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Picard iteration

  1. Jun 28, 2005 #1
    [SOLVED] picard iteration

    I tried reading beyond my level,but i am having some problems.i do not understand the picard iteration,who can explain this to a boy of 16.
  2. jcsd
  3. Jun 28, 2005 #2


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    http://www.tmt.ugal.ro/crios/Support/ANPT/Curs/math/s1/s1pic/s1pic.html [Broken] explains it. If you have a DE

    y' = f(x,y)

    then y is that function which satisfies:

    [tex]\mathbf{y}(x) = y(x_0) + \int _{x_0} ^x f(\xi , \mathbf{y}(\xi ))d\xi [/tex]

    Notice how I boldfaced the "y"s. Pick functions f and w "randomly," and plug them into the right side of the equation. Compute it, and you'll get function of x, call it z(x). Is this z(x) the same as your w(x)? Probably not. Picard iteration gives you a technique so that even if you're starting out with some function y_0 that doesn't satisfy the differential equation, you can iteratively get new functions, and approach the actual function in the limit. Suppose you're looking for the smallest non-negative number, and you start with 2. That's not small enough, so you divide by 2 and get 1. Still not small enough, so you keep dividing by 2, and in the limit you will approach the answer, 0. Picard's technique is a similar idea. It gives you an iterative procedure to find a sequence of functions whose limit is the function that satisfies the DE. This process works because the functions in question form a complete metric space, so you can prove that this sequence has a limit, and that the limit is a fixed point if you try to apply the iterative function to it. Our function w was not a fixed point, because we put it in the right side, and got a different function, z, coming out. But remember how I put "y" in bold. You can see that y is a fixed point. You put y in the right side, and you get y back.
    Last edited by a moderator: May 2, 2017
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