ANALYSIS AND COMPUTATION OF TRAVELING WAVE SOLUTIONS OF BISTABLE DIFFERENTIAL-DIFFERENCE EQUATIONS CHRISTOPHER E. ELMER
AND ERIK S. VAN VLECK
y
Abstract. We consider traveling wave solutions to a class of di erential-di erence equations. Our interest is in understanding propagation failure, directional dependence due to the discrete Laplacian, and the relationship between traveling wave solutions of the spatially continuous and spatially discrete limits of this equation. The di erential-di erence equations that we study include damped and undamped nonlinear wave and reaction-di usion equations as well as their spatially discrete counterparts. Both analytical and numerical results are given.
1. Introduction. We consider analytically and numerically obtained traveling wave solutions of lattice-di erential equations of the form u_ (; t) + u(; t) = u(; t) + LD u(; t) ? f (u(; t)); P where u(; t) maps IRn IR ! IR, ; ; 2 IR+ [ 0, is the continuous Laplacian operator ni=1 Di ;i , and \_" denotes di erentiation with respect to t. We let LD be a discrete Laplacian operator de ned
(1.1)
with respect to an integer lattice of the form
LD u(; t) =
n X k=1
"k [u( + ek ; t) + u( ? ek ; t) ? 2u(; t)];
where "k 2 IR+ [ 0 and ek is the unit vector whose kth element equals 1, for k = 1; :::; n. Using an "k for each dimension allows us to vary the directional dependence, and using ek for our forward and backward shifts on an integer lattice is equivalent to using a nearest neighbor representation for LD . We are interested in bistable systems, systems where f (u) can be represented by a cubic-like nonlinear function with \stable" zeros at 0 and 1, and an \unstable" zero at the detuning parameter a 2 (0; 1): In this work, we use a piecewise linear representation for f de ned as
u(; t) u(; t) < a; u(; t) ? 1 u(; t) > a; 0 < a < 1: P Let be a vector in IRn such that ni=1 i2 = 1. A solution u(; t) = '( ? ct) of (1.1), where ' : IR ! IR, \ " denotes the Euclidean dot product, and c 2 IR is the unknown wave speed, is called
(1.2)
fa (u(; t)) =
a traveling wave solution. Substituting the traveling wave ansatz into (1.1), we obtain the spatially discrete ordinary di erential equation
?c '0 ( ) + c2 '00 ( ) = '00 ( ) + LT '( ) ? f ('( )); or
?c'0 ( ) ? c '00 ( ) = LT '( ) ? f ('( )); where c = ? c2 , c = c , = ? ct, and n X LT '( ) = "k ['( + ek ) + '( ? ek ) ? 2'( )]:
(1.3)
k=1
Department of Mathematical and Computer Sciences, Colorado School of Mines, Golden, CO 80401 USA (
[email protected]). y Department of Mathematical and Computer Sciences, Colorado School of Mines, Golden, CO 80401 USA (
[email protected]). The work of this author was supported in part by NIST contract # 43NANB714674 and NSF grant # DMS-9505049. 1
We present traveling wave solutions for c nonnegative when c 6= 0 and c positive when c = 0. The unit vector is normal to the wavefront and indicates the direction of the traveling wave with respect to the lattice. Equation (1.1) includes the continuous and spatially discrete nonlinear wave, damped wave, and reaction-di usion equations. We present a description of the family of traveling wave solutions de ned by (1.1), equivalently (1.3), with nonlinearity (1.2). The parameters in (1.1) allow for continuation between the discrete and continuous Laplacian terms. Throughout the modeling community interest is increasing in models of time dependent di erential equations that are discrete in space. In many areas of science, systems exhibit a spatially discrete structure. This spatially discrete structure is often the result of a underlying spatial lattice which a ects the evolution and dynamics of the system. The representation in (1.1) allows the microscopic lattice structure of a system to have macroscopic e ects on the solution behavior, in particular we note that the lattice Laplacian allows the phenomena of anisotropy and propagation failure (phenomena that continuous Laplacian models fail to represent). Some of the elds of research in which we see the e ects of a spatially discrete structure and can be modeled with equations of the form of (1.1) include material science (crystal growth and liquid-solid materials) [8],[18],[32], chemical reaction theory (chemical reactors) [24], [38], image processing and pattern recognition (cellular neural networks and optical memory) [27],[15],[16],[17],[45], [56], physics (relativistic quantum mechanics) [49], [50], [51], [55], and biology (myelinated nerve axons and myocardium) [5],[22],[36],[37],[58], [6],[23],[35]. If for some k = 1; :::; n, "k 6= 0 and k 6= 0, with either c 6= 0 or c 6= 0, then (1.3) is a delay equation with both forward and backward delays, a functional di erential equation of mixed type. Very little work has been done on general systems of such mixed functional equations, except for the work of Rustichini [47], [48] and Mallet-Paret [41]. Propagation failure for one dimension spatially discrete reaction-di usion equations was studied by Keener [35], [36]. Zinner provided existence and stability of traveling wave solutions for spatially discrete reaction-di usion equations [60], [61], [62]. In [29], Gao looked at varying the "k in LT and LD . In [10], Cahn, Mallet-Paret, and Van Vleck provide propagation failure and lattice anisotropy results for traveling wave solutions of two dimensional spatially discrete reaction-di usion equations with (1.2), see also the analysis of Fath [25] for one dimension spatially discrete reaction-di usion equations. Mallet-Paret in [42] looks at the global structure of solutions to lattice-di erential reaction-di usion equations. Shen's work includes (1.1) with = 0 and nonlinearity f (u; t) almost periodic in t [53], [54]. In [33] and [34], Johnston studies bifurcation phenomena for several bistable nonlinearities. Other investigations into the solutions of mixed equations of the form (1.1) include the works [61], [31], [62], [60], and [14]. Our physical motivation for considering (1.1) is to incorporate both microscopic and macroscopic features into the model. This could be due to having di usive interactions at a given length scale within a material while simultaneously having di usive interactions at an arbitrary small length scale. This results in having both the discrete Laplacian operator, LD , and the spatially continuous Laplacian operator, . By considering both operators we are able determines which dynamic behaviors are due to LD and which are due to . For instance, we nd that for > 0 there is no propagation failure but there may still be directional dependence of the wave speed due to the discrete term LD . Our goal is to gain insight into the dynamics of traveling waves solutions, with spatially discrete and/or spatially continuous terms, as characterized by dependence between the detuning parameter, a, and the wave speed, c, and by graph of the wave form. We present both analytically and numerically obtained solutions implementing each solution method independent of the other. The analytically obtained solutions are produced using classical methods. Using relaxation, continuation, and xed point iteration, the numerically obtained solutions are produced with a standard boundary value problem solver. Since the phenomena uncovered is veri ed with both solution methods, we have a natural basis for comparison of the two solution techniques. From our analytically obtained solution, we derive a relation between the detuning parameter a, in (1.2), and the wave speed c, from which we generate a(c) plots to help illustrate phenomena we discuss. For all our results, we incorporate the classical phase condition '(0) = a. 2
For c 2 IR, c 0, with c and c not both equal to zero, we show that a solution '( ) of (1.3), with nonlinearity (1.2) and boundary conditions '(?1) = 0; '(1) = 1, exists for all 2 IR and is strictly monotone increasing on IR. The solution '() 2 C (IR) with '() 2 C 1 (IR) if and only if c > 0: The discontinuity in '0 when c = 0 comes from the discontinuity in the nonlinearity that we are using and produces \kinks" in the numerical solution plots. Because of the spatially discrete Laplacian term in (1.1) and (1.3), it is possible for the wave speed c to equal zero for a nontrivial interval of the detuning parameter a (the interval of propagation failure). In fact, there exists of propagation failure P an interval if and only if the coecient on the spatially discrete Laplacian, nk=1 "k ,P is not zero and the coecient on the spatially continuous Laplacian, , is zero. Also, for = 0, as nk=1 "k ! 0, the interval of propagation failure expands to (0; 1). Information about the interval of propagation failure comes from letting c go to zero in the a(c) relation. We also study the a(c) relation in general, when we are not taking c ! 0. The relation a(c) is odd and analytic in c for all c if c > 0, and odd and analytic in c for all c 6= 0 if c = 0. The relation a(c) = 1=2 holds if and only if c = 0, thus a(c) = 1=2 when = 0. For c 6= 0, the a(c) relation is strictly increasing. If we are solving the reaction-di usion equation, = 0, then c 2 (?1; 1) and a 2 (0; 1), but if we are solving the damped wave equation, > 0, then c is restricted to a proper subinterval of IR and through the a(c) relation, we see that a is restricted to a proper subinterval of (0; 1). In fact, the range of a for the damped wave equation is symmetric about a = 1=p 2 and the lowest upper and greatest lower boundspcan be found with the a(c) relation by setting cp= = for the upper bound and by setting c = ? = for the lower bound. Notice that as jcj ! = the spatially mixed damped wave equation approaches the spatially discrete reaction-di usion equation. This indicates that these lowest upper and greatest lower bounds on a for the spatially mixed damped wave equation can be found using the a(c) relation for the spatially discrete reaction-di usion equation with the same ; "k 's, and values. Lets suppose c ! 0, c ! 0, and that the elements of the direction vector are rationally related. Then the solution ' approaches a step function in the limit. For our phase condition a jump discontinuity between steps occurs at = 0. The step to the right of = 0 is equal to the least upper bound of the interval of propagation failure and the step to the left of = 0 is equal to the greatest lower bound of the interval of propagation failure. Hence the jump between steps at = 0 is equal in length to the interval of propagation failure. In addition to the above results, the numerical plots of the solution demonstrate the di erences in traveling wave solutions of the reaction-di usion, damped wave, and the undamped wave equation, and the impact of the spatially discrete and the spatially continuous Laplacian terms. We also present contour plots in the c versus c parameter space and the versus parameter space which show the level set of a. Section 2 consists of our analytical derivation of a solution along with our analytically derived results for (1.3). First we complete the de nition the problem. This includes discussing boundary conditions and a phase condition. Next we derive a solution to (1.3) when c and c are not both zero, including results about monotonicity and smoothness. Then we present the a(c) relation and use it to derive results about the phenomenon of propagation failure and the restriction of the interval for a. We also provide plots of the a(c) relation. For completeness of presentation, a brief outline of results when c and c both approach zero is included. The importance of this case is that our traveling wave equation becomes a di erence equation as c and c both approach zero. We also show a relation between the function that is produced in the limit and the interval of propagation failure. Section 3 is dedicated to the numerical solutions of our traveling wave equation. We rst de ne the numerical method. We then present plots of numerically obtained waveforms. This includes a presentation of the impact of the spatially discrete versus the spatially continuous Laplacian, and a presentation of the solutions in terms of the three classi cations, reaction-di usion, damped wave, and undamped wave. We nish this section with a summary of how the numerical and analytical solutions support one another. Section 4 consists of conclusions with present and future directions.
2. Analytical Results. We complete the de nition of the traveling wave equation (1.3), nd a an analytical solution to the traveling wave equation, and investigate the phenomena of propagation failure, solution \kinks", restrictions on the wave speed and the detuning parameter, and lattice induced 3
anisotropy.
2.1. Background. Because ' = 0 and ' = 1 are zeros of fa, in (1.2), it is natural to impose the
boundary conditions
'(?1) = 0; '(1) = 1
(2.1)
to (1.3). Traveling wave solutions to (1.3) with (2.1) are orbits which connect the points ('; '0 ) = (0; 0) and ('; '0 ) = (1; 0). Solutions to (1.3) with (2.1) are only unique up to a phase shift, i.e., if '( ) is a solution to (1.3) with (2.1) then so is '( + b), thus we need to choose a phase for uniqueness. Assuming that ' is strictly monotone increasing, '( ) = a for only one value of , call it 0 . Without loss of generality, we may assume 0 = 0. We will later justify these assumptions and our choice of a phase condition. This choice of phase condition implies
'( ) < a;
< 0;
and
Thus, for the Heaviside function de ned as
h(x) =
'( ) > a;
> 0:
1 x > 0; 0 x < 0;
h('( ) ? a) = h( ) for 6= 0. Therefore, fa ('( )) in (1.2) becomes (2.2) f ('( )) = '( ) ? h('( ) ? a) = '( ) ? h( )
for 6= 0:
This form of the nonlinearity incorporates the phase condition into our problem. Thus for any function ' the right-hand side of (1.3) becomes LT '( ) ? '( )+ h( ): Taking the limits from the left and right of as it approaches 0 we see that lim L '( ) ? '( ) + h( ) = LT '(0) ? '(0) 6= LT '(0) ? '(0) + 1 = ! lim0+ LT '( ) ? '( ) + h( ): !0? T This in turn implies for any function ' that satis es (1.3) with (2.2) and (2.1) that (2.3)
lim (?c '0 ( ) ? c '00 ( )) 6= ! lim0+(?c '0 ( ) ? c '00 ( )):
!0?
2.2. An Analytic Solution and the c(a) Relation. We now derive the solution of the continuous/discrete traveling wave equation (1.3) with c = c , c = ? c2 , c 0 when c 6= 0, c > 0 when c = 0, the boundary conditions (2.1), and the piecewise linear function f (2.2). From our derived solution, we obtain a relationship between the detuning parameter, a, and the wave speed, c. Also, from this solution and the a(c) relation, we present results concerning propagation failure. In order to construct our solution, we use the Fourier transform. This is the reason for using a \linearized" nonlinearity f . This section begins with the construction of a solution, '( ), for (1.3). Our construction follows the construction in [10] and [42] for traveling wave solutions to spatially discrete reaction-di usion equations. The construction consists of de ning a function '" ( ) = e?" '( ) and substituting into Equation (1.3). We next show that the solution '" ( ) of the resulting equation approaches zero in an asymptotic fashion as j j ! 1. This allows the us to apply the integral Fourier transform to '" ( ) (the integral exists), and solve the equation for '" in Fourier space. The Fourier inversion theorem then gives '( ) = e" '" ( ). We then show that this solution, '( ), is continuous in both and c , and is analytic in c for c 6= 0. We also show that Dc '( ) is bounded. From the formula for '( ), we next derive a relation between a and c, called ?. The function ? is odd in c , analytic in c, and discontinuous at c = 0 if and only if = 0. When = 0, the discontinuity in ? is a jump discontinuity and is called the range of propagation failure. We then show that '( ) is strictly increasing on the real line. The last key result of this section is that c depends analytically on a for a outside the interval of propagation failure. This is not true for a in the interval of propagation failure. 4
We begin our derivation of a solution with a preliminary lemma which we will use to show that
'" ( ) ! 0 as j j ! 1. Lemma 2.1. Let ' be a solution of (1.3) for c 6= 0; c 0 or c = 0; c > 0. Then there exists an "0 > 0 such that, for some K > 0, (2.4) j'( )j Ke"0 for 0:
Proof.
The case when c = c, c = 0, and the dimension n = 2, is restatement of Lemma 4.1 in [10], and the case for general n follows directly from this lemma. The proof of the case when c 6= 0 follows the method of proof of Lemma 4.1 in [10] with Equation (4.6) replaced by
?c 00 ( ) = ?c 0 ( ) + LT ( ) ? ( ) for all > 0 and Equations (4.8) replaced by
P (s) = c s ? c s2 + 1 ? H; and
q(s) = (c ? c s) (0) ? c 0 (0) ? J (s); Pn " (eek + e?ek ? 2); with H = P k=1 k R R and J (s) = nk=1 "k (eek 0ek e?s ( )d ? e?ek ?0ek e?s ( )d ): When c = 0 care is needed in
the integrations. Now we are ready to construct our solution to the traveling wave equation (1.3) with boundary conditions '(?1) = 0 and '(+1) = 1, and reaction term f given by (2.2). First, let '" ( ) = e?" '( ), with " > 0, " small. By Lemma 2.1, if ' is a solution to (1.3), then, as ! ?1;
'" ( ) = e?" '( ) je?" jj'( )j jKe("0 ?") j ! 0 for 0 < " < "0 :
Also as ! 1;
'" ( ) = e?" '( ) ! 0 since '(+1) = 1. This implies that '" (?1) = 0 and '" (+1) = 0. Substituting '( ) = e" '" ( ) into
(1.3), we obtain
?ce" ("'" ( ) + '0" ( )) = c e" ("2 '" ( ) + 2"'0" ( ) + '00" ( )) + e" L" '" ( ) ? e" '" ( ) + h( ); which simpli es to (2.5)
(?c ? 2c")'0" ( ) = c'00" ( ) + L" '" ( ) ? (1 ? c" ? c "2 )'" ( ) + e?" h( );
P
where L" '" ( ) = nk=1 "k (e"ek '" ( + ek ) + e?"ek '" ( ? ek ) ? 2'" ( )): Following the Fourier solution technique presented in [10], we apply the Fourier transform
'^" (s) =
Z1
?1
e?is '" ( )d;
where " > 0 is suciently small, to both sides of (2.5) and solve for '^" (s). We then apply the Fourier inversion theorem Z +1 1 " e(is+") '^" (s)ds; '( ) = e '" ( ) = 2 ?1 5
which is absolutely convergent, to obtain '( ). This technique produces the following formula for the solution of (1.3):
Z 1 A(s) sin s Z 1 cos s 1 1 c (2.6) '( ) = 2 + ds; ds + 0 s(A2 (s) + 2c s2 ) 0 A2 (s) + 2c s2 Z 1 sin s 1 1 '( ) = 2 + ds; 0 sA(s) where
A(s) = 1 + cs2 + 2
(2.7)
n X k=1
for c 6= 0; c 0; for c = 0; c > 0;
"k [1 ? cos(sek )]:
Remark 2.1.
i) Noting that
Z 1 ?A(s) sin(s ) Z 1 cos(s ) 1 ? 1 c = '(; c ); ds ? 1 ? '(?; ?c) = 2 ? 0 s(A2 (s) + 2c s2 ) 0 A2 (s) + 2c s2
we see that, for '( ) = '(; c ), we have the symmetry property '(; c ) = 1 ? '(?; ?c). ii) Recalling that
Z ?i"+1 eis 1 ds: '( ) = 2i ?i"?1 sR(s)
we can see that as ! ?1, '( ) ! 0. Our symmetric property then gives us that '(+1) = 1: This con rms our choice of boundary conditions. Proposition 2.1. Let '(; c ) denote the solution function (2.6). Then '(; c ) is continuous in both and c , and is analytic in c, for c 6= 0. In addition, for c 6= 0, the derivative of '(; c ) with respect to c,
( ) = (; c ) = Dc '(; c ); is bounded as ! 1: For 6= 0; ( ) satis es the equation
?'0 ( ) + 2 c '00 ( ) + (c2 ? ) 00 ( ) ? c 0 ( ) = LT ( ) ? ( );
which is the derivative of the equation with respect to c. Proof. The proof follows the proof of Proposition 4.2 in [10] with the derivative Dc replacing Dc and
?'0 ( ) + 2 c '00 ( ) + (c2 ? ) 00 ( ) ? c 0 ( ) = LT ( ) ? ( );
replacing Equation (4.22). Making the traveling wave solution assumption for (1.1) and substituting the traveling wave ansatz u(; t) = '( ? ct) actually gives us two unknown functions to solve for, ' and c, which depend on the detuning parameter a. When using the Fourier transform technique, we assume that c is known and a is the unknown. We never see a explicitly appear in our derivation of an analytic solution ' because of our particular choice of nonlinearity, fa . Using (2.6) we now have an explicit representation of the relation between the wave speed c and the detuning parameter a. Remark 2.2.
6
Recall that we solved for the translate de ned by setting '(0) = a. This then gives us the following relation between a and c: Z1 ds '(0) = a = 21 + c (2.8) 2 A ( s ) + 2c s2 ; when c 6= 0; 0 and '(0) = a = 12 when c = 0:
We assumed that our solution was strictly increasing thus allowing us to pick a unique translate of the traveling wave solution. We now justify that assumption. Lemma 2.2. The solution ', (2.6), of (1.3) with (2.2) and (2.1), is strictly increasing for 2 IR. Proof. The proof of the lemma follows the proofs of Corollary 4.5 and Theorem 4.6 from [10] if we recall that '00 ( ) = lim!0 (1=)['( + ) + '( ? ) ? 2'( )], and if we recall that ?c '0 ( ) = 0 when c = 0. A drawback of using the piecewise linear nonlinearity fa , (1.2), is that the solution '( ) is not smooth at ' = a. Lemma 2.3. For c 6= 0 and c 0 the derivative '0 ( ) of (2.6) is continuous if and only if c > 0. In addition, if c > 0 lim '00 ( ) ? lim '00 ( ) = 1 ; !0?
and if c = 0
!0+
c
lim '0 ( ) ? ! lim0+ '0 ( ) = 1 :
!0?
c
Proof. We note that by (1.3) with (2.2), any diculty in smoothness occurs at = 0. Taking the di erence of (1.3) with (2.2) as approaches zero from the left and from the right, we obtain (2.9) lim (? '0 ( ) ? c '00 ( )) ? ! lim0+(?c'0 ( ) ? c '00 ( )) = ?1 !0? c since ' is continuous on IR. Suppose c = 0. Then (2.9) implies
lim '0 ( ) ? ! lim0+ '0 ( ) = 1 :
!0?
c
Thus '0 is discontinuous at = 0. Now suppose c > 0. Taking the derivative of '( ) with respect to in (2.6), we see that lim!0? '0 ( ) = lim!0+ '0 ( ). Thus (2.9) implies that lim '00 ( ) ? lim '00 ( ) = 1 : !0?
!0+
c
We now address exactly what is meant by a solution to (1.3). To solve (1.3), it is sucient to regard both f ('), (2.2), and h( ) as set valued functions, with f and h singleton sets for ' 6= a and 6= 0, and f (a) = [a ? 1; a] and h(0) = [0; 1]. This amounts to lling in the jump discontinuities in the graphs of f and h. By a solution to (1.3) we mean that ' 2 C (IR) and the di erential inclusion ?c'0 ( ) ? c'00 ( ) 2 LT '( ) ? '( ) + h( ): holds for 2 IR. 7
2.3. Propagation Failure. We now analyze the relationship between the wave speed c and the detuning parameter a. We are particularly interested in the phenomenon of propagation failure, the existence of a nontrivial interval for a such that the wave speed is zero. Throughout this subsection we include plots of the a(c) relation calculated from (2.8) using the three point Gaussian quadrature algorithm adapt [52]. Set Z1 1 1 ds c ?(c ; c) := '(0) ? 2 = a ? 2 = (2.10) ; c 6= 0: 2 0 A (s) + 2c s2 where n X 2 A(s) = 1 + cs + 2 "k [1 ? cos(sek )]: k=1
(2.11)
For a change in variables in (2.10) we obtain
Z1 ds sgn( ) c ; ?(c ; c ) = 2 A ( s= c ) + s2 0
(2.12)
c 6= 0:
Remark 2.3. i) ?(c ; c ) with c 6= 0 is nonzero. ii) For k 6= 0, ?(c ; c ) ! 0 as "k ! 1, k = 1; :::; n. iii) Recall a ? 1=2 = 0 when c = 0.
iv) For = 0, the c versus a curve is a(c) = 1=2 for all c. Lemma 2.4. For c 6= 0, the function ?(c ; c ) in (2.10) is strictly increasing in c . Proof. The case when c = 0 is proven in Theorem 4.7 of [10]. For c > 0, the application of the Mel'nikov method in the style of Theorem 4.7 of [10] proves the lemma. In this application it is important to recall that
lim (?c'0 ( ) ? c '00 ( )) 6= ! lim0+(?c'0 ( ) ? c '00 ( ))
!0?
and that '0 ( ) is continuous at = 0. We now show that ? is bounded and we show that ? is odd and analytic in c . The proof of the following theorems appear at the end of this section. Theorem 2.1. Let '( ) be a solution to (1.3) with (2.2) and (2.1). Thus ?(c ; c) is de ned by (2.10) with c 6= 0. Then ? is odd and analytic in c . In addition, if = 0 (2.13) j ?(c ; c)j = 12 : j?(c ; c)j < 21 and j lim c j!1 If > 0 then c is restricted to the interval [?d; d],
j?(c; c )j ?(d; 0);
(2.14)
lim j?(c ; c )j = ?(d; 0);
and
jc j!d
p
where d = = . Remark 2.4.
Lets consider the damped wave equation, i.e. > 0 and > 0. Lets pick jcj as large as possible p p for this case, let c = ? = and c = = . For these values of c, c = 0, (1.3) becomes the traveling wave form of a spatially discrete reaction-di usion equation, and j?j = ?(d; 0), the ? of the spatially discrete reaction-di usion equation. Thus the range of a values for the damped wave equation is [1=2 ? ?(d; 0); 1=2 + ?(d; 0)]. (see Figures 2.2, 2.3, 2.4, and 2.5). 8
a 1 Curve Curve Curve Curve Curve
0.9 0.8
1 2 3 4 5
0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 -10
-8
-6
-4
-2
0
2
4
6
8
10
c
Curve 1 2 3 4 5 ? 1 ? 2 5 1 10 10 10?5 Fixed ("1 ; "2 ; "3 ) Parameters 0 10 (1; 0; 0) (1; 0; 0) . An a(c) plot for the Reaction-Di usion where we vary .
Fig. 2.1
We see in Figures 2.1, 2.2, and 2.3 that as increases, j?(c ; c)j increases toward its upper bound of 1=2. This agrees with both (2.13) and (2.14). Figures 2.1, 2.2, 2.6, and 2.7 illustrate (2.13). Figures 2.3, 2.4, and 2.5 illustrate (2.14). Theorem 2.2. Let ?(c ; c ) be de ned by (2.10) with c 6= 0 and let %(c ) limjcj!0+ j?(c ; c )j for c 0: Then (2.15)
%(c ) =
(
R limT !1 21T 0T Ads(s) 0
c = 0; c > 0:
Remark 2.5. i) Since c ! 0+ and c ! as c ! 0+, when = 0, (1=2 ? %(0); 1=2 + %(0)) is the range for the
detuning parameter a for which the solutions to (1.3) exhibit propagation failure. ii) The interval [1=2 ? ?(d; 0); 1=2 + ?(d; 0)], in Remark 2.4, is equal to the interval of propagation failure when = 0 (see Figures 2.2, 2.3, 2.4, and 2.5). Lets take a close look at when c ! 0+ because ! 0+. Suppose = 0 and we are solving the reaction-di usion equation. When 6= 0, ja ? 1=2j = %(c ) = %( ) = 0 as ! 0+ for all c. Figure 2.1 is an a(c) plot showing how a ! 1=2 as ! 0+ in the mixed (spatially continuous/discrete) reactiondi usion model. When = 0, ja ? 1=2j = %(0) > 0 as ! 0+ for all c. Figure 2.2 is an a(c) plot showing 9
a 1 Curve 1 Curve 2 Curve 3
0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 -10
-8
-6
-4
-2
0
2
4
6
8
10
c
Curve
1 2 3 1 0:1 0:01 Fixed ("1 ; "2 ; "3 ) Parameters 0 0 (1; 0; 0) (1; 0; 0) . An a(c) plot for the Spatially Discrete Reaction-Di usion where we vary .
Fig. 2.2
how a ! 1=2 %(0) as ! 0+ in the spatially discrete reaction-di usion model. We have propagation failure in this case. The interval of propagation failure is the interval of the jump discontinuity in Figure 2.2. Now suppose > 0 and > 0 and wepare solving the damped wave equation. When > 0, c 0, we have seen that the bound for c, jcj = , impliespa bound on a. Figure 2.3, for a mixed damped wave model, is an a(c) plot where the bound on jcj is = = 10. Figure 2.3 also shows the upper (at c = 10) and lower (at c = ?10) bounds on the detuning parameter a for di erent values of . Now notice in Figure 2.2 the values of a at c = ?10 and c = 10 for the same values of . We see the two set of a values (in Figure 2.2 and Figure 2.3) are the same. Notice p how the range of a depends p in Figure 2.3 on . As ! 0+ the a(c) curve approaches 1=2 for jcj < = and a( = ) approaches 1=2 %(0). Figure 2.4 also shows this result for various values of . Figure 2.5 shows explicitly the restriction of the range of a. Here we present a(c) plots for various (the sigmoidal curves) along with the a(c) curve for the spatially discrete reaction di usion equation with the same , ("1 ; "2 ; "3 ), and . Suppose = 0, so c = 0 and we have the undamped wave equation. Then regardless of the choice of the other parameters a(c) = 1=2 for all c 2 IR. Proof.(of Theorem 2.1) The oddness of ? follows directly from (2.10). Analyticity follows from Proposition 2.1. Let c 6= 0. P Suppose = 0. Then A(s=c ) = [1 + c =2c s2 + 2 nk=1 "k (1 ? cos((s=c)ek )] 1 and 10
a 1 Curve Curve Curve Curve Curve Curve Curve
0.9 0.8
1 2 3 4 5 6 7
0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 -10
-8
-6
-4
-2
0
2
4
6
8
10
c
Curve 1 2 3 4 5 6 7 ? 1 ? 1 ? 2 ? 2 5 1 5 10 10 5 10 10 10?5 Fixed ("1 ; "2 ; "3 ) Parameters 0.1 10 (1; 0; 0) (1; 0; 0) . An a(c) plot for the Damped Wave where we vary .
Fig. 2.3
P limjcj!1 A(s=c ) = limjc j!1 [1 + c =2c s2 + 2 nk=1 "k (1 ? cos((s=c )ek )] = 1 for s > 0. Thus Z1 Z 1 ds 1 1 ds 1 = j?(c ; c)j = 2 2 2 0 1+s 2 0 A (s=c ) + s
and
1Z 1 1 Z 1 ds = 1 : ds lim j?(c ; c )j = lim = A2 (s=c) + s2 0 1 + s2 2 c !1 0 jcj!1 p p Now suppose > 0. This implies, since c 0, that jcj = j cj = . Let d = = . P P n n Then A(s=c) = [1 + c=2c s2 + 2 k=1 "k (1 ? cos((P s=c )ek )] [1 + 2 k=1 "k (1 ? cos((P s=c )ek )] and limjc j!d A(s=c ) = limjc j!d [1 + c =2c s2 + 2 nk=1 "k (1 ? cos((s=c)ek )] = [1 + 2 nk=1 "k (1 ? cos((s=d)ek )] for s > 0. Thus
j?(c ; c )j j?(c; 0)j
and
lim j?(c ; c)j = ?(d; 0):
jc j!d
By Lemma 2.4, if jc j d then j?(c ; c )j ?(d; c ). This and the inequality above gives us
j?(c ; c )j ?(d; c ) ?(d; 0): 11
a 1 Curve 1 Curve 2 Curve 3 0.8
0.6
0.4
0.2
0 -10
-5
0
5
10
c
Curve
1 2 3 10 1:6 0:1 Fixed ("1 ; "2 ; "3 ) p p ? 6 Parameters 10 0:1 (1; 1; 0) ( 22 ; 22 ; 0)
a
. An a(c) plot for the Wave equation where we vary .
Fig. 2.4
1 Curve Curve Curve Curve
1 2 3 4
0.8
0.6
0.4
0.2
0 -10
-5
0
Curve
1
5
2
3
10 1:6 0:1 Fixed ("1 ; "2 ; "3 ) Parameters 0:1 0:1 (1; 0; 0) (1; 0; 0) Curve 4 = 0 = 0 = 0:1 ("1 ; "2 ; "3 ) = (1; 0; 0) = (1; 0; 0) . An a(c) plot for the Damped Wave equation where we vary . 12
Fig. 2.5
10
c
a 1 Curve Curve Curve Curve Curve
0.9 0.8
1 2 3 4 5
0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 -10
-8
Curve
-6
-4
-2
0
2
4
6
8
10
c
1 4p p 2p p p3 p p p5 p (1; 0; 0) ( 22 ; 22 ; 0) ( 33 ; 33 ; 33 ) ( 12 ; 12 ; 22 ) ( 22 ; 33 ; 66 ) Fixed ("1 ; "2 ; "3 ) Parameters 1 0 0 (1; 1; 1)
. An a(c) plot for the Spatially Discrete Reaction-Di usion equation where we vary .
Fig. 2.6
Proof.(of Theorem 2.2)
We have that 1 + cs2 =2c A(s=c), which implies
(2.16)
Z1 1 ds p ?(c ) 1 = ; 2 2 2 2 2 1 + 4c =2c 0 (1 + c s =c ) + s
for c small and c =2c 0. As c ! 0+, c =2c ! 1 for c > 0. So (2.16) implies that as c ! 0+, ?(c ) is bounded above by zero when c > 0.PRecall that for c > 0, ?(c ) > 0: Now suppose c = 0. Then A(s) = 1 + 2 nk=1 "k [1 ? cos(sek )] and 1 %(0) = Tlim !1 2T
Z T ds 0 A(s)
follows directly from Theorem 4.3 in [10]. Remark 2.6.
R
i) The quantity limT !1 21T 0T Ads(s) always exists and is positive when c = 0. ii) Since c 0, when c 6= 0, and 0, = 0 implies c = ? c2 = 0. Thus lim !0 %(c ) = limc !0 %(c ). iii) For 6= 0 as c ! 0, there exists a nontrivial interval for the detuning parameter if and only if
= 0: In other P words, we have propagation failure if and only if = 0. 0, %(0) ! 1=2: This shows that the interval of propagation failure can expand to iv) As nk=1 "k !P the interval (0; 1). As nk=1 "k ! 1, %(0) ! 0. In our previous discussions, we presented a(c) relations where and ("1 ; "2 ; "3 ) are xed. In Figures 2.6 and 2.7 we see the e ects of varying , lattice induced anisotropy. In both plots we see the direct 13
a 1 Curve Curve Curve Curve Curve
0.9 0.8
1 2 3 4 5
0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 -10
-8
-6
Curve
-4
-2
0
2
4
6
8
10
c
1 4p p 2p p p3 p p p5 p (1; 0; 0) ( 22 ; 22 ; 0) ( 33 ; 33 ; 33 ) ( 12 ; 12 ; 22 ) ( 22 ; 33 ; 66 ) Fixed ("1 ; "2 ; "3 ) Parameters 1 0 10?4 (1; 1; 1) . An a(c) plot for the Reaction-Di usion equation where we vary .
Fig. 2.7
dependence of the a(c) relation on the direction parameter . Figure 2.7 shows how the a(c) relation becomes discontinuous, how the interval of propagation failure develops, as ! 0+. We discuss Figures 2.6 and 2.7 further in Section 3. 2.4. c and c Equal Zero. In this section, we consider solutions to (1.3) with c = 0 = c. In this case we are only concerned with values '( ) for 2 D, where D IR is the countable set D = f lj 2 ZZ n g: This is because with c = 0 = c , (1.3) becomes a di erence equation. Below is a summary of results, but rst some notation. If for any j 6= k, k and j = 1; :::; n, j and k are rationally independent, then we call rationally independent, and is called rationally dependent otherwise. If is rationally independent, then D is a dense subset of IR, while if is rationally dependent, then D = fm jm 2 ZZ g and D is a discrete subset of IR. Both forms of D have measure zero. Next we take the limit as c ! 0 together with the limit c ! 0 of the solution found in the last section. Let ' ( ) be de ned as the function such that '(; (c )n ; (c )n ) ! ' ( ) as the sequences (c )n ! 0 and (c )n ! 0 with (c )n < j(c )n j for n greater than some N 2 ZZ . ' ( ) is de ned almost everywhere on IR, but can contain jump discontinuities. Let '+ equal ' everywhere ' is de ned and equal to the right-hand limit of ' elsewhere. Let '? equal ' everywhere ' is de ned and equal to the left-hand limit of ' elsewhere. Let ' be de ned as the linear combination
' ( ) = '+ ( ) + (1 ? )'? ( ): The following is a summary of results for c = 0 = c . Remark 2.7.
i) The equation
(2.17)
0 = LT ' ( ) ? ' ( ) + h ( ); 14
holds for all 2 IR. ii) The function ' is monotone. iii) The boundary conditions ' (?1) = 0, and ' (+1) = 1, which implies that ' satis es the boundary conditions of (1.3). iv) Let 2 [0; 1]. If is rationally independent, then the function ' is strictly increasing on IR. If is rationally dependent, then ' ( ) < ' ( + ) for all 2 IR, where is such that D = fm jm 2 ZZ g. In any case, if we de ne
(2.18)
' 0 ( ) = ' ( ? 0 );
then for any 2 [0; 1] and 0 2 IR, the function ' 0 () restricted to the set D is strictly increasing. v) Let ; 0 2 IR. Then a bounded function ' : D ! IR satis es equation (2.17) for all 2 D if and only if '( ) = ' 0 ( ) for all 2 D: vi) Assume that is rationally dependent, with as in D = fj jj 2 ZZ g: Then for each j 2 ZZ , the function ' is constant on the interval (j; (j + 1) ). vii) Let be rationally dependent with as in D = fj jj 2 ZZ g, let 2 [0; 1], and let %(0) be de ned as in Theorem 2.2. Then
8 1 < 2 + %(0); 2 (0; ); ' ( ) = : 1 ? %(0); 2 (?; 0): 2
In addition,
8 1 < 2 + %(0); = 1; ' (0) = : 1 2 ? %(0); = 0;
viii) Let be rationally dependent. From the de nition of ' and by Remark 2.7(vi), ' ( ) for 62 D consists of a discrete set of values in [0; 1]. These are the values of ' on the constant intervals. Remark 2.7(vii) gives us two consecutive values from this set. Using 0 = LT ' ( ) ? ' ( ) + h ( ), we
can nd all the values of this discrete set. 3. Numerical Results. In this section we present our numerical results for equation (1.3) with n = 3. We start by de ning the numerical method used. After, we present traveling wave solutions across a wide range of the parameters. We also show how the various solution curves are related, i.e. how we can continue from the solution to one set of parameters to the solution with another set of parameters. The solution behavior we present ranges from smooth solutions to solutions with kinks and steps. We nish this section with a survey of the relationship between the numerical and analytical solutions to (1.3). 3.1. The Numerical Method. We solve (1.3) with nonlinearity (2.2) on the truncated interval 2 [?T; T ] and use asymptotic boundary conditions to match the solution on this truncated problem with the boundary conditions (2.1) [7], [19]. We do not use the analytical solution (2.6), or the a(c) relation (2.8) derived from it, in our numerical computations. In (1.3) with (2.2) the wave speed parameter c appears, but the detuning parameter a does not. Thus we only need to provide the wave speed c, not a or the a(c) relation. Our intent is to develop a robust algorithm that we can use to numerically solve equations with nonlinearities f such that analytical solutions are unknown. Our approach is to use a standard boundary value problem solver and to use continuation techniques to go from one set of parameters to another. The most obvious concerns that appear using this approach are the forward and backward delay terms of LT ' in (1.3). We handle the delay terms by obtaining their values from a previous solution in the continuation sequence. We then perform xed point iteration, updating the 15
0:5
0:55 0:6 0:65 0:70
0:75
0:8
0:85
0:9
9
8
7
6
c
0:95
5
4
3
2
1
0
0
1
2
3
Curve
4
|{
5
c
6
p2 {p3{ p6
7
8
9
? ?
(1; 0; 0) ( 2 ; 3 ; 6 ) (1; 0; 0) (1; 1; 1) (1; 1; 1) ( 12 ; 0; 0) . This contour plot contains level sets of a with respect to the parameters c and c . For each value of a, ("1 ; "2 ; "3 )
Fig. 3.1
shown along the top and right side of the plot, there are three contours.
delay terms along the way. Reformulating our traveling wave equation with this relaxation gives us
?c'0m+1 ( ) ? c '00m+1 ( ) + (1 + 2 (3.1)
=
n X k=1
n X k=1
"k )'m+1 ( )
"k ['m ( + ek ) + 'm ( ? ek )] + h( );
where '0 ( ) is the solution to (1.3) at a previous choice of parameters. We perform the iteration in (3.1) until the residual generated by substituting 'm into (1.3) is suciently small. We now present results concerning convergence of the method de ned by (3.1). Let G('m ( )) = ?c'0m ( ) ? c '00m ( ) + (1 + P P n n 2 k=1 "k )'m ( ), let F ('m ( )) = k=1 "k ['m ( + ek ) + 'm ( ? ek )]. Lemma 3.1. Let ' satisfy G('( )) = F ('( )) + h( ) on IR with boundary conditions '(?1) = 0 and '(1) = 1. Let 1 ( ) = 1 ? b1 e?c1 for 0 and let 2 ( ) = b2 ec2 for 0 with b1 ; c1 ; b2; c2 > 0. Let T? and T+ 2 IR+ . Then b1 ; c1 ; b2 ; c2 may be chosen such that j'( ) ? 1 ( )j b1 e?c1 b1e?c1 T+ for T+ and j'( ) ? 2 ( )j b2 ec2 b2 e?c2T? for ?T?. 16
' 1 Curve Curve Curve Curve Curve
0.9 0.8
1 2 3 4 5
0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 -8
-6
-4
-2
Curve
1 p2 2p2 (1; 0; 0) ( 2 ; 2 ; 0) Fixed c c Parameters 10?5 0.01
0
2
p3 p33 p3
(3; 3; 3)
1 0 10?5
4
6
4
8
5
p p p p ( 12 ; 12 ; 22 ) ( 22 ; 33 ; 66 ) ("1 ; "2 ; "3 ) c
(1; 1; 1)
0.01
. An '() plot approaching the spatially discrete limit for the reaction-di usion equation where we vary .
Fig. 3.2
Proof. Let H (') = 1c [c'00 + c'0 + LT ' ? ' + h()]: Then for > 0, H ( 1 ( )) = H (1 ? b1 e?c1 ) = 1 [?cc21 + 2 c
n X
k=1
"k (1 ? cosh[c1 (ek )]) + c c1 + 1]b1 e?c1 :
We want 1 to be a subsolution for T+ , so we set H ( 1 ( )) 0 for T+ and obtain the upper bound on c1 de ned by 2
n X k=1
"k + 1 c c21 + ?c c1 + 2
n X k=1
"k cosh[c1 (ek )]:
Letting p1 = maxc1 2IR [H (1 ? b1e?c1 ) 0] for T+ we see that c1 2 (0; p1 ] for 1 to be a subsolution. Since we want 1 to be a subsolution we also need 1 (T+ ) '(T+ ) and 1 (1) '(1): Since both 1 (1) = 1 and '(1) = 1 the second inequality is satis ed. The inequality 1 (T+ ) '(T+ ) is satis ed when b1 (1 ? '(T+ ))ec1 T+ once c1 is chosen. Thus for b1 and c1 satisfying the above conditions, 1 is a subsolution for T+ . Now let w1 ( ) = 1 for T+ . Then H (w1 ) 0, w1 (T+ ) '(T+ ), and w1 (1) '(1). Thus w1 is a supersolution for T+ . Hence for T+ , j'( ) ? 1 ( )j jw1 ( ) ? 1 ( )j = b1e?c1 b1 e?c1 T+ : 17
' 1 Curve Curve Curve Curve Curve
0.9 0.8
1 2 3 4 5
0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 -8
-6
Curve
-4
-2
1 p2 2p2 (1; 0; 0) ( 2 ; 2 ; 0) Fixed c c Parameters 10?5 0.1
0
p3 p33 p3
2
(3; 3; 3)
1 0 10?5
4
6
4
8
5
p p p p ( 12 ; 12 ; 22 ) ( 22 ; 33 ; 66 ) ("1 ; "2 ; "3 ) c
(1; 1; 1)
0.1
. An '() plot for the reaction-di usion equation where we vary .
Fig. 3.3
The proof that for ?T?,
j'( ) ? 2 ( )j b2 ec2 b2 e?c2 T? ;
is similar.
Remark 3.1.
When "1 = "2 = ::: = "n = 0, there exist b1 ; c1 ; b2 ; c2 such that for T+ , j'( ) ? 1 ( )j = 0, and for ?T?, j'( ) ? 2 ( )j = 0. The Lemma 3.1 allows us to consider convergence of (3.1) on in nite intervals with the use of asymptotic boundary conditions. Theorem 3.1. Let c > 0 and let ' satisfy (3.2) G('( )) = F ('( )) + h( ) on IR with boundary conditions '(?1) = 0 and '(1) = 1. De ne the sequence of functions f'm ( )g1 m=0 by the iteration (3.3) G('m+1 ( )) = F ('m ( )) + h( ) in (3.1). De ne k k = sup2IR j j, em ( ) = 'm ( ) ? '( ), and let '0 be such that e0 is continuously di erentiable, '0 (?1) = 0, and '0 (1) = 1. Then limm!1 kem k = 0. 18
' 1 Curve Curve Curve Curve
0.9
1 2 3 4
0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 -8
-6
-4
Curve
-2
1
("1 ; "2 ; "3 ) ( 12 ; 21 ; 0) Fixed c c Parameters 10?5 0
0
2
2
( 21 ; 32 ; 0)
4
3 4 (1; 2; 0) (2; 2; 0)
6
8
c
p p 1 0 10?5 ( 22 ; 22 ; 0) 0
. An '() plot approaching the spatially discrete limit for the reaction-di usion equation where we vary "k .
Fig. 3.4
Proof. We start o by establishing two inequalities. Let = (2 Pnk=1 "k )=(1 + 2 Pnk=1 "k ). Then for kem k = 6 0, kF (em)k = sup j
n X
2IR k=1
(3.4)
= (1 + 2
"k (em ( + ek ) + em( ? ek ))j 2
n X k=1
n X
"k kemk
k=1 n X
"k )kemk k ? c e0m ? c e00m + (1 + 2
k=1
"k )em k = kG(em )k;
The last inequality in (3.4) is established by choosing 0 2 (?1; +1) such that jem (0 )j = kem k. 0m (0 ) = 0, ?c e00m(0 ) 0 when (1 + 2 Pn "k )em (0 ) > 0, and Then since em(1) = em(?1 ) = 0, e k =1 ?c e00m(0 ) 0 when (1 + 2 Pnk=1 "k )em (0 ) < 0, which implies (1 + 2
n X k=1
"k )kemk j ? c e0m( ) ? c e00m ( ) + (1 + 2
n X k=1
"k )em ( )j:
Applying (3.2) and (3.3) to (3.4) we get the two inequalities (3.5)
kG(em)k m kG(e0 )k and kF (em)k m kF (e0 )k: 19
' 1 Curve Curve Curve Curve Curve
0.9 0.8
1 2 3 4 5
0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 -8
-6
-4
-2
Curve
1
0
2
2
3
4
4
6
8
5
c = c 0 ?0:1 ?0:5 ?0:8 ?1 Fixed c ("1 ; "2 ; "3 ) p p p Parameters 10?5 1 0 10?5 (1; 1; 1) ( 22 ; 33 ; 66 ) . An '() plot approaching the spatially discrete reaction-di usion equation where we vary c.
Fig. 3.5
' 1
Curve Curve Curve Curve
1 2 3 4
0.8
0.6
0.4
0.2
0 -8
-6
-4
-2
Curve
0
1
2
2
3
4
6
8
4
= c2 0 0:25 0:50 1 Fixed c ("1 ; "2 ; "3 ) Parameters 10?5 0.5 (1; 0; 0) (1; 0; 0) Fig. 3.6. An '( ) plot of the damped wave equation approaching the spatially discrete reaction-di usion equation where we vary and c.
20
' 1 Curve Curve Curve Curve
1 2 3 4
0.8
0.6
0.4
0.2
0 -8
-6
-4
-2
Curve
0
1
2
2
3
4
6
8
4
= c2 0 0:25 0:50 1 Fixed ("1 ; "2 ; "3 ) Parameters 0 0.5 (1; 0; 0) (1; 0; 0) Fig. 3.7. An '( ) plot of the damped wave equation approaching the spatially discrete reaction-di usion equation where we vary and c.
Thus limm!1 kG(em ( ))k = 0 and limm!1 kF (em ( ))k = 0. Using the de nition of G we have
Pn e00m ( ) + c e0m ( ) = ? 1 G(em ( )) + (1 + 2 k=1 "k ) em( ): c c c P P n Let C1 = 1=(1 + 2 k=1 "k ) > 0: Since limm!1 kF (em( ))k = 0, limm!1 j nk=1 "k (em ( + ek ) + em ( ? ek ))j = 0, and as m ! 1 the local maximums of em( ) are positive and the local minimums are negative. Let 1 be a value of such that em(1 ) > 0 is a local maximum of em( ). Suppose that there exist an M1 such that for m > M1 , em(1 ) > C1 jG(em (1 ))j. But if em (1 ) is a local maximum, e00m (1 ) + cc e0m (1 ) < 0. This violates equation (3.6) and hence em (1 ) C1 G(em (1 )): Using the same argument for local minimum, we obtain jem ( )j C1 jG(em ( ))j whenever em( ) has a local extrema. This in turn implies that jem ( )j C1 kG(em )k and kemk C1 kG(em)k.
(3.6)
Remark 3.2.
Recall that when c > 0, the solution ' is continuously di erentiable on IR. This implies that our initial guess '0 should also be at least continuously di erentiable on IR. In fact, any continuously di erentiable function '0 such that '0 (?1) = 0 and '0 (1) = 1 will work. The reformulation (3.1) is an ordinary di erential equation, allowing us to use the standard boundary value solver colmod, from the family of collocation boundary value solvers which include colsys and colnew [2], [3], [4], [11], [12]. Our choice of a particular solver is based on several criteria. We want solutions to our system for a wide range of the parameters ; ; ; ("1 ; "2; "3 ) and wave speed c. We also want a code in which we can perform continuation. Lastly, we want to solve this problem for a family of bistable nonlinearities, which include the cubic f (') = '(' ? a)(' ? 1) for example. We have found that while the continuation feature of colmod is quite robust in our application, xed point iteration is needed if we wish to continue from any solution to any other solution. A complete analysis of the homotopy and convergence properties of our method will appear in future works. The absolute error tolerances for the 21
wave solutions is approximately 10?8 except for the curves where c is close to zero. For our numerical method, this is a perturbation limit thus we used the absolute error tolerance of approximately 10?6 to limit how small a c ( 10?5) we could use. The a(c) curves that appear in the previous section are computed using equation (2.10) and the quadrature algorithm adapt [52]. The algorithm adapt uses three point Gaussian quadrature to estimate the integrals and the seven point Kronrod rule to estimate errors. Absolute and relative errors at each quadrature step are set to 10?12 and the estimated error of the tail is set to 10?16. The quadrature steps are of length 10. Since we calculate the a(c) relation using (2.10), the a(c) plots are generated from the analytically derived traveling wave solution (2.6). Our numerical experiments were performed on a Silicon Graphics Indigo2. 3.2. Numerical Solution Results. We now present numerically obtained traveling wave solutions, for a wide range of the parameters, in the form of solution plots which show the range and richness of solutions to (1.3). Recall, for a particular choice of c 0, c, a 2 (0; 1), and "1 ; "2 ; "3 0, we have shown that there exists a traveling wave solution to (1.3) if c and c are not both zero. Let and ("1 ; "2 ; "3 ) be xed. The c; c parameter map in Figure 3.1 exist for each choice of
and ("1 ; "2; "3 ) with c bounded from above by . Each point in the upper half plane of Figure 3.1, excluding (0; 0), corresponds to a family of solutions to (1.3) parameterized by ; ; and c. Since '(?; ?c ) = 1 ? '(; c ) (Section 2.2) we focus only on the c ; c 0 quadrant (c > 0). Recall that c = ? c2 and c = c . Thus for each xed (c ; c), in rst quadrant of Figure 3.1 the parameters and have the relation 2c = 2 ( ? c), a parabolic curve parameterized by c. In fact all solutions along the parabola for xed ; ("1 ; "2 ; "3 ); c ; c, have the same a value and thus the ; map is a map of level sets of the detuning parameter a. The location along a particular parabola is determined by c. As c ! 1, ( ; ) ! (0; 0) and as c ! 0+, ( ; ) ! (1; 1). The parameter map c ; c also exhibits level sets of a. Figure 3.1 contains three di erent contour plots with respect to a for di erent values of and ("1 ; "2 ; "3 ). Looking at the two contour plots where ("1 ; "2; "3 ) = (1; 1; 1) we see that as the magnitudes of c and c increase, the direction has a decreasing e ect on the value of a. We see the e ect of , and in fact the spatially discrete Laplacian, is greatest when c = 0. Now looking at the two contour plots where = (1; 0; 0) we see the e ect of varying ("1 ; "2; "3 ). For = (1; 0; 0), the problem with ("1 ; "2 ; "3 ) = (1; 1; 1) is equivalent to the one with ("1 ; "2; "3 ) = (1; 0; 0). We now discuss how the parameters "k and k e ect the c versus a relation and how they e ect the solution wave forms. p p Figure 3.4 is a set of solutions plots in which we vary "k . Note that = ( 2=2; 2=2; 0) for all three pplots. pFor this choice of the pproblem becomes equivalent to the two dimension problem with = ( 2=2; 2=2). With 1 =P2 = 2=2,Pchanges in the solution curves and thePa(c) curves depend directly on the magnitude of 3k=1 "k . If 3k=1 "k for one curve is greater than 3k=1 "k for a second curve, with all other parameters xed, then the rst curve will lie inside (closer to a for all 6= 0 or closer to 1/2 for all c 6= 0) the second curve. Figures 3.2 and 3.3 provide a detailed example of lattice induced anisotropy for (1.3). These plots allow us to see what happens as c ! 0; c ! 0 for solutions of di erent direction. Figures 3.2 and 3.3 are sets of solution curves '( ) where we vary . In Figures 3.2 and 3.3, Curves 1, 2, and 3 have 's that are rationally dependent. Recall from Remark 2.7(vi) that for rationally dependent and c = 0, as c ! 0, ' 0 becomes constant on the intervals (j; (j + 1) ) for all j 2 ZZ , with = GCD(1 ; 2 ; 3 ) as in D = fj jj 2 ZZ g and ' = ' 0 on D. In Figures 3.2 and 3.3, Curves 4 and 5 have s that are as close to rationally independent as computer numbers allow. Recall from Remark 2.7(iv) that for rationally independent and c = 0, as c ! 0, ' 0 is strictly increasing. Figures 2.6 and 2.7 are sets of a(c) curves where we vary . In these plots we see how the interval of propagation failure depends on the direction of the wave, . We now present and P3 discuss solutions to (1.3) for = 0 which implies c = . Depending on the magnitudes P of and k=1 "k , (1.3) represents the Ptraveling wave equation for the spatially discrete ( = 0, 3k=1 "k 6= 0) to the spatially continuous ( 3k=1 "k = 0, 6= 0) reaction-di usion equations. 22
Since we are using a code to solve a second order ordinary di erential equation with as the coecient on the second order term, we treat as a perturbation parameter (taking as small as possible) when we want to look at the solution behavior for the spatially discrete reaction-di usion equation. Looking at a(c) curves for = 0, Figures 2.1 and 2.2, we see that for c = c > 0 as ! 0+ the a(c) curve goes to a(c) = 1=2 and for c = c = 0 as ! 0+ the a(c) curve goes to a(c) = 1=2 %(0). We now consider solution plots where we approach the spatially discrete limit. In Figure 3.5 we vary c and take as small as possible ( 10?5). The feature we wish to illustrate with this plot is the \kink" which forms in the solution curves at '(0). Recall from (2.3) that c = = 0 implies for c 6= 0 that lim!0+ '0 ( ) 6= lim!0? '0 ( ), hence the \kink". Notice that in both these plots the larger the magnitude of c , the farther a is from 1=2. The rst of the ("1 ; "2 ; "3 ) plots, Figure 3.4, and Figures 3.2 and 3.3 are also examples of the solutions to the reaction-di usion equation as we approach the spatially discrete limit. We now present and discuss solutions to (1.3) for ; 6= 0. Figure 3.6 is example of solutions to the damped wave equation for > 0 and ! =c2. Here we vary = c2 . Notice that these are the same solutions as for the spatially discrete reaction-di usion equation since c = ? c2 ! 0. We now present and discuss solutions to (1.3) for c = = 0 and > 0. Figure 3.7 is an example of solutions to the wave equation. The lack of a damping term xes a = 1=2 and disallows the possibility of the \kink" e ect. Comparing the formation of steps for the wave equation, with the formation of steps for the reaction-di usion or damped wave equation, Figures 3.2, 3.3, and 3.6 we see that although all the step solutions are approaching the same solution, c ! 0 and c ! 0, the formation of kinks depend on how we approach these solutions.
4. Conclusions. We present an investigation of traveling wave solutions to the bistable di erentialdi erence equation (1.3) with boundary conditions (2.1) and the piecewise linear nonlinearity (2.2), incorporating the phase condition '(0) = a. We provide two methods of solution, one analytical and one numerical. The analytic solution method consists of solving the problem in Fourier transform space. The Fourier transform technique works because the transformed equation possesses no singularities on the imaginary axis. This is in some sense equivalent to the fact that the characteristic equation associated with (1.3) has no purely imaginary roots. Finding the traveling wave solution of (1.3) involves nding the speed c of the wave as well as the solution curve '. Our choice of nonlinearity (2.2) not only allows us to apply the Fourier transform, it allows us to specify c and solve for ' when the phase condition '(0) = a is used. From the analytical solution we derived the relation between the wave speed c and the detuning parameter a. For c 6= 0; c 0, we prove that a traveling wave solution '( ) exists, is continuous, and is strictly increasing on the real line, results that can be seen in all of the solution plots. The derivative of the solution, '0 ( ), is continuous on the real line if and only if c > 0. When c 6= 0; c = 0 the discontinuous nonlinearity f produces the discontinuity in '0 . In all of the solution, ', plots we see \kinks" form when P n c is small. For k=1 "k 6= 0, there exists a nontrivial interval for a of propagation failure if and only if = 0. In other words, propagation failure exhibited by the spatially discrete model is lost when the spatially continuous di usion term is introduced into the equation. In Figures 2.2, 2.5, and 2.6, where
= 0, we see explicitly the intervals of propagation failure. In the remaining a(c) plots, Figures 2.1, 2.3, 2.4, and 2.7, where 6= 0, we see there is no such interval. Figures 3.6 and 3.7 illustrates what happens to the solution curves as ! 0. The spatially discrete model also exhibits lattice anisotropy, and this phenomenon is not lost with the introduction of the spatially continuous di usion term, even though the spatially continuous model does not exhibit directional dependence. The a(c) plots in Figures 2.6 and 2.7 as well as the solution plots in Figures 3.2 and 3.3 illustrate this analytical result. For the reactiondi usion equation the detuning parameter varies with the wave speed c, for the damped wave equation the wave speed c is restricted to a proper subinterval of IR, and for the undamped wave equation a = 1=2 for all c. If we set the wave speed c = 0, = 0, and pick the direction normal so that its elements are rationally related, then, from the least upper bound and the greatest lower bound of the interval of propagation failure, we can construct a set of discrete ' values that satisfy (1.3) with (2.1). Since c = 0 23
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