# Note on propagation speed of travelling waves for a weakly coupled parabolic system

## Note on propagation speed of travelling waves for a weakly coupled parabolic system

Nonlinear Analysis 44 (2001) 239 – 246 www.elsevier.nl/locate/na Note on propagation speed of travelling waves for a weakly coupled parabolic system...

Nonlinear Analysis 44 (2001) 239 – 246

www.elsevier.nl/locate/na

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–di usion 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.  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 . In Section 3, we shall state the proof of the above theorem by using the method of moving planes which was proposed in .

2. Application In this section, as an application of Theorem 1.1, we consider a Lotka–Volterra competition model with di usion 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).

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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(21 − ) 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(2ˆ − ) 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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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))

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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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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(20 −  ; 0 ) o 0 holds because of (3.6), and since U0 (; ) = 2u0 () ≺o 0 holds for any  ∈ Int Ju ∩ Jv , we

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see that there exists  ∈ (20 − + ; 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 (20 − ) = ¡ 0 otherwise; we have 0 ≤ U 00 (; 0 ) = s0 u0 (20 − ) − s0 u0 () + f(u(20 − )) − f(u()) ( = 0 if  = 0 = − u ; = 2s0 u0 (20 − ) − C4 V (; 0 ) = ¡ 0 otherwise; where

Z C4 =

0

1

fv ( u() + (1 − )u(20 − )) 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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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  R.A. Gardner, Existence and stability of travelling wave solutions of competition models: a degree theoretic approach, J. Di erential Equations 44 (1982) 343–364.  B. Gidas, W.M. Ni, L. Nirenberg, Symmetry and related properties via the maximum principle, Commun. Math. Phys. 68 (1979) 209–243.  Y. Hosono, Singular perturbation analysis of travelling waves for di usive Lotka–Volterra competition models, Numerical and Applied Mathematics Part II, Baltzer, Basel, 1989, pp. 687– 692.  Y. Kan-on, A note on the propagation speed of travelling waves for a Lotka–Volterra competition model with di usion, J. Math. Anal. Appl. 217 (1998) 693–700.  M.M. Tang, P.C. Fife, Propagating front for competing species equations with di usion, Arch. Rational Mech. Anal. 73 (1980) 69–77.  A.I. Volpert, V.A. Volpert, Traveling wave solutions of parabolic systems, American Mathematical Society, Providence, RI, 1994.