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Does my proof make sense?

  1. Nov 10, 2011 #1
    Here's the problem:
    Show that if we have two independent solutions to a 3rd order homogenous linear differential equation y'''+a2y''+a1y'+a0=0 we can always find a third solution which is independent from the two given solutions.

    Here is my proof:
    since y2 and y1 are linearly independent, there exists a non-constant function u such that y2=uy1.
    Now, If we plug y2 into the equation, we can reduce the order of the original equation to a 2nd order linear ODE of the form a(x)v+b(x)v'+c(x)'' = 0 where:
    [tex]u=y_2/y_1=\int{v(x)dx}[/tex]

    Now, If I set v=wy3, and I suppose t=w', then doing the same process I can reduce the ODE a(x)v+b(x)v'+c(x)v''=0 into a first order linear ODE of the form r(x)t+s(x)t'=0.
    This ODE is always solvable and the solution to this ODE is:
    [tex]t=Ce^{\int{-r(x)/s(x)dx}}[/tex]
    so, w would be:
    [tex]w=C\int{e^{\int{-r(x)/s(x)dx}}}[/tex]

    Having w and v, we can find y3=v/w.

    This shows us that y3 can be found, but I don't know how to show that y3 is necessarily independent from y1 and y2.

    Any idea that completes my proof would be highly appreciated.
     
  2. jcsd
  3. Nov 12, 2011 #2
    Did you try using the wronskian?
     
  4. Nov 12, 2011 #3
    I used Wronskian of y1 and y2 to prove that v is not a constant function, but I don't know how to use Wronskian to prove that y3 is linearly independent from y1 and y2. You mean that I should form a 3 by 3 determinant that each row is a vector whose components are yi yi' and yi'' (i=1,2,3)? then I'll have to show that the Wronskian isn't equal to 0? Is that what you meant?
     
  5. Nov 12, 2011 #4
    Yeah, you have to take a 3x3 matrix determinant and show it's not zero.
     
  6. Nov 12, 2011 #5
    but How could I form that determinant? I mean I don't precisely have y3, in case of y1 and y2 I got some information that they are linearly independent, but regarding y3 I got no information explained precisely. would you give me a hint?
     
  7. Nov 12, 2011 #6
    You just gave me y3=y1/y2.

    Just use quotient rule twice on y3 and form the wronskian.

    See http://en.wikipedia.org/wiki/Wronskian

    You want to know how to take 3x3 determinant, use the Sarrus rule.

    Am I misunderstanding?
     
  8. Nov 12, 2011 #7
    If you have an expression for y3, all the hard work is done.
     
  9. Nov 12, 2011 #8
    I hope I haven't done such a mistake in typing my proof, because y3 isn't necessarily equal to y1/y2, u = y1/y2, not y3.
    y3=v/w. but v is a solution of the 2nd order linear ODE, so I don't precisely have it.

    I doubt I could find a precise expression for y3.
     
  10. Nov 12, 2011 #9
    I think there is a different way of doing this problem. Construct the characteristic equation:
    [tex]z^3+a_2z^2+a_1z+a_0.[/tex]
    Since this is a cubic, there are 3 types of zeros (by the fundamental theorem of algebra):
    1. 3 distinct real zeros
    2. 2 repeated and 1 distinct real zeros
    3. 1 real zero and two conjugate pairs of zeros

    Observe that there cannot be three complex roots because complex roots come in pairs.

    Now show that in each case, if there are two linearly independent zeros, then the third has to be linearly independent.
     
  11. Nov 12, 2011 #10
    I believe you have insufficiently shown how to go from a second-order to a first-order equation. Also, when you get something you think is right, would be nice if you could check it with a real-live problem that you know the answer to and you can do that check numerically. That is, get two solutions, then numerically compute whatever difficult expression for the third and then see if it matches (numerically at least), what you know is correct. For example, does your expression above for y_3 agree with an actual set of solutions for a third-order ODE?
     
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