A Differential Equation

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

 

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?

 

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.

 

(1) I think you have sign errors.
(2) You could cross multiply, at least.
(3) I don't see how to finish it off either.

 

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

 

Here is your sign error.

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]

 

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

 

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?

 

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?

 

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.

 

I've gotten a consistency check:

I tried this change of variables:

y = 2 p q
x = p² - q²

In the resulting DE, if I make the hypothesis that either p or q is really large, and kill off the lower order terms, then you can solve the asymptotic form to get pq = K, or that y is again just some arbitrary constant.

 

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

 

Well, I have no proof that these are not spurious solutions. It takes quite a bit of work to show by this method that in y'=-y/x that y→0 as x→∞.

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.