Solving ODE: $\frac{d^2x}{dt^2} = -\frac{L}{x}$

[gulsen]

I've been trying to deterimine the path of a particle. And somewhere in the solution, I need to solve this:
$$\frac{d^2x}{dt^2} = -\frac{L}{x}$$
where L is a positive number (actually length). And ideas how this can be solved? I was just reluctant to try series expansion (frob., etc) since the problem isn't finished with the solution of this, and it'd be better if I had finite terms to go on.

 

[quantumdude]

How about integrating twice?

 

[gulsen]

How about reading the question?

In case you haven't noticed, the variable on the right side is x and not t.

 

[quantumdude]

Whoops.

I haven't finished solving it yet, but I have started by moving the $x$ over to the left side and taking the Laplace transform. It requires some clever integration by parts, but I think it is going to come out.

 

[gulsen]

Thanks for taking time :)

 

[quantumdude]

The Laplace transform method doesn't work. $X(s)$ cancels out entirely.

I then tried substituting $v=\frac{dx}{dt}$, which gives the following:

[tex]\frac{d^2x}{dt^2}=\frac{dv}{dt}=\frac{dv}{dx}\frac{dx}{dt}=v\frac{dv}{dx}[/itex]<br /> <br /> This works <i>in principle</i> because it gives you a seperable first order ODE in $v(t)$. But the integration to get from $v$ to $x$ looks nasty, and I don't know if it can be done <i>in practice</i>.[/tex]

 

[gulsen]

I too have tried solving the equation using the same trick.
$$\frac{dv}{dt} \frac{dx}{dx} = vdv\frac{1}{dx}$$
and
$$vdv = -L\frac{dx}{x}$$

$$\frac{v^2}{2} = -L \ln(x) + C_0$$
$$v = \frac{dx}{dt} = \sqrt{-2L \ln(x) + C_0}$$
$$\frac{dx}{\sqrt{-2L \ln(x) + C_0}} = dt$$
Which seemed to be worse...
But at least, it's first order!

(note: negative v won't make sense, so I just ignored it)

 

[da_willem]

I've been trying to deterimine the path of a particle. And somewhere in the solution, I need to solve this:
$$\frac{d^2x}{dt^2} = -\frac{L}{x}$$
where L is a positive number (actually length). And ideas how this can be solved? I was just reluctant to try series expansion (frob., etc) since the problem isn't finished with the solution of this, and it'd be better if I had finite terms to go on.

I guess x and t are not position and time as then the equation is dimensionally incorrect. Or is there still some coefficient missing?

 

[dextercioby]

Yes, it all means that x(t) is not expressible in terms of elementary functions, and possibly neither in terms of known special functions...

Daniel.

 

[Clausius2]

aaahhh nonlinearity...makes a simple equation to be a very hard one. If you assume a series solution sometimes you come out with the series expansion of an analytical function. It's only an idea.

 

[gulsen]

I guess x and t are not position and time as then the equation is dimensionally incorrect. Or is there still some coefficient missing?

^-^' actually, I've just merged them into L to keep it simple, but L is still positive.

BTW, integrator has given this solution:
$$\int \frac{dx}{\sqrt{-2L \ln(x) + C_0}} = e^{\frac{C_0}{2L}} \sqrt{\frac{\pi}{2L}} erf(\sqrt{\frac{-2L \ln(x) + C_0}{2L}})$$