Existence of Hamiltonian for Planar Ordinary Differential Equations

Question

Consider the following planar ODE $$ \begin{cases} \dot x = f(x,y) \\ \dot y = g(x,y) \end{cases} $$ and suppose $ \frac{\partial f }{\partial x} + \frac{\partial g }{\partial y} = 0$. Is this a sufficient and necessary condition for the existence of a Hamiltonian (or any first integral) for the system?

Answer 1

Say that we are on a rectangle (for simplicity) and that $f$ and $g$ are $C^1$.

Integrating $f=\partial H/\partial y$ with respect to $y$ gives $$ \int_{y_0}^y f(x,s) \, d s =H(x,y)-H(x,y_0). $$ Similarly, integrating $g=-\partial H/\partial x$ with respect to $x$ for $y=y_0$ gives $$ \int_{x_0}^x g(s,y_0) \, d s =-H(x,y_0)+H(x_0,y_0). $$ Subtracting the two yields the identity $$ H(x,y)=H(x_0,y_0)+\int_{y_0}^y f(x,s) \, d s-\int_{x_0}^x g(s,y_0) \, d s. $$

So we are led to consider $$ H(x,y)=\int_{y_0}^y f(x,s) \, d s-\int_{x_0}^x g(s,y_0) \, d s $$ (the constant doesn't matter).

Note that indeed $\partial H/\partial y=f$ and $$ \begin{split} \frac{\partial H}{\partial x}(x,y) &=\int_{y_0}^y \frac{\partial f}{\partial x}(x,s) \, d s-g(x,y_0)\\ &=-\int_{y_0}^y \frac{\partial g}{\partial y}(x,s) \, d s-g(x,y_0)\\ &=-g(x,y). \end{split} $$

The other direction is simpler (since $f$ and $g$ are $C^1$, the Hamiltonian $H$ is $C^2$).