# 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?

• Energy is a complete integral of motion, but it does not mean that it's easy to compute the canonical transformation into action-angle variables. See: Liouville integrability. – Ryan Thorngren Oct 5 '18 at 14:36
• Thank You @RyanThorngren, however isn't there a qualitative difference between "integrability by quadratures" (i.e. Liouville integrability) and nonintegrability due to impossibility to find an integrating factor in a planar system (e.g. consider a dissipative one)? – Lo Scrondo Oct 6 '18 at 13:16

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.

• Marvelous answer! In so much textbooks I've looked into, the generic planar differential system is shown as an example of NONintegrable system! – Lo Scrondo Oct 5 '18 at 8:10
• Just to be sure @Qmechanic, the local integrability stands also in the case $A$ and $B$ in my example are nonlinear? – Lo Scrondo Oct 5 '18 at 14:22
• $\uparrow$ Yes. – Qmechanic Oct 5 '18 at 14:26