- #1

StretchySurface

- 6

- 0

- Homework Statement
- Find General Solution of Heat Equation on a Ring.

- Relevant Equations
- See below

I'm solving the heat equation on a ring of radius ##R##. The ring is parameterised by ##s##, the arc-length from the 3 o'clock position. Using separation of variables I have found the general solution to be:

$$U(s,t) = S(s)T(t) = (A\cos(\lambda s)+B\sin(\lambda s))*\exp(-\lambda^2 kt)$$.

Here is my confusion:

Periodicity demands that ##U(s',t) = U(s' + 2\pi R, t).## This means that

$$\cos(\lambda s') = cos(\lambda s' + \lambda*2\pi R)$.$

This finally gives condition that ##\lambda = k/R%## where ##k## is an integer. This is the correct answer as told by my professor/textbook.

However, periodicity can also demands ##U(s',t) = U(s' + 2\pi nR, t)##, i.e revolving around the ring ##n## complete times. But this leads to the condition ##2\pi \lambda nR = 2 \pi m## where ##m## and ##n## are arbitrary integers. This gives us ##\lambda = m/(nR)##. This doesn't seem to be the correct answer but I don't understand why and rational number divided by ##R## is not an eigenvalue.

$$U(s,t) = S(s)T(t) = (A\cos(\lambda s)+B\sin(\lambda s))*\exp(-\lambda^2 kt)$$.

Here is my confusion:

Periodicity demands that ##U(s',t) = U(s' + 2\pi R, t).## This means that

$$\cos(\lambda s') = cos(\lambda s' + \lambda*2\pi R)$.$

This finally gives condition that ##\lambda = k/R%## where ##k## is an integer. This is the correct answer as told by my professor/textbook.

However, periodicity can also demands ##U(s',t) = U(s' + 2\pi nR, t)##, i.e revolving around the ring ##n## complete times. But this leads to the condition ##2\pi \lambda nR = 2 \pi m## where ##m## and ##n## are arbitrary integers. This gives us ##\lambda = m/(nR)##. This doesn't seem to be the correct answer but I don't understand why and rational number divided by ##R## is not an eigenvalue.