# Basic Idea

The Picard iterative process is an iterative method of solving an initial value problem.

Let's say we have this differential equation.
$$y'(x) = f(x, y), y(x_0) = y_0$$
We integrate both sides from $x_0$ to $x$.
$$\int_{x_0}^x y'(x)\,\mathrm{d}x = \int_{x_0}^x f(t,y(t))\,\mathrm{d}t$$
$$\int_{x_0}^x y'(x)\,\mathrm{d}x = \int_{x_0}^x f(t,y(t))\,\mathrm{d}t$$
$$y(x) - y(x_0) = \int_{x_0}^x f(t,y(t)\,\mathrm{d}t$$
$$y(x) = y(x_0) + \int_{x_0}^x f(t,y(t))\,\mathrm{d}t$$
This is the associated integral equation to this differential equation.
We can make this iterative however. We write this:
$$y_{n+1}(x) = y_0 + \int_{x_0}^x f(t,y_n(t))\,\mathrm{d}t$$

## A Little Bit of Intuition

$$y_{n+1}(x) = y_0 + \int_{x_0}^x f(t,y_n(t))\,\mathrm{d}t$$
The purpose of this iterative process is to eventually get closer and closer to the true value of y(x).
So, $y(x) \approx y_{1001}(x) = y_0 + \int_{x_0}^x f(t,y_{1000}(t))\,\mathrm{d}t.$

We know this is true, because as $n \to \infty$, $y_n \to y$, and $y(x) = y_0 + \int_{x_0}^x f(t,y(t))\,\mathrm{d}t$, so we do eventually approach a solution.

We know this is true, because as $n \to \infty$, $y_n \to y$, and $y(x) = y_0 + \int_{x_0}^x f(t,y(t))\,\mathrm{d}t$, so we do eventually approach a solution.

## Example to Setup

$$y'(x) = sin(x + y(x), y(\pi) = 0$$
Our initial step is to write a $y_0(x)$. The best thing to do is define $y(x) = y(\pi) = 0$.
Now, we write the general equation
$$y_{n+1}(x) = 0 + \int_{\pi}^{x}\sin(t + y_n(t))\, \mathrm{d}t$$
$$y_{1}(x) = \int_{\pi}^{x}\sin(t)\, \mathrm{d}t = -\cos(x) - 1$$
At this point, it's difficult to continue, because we have cos inside of sin.

## Example to Completion

$$y' = -y, y(0) = 2$$
We know from observation our answer should be $y = 2e^{-x}$. Let's verify this using Picard iterations.
$$y_1(x) = 2 + \int_0^x -2\, \mathrm{d}x = 2 - 2x$$
$$y_2(x) = 2 + \int_0^x -2+2x\, \mathrm{d}x = 2 - 2x + x^2 = 2 - 2x + \dfrac{2x^2}{2!}$$
$$y_3(x) = 2 + \int_0^x -2 + 2x - x^2\, \mathrm{d}x = 2 - 2x + x^2 - \frac{x^3}{3} $$
$$= 2\left(1 + (-x) + \frac{(-x)^2}{2!} + \frac{(-x)^3}{3!} \right) \approx 2e^{-x}$$
We can see that this expression approaches the Taylor series for $e^{-x}$.

David Witten