Lyapunov Stability

This tech note summarizes Chapter 4 of “Nonlinear Systems” by Hassan K. Khalil (3rd edition). I have tried to be as precise as possible, but one should refer to the textbook for rigourous definitions. The proofs are also not repeated here.

Autonomous Systems:

Consider the system

\[\dot x = f(x)\]

where

\(f: D \rightarrow R^n\) is a locally Lipschitz map from \(D \sub R^n\) to \(R^n\)
\(x=0\) is an equilibium point, ie \(f(0) =0\)

Definitions: The equilibrium point \(x=0\) is

unstable if not stable
stable if, for every \(\epsilon > 0\) there exists a \(\delta = \delta(\epsilon) > 0\) such that

\[|| x(0) || < \delta \implies || x(t) || < \epsilon, \quad \forall t\geq 0\]

asymptotically stable if \(x=0\) is stable, and \(\delta\) can be chosen such that

\[||x(0) || < \delta \implies \lim_{t\rightarrow 0} x(t) = 0\]

exponentially stable if \(x=0\) is stable, and constants \(L, \gamma\) can be chosen such that

\[||x(0) || < \delta \implies ||x(t)|| < L \|x(0)\| e^{-\gamma t}, \quad \forall t\geq 0\]

Note: locally Lipschitz does not imply that a solution is defined for \(t\geq0\), but the additional conditions involved in Lyapunov stability guarantee this.

Theorem 4.1: Lyapunov’s theorem

Let \(x=0\) be an equilibrium point
Let \(D \sub R^n\) be a domain containing \(x=0\)
Let \(V : D \rightarrow R\) be a continously differentiable function such that \(V\) is positive definite.

If \(\dot V\) is negative semi-definite, then \(x = 0\) is stable.

If \(\dot V\) is negative definite, then \(x=0\) is asymptotically stable.

Proof: Construct suitable subsets, and apply the definitions of stability etc.

Some useful Lyapunov Functions

Same sector nonlinearity:

Suppose we have

\[\dot x = - g(x)\]

where \(g(0) = 0\) and \(x g(x) > 0\) for all \(x \in D\) and \(x \neq 0\). Essentially looks like a feedback system.

So we choose

\[V(x) = \int_0^x g(y) dy\]

(which is positive definite) and now \(\dot V = g(x) \dot x = - g(x)^2\) and so the Lyapunov conditions are satisfied.

Variable Gradient Method:

For some \(V(x)\) let

\[g(x) = (dV/dx)^T\]

Then \(\dot V = g(x)^T f(x)\) and we need the jacobian \(dg/dx\) to be symmetric. So choose a \(g(x)\) (based on what \(f\) looks like) such that \(g(x)^T f(x)\) is negative definite.

Then compute

\[V(x) = \int_0^x g(y)^T dy = \int_0^{x_1} g_1(y_1, 0, ..., 0) dy_1 + \int_0^{x_2} g_2(0, y_2, 0, ..., 0) dy_2 + ...\]

If we chose \(g\) with some unknown parameters, we can now choose those parameters such that the resulting \(V\) is positive definite. (See example 4.5 of Khalil to see this in practice).

Theorem 4.2

Let \(x=0\) be an equilibrium point
Let \(V : R^n \rightarrow R\) be a continously differentiable function such that \(V\) is positive definite on the entire domain.
Let \(\|x\| \rightarrow \infty \implies V(x) \rightarrow \infty\) (ie is radially unbounded)
Let \(\dot V\) be negative definite

Then \(x=0\) is globally asymptotically stable

Globally just means that the property holds for all \(x \in R^n\).

Theorem 4.3: Chetaev’s theorem: (Converse Theorem)

Let \(V: D \rightarrow R\) be a continously differentiable function (not necessarily positive definite) and \(D\) contains the origin, and \(V(0) = 0\)
Suppose there exists a \(x_0\) arbitrarily close to the origin such that \(V(x_0) > 0\).
Choose \(r> 0\) such that \(B_r = \{x : \|x\| < r\}\) is contained in \(D\).
Let \(U = \{ x \in B_r : V(x) > 0\}\)
Suppose that \(\dot V(x) > 0\) for all \(x \in U\)

Then \(x=0\) is unstable

Sketch of Proof: Notice, by construction, there exists \(x_0 \in U\) where \(\dot V(x_0) > 0\). Since \(x_0\) can be chosen arbitrarily close to the origin, there exists an arbitrarily small ball (of radius \(r = \|x_0\|\)) such that the system must leave this ball. Thus, the origin cannot be stable.

Definitions

A set \(M\) is said to be invariant with respect to \(\dot x = f(x)\) if \(x(0) \in M \implies x(t) \in M \forall t\).
A set \(M\) is positively invariant if (above) for all \(t\geq 0\).
A system approaches a set \(M\) if, for each \(\epsilon > 0\) there exists a \(T = T(\epsilon) > 0\) such that \(dist(x(t), M) < \epsilon\) for all \(t \geq T\).

Theorem 4.4: La Salle’s Theorem

Let \(\Omega \in D\) be a compact set that is positively invariant with respect to \(\dot x = f(x)\).
Let \(V : D \rightarrow R\) be a continously differentiable function
Let \(\dot V \leq 0\) for all \(x \in \Omega\) (no need for positive definite \(V\))
Let \(E = \{ x \in \Omega : \dot V = 0\}\)
Let \(M\) be the largest invariant set in \(E\)

Then every solution starting in \(\Omega\) approaches \(M\) as \(t \rightarrow \infty\).

Note, if the only invariant point is \(M = \{ 0 \}\) then the system must be asymptotically stable.

If you use these theorems to prove stability, you have used Lyapunov’s direct method. The indirect method is based on linearization:

Theorem 4.5: The equilibrium point \(x=0\) of \(\dot x = A x\) is

stable if, and only if,
- eigenvalues of \(A\) satisfy
\[Re(\lambda_i) \leq 0\]
- AND for any eigenvalue with \(Re(\lambda_i) = 0\) with algebraic multiplicity \(q_i \geq 2\), we have
\[\text{dim} N(A-\lambda_i I) = q_i\]
ie all the eigenvectors are of rank 1, (ie no generalised eigenvectors).
asymptotically stable, if and only if

\[Re(\lambda_i) < 0\]

for all eigenvectors. That is, \(A\) is a Hurwitz matrix.

Theorem 4.6

A matrix \(A\) is Hurwitz, if, and only if, for any positive definite symmetric matrix \(Q\) there exists a positive definite symmetric matrix \(P\) such that

\[PA + A^T P = -Q\]

Moreover, \(P\) is unique.

Proof: To prove sufficiency, let \(V(x) = x^T P x\) work through \(\dot V\), and define \(Q\) as needed. To prove necessity, define \(P = \int_0 ^\infty \exp(A^T t) Q \exp(At) dt\), which must exist since \(A\) is stable. And then prove that \(P\) is positive definite. Substitute \(P\) into the Lyapunov equation, and show that it can be simplified to \(-Q\).

Note \(Q\) can be relaxed to positive semi-definite, if \(Q = C^T C\) and \((A, C)\) is observable.

The key benefit of this method is that we can make the following claim:

Theorem 4.7

Let \(x=0\) be an equilibrium point for the nonlinear system \(\dot x = f(x)\).
Let

\[A = \frac{df}{dx}(x) |_{x=0}\]

If \(A\) is Hurwitz, then the nonlinear systme is asymptotically stable.

If \(Re(\lambda_i) \geq 0\) for one or more eigenvalues, then the origin is unstable.

Proof: essentially you can show that there is a small enough neighborhood of the origin that is dominated by the linear term \(\dot x = Ax\).

Note, if \(Re(\lambda_i) = 0\) for some \(\lambda_i\), the system could be stable, asymptotically stable, or unstable! - there is no way to know, without greater analysis

Non-Autonomous Systems

Now we extend these ideas to systems of

\[\dot x = f(t, x)\]

where

\[f : [0, \infty) \times D \rightarrow R\]
\(f(t, 0) = 0\) for all \(t \geq 0\). (this can be assumed, since we can change coordinates: \(x = y - \bar y\), where \(\bar y\) is a solution to the above equation)
\(D\) contains \(x=0\).
\(f\) is piecewise continous in \(t\)
\(f\) is locally Lipschitz in \(x\) on \([0, \infty) \times D\).

Definitions: The equilibrium point \(x=0\) for the non-autonomous system is

unstable if it is not stable.
stable, if, for every \(\epsilon > 0\) there exists a \(\delta = \delta(\epsilon, t_0) > 0\) such that

\[\|x(t_0)\| < \delta(\epsilon, t_0) \implies \| x(t) \| < \epsilon, \quad \forall t \geq t_0 \geq 0\]

uniformly stable if for each \(\epsilon > 0\), there is a \(\delta = \delta(\epsilon) >0\), independent of \(t_0\), such that

\[\|x(t_0)\| < \delta(\epsilon) \implies \| x(t) \| < \epsilon, \quad \forall t \geq t_0 \geq 0\]

asymptotically stable if it is stable, and there is a positive constant \(c = c(t_0) > 0\) such that

\[\lim_{t\rightarrow\infty} x(t) = 0, \quad \forall \| x(t_0) \| < c(t_0)\]

uniformly asymptotically stable if it is uniformly stable, and there is a \(c >0\), independent of \(t_0\), such for all \(\|x(t_0) \| < c\),

\[x(t) \rightarrow 0 \text{ as } t \rightarrow \infty\]

That is, for each \(\eta > 0\), there is a \(T = T(\eta) > 0\) such that

\[\| x(t) \| < \eta, \quad \forall t \geq t_0 + T(\eta), \quad \forall \|x(t_0)\| < c\]

globally uniformly assymptotically stable if
1. it is uniformly stable,
2. \(\delta(\epsilon)\) can be chosen to statisfy \(\lim_{\epsilon \rightarrow \infty} \delta(\epsilon) = \infty\)
3. for each pair of positive numbers \(\eta, c > 0\) there exists a \(T(\eta, c) > 0\) such that
\[\| x(t) \| < \eta, \quad \forall t\geq t_0 + T(\eta, c), \quad \forall \|x(t_0)\| < c\]

Lemma 4.5 The equilibrium point \(x=0\) is

uniformly stable if, and only if, there exists a class \(K\) function \(\alpha\) and a positive constant \(c>0\) independent of \(t_0\) such that

\[\| x(t) \| < \alpha(\| x(t_0) \|), \quad \forall t \geq t_0 \geq 0, \quad \forall \| x(t_0) \| < c\]

uniformly asymptotically stable if, and only if, there exists a class \(KL\) function \(\beta\) and a positive constant \(c>0\) independent of \(t_0\), such that

\[\| x(t) \| < \beta ( \| x(t_0)\| , t- t_0), \quad \forall t \geq t_0 \geq 0, \quad \forall \| x(t_0) \| < c\]

exponentially stable if there exist positive constants \(c, k, \lambda > 0\) such that

\[\| x(t) \| < k \| x(t_0)\| e^{-\lambda (t-t_0)}, \quad \forall \| x(t_0) \| < c\]

globally exponentially stable if the system is exponentially stable for all \(x(t_0)\).

Theorem 4.8 Lyapunov’s theorem for non-autonomous systems:

Let \(x=0\) be an equilibrium point
Let \(D \sub R^n\) be a domain containing \(x=0\).
Let \(W_1(x), W_2(x)\) be continous positive definite functions on \(D\).
Let \(V : D \rightarrow R\) be a continously differentiable function such that

\[W_1(x) \leq V(t, x) \leq W_2(x)\] \[\frac{\partial V}{\partial t} + \frac{\partial V}{\partial x} f(t, x) \leq 0\]

for all \(t \geq 0\) and for all \(x \in D\).

Then \(x=0\) is uniformly stable.

Theorem 4.9 extends this: if we also have

\[\frac{\partial V}{\partial t} + \frac{\partial V}{\partial x} f(t, x) \leq -W_3(x)\]

where \(W_3\) is a positive definite function on \(D\), then \(x=0\) is uniformly asymptotically stable.

If \(D = R^n\) and \(W_1\) is radially unbouded, then \(x=0\) is globally uniformly asymptotically stable.

Theorem 4.10 says if there exists \(k_1, k_2, k_3, a > 0\) such that it is uniformly asymptotically stable with \(W_1(x) = k_1 \|x\| ^ a, W_2(x) = k_2\|x\|^a, W_3(x) = k_3\|x\| ^a\), then the origin is exponentially stable.

If this holds globally, it is globally exponentially stable.

todo

application of linearization to non-autonomous systems, and the Lyapunov differential equations
converse theorems (these essentially say that stability of some form exist, a Lyapunov function, satisifying the corresponding conditions exists)
boundedness and ultimate boundedness (which is a relaxation on stability requirements)
Input to state stability (which leads to \(\mathcal{L}\) stability, as in Chapter 5)