Differential Equations/Successive Approximations
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y'=f(x,y) has a solution y satisfying the initial condition y(x0)=y0, then it must satisfy the following integral equation:
y=y0+
Now we will solve this equation by the method of successive approximations.
Define y1 as:

And define yn as

We will now prove that:
- If f(x,y) is bounded and the Lipschitz condition is satisfied, then the sequence of functions converges to a continuous function
- This function satisfies the differential equation
- This is the unique solution to this differential equation with the given initial condition.
[edit] Proof
First, we prove that yn lies in the box, meaning that
. We prove this by induction. First, it is obvious that
. Now suppose that
. Then
so that
. This proves the case when x0 < x, and the case when x < x0 is proven similarily.
We will now prove by induction that
. First, it is obvious that | y1(x) − y0 | < M(x − x0). Now suppose that it is true up to n-1. Then
due to the Lipschitz condition.
Now,
.
Therefore, the series of series
is absolutely and uniformly convergent for
because it is less than the exponential function.
Therefore, the limit function
exists and is a continuous function for
.
Now we will prove that this limit function satisfies the differential equation.