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!

Question on the Wronskian

  1. Feb 16, 2012 #1
    1. The problem statement, all variables and given/known data
    Given the two functions:
    [itex]f(t) = t[/itex]
    [itex]g(t) = |t|[/itex]

    Use the Wronskian to determine if the two functions are dependent or independent.

    2. The attempt at a solution
    I have already found the correct answer to this, which is that it is independent but I have some questions as to how this is. When I first tried to solve this I found that it was dependent, based on the following reasoning:

    [itex]W[f(t),g(t)] = \left| {\begin{array}{cc}
    t & |t| \\
    1 & \pm 1 \\
    \end{array} } \right|

    Calculating the determinant:
    [itex] W[f(t),g(t)] = (t)(\pm 1) - (|t|)(1)[/itex]
    [itex] W[f(t),g(t)] = (t)(\pm 1) - |t|[/itex]

    Given that [itex]t<0[/itex] and [itex]0<t[/itex] will determine the sign of [itex]g'(t) = \frac{d}{dt} |t|[/itex], then utilizing the following conditions:

    If [itex]t = +1[/itex], then [itex]g'(t) = +1[/itex].
    If [itex]g'(t) = +1[/itex], then:
    [itex] W[f(t),g(t)] = (t)(\pm 1) - |t|[/itex]
    [itex] W[f(1),g(1)] = (1)(+1) - |1| = 0[/itex]

    Likewise, if [itex]t = -1[/itex], then [itex]g'(t) = -1[/itex].
    If [itex]g'(t) = -1[/itex], then:
    [itex] W[f(t),g(t)] = (t)(\pm 1) - |t|[/itex]
    [itex] W[f(-1),g(-1)] = (-1)(-1) - |-1| = 1 - 1 = 0[/itex]

    Thus, based on the above reasoning, I thought the answer was dependent, but in fact, it is independent; can someone point out my mistake?
  2. jcsd
  3. Feb 16, 2012 #2


    User Avatar
    Science Advisor

    the Wronskian is undefined at t = 0, and is therefore not identically 0 everywhere.

    indeed suppose that there existed a,b in R, with af(t) + bg(t) identically 0, for all t.

    for t > 0, we get that at + bt = 0, and since t ≠ 0, a + b = 0.

    for t < 0, we have that at - bt = 0, so that a - b = 0.

    therefore, 2a = 0, so a = 0, and thus b = 0, so f and g are linearly independent over R.
  4. Feb 16, 2012 #3


    User Avatar
    Science Advisor
    Homework Helper

    I would say you can't really apply the Wronskian if the interval you are testing on contains 0. |t| isn't differentiable at t=0. But they are linearly independent since there are no constants such that a*t+b*|t|=0 for all t except for a=b=0. On the other hand as your Wronskian suggests, if you are working on an interval of nonnegative numbers, 1*t+(-1)*|t|=0 or an interval of nonpositive numbers, 1*t+1*|t|=0. So they are linearly dependent over some intervals.
    Last edited: Feb 16, 2012
  5. Feb 16, 2012 #4


    User Avatar
    Science Advisor

    i think it fair to point out that the Wronskian test is a negative test:

    Wronskian not identically 0 => independent functions.

    Wronskian identically 0: might be independent, might not.

    (counter-example: f(t) = t2, g(t) = t|t|).
  6. Feb 16, 2012 #5


    User Avatar
    Science Advisor
    Homework Helper

    Right. That's why I was careful to say "Wronskian suggests" instead of "Wronskian shows". Good point to make explicit though.
  7. Feb 16, 2012 #6


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    To add a little more, W(f,g) identically 0 does imply dependence if f and g are solutions of a 2nd order linear DE. To get away from the derivative doesn't exist example, consider {t,t2} on [-1,1]. Here their Wronskian is W = t2. Since that doesn't satisfy that it is either never 0 or identically 0 on the interval, what it shows, besides that they are independent, is that they aren't solutions to a second order linear DE.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Discussions: Question on the Wronskian
  1. Wronskian question (Replies: 3)

  2. Find the Wronskian? (Replies: 2)