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

General Solutions of 2nd Order Linear Homogeneous Ordinary D.E.s

  1. Oct 17, 2004 #1

    cepheid

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    A theorem in my textbook is confusing me:

    For the functions [itex] p(t) \ \ \text{and} \ \ q(t) [/itex] continuous on an open inteval [itex] I [/itex] defined by [itex] \alpha < t < \beta [/itex]:

    We have differential equation [itex] L[y] = 0 [/itex] where

    [tex] L = (\frac{d^2}{dt^2} + p\frac{d}{dt} + q) [/tex]

    The theorem attempts to prove that the general solution can be expressed in the form

    [tex] y = c_{1} y_{1} + c_{2} y_{2} [/tex]


    It does so by looking at an arbitrary solution [itex] y = \phi (t) [/itex]. Let's say that at some point [itex] t_0 [/itex], we let [itex] \phi (t_0) = y_0 \ \ \text{and} \ \ \phi^{\prime} (t_0) = y^{\prime}_0 [/itex].

    Then [itex] \phi (t) [/itex] is certainly a solution to the initial value problem:

    [tex] y^{\prime\prime} + py^{\prime} + qy = 0 [/tex]
    [tex] y(t_0) = y_0 [/tex]
    [tex] y^{\prime}(t_0) = y^{\prime}_0 [/tex]

    But, as proven by another theorem, as long as the Wronskian of [itex] y_1 \ \ \text{and} \ \ y_2 [/itex] is not zero at [itex] t_0 [/itex], then we can find constants c1 and c2 such that

    [tex] y = c_{1} y_{1} + c_{2} y_{2} [/tex]

    is also a solution to this initial value problem. By the uniqueness theorem (which the textbook declines to prove), there is only one solution to the initial value problem i.e.:

    [tex] y = \phi (t) = c_{1} y_{1} + c_{2} y_{2} [/tex]

    In my mind, this constitutes an actual proof of the theorem if and only if it is true for all possible initial value problems i.e. if I can show this for every possible point [itex] t_0 [/itex] that I care to choose in [itex] I [/itex]. That condition demands that the Wronskian of [itex] y_1 \ \ \text{and} \ \ y_2 [/itex] is not zero at any point in the interval! Not only does the textbook not make this point clear, it says something entirely different!

    "Not everywhere zero" is not the same thing as "nonzero everywhere"! As I've already said, I think it should be the latter. The former implies that it can be zero in certain places, as long as it isn't zero everywhere.

    So which is wrong, my interpretation of the proof, or the statement in the textbook? :confused:
     
    Last edited: Oct 17, 2004
  2. jcsd
  3. Oct 17, 2004 #2
    Remember that these conditions are all local. If [itex]W[y_{1},y_{2}](t_{0})\neq 0[/itex], means that you can build a solution in the form [itex]c_{1}y_{1}+c_{2}y_{2}[/itex] locally (in a vecinity of [itex]t_{0}[/itex], not in the whole interval). The curve [itex]\phi(t)[/itex] happens to be the graph of that solution, but only locally. To extend the solution to the whole interval, you need to be more carefull.

    Im sure you can find a proof for the existence and uniqueness theorem for 2 order ode in Brown or Coddington book.

    If you cant find the proof, give me some time and ill post a detailed one.
     
    Last edited: Oct 17, 2004
  4. Oct 18, 2004 #3

    cepheid

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    I don't mind the lack of proof for the existence/uniqueness theorem at the moment. It's just this theorem for now that still baffles me. I know now that I'm wrong and that the W need only be nonzero at one point, but I have no idea how this proves the theorem. Couldn't I equally well pick a point at which the Wronskian is zero for sure, set up an initial value problem to which phi is a solution, and show that phi cannot be expressed in the form c1y1 + c2y2?
     
  5. Oct 18, 2004 #4

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    It can be shown that, if y1 and y2 are both solutions to a second order, linear, homogeneous differental equation, then the Wronskian can be written as an exponential. It is either identically 0 (so y1 and y2 are dependent) or never 0 (so y1 and y2 are independent) on an interval.
     
  6. Oct 18, 2004 #5
    also, you migh take in mind that if [itex]W[y_{1},y_{2}](t)=0[/itex] and y is non trivial

    [tex]y_{1}\dot{y}_{2}-\dot{y}_{1}y_{2}=0[/tex]

    if [itex]y_{2}\neq 0[/itex] (if it is 0 then take [itex]y_{1}[/itex])

    [tex]\frac{y_{1}\dot{y}_{2}-\dot{y}_{1}y_{2}}{{y_{2}}^{2}}}=0[/tex]

    [tex]\frac{d}{dt}(\frac{y_{1}}{y_{2}})=0[/tex]

    integrating

    [tex]\frac{y_{1}}{y_{2}}=\lambda[/tex]

    [tex]y_{1}=\lambda y_{2}[/tex]

    so [itex]y_{1}[/itex] and [itex]y_{2}[/itex] are lineary dependent.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: General Solutions of 2nd Order Linear Homogeneous Ordinary D.E.s
Loading...