1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

How many solutions?

  1. Aug 19, 2007 #1
    Hello, this is my first post so I am not familiar with how to write maths in here.

    I have a third order ode of the form

    y'''+A/y=0 where y=y(x) (A is just a constant) with well defined boundary conditions.

    I believe there is no analytic solution to an ode of this form, but I can (by imposing a further condition) solve this numerically for y.

    My question is this; I solve this numerically in Maple and one answer pops out. Is there any way I can easily prove this is the only real, non-negative solution?
     
  2. jcsd
  3. Aug 19, 2007 #2

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    I can't think of any way to answer your question because I can't imagine why you would "I believe there is no analytic solution to an ode of this form". It seems to me pretty obvious that there exist a solution as long as the boundary conditions do not require y= 0. Perhaps you could explain a little more exactly what you mean.

    You don't say WHAT boundary conditions you have. Whether or not a boundary value problem has a unique solution depends strongly on the boundary conditions themselves.

    For example, the problem y"+ y= 0, with boundary conditions, y(0)= 0, y(1)= 0 has a unique solution. The same equation with boundary conditions y(0)= 0, [itex]y(\pi)= 1[/itex] has no solution. The same equation with boundary conditions y(0)= 0, [itex]y(\pi)= 0[/itex] has an infinite number of solutions.
     
    Last edited: Aug 20, 2007
  4. Aug 20, 2007 #3
    Thanks for the reply,

    my bcs are y(-1)=0, y(1)=0, y'(-1)=beta1, y'(1)=(beta1^2+4A)^(1/2) and y must also satisfy (integral y dx, x=-1..1)=2/3.

    I have a third order ode, 2 unknowns and 5 conditions so I have enough for a solution.

    This does mean y''' is singular at x=-1, 1 but that is ok.

    As far as I can see this cannot be solved analytically, but I can find one numerical solution, can I know there are no other solutions that are everywhere real and positive?
     
  5. Aug 21, 2007 #4
    Hm. All I can think of is that your equation may be derived from

    [tex]y y^{\prime \prime} + A x + B - \frac{y^{\prime 2}}{2} = 0[/tex]

    where B is a constant. But

    [tex]y y^{\prime \prime} - \frac{y^{\prime 2}}{2} = 2 y^{\frac{3}{2}} (\sqrt{y})^{\prime \prime}[/tex]

    Therefore you can re-write your equation

    [tex]u^{\prime \prime} + \frac{Ax + B}{2 u^{ 3}} = 0[/tex]

    with the substitution [tex]u = \sqrt{y}[/tex]

    ...and with a suitable sub for x, you should get a standard Emden-Fowler equation. See here. (EqWorld)
     
    Last edited: Aug 21, 2007
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: How many solutions?
Loading...