LCS Tutorial: Proof of flux theorem

This section of the tutorial contains the proof of the flux theorem listed in the previous page. The proof is broken into intermediate results which are listed in theorems, lemmas, etc., which lead to the proof of the flux theorem. Since this page is a little more dense than the others, one can skip this page on a first reading and come back to it later.

Since our flow is Lipschitz continuous, cf. Thm. 1.4 of Verhulst (1996), it satisfies the following condition:

There is a positive constant k such that

∥ t ∥ ∥ dφ t0(x) ∥ k ∣t- t ∣ ∥ ----------∥ ≤ e 0 , ∥ dx ∥

(26)

Theorem 6.2 The finite-time Lyapunov exponent becomes constant along trajectories for large integration times T.

PROOF. We compare the value of the finite-time Lyapunov exponent computed at two different points of the same trajectory. Without loss of generality, we assume that the initial time is t₀= 0. Let

where we have used properties of the flow map given in Eq. (3) and the maximum exponential stretching hypothesis of Eq. (26). Similarly,

( ) ∥ s+T ∥ ∥ T ∥ T T ∥ dφ-s----(y)-∥ ∥ d-φ0-(x)-∥ ∣T ∣ σ s (y) - σ 0 (x) = ln ∥ dy ∥ - ln ∥ dx ∥ ∥ s+T T 0 ∥ ∥ T ∥ ∥ d(-φT----(φ0-(φ-s(y))))-∥ ∥ -d(φ-0 (x))-∥ = ln ∥ dy ∥ - ln ∥ dx ∥ ∥ ∥ ∥ ∥ ∥ dφs+T (ˆx) d φT (x) dφ0 (y) ∥ ∥ d φT (x) ∥ = ln ∥ ---T-----------0---------s----∥ - ln ∥ ----0----∥ d ˆx dx dy dx ( ∥ ∥ ∥ ∥ ∥ ∥ ) ∥ ∥ ∥ d φs+T (ˆx) ∥ ∥ dφT (x)∥ ∥ d φ0 (y) ∥ ∥ d φT (x) ∥ ≤ ln ∥ ---T--------∥ ∥ ----0----∥ ∥ ---s-----∥ - ln ∥ ----0----∥ d ˆx dx dy dx ∥ s+T ∥ ∥ ∥ ∥ dφ-T----(ˆx)-∥ ∥ -dφ0s(y)-∥ = ln ∥ d ˆx ∥ + ln ∥ dy ∥ ≤ 2k ∣s ∣ ,

so we have

∥ ∥ ∣s ∣ ∥ T T ∥ ----- σ 0 (x) - σ s (y) ≤ 2k . ∣T ∣

(27)

Therefore

∥ ∥ ∥ T T ∥ ∥ d σT (x) ∥ ∥ σ t+s(y) - σ t (x) ∥ 2k ∥ -----t------∥ = lim -------------------------------≤ ----- = 𝒪(1 ∕ ∣T ∣) . ∥ dt ∥ s →0 ∣s ∣ ∣T ∣

(28)

Taking the limit as

∣T ∣ → ∞

gives

∥ ∥ ∥ d σT (x) ∥ lim sup ∥ -----t-------∥ = 0 , ∥ ∥ ∣T ∣→ ∞ dt

(29)

which implies

∥ T ∥ ∥ d σ t (x) ∥ lim ∥ -------------∥ = 0 , ∣T ∣→ ∞ ∥ dt ∥

(30)

Corollary 6.1 We have
$∂-∇--σ-- * = - J ∇ σ - Σv + 𝒪 (1 ∕ ∣T ∣) , ∂ t$ (31)

where J is the Jacobian matrix of the velocity field $v$ .

As a result,

∂--σ- = - 〈v, ∇ σ 〉 + 𝒪 (1 ∕∣T ∣) . ∂ t

(32)

Taking the (spatial) derivative of this equation yields

∂ ∇ σ ∂ σ -------- = ∇ ----- = - J * ∇ σ - Σv + 𝒪 (1 ∕ ∣T ∣) , ∂ t ∂ t

(33)

which is reasonable since

∇ σ

is smooth since the flow is assumed to be smooth.

Let

Λ

be the Hessian of

L

and note the following properties of

Λ

and

Σ

Lemma 6.1 $Σ$ and $Λ$ are self-adjoint.

PROOF. This result holds due to the symmetry of mixed partials. For example, from

Σ(u, v) = Σ(v, u)

, we deduce immediately that

〈u, Σv 〉 = 〈v, Σu 〉 = 〈Σu, v〉

because the derivatives are necessarily real numbers.

Theorem 6.3 For $L = 0$ , we have

〈 〉 〈 〉 ˆt, Σ ˆn = ˆn, Σ ˆt = 0

PROOF. From Def. 5.1, SR2 implies that

∇L = ˆn

is an eigenvector of

Σ

. Hence,

since by definition the two vectors are orthogonal, where

λ (Σ) min

is the smallest eigenvalue of

Σ

Corollary 6.2 Along $L = 0$ and for an arbitrary vector v, we have

〈ˆn, Σv 〉 = 〈ˆn, Σ ˆn 〉〈ˆn, v 〉

PROOF. Developing v in the orthonormal basis

(ˆt, ˆn)

gives

〈 〉 ˆ ˆ v = t, v t + 〈 ˆn, v 〉 ˆn .

(34)

A direct computation of

〈nˆ, Σv 〉

, and applying Thm. 6.3, gives the desired result.

Lemma 6.2 $Λˆn = 0$ everywhere in U.

PROOF. Everywhere in U, L is smooth and the gradient

∇L

is well-defined. In particular, from Eq. (19)

∥∇L ∥ = 1

, therefore

( ) 2 0 = ∇ ∥ ∇L ∥ = 2 Λ ∇L = 2 Λ nˆ .

(35)

Fig. 10: Geometry of quantities discussed in Lemma 6.3.

Lemma 6.3 On the LCS, i.e. for L = 0,
$∂ L ∂ 〈ˆn, ∇ σ 〉 ----- ---------------- 〈 ˆn, Σ ˆn 〉 = . ∂ t ∂ t$ (36)

PROOF. Take x on the LCS at time t, i.e.

L(x, t) = 0

. Define

y = x + α( δt)ˆn

such that

L(y, t + δt) = 0

. In other words, y is at the intersection of the LCS at time

t + δt

and the line starting at x, orthogonal to the LCS at time t (see Fig. 10). Since we require y = x for

δt = 0

, it follows that

α(δt)

𝒪( δt)

. Expanding L to second order in

δt

gives the following (where all derivatives on the right-hand side of Eqs. (37)–(41) are evaluated at x and t unless otherwise specified):

0 = L(y, t + δt) = L(x, t) + α + ∂L--δt + 𝒪( δt2) , ∂t = α + ∂L--δt + 𝒪( δt2) . ∂t

(37)

Now expanding

∇L

, and plugging in Lem. 6.2, gives

∂ ∇L ∇L ∣ = ∇L + --------δt + 𝒪( δt2) . y,t+ δt ∂ t

(39)

From Eqs. (19) and (38) we have

∂ L ∂ ∇ σ ∇ σ ∣ = ∇ σ - -----Σ ∇L δt+ --------δt+ 𝒪( δt2) . y,t+ δt ∂ t ∂ t

(40)

Since y is on the LCS at time

t + δt

, we must have

〈 〉 0 = ∇L ∣y,t+ δt , ∇ σ ∣y,t+ δt ( ) ∂L-- ∂〈∇L,--∇--σ〉- 2 = 〈∇L, ∇ σ 〉 + δt - 〈∇L, Σ ∇L 〉 + + 𝒪( δt ) ( ∂t ) ∂t ∂L ∂〈ˆn, ∇ σ 〉 2 = δt - 〈ˆn, Σ ˆn 〉 ---- + ----------- + 𝒪( δt ) . ∂t ∂t

(41)

We now have all the precursors to prove Thm. 6.1, which we have restated below for convenience.

Theorem 6.1 (restated) For L = 0, we have
$〈 〉 〈 〉 dL ˆt, ∇ σ ∂ nˆ ----- = ------------- ˆt, ----- - J nˆ + 𝒪 (1 ∕ ∣T ∣) . dt 〈 ˆn, Σ nˆ 〉 ∂ t$ (42)