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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|
    [/itex]

    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

    Deveno

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    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

    Dick

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    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

    Deveno

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    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

    Dick

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    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

    LCKurtz

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    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.
     
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