Example 4.1

Consider the system \begin{align*} \dot{x} &= x + y - 4x(x^2 +y^2) \\ \dot{y} &= -x + y - 4y(x^2 +y^2) \end{align*} The stationary Point \((0,0)\) is obvious. Are there any others?

From the first equation \[ 4x(x^2+y^2) = x+y \quad \text{ i.e. } 4xy(x^2+y^2) = xy + y^2 \] and from the the second equation \[ 4y(x^2+y^2) = -x+y \quad \text{ i.e. } 4xy(x^2+y^2) = -x^2 + xy \] Therefore, \[ 4xy(x^2+y^2) = xy + y^2 = -x^2 + xy, \] which implies that \(x^2+y^2=0\), or there is no solutions other than \((0,0).\)

Now consider \(V(x,y)=\frac{1}{2}x^2 + \frac{1}{2}y^2\) (can your think of any motivation for this choice?). Then \begin{align}\label{eq:itV} \dot{V} &= x \dot{x} + y \dot{y} \cr &= x(x+y-4x(x^2+y^2)) + y (-x + y -4y(x^2+y^2)\cr &= x^2 + y^2 -4(x^2+y^2)^2. \end{align} Define \[ \mathcal{D}=\left\{ (x,y) ~|~ \frac{1}{8} \le x^2 + y^2 \le 1 \right\}. \] Then \(\dot{V}>0\) on \( x^2 + y^2 =\frac{1}{8}\) (which is the set \(V(x,y)=\frac{1}{16}\)) i.e. solutions enter \(\mathcal{D}\) across this boundary; \(\dot{V}<0\) on \( x^2 + y^2=1\) (which is the set \(V(x,y)=\frac{1}{2}\)) i.e. solutions enter \(\mathcal{D}\) across this boundary. So solutions enter and do not leave \(\mathcal{D}\) which is closed, bounded and contains no stationary points, and hence there is at least one periodic orbit in \(\mathcal{D}\). (Note that any outer boundary bigger than \(\frac{1}{4}\) and lower boundary less than this will do.)

Example 4.2 Prove existence of a periodic orbit for \begin{align}\label{eq:diffpode} \dot{x} = y, \qquad \dot{y} = -x + y (1-3x^2 -6y^2) . \end{align} First consider stationary points: from \(\dot{x}=0\) we find \(y =0\) and then \(\dot{y} = 0\) implies \(x = 0\) so \((0,0)\) is a unique stationary point. Try \(V(x,y)=x^2 + y^2\), then \begin{align*} \dot{V} = 2x \dot{x} + 2 y \dot{y} = 2xy + 2y \left[ -x + y (1-3x^2 -6y^2)\right] = 2y^2 (1-3x^2 -6y^2). \end{align*} So if \(1-3x^2 -6y^2 <0\) (i.e. in the neighbourhood of the origin), \(\dot{V} \le 0\), and if \(1-3x^2 -6y^2>0\) (i.e. far away from the origin), \(\dot{V} \ge 0\). To apply the Poincar\'e-Bendixson Theorem, we have to understand how the geometry of these ellipses \(\{(x,y)\mid 1-3x^2-6y^2=0\}\) describing the sign of \(\dot V\) interacts with the geometry of the circles of constant \(V(x,y)=x^2+y^2\) that will be used to define \(\mathcal{D}\). Moreover, every such circle contains some points on the \(x\)-axis (with \(y=0\)), and \(\dot V=0\) on some parts of \(V=c\). As a result, the simple geometric arguments used above will not hold (we need \(\dot{V}\) strictly less or greater than zero. How does the \(\dot V\) equation relate to the circles \(V=c\)? First, \[\dot{V}= 2y^2(1-3x^2-6y^2) \le 2y^2 (1-3x^2-3y^2) \] So if \(x^2 + y^2 = V \ge \frac{1}{3}\), then \(\dot{V} \le 0 \). Let the outer boundary of \(\mathcal{D}\) be \(x^2 +y^2 = \frac{1}{3}\), call this \(\partial \mathcal{D}_+\). Then if \(p=(x_0,y_0)\) with \(|p| =\sqrt{x_0^2+y_0^2} \le \frac{1}{3}\) and \(V(p) \le \frac{1}{3}\), we claim \(V(\varphi_t(p))=|\varphi_t(p)|^2 \le \frac{1}{3}\) for all \(t \ge 0\). Here \(\varphi_t(p)\) is the solution to~\eqref{eq:diffpode} starting from \(p\) at time \(t=0\) (this form is used to emphasise the dependence of the initial condition on \(p\)). Suppose this claim is not true, then there exist two times \(t_0\) and \(t_1\), such that \[ V(\varphi_{t_0}(p)) = \frac{1}{3} \] and \[ V(\varphi_t(p)) > \frac{1}{3} \quad \text{ for~ all}\quad t \in (t_0, t_1].\] In particular \[ V(\varphi_{t_1}(p))- V(\varphi_{t_0}(p)) >0. \] Hence by the Mean Value Theorem, \[ V(\varphi_{t_1}(p, t_1))- V(\varphi_{t_0}(p))=\dot V(\varphi_t(p))(t_1-t_0) \quad \text{for~some}\quad t\in [t_0,t_1]. \] On the other hand, \[ \frac{dV(\varphi_t(p)}{dt} \leq 0 \text{ for all } t \in \left[ t_0, t_1 \right]. \] Therefore, \(0 < V(\varphi_{t_1}(p, t_1))- V(\varphi_{t_0}(p))=\dot{V}(\varphi_t(p))(t_1-t_0)\leq 0\), which is a contradiction. Now we consider the inner boundary. The fact \[ \dot{V}=2y^2 (1-3x^2 - 6y^2) \ge 2 y^2 (1-6x^2 - 6y^2)\] implies that \(\dot{V}\geq 0\) on the set \(\{(x,y)\mid x^2+y^2 \leq 1/6\}\), and motivates the choice of the inner boundary \[ \partial \mathcal{D}_{-}=\left\{ (x,y) \mid x^2 + y^2=\frac{1}{6}\right\}. \] We claim that if \(V(p) \ge \frac{1}{6}\) then \(V(\varphi_t(p)) \ge \frac{1}{6}\) for all \(t \ge 0\). The proof is almost exactly the same as the above case (try it!). So if we choose \(\mathcal{D}=\left\{ (x,y) \mid \frac{1}{6} \le V(x,y) \le \frac{1}{3} \right\}\), a solution that starts in \(\mathcal{D}\) stays in \(\mathcal{D}\). Moreover, \(\mathcal{D}\) contains no stationary points and hence by the Poincar\'e-Bendixson Theorem there is at least one periodic orbit in \(\mathcal{D}\).

Example 4.3 (Glycolysis oscillation) In this system of ODEs modelling the process how the human body converts glucose (sugar) into energy, \(x\) is the ADP concentration and \(y\) is the F6P (fructose-t-phosphate) concentration \((a>0)\): \[ \dot{x}= -x + ay +x^2 y, \qquad \dot{y}= \frac{1}{2} - ay -x^2 y. \] We want to show that there are oscillations (periodic orbits) if the positive parameter \(a\) is sufficiently small. Start by considering stationary points and their stability. \emph{Stationary Points}: Solving the system \[ x^2 y = x-ay,\qquad x^2 y = \frac{1}{2}-ay, \] we get \(x^*= \frac{1}{2}\) and \( y^* = \frac{1}{2(a+x^2)}= \frac{2}{4a+1}\). So \(\left(\frac{1}{2}, \frac{2}{4a+1}\right)\) is the only stationary point. \emph{Stability}: The Jacobian matrix is \[ J= \begin{pmatrix} -1 + 2xy & x^2 + a \\ -2xy & -(x^2+a) \end{pmatrix}.\] At the stationary point \(\left(\frac{1}{2}, \frac{2}{4a+1}\right)\), \[ J(x^*, y^*) = \begin{pmatrix} -1 + y^* & \frac{1}{2y^*} \\ -y^* & -\frac{1}{2y^*} \end{pmatrix}.\] Therefore, we get the determinant and the trace \[ \det J = \frac{1}{2y^*} >0,\quad \mbox{tr} J = -1+y^*-\frac{1}{2y^*}. \] Since the product of the roots (\({\rm det}J\)) is positive the stationary point is not a saddle, and so it is stable if \(\text{ Tr } J < 0\) and unstable if \(\text{ Tr } J> 0\). In other words, the stationary point is unstable if \(1+\frac{1}{2y^*} - y^* < 0\) or \[1+\frac{4a+1}{4} - \frac{2}{4a+1} < 0.\] This equation can be written as \[ 4(4a+1) + (4a+1)^2 - 8 <0 \qquad \mbox{or}\qquad 16a + 4 + 16a^2 + 8a + 1 - 8 <0 \] which is simplified as \[3 - 24a - 16a^2> 0 .\] Thus if \(a\) is sufficiently small, then the fixed point is unstable and there is a small closed curve containing the fixed point that solutions cross outwards.

Constructing a PB region (when \(a>0\) is sufficiently small): As noted earlier, if \(a\) is sufficiently small then the stationary point \((\frac{1}{2}, \frac{2}{1+4a})\) is unstable and so we can use the Lyapunov function constructed from adjoint eigenvectors in reverse time to construct an inner boundary that trajectories cross outwards. In what follows \(\epsilon >0\) is a small constant. The remainder of the Poincar\'e-Bendixson region will be constructed using four straight lines (see Figure 4.2): the \(x\)-axis, a vertical line near the \(y\)-axis, a horizontal line at larger \(y\) and a part of \(x+y=c\), with \(c\) chosen later. The first two are obvious. If \(y=0\) (the \(x\)-axis) then \(\dot y= 1/2>0\) and so trajectories cross the \(x\)-axis upwards into \(y>0\). If \(x=-\epsilon<0\) (near the \(y\)-axis) then \(\dot x=\epsilon +ay\) and so \(\dot x> 0\) if \(y>0\). Thus solutions cross the \(x\)-axis and a horizontal line near the \(y\)-axis into the positive quadrant formed by the two lines. (The use of \(\epsilon\) is so that we don't have to worry about the behaviour at the origin, where \(\dot x=0\) had we chosen the two coordinate axes.) To think about the remaining straight lines of the boundary, draw the two nullclines: \begin{align*} y = \frac{x}{a+x^2} \quad \mbox{ on which } \dot{x}=0,\\ y = \frac{1}{2(a+x^2)} \quad \mbox{ on which } \dot{y}=0. \end{align*} In other words, the direction field (or the vector field) is vertical on the line \(y=x/(a+x^2)\) and horizontal on \(y=1/2(a+x^2)\). Thus the nullcline \(y=1/2(a+x^2)\) for \(\dot{y}\) has a maximum at \(y=\frac{1}{2a}\); if \(y>\frac{1}{2a}\), then \(\dot y<0\) and trajectories have a component downwards. This suggests using the line \[ y=\frac{1}{2a}+\epsilon \] which all solutions cross downwards. But we can not choose a vertical line \(x=M\) for some large \(M\) to complete the closed region, because of the term \(x^2y\) ( \(\dot{x}\) may not be negative on this vertical line). However, we note that the sign for \(\dot{x}+\dot{y}=-x+1/2\) is much simpler (negative when \(x>1/2\)). This motivate the choice of straight lines of the form \(x+y=c\) for some \(c\). If we choose \(F(x,y)=x+y-c\) and define the region \(D=\{ (x,y)\mid F(x,y)<0\), then \(\dot{F}<0\) if \(x>1/2\). In other words, trajectories on the line \(x+y=c\) cross inwards to \(D\). The choice of \(c\) is determined by closing the outer boundary as simply as possible: choose it so that \(x+y=c\) intersects \(y=\frac{1}{2a}+\epsilon\) when \(x=\frac{1}{2}\), i.e. \[ c=\frac{1}{2}+\frac{1}{2a}+\epsilon . \] So consider region (see Figure 4.2) bounded by

• \(\displaystyle x=0, \qquad 0 \le y \le \frac{1}{2a}+\epsilon \)
• \(\displaystyle y=0, \qquad 0 \le x \le \frac{1}{2}+\frac{1}{2a}+\epsilon \)
• \(\displaystyle y=\frac{1}{2a}+\epsilon , \qquad 0 \le x \le \frac{1}{2}\)
• \(\displaystyle x+y-(\frac{1}{2}+\frac{1}{2a}+\epsilon )=0, \qquad \frac{1}{2} \le x \le \frac{1}{2}+\frac{1}{2a}+\epsilon\)

Solutions enter but do not leave this region bounded by these curves. Together with the small closed curve about the unstable (\(a\) sufficiently small) stationary point which solutions cross outwards, this forms a Poincar\'e-Bendixson region. Hence there exists a periodic orbit.