Bendixson–Dulac theorem

Template:Multiple issues In mathematics, the Bendixson–Dulac theorem on dynamical systems states that if there exists a $C^1$ function $\phi (x,y)$ (called the Dulac function) such that the expression

File:Dulac.svg

According to Dulac theorem any 2D autonomous system with a periodic orbit has a region with positive and a region with negative divergence inside such orbit. Here represented by red and green regions respectively

$\frac{\partial (\phi f)}{\partial x}+\frac{\partial (\phi g)}{\partial y}$

has the same sign ($\ne 0$) almost everywhere in a simply connected region of the plane, then the plane autonomous system

$\frac{dx}{dt}=f(x,y),$

$\frac{dy}{dt}=g(x,y)$

has no nonconstant periodic solutions lying entirely within the region. "Almost everywhere" means everywhere except possibly in a set of measure 0, such as a point or line.

The theorem was first established by Swedish mathematician Ivar Bendixson in 1901 and further refined by French mathematician Henri Dulac in 1923 using Green's theorem.

Proof

Without loss of generality, let there exist a function $\phi (x,y)$ such that

$\frac{\partial (\phi f)}{\partial x}+\frac{\partial (\phi g)}{\partial y}>0$

in simply connected region $R$. Let $C$ be a closed trajectory of the plane autonomous system in $R$, meaning $f(t+T,x,y)=f(t,x,y)$ and $g(t+T,x,y)=g(t,x,y)$ for all $t\ge 0$ and some $T>0$ on the curve $C$. Let $D$ be the interior of $C$. Then by Green's theorem,

$\begin{array}{rl}& \underset{D}{\iint }(\frac{\partial (\phi f)}{\partial x}+\frac{\partial (\phi g)}{\partial y})dxdy=\underset{D}{\iint }(\frac{\partial (\phi \dot{x})}{\partial x}+\frac{\partial (\phi \dot{y})}{\partial y})dxdy\\ =& {\oint }_C\phi (-\dot{y}dx+\dot{x}dy)={\oint }_C\phi (-\dot{y}\dot{x}+\dot{x}\dot{y})dt=0\end{array}$

with $\dot{x}=dx/dt$ and $\dot{y}=dy/dt$ being Newton's notation and the first equality following from the plane autonomous system. Because of the constant sign, the left-hand integral in the previous line must evaluate to a positive number. But on $C$, $dx=\dot{x}dt$ and $dy=\dot{y}dt$, so the bottom integrand is in fact 0 everywhere and for this reason the right-hand integral evaluates to 0. This is a contradiction, so there can be no such closed trajectory $C$.

See also

• Template:Slink
• Liouville's theorem (Hamiltonian), similar theorem with $\frac{dq}{dt}=\frac{\partial H(q,p)}{\partial p}(=f(q,p)),\frac{dp}{dt}=-\frac{\partial H(q,p)}{\partial q}(=g(q,p))$

References

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