Timeline for Systems of (hyperbolic) 2nd order PDEs with lower order constraints
Current License: CC BY-SA 4.0
13 events
| when toggle format | what | by | license | comment | |
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| Apr 9 at 17:12 | history | edited | Robert Bryant | CC BY-SA 4.0 |
Added some more detail about how to get to the Frobenius case
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| Apr 9 at 16:47 | comment | added | Robert Bryant | @DanielCastro: I'll edit my answer above to explain what to do in this case. | |
| Apr 9 at 14:42 | comment | added | Daniel Castro | Indeed $H$ does not identically vanish. The space of solutions of the system $H_u=H_v=0$ plus $(1)$ is 'bigger' than the original system, because we are essentially setting $H$ to be any constant whereas it has to be zero, is it ? Besides that, that's a 'doubly overdetermined' system - 4 equations for a single function $\gamma$, which seems even more difficult to handle than just $(1)$. Is there in the literature a similar case-study to educate myself about what to do ? | |
| Apr 9 at 13:59 | comment | added | Robert Bryant | @DanielCastro: I think you meant to write "$H$ does not vanish (identically)" in your comment above. So it seems that you need to add $H_u=H_v=0$ to (1) to get a complete system. This 'prolonged' system of 4 third-order equations will probably allow you to solve for all 4 of the third derivatives of $\gamma$ in terms of lower derivatives. In this case, you are now reduced to a system that can be studied by Frobenius' method, and there is at most a finite dimensional space of solutions $\gamma$. | |
| Apr 9 at 13:05 | comment | added | Daniel Castro | In this case and similar ones of interest indeed $A_0=A_3=0$ identically, but $H$ does vanish and $(1)$ cannot be written as linear combinations of its derivatives. So $H=0$ is a second order determined equation for $\gamma$ but it does not necessarily imply $(1)$. | |
| Apr 6 at 14:23 | vote | accept | Daniel Castro | ||
| Apr 3 at 23:02 | history | bounty ended | CommunityBot | ||
| Mar 29 at 9:42 | history | edited | Robert Bryant | CC BY-SA 4.0 |
Added the missing term A_4, which was pointed out by Daniel Castro
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| Mar 29 at 9:38 | comment | added | Robert Bryant | @DanielCastro: Of course, yes you are correct. I overlooked that term when I was writing it up, but it doesn't substantially change the story. I'll fix that. | |
| Mar 28 at 9:41 | comment | added | Daniel Castro | Thank you. Also, in $(2)$ shouldn't we have some $A_4$ containing (only) second and lower derivatives ? | |
| Mar 27 at 20:18 | comment | added | Robert Bryant | @DanielCastro: But $(\tau_u)_v = (\tau_v)_u$ is an identity, no matter what expression you plug in (as long as $\tau$ is $C^2$), so plugging in the expression for $\tau$ as above will only give you the trivial third-order equation on $\gamma$. | |
| Mar 27 at 19:27 | comment | added | Daniel Castro | Thanks for the guidance. A preliminary question is, shouldn't we consider a third equation? I mean, in the original system we take $\tau$ from the Gauss (third equation) and replace in the first two (Codazzi) equations, thus obtaining the two in $(1)$. But taking cross derivatives of the Codazzi's, $(\tau_u)_v=(\tau_v)_u$, puts into another equation containing $\gamma_{uvv}$ and $\gamma_{uuv}$. | |
| Mar 27 at 18:26 | history | answered | Robert Bryant | CC BY-SA 4.0 |