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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


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    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
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