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Let $S(x)$ be the area of the yellow curvilinear triangle. I'd like to find a graph for which $S(x)=H(x)$ where $H $ is some prescribed function (small, smooth, vanishing near the endpoints to any order you wish, etc.). Is it always possible or there are some non-obvious hidden restrictions?

alt text http://www.artofproblemsolving.com/Forum/latexrender/pictures/b/4/6/b46dbb397d96f17cce1e421bfaf29b9a80d26c81.png

The question comes from the infamous t-section problem (if you know the areas of all sections of a symmetric convex body by the hyperplanes at some fixed small distance $t$ from the origin (so small that all sections are non-empty), can you recover the body?). The problem is open even on the plane. I do not say that this toy question is directly relevant here but an answer to it will certainly make a few things clearer for me.

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  • $\begingroup$ What happens when you try the obvious, which to me is adding dx to x, and then figuring the incremental shape of the curve based on the difference in the area of the triangles? Or is 45 not 45 degrees but an angle that depends on x? Gerhard "Ask Me About System Design" Paseman, 2010.05.31 $\endgroup$
    – Gerhard Paseman
    Commented Jun 1, 2011 at 2:16
  • $\begingroup$ This strikes me as something that deserves a 'calculus of variations' tag. Gerhard "Ask Me About System Design" Paseman, 2011.05.31 $\endgroup$
    – Gerhard Paseman
    Commented Jun 1, 2011 at 2:32
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    $\begingroup$ The answer to the first question is "nothing good". You get a delayed differential equation, which is highly unstable. As to the tag, I'll add it :). $\endgroup$
    – fedja
    Commented Jun 1, 2011 at 4:32

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There are restrictions.

At most points, $S'(x)$ is the length of the right leg minus the length of the left leg of the curvilinear triangle, perhaps with exceptions on a null set where there are tangencies. If $S(0)=S(1)=0$ then the lengths of these legs are at most $\sqrt{2}(1-x)$ and $\sqrt{2}x$. For almost all $0\le x \le 1$, $S'(x)$ satisfies $-\sqrt{2}\le -\sqrt{2} x \lt S'(x) \lt \sqrt(2) (1-x) \le \sqrt{2}$. This is an extra condition on $H$ which rules out some smooth small functions which have large derivatives near some points, such as $10^6 \exp(-1/(x (1-x))^2)$ for $0\lt x \lt 1$, which has a derivative of $1.132$ at $x=0.436$ although the value of the function is small.

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  • $\begingroup$ Thanks a lot! I upvoted it but, unfortunately, it still fits under the "trivial restriction" category. To be more precise, you can assume $H$ to be small in any function space you desire (say, $C^k$ with any $k$). What I mean is that "roughly speaking" $S(x)=\frac f(x)^2$ and cannot change too much when $x$ goes by $f(x)$ to the right or to the left where $f$ is the function whose graph gives $S$, so if you have hard time with the square root or the rate of change is too big, you have trouble for sure. But what if you have $H(x)=10^{-100}[x(1-x)]^{10}$, say? $\endgroup$
    – fedja
    Commented Jun 5, 2011 at 17:40
  • $\begingroup$ I wasn't sure if this was supposed to be ruled out by the condition that $H$ is small. One of the things I considered was whether you meant that if $H$ is smooth and vanishes to all orders, then there is some function so that $S = cH$ for some $c \gt 0$. $\endgroup$
    – Douglas Zare
    Commented Jun 5, 2011 at 18:03
  • $\begingroup$ Yes, that would be another good way to formalize it (provided that you do not start playing with coming very close to $0$ and then lifting off infinitely many times). The bad thing is that I don't know myself what effect I'm looking for. I rather know a long list of trivial effects that I do not care much about. The particular function I wrote would have none of them though, so if you can show that it is impossible, it'll tell me something new. $\endgroup$
    – fedja
    Commented Jun 6, 2011 at 1:35

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