Ordinary Differential Equations/Successive Approximations

$$y'=f(x,y)$$ 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
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.