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This is an economics problem, but I'm pretty sure this kind of thing comes up elsewhere. I've used dynamic programming to find the optimal path of a system (law of motion), which is:

$k_{t+1}=\beta\alpha k_t^\alpha$,

where $\alpha,\beta \in R(0,1)$

now I want to find out if this expression converges to a steady state, defined as

$\lim_{t\to\infty} k_{t+1}=k_t=k^*.$

I thought about trying to tackle it as an eigenvalue problem, of the form

$\left[ \begin{matrix} \log k(t+1)\\ \log k(t+2) \end{matrix} \right] = \left[ \begin{matrix} 0 &1\\ 0 &\alpha \end{matrix}\right] \left[ \begin{matrix} \log k(t)\\ \log k(t+1)\end{matrix} \right] + \left[ \begin{matrix} 0\\ \log\beta + \log\alpha \end{matrix} \right]$


$U_{t+1} = AU_t +B$

My question is this: will the system converge as long as the eigenvalues of A are inside the unit circle? I'm concerned about how the contribution of B behaves over time. Am I on the right track with this thing at all?

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up vote 4 down vote accepted

This dynamical system is explicitly solvable.

Let $x_t=\log k_t$, then $x_{t+1}=\alpha x_t+b$ with $b=\log(\alpha\beta)$. Since $\alpha\ne1$, this is equivalent to $x_{t+1}-x^*=\alpha (x_t-x^*)$ with $x^*=b/(1-\alpha)$, hence $x_t=x^*+\alpha^t(x_0-x^*)$ is the explicit solution of your dynamical system. Since $|\alpha|<1$, $x_t\to x^*$ for every starting point $x_0=\log k_0$, which means that $k_t\to k^*$ with $k^*=\exp(x^*)=(\alpha\beta)^{1/(1-\alpha)}$.

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Thank you very much. I think $a=\alpha$ though, not $\log\alpha$. – jefflovejapan May 20 '11 at 1:55
Yes. Corrected. – Did May 20 '11 at 5:24

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