Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

A Integral equations -- Picard method of succesive approximation

  1. Nov 14, 2017 #1
    Equation
    [tex]\varphi(x)=x+1-\int^{x}_0 \varphi(y)dy[/tex]
    If I start from ##\varphi_0(x)=1## or ##\varphi_0(x)=x+1## I will get solution of this equation using Picard method in following way
    [tex]\varphi_1(x)=x+1-\int^{x}_0 \varphi_0(y)dy[/tex]
    [tex]\varphi_2(x)=x+1-\int^{x}_0 \varphi_1(y)dy[/tex]
    [tex]\varphi_3(x)=x+1-\int^{x}_0 \varphi_2(y)dy[/tex]
    ...
    Then solution is given by
    [tex]\varphi(x)=\lim_{n \to \infty}\varphi_n(x)[/tex].
    When I could say that this sequence will converge to solution of integral equation. How to see if there is some fixed point? I know how to use this method, but I am not sure from the form of equation, when I can use this method. Thanks for the answer.
     
  2. jcsd
  3. Nov 14, 2017 #2

    mathman

    User Avatar
    Science Advisor
    Gold Member

    [tex]\phi_0(x)=1\ results \ in \ \phi_1(x)=1[/tex], you're done!
     
  4. Nov 15, 2017 #3
    This is not my question. I know how to solve this. I am not sure when I can use this method. When sequence of functions ##\varphi_0(x)##, ##\varphi_1(x)##... will converge to ##\varphi(x)##.
     
  5. Nov 15, 2017 #4

    mathman

    User Avatar
    Science Advisor
    Gold Member

    Since all [tex]\phi_n(x)=1[/tex] are the same, the sequence trivially converges to [tex]\phi(x)=1.[/tex] I am not sure what you are looking for.
     
  6. Nov 15, 2017 #5

    WWGD

    User Avatar
    Science Advisor
    Gold Member

    I think s/he is looking for general conditions for convergence, not just for this particular problem.
     
  7. Nov 16, 2017 #6
    Yes. Thanks.
     
  8. Nov 16, 2017 #7

    MathematicalPhysicist

    User Avatar
    Gold Member

    You can use this method when you have: ##\int \lim_{n\to\infty} \varphi_n(y)dy = \lim_{n\to \infty} \int \varphi_n(y)dy##.
     
  9. Nov 16, 2017 #8

    WWGD

    User Avatar
    Science Advisor
    Gold Member

    Isn't this equivalent to dominated or monotone convergence?
     
  10. Nov 16, 2017 #9

    mathman

    User Avatar
    Science Advisor
    Gold Member

    Dominated convergence is a sufficient condition, but not necessary.
     
  11. Nov 16, 2017 #10

    WWGD

    User Avatar
    Science Advisor
    Gold Member

    Ah, yes, Dominated, no reason for Monotone here. Need some caffeine.
     
  12. Nov 17, 2017 #11
    Yes but if I have for example equation in the form
    [tex]\varphi(x)=f(x)+\lambda \int^x_0K(x,t)\varphi(t)dt[/tex]
    could I see this just for looking in kernel ##K(x,t)## and parameter ##\lambda##?
     
  13. Nov 17, 2017 #12

    MathematicalPhysicist

    User Avatar
    Gold Member

    @LagrangeEuler in your last post this is an eigenvalue problem: if we denote by: ##K\varphi(x) = \int_0^x K(x,t)\varphi(t)dt##

    Then you want to solve the equation: ##(I-\lambda K)\varphi = f##; you need to solve the equation ##\det |I-\lambda K| \ne 0 ## and then you have a solution: ##\varphi(x) = (I-\lambda K)^{-1}f(x)##; how to find the inverse, check any functional analysis textbook or Courant's and Hilbert's first volume.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Integral equations -- Picard method of succesive approximation
Loading...