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I'm sorry, I'm having such a hard time with this question in my text:

Determine the equilibrium temperature distribution for the thin circular ring, i.e. the steady state solution for the heat equation

$\dfrac{\partial{u}}{\partial{t}} = k\dfrac{\partial^2{u}}{\partial{x^2}}$

with periodic boundary conditions

$u(-L, t) = u(L, t)$ and $\dfrac{\partial{u}}{\partial{x}}(-L, t) = \dfrac{\partial{u}}{\partial{x}}(L, t)$

and initial condition

$u(x, 0) = f(x)$

Solve from the equilibrium solution and by computing the limit ${t \to \infty}$ of the time dependent problem.

I really am unsure of where to even start on this problem. All I know is the steady state solution is $\lim_{t \to \infty} u(x, t) = A_0 = \dfrac{1}{L} \int_0^L f(x)\,dx $.

If I could get any tips, pointers, etc. on starting the problem, that would be great. I'm not looking for an answer, just a nudge in the right direction.

Brooklyn Tanner
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  • To begin with, the equation doesn't match the geometry. You provided a 1D heat equation, but the ring is a circle in 2D. – Ron Gordon Sep 16 '15 at 20:39
  • I can only assume because the ring is "thin" then we can neglected variation across the ring. Instead we have periodic boundary as it is connected. However, i could be way off mark. OP clarify please. – Chinny84 Sep 16 '15 at 20:46
  • @Chinny84 In this situation the ring should satisfy a one dimensional heat equations where the distance is the arc length along the ring. The temperature in the ring is constant along the cross sections because it's so thin. – Brooklyn Tanner Sep 16 '15 at 20:52
  • I thought as much. But you should solve the Pde with the boundary conditions given and then find steady state from there. – Chinny84 Sep 16 '15 at 20:56
  • @Chinny84 I see, okay thank you very much! – Brooklyn Tanner Sep 16 '15 at 21:01

1 Answers1

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let $u_s(x)\equiv \lim_{t \to \infty}u(x,t)$ ( steady state is time independent)

clearly $\dfrac{\partial{u_s}}{\partial{t}} = k\dfrac{\partial^2{u_s}}{\partial{x^2}}=0$

so the only solutions take the form $u_s(x)=ax+b$ which can only satisfy the boundary conditions if $a=0$

so $$u_s(x)=\text{Const.}$$

The specific value you have for the constant makes perfect sense for $f(x)=\text{Const.}$

You may be able to justify it in general by appealing to conservation of molecular kinetic energy.

Otherwise you might need to revisit your boundary conditions, if they are truly periodic they should read something like $$u(x,t)= u(x+L, t)$$

WW1
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