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Integrability (meaning)

by anthony2005
Tags: integrability, meaning
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anthony2005
#1
Jan18-13, 09:58 AM
P: 24
The title is self-explanatory. What is it meant in the physics and maths community by the words integrability and integrable system?
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Naty1
#2
Jan18-13, 10:18 AM
P: 5,632
Have you seen an explanation like this:

http://en.wikipedia.org/wiki/Integration_(mathematics)

where they first discuss integrating a smooth function.....

or is this what really interests you:

http://en.wikipedia.org/wiki/Integrable_system

like maybe one of the systems listed at the end of the article??
anthony2005
#3
Jan18-13, 11:23 AM
P: 24
So, is it correct to state in general that: "an integrable system is a system which thanks to certain properties its dynamics is exactly solvable" ?

Naty1
#4
Jan18-13, 12:39 PM
P: 5,632
Integrability (meaning)

You should wait for someone who is more up to date on math and current terminology than I....but I'll give you my 2 cents:

first, you posted this under Quantum Physics,so if you are looking for a specific answer, check here in the Wikipedia article:

Quantum integrable systems

that seems different from you latest post.

second, You may have to define what 'solvable' means to you because the section in Wikipedia says this:

General dynamical systems

...The distinction between integrable and nonintegrable dynamical systems thus has the qualitative implication of regular motion vs. chaotic motion and hence is an intrinsic property, not just a matter of whether a system can be explicitly integrated in exact form.
and under the 'Chaos' link

... the deterministic nature of these systems does not make them predictable.[
Bill_K
#5
Jan18-13, 01:24 PM
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So, is it correct to state in general that: "an integrable system is a system which thanks to certain properties its dynamics is exactly solvable" ?
No, integrability means: can a given relationship between derivatives be integrated to yield a relationship between functions. For example, given the system

∂f/∂x = F(x,y)
∂f/∂y = G(x,y)

does f(x,y) exist? Answer, only if an integrability condition is satisfied: ∂2f/∂x∂y = ∂2f/∂y∂x,

that is, ∂F/∂y = ∂G/∂x.
andrien
#6
Jan19-13, 02:35 AM
P: 1,020
it is more suited with classical section,integrability of system is classified according to it's holonomicity.In classical dynamics a system which is non holonomic has at least one non-integrable eqn.they look like
Ʃaidqi +atdt=0
this eqn should not be a total differential(or can be converted).there are many examples of it.One simple and particular is rolling of a sphere on a rough surface.Point of contact satisfy a non integrable relation.
vanhees71
#7
Jan19-13, 04:10 AM
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P: 2,458
Quote Quote by Bill_K View Post
No, integrability means: can a given relationship between derivatives be integrated to yield a relationship between functions. For example, given the system

∂f/∂x = F(x,y)
∂f/∂y = G(x,y)

does f(x,y) exist? Answer, only if an integrability condition is satisfied: ∂2f/∂x∂y = ∂2f/∂y∂x,

that is, ∂F/∂y = ∂G/∂x.
That's only half of the truth! Your integrability conditions are sufficient only for simply connected regions in the [itex](x,y)[/itex] plane, where [itex]F[/itex] and [itex]G[/itex] are free of singularities and smoothly differentiable.

A simple but eluminating example is the potential curl
[tex]\vec{F}(\vec{x})=\frac{-y \vec{e}_x+x \vec{e}_y}{r^2}.[/tex]
It's everywhere curl free, except in the origin, i.e.,
[tex]\partial_x F_y-\partial_y F_x=0,[/tex]
but it does not have a unique potential in every region in the plane that contains the origin, where the singularity sits.

Indeed, integrating the vector field along any circle around the origin gives [itex]2 \pi[/itex].

To make the potential unique, one has to cut the plane by a ray starting from the origin. A standard choice is the negative [itex]x[/itex]-axis. I.e., you take out the points [itex](x,0)[/itex] with [itex]x \leq 0[/itex].

It's most easy to find the corresponding potential by introducing polar coordinates. Here, we use
[tex](x,y)=r (\cos \varphi,\sin \varphi)[/tex]
with [tex]\varphi \in (-\pi,\pi),[/tex]
which automatically excludes the negative x axis. The function [itex]\vec{F}[/itex] then reads
[tex]\vec{F}=\frac{\vec{e}_{\varphi}}{r}.[/tex]
The potential thus can be a function of only [itex]\varphi[/itex], and the gradient reads
[tex]\vec{F} \stackrel{!}{=}-\vec{\nabla} V(\varphi)=-\frac{1}{r} V'(\varphi).[/tex]
This gives, up to a constant
[tex]V(\varphi)=-\varphi.[/tex]
The potential is indeed unique everywhere except along the negative [itex]x[/itex] axis, along which it has a jump
[tex]V(\varphi \rightarrow \pi-0^+)=-\pi, \quad V(\varphi \rightarrow -\pi + 0^+)=+\pi.[/tex]
In Cartesian Coordinates this potential is given by
[tex]V(\vec{x})=-\mathrm{sign} y \arccos \left (\frac{x}{\sqrt{x^2+y^2}} \right ).[/tex]


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