MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
What are the boundary conditions? – S. Carnahan Jun 23 '10 at 1:00
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. – sigoldberg1 Jun 23 '10 at 1:09
I guess to more specifically respond, the boundary conditions are free, with forcing at theta = 0, on a line orthogonal to the midline circle. – sigoldberg1 Jun 23 '10 at 1:19
Force is harmonic (sinusoidal), to be applied normal to the surface. – sigoldberg1 Jun 23 '10 at 1:24
up vote 6 down vote accepted

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.

share|cite|improve this answer
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? – sigoldberg1 Jun 25 '10 at 0:32
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. – S. Carnahan Jun 29 '10 at 21:38

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.

share|cite|improve this answer
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. – sigoldberg1 Jun 23 '10 at 15:05

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.