The Bendixson-Dulac Criterion

The Bendixson-Dulac criterion provides a sufficient condition for the non-existence of periodic orbits, complementing the Poincare-Bendixson existence theorem.

Statement

Theorem6.6Bendixson-Dulac theorem

Let $D \subseteq \mathbb{R}^2$ be a simply connected domain and $\mathbf{F} = (f, g) \in C^1(D)$ . If there exists a $C^1$ function $B: D \to \mathbb{R}$ (called a Dulac function) such that

$\frac{\partial(Bf)}{\partial x} + \frac{\partial(Bg)}{\partial y}$

does not change sign in $D$ and is not identically zero on any open subset, then the system $x' = f(x,y)$ , $y' = g(x,y)$ has no periodic orbits entirely contained in $D$ .

Proof

Suppose for contradiction that $\gamma$ is a periodic orbit in $D$ . Let $R$ be the region enclosed by $\gamma$ (which exists since $D$ is simply connected). By Green's theorem:

$\iint_R \left[\frac{\partial(Bf)}{\partial x} + \frac{\partial(Bg)}{\partial y}\right]dA = \oint_\gamma (Bf\,dy - Bg\,dx).$

On $\gamma$ : $dx = f\,dt$ and $dy = g\,dt$ , so

$\oint_\gamma (Bf\,dy - Bg\,dx) = \oint_\gamma B(fg - gf)\,dt = 0.$

But the double integral is nonzero (since the integrand has one sign and is not identically zero). Contradiction. $\blacksquare$

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Examples

ExampleSimple application

For $x' = -y + x(x^2+y^2-1)$ , $y' = x + y(x^2+y^2-1)$ : take $B = 1$ .

$\frac{\partial f}{\partial x} + \frac{\partial g}{\partial y} = (3x^2+y^2-1) + (x^2+3y^2-1) = 4(x^2+y^2) - 2.$

This changes sign (positive for $x^2+y^2 > 1/2$ , negative for $x^2+y^2 < 1/2$ ), so $B = 1$ does not work directly. A different Dulac function is needed.

However, in polar coordinates: $r' = r(r^2-1)$ has a limit cycle at $r = 1$ , so the criterion correctly cannot rule it out.

ExampleChoosing a Dulac function

For $x' = y$ , $y' = -x - y + x^2$ : try $B(x,y) = e^{2y}$ .

$\frac{\partial(Be^{2y}\cdot y)}{\partial x} + \frac{\partial(Be^{2y}(-x-y+x^2))}{\partial y} = 0 + e^{2y}[2(-x-y+x^2) + (-1)] = e^{2y}(2x^2-2x-2y-1).$

For the region where $2x^2-2x-2y-1 < 0$ , no periodic orbits exist.

RemarkGradient and Hamiltonian systems

Gradient systems $\mathbf{x}' = -\nabla V(\mathbf{x})$ never have periodic orbits (along any trajectory, $\dot{V} = -|\nabla V|^2 \leq 0$ , so $V$ decreases and cannot return to its initial value). Hamiltonian systems $x' = \partial H/\partial y$ , $y' = -\partial H/\partial x$ satisfy $\text{div}(\mathbf{F}) = 0$ (Bendixson with $B=1$ ), but this does not rule out periodic orbits — instead, Hamiltonian systems typically have families of periodic orbits.