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Derivative of a Wronskian

  1. Oct 7, 2008 #1
    In general, the question is how do you take the derivative of the determinant of a matrix of functions, but more specifically how does one do this for a Wronskian?

    I've read a remark that seemed to say that the derivative for an nth order Wronskian is the determinant of a sum of n matrices, each of which is made by differentiating one row of the Wronskian. Is that right?

    In linear algebra texts derivatives of matrices of functions are discussed but it's been so long that the language of the latter chapters of those texts is no longer accessible to me. Is there a way to understand this without adjoints etc?

    Thank you,
    Genya
     
  2. jcsd
  3. Oct 8, 2008 #2
    A determinant of a matrix NxN is a sum of N! products of N matrix elements. Each product contains exactly one element from each row and each column, if you draw it on the matrix it looks like 'lightning'. So the determinant is the sum of all possible 'lightnings'. When you differentiate that sum, the result is the sum of the derivatives of each 'lightnings'. The derivative of each lightning, by product rule, is sum of N products, in each product only one element of the lightning is differentiated.

    That's why the derivative of the determinant is a sum of N determinants of N matrices, each matrix obtained from the original one by differentiating only one row (or column if u prefer to work with the columns). The one differentiated row corresponds to only one element differentiated in each lightning.
     
  4. Apr 5, 2009 #3

    thr

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    so how do you prove that? it's not hard to prove for n=2 that the derivative of the wronskian

    (det(f,g))'(t) = det(f'(t),g(t)) + det(f(t),g'(t))

    but how to go on from here? it smells a bit of induction... but then again it doesn't seem to be the best way.
     
  5. Apr 6, 2009 #4
    I think to prove it they make use of the definition of determinant by permutation (can't recall it exactly). The definition is something like this

    det(A)=[tex]\sum (-1)^\sigma a_{1i_1} a_{2i_2} ... a_{ni_n}[/tex]

    where the sum is over all permutation and [tex]\sigma[/tex] may be related to odd or even permutation (can't remember).

    If we differentiate det(A) then I think we should be able to obtain the result.
     
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