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For the difference equation

$k_{t+1}=\sqrt{k_t}-\frac{k_t}{2}$

one has to find all "fixed points" and determine whether they are locally or globally asymptotically stable.

Now I'm not quite sure what "fixed point" means in this context. Is it the same as "equilibrium point" (i.e., setting $\dot{k}=0$ , and calculate $k_{t+1}=k+\dot{k}=k+0$ from there)? Or something different?

I feel confident in solving such types of DE, just not sure what "fixed point" is supposed to mean here. Thanks for providing some directions!

Harry Peter
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bonifaz
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2 Answers2

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(I deleted my first answer after rereading your question; I thought perhaps I gave more info than you wanted.)

Yes, you can read that as "equilibrium points". To find them, just let $k_t = \sqrt{k_t} - \frac{1}{2}k_t$. Solving for $k_t$ will give you a seed $k_0$ such that $k_0 = k_1$. As you wrote, letting $k_0 = 0$ is one such value. However, there's another $k_0$ that will behave similarly. Furthermore, it's obvious what happens if $k_0 < 0$. Does the same thing happen for any other $k_0$? Now that you've found all the totally uninteresting seeds and the really weird ones, what about the others?

dmk
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A fixed point in a difference equation is similar to a fixed point in a differential equation: it is a value $k_1$ for which $k_1=k_2$.

Unfortunately, you don't have a $k_t$ on the RHS of this equation, so if you would like more specific help then you'll need to update the question.

Eric Stucky
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