Timeline for Analytic solution of a system of linear, hyperbolic, first order, partial differential equations
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
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| Jun 18, 2014 at 12:56 | comment | added | FraSchelle | Yes :-), a Taylor expansion. Sorry to have missed the point when I said series. It is always good to be precise indeed. | |
| Jun 18, 2014 at 12:52 | comment | added | Robert Bryant | It's hard to tell until you specify what you mean by 'series expansion'. For example, since $\mathbf{B}$ is $2\pi$-periodic, you might mean Fourier series expansion, which every continuous, differentiable, $2\pi$-periodic function has. What I mean by 'analytic' is that the function equals its Taylor series expansion (which is assumed to have a positive radius of convergence) at every point. Is that what you mean? | |
| Jun 18, 2014 at 12:17 | comment | added | FraSchelle | Well analytic in my sense means continuous, differentiable, having series expansion... In short, the $t$-dependency of the B-matrix is $B \propto e^{it}$. Is that analytical enough ;-) ? | |
| Jun 18, 2014 at 11:59 | comment | added | Robert Bryant | The Cauchy-Kowalewski Theorem will still apply if you allow $\mathbf{u}$ and $\mathbf{B}$ to be complex valued, as long as they are real-analytic functions of their arguments. If you only have smoothness (or weaker regularity) of the coefficients, then the C-K Theorem may not apply. By the way, I now realize that you are using 'analytic' in a different sense from what I mean when I use 'analytic' (which is really 'real-analytic'), you probably mean 'analytic' to be some concept that is in opposition to 'numerical' and somehow related to 'explicit'. Is that right? | |
| Jun 18, 2014 at 10:32 | comment | added | FraSchelle | Thanks a lot for your answer. I believed my problem is indeed Cauchy's, since the initial data are non-characteristic. But I didn't care so much about all the details. Actually, B is Hermitian (still continuous, differentiable in t, ... all the good physical properties of a Hamiltonian somehow), does your theorem still apply ? I guess so, since the eigenvalues are then real (and analytic, ...) or is there a trick I don't see ? Thanks again. | |
| Jun 17, 2014 at 19:24 | history | answered | Robert Bryant | CC BY-SA 3.0 |