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Let's say I have something like[tex]d^2 u / d x^2 = u + u^2 /

  1. Jul 21, 2012 #1
    Let's say I have something like

    [tex]d^2 u / d x^2 = u + u^2 / a[/tex], where [tex]u = 1/r[/tex]

    How do I integrate to solve for x in terms of r?
  2. jcsd
  3. Jul 21, 2012 #2
    Re: derivatives

    If [itex]u=1/r[/itex], then you are going to have to use the chain rule on the LHS to get something like:

    [tex]\frac{2}{r^3}\big(\frac{dr}{dx}\big)^2-\frac{1}{r^2}\frac{d^2r}{dx^2}= \frac{1}{r}+\frac{1}{ar^2}[/tex]

    Unless I have just really screwed something up, I don't see how this ODE can be solved, since each [itex]r[/itex] is (presumably) a function of [itex]x[/itex]

    I should say that I don't know how to solve it.
  4. Jul 21, 2012 #3
    Re: derivatives

    Dang, I don't know how you did that, so I should have just posted the actual equation I am looking for, although I didn't realize until now that it would reduce to mostly constants. It is

    [tex]d^2u / dθ^2 = a u^2 - u + b[/tex]

    where u = 1/r and a and b are constants. θ isn't a function of r except as related in the equation after the integration as far as I can tell.

    From what you have, (dr / dx)^2 would be exactly what I am looking for. It still includes a term for a second derivative of [tex]d^2r / dx^2[/tex], though, but if we just solved for u first, could we susbstitute back in for the second derivative of [tex]d^2u / dx^2[/tex] from the original equation, then integrate using u, then just change u to 1/r in the final form? I suppose just solving for du / dθ with the equation I just gave would be easiest, and I could take it from there.
    Last edited: Jul 21, 2012
  5. Jul 21, 2012 #4
    Re: derivatives

    Well, if [itex]u[/itex] is a function of [itex]\theta[/itex] then this is just a Non-homogeneuous ODE (which is incredibly tough to solve, I think.) However, [itex]u[/itex] is a function of [itex]r[/itex] which means you must use the chain rule to get the second derivative of [itex]u[/itex] with respect to [itex]\theta[/itex]. Then, you get a DE that I have no idea how to solve (and I suspect there is no solution.)
  6. Jul 21, 2012 #5
    Re: derivatives

    Okay, thank you, Robert1986. I'll see if I can find something else that approximates it.
  7. Jul 21, 2012 #6
    Re: derivatives

    Hold on; I had a minor malfunction.

    Yes, I believe you can solve this for [itex]u(x)[/itex] and then [itex]r(x)=1/u(x)[/itex]. However, this is a very hard ODE to solve.
  8. Jul 22, 2012 #7


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    Re: derivatives

    By change of variables, you can massage it into the form
    y'' = y2+c [1]
    where |c| = 1, and the sign of c is that of ab-1/4.
    Consider instead
    y'' = y2 [2]
    This has solutions y = 6/(x-α)2. (There must be more general solutions too, but I haven't found them. In particular, there should be a solution through any prescribed point and any prescribed slope through that point.) For large y, [1] and [2] must behave much the same, so [1] has vertical asymptotes.
    For c = -1, there is also the solution y = 1. Nearby solutions diverge from this in both x-directions.
    For c = +1, [1] cannot have a horizontal asymptote. So I would think it must have multiple vertical asymptotes. The gap between the asymptotes need not be constant. Between any pair, the curve is symmetric: it descends from +∞, bottoms out somewhere, possibly y < 0, then reascends. The exact path is independent between each pair of asymptotes. It is completely determined by the size of the gap (or equivalently, by the value of y at y' = 0). It would be interesting to plot how the gap size depends on the minimum y.
    Last edited: Jul 22, 2012
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