MHB Approximate solution for non-linear ODE

AI Thread Summary
The discussion centers on finding an interval over which the acceleration of an object is approximately constant in a non-linear ordinary differential equation (ODE). The proposed method involves dropping an object from a height and using a Taylor series expansion of the exponential function to derive an interval for the distance where acceleration remains roughly constant. Feedback from another participant suggests an alternative approach by analyzing the equation to derive a relationship between acceleration and the function's derivatives. Ultimately, the conversation highlights the complexity of the problem and the need for a suitable approximation method to achieve the desired results. The participants express appreciation for the insights shared, indicating a collaborative effort to refine the approach.
topsquark
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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
 
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Hey Dan!

That sounds pretty complicated.

Just brainstorming here, but if I look at:
topsquark said:
[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:
$$x \approx \frac 1 {2c}\ln\frac{A}{-ca}$$
 
I like Serena said:
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
 
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