Picard Iteration Help - Get Answers Now

  • Thread starter ravicplk
  • Start date
In summary, Picard iteration is a method for approximating a solution for a differential equation by starting with a constant function and then continually integrating and setting initial conditions to get higher order Taylor Polynomials. It is not meant as a method for actually solving a differential equation, but rather for establishing a solution for use in Picard's "existence and uniqueness" theorem.
  • #1
ravicplk
6
0
can anyone help me on the concept picard iteration?

thanks in advance
 
Physics news on Phys.org
  • #2
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 ex, the actual solution to y'= y, y(0)= 1.
 
  • #3
HallsofIvy said:
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 ex, the actual solution to y'= y, y(0)= 1.
thanks bro.
please give me a full course if u can.
 
  • #4
Sorry, I'm out of that business now!
 
  • #5
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
 
  • #6
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.
 

FAQ: Picard Iteration Help - Get Answers Now

What is Picard Iteration and how does it work?

Picard Iteration is a numerical method used to approximate the solution of a differential equation. It works by breaking down the differential equation into smaller, simpler equations and then solving them iteratively until the solution converges to the desired accuracy.

What types of problems can be solved using Picard Iteration?

Picard Iteration can be used to solve a wide range of differential equations, including ordinary differential equations (ODEs), partial differential equations (PDEs), and integral equations. It is particularly useful for solving non-linear problems that cannot be solved analytically.

What are the advantages of using Picard Iteration?

Picard Iteration is a simple and efficient method for solving differential equations. It is easy to implement and does not require a lot of computational resources. It also provides a good approximation of the solution, even for complex non-linear problems.

What are the limitations of Picard Iteration?

Picard Iteration may not converge for all types of differential equations. In some cases, the solution may oscillate or diverge instead of converging. It also requires a good initial guess to start the iteration process, which can be difficult to determine for certain problems.

Are there any other methods similar to Picard Iteration?

Yes, there are other methods similar to Picard Iteration, such as Newton's method and Runge-Kutta methods. These methods also use iterative processes to approximate the solution of a differential equation, but they may converge faster or be more suitable for certain types of problems.

Similar threads

Back
Top