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Note on propagation speed of travelling waves for a weakly coupled parabolic system Yukio Kan-On Department of Mathematics, Faculty of Education, Ehime University, Matsuyama 790-8577, Japan Received 15 October 1997; accepted 26 January 1999

Keywords: Weakly coupled parabolic system; Propagation speed; Method of moving planes

1. Introduction One interesting phenomenon in various ÿelds is the appearance of travelling waves, and a well-used model to explain the phenomenon is the system of reaction–diusion equations wt = Dwxx + f(w);

x ∈ R; t ¿ 0;

(1.1)

where w and f are n-dimensional vectors, and D is a diagonal matrix whose elements are positive. In this paper, we treat travelling waves in the form w(t; x)=u(); =x−st, where u() is of C 2 -class, and s is the so-called propagation speed. Hence, such waves necessarily satisfy the system of ODEs 0 = Du00 + su0 + f(u);

∈R

(1.2a)

for some s ∈ R, where 0 = d=d. In order to determine the propagation speed, we lay the boundary condition u(−∞) = u(+∞) = u0 (∈ Rn )

(1.2b)

on the wave. It follows from u0 ()= = −D−1 f(u0 ) + o(1)

as || → +∞

that u0 must be a zero of f(u). Here we assume that (H.1) the real part of every eigenvalue of the matrix fu (u0 ) is negative, which implies that u0 is an exponentially stable equilibrium point of ut = f(u). 0362-546X/01/$ - see front matter ? 2001 Elsevier Science Ltd. All rights reserved. PII: S 0 3 6 2 - 5 4 6 X ( 9 9 ) 0 0 2 6 1 - 8

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For the case n = 1, since (H.1) leads to fu (u0 ) ¡ 0, we can easily check that every solution of (1:2) exponentially decays to u0 as || → +∞, and then obtain Z Z s u0 ()2 d = − (Du00 () + f(u()))u0 () d = 0: R

R

The above equation says that if (1:2) has a nonconstant solution for s ∈ R, the propagation speed s must satisfy s = 0. As such a property does not always hold in case of n ≥ 2, one important problem for the travelling wave is the qualitative study on the propagation speed. There are many studies of travelling waves of (1.1) with a variety of boundary conditions for n ≥ 2 (for instance, see Volpert et al. [6] and its references). However so far, we have not yet known enough results on the propagation speed. In this paper, as a ÿrst step to approach the propagation speed, we restrict our discussion to the case n = 2, where u = (u; v); u0 = (u0 ; v0 ); D = diag(1; d); d ¿ 0, and f(u) = (f; g)(u). In addition to (H.1), for the nonlinearity f, we assume (H.2) fv (u) ¡ 0 and gu (u) ¡ 0 for any u, (H.3) f(u; v0 )(u − u0 ) ¡ 0 for any u 6= u0 , and g(u0 ; v)(v − v0 ) ¡ 0 for any v 6= v0 . Here is the main result of this paper. Theorem 1.1. Suppose (H:1)–(H:3). If (1:2) has a nonconstant solution for s=s0 ∈ R; then s0 = 0 holds. The above theorem is an extension of Theorem 1.1 in [4]. In Section 3, we shall state the proof of the above theorem by using the method of moving planes which was proposed in [2].

2. Application In this section, as an application of Theorem 1.1, we consider a Lotka–Volterra competition model with diusion which describes the dynamics of the population u of two competing species, where f(u) = uf0 (u); g(u) = vg0 (u), and f 0 (u) = (f0 ; g0 )(u) is a C 2 -class function in u. As u indicates the population density, we only consider positive solutions of (1:2) with u0 ¿ 0 and v0 ¿ 0, where (u; v)() is called positive if u() ¿ 0 and v() ¿ 0 are satisÿed for any ∈ R. For the existence of travelling waves, we refer to [5,1,3], for instance. It is easy to check that (H.2) is rewritten as f0v (u) ¡ 0 and g0u (u) ¡ 0 for any u in the ÿrst quadrant, and that (H.3) holds in the ÿrst quadrant when (H.4) f0u (u) ¡ 0 and g0v (u) ¡ 0 are satisÿed for any u in the ÿrst quadrant. Furthermore, we see that the classical model with f0 (u) = au − bu u − cu v;

g0 (u) = av − bv u − cv v

is an example satisfying (H.4), where a ; b and c are positive constants ( = u; v). Throughout this section, we assume (H.1), (H.2) and (H.4).

Y. Kan-On / Nonlinear Analysis 44 (2001) 239 – 246

241

Let u() = (u; v)() be an arbitrary nonconstant positive solution of (1:2) for s = s0 ∈ R. By Theorem 1.1, we obtain s0 = 0. Let ∈ R be an arbitrary point satisfying u0 () = 0. It is easy to check that U()(=(U; V )()) = u() − u(2 − ) is a solution of 0 = DU00 + A()U;

U () = 0; U00 () = 0;

U() = 0; where

Z A() =

0

1

∈ R;

0

fu (u() + (1 − )u(2 − )) d:

Let us derive a contradiction by assuming V 0 () 6= 0. By U(+∞) = 0 and U 000 ()V 0 () = −fv (u())V 0 ()2 ¿ 0; we can take 1 (∈ (; +∞]) as satisfying U (1 )V (1 ) = 0 and U ()V () ¿ 0 for any ∈ (; 1 ). For the case U (1 ) = 0, we have Z 1 (u00 () + f(u()))u(21 − ) d = u(1 )U 0 (1 ) 0=

+

Z 1Z

0

1

f0u (u() + (1 − )u(2 − ))U()u()u(2 − ) d d 6= 0

by using integration by parts. Similarly we can derive a contradiction in case of V (1 )= 0. Hence, we obtain V 0 ()=0. By uniqueness, we have u()=u(2−) for any ∈ R. ˆ = 0 is satisÿed for some ˆ (¡ ). Since u() = We consider the case where u0 () u(2ˆ − ) holds for any ∈ R, we obtain u() = u( + 2j( − )) ˆ for any j ∈ Z, which implies u() = u0 . By u0 () = 0 and uniqueness, we have u() = u0 for any ∈ R. This contradiction implies u0 () 6= 0 for any ¡ . Similarly, we can prove v0 () 6= 0 for any ¡ . When u0 () ¿ 0 and v0 () ¿ 0 hold for any ¡ , we have u() ¿ u0 and v() ¿ v0 for any ≤ , and then obtain 0 ¿ f(u()) = −u00 () ≥ 0 because of (H.2) and (H.4). Similarly we can derive a contradiction by assuming u0 () ¡ 0 and v0 () ¡ 0 for any ¡ . Hence we obtain u0 ()v0 () ¡ 0 for any ¡ . Theorem 2.1. Suppose (H:1); (H:2) and (H:4). Let u() = (u; v)() be an arbitrary nonconstant positive solution of (1:2) for s = s0 ∈ R. Then s0 = 0 holds and there exists ∈ R such that u() satisÿes u() = u(2 − ) and u0 ()v0 () ¡ 0 for any ¡ . 3. Proof Contrary to the conclusion, we assume that (1:2) has a nonconstant solution u() = (u; v)() for s = s0 ∈ R (6= 0). Since u(2 − ) is a nonconstant solution of (1:2) with s = −s0 for each ∈ R, we may assume s0 ¿ 0 without loss of generality.

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Y. Kan-On / Nonlinear Analysis 44 (2001) 239 – 246

We deÿne the order relations s and o in the following manner: (u1 ; v1 ) s (u2 ; v2 ) ⇔ u1 ≤ u2 ; v1 ≤ v2 ; (u1 ; v1 ) o (u2 ; v2 ) ⇔ u1 ≤ u2 ; v1 ≥ v2 : Furthermore, we denote by ≺s and ≺o the relation obtained from the above deÿnition by replacing ≤ with ¡. By (H.2) and (H.3), we have s 0 for any u s u0 with u 6= u0 ; (3.1) f(u) ≺s 0 for any u s u0 with u 6= u0 : From (H.1) and (H.2), we easily obtain fu (u0 ) ¡ 0;

fv (u0 ) ¡ 0;

gu (u0 ) ¡ 0;

gv (u0 ) ¡ 0;

det fu (u0 ) ¿ 0:

(3.2)

The linearized operator of (1.2a) around u = u0 can be represented as ! d ; s) fv (u0 ) pu ( d ; d pv ( d ; s) gu (u0 ) where pu ( ; s) = 2 + s + fu (u0 ) and pv ( ; s) = d 2 + s + gv (u0 ). It follows from (3.2) that pu ( ; s) has two real zeros − (s) and + (s) satisfying − (s) ¡ 0 ¡ + (s) for any s ∈ R. Setting R( ; s) = pu ( ; s)pv ( ; s) − fv (u0 ) gu (u0 ); we see from R( ± (s); s) = −fv (u0 )gu (u0 ) ¡ 0; that R( ; s) has four real zeros

± 1 (s)

R(0; s) = det fu (u0 ) ¿ 0

and ± 2 (s) satisfying

− + +

− 1 (s) ¡ − (s) ¡ 2 (s) ¡ 0 ¡ 2 (s) ¡ + (s) ¡ 1 (s);

± pu ( ± 2 (s); s) ¡ 0 ¡ pu ( 1 (s); s);

− R ( + 2 (s); s) ¡ 0 ¡ R ( 2 (s); s)

(3.3)

for any s ∈ R. By ± ± ± Rs ( ± 2 (s); s) = 2 (s)(pu ( 2 (s); s) + pv ( 2 (s); s)); ± pu ( ± 2 (s); s)pv ( 2 (s); s) = fv (u0 ) gu (u0 ) ¿ 0;

− − + we obtain Rs ( + 2 (s); s) ¡ 0 ¡ Rs ( 2 (s); s) for any s ∈ R. By 2 (0) + 2 (0) = 0 and

Rs ( + Rs ( − d − 2 (s); s) 2 (s); s) [ 2 (s) + + − (s)] = − ¡0 2 − ds R ( 2 (s); s) R ( + 2 (s); s)

for any s ∈ R;

+ we have 0 ≡ − 2 (s0 ) + 2 (s0 ) ¡ 0. Setting

ej± = (−fv (u0 ); pu ( ± j (s0 ); s0 ));

j = 1; 2;

we obtain e1± s 0 and e2± o 0 because of (3.2) and (3.3). Since u() is nonconstant, we have ∓

u() = u0 + C1± e1∓ e 1

(s0 )

∓

(1 + o(1)) + C2± e2∓ e 2

(s0 )

(1 + o(1))

Y. Kan-On / Nonlinear Analysis 44 (2001) 239 – 246

243

as → ±∞, where C1± and C2± are suitable constants satisfying (C1− ; C2− ) 6= 0 and (C1+ ; C2+ ) 6= 0. By replacing the roles between u and v if need be, we may assume C2− ≥ 0 without loss of generality. − ± + + Lemma 3.1. If there exist − 2 and 2 with −∞ ≤ 2 ¡ 2 ≤ +∞ such that u(2 ) = ± u0 (respectively; v(2 ) = v0 ) and u() 6= u0 (respectively; v() 6= v0 ) for any ∈ − + + (− 2 ; 2 ) are satisÿed; then H () ¡ 0 holds for some ∈ (2 ; 2 ); where H ()=(u()− u0 ) (v() − v0 ).

Proof. We only show the proof for the case where u(± 2 ) = 0 and u() ¿ u0 for + ; ) are satisÿed. Since the others can be proved in a similar manner. any ∈ (− 2 2 + Contrary to the conclusion, we assume v() ≥ v0 for any ∈ (− 2 ; 2 ). By (3.1), we − + have f(u()) ≺s 0 for any ∈ (2 ; 2 ). Since u() attains a local maximum at some + 0 = 3 ∈ (− 2 ; 2 ), we obtain u (3 ) = 0 and 0 ≥ u00 (3 ) = −s0 u0 (3 ) − f(u(3 )) ¿ 0: This contradiction implies that the desired results holds. When C2− = 0 is satisÿed, we see from C1− 6= 0 and u(+∞) = u0 that there exists 4 ∈ (−∞; +∞] such that H (4 ) = 0 and H () ¿ 0 for any ∈ (−∞; 4 ) hold. This contradicts the fact of Lemma 3.1. Hence, we have C2− ¿ 0, which implies u() o u0 for any in a neighborhood of = −∞. Analogously, we can prove C2+ 6= 0. By

0 ¡ 0, we obtain u(2 − ) − u0 1 − 0 (1 + o(1)) ¡ 1; u() − u0 = C3 e (3.4) v(2 − ) − v0 = C32 e− 0 (1 + o(1)) ¡ 1 v() − v0 as → −∞ for any ÿxed ∈ R, where C + e2 − C + p ( − (s ); s )e2 − (s0 ) 2 2 2 (s0 ) 2 u 2 0 0 1 2 C3 = : ; C3 = C2− C2− pu ( + (s ); s ) 0 0 2 We set − = min(u ; v ); + = max(u ; v ), u = sup{ | u() ≥ u0 for any ≤ };

v = sup{ | v() ≤ v0 for any ≤ }:

From Lemma 3.1, we see that u ¡+∞ is equivalent to v ¡+∞, and obtain u() ≤ u0 for any ∈ (u ; + ] and v() ≥ v0 for any ∈ (v ; + ]. We consider the case where u(5 ) = u0 is satisÿed for some 5 (¡ − ). By u0 (5 ) = 0; 0 ≤ u00 (5 ) = −f(u(5 )) and (H.3), we have v(5 ) ≥ v0 . By the deÿnition of − , we obtain v(5 ) = v0 and v0 (5 ) = 0. By uniqueness, we have u() = u0 for any ∈ R. This contradiction implies u() ¿ u0 for any ¡ − . In a similar manner, we can prove v() ¡ v0 for any ¡ − . Consequently we have u() o u0

for any ¡ − :

(3.5)

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Y. Kan-On / Nonlinear Analysis 44 (2001) 239 – 246

Case 1: u ¡ + ∞. Clearly we have v ¡ + ∞ and u0 (u ) ≤ 0 ≤ v0 (v ). We consider the case u0 (u ) = 0. From the deÿnition of u , we have 0 = u00 (u ) = −f(u(u )). By (H.3), we obtain v(u ) = v0 and H (u ) = 0. Since u() is nonconstant, we see that v0 (u ) 6= 0 must be satisÿed. From u(+∞) = u0 and u000 (u )v0 (u ) = −fv (u(u ))v0 (u )2 ¿ 0; it follows that there exists 6 (¿ u ) such that H (6 ) = 0 and H () ¿ 0 for any ∈ (u ; 6 ) hold. This contradicts the fact of Lemma 3.1. Hence we obtain u0 (u ) ¡ 0. In a similar manner, we can prove v0 (v ) ¿ 0. + − + Let Ju = [− u ; u ] (respectively, Jv = [v ; v ]) be the maximal extended interval satisfying u ∈ Ju (respectively, v ∈ Jv ) and u0 () ≤ 0 (respectively, v0 () ≥ 0) for any ∈ Ju (respectively, ∈ Jv ). For the case u ≤ v , we have f(u()) s 0, Z 0 s(u −) 0 u u ( ) − es(−) f(u()) d ¡ 0; u () = e u

0

s(v −)=d 0

v () = e

u

v ( ) −

Z

v

es(−)=d g(u()) d ¿ 0

for any ∈ [u ; v ], and then obtain [u ; v ] ⊂ Ju ∩ Jv . Analogously for the case v ≤ u , we can prove [v ; u ] ⊂ Ju ∩ Jv and u0 () ≺o 0 for any ∈ [v ; u ]. Consequently, we have [− ; + ] ⊂ Ju ∩ Jv and u0 () ≺o 0

for any ∈ [− ; + ]:

The above fact and (3.5) lead to ( ¿ u0 for any ∈ (−∞; u ); u() = ¡ u0 for any ∈ (u ; + ]; ( v() =

¡ v0 for any ∈ (−∞; v ); ¿ v0 for any ∈ (v ; + ]:

(3.6)

We consider the case where u0 (7 ) = 0 holds for some 7 ∈ Int Ju ∩ Jv . Then we have u00 (7 ) = 0. Since 0 ≥ u000 (7 ) = −fv (u(7 ))v0 (7 ) ≥ 0 is satisÿed because of (H.2), we have v0 (7 ) = 0 and v00 (7 ) = 0. We see from uniqueness that u() must be a constant function. This contradiction implies u() ¡ 0 for any ∈ Int Ju ∩ Jv . Similarly we can prove v0 () ¿ 0 for any ∈ Int Ju ∩ Jv . Setting U(; )(=(U; V )(; )) = u() − u(2 − ) and 0 = inf { | U(; ) o 0 for any ∈ [; + ] and ∈ [2 − + ; ]}; we have 0 ≤ + . It follows from the deÿnition of Ju that there exists (¡ − u ) such 0 0 ). By U (; ) = 0; U (; ) = 2u () and the that u0 () ¿ 0 holds for any ∈ (; − u − . In a similar manner, we can prove deÿnition of 0 , we obtain 0 ≥ − 0 ≥ v . u − − + Consequently, we obtain 0 ≥ 0 = max(u ; v ). Since U(20 − ; 0 ) o 0 holds because of (3.6), and since U0 (; ) = 2u0 () ≺o 0 holds for any ∈ Int Ju ∩ Jv , we

Y. Kan-On / Nonlinear Analysis 44 (2001) 239 – 246

245

see that there exists ∈ (20 − + ; 0 ] such that (U; U 0 )(; 0 ) = 0

and=or

(V; V 0 )(; 0 ) = 0

(3.7)

are satisÿed. Case 2: u = +∞. Clearly, we have u() o u0 for any ∈ R. Setting 0 = − max(− u ; v ), 0 − u = inf { | u () ≤ 0 for any ≥ };

0 − v = inf { | v () ≥ 0 for any ≥ };

we have u0 () ≺o 0 for any ¿ 0 in the similar manner with the argument for Case 1. Since U(; ) = 0 and U0 (; ) = u0 () + u0 (2 − ) ≺o 0 are satisÿed for any ≥ 0 and ∈ (0 ; ), we have U(; ) o 0

for any ≥ 0 and ∈ [0 ; ):

It follows from (3.4) that there exists 8 (≤ 0 ) such that U(; 0 ) o 0 holds for any ≤ 8 . Since U (; ) = −2u0 (2 − ) o 0 is satisÿed for any ≥ 0 and ¡ , we have U(; ) o 0

for any ≥ 0 and ≤ 8 :

0

By u () ≺o 0 for any ¿ 0 , we can take 9 (≥ 0 ) as satisfying u(9 ) o u() for any ∈ [8 ; 0 ], and then obtain U(; ) o u() − u(9 ) o 0

for any ≥ 9 and ∈ [8 ; 0 ]:

Hence we arrive at U(; ) o 0 for any ≥ 9 and ≤ . Setting 0 = inf { | U(; ) o 0 for any ≥ and ≤ }; we have 0 ≤ 9 and U(; 0 ) o 0 for any ≤ 0 . In the similar manner with the argument for Case 1 we can obtain 0 ≥ 0 , and ÿnd that there exists ∈ [8 ; 0 ] such that (3.7) holds. Proof of Theorem 1.1. Let us consider the case (U; U 0 )(; 0 ) = 0. By ( = 0 if = 0 = − u ; u0 (20 − ) = ¡ 0 otherwise; we have 0 ≤ U 00 (; 0 ) = s0 u0 (20 − ) − s0 u0 () + f(u(20 − )) − f(u()) ( = 0 if = 0 = − u ; = 2s0 u0 (20 − ) − C4 V (; 0 ) = ¡ 0 otherwise; where

Z C4 =

0

1

fv ( u() + (1 − )u(20 − )) d(¡ 0):

00 0 Hence we obtain = 0 = − u and U (0 ; 0 ) = 0. By deÿnition, we have u (0 ) = 00 0 0; u (0 ) ≤ 0 and v (0 ) ≥ 0. The inequality

0 ≥ U 000 (0 ; 0 )=2 = u000 (0 ) = −s0 u00 (0 ) − fv (u(0 ))v0 (0 ) ≥ 0

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Y. Kan-On / Nonlinear Analysis 44 (2001) 239 – 246

says that u00 (0 ) = 0 and v0 (0 ) = 0 are satisÿed. By deÿnition, we obtain v00 (0 ) ≥ 0. From V (0 ; 0 ) = 0;

V 0 (0 ; 0 ) = 0;

V 00 (0 ; 0 ) = 0;

0 ≤ dV 000 (0 ; 0 )=2 = dv000 (0 ) = −s0 v00 (0 ) ≤ 0; we have v00 (0 ) = 0, which implies u0 (0 ) = 0 and u00 (0 ) = 0. By uniqueness, we see that u() must be a constant function. This is a contradiction. Analogously, we can derive a contradiction when (V; V 0 )(; 0 ) = 0 holds. Hence we arrive at s0 = 0. Thus the proof of Theorem 1:1 is completed. References [1] R.A. Gardner, Existence and stability of travelling wave solutions of competition models: a degree theoretic approach, J. Dierential Equations 44 (1982) 343–364. [2] B. Gidas, W.M. Ni, L. Nirenberg, Symmetry and related properties via the maximum principle, Commun. Math. Phys. 68 (1979) 209–243. [3] Y. Hosono, Singular perturbation analysis of travelling waves for diusive Lotka–Volterra competition models, Numerical and Applied Mathematics Part II, Baltzer, Basel, 1989, pp. 687– 692. [4] Y. Kan-on, A note on the propagation speed of travelling waves for a Lotka–Volterra competition model with diusion, J. Math. Anal. Appl. 217 (1998) 693–700. [5] M.M. Tang, P.C. Fife, Propagating front for competing species equations with diusion, Arch. Rational Mech. Anal. 73 (1980) 69–77. [6] A.I. Volpert, V.A. Volpert, Traveling wave solutions of parabolic systems, American Mathematical Society, Providence, RI, 1994.

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