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A Implicit function parameterization

  1. Dec 30, 2016 #1

    I've been trying to get a hang of parameterizing a function (explicit or implicit).
    The main view seems to be that there is no general way of doing this, but this document seems to say that you can get solutions using differential equations?
    http://www.mathematics-and-its-applications.com/preview/june2013/data/art 10.pdf
    I'm just not so good at differential equations :) Can someone demonstrate the method for some simple function, like y=x^2?

    Have I got the idea of parameterization correct?
    For an implicit form function fi(x1, x2, ..., xn) = 0:
    - Parameterization will let you introduce p new variables (where p might be anything from 1 to n-1) that does not live in any of the n dimensions of the function but that n-p variables of the function will depend on.
    - Parameterization will create an n-dimensional vector of relations tracing out the function's contour of points.
    - The "default" parameterization is (t1, t2, ..., tp, fe(t1, t2, ..., tp)) where ti is the i'th new variable/parameter and fe() is the explicit form function of xn.
    - The ideal parametrization is when each parameter represents the arc length (at unit velocity) trace along a coordinate axis of the function surface (according to some coordinate base in the surface)
    Last edited: Dec 30, 2016
  2. jcsd
  3. Dec 30, 2016 #2
    What do you mean? Please elaborate.
  4. Jan 3, 2017 #3

    Stephen Tashi

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    That one dimensional case isn't treated by the paper. The first case considered by the paper (section 2) is the "two dimensional case" is described by:

  5. Jan 3, 2017 #4

    I thought y=x^2 was like an implicit 2D function g(x,y) = x^2-y ? The contour space is 1D but you need to visualize it in 2D, so it is a 2D case?
    I don't get the quoted part, should I add the absolute value of g(x,y) with the length of the gradient? What does C^1 mean?
  6. Jan 3, 2017 #5

    Stephen Tashi

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    It is, if you include the additional stipulation that g(x,y) = 0.

    Yes, it is.
    ##C^1(\Omega)## is usually the notation for the set of functions whose first derivative exists and is continuous at each point on the set ##\Omega##.

    The statement ##|g(x,y)| + |\nabla g(x,y)| > 0 ## isn't an instruction for you to do something. It is an inequality. It amounts to saying that ##g(x,y)## and ##\nabla g(x,y)## are not simultaneously zero at any point in ##\Omega##.
  7. Jan 3, 2017 #6
    So if my function meet these conditions I can find parametric solutions using some differential equations?
  8. Jan 3, 2017 #7

    Stephen Tashi

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    We have to read the paper to find out! Pick a set for ##\Omega## and try your example on the equations in the paper.
  9. Jan 3, 2017 #8
    hehe ;) I have looked through it and concluded I don't get it :)

    But ok, a set is the input? I would like for it to work on the whole curve, but if I need to pick something, lets say 0 <= y <= 4 and 0 <= x <= 2.
    But that would mean the inequality condition fails at (0,0)? |g(0,0)| = 0 and |(dg/dx(0,0), dg/dy(0,0))| = 0, if I understand it correctly?
  10. Jan 3, 2017 #9

    Stephen Tashi

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    Why do you think ##\frac{dg}{dy} ## is zero at ##(0,0)## ?
  11. Jan 4, 2017 #10
    Because I'm wrong :)
    Ok, Then:

    x'(t) = -dg/dy(x(t), y(t)) = -(-1)
    y'(t) = dg/dx(x(t), y(t)) = 2*x(t) ?
  12. Jan 4, 2017 #11
    That would lead to:
    x(t) = -t
    y(t) = -t^2
    Which is just the negative "default" parameterization.. oh well, don't know what I was expecting anyway :]
  13. Jan 5, 2017 #12
    In the case of
    g(x,y) = x^2 + y^2 - r^2 = 0,
    is this correct?

    x'(t) = -dg/dy(x(t), y(t)) = -2*y(t)
    y'(t) = dg/dx(x(t), y(t)) = 2*x(t)

    Wolfram Alpha gives me:
    https://www4f.wolframalpha.com/Calculate/MSP/MSP2481c65ih2fhc6feci600003e87c45c61da9i1c?MSPStoreType=image/gif&s=57 [Broken]
    https://www4f.wolframalpha.com/Calculate/MSP/MSP2501c65ih2fhc6feci60000602873g0fg73gc5b?MSPStoreType=image/gif&s=57 [Broken]

    The document says:
    "initial condition x(0) = x0, y(0) = y0 where g(x0,y0) = 0 is assumed."

    Then I'm free to choose c1 and c2 as I wish (to set the point of t=0) according to:
    g(x0,y0) = (c1)^2 + (-c2)^2 - r^2 = 0?

    I think what I really want is to involve the velocity in the differential equations.
    In this case I get a speed of 2 length units per one time unit?
    In the first case it's not as obvious what the speed is/will be. Is it possible to involve some expression for the arc length in the first equations before soving them, so that you get arc length parameterization, or is it impossible to get such a parameterization for the parabola?
    Last edited by a moderator: May 8, 2017
  14. Jan 5, 2017 #13

    Stephen Tashi

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    The images don't show up. Do we get ?:

    ##y''(t) = 2 x'(t)##
    ## x'(t) = y''(t)/2##

    ##y''(t)/2 = -2 y'(t)##
    ##y''(t)/2 + 2y'(t) = 0##

    ##y'(t) = C_1e^{-4t} ##
    ##y(t) = (-1/4)C_1e^{-4t} + C_2##
    ##x(t) = -2y(t) = (1/2)C_1e^{-4t} - C_2##

    Yes, if you choose things so ##(x_0,y_0) \in \Omega##.

    It seems to me that once you find a parametrization ## g(x(t),y(t)) = 0## then you can adjust the "speed" of ##(x(t),y(t))## by rescaling it to ##(x(h(t)),y(h(t)))##. You don't have to include the problem of finding ##h(t)## in the problem of finding the initial parameterization.
  15. Jan 5, 2017 #14
    No, but interesting solution. Wolfram gave:
    x(t) = c1*cos(2t) - c2*sin(2t)
    y(t) = c1*sin(2t) + c2*cos(2t)

    Ok. I found a formula for the arc length of the parabola, but the problem was that it couldn't be inversed to substitute t in the parameterization.
    Since the general formula for arc length sums small pieces of sqrt(dx^2+dy^2), I thought it might be integrated in the differential equation system somehow.

    For g(x,y) = x^2 - y = 0:
    x'(t) = 1
    y'(t) = 2*x(t)

    For the arc length s(t), would this be true?
    s'(t)^2 = x'(t)^2 + y'(t)^2

    Not sure where I'm going with that, but it's interesting to try stuff.. :)
    Maybe one could find ds/dx and ds/dy, invert them and integrate to get x(s) and y(s)?
  16. Jan 6, 2017 #15
    Cool.. try draw this with b=~100 and a=~7.1
    must be a limit where it coincides with y=x^2?
    (a*ln(2t + sqrt(1 + 4t²)), b*sqrt(1 + 4t²) - b)
  17. Feb 1, 2017 #16
    How can I check if my parameterization is arc length parameterized?
    for f(x,y) = x^2-y = 0, I want to check this parameterization:
    x(t) = t / sqrt(t²) sqrt(sqrt(t²))
    y(t) = sqrt(t²)
    Last edited: Feb 1, 2017
  18. Feb 1, 2017 #17
    I think this is closer?
    x(t) = k*t/(t^2)^(1/4)
    y(t) = (k*t/(t^2)^(1/4))^2
    where k = sqrt( (sqrt(5)-1)/2 )
    Last edited: Feb 1, 2017
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