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Limit Sets
Consider an autonomous system of the form ˙x = f (x) for which solutions are defined for all time in an open set D. For x ∈ D, the ω−limit set of x, denoted ω(x) is the set of points y such that there is a sequence t 1 < t 2 <... with ti → +∞ as i → ∞, and φ(ti, x) → y. Similarly, the α−limit set of x is the set of points y for which there is a sequence t 1 > t 2 >... with ti → −∞ as i → ∞ and φ(ti) → y as i → ∞. The ω−limit set of x is denoted ω(x) and the α−limit set of x is denoted α(x). It is easy to show that these are closed subsets of D. They may be empty. One can define ω(x) in the case that φ(t, x) exists for all t > t 0 for some t 0. A similar statement holds for α(x), if φ(t) exists for t < t 0. A point x 0 for which f (x 0 ) = 0 is called an equilibrium or stationary point or critical point of ˙x = f (x). We also say that a function f : D → Rn^ is a vector field in D. Thus, being given a vector field in D is the same as being given an autonomous differential equation in D. An invariant set for the differential equation ˙x = f (x) or for the vector field f in D is a subset Ω of D such that if x ∈ Ω and φ(t, x) is a solution to x˙ = f (x) with φ(0) = x, then φ(t, x) ∈ Ω for all t. Facts:
- Suppose the f is a C^1 vector field in D ⊂ Rn^ and x ∈ D has the property that the orbit φ(t, x) of x remains in a compact subset F of D for t ≥ 0. Then, ω(x) is a compact invariant connected subset of F.
- Any orbit is an invariant set.
Vector Fields as Differential operators
Recall that an autonomous differential equation ˙x = f (x) is given by simply giving a function f : D → Rn^ from a domain D in Rn. Suppose that f is Ck^ for k ≥ 1. Let Ck[D, R] be the space of Ck^ real-valued functions defined on D. We can use f to define an operator Lf from Ck+1[D, R] to Ck[D, R] in the following way. For x ∈ D, let φ(t, x) be the solution to x˙ = f (x), φ(0, x) = x. For, ψ ∈ Ck+1[D, R], let (Lf ψ)(x) = (^) dtd ψ(φ(t, x)) |t=0. This defines a mapping
from Ck+1[D, R] to Ck[D, R] which satisfies the following two properties.
- (linearity). L is a linear mapping; i.e., Lf (αψ+βη) = αLf (ψ) = βLf (η) for any two functions ψ, η and scalars α, β.
- (derivation). For ψ, η ∈ Ck+1[D, R],
Lf (ψ · η) = Lf (ψ) · η + ψLf (η).
The operator Lf is called the Lie derivative operator. It maps Ck+ functions to Ck^ functions. Let πi : x → xi be the projection of a vector onto its i−th coordinate as a function on Rn. Facts.
- The value of the function Lf (ψ) can be computed from knowledge of the vector field and the partial derivative functions of ψ by the formula
Lf (ψ)(x) =
∑^ n
i=
∂ψ ∂xi
(x)(πi ◦ f )(x). (1)
- It follows from this formula that the component functions of the vector field f are equal to the functions Lf (πi). Indeed, if f (x) = (f 1 (x), f 2 (x),... , fn(x)) = (π 1 ◦ f (x),... , πn ◦ f (x)), then
Lf (πi) = fi.
Thus, the operator Lf and the vector field f completely determine each other, and we can think of vector fields as differential operators on real- valued functions or as assignments of vectors at each point in a domain D.
- The function ψ is constant along solution curves of ˙x = f (x) if and only if Lf (ψ) is the zero function in D.
Let ei be the unit vector in Rn^ whose i−th coordinate is 1 and whose other coordinates are 0. It is common to write (^) ∂x∂i for the operator Lf where f (x) = ei is the constant vector field whose value at each x is ei.
is then transverse to the vector field f (x 0 ) at x 0. By the local continuity of solutions to ˙x = f (x) on initial conditions and the continuity of f , there are a neighborhood V 1 of x 0 in H and an interval I about 0 in R such that if x ∈ V 1 , then φ(t, x) is defined on all of I and meets H only for t = 0. For (y 2 ,... , yn) = y near 0 in Rn−^1 , we have an associated point η(y) = x 0 +
∑ j yj^ vj^ ∈^ H.^ Write (y^1 , y) for the point (y^1 , y^2 ,... , yn) in^ R n
with y ∈ Rn−^1. We define a mapping ρ(y 1 , y) by
ρ(y 1 , y) = φ(y 1 , η(y)).
We claim that this transformation ρ is the required change of coordinates. First, note that ρ is a Ck^ mapping of the variables (y 1 , y). To prove that ρ is a change of coordinates, it suffices to show that its jacobian determinant at 0 is not zero and use the implicit function theorem. Now, at (y 1 , y) = 0, the first column of the jacobian matrix of ρ, (^) ∂y∂ρ 1 is
just f (x 0 ), while the j −th column is just (^) ∂y∂ρj is vj (exercise). By the choice of
the v j′ s, these vectors are linearly independent. Thus, the required jacobian determinant is not zero. Finally, we have to show that the mapping ρ carries solutions to (^) ∂x∂ 1 to those of f. A solution to the constant vector field (^) ∂x∂ 1 is simply a function (t, (y 1 , y)) → (t + y 1 , y). Transforming this by ρ gives the function (t, (y 1 , y)) → ρ(t + y 1 , y) = φ(t + y 1 , η(y)). But, as we saw in the proof of the local flow property of autonomous systems, if φ(t, z) is a solution, then so is φ(t + s, z). Thus, the function t → φ(t + y 1 , η(y)) is a solution to the equation ˙x = f (x). Suppose the f is a C^1 vector field defined in an open set D ⊂ Rn. Definition. An invariant set K is called a minimal set if it is compact, non-empty, and does not properly contain another compact, non-empty, in- variant set. Proposition. Any compact invariant set contains a minimal set. Proof. Let K be a compact invariant set. The set C of non-empty compact invariant subsets of K is partially ordered by inclusion A ≺ B if and only if A ⊇ B. Each totally ordered subset has an upper bound, so by Zorn’s lemma, C contains a maximal element, say Σ. Then, Σ is a minimal set. QED. Example and Remark.
- A critical point or periodic orbit is a minimal set.
- It is remarkable fact that in the plane for a C^1 autonomous vector field, there are no other minimal sets.
- In Rn, n > 2, there are many examples of non-trivial minimal sets. We will see this later.
Proposition. Suppose f is a C^1 vector field in an open set D ⊂ Rn^ and there is a closed non-empty ball B ⊂ D such that f is non-zero and nowhere tangent on the boundary of B. Then, f possesses a critical point in B. Proof. Let φ(t, x) be the local flow of f. Since, f is non-zero and not tangent to the boundary of B, orbits at the boundary either flow into or outof B. We suppose they flow into B. In the other case, replace f by −f. For x ∈ B, the solution φ(t, x) is defined and remains in B for all t > 0. Let m > 0 be a positive integer, and consider the mapping x → φ (^) m 1 (x). This is a continuous self-map of the closed ball B to itself. By the Brouwer fixed point theorem, it has a fixed point, say xm. Since B is compact, the sequence xm has a subsequence xmk which converges, say to the point y as k → ∞. Let us show that f (y) = 0. If not, then by the flow box theorem, there are a neighborhood U of y in D and an interval I = [−, ] about 0 in R such that,
(*) for z ∈ U , the solution φ(t, z) is defined for all t ∈ [−, ]
(**) φ(t 1 , z) 6 = φ(t 2 , z) for t 1 6 = t 2 ∈ I
But, if k is large enough, then xmk ∈ U , and (^) m^1 k < . But then, φ 1 m (xmk ) 6 = xmk by (**) which contradicts the definition of xmk. QED.
