# Optimal control, part 2

In the first part (here), we introduced Bellman’s idea of backward induction. But what if we consider now infinite time horizon ? Actually, the maths will be even more simple… and we will be able to use fixed pointed theorem to derive solutions.

• The mathematical framework

Here, consider the following value function and define A sequence is said to be an admissible solution for starting point x, if and . If we reformulate the dynamic programming idea, we obtain that if is a solution to problem , then for all , sequence is a solution to problem . It comes that function v is a solution of Bellman’s equation Note that is can be ssen as some fixed point resul, since i.e. So far, it shouldn’t be so hard….

• Frank Ramsey’s model (discrete version)

In 1928, Frank Ramsey wanted to understand the amount of savings in a dynamic perspective (in how much of its income should a nation save). Consider the following infinite horizon problem, where some planifier wants to maximize subject to constraints  and .
Assuming concavity of utility function and production function, we should look only for interior solutions. Define the Lagrangian as Thus, the first order conditions are then given by and Assume further some terminal condition, e.g. (also called transversality condition). If we combine those two conditions, and assume that the first constraint is saturated, we obtain the so-called Euler equation, It is also possible to use Bellman’s equation: given the dynamic of the capital  The first order condition states But since v is unknown, so is its derivative. But from the enveloppe theroem, we obtain something like where We can then write i.e. and finally which is, Euler’s equation.

• A specified model, with calculations

As in the previous post (here), consider a log utility function, and a power production function, and . The dynamic is then and Note that fixed points are here and Recall that the value function is defined as A natural idea to derive the value function can be to iterate, i.e. starting with a simple function, e.g the null function, at step 0. At step n=1 thus At step n=2, i.e. The first order condition is then and thus, we obtain that can be plugged in the previous equation, i.e. At step 3, we start from that new expression, derive the first order condition, and we get and and so on…
And finally, we can prove that i.e. . Assuming that actually, we can prove that (and has a form that can be explicited).