classical mechanics - Is *every* planar/2D system integrable?

Consider the generic following planar/2D system:

$$\begin{cases} \frac{dx}{dt} = A(x,y)\\ \\ \frac{dy}{dt} = B(x,y), \end{cases}$$

where $A,B$ are two functions. Reading Classical Mechanics by Joseph L. McCauley I found the following statements:

Every two dimensional flow, $$dx/dt = A(x,y), \qquad dy/dt = B(x,y),$$ whether dissipative or conservative, has a conservation law,

and, if we rewrite the system equations as $dt=dx/A=dy/B$,

every differential form $B(x,y)dx-A(x,y)dy=0$ in two variables either is closed or else has an integrating factor $M(x,y)$ that makes it integrable.

So is really every planar system integrable, or have I missed some detail?

Answer

A global integrability statement for general 2D systems does not hold, but a local integrability statement is true. Let us reformulate OP question as follows.

Suppose that we are given a two-dimensional first-order problem $$ \dot{x}~=~f(x,y), \qquad \dot{y}~=~g(x,y), \tag{1}$$ where $f$ and $g$ are two given smooth functions. Is eq. (1) a Hamiltonian system $$ \dot{x}~=~\{x,H\}, \qquad \dot{y}~=~\{y,H\}, \tag{2}$$ with a symplectic structure $\{\cdot,\cdot\}$ and Hamiltonian $H(x,y)$?

The answer is, perhaps surprisingly: Yes, always, at least locally. The Hamiltonian $H$ is the sought-for integral of motion/first integral.

1. 
   

   Proof: In two dimensions, a Poisson bracket is completely specified by the fundamental Poisson bracket relations $$ \{x,y\} ~=~B(x,y)~=~-\{y,x\}, \qquad \{x,x\}~=~0~=~\{y,y\}, \tag{3} $$ where $B$ is some function that doesn't take the value zero. [Exercise: Check that eqs. (3) automatically satisfy the Jacobi identity.] The Hamilton's eqs. (2) become $$ \dot{x}~=~B\frac{\partial H}{\partial y}, \qquad \dot{y}~=~-B\frac{\partial H}{\partial x}.\tag{4} $$ Next consider the one-form $$ \eta ~:=~ f{\rm d}y -g{\rm d}x, \tag{5}$$ which is possibly an inexact differential. However, it is known from the theory of PDE's, that there locally exists an integrating factor $\frac{1}{B}$, so that the one-form $$ \frac{1}{B}\eta~=~{\rm d}H \tag{6} $$ is locally an exact differential given by some function $H$. It is straightforward to check that one can use $B$ as the Poisson structure (3) and $H$ as the Hamiltonian. $\Box$

   
   


2. 
   

   Remark. The existence of a pair of canonical variables $q(x,y)$ and $p(x,y)$, with $\{q,p\}=1$, are, in turn, guaranteed locally by Darboux' Theorem.

   
   


3. 
   

   Example: 1D system with friction force: Eqs. of motion: $$m\dot{v}~=~-kv-V^{\prime}(x), \qquad \dot{x}~=~v. \tag{7} $$ According to the theorem there in principle exists locally a Hamiltonian formulation without explicit time dependence, however it is not possible to give a general formula. If we are allowed to have explicit time dependence (which is outside the main topic of this answer), then there is a simple solution: Define $e(t):=\exp(\frac{kt}{m})$. Lagrangian: $L=e(t)L_0$, where $L_0=\frac{m}{2}v^2-V(x)$. Momentum: $p=e(t)mv$. Hamiltonian: $H=\frac{p^2}{2me(t)}+e(t)V(x)$.

   
   
   


4. 
   

   Example: Math.SE q1577274.

   
   


5. 
   

   Counterexample. Solutions to diff. eqs. exist in general only locally. Consider $$f(q,p)~=~\frac{q}{q^2+p^2}\quad\text{and}\quad g(q,p)~=~\frac{p}{q^2+p^2}\tag{8}$$ in the domain $D=\mathbb{R}^2\backslash\{(0,0)\},$ which is not contractible. It is relatively straightforward to check that $$\eta~=~f\mathrm{d}p-g\mathrm{d}q ~=~\frac{q\mathrm{d}p-p\mathrm{d}q}{q^2+p^2} \tag{9}$$ is a closed $1$-form, and there doesn't exist a globally defined Hamiltonian $H$ on $D$ such that eqs. (2) are satisfied. The best one can do is to put $H$ equal to a single-valued branch of ${\rm arg}(q+ip)$, which is not globally defined.

   
   


6. 
   

   Counterexample: Contractible domain without global solution: Math.SE q2710698.