Spurious Chaotic Solutions of Dierential. Equations. Sigitas Keras. September Department of Applied Mathematics and Theoretical Physics

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1 UNIVERSITY OF CAMBRIDGE Numerical Analysis Reports Spurious Chaotic Solutions of Dierential Equations Sigitas Keras DAMTP 994/NA6 September 994 Department of Applied Mathematics and Theoretical Physics Silver Street Cambridge CB3 9EW England

2 Spurious Chaotic Solutions of Dierential Equations Sigitas Keras In this paper we present results concerning a chaotic behaviour of dierence equations. It is proved that there exists a horseshoe for the dierence equation y n+k+ = kx i= i y n+i + mx i=l i g(y n+i ); l m < k P m whenever is suciently large, provided that i=l i 6=, the function g is continuous and changes sign at least twice. This equation is of great importance in numerical analysis, since many well known numerical methods for ODEs, in particular multistep methods, can be rendered in this form. Some results concerning the non-chaotic case are also presented.. Introduction When we solve a dierential equation numerically, the accuracy of the numerical solution is our main but not the only concern. We also want the solution to preserve important qualitative features of the initial system: the number of stable and unstable steady states, the global behaviour of the trajectories, etc. This can easily be lost even in the simplest cases, as the following example illustrates. Let us consider a nonlinear second order equation u + u(? u) = u() = ; u () = v (.) It is known (?) that the solution of this equation can exhibit three dierent types of behaviour: u is periodic if jvj < v, with period T! as jvj! v ; 2 u! if jvj = v ; 3 u! if jvj > v. Department of Applied Mathematics and Theoretical Physics, University of Cambridge, England ( S.Keras@amtp.cam.ac.uk)

3 2 Spurious Chaotic Solution Fig.. The solution of the equation :2 with h 2 = : and v = :8 (v < v h ) Fig. 2. The solution of the equation :2 with h 2 = : and v = :9 (v > v h ) Let us now consider a simple numerical scheme for the equation (??) u n+? 2u n + u n? + u h 2 n (? u n ) = (.2) u? u u = ; = v h If h 2 is small (h 2 ), the solution of the equation (??) has the same structure as the continuous solution, i.e. it is periodic for v < v h and unbounded for v > v h with a critical value in between (Figures??,??). However, when h 2, the behaviour of the solution becomes much more complicated. While there still exist periodic solutions (Figure??), the set of initial values v for which the solution is periodic does not form an interval as in the continuous case. What is even more interesting, there are bounded solutions which are not periodic and clearly do not approximate any periodic continuous solution (Figure??). A closer inspection reveals that most of the solutions sooner or later become unbounded and this happens even if the initial value v is close to. In order to investigate this we will prove the following Lemma. Then u n! as n! Let us assume that the following inequality holds u < (.3) u k > for some k > (.4) Proof. Let i be the smallest number such that u i > u i?. It is easy to verify that, under the aforementioned assumptions, u i+ > u i. It follows by induction that u n monotonically increases if n i, hence, there exists a limit u of the sequence fu n g which may be nite or innite. If it is nite then u is a solution of the equation (u? 2u + u )=h 2 + u (? u ) =

4 Sigitas Keras Fig. 3. The solution of the equation :2 with h 2 = and v = 9: Fig. 4. The solution of the equation :2 with h 2 = and v = 8: The only solutions for the equation above are u = ;, which leads to contradiction. Thus, u =. 2 Let D n denote the set of initial values for which u i < ; i = ; : : :; n. Figure?? displays the sets D 4 ; D 5 and D 6. Since D i contracts very fast as i!, for the purpose of a better visualisation we present only a half of the set D 5 at the interval [?:2; :2] and a quarter of the set D 6 at the interval [?:2; :6]. The whole set D 5 consists of 8 disjoint intervals and the whole set D 6 consists of 6 disjoint intervals. Numerical experiments suggest that in general D n consists of 2 n?2 intervals, so that D = \ D i= i is a Cantor set. Nothing similar to the continuous case, where the set D was a single interval! The fact that the equation (??) has a dierent structure of bounded solutions is not the only reason for our studies. Let us consider equation (??) as a boundary value problem with boundary values u() = u() =. It is known (?) that, if n 2 < (n + ) 2, the aforementioned equation has 2n? steady solutions. However, if h is not suciently small, the discretization (??) has 2 N?2 dierent steady solutions where N is the number of discretization points and is independent of. Figure?? shows a bifurcation diagram to the boundary value problem. While it is known (?) that in a continuous case all steady solutions bifurcate from a zero solution, one can notice that discretization produces spurious bifurcations. Since the steady states play an important role in a global dynamical behaviour of the nonlinear parabolic equation u t = u xx + u(? u); one may expect a dierent global behaviour of the solution of the equation above whenever h 2 is large. Spurious solutions as in Figure?? are examples of chaos in dierence equations and the equation (??) can be described by what is known in the theory of dynamical systems as a Smale's horseshoe map (?). Although very simple in appearance, Smale's horseshoe maps are dicult to investigate. The theory of spurious solutions for ODEs has been discussed

5 4 Spurious Chaotic Solution n=4.6.4 y n= n=6 x lambda*h^2 Fig. 5. The sets D 4, D 5 and D 6 for the equation :2 with = Fig. 6. A bifurcation diagram for the equation :2 with N = 6 in (?,?,?,?) and chaotic spurious solutions have been found by (?,?,?). In our paper we will prove the existence of a horseshoe for a wide class of dierence equations which are of great importance in numerical analysis. In the next section we will present a background material from dynamical systems theory and prove the main results of this paper. Section 3 is devoted to some cases of the non-chaotic dierence equations. 2. Smale's horseshoe and dierence equations We commence this section by introducing several denitions from the theory of nonlinear dynamical systems. The original horseshoe map as dened by Stephen Smale (?) was a map of a square, where the image was a gure vaguely reminiscent of a horseshoe (Figure??). Smale considered the set D n of the initial values, for which the nth iterate of the map remains in the square and proved that D = \ D i= i is a Cantor set. It is possible to describe this behaviour using symbolic dynamics on binary sequences. It is proved in (?) that D n consists of 2 n disjoint sets D n (i), all of them in two vertical strips (Figure??), which is in fact the set D. We can assign a nite binary sequence of length n to each set D n (i), where the kth element of the sequence is either or depending on whether the kth iterate of the set D n (i) is in the upper or lower part of the \horseshoe". It is easy to prove (?) that in this case k iterations on a set correspond to the shift by k symbols in the corresponding binary sequence which leads to the following formal denition of the horseshoe map. Denition 2. Let S be some (nite or innite) set of symbols with more than one element and let I S = fi s js 2 Sg be a set of disjoint intervals in R. We say that the map : IS k 7! R has a horseshoe if for every sequence of symbols fs ig, where s i= i 2 S, there exists a sequence fy si g, y i= s i 2 I si such that (y si?k+ ; : : :; y si ) = y si+ The simplest case occurs when S has only two elements, S = f; g say. Our numerical example, as given in the introductory part of this paper, corresponds to this case.

6 Sigitas Keras 5 D f(d ) Fig. 7. The horseshoe map, and sets D and f(d ) Now we are able to formulate our main results Theorem 2. Consider the dierence equation y n+k+ = kx i= i y n+i + g(y n+j ); j k; (2.) where the function g is continuous and changes sign at least twice. Then for suciently large there exist two intervals I and I such that for every innite binary sequence fs i g i= there exists a sequence fy ng such that y n 2 I sn and y n satises the equation (??) Remark. The number of points where function g changes sign corresponds to the number of elements in S from our denition. This explains why we require the function g to change sign at least twice. Proof. We prove the theorem by induction. Without loss of generality we assume that g changes sign at the points and. Let I = [?; 2]. We dene M = max y n+i 2I X k i y n+i : i= Since g is continuous and changes sign at the points and, for small enough there exist intervals I and I such that the image of I i ; i = ; under g is [?; ] and I i does not contain other zeroes of the function g. Taking = (M + 2)= we deduce that the image of intervals I i under g is [?(M + 2); M + 2] In other words, for arbitrary values of y n+ ; : : :; y n+j? ; y n+j+ ; : : :; y n+k from interval I, when y n+j spans either of the intervals I, I, y n+k+ spans both intervals I and I. Now we can formulate our inductive assumption: A n Let S n be a binary sequence of length n, S n = fs i g n i=, and s i 2 f; g. Then there exists a compact set D n = D n (S n ) of initial values for equation (??) such that y i 2 I si

7 6 Spurious Chaotic Solution for i = ; : : :; n + j? and when y ; : : :; y k span D n then y n+j ; : : :; y n+k span both I and I. A is obviously true, since j k and we are free to chose initial values for the y ; : : :; y k. Suppose that A n is true. Without loss of generality we may assume that the last element of S n+ is. Let D n+ be a set of initial values such that y n+j 2 I. Under the assumption A n, D n+ D n is not an empty set, and, from our denition of the interval I, when y n+j spans I, y n+k+ spans both intervals I and I. Since I is compact and g is continuous, D n+ is compact. It follows now that D = \ i= D i is nonempty. 2 Theorem 2.2 Consider the following recurrence relation y n+k+ = mx i=l kx i= i y n+i + mx i=l i g(y n+i ); l m < k; (2.2) i 6= ; (2.3) where g is continuous and changes sign at least twice. Let j l j K = P mi=l+ j i j : Then If K > ; then for suciently large there exist two intervals I and I such that for every innite binary sequence fs i g there exists a sequence fy i= ng where y n 2 I sn and y n satises the equation (??) for all n = ; ; 2; : : :; 2 If K ; then for every nite binary sequence fs i g N there exist and two intervals I i= and I such that there exists a sequence fy n g N where y n= n 2 I sn and y n satises the equation (??) for all n = ; ; 2; : : :; N. Proof. The proof is similar to the previous one. Since the sum of coecients i is nonzero and our function g changes sign at least twice at the points x ; x, it is easy to see that for suciently large equation (??) has at least two equilibrium solutions z and z : z = kx i= i z + mx i=l i g(z ); = ; ; and z! x ; = ; as!. Without loss of generality we may assume that x =,

8 x =. We dene and rewrite equation (??) as f(; x ; : : :; x k ) = Sigitas Keras 7 kx i= i x i + mx i=l+ i g(x i ) y n+k+ = l g(y n+l ) + f(; y n+ ; : : :; y n+k ): Since g is continuous and changes sign twice, for small enough there exist distinct intervals I ; = ; such that Function g maps I to the interval [?; ] 2 Intervals I do not contain other zeroes of the function g except for x and x Let us dene constants M and M 2 M = max x ;:::;x k 2I [I M 2 = max x2i [I jxj: X k i x i ; i= If K >, we can take > (M +M 2 )=(K?) which results in the following inequalities: max l g(x) x2i > f(; x ; : : :; x k ) + M 2? min l g(x) x2i > f(; x ; : : :; x k ) + M 2 where x ; : : :; x k 2 I [ I. Similar inequalities can be obtained for the interval I. This means that when y n+l spans either of the intervals I, I, y n+k+ spans both intervals I and I for arbitrary values of y n ; : : :; y n+l? ; y n+l+ ; : : :; y n+k from I or I. Hence we may formulate an inductive assumption similar to A n from Theorem : A n Let S n be a sequence of n s or s, S n = fs i g n i=, and s i 2 f; g. Then there exists a compact set D n = D n (S n ) of initial values for equation (??) such that y i 2 I si for i = ; : : :; n + l and when y j ; j = ; : : :; k span D n, then y n+l+ ; : : :; y n+j span both I and I A is true since we are free to choose initial values for our equation (??) and (4:4{5) shows that A n yields A n+. Since D n D n? and D n is compact, D = \D n 6= ;. Consider now the case K. Let, I, I, M, M 2 be as in the rst part of the theorem We will need the following lemma Lemma 2. Let us x N. Then, if is suciently large, there exists a sequence of intervals I (n) ; = ; ; n = ; : : :; N? ; satisfying the following conditions. I () I ; I (n+) I (n) ; I (n+) 6= ;

9 8 Spurious Chaotic Solution Proof. max x2i (n) min x2i (n) l g(x) > max jf(; x ; : : :; x k )j + M 2 ; x ; : : :; x k 2 I (n+) [ I (n+) ; l g(x) <? max jf(; x ; : : :; x k )j? M 2 ; x ; : : :; x k 2 I (n+) [ I (n+) : Let us choose such that We dene c n recursively as follows It is easy to prove by induction that = M + M 2 N? X i= K i X N? c = = (M + M 2 ) K i i= c n = (c n?? M? M 2 )K X N?n c n = (M + M 2 ) K ; i which means that c n M + M 2 for all n N. Choose I (n) so that l g(x) maps I (n) to the interval I cn = (?c n ; c n ). The rst two conditions are satised since c n+ < c n and c = C. Inequality i= max jf(; x ; : : :; x k )j + M 2 Kc n+ + M + M 2 = c n ; yields that the last two conditions are also satised. 2 Now we can proceed as in the rst part of the theorem. x ; : : :; x k 2 I (n+) [ I (n+) A n Let S n be a sequence of n s or s, S n = fs i g n i=, and s i 2 f; g. Then there exist a set D n = D n (S n ) of initial values for equation (??) such that y i 2 I s (i) i for i = ; : : :; n + l and when y j ; j = ; : : :; k span D n, then y n+l+ ; : : :; y n+j span both I (n+l+) and I (n+l+), where I s (i) i 2 I si A is true and A follows from n+ A n if we choose large enough and dene intervals I (i) as in Lemma 2: Non-chaotic case One may ask what happens when the function g has less then two real zeroes. Unfortunately, no general results for dierence equations of type (??) or (??) are available. The following two theorems address themselves to a much simpler case. Nevertheless, they are of considerable interest in numerical analysis, because of their relevance to two-step methods for second order nonlinear ODEs.

10 Theorem 3. Sigitas Keras 9 Let us consider the following dierence equation y n+ = 2y n? y n?? g(y n ) (3.) Suppose that g does not change sign. Then ^y = lim n! y n exists on the extended real line and, unless g(^y) =, it is true that j^yj = Proof. Indeed, from (??) we have by induction y n+? y n = y n? y n?? g(y n ) = = y? y? nx i= g(y i ) (3.2) Without loss of generality we may assume that the function g is nonpositive. Then the expression on the right hand side of (??) is monotonically increasing as n! and changes its sign at most once. This means that for suciently large n the expression y n+? y n has a constant sign, i.e. y n is monotonic and hence tends to a limit. Let ^y = lim n! y n and suppose that g(^y) 6=. Then the expression on the right hand side of (??) has an innite limit and so does y n. 2 Theorem 3.2 Suppose that g changes sign at the point x and that g(x) > for x < x and g(x) < for x > x. Then ^y = lim n! y n exists on the extended real line and, unless g(^y) =, it is true that j^yj = Proof. Without loss of generality we may assume that x = and that y < y. Let us recall equation (??). It is easy to see that if y >, then y n! +. If y < then y i+? y i is decreasing and y i is increasing, so that for some n either y n > ; y n? y n? >, with following monotonous convergence to +, or y n < ; y n? y n? < with following monotonous convergence to?, or y n+? y n! ; y n! ^y as n!. In the latter case we deduce again that the sum on the right hand side of (??) has a nite limit, hence = lim n! g(y n ) = g(^y) Acknowledgments The author would like to thank Dr Arieh Iserles, Prof. Mike Powell and Prof. Herb Keller for many useful comments on this work. The author was supported by Leslie Wilson Scholarship from Magdalene College, Cambridge and ORS award. REFERENCES Chafee, N. & Infante, E. (974), `A bifurcation problem for a nonlinear parabolic equation', Jnl Appl. Anal. 4, 7{37. Davis, H. T. (962), Introduction to Nonlinear Dierential and Integral Equations, Dover Publications, Inc, New York. Griths, D. F. & Mitchell, A. R. (988), `Stable periodic bifurcations of an explicit discretization of a nonlinear partial dierential equation in reaction diusio', IMA J. Num. Anal. 8, 435{454. Henry, D. (98), Geometric theory of semilinear parabolic equations, Springer-Verlag, Berlin, New York.

11 Spurious Chaotic Solution Humphries, A. R. (993), `Spurious solutions of numerical methods for initial value problems', IMA J. Num. Anal. 3, 263{29. Iserles, A. (99), `Stability and dynamics of numerical methods for nonlinear ordinary dierential equations', IMA J. Num. Anal., {3. Iserles, A., Peplow, A. T. & Stuart, A. M. (99), `A unied approach to spurious solutions introduced by time discretisation. Part : Basic theory', SIAM J. Num. Anal. 28, 723{75. Keener, J. P. (987), `Propagation and its failure in coupled systems of discrete excitable cells', SIAM Jnl Appl. Maths 47, 556{557. Mitchell, A. R. & Bruch, J. C. (985), `A numerical study of chaos in a reaction-diusion equation', Num. Meth. PDE, 3{23. Smale, S. (967), `Dierentiable dynamical systems', Bull. Amer. Math. Soc. 73, 747{87. Stuart, A. M. & Peplow, A. T. (99), `The dynamics of the theta method', SIAM J. Sci.Stat. Comp 2, 35{372.

Waveform Relaxation Method with Toeplitz. Operator Splitting. Sigitas Keras. August Department of Applied Mathematics and Theoretical Physics

UNIVERSITY OF CAMBRIDGE Numerical Analysis Reports Waveform Relaxation Method with Toeplitz Operator Splitting Sigitas Keras DAMTP 1995/NA4 August 1995 Department of Applied Mathematics and Theoretical