# On Using Curvature to Demonstrate Stability

Authors: C. Connell McCluskey
Journal: Differential Equations and Nonlinear Mechanics
 (2008)

---

## Abstract

A new approach for demonstrating the global stability of ordinary differential equations
is given. It is shown that if the curvature of solutions is bounded on some set, then
any nonconstant orbits that remain in the set, must contain points that lie some minimum
distance apart from each other. This is used to establish a negative-criterion for periodic
orbits. This is extended to give a method of proving an equilibrium to be globally stable. The
approach can also be used to rule out the sudden appearance of large-amplitude periodic orbits.


---

## Body

## 1. Introduction

A key issue in the analysis of a system of ordinary
differential equations is to determine the long-term dynamics. Generally, it is
desirable, if possible, to   know whether a system settles down to an
equilibrium, approaches a limit cycle, or exhibits some other long-term
behavior. In this paper, we give a method that uses the curvature of solutions
to help show that a system approaches an equilibrium.A nice feature of considering the curvature is that
the calculation only involves local information. Furthermore, it is not
necessary to calculate solutions to the differential
equations since the curvature calculation uses the
derivative of a solution, and this is given by the vector field that defines
the flow.On any set for which the curvature is bounded, there
is a minimum length of a closed orbit (not including equilibria). Also, in such
a set, any closed orbit must contain points that are a certain distance away
from each other. The smaller the curvature, the farther apart certain points on
the orbit must be from each other. This idea is used to formulate a negative
criterion for periodic orbits. The method is then extended to give a criterion
that shows that all forward trajectories leave the set.Unfortunately, if the eigenvalues of the Jacobian at
an equilibrium are not all equal, then the curvature near the equilibrium is
unbounded. This makes it necessary to deal with the behavior near an
equilibrium separately. For instance, one may use a Lyapunov function near an
equilibrium to show that a certain neighborhood is contained in the
equilibrium's basin of attraction. Then, analysis of the curvature can be used
to show that all other solutions enter this set (by showing that they leave the
complement), making the stability global.This approach can also be useful in ruling out
nonlocal bifurcations as parameters are varied. It is often possible to show
that an equilibrium is locally asymptotically stable for all parameter values,
and that it is globally stable for a particular set of parameter values. As
parameters are varied, the continued local stability precludes the possibility
of local bifurcations such as Hopf. Nonlocal bifurcations, however, may still
occur. The method presented here can be used to preclude the existence (and
sudden appearance) of large-amplitude periodic orbits and, therefore, help rule
out nonlocal bifurcations.We now give a brief outline of the paper. In Section2, the curvature of parametric curves is discussed. This is
applied to periodic solutions of differential equations in Section 3. In Section 4, results are given for
omega limit points. Section 5 stability gives stability theorems. A discussion of how to apply the results is included in Section 6.


## 2. Curvature Calculations

Consider a curveΓ given by the function r:ℝ→ℝn,
where r is C2 and r′ is nonzero for all t∈ℝ.
In this paper, we use the convention that (′) represents differentiation with respect to
time.Lets be the arclength along Γ from some reference point, that is, s(t)=∫t0t∥r′(τ)∥dτ.
We use the convention that when the arguement of a function is changed to s,
the function is considered to be reparameterized in terms of arc length. Note
that ds/dt=∥r′(t)∥,
and so d/ds=(d/dt)(dt/ds)=(1/∥r′(t)∥)(d/dt).For eacht,
the unit tangent at the point r(t) is(2.1)T(t)=r′(t)∥r′(t)∥,and so(2.2)T′(t)=∥r′∥r′′−((1/∥r′∥)r′⋅r′′)r′∥r′∥2=∥r′∥2r′′−(r′⋅r′′)r′∥r′∥3.The curvature [1] of Γ is defined to be K=∥dT/ds∥ and is given by(2.3)K=∥T′(t)∥∥r′(t)∥.Using (2.2) gives(2.4)K=∥∥r′∥2r′′−(r′⋅r′′)r′∥∥r′(t)∥4.LetT0 be a unit vector in ℝn.
Let θ(t) be the angle between T(t) and T0 :
Proposition 2.1.
(2.5)|dθds|≤K.Proof.
Noting thatcos(θ)=T0⋅T,
we have −sin(θ)(dθ/ds)=T0⋅dT/ds.
Letting β be the angle between dT/ds and T0,
then, T0⋅dT/ds=∥dT/ds∥cos(β).
Thus,(2.6)dθds=−1sin(θ)T0⋅dTds=−1sin(θ)∥dTds∥cos(β)=−cos(β)sin(θ)K.We now show that |cos(β)/sin(θ)| is at most one. Since T(s) is a unit vector for all s,
it follows that dT/ds is orthogonal to T.
Since T0 is a unit vector separated from T by angle θ,
the component of T0 normal to T has magnitude sin(θ).
Thus, the component of T0 in the direction of dT/ds has magnitude less than or equal to sin(θ).
On the other hand, this magnitude is given by |cos(β)|,
showing that |cos(β)/sin(θ)|≤|sin(θ)/sin(θ)|=1.
Combining with (2.6) yields the proposition.Now, supposeΓ is a simple closed curve given by the C2ω-periodic function r:ℝ→ℝn,
where r′ is nonvanishing. Let x0∈Γ.
Choose t0∈ℝ such that r(t0)=x0,
and let T0=T(t0),
that is, T0 is the unit tangent to Γ at x0.Sincer is ω-periodic, it follows that 0=r(t+ω)−r(t)=∫tt+ωr′(τ)dτ for all t.
Hence, 0=∫tt+ω(r′(τ)⋅T0)dτ.
Combining this with the fact that (r′(t)⋅T0) is ω-periodic, continuous and positive at t=t0,
we see that (r′(t)⋅T0) oscillates in sign and, therefore, must have
at least two zeros in every interval of length ω.
Let (t−,t+) be the maximal interval containing t0 such that (r′(t)⋅T0) is positive. Note that θ(t−)=θ(t+)=π/2.Letd(x,y) be the Euclidean distance between points x,y∈ℝn.
We now work toward finding a lower bound on the distance between the points r(t0) and r(t+) :(2.7)(r(t+)−r(t0))⋅T0=∫t0t+(r′(t)⋅T0)dt=∫t0t+∥r′(t)∥cos(θ(t))dt.Let KΓ be the least upper bound of K on Γ.
Then, recalling that ds/dt=∥r′(t)∥ gives(2.8)(r(t+)−r(t0))⋅T0=∫s(t0)s(t+)cos(θ(s))ds=∫s(t0)s(t+)KΓKΓcos(θ(s))ds≥1KΓ∫s(t0)s(t+)dθdscos(θ(s))ds=1KΓ∫θ(t0)θ(t+)cos(θ)dθ=1KΓ∫0π/2cos(θ)dθ=1KΓ,where the inequality follows
from Proposition 2.1. Similarly, (r(t+)−r(t0))⋅T(t+)≥1/KΓ.
Since T0 and T(t+) are orthogonal, it follows that d(r(t+),r(t0))≥2/KΓ.
Similarly, d(r(t−),r(t0))≥2/KΓ and d(r(t−),r(t+))≥2/KΓ.
This leads to the following result.Theorem 2.2.
LetΓ be a simple closed curve and let K⋆ be an upper bound for the curvature on Γ.
Then, given a point x0∈Γ,
there exist points x−,x+∈Γ such that d(x0,x−),d(x0,x+)≥2/K⋆ and d(x−,x+)≥2/K⋆.
Furthermore, the tangents to Γ at x− and x+ are orthogonal to the tangent at x0.By not requiring that the curveΓ be closed, we obtain the following result,
which will be useful in Section 4.Theorem 2.3.
LetK⋆ be an upper bound for the curvature on a curve Γ which is given by r:ℝ→ℝn,
with r′ nonvanishing. Suppose further that there exist t−,t0,t+∈ℝ,
with t−<t0<t+ such that the tangent at x0=r(t0) is orthogonal to the tangents at x−=r(t−) and x+=r(t+),
and r′(t)⋅r′(t0)>0 for all t∈(t−,t+).
Then, d(x0,x−),d(x0,x+)≥2/K⋆ and d(x−,x+)≥2/K⋆.


## 3. Periodic Solutions of Differential Equations

Consider the differential equation(3.1)x′=f(x),where f:ℝn→ℝn is C1.
We denote by x(t;x0) the solution to (3.1) which passes through x0 at time t=0.
Suppose x(t;x0) is a nonconstant solution to (3.1). Noting that x′′=d(x′)/dt=(∂(x′)/∂x)(dx/dt)=(∂f/∂x)f(x),
it follows from (2.4) that the curvature at x is(3.2)K(x)=∥∥f(x)∥2x′′−(f(x)⋅x′′)f(x)∥∥f(x)∥4=∥∥f∥2(∂f/∂x)f−(fT(∂f/∂x)f)f∥∥f∥4.Thus, as long as the point x is not an equilibrium of (3.1), the curvature at x of the solution through x can be precisely calculated without any
explicit knowledge of the solution.We now relate Theorem2.2 to periodic
solutions of ordinary differential equations.Theorem 3.1.
LetΓ be the simple closed curve traced out by a
nontrivial periodic solution of (3.1) and suppose
that Γ lies entirely in a region, where the curvature
of solutions is bounded above by K⋆.
Then, the conclusions of Theorem 2.2 hold.LetΔ⊆ℝn.
Define(3.3)KΔ=supx∈ΔK(x).Then, any periodic solution
to  (3.1) which lies
entirely in Δ has maximum curvature less than or equal to KΔ.Corollary 3.2.
IfΔ contains no simple closed curve for which the
maximum curvature is less than or equal to KΔ,
then there are no nontrivial periodic orbits of
(3.1) which lie entirely in Δ.Property (The Negative Property).
LetK⋆ be the maximum curvature of f on a set which contains no equilibria. The set
is said to have the negative property if it does not contain a point x0 such that there are two corresponding points x− and x+ satisfying d(x0,x−),d(x0,x+)≥2/K⋆ and d(x−,x+)≥2/K⋆,
with f(x−) and f(x+) orthogonal to f(x0).Corollary 3.3.
IfΔ has the negative property, then there are no
periodic solutions to (3.1) contained entirely in Δ.


## 4. Omega Limit Points

Theorem 4.1.
SupposeΔ⊆ℝn has the negative property, and that φ(t) is a forward trajectory of (3.1) which is
contained entirely in Δ.
Then, the omega limit set of φ consists entirely of equilibria which lie in
the closure of Δ,
but not in Δ.Proof.
Supposex¯ is an omega limit point of φ(t),
where φ(t)∈Δ for all t≥0,
and that f(x¯)≠0.
Then, there is an increasing sequence of times {tn} which limits to ∞ such that limn→∞φ(tn)=x¯.
Then,(4.1)0=limn→∞(φ(tn+1)−φ(tn))=limn→∞∫tntn+1φ′(t)dt=limn→∞∫tntn+1f(φ(t))dt.Fix ϵ,δ>0 such that f(y)⋅f(z)>2ϵ if y,z∈Nδ,
the open ball centred at x¯ with radius δ.
Choose τ>0 so that f(x(t;y))⋅f(z)>ϵ if y,z∈Nδ and |t|<τ.
Taking the dot product of each side of (4.1) with a vector v∈f(Nδ) gives(4.2)limn→∞∫tntn+1(f(φ(t))⋅v)dt=0.For sufficiently large n,
we have φ(tn)∈Nδ and, therefore, f(φ(t))⋅v>ϵ for |t−tn|<τ.
Thus,(4.3)∫tntn+τ(f(φ(t))⋅v)dt>ϵτ>0.Hence, the integrand must
oscillate between positive and negative for large enough t,
in order that (4.2) be satisfied. Thus, for large enough n,
there exist tn− and tn+ with tn−<tn<tn+,
such that f(φ(tn−)) and f(φ(tn+)) are orthogonal to v.
Furthermore, tn− and tn+ can be chosen so that f(φ(t))⋅v is positive on the interval with these
endpoints.
The only restriction onv,
thus far, is that v∈f(Nδ).
In order for n to be sufficiently large for tn− and tn+ to be defined, it is only necessary that φ(tn)∈Nδ.
(For large n,
a change in the choice of v will change the values of tn− and tn+ but will not change the fact that they exist.)
Now, fix n to be sufficiently large that φ(tn)∈Nδ,
and choose v=f(φ(tn)).
Applying
Theorem2.3, with (x0,x−,x+)=(φ(tn),φ(tn−),φ(tn+)),
yields d(x0,x−),d(x0,x+)≥2/KΔ and d(x−,x+)≥2/KΔ.
This contradicts the negative property, and so if x¯ is an omega limit point, then we must have f(x¯)=0.
SinceΔ satisfies the negative property, there are no
equilibria in Δ and, therefore, x¯ is not an element of Δ.
However, since x¯ is an omega limit point of φ,
it must be an element of the closure of Δ,
completing the proof.Corollary 4.2.
IfΔ⊆ℝn is a compact set which has the negative
property, then there are no forward trajectories contained entirely in Δ.Proof.
SinceΔ is compact, any forward trajectory contained
entirely in Δ has a nonempty omega limit set Ω⊆Δ.
By Theorem 4.1, Ω consists entirely of equilibria, but by the
negative property Δ contains no equilibria, so Δ cannot contain an entire forward trajectory.Corollary 4.3.
Letd be the diameter of a compact set Δ which contains no equilibria. If dKΔ<2,
then any orbit starting in Δ leaves Δ.


## 5. Stability Theorems

SupposeD⊆ℝn is positively invariant under (3.1). If Δ⊆D is compact and has the negative property, then
by Corollary 4.2  there are no forward trajectories
contained entirely in Δ.
If it is known that all solutions intersecting D∖Δ limit to an equilibrium, then we can conclude
that all solutions in D limit to an equilibrium giving the following
result.Theorem 5.1.
SupposeD is positively invariant, and Δ⊆D is a compact set which has the negative
property. If all solutions in D∖Δ limit to an equilibrium, then all solutions in D limit to an equilibrium.Corollary 5.2.
SupposeD is positively invariant and contains a unique
equilibrium x¯.
Suppose Δ⊆D is a compact set which has the negative
property. If all solutions in D∖Δ limit x¯,
then x¯ is globally stable in D.We now consider the effect of making a change of
variables. Suppose(5.1)x=α(y),where α:ℝn→ℝn is a diffeomorphism. Then, x′=(∂α/∂y)y′,
and so(5.2)y′=[∂α∂x]−1f(x)=[∂α∂x]−1f(α(y)).If the set α−1(Δ) is compact and has the negative property
for (5.2), then Corollary 4.2 implies solutions of (5.2) leave α−1(Δ),
which then implies solutions of (3.1) leave Δ.
This allow Theorem 5.1 to be generalized as
follows.
Theorem 5.3.
SupposeD is positively invariant under (3.1), Δ⊆D is compact and that all solutions in D∖Δ limit to an equilibrium. If there exists a
diffeomorphism α:ℝn→ℝn such that α−1(Δ) has the negative property for (5.2), then all
solutions in D limit to an equilibrium.


## 6. Discussion

If it is known that the curvature of a vector field is
bounded on a set, then it follows that the rate at which trajectories turn is
bounded. A consequence of this is that a closed orbit has a minimum size. In
particular, there exist points on the orbit which are separated by a distance
at least twice as large as the reciprocal of the
bound on the curvature.This approach can be used to rule out periodic
trajectories of a differential equation and to show that solutions which remain
in a given set must limit to sets consisting of equilibria.It may appear that this alone could be used to show
global stability of an equilibrium in an invariant compact set, but there is a
problem. Near an equilibrium the curvature is usually unbounded. This is true
whenever the linearization at the equilibrium has at least two distinct
eigenvalues.Thus, in determining the global stability of an
equilibrium, it would be necessary to deal with the behavior near the
equilibrium in some other manner. For instance, one could use the linearization
to obtain a quadratic form that could be used to show that a given neighborhood
limits to the equilibrium. Then, by showing that the remainder of the compact
set has the negative property, it would follow that the equilibrium is globally
stable.



---

*Source: 745242-2008-05-12.xml*