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comment Does the Legendre-Hadamard condition imply a generalized Gårding inequality?
Terry, thanks again for your help. As Denis' answer shows, the tangential Fourier transform does indeed provide the answer.
1d
comment Does the Legendre-Hadamard condition imply a generalized Gårding inequality?
Many thanks! I haven't had a chance to study your paper carefully, but it looks good. I've learned that elliptic systems of PDE's can have much more subtle behavior than scalar elliptic PDE's.
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accepted Does the Legendre-Hadamard condition imply a generalized Gårding inequality?
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Jan
23
comment Does the Legendre-Hadamard condition imply a generalized Gårding inequality?
Terry, perhaps I'm trying to do the reflection too naively but I can't get it to work. If I reflect the function about the hyperplane $x^n = 0$, then the argument seems to work only if $A^{in}_{ab} = 0$ for $1 \le i \le n-1$. Or if $u=0$ along the hyperplane.
Jan
22
comment Does the Legendre-Hadamard condition imply a generalized Gårding inequality?
Terry, thanks! Obvious after you said it. I'll try that.
Jan
21
revised Does the Legendre-Hadamard condition imply a generalized Gårding inequality?
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Jan
21
asked Does the Legendre-Hadamard condition imply a generalized Gårding inequality?
Jan
20
comment The Minkowski sum of two curves
In the generic situation it should be possible to approximate by smooth curves and take a limit. But there will definitely be singularities, where the sum is not open. I don't know how to identify this when the curves are not differentiable.
Jan
20
comment The Minkowski sum of two curves
It's instructive to look at some specific examples. Setting $\gamma $ to $y = 0$ and $y = 1$ are instructive. More generally, if you parameterize both $\gamma$ and $\gamma'$ with two different parameters, say, $s$ and $t$, then you get a map from $\mathbb{R}^2$ to the Minkowski sum, and you can compute the rank of this map.
Jan
18
comment Gauß Bonnet operator
I would say that the classical definition has won and $\Delta$ usually denotes the negative operator. But you always need to check.
Jan
17
comment Gauß Bonnet operator
Well, is it a positive or negative operator?
Jan
17
comment Gauß Bonnet operator
This really should be asked on math.stackexchange.com, since it's not a research level question. Nevertheless, here's the answer: There is an ambiguity in how $\Delta$ is defined. Overall, people in PDE theory define it to be the negative operator, but some (but not all) people in differential geometry prefer to define it to be the positive operator. So you always have to watch out for this when learning or doing anything involving a Laplacian.
Jan
16
comment Immersability and applications of a particular Riemannian metric
The calculation of curvature is probably easier using moving frames. You simply start with a moving frame $e_1, \dots, e_n$ of the flat metric with dual frame $\omega^1, \cdots \omega^n$, where $e_n = N$. There are corresponding connection $1$-forms $\omega^i_j$, and the Maurer-Cartan equations give the curvature. Now define a new set of forms $\eta^1 = \omega^1, \dots, \eta_{n-1} = \omega^{n-1}, \eta^n = (1+\alpha)\omega^n$. This is an orthonormal frame for $G$. Now solve for the connection $1$-forms for $G$ and compute its curvature.
Jan
14
revised Cauchy problem for an overdetermined system of PDE
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Jan
14
comment Cauchy problem for an overdetermined system of PDE
Sorry but if $E_1$ and $E_2$ are linearly independent, you need to solve both equations. In particular, if $p + q = 1$, then $E_1 + pE_3$ and $E_1 + qE_3$ are linearly dependent and therefore do not imply solutions to $E_1$ and $E_2$. So you have to assume that $p + q \ne 1$ in order to get two linearly independent equations.
Jan
13
revised Cauchy problem for an overdetermined system of PDE
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Jan
13
comment Cauchy problem for an overdetermined system of PDE
Yes, this case is handled in section 1.7 of "Involutive Hyperbolic Differential Systems" (Memoirs of the AMS, #370). In the generic case, the so-called reduced Cartan characters are, I believe, $3, 2, 2$. Theorem (1.26) gives an invariant way to identify whether the procedure yields a hyperbolic system of PDE's or not. Alas, as proud as I am of this paper, it's rather difficult to read. My advice is to play around with the calculations described above. I've appended some additional information.
Jan
13
comment Cauchy problem for an overdetermined system of PDE
I may have misspoken. I'm not sure an invariant way has been worked out. If it has, it would be in the reference cited by Ben McKay above (yes, I wrote it but I don't have a copy handy and I don't remember whether I dealt with this case). I'll look at it when I get a chance.