# 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).