Approximate solution for non-linear ODE

[topsquark]

I've decided to finish off this stage of my GR problem by finding an interval over which the acceleration of the object is "roughly" constant. I don't need help with the Math per se, but I would like your opinion on the method I am proposing. The Math is sufficiently ugly that I'd like some feedback before I get started on a path that will take me too long to realize is fruitless. So here we go:

The equation is this:
[math]\frac{d^2 x}{d \tau ^2} + \frac{c}{2} \left ( \frac{d x}{d \tau } \right ) ^2 = -A e^{-2 c x}[/math]
where c and A are constant. I am convinced that this has no non-trivial exact solutions.

What I would like to do is find an interval in x such that [math]\frac{d ^2 x}{d \tau ^2} \approx \text{constant} \equiv a[/math].

I am proposing the following strategy: I am going to drop an object from a height H, so [math]x_0 = H[/math] from rest, so [math]v_0 = \left . \frac{dx}{d \tau} \right |_0 = 0[/math]. That means that I'm going to set [math]x = H - \frac{1}{2}a \tau ^2[/math]. My thought from here is to expand the exponential function on the RHS as a Taylor series to zeroth order, with remainder: this will give me a range to solve for [math]\Delta h[/math], which is an interval over the distance x where the acceleration is roughly constant. It will pretty much go like this:
[math]\frac{d^2 x}{d \tau ^2} + \frac{c}{2} \left ( \frac{d x}{d \tau } \right ) ^2 = -A e^{-2 c x}[/math]

[math]\frac{d^2x}{d \tau^2} = a, ~~ \frac{dx}{d \tau} = a \tau, ~~ x = \frac{1}{2}a \tau ^2[/math]

[math]a + \frac{c}{2} \cdot ( a \tau ) ^2 \approx -A e^{-2 c x -2c x_0} [/math]

[math]a + \frac{1}{2}c a^2 \tau ^2 \approx -A \left ( e^{-2cH} \cdot e^{c a \tau ^2} - 2ce^{-2 c \xi } \Delta h \right )[/math]
Where the last term is the remainder written in the Lagrange form, [math]\Delta h = x - x_0[/math], and [math]x_0 < \xi < x[/math].

How do I finish this? I want an interval on x (which would be [math]\Delta h[/math]) where the acceleration is roughly constant... I want to solve for [math]\Delta h[/math] in terms of [math]\tau[/math]. Do I vary the value of [math]\xi[/math] to get this?

Is this method even appropriate? Is there another approximation scheme I should try instead?

Thanks in advance!

-Dan

 

[I like Serena]

Hey Dan!

That sounds pretty complicated.

Just brainstorming here, but if I look at:

[math]\frac{d^2 x}{d \tau ^2} + \frac{c}{2} \left ( \frac{d x}{d \tau } \right ) ^2 = -A e^{-2 c x}[/math]
where c and A are constant. I am convinced that this has no non-trivial exact solutions.

What I would like to do is find an interval in x such that [math]\frac{d ^2 x}{d \tau ^2} \approx \text{constant} \equiv a[/math].

I would first think like:
$$\ddot x + \frac c 2 \dot x^2 = f(x) \Rightarrow \ddot x=f(x) - \frac c 2 \dot x^2$$

If we want $\ddot x \approx a$, that means that:
$$\d {}\tau \ddot x \approx 0 \Rightarrow f'(x)\dot x - c \dot x \ddot x \approx 0 \Rightarrow f'(x) \approx ca$$

If I'm not mistaken, that comes out as:
$$x \approx \frac 1 {2c}\ln\frac{A}{-ca}$$

 

[topsquark]

Hey Dan!

That sounds pretty complicated.

Just brainstorming here, but if I look at:I would first think like:
$$\ddot x + \frac c 2 \dot x^2 = f(x) \Rightarrow \ddot x=f(x) - \frac c 2 \dot x^2$$

If we want $\ddot x \approx a$, that means that:
$$\d {}\tau \ddot x \approx 0 \Rightarrow f'(x)\dot x - c \dot x \ddot x \approx 0 \Rightarrow f'(x) \approx ca$$

If I'm not mistaken, that comes out as:
$$x \approx \frac 1 {2c}\ln\frac{A}{-ca}$$

Actually it comes out to be
[math]x \approx -\frac{1}{2c} ln \left ( \frac{a}{2A} \right )[/math]

Which implies that, since a < 0 (which I should have noted earlier!) that A < 0 which implies c < 0 as well. (I never gave the form of A.)

I think this concept is better than the one I was trying to derive. I was trying to find what intervals [math]\Delta h[/math] of x would be related to a roughly constant acceleration, but I just couldn't figure out how to formulate it. But this form is actually better for what I was trying to figure out. Thanks for the help!

-Dan