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\begin{document}
\centerline{Math 322 lecture for Monday, Week 9}
\bigskip

\centerline{Liapunov functions and stability}
\medskip

\textbf{Definition.} An equilibrium point~$x_0$ for a system~$x'=f(x)$ is
{\em stable}  if for each open neighborhood~$U$ of~$x_0$, there exists another
open neighborhood~$W$ of~$x_0$ such that if~$p\in W$, then~$\phi(t,p)\in U$ for
all~$t\geq0$.  Otherwise,~$x_0$ is {\em unstable}.  We say~$x_0$ is {\em
asymptotically stable} if it has an open neighborhood~$W$ such that
\[
  \lim_{t\to\infty}\phi_t(p)=x_0
\]
for all~$p\in W$.
\bigskip

{\bf Facts.}
\begin{enumerate}
  \item Surprisingly, an equilibrium point can be both unstable and
    asymptotically stable! We'll see an example in the homework.
  \item Suppose~$x_0$ is a hyperbolic equilibrium point, i.e., it's linearized system has
  no eigenvalues with real part equal to~$0$. To analyze the stability of~$x_0$, we use
  Hartman-Grobman to replace the system~$x'=f(x)$ with its
  linearization~$x'=Df_{x_0}(x)$ at~$x_0$.  If all eigenvalues of~$Df_{x_0}$
  have
  negative real part, then~$x_0$ is stable and asymptotically stable, and the
  approach of a trajectory to~$x_0$ is exponential in time.
  Otherwise, some eigenvalue has positive real part, and~$x_0$ is unstable.
  \item In any case, it turns out that if an equilibrium point~$x_0$ is stable,
  then no eigenvalue of~$Df_{x_0}$ has positive real part (even in the
  non-hyperbolic case).
\end{enumerate}
\medskip


{\bf Liapunov functions.}  Let~$x_0$ be an equilibrium point.  Suppose
there is a way to assign a smoothly changing ``temperature'' to each point
in~$E$ such that: (i) the temperature at~$x_0$ is~$0$, (ii) the temperature at
every other point is positive.  Could we determine stability only knowing the
temperatures along trajectories?  This is the idea behind the notion of a
Liapunov function. (Below, we label the temperature function by~$V$.)
\medskip

Given~$V\colon E\to\R$ and~$p\in E$, we define
\[
  \dot{V}(p)=\frac{d}{dt}V(\phi_t(p))\bigg|_{t=0}.
\]
Thus~$\dot{V}(p)$ tells us how fast the temperature is changing along the
solution trajectory as it passes through~$p$.
\medskip

\textbf{Theorem.} Let~$f\in C^1(E)$ and~$f(x_0)=0$.  Let~$V\colon E\to\R$ also be~$C^1$ (continuously
differentiable).  Suppose that~$V(p)\geq 0$ and~$V(p)=0$ if and only
if~$p=x_0$. Then:
\begin{enumerate}
  \item If~$\dot{V}$ is negative semidefinite ($\dot{V}(p)\leq0$ for
    all~$p\in E\setminus\left\{ x_0 \right\}$)  then~$x_0$ is stable.
  \item If~$\dot{V}$ is negative definite ($\dot{V}(p)<0$ for all~$p\in
    E\setminus\left\{ x_0 \right\}$)
    then~$x_0$ is asymptotically stable.
  \item If~$\dot{V}$ is positive definite ($\dot{V}(p)>0$ for all~$p\in
    E\setminus\left\{ x_0 \right\}$), then~$x_0$ is unstable.
\end{enumerate}
\medskip

\textbf{Definition.} A function satisfying the hypotheses of the previous
theorem is called a {\em Liapunov function}.
\medskip

Happily, thanks to the chain rule, the conditions on~$\dot{V}$ in the
theorem can be verified \emph{without solving the system}:
\medskip

\textbf{Proposition.}  With~$V$ as above, 
\[
  \dot{V}(p)=\nabla V(p)\cdot f(p).
\]
\begin{proof}
 Let~$\psi(t):=\phi_t(p)$.  Apply the chain rule:
 \begin{align*}
   J(V\circ\psi)(0)
   &= JV(\psi(0))J\psi(0)\\
   &= JV(p)J\psi(0)\\
   &= \left(\begin{array}{ccc}
       \frac{\partial V}{\partial x_1}(p)&\dots&\frac{\partial V}{\partial
       x_n}(p)
   \end{array} \right)
   \left(\begin{array}{c}
       \psi_1'(0)\\\vdots\\\psi_n'(0)
   \end{array} \right)\\
   &= \nabla V(p)\cdot\psi'(0).
 \end{align*}
 Now,~$\psi$ is the solution to the system~$x'=f(x)$ with initial condition~$p$.
 Therefore,~$\psi'(t)=f(\psi(t))$, and~$\psi'(0)=f(\psi(0))=f(p)$.  The result follows.
\end{proof}

{\bf Example.} Consider the system
\begin{align*}
  x'=-y^3\\
  y'=x^3.
\end{align*}
The origin is a non-hyperbolic equilibrium point and
\[
  V(x,y)=x^4+y^4
\]
is a Liapunov function for that point. (Clearly,~$V$ is smooth and~$V(x,y)\geq0$
with equality only at the origin.)  For any trajectory~$(x,y)=(x(t),y(t))$, we have
\[
  \dot{V}(x,y)=4x^3x'+4y^3y'=4x^3(-y^3)+4y^3(x^3)=0.
\]
Hence, the origin is stable.  In fact, our calculation shows
that~$V(\phi_t(p))$ is a constant as a function of~$t$.
In other words, trajectories (solutions) sit on level sets for~$V$, as seen in
the following:
\begin{center}
  \begin{tikzpicture}
    \node at (0,0) {\includegraphics[height=2.0in]{09mon1.png}};
      \node at (9,0) {\includegraphics[height=2.0in]{09mon2.png}};
  \end{tikzpicture}
\end{center}
\begin{proof}[Proof of theorem]
  We may assume~$x_0=0\in\R^{n}$ is the equilibrium point.

  (1) Suppose that~$\dot{V}(p)\leq 0$ for all~$p\in E\setminus\left\{ x_0
  \right\}$.  Choose~$\ve>0$ such that the open ball~$B_{\ve}(x_0)$ of
  radius~$\ve$ centered at~$x_0$ is contained in~$E$.  Let
  \[
    \overline{B_{\ve}(x_0)}:=\left\{ x\in\R^n:|x-x_0|\leq\ve \right\}.
  \]
  Replacing~$\ve$ by~$\ve/2$, if necessary, we may
  assume~$\overline{B_{\ve}(x_0)}\subset E$.  Let
  \[
    \alpha:=\min_{|x|=\ve} V(x),
  \]
  the minimum of~$V$ on the boundary of~$\overline{B_{\ve}(x_0)}$.  The
  function~$V$ achieves its minimum on the boundary since~$V$ is continuous and
  the boundary is compact (closed and bounded).  Since the minimum is achieved at
  some point on the boundary and~$V$ is strictly greater than~$0$ away from the
  origin, we have~$\alpha>0$.  

  Define
  \[
    W:=\left\{ x\in B_{\ve}(x_0):V(x)<\alpha \right\}.
  \]
  We think of~$W$ as the set of points in~$B_{\ve}$ whose ``temperature'' is less
  that~$\alpha$, the minimum temperature on the boundary of~$B_{\ve}$. Then~$W$
  is an open\footnote{The set~$W$ is open since~$W=V^{-1}((-\infty,\alpha))$, and
    by definition of continuity, the inverse image of an open subset under a
  continuous function is continuous.} neighborhood of the origin, and no solution
  starting at a point in~$W$ can leave~$W$ since~$V$ is nonincreasing on solution
  curves. Thus~$x_0$ is stable.
  \medskip

  (2) Suppose now that~$\dot{V}(p)<0$ for all~$p\in E\setminus\left\{ x_0
  \right\}$.  As in the proof for part~(1), we choose~$\ve>0$ so
  that~$\overline{B_{\ve}(x_0)}\subset E$.  We let
  \[
    \alpha:=\min_{|x|=\ve} V(x),
  \]
  and take
  \[
    W:=\left\{ x\in B_{\ve}(x_0):V(x)<\alpha \right\}.
  \]
  Since~$\dot{V}(p)<0$ for all~$p\in E\setminus\left\{ x_0 \right\}$, we saw in
  the proof of part~(1) that solution trajectories starting in~$W$ never
  leave~$W$.  We would like to show that~$\lim_{t\to\infty}\phi_t(p)=0$ for
  all~$p\in W$.  Pick any sequence~$t_1<t_2<\dots$ such that~$t_n\to\infty$, and
  consider the sequence
  \[
    \left\{ \phi(t_n,p) \right\}.
  \]
  By part (1), this sequence never leaves~$W$, and hence it is contained in the
  closure~$\overline{W}\subseteq\overline{B_{\ve}(x_0)}$, which is compact.  So by
  the Bolzano-Weierstrass theorem, there exists a convergent subsequence.  This
  means that there is a subsequence~$t_{n_k}$ such that
  \[
    \lim_{k\to\infty}\phi(t_{n_{k}},p)=q
  \]
  for some~$q\in\overline{W}$.  For ease of writing, replace our original sequence
  with the subsequence~$\left\{ t_{n_k} \right\}_k$.  We then have
  \[
    \lim_{n\to\infty}\phi(t_n,p)=q.
  \]

  We would like to show that~$q=x_0=0$, and we will do this by contradiction.
  Suppose that~$q\neq 0$.  Then~$V(q)>0$. Also since~$V$ is strictly decreasing
  along trajectories, we have
  \[
    V(q)>V(\phi(1,q)).
  \]
  Since~$\lim_{n\to\infty}\phi(t_n,p)=q$, by continuity of solutions with respect
  to both time and
  initial conditions, and by continuity of~$V$, there exists an integer~$N$ large
  enough so that~$\phi(t_N,p)$ is close enough to~$q$ so that~$V(\phi(1,\phi(t_N,p)))$
  is close enough to~$V(\phi(1,q))$ so that
  \[
    V(\phi(1+t_N,p)) = V(\phi(1,\phi(t_N,p))<V(q).
  \]
  Since~$t_n\to\infty$, we can find~$M$ such that~$t_{M}>1+t_N$. Then,
  since~$V$ is strictly decreasing along trajectories, we have
  \[
    V(q)>V(\phi(1+t_N,p))>V(\phi(t_M,p))
  \]
  This is a problem: since~$V$ stricty decreases along trajectories
  and~$V$ is continuous, we have that the sequence~$\left\{ V(\phi(t_n,p)
  \right\}$ is strictly decreasing and converges to~$V(q)$.  So in
  contradiction to the inequalities displayed above,
  \[
    V(\phi(t_M,p)>V(q).
  \]

  We have shown that~$q=0$ and that there is a sequence~$\{t_{n}\}$ such
  that $\lim_{n \to \infty} \phi(t_{n},p)=q=0$.  We now need to
  show~$\lim_{t\to\infty}\phi(t,p)=x_0=0$.  If not, there exists an~$\eta>0$
  such that for all~$n$, there exists~$s_n>n$ such that
  \begin{equation}
    |\phi(s_n,p)|\geq\eta>0.
    \label{eqn: 09wed1}
  \end{equation}
  We may assume that the sequence~${s_n}$ is increasing.  However, by
  Bolzano-Weierstrass, there again exists a subsequence~$\left\{ s_{n_k}
  \right\}$ of~$\left\{ s_n \right\}$ such that~$\phi(s_{n_k},p)$ converges, and
  as we have seen, it must converge to~$0$.  But that's impossible in light
  of~\eqref{eqn: 09wed1}.

  (3) Finally, now suppose that~$\dot{V}(p)>0$ for all~$p\in E\setminus\left\{ x_0 \right\}$.  
  Choose~$\ve>0$ such that~$\overline{B_{\ve}(0)}\subset E$.  We'll show
  that given any point~$p\in E$, we have that~$\phi_t(p)$
  leaves~$B_{\ve}(0)$ at some point, i.e., there exists~$t\geq 0$
  such that~$|\phi_t(p)|>\ve$.  Hence,~$x_{0}$ is unstable.

  Given~$p\in E\setminus\left\{ 0 \right\}$, since~$V$ is strictly increasing on trajectories,
  \[
    V(\phi_t(p))>V(\phi_0(p))=V(p)>0
  \]
  for all~$t>0$.  Thus,~$\phi_t(p)$ is bounded away from~$0$.
  Say~$|\phi_t(p)|\geq\eta>0$ for all~$t\geq0$.  If~$\eta\geq\ve$, then we are
  done since~$|p|=|\phi_0(p)|\geq\eta>\ve$, which says~$p$ is already out
  of~$B_{\ve}(x_0)$.  Otherwise, define
  \[
    m := \min_{y:\eta\leq|y|\leq\ve}\dot{V}(y),
  \]
  which exists since~$\dot{V}$ is continuous and~$y$ is restricted to a compact
  set. In fact, for that same reason,~$m=\dot{V}(q)$ for some point in the set over
  which we are minimizing.  Therefore,~$m>0$.  Supposing for contradiction that~$\phi_t(p)$ stays
  inside~$B_{\ve}(x_0)$ for all~$t\geq0$, we have~$\dot{V}(\phi_t(p))\geq m$ for
  all~$t\geq0$.  Hence,
  \[
    V(\phi_t(p))-V(p)=V(\phi_t(p))-V(\phi_0(p))=\int_{s=0}^t\dot{V}(\phi_s(p))\,ds\geq mt\to\infty
  \]
  as~$t\to\infty$.  But since~$V$ is continuous, it achieves a maximum
  on~$\overline{B_{\ve}(x_0)}$---a contradiction.
  \hfill$\Box$
  \medskip

  {\bf Example.}  Consider the system
  \begin{align*}
    x'&= -2y+yz\\
    y'&= x-xz\\
    z'&= xy.
  \end{align*}
  The Jacobian at the origin is
  \[
    J(0)=
    \left(\begin{array}{ccc}
	0&-2&0\\
	1&0&0\\
	0&0&0
    \end{array} \right).
  \]
  The characteristic polynomial is
  \[
    \det
    \left(\begin{array}{ccc}
	-x&-2&0\\
	1&-x&0\\
	0&0&-x
    \end{array} \right)
    =-x^3-2x=-x(x^2+2).
  \]
  So the eigenvalues are~$0,\pm\sqrt{2}i$.  So the origin is a nonhyperbolic
  equilibrium point.  To determine stability, we look for a suitable Liapunov function.
  We guess a function of the form
  \[
    V = ax^2+by^2+cz^2
  \]
  with positive constants~$a,b,c$. We have
  \begin{align*}
    \dot{V}&= 2axx'+2byy'+2czz'\\
	   &= 2ax(-2y+yz)+2by(x-xz)+2cz(xy)\\
	   &= 2(-2a+b)xy+2(a-b+c)xyz.
  \end{align*}
  Take~$a=c=1$ and~$b=2$, and we get~$V=x^2+2y^2+z^2$ with~$\dot{V}=0$.  This
  means that trajectories stay on the ellipsoids that are level sets of~$V$.
\end{proof}
\end{document}


