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I'm asked to find the following difference equation's fixed points and corresponding eigenvalues:

${u_{t + 1}} = f({u_t})$ where $f({u_t}) = \exp \left[ {\gamma \left( {1 - \frac{{{u_t}}}{K}} \right)} \right]$ and $K > 0,\gamma > 0$.

I was taught that a fixed point of a difference equation of this form is an initial condition ${u_0}$ such that $f({u^*})={u^*}$.

Thus, it is necessary to solve the equation $\exp \left[ {\gamma \left( {1 - \frac{{{u^*}}}{K}} \right)} \right] - {u^*} = 0$ for $u^*$ in terms of $\gamma$ and $K$. However, when solving this equation in WolframAlpha I obtain a solution in terms of the product-log function which we never saw in class. What am I missing?

Note: I am adding the differential-equations tag because this part of my course (difference equations) was immediately preceded by an extensive treatment of ordinary differential equations, so maybe the solution to this problem is related in some way to ODEs?

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Welcome to the world of Lambert function !

The only analytical solution of equation $$\exp \left[ {\gamma \left( {1 - \frac{{{u}}}{K}} \right)} \right] - {u} = 0$$ is effectively given by $$u=\frac{K }{\gamma }W\left(\frac{ \gamma }{K}e^{\gamma }\right)$$ The Wikipedia page will show you, for many examples, the steps to arrive to this expressions as well as series expansions for its evaluation.

If you cannot use Lambert function, only numerical methods would do the work.

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  • $\begingroup$ Is it at least known without numerics in which range $W(z) > 1$? $\endgroup$ – Evgeny Dec 4 '17 at 5:35
  • $\begingroup$ @Evgeny. $W(x)$ is an increasing function and $W(e)=1$ $\endgroup$ – Claude Leibovici Dec 4 '17 at 5:39
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    $\begingroup$ Perfect! That would be enough for deducing stability then. $\endgroup$ – Evgeny Dec 4 '17 at 5:41
  • $\begingroup$ Thank you for your insight! $\endgroup$ – Maximiliano Santiago Dec 5 '17 at 6:37
  • $\begingroup$ @MaximilianoSantiago. You are welcome ! This is a just fascinating function with more and more applications in many areas. Just search for Lambert in this site; you will see plenty of them. Cheers. $\endgroup$ – Claude Leibovici Dec 5 '17 at 6:49

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