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Existence (and uniqueness) of parameterization

  1. Aug 12, 2009 #1


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    This topic has been masterfully avoided in my classes, but several proofs of theorems in multivariate calculus use the existence of a parametrization like this:

    Let [tex]f:\mathbb{R}^2\to\mathbb{R}[/tex]. Then we can write: [tex]f(x,y)=g(t)=f(x(t),y(t))[/tex]

    And from this, we can get some interesting results, like the Lagrange multiplier theorem. However, an existence proof has eluded me online and my professor's explanation was essentially that whenever the level curves of f are smooth, closed curves, this parametrization "can be done". Can anyone help me with this, or direct me to a source (maybe even on these forums) that has the proof?

  2. jcsd
  3. Aug 12, 2009 #2
    It's not a formal proof but it's the idea behind this. If you want a level curve, you need this condition:

    [tex]f(x(t),y(t)) = k[/tex]


    [tex]\frac{d}{dt} f(x(t),y(t)) = \frac{df}{dx}\frac{dx}{dt} + \frac{df}{dy}\frac{dy}{dt}= \nabla f \cdot \frac{d \vec{x}}{dt} = 0[/tex]


    [tex]\frac{d \vec{x}}{dt}=(\frac{dx}{dt}, \frac{dy}{dt})[/tex]

    This means that the direction of your curve is perpendicular to the direction of ascent of the function. If the gradient is not 0, there's only one path that you can take. The length of the vector [tex]d\vec{x} / dt[/tex] will only define the speed at which you travel, but the path will stay the same. By always taking the same direction, you will always find a path ( not necessarily closed for some special functions ) that correspond to the level curve you need.

    But the parameterization is not unique. Like I said, you can go slower or faster on certain part of the curve, resulting in a different parameterization. Also, the curve is not unique, it depends on the first point you choose to begin your curve.

    I need to add that if the has to intersect itself, it will necessarily create a closed curve (possibly smooth by carefully choosing the speed at which you travel at the beginning and at the end). Since you always choose the same direction to move, you can never go back, ensuring that the velocity at the end is not in the opposite direction. And since only a one dimensional path is possible, the two velocities will be in the same direction, forming a smooth closed curve.
    Last edited: Aug 12, 2009
  4. Aug 14, 2009 #3


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    Thank you for the explanation, GPPaille, but I am still in search of a Hypothesis-Conclusion-Formal Proof exposition on this topic.

    The part that you explained about orthogonality and different speeds => non-uniqueness, which make sense but still require a proof in terms of other calculus theorems or axioms.. I'd like to be able to do this myself, but I'm still a junior analyst and am stuck.
  5. Aug 18, 2009 #4
    I'm having a bit of trouble understanding exactly what your question is. As far as I can tell, however, what you want is a proof that the equation [tex] f(x,y) = k [/tex] determines a curve in the [tex] xy [/tex]-plane which has some smooth parametrization [tex] \gamma : t \mapsto (x(t), y(t)) [/tex]. Actually, the issue of which subsets of [tex] \mathbb{R}^n [/tex] defined by equations (like [tex] f(x,y) = k [/tex]) admit "parametrizations" is rather complicated, not least of all because sensibly defining what constitutes a "parametrization" is much trickier than one might think. Given that your function [tex] f [/tex] is well-behaved (smooth, with nonvanishing gradient), it is certainly true that [tex] f(x,y) = k [/tex] determines a smooth, nonsingular curve. However, even the circle cannot be parametrized unless one makes allowances for "trouble spots" of volume zero (e.g., the usual parametrization of the circle fails to be one-to-one at the point (1,0)). Once this concession is made, there is a theorem guaranteeing that all compact smooth manifolds have parametrizations; it's rather technical, and I actually don't know the proof. (This would, however, suffice to show that all non-stupid closed level curves are parametrizable, as your professor claims.) This theorem is indispensable in integration theory, because essentially the entire theory of integration over smooth manifolds is built on the assumption that the given manifold can be parametrized. (This is also the source of the "volume zero" caveat mentioned above: regions of volume zero can be safely ignored when performing integrals.)

    However, from what you've said so far, I would guess that all you really need is a proof of the existence of local parametrizations of your level curve, i.e., a set of finitely many smooth maps that, together, parametrize some neighborhood of every point on your curve. For this, the Implicit Function Theorem does the job perfectly well. Briefly, the IFT states that if [tex] F : \mathbb{R}^n \to \mathbb{R}^m [/tex] is a smooth function with surjective Jacobian at a point [tex] \mathbf{p} \in \mathbb{R}^n [/tex], then the equation [tex] F(\mathbf{x}) = \mathbf{0} [/tex] may be "solved" in terms of the pivot variables in some neighborhood of [tex] \mathbf{p} [/tex]. In your case, this means that if the gradient of [tex] f [/tex] is never equal to the zero vector, then in some neighborhood of every point satisfying [tex] f(x,y) - k = 0 [/tex], you can either solve for [tex] y [/tex] in terms of [tex] x [/tex], or vice-versa. This argument suffices to cover the level curve with finitely many open, smoothly-parametrized curve segments (graphs of functions, really), which should be enough to prove the Lagrange multiplier theorem on constrained local maxima and minima.

    The proof of the Implicit Function Theorem is actually extremely cool (in my opinion, at least). Basically, you first prove another theorem called the Inverse Function Theorem, which is an obvious special case of the IFT. There is then a sneaky way of showing that the IFT is, in fact, a special case of the InvFT. This post is already too long, so if you want details, PM me (or just google Implicit Function Theorem).
  6. Aug 18, 2009 #5


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    Thank you for your post VKint! The local parametrization was the question I intended to pose; it seems to me that I need to do some more foundational work before I can fully grasp the outline of this proof (happily, this is exactly what I'll be taking at uni. next school semester). Thanks again VKint!
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