Riassunti Biomatematical modeling

Short notes on dynamical systems

Definition 0.1

Let X be a set and (T, +) an abelian group with identity element e. A dynamical system on X (with time T) is a map φ: T × X → X such that:

For all x ∈ X: φ(e, x) = x;

For all s, t ∈ T, x ∈ X: φ(s, φ(t, x)) = φ(s + t, x).

A dynamical system on T = Z is called discrete; a dynamical system with T = ℝ is called continuous.

From the definition, it is clear that φ(s, φ(t, x)) = φ(t, φ(s, x)), since they are both equal to φ(s + t, x). The function t → φ(t, x) is called trajectory or orbit through x.

Definition 0.2

An x ∈ X such that for all t₁, t₂ ∈ T: φ(t₁, x) = φ(t₂, x) is called an equilibrium position or equilibrium point for the system.

Clearly, taking t₁ = t₂ ∈ T and t₂ = e, if x is an equilibrium position it holds for all t ∈ T: φ(t, x) = φ(e, x) = x.

Something more can be said if X is a vector space, or a difference space (like the Euclidean space). Then x - y is a vector on some vector space and then the quotient (φ(t + ε, x) - φ(t, x))/ε is well-defined. Then, if the limit of this quotient exists,

\[ \frac{d}{dt} \phi(t + \epsilon, x) - \phi(t, x) = \lim_{\epsilon \to 0} \frac{\phi(t + \epsilon, x) - \phi(t, x)}{\epsilon} \]

is clearly a function F of φ(t, x) only. One therefore has φ'(t, x) = F(φ(t, x)).

Therefore, differentiable continuous dynamical systems obey a “differential equation”. More or less equivalent notations are u(t) for φ(t, x) (but forgetting the fact that u(0) = x, or S(t)x). The maps S(t) form then a group with respect to composition with S(0) as identity element, since S(t + s) = S(t)S(s) and S(0) = Id (remember that the composition is always associative). Notice also that, since S(-t)S(t) = Id, that S(-t) = S(t)^-1.

In some cases, the set T is not a group but a semigroup, like ℝ₊. In this case, the maps S(t) form a semigroup with respect to composition.

Conversely, if a function t → u(t) on a vector or difference space X which allows derivation (i.e., has a topological structure) verifies:

u'(t) = F(u(t)), u(0) = x

then the function (t, x) → φ(t, x) = u(t) is a dynamical system on X with time T = ℝ.

First, obviously φ(0, x) = u(0) = x and second, it is easy to verify that for every s the function v: t → u(t + s) verifies:

v'(t) = F(v(t)) where v(0) = u(s).

Then, v(t) = u(t + s) = φ(t + s, x) but since F remains the same, v(t) = φ(t, u(s)) = φ(t, φ(s, x)), which is the second property of (1).

From now on, we will only consider the case of discrete or continuous dynamical systems.

Definition 0.3

Let X be a topological space and let 𝒩_x be the set of all neighbourhoods of x ∈ X. An equilibrium position x of a dynamical system is said to be stable if:

For all V ∈ 𝒩_x, there exists U ∈ 𝒩_x such that for all x ∈ U and all t ∈ T: φ(t, x) ∈ V, otherwise, it is said to be unstable.

Instability means in general unobservability. Stable equilibrium positions may be observable if the neighbourhood U is not too small. Stability is very difficult to check if the behaviour of φ is not explicitly known and if X is a general topological space; moreover, stability properties may change if the topology changes.

A simpler situation appears when X = ℝⁿ and T = ℝ, i.e., the case of ordinary differential equations.

Theorem 0.1

Let t → u(t) be a solution of (2) and suppose that it exists and is unique for all t ∈ ℝ. Let x be an equilibrium position of the system and set:

\[ A = \frac{\partial F}{\partial u} \bigg|_{x} \]

If all the eigenvalues of A have strictly negative real parts, then x is stable. If there exists at least one eigenvalue of A with a strictly positive real part, then x is unstable.

Example 0.1

Take X = ℝ², ω = 0, and the system:

• x' = y
• y' = -ω²sin(x)

which is the system associated with the simple pendulum. Then x = (π, 0) is an equilibrium position and:

\[ A = \begin{pmatrix} 0 & 1 \\ -ω^2 & 0 \end{pmatrix} \]

whose eigenvalues are ±iω. Since one of them is strictly positive, x is unstable. This theorem works in most cases but with one important exception: when there exists at least one eigenvalue with zero real part. In that case, there may be stability as well as instability. In some of these cases, the concept of a Lyapunov function may be very useful.

Definition 0.4

Let φ be a dynamical system on X = ℝⁿ having x as an equilibrium point and let W: X → ℝ be a continuous function having a strict minimum in x. W is said to be a Lyapunov function relative to x if there exists a neighbourhood Z of x such that:

For all x ∈ Z, t₁ ≤ t₂, t₁ ≥ 0: W(φ(t₂, x)) ≤ W(φ(t₁, x)).

This is equivalent to saying that W has a strict minimum in x and the function t → W(φ(t, x)) (or t → W(u(t)) with u(0) = x) is nonincreasing.

Theorem 0.2

If a dynamical system on X = ℝⁿ which is continuous in time has an equilibrium point x which admits a Lyapunov function relative to it, then x is stable.

Proof. Let Z be a neighbourhood of x such that W(x) > W(x) for all x ∈ Z. Clearly, Z = Z ∩ Z, (7) holds true also for the function W - W(x), so it is not restrictive to suppose W(x) = 0 and W(x) > 0 for all x ∈ Z different from x. Let now V be an arbitrary neighbourhood of x and let B(x, r) be a ball centered in x and contained in V ∩ Z. Clearly, x ∉ ∂B(x, r), so W is strictly positive on the closed set ∂B(x, r). Since W is continuous, W has a minimum w > 0 on ∂B(x, r). Let U = W^-1(]0, w[) which, by continuity, is a neighbourhood of x and let U = U ∩ B(x, r). If x ∈ U, then, by (7) W(φ(t, x)) ≤ W(φ(0, x)) = W(x) < w for all t ≥ 0. Suppose by contradiction that x is unstable. Then for every neighbourhood U of x there exist x ∈ U and t > 0 such that φ(t, x) ∉ V. But then ||φ(t, x)|| > r and by the continuity of φ in time and of the norm there exists t such that φ(t, x) ∈ ∂B(x, r), hence W(φ(t, x)) ≥ w, which is a contradiction.

Lyapunov functions are in general not easy to find. If this happens, especially when they are constant on trajectories, they often carry important information on the system. In holonomic systems, for example, W is often the total mechanical energy. In some cases, however, there is no information, as in the case of the Lotka-Volterra system (see XXX).

Example 0.2

Consider the equilibrium position (0, 0) in (3). In this case:

\[ A = \begin{pmatrix} 0 & 1 \\ -ω^2 & 0 \end{pmatrix} \]

so that its eigenvalues are ±iω and theorem 0.1 is not applicable. The function:

\[ W(x, y) = \frac{ω^2}{2}(1 - \cos(x)) + \frac{y^2}{2} \]

has clearly a strict minimum in (0, 0) and moreover:

\[ \frac{d}{dt} W(x(t), y(t)) = ω^2 \sin(x)x' + yy' = 0 \]

along trajectories of (3) and therefore (0, 0) is stable.

If one wants to incorporate (or is in the presence of) dissipative effects, the notion of asymptotic stability is more appropriate.

Definition 0.5

Let φ be a dynamical system on a topological space X having x as an equilibrium point. x is said to be attractive if there exists a neighbourhood U of x such that for all x ∈ U: lim_t→+∞ φ(t, x) = x.

If X is a metric space, a set Y ⊂ X is attractive if the same holds with dist(φ(t, x), Y), where dist(x, Y) = inf_y∈Y d(x, y).

Definition 0.6

Let φ be a dynamical system on a topological space X having x as an equilibrium point. If x is stable and attractive, then it is said to be asymptotically stable.

Example 0.3

Attractivity does not imply stability, as the following example shows. Take X = ℝ² and the system written in polar coordinates as:

• r' = r(1 - r)
• θ' = r(1 - cos(θ))

It can be checked that the right-hand side is continuous and Lipschitz on the whole of ℝ². Then (r = 1, θ = 0) ((1, 0) in Cartesian coordinates) is an equilibrium point. It is easy to verify that if r(0) = 1 and θ(0) = θ₀ ≠ 0, then the solution is given by:

θ(t) = θ₀ + 2 arctan(cot(θ₀/2 + t))

and therefore the orbit goes away from zero for some time and then, as θ(t) approaches 2π, it slows down towards the equilibrium point.

Next, integration of the equation in r shows that:

r(t) = r₀ e^t / (1 + r₀ (e^t - 1))

which is a bounded function tending to 1 as t → +∞. Therefore, the corresponding θ(t) is increasing in a way similar to the equation which can be computed explicitly.

Therefore, taken a sufficiently small neighbourhood U of the equilibrium point r = 1, θ = 0, every solution from U tends to r = 1, θ = 0 = 2π and this position is attractive. However, even if U is chosen arbitrarily small, the solution from θ₀ > 0 will not remain in it for all times; therefore, the equilibrium is unstable, though being attractive.

Remark 1

Regular dynamical systems, however, cannot describe a system evolving from x and arriving at a position y and there stopping. For, if t → u(t) were a solution of u'(t) = F(u(t)), then if there is a sequence (t_k) of instants tending to t such that lim_k→∞ u'(t_k) = 0 and u'(t) = 0 for t > t, then it would first be F(y) = 0 by continuity, since:

F(y) = F(lim_k→∞ u(t_k)) = lim_k→∞ F(u(t_k)) = lim_k→∞ u'(t_k) = 0

and second, the function v(t) = u(t) would satisfy v'(t) = F(v(t)) = G(v(t)) where G has the same regularity as F. Hence, uniqueness holds also for v, but there would be two solutions starting from y: the constant trajectory v(t) = y and the trajectory given by v(t) = u(t), which is impossible.

Asymptotic stability holds both via linearization and an enforcement of Lyapunov’s theorem.

Theorem 0.3

Let t → u(t) be the solution of (2) and suppose that it exists for all t ∈ ℝ. Let x be an equilibrium position of the system and set:

\[ A = \frac{\partial F}{\partial u} \bigg|_{x} \]

If all the eigenvalues of A have strictly negative real parts, then x is asymptotically stable.

Definition 0.7

Let φ be a dynamical system on X = ℝⁿ having x as an equilibrium point and let W: X → ℝ be a continuous function having a strict minimum in x. W is said to be a strict Lyapunov function relative to x if there exists a neighbourhood Z of x such that:

For all x ∈ Z, x ≠ x, t₁ ≥ 0, t₁ < t₂ implies W(φ(t₂, x)) < W(φ(t₁, x)).

For all x ∈ Z, if for all t₁, t₂ ≥ 0: W(φ(t₂, x)) = W(φ(t₁, x)) implies x = x.

In other words, in a neighbourhood of the equilibrium point, the function W strictly decreases along every non-constant trajectory.