Timeline for Sufficient condition for the unique solvability of Dirichlet problem of Hamilton-Jacobi equation
Current License: CC BY-SA 3.0
19 events
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| May 19, 2020 at 7:54 | history | edited | leo monsaingeon |
added the [viscosity-solutions] tag
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| Sep 3, 2016 at 9:37 | comment | added | kenneth | @Jeff Thanks for the interesting paper. | |
| Sep 2, 2016 at 11:57 | comment | added | Jeff | @kenneth Right, uniqueness is still open for the general Dirichlet problem for second order equations. There are some uniqueness results, but they require the sub- and supersolutions to be continuous and require a lot from the PDE, so they are not usually very useful (there is one in the user's guide, for example). An interesting result by Barles and Da Lio in the appendix of the Kohn-Serfaty paper (math.nyu.edu/faculty/kohn/papers/kohn-serfaty-cpam.pdf) establishes uniqueness for curvature flow equation on star shaped domains. I think these techniques are a bit more general. | |
| Sep 2, 2016 at 9:47 | comment | added | kenneth | @Jeff I agree with you. Generalized solution makes our life easier from the existence issue (relieving us to check boundary), but not the other way around from uniqueness. | |
| Aug 31, 2016 at 21:18 | comment | added | Jeff | @kenneth You always have problems like this for first order equations and there is no way around it. The Dirichlet problem is in general over-determined (as soon as a characteristic curve intersects the boundary twice, it is impossible to specify arbitrary Dirichlet conditions). The generalized Dirichlet problem is the "right way" to interpret first order (and second order) equations with Dirichlet boundary conditions. It relieves you of the duty of analyzing the geometry of the projected characteristics, which can be quite complicated. Unfortunately, the theory is still incomplete though. | |
| Aug 31, 2016 at 21:11 | comment | added | Jeff | @kenneth Probably the best explanation is given in the paper. The boundary conditions are incompatible with the PDE at the boundary. If you changed the boundary conditions to $u(\pm 1)=-1$, then the second equation would be fine (solution is $u=-1$). If you think back to the method of characteristics (e.g., in Evans book), there are no admissible triples for the second equation at either boundary point, so local solvability must fail. | |
| Aug 29, 2016 at 15:05 | comment | added | kenneth | @Jeff By the way, there is another possible explanation in Example 14 of the paper arxiv.org/pdf/1602.06109.pdf | |
| Aug 29, 2016 at 15:03 | comment | added | kenneth | @Jeff Thanks for your reply. It explains the situation and thanks for the nice notes on the viscosity solution. | |
| Aug 28, 2016 at 2:40 | comment | added | Jeff | @kenneth There probably is no good answer to this question. It is very easy to write down PDE for which there do not exist continuous solutions. For [ex3] you may as well be asking why $|u'(x)| + 1=0$ has no solution. That being said, there are some partial answers. If the equation is first order and the Hamiltonian is coercive and nonnegative (in a certain sense), then the Dirichlet problem is solvable. See the definitions at the start of Section 5 here: math.umn.edu/~jwcalder/222BS16/viscosity_solutions.pdf. [ex2] satisfies the nonnegativity condition, while [ex3] does not. | |
| Jun 30, 2016 at 6:40 | history | edited | kenneth | CC BY-SA 3.0 |
added 211 characters in body
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| Jun 28, 2016 at 16:14 | comment | added | kenneth | @WillieWong I got you. The other case is trivially NO. It's interesting | |
| Jun 28, 2016 at 16:10 | comment | added | kenneth | @WillieWong Thanks for checking. I've also double checked the solutions, and it seems correct to me. | |
| Jun 28, 2016 at 16:09 | comment | added | Willie Wong | With the other sign the second equation has no (continuous) solutions. $|u'| \geq 0$ but by boundary data assumption $u-1 < 0$ in a neighborhood of $\pm1$. (In fact that's why I queried in my first comment; I've mistakenly thought you were in this case.) | |
| Jun 28, 2016 at 16:03 | comment | added | Willie Wong | Sorry, you are right about the minus sign. Mea culpa (I dropped a minus sign in checking your computations.) // I'll let someone more familiar with the method answer the technical part of your question. | |
| Jun 28, 2016 at 16:02 | comment | added | kenneth | @WillieWong I am also curious about the case without leading minus sign. I believe the unique solvability shall be true? | |
| Jun 28, 2016 at 15:58 | comment | added | kenneth | @WillieWong I am not familiar in PDE, so I am not confident. Anyway, by the definition of [User's guide], it may be regarded as elliptic, but degenerate. In [ex2], I do really mean $- \inf b u'$, which is indeed $|u'|$ as you mentioned. But, the result is the same if the $\inf$ is over $[-1, 2]$, by only changing a bit on explicit form of the solution. Thanks. | |
| Jun 28, 2016 at 15:45 | comment | added | Willie Wong | Regarding your tag choice: [ex1] is not elliptic in any way you look at it. The principal part of [ex2] is in fact just a complicated way to write $|u'|$ which I guess may be thought of as being elliptic due to coercivity? (Incidentally, do you really mean $-\inf bu'$? Maybe you want it without the leading minus sign?) | |
| Jun 28, 2016 at 15:28 | history | edited | kenneth | CC BY-SA 3.0 |
fixed typos
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| Jun 28, 2016 at 14:55 | history | asked | kenneth | CC BY-SA 3.0 |