# Heat Equation with Periodic Boundary Conditions

StretchySurface
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.

Homework Helper
Something with a period of ##4\pi## does not necessarily satisfy the condition that ##f(\theta)=f(\theta+2\pi)## !

StretchySurface
Does that mean the following is untrue: We require ##U(s′,t)=U(s′+4πR,t)##?

Homework Helper
It is a true statement, but:
It's a necessary but insufficient condition.

StretchySurface
Okay, so I don't really see the point you're trying to convey and how it relates to my question. Below, I'll write out a set steps in reasoning to see if you can pinpoint the error I'm making.

Let's only consider the case ##n = 2## i.e traveling around the ring twice.
Periodicity demands that ##U(s',t)=U(s'+4πR,t)##.
This means that ##2\lambda R = k##, for any integer ##k##.
Thus ##\lambda_{k} = k/(2R)## which includes some values not mentioned in the notes I'm working from.

Homework Helper
some values not mentioned
Yes. And those do NOT satisfy the condition ##U(s′,t)=U(s′+2\pi R,t)##. Do you propose to drop that condition ?

Staff Emeritus
Homework Helper
Gold Member
Put slightly differently, 2pi periodic functions are automatically 4pi periodic but 4pi periodic functions are not necessarily 2pi periodic.

StretchySurface
Ah okay, now I understand! Thank you very much!

Gold Member
Btw if ##\lambda## is the inverse of a length, and ##k## is an integer (I suppose dimensionless), that exponential doesn't look good to me.

Edit: ok the ##k## in the exponential is the diffusion coefficient (I suppose)... my bad

Staff Emeritus