Timeline for Existence of second order potential for PDE
Current License: CC BY-SA 3.0
14 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 12, 2017 at 11:29 | comment | added | Yegor | Thank you for the pointers. Let me check, if a more naive approach is possible. General solution to the original problem above involves one arbitrary scalar function $T$ of all the coordinates. I guess it means that the original system of PDEs is underdetermined. Could one just try to count, how many arbitrary functions of all coordinates should parametrize general solution? In simple examples this could provide some guesses, what possible potentials could be. However I am not able to do the counting even for this simple problem. How can one see that everything is encoded in just one function? | |
| Dec 12, 2017 at 10:52 | comment | added | Igor Khavkine | ... A comprehensive modern treatment can be found in the book Involution by W. Seiler. Unfortunately, the theory is quite involved. In some cases, if you write out all the operators in coordinates, some computer algebra packages may do the calculations for you (like for instance the Janet package for Maple). | |
| Dec 12, 2017 at 10:50 | comment | added | Igor Khavkine | @Yegor, if you have a linear differential operator $D$ and a linear operator $C$ such that $CD=0$, with the additional property that any other operator $E$ such that $ED=0$ must factor through $C$, $E=E'C$ for some $E'$, then $C$ is a (universal/complete/generating) compatibility operator for $D$. Your problem with $LS=0$ is mapped to the one I described by taking formal adjoints, $D=L^*$, $C=S^*$ and vice versa. As Robert Bryant said, compatibility operators exist under general conditions, and can be analyzed with Spencer cohomologies. ... | |
| Dec 12, 2017 at 9:00 | comment | added | Robert Bryant | @Yegor: Yes, these questions can be answered. There are effective tests for when the kernel of a given differential operator can be written as the image of another. This can even be generalized to cover nonlinear operators in some cases. I should say right away that, most of the time, this cannot be done, but when it can, one can usually employ it to good effect. I cannot go into details here, but the most important cases can be described via the machinery of Spencer sequences, resolutions, and cohomology. The Poincaré sequence is the most famous, of course, but there are many others. | |
| Dec 12, 2017 at 8:28 | comment | added | Yegor | It looks to me like some kind of cohomology here. In electromagnetism we have that $dF=0$, so we are looking for an operator which lies in the kernel of $d$, which is $d$ itself, hence $F=dA$. So more generally, given $L(u)=0$, we should be looking for differential operators S lying in the kernel of L, i.e. $L S =0$? Can such questions be answered? Does it even make sense? | |
| Dec 12, 2017 at 8:24 | comment | added | Yegor | These were indeed useful details. May I ask a related question? The technique you are using is very powerful in proving that certain tensor can be expressed in certain way, say, in this case it is shown that second rank tensor (satisfying certain condition) can be written in the Hessian like form in unique way. But is there a method to figure out which kind of a potential can one use to represent given tensor? Concretely: why in the electromagnetism we represent the field strength $F$ using 'first order potential' $F=dA$, whereas in the case above we can get even 'second order potential'? | |
| Dec 12, 2017 at 8:07 | history | edited | Robert Bryant | CC BY-SA 3.0 |
Rewrote in response to the OP's request for a more explicit description of the solution
|
| Dec 12, 2017 at 7:49 | comment | added | Robert Bryant | @Yegor: I'm alluding to the fact that the underlying manifold is simply connected and that the affine sheaf of solutions is 'flat'. However, I can rewrite so as to avoid this, so I will. Maybe that will be more useful to you. | |
| Dec 11, 2017 at 15:03 | comment | added | Yegor | sorry, my background isn't in PDEs, rather in physics. What is the "usual cohomological argument" you refer to in the last paragraph? | |
| Dec 6, 2017 at 15:40 | vote | accept | Yegor | ||
| Dec 5, 2017 at 12:28 | comment | added | Robert Bryant | @Yegor: Yes, it does. It is determined by the (constant) sectional curvature. For an explanation of where the second equation comes from, see the other MO question that you referenced. | |
| Dec 5, 2017 at 12:02 | comment | added | Yegor | How do you come up with the second equation in (1)? In particular, does the coefficient in front of $f \delta_{ij}$ term matter? | |
| Dec 5, 2017 at 10:19 | history | edited | Robert Bryant | CC BY-SA 3.0 |
added 25 characters in body
|
| Dec 5, 2017 at 10:12 | history | answered | Robert Bryant | CC BY-SA 3.0 |