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Entry  Sat Jun 28 21:59:11 2014, Sam Moore, Optics, General, Difficulty with the COMSOL stationary module; Test Cases 6_27_14.pdf
    Reply  Sun Jun 29 15:37:18 2014, Sam Moore, Optics, General, Difficulty with the COMSOL stationary module; Test Cases 6_29_14.pdf
    Reply  Sun Jun 29 20:25:44 2014, Koji, Optics, General, Difficulty with the COMSOL stationary module; Test Cases 
Message ID: 90     Entry time: Sun Jun 29 20:25:44 2014     In reply to: 88
Author: Koji 
Type: Optics 
Category: General 
Subject: Difficulty with the COMSOL stationary module; Test Cases 

What about this example? The result is easier to understand intuitively.

Consider a bar with the length of L.

Let's say there is no body heat applied, but the temperature of the bar at x=L is kept at T=0
and at x=0 is kept at T=T0 Exp[I w t].

The equation for the bar is

...(1)

Consider the solution with the form of T(x, t) = T(x) T0 Exp(I w t), where T(x) is the position dependent transfer function.
T(x) is a complex function.

Eq.1 is modified with T(x) as

With the boundary condition of

This can be analytically solved in the following form

where alpha is defined by

So kappa/Cp is the characteristic (angular) frequency of the system.
Here is the example plot for L=1 and alpha = 1 (red), 10 (yellow green), 100 (turquoise), 1000 (blue)

If the oscillation is slow enough, the temperature decay length is longer than the bar length and thus the temperature is linear to the position.
If the oscillation is fast, the decay length is significantly shorter  than the bar length and the temperature dependence on the position is exponential.

Now what we need is to solve this in COMSOL

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