Traveling wave solutions for some discrete quasilinear parabolic equations

Nonlinear Analysis 48 (2002) 1137 – 1149

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Traveling wave solutions for some discrete quasilinear parabolic equations Sheng-Chen Fu, Jong-Shenq Guo ∗ , Shang-Yau Shieh Department of Mathematics, National Taiwan Normal University, 88, Sec. 4, Ting-Chou Road, Taipei 117, Taiwan Received 7 September 1999; accepted 9 July 2000

Keywords: Traveling wave solution; Lattice ordinary di2erential equations; Quasilinear parabolic equation; Fisher-type source term

1. Introduction In this paper, we study the existence of traveling wave solutions of the following lattice ordinary di2erential equations: u˙ n = d[g(un−1 ) − 2g(un ) + g(un+1 )] + f(un );

(1.1)

where n ∈ Z; d ¿ 0. Note that (1.2) is the discrete analogue of the following quasilinear parabolic equation: ut = d(g(u))xx + f(u):

(1.2)

In particular, we shall focus our attention on the case when g(u) = um ; m ≥ 1;

f(u) = u(1 − u);

(1.3)

which is the so-called standard heat equation (for m = 1) or porous medium equation (for m ¿ 1) with Fisher-type source term f. The case for m = 1 was
0362-546X/02/$ - see front matter ? 2002 Elsevier Science Ltd. All rights reserved. PII: S 0 3 6 2 - 5 4 6 X ( 0 0 ) 0 0 2 4 2 - X

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A solution {un }n∈Z of (1.1) is called a traveling wave solution with speed c ¿ 0 if there exists a di2erentiable function : R → [0; 1] such that (−∞) = 0; (∞) = 1, and un () = (n + c);

∀n ∈ Z ; ∀ ∈ R:

In [16], they studied the existence of traveling wave solutions of an in
∀x ∈ [0; 1]:

(1.4)

However, condition (1.4) does not hold in our case. Note that in our case g(x) is nondecreasing for x ≥ 0 and the lower solution we constructed is nonnegative. In [19], the authors used the continuation and comparison methods to obtain the existence and nonexistence of traveling wave solutions of the discrete Fisher equation, i.e., the case m=1. Their result is quite sharp in the sense that given any c ¿ 0 there is a positive constant d∗ such that a traveling wave solution exists if and only if d ≤ d∗ . It is well known that the monotone iteration method is a very powerful tool in solving di2erential equations. It is simple, although the main diGculty is to
rc − 1 : 4 sinh2 (mr=2)

This is the subject of Section 3. We remark that indeed the above result holds for any m ¿ 1. But, for the case m ≥ 2, the result obtained by monotone iteration method is better than the method of Zinner et al. In fact, we have dK m ¡ 1=[4 sinh2 (m=(2c))] for a given c ¿ 0. Another well-known and related equation is the Nagumo equation [15] where the source term f is replaced by f(u) = u(u − a)(1 − u) for some a ∈ (0; 1). For results on partial di2erential equations case, we refer the readers to, e.g., [7,10], and the references cited in these papers. For lattice ordinary di2erential equations and its discrete analogue, we refer the readers to [3–5,9,11–14,17,18], especially the nice survey paper [3]. This paper is organized as follows. In Section 2, we study the case for m ≥ 2 using the monotone iteration method. The case for 1 ¡ m ¡ 2 is treated in Section 3 by the method of [19]. Finally, the case m = 1 is considered in Section 4. Using the monotone iteration method, we are able to recover the known results in [19] for the existence of

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traveling wave solutions of (1.1) for the case m = 1 which were proved by a di2erent method. 2. The case m ≥ 2 In this section, we study the following lattice ordinary di2erential equations: u˙ n = d[g(un−1 ) − 2g(un ) + g(un+1 )] + f(un ); n ∈ Z;

(2.1)

m

where d ¿ 0; g(u)=u ; m ¿ 1, and f(u)=u(1−u). Our goal is to prove the existence of nondecreasing traveling wave solutions. We
(−∞) = 0; (∞) = 1, and un () = (n + c); ∀n ∈ Z ; ∀ ∈ R. It is easy to see that (2.1) has a nondecreasing traveling wave solution with speed c if and only if the equation ˙ = d[g( (t − 1)) − 2g( (t)) + g( (t + 1))] + f( (t)) c (t) (2.2) has a nondecreasing solution de
H ( )(t) =  (t) +

(2.3)

and the set O = { | (−∞) = 0; (∞) = 1; and is nondecreasing}: The proof of the following proposition is similar to the proof of [16, Proposition 3:1] and we omit it. Proposition 2.1. Suppose that ∈ O and is a function from R to R such that

≤ ≤ 1. Then H ( ) is nondecreasing and H ( )(t) ≤ H ( )(t); ∀t ∈ R. Denition 2.2. A continuous function : R → [0; 1] is called an upper solution of (2.2) if it is di2erentiable almost everywhere and satis
(s) (2.5)

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Multiplying (2.5) by es and integrating it from −∞ to t, we obtain that
t

(t) = e−t es H ( )(s) ds; t ∈ R: −∞

(2.6)

Conversely, any solution of the integral equation (2.6) is also a solution of (2.2). For + , we de
1 (t) = e−t es H ( + )(s) ds; t ∈ R: −∞

Then we have Proposition 2.3. The inequalities ˙ 1 (t) ≥ 0 and − (t) ≤ 1 (t) ≤ + (t) hold; ∀t ∈ R. Proof. Since




˙ 1 (t) = −e−t = e−t

t

−∞




t

−∞

es H ( + )(s) ds + H ( + )(t)

es [H ( + )(t) − H ( + )(s)] ds;

it follows that ˙ 1 (t) ≥ 0. Since + is an upper solution of (2.2), it follows from (2.4) that
t

+ (t) ≥ e−t es H ( + )(s) ds: −∞

Hence by the de
− (t) ≤ e−t es H ( − )(s) ds ≤ e−t es H ( + )(s) ds = 1 (t): −∞

−∞

The proof it is now complete. Now, for any k ∈ N , we de
k+1 (t) = e−t es H ( k )(s) ds; t ∈ R: −∞

(2.7)

Then by a similar argument as above we obtain that ˙ k+1 (t) ≥ 0 and

− (t) ≤ k+1 (t) ≤ k (t) ≤ + (t);

∀t ∈ R

(2.8)

for all k ∈ N . It follows from (2.8) that the limit

(t) = lim k (t) k→∞

(2.9)

exists for all t ∈ R such that − ≤ ≤ + and is nondecreasing. Hence (−∞) and (∞) exist. Moreover, (−∞) = 0. Indeed, the function de
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a nondecreasing solution of (2.2) with (−∞) = 0 and (∞) = 1. To see this, we apply the Lebesgue dominated convergence theorem to (2.7) and obtain that satis
(∞) = 0 or (∞) = 1. But, we assume that − ≡ 0. We can
− (t) ¿ 0. Thus, by the monotonicity of , (∞) ≥ (t) ≥ − (t) ¿ 0. We conclude that (∞) = 1. It remains to
+ (t) = min{et=c ; 1} is an upper solution of (2:2). Proof. It suGces to verify inequality (2.4). For t ≥ 0, it is trivial. Now, suppose that t ¡ 0. Recall that 4 sinh2 (m=(2c)) = e−m=c − 2 + em=c . Note that + c ˙ (t) = + (t) for t ¡ 0. It follows from the de
∀x ∈ [0; 1]:

(2.10)

But, by the assumptions on d and m, we have 4d sinh2 (m=(2c))g(x) + f(x) − x = x2 [4d sinh2 (m=(2c))xm−2 − 1] ≤ 0 for all x ∈ [0; 1]. Condition (2.10) holds. Therefore, + is an upper solution of (2.2). Proposition 2.5. Suppose that m ¿ 1. For 0 ¡  ¡ min{m − 1; 1} and a su2ciently large M; the function

− (t) = max{0; (1 − M et=c )et=c } is a lower solution of (2:2). Proof. First, by the assumption on , we have A(t) ≡ e−(+1)t=c (dpemt=c + e2t=c ) → 0 as t → −∞, where p = e−m=c + 2 + em=c . Therefore, there is a t0 ¡ 0 such that A(t) ¡ 

(2.11) −t0 =c

for all t ≤ t0 . Take M = e . For t ≥ t0 , since − (t) = 0, it is trivial that the inequality −

c ˙ (t) ≤ d[g( − (t − 1)) − 2g( − (t)) + g( − (t + 1))] + f( − (t))

(2.12)

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holds. For t ¡ t0 , we have d[g( − (t − 1)) − 2g( − (t)) + g( − (t + 1))] + f( − (t)) ≥ −{d[g( − (t − 1)) + 2g( − (t)) + g( − (t + 1))] + ( − (t))2 } + − (t) ≥ −[dpemt=c + e2t=c ] + − (t) ≥ −e(+1)t=c + − (t) − ≥ c ˙ (t):

Hence − is a lower solution of (2.2). Note that the functions + and − were found to satisfy the conditions: 0 ≤ − (t) ≤ +

(t) ≤ 1, ∀t ∈ R, − ≡ 0, and + (−∞) = 0; + (∞) = 1; ˙ (t) ≥ 0, a.e. in R. Hence we have proved the following main theorem of this section. +

Theorem 2.6. Given any c ¿ 0. For m ≥ 2; there is a nondecreasing traveling wave solution of (2:1) with speed c; if d ≤ 1=[4 sinh2 (m=(2c))]. 3. The case 1 ¡ m ¡ 2 In this section we shall study the existence of traveling wave solutions for the system m m u˙ n = d(un−1 − 2unm + un+1 ) + f(un );

n ∈ Z;

(3.1)

where d ¿ 0, m ¿ 1, and f(x) = x(1 − x). First, we state the main theorem of this section as follows. Theorem 3.1. For m ¿ 1; (3:1) has a traveling wave solution with speed c ¿ 0 if d ¡ dK m . Given any  ¿ 0, let dK  , supr¿0 [(1 + )1−m rc − 1]=[4 sinh2 (mr=2)]. To prove Theorem 3.1, we will
un (0) = x n ;

where u−N −1 ≡ x−N ; uN +1 ≡ 1; and x , {x n : n ∈ ZN }.

n ∈ ZN ;

(P)

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Denition 3.4. Let  , 1=c ¿ 0 and let F : R+ × R2N +1 → R2N +1 be de
xn0 = v(n);

n ∈ ZN ;

where v is the solution of the initial value problem v (t) = f(v(t)); v(0) = 1=2: For d = 0, we recall from [19, Lemma 2:6] that Dx F(0; x0 ) is nonsingular. We de
(3.2)

Proof. Since limr→∞ (p1−m rc − 1)=4 sinh2 (mr=2) = 0, there exists an r ¿ 0 such that p1−m rc − 1 = dK  : 4 sinh2 (mr=2)

(3.3)

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Since d ≤ dK  , we have p1−m rc = dK  (4 sinh2 (mr=2)) + 1 ≥ d(e−mr − 2 + emr ) + 1:

(3.4)

We divide it into four cases. For z ≤ −1, multiplying both sides of (3.4) by pm erz , we obtain pcrerz ≥ erz d(pm e−mr − 2pm + pm emr ) + pm erz ≥ e−mrz+rz d[(per(z−1) )m − 2(perz )m + (per(z+1) )m ] + perz ; since pm erz ≥ perz : Also, since x ¿ f(x) for all x = 0 and e−mrz+rz = e(1−m)rz ≥ 1, c(prerz ) ¿ d[(per(z−1) )m − 2(perz )m + (per(z+1) )m ] + f(perz ):

(3.5)

Hence (3.2) is satis
for some n ∈ ZN \{N };

m m − 2xnm + xn+1 ) + f(x n ) = 0 d(xn−1

(3.6) for some n ∈ ZN ;

(3.7)

xN = 1;

(3.8)

x−N = 0:

(3.9)

A similar argument as in the proof of [19, Lemma 2:12] shows that none of (3.6) – (3.8) can hold. We may assume that d ¿ 0.

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Suppose that (3.9) holds, i.e., x−N = 0. Choose r ¿ 0 as in Lemma 3.7 so that V (z) = min{perz ; p} satis
if

j ≤ z ≤ j + 1;

j ∈ ZN ;

(3.10)

where uN; j is the solution of (P) with initial value x ∈ Ud . For j ≤ z ¡ j + 1, cU˙ N (z) m m m = d[uN; j−1 ((z − j)) − 2uN; j ((z − j)) + uN; j+1 ((z − j))] + f(uN; j ((z − j)))

= d[UN (z − 1)m − 2UN (z)m + UN (z + 1)m ] + f(UN (z)):

(3.11)

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The equivalent integral equation is
t d[UN (z − 1)m − 2UN (z)m + UN (z + 1)m ] d z cUN (t) = cUN (0) +
+

0

0

t

f(UN (z)) d z;

where t ∈ [ − N; N + 1). Since UN (z) is bounded by 1; ∀N ≥ N0 ; ∀z ∈ [ − N; N + 1) and cUN is bounded by 4d + f∞ , from (3.11) we conclude that {UN }∞ N =N0 is equicontinuous. It follows from the Arzela-Ascoli Theorem that the sequence {UN }∞ N =N0 contains a subsequence which converges uniformly on any
0

0

t

f(U (z)) d z;

∀t ∈ R

and so (1.4) holds. The following lemma will be used to show that the limit function U (t) satis

=

0



m m d[uk−1 (s) − 2ukm (s) + uk+1 (s)] ds +


0



f(uk (s)) ds;

(3.12)

we obtain m m − xk+1 ) + +; xk+1 − xk ≥ d(xk−1

(3.13)

where + = *(1 − *) = min{f(x): * ≤ x ≤ 1 − *}. Also, xk+1 ¿ xk−1 and m ¿ 1 imply m m that (xk+1 − xk−1 ) ≥ (xk+1 − xk−1 )m . Together with (3.13), we get m m m m (xk+1 − xk−1 )1=m ≥ d(xk−1 − xk+1 ) + +:

Set Q ,

m xk+1

−

m xk−1 .

Note that Q

1=m

≥ Q, since 1=m ¡ 1 and 0 ¡ Q ¡ 1.

(3.14)

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If d ≥ 1, then (3.14) implies that + ≤ Q1=m + dQ ≤ d(Q1=m + Q) ≤ d(Q1=m + m m )1=m ≥ +=(2d). Q1=m ). Thus Q1=m ≥ +=(2d), i.e. (xk+1 − xk−1 If d ¡ 1, then (3.14) implies that + ≤ Q1=m + dQ ¡ Q1=m + Q ≤ Q1=m + Q1=m . m m − xk−1 )1=m ¿ +=2. Thus Q1=m ¿ +=2, i.e. (xk+1 From the de
≤ de-t (e- + e−- − 2) + e-t − c-e-t = 0: Hence (2.4) holds for all t ∈ R and + is an upper solution of (2.2).

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Next, for the same d and -, take a constant  with 0 ¡  ¡ min{-∗ − -; -}. Then for some M ¿ 0 the function − (t) = max{0; e-t − M e(-+)t } is a lower solution of (2.2). We give a proof as follows. First, by the choice of , e(-−)t → 0 as t → −∞. Hence there is a t0 ¡ 0 such that   c(- + ) − 1 −(-+) (-−)t -+ e − 2] -+ ≤ [e + e − d ; ∀t ≤ t0 : e + e−(-+) − 2 We note that the function (cr − 1)=(er + e−r − 2) is strictly increasing in (1=c; -∗ ). Take M = e−t0 . We are in a position to verify inequality (2.12) for m = 1. The case for t ≥ t0 is trivial. For t ¡ t0 , we have d[ − (t − 1) − 2 − (t) + − (t + 1)] + f( − (t)) ≥ e-t [d(e−- + e- − 2) + 1] − e(-+)t M [d(e−(-+) + e-+ − 2) + 1] − − (t)2 ≥ c-e-t − M e(-+)t {[d(e−(-+) + e-+ − 2) + 1] + e(-−)t } ≥ c-e-t − M e(-+)t c(- + ) − = c ˙ (t):

Hence (2.12) is satis
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[4] S.-N. Chow, W. Shen, Dynamics in discrete Nagumo equation: spatial topological chaos, SIAM J. Appl. Math. 55 (1995) 1764. [5] S.-N. Chow, W. Shen, Stability and bifurcation of traveling wave solutions in coupled map lattices, Dynamic Systems Appl. 4 (1995) 1. [6] A. De Pablo, J.L. Vazquez, Travelling waves and