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

[grav-universe]

Let's say I have something like

$$d^2 u / d x^2 = u + u^2 / a$$, where $$u = 1/r$$

How do I integrate to solve for x in terms of r?

 

[Robert1986]

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

$$\frac{2}{r^3}\big(\frac{dr}{dx}\big)^2-\frac{1}{r^2}\frac{d^2r}{dx^2}= \frac{1}{r}+\frac{1}{ar^2}$$

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

EDIT:
I should say that I don't know how to solve it.

 

[grav-universe]

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

$$\frac{2}{r^3}\big(\frac{dr}{dx}\big)^2-\frac{1}{r^2}\frac{d^2r}{dx^2}= \frac{1}{r}+\frac{1}{ar^2}$$

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

EDIT:
I should say that I don't know how to solve it.

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

$$d^2u / dθ^2 = a u^2 - u + b$$

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 $$d^2r / dx^2$$, though, but if we just solved for u first, could we susbstitute back in for the second derivative of $$d^2u / dx^2$$ 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.

 

[Robert1986]

Well, if $u$ is a function of $\theta$ then this is just a Non-homogeneuous ODE (which is incredibly tough to solve, I think.) However, $u$ is a function of $r$ which means you must use the chain rule to get the second derivative of $u$ with respect to $\theta$. Then, you get a DE that I have no idea how to solve (and I suspect there is no solution.)

 

[grav-universe]

Okay, thank you, Robert1986. I'll see if I can find something else that approximates it.

 

[Robert1986]

Hold on; I had a minor malfunction.

Yes, I believe you can solve this for $u(x)$ and then $r(x)=1/u(x)$. However, this is a very hard ODE to solve.

 

[haruspex]

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.
HTH