
#1
Jan1813, 09:58 AM

P: 24

The title is selfexplanatory. What is it meant in the physics and maths community by the words integrability and integrable system?




#2
Jan1813, 10:18 AM

P: 5,634

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?? 



#3
Jan1813, 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" ?




#4
Jan1813, 12:39 PM

P: 5,634

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 



#5
Jan1813, 01:24 PM

Sci Advisor
Thanks
P: 3,856

∂f/∂x = F(x,y) ∂f/∂y = G(x,y) does f(x,y) exist? Answer, only if an integrability condition is satisfied: ∂^{2}f/∂x∂y = ∂^{2}f/∂y∂x, that is, ∂F/∂y = ∂G/∂x. 



#6
Jan1913, 02:35 AM

P: 983

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 nonintegrable eqn.they look like
Ʃa_{i}dq_{i} +a_{t}dt=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. 



#7
Jan1913, 04:10 AM

Sci Advisor
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P: 2,147

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 \pi0^+)=\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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