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cefarix

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- Thread starter cefarix
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cefarix

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- #2

Hurkyl

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I would try a change of variable into polar coordinates (based on the x²+y² term)

- #3

cefarix

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- #4

Hurkyl

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can we replace y' by tan theta?

It's not quite that simple. First, you need to realize that everything is now a function of θ, so you're going to have to do things like the chain rule.

Notice, for example, that:

[tex]

\frac{dx}{d\theta} = \frac{d}{d\theta} (r \cos \theta)

= -r \sin \theta + \frac{dr}{d\theta} \cos \theta

[/tex]

Can you figure out a way to rewrite dy/dx now?

- #5

quantumdude

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cefarix said:Btw, its not a homework prob... thanks

The idea behind moving homework questions to the Homework section is to clear up the Math section for discussion of mathematical topics, without having this kind of step-by-step help discussions hanging around. While this may not be a homework problem that was assigned to you, it looks just like one, and it is handled just like one, so it really should be posted there.

- #6

cefarix

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Plugging that into [tex]\frac{dy}{dx} = - \frac{\sin \theta}{r^2}[/tex] gets me the equation [tex]\frac{r \cos \theta + \frac{dr}{d\theta} \sin \theta}{r \sin \theta + \frac{dr}{d\theta} \cos \theta} = \frac{\sin \theta}{r^2}[/tex] which I have no idea how to solve

- #7

Hurkyl

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(2) You could cross multiply, at least.

(3) I don't see how to finish it off either.

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cefarix

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- #9

Hurkyl

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But what you've written in post #6 is not (dy/d&theta) / (dx/dθ) because of the sign error.

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cefarix

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what sign error?

- #11

quantumdude

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cefarix said:[tex]\frac{dy}{dx} = - \frac{r \cos \theta + \frac{dr}{d\theta} \sin \theta}{r \sin \theta + \frac{dr}{d\theta} \cos \theta}[/tex]

Should be:

[tex]\frac{dy}{dx} = \frac{r \cos \theta + \frac{dr}{d\theta} \sin \theta}{-r \sin \theta + \frac{dr}{d\theta} \cos \theta}[/tex]

- #12

saltydog

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Using Hurkyl's substitution, I can get it down to:

[tex](r-rCos(2\theta)-2r^3Cos(\theta))d\theta-(2r^2Sin(\theta)+Sin(2\theta))dr=0[/tex]

but no further.

Edit: corrected first term: r^2 to r

[tex](r-rCos(2\theta)-2r^3Cos(\theta))d\theta-(2r^2Sin(\theta)+Sin(2\theta))dr=0[/tex]

but no further.

Edit: corrected first term: r^2 to r

Last edited:

- #13

quantumdude

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Tom Mattson said:The idea behind moving homework questions to the Homework section is to clear up the Math section for discussion of mathematical topics, without having this kind of step-by-step help discussions hanging around. While this may not be a homework problem that was assigned to you, it looks just like one, and it is handled just like one, so it really should be posted there.

You know what? I think I made a mistake moving this thread--sorry. When I initially looked at it I assumed that it would have a quick, closed-form solution. But it's clear now that that is not the case, and that this could lead to an interesting mathematical discussion. So, I'm sending it to Differential Equations.

So, cefarix, if we can't get a closed form solution to this, then do you have any initial conditions by which we might get a numerical solution? And just out of curiosity, how did you come upon this equation?

- #14

cefarix

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I'm trying to make a model of the path a light beam would take by a non-moving point mass, according to the principles of my own unified theory. Basically, I am treating spacetime as a 4-dimensional medium with varying density, and masses path's are bent by density changes. Taking an at-rest point mass and using a light beam simplifies calculations by making the problem symmetrical, and having to deal with only a single plane within space and being able to ignore the time-like density changes (akin to ignoring time-like curvature in GR). I figured if I modeled the acceleration as inversely proportional to the square of the distance, and pointing towards the mass, I could write up a differential equation and integrate it to get the path of the light beam. The different solutions would correspond to different initial starting positions and angles of the beam. According to the principles of the theory, the light beam would have an initial magnitude of velocity inversely proportional to the magnitude of the acceleration at that point. I'm not explaining why here, as this probably isn't the right place to do so, but I could explain to you elsewhere if you like.

- #15

saltydog

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Well if no one can offer an analytical approach, I have some questions about the qualitative behavior of the numerical solution.

[tex]y^{'}=\frac{-y}{(x^2+y^2)^{3/2}};\quad y(0)=y_0[/tex]

The plots below are for [itex]y_0[/itex]=2, 1.5, and 1 respectively. Just looking at the ODE from a qualitative perspective, the plots exhibit an expected behavior:

The slope is initially negative because of the numerator and thus the plot decreases. But eventually the squares in the denominator build up making the slope smaller and smaller causing the value of y to change less and less. As x goes to infinity, the slope monotonically goes to zero. These are my questions:

1. Is the x-axis an asymptote for these cases or does it reach an asymptote before this?

2. Is the answer to (1) dependent on the initial value of y?

3. Are there some initial (positive) values of y which cause the solution to dip below the x-axis?

I don't know how to answer definitively (1) and (2). For [itex]y_0[/itex] small, say 0.1 or 0.2, the numerical plots appear to dip below the x-axis but I'm skeptical and would like to see a more rigorous answer.

Interested?

[tex]y^{'}=\frac{-y}{(x^2+y^2)^{3/2}};\quad y(0)=y_0[/tex]

The plots below are for [itex]y_0[/itex]=2, 1.5, and 1 respectively. Just looking at the ODE from a qualitative perspective, the plots exhibit an expected behavior:

The slope is initially negative because of the numerator and thus the plot decreases. But eventually the squares in the denominator build up making the slope smaller and smaller causing the value of y to change less and less. As x goes to infinity, the slope monotonically goes to zero. These are my questions:

1. Is the x-axis an asymptote for these cases or does it reach an asymptote before this?

2. Is the answer to (1) dependent on the initial value of y?

3. Are there some initial (positive) values of y which cause the solution to dip below the x-axis?

I don't know how to answer definitively (1) and (2). For [itex]y_0[/itex] small, say 0.1 or 0.2, the numerical plots appear to dip below the x-axis but I'm skeptical and would like to see a more rigorous answer.

Interested?

Last edited:

- #16

Hurkyl

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I don't really know if this approach is kosher, but...

I made the change of variable:

y → y(t) / z(t)

x → x(t) / w(t)

from which I get the differential equation:

[tex]

(y' z - y z') (x^2 z^2+ y^2 w^2)^{3/2} = (x'w - xw') z^4 y w

[/tex]

where differentiation is with respect to t.

Now, notice that when w(a) = 0 (corresponding to infinite x in the original problem), I can satisfy the differential equation

y'z - yz' = 0

Or equivalently,

y = Kz

meaning that the original y(t) is roughly the constant K when t is near a.

- #17

Hurkyl

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I tried this change of variables:

y = 2 p q

x = p² - q²

In the resulting DE, if I make the hypothesis that either

- #18

saltydog

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Hurkyl said:It doesn't look like the x-axis is the asymptote for those plots.

I don't really know if this approach is kosher, but...

I made the change of variable:

y → y(t) / z(t)

x → x(t) / w(t)

from which I get the differential equation:

[tex]

(y' z - y z') (x^2 z^2+ y^2 w^2)^{3/2} = (x'w - xw') z^4 y w

[/tex]

where differentiation is with respect to t.

Now, notice that when w(a) = 0 (corresponding to infinite x in the original problem), I can satisfy the differential equation

y'z - yz' = 0

Or equivalently,

y = Kz

meaning that the original y(t) is roughly the constant K when t is near a.

Thanks Hurkyl. That's an interesting approach. Seems though the actual asymptote cannot be determined analytically without the analytical solution in hand.

- #19

Hurkyl

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Incidentally, the motivation is to pass from the real line to the projective line, so that infinity is an actual point. The usual way to coordinatize the real line is by pairs (u : v) (not both zero), and define that (u : v) = (ku : kv) for any nonzero k.

So going back to the real line, when v is not zero, you can map (u : v) → u/v... when v is zero, that corresponds to the point at infinity. That is what motivated my transformation.

- #20

cefarix

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- #21

saltydog

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cefarix said:

Not sure what you mean. I posted three particular solutions to the first equation above (numerically determined but particular nevertheless). What's with the x'? With respect to y or a coupled system with respect to t?

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arildno

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- #23

cefarix

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Now that I think about it, I think it would be best modeled as dy/dt and dx/dt.

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arildno

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On further reflection, just find the equation for dy/dx.

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cefarix

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- #26

arildno

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If you're not satisfied with that, do the polar coordinate substitution of the dependent variables.cefarix said:

you'll readily find that the radial variable decreases linearly with time, and the angle is constant. This is, of course, under assumption that r(0) is different from zero.

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