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I'm learning about the 1-dimensional diffusion/heat equation with certain boundary conditions, and currently I'm stuck at the last step, which is normalization at the initial time t=0.

The problem is to find the function $\phi(s,s',t)$, with the spatial variable $s\in (0,1)$, some fixed position $s'\in (0,1)$, and for all time $t\geq 0$. It must satisfy the heat equation

$\frac{\partial \phi (s,s',t)}{\partial t} = \frac{\partial^2 \phi(s,s',t)}{\partial s^2}$

under the boundary conditions

$\frac{\partial \phi(s,s',t)}{\partial s}\Big |_{s=1} = -2\mu \phi(1,s',t)$

$\frac{\partial \phi(s,s',t)}{\partial s}\Big |_{s=0} = 2\mu \phi(0,s',t)$

and the initial shape

$\phi(s,s',0) = e^{-2\mu |s-s'|}$

There is a fixed real constant $\mu > 0$. So far I could work out the solution up to the normalization step:

$e^{-2\mu |s-s'|} = \sum_{n=1}^{\infty} A_n \cos [\alpha_n (2s-1)] + B_n \sin [\beta_n (2s-1)]$

The Fourier frequencies are the solutions of these equations:

$\alpha_n \tan \alpha_n = \mu\\ \beta_n \cot \beta_n = -\mu$

In the Doi and Edwards article http://www.physics.emory.edu/faculty/roth/polymercourse/historical/Doi-Edwards_JCS-FT78.pdf Equation 4.37, the exact amplitudes are given as

$A_n = \frac{ \mu}{\mu^2 + \alpha_n^2 + \mu }\cos(\alpha_n (2s'-1))\\ B_n = \frac{ \mu}{\mu^2 + \beta_n^2 + \mu }\sin(\beta_n (2s'-1))$

without explanation and I was wondering how can one derive these expressions. If the frequencies where the usual $\alpha_n = \pi n$ etc., we could simply use the orthogonality of the trigonometric functions, but in the current situation I have no clue how to proceed.

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  • $\begingroup$ You should put in two more groups of parentheses in the first displayed formula. Could you give more background on the governing equation? Fourier series can directly be applied to the pure heat equation. You already told us that this is not possible in your example. $\endgroup$ – Tobias Dec 30 '16 at 22:00
  • $\begingroup$ I've added the missing info. Let me know if there's anything else I can do! $\endgroup$ – Airidas Korolkovas Dec 30 '16 at 23:05
  • $\begingroup$ Dear Airidas, I think this qualifies as a question for mathoverflow.net. $\endgroup$ – Tobias Jan 1 '17 at 9:38

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