Picard Iteration Help - Get Answers Now

[ravicplk]

can anyone help me on the concept picard iteration?

thanks in advance

 

[HallsofIvy]

What do you want? A full course in the subject?

Here is a simple example: To solve the equation y'= y, y(0)= 1 using Picard iteration, start by approximating y by a constant. Since we know y(0)= 1, "1" is a good choice. Then the equation becomes y'= 1 and, integrating, y= x+ C. When x= 0, y(0)= C= 1 so the first "iteration" gives y= x+ 1.

Now the equation is y'= x+ 1. Integrating, y= (1/2)x²+ x+ C and, setting x= 0, y(0)= C= 1 so y= (1/2)x²+ x+ 1.

Now the equation is y'= (1/2)x²+ x+ 1. Integrating, y= (1/6)x³+ (1/2)x+ x+ C and, setting x= 0, y(0)= C= 1 so y= (1/6)x³+ (1/2)x²+ x+ 1.

Now the equation is y'= (1/6)x³+ (1/2)x²+ x+ 1. Integrating, y= (1/24)x⁴+ (1/6)x³+ (1/2)x²+ x+ C and, letting x= 0, y(0)= C= 1 so y= y= (1/24)x⁴+ (1/6)x³+ (1/2)x²+ x+ 1.

Continuing the iteration will give higher and higher powers of x. It should be clear now that we are getting terms of the form (1/n!)xⁿ and that this is giving higher and higher order Taylor Polynomials for e^x, the actual solution to y'= y, y(0)= 1.

 

[ravicplk]

What do you want? A full course in the subject?

Here is a simple example: To solve the equation y'= y, y(0)= 1 using Picard iteration, start by approximating y by a constant. Since we know y(0)= 1, "1" is a good choice. Then the equation becomes y'= 1 and, integrating, y= x+ C. When x= 0, y(0)= C= 1 so the first "iteration" gives y= x+ 1.

Now the equation is y'= x+ 1. Integrating, y= (1/2)x²+ x+ C and, setting x= 0, y(0)= C= 1 so y= (1/2)x²+ x+ 1.

Now the equation is y'= (1/2)x²+ x+ 1. Integrating, y= (1/6)x³+ (1/2)x+ x+ C and, setting x= 0, y(0)= C= 1 so y= (1/6)x³+ (1/2)x²+ x+ 1.

Now the equation is y'= (1/6)x³+ (1/2)x²+ x+ 1. Integrating, y= (1/24)x⁴+ (1/6)x³+ (1/2)x²+ x+ C and, letting x= 0, y(0)= C= 1 so y= y= (1/24)x⁴+ (1/6)x³+ (1/2)x²+ x+ 1.

Continuing the iteration will give higher and higher powers of x. It should be clear now that we are getting terms of the form (1/n!)xⁿ and that this is giving higher and higher order Taylor Polynomials for e^x, the actual solution to y'= y, y(0)= 1.

thanks bro.
please give me a full course if u can.

 

[HallsofIvy]

Sorry, I'm out of that business now!

 

[aroosak]

I was reading your reply,very helpful.
It is just one thing: how do you know you have to start from a constant for your picard iteration, why not some polynomial for example?

thanks

 

[HallsofIvy]

Because that's what "Picard iteration" means!

Given a problem like dy/dx= f(x,y), y(x₀)= y₀, start with the constant function y(x)= y₀.

You could, if you like, start with some polynomial, some exponential, etc. but then it would be harder to say what function to start with.

In any case, Picard's iteration was never meant as a method for actually solving a differential equation. It was a method for establishing how to write a solution for use in Picard's "existence and uniquness" theorem.