is created by David Witten, a mathematics and computer science student at Vanderbilt University. For more information, see the "About" page.

Picard Iterations

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.

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

Homogeneous Equations with Constant Coefficients

Integrating Factor