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Math Help - Wronskian and Linear Independence

  1. #1
    Newbie Phantasma's Avatar
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    Wronskian and Linear Independence

    Suppose y_1 and y_2 are solutions of y''+p(x)y'+q(x)y=r(x), with p, q, and r being continuous functions of x. If y_1 and y_2 are linearly independent on an interval, is it possible to still get a Wronskian of 0 (id est x^2 and x|x|)?
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    Re: Wronskian and Linear Independence

    My guess is that the answer is no, but I would like someone to confirm this.
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    Re: Wronskian and Linear Independence

    Quote Originally Posted by Phantasma View Post
    My guess is that the answer is no, but I would like someone to confirm this.
    At first I though you were correct, it will not vanish as that would show that they are in fact linearly dependent.


    A common misconception is that W = 0 everywhere implies linear dependence, but Peano (1889) pointed out that the functions x2 and |x|x have continuous derivatives and their Wronskian vanishes everywhere, yet they are not linearly dependent in any neighborhood of 0.
    -wikipedia Wronskian - Wikipedia, the free encyclopedia
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    Re: Wronskian and Linear Independence

    Wronskian=0 implies that the functions are dependent if they are analytic on some intervals. The abs(x) is not analytic on the neighborhood of zero. So x^2 and |x|x are linearly dependent on any open interval not containing zero.So even though y1 & y2 seems linearly independent, they are actually not on any open interval not containing zero .
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  5. #5
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    Re: Wronskian and Linear Independence

    Hi,
    The attachment does not answer your question, and you may will know everything I say. But it does give a partial solution:

    Wronskian and Linear Independence-mhfdiffeqs.png
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    Re: Wronskian and Linear Independence

    Quote Originally Posted by Phantasma View Post
    Suppose y_1 and y_2 are solutions of y''+p(x)y'+q(x)y=r(x), with p, q, and r being continuous functions of x. If y_1 and y_2 are linearly independent on an interval, is it possible to still get a Wronskian of 0 (id est x^2 and x|x|)?
    The problem with your example is that x|x|, not being differentiable on any interval containing 0, cannot be a solution to such an equation. And if we are restricted to an interval that does NOT contain 0, we have all positive numbers, in which x|x|= x^2 or all negative numbers in which x|x|= -x^2, in either case not independent of x^2.
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