On 10/11/2009 7:50:32 PM, jwrudn wrote:
>Tom, thanks for your comment. I agree, you are correct. When I said that the equations are stable, I meant that the exact solution of the equations is stable (decays exponentially) for the given initial conditions used in the worksheet.
This isn't the concept of stability. If a magnitude grows without limit (eventually exponentially) have a physical sense: something go to broken, and this not have relation with the stability of the solutions. In your case: if the pressure grows too much, then sure the material under this pressure go to have a fracture. And then your model must to change, because now is not true or good for the new situation. Same thing with the temperature, let say that the material could burn or change the state, or something that make the model inadequate and need to set to another model.
Also this not meaning that the model is not useful: only says that you need to check the material properties to know the time limit that could be exposed to this situation before it go to be broken.
For a rigourous definition for the 'stability in the future as Lyapunov' you can see
http://www.fing.edu.uy/~eleonora/dvi/cual4.pdfin spanish, but for odes, not pdes.
Stability is usually the study of small variations of the parameters in the diff equations, and is usual taking initials conditions as parameters, but eventually could be anything. For example, c_thermical or c_hydraulic could be the parameters to study the stability. In your case seems that the stability that you want to study is precisely this 'stability in the future', this is, for t big (4 or 5 seconds are enough big with the others parameters fixed). So, your parameter is the time, and the 'small variations' about a fixed point is the grow of the time without limit. As you can see in the paper, particular solutions (which depends on initials conditions) must to be taken in a 'maximal interval', to ensure that the solution is stable also for small variations around the initials conditions. So, this stability must to study first.
In your case, for N>Ncrit=10.972 what you have is not stability, only solutions that not diverges to inf as t diverges to inf. This meaning only that your material not burns or fissure, but not implies that the analytical solutions are stables. Notice that for N=6, for example, analytic and pdsolve solutions are the same.
For numerical solutions the situation is more complicated: you need that the analytic solution must to be stable, but also needs to add numerical criterion to the 'stability' of the numerical solutions, but this depends on the particular algorithm, and mathcad don't give information about their numerical algorithms to do this.
Regards. Alvaro.