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<p style="font-size: 18px;"><a href="http://people.csail.mit.edu/russt/">Russ Tedrake</a></p>
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<chapter style="counter-reset: chapter 8"><h1>Lyapunov
Analysis</h1>
<p>Optimal control provides a powerful framework for formulating control
problems using the language of optimization. But solving optimal control
problems for nonlinear systems is hard! In many cases, we don't really care
about finding the <em>optimal</em> controller, but would be satisfied with any
controller that is guaranteed to accomplish the specified task. In many
cases, we still formulate these problems using computational tools from
optimization, and in this chapter we'll learn about tools that can provide
guaranteed control solutions for systems that are beyond the complexity for
which we can find the optimal feedback.</p>
<p>There are many excellent books on Lyapunov analysis; for instance
<elib>Slotine90</elib> is an excellent and very readable reference and
<elib>Khalil01</elib> can provide a rigorous treatment. In this chapter I
will summarize (without proof) some of the key theorems from Lyapunov
analysis, but then will also introduce a number of numerical algorithms...
many of which are new enough that they have not yet appeared in any mainstream
textbooks.</p>
<section><h1>Lyapunov Functions</h1>
<p>Let's start with our favorite simple example. </p>
<example><h1>Stability of the Damped Pendulum</h1>
<center><img width="25%" src="figures/simple_pend.svg"/></center>
<p>Recall that the equations of motion of the damped simple pendulum are
given by \[ ml^2 \ddot{\theta} + mgl\sin\theta = -b\dot{\theta}, \] which
I've written with the damping on the right-hand side to remind us that it
is an external torque that we've modeled.</p>
<p>These equations represent a simple second-order differential equation;
in chapter 2 we discussed at some length what was known about the
solutions to this differential equation--in practice we do not have a
closed-form solution for $\theta(t)$ as a function of the initial
conditions. Since we couldn't provide a solution analytically, in chapter
2 we resorted to a graphical analysis, and confirmed the intuition that
there are fixed points in the system (at $\theta = k\pi$ for every integer
$k$) and that the fixed points at $\theta = 2\pi k$ are asymptotically
stable with a large basin of attraction. The graphical analysis gave us
this intuition, but can we actually prove this stability property? In a
way that might also work for much more complicated systems?</p>
<p>One route forward was from looking at the total system energy (kinetic
+ potential), which we can write down: \[ E(\theta,\dot{\theta}) =
\frac{1}{2} ml^2\dot{\theta}^2 - mgl \cos\theta. \] Recall that the
contours of this energy function are the orbits of the undamped
pendulum.</p>
<center><img width="50%" src="figures/pend_undamped_phase.svg"/></center>
<p>A natural route to proving the stability of the downward fixed points
is by arguing that energy decreases for the damped pendulum (with $b>0$)
and so the system will eventually come to rest at the minimum energy, $E =
-mgl$, which happens at $\theta=2\pi k$. Let's make that argument
slightly more precise. </p>
<p>Evaluating the time derivative of the energy reveals \[ \frac{d}{dt} E
= - b\dot\theta^2 \le 0. \] This is sufficient to demonstrate that the
energy will never increase, but it doesn't actually prove that the energy
will converge to the minimum when $b>0$ because there are multiple
states(not only the minimum) for which $\dot{E}=0$. To take the last
step, we must observe that set of states with $\dot\theta=0$ is not an
invariant set; that if the system is in, for instance
$\theta=\frac{\pi}{4}, \dot\theta=0$ that it will not stay there, because
$\ddot\theta \neq 0$. And once it leaves that state, energy will decrease
once again. In fact, the fixed points are the only subset the set of
states where $\dot{E}=0$ which do form an invariant set. Therefore we can
conclude that as $t\rightarrow \infty$, the system will indeed come to
rest at a fixed point (though it could be any fixed point with an energy
less than or equal to the initial energy in the system, $E(0)$).</p>
</example>
<p>This is an important example. It demonstrated that we could use a
relatively simple function -- the total system energy -- to describe
something about the long-term dynamics of the pendulum even though the
actual trajectories of the system are (analytically) very complex. It also
demonstrated one of the subtleties of using an energy-like function that is
non-increasing (instead of strictly decreasing) to prove asymptotic
stability.</p>
<p>Lyapunov functions generalize this notion of an energy function to more
general systems, which might not be stable in the sense of some mechanical
energy. If I can find any positive function, call it $V(\bx)$, that gets
smaller over time as the system evolves, then I can potentially use $V$ to
make a statement about the long-term behavior of the system. $V$ is called
a <em>Lyapunov function</em>. </p>
<p>Recall that we defined three separate notions for stability of a
fixed-point of a nonlinear system: stability i.s.L., asymptotic stability,
and exponential stability. We can use Lyapunov functions to demonstrate
each of these, in turn.</p>
<theorem><h1>Lyapunov's Direct Method (for local stability)</h1>
<p>Given a system $\dot{\bx} = f(\bx)$, with $f$ continuous, and for some
region ${\cal D}$ around the origin (specifically an open subset of
$\mathbf{R}^n$ containing the origin), if I can produce a scalar,
continuously-differentiable function $V(\bx)$, such that \begin{gather*}
V(\bx) > 0, \forall \bx \in {\cal D} \setminus \{0\} \quad V(0) = 0,
\text{ and} \\ \dot{V}(\bx) = \pd{V}{\bx} f(\bx) \le 0, \forall \bx \in
{\cal D} \setminus \{0\} \quad \dot{V}(0) = 0, \end{gather*} then the
origin $(\bx = 0)$ is stable in the sense of Lyapunov (i.s.L.). [Note: the notation $A \setminus B$ represents the set $A$ with the elements of $B$ removed.]</p>
<p>If, additionally, we have $$\dot{V}(\bx) = \pd{V}{\bx} f(\bx) < 0,
\forall \bx \in {\cal D} \setminus \{0\},$$ then the origin is (locally)
asymptotically stable. And if we have $$\dot{V}(\bx) = \pd{V}{\bx} f(\bx)
\le -\alpha V(x), \forall \bx \in {\cal D} \setminus \{0\},$$ for some $\alpha>0,$
then the origin is (locally) exponentially stable. </p>
</theorem>
<p>Note that for the sequel we will use the notation $V \succ 0$ to denote a
<em>positive-definite function</em>, meaning that $V(0)=0$ and $V(\bx)>0,$
for all $\bx \ne 0$. We can use "positive definite in ${\cal D}$" if this
applies $\forall x \in {\cal D}$. Similarly, we'll use $V \succeq 0$ for
positive semi-definite, $V \prec 0$ for negative-definite functions,
etc.</p>
<p>The intuition here is exactly the same as for the energy argument we made
in the pendulum example: since $\dot{V}(x)$ is always zero or negative, the
value of $V(x)$ will only get smaller (or stay the same) as time progresses.
Inside the subset ${\cal D}$, for every $\epsilon$-ball, I can choose a
$\delta$ such that $|x(0)|^2 < \delta \Rightarrow |x(t)|^2 < \epsilon,
\forall t$ by choosing $\delta$ sufficiently small so that the sublevel set
of $V(x)$ for the largest value that $V(x)$ takes in the $\delta$ ball is
completely contained in the $\epsilon$ ball. Since the value of $V$ can only
get smaller (or stay constant) in time, this gives stability i.s.L.. If
$\dot{V}$ is strictly negative away from the origin, then it must eventually
get to the origin (asymptotic stability). The exponential condition is
implied by the fact that $\forall t>0, V(\bx(t)) \le V(\bx(0)) e^{-\alpha
t}.$</p>
<p> Notice that the system analyzed above, $\dot{\bx}=f(\bx)$, did not have
any control inputs. Therefore, Lyapunov analysis is used to study either the
passive dynamics of a system or the dynamics of a closed-loop system (system
+ control in feedback). We will see generalizations of the Lyapunov
functions to input-output systems later in the text. </p>
<subsection><h1>Global Stability</h1>
<p>The notion of a fixed point being stable i.s.L. is inherently a local
notion of stability (defined with $\epsilon$- and $\delta$- balls around
the origin), but the notions of asymptotic and exponential stability can
be applied globally. The Lyapunov theorems work for this case, too, with
only minor modification.</p>
<theorem><h1>Lyapunov analysis for global stability</h1>
<p>Given a system $\dot{\bx} = f(\bx)$, with $f$ continuous, if I can
produce a scalar, continuously-differentiable function $V(\bx)$, such
that \begin{gather*} V(\bx) \succ 0, \\ \dot{V}(\bx) = \pd{V}{\bx}
f(\bx) \prec 0, \text{ and} \\ V(\bx) \rightarrow \infty \text{ whenever
} ||\bx||\rightarrow \infty,\end{gather*} then the origin $(\bx = 0)$ is
globally asymptotically stable (G.A.S.).</p>
<p>If additionally we have that $$\dot{V}(\bx) \preceq -\alpha V(\bx),$$
for some $\alpha>0$, then the origin is globally exponentially
stable.</p>
</theorem>
<p>The new condition, on the behavior as $||\bx|| \rightarrow \infty$ is
known as "<a
href="https://en.wikipedia.org/wiki/Radially_unbounded_function">radially
unbounded</a>", and is required to make sure that trajectories cannot
diverge to infinity even as $V$ decreases; it is only required for global
stability analysis.</p> <!-- note: the first exercise below has an example -->
</subsection> <!-- global stability -->
<subsection id="lasalle"><h1>LaSalle's Invariance Principle</h1>
<p>Perhaps you noticed the disconnect between the statement above and the
argument that we made for the stability of the pendulum. In the pendulum
example, using the mechanical energy resulted in a Lyapunov function time
derivative that was only negative semi-definite, but we eventually argued
that the fixed points were asymptotically stable. That took a little
extra work, involving an argument about the fact that the fixed points
were the only place that the system could stay with $\dot{E}=0$; every
other state with $\dot{E}=0$ was only transient. We can formalize this
idea for the more general Lyapunov function statements--it is known as
LaSalle's Theorem.</p>
<theorem><h1>LaSalle's Theorem</h1>
<p>Given a system $\dot{\bx} = f(\bx)$ with $f$ continuous. If we can
produce a scalar function $V(\bx)$ with continuous derivatives for
which we have $$V(\bx) \succ 0,\quad \dot{V}(\bx) \preceq 0,$$ and
$V(\bx)\rightarrow \infty$ as $||\bx||\rightarrow \infty$, then $\bx$
will converge to the largest <i>invariant set</i> where $\dot{V}(\bx) =
0$.</p>
</theorem>
<p>To be clear, an <em>invariant set</em>, ${\cal G}$, of the dynamical
system is a set for which $\bx(0)\in{\cal G} \Rightarrow \forall t>0,
\bx(t) \in {\cal G}$. In other words, once you enter the set you never
leave. The "largest invariant set" need not be connected; in fact for the
pendulum example each fixed point is an invariant set, so the largest
invariant set is the <em>union</em> of all the fixed points of the
system. There are also variants of LaSalle's Theorem which work over
a region.</p>
<p>Finding a Lyapunov function which $\dot{V} \prec 0$ is more difficult
than finding one that has $\dot{V} \preceq 0$. LaSalle's theorem gives us
the ability to make a statement about <i>asymptotic</i> stability even in
this case. In the pendulum example, every state with $\dot\theta=0$ had
$\dot{E}=0$, but only the fixed points are in the largest invariant
set.</p>
<example id="cartpole_swingup"><h1>Swing-up for the Cart-Pole System</h1>
<p>Recall the <a href="acrobot.html#cartpole_swingup">example of using partial-feedback linearization to generate a swing-up controller for the cart-pole system</a>. We first examined the dynamics of the pole (pendulum) only, by writing it's energy: $$E(\bx) = \frac{1}{2}\dot\theta^2 - \cos\theta,$$ desired energy, $E^d = 1$, and the difference $\tilde{E}(\bx) = E(\bx) - E^d.$ We were able to show that our proposed controller produced $$\dot{\tilde{E}} = -k \dot\theta^2 \cos^2\theta \tilde{E},$$ where $k$ is a positive gain that we get to choose. And we said that was good!</p>
<p>Now we have the tools to understand that really we have a Lyapunov function $$V(\bx) = \frac{1}{2}\tilde{E}^2(\bx),$$ and what we have shown is that $\dot{V} \le 0$. By LaSalle, we can only argue that the closed-loop system will converge to the largest invariant set, which here is the entire homoclinic orbit: $\tilde{E}(\bx) = 0$. We have to switch to the LQR controller in order to stabilize the upright.</p>
<figure>
<iframe id="igraph" scrolling="no" style="border:none;" seamless="seamless" src="data/cartpole_swingup_V.html" height="350" width="100%"></iframe>
<figcaption>Lyapunov function: $V(\bx) = \frac{1}{2}\tilde{E}^2(\bx).$</figcaption>
</figure>
<figure>
<iframe id="igraph" scrolling="no" style="border:none;" seamless="seamless" src="data/cartpole_swingup_Vdot.html" height="350" width="100%"></iframe>
<figcaption>Time-derivative of the Lyapunov function: $\dot{V}(\bx).$</figcaption>
</figure>
<p>As you can see from the plots, $\dot{V}(\bx)$ ends up being a quite non-trivial function! We'll develop the computational tools for verifying the Lyapunov/LaSalle conditions for systems of this complexity in the upcoming sections.</p>
</example>
</subsection> <!-- lasalle -->
<subsection id="HJB"><h1>Relationship to the Hamilton-Jacobi-Bellman
equations</h1>
<p> At this point, you might be wondering if there is any relationship
between Lyapunov functions and the cost-to-go functions that we discussed
in the context of dynamic programming. After all, the cost-to-go
functions also captured a great deal about the long-term dynamics of the
system in a scalar function. We can see the connection if we re-examine
the HJB equation \[ 0 = \min_\bu \left[ \ell(\bx,\bu) +
\pd{J^*}{\bx}f(\bx,\bu). \right] \]Let's imagine that we can solve for the
optimizing $\bu^*(\bx)$, then we are left with $ 0 = \ell(\bx,\bu^*) +
\pd{J^*}{\bx}f(\bx,\bu^*) $ or simply \[ \dot{J}^*(\bx) = -\ell(\bx,\bu^*)
\qquad \text{vs} \qquad \dot{V}(\bx) \preceq 0. \] In other words, in
optimal control we must find a cost-to-go function which matches this
gradient for every $\bx$; that's very difficult and involves solving a
potentially high-dimensional partial differential equation. By contrast,
Lyapunov analysis is asking for much less - any function which is going
downhill (at any rate) for all states. This can be much easier, for
theoretical work, but also for our numerical algorithms. Also note that
if we do manage to find the optimal cost-to-go, $J^*(\bx)$, then it can
also serve as a Lyapunov function so long as $\ell(\bx,\bu^*(\bx)) \succeq
0$.</p>
</subsection> <!-- relationship to HJB -->
<todo>Include instability results, as in Briat15 Theorem 2.2.5</todo>
</section> <!-- Lyapunov definitions -->
<section><h1>Lyapunov analysis with convex optimization</h1>
<p>One of the primary limitations in Lyapunov analysis as I have presented
it so far is that it is potentially very difficult to come up with suitable
Lyapunov function candidates for interesting systems, especially for
underactuated systems. ("Underactuated" is almost synonymous with
"interesting" in my vocabulary.) Even if somebody were to give me a
Lyapunov candidate for a general nonlinear system, the Lyapunov conditions
can be difficult to check -- for instance, how would I check that $\dot{V}$
is strictly negative for all $\bx$ except the origin if $\dot{V}$ is some
arbitrarily complicated nonlinear function over a vector $\bx$?</p>
<p>In this section, we'll look at some computational approaches to verifying
the Lyapunov conditions, and even to searching for (the coefficients of) the
Lyapunov functions themselves.</p>
<p>If you're imagining numerical algorithms to check the Lyapunov conditions
for complicated Lyapunov functions and complicated dynamics, the first
thought is probably that we can evaluate $V$ and $\dot{V}$ at a large number
of sample points and check whether $V$ is positive and $\dot{V}$ is
negative. <a
href="https://github.com/RobotLocomotion/drake/blob/master/systems/analysis/test/lyapunov_test.cc">This
does work</a>, and could potentially be combined with some smoothness or
regularity assumptions to generalize beyond the sample points. <todo>Add
python bindings for the pendulum energy lp lyapunov example.</todo> But in
many cases we will be able to do better -- providing optimization algorithms
that will rigorously check these conditions <em>for all $\bx$</em> without
dense sampling; these will also give us additional leverage in formulating
the search for Lyapunov functions. </p>
<subsection><h1>Lyapunov analysis for linear systems</h1>
<p>Let's take a moment to see how things play out for linear systems.</p>
<theorem><h1>Lyapunov analysis for stable linear systems</h1>
<p>Imagine you have a linear system, $\dot\bx = {\bf A}\bx$, and can
find a Lyapunov function $$V(\bx) = \bx^T {\bf P} \bx, \quad {\bf P} =
{\bf P^T} \succ 0,$$ which also satisfies $$\dot{V}(\bx) = \bx^T {\bf
PA} \bx + \bx^T {\bf A}^T {\bf P}\bx \prec 0.$$ Then the origin is
globally asymptotically stable.</p>
</theorem>
<p>Note that the radially-unbounded condition is satisfied by ${\bf P}
\succ 0$, and that the derivative condition is equivalent to the matrix
condition $${\bf PA} + {\bf A}^T {\bf P} \prec 0.$$</p>
<p>For stable linear systems the existence of a quadratic Lyapunov
function is actually a necessary (as well as sufficient) condition.
Furthermore, a Lyapunov function can always be found by finding the
positive-definite solution to the matrix Lyapunov equation
\begin{equation}{\bf PA} + {\bf A}^T{\bf P} = - {\bf
Q},\label{eq:algebraic_lyapunov} \end{equation} for any ${\bf Q}={\bf
Q}^T\succ 0$.</p>
<todo> add an example here. double integrator? re-analyze the LQR output?
</todo>
<p>This is a very powerful result - for nonlinear systems it will be
potentially difficult to find a Lyapunov function, but for linear systems
it is straight-forward. In fact, this result is often used to propose
candidates for non-linear systems, e.g., by linearizing the equations and
solving a local linear Lyapunov function which should be valid in the
vicinity of a fixed point.</p>
</subsection> <!-- lyapunov for linear -->
<subsection><h1>Lyapunov analysis as a semi-definite program
(SDP)</h1>
<p>Lyapunov analysis for linear systems has an extremely important
connection to convex optimization. In particular, we could have also
formulated the Lyapunov conditions for linear systems above using
<em>semi-definite programming</em> (SDP). Semidefinite programming is a
subset of convex optimization -- an extremely important class of problems
for which we can produce efficient algorithms that are guaranteed find the
global optima solution (up to a numerical tolerance and barring any
numerical difficulties).</p>
<p>If you don't know much about convex optimization or want a quick
refresher, please take a few minutes to read the optimization
preliminaries in the appendix. The main requirement for this section is
to appreciate that it is possible to formulate efficient optimization
problems where the constraints include specifying that one or more
matrices are positive semi-definite (PSD). These matrices must be formed
from a linear combination of the decision variables. For a trivial
example, the optimization $$\min_a a,\quad \subjto \begin{bmatrix} a & 0
\\ 0 & 1 \end{bmatrix} \succeq 0,$$ returns $a = 0$ (up to numerical
tolerances).</p>
<p>The value in this is immediate for linear systems. For example, we
can formulate the search for a Lyapunov function for the linear system
$\dot\bx = {\bf A} \bx$ by using the parameters ${\bf p}$ to populate a
symmetric matrix ${\bf P}$ and then write the SDP: \begin{equation}
\find_{\bf p} \quad \subjto \quad {\bf P} \succeq 0, \quad {\bf PA} +
{\bf A}^T {\bf P} \preceq 0.\label{eq:lyap} \end{equation} Note that you
would probably never use that particular formulation, since there
specialized algorithms for solving the simple Lyapunov equation which are
more efficient and more numerically stable. But the SDP formulation does
provide something new -- we can now easily formulate the search for a
<i>"common Lyapunov function"</i> for uncertain linear systems.</p>
<example id="common-lyapunov-linear"><h1>Common Lyapunov analysis for
linear systems</h1>
<p>Suppose you have a system governed by the equations $\dot\bx = {\bf
A}\bx$, where the matrix ${\bf A}$ is unknown, but its uncertain
elements can be bounded. There are a number of ways to write down this
uncertainty set; let us choose to write this by describing ${\bf A}$ as
the convex combination of a number of known matrices, $${\bf A} =
\sum_{i} \beta_i {\bf A}_i, \quad \sum_i \beta_i = 1, \quad \forall i,
\beta_i > 0.$$ This is just one way to specify the uncertainty;
geometrically it is describing a polygon of uncertain parameters (in the
space of elements of ${\bf A}$ with each ${\bf A}_i$ as one of the
vertices in the polygon.</p>
<figure> <img width="50%"
src="data/affine_combination_of_matrices.svg"/> </figure>
<p>Now we can formulate the search for a common Lyapunov function using
\[ \find_{\bf p} \quad \subjto \quad {\bf P} \succeq 0, \quad
\forall_i, {\bf PA}_i + {\bf A}_i^T {\bf P} \preceq 0.\] The solver
will then return a matrix ${\bf P}$ which satisfies all of the
constraints, or return saying "problem is infeasible". It can easily be
verified that if ${\bf P}$ satisfies the Lyapunov condition at all of
the vertices, then it satisfies the condition for every ${\bf A}$ in the
set: \[ {\bf P}(\sum_i \beta_i {\bf A}_i) + (\sum_i \beta_i {\bf
A}_i)^T {\bf P} = \sum_i \beta_i ({\bf P A}_i + {\bf A}_i^T {\bf P})
\preceq 0, \] since $\forall i$, $\beta_i > 0$. Note that, unlike the
simple Lyapunov equation for a known linear system, this condition being
satisfied is a sufficient but not a necessary condition -- it is
possible that the set of uncertain matrices ${\bf A}$ is robustly
stable, but that this stability cannot be demonstrated with a common
quadratic Lyapunov function.</p>
<p> You can try this example for yourself in <drake></drake>.</p>
<script>document.write(notebook_link('lyapunov'))</script>
<p>As always, make sure that you open up the code and take a look.</p>
<p>There are many small variants of this result that are potentially of interest. For instance, a very similar set of conditions can certify "mean-square stability" for linear systems with multiplicative noise (see e.g. <elib>Boyd94</elib>, § 9.1.1).</p>
</example>
<todo>Add example or exercise based on e.g. Briat15 (LPV) example 1.3.1, showing off how powerful this can be. Also perhaps the parameter-dependent robust stability of his Def 2.3.5.</todo>
<p>This example is very important because it establishes a connection
between Lyapunov functions and (convex) optimization. But so far we've
only demonstrated this connection for linear systems where the PSD
matrices provide a magical recipe for establishing the positivity of the
(quadratic) functions for all $\bx$. Is there any hope of extending this
type of analysis to more general nonlinear systems? Surprisingly, it
turns out that there is.</p>
</subsection>
<subsection><h1>Lyapunov analysis for polynomial systems</h1>
<p><a href="optimization.html#sums_of_squares">Sums of squares
optimization</a> provides a natural generalization of SDP to optimizing
over positive polynomials (if you are not familiar, take a moment to <a
href="optimization.html#sums_of_squares">read the appendix</a>). This
suggests that it may be possible to generalize the optimization approach
using SDP to search for Lyapunov functions for linear systems to searching
for Lyapunov functions for at least the polynomial systems: $\dot\bx =
f(\bx),$ where $f$ is a vector-valued polynomial function. If we
parameterize a <em>fixed-degree</em> Lyapunov candidate as a polynomial
with unknown coefficients, e.g., \[ V_\alpha(\bx) = \alpha_0 + \alpha_1
x_1 + \alpha_2 x_2 + \alpha_3 x_1x_2 + \alpha_4 x_1^2 + ..., \] then the
time-derivative of $V$ is also a polynomial, and I can formulate the
optimization: \begin{align*} \find_\alpha, \quad \subjto \quad&
V_\alpha(\bx) \sos \\ & -\dot{V}_\alpha(\bx) = -\pd{V_\alpha}{\bx} f(\bx)
\sos. \end{align*} Because this is a convex optimization, the solver will
return a solution if one exists.</p>
<example><h1>Verifying a Lyapunov candidate via SOS</h1>
<p>This example is example 7.2 from <elib>Parrilo00</elib>. Consider
the nonlinear system: \begin{align*} \dot{x}_0 =& -x_0 - 2x_1^2 \\
\dot{x}_1 =& -x_1 - x_0 x_1 - 2x_1^3,\end{align*} and the <i>fixed
</i> Lyapunov function $V(x) = x_0^2 + 2x_1^2$, test if $\dot{V}(x)$ is
negative definite.</p>
<p>The numerical solution can be written in a few lines of code, and
is a convex optimization.</p>
<script>document.write(notebook_link('lyapunov'))</script>
</example>
<example><h1>Cart-pole swingup (again)</h1>
<p>In the <a href="#cartpole_swingup">cart-pole swingup example</a>, we
took $$V(\bx) = \frac{1}{2} \tilde{E}^2(\bx) = \frac{1}{2}
(\frac{1}{2}\dot\theta^2 - \cos\theta - 1)^2.$$ This is clearly a sum
of squares. Furthermore, we showed that $$\dot{V}(\bx) = -
\dot\theta^2 \cos^2\theta \tilde{E}^2(\bx),$$ which is also a sum of
squares. So the proposal of using sums-of-squares optimization is not so different, actually, than the recipes that nonlinear control theorists have been using (on pen and paper) for years. In this case, I would need a basis vector that includes many more monomials: $[1, \dot\theta, \cos\theta, ..., \dot\theta^3, ... ]^T,$ and setting up the equality constraints in an optimization requires more complicated term matching when trigonometric functions are involved, but the recipe still works.
</p>
</example>
<example><h1>Searching for a Lyapunov function via SOS</h1>
<p>Verifying a candidate Lyapunov function is all well and good, but
the real excitement starts when we use optimization to <i>find</i> the
Lyapunov function. In the following code, we parameterize $V(x)$ as
the polynomial containing all monomials up to degree 2, with the
coefficients as decision variables: $$V(x) = c_0 + c_1x_0 + c_2x_1 +
c_3x_0^2 + c_4 x_0x_1 + c_5 x_1^2.$$ We will set the scaling
(arbitrarily) to avoid numerical issues by setting $V(0)=0$, $V([1,0])
= 1$. Then we write: \begin{align*} \find_{\bc} \ \ \subjto \ \ &
V\text{ is sos, } \\ & -\dot{V} \text{ is sos.}\end{align*}</p>
<script>document.write(notebook_link('lyapunov'))</script>
<p>Up to numerical convergence tolerance, it discovers the same
coefficients that we chose above (zeroing out the unnecessary
terms).</p>
</example>
<p>It is important to remember that there are a handful of gaps which make
the existence of this solution a sufficient condition (for proving that
every sublevel set of $V$ is an invariant set of $f$) instead of a
necessary one. First, there is no guarantee that a stable polynomial
system can be verified using a polynomial Lyapunov function (of any
degree, and in fact there are known counter-examples
<elib>Ahmadi11a</elib>) and here we are only searching over a fixed-degree
polynomial. Second, even if a polynomial Lyapunov function does exist,
there is a gap between the SOS polynomials and the positive
polynomials.</p>
<p>Despite these caveats, I have found this formulation to be surprisingly
effective in practice. Intuitively, I think that this is because there is
relatively a lot of flexibility in the Lyapunov conditions -- if you can
find one function which is a Lyapunov function for the system, then there
are also many "nearby" functions which will satisfy the same
constraints.</p>
</subsection>
</section> <!-- computing Lyapunov -->
<section><h1>Lyapunov functions for estimating regions of
attraction</h1>
<p>There is another very important connection between Lyapunov functions and
the concept of an invariant set: <em>any sublevel set of a Lyapunov
function is also an invariant set</em>. This gives us the ability to use
sublevel sets of a Lyapunov function as approximations of the region of
attraction for nonlinear systems.</p>
<theorem><h1>Lyapunov invariant set and region of attraction
theorem</h1>
<p>Given a system $\dot{\bx} = f(\bx)$ with $f$ continuous, if we can find
a scalar function $V(\bx) \succ 0$ and a sublevel set $${\cal G}: \{ \bx
| V(\bx) \le \rho \}$$ on which $$\forall \bx \in {\cal G}, \dot{V}(\bx)
\preceq 0,$$ then ${\cal G}$ is an invariant set. By LaSalle, $\bx$ will
converge to the largest invariant subset of ${\cal G}$ on which
$\dot{V}=0$.<p>
<p>Furthermore, if $\dot{V}(\bx) \prec 0$ in ${\cal G}$, then the origin
is locally asymptotically stable and the set ${\cal G}$ is inside the
region of attraction of this fixed point. Alternatively, if $\dot{V}(\bx)
\preceq 0$ in ${\cal G}$ and $\bx = 0$ is the only invariant subset of
${\cal G}$ where $\dot{V}=0$, then the origin is asymptotically stable and
the set ${\cal G}$ is inside the region of attraction of this fixed point.
</p>
</theorem>
<example><h1>Region of attraction for a one-dimensional system</h1>
<p> Consider the first-order, one-dimensional system $\dot{x} = -x + x^3.$
We can quickly understand this system using our tools for graphical
analysis.</p> <figure> <img width="80%"
src="figures/cubicPolynomialExample.svg"/> </figure> <p>In the vicinity of
the origin, $\dot{x}$ looks like $-x$, and as we move away it looks
increasingly like $x^3$. There is a stable fixed point at the origin and
unstable fixed points at $\pm 1$. In fact, we can deduce visually that
the region of attraction to the stable fixed point at the origin is $\bx
\in (-1,1)$. Let's see if we can demonstrate this with a Lyapunov
argument.</p>
<p>First, let us linearize the dynamics about the origin and use the
Lyapunov equation for linear systems to find a candidate $V(\bx)$. Since
the linearization is $\dot{x} = -x$, if we take ${\bf Q}=1$, then we find
${\bf P}=\frac{1}{2}$ is the positive definite solution to the algebraic
Lyapunov equation (\ref{eq:algebraic_lyapunov}). Proceeding with $$V(\bx)
= \frac{1}{2} x^2,$$ we have $$\dot{V} = x (-x + x^3) = -x^2 + x^4.$$
This function is zero at the origin, negative for $|x| < 1$, and positive
for $|x| > 1$. Therefore we can conclude that the sublevel set $V <
\frac{1}{2}$ is invariant and the set $x \in (-1,1)$ is inside the region
of attraction of the nonlinear system. In fact, this estimate is
tight.</p>
</example>
<subsection id="common-lyapunov"><h1>Robustness analysis using "common Lyapunov functions"</h1>
<p>While we will defer most discussions on robustness analysis until <a
href="robust.html">later in the notes</a>, the idea of a <em>common
Lyapunov function</em>, which we introduced briefly for linear systems in
the <a href="#common-lyapunov-linear">example above</a>, can be readily
extended to nonlinear systems and region of attraction analysis. Imagine
that you have a model of a dynamical system but that you are uncertain
about some of the parameters. For example, you have a model of a
quadrotor, and are fairly confident about the mass and lengths (both of
which are easy to measure), but are not confident about the moment of
inertia. One approach to robustness analysis is to define a bounded
uncertainty, which could take the form of $$\dot{\bx} = f_\alpha(\bx),
\quad \alpha_{min} \le \alpha \le \alpha_{max},$$ with $\alpha$ a vector
of uncertain parameters in your model. Richer specifications of the
uncertainty bounds are also possible, but this will serve for our
examples.</p>
<p>In standard Lyapunov analysis, we are searching for a function that
goes downhill for all $\bx$ to make statements about the long-term
dynamics of the system. In common Lyapunov analysis, we can make many
similar statements about the long-term dynamics of an uncertain system if
we can find a single Lyapunov function that goes downhill <em>for all
possible values of $\alpha$</em>. If we can find such a function, then we
can use it to make statements with all of the variations we've discussed
(local, regional, or global; in the sense of Lyapunov, asymptotic, or
exponential).</p>
<example><h1>A one-dimensional system with gain uncertainty</h1>
<p>Let's consider the same one-dimensional example used above, but add
an uncertain parameter into the dynamics. In particular, consider the
system: $$\dot{x} = -x + \alpha x^3, \quad \frac{3}{4} < \alpha <
\frac{3}{2}.$$ Plotting the graph of the one-dimensional dynamics for a
few values of $\alpha$, we can see that the fixed point at the origin is
still stable, but the <em>robust region of attraction</em> to this fixed
point (shaded in blue below) is smaller than the region of attraction
for the system with $\alpha=1$.</p>
<figure> <img width="80%"
src="figures/cubicPolynomialExample_gain_uncertainty.svg"/>
</figure>
<p>Taking the same Lyapunov candidate as above, $V = \frac{1}{2} x^2$,
we have $$\dot{V} = -x^2 + \alpha x^4.$$ This function is zero at the
origin, and negative for all $\alpha$ whenever $x^2 > \alpha x^4$, or
$$|x| < \frac{1}{\sqrt{\alpha_{max}}} = \sqrt{\frac{2}{3}}.$$ Therefore,
we can conclude that $|x| < \sqrt{\frac{2}{3}}$ is inside the robust
region of attraction of the uncertain system.</p>
</example>
<p>Not all forms of uncertainty are as simple to deal with as the gain
uncertainty in that example. For many forms of uncertainty, we might not
even know the location of the fixed points of the uncertain system. In
this case, we can often still use common Lyapunov functions to give some
guarantees about the system, such as guarantees of <em>robust set
invariance</em>. For instance, if you have uncertain parameters on a
quadrotor model, you might be ok with the quadrotor stabilizing to a pitch
of $0.01$ radians, but you would like to guarantee that it definitely does
not flip over and crash into the ground.</p>
<example><h1>A one-dimensional system with additive uncertainty</h1>
<p>Now consider the system: $$\dot{x} = -x + x^3 + \alpha, \quad
-\frac{1}{4} < \alpha < \frac{1}{4}.$$ Plotting the graph of the
one-dimensional dynamics for a few values of $\alpha$, this time we can
see that the fixed point is not necessarily at the origin; the location
of the fixed point moves depending on the value of $\alpha$. But we
should be able to guarantee that the uncertain system will stay near the
origin if it starts near the origin, using an invariant set
argument.</p>
<figure>
<img width="80%" src="figures/cubicPolynomialExample_additive_uncertainty.svg"/>
</figure>
<p>Taking the same Lyapunov candidate as above, $V = \frac{1}{2} x^2$,
we have $$\dot{V} = -x^2 + x^4 + \alpha x.$$ This Lyapunov function
allows us to easily verify, for instance, that $V \le \frac{1}{3}$ is a
<em>robust invariant set</em>, because whenever $V = \frac{1}{3}$, we
have $$\forall \alpha \in [\alpha_{min},\alpha_{max}],\quad
\dot{V}(x,\alpha) < 0.$$ Therefore $V$ can never start at less than
one-third and cross over to greater than one-third. To see this, see
that $$ V=\frac{1}{3} \quad \Rightarrow \quad x = \pm \sqrt{\frac{2}{3}}
\quad \Rightarrow \quad \dot{V} = -\frac{2}{9} \pm \alpha
\sqrt{\frac{2}{3}} < 0, \forall \alpha \in
\left[-\frac{1}{4},\frac{1}{4} \right]. $$ Note that not all sublevel
sets of this invariant set are invariant. For instance $V <
\frac{1}{32}$ does not satisfy this condition, and by visual inspection
we can see that it is in fact not robustly invariant.</p>
</example>
<todo>robust quadrotor example</todo>
</subsection> <!-- end common lyap -->
<subsection><h1>Region of attraction estimation for polynomial systems</h1>
<p>Now we have arrived at the tool that I believe can be a work-horse for
many serious robotics applications. Most of our robots are not actually
globally stable (that's not because they are robots -- if you push me hard
enough, I will fall down, too), which means that understanding the regions
where a particular controller can be guaranteed to work can be of critical
importance.</p>
<p>Sums-of-squares optimization effectively gives us an oracle which we
can ask: is this polynomial positive for all $\bx$? To use this for
regional analysis, we have to figure out how to modify our questions to
the oracle so that the oracle will say "yes" or "no" when we ask if a
function is positive over a certain region which is a subset of $\Re^n$.
That trick is called the S-procedure. It is closely related to the
Lagrange multipliers from constrained optimization, and has deep
connections to "Positivstellensatz" from algebraic geometry.</p>
<subsubsection><h1>The S-procedure</h1>
<p>Consider a scalar polynomial, $p(\bx)$, and a semi-algebraic set
$g(\bx) \preceq 0$, where $g$ is a vector of polynomials. If I can
find a polynomial "multiplier", $\lambda(\bx)$, such that \[ p(\bx) +
\lambda^T(\bx) g(\bx) \sos, \quad \text{and} \quad \lambda(\bx) \sos,
\] then this is sufficient to demonstrate that $$p(\bx)\ge 0 \quad
\forall \bx \in \{ \bx | g(\bx) \le 0 \}.$$ To convince yourself,
observe that when $g(\bx) \le 0$, it is only harder to be positive, but
when $g(\bx) > 0$, it is possible for the combined function to be SOS
even if $p(\bx)$ is negative. We will sometimes find it convenient to
use the short-hand: $$g(\bx) \le 0 \Rightarrow p(\bx) \ge 0$$ to denote
the implication certified by the S-procedure (e.g. "whenever $g(\bx) \le 0$, we have $p(\bx) \ge 0$"). </p>
<p>We can also handle equality constraints with only a minor
modification -- we no longer require the multiplier to be positive. If
I can find a polynomial "multiplier", $\lambda(\bx)$, such that
\[p(\bx) + \lambda^T(\bx) g(\bx) \sos \] then this is sufficient to
demonstrate that $$p(\bx)\ge 0 \quad \forall \bx \in \{ \bx | g(\bx) = 0
\}.$$ Here the intuition is that $\lambda(x)$ can add arbitrary
positive terms to help me be SOS, but those terms contribute nothing
precisely when $g(x)=0$.</p>
</subsubsection>
<subsubsection><h1>Basic region of attraction formulation</h1>
<p>The S-procedure gives us the tool we need to evaluate positivity
only over a region of state space, which is precisely what we need to
certify the Lyapunov conditions for a region-of-attraction analysis.
Let us start with a positive-definite polynomial Lyapunov candidate,
$V(\bx) \succ 0$, then we can write the Lyapunov conditions:
$$\dot{V}(\bx) \prec 0 \quad \forall \bx \in \{ \bx | V(\bx) \le \rho
\},$$ using sums-of-squares and the S-procedure: $$-\dot{V}(\bx) +
\lambda(\bx)(V(\bx) - \rho) \text{ is SOS,} \quad \text{and} \quad
\lambda(\bx) \text{ is SOS},$$ where $\lambda(\bx)$ is a multiplier
polynomial with free coefficients that are to be solved for in the
optimization.</p>
<p>I think it's easiest to see the details in an example.</p>
<example id="roa_cubic_system"><h1>Region of attraction for the
one-dimensional cubic system</h1>
<p>Let's return to our example from above: \[ \dot{x} = -x + x^3 \]
and try to use SOS optimization to demonstrate that the region of
attraction of the fixed point at the origin is $x \in (-1,1)$, using
the Lyapunov candidate $V = x^2.$ </p>
<p>First, define the multiplier polynomial, \[ \lambda(x) = c_0 + c_1
x + c_2 x^2. \] Then define the optimization
\begin{align*} \find_{\bf c} \quad & \\ \subjto \quad& - \dot{V}(x) +
\lambda(x) (V(x)-1) \sos \\ & \lambda(x) \sos \end{align*} </p>
<p>You can try this example for yourself in <drake></drake>.</p>
<script>document.write(notebook_link('lyapunov'))</script>
</example>
<p>In this example, we only verified that the one-sublevel set of the
pre-specified Lyapunov candidate is negative (certifying the ROA that
we already understood). Even more useful is if you are able to search
for the largest $\rho$ that can satisfy these conditions.
Unfortunately, in this first formulation, optimizing $\rho$ directly
would make the optimization problem non-convex because we would have
terms like $\rho c_0$, $\rho c_1 x$, ... which are bilinear in the
decision variables; we need the sums-of-squares constraints to be only
linear in the decision variables.</p>
<p>Fortunately, because the problem is convex with $\rho$ fixed (and
therefore can be solved reliably), and $\rho$ is just a scalar, we can
perform a simple line search on $\rho$ to find the largest value for
which the convex optimization returns a feasible solution. This will
be our estimate for the region of attraction.</p>
<p>There are a number of variations to this basic formulation; I will
describe a few of them below. There are also important ideas like
rescaling and degree-matching that can have a dramatic effect on the
numerics of the problem, and potentially make them much better for the
solvers. But you do not need to master them all in order to use the
tools effectively.</p>
<example><h1>Region of Attraction codes in Drake</h1>
<p>In <drake></drake>, we have packaged most of the work in setting up
and solving the sums-of-squares optimization for regions of attraction
into a single method <code>RegionOfAttraction(system, context,
options)</code>. This makes it as simple as, for instance:</p>
<pre><code class="python">x = Variable("x")
sys = SymbolicVectorSystem(state=[x], dynamics=[-x+x**3])
context = sys.CreateDefaultContext()
V = RegionOfAttraction(sys, context)
</code></pre>
<script>document.write(notebook_link('lyapunov'))</script>
<p>Remember that although we have tried to make it convenient to call
these functions, they are not a black box. I highly recommend
opening up the <code>RegionOfAttraction</code> method and
understanding how it works. There are lots of different options /
formulations, and numerous numerical recipes to improve the numerics
of the optimization problem.</p>
</example>
<!--
<example><h1>Region of attraction for the time-reversed Van der Pol oscillator</h1>
<pre><code class="matlab" testfile="testVanDerPol">
cd(fullfile(getDrakePath,'examples'));
VanDerPol.run();
</code></pre>
</example>
-->
</subsubsection>
<subsubsection><h1>The equality-constrained formulation</h1>
<p>Here is one important variation for finding the level set of a
candidate region of attraction, which turns an inequality in the
S-procedure into an equality. This formulation <i>is</i> jointly
convex in $\lambda(\bx)$ and $\rho$, so one can optimize them in a
single convex optimization. It appeared informally in an example in
<elib>Parrilo00</elib>, and was discussed a bit more in
<elib>Shen20</elib>.</p>
<p>Under the assumption that the Hessian of $\dot{V}(\bx)$ is
negative-definite at the origin (which is easily checked), we can write
\begin{align*} \max_{\rho, \lambda(\bx)} & \quad \rho \\ \subjto &
\quad (\bx^T\bx)^d (V(\bx) - \rho) + \lambda(\bx) \dot{V}(\bx) \text{
is SOS},\end{align*} with $d$ a fixed positive integer. As you can
see, $\rho$ no longer multiplies the coefficient of $\lambda(\bx)$.
But why does this certify a region of attraction?</p>
<p>You can read the sums-of-squares constraint as certifying the
implication that whenever $\dot{V}(x) = 0$, we have that either $V(\bx)
\ge \rho$ OR $\bx = 0$. Multiplying by some multiple of $\bx^T\bx$ is
just a clever way to handle the "or $\bx=0$" case, which is necessary
since we expect $\dot{V}(0) = 0$. This implication is sufficient to
prove that $\dot{V}(\bx) \le 0$ whenever $V(\bx) \le \rho$, since $V$
and $\dot{V}$ are smooth polynomials; we examine this claim in full detail in <a href="#van_der_pol_roa">one of the exercises</a>.</p>
<p>Using the S-procedure with equalities instead of inequalities also
has the potential advantage of removing the SOS constraint on
$\lambda(\bx).$ But perhaps the biggest advantage of this formulation
is the possibility of dramatically simplifying the problem using the
quotient ring of this algebraic variety, and in particular some recent
results for exact certification using sampling
varieties<elib>Shen20</elib>.</p>
</subsubsection>
<subsubsection><h1>Searching for $V(\bx)$</h1>
<p>The machinery so far has used optimization to find the largest
region of attraction that can be certified <i>given a candidate
Lyapunov function</i>. This is not necessarily a bad assumption. For
most stable fixed points, we can certify the local stability with a
linear analysis, and this linear analysis gives us a candidate
quadratic Lyapunov function that can be used for nonlinear analysis
over a region. Practically speaking, when we solve an LQR problem, the
cost-to-go from LQR is a good candidate Lyapunov function for the
nonlinear system. If we are simply analyzing an existing system, then
we can obtain this candidate by solving a Lyapunov equation (Eq
\ref{eq:algebraic_lyapunov}).</p>
<p>But what if we believe the system is regionally stable, despite
having an indefinite linearization? Or perhaps we can certify a larger
volume of state space by using a Lyapunov candidate with degree greater
than two. Can we use sums-of-squares optimization to find that in the
region of attraction case, too?</p>
<p>To accomplish this, we will now make $V(x)$ a polynomial (of some
fixed degree) with the coefficients as decision variables. First we
will need to add constraints to ensure that $$V(0) = 0 \quad {and}
\quad V(\bx) - \epsilon \bx^T\bx \text{ is SOS},$$ where $\epsilon$ is
some small positive constant. This $\epsilon$ term simply ensures that
$V$ is strictly positive definite. Now let's consider our basic
formulation: $$-\dot{V}(\bx) + \lambda(\bx)(V(\bx) - 1) \text{ is
SOS,} \quad \text{and} \quad \lambda(\bx) \text{ is SOS}.$$ Notice
that I've replaced $\rho=1$; now that we are searching for $V$ the
scale of the level-set can be set aribtrarily to 1. In fact, it's
better to do so -- if we did not set $V(0)=0$ and $V(\bx) \le 1$ as the
sublevel set, then the optimization problem would be underconstrained,
and might cause problems for the solver.</p>
<p>Unfortunately, the derivative constraint is now nonconvex (bilinear)
in the decision variables, since we are searching for both the
coefficients of $\lambda$ and $V$, and they are multiplied together.
Our equality-constrained formulation doesn't get us around this one,
either (since the coefficients of $V$ also appear in $\dot{V}$). Here
we have to resort to some weaker form of optimization. In practice, we
have had good practical success using bilinear alternations: start with
an initial $V$ (e.g. from LQR), and search for $\lambda$; then fix
$\lambda$ and search for $V$ and repeat until convergence (see, for
instance <elib>Tedrake10+Majumdar16a</elib>). A typical choice for the
objective is to maximize some convex surrogate for the volume of the
certified region of attraction in state space; typically we use the
determinant of the quadratic form describing a contained ellipse.</p>
<p>Barring numerical issues, this algorithm is guaranteed to have
recursive feasibility and tends to converge in just a few alternations,
but there is no guarantee that it finds the optimal solution.</p>
<example><h1>Searching for Lyapunov functions</h1>
<p>(Details coming soon...)</p>
<script>document.write(notebook_link('lyapunov'))</script>
</h1></example>
</subsubsection>
<subsubsection><h1>Convex outer approximations of the ROA</h1>
<p>All of our region of attraction approximations so far have been
"inner approximations" -- the certified region is guaranteed to be
contained within the true region of attraction for the system. This is
what we want if we are to give some guarantee of stability. As we
increase the degree of our polynomial and multipliers, we expect that
the estimated regions of attraction will converge on the true
regions.</p>
<p>It turns out that if we instead consider outer approximations, which
converge on the true ROA from the other side, we can write formulations
that enable searching for the Lyapunov function directly as a convex
optimization. As we will see below, this approach also allows one to
write convex formulations for controller design. These methods have
been explored in a very nice line of work by Henrion and Korda (e.g.
<elib>Henrion14</elib>), using the method of moments from
<elib>Lasserre10</elib> also called "occupation measures". Their
treatment emphasizes infinite-dimensional linear programs and
heirarchies of LMI relaxations; these are just dual formulations to the
SOS optimizations that we've been writing here. I find the SOS form
more clear, so will stick to it here. Some of the occupation measure
papers include the dual formulation of the infinite-dimension linear
program which will look quite similar; my coauthors and I typically
call out the sums-of-squares version in our papers on the topic (e.g.
<elib>Posa17</elib>).</p>
<p>To find an outer approximation, instead of solving for a Lyapunov
function that certifies convergence to the origin, we use a very
related set of conditions to search for a Lyapunov-like "barrier
certificate", $\mathcal{B}(\bx).$ Like a Lyapunov function, we'd like
$\dot{\mathcal{B}}(\bx) \leq 0$; this time we'll ask for this to be
true everywhere (or at least in some set that is sufficiently larger
than the ROA of interest). Then we will set $\mathcal{B}(0) > 0$. If
we can find such a function, then certainly any state for which
$\mathcal{B}(\bx) < 0$ is outside of the region of attraction of the
fixed point -- since $\mathcal{B}$ cannot increase it can never reach
the origin which has $\mathcal{B} > 0$. The <i>super</i>level set
$\{ \bx | \mathcal{B}(\bx) \ge 0\}$ is an outer approximation of the