# Nonlinear first-order ODE?

• WendysRules
In summary, the conversation is about setting up a problem involving conservation of energy for an object in a gravitational potential. The equations are manipulated to find a solution for r(t) and theta(t), but it is noted that there is no closed form solution for these functions. Care must also be taken in choosing the correct root when converting between different forms of the equations.

## Homework Statement

Let $$\frac{1}{2}\dot{r}^2=e+\frac{m}{r}-\frac{L^2}{2r^2}$$ where L is angular moment, and e is energy (so I guess I'll take as constants for now...)

## Homework Equations

Not sure for now.

## The Attempt at a Solution

So, if I let $$u=\frac{1}{r}$$ then my equation becomes $$\frac{1}{2}\dot{r}^2=e+mu-\frac{L^2u^2}{2}$$ However, I'm not sure if I should also get my differential in terms of u as well. IE $$r=\frac{1}{u} \rightarrow dr=-\frac{1}{u^2} du$$ or just continue down the first path.

If I don't do the differential change, then I'd get $$\frac{1}{2}\dot{r}^2=e+mu-\frac{L^2u^2}{2} \rightarrow \dot{r}^2 = 2e+2mu-\frac{2L^2u^2}{2}$$ finally giving us... $$\dot{r} =\sqrt{2e+2mu-L^2u^2}$$ which is just... yuck.

If I do the substitution, i'll get... $$\frac{1}{2}\dot{r}^2=(\frac{dr}{2dt})^2 \rightarrow (\frac{-\frac{1}{u^2} du}{2dt})^2 \rightarrow (\frac{-du}{2u^2dt})^2$$ From here, we set it equal to the righthand side, and then see that $$\frac{-du}{2u^2dt} = \sqrt{e+mu-\frac{L^2u^2}{2}} \rightarrow \frac{du}{u^2} =-2\sqrt{e+mu-\frac{L^2u^2}{2}} dt$$ which is also, bad looking. So before I go ahead and solve either, I was looking to maybe see if there is an easier way to look at these since the next one is even *worse* (cubic in r..)

Could it also be that I'm not seeing the simplicity of it all? I'm not sure if r *depends* on t really, since this comes from our metric that ##ds^2=dr^2+r^2d\phi^2##(working in equatorial plane thus, ##\sin\theta = 1## and ##d\theta = 0##) Can go more into this if needed. So, it could be that the right hand side on both is simple because none of the terms depend on t.

The equation you started with is conservation of energy for an object in a gravitational potential, so you should expect ##r## and ##\phi## to vary with time in general. What's your end goal for this problem? Finding ##r(t)## and ##\phi(t)##, or finding ##r(\phi)##? If it's the latter, you can use the fact that angular momentum is conserved to relate ##\dot r## with ##dr/d\phi##.

vela said:
The equation you started with is conservation of energy for an object in a gravitational potential, so you should expect ##r## and ##\phi## to vary with time in general. What's your end goal for this problem? Finding ##r(t)## and ##\phi(t)##, or finding ##r(\phi)##? If it's the latter, you can use the fact that angular momentum is conserved to relate ##\dot r## with ##dr/d\phi##.

The end goal is something I already know, but I've never explicitly shown it. Essentially, I want to find r(t), and then compare it to the relativistic case where $$\frac{1}{2}\dot{r}^2= e+\frac{m}{r}-\frac{L^2}{2r^2}+\frac{L^2m}{r^3}$$ and then compare orbits. I know I can compare these with the potentials (since in both cases we have ##\frac{1}{2}\dot{r}^2= const-potential## if we factor out a - from both V), but I should also explicitly solve for r(t) in both cases as I've never done it. And after seeing the integrals being done, I realize why I have just accepted the hand-wavy case!

If I've set everything up right, then I guess I should just chug through the integral.

WendysRules said:

## Homework Statement

Let $$\frac{1}{2}\dot{r}^2=e+\frac{m}{r}-\frac{L^2}{2r^2}$$ where L is angular moment, and e is energy (so I guess I'll take as constants for now...)

## Homework Equations

Not sure for now.

## The Attempt at a Solution

So, if I let $$u=\frac{1}{r}$$ then my equation becomes $$\frac{1}{2}\dot{r}^2=e+mu-\frac{L^2u^2}{2}$$ However, I'm not sure if I should also get my differential in terms of u as well. IE $$r=\frac{1}{u} \rightarrow dr=-\frac{1}{u^2} du$$ or just continue down the first path.

If I don't do the differential change, then I'd get $$\frac{1}{2}\dot{r}^2=e+mu-\frac{L^2u^2}{2} \rightarrow \dot{r}^2 = 2e+2mu-\frac{2L^2u^2}{2}$$ finally giving us... $$\dot{r} =\sqrt{2e+2mu-L^2u^2}$$ which is just... yuck.

If I do the substitution, i'll get... $$\frac{1}{2}\dot{r}^2=(\frac{dr}{2dt})^2 \rightarrow (\frac{-\frac{1}{u^2} du}{2dt})^2 \rightarrow (\frac{-du}{2u^2dt})^2$$ From here, we set it equal to the righthand side, and then see that $$\frac{-du}{2u^2dt} = \sqrt{e+mu-\frac{L^2u^2}{2}} \rightarrow \frac{du}{u^2} =-2\sqrt{e+mu-\frac{L^2u^2}{2}} dt$$ which is also, bad looking. So before I go ahead and solve either, I was looking to maybe see if there is an easier way to look at these since the next one is even *worse* (cubic in r..)

Could it also be that I'm not seeing the simplicity of it all? I'm not sure if r *depends* on t really, since this comes from our metric that ##ds^2=dr^2+r^2d\phi^2##(working in equatorial plane thus, ##\sin\theta = 1## and ##d\theta = 0##) Can go more into this if needed. So, it could be that the right hand side on both is simple because none of the terms depend on t.

You need to be careful: passing from a formula for ##(dr/dt)^2## to a formula for ##dr/dt## requires you to choose the correct root. In your ##r(t)## DE you need to use ##+\sqrt{\cdot}## when ##r(t)## is increasing and ##-\sqrt{\cdot}## when ##r(t)## is decreasing. So, you have one DE when ##r## is moving away from ##0## and another when it is moving towards ##0##. An opposite choice must be made in your ##u(t)## DE.

The need for such choices can create headaches when it comes to solving the DE numerically, using a standard DE-solving package.

WendysRules