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ODE Proof (2nd order linear homogeneous equations)

  1. Feb 7, 2013 #1
    1. The problem statement, all variables and given/known data
    Suppose u, v are two linearly independent solutions to the differential equation u''+p(x)u'+q(x)v=0. If x0,x1 are consecutive zeros of u, then v has a zero on the open interval (x0,x1)


    2. Relevant equations



    3. The attempt at a solution
    I'm trying to use the Wronskian(u,v;x) to arrive at a contradiction.

    I know that the W = uv'-u'v is never 0 since u,v are linearly independent. Furthermore, since it is a combination of C2 functions it must also be continuous on [x0,x1] = I. Also if I suppose that v is never zero on I, then it must be always positive or always negative. Same with u, since we know x0,x1 are consecutive zeros. From this I want to show that the Wronskian is positive at some point in I, and negative at another point, which would lead to a contradiction. Is this the right idea?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Feb 7, 2013 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Surely you don't mean to have both u and v in the equation itself? Since u and v are to be specific solutions, it would be better to say
    "Suppose u, v are two linearly independent solutions to the differential equation y''+p(x)y'+q(x)y=0.


    Sounds like a good plan. How are you going to implement it?

     
  4. Feb 7, 2013 #3

    pasmith

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    Homework Helper

    I think there is a direct proof: You have
    [tex]W(x_0) = u(x_0)v'(x_0) - u'(x_0)v(x_0) = -u'(x_0)v(x_0)[/tex]
    and similarly
    [tex]W(x_1) = -u'(x_1)v(x_1)[/tex]
    and, since [itex]W(x)[/itex] vanishes nowhere, [itex]W(x_0)[/itex] and [itex]W(x_1)[/itex] must have the same sign. Thus
    [tex]W(x_0)W(x_1) = u'(x_0)v(x_0)u'(x_1)v(x_1) > 0.[/tex]

    Now use the fact that [itex]x_0[/itex] and [itex]x_1[/itex] are consecutive zeroes of [itex]u[/itex] to show that [itex]u'(x_0)u'(x_1) < 0[/itex].

    What does that require of [itex]v(x_0)v(x_1)[/itex] if the condition on [itex]W(x_0)W(x_1)[/itex] is to hold?
     
  5. Feb 8, 2013 #4
    Yeah thanks guys, I think I get it now.

    v(x0)v(x1) would have to be negative. And since v is cts it must have an 0 between x0 and x1.
     
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