I have the following curve on the usual $x$-$y$ Euclidean plane: \begin{align} (x^2-y^2+2x)^2 = 4\alpha({x^2-y^2}) \end{align} for some $\alpha > 1$. You may plot the curve here:


by entering for example, ((x^2)-y^2+2*x)^2=4*12*(x^2-y^2), to check out the case for $\alpha = 12$.

For $\alpha = 6$, say I get one level set, and for $\alpha = 6+\epsilon$ for some small $\epsilon$, I get another level set. I want to show that in the right half plane, the area between the two level sets will be bounded by $\epsilon$ times a constant. Of course, I would like to show that this property holds for any $\alpha$ and sufficiently small $\epsilon$ (In this case the constant multiplier may depend on $\alpha$).

This curve with the equation above is just one example. The main question is how can I do this for any curve (assuming that curve has some certain properties)? (Although it would also be nice to see an answer for the particular case above). Answers in the form "Read this noob" would also be greatly appreciated.

$\textbf{Edit:}$ To clarify, formally, we consider the solutions of the equation $f(x_1,\ldots,x_n) - \alpha g(x_1,\ldots,x_n) = 0$, where $x_1,\ldots,x_n$ are real coordinates, $f$ and $g$ are (say) polynomials. Let $\mathcal{S}_{\alpha} = \{(x_1,\ldots,x_n)\in\mathbb{R}^n:f(x_1,\ldots,x_n) - \alpha g(x_1,\ldots,x_n) = 0\}$. Roughly, I would like to show something like ($\mu$ is the Lebesgue measure) \begin{align} \forall\alpha>0,\,\forall\epsilon>0,\,\mu\left(\bigcup_{\beta\in[\alpha,\alpha+\epsilon]}\mathcal{S}_{\beta}\right)\leq h(\alpha)\epsilon, \end{align} for some $0<h(\alpha)<\infty$ whenever it is possible to do so. In the definition of $\mathcal{S}_{\alpha}$, we may restrict the solutions to e.g. only positive coordinates as in the example above.

$\textbf{Postmortem:}$ I have got two great answers, none of which however were able to completely resolve the problem. Robert Israel's argument is quite general, but to my reading, does not simply the problem as much as I was expecting. Anıl Başeski's argument is very specific, but it at least solves the particular case as posed in the question. The particular case was related to my own research, and the existence of a parametrization and a closed form solution for the area is quite helpful. I have therefore decided to award the bounty to Anıl. I would also like to thank all who have participated in the dicussion. The general question however is still open, and hence there are no "accepted answers."

  • $\begingroup$ Wow, you've put almost all your money on this question! I hope someone will produce an answer you like. Just a remark: it will be difficult to get such a generalization. Say $f(x,y)=x$ and $g(x,y)=y$. Then every union of $S_\beta$ over a nontrivial interval has infinite measure as it is some doubly infinite nontrivial cone. Same if you restrict to positve coordinates, and $\alpha>0$. $\endgroup$ – Julien Apr 10 '13 at 20:24
  • $\begingroup$ @julien Hehe, yes, if I had more points I would put them too :) As you have said, it will not be possible to do this for any curve. I was thinking that there will be a "discriminant-like" function (or functions) that will tell us whether or not the sets will have finite measure (And whenever they say the measure will be finite, something else will tell us "This bound holds.") $\endgroup$ – Lord Soth Apr 10 '13 at 20:27
  • $\begingroup$ Well, this set is somehow, in the generic case, a thickened $n-1$ dimensional object. So you need at least your levels sets to be compact (i.e. bounded, as they are closed) if you don't want this to explode. I am not aware of an algebraic characterization that guarantess compactness of such algebraic curves. But I'm atrociously ignorant in algebraic geometry. I would know what to do with quadratic forms, but that's about it. $\endgroup$ – Julien Apr 10 '13 at 20:34
  • $\begingroup$ @julien It is great if you know what to do if $f$ and $g$ are something like in the example, and I would love to see it (But if you meant the case of conic sections, I guess that can (perhaps) be handled "by hand.") $\endgroup$ – Lord Soth Apr 10 '13 at 20:39
  • $\begingroup$ @julien, The example is quite special actually, if you rotate everything by $-45^{\circ}$ you can solve for $y$ given $x$, but even for that case the derivatives, etc, become extremely complicated, I am hoping there is a simple way to do these things. $\endgroup$ – Lord Soth Apr 10 '13 at 20:42

For your specific equation the curve in the right half plane can be parameterized by $y=t\ x(t)$ to give $$x(t)=\frac{2\times \sqrt{\alpha (1-t^2)}-2}{1-t^2}$$ $$y(t)=t\ x(t)=t\frac{2\times\sqrt{\alpha (1-t^2)}-2}{1-t^2}$$ where $-\beta\lt t\lt\beta$ and $\beta=\frac{\sqrt{\alpha-1}}{\sqrt{\alpha}}$

The area can be formulated as $$A=\frac{1}{2}\int_{-\beta}^{\beta}\bigg(x(t)\frac{dy(t)}{dt}-y(t)\frac{dx(t)}{dt}\bigg)dt $$ Due to fact that $y=t\ x(t)$ $$A=\frac{1}{2}\int_{-\beta}^{\beta}\bigg(x(t)^2\bigg)dt =F(\beta)-F(-\beta)$$ and $$F=\frac{t \bigg( 8 \sqrt{\alpha (1-t^2)}-2\bigg ) + 2 (1 + 2 \alpha) (-1 + t^2) \text{arctanh}(t)}{t^2-1}$$ which gives following formulation for $A$ $$A=2 (1 + 2 \alpha) \text{arctanh}\bigg(\frac{\sqrt{\alpha-1}}{\sqrt{\alpha}}\bigg)-6\sqrt{\alpha^2-\alpha} $$ By Taylor expansion $$dA=A(\alpha+\epsilon)-A(\alpha)=\frac{dA}{d\alpha}\epsilon+O(\epsilon^2)$$ and by neglecting higher order terms $$dA= 4\bigg( \text{arctanh}\bigg(\frac{\sqrt{\alpha-1}}{\sqrt{\alpha}}\bigg)- \frac{\sqrt{\alpha-1}}{\sqrt{\alpha}}\bigg)\epsilon $$

PS: Since equations are too long and time is too short I recommend you to check for typos.

  • $\begingroup$ Great! I was wondering how were you able to "see" that parametric formulation? $\endgroup$ – Lord Soth Apr 15 '13 at 21:08
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    $\begingroup$ I have verified that your formulas are correct. One small thing is that all the areas should be divided by $2$. I have edited the answer. I have also changed $O^2$ to $O(\epsilon^2)$. $\endgroup$ – Lord Soth Apr 16 '13 at 1:25
  • $\begingroup$ @LordSoth thanks for the edits. In parametric formulation I was just lucky i guess $\endgroup$ – AnilB Apr 16 '13 at 5:05

If $S_\alpha$ is a compact $n-1$-manifold you should get $h(\alpha) = \int_{S_\alpha} \dfrac{1}{\|\nabla (f/g)\|}$, when that is finite. A counterexample would be the curve $(x^2 + y^2 - 1)^3 = \alpha$, which is a circle of radius $1 + \alpha^{1/3}$. The area between $S(0)$ and $S(\epsilon)$ is approximately $2 \pi \epsilon^{1/3}$ and thus is not $O(\epsilon)$ as $\epsilon \to 0$.

  • $\begingroup$ Thank you for the answer. One of the main difficulties I am facing is the characterization of $\mathcal{S}_{\alpha}$ itself. Can we say anything about when $S_{\alpha}$ becomes a "compact $n-1$-manifold" or when that integral becomes finite. It would also be great if you could explain how you obtain the integral. Does it guarantee $\leq h(\alpha)\epsilon$ for all positive (or sufficiently small) $\epsilon$ or is it more like $\sim h(\alpha)\epsilon$ for small $\epsilon$? $\endgroup$ – Lord Soth Apr 10 '13 at 20:51
  • $\begingroup$ Let me reput it this way. Can I just (relatively quickly) obtain what I want from your formula for my example in the question? $\endgroup$ – Lord Soth Apr 10 '13 at 20:58
  • $\begingroup$ Actually I was using my Samsung Note; and I pressed "downvote". Now I can't get it back. Sorry... If you edit your answer I will be able to correct my mistake. $\endgroup$ – AnilB Apr 16 '13 at 18:01
  • $\begingroup$ If $\nabla(f/g)$ is nonzero on $S_\alpha$, then the Implicit Function Theorem says $S_\alpha$ is an $n-1$-manifold. It is also closed, so it is compact iff it is bounded. In your example the complete $S_\alpha$ is unbounded, but you're only taking the part in the right half-plane, which is bounded. The singularity of $f/g$ at $(0,0)$ makes matters somewhat more complicated. $\endgroup$ – Robert Israel Apr 17 '13 at 6:56

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