1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Difference equations

  1. Mar 14, 2010 #1
    1. The problem statement, all variables and given/known data
    Suppose that {uk} and {vk} are sequences satisfying uk = Auk-1 k = 1, 2, ... and vk = Avk-1 k = 1, 2,... Show that if u0 = v0 then ui = vi for all values of i.

    2. Relevant equations

    uk --> is u subscript k
    u0 --> is u subscript 0
    uk-1 --> is u subscript k-1
    ui --> is u subscript i

    3. The attempt at a solution

    Well so far I have...
    uk = A^k(u0)
    = A^k(a1u1 + a2u2 +...+anun)
    = a1(A^k)u1 + a2(A^k)u2 +...+ an(A^k)un
    = a1(lambda1^k)u1 + a2(lambda2^k) +...+ an(lambdan^k)un

    But since u0 = v0 A^k(u0) = A^k(v0)

    But after there I get uncertain cause I think my next steps would be:
    = A^k(a1v1 + a2v2 +...+ anvn)
    = a1(A^k)v1 + a2(A^k)v2 +...+ an(A^k)vn
    = a1(lambda1^k)v1 + a2(lambda2^k)v2 +...+ an(lambdan^k)vn)

    Then conclude uk = vk?
    Is this correct?

    Thanks! :)
  2. jcsd
  3. Mar 14, 2010 #2


    User Avatar
    Science Advisor
    Homework Helper

    Hi simmonj7! :smile:

    (try using the X2 tag just above the Reply box :wink:)
    (what are the as and the lambdas? :confused: anyway …)

    Hint: put wk = uk - vk, for all k. :wink:
  4. Mar 14, 2010 #3
    When solving a difference equation all the way out you have to find the eigen values, then find the corresponding eigen vectors and then have to find a relationship between those eigen vectors and the vector x0. Thus x0 = a1u1 + a2u2 +...+ anun is the relationship between all the eigen vectors.

    Then you just plug in that relationship for x0 into the equation xk = (A^k)x0. From there you distribute the A^k through and then (because (A)x = (lambda)x then (A^k)x = (lambda^k)x where lambda is an eigen vector of A) you substitute that back into the equation and that is what all the a's and lambda's are.
  5. Mar 14, 2010 #4


    User Avatar
    Science Advisor
    Homework Helper

    But there's only one eigenvalue here (and it's A). :confused:
  6. Mar 14, 2010 #5
    Where are you getting that there is only one eigen value? There is no way to determine how many eigen values there are of A...
  7. Mar 15, 2010 #6


    User Avatar
    Science Advisor
    Homework Helper

    The eigenvalues of A don't matter …

    all that matters is the roots of the characteristic equation of this recurrence relation which in this case is the single root, A.

    Try using the wk I mentioned earlier. :smile:
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook