26

• Concerning question 2, you might want to take a look at Simulation of cubical particle packing under mechanical vibration (2016). The precise effect mentioned in the 2017 paper is not considered in that study, but a variety of ordering mechanisms are examined.
• Concerning question 1, the complexity of this mechanical problem (in the sense of many ...

answered Dec 16 '17 at 16:42

Carlo Beenakker

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18

Stable sheaves are simple, i.e., $\textrm{End}E\simeq \mathbb{C}$. One thing that you want to avoid is the jumping of the automorphism group in a family.
A classical example is to consider a hyperelliptic curve $X$, and $[L]\in\textrm{Pic}^{g-1}X$.
If $\pi:X\to \mathbb{P}^1$ is the $g^1_2$, then Grothendieck-Riemann-Roch plus Riemann-Hurwitz tell you that
$...

18

I doubt that a mathematically rigorous explanation of the phenomenon discovered in that paper exists using today's technology. While mathematical statistical mechanics is a well-developed field of study, there is still much to be done before the applications of statistical mechanics by physicists can be made rigorous.
Any explanation should include a ...

13

The physical reason is that the cubically packed state has lower gravitational potential energy than the jammed random state. The overall process is analogous to annealing, although the reduction in energy in the latter case is to do with atomic bonding, not gravity.
For the process to work, it is necessary for the jostling to be enough to ease the elements ...

12

Now that your comment has clarified your question, we can answer it: The answer is 'no'. There is the following well-known example:
Consider the following family of circles: $C_\lambda$ is defined as $x^2+y^2 = 1$ and $z = \lambda$. Let $\lambda>0$ be fixed and orient $C_{-\lambda}$ counterclockwise and orient $C_\lambda$ clockwise. Then for $\...

answered Nov 21 '14 at 11:09

Robert Bryant

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10

To make GIT work is not a good reason. There are constructions of moduli spaces that do not use GIT (mainly, because in certain situations GIT does not work). Boundedness as Donu notes is a better reason. In other words it's not that unstable (in itself) is "bad", but that (semi-)stable is "good". Yet in other words, one "bad" property of unstable sheaves is ...

answered Oct 10 '12 at 15:18

Sándor Kovács

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8

Probably the main reason people avoid singular varieties is because of boundedness. Arguments using induction and, as you say, spectral sequences need boundedness of the complexes in play in one direction or the other (and sometimes both).
Related to this, is the failure of Serre duality, which lies at the heart of many derived arguments. (the fact that the ...

7

One possible answer could be: by opening any book on vector bundles, and looking at the proposition right after the definition of stable vector bundle. :)
Another possible answer is as follows.
What you ask is valid in much more generality on any compact Kähler manifold.
Proposition. Let $\mathcal F$ and $\mathcal G$ two torsion free coherent sheaves over ...

6

If $p=2$ and $\sigma_1, \sigma_2 \in L^\infty(\Omega)$ one has the estimate
$$\|\nabla u_1 - \nabla u_2\|_{L^2(\Omega)} \leq C \|f\|_{H^{1/2}(\partial\Omega)} \|\sigma_1 -\sigma_2\|_{L^\infty(\Omega)}$$ where $C = C(\Omega,c)$.
To see this, write $0 = (u_1-u_2) \operatorname{div}(\sigma_1 \nabla u_1 - \sigma_2 \nabla u_2)$ and integrate by parts, using the ...

6

In general, reductiveness of the lie algebra of holomorphic vector fields is an obstruction to the existence of cscK metrics (constant scalar curvature Kähler), not just Kähler-Einstein metrics. In particular, the blow up of $\mathbb{P}^2$ in $1$ or $2$ points cannot admit a cscK metric in any Kähler class. I'm not sure of a good reference for this, I can't ...

6

Solutions with $|x|=1$ are given by $x=e^{i\theta}$ and $\alpha=2\sin(\theta/2)$ with $\theta=\pi/(2\tau+1)$. From this we get $1/\alpha=(2\tau+1)/\pi + O(1/\tau)$.

answered May 9 '19 at 13:45

Neil Strickland

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5

A stable theory is dp-minimal if and only if all 1-types have weight 1. (See "On dp-minimality, strong dependence and weight" by A. Onshuus and A. Usvyatsov.)
A differentially closed field has 1-types of arbitrary large finite weight, hence neither $DCF_0$ nor $DCF_p$ are dp-minimal. Similarly, the generic type of the free group has infinite weight---see ...

5

First, in the standard definition $D = K_X$, so I will give an example in this case. Let $X$ be a curve of genus 2 and $E = O \oplus O(P)$ for a point $P \in X$. Clearly $E$ is unstable with $O(P)$ being the only destabilizing subbundle. Define $\phi$ to be the composition
$$
O \oplus O(P) \to O(P) \to O(K_X) \to O(K_X) \oplus O(K_X + P),
$$
where the first ...

5

A longish bunch of remarks: there's a big leap that doesn't make sense between your question and your motivation.
On the maximally extended Schwawrzschild solution there is no decay to the wave equations: the space-time "ends" in finite proper time at the singularity, and there's not enough time (compare to the not-enough-space scenario you ...

5

I suggest looking at Lyapunov's direct method and possibly LaSalle's principle and their nonsmooth extensions.
http://www4.ncsu.edu/~schecter/ma_532_fa12/lasalle.pdf
http://ieeexplore.ieee.org/abstract/document/317122/

reference-request ca.classical-analysis-and-odes ds.dynamical-systems differential-equations stability

5

A simple counterexample can be constructed as follows: take
$$ A(t) = \pmatrix{0&\tfrac{\pi}{2}\\-\tfrac{\pi}{2}&0} \qquad \text{for } t \in [2n, 2n+1) ,$$
and
$$ A(t) = \pmatrix{-\alpha&0\\0&-\beta} \qquad \text{for } t \in [2n+1, 2n+2) ,$$
where $0 < \alpha \leqslant \beta$.
The equation $x'(t) = c A(t) x(t)$ describes then the ...

answered Sep 11 '18 at 19:53

Mateusz Kwaśnicki

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5

This is more a long comment than an answer, but I know that a similar problem, also originated from (macromolecular) biology, was studied and solved by Gaetano Fichera, Maria Adelaide Sneider and Jeffreys Wyman. The system of ODEs they studied was the following one
$$
\left\{
\begin{split}
x_1' &= Lc_1^2 - (\rho_1 + 2Lc_1 + Qc_2)x_1 + (P - 2Lc_1)x_2 — ...

5

Rather incredibly, your (corrected) system does have a closed-form solution, which I found with Maple's help.
$$ x(t) = 1+4\,\cos \left( 2\,t \right) +3\,\sqrt {8\,\cos \left( 2\,t \right) +
17}$$
$y(t)$ is a rather complicated beast that I'll just write in text rather than LaTeX: when pretty-printed, it still doesn't look pretty.
-16384/(5-(16*cos(t)^2+9)^(...

4

The standard explanation is that if we want to work on the level of the coarse moduli space and not on the level of the stack itself, we need to tame the automorphism groups of objects (as Peter pointed out) and have some control over the maps between objects in your moduli. Semi-stable objects, Harder-Narasimhan and Jordan-Holder filtrations are exactly the ...

4

First, since the sets of equilibrium points $A = \{ 0 \} \times \mathbb{R}_+^3$ and $B = \mathbb{R}_+ \times \{ (0,0,0) \}$ are three and one-dimensional respectively, you should expect at least 3 and 1 eigenvalues zero for the Jacobian at points in these sets. If the other eigenvalues have negative real part for all points in the set ($A$ or $B$), then the ...

4

It seems relatively easy to construct such a bundle by induction. Namely:
Step 0 If $r=1$, there are clearly line bundles of given degree $d$ on $X$, since $k$ is assumed to be algebraically closed.
Step 1 Choose $(r',d')$ such that $0<r'<r$, $d'/r'>d/r$, and there are no integral points within (or on the edges) of the triangle with vertices $(0,0)...

4

This should be seen as a comment, but it is too long for a comment:
One aspect, that has not been addressed in the answers, is the influence of the container geometry. In the experiment, a cylindrical container has been subjected to two different kinds of motions: a coaxial rotation with alternating directions and tapping to the side of side of the ...

4

Given a realization of the Ornstein-Uhlenbeck process $X_t$, the SDE $$
d Y_t = Y_t (1- Y_t) X_t (dt + d V_t) \tag{1}
$$ is scalar, nonautonomous, and nonlinear. Note that (1) has two fixed points at $0$ and $1$, which are asymptotically stable in the following sense.
Theorem. For almost all $Y_0 \in (0,1)$, we have $\lim_{t \to \infty} Y_t \in \{0, 1\}$ ...

3

I believe it's the case that if you have $X \subseteq \mathbb P^n$ with an action of $G$ on $X$ and $\mathbb P^n$ you can understand the difference between stable, semistable, and unstable using the affine cone on $X$ in $\mathbb A^{n+1}$.
Semistable ones are the points such that, for any nonzero point on the corresponding line in the affine clone, its ...

ag.algebraic-geometry rt.representation-theory sg.symplectic-geometry geometric-invariant-theory stability

3

Proposition: Let $X$ be a smooth projective surface for which $H^{0}(\omega_{X})=0.$ Then there exists a vector bundle $E$ on $X$ satisfying the property that for any ample divisor $H$ on $X,$ $E$ is slope-semistable with respect to $H$ but not Gieseker-semistable with respect to $H.$
Proof: Let $x \in X$ be a closed point, and consider the exact ...

3

Another approach would be to expand in series form around some equilibrium point, and observe whether the lower order terms resemble some "known" behaviour, which is known as a normal form on the center manifold: https://en.wikipedia.org/wiki/Center_manifold
First of all, to simplify the system, observe $E-Z S=c$, where the constant $c$ could be obtained ...

3

I will assume that "strict negativity of the real parts" means that there is a constant $c>0$ such that all eigenvalues of all matrices $A(t)$ have a real part less than $-c$.
(i) if $A(t)$ converges to a limit which is Hurwitz, i.e. for which all eigenvalues have negative real part, then you can use a quadratic Lyapunov function for that limit, which ...

3

Stability
For all $k$ up to the total number of blocks $n$: the center of mass of the top $k$ blocks must lie above the surface of the $k+1$ block supporting them.

3

I do not know if this is what you are after. The following shows that the Jordan form can be replaced by the Schur form. This feels a bit like cheating. Is this enough? The trickier parts are the statements about the real parts of the
eigenvalues. It is of course a bit harder to say useful things about
eigenvalues when we are not allowed to speak about ...

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