# Number of limit cycles: Counterexample of the extended Bendixson-Dulac criterion?

The problem concerns the number of limit cycles in the vector field of coupled differential equations (ODEs) in two dimensions, i.e. $$\ \dot{x} = X(x,y)\\ \dot{y} = Y(x,y)$$ Specifically, let $$\ X = \frac{2}{15}\left(-99-\frac{100(-66.3357 + x)}{(1 + 0.00491694 e^{-0.136986 x})^3\left(1+\frac{ 4684.43e^{0.04 x}}{\left(-1 + \frac{1}{y} \right)^{2.336}}\right)} - 2 x - 150 y^2 (85 + x)\right) \\ \\ Y = 0.00653522 + y (0.0118225 + 0.000846776 x) + 0.0000821902 x$$ The functions $X$ and $Y$ reflect a reduced Hodgkin-Huxley model of neuronal firing. I have considered a region $D$ in the plane such that -70$<x<$20, 0.04$<y<$0.16 and $M(x,y)>$0. A function $f:D\rightarrow R$ that I believe is a Dulac function candidate in $D$ is $$\ f(x,y) =\frac{\log ((x+71)y)}{X(x,y)}$$ Cumputing
$$\ M(x,y)=\frac{\partial }{\partial x}(f X)+\frac{\partial }{\partial y}(f Y) > 0$$ yields that $D$ is a multiple connected region, however there are two limit cycles in the plane and only one "hole" (see figure below, one unstable limit cycle and one stable, not fully shown). By this paper, one bounded complement should limit the number of limit cycles to one. Or is the region outside $D$ also a complement region that should be counted (obviously it is not in $D$, although not encircled by the limit cycles)? The proof for this theorem in the linked paper only concerns complement regions contained by limit cycles..

Well, I might also be missing something fundamental.

I realized that the paper I refer to requires subscription. Here is the theorem and the proof of it (N.G. Lloyd (1979) A note of the number of limit functions in certain two-dimensional systems, J. London Mat. Soc. (2), 20:277-286). It is also given as Theorem 2.1 in this freely available paper.

THEOREM. Suppose that $D$ is an open, connected subset of the plane and that $X$ and $Y$ have continuous first order partial derivatives in D. Suppose further that there is a continuously differentiable function $f:D\rightarrow R$ such that

$$\ \frac{\partial }{\partial x}(f X)+\frac{\partial }{\partial y}(f Y) \neq 0$$

in D. If $\mathcal{c}D$ (the complement of $D$) has $k$ bounded components, then the [system] has at most $k$ limit cycles entirely contained in $D$.

Proof. Let the bounded components of $\mathcal{c}D$ be $D_1,\ldots ,D_k$; the argument does infact apply when $k=0$ (which is, of course, the configuration of Bendixson's theorem). We show that if F is a limit cycle entirely contained in $D$, then the interior domain,$\Gamma_{\circ}$ say, of $\Gamma_{\circ}$ contains at least one $D_i$ which is adjacent to $\Gamma_{\circ}$ in the sense that $D_i$ is encircled by no other limit cycle of [the system] which is contained in $\Gamma_{\circ}$. Having proved this we can immediately conclude that there are at least as many $D_i$, as there are limit cycles entirely contained in $D$; hence there are at most $k$ limit cycles entirely contained in $D$.

The proof is by contradiction. Let $\Gamma$ be a closed orbit such that $D_1,\ldots,D_r$, say, are contained in $\Gamma_{\circ}$. Suppose, if possible, that every $D_i$ is encircled by a closed orbit. Let $\Gamma_{1},\ldots,\Gamma_{s}$, be closed orbits entirely contained in $\Gamma_{\circ}$ which are so chosen that together they encircle $D_1,\ldots,D_r$, in the sense that $$D_{i} \subset \bigcup_{j=1}^{s}\Gamma_{j}^{\circ} \;\;\; (j=1,\ldots,r)$$ (wher $\Gamma_{j}^{\circ}$ is the interior domain of $\Gamma_{j}$), and furthermore are such that the $\Gamma_{j}^{\circ}$ are mutually disjoint. Then $s \leq r$, and the set $$R=\Gamma^{\circ} \backslash \bigcup_{i=1}^{s} (\Gamma_{i}^{\circ} \cup \Gamma_{i})$$ is an open, connected subset of $D$. We now apply Green's theorem to $R$. We have $$\int \int_R \left[\frac{\partial }{\partial x}(f X)+\frac{\partial }{\partial y}(f Y)\right]=\int_\Gamma f(-Ydx+Xdy)-\sum_{i=1}^{s}\int_{\Gamma_i}f(-Ydx+Xdy).$$ Since $\Gamma$ and $\Gamma_{i}(i=1,\ldots,s)$ are orbits of the system, every term on the right hand side is zero. But by the [inequality above] the left hand side is non-zero. This is the contradiction we sought.

• The papers you mentioned requires subscription for viewing. – user137035 Aug 18 '14 at 7:19
• Thanks for pointing that out. Theorem added. – Hugo Aug 18 '14 at 13:17
• The pictures, as shown, contradict the divergence theorem rather directly (without need for referencing the paper): Let $\Omega$ be the doubly connected region between the periodic orbits; then $[fX,fY]$ is a vector field with positive divergence in $\Omega$ but with zero flux across the boundary (since it's tangent to $\partial \Omega$). So... there must be a mistake somewhere. For one thing, are you sure you can divide by $X(x,y)$? This function turns into zero in places, namely where the vector field is vertical. – user147263 Aug 19 '14 at 23:54
• Yes, there is an N-shaped nullcline of $X$ passing through $D$, so there is a division with zero along this line and consequently $M$ vanish to infinity. However, the sign is still positive.. I assume that this might relate to the notion of "almost everywhere" (i.e. $M$ vanishing only on a set of zero Lebesgue measure). Given that the nullcline is a line the Lebesgue measure is one, I assume. This might be the problem, I agree, although I'm not certain why. – Hugo Aug 20 '14 at 8:02

I believe I have found my mistake (thanks to the comment above). The division with $X(x,y)$ violate the assumption that $f$ is a $C^1$ function in $D$ since the derivative is not continuous, approaching infinity from two directions around the nullcline of $X(x,y)$. Therefore one of the assumption of the Bendixon-Dulac theorem does not hold and a conclusion regarding the number of limit cycles cannot be made.