# Which plane convex arcs have the smallest maximum curvature?

Let $p$ and $q$ be positive real numbers with $p \leq q$. Suppose that $H(p,q)$ is the class of all convex arcs $c$ in the Cartesian $x-y$ plane which satisfy the following conditions:

(1)The $y$-axis is an axis of symmetry of $c$.

(2)The points $(-p,0)$ and $(p,0)$ are the end-points of $c$ and the point $(0,q)$ is also a point of $c$.

(3)The curvature of $c$ is defined and continuous at each point of $c$ and never changes sign (which can always be taken to be non-negative).

QUESTION: Given $p$ and $q$ is there a formula for the greatest lower bound of the maximum curvature that an arc $c$ belonging to the class $H(p,q)$ can have? Is there a particular arc in this class which actually attains this greatest lower bound?

The (upper half of the) ellipse whose equation is $\frac{x^2}{p^2}+\frac{y^2}{q^2}=1$ has a maximum curvature of $\frac{q}{p^2}$ and is an arc belonging to the class $H(p,q)$, but I cannot prove-and do not think- that $\frac{q}{p^2}$ is actually a greatest lower bound for the whole class (except in the case $p=q$ which is specifically excluded).

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The end of your post is missing; if I were you, I would look at a stadium curve (union of half circles and lines). – Benoît Kloeckner Apr 26 '10 at 14:57
If the circle arc through the three points in question is an admissible curve, it is the answer. – fedja Apr 26 '10 at 15:01
What is a "convex arc"? If it is anything with curvature of constant sign, then the lower bound for the max curvature is zero. If it is an arc that lies on the boundary of its own convex hull, then the answer is a circle arc. If it is a graph of a convex function, the answer is either circle arc or a stadium curve depending on which of p and q is larger. – Sergei Ivanov Apr 26 '10 at 15:19
Thanks for your answer, Sergei. The convex arcs I am considering are always graphs of convex functions and I should have made this clear. I am especially interested in cases where p is "very small" and q is "much greater" than p. Your answer then indicates that the greatest lower bound would be attained by a stadium curve-call it "s(p,q)". But I would still like to know whether, given p and q, there is a formula expressing the maximum curvature of s(p,q) as a function of p and q. Does the curvature of stadium curves have discontinuities? If so, they could not be in the class H(p,q). – Garabed Gulbenkian Apr 29 '10 at 20:47
If $q>p$, the answer is $1/p$. This is the greatest lower bound which is not attained. The stadium curve where it would be attained is not admissible (it is not a graph of a function, and its curvature is discontinuous). But is can be approximated by admissible curves with slightly larger maximum curvature. – Sergei Ivanov May 4 '10 at 14:15