- #1
dRic2
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Hi, I'm reading a book about numerical models for PDE and it says that a method is said to be stable if this condition holds:
$$|| \mathbf u^{n+1} || \le c_t || \mathbf u^{n} ||$$
where ##c_t## is a constant greater than zero, and ##u## is the numerical solution to the problem. (In particular I'm studying transport equations). I think that this condition is equivalent to the following:
$$||\mathbf {\Phi}|| < c_t$$
where ##\Phi## is the "iteration matrix" (##\mathbf u^{n+1} = \mathbf {\Phi} \mathbf u^{n}##).
There is something that bothers me very much in this definition... Here it seems to me that any norm will do the job. I mean, I know that in ##R^N## all the norms are equivalent so if the condition happens to be true for a particular norm then it will be true for all the other norms... But in the book the author says that stability is linked to the norm I chose to evaluate it; in particular a problem could be stable with respect to a certain norm A, but not B.
What am I missing ?
PS: I hope I explained myself properly, otherwise let me know if you have problems to understand
$$|| \mathbf u^{n+1} || \le c_t || \mathbf u^{n} ||$$
where ##c_t## is a constant greater than zero, and ##u## is the numerical solution to the problem. (In particular I'm studying transport equations). I think that this condition is equivalent to the following:
$$||\mathbf {\Phi}|| < c_t$$
where ##\Phi## is the "iteration matrix" (##\mathbf u^{n+1} = \mathbf {\Phi} \mathbf u^{n}##).
There is something that bothers me very much in this definition... Here it seems to me that any norm will do the job. I mean, I know that in ##R^N## all the norms are equivalent so if the condition happens to be true for a particular norm then it will be true for all the other norms... But in the book the author says that stability is linked to the norm I chose to evaluate it; in particular a problem could be stable with respect to a certain norm A, but not B.
What am I missing ?
PS: I hope I explained myself properly, otherwise let me know if you have problems to understand