TRAVELING WAVE SOLUTIONS IN CELLULAR NEURAL ... - AIMS

DISCRETE AND CONTINUOUS DYNAMICAL SYSTEMS Supplement Volume 2005

Website: http://AIMsciences.org pp. 410–419

TRAVELING WAVE SOLUTIONS IN CELLULAR NEURAL NETWORKS WITH MULTIPLE TIME DELAYS

Cheng-Hsiung Hsu and Suh-Yuh Yang Department of Mathematics, National Central University Chung-Li 32054, Taiwan Abstract. This work investigates the existence of traveling wave solutions of the cellular neural network distributed in Z1 with multiple time delays. Applying the method of step with the help of the characteristic function, we can figure out an analytic solution in an explicit form with many parameters. We then focus on the mechanism for producing the so-called camel-like traveling wave solutions and study the effect of delays on the shape of solutions. Some numerical results are also provided to demonstrate the theoretical analysis.

1. Introduction. The aim of this work is to study the existence of traveling wave solutions of the following cellular neural network (CNN) with multiple time delays: m X dxi (t) = −xi (t) + af (xi (t)) + αj f (xi−j (t − τj )), dt j=1

(1)

for all i ∈ Z1 and t ∈ R, where m ∈ N, 0 ≤ τ1 ≤ τ2 ≤ · · · ≤ τm are time delays; the nonlinearity f is a piecewise-linear output function given by 1 for x ∈ R; (2) f (x) = (|x + 1| − |x − 1|) 2 the coefficients a and αj (j = 1, 2, · · · , m) of f are real constants which measure the synaptic weights of self-feedback and left neighborhood interaction. In recent years, the study of CNNs has been receiving increasing attention in both the engineering and mathematics communities owing to the advantage that they can be applied to a broad scope of applications in, for example, image and video signal processing, robotic and biological versions, and higher brain functions (see [5, 6, 7] for more details). In models of electronic neural networks, time delays are unavoidably encountered due to the finite switching speed of amplifiers. Therefore, it is crucial to take time delays into consideration when studying CNNs. In the present paper we consider the system of delay differential equations (1) in which the dynamics of each given cell xi depends on itself with instantaneous feedback and its nearest m left neighbors xi−j (j = 1, 2, · · · , m) with delayed interaction. In the following, we focus on the existence of the traveling wave solutions of system (1) and investigate the effect of delays on the shape of solutions. A C 1 2000 Mathematics Subject Classification. Primary: 34K10, 34K28; Secondary: 34B15. Key words and phrases. lattice differential equations, cellular neural networks, time delays, method of step, camel-like traveling waves. This work was partially supported by the National Science Council of Taiwan and the Brain Research Center of University System of Taiwan. 410

TRAVELING WAVES IN CNN WITH DELAYS

411

function ϕ : R → R such that xi (t) = ϕ(i − ct) for all i ∈ Z1 and t ∈ R is a solution of system (1) is called a traveling wave solution, where s := i − ct ∈ R is the moving coordinate for a given nonzero wave speed c ∈ R. Substituting this traveling wave ansatz into (1) with α0 := a and τ0 := 0, we obtain the following profile equation m X −cϕ′ (s) = −ϕ(s) + αj f (ϕ(s − j + cτj )), (3) j=0

which is a functional differential equation with retarded terms for c < 0. FurtherPm more, assumption (1) has three homogeneous j=0 αj > 1 ensures that system Pm stationary solutions given by x0 := 0 and x± := ± j=0 αj , and we are interested in finding traveling wave solutions connecting x0 with x+ , that is, lim ϕ(s) = x0

s→−∞

and

lim ϕ(s) = x+ .

s→∞

(4)

The study of traveling wave solutions for lattice differential equations has drawn considerable attention in the past decade (see, e.g., [1]-[4], [8], [9], [11]-[17] and many references contained therein). Most existing results in above-mentioned references focus on the existence of monotonic traveling waves for equations without time delays. However, it is shown in [10] that, in the absence of time delays, system (1) possesses non-monotonic traveling wave solutions. In particular, the existence of the so-called camel-like traveling waves is investigated there. In this work, we will generalize the results in [10] to system (1) with multiple time delays. We will follow the notation given in [10]. A solution ϕ of (3) with the asymptotic boundary condition (4) is called a traveling wave solution of (1). Such a solution is said to be of type n if it has exactly n critical points with strictly local extreme values. Furthermore, if the maximum of an n-type traveling wave ϕ is greater or equal to x+ , then it is said to be of type n+ ; otherwise, it is said to be of type n− . Due to the shape of solution, we also call an n-type solution (n ≥ 1) as a camel-like traveling wave solution. By means of the method of step with the aid of the positive root of the corresponding characteristic function, we obtain the following results: Pm Theorem 1. Assume that m is even, α0 := a ≥ 1, τ0 := 0, j=0 αj > 1 and c < 0. (I) If αj > 0 for 1 ≤ j ≤ m then for each c < 0 there exists a monotonic increasing solution of (3) and (4). (II) Let αj P < 0 for 1 ≤ j ≤ m and αj ≥ αj+1 for 1 ≤ j ≤ m − 1. m (i) If j=0 αj eτj ≥ 0 then there exists cˆ < 0 such that (3) and (4) possess a monotonic increasing solution if c ≤ cˆ and a 1+ -type solution when c > cˆ. Pm (ii) If j=0 αj eτj < 0 then for each c < 0 there exists a 1+ -type solution of (3) and (4). (III) Let the signs of {αj }m j=1 be alternating with |αj | ≥ |αj+1 | for 1 ≤ j ≤ m − 1. (i) If α1 > 0 (resp., 1 − α0 ≤ α1 < 0) then there exists c∗ < 0 such that (3) and (4) have an (m − 1)-type solution (resp., m-type solution), provided c∗ P < c < 0. Pm n−1 (ii) If j=0 αj e(τj −τn ) + j=n αj e−γ(τj −τn ) > 0 for n = 1, 2, · · · , m, then there exists c∗ < 0 with sufficiently large magnitude |c∗ | such that (3) and (4) possess a monotonic increasing solution for c < c∗ , where γ is defined Pn−1 Pm in (20). Otherwise, if j=0 αj e(τj −τn ) + j=n αj e−γ(τj −τn ) < 0 for some 1 ≤ n ≤ m, then there exists c∗ < 0 with sufficiently large magnitude |c∗ | such that (3) and (4) possess a non-monotonic solution for c < c∗ .

412

CHENG-HSIUNG HSU AND SUH-YUH YANG

Similar results can be obtained when m is odd, but we omit the details. Based on a comparison of Theorem 1 and the Main Theorem in [10], we can conclude that in the absence of time delays, by the method of step presented in this paper, system (1) has monotonic increasing traveling wave solutions whenever the magnitude of the wave speed c < 0 is sufficiently large and the monotonic solutions, however, might lose their monotonicity when the time delays are presented and large enough. This is exactly the effect of delays on the shape of solutions. This paper is organized as follows. In Section 2, we derive an analytic solution of (3) and (4) in an explicit form by using the method of step with the help of the characteristic function. Some preliminary results of the solution are introduced in Section 3. We then give the proof of the Theorem 1 in Section 4. Finally, in Section 5, we examine the solution structure for some specific parameters. 2. An analytic solution of (3) and (4) with negative wave speed. In this section, we will construct an analytic solution of (3) and (4) with c < 0 by the method of step. Since the traveling waves are translation invariant, we may normalize the solution at s = 0 by letting ϕ(0) = 1 and assume in advance that ϕ(s) > 1 for s ∈ (0, ∞) and 0 < ϕ(s) < 1 for s ∈ (−∞, 0). Later, one can verify that the solution under consideration satisfies such assumptions (cf. (iv) and (vi) of Lemma 2). Now, if we partition the s-axis into m + 2 subintervals by I−∞ In

= =

(−∞, 0], [n − cτn , n + 1 − cτn+1 ] for n = 0, 1, · · · , m − 1,

I∞

=

[m − cτm , ∞),

then equation (3) can be rewritten as  m X    (α − 1)ϕ(s) + αj ϕ(s − j + cτj ) 0     j=1 m n X X −cϕ′ (s) =  αj ϕ(s − j + cτj ) α + −ϕ(s) +  j    j=n+1 j=0   −ϕ(s) + x+

if s ∈ I−∞ , if s ∈ In ,

(5)

if s ∈ I∞ ,

where n = 0, 1, · · · , m − 1. Since (5) is a functional differential equation with retarded terms, if we know the solution form ϕ(s) on I−∞ then we may solve (5) for s ∈ (0, ∞) piece by piece. To this aim, suppose ϕ(s) = eσs for s ∈ I−∞ , where σ is some positive constant that need to be determined. Then ϕ(s) → x0 as s → −∞ and σ must be a positive root of the following so-called characteristic function of (3) at x0 = 0: m X αj e(−j+cτj )σ . (6) ∆(σ, c; x0 ) = −cσ + 1 − j=0

By inspection, for any fixed c < 0 we always have ∆(0, c; x0 ) < 0 and ∆(σ, c; x0 ) > 0 as σ → ∞. Hence, there actually exists a σ = σ(c) > 0 such that ∆(σ, c; x0 ) = 0. Thus, we may set ϕ(s) = eσ(c)s for s ∈ I−∞ . (7) Now, a solution of (5) for s ∈ (0, ∞) can be constructed step by step. First, it is obvious that 1

ϕ(s) = (ϕ(m − cτm ) − x+ )e c (s−m+cτm ) + x+

for s ∈ I∞ ,

(8)


413

and ϕ(s) → x+ as s → ∞. Proposition 1. Let c < 0 and σ := σ(c) be a positive root of ∆(σ, c; x0 ). Assume that (7) holds. Then we have n n X 1 cσ X (s−j+cτj ) c + αj e αj + Jn (s) ϕ(s) = 1 − cσ j=0 j=0

(9)

Pm for s ∈ In with 0 ≤ n ≤ m − 1, where Jn (s) := ( j=n+1 αj eσ(s−j+cτj ) )/(1 − cσ). Proof. If s ∈ In then ϕ(s − j + cτj ) = eσ(s−j+cτj ) for j ≥ n + 1, and we have −cϕ′ (s) = −ϕ(s) +

n X

αj +

m X

αj eσ(s−j+cτj ) .

(10)

j=n+1

j=0

For (10) with the initial value ϕ(n − cτn ), one can verify that n n 1 X X αj − Jn (n − cτn ) e c (s−n+cτn ) + Jn (s) + ϕ(s) = ϕ(n − cτn ) − αj . j=0

(11)

j=0

If n = 0, then the solution form (11) is just the same with (9). Next, assume that ϕ(s) has the form (9) on Ik , then ϕ(k + 1 − cτk+1 ) =

k 1 cσ X αj e c (k+1−j+cτj −cτk+1 ) 1 − cσ j=0

+

k X

αj + Jk (k + 1 − cτk+1 ).

(12)

j=0

Thus, if s ∈ Ik+1 then combining (11) with (12), we obtain

ϕ(k + 1 − cτk+1 ) −

=

k+1 X j=0

cσ 1 − cσ

k+1 X

1 αj − Jk+1 (k + 1 − cτk+1 ) e c (s−k−1+cτk+1 )

1

αj e c (s−j+cτj ) ,

(13)

j=0

which implies (9). This completes the proof. 3. Some preliminary results. In this section, some fundamental properties of the solution ϕ(s) given by (7), (8), and (9) are further studied. Throughout the remainder of this paper, we always assume that m is even. Lemma 1. For n = 1, 2, · · · , m, if ϕ(n − cτn ) < (resp., =, >) x+ then ϕ′ (n − cτn ) > (resp., =, ) x+ then n−1 m X 1 cσ X αj e c (n−cτn −j+cτj ) < (resp., =, >) αj − Jn−1 (n − cτn ), 1 − cσ j=0 j=n

(14)

414


which implies that ′

ϕ (n − cτn ) = > =

n−1 1 σ X αj e c (n−cτn −j+cτj ) + σJn−1 (n − cτn ) 1 − cσ j=0

(resp., =, 0 be a solution of ∆(σ, c; x0 ) = 0. Then (i) limc→−∞ σ(c) = 0. (ii) If α1 > 0 (or 1 − α0 ≤ α1 < 0) and α0 ≥ 1 then limc→0− σ(c) = ∞. (iii) If α1 > 0 then ϕ′ (1 − cτ1 ) > 0 and ϕ(s) is monotonic increasing on I0 . (iv) If α1 > 0 and ϕ(s) has a critical point at sn ∈ In , 1 ≤ n ≤ m − 1, then ϕ(s) is convex (resp., concave) at sn provided n is even (resp., odd). Moreover, if ϕ(s) is convex at sn then ϕ(sn ) > 1. (v) If α1 > 0, ϕ(m − cτm ) > x+ and ϕ(s) is non-monotonic, then the number of critical points for ϕ(s) is odd and at most m − 1. (vi) If α1 < 0 and ϕ(s) has a critical point at sn ∈ In , 0 ≤ n ≤ m − 1, then ϕ(s) is convex (resp., concave) at sn provided n is odd (resp., even). Moreover, if ϕ(s) is convex at sn with 1 − α0 ≤ α1 < 0 then ϕ(sn ) > 1. (vii) If α1 < 0, ϕ(m − cτm ) < x+ and ϕ(s) is non-monotonic, then the number of critical points for ϕ(s) is even and at most m. Pm Proof. First, define g(σ) := −1 + α0 + ℓe−σ with ℓ := j=1 αj e(1−j+cτj )σ . Then g(0) = x+ − 1 > 0. Since −cσ(c) = g(σ(c)), we have (i) directly. Now, one can check that limσ→∞ g(σ) = α0 − 1 ≥ 0 and m X αj e(−j+cτj )σ > 0 for σ ∈ [0, ∞). g(σ) = −1 + α0 + j=1

Thus, assertion (ii) follows immediately. By solution form (9), we obtain 1 σ (15) α0 e c (1−cτ1 ) + ℓe−cστ1 . 1 − cσ Since α1 > 0 and |αj | ≥ |αj+1 | for 1 ≤ j ≤ m − 1 with alternating signs, m α2j−1 e(−2j+2+cτ2j−1 )σ + α2j e(−2j+1+cτ2j )σ > 0 for j = 1, 2, · · · , . (16) 2 Hence, ℓ > 0 which combining with (15) implies ϕ′ (1 − cτ1 ) > 0. Next, assume ϕ(s) is non-monotonic with ϕ′ (s0 ) = 0 for some s0 ∈ int I0 . Then we have α0 σ 1 s0 (17) e c < 0, ϕ′′ (s0 ) = c i.e., ϕ(s) is concave at s0 . But this contradicts the facts that ϕ′ (0) = σ > 0 and ϕ′ (1 − cτ1 ) > 0. Hence ϕ′ (s) > 0 for s ∈ I0 , which completes the proof of (iii). ϕ′ (1 − cτ1 ) =


415

For proving (iv) and (vi), we apply solution form (9) again, n 1 σ X αj e c (sn −j+cτj ) + σJn (sn ) = 0, ϕ′ (sn ) = 1 − cσ j=0

(18)

n m X 1 σ σ X αj e c (sn −j+cτj ) + σ 2 Jn (sn ) − αj eσ(sn −j+cτj ) . c(1 − cσ) j=0 c j=n+1 Pm Since m is even, if α1 > 0 then j=n+1 αj eσ(sn −j+cτj ) > 0 (resp., < 0) provided n is Pm even (resp., odd). Similarly, if α1 < 0 then j=n+1 αj eσ(sn −j+cτj ) < 0 (resp., > 0) provided n is even (resp., odd). Hence, we have the convexity results of ϕ(sn ). Assume ϕ is convex at sn . Then (18) implies n m X X αj , αj eσ(sn −j+cτj ) + ϕ(sn ) =

ϕ′′ (sn ) =

j=0

j=n+1

and one can check that ϕ(sn ) > 1 provided α1 > 0 or 1 − α0 ≤ α1 < 0. This completes the proof of (iv) and (vi). Finally, we will prove part (v), and the proof of part (vii) can be achieved in a similar way. Since ϕ(m − cτm ) > x+ , we have ϕ′ (m − cτm ) < 0. If α1 > 0, by (iii) and (iv), the number of critical points of ϕ(s) is at most m − 1. Assume the number of critical points is even and denoted by {s1 , · · · , sn } with 0 < s1 < · · · < sn < m − cτm . Since ϕ′ (1 − cτ1 ) > 0, we have ϕ′′ (s1 ) < 0 and ϕ′′ (sn ) > 0. Hence ϕ′ (m − cτm ) > 0 and this is a contradiction. 4. Proof of Theorem 1. We are now in the position to prove Theorem 1. (I). αj > 0 for 1 ≤ j ≤ m: For s ∈ In , 0 ≤ n ≤ m − 1, we have ϕ′ (s) =

n m X 1 σ σ X αj e c (s−j+cτj ) + αj eσ(s−j+cτj ) > 0. 1 − cσ j=0 1 − cσ j=n+1

(19)

Hence ϕ(s) is monotonic increasing on (−∞, m − cτm ]. Furthermore, if ϕ(m − cτm ) ≥ x+ then (8) implies ϕ′ (m − cτm ) ≤ 0 which contradicts (19). Therefore, ϕ(m − cτm ) < x+ which combining with Proposition 1 implies ϕ′ (s) > 0 on I∞ and ϕ(s) → x+ as s → ∞. This completes the proof of part (I) of Theorem 1. See some numerical results for m = 4 given in Figure 1 and Figure 2.

3.5

3

φ(s)

2.5

4

a = 0.5 α1 = 0.5 α2 = 0.5 α3 = 0.5 α4 = 0.5 x+ = 2.5 c = −1.5 σ = 0.173136 τ1 = 1.0 τ2 = 2.0 τ3 = 3.0 τ4 = 4.0

3

2.5

2

2

1.5

1.5

1

1

0.5

0.5

0

0

a = 0.5 α1 = 0.5 α2 = 0.5 α3 = 0.5 α4 = 0.5 x+ = 2.5 c = −0.5 σ = 0.333239 τ1 = 1.0 τ2 = 2.0 τ3 = 3.0 τ4 = 4.0

3.5

φ(s)

4

2

4

6

8

10

s − axis

Figure 1: A 0-type traveling wave

0 −1

0

1

2

3 s − axis

4

5

6


7

416


(II). αj < 0 and αj ≥ αj+1 : (II-i). We first claim that there exists cˆ < 0 such that ϕ(m − cτm ) = x+ . Let g(σ) be that defined in the proof of Lemma 2. If αj < 0 for j = 1, 2, · · · , m, then g(σ) is increasing in σ which implies σ(c) is increasing with respect to c, lim σ(c) = ∞, and c→0−

lim −cσ(c) = γ

c→−∞

for some γ > 0 satisfying γ = −1 +

m X

αj e−γτj .

(20)

j=0

Hence, we have lim− ϕ(m − cτm ) − x+

c→0

lim ϕ(m − cτm ) − x+

c→−∞

αm − αm > 0, α0 m γ X = − αj eτj −τm < 0. 1 + γ j=0

=

Therefore, the claim follows from the intermediate value theorem. Next, we show that such cˆ is unique by proving ϕ(m−cτm ) is increasing with respect to c. Suppose false. Let c∗ > cˆ be such that ϕ(m−cτm ; c∗ ) ≤ ϕ(m−cτm ; cˆ). Since σ(ˆ c) = ϕ′ (0; cˆ) < ′ ∗ ∗ ∗ ∗ ϕ (0, c ) = σ(c ), there must exist s ∈ (0, m − cτm ) such that ϕ(s ; cˆ) = ϕ(s∗ , c∗ ) and ϕ′ (s∗ ; cˆ) > ϕ′ (s∗ , c∗ ). However, by (5), one can prove that ϕ′ (s∗ ; cˆ) < ϕ′ (s∗ , c∗ ). This is a contradiction, and therefore cˆ is unique. Hence, we have ϕ(m − cτm ) ≤ x+ if c ≤ cˆ and ϕ(m − cτm ) > x+ for c > cˆ. Now, suppose ϕ(s) has a critical point sn ∈ int In with n ≤ m − 1 then ϕ′′ (sn ) = −

m σ X αj eσ(sn −j+cτj ) < 0. c j=n+1

(21)

Hence, if c ≤ cˆ then ϕ′ (0) = σ > 0 and ϕ′ (m − cτm ) > 0, i.e., ϕ(s) is monotonic increasing on [0, m − cτm ]. On the other hand, if c > cˆ then ϕ′ (m − cτm ) < 0 and ϕ′′ (sn ) < 0. This implies that ϕ(s) is a non-monotonic 1+ -type traveling wave. (II-ii). By the similar arguments as that in the proof of (i), we can derive that ϕ(m − cτm ) > x+ for all c < 0. Thus (21) shows that the solution is of type 1+ . Some numerical results for m = 4 are illustrated in Figure 3 and Figure 4. 4

3

φ(s)

2.5

3.5

3

2.5

φ(s)

3.5

4 a = 3.8 α1 = −0.25 α2 = −0.3 α3 = −0.35 α4 = −0.4 x+ = 2.5 c = −3.0 σ = 0.871153 τ1 = 0.01 τ2 = 0.02 τ3 = 0.03 τ4 = 0.04

2

2

1.5

1.5

1

1

0.5

0.5

0 −1

0

1

2 s − axis

3

4


5

a = 3.8 α1 = −0.25 α2 = −0.3 α3 = −0.35 α4 = −0.4 x+ = 2.5 c = −0.5 σ = 5.598191 τ1 = 0.01 τ2 = 0.02 τ3 = 0.03 τ4 = 0.04

0 −1

0

1

2 s − axis

+

3

4

Figure 4: A 1 -type traveling wave

5


417

(III). |αj | ≥ |αj+1 | for 1 ≤ j ≤ m − 1 with alternating signs: (III-i). Define the real-valued function G in variable c by G(c) =

n−1 X

1

αj e c (n−j−cτn +cτj ) +

m X

αj eσ(n−j−cτn +cτj ) .

(22)

j=n

j=0

According to Proposition 1, for n = 1, 2, · · · , m, we have σ ϕ′ (n − cτn ) = G(c). 1 − cσ

(23)

Then (ii) of Lemma 2 implies that lim− cσ = 1 − α0

c→0

and

lim− G(c) = αn +

c→0

m X

αj e(1−α0 )(τj −τn ) .

j=n+1

Thus, lim G(c) < 0 (> 0) if n is even and α1 > 0 (1 − α0 ≤ α1 < 0),

(24)

lim G(c) > 0 (< 0) if n is odd and α1 > 0 (1 − α0 ≤ α1 < 0).

(25)

c→0− c→0−

Therefore, if α1 > 0 then there exists {ˆ cn }m ˆn < 0 such that if c > cˆn then n=1 with c ′ ϕ (n − cτn ) > 0 (resp., < 0), provided n is odd (resp., even). Let c∗ := max{ˆ cn }m n=1 . ∗ Since ϕ(s) is increasing on I0 , if c > c then ϕ(s) is a traveling wave solution of type (m − 1). On the other hand, if 1 − α0 ≤ α1 < 0 and c > c∗ then ϕ′ (n − cτn ) > 0 (resp., < 0), provided n is even (resp., odd). By the same arguments, ϕ(s) is an m-type traveling wave. (III-ii). According to part (i) of Lemma 2, we know that lim G(c) =

c→−∞

n−1 X

αj e(τj −τn ) +

m X

αj e−γ(τj −τn ) > 0.

j=n

j=0

Hence, for sufficiently large |c|, we have ϕ′ (n − cτn ) > 0 for n = 0, 1, · · · , m. By (iv) and (vi) of Lemma 2, there exists c∗ < 0 such that if c < c∗ then (3) and (4) possess a monotonic increasing solution. This completes the proof. Some numerical results for m = 4 are illustrated in Figure 5 and Figure 6. 4

3

φ(s)

2.5

3.5

3

2.5

φ(s)

3.5

4 a = 2.3 α1 = 0.5 α2 = −0.4 α3 = 0.3 α4 = −0.2 + x = 2.5 c = −0.15 σ = 8.666823 τ1 = 1.0 τ2 = 2.0 τ3 = 3.0 τ4 = 4.0

2

2

1.5

1.5

1

1

0.5

0.5

0 −1

0

1

2 s − axis

3

4

5

Figure 5: A 3+ -type traveling wave

a = 2.7 α1 = −0.5 α2 = 0.4 α3 = −0.3 α4 = 0.2 + x = 2.5 c = −0.15 σ = 11.333326 τ1 = 1.0 τ2 = 2.0 τ3 = 3.0 τ4 = 4.0

0 −1

0

1

2 s − axis

3

4

5


418


5. |αj | = |αj+1 | for 1 ≤ j ≤ m − 1 with alternating signs. In this section, two specific examples of part (III) of Theorem 1 will be investigated in detail. Proposition 2. Assume that |αj | = |αj+1 | for 1 ≤ j ≤ m − 1. (I) Let α0 ≥ 1 and αj = (−1)j+1 α with α > 0 for 1 ≤ j ≤ m. Then (i) If ϕ(s) has a critical point sn ∈ In , 0 ≤ n ≤ m − 1, then ϕ(sn ) > x+ . (ii) For 1 ≤ n ≤ m, if ϕ(n − cτn ) ≤ x+ then ϕ′ (n − cτn ) > 0. (iii) For 1 ≤ n ≤ m − 1, if n is even and ϕ(n − cτn ) < x+ then ϕ′ (n + 1 − cτn+1 ) > 0. (iv) For 1 ≤ n ≤ m − 2, we have ϕ′ (n − cτn ) > ϕ′ (n + 2 − cτn+2 ). (v) If ϕ(m − cτm ) ≤ x+ then ϕ(s) is monotonic increasing. (vi) If there exists s˜ ∈ (0, m − cτm ) such that ϕ(˜ s) = x+ then ϕ(s) > x+ for s > s˜. (II) Let α0 ≥ 1 and αj = (−1)j α with α > 0 for 1 ≤ j ≤ m. Then (i) If ϕ(s) has a critical point sn ∈ In , 0 ≤ n ≤ m − 1, then ϕ(sn ) < x+ . (ii) For 1 ≤ n ≤ m−1, if n is odd and ϕ(n−cτn ) < x+ then ϕ′ (n+1−cτn+1 ) > 0. (iii) ϕ(s) < x+ for all c < 0 and s ∈ (−∞, ∞). (iv) If ϕ(s) is non-monotonic then ϕ(s) has even and at most m critical points. Proof. For simplicity, we only prove the results of part (I). (i). Since m is even, by Proposition 1, it is easy to check that m X ϕ(sn ) − x+ = α (−1)j+1 (eσ(sn −j+cτj ) − 1) > 0. j=n+1

+

(ii). If ϕ(n − cτn ) ≤ x then by Lemma 1 we have m αX (−1)j+1 (1 − eσ(n−j+cτj ) ) > 0. ϕ′ (n − cτn ) > c j=n (iii). Since n is even and ϕ(n − cτn ) < x+ then Jn (n − cτn ) > 0 and σJn (n + 1 − cτn+1 ) > 0. By Proposition 1, we have 1

ec (ϕ(n − cτn ) − x+ − Jn (n − cτn )) + σJn (n + 1 − cτn+1 ) > 0. ϕ (n + 1 − cτn+1 ) = c (iv). By Proposition 1 again, we have ′

ϕ′ (n + 2 − cτn+2 ) − ϕ′ (n − cτn ) n+2 n+1 n n σ α0 (e c −τn+2 − e c −τn ) + α(e c −τn+1 − e c −τn = 1 − cσ +eσ(n−m−cτn +cτm ) − eσ(n+1−m−cτn+1 +cτm ) ) < 0. (v). Suppose ϕ(m − cτm ) ≤ x+ , by Proposition 1 we have ϕ′ (m − cτm ) > 0. Since ϕ′ (0) = σ > 0, if ϕ(s) is not monotonic on [0, m − cτm ] then there exists s¯ such that ϕ′ (¯ s) = 0 and ϕ(¯ s) < x+ , but this contradicts the result of part (i). (vi). By the result of part (v), we know that ϕ(m − cτm ) > x+ . If there exists s˜1 ∈ (˜ s, m − cτm ) such that ϕ(˜ s1 ) < x+ then we have ϕ′ (˜ s2 ) = 0 and ϕ(˜ s2 ) < x+ for some s˜2 ∈ (˜ s, m − cτm ), but this contradicts the result of part (i). Some numerical results for m = 4 are depicted in Figure 7 and Figure 8.

TRAVELING WAVES IN CNN WITH DELAYS 4

4

3

φ(s)

2.5

a = 2.5 α1 = 0.5 α2 = −0.5 α3 = 0.5 α4 = −0.5 x+ = 2.5 c = −0.2 σ = 7.500308 τ1 = 1.0 τ2 = 2.0 τ3 = 3.0 τ4 = 4.0

3

2.5

2

2

1.5

1.5

1

1

0.5

0.5

0 −1

0

a = 2.5 α1 = −0.5 α2 = 0.5 α3 = −0.5 α4 = 0.5 x+ = 2.5 c = −0.3 σ = 4.997490 τ1 = 1.0 τ2 = 2.0 τ3 = 3.0 τ4 = 4.0

3.5

φ(s)

3.5

419

1

2 s − axis

3

4

5

0 −1

0

1

2

3

4

5

6

s − axis


Figure 8: A 4− -type traveling wave

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Received August, 2004; revised March, 2005. E-mail address: [email protected] E-mail address: [email protected]