Timeline for Differences between the Poisson's and elliptic Monge-Ampere equations?
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
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| Aug 29, 2013 at 0:58 | comment | added | timur | Try solving the both equations for $\rho$ spherically symmetric and see if they are different. | |
| Feb 7, 2012 at 8:22 | comment | added | Dan Fox | Igor, my point was that the local geometry naturally associated to the two operators is different, so already in the case where the domain is the entire space and $\rho$ is a constant, or for the Dirichlet problem, the equations/solutions behave quite differently. The Monge-Ampere equation is more difficult than the Laplacian. As Deane Yang has pointed out above, the nonlinearity/linearity is one reason. The local geometry is another. On the other hand, by the arithmetic/geometric mean inequality there is some relation. In this regard see papers of Dean/Glowinski about the M-A case. | |
| Feb 6, 2012 at 23:22 | comment | added | Igor Markov | Hmm... I am not sure how to exploit such invariance in my case, since the domain is bounded and $\rho$ seems unlikely to be invariant. | |
| Feb 6, 2012 at 17:28 | history | edited | Igor Markov | CC BY-SA 3.0 |
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| Feb 6, 2012 at 17:12 | vote | accept | Igor Markov | ||
| Feb 6, 2012 at 17:11 | history | edited | Igor Markov | CC BY-SA 3.0 |
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| Feb 6, 2012 at 15:56 | answer | added | Andrew T. Barker | timeline score: 1 | |
| Feb 6, 2012 at 14:40 | comment | added | Dan Fox | A basic difference is that the Laplacian is invariant under Euclidean motions, while the Hessian determinant is invariant under unimodular affine motions. | |
| Feb 6, 2012 at 10:26 | comment | added | Deane Yang | Your reasons for wanting to consider the Monge-Ampere equation are rather vaguely stated, and your questions about the equation are quite broad. The differences are sufficiently big that you should not pursue this unless the application really does call for the Monge-Ampere equation, which is nonlinear, not necessarily elliptic, and therefore much more difficult to work with numerically. I also see that you are at Michigan, so I encourage you to consult colleagues in the math department and maybe even others in your own department about this. | |
| Feb 6, 2012 at 9:56 | comment | added | Igor Markov | Thinking aloud: for #1, we may want to consider two cases (1) $\rho$ with well-articulated saddle points and (2) $\rho$ without saddle points. | |
| Feb 6, 2012 at 9:40 | history | asked | Igor Markov | CC BY-SA 3.0 |