Timeline for Local solvability and Cauchy-Kovalevskaya theorem for PDEs
Current License: CC BY-SA 4.0
18 events
| when toggle format | what | by | license | comment | |
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| Aug 2, 2020 at 23:31 | answer | added | Deane Yang | timeline score: 1 | |
| Jul 31, 2020 at 22:12 | comment | added | Igor Khavkine | @Michele There is no reason to be hypothetical. All of the tools that you have mentioned are in fact already being used in the relevant literature, without any compunctions about using $\infty$-jets when convenient. This discussion seems to be veering more into philosophy. I'm not sure what other practical information I could add, besides what I already wrote. | |
| Jul 31, 2020 at 15:37 | comment | added | Michele | @Igor. Just to clarify, my direction is merely pragmatical. In particular, again restricting to the polynomial case, being able to use certain well-known algebraic-geometric notions (notably, Groebner bases) for algorithmic purposes. Just to make a simple example, under suitable conditions, the set of invariant polynomials of $\Delta$ --- those identically 0 on solutions of $\Delta$ --- coincide with the ideal generated by $\Delta$, on $n$-order jets. This in turn may be useful in direct methods to search for conservation laws. | |
| Jul 31, 2020 at 14:50 | comment | added | Igor Khavkine | @Michele There is no argument against both subjective and temporal fluctuations in what is or is not a good definition. There are many ways in which a definition may be modified to suit one's taste. Is that the philosophical direction you intended your question to go in? The gap between $\infty$-jets and finite order data is closed by the notion of an involutive form of a PDE, about which many volumes have been written. Your last reference is an example. Another notable reference is Involution by W. Seiler (Springer, 2010). | |
| Jul 31, 2020 at 6:55 | comment | added | Michele | @Igor. This fills the gap between $S(\Delta)$ and $V(\Delta)$, at the cost of forcing infinite jets in the definition of $V(\Delta)$, thus spoiling its finitary, 'algebraic' nature. For instance, suppose $\Delta$ is polynomial, then it may be useful to reason on $V(\Delta)$ in algebraic-geometric terms, that is as the affine variety induced by $\Delta$ on $n$-th order jets. Note that local solvability (Olver style) is central also in differential-completion algorithms for PDEs, like Cartan-Kuranishi: see e.g. p. 15 [link] (maths.dundee.ac.uk/plin/SiAM_ReidLinWittkopf_01.pdf ). | |
| Jul 30, 2020 at 15:03 | comment | added | Igor Khavkine | @Michele To be specific, replace $V(\Delta)$ by the locus of $\infty$-jets cut out by $\Delta=0$ and all of its differential consequences, at the same time replace $S(\Delta)$ by the locus of $\infty$-jets of locally defined smooth/analytic (as you prefer) solutions. Then $S(\Delta) = V(\Delta)$ is a good definition of local solvability. Sure the ext-K form gives you lots of information, but also including the ability to specify formal solutions by finite amounts of data (the free initial data), as I wrote earlier. The hypotheses of the Cartan-Khäler theorem do that in even more generality. | |
| Jul 29, 2020 at 22:23 | comment | added | Michele | @Igor. Ok, so to be specific, you propose to replace the definition of $S(\Delta)$ given in my question with $S(\Delta):=...$? As an aside, concerning the extended Kovalevskaya form, my understanding is that it additionally guarantees analiticity (given analytic initial data). If one is interested in formal power series solutions, disregarding analyticity, there are more generous formats. See e.g. here link. | |
| Jul 29, 2020 at 19:49 | comment | added | Igor Khavkine | I would second Deane's comment that Olver's defnition is not the most common/reasonable one. In fact, it reads a little "lazy," which surprises me. Here's a better definition. Locally solvable: every formal power series solution (aka an $\infty$-jet at a point) is the Taylor series of a smooth solution (locally extends to a smooth solution). Dropping this definition into your example removes any contradictions. Note that the extended Kovalevskaya form precisely guarantees a unique formal solution for any free initial data (of appropriate order for $u$ and $v$). | |
| Jul 29, 2020 at 18:36 | history | edited | Michele | CC BY-SA 4.0 |
Pasted Olver's original definition.
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| Jul 29, 2020 at 18:34 | comment | added | Michele | I have pasted the original definition into my question. In short, for l.s. we require $V(\Delta)=S(\Delta)$ on the 2nd order jet, which does not seem to be the case for my 2-equations system. Of course it is the case for your 3-equations system, but this does not solve my doubt about Olver's statement. | |
| Jul 29, 2020 at 18:29 | history | edited | Michele | CC BY-SA 4.0 |
added 147 characters in body
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| Jul 29, 2020 at 18:27 | comment | added | Michele | It may not be the 'standard definition' of local solvability, but it is the way it is defined Olver's textbook. I copy it below: | |
| Jul 29, 2020 at 16:28 | comment | added | Deane Yang | Here, it appears that local solvability means that any second order jet that solves the system can be extended to a local smooth solution (which is not the standard definition of local solvability). To be a second order solution, the three equations and the first partial derivatives with respect to both $x$ and $t$ of the first two equations must hold. In particular, if a 2-jet is a solution at $(x_0,t_0)$, $u_{tx} = v_x$ then holds at $(x_0,t_0)$. So in fact this equation is a consequence of the jet being in $S(\Delta)$. | |
| Jul 29, 2020 at 7:21 | comment | added | Michele | This is in the definitiion of $S(\Delta)$, but not in that of $V(\Delta)$. That is, there are 2nd order differential consequences (e.g. $u_{tx}=v_x$) that are not algebraic consequences. | |
| Jul 28, 2020 at 22:33 | comment | added | Deane Yang | Why is this system not locally solvable? Doesn’t $u_{0,tx} = v_{0,x}$ follow from the assumption that the equations hold at $(x_0,t_0)$ up to second order. | |
| Jul 28, 2020 at 15:35 | comment | added | Michele | Your system is "equivalent" to the old one in terms of solutions. However the point here is not finding the solutions, but solving the apparent contradiction I have pointed out. Cheers | |
| Jul 28, 2020 at 14:37 | comment | added | Deane Yang | Couldn't you find all solutions by solving the first order system \begin{align*} u_t &= v\\ v_t &= w\\ w_t &= u_x \end{align*} with initial data $(u,v,w)$ along $t = 0$? | |
| Jul 28, 2020 at 13:18 | history | asked | Michele | CC BY-SA 4.0 |