The Picard Iteration
Now we can start actually closing in on a solution to our initial value problem. Recall the setup:
The first thing we’ll do is translate this into an integral equation. Integrating both sides of the first equation and using the second equation we find
Conversely, if satisfies this equation then clearly it satisfies the two conditions in our initial value problem.
Now the nice thing about this formulation is that it expresses as the fixed point of a certain operation. To find it, we will use an iterative method. We start with and define the “Picard iteration”
This is sort of like Newton’s method, where we express the point we’re looking for as the fixed point of a function, and then find the fixed point by iterating that very function.
The one catch is, how are we sure that this is well-defined? What could go wrong? Well, how do we know that is in the domain of ? We have to make some choices to make sure this works out.
First, let be the closed ball of radius centered on . We pick so that satisfies a Lipschitz condition on , which we know we can do because is locally Lipschitz. Since this is a closed ball and is continuous, we can find an upper bound for . Finally, we can find a , and the interval . I assert that is well-defined.
First of all, for all , so that’s good. We now assume that is well-defined and prove that is as well. It’s clearly well-defined as a function, since by assumption, and is contained within the domain of . The integral makes sense since the integrand is continuous, and then we can add . But is ?
So we calculate
which shows that the difference between and has length smaller than for any . Thus , as asserted, and the Picard iteration is well-defined.