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

Solving this type of recurrence equation

  1. Aug 3, 2011 #1
    Hi,

    The problem is to solve
    [tex](1-g_{i+1})P_{i+1}-P_{i}+g_{i-1}P_{i-1}=0[/tex]
    for [tex]P_{i}[/tex]
    with boundary condition
    [tex]P_{i}=P_{i+L}, g_{i}=g_{i+L}[/tex]
    Can anyone provide any guide of solving this type of recurence equation?
    Thank you!
     
    Last edited: Aug 3, 2011
  2. jcsd
  3. Aug 4, 2011 #2

    Stephen Tashi

    User Avatar
    Science Advisor

    I don't know of any general methods for solving recurrence relations in several unknown functions.

    Since setting g and P to be constant sequences gives a solution to the problem, I suggest that you state additional conditions that g and P must satistfy if having them constant isn't what you are after.
     
  4. Aug 5, 2011 #3
    Thanks for your reply.

    In my problem, [tex]g_i[/tex] is given and arbitrary, so in general [tex]g_i[/tex] is not a constant sequence.
     
  5. Aug 5, 2011 #4

    Stephen Tashi

    User Avatar
    Science Advisor

    Then, as I interpret the problem, it amounts to solving [itex] L [/itex] simultaneous linear equations with constant coefficients and unknowns [itex] P_1,P_2,..P_L [/itex].

    Are you looking for a closed form symbolic answer instead of a numerical one?

    ( The wikipedia article on "recurrence relation" says that linear constant coefficient difference equations can be solved with z-transforms. I, myself, have never done that. )
     
  6. Aug 5, 2011 #5
    Because both [itex]\lbrace P_{i}\rbrace[/itex] and [itex]\lbrace g_{i}\rbrace[/itex] are period with a common period of L, you should use the discrete Fourier transform:

    [tex]
    P_{i} = \frac{1}{\sqrt{L}} \, \sum_{k = 0}^{L - 1}{\tilde{P}_{k} \, \exp{\left(\frac{2 \pi j k \, i}{L}\right)}}
    [/tex]
    and similarly for [itex]g_{i}[/itex]. Then you will use the convolution theorem for the products.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook