Question

State and prove existence and uniqueness theorem

Answer 1

Existence and uniqueness theorem is the tool which makes it possible for us to conclude that there exists only one solution to a first order differential equation which satisfies a given initial condition.

Theorem. 

Let f(x,y) be a real valued function which is continuous on the rectangle 

$R=$ {$(x,y); |x-x_{o}|\leq a, |y-y_{o}|\leq b$ }. Assume f has a partial derivative with respect to y and that $\dfrac{\delta f}{\delta y}$ is also continuous on the rectangle R. Then there exists an interval $I = [ x_{o}-h,x_{o}+h]$ (with $h\leq a$ ) such that the initial value problem

$\begin{cases} y'=f(x,y), &\\ y(x_{o})=y_{o} & \end{cases}$

has a unique solution y(x) defined on the interval I.

Note that the number h may be smaller than a. In order to understand the main ideas behind this theorem, assume the conclusion is true. Then if y(x) is a solution to the initial value problem, we must have

$$y(x)=y_{o}+\int^{x}_{x_{o}} f(t,y(t))dt$$

It is not hard to see in fact that if a function y(x) satisfies the equation (called functional equation)

$$y(x)=y_{o}+\int^{x}_{x_{o}} f(t,y(t))dt$$

on an interval I, then it is solution to the initial value problem

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

$$y(x_{o})=y_{o}$$

Picard was among the first to look at the associated functional equation. The method he developed to find y is known as the method of successive approximations or Picard's iteration method. This is how it goes:

Step 1. Consider the constant function

$$y_{o}(x)=y_{o}$$

Step 2. Once the function $y_{n}(x)$  is known, define the function

$$y_{n+1}(x)=y_{o}+\int^{x}_{x_{o}} f(t,y(t))dt$$

Step 3. By induction, we generate a sequence of functions {$y_{n}(x)$ }  which, under the assumptions made on f(x,y), converges to the solution y(x) of the initial value problem

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

$$y(x_{o})=y_{o}$$