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Back to basics : ODEs Lyapunov stability

Caricature of Camille Jordan (1838 - 1922) by one of his students at École Polytechnique, during a lecture on differential calculus : « J'ai mis ∂u mais c'est une faute d'impression ».
Caricature of Camille Jordan (1838 - 1922) by one of his students at École Polytechnique, during a lecture on differential calculus : « J'ai mis ∂u mais c'est une faute d'impression ».

This tiny post is about stability in the sense of Lyapunov of stationary points of autonomous ordinary differential equation (ODE) in Rn given by x(t)=f(x(t)),x(0)=x0 where f:ORnRn is a C1 vector field. This includes two important examples:

  • Gradient flows : f=E where E:OR is C2
    Geometrically, the system descents down the levelsets of E
  • Hamiltonian systems : f=ΩH with Ω skew-symmetric and H:OR is C2
    Geometrically, the system remains in the levelset of H where it started

The C1 regularity of f ensures, via the Cauchy-Lipschitz or Picard–Lindelöf theorem, the local existence and uniqueness of solutions for all x0O. As a consequence, the solution that takes the value x at initial time 0, denoted tφt(x), is uniquely defined on a maximal time interval ending up at Tmax(x), and blows as tTmax(x) if Tmax(x)<. We say that φt:OO is the flow of the ODE. The phase space O is partitioned into disjoint integral curves associated to the solutions. We are interested in the long time behavior of solutions near stationary points.

Stability of stationary points. Suppose that x0 is stationary : f(x0)=0. Then x(t)=x0 for all t. We say that this stationary point is stable in the sense of Lyapunov when for all ε>0, there exists η>0 such that for all x¯B(x0,η),
Tmax(x)=+andφt(x)x0ε for all t0, the solution remains as close to x0 as we want and for all time provided that we start close enough. We say that the stationary point x0 is unstable otherwise. Moreover, we say that a stable stationary point x0 is asymptotically stable if for some η0>0 and all x¯B(x0,η0),
limtφt(x)=x0. Furthermore, we say that it is exponentially stable if there exist positive constants α,c,C>0 such that forall x¯B(x0,η0) and all t0,
φt(x)x0Ceαtxx0.

Lyapunov function. It is a C2 function Φ:OR such that for all xO,
Φ(x),f(x)0. This is equivalent to say that tΦ(φt(x)) is non-increasing for all x, since φ0(x)=x and
tΦ(φt(x))=Φ(φt(x),φt(x)=Φ(φt(x)),f(φt(x))0. For a gradient flow f=Φ, the function Φ is a Lyapunov function and in this case
tΦ(φt(x))=Φ(φt(x))2.

Prime integral. It is a C2 function H:OR such that for all xO,
H(x),f(x)=0. This is equivalent to say that both H and H are Lyapunov functions, and also to say that tH(φt(x)) is constant for all t and x. For a Hamiltonian system f=ΩH, the Hamiltonian H is a prime integral, indeed thanks to the skew-symmetry of Ω,
tH(φt(x))=H(φt(x)),f(φt(x))=H(φt(x),ΩH(φt(x))=0.

Criterion for stability.. If x0 is a strict local minimum of a Lyapunov function Φ then x0 is stationary and stable. Actually the proof shows that we only need to know that Φ is locally a Lyapunov function, namely Lyapunov on a small enough ball centered at x0 included in O, this modification of the open set on which we define the ODE has potentially an impact on Tmax, but at the end, we conclude that Tmax=+ and it does not matter.

Proof. We use a reasoning by continuity. Since x0 is a strict minimum, for ε>0 small enough, ¯B(x0,ε)O and Φ(x0)<Φ(x) for all x¯B(x0,ε){x0}. Since the sphere ¯S(x0,ε) is compact because the dimension is finie, and Φ is continuous, there exists x1S(x0,ε) such that m:=minS(x0,ε)Φ=Φ(x1), and since Φ(x1)>Φ(x0), we get m>Φ(x0). On the other hand, since Φ is continuous at x0 and Φ(x0)<m, there exists 0<η<ε such that the maximum M of Φ on ¯B(x0,η) is <m. Now, for all x¯B(x0,η), since φ0(x)=x and tΦ(φt(x)) is non-increasing because Φ is a Lyapunov function, the function Φ remains M<m along the trajectory, hence the trajectory cannot reach the sphere S(x0,ε), thus it remains in B(x0,ε). Therefore x0 is stable. We have proved stability before stationarity. To show that it is stationary, it suffices to linearize the flow as φt(x)x0=tf(x)+ot0(t) and to take η=tη. Finally, since the trajectory is bounded and the dimension is finite, we have necessarily Tmax(x)=+.

Criterion for being exponentially stable.. If x0 is stationary and
maxspec(Jac(f)(x0))<0,then x0 is exponentially stable. This is a spectral criterion by linearization of the ODE.

Proof. Let A:=Jac(f)(x0) be the Jacobian matrix of f at x0, and its Jordan reduction
A=P1(D+E)P,D:=Diag(λ1,,λn),PGLn(C),EMn(R). The vector x0 (and x below) and the matrix A are real but P and D are complex (E is real and DE=ED but we do not use that). We can always assume that E is arbitrary small since the Jordan blocs can be conjugated by diagonal matrices, at the price of making P bigger:
[λ1000λ10000λ10000λ]=Diag(1,ε,ε2,)[λε000λε0000λε0000λ]Diag(1,ε1,ε2,). We have α:=maxspec(Jac(f)(x0))=max1knλk<0. Now
Φ(x):=P(xx0)2=¯PP(xx0),xx0, hence, using the fact that xx0 is real for the first equality,
Φ(x)=2(¯PP)(xx0)and in particularΦ(x)=O(xx0). On the other hand, f(x0)=0 since x0 is stationary and thus
f(x)=A(xx0)+o(xx0). We have then, using the fact that A and xx0 are real for the first equality,
Φ(x),f(x)=2¯PP(xx0),A(xx0)+o(xx02)=2P(xx0),(D+E)P(xx0)+o(xx02)=2z,Dz+2z,Ez+o(xx02) where z:=P(xx0). But now by the Cauchy--Schwarz inequality
2z,Ez2Ez2, while
z,Dz=z,(D)ziz,(D)z=k|zk|2Dk,kik|zk|2Dk,k gives
2z,Dz2αz2. Thus, for β<2α, E small enough, and x¯B(x0,η0), with η0>0 small enough, we get
Φ(x),f(x)2αP(xx0)2+2EP(xx0)2+o(xx02)βP(xx0)2=βΦ(x). In particular Φ is a Lyapunov function, on the restricted set O0:=B(x0,η0)O, with a strict minimum at x0 because P is invertible, hence x0 is stable. Moreover tΦ(φt(x))βΦ(φt(x)) for x¯B(x0,η0), and the Grönwall lemma below gives
Φ(φt(x))Φ(φ0(x))eβt=Φ(x)eβt. Next, since P is invertible, there exists a positive constant C>0 such that
C1xx02Φ(x)Cxx02, hence for x¯B(x0,η0) and all t0,
φt(x)x0C2eβ2txx0 which means that x0 is exponentially stable.

Grönwall lemma.. If u:[a,b]R is continuous on [a,b] and differentiable on (a,b), and if
u(t)α(t)u(t) for all t(a,b), with α:[a,b]R continuous, then for all t[a,b],
u(t)u(a)exp(taα(s)ds). The equality case u(t)=α(t)u(t) is a one-dimensional non-autonomous linear ODE. To see it, we observe that the solution of the equality case is positive : v(t):=etaα(s)ds>0 for all t[a,b], and v(t)=α(t)v(t) for all t(a,b), while v(a)=1, hence
(u(t)v(t))=u(t)v(t)u(t)v(t)v(t)2=(u(t)α(t)u(t))v(t)v(t)20 for all t(a,b), therefore u(t)/v(t)u(a)/v(a)=u(a) for all t[a,b].

It seems that it was already known to Alexandre Lyapunov (1857 - 1918) as well as to Henri Poincaré (1854 - 1912) that the solution of a linear non-autonomous ODE x(t)=A(t)x(t) in dimension n2 can perfectly diverge as t while A(t) has all eigenvalues <0 for all t.

Note. This tiny post is taken from a course on topology and differential calculus, that I have the pleasure of teaching, succeeding to my former colleague Dmitry Chelkak.

Further reading.

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