Given a mobius strip, what do the solutions of the wave equation look like qualitatively? How do they differ from solutions on the equivalent strip glued together as a cylinder? Any refs, particularly to symmetry?
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$\begingroup$ books.google.com/books?id=t0D_YmZpTucC&pg=PA333 $\endgroup$– Steve HuntsmanCommented Jun 23, 2010 at 0:53
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$\begingroup$ What are the boundary conditions? $\endgroup$– S. Carnahan ♦Commented Jun 23, 2010 at 1:00
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$\begingroup$ Well, say I'm trying to excite it with vibrations near the fundamental frequency of the corresponding orientable cylindrical surface. Is the resonant frequency (eigenvalue) of the mobius strip shifted? I know this doesn't exactly answer your question, but you see what I'm imagining. $\endgroup$– sigoldberg1Commented Jun 23, 2010 at 1:09
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$\begingroup$ I guess to more specifically respond, the boundary conditions are free, with forcing at theta = 0, on a line orthogonal to the midline circle. $\endgroup$– sigoldberg1Commented Jun 23, 2010 at 1:19
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$\begingroup$ Force is harmonic (sinusoidal), to be applied normal to the surface. $\endgroup$– sigoldberg1Commented Jun 23, 2010 at 1:24
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2 Answers
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Any solution to the wave equation on a Möbius strip lifts to a solution on its orientation double cover, which is a cylinder of equal width but twice the circumference. In order for a solution on the cylinder to descend to the Möbius strip, it is necessary and sufficient that it be invariant under a certain order two symmetry. If the cylinder has coordinates given by $[0,\pi a]$ in the free direction and $[0, 2\pi L]$ in the loop direction, then the solutions are linear combinations of products $AB$, where $A$ has the form $\cos (kn_a t)\cos(\frac{n_a}a x)$ or $\sin (kn_a t)\cos(\frac{n_a}a x)$, and $B$ has the form $\cos(kn_Lt)\cos(\frac{n_L}{L}y)$, $\sin(kn_Lt)\cos(\frac{n_L}{L}y)$, $\cos(kn_Lt)\sin(\frac{n_L}{L}y)$, or $\sin(kn_Lt)\sin(\frac{n_L}{L}y)$. Here, $n_a$ and $n_L$ are nonnegative integers, and $k$ is a constant. Invariance under the symmetry is equivalent to $n_a + n_L$ being an even number. In contrast, solutions on the cylinder of the same dimensions correspond to solutions on the double cover such that only $n_L$ is even.
answered Jun 23, 2010 at 16:40
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$\begingroup$ A surprising and lovely answer. Can you say a little more about a) the order two symmetry b) the proof that n_sub_a + n_sub_L must be even, or refer us to a reference? $\endgroup$ Commented Jun 25, 2010 at 0:32
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1$\begingroup$ The order two symmetry is defined by $(x,y) \mapsto (\pi a - x, \pi L + y)$, where $x$ is the coordinate in the free direction, $y$ is the coordinate in the loop direction, and the $y$ coordinates are considered modulo $2 \pi L$. The proof of invariance comes from examining the eigenfunctions I listed above, and seeing which ones are unchanged under the transformation. This reduces to the fact that cosine and sine get multiplied by $-1$ when the domain is translated by $\pi$. I do not know a reference. $\endgroup$– S. Carnahan ♦Commented Jun 29, 2010 at 21:38
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The following paper by: Kousuke Yakubo,Yshai Avishai,and Doron Cohen describes in section II the solution of the Helmholtz equation on a flat rectangular surface having the topology of a Mobius strip. The solution is given in terms of the admissible wave numbers.
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$\begingroup$ While not directly on point (answers a slightly different question, with different boundary conditions), this suggestion materially advances my understanding of the subject. I would greatly appreciate further contributions by others as well. $\endgroup$ Commented Jun 23, 2010 at 15:05