# Integrability (meaning)

1. Jan 18, 2013

### anthony2005

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

2. Jan 18, 2013

### Naty1

Have you seen an explanation like this:

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

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??

Last edited by a moderator: May 6, 2017
3. Jan 18, 2013

### anthony2005

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. Jan 18, 2013

### Naty1

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. Jan 18, 2013

### Bill_K

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.

Last edited: Jan 18, 2013
6. Jan 19, 2013

### andrien

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.

7. Jan 19, 2013

### vanhees71

That's only half of the truth! Your integrability conditions are sufficient only for simply connected regions in the $(x,y)$ plane, where $F$ and $G$ are free of singularities and smoothly differentiable.

A simple but eluminating example is the potential curl
$$\vec{F}(\vec{x})=\frac{-y \vec{e}_x+x \vec{e}_y}{r^2}.$$
It's everywhere curl free, except in the origin, i.e.,
$$\partial_x F_y-\partial_y F_x=0,$$
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 $2 \pi$.

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

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