3 Linearisation and equilibria

3.4 Stability and Lyapunov functions

We have seen that if \(\mathrm{Re} \lambda_i < 0\) for all eigenvalues \(\lambda_i\) of \(A\), solutions of linear system \(\dot{x}=A x\) converge to the origin. In fact, if \(A\) can be diagonised with the eigenpairs \((\lambda_i,e_i)\), then the solution can be written as \[ x(t)=\sum_{i=1}^n c_i e^{\lambda_i t} e_i, \] while the prefactor \(e^{\lambda_i t}\) goes to zero as \(t\) goes to infinity. Is this true of the corresponding nonlinear system near the stationary point, when the solution can not be obtained in explicit form? This is part of a much more general question about stability of stationary points, so first let us introduce some definitions. The main questions to be answered in this subsection are: (1) How do stability results for linear systems carry over to nonlinear systems locally? (2) How about the boundedness or stability of solutions for \(\dot{x}=f(x), x \in \mathbb{R}^n\).

A stationary point \(x^*\) of an autonomous system \(\dot{x}=f(x)\) is asymptotically stable iff there exists an open neighbourhood \(U\) of \(x^*\) such that \(\varphi_t(x_0) \to x^*\) as \(t \to \infty\) for all \(x_0 \in U\).

If all eigenvalues \(\lambda_i\) are negative, then the solution starting at any \(x_0\) converges to the origin. If \(A=\mbox{diag}(\lambda_1,\lambda_2,\cdots,\lambda_n)\) is diagonal, this asymptotical stability can also be shown alternatively by considering the (squared) distance \(V(x)=| x|^2\) between the solution \(x(t)\) and the origin. Since \[ \frac{d}{dt} V(x)=2 x\cdot \dot{x}=2 \sum_{i=1}^n \lambda_i x_i^2\leq 0, \] the square distance \(|x(t)|^2\) is strictly decreasing, until \(x(t)\) reaches the origin. This means that \(V(x(t))=|x(t)|^2\) converges to zero .

Figure 3.13 (a) Asymptotic stability; (b) Lyapunov stability.

This means that if a solution starts sufficiently close to \(x^*\), itsfisolution eventually becomes arbitrary close to \(x^*\). But it does not implyfithe solution stays within \(U\) for all \(t>0\).

A stationary pointfi\(x^*\) of an autonomous ODE is Lyapunov stable iff for every openfineighbourhood \(U\) of \(x^*\) there exists an open neighbourhood \(W \subset U\) such that \(x_0 \in W\) implies \(\varphi_t(x_0) \in U\) for all \(t \ge 0\).

In other words solutions stay close to \(x^*\) if they start closefienough to \(x^*\). Note that Lyapunov stability does not imply asymptoticfistability (think of a linear centre). We have been thinking about linearisation,fiwhich motivates our final definition.

A stationary point \(x^*\) of anfiautonomous ODE is linearly stable iff the real parts of everyfieigenvalue of \({D}f(x^*)\) is negative.

For linear systemsfi\(\dot{x}=Ax\), if the eigenvalues of \(A\) have negative real parts thenfi\(|x(t)|\to 0\) as \(t \to \infty\), i.e. solutions are asymptotically stablefi(we have shown this in the special case of distinct eigenvalues). We will take a geometric view of stability (see the textbook byalues). Meiss for a more analytic treatment). The geometric approach is via motivatingalues). example for showing solutions are bounded (often an important first step inalues). The proof of the lemma uses some results from calculus. First recall the chain rule for the derivative of a function of a function. In one dimension \[ \frac{\mathrm{d}}{\mathrm{d} t}V(x(t)) =\frac{\mathrm{d} V(x(t))}{\mathrm{d} x}\frac{\mathrm{d} x(t)} {\mathrm{d} t} \] and in higher dimensions (\(x(t)\in\mathbb{R}^n\)) \[ \frac{\mathrm{d}}{\mathrm{d} t}V(x(t))=\nabla V(x(t)) \cdot \frac{\mathrm{d} x(t)}{\mathrm{d} t} =\nabla V(x)\cdot \dot{x}. \] So if \(x\) satisfies the differential equation \(\dot x=f(x)\) then \[ \frac{\mathrm{d}}{\mathrm{d} t}V(x(t))=\nabla V(x) \cdot f(x) . \]

Bounding Lemma Consider \(\dot{x}=f(x)\), \(x \in \mathbb{R}^n\), \(f\) smooth. Suppose there exists a compact set \(U \subset \mathbb{R}^n\), \(\epsilon >0\) and a continuously differentiable function \(V:U\to \mathbb{R}\) such that

then for all \(x_0 \in \mathbb{R}^n\) there exists \(t_0> 0 \) such that \(\varphi_t(x_0) \in S_{c_0}\) for all \(t \ge t_0.\)

Proof We first show that \(x(t)\) enters \(S_{c_0}\) at some time, even initially \(x_0\) is not in \(S_{c_0}\). If \(x_0 \notin S_{c_0}\), then \(\dot{V} < - \varepsilon\) and \(V(x_0)>c_0\), so \begin{equation*} V(\varphi_t(x_0)) < V(x_0) - \varepsilon t, \end{equation*} and so there exists \(t_0 < \frac{V(x_0) - c_0}{\varepsilon}\) such that \begin{equation*} V(\varphi_{t_0}(x_0))=c_0, \qquad \text{ i.e. } \varphi_{t_0}(x_0) \in S_{c_0}. \end{equation*}

Figure 3.14 What if \(\varphi_t(x_0)\) leaves \(S_{c_0}\)?

Once \(x(t)\) is in \(S_{c_0}\), we show that it stays in \(S_{c_0}\). To make this argument formally, suppose \(\varphi_{t^*}(x_0) \in \partial S_{c_0}\) and for \(\delta>0\) sufficiently small \(\varphi_t(x_0) \notin S_{c_0}\), for any \(t \in (t^*,t^*+\delta)\) (see Figure 3.14). Then \( V(\varphi_{t^*+\delta}(x_0))>c_0 = V(\varphi_{t^*}(x_0)) \) and \( \dot{V}(\varphi_t(x_0))<-\epsilon \) for any \( t \in (t^*,t^*+\delta) \). Then \[ 0< V(\varphi_{t^*+\delta}(x_0)) - V(\varphi_{t^*}(x_0))=\int_{t^*}^{t^*+\delta} \frac{\mathrm{d}}{\mathrm{d} t} V(\varphi_t(x_0))\mathrm{d} t < \int_{t^*}^{t^*+\delta} (-\epsilon)\mathrm{d} t=-\delta \epsilon. \] So we get a contradiction.

Example 3.6 ( Lorenz Equations) The Lorenz equations are three coupled ODEs introduced as a simple model of the weather. They are one of the earliest examples which appeared to behave chaotically. The equations are \begin{equation*} \dot{x} = \sigma (y-x), \qquad \dot{y} = rx-y-xz, \qquad \dot{z} = -bz + xy \end{equation*} with \(\sigma,r,b \ge 0\).ons are We wish to show that all solutions are bounded. Need to 'create' a positive function \(V(x,y,z)\) withounded. the properties of the lemma. The obvious thing to do is to choose quadraticounded. function, slightly less obvious is which quadratic function. The followingounded. approach is not optimal but works! Set \begin{equation*} V(x,y,z)=Ax^2 + By^2 + C(z-2r)^2 \end{equation*} and we want to choose the positive constants A,B,C to satisfy the lemma. So \begin{align*} \frac{1}{2}\dot{V} &= Ax \dot{x} + B y \dot{y} + C \dot{z} w(z-2r)\cr &= A \sigma x (y-x) + By(rx-y-xz) + C(-bz +xy)(z-2r) \end{align*} wwhere we have used the ODE to replace \(\dot{x}, \dot{y},\) and \(\dot{z}\). We want \(\dot{V}<0\) for large enough \(x,y\) and \(z\), so the `difficult' terms are cross multiplications like \(xy\) and \(xyz\) (good terms are quadratic terms, with negative coefficients). Choose the constants to remove them i.e. \[\begin{array}{rlll} B - C &=0 &~~~~~~ & (xyz \text{ terms vanish }) \\ A \sigma + B r -2Cr &=0 & & (xy \text{ terms vanish}) \end{array}\] So set \(B=C\) wand \(A\sigma=Br\), e.g. \(A=r, B=C=\sigma\) (we have one degree of freedom to choose these constants: \(A=r/\sigma,B=C=1\) works as well), then \[ V(x,y,z)=\sigma rx^2 + \sigma y^2 + \sigma (z-2r)^2 \] and \[ \frac{1}{2}\dot{V}(x,y,z)=- {\sigma}r x^2 - \sigma y^2 - \sigma (bz^2 -2rbz) {=-\sigma rx^2 -\sigma y^2 - \sigma b(z-r)^2 + \sigma br^2. } \] Notice it is { \((z-r)^2\)} on the right hand side, instead of { \((z-2r)^2\)} as in \(V\). The specific form of \(\dot{V}/2\) also tells us which set to look at. So provided that \((x,y,z)\) lies outside the set \begin{equation*} \tilde{U}=\big\{(x,y,z)| {\sigma}rx^2 + y^2 + \sigma b (z-r)^2 \le b r^2 \big\} \qquad \text{ (an ellipsoid)} \end{equation*} then \(\dot{V}<0\). It is possible to modify the Bounding Lemma to deal with this, but the actual conditions of the bounding lemma require \(\dot V<-\epsilon <0\). Thus we choose \(U\) a little larger: pick \(\epsilon>0\) and set \begin{equation*} {U} = \{(x,y,z)|{\sigma} rx^2 + \sigma y^2 + \sigma b (z-r)^2 \le \sigma b r^2+{\epsilon}\} \end{equation*} and so in \(\mathbb{R}^3\backslash U\) \[ \dot V= - r\sigma x^2 - \sigma y^2 - \sigma b(z -r) + \sigma b r^2\le -2\epsilon<-\epsilon . \] Now just choose \(c_0\) sufficiently large that \begin{equation*} S_{c_0}=\{(x,y,z)| rx^2 + \sigma y^2 + \sigma b (z-2r)^2 \le c_0 \} \end{equation*} (another ellipsoid) contains \(U\).
You can use your mouse to drag one trajectory of the Lorenz ODE with the parameters \( \sigma =10, r= 8/3\) and \(b = 28 \). Notice how the solution moves from one parts to the other parts.
Remark We can relate \(\dot{V}\) to \(V\) in a differential inequality, to induce some information about the behaviour of \(V\). More precisely, the above expression for \(\dot{V}\) implies that \[ \dot{V}(x,y,z) = - 2r\sigma x^2 -2\sigma y^2 - \sigma b(z -2r)^2 -\sigma bz^2+ 4\sigma b r^2 \leq - \mu V(x,y,z) + 4\sigma br^2. \] with \(\mu=\min(2\sigma,2,b)\). The advantage here is that we can ``solve'' this differential inequality. Since \[ \frac{\mathrm{d}}{\mathrm{d} t} \big(e^{\mu t} V(x,y,z)\big) = e^{\mu t} \big( \dot{V} + \mu V\big) \leq 4\sigma br^2 e^{\mu t}. \] Integrating both sides for time from zero to \(t\), then \[ e^{\mu t} V(x,y,z) - V(x_0,y_0,z_0) \leq 4\sigma b r^2 \int_0^t e^{\mu\tau}\mathrm{d}\tau =\frac{4\sigma b r^2}{\mu} (e^{\mu t} - 1). \] { That is \[ V(x,y,t) \leq e^{-\mu t} V(x_0,y_0,z_0) + \frac{4\sigma br^2}{\mu}\left(1-e^{-\mu t}\right) \leq e^{-\mu t} V(x_0,y_0,z_0) + \frac{4\sigma br^2}{\mu}, \] which is bounded for any \(t\geq 0\). We can estimate the time when the trajectories enter the set (different from the above one) \[ \bar{U} = \left\{ (x,y,z) | V(x,y,z) \leq \frac{4\sigma br^2}{\mu}+\epsilon \right\}, \] that is \(t \geq \frac{1}{\mu} \log \frac{\epsilon}{V(x_0,y_0,z_0)}\).
Remark The function \(V\) above is not unique, and we can choose alternative ones (along with many others) like \[ V(x,y,z) = x^2+y^2+(z-r-\sigma)^2. \] Then \[ \frac{\mathrm{d} V}{\mathrm{d} t} = -2\sigma x^2 - 2y^2 - b(z-r-\sigma)^2 -bz^2 + b(r+\sigma)^2 \leq -\alpha V + b(r+\sigma)^2, \] where \(\alpha= \min(2\sigma,2,b)\). Then we can choose the set to be \[ \tilde{U} = \left\{(x,y,z) \mid x^2 + y^2 + (z-\sigma-r)^2 \leq \frac{b(r+\sigma)} {\alpha} \right\}, \] or relax the right hand side by changing it to \(b(r+\sigma)^2/\alpha + \epsilon\).

The same basic idea works for stationary points, leading to the so called Lyapunov functions.

A function \(V:U\to \mathbb{R}\) is called a Lyapunov function on \(U \subseteq \mathbb{R}^n\) iff it is continuously differentiable, \(V(x) \ge 0\) on \(U\) and \(\dot{V}\le 0\) on \(U\).

Lyapunov's Stability Theorem Suppose \(x^* \in \mathbb{R}^n\) is a)stationary point of \(\dot{x}=f(x)\) with \(f\) smooth. Let \(U\) be an open neighbourhood of \(x^*\) and suppose there exists a Lyapunov function \(V:U \to \mathbb{R}\) such that \(V(x) >0\) on \(U \setminus \{x^*\}\) and \(V(x^*)=0\). Then \(x^*\) is Lyapunov stable. If in addition \(\dot V<0\) in \(U \setminus \{x^*\}\) then \(x^*\) is asymptotically stable.

Proof. Choose \(\varepsilon>0\) small enough so that \(\{ x \mid |x-x^*|)\le \varepsilon \}\) lies entirely in \(U\), and let \(\displaystyle c_0= \min_{|x-x^*| = \varepsilon} V(x)\) which exists as \(|x-x^*|=\varepsilon\) is compact (closed and bounded), and \(c_0>0\) as \(x^* \notin \{ x \mid | x-x^*| =\varepsilon\}\). Let \(\mathcal{B}_\varepsilon(x^*) = \{ x \mid |x-x^*| <\varepsilon \}\). Now \(V\) is continuous, and \(V(x^*)=0\), so there exists \(\delta>0\) such that for all \(x \in \mathcal{B}_\delta(x^*), V(x) <\frac{1}{2} c_0.\) Consider \(x_0 \in \mathcal{B}_\delta (x^*).\) Since \(\dot{V} \le 0\) in \(U\), \(V(\varphi_t(x_0)) \le V(x_0) < \frac{1}{2} c_0\) for all \(t\) such that \(\varphi_t(x_0) \in U\), and hence \(V(\varphi_t(x_0))<\frac{1}{2}c_0 < c_0\), the minimum on \(|x-x^*|=\varepsilon\). Hence \(\varphi_t(x_0)\in \mathcal{B}_\varepsilon(x^*)\) for all \(t>0\).

Figure 3.15 Regions defined in proof.

Suppose in addition that \(\dot{V}<0\) if \(x\in U \backslash \{x^*\}\), (note that \(\dot{V}(x^*)=0\) since \(x^*\) is stationary). Then if \(x_0 \in \mathcal{B}_\delta(x^*)\) as before, \(V(\varphi_t(x_0))\) is strictlyfidecreasing and hence tends to a limit \(\bar{V}=\lim_{t\to\infty} V(\varphi_t(x_0))\). At the limit \(\dot{V}=0\) hence the limit must be \(x^*\).

Example 3.7 Consider the Lorenz equations with \(r \in (0,1)\). That is \begin{equation*} \dot{x}=\sigma (y-x), \qquad \dot{y}=rx-y-xz, \qquad \dot{z}=-bz + xy \end{equation*} with \(\sigma,b \ge 0\) and \( 0 < r < 1\). Try a Lyapunov function of the form \begin{equation*} V(x,y,z)=Ax^2 + By^2 +Cz^2, \end{equation*} with \(\frac{1}{2} \dot{V}=A \sigma x(y-x) + By(rx-y-xz) +Cz(-bz +xy)\). Choose \(B=C\) to remove the \(xyz\) terms: \begin{equation*} \frac{1}{2} \dot{V}=A \sigma (-x^2 + xy) + B(rxy-y^2-bz^2) \end{equation*} We deal with the \(xy\) terms to ensure \(\dot{V}\) is negative, by choosing \(A\) and \(B\) appropriately (matching the square!). Set \(B=\sigma\), \(A=r\) so all terms together \begin{align*} \frac{1}{2} \dot{V} &=r \sigma (-x^2 + 2xy) - \sigma y^2 -\sigma bz^2 \\ &=r \sigma (-(x - y)^2) - (\sigma - r \sigma) y^2-\sigma bz^2 \\ &=-r \sigma (x - y)^2 - \sigma(1 - r ) y^2 - \sigma bz^2 \\ &<0 \qquad \qquad \text{ if } (x,y,z) \neq (0,0,0) \end{align*} if \(r\) is between \(0\) and \(1\). Hence the origin is asymptotically stable when \(r \in (0,1)\).
Example 3.8 Let us consider a degenerate node with repeated eigenvalues \(\lambda=-2\): \begin{align*} \dot{x} = -2x + y,\quad \dot{y} = -2y. \end{align*} Set \(V(x,y)=x^2+By^2\) and \(B\) is chosen later. Then \begin{align*} \frac{1}{2} \dot{V} &= x(-2x+y) + By(-2y) = -2x^2 +xy -2By^2 = -2(x-\frac{1}{4} y)^2 + \frac{1}{8} y^2 -2By^2. \end{align*} So for any \(B>\frac{1}{16}\) the function is a Lyapunov function and the origin is asymptotically stable.

In many practical examples, the stationary point \(x^*\) can still be asymptotic stable, when the condition \(\dot{V} < 0\) in \(U\subset \{x^*\}\) in Lyapunov's Stability Theorem can be relaxed to \(\dot{V} \leq 0\), provided that \(x^*\) is the only fixed point.

Theorem (LaSalle's Invariance Principle) Suppose that \(V: U\subset \mathbb{R}^n \to \mathbb{R}\) is a Lyapunov function for the system \(\dot{x}=f(x)\). If the set \(\{x\in U \mid \dot{V}(x)=0\}\) contains only one fixed point \(x^*\), then \(x^*\) is asymptotically stable.

Example 3.9 Consider the equation \(\ddot{x} + \mu \dot{x} + \omega^2 x=0\) describing the motion of a harmonic oscillator with friction (\(\mu >0\)), or equivalently \(\dot{x} = y,\quad \dot{y} = -\mu y -\omega^2x\). If we choose \(V(x,y) = \omega^2 x^2 + y^2\), then \(\dot{V} = -2\mu y^2 \leq 0\). Since \((0,0)\) is the only fixed point in the set \(\{(x,y) \mod \dot{V}(x,y)=0\} =\{ (x,y) \mid y=0\}\), by Lasalle's invariance principle, the origin is asymptotically stable. In other words, the harmonic oscillator will eventially stop moving.