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Wronskian ODEs

  1. Sep 6, 2010 #1
    I'm reading Ince on ODEs, and I'm in the section (in Chapter 5) where he talks about the Wronskian. There are quite a few things here that I don't quite understand or follow.

    I'm not going to get into all the details, but briefly, suppose we have the Wronskian of k functions:

    [tex]W = \left|\array{cccc} u_1 & u_2 & ... & u_k \\
    u_1' & u_2' & ... & u_k' \\
    ... & ... & ... & ... \\
    u_1^{(k-1)} & u_2^{(k-1)} & ... & u_k^{(k-1)} \endarray\right|[/tex]

    and we designate by [tex]U_1, U_2, ... U_k[/tex] the minors of the elements in the last line of the Wronskian.

    So far so good.

    Then there is quite a lot of derivation that I don't quite follow, but my main question is this: eventually he gets to the point where he has the following equation:

    [tex]U_1'U_k - U_k' U_1 = 0[/tex]

    Ince then tersely states that since [tex]U_k[/tex] is not zero, it follows that:

    [tex]U_1 = -c_1U_k[/tex]

    or in other words, that U_1 is a constant multiple of U_k.

    That's the part I don't quite follow.

    All things being equal, his expression itself seems to be "like" a Wronskian, except that it involves the minors of a determinant, rather than "functions" (or, solutions to a differential equation). That is, we could re-write his initial equation:

    [tex]\left|\array{cc}U_1 & U_k \\ U_1' & U_k'\endarray\right| = 0[/tex]

    But I don't see how linear dependence follows from that equation.

    In do see how if U_1 and U_k are linearly dependent, then we can conclude that the "Wronkian" must be zero.. but isn't it a standard result from linear algebra that if the Wronskian is zero, we can't necessarily conclude that the functions are linearly dependent?

    Does this restriction not apply in this case, since we are working w/ the minors of a determinant, rather than "real" functions?
  2. jcsd
  3. Sep 6, 2010 #2
    Re: Wronskian

    The way I understood this when I learned it is:
    If U_1 and U_k are both not identically zero, we can divided W by (U_1)^2 which yields:
    [tex] \frac{W}{U_{1}^{2}}=\frac{U_1'U_k - U_k' U_1}{U_{1}^{2}}=(\frac{U_k}{U_1})'=0[/tex]

    [tex](\frac{U_k}{U_1})'=0[/tex] suggests U_1=const x U_k
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