the fixed points, $\pm\sqrt{a/b}$ (if real) and $\pm\infty$. The latter are probably good places for your continuum approximation.

Then, there is strong contraction when the derivative of the RHS of the first equation is zero, ie $0$ and $2b/3$. Depending on the parameters this may lead to stable periodic orbits, and/or may determine behaviour of chaotic orbits - read up on kneading theory.

But I doubt if there is a closed form solution, for either the original equation or for the desired inverse.

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  • $\begingroup$ Wouldn't you need to cubic root the RHS and then take the derivative to analyse whether it's stable or not? $\endgroup$
    – user88595
    Commented Aug 5, 2013 at 9:26
  • $\begingroup$ For the fixed points (first paragraph of my answer), you have the right approach. For the periodic orbits (second paragraph), if a critical point is periodic it is superstable (has zero derivative), and there will be a range of parameters near it where there is a stable periodic orbit. But the exact location is a solution of a high degree polynomial and so more difficult to analyse exactly. $\endgroup$
    – user25199
    Commented Aug 19, 2013 at 8:16

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