Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Frobenius method?

  1. Nov 29, 2005 #1
    for the forbenius method,
    if the roots to the indical equation differ by an integer,
    why do you always have to take the larger root to find the smaller root?
  2. jcsd
  3. Nov 29, 2005 #2


    User Avatar
    Science Advisor

    ?? I'm sorry, this makes no sense to me at all. You find the roots of the indicial equation by solving it! It is not necessary to "take the larger root to find the smaller"!

    I expect you mean that you find a solutions by "plugging in" the larger root in order to find the coefficients of the power series expansion and the second, independent solution, will involve that first solution as well as the second, smaller root.

    Specifically, if y1(x) is the series solution to the equation using r1, the larger root, then another, independent, solution is of the form
    [tex]ay_1(x)ln x+ x^{r_2}\left[1+ \Sigma_{n-1}^\infty c_n x^n\right][/tex]
    That is from Elementary Differential Equations and Boundary Value Equations by Boyce and DiPrima. Unfortunately they say "For this case the derivation is considerably more complicated and will not be given here"! They do note that a derivation is given in An Introduction to Ordinary Differential Equations by E.A. Coddington.
  4. Nov 29, 2005 #3
    yes, that's what i mean... but why does that happen?
  5. Nov 3, 2008 #4
    How it can be decided if the natural logarithm function will be used for the second solution(for the case 3 in which the roots of indicial eqn. differ by an integer)?

    Here is a quotation from Erwin Kreyszig, Advanced Eng. Mathematics 9th ed.,
    "Indicial equation r(r - 1) + r - 1 = 0. Hence r1 = 1, r2 = -1. These roots differ
    by an integer; this is Case 3. It turns out that no logarithm will appear."
  6. Nov 4, 2008 #5


    User Avatar
    Science Advisor

    Well, that says "this is Case 3". What are "Case 1", "Case 2", and "Case 3"?

    The simplest way to get the idea, though not the proof, is to look at the Euler type equation, which is, in a sense, the "critical case" here.

    An Euler type equation, also called "equipotential" has xn as coefficient of the nth derivative. It can be shown that the substitution t= ln(x) converts an Euler type equation, in x, to an equation with constant coefficients, in t. The two equations have the same "characteristic equation". Of course, if an equation with constant coefficients has a double root, say r as double root, then two independent solutions to the differential equation are ert and tert. Replacing t with ln(x) gives erln(x)= xr and ln(x)xr as independent solutions to the original Euler type equation.
  7. Nov 11, 2009 #6
    i don't remember the proof, but in practice when you work out the problems with the small root normally the ak in recursion relation diverge. for example the bessell solution is like

    ak+2=- ak 1/((k+2)(2n+k+1) in the case that the root are n and -n both integer and n is a natural number.

    if we use -n, then for the case a2n-3=a2n-1 1/((2n-3)(-2n +(2n-1)+1)) then a2n-3 is not define.

    the proof of why for the larger root is always possible, should be a special case of the fuch' s theorem.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook