Welcome back. In today’s blog post, I am excited to discuss Lyapunov functions, which can prove the stability of equilibrium points of a broad class of dynamical systems, including nonlinear systems. Lyapunov functions generalize the concept of energy from mechanical systems. If a pendulum is damped, then its downward position is a stable equilibrium point, and its energy will decrease as the pendulum approaches its stable equilibrium position. Similarly, if we take a Lyapunov function and evaluate it along trajectories near a stable equilibrium point of any dynamical system, then the value of the Lyapunov function will decrease as the trajectory approach the stable equilibrium. In the following paragraphs, we will explore the basic properties of Lyapunov functions, and then we will focus on linear time-invariant systems to see how the theory can be applied in this basic but important case. For simplicity, we will restrict our entire discussion to autonomous systems, but our principles can be generalized to time-varying systems.
Suppose we have an autonomous nonlinear system of the form:
dx/dt = f(x),
where x ∈ ℝn is an n-dimensional state vector and f: ℝn → ℝn is an n-dimensional vector field. Let’s also suppose that xeq is an equilibrium point of the system (that is to say, f(xeq) = 0). Without loss of generality, we can assume that xeq = 0 (if xeq ≠ 0, then we can shift our coordinate system to make xeq coincide with the origin).
A big theorem then tells us that, xeq = 0 is locally stable in the sense of Lyapunov if and only if there exists a function V: ℝn → [0, ∞) and a neighborhood D of the origin such that:
(1) V(0) = 0
(2) V > 0 for all x ∈ D\{0}
(3a) dV/dt = [∇V]T f(x) ≤ 0 for all x ∈ D\{0}.
By stable in the sense of Lyapunov, we mean that trajectories starting close to xeq = 0 will remain close to xeq for all time. Therefore, a linear center would be classified as stable in the sense of Lyapunov. If an equilibrium point is not stable, then it is unstable.
The theorem goes on to say that xeq = 0 is asymptotically stable if and only if properties (1) and (2) are satisfied and:
(3b) dV/dt = [∇V]T f(x) < 0 for all x ∈ D\{0}.
The difference between stability and asymptotic stability is the strict inequality in (3b). If xeq is asymptotically stable, then trajectories starting close will not only stay close, but also converge to xeq = 0 as t → ∞.
These stability definitions in terms of Lyapunov functions are useful because they bypass potentially costly computations of the Jacobian of f around xeq and its eigendecomposition. These Lyapunov functions also give us an idea of the region of attraction. If the domain D in the theorem above is very large, then xeq has a large basin of attraction. If D is very small, then trajectories must start extremely close to xeq in order to stay close. If D = ℝn, then xeq = 0 is globally stable, asymptotically stable, or unstable.
We can specialize our theory of Lyapunov functions to linear time-invariant systems of the form:
dx/dt = Ax,
where A ∈ ℝn ✕ n is a constant matrix. Since f(x) = Ax is linear, valid Lyapunov functions can be constructed as quadratic forms, V(x) = xT P x, where P is a symmetric matrix. One very important condition for P is that it is positive definite, as then it satisfies properties (1) and (2) by definition:
(1) V(0) = 0
(2) V(x) = xT P x > 0 for x ∈ ℝn\{0}.
If we want to show that xeq = 0 is globally asymptotically stable, then we need dV/dx < 0 for x ∈ ℝn\{0}. For a quadratic Lyapunov function, we can ensure this property if dV/dx = -xT Q x for some symmetric positive-definite matrix Q (recall that if Q is positive-definite, then -Q is negative-definite). From V(x) = xT P x, we already know that
dV/dt = (dx/dt)T P x + xT P (dx/dt) =
= xT AT P x + xT P A x = xT (AT P + P A) x.
Hence, to construct a Lyapunov function for dx/dt = Ax, we can solve the following algebraic Lyapunov equation for P:
AT P + P A = -Q.
In fact, there is a theorem that says that for a linear time-invariant system dx/dt = Ax, the origin is globally asymptotically stable if and only if for any symmetric positive-definite matrix Q = QT > 0, there exists a symmetric positive-definite P = PT > 0 such that
AT P + P A = -Q.
Next week we will explore dynamical systems further and potentially see how Lyapunov function theory can be applied to fluid mechanics. Until then, please take care.
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