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APPEARED IN BULLETIN OF THE

arXiv:math/9210217v1 [math.DS] 1 Oct 1992APPEARED IN BULLETIN OF THEAMERICAN MATHEMATICAL SOCIETYVolume 27, Number 2, October 1992, Pages 298-303A SHOOTING APPROACH TO THE LORENZ EQUATIONSS. P. Hastings and W. C. TroyAbstract.

We announce and outline a proof of the existence of a homoclinic orbitof the Lorenz equations. In addition, we develop a shooting technique and two keyconditions, which lead to the existence of a one-to-one correspondence between a setof solutions and the set of all infinite sequences of 1’s and 3’s.1.

IntroductionThe system of equations discovered by Lorenz [6] is found in computer simula-tions to have chaotic behavior, by practically any definition of that term. A surveyappears in [7].

However, aside from local results and various kinds of bifurcationanalysis, little has been proved about these equations.We have now been able to prove the existence of a homoclinic orbit for an openset of parameter values. That is, there is a nonconstant solution tending to thesame equilibrium point, in this case (0,0,0), at ±∞.

Such a solution has long beenconjectured to exist and is recognized as an important feature of these equations.Further, we have proved a theorem, which reduces the question of whether thereare “chaotic” orbits to one which, in principle, can be handled for an open set ofparameter values with the techniques of rigorous numerical analysis, such as intervalarithmetic [1].In other words, the amount of computation required for rigorousverification of the hypotheses of our second theorem is finite; whether it is practicalremains an open question at this time. Computer assisted proofs generally leave agap in understanding, but with Theorems 1 and 2, we believe the gap is smallerthan before.The use of a “shooting” technique to obtain a homoclinic orbit is clearly sug-gested by numerical integrations.

However, its analytic implementation is difficultand has not been done before, to our knowledge. The proof of the second the-orem is easier.

Its conclusion is interesting because it discusses the existence ofchaos without reference to any bifurcation phenomena. The hypotheses appear tobe true (based on standard numerical integrations) for the “classical” parametervalues used for the Lorenz equations.Our approach is more elementary than some other approaches to these equations,such as geometric models, which have given deep insights about chaotic behaviorbut have not been shown to apply to the system which motivated all this work.Instead of Poincar´e maps we use simple one parameter “shooting.” Therefore, thisis a further application of methods begun in [5] and continued in [3, 4].

Techniquesfrom [8] are also important.1991 Mathematics Subject Classification. Primary 58F15, 58F13.Received by the editors January 7, 1992 and, in revised form, April 20, 1992c⃝1992 American Mathematical Society0273-0979/92 $1.00 + $.25 per page1

2S. P. HASTINGS AND W. C. TROYTheorem 1.

For each (s, q) in some neighborhood of the point (10, 1) there is anR in the interval (1, 1000) such that the system of equations(1)x′ = s(y −x),y′ = Rx −y −xz,z′ = xy −qzhas a homoclinic orbit.We chose q = 1 as the point to do our analysis instead of the usual value of 8/3,purely for convenience in the many numerical calculations. The lack of precision inthe R value could easily be reduced dramatically with computer assistance.

Also,the range of parameters (s, q) for which an R exists could be expanded greatly; butit would be more desirable to find a method that would reveal the set of such (s, q)analytically, by proving an extension of Lemma 2. Further homoclinic orbits for agiven (s, q) could, in principle, also be found by our method.2.

Outline of proof that there is a homoclinic orbitFor any R > 1, the equilibrium point (0,0,0) is unstable and has a one-dimensionalunstable manifold, which we denote by γ. We analyze the behavior of this manifoldfor R close to 1 and for R = 1000.

We prove two lemmas, which deal with a solutionp(t) = (x(t), y(t), z(t)) on the “positive” branch γ+ of the unstable manifold, sothat p(t) →(0, 0, 0) as t →−∞, and x(t), y(t), and z(t) are all positive for largenegative t.Lemma 1. For (s, q) in some open neighborhood of (10, 1) and R −1 positive butsufficiently small, x(t), y(t), and z(t) are all positive on the entire line −∞< t < ∞.Lemma 2.

For (s, q) in some open neighborhood of (10, 1) and R = 1000, x(t) hasat least one zero, t1, and x′ changes sign exactly once in (−∞, t1].Lemma 1 is easy to prove analytically, but Lemma 2 is much more difficult.With these two results, the existence of a homoclinic orbit is seen fairly quickly.We let R∗be the infimum of all values of R > 1 for which the behavior in Lemma2 occurs. For R = R∗and (x(t), y(t), z(t)) on γ+, we can show that either (a) xand x′ vanish simultaneously, or (b) after τ1, the first zero of x′, x decreases as longas the solution exists but never becomes negative.

This requires eliminating otheroptions, such as the bifurcation from some finite point of new zeros of x′ at R∗.In case (a), x = y = 0 at some finite time t, and then the uniqueness theory forinitial value problems implies that x and y are identically zero. This is impossibleon γ+.

In case (b), it is clear that (x, y, z) must approach an equilibrium point ast →∞. Only a little more effort is needed to show that this is the origin and theorbit is homoclinic.The proof of Lemma 2 requires a careful study of γ+.

The value R = 1000 is,of course, rather arbitrary. It is not hard to prove that on some initial interval(−∞, t0], x, y, and z increase monotonically, reaching a point where y = 1, 0.096 ≤x ≤0.1, and x2/20 < z < 0.1.

From this point the result would follow easilywith computer assistance; but in our opinion, ingenuity, and considerable effort,is required to follow the solution analytically. We show that x, y, and z continueto rise at least until they reach the levels z = 1000, 126.4 < x < 135.6, and

A SHOOTING APPROACH TO THE LORENZ EQUATIONS3798 < y < 1000. From this region y, and eventually x and z, begin to fall.

It isapparent that at a point where y becomes negative, z must be greater than R. Infact, we show that y = 0 at a point where 155 < x < 189 and z > 10.4x.To obtain these inequalities we use (1) for initial estimates over suitable intervals,and then iterate to obtain better bounds. Use is made of the functions S = 12(y2 +z2) −50x2 and Q = z −x2/20.

A continuation of this process into the region wherey is negative requires some “tricks” but finally yields the result. The details havebeen submitted elsewhere.3.

Criteria for the existence of complicated solutionsIn this section we give a theorem with the conclusion that for some values of(s, q, R), equation (1) has solutions with very complicated behavior, in a sense tobe made precise. The hypotheses of this theorem seem difficult to check analytically;however, the result seems to us to be an improvement over previous work becausethese hypotheses can be confirmed by examining only a compact segment of γ+,together with a set of solutions p such that p(0) is in a compact line segment.

Thisline segment lies in the intersection of the two planes x = y and z = R −1. Thesolutions only have to be followed over compact time intervals, suggesting that thehypotheses can be checked rigorously with computer assistance.Standard numerical analysis indicates that the hypotheses are satisfied, for someparameters, in a robust fashion so that the errors in the numerical analysis shouldnot be so great that the result is false.

This reinforces our hope that the theoremcan be shown to apply to (1).We have two principle hypotheses for Theorem 2. The first is an extension ofLemma 2.Condition A.

If p is a solution of (1) with p(0) ∈γ+, then x′ has at least fivesign changes and x has at least one zero. If τ1 < τ2 < τ3 < τ4 < τ5 are the first fivesign changes of x′, while t1 < t2 are the first two zeros of x, then τ1 < t1 < τ2 <τ3 < τ4 < τ5 < t2.

(If x does not have a second zero, set t2 = ∞. )This condition is obviously more restrictive than the conclusion of Lemma 2.When (s, q) = (10, 1), standard numerical integrations suggest that it holds forR approximately in the range (8.2, 17.2).

If q = 8/3, then the R range becomesabout (14, 46.6). Unpredictable behavior exists outside of this range, and a straightforward extension of our theorem would partly explain this, but be harder to checkrigorously.Before stating our second hypothesis, we must describe the “shooting” procedureused to obtain the complicated solutions.

The method is to choose initial conditionsp(0) in a certain line segment in the plane x = y and give an inductive procedurefor varying p(0) to obtain more and more complex behavior.To specify this line segment, suppose that Condition A is satisfied. Then thebranch γ+ of γ first crosses the plane x = y at some point p1, which can be shownto lie in the region z > R −1.

Also, since R > 1, there is an equilibrium point p0of (1) in this plane in the positive octant. At p0, z = R −1.

Our shooting set isthe line segment L connecting p0 and p1.The idea of our second condition is, roughly, that solutions starting on L donot gain or lose sign changes of x′ by bifurcation as the initial point changes on

4S. P. HASTINGS AND W. C. TROYL.Stated that way, however, it appears necessary to follow these solutions on0 ≤t < ∞, clearly not possible numerically.

Instead, we consider solutions startingalong the line where zeros of x′ bifurcate and follow these backwards. This is theline M defined by the equations x = y, z = R −1.

Note that this line intersects Lonly at p0. We will explain below why this should require only a finite amount ofcomputation.Condition B.

Suppose p is a nonconstant solution of (1) such that p(0) ∈M. Thenat least one of the following is true.

(2a) p(t) /∈L for t < 0(2b) In some interval containing t = 0, x ̸= 0 and x′ changes sign four times.To check Condition A numerically, it is first necessary to give estimates thatshow that γ+ intersects a specific planar rectangle close to, but not including, theorigin. It must then be shown that every solution starting in this rectangle behavesas described in the condition.

This process is difficult, because current methods ofinterval arithmetic lose about 10 decimal places of accuracy for every time unit ofintegration for this system; however it has been successfully carried out by Hassardand Zhang [2].To check Condition B, we suggest the use of the well-known result [7] that theellipsoid E defined by the inequalityx2 + 10R y2 + 10R (z −2R)2 ≤40Ris a positively invariant set for (1) for a range of values of q and s. The line segmentL lies in E. We consider initial points p(0) on the (different) line segment M ∩E.Assuming that p is not constant and (2b) cannot be verified, we would integratefrom p(0) backwards in t and show that the solution leaves E before intersectingthe line segment L. Once p leaves E as t decreases, it cannot reenter E at a lowert value.By its nature, a single integration using interval arithmetic can verifyCondition B for an interval of initial conditions on M around p(0). The practicaldifficulty, which up to now has prevented us from completing this step, is that thelength of these intervals is quite small, so that several thousand initial conditionsmust be considered.

A local analysis around p0 results in a bound on the length ofthe time intervals in our integrations.We can now state our second theorem.Theorem 2. Suppose that Conditions A and B hold for some (s, q, R).

Supposealso that two of the eigenvalues of the linearized system around p0 are complex.Moreover, suppose that {Mj} is any infinite sequence of 1’s and 3’s. Then thereis a solution p = (x, y, z) of (1) such that x has an infinite number of zeros in0 < t < ∞, and if {ti} is the sequence of consecutive zeros of x in [0, ∞) and σi isthe number of sign changes of x′ in (ti, ti+1), then σi = Mi for 1 ≤i < ∞.Outline of Proof.

Parametrize L by setting pα(0) = αp0 + (1 −α)p1, for 0 ≤α ≤1.The proof proceeds by induction, choosing a sequence of α’s giving more and moreof the prescribed numbers of critical points between zeros of x.Because two of the eigenvalues of the linearized system around p0 are complex,it follows that if α is close to 1, then p crosses the plane y = x in 0 < t < ∞beforeany possible zero of x. On the other hand, for small α, x decreases monotonically

A SHOOTING APPROACH TO THE LORENZ EQUATIONS5to below 0, after which x′ changes sign at least four times before x = 0 a secondtime. Therefore, the first positive zero of x, t1(α), is defined and continuous onsome maximal interval of the form [0, ¯α), where ¯α < 1.If xα has at least n positive zeros t1(α), .

. .

, tn(α), let tn+1(α) denote the (n+1)stpositive zero of x if this exists, or else tn+1(α) = ∞. Also, for 1 ≤i ≤n, letσj = σj(α) denote the number of sign changes of x′ in [ti, ti+1).Suppose that tn( ) is continuous on some interval In ⊂[0, ¯α).

We define threesubsets of In as follows, where the dependence of the tj and σj on α is againunderstood:An(In) = {α ∈In|tn+1 < ∞, σn = 1 and σn+1 ≥4},Bn(In) = {α ∈In|tn+1 < ∞, σn = 3 and σn+1 ≥4},Cn(In) = {α ∈In|tn+1 < ∞, σn ≥4 and σn+1 ≥4}.We prove the following, which imply the theorem. (i) Let I1 = (0, ¯α).

Then A1(I1), B1(I1), and C1(I1) are all nonempty. (ii) If, for some n and some In, An = An(In), Bn, and Cn are all nonempty,then there are intervals In+1 ⊂In and I′n+1 ⊂In such that(a) tn+1( ) is continuous on In+1 and on I′n+1;(b) In+1 ∩An and I′n+1 ∩Bn are nonempty;(c) The sets An+1(In+1), Bn+1(In+1), Cn+1(In+1), An+1(I′n+1), Bn+1(I′n+1),and Cn+1(I′n+1) are all non-empty.We do not have space for the details here, and they have been submitted else-where.

Condition B is used to show that as α varies, the number of sign changes ofx′α between consecutive zeros of x can decrease from four or more to three, or fromtwo or three to one, only when the fourth or second of these sign changes tends toinfinity on the t axis. The number of sign changes of x′ cannot jump directly fromfour or more to one without passing through an open set of α’s where there arethree.These theorems have a few simple corollaries, which we will mention when detailsof the proofs are published.Since the divergence of the Lorenz vector field isnegative, volumes are reduced by the flow.

We hope to investigate whether ourresults have any consequences about the existence of “strange attractors.”References1. O. Aberth, Precise numerical analysis, William C. Brown Publishers, Dubuque, IA, 1988.2.

B. Hassard and J. Zhang (to appear).3. S. Hastings and J.

B. McLeod, On the periodic solutions of a forced second-order equation,Nonlinear Science 1 (1991), 225–245.4. S. Hastings and J.

B. McLeod, On the chaotic motion of a forced pendulum, Amer. Math.Monthly (to appear).5.

S. Hastings and W. Troy, Oscillating solutions of the Falkner-Skan equation for positiveβ, J. Differential Equations 71 (1988), 123–144.6. E. N. Lorenz, Deterministic non-periodic flow, J. Atmospheric Sci.

20 (1963), 130-141.7. C. Sparrow, The Lorenz equations: bifurcations, chaos, and strange attractors, AppliedMath.

Sci., vol. 41, Springer-Verlag, Berlin and New York, 1982.8.

W. Troy, The existence of bounded solutions of the Kuramoto-Sivashinskii equations, J.Differential Equations 82 (1989), 269–313.Department of Mathematics, University of Pittsburgh, Pittsburgh, Pennsylvania15260

6S. P. HASTINGS AND W. C. TROYE-mail address: SPH@ MTHSN4.Math.P.H.eduE-mail address :Troy@ VMS.CIS.PITT.EDU

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