Implication of stability of Van der Pol oscillator.

Question

Consider the homogeneous Van der Pol equation, $\ddot{x} + \mu (x^2-1)\dot{x} + x = 0$, with $\mu>0$. We convert it into a dynamical system, $$\dot{\bf x} = (y, -(x+\mu(x^2-1)y), \ \mathbf{x} \equiv (x,y), \ \text{where} \ y = \dot{x}.$$ It is given that there exists $\epsilon > 0$ such that for any $||\bf x(0) - a||$, there exists a positive $M$ such that $||\mathbf{x}(t) - \mathbf{a}|| < \epsilon$ for all $t > M$.

The problem I am trying to solve is to prove that this property does not imply that the equilibrium point at the origin is not stable, according to this definition:

an equilibrium point $\bf a$, i.e $\bf \dot{x}(a) = \bf 0$, is defined to be stable if $\forall \epsilon > 0$, $\exists \delta > 0$ such that $||\bf x(0) - a|| < \delta \implies ||\bf x(t) - a|| < \epsilon$.

The following counter example is proposed,

Let $\delta > 0$, and $\mathbf{y} = (0, \delta)$. Now, $\bf \dot{x}|_{x =y} = (\delta,\mu \delta)$. We can see that the rate of change of $x$ and $y$ are both positive at this point. Let $\epsilon = \frac{1}{2}$. We choose $\mu = \frac{1}{\delta}$ to ensure the change in $y$ is large. This change will be positive so long as $$-(x + \mu (x^2 - 1)y) = (1-x^2)\mu y - x > 0.$$ Notice that for $0 \leq x \leq 0.5$, $1-x^2 \geq x \implies y \geq \delta$ since $\dot{y}_{y=\delta} = 1$. As $x \to \frac 12$, $$\dot{y} \to -(0.5 - 0.75 \mu y) = 0.75 \frac{y}{\delta} - 0.5 > 0$$. As both $x$ and $y$ are increasing until $x$ reaches $0.5$, this implies $y > \delta$ and hence there exists $t$ such that $||\bf x||$$> 0.5 = \epsilon$. Note that this holds for all $\delta < 0.5$, so no $\delta$ satisfies the definition of stability and since the definition does not hold for a particular $\epsilon$, the Van der Pol oscillator is not stable.

Is this argument valid? If not, how could it be fixed?

Answer 1

For $(x,y)\approx (0,0)$, the equation is almost linear, the linear version is $$ \ddot x-μ\dot x+x=0. $$ This is a linear oscillator with "anti-friction", the faster it moves the faster it moves.

Using the levels of $V=x^2+y^2$ to measure the motion one finds that $\dot V = 2μy^2\ge 0$, so that any motion that is not at rest will almost always increase the distance to the origin.

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Using $V=2x^2+y^2+(y-μx)^2$ even gives $\dot V=2μ(x^2+y^2)>0$ for $(x,y)\ne (0,0)$ so that any non-zero solution of the linear equation crosses the elliptical level sets of $V$ outwards at any time.

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Now one could check how these same functions $V$ look for the non-linear equation. \begin{align} \frac12\dot V &= (2x-μ(y-μx))\dot x + (y+(y-μx))\dot y \\ &=((2+μ^2)x-μy)y + (2y-μx)[(1-x^2)\mu y-x] \\ &=μx^2 + μy^2 -μ(2y-μx)x^2y \\ &\geμ(x^2+y^2)-\frac{\mu}2(x^2+y^2)^2-\frac{3\sqrt3μ^2}{16}(x^2+y^2)^2 \end{align} as by the inequality of geometric and quadratic mean $x^3(\sqrt3y)\le \left(\frac{3\cdot x^2+(\sqrt3y)^2}{4}\right)^2$. Thus $\dot V>0$ for $$x^2+y^2<\frac2{1+\frac{3\sqrt3}8μ}.$$ The solutions will escape thus any level set of $V$ that is inside the circle of this radius.

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As to the Counter-Example

Note that $μ$ is a fixed constant from the start. You need to find some $ϵ$ so that independent of how small $δ$, any solution starting in the $δ$ ball (or for convenience the ball of double that radius) leaves the $ϵ$ ball. You propose that one can show this using $ϵ=\frac12$. Then for any given $0<δ<ϵ$ you consider the solution starting in ${\bf y}=(0,δ)$ and claim that the solution moves to the right.

Check for $μ=10$, $0<δ<ϵ=0.5$ so indeed the solutions seem to leave this initial position almost linearly into the first quadrant.

Obviously one can not use the claim $μδ=1$ if $δ$ is variable and arbitrarily small. It might be possible to show that $(1-x^2)\mu y-x\ge\frac{\mu}2y$ which implies $y\ge δ+\frac{\mu}2x$ inside that region.