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Ordinary 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+∫x0xf(t,y(t))dt

Now we will solve this equation by the method of successive approximations.

Define y1 as:

y1=y0+∫x0xf(t,y0)dt

And define yn as

yn=y0+∫x0xf(t,yn−1)dt

We will now prove that:

  1. If f(x,y) is bounded and the Lipschitz condition is satisfied, then the sequence of functions converges to a continuous function
  2. This function satisfies the differential equation
  3. This is the unique solution to this differential equation with the given initial condition.

Proof

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First, we prove that yn lies in the box, meaning that |yn(x)−y0|<12h. We prove this by induction. First, it is obvious that |y1(x)−y0|≤12h. Now suppose that |yn−1(x)−y0|≤12h. Then |f(t,yn−1(t))|≤M so that

|yn(x)−y0|≤∫x0x|f(t,yn−1(t))|dt≤M(x−x0)≤12Mw≤12h. This proves the case when x0<x, and the case when x<x0 is proven similarily.

We will now prove by induction that |yn(x)−yn−1(x)|<MKn−1n!(x−x0)n. First, it is obvious that |y1(x)−y0|<M(x−x0). Now suppose that it is true up to n-1. Then

|yn(x)−yn−1(x)|≤∫x0x|f(t,yn−1(t))−f(t,yn−2(t))|dt<∫x0xK|yn−1(t)−yn−2(t)|dt due to the Lipschitz condition.

Now,

|yn(x)−yn−1(x)|<MKn−1(n−1)!∫x0x||u−x0|n−1du=MKn−1n!|x−x0|n.

Therefore, the series of series y0+∑n=1∞(yn(x)−yn−1(x)) is absolutely and uniformly convergent for |x−x0|≤12w because it is less than the exponential function.

Therefore, the limit function y(x)=y0+∑n=1∞(yn(x)−yn−1(x))=limn→∞yn(x) exists and is a continuous function for |x−x0|≤12w.

Now we will prove that this limit function satisfies the differential equation.