Ordinary Differential Equations/Successive Approximations

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y'=f(x,y) has a solution y satisfying the initial condition y(x_0)=y_0, then it must satisfy the following integral equation:

y=y_0+\int_{x_0}^x f(t, y(t))dt

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

Define y_1 as:

y_1=y_0+\int_{x_0}^x f(t,y_0)dt

And define y_n as

y_n=y_0+\int_{x_0}^x f(t,y_{n-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[edit]

First, we prove that y_n lies in the box, meaning that |y_n(x)-y_0|<\frac{1}{2}h. We prove this by induction. First, it is obvious that |y_1(x)-y_0|\le\frac{1}{2}h. Now suppose that |y_{n-1}(x)-y_0|\le\frac{1}{2}h. Then |f(t,y_{n-1}(t))|\le M so that

|y_n(x)-y_0|\le\int_{x_0}^x |f(t,y_{n-1}(t))|dt\le M(x-x_0)\le \frac{1}{2}Mw\le \frac{1}{2}h. This proves the case when x_0<x, and the case when x<x_0 is proven similarily.

We will now prove by induction that |y_n(x)-y_{n-1}(x)|<\frac{MK^{n-1}}{n!}(x-x_0)^n. First, it is obvious that |y_1(x)-y_0|<M(x-x_0). Now suppose that it is true up to n-1. Then

|y_n(x)-y_{n-1}(x)|\le\int_{x_0}^x |f(t,y_{n-1}(t))-f(t,y_{n-2}(t))|dt<\int_{x_0}^x K|y_{n-1}(t)-y_{n-2}(t)|dt due to the Lipschitz condition.

Now,

|y_n(x)-y_{n-1}(x)|<\frac{MK^{n-1}}{(n-1)!}\int_{x_0}^x ||u-x_0|^{n-1}du=\frac{MK^{n-1}}{n!}|x-x_0|^n.

Therefore, the series of series y_0+\sum_{n=1}^\infty (y_n(x)-y_{n-1}(x)) is absolutely and uniformly convergent for |x-x_0|\le\frac{1}{2}w because it is less than the exponential function.

Therefore, the limit function y(x)=y_0+\sum_{n=1}^\infty (y_n(x)-y_{n-1}(x))=\lim_{n\rightarrow\infty}y_n(x) exists and is a continuous function for |x-x_0|\le\frac{1}{2}w.

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