Traveling wave to a reaction-hyperbolic system for axonal transport

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ScienceDirect J. Differential Equations ••• (••••) •••–••• www.elsevier.com/locate/jde

Traveling wave to a reaction-hyperbolic system for axonal transport ✩ Feimin Huang a,b , Xing Li c,∗ , Yinglong Zhang d a College of Mathematics and Computer Science, Hunan Normal University, China b Academy of Mathematics and Systems Science, Chinese Academy of Sciences, China c The College of Mathematics and Statistics, Shenzhen University, China d Department of Mathematical Sciences, Seoul National University, Republic of Korea

Received 8 October 2016; revised 10 February 2017

Abstract In this paper, we study a class of nonlinear reaction-hyperbolic systems modeling the neuronal signal transfer in neuroscience. This reaction-hyperbolic system can be regarded as n × n (n ≥ 2) hyperbolic system with relaxation. We first prove the existence of traveling wave by Gershgorin circle theorem and mathematically describe the neuronal signal transport. Then for a special case n = 2, we show the traveling wave is nonlinearly stable, and obtain the convergence rate simultaneously by a weighted estimate. © 2017 Elsevier Inc. All rights reserved. Keywords: Reaction-hyperbolic system; Axonal transport; Traveling wave; Convergence rate

1. Introduction In this paper, we consider a class of reaction-hyperbolic systems in one space dimension as follows ✩

F. Huang’s research was supported in part by NSFC Grant No. 11371349 and the Construct Program of the Key Discipline in Hunan Province of China. X. Li’s research was supported in part by NSFC Grant No. 11301344. * Corresponding author. E-mail addresses: [email protected] (F. Huang), [email protected] (X. Li), [email protected] (Y. Zhang). http://dx.doi.org/10.1016/j.jde.2017.02.033 0022-0396/© 2017 Elsevier Inc. All rights reserved.

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⎧ 1 ⎪ ⎪ = f1 (u1 , u2 ), u1t +λ1 u1x ⎪ ⎪  ⎪ ⎪ ⎪ ⎪ ⎪ 1 1 ⎪ ⎪ = − f1 (u1 , u2 ) + f2 (u2 , u3 ), u2t +λ2 u2x ⎪ ⎪   ⎪ ⎨ ...... ⎪ ⎪ ⎪ ⎪ 1 1 ⎪ n−2 n−1 ⎪ +λn−1 un−1 , u ) + fn−1 (un−1 , un ), un−1 ⎪ t x = − fn−2 (u ⎪   ⎪ ⎪ ⎪ ⎪ ⎪ 1 ⎪ ⎩ unt +λn unx = − fn−1 (un−1 , un ), n ≥ 2, 

(1.1)

which model the axonal transport in neuroscience, see Reed and Blum [16]. Here ui represents the i-th subpopulations, the term λi uix accounts for the transport of the i-th subpopulation with constant velocities λi satisfying λ1 < λ2 < · · · < λn . Each fi is a smooth function describing the biochemical processes of the constituents. The small parameter  characterizes the fact that the biochemical process is much faster than the transport one. The axonal transport is important for the maintenance and functions of nerve cells. The system (1.1) was introduced by Reed and Blum to model the axonal transport mathematically, see [2, 3,14–16]. It was observed in experiments that the neuronal signal can be transferred through a jump of concentration of subpopulations, see [1,9]. This phenomenon can be explained in mathematics by the wave-like solution of the system (1.1). It was also observed by numerical simulations in [2,3,14–16] that these models exist wave-like solutions moving at a more or less constant velocity both in linear and nonlinear case, which is consistent with the experimental observation in neuroscience. However theoretically there is no traveling wave for the linear case, see [17]. Instead an approximate traveling wave was derived in [17], see also [3,7,8]. To our best knowledge, except the special case n = 2, there is no theoretical result on the traveling wave for the nonlinear cases which are more important in neuroscience, that is, in the system (1.1) fi is nonlinear. For other interesting works, see [5,11] and the references therein. In this paper, we focus on the traveling wave to the system (1.1). We first show the existence of traveling wave motivated by the kinetic theory for the Boltzmann equation. Precisely speaking, we observe that the equilibrium equation of the system (1.1) as  → 0 is scalar conservation law in which there naturally has a shock wave. Based on this, we can expect a traveling wave exists for the system (1.1) with the same speed of the shock wave to the scalar conservation law, determined by the Rankine–Hugoniot condition. Before formulating our main results, we need the following assumption: (A) The function fi (ui , ui+1 ) is strictly decreasing with respect to ui and strictly increasing ∂fi ∂fi with respect to ui+1 , i.e. ∂u i < 0 and ∂ui+1 > 0. In addition fi (0, 0) = 0. The assumption (A) means that the chemical reactions form a chain and un is made from which is made from un−2 . Note that the system (1.1) can be rewritten as

un−1 ,

⎧ (u1t +λ1 u1x )=f1 (u1 , u2 ), ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ (u2 +λ2 u2 )= − f1 (u1 , u2 ) + f2 (u2 , u3 ), t x ⎪ ...... ⎪ ⎪ ⎪ ⎪ ⎩ (unt +λn unx )= − fn−1 (un−1 , un ), n ≥ 2,

(1.2)

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which formally takes a limit, as  → 0, that fi (ui , ui+1 ) = 0 for all i = 1, 2, · · · , n − 1 (the limit is actually proved in [4] and [20] for special case). The state (u1 , · · · , un ) is called equilibrium state if it satisfies fi (ui , ui+1 ) = 0 for all i. Then we obtain the following equilibrium system corresponding to system (1.1) ⎧ n n   ⎪ ⎪ ⎪ ui )t + ( λi ui )x = 0, ( ⎪ ⎪ ⎪ ⎪ i=1 ⎪ ⎨ i=1 1 2 f1 (u , u ) = 0, ⎪ ⎪ ⎪ ⎪ ······ ⎪ ⎪ ⎪ ⎪ ⎩ fn−1 (un−1 , un ) = 0,

(1.3)

where the first equation is derived by adding all equations of (1.1). In view of assumption (A), there exists a unique globally-defined function hi (ui ) such that fi (ui , ui+1 ) = 0 iff ui+1 = hi (ui ), h (ui ) > 0, i = 1, 2, · · · , n − 1.

(1.4)

Thus the equilibrium system (1.3) can be written as ⎧ n n   ⎪ ⎪ ⎨( ui )t + ( λi ui )x = 0, ⎪ ⎪ ⎩

i=1

(1.5)

i=1

ui = hi−1 (ui−1 ), i = 2, · · · , n.

Let ω = u1 + h1 (u1 ) + · · · + hn−1 ◦ hn−2 ◦ · · · ◦ h1 (u1 ). Notice that since hi > 0, there exists an increasing function g(ω) such that u1 = g(ω). Let F (ω) = λ1 g(ω) + λ2 h1 ◦ g(ω) + · · · + λn hn−1 ◦ · · · ◦ h1 ◦ g(ω), then the equilibrium system (1.5) is reduced to a scalar conservation law, i.e. ωt + F (ω)x = 0.

(1.6)

By a direct calculation, we get F  (ω) =

λ1 + λ2 h1 + · · · + λn hn−1 · · · h1 1 + h1 + · · · + hn−1 · · · h1

,

(1.7)

where hi−1 = −

(fi−1 )ui−1 , (fi−1 )ui

i = 2, · · · , n.

(1.8)

Next we consider the following Riemann initial data  ω|t=0 =

ω− , x < 0, ω+ , x > 0,

(1.9)

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where ω± = u1± + h1 (u1± ) + · · · + hn−1 ◦ · · · ◦ h1 (u1± ). In the present paper we investigate the nonlinear case, that is, F  (ω) = 0. In particular, F  (ω+ ) = 0. If F  (ω+ ) > 0, then there exists a small neighborhood U of ω+ such that F  (ω) > 0, ∀ω ∈ U .

(1.10)

ω− ∈ U and ω− > ω+ ,

(1.11)

Assume that

the system (1.6) admits a shock wave satisfying  ω− , x < σ t, ω(x, t) = ω+ , x > σ t,

σ=

F (ω+ ) − F (ω− ) , ω + − ω−

(1.12)

where σ is the propagation speed of shock wave, given by the Rankine–Hugoniot condition, see [18]. If F  (ω+ ) < 0, then the conditions (1.10) and (1.11) can be replaced by ω− ∈ U and ω− < ω+ , F  (ω) < 0, ∀ω ∈ U ,

(1.13)

and the system (1.6) still admits a shock wave with the formula (1.12), see [18]. Therefore, we always assume F  (ω+ ) > 0 in what follows. Note that the system (1.1) can be regarded as n × n (n ≥ 2) one dimension hyperbolic system with relaxation, which takes the form 1 Ut + F (U )x = R(U ), 

(1.14)

where  is the relaxation time, see [6,12,19]. It is straightforward to check from (1.7) that the characteristic speed F  (ω) satisfies λ1 < F  (ω) < λn .

(1.15)

Thus the so-called sub-characteristic condition is satisfied, see [12]. As pointed out by [12], the sub-characteristic condition (1.15) is important to the stability of solution to (1.14). On the other hand, if we consider a viscous perturbation of the scalar equation (1.6), i.e., ωt + F (ω)x = ωxx ,

(1.16)

t there exists a viscous traveling wave ω( x−σ  ) of the viscous conservation law (1.16) with scalt x   ing t =  , x =  , see [18] for details. Similar to the viscous conservation law (1.16), the system (1.1) can be regarded as a perturbation of the scalar equation (1.6) when  is small, one natut rally expects that the system (1.1) admits a traveling wave (U 1 , · · · , U n )( x−σ  ) satisfying the boundary condition

lim U i (ξ ) = ui± = hi−1 ◦ · · · ◦ h1 (u1± ),

ξ →±∞

where ξ =

x−σ t  ,

and the entropy condition

(1.17)

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F  (ω+ ) < σ < F  (ω− ).

5

(1.18)

From (1.15), there always exists 1 < k < n such that λk ≤ F  (ω+ ) < λk+1 ,

(1.19)

which, together with (1.10), (1.11) and (1.18), implies that λk < σ < λk+1 .

(1.20)

Now we are ready to state the first result of present paper in the following. Theorem 1.1. Under the assumption (A) and the condition (1.11), there exists a unique traveling t wave (U 1 , U 2 , · · · , U n )( x−σ  ) of system (1.1) up to a shift. In particular, when n = 2, we can show that the above traveling wave is nonlinearly stable, and obtain the convergence rate simultaneously. Let u := u1 and v := u2 . Then the system (1.1) becomes ⎧ 1 ⎪ ⎪ ⎨ ut + λ1 ux = f (u, v),  ⎪ 1 ⎪ ⎩ vt + λ2 vx = − f (u, v), 

(1.21)

which can be reduced into a simple system 

ut = f (u, v), vt + vx = −f (u, v),

(1.22)

that corresponds to  = 1, λ1 = 0 and λ2 = 1. In fact, through a coordinate transformation t  = t , x  = x , the system (1.21) is changed into 

ut + λ1 ux = f (u, v), vt + λ2 vx = −f (u, v),

(1.23)

where we still denote (x  , t  ) by (x, t). Again using the coordinate transformation t  = t, x  = x−λ1 t λ2 −λ1 , we get (1.22). So we always consider the simple system (1.22) for n = 2. We assume the initial data satisfies  (u0 (x) + v0 (x) − U (x) − V (x))dx = 0,

(1.24)

R t where (U, V )( x−σ  ) is the traveling wave. If (1.24) is not satisfied, it is easy to find a unique shift s0 determined by the initial perturbation (cf. [13]) so that

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 (u0 (x) + v0 (x) − U (x + s0 ) − V (x + s0 ))dx = 0.

(1.25)

R

Set x (x, t) :=

[u(y, t) + v(y, t) − U (y − σ t) − V (y − σ t)]dy,

(1.26)

−∞

with initial data 0 (x) := (x, 0), where σ is the shock speed given in (1.12). Define a weight 1 function as x = (1 + x 2 ) 2 and for α ≥ 0, denote L2α (R) the weighted Sobolev space L2α on R 1 with the norm |f |α = ( x α |f |2 dx) 2 . Then we give the second result of present paper in the following. Theorem 1.2. Let (U, V )(x − σ t) be the traveling wave of system (1.22). Let δ ≡ |(u+ − u− , v+ − v− )| be the strength of the traveling wave. Then there exists a positive constant δ0 such that if the initial data 0 satisfies 0 H 2 ∩L2α + v0 − V H 1 ∩L2α + δ ≤ δ0 , the system (1.22) admits a unique global solution (u, v)(x, t) satisfying lim sup |(u, v) − (U, V )| ≤ CEα (1 + t)−

[α] 2

(1.27)

t→∞x∈R

where C is a positive constant independent of time t and δ. The outline of this paper is organized as follows. In section 2, we prove the existence of the traveling wave. In section 3, we show the nonlinear stability of traveling wave and obtain the convergence rate when n = 2 by a weighted estimate. Before closing this part, we give some notations. Notations. Throughout this paper, several positive generic constants which are independent of time t and δ are denoted by C without confusions. For function spaces, H l (R) denotes the l-th order Sobolev space with its norm l  1 j f l = ( ∂x f 2 ) 2 ,

when  ·  :=  · L2 (R) .

j =0

2. Existence of traveling wave t n x−σ t This section is devoted to the existence of traveling wave (U 1( x−σ  ), · · · , U (  )) of system (1.1) with the boundary conditions

lim (U 1 , U 2 , · · · , U n )(ξ ) = (u1± , u2± , · · · , un± ),

ξ →±∞

satisfying (1.11). Let ξ =

x−σ t  ,

then we have

(2.1)

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⎧ ⎪ (λ1 − σ )Uξ1 = f1 (U 1 , U 2 ), ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ (λ2 − σ )Uξ2 = −f1 (U 1 , U 2 ) + f2 (U 2 , U 3 ), ⎪ ⎪ ⎨ ······ ⎪ ⎪ ⎪ ⎪ ⎪ (λn−1 − σ )Uξn−1 = −fn−2 (U n−2 , U n−1 ) + fn−1 (U n−1 , U n ), ⎪ ⎪ ⎪ ⎪ ⎩ (λ − σ )U n = −f (U n−1 , U n ). n n−1 ξ

(2.2)

∂fi i ∂fi i+1 i+1 i Let ∂u i (u+ , u+ ) = −ai , ∂ui+1 (u+ , u+ ) = bi+1 . In terms of assumption (A), ai and bi are

positive constants. Since we only consider the weak traveling wave, that is, ni=1 |ui+ − ui− | is small, we linearize the system (2.2) at (u1+ , · · · , un+ ) and obtain

w = −1 A+ w,

(2.3)

where w = (U 1 − u1+ , U 2 − u2+ , · · · , U n − un+ )T , −1 and A+ are n × n matrices satisfying ⎞ ··· 0 ⎟ ⎜ 1 ··· 0 ⎟ ⎜ 0 λ2 −σ −1 = ⎜ ⎟ ⎝. . . . . . . . . . . . . . . . . . . . . . . .⎠ 0 0 · · · λn1−σ ⎛

1 λ1 −σ

0

and ⎞ b2 0··· 0 0 0 −a1 ⎜ a1 −(a2 + b2 ) b3 · · · 0 0 0 ⎟ ⎟ ⎜ ⎟ ⎜ 0 a −(a + b ) · · · 0 0 2 3 3 ⎟ ⎜ A+ = ⎜ ⎟ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ⎟ ⎜ ⎝ 0 0 0··· an−2 −(an−1 + bn−1 ) bn ⎠ 0 0 0··· 0 an−1 −bn ⎛

with rank(A+ ) = n − 1. Note that from (1.20), −1 has k negative eigenvalues and n − k positive ones. For matrix −1 A+ , we have Lemma 2.1. −1 A+ has at least n − k eigenvalues with non-positive real parts and at least k eigenvalues with non-negative real parts. Moreover in the imaginary axis, there is no other eigenvalue except 0. We use Gershgorin circle theorem to prove Lemma 2.1. For a n × n complex matrix

B= (bi,j ), denote by Di (B) the closed disc in C with center at bi,i and radius Ri (B) = | bi,j |. j =i

We call Di (B) the Gershgorin disc of B. Theorem 2.2 (Gershgorin circle theorem). Let B be an n × n complex matrix, then (i) Each eigenvalue of B lies within at least one of the Gershgorin discs. (ii) If the union of k discs is disjoint from the union of the other n − k discs then the former union contains exactly k and the latter n − k eigenvalues of B.

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Proof of Lemma 2.1. It is easy to check that −1 A+ is similar to A+ −1 , so we just need to discuss the eigenvalues of A+ −1 , here ⎛

−a1 λ1 −σ ⎜ a1 ⎜ λ −σ ⎜ 1

b2 λ2 −σ −(a2 +b2 ) λ2 −σ a2 λ2 −σ

0

···

0

0

b3 λ3 −σ −(a3 +b3 ) λ3 −σ

···

0

0

0

⎞

⎟ 0 ⎟ ⎟ ⎜ ⎟ ··· 0 0 0 ⎟ ⎜ 0 −1 A+  = ⎜ ⎟. ⎜. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .⎟ ⎜ ⎟ −(an−1 +bn−1 ) an−2 bn ⎟ ⎜ 0 0 0 · · · λn−2 −σ λn−1 −σ λn −σ ⎠ ⎝ an−1 −bn 0 0 0 ··· 0 λn−1 −σ λn −σ Let A+ −1 = (ai,j ), D = (di,j ) be the diagonal matrix with di,i = ai,i , B(t) = (1 − t)D + tA+ −1 , 0 ≤ t ≤ 1 and Di (t) = Di (B(t)T ), where B(t)T is the transpose of B(t). By (1.20), ai,i is positive real number for i ≤ k and negative one for i ≥ k + 1. Note that in 1 1 + λ1a−σ = 0. Then A+ −1 , the sum of each column is zero, for instance in the first column λ−a 1 −σ the radius of Di (t) is t|ai,i | with the center at ai,i . Thus for any 0 ≤ t < 1, the disc Di (t) is located at half plane Reξ > 0 for i ≤ k and located at half plane Reξ < 0 for k < i ≤ n. By the Gershgorin circle theorem, we conclude that for any 0 ≤ t < 1, there have n − k eigenvalues of B(t)T with negative real part and k eigenvalues with positive real part. Note that A+ −1 = B(1) and the radius of Di (B(1)T ) is |ai,i | with the center at ai,i . Since ai,i = 0 is located at real axis, the disc Di (B(1)T ) only intersects with imaginary axis at the original point. By the Gershgorin circle theorem, there is no eigenvalue of B(1) at the imaginary axis except the original point. From the fact that rank(B(1)) = n − 1, we conclude that only one eigenvalue can be on the imaginary axis and this eigenvalue is 0. Therefore, the conclusion of Lemma 2.1 follows from the following facts: (i) B(t) has the same eigenvalues with B(t)T , (ii) the i-th eigenvalue μi (t) of B(t) is a continuous function of t , (iii) A+ −1 = B(1), (iv) rank(A+ ) = n − 1. 2 Integrating (2.2) with respect to ξ over (ξ, +∞), we can get that U n (ξ ) =

−(λ1 − σ )U 1 − · · · − (λn−1 − σ )U n−1 + a , λn − σ

(2.4)

where a = (λ1 − σ )u1+ + · · · + (λn − σ )un+ . Then the system (2.2) is changed into the following (n − 1) × (n − 1) ODE system ⎧ ⎪ (λ1 − σ )Uξ1 = f1 (U 1 , U 2 ), ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ (λ2 − σ )Uξ2 = −f1 (U 1 , U 2 ) + f2 (U 2 , U 3 ), ⎪ ⎪ ⎪ ⎪ ⎨ ······ ⎪ n−1 ⎪ (λ ⎪ = −fn−2 (U n−2 , U n−1 ) + fn−1 (U n−1 , U n ), n−1 − σ )Uξ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ −(λ1 − σ )U 1 − · · · − (λn−1 − σ )U n−1 + a ⎪ ⎪ ⎩ U n (ξ ) = . λn − σ

(2.5)

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Linearize (2.5) at (u1+ , · · · , un−1 + ), we have the following linear system w0 = A˜ + w0 ,

(2.6)

T where w0 = (U 1 − u1+ , · · · , U n−1 − un−1 + ) ,

⎛

−a1 λ1 −σ ⎜ a1 ⎜ ⎜ λ2 −σ

b2 λ1 −σ −(a2 +b2 ) λ2 −σ a2 λ3 −σ

0

···

0

0

0

b3 λ2 −σ −(a3 +b3 ) λ3 −σ

···

0

0

0

⎞

⎟ ⎟ ⎟ ⎜ ⎟ ⎜ ⎟ · · · 0 0 0 0 A˜ + = ⎜ ⎟ ⎜. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .⎟ ⎜ ⎟ −(an−2 +bn−2 ) bn−1 ⎟ an−3 ⎜ 0 0 0 · · · λn−2 ⎝ −σ λn−2 −σ λn−2 −σ ⎠ c1 c2 c3 ··· cn−3 cn−2 cn−1 is a (n − 1) × (n − 1) matrix with ⎧ −(λi − σ )bn ⎪ ⎪ , i = 1, · · · , n − 3, ci = ⎪ ⎪ (λ n−1 − σ )(λn − σ ) ⎪ ⎪ ⎪ ⎨ an−2 (λn−2 − σ )bn − , cn−2 = ⎪ λn−1 − σ (λn−1 − σ )(λn − σ ) ⎪ ⎪ ⎪ ⎪ ⎪ −(an−1 + bn−1 ) bn ⎪ ⎩ cn−1 = − . λn−1 − σ λn − σ For the matrix −1 A+ and matrix A˜ + , we have Lemma 2.3. −1 A+ ∼



A˜ + 0

lT 0

 ,

bn where l = (0, · · · , 0, λn−1 −σ ).

Proof. Let ⎞ 1 0 0 ··· 0 0 0 ⎜ 0 1 0 ··· 0 0 0⎟ ⎟ ⎜ ⎜ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .⎟ P =⎜ ⎟, ⎝ 0 0 0 ··· 0 1 0⎠ λn−1 −σ −σ λ3 −σ λ2 −σ λ1 −σ · · · λλn−2 1 λn −σ λn −σ λn −σ λn −σ n −σ ⎛

which gives that P

−1

A+ P

−1



A˜ + = 0

lT 0



We are now ready to give the following proposition of A˜ + .

.

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Proposition 2.4. Under the assumption (A), (1.10) and (1.11), the matrix A˜ + has n − k numbers of eigenvalues with negative real part. Proof of Proposition 2.4. By Lemma 2.1 and the fact that rank(A+ ) = n − 1, −1 A+ has only one zero eigenvalue, at least n − k − 1 eigenvalues with negative real part and at least k − 1 ones with positive real part. Moreover, by Lemma 2.3, all the non-zero eigenvalues of −1 A+ and A˜ + are the same. So the matrix A˜ + has at least n − k − 1 eigenvalues with negative real part and at least k − 1 eigenvalues with positive real part. On the other hand, a direct calculation yields that |A˜ + | =

(−1)n−k−1 η , |λ1 − σ | · · · |λn − σ |

(2.7)

where η = (λn − σ )a1 a2 · · · an−1 + (λn−1 − σ )a1 a2 · · · an−2 bn + · · · + (λ1 − σ )b2 · · · bn . By the entropy condition (1.18), we have (λ1 − σ ) + (λ2 − σ )

a1 an−1 a1 + · · · + (λn − σ ) ··· < 0, b2 bn b2

(2.8)

which is equivalent to η < 0. Then we can get that (1) if n − k − 1 is odd, |A˜ + | > 0, then A˜ + has n − k eigenvalues with negative real part; (2) if n − k − 1 is even, |A˜ + | < 0, then A˜ + has n − k eigenvalues with negative real part.

2

In the same way, linearize the system (2.5) at (u1− , · · · , un−1 − ), we have w0 = A˜ − w0 ,

(2.9)

T 1 ˜ where w0 = (U 1 − u1− , · · · , U n−1 − un−1 − ) and A− is the corresponding matrix at (u− , · · · , n−1 T u− ) . As in Proposition 2.4, we can obtain

Proposition 2.5. Under the assumption (A), (1.10) and (1.11), the matrix A˜ − has k numbers of eigenvalues with positive real part. Proof of Theorem 1.1. By Proposition 2.4, the matrix A˜ + has n − k eigenvalues with negative real part. Thus there exists a stable manifold M+ ⊂ R n−1 with dimM+ = n − k in a small neighborhood of (u1+ , · · · , un−1 + ) for the nonlinear system (2.5). Similarly, from Proposition 2.5, there exists a unstable manifold M− ⊂ R n−1 with dimM− = k in a small neighborhood of (u1− , · · · , un−1 − ) for the system (2.5). Since the manifold M− continuously depends on n−1 1 (u− , · · · , u− ), the radius of the manifold M− is almost the same when (u1− , · · · , un−1 − ) is close  ). Thanks to dimM + dimM = n > n − 1, we have M =  ∅. Therefore M to (u1+ , · · · , un−1 + − + − + there exists a unique traveling wave of system (1.1) connecting (u1− , · · · , un− ) and (u1+ , · · · , un+ ) satisfying (2.1) up to a shift. 2 3. Convergence rate toward the traveling wave This section is devoted to the stability of traveling wave with convergence rate for a special case n = 2 by a weighted estimate introduced in [10]. As discussed in (1.21)–(1.23), we only need to consider the system (1.22), that is,

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11

∂t u = f (u, v)

(3.1)

(∂t + ∂x )v(x, t) = −f (u, v). By (1.10) and (1.11), we have Lemma 3.1. Let (U, V )(x − σ t) be the traveling wave of (3.1), then U  < 0.

Proof. As in (1.6), there exists h(u) such that f (u, h(u)) = 0 and w = u +h(u). Then from (1.7), we have F  (w) =

h (u) h (u) and F  (w) =  1 + h (u) (1 + h (u))3

(3.2)

which together with (1.10) implies h (u) > 0. On the other hand, we can get 1 σU − a U  = − f (U, ), σ 1−σ

(3.3)

where a = σ u− − (1 − σ )v− , v± = h(u± ). Because h is convex and u+ < u− , σU − a h(u+ ) − h(u− ) (U − u− ) > h(U ), = h(u− ) + 1−σ u+ − u− where

σ 1−σ

=

h(u+ )−h(u− ) . u+ −u−

(3.4)

Note that fv > 0, we obtain

1 σU − a 1 U  = − f (U, ) < − f (U, h(U )) = 0. σ 1−σ σ

2

(3.5)

Now let G(x) = (1 − σ )f u + σ f v , f = f (U, V ), f u = fu (U, V ), f v = fv (U, V ). By f (u, h(u)) = 0 and h > 0, a direct calculation gives    fuu fuv 1   fv h = −(1, h ) > 0, fuv fvv h which, together with the fact that |u+ − u− | << 1, implies G (x) = ((1 − σ )f uu + 2σ f uv +

σ2 f )Ux > 0. 1 − σ vv

(3.6)

On the other hand, the equality f (u, h(u)) = 0 yields that fu (u± , v± ) + fv (u± , v± )h (u± ) = 0,

(3.7)

which, from u+ < u− and h > 0, gives G(−∞) = (1 − σ )(fu (u− , v− ) + fv (u− , v− )

h(u+ ) − h(u− ) ) u+ − u−

< (1 − σ )(fu (u− , v− ) + fv (u− , v− )h (u− )) = 0. Similarly we have G(+∞) > 0. So there exists a unique point x∗ such that G(x∗ ) = 0.

(3.8)

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Lemma 3.2. There is a positive constant C0 such that for any β ∈ [0, α], 1 x − x∗ Fβ (x) = {G x − x∗ + βG(x) } ≥ βC0 . 2 x − x∗

(3.9)

Proof. From (3.6), G(x) is an increasing function of x ∈ R and G(x∗ ) = 0, G (x∗ ) > 0, so  1  x − x∗ G (x∗ )(x − x∗ )2 , G(x) ≥ 2 x − x∗ c,

|x − x∗ | << 1, otherwise,

where c is a positive constant. On the other hand, x − x∗ G (x) > Thus the lemma is proved. 2

G (x ∗ ) 2

when |x − x∗ | << 1.

Define w = u + v, then the system (3.1) can be re written as 

wt + (w − u)x = 0, ut = f (u, w − u).

(3.10)

We change the coordinates (x, t) to (ξ = x − σ t, t) (still denoted by (x, t)), then we have 

wt + (1 − σ )wx − ux = 0, ut − σ ux = f (u, w − u),

(3.11)

with the following initial data (w, u)(x, 0) = (w0 , u0 )(x) = (u0 + v0 , u0 ).

(3.12)

Without loss of generality, we assume the initial data w0 satisfies  (w0 − W )(x)dx = 0,

(3.13)

R

where W = U + V . Let x (x, t) =

w(y, t) − W (y)dy,  = u(x, t) − U (x).

(3.14)

−∞

Substituting the above equality into (3.11), we get 

xt + (1 − σ )xx − x = 0, t − σ x = f (u, w − u) − f (U, W − U ),

(3.15)

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13

which gives  = t + (1 − σ )x .

(3.16)

Then we obtain a wave equation tt + (1 − 2σ )xt − σ (1 − σ )xx = f ( + U, x −  + W − U ) − f (U, W − U )

(3.17)

with the corresponding initial data ⎧ x ⎪ ⎪ ⎪ ⎨ (x, 0) = 0 = (w0 − W )(y)dy, ⎪ ⎪ ⎪ ⎩

−∞

(3.18)

t (x, 0) = t0 = 0 − (1 − σ )0x , 0 = u0 − U (x).

We will seek the solution in the following space X(0, T ) = { ∈ C([0, T ]; H 2 ) ∩ C 1 ([0, T ]; H 1 ), (x , t ) ∈ L2 ([0, T ]; H 1 )}. Theorem 3.3. Suppose that 0 ∈ H 2 ∩ L2α , (0x , 0t ) ∈ H 1 ∩ L2α for some α ≥ 0. Then there exist positive constants δ0 and C such that if Eα + δ ≤ δ0 , the Cauchy problem (3.17) and (3.18) has a unique solution  ∈ X(0, ∞) satisfying t (1 + t)

γ

(2H 2

+ t 2H 1 ) +

(1 + τ )γ x , t 2H 1 dτ ≤ CEα2 ,

(3.19)

0

where Eα = 0 , 0t , 0x α + 0 H 2 + 0t H 1 and 0 ≤ γ ≤ [α]. Proof of Theorem 1.2. Once we have Theorem 3.3, then (1.27) can be directly derived from (3.19) by Sobolev’s inequality. 2 The local existence of solution can be proved by fixed point theorem and the proof is omitted. To prove Theorem 3.3, it remains to close the following a priori estimates. Proposition 3.4 (A priori estimates). Let T be any positive constant. Suppose the problem (3.17) and (3.18) has a solution  ∈ X(0, T ) satisfying (, t , x ) ∈ C([0, T ]; L2α ), (t , x ) ∈ L2 ([0, T ]; L2α ) for some α ≥ 0. Let δ = |w+ − w− | be the strength of traveling wave and N (T ) = sup (H 2 , t H 1 ) t∈[0,T ]

then there exists a small positive constant δ0 such that if N (T ) + δ ≤ δ0 , it holds that for t ∈ [0, T ],

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14

t (1 + t)

γ

(2H 2

+ t 2H 1 ) +

(1 + τ )γ  x , t 2H 1 dτ ≤ CEα2 ,

(3.20)

0

where 0 ≤ γ ≤ [α], C is a constant independent on T and γ . Proposition 3.4 follows from some lemmas below. Lemma 3.5. For any β, γ ∈ [0, α], there is a positive constant C independent of T , β and γ such that t (1 + t)

γ

 , x , t 2β

+

(1 + τ )γ  x , t 2β dτ 0

t +β

(1 + τ )γ   2β−1 dτ 0

t ≤ C{0 , 0x , 0t 2β

+γ

(1 + τ )γ −1 , x , t 2β dτ

(3.21)

0

t +β

t (1 + τ ) x , t  dτ + (N (T ) + δ) γ

(1 + τ )γ x , t 2β dτ }.

2

0

0

Proof. Note that f (ψ + U, x − ψ + W − U ) − f (U, W − U ) = (f u − f v )t + ((1 − σ )f u + σ f v )x + O(2x + ψ 2 ).

(3.22)

The proof is divided into two steps. Step 1: Multiply (3.17) by (1 + t)γ x − x∗ β , we have d 1 { (f − f u )(1 + t)γ x − x ∗ β 2 + (1 + t)γ x − x ∗ β t } dt 2 v + σ (1 − σ )(1 + t)γ x − x ∗ β 2x + (1 + t)γ x − x ∗ β−1 Fβ (x)2 = (1 + t)γ x − x ∗ β 2t + γ (1 + t)γ −1 x − x ∗ β t + β(1 − 2σ )(1 + t)γ x − x ∗ β−2 (x − x ∗ )t + (1 − 2σ )(1 + t)γ x − x ∗ β x t − βσ (1 − σ )(1 + t)γ x − x ∗ β−2 (x − x ∗ )x γ − (f u − f v )(1 + t)γ −1 x − x ∗ β 2 + (1 + t)γ x − x ∗ β O(2x +  2 ) 2 + {· · ·}x

(3.23)

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15

where {· · ·}x denotes the term which disappears after integrating with respect to x. Integrate (3.23) over R × (0, t), we get t 2β

(1 + t)

γ

+

t (1 + τ )

γ

x 2β dτ

+β

(1 + τ )γ 2β−1 dτ

0

0

t ≤ C1 {0 , 0t 2β

+ (1 + t)

γ

t 2β

+

t (1 + τ )

γ

t 2β dτ

+γ

0

t 

0

t 

(1 + τ )γ −1 x − x ∗ β |t |dxdτ + β

+γ 0 R

t  +

(1 + τ )γ −1 2β dτ

(1 + τ )γ x − x ∗ β−1 |t |dxdτ

0 R

(1 + τ )γ x − x ∗ β |x t |dxdτ + β

0 R

t 

(1 + τ )γ x − x ∗ β−1 |x |dxdτ

0 R

t + N(T )

(1 + τ )γ t , x 2β dτ },

(3.24)

0

where we have used Lemma 3.2 and the fact that ∞ ≤ C0 N (t). By using the Cauchy inequality, it is straightforward to check that t 

(1 + τ )γ x − x ∗ β |x t |dxdτ

C1 0 R t

1 ≤ 4

(1 + τ )

γ

x 2β dτ

+ C˜ 2

0

t (1 + τ )γ t 2β dτ.

(3.25)

0

Then inequality (3.24) is changed to t (1 + t)

γ

2β

+

t (1 + τ )

γ

x 2β dτ

+β

0

(1 + τ )γ 2β−1 dτ 0

t ≤ C1 {0 , 0t 2β

+ (1 + t)

γ

t 2β

+

t (1 + τ )

γ

t 2β dτ

0

t  +β

(1 + τ )γ x − x ∗ β−1 |t |dxdτ + β

0 R

+γ

(1 + τ )γ −1 , t 2β dτ

0

t 

(1 + τ )γ x − x ∗ β−1 |x |dxdτ

0 R

t + N(T )

(1 + τ )γ x , t 2β dτ }. 0

(3.26)

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16

Step 2: Multiply (3.17) by (1 + t)γ x − x∗ β t , we have d 1 1 { (1 + t)γ x − x ∗ β 2t + σ (1 − σ )(1 + t)γ x − x ∗ β 2x } dt 2 2 + (f v − f u )(1 + t)γ x − x ∗ β 2t γ 1 (1 + t)γ −1 x − x ∗ β 2t + β(1 − 2σ )(1 + t)γ x − x ∗ β−2 (x − x ∗ )2t 2 2 γ + σ (1 − σ )(1 + t)γ −1 x − x ∗ β 2x − βσ (1 − σ )(1 + t)γ x − x ∗ β−2 (x − x ∗ )t x 2

=

((1 − σ )f u + σ f v )(1 + t)γ x − x ∗ β t x + (1 + t)γ x − x ∗ β O(2x +  2 )t + {· · ·}x .

(3.27)

Since δ =| w+ − w− | 1, we have |

σ − h (u+ ) |= O(1)δ 1−σ

which together with f (u, h(u)) = 0, yields |(1 − σ )f u + σ f v | σ = (1 − σ )|f u + f − fu (u+ , v+ ) − fv (u+ , v+ )h (u+ )| 1−σ v = O(1)δ.

(3.28)

Thus integrate (3.27) over R × (0, t), we get t (1 + t)

γ

x , t 2β

+

(1 + τ )γ t 2β dτ 0

t ≤ C2 {Eβ2

+γ

γ −1

(1 + τ )

t  t , x 2β dτ

+β 0 R

0

t



+β

(1 + τ )γ x − x ∗ β−1 |t |2 dxdτ

(1 + τ )γ x − x ∗ β−1 |x t |dxdτ

0 R

t (1 + τ )γ t , x 2β dτ }.

+ (N (T ) + δ)

(3.29)

0

Multiply (3.29) by a large constant C0 , and add the result to the inequality (3.26), we can obtain

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17

t (1 + t)

γ

 , x , t 2β

+

(1 + τ )γ  x , t 2β dτ 0

t +β

(1 + τ )γ   2β−1 dτ 0

t ≤ C3 {0 , 0x , 0t  + γ 2

(1 + τ )γ −1 , x , t 2β dτ

(3.30)

0

t



+β

∗ β−1

(1 + τ ) x − x γ

t  |t | dxdτ + β 2

0 R

0 R

t  +β

(1 + τ )γ x − x ∗ β−1 |x t |dxdτ

∗ β−1

(1 + τ ) x − x γ

t  |x |dxdτ + β

0 R

(1 + τ )γ x − x ∗ β−1 |t |dxdτ

0 R

t + (N (T ) + δ)

(1 + τ )γ x , t 2β dτ }.

(3.31)

0

Now we are ready to estimate the right hand side of (3.30). We have t  C3 β

(1 + τ )γ x − x ∗ β−1 |x |dxdτ

0 R

β ≤ 4

t (1 + τ )

γ

2β−1 dτ

+ C˜ 3 β

t (1 + τ )

γ

2β−1 dτ

+ C˜ 3 β

t



(1 + τ )γ x − x ∗ β−1 2x dxdτ

0 |x−x ∗ |≤A

0

+ C˜ 3 β

(1 + τ )γ x − x ∗ β−1 2x dxdτ

0 R

0

β ≤ 4

t 

t



(1 + τ )γ x − x ∗ β−1 2x dxdτ

0 |x−x ∗ |≥A

β ≤ 4

t (1 + τ )

γ

2β−1 dτ

0

+

1 4

+ C˜ 3 Aβ−1 β

t (1 + τ )γ x 2 dτ 0

t (1 + τ )γ x 2β dτ 0

(3.32)

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18

where we choose A large such that C˜ 3 β x − x ∗ −1 ≤ obtain t  C3 β

1 4

when |x − x ∗ | ≥ A. Similarly, we can

(1 + τ )γ x − x ∗ β−1 |t |dxdτ

0 R

β ≤ 4

t (1 + τ )

γ

2β−1

+ C˜ 3 A

β−1

t

0

+

1 4

(1 + τ )γ t 2 dτ

β 0

t (1 + τ )γ t 2β dτ

(3.33)

0

and t  C3 β

(1 + τ )γ x − x ∗ β−1 |x t |dxdτ

0 R

t 

≤ C˜ 3 β

(1 + τ )γ x − x ∗ β−1 (|x |2 + |t |2 )dxdτ

0 R

≤ C˜ 3 Aβ−1 β

t

1 (1 + τ ) x , t  dτ + 4

t (1 + τ )γ x , t 2β dτ.

2

γ

0

(3.34)

0

Inserting the above inequalities into (3.30) yields Lemma 3.5.

2

Lemma 3.6. For any γ ∈ [0, α], there is a positive constant C independent on T and γ such that t (1 + t)

γ

(x 2H 1

+ xt  ) +

(1 + τ )γ xx , xt 2 dτ

2

0

t ≤ C{0x 2H 1

+ 0xt  + γ 2

(1 + τ )γ −1 (x 2H 1 + xt 2 )dτ

0

t +

(1 + τ )γ x , t 2 dτ }.

(3.35)

0

Proof. Step 1: Differentiate (3.17) with respect to x and multiply the result by (1 + t)γ x , we obtain

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19

d {(1 + t)γ x xt } − (1 + t)γ 2xt − γ (1 + t)γ −1 x xt dt − (1 − 2σ )(1 + t)γ xx xt + σ (1 − σ )(1 + t)γ 2xx = (f (u, w − u) − f (U, W − U ))x (1 + t)γ x + {· · · }x .

(3.36)

A direct calculation gives (f (u, w − u) − f (U, W − U ))x (1 + t)γ x d 1 γ { (f − f u )(1 + t)γ 2x } + (f v − f u )(1 + t)γ −1 2x dt 2 v 2 1 − (f v − f u )x (1 + t)γ t x + (1 + t)γ ((1 − σ )f u + σ f v )x 2x 2

=−

+ O(2x +  2 )xx + {· · · }x .

(3.37)

Substitute (3.37) into (3.36) and integrate the results over R × [0, t], we obtain t (1 + t)  x  +

(1 + τ )γ xx 2 dτ

2

γ

0

t ≤ C4 {0x , 0xt  + (1 + t) xt  + 2

(1 + τ )γ xt 2 dτ

2

γ

0

t +γ

(1 + τ )γ −1 x , xt 2 dτ +

0

t (1 + τ )γ x , t 2 dτ } 0

t + N (T )

(1 + τ )γ xx 2 dτ.

(3.38)

0

Step 2: Differentiate (3.17) with respect to x and multiply the result by (1 + t)γ xt , we have t (1 + t) xx , xt  +

(1 + τ )γ xt 2 dτ

2

γ

0

t ≤ C4 {0xx , 0xt  + +γ 2

(1 + τ )γ −1 xx , xt 2 dτ

0

t +

t (1 + τ ) t , x  dτ } + (N (T ) + δ) γ

0

(1 + τ )γ xx , xt 2 dτ.

2

0

(3.39)

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20

Multiply (3.39) by a large constant C0 and add the result to the inequality (3.38), the lemma is proved. 2 Let β = 0, γ = 0 in Lemma 3.5 and Lemma 3.6, we have Lemma 3.7. There exists a positive constant C independent on T such that if N (T ) + δ ≤ δ0 , Proposition 3.4 holds with γ = α = 0 for t ∈ [0, T ], t 2H 2

+ t 2H 1

+

x , t 2H 1 dτ ≤ CE02 .

(3.40)

0

Lemma 3.8. Let γ ∈ [0, α] ∩ Z, there exists a positive constant C independent on T and γ such that if N(T ) + δ ≤ δ0 , then t (1 + t)

γ

, t , x 2α−γ

+

(1 + τ )γ x , t 2α−γ dτ 0

t + (α − γ )

(1 + τ )γ 2α−γ −1 dτ ≤ CEα2 .

(3.41)

0

Moreover for any 0 ≤ γ ≤ [α], t (1 + t)

γ

(2H 1

+ t  ) +

(1 + τ )γ x , t 2 dτ ≤ CEα2 .

2

(3.42)

0

Proof. Following [10], let β = α, γ = 0 in inequality (3.21), we have inequality (3.41) with γ = 0. Let β = 0, γ = 1 and use inequality (3.41) with γ = 0, we have inequality (3.42) with γ = 1. Repeating the same procedure, the lemma is complete. 2 Similarly, we can get Lemma 3.9. Let γ ∈ [0, α] ∩ Z, there exists a positive constant C independent on T and γ such that if N (T ) + δ ≤ δ0 , then t (1 + t)

γ

(x 2H 1

+ xt  ) +

(1 + τ )γ xx , xt 2 dτ ≤ CEα2 .

2

(3.43)

0

Proof of Proposition 3.4. Proposition 3.4 is complete due to Lemma 3.7, 3.8 and 3.9.

2

Acknowledgment The authors are very appreciated for the anonymous referee whose suggestions greatly improve the current manuscript.

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