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Let $T\subset \mathbb{R}^2$ be any triangle and $T^t$ a deformation of $T$. Call $l_1,l_2,l_3$ the squares of the lengths of the sides of $T$ and $l_1^t,l_2^t,l_3^t$ the squares of the lengths of the corresponding sides of $T^t$. Let $\phi_t:T\rightarrow T^t$ be the affine map which sends sides to corresponding sides.

This is the problem I'd like to solve:

Find triangles $T$ such that the following equality is verified for any deformation $T^t$: $$Max\{\frac{l_1}{l_1^t},\frac{l_2}{l_2^t},\frac{l_3}{l_3^t}\}=Lip(\phi_t)^2$$

where $Lip(\phi_t)$ is the Lipschitz constant of $\phi_t$.

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The equality is never satisfied.

We send $T$ to $T^t$ by an affine map. This sends the unit circle to an ellipse. The bilipschitz constant is either the half of the major axis or the reciprocal of the half of the minor axis. When we look at how the sides of a triangle are distorted, we "measure the ellipse" in three directions only. There is no guarantee that one of these directions corresponds to the maximum or minimum distortion.

More concretely, for every triangle you can keep the length of one side and move the opposite vertex in a direction perpendicular to that side. The maximum distortion is along this perpendicular and is bigger than the distortion along any of the sides.

On the other hand, one can ask for an estimate of the bilipschitz constant in terms of the distortions along the sides of a triangle...

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  • $\begingroup$ It works in the first order case as well, although I agree that it is vague. The concrete example I gave should leave no doubts: when the square of the height $h_1$ is increased by $t$, the squares of $l_2$ and $l_3$ are also increased by $t$. Since $l_2, l_3 > h_1$ (we assume there is no right angle at the side $l_1$), the first-order relative distortions of $l_2$ and $l_3$ are smaller than that of $h_1$. $\endgroup$ – Ivan Izmestiev Nov 17 '16 at 14:45
  • $\begingroup$ As for the last sentence in the answer, I mean an upper estimate of the right hand side of the conjectured equation in terms of the left hand side. There should be something in terms of the angles of the triangle. $\endgroup$ – Ivan Izmestiev Nov 17 '16 at 14:47
  • $\begingroup$ could you elaborate a little more your last comment? I'm really interested $\endgroup$ – user101163 Nov 17 '16 at 15:13
  • $\begingroup$ I don't have an exact answer, but for example for a regular triangle the worst ratio of the RHS to LHS should be attained when the deformation is as described in my previous comment. In the general case I'd suggest to use the barycentric coordinates for computation. $\endgroup$ – Ivan Izmestiev Nov 17 '16 at 15:42

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