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A Differential Equation

  1. Sep 25, 2005 #1
    How would I solve the differential equation y' = - y / ( (x^2 + y^2)^1.5 ) ? Btw, its not a homework prob... thanks
     
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  3. Sep 25, 2005 #2

    Hurkyl

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    I would try a change of variable into polar coordinates (based on the x²+y² term)
     
  4. Sep 25, 2005 #3
    I'm very new to differential equations...could you show me how that would go along? So far I've figured out it would be of the form y' = ( sin theta ) / r^2, but can we replace y' by tan theta?
     
  5. Sep 25, 2005 #4

    Hurkyl

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    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?
     
  6. Sep 25, 2005 #5

    Tom Mattson

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    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.
     
  7. Sep 25, 2005 #6
    I get [tex]\frac{dy}{dx} = - \frac{r \cos \theta + \frac{dr}{d\theta} \sin \theta}{r \sin \theta + \frac{dr}{d\theta} \cos \theta}[/tex]
    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 :cry:
     
  8. Sep 25, 2005 #7

    Hurkyl

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    (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. :frown:
     
  9. Sep 25, 2005 #8
    Well if [tex]\frac{dx}{d\theta} = \frac{d}{d\theta}r \cos \theta[/tex] then [tex]\frac{dy}{d\theta} = \frac{d}{d\theta}r \sin \theta = r \cos \theta + \frac{dr}{d\theta} \sin \theta[/tex]. That gives me [tex]\frac{dy}{dx} = \frac{dy}{d\theta}\frac{d\theta}{dx}[/tex]. I assume that [tex]\frac{d\theta}{dx} = \frac{1}{\frac{dx}{d\theta}}[/tex] or would it rather be the negative reciprocal? Also I tried cross multiplying on paper, and couldn't make sense out of the resulting mess either :frown:
     
  10. Sep 26, 2005 #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.
     
  11. Sep 26, 2005 #10
    what sign error?
     
  12. Sep 26, 2005 #11

    Tom Mattson

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    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]
     
  13. Sep 26, 2005 #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
     
    Last edited: Sep 26, 2005
  14. Sep 26, 2005 #13

    Tom Mattson

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    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?
     
  15. Sep 26, 2005 #14
    Oh I see the sign error...I was accidentally multiplying the whole thing by -1 instead of just the r sin theta term...As for how I come upon the equation:

    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 modelled 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.
     
  16. Sep 26, 2005 #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?
     

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    Last edited: Sep 26, 2005
  17. Sep 26, 2005 #16

    Hurkyl

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    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.
     
  18. Sep 26, 2005 #17

    Hurkyl

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    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.
     
  19. Sep 27, 2005 #18

    saltydog

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    Thanks Hurkyl. That's an interesting approach. Seems though the actual asymptote cannot be determined analytically without the analytical solution in hand.
     
  20. Sep 27, 2005 #19

    Hurkyl

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    Well, I have no proof that these are not spurious solutions. :frown: 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.
     
  21. Sep 27, 2005 #20
    Can someone show me a particular solution of the two equations y' = -y/((x^2+y^2)^1.5) and x' = -x/((x^2+y^2)^1.5)?
     
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