Results 1 to 2 of 2

Math Help - Methods for Inhomogeneous Difference Equations

  1. #1
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,675
    Thanks
    302
    Awards
    1

    Methods for Inhomogeneous Difference Equations

    I believe that the term "inhomogeneous difference equation" describes the following problem:
    a_{n + 2} - 2a_{n + 1} + a_n = 4 with the conditions a_0 = 1 and a_1 = 2. (This problem has no relevance to anything, I just made it up.)
    ====================================
    TPH gave a method of solution for this.
    The homogeneous version of this difference equation is
    a_{n + 2} - 2a_{n + 1} + a_n = 0
    and has the characteristic equation
    m^2 - 2m + 1 = 0

    (m - 1)^2 = 0

    So m = 1 is a double root. Thus
    a_n = A(1)^2 + Bn(1)^2 = Bn + A
    as the homogeneous solution.

    We need a particular solution. So let a_n = C. Then a_{n + 2} = C, a_{n + 1} = C and substitution into the original equation gives:
    C - 2C + C = 4
    which is impossible.
    ====================================

    Soroban also gave a method. Consider the related difference equation:
    a_{n + 3} - 2a_{n + 2} + a_{n + 1} = 4

    When we subtract the original equation
    a_{n + 2} - 2a_{n + 1} + a_n = 4
    from this we get the homogeneous difference equation:
    a_{n + 3} - 3a_{n + 2} + 3a_{n + 1} - a_n = 0

    The characteristic equation here is
    m^3 - 3m^3 + 3m - 1 = 0

    (m - 1)^3 = 0
    and has m = 1 as a triple root. So the solution of the homogeneous equation is
    a_n = A(1)^n + Bn(1)^n + Cn^2(1)^n = Cn^2 + Bn + A

    Now we substitute this solution into the original equation:
    a_{n + 2} - 2a_{n + 1} + a_n = 4

    [C(n + 2)^2 + B(n + 2) + A] - 2[C(n + 1)^2 + B(n + 1) + A] + [Cn^2 + Bn + A] = 4

    After a number of cancellations:
    2C = 4

    So C = 2.

    Thus
    a_n = 2n^2 + Bn + A

    Application of the conditions a_0 = 1 and a_1 = 2 gives (respectively)
    a_0 = A = 1 ==> a_n = 2n^2 + Bn + 1
    and
    a_1 = 2(1)^2 + B(1) + 1 = 2

    2 + B + 1 = 2

    B + 3 = 2

    B = -1

    Which implies
    a_n = 2n^2 - n + 1 <-- This correctly defines the solution.
    ====================================

    My question (long in coming) is why don't both of these methods work? Is there an assumption in TPH's method that makes it unviable for this specific problem? And, if so, is Soroban's the most general method?

    -Dan
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,675
    Thanks
    302
    Awards
    1
    Oooh! Wait a minute!

    In TPH's method we have that a_n = Bn + A as a solution to the homogeneous equation. Does that mean when I want to pick a particular solution, since constant and linear terms are already part of the homogeneous solution that I need to pick a_n = Cn^2 as the particular solution? That would make sense with Soroban's solution...

    -Dan
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Homogeneous and Inhomogeneous differential equations
    Posted in the Differential Equations Forum
    Replies: 4
    Last Post: April 24th 2011, 07:03 AM
  2. Finite difference methods boundary conditions
    Posted in the Business Math Forum
    Replies: 0
    Last Post: March 29th 2011, 05:23 AM
  3. Replies: 0
    Last Post: October 29th 2010, 08:12 AM
  4. Replies: 8
    Last Post: August 30th 2010, 07:00 AM
  5. Replies: 5
    Last Post: October 20th 2008, 02:55 AM

Search Tags


/mathhelpforum @mathhelpforum