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Apparently this 2nd-order ODE has 3 solutions?

  1. Sep 30, 2004 #1
    Apparently this 2nd-order ODE has 3 solutions??

    The following apparently has 3 solultions:

    [tex]
    \frac {d^2u}{d\theta^2} + u = -\frac {1}{ml^2u^2}f(u^{-1})
    [/tex]

    where:

    u = 1/r
    m = mass
    l = angular momentum

    One of the solutions is:

    [tex]r=r_0e^{k\theta} \text{ where } \theta \text { varies logarithmically with time}[/tex]

    Apparently there are also 2 additional solutions (depending on the value of the constant [tex]\alpha[/tex])
    that could be in the form of:

    [tex]r=Ae^{\sqrt{\alpha x}}+Be^{-\sqrt{\alpha x}} \text{ or }[/tex]
    [tex]r=A\theta + B \text{ or }[/tex]
    [tex]r=Asin({\sqrt{\alpha x}})+Bcos({\sqrt{\alpha x}})[/tex]

    So, knowing:

    [tex]
    \frac {d^2u}{d\theta^2} + u = -\frac {1}{ml^2u^2}f(u^{-1})
    [/tex]

    and

    [tex]r=r_0e^{k\theta}[/tex]

    How does one specifically determine the equations of the additional solutions?

    Thanks!
     
  2. jcsd
  3. Sep 30, 2004 #2

    NateTG

    User Avatar
    Science Advisor
    Homework Helper

    I saw your other post, and really wasn't able to understand the equation. What function is [tex]f[/tex]?

    Something like:

    [tex]
    \frac {d^2}{d\theta^2}u(\theta)^2+ u(\theta) = -\frac {1}{ml^2u(\theta)^2}
    [/tex]

    Makes more sense, but if you plug in [tex]u(\theta)=\cos \theta[/tex] which represents a solution to your original problem you get:
    [tex]-\cos \theta + \cos \theta = k \sec(\theta)^2[/tex]
    or
    [tex]0 = k \sec(\theta)^2[/tex]
    which is...problematic.

    Perhaps you mean that [tex]f(u^{-1})=f(r)=-\frac{1}{r^3}=-u^3[/tex]
    then your eqation simplifies to:
    [tex]
    \frac {d^2}{d\theta^2}u(\theta)^2+ u(\theta) = \frac {u(\theta)}{ml^2}
    [/tex]
    or
    [tex]\frac{d^2}{d\theta^2}u(\theta)^2 + (1-\frac{1}{ml^2}) u(\theta)=0[/tex]
    which has solutions of the form
    [tex]u(\theta)=k_1e^{i(\theta)\sqrt{1-\frac{1}{ml^2}}}[/tex]
    so
    [tex]r(\theta)=k e^{-i(\theta) \sqrt{(1-\frac{1}{ml^2}}}[/tex]
    It's relatively easy to see that if [tex]1-\frac{1}{ml^2} < 0 [/tex] the functions are not orbits.
     
  4. Oct 3, 2004 #3
    I think you've addressed this NateTG, but I still
    have some digesting to do...

    The following is basically Newton's DE of motion using two-dimensional polar coordinates instead of three:

    [tex]
    \frac {d^2u}{d\theta^2} + u = -\frac {1}{ml^2u^2}f(u^{-1})
    [/tex]

    where:

    u = 1/r
    m = mass
    l = angular momentum
    [tex]f(u^{-1}) \text { = the central force} [/tex]

    And, yes, l = angular momentum per unit mass (thanks ehild!).

    There is a fair amount of derivation provided in the text. The DE is labled as "the differential equation of the orbit of a particle moving under a central force."

    Setting u = 1/r you thus have:

    [tex]u={r_0}^{-1}e^{-k\theta}[/tex]

    Referring back to the DE and using

    [tex] \text{u, } \frac {d^2u}{d\theta^2}[/tex]

    then simply re-arranging and solving algebraically
    for [tex]f(u^{-1} )[/tex] you get

    [tex]f(r)=-(k^2+1)ml^2\cdot \frac {1}{r^3}[/tex]

    which shows that the central field corresponds to an inverse-cube force.

    I'm stuck at the proposition of having to find two other types of possible orbits (and their equations). I was thinking it was a simple matter of expressing the solution to the DE in basic alternative forms -- but maybe not??

     
  5. Oct 3, 2004 #4
    If the diff. equation is of the form au'' + bu' + cu = 0, then shouldn't it only have two possible solutions? Unless of course you are cosidering u = 0 a solution.
     
  6. Oct 3, 2004 #5

    cj

    User Avatar

    I was thinking the same thing, but I also seem to
    remember that DE of this form have a bunch of solutions
    based on the sign of a combination of constants.

    I guess I need to refresh my working knowledge of DE.
     
  7. Oct 4, 2004 #6
    Your problem is incomplete. You need two boundary conditions in order to set a unique solution for the ode. Also you need to consider under wich circumstances au'' + bu' + cu = 0 + (boundary conditions) has a solution and if this solution is unique. (Existence and Uniqueness Theory).

    In the case that au'' + bu' + cu = 0 + (boundary conditions) satisfies the uniqueness and existence theorem, the there is only two l.i. solutions to the ode.

    On relation to 1 If you already founded a solution, use the reduction of order method to find the second solution ie
    [tex]y_{2}(x)=v(x)y_{1}(x)[/tex]
     
    Last edited: Oct 4, 2004
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