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Math Help - Picard's Iteration

  1. #1
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    Picard's Iteration

    Guys I really need help with understanding how to use this method for my O.D.E. class. The main part I don't understand is, I don't get what f(t,Yn(t)) is supposed to be. I also do not understand how to do the next iteration after the first and the one after that and so on. I know their is a pattern but could someone show me how to do all this stuff to help me prepare for a Test? For instance

    dy/dx = (x^2)(y)
    y(0) = 1

    This is from my notes, Why is f(t,Yn(t)) = to [t^2 - 1]? What procedure is there for figuring this out? Also if possible, could you tell it to me in a way for which it will work for any Picard Iteration problem the professor gives us? I just need to know how f(t,Yn(t)) becomes what it is.
    Last edited by JonathanEyoon; February 13th 2009 at 07:33 PM.
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  2. #2
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    Here's a procedure that should always work. Start with the equation \tfrac{dy}{dx} = x^2y. For convenience, replace the variable x by t, so that y is a function of t: \tfrac{dy}{dt} = t^2y(t). Integrate both sides from 0 to x: y(x)-y(0) = \int_0^xt^2y(t)\,dt, or y(x) = 1 + \int_0^xt^2y(t)\,dt.

    The idea now is to start with an approximate solution y_0(t) which you substitute into the integral, getting a new function y_1(x) which you then substitute into the integral to get another function y_2(x), and so on.

    The first approximation y_0(t) is just the constant function 1 (because this is given as the value of y at x=0). Then
    y_1(x) = 1 + \int_0^xt^2y_0(t)\,dt = 1 + \int_0^xt^2\,dt = 1+x^3/3. So y_1(t) = 1+t^3/3. Next,
    y_2(x) = 1 + \int_0^xt^2y_1(t)\,dt = 1 + \int_0^xt^2(1+t^3/3)dt = \ldots, and so on.
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  3. #3
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    Quote Originally Posted by JonathanEyoon View Post
    Guys I really need help with understanding how to use this method for my O.D.E. class. The main part I don't understand is, I don't get what f(t,Yn(t)) is supposed to be. I also do not understand how to do the next iteration after the first and the one after that and so on. I know their is a pattern but could someone show me how to do all this stuff to help me prepare for a Test? For instance

    dy/dx = (x^2)(y)
    y(0) = 1

    This is from my notes, Why is f(t,Yn(t)) = to [t^2 - 1]? What procedure is there for figuring this out? Also if possible, could you tell it to me in a way for which it will work for any Picard Iteration problem the professor gives us? I just need to know how f(t,Yn(t)) becomes what it is.
    It isn't equal to t^2- 1. Where did you get that? You must be looking at an example that has nothing to do with this given problem. f(t, Yn(t)) is simply f applied to t and whatever you have just calculated Yn(t) to be!

    In this case dy/dx= x^2y. Since we are given that y(0)= 1, a reasonable starting approximation for y is the constant y= 1; Then we have dy/dx= x^2(1) and, integrating y= (1/3)x^3+ C. In order to have y(0)= 1, we must have y(0)= (1/3)0+ C= 1 so C= 1 and y= (1/3)x^3+ 1. Replacing the y in the differential equation with that, dy/dx= x^2((1/3)x^3+ 1)= (1/3)x^5+ x^2 and integrating, y= (1/18)x^6+ (1/3)x^3+ C. Again, y(0)= 1 gives C= 1. Take dy/dx= f(x,y(x))= x^2((1/18)x^3+ (1/3)x^3+ 1) and integrate again.
    Last edited by HallsofIvy; February 14th 2009 at 08:32 AM.
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  4. #4
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    I just figured out that the example the professor showed us in class was wrong. You are right, it should most definitely not be what I have above. Thanks guys for your input~! It really helped me get this.


    Quote Originally Posted by HallsofIvy View Post
    It isn't equal to t^2- 1. Where did you get that? You must be looking at an example that has nothing to do with this given problem. f(t, Yn(t)) is simply f applied to t and whatever you have just calculated Yn(t) to be!

    In this case dy/dx= x^2y. Since we are given that y(0)= 1, a reasonable starting approximation for y is the constant y= 1; Then we have dy/dx= x^2(1) and, integrating y= (1/3)x^3+ C. In order to have y(0)= 1, we must have y(0)= (1/3)0+ C= 1 so C= 1 and y= (1/3)x^3+ 1. Replacing the y in the differential equation with that, dy/dx= x^2((1/3)x^3+ 1)= (1/3)x^5+ x^2 and integrating, y= (1/18)x^6+ (1/3)x^3+ C. Again, y(0)= 1 gives C= 1. Take dy/dx= f(x,y(x))= x^2((1/18)x^3+ (1/3)x^3+ 1) and integrate again.
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