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Breast cancer chemical structures and their partition resolvability


  • Cancer is a disease that causes abnormal cell formation and spreads throughout the body, causing harm to other organs. Breast cancer is the most common kind among many of cancers worldwide. Breast cancer affects women due to hormonal changes or genetic mutations in DNA. Breast cancer is one of the primary causes of cancer worldwide and the second biggest cause of cancer-related deaths in women. Metastasis development is primarily linked to mortality. Therefore, it is crucial for public health that the mechanisms involved in metastasis formation are identified. Pollution and the chemical environment are among the risk factors that are being indicated as impacting the signaling pathways involved in the construction and growth of metastatic tumor cells. Due to the high risk of mortality of breast cancer, breast cancer is potentially fatal, more research is required to tackle the deadliest disease. We considered different drug structures as chemical graphs in this research and computed the partition dimension. This can help to understand the chemical structure of various cancer drugs and develop formulation more efficiently.

    Citation: Qingqun Huang, Adnan Khalil, Didar Abdulkhaleq Ali, Ali Ahmad, Ricai Luo, Muhammad Azeem. Breast cancer chemical structures and their partition resolvability[J]. Mathematical Biosciences and Engineering, 2023, 20(2): 3838-3853. doi: 10.3934/mbe.2023180

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  • Cancer is a disease that causes abnormal cell formation and spreads throughout the body, causing harm to other organs. Breast cancer is the most common kind among many of cancers worldwide. Breast cancer affects women due to hormonal changes or genetic mutations in DNA. Breast cancer is one of the primary causes of cancer worldwide and the second biggest cause of cancer-related deaths in women. Metastasis development is primarily linked to mortality. Therefore, it is crucial for public health that the mechanisms involved in metastasis formation are identified. Pollution and the chemical environment are among the risk factors that are being indicated as impacting the signaling pathways involved in the construction and growth of metastatic tumor cells. Due to the high risk of mortality of breast cancer, breast cancer is potentially fatal, more research is required to tackle the deadliest disease. We considered different drug structures as chemical graphs in this research and computed the partition dimension. This can help to understand the chemical structure of various cancer drugs and develop formulation more efficiently.



    In the present paper, we focus on the nonlocal dispersal equation with spatially dependent bistable nonlinearity

    ut(x,t)=Ju(x,t)u(x,t)+f(x,u),x,tR, (1)

    where the nonlinearity f(x,u) and kernel function J(x) satisfy the following assumptions.

    (F) The nonlinearity f satisfies that

    {f(x,u)=f1(u)forx0,u[0,1],f(x,u)=f2(u)forxL,u[0,1],f2(u)f(x,u)f1(u),fu(x,u)<1forx[L,0],u[0,1], (2)

    where L>0, f1 and f2 are two given bistable nonlinearities satisfying that

    fiC1,1([0,1]),fi(0)=fi(1)=0,fi(0),fi(1)<0,fi<0in(0,θi),fi>0in(θi,1),0<θ1<θ2<1,and10fi(s)ds>0,i=1,2.

    (J) The kernel function JC1(R) satisfies that

    {J(x)=J(x),J(x)0,RJ(y)dy=1,R|J(y)|dy<,RJ(y)eλydy<for allλ>0. (3)

    A typical example is f(x,u)=u(1u)(a(x)u).

    The nonlocal dispersal equation has attracted so much attention because of its extensive applications to account for diffusion phenomena involving jumps in biology, physics and chemistry [1,12]. Traveling waves as one kind of special solutions with invariant profile and fixed speed as well as entire solutions have been adequately investigated. It is well-known that the importance of the study of entire solutions of reaction-diffusion equations (nonlocal dispersal equations) is frequently recalled in the literature. Since the pioneering works of Hamel and Nadirashvili [13,14], there have been tremendous advances in studying the existence of entire solutions for various models. In particular, when the nonlinearity is homogeneous (i.e. L=0 and f1=f2), the theories of traveling waves and entire solutions for nonlocal dispersal equation 1 with various types of nonlinearities have been well established, the related results refer to [2,6,7,8,9,15,18,23,24,26,27] and references therein. Specifically, when the kernel J(x) is compactly supported, Sun et al. [23] constructed a two-dimensional manifold of entire solutions which behave as two traveling wave solutions coming from both directions for bistable nonlocal dispersal equation.

    However, for the inhomogeneous nonlinearity, several works are devoted to transition fronts (see [3,19]) of nonlocal dispersal equations [17,20] and forced waves in shifting habitats [25]. Traveling waves and spreading speed of monostable nonlocal dispersal equations with space periodic nonlinearity were studied in [21]. Li et al. [16] further obtained the existence of entire solutions for space periodic nonlinearity. Particularly, Eberle [10,11] constructed a heteroclinic orbit connecting two traveling waves for bistable local dispersal equation 1 in cylinders. Meanwhile, Berestycki and Rodríguez [5] considered a bistable nonlocal dispersal equation with a gap in one dimension. As far as we know, there is no result on entire solutions for the nonlocal dispersal equation 1 with inhomogeneous nonlinearity f(x,u) satisfying (F).

    In this paper, we aim to construct an entire solution connecting traveling waves with the two different nonlinearities, motivated by [10,11]. By constructing suitable sub- and super-solutions, we establish the existence and uniqueness of the entire solution behaving as the traveling wave coming from one side and eventually going to the other side, which is different with the one constructed in [23] for the case L=0 and f2=f1. It should be pointed out that there are many differences to [4], though the similar method is applied. More precisely, since the nonlocal operator is not a compact operator as the Laplacian operator in [4], we have to establish the Lipschitz continuity of entire solutions in order to prove its uniqueness as in [23]. Compared to [23], we drop the assumption the kernel J(x) is compactly supported.

    The crucial part of this paper is to figure out the long time asymptotic behavior of the entire solution. Since the lack of compactness of the nonlocal operator, we can not use the Lyapunov function argument as [10,11] to show that the entire solution converges to a translation of the other traveling wave as time goes to positive infinity. However, inspired by the idea in [7], we use a "squeezing" technique to address this issue. Furthermore, we also apply sub- and super-solutions method with comparison principle to establish Lyapunov stability of the entire solution. This can be done because that we can obtain a positive estimate on the derivative of the entire solution with respect to t when it is not so close to 0 as well as 1 and t is large enough.

    Now we state the main results of this paper as follows.

    Theorem 1.1. Let assumptions (F) and (J) hold. Then there exists a unique entire solution u(x,t) of 1 with 0<u(x,t)<1, ut(x,t)>0 for all tR and xR such that

    u(x,t)ϕ1(x+c1t)0astuniformly inxR

    and

    u(x,t)ϕ2(x+c2t+β)0ast+uniformly inxR

    for some βR, where (ϕi,ci)(i=1,2) are the traveling wave solutions solving

    {Jϕiϕi+fi(ϕi)ciϕi=0,ϕi()=0,ϕi(+)=1,0<ϕi<1. (4)

    Theorem 1.2. The entire solution u(x,t) constructed in Theorem 1.1 is Lyapunov stable in the following sense: For any given ϵ>0, there exists δ>0 such that for any uniformly continuous v0(x)[0,1] with supxR|v0(x)u(x+a,t0)|<δ, the solution v(x,t,v0(x)) of 1 with initial value v0(x) satisfies

    |v(x,t,v0)u(x+a,t+t0)|<ϵ

    for xR and t0, where a,t0R are two constants.

    The rest of this paper is organized as follows. In Section 2, we recall some results of the bistable traveling waves for homogeneous nonlinearities and prove the comparison principle for 1. Section 3 is devoted to constructing the unique entire solution. In Section 4, we study the asymptotic profile as time goes to positive infinity. Finally, we establish Lyapunov stability of the entire solution in Section 5.

    In this section, some known results on the traveling waves of 4 are outlined and the comparison principle is established.

    It follows from Theorem 2.7 in [2] and Theorem 2.7 in [23] that 4 admits a solution ϕ1(z) satisfying

    {β0eλ0zϕ1(z)α0eλ0z,z0,β1eλ1z1ϕ1(z)α1eλ1z,z>0, (5)

    where α0,α1,β0 and β1 are positive constants, λ0 and λ1 are the positive roots of

    cλ0=RJ1(y)eλ0ydy1+f1(0)andcλ1=RJ1(y)eλ1ydy1+f1(1),

    respectively. Moreover, we have

    {˜β0eλ0zϕ1(z)˜α0eλ0z,z0,˜β1eλ1zϕ1(z)˜α1eλ1z,z>0 (6)

    for some constants ˜α0,˜α1,˜β0 and ˜β1>0. At last, note that f1C1,1([0,1]), there exists some Lf>0 such that

    |f1(u+v)f1(u)f1(v)|Lfuvfor0u,v1.

    We show that the following comparison principle holds by a contradiction argument.

    Proposition 1. Suppose that assumptions (F) and (J) hold. Furthermore, let u(x,t), v(x,t) be continuous and bounded functions on R×[0,T] for some T>0 satisfy

    {u(x,t)t(RJ(xy)[u(y,t)u(x,t)]dy)f(x,u(x,t))0,(x,t)R×(0,T],u(x,0)0,xR,
    {v(x,t)t(RJ(xy)[v(y,t)v(x,t)]dy)f(x,v(x,t))0,(x,t)R×(0,T],v(x,0)0,xR,

    respectively. Then,

    u(x,t)v(x,t)inR×[0,T].

    Furthermore, if u(x,0)v(x,0) for xR, then u(x,t)>v(x,t) for xR,t[0,T].

    Proof of Proposition 1. Let ˉw(x,t)=u(x,t)v(x,t). In fact, it is sufficient to show ˉw(x,t)0 for (x,t)R×[0,ϵ0T] with ϵ0(0,1). Otherwise, suppose that infR×[0,ϵ0T]ˉw(x,t)<0. Denote ˇw(x,t)=eZtˉw(x,t), where Z=fu(x,u)+1. It follows that infR×[0,ϵ0T]ˇw(x,t)<0. Then there exists a sequence (xn,tn)R×(0,ϵ0T] such that

    limn+ˇw(xn,tn)=infR×[0,ϵ0T]ˇw(x,t)<0.

    Observe that

    ˇwt(x,t)=Zˇw(x,t)+eZtˉwt(x,t)Zˇw(x,t)+eZtRJ(xy)[ˉw(y,t)ˉw(x,t)]dy+eZt[f(x,u(x,t))f(x,v(x,t))]=Zˇw(x,t)+RJ(xy)[ˇw(y,t)ˇw(x,t)]dy+fu(x,uθ(x,t))ˇw(x,t).

    where uθ(x,t) is between v(x,t) and u(x,t). This implies that

    ˇw(xn,tn)ˇw(xn,0)tn0[Jˇw(xn,s)ˇw(xn,s)+Zˇw(xn,s)+fu(x,uθ(x,s))ˇw(xn,s)]dstn0[Jˇw(xn,s)+(fu(x,u)fu(x,uθ(x,s)))infR×[0,ϵ0T]ˇw(x,t)]ds.

    Letting n converge to infinity, we have

    infR×[0,ϵ0T]ˇw(x,t)(1+fu(x,u)minR×[0,1]|fu(x,uθ(x,s))|)ϵ0TinfR×[0,ϵ0T]ˇw(x,t).

    Since infR×[0,ϵ0T]ˇw(x,t)<0, we can choose ϵ0 sufficiently small such that

    infR×[0,ϵ0T]ˇw(x,t)(1+fu(x,u)minR×[0,1]|fu(x,uθ(x,s))|)ϵ0TinfR×[0,ϵ0T]ˇw(x,t)>infR×[0,ϵ0T]ˇw(x,t).

    Thus, we get a contradiction. Therefore, we obtain u(x,t)v(x,t) for xR and t[0,T]. The remaining part of this proposition can be proved similarly by replacing the auxiliary function ˇw(x,t)=eZtˉw(x,t) with ˇw(x,t)=eZtˉw(x,t)ϵt for small ϵ>0. In fact, we can similarly obtain

    infR×[0,ϵ0T]ˇw(x,t)(1+fu(x,u)minR×[0,1]|fu(x,uθ(x,s))|ϵinfR×[0,ϵ0T]ˇw(x,t))ϵ0TinfR×[0,ϵ0T]ˇw(x,t).

    Here, we can also choose ϵ0 such that

    (1+fu(x,u)minR×[0,1]|fu(x,uθ(x,s))|ϵinfR×[0,ϵ0T]ˇw(x,t))ϵ0T<1.

    This finishes the proof.

    In this section, we focus on the construction of the entire solution which behaves like a traveling wave approaching from infinity. The main idea is to establish suitable sub- and super-solutions, which are defined as follows.

    W(x,t)={ϕ1(x+c1tξ(t))ϕ1(x+c1tξ(t)),x0,0,x<0,

    and

    W+(x,t)={ϕ1(x+c1t+ξ(t))+ϕ1(x+c1t+ξ(t)),x0,2ϕ1(c1t+ξ(t)),x<0,

    here ξ(t) is the solution of the following equation

    ˙ξ(t)=Meλ(c1t+ξ),t<T,ξ()=0,

    where M,λ and T are positive constants to be specified later. A direct calculation yields that

    ξ(t)=1λln11c11Meλc1t.

    For the function ξ(t) to be defined, one must have 1c11Meλc1t>0. Besides, there is

    c1t+ξ(t)0for<tT,

    where T:=1λc1lnc1+Mc1.

    Now we verify that W are sub- and super-solutions of 1 for tT. Define

    Lu=ut(Juu)f(x,u).

    We first deal with the sub-solution W(x,t). Since it is obvious for x<0, we only consider the case x0. A straightforward calculation yields that

    LW(x,t)=(c1˙ξ(t))[ϕ1(x+c1tξ(t))ϕ1(x+c1tξ(t))]+0J(xy)[ϕ1(y+c1tξ(t))ϕ1(y+c1tξ(t))]dy+[ϕ1(x+c1tξ(t))ϕ1(x+c1tξ(t))]f(x,W)=˙ξ(t)[ϕ1(x+c1tξ(t))ϕ1(x+c1tξ(t))]+0J(xy)[ϕ1(y+c1tξ(t))ϕ1(y+c1tξ(t))]dy+f1(ϕ1(x+c1tξ(t)))f1(ϕ1(x+c1tξ(t)))f(x,W).

    Recall that f(x,u)=f1(u) for x0 and ϕ1>0, it follows that

    LW(x,t)˙ξ(t)[ϕ1(x+c1tξ(t))ϕ1(x+c1tξ(t))]+f1(ϕ1(x+c1tξ(t)))f1(ϕ1(x+c1tξ(t)))f1(ϕ1(x+c1tξ(t))ϕ1(x+c1tξ(t))).

    Then we continue to show LW(x,t)0 in two cases 0<x<c1t+ξ(t) and xc1t+ξ(t).

    Case 1. For 0<x<c1t+ξ(t), similar to [4], we have

    ϕ1(x+c1tξ(t))ϕ1(x+c1tξ(t))>m[ϕ1(x+c1tξ(t))ϕ1(x+c1tξ(t))]

    for some m>0. Consequently,

    LW(x,t)˙ξ(t)[ϕ1(x+c1tξ(t))ϕ1(x+c1tξ(t))]+Lfϕ1(x+c1tξ(t)))[ϕ1(x+c1tξ(t))ϕ1(x+c1tξ(t))][ϕ1(x+c1tξ(t))ϕ1(x+c1tξ(t))][mMeλ(c1t+ξ(t))+Lfα0eλ0(x+c1tξ(t))]=[ϕ1(x+c1tξ(t))ϕ1(x+c1tξ(t))]eλ(c1t+ξ(t))[mM+Lfα0eλ0(x2ξ(t))].

    Therefore, if we choose MLfα0m then LW(x,t)0.

    Case 2. For xc1t+ξ(t), if λ0λ1, then we obtain

    LW(x,t)Meλ(c1t+ξ(t))[~β1eλ1(x+c1tξ(t))˜αeλ0(x+c1tξ(t))]+Lfα0eλ0(x+c1tξ(t))[α1eλ1(x+c1tξ(t))β0eλ0(x+c1tξ(t))]eλ0(x+c1tξ(t))[M~β1e(λ0λ1)xe(λ0+λ1λ)c1t+(λ+λ0+λ1)ξ(t))M˜αeλ(c1t+ξ(t))Lfα0].

    Since c1t+ξ(t)<0 for t<T and λ<min{λ0,λ1}, LW(x,t)0 provided M2Lfα0~β1.

    For λ0<λ1, which means f1(0)>f1(1), we have

    LW(x,t)Meλ(c1t+ξ(t))ϕ1(x+c1tξ(t))+f1(ϕ1(x+c1tξ(t)))f1(ϕ1(x+c1tξ(t)))f1(ϕ1(x+c1tξ(t))ϕ1(x+c1tξ(t))).

    Moreover, if x+c1tξ(t)>L for some L1 such that ϕ1(x+c1tξ(t))[1σ,1] and ϕ1(x+c1tξ(t))[0,σ]. It follows that

    LW(x,t)Meλ(c1t+ξ(t))ϕ1(x+c1tξ(t))+[f1(1)f(0)]ϕ1(x+c1tξ(t))+o(ϕ1(x+c1tξ(t)))eλ0(x+c1tξ(t))[M˜α0eλ(c1t+ξ(t))+(f1(1)f(0))o(1)].

    Since t<T such that c1t+ξ(t)1, we have LW(x,t)0. Finally, if 0<x+c1tξ(t)L, then, from the above case λ0λ1, there holds

    LW(x,t)eλ0(x+c1tξ(t))[M~β1e(λ0λ1)xe(λ0+λ1λ)c1t+(λ+λ0+λ1)ξ(t))M˜αeλ(c1t+ξ(t))Lfα0]eλ0(x+c1tξ(t))[M~β1e(λ0λ1)Le(λ2λ0)c1t+(λ+2λ0)ξ(t))M˜αeλ(c1t+ξ(t))Lfα0].

    Thanks to t<T, λ<min{λ0,λ1} and M>2Lfα0~β1e(λ1λ0)L, we obtain

    LW(x,t)0.

    We intend to testify the super-solution W+(x,t) in two steps.

    Step 1. W+(x,t)=2ϕ1(c1t+ξ(t)) for x<0. Following a direct calculation, we have

    LW+(x,t)=2(c1+˙ξ(t))ϕ1(c1t+ξ(t))RJ(xy)W+(y,t)dy+2ϕ1(c1t+ξ(t))f(x,2ϕ1(c1t+ξ(t)))2(c1+˙ξ(t))ϕ1(c1t+ξ(t))f1(2ϕ1(c1t+ξ(t)))+0J(xy)[ϕ1(y+c1t+ξ(t))+ϕ1(y+c1t+ξ(t))]dy.

    Denote

    I=+0J(xy)[ϕ1(y+c1t+ξ(t))+ϕ1(y+c1t+ξ(t))]dy.

    Then, it follows that

    I=c1tξ(t)0J(xy)[ϕ1(y+c1t+ξ(t))+ϕ1(y+c1t+ξ(t))]dy++c1tξ(t)J(xy)[ϕ1(y+c1t+ξ(t))+ϕ1(y+c1t+ξ(t))]dyα0eλ0(c1t+ξ(t))+0J(xy)[eλ0y+eλ0y]dy+2+c1tξ(t)J(xy)dy.

    Let

    Jλ0=+0J(y)[eλ0y+eλ0y]dy.

    It follows from assumption (J) that there exists KJ>0 such that

    J(x)KJe2λ0xforx0.

    Furthermore, there hold

    IJλ0α0eλ0(c1t+ξ(t))+2KJ+c1tξ(t)e2λ0(xy)dyJλ0α0eλ0(c1t+ξ(t))+2KJλ0e2λ0(c1t+ξ(t)).

    Therefore, according to t<T such that ϕ1(c1t+ξ(t))1, we get

    LW+(x,t)2(c1+˙ξ(t))ϕ1(c1t+ξ(t))Jλ0α0eλ0(c1t+ξ(t))2KJλ0e2λ0(c1t+ξ(t))f1(2ϕ1(c1t+ξ(t)))(2c1~β0α0Jλ02f1(0)β0o(1))eλ0(c1t+ξ(t))+(2M~β02KJλ0)e2λ0(c1t+ξ(t)).

    Since

    M>KJ~β0λ0andJλ0<2c1~β0+2|f1(0)|β0α0,

    consequently, we have LW+(x,t)0 for t<T.

    Step 2. For x0, W+(x,t)=ϕ1(x+c1t+ξ(t))+ϕ1(x+c1t+ξ(t)). Then we can obtain

    LW+(x,t)=(c1+˙ξ(t))[ϕ1(x+c1t+ξ(t))+ϕ1(x+c1t+ξ(t))]RJ(xy)W+(y,t)dy+[ϕ1(x+c1t+ξ(t))+ϕ1(x+c1t+ξ(t))]f1(W+)=˙ξ(t)[ϕ1(x+c1t+ξ(t))+ϕ1(x+c1t+ξ(t))]+0J(xy)[ϕ1(y+c1t+ξ(t))+ϕ1(y+c1t+ξ(t))2ϕ1(c1t+ξ(t))]dy+f1(ϕ1(x+c1t+ξ(t)))+f1(ϕ1(x+c1t+ξ(t)))f1(ϕ1(x+c1t+ξ(t))+ϕ1(x+c1t+ξ(t))).

    Denote

    II=0J(xy)[ϕ1(y+c1t+ξ(t))+ϕ1(y+c1t+ξ(t))2ϕ1(c1t+ξ(t))]dy.

    We consider two cases.

    Case 1. 0xc1tξ(t). Without loss of generality, let ϕ1(0)=θ1. Since ϕ1>0 and t<T such that ϕ1(c1t+ξ(t))<12θ1, we have

    II=0c1t+ξ(t)J(xy)[ϕ1(y+c1t+ξ(t))+ϕ1(y+c1t+ξ(t))2ϕ1(c1t+ξ(t))]dy+c1t+ξ(t)J(xy)[ϕ1(y+c1t+ξ(t))+ϕ1(y+c1t+ξ(t))2ϕ1(c1t+ξ(t))]dy0c1t+ξ(t)J(xy)Cλ0eλ0(c1t+ξ(t))[eλ0y+eλ0y2(2+e(λ0+η)y+e(λ0+η)y)Cηeη(c1t+ξ(t))]dy(2+Jη)Cλ0Cηe(λ0+η)(c1t+ξ(t)).

    The second inequality follows from

    |ϕ1(x)Cλ0eλ0x|Cηe(λ0+η)xforx0and some0<η<λ0,

    which can be easily obtained by 5 and 6. As a consequence,

    LW+(x,t)Me(λ+λ0)(c1t+ξ(t))~β0(eλ0x+eλ0x)(2+Jη)Cλ0Cηe(λ0+η)(c1t+ξ(t))Lfα20e2λ0(c1t+ξ(t)))e(λ+λ0)(c1t+ξ(t))[2M~β0(2+Jη)Cλ0Cηe(ηλ)(c1t+ξ(t))Lfα20e(λ0λ)(c1t+ξ(t))]e(λ+λ0)(c1t+ξ(t))[2M~β0(2+Jη)Cλ0CηLfα20].

    The last inequality holds since λ<min{λ0,λ1,η}. Thus LW+(x,t)0, provided that

    M(2+Jη)Cλ0Cη+Lfα202~β0.

    Case 2. Here x>c1tξ(t). From Case 1, we know

    II0c1t+ξ(t)J(xy)Cλ0eλ0(c1t+ξ(t))(2+e(λ0+η)y+e(λ0+η)y)Cηeη(c1t+ξ(t))dyCλ0CηKJe(λ0+η)(c1t+ξ(t))0c1t+ξ(t)e2λ0(xy)(3+e(λ0+η)y)dyCλ0CηKJeλ0(x+c1t+ξ(t))λ0x(3λ0eη(c1t+ξ(t))+1η).

    Moreover, if λ1λ0, then

    LW+(x,t)Meλ(c1t+ξ(t))(~β1eλ1(x+c1t+ξ(t))+~β0eλ0(x+c1t+ξ(t)))Lfα0eλ0(x+c1t+ξ(t))Cλ0CηKJeλ0(x+c1t+ξ(t))(3λ0eη(c1t+ξ(t))+1η)eλ(c1t+ξ(t))+λ0(x+c1t+ξ(t))[M~β1e(λ0λ1)x(λ0+λ1)(c1t+ξ(t))3Cλ0CηKJλ0e(ηλ)(c1t+ξ(t))(Lfα0+Cλ0CηKJη)eλ(c1t+ξ(t))].

    Remember that λ<min{λ0,λ1,η} and t<T. As a matter of fact, take M sufficiently large such that

    M~β13Cλ0CηKJλ0(Lfα0+Cλ0CηKJη)0. (7)

    This yields that LW+(x,t)0.

    If λ0<λ1, we have f1(0)>f1(1). In addition, for x>c1tξ(t)+M with M1 such that ϕ1(x+c1t+ξ(t))[1σ,1], we have

    f1(ϕ1(x+c1t+ξ(t)))+f1(ϕ1(x+c1t+ξ(t)))f1(W+)=[f1(0)f1(1)]ϕ1(x+c1t+ξ(t))+o(ϕ1(x+c1t+ξ(t))).

    Meanwhile,

    LW+(x,t)eλ(c1t+ξ(t))+λ0(x+c1t+ξ(t))(M~β1e(λ0λ1)x(λ0+λ1)(c1t+ξ(t))3Cλ0CηKJλ0e(ηλ)(c1t+ξ(t)))+([f1(0)f1(1)]β0Cλ0CηKJηeλ0x)eλ0(x+c1t+ξ(t)).

    Then LW+(x,t)0 by 7 and x<c1t+ξ(t)M1 such that

    [f1(0)f1(1)]β0Cλ0CηKJηeλ0x0.

    When c1tξ(t)<xc1tξ(t)+M, it follows from the case λ0λ1 that

    LW+(x,t)eλ(c1t+ξ(t))+λ0(x+c1t+ξ(t))[M~β1e(λ0λ1)x(λ0+λ1)(c1t+ξ(t))3Cλ0CηKJλ0e(ηλ)(c1t+ξ(t))(Lfα0+Cλ0CηKJη)eλ(c1t+ξ(t))]eλ(c1t+ξ(t))+λ0(x+c1t+ξ(t))[M~β1e(λ0λ1)M2λ0(c1t+ξ(t))3Cλ0CηKJλ0e(ηλ)(c1t+ξ(t))(Lfα0+Cλ0CηKJη)eλ(c1t+ξ(t))].

    Similar to the case λ1λ0, we can obtain LW+(x,t)0 for t<T by letting M be sufficiently large.

    Proof of Theorem 1.1. Let un(x,t) be the unique solution of the Cauchy problem

    {(un)t(x,t)=Jun(x,t)un(x,t)+f(x,un),xR,t>n,un(x,n)=W(x,n),xR.

    Since W(x,n)=un(x,n)W+(x,n), the comparison principle yields for n>T

    W(x,t)un(x,t)W+(x,t)forxR,t[n,T].

    Moreover, since m:=minR×[0,1](1fu(x,u))>0 by the assumption fu(x,u)<1, it is not difficult to see that δun(x,t)=un(x+η,t)un(x,t) with ηR is a sub-solution of

    {v(t)=Lηmv(t),t>n,v(n)=Mη,

    where LR|J(y)|dy and M is some positive constant. It is then similar to the proof of Proposition 2.4 in [23] that un(x,t) and (un)t(x,t) are Lipschiz continuous with respect to x. Besides, a direct calculation gives that |un(x,t)|<C and |(un)t(x,t)|<C for some positive constant C. Then applying Arzela-Ascoli Theorem, there exists a subsequence of {un}+n=1 and {(un)t}+n=1, still denoted by {un}+n=1 and {(un)t}+n=1, such that {un}+n=1 and {(un)t}+n=1 converge to a function u(x,t) and ut(x,t). Now we have obtained the entire solution. In fact, the regularity of this entire solution is not so well. But we can show the entire solution is Lipschitz continuous in xR, which is important to show the uniqueness of the entire solution.

    Proposition 2. Let u(x,t) be the entire solution in Theorem 1.1. Then u(x,t) satisfies

    |u(x+η,t)u(x,t)|Mη,

    and

    |u(x+η,t)tu(x,t)t|M"η

    for M,M">0.

    The proof is similar to that of Proposition 3.1 in [23] by virtue of the fact fu(x,u)<1 for xR and u[0,1], so we omit it here.

    As for the uniqueness, since it is easy to see u(x,t)ϕ1(x+c1t) as t by the sub- and super-solutions established above, we can similarly show that the conclusion of Lemma 3.1 in [4] also holds true here. Then referring to the process of Section 3 in [4], we can obtain the uniqueness of the entire solution by constructing similar sub- and super-solutions. Thus we have shown the first part of Theorem 1.1.

    In this section, we are going to show the entire solution established in previous section converges to a shift of ϕ2(x+c2t) as t+. The main idea here is first to construct suitable sub-and super-solutions taking advantage of the traveling wave ϕ2(x+c2t) to get proper lower and upper bounds, which is a classical and regular method in study of traveling waves. Then we apply a "squeezing" technique which is used to prove the asymptotic stability of the traveling waves for nonlocal dispersal equation in [7] to obtain the convergence of the entire solution. Here, we assume that for some X>0,ω,σ>0 and

    ω<min{|f1(0)|4,|f2(0)|4,|f1(1)|4,|f2(1)|4},

    there hold

    ϕ1(x),ϕ2(x)σ2forxXandϕ1(x),ϕ2(x)1σ2forxX,

    besides,

    f1(s),f2(s)ωfors[0,σ][1σ,1].

    Theorem 4.1. Let assumptions (F) and (J) hold, and u(x,t) be the unique entire solution of 1 with 0<u(x,t)<1, ut(x,t)>0 for all tR and xR such that

    u(x,t)ϕ1(x+c1t)0astuniformly inxR. (8)

    Then

    u(x,t)ϕ2(x+c2t+β)0ast+uniformly inxR

    for some βR.

    Lemma 4.2. Suppose assumptions (F) and (J) hold. Then there exist T, β, δ,ω>0 such that

    u(x,t)max{ϕ2(x+c2tβ)δeωt,0}

    for xR and tT.

    Proof of Lemma 4.2. In fact, it follows from 8 that there exists tR such that

    |u(x,t)ϕ1(x+c1t)|ϵ2

    for any ϵ>0 and xR. Denote

    ˜u(x,t)=u(x,t+t)andu_(x,t)=max{ϕ2(ξ(x,t))Cveωt,0},

    where

    ξ(x,t)=x+c2(t+t)β0+CveωtCvfort0,xR.

    We first show ˜u(x,0)u_(x,0) for xR. Then there exists x1>0 such that

    ϕ1(x+c1t)1ϵ2andϕ2(x+c2t)1ϵ2forxx1.

    As ut>0 and ϕi>0 (i=1,2), we have

    1ϵ2u(x,t)1and1ϵ2ϕi1,i=1,2

    for tt,xx1. Then,

    |ϕiu(x,t)|ϵ,i=1,2fortt,xx1.

    Meanwhile, there exists x2R such that

    0<ϕi(x+cit)ϵ2,i=1,2forxx2.

    Particularly, there holds

    |u(x,t)ϕ2(x+c2t)|ϵforxR(x1,x2).

    In addition, minx[x1,x2]u(x,t)>0 since 0<u(x,t)<1. Then we can choose β0>0 such that

    ϕ2(x+c2tβ0)u(x,t)forx[x1,x2].

    From all the discussion above we obtain ˜u(x,0)u_(x,0). Next, we prove

    Lu_(x,t)=u_t(x,t)(Ju_(x,t)u_(x,t))f(x,u_)0fort0,xR.

    In fact, we only need to consider u_(x,t)=ϕ2(ξ(x,t))Cveωt because 0 is obviously a sub-solution of 1. Hence,

    Lu_(x,t)=(c2Cvωeωt)ϕ2(ξ(x,t))+CvωeωtRJ(xy)ϕ2(ξ(y,t))dy+ϕ2(ξ(x,t))f(x,u_(x,t))Cvωeωtϕ2(ξ(x,t))+f2(ϕ2(ξ(x,t)))f2(ϕ2(ξ(x,t))Cveωt)+Cvωeωt.

    In the following, we continue the proof in three cases.

    Case ξ(x,t)X. In this case,

    ϕ2(ξ)Cveωtϕ2(ξ)σ.

    Then, by ω<|f2(0)|4,

    Lu_(x,t)(f2(0)+2ω)Cveωt0.

    Case ξ(x,t)X. For this case, by 2Cv<σ,

    1σ1σ2Cvϕ2(ξ)Cveωtϕ2(ξ)<1.

    Then, since ω<|f2(1)|4,

    Lu_(x,t)(f2(1)+2ω)Cveωt0.

    Case ξ[X,X]. Let τ0=minξ[X,X]ϕ2(ξ). Then,

    Lu_(x,t)τ0Cvωeωt+f2Cveωt+Cvωeωt(f2Cv++Cvωτ0Cvω)eωt.

    Choose Cv=τ10ω10(f2Cv++Cvω). This gives Lu_(x,t)0. We finish the proof by letting β=β0+Cv and δ=Cv.

    Similarly, we can prove the following lemma.

    Lemma 4.3. Under the assumptions of Lemma 4.2, there exist T+>T, β+, C+,ω>0 such that

    min{ϕ1(x+c1t+β+)+C+eωt,1}u(x,t)

    for xR,tT+, where ω,T are defined in Lemma 4.2.

    Now we are in a position to establish an important lemma.

    Lemma 4.4. Under the assumptions of Lemma 4.2, there exist T>max{T+,T}, β,C,ω>0 such that

    min{ϕ2(x+c2t+β)+Ceωt,1}u(x,t)

    for xR,tT, where ω,T,T+ are defined in Lemmas 4.2 and 4.3.

    Proof of Lemma 4.4. Define ˆu(x,t)=u(x,t+T) and ¯u(x,t)=min{ϕ2(ξ+(x,t))+v(t),1}, where

    ξ+(x,t)=x+c2(t+T)+β++V(t),V(t)=KVt0v(s)ds,

    and

    v(t)=(ϵ1+C3)eωtC3eλ1c2t

    for 0<ϵ1,C3<σ4 and t0. Here, λ1 is the positive root of

    c2λ1=RJ(xy)eλ1ydy1+f2(1).

    Then

    v(0)=ϵ1,0<v(t)<σ2fort0

    and

    ˙v(t)=ωv(t)(wλ1c2)C3eλ1c2tfort0.

    Now we intend to show ˆu(x,0)¯u(x,0). Recall the fact that |1ϕ2(x)|C2eλ1x for xR,C2>0. Meanwhile, we assume that 0<ω<λ1c2. Additionally, by the assumptions of f(x,u), there exists C0>0 such that

    |f1(s)f2(s)|C0|1s|fors[0,1].

    Since ϕi()=0 and ϕi(+)=1, there exists x3<0 such that

    ϕ1(x+c1T+β+)<ϵ12andϕ2(x+c2T)<ϵ1forx<x3

    as well as x4>0 such that

    ϕ2(x+c2Tβ)1ϵ1andϕ2(x+c2T)1ϵ1

    for x>x4. Note that we can take T>0 sufficiently large such that

    max{C+,δ}eωT<ϵ12andXc2Tx0.

    Therefore, it is obvious that

    u(x,T)min{ϕ2(x+c2T)+ϵ1,1}¯u(x,0)forxR(x3,x4).

    Furthermore, since maxx[x3,x4]u(x,T)<1, one can choose β+>0 such that

    ϕ2(x+c2T+β+)u(x,T)forx[x3,x4].

    This yields that ˆu(x,0)¯u(x,0).

    In the following, we are going to prove

    L¯u(x,t)=¯ut(x,t)(J¯u(x,t)¯u(x,t))f(x,¯u)0fort0,xR.

    Here we only need to consider ¯u(x,t)=ϕ2(ξ+(x,t))+v(t) since 1 is obviously a super-solution of 1. A direct calculation gives that

    L¯u(x,t)=˙ξ+(x,t)ϕ2(ξ+(x,t))+˙v(t)RJ(xy)ϕ2(ξ+(y,t))dy+ϕ2(ξ+(x,t))f(x,ϕ2(ξ+(x,t))+v(t))
    =˙V(t)ϕ2(ξ+(x,t))+˙v(t)+f2(ϕ2(ξ+(x,t)))f(x,ϕ2(ξ+(x,t))+v(t)){˙V(t)ϕ2(ξ+(x,t))+˙v(t)+f2(ϕ2(ξ+(x,t)))f2(ϕ2(ξ+(x,t))+v(t))forx<L,˙V(t)ϕ2(ξ+(x,t))+˙v(t)+f2(ϕ2(ξ+(x,t)))f1(ϕ2(ξ+(x,t))+v(t))forxL.

    We continue to prove L¯u(x,t)0 in three cases.

    Case 1. If ξ+>X and xL, then

    L¯u(x,t)˙V(t)ϕ2(ξ+(x,t))+˙v(t)+f1(ϕ2(ξ+(x,t)))f1(ϕ2(ξ+(x,t))+v(t))C0|1ϕ2(ξ+(x,t))|˙V(t)ϕ2(ξ+(x,t))+˙v(t)(f1(1)+ω)˙v(t)C2C0eλ1(x+c2(t+T)+β++V(t))˙V(t)ϕ2(ξ+(x,t))+˙v(t)(f1(1)+ω)˙v(t)C2C0eλ1c2teλ1(Lc2T).

    Particularly, choose C3 such that

    ˙v(t)=ωv(t)+C2C0eλ1c2teλ1(Lc2T),

    which means

    C3=C2C0eλ1(Lc2T)λ1c2ω.

    As a consequence,

    L¯u(x,t)(f1(1)+2ω)v(t)0.

    In addition, if ξ+>X and x<L, then

    L¯u(x,t)ωv(t)(wλ1c2)C3eλ1c2t(f2(1)+ω)v(t)(f2(1)+2ω)v(t)0.

    Case 2. If ξ+<X, then x<L. Similar to Case 1 with x<L, we can obtain L¯u(x,t)0.

    Case 3. For ξ+[X,X], we know x<L. Therefore,

    L¯u(x,t)˙V(t)ϕ2(ξ+(x,t))+˙v(t)f2v(t)˙V(t)τ0˙V(t)ωv(t)f2v(t)(KVτ0ωf2)v(t),

    where τ0 is defined as in Lemma 4.2. It follows from

    KVω+f2τ0

    that L¯u(x,t)0. This finishes the proof by letting β=β++ϵ1+C3 and C=C3+ϵ1.

    Now we are ready to show that the entire solution converges to a shift of ϕ2(x+c2t) using a "squeezing" technique [7]. We should declare that assumption (A) (see (B2) in [7]) holds.

    (A) There exists a positive nonincreasing function ζ(n) defined on [1,+) such that for any u1(x,t),u2(x,t) satisfying 1u1(x,t),u2(x,t)2, \mathcal{L}u_1(x,t)\geq0 , \mathcal{L}u_2(x,t)\leq0 and u_1(x,0)\geq u_2(x,0) , there holds

    \begin{equation*} \min\limits_{x\in[-n,n]}\{u_1(x,1)-u_2(x,1)\}\geq\zeta(n)\int_0^1[u_1(y,0)-u_2(y,0)]dy \; \text{for}\; m\geq1. \end{equation*}

    In fact, it follows from the comparison principle that u(x,t)\geq v(x,t) for x\in\mathbb{R} and t\geq0. Furthermore, denote \breve{w}(x,t) = e^{Kt}[u(x,t)-v(x,t)]\geq0 with K\geq1+\max_{s\in[-1,2]}|f_s(x,s)| . Then

    \begin{equation*} \begin{split} \breve{w}_t(x,t)\geq&K\breve{w}(x,t)+J*\breve{w}(x,t)-\breve{w}(x,t)+f(x,u(x,t))-f(x,v(x,t))\\ \geq&J*\breve{w}(x,t)+[K-(1+\max\limits_{s\in[-1,2]}|f_s(x,s)|)]\breve{w}(x,t)\\ \geq&J*\breve{w}(x,t), \end{split} \end{equation*}

    which implies that \breve{w}(x,t)\geq\breve{w}(x,0) for t\geq0 . Therefore, \breve{w}_t(x,t)\geq J*\breve{w}(x,0) and \breve{w}(x,t_*)\geq t_*J*\breve{w}(x,0) . Repeating the same progress on [t_*,2t_*],\; ...,\; [(N-1),Nt_*] , we have \breve{w}(x,Nt_*)\geq t^N_*J*\breve{w}(x,0) for N\geq1 . In addition, since J(x)\geq0 and \int_{\mathbb{R}}J(y)dy = 1 , there exists N_0 = N_0(M)\geq1 such that by letting t_* = \frac{T}{N_0} we have

    \begin{equation*} \breve{w}(x,T)\geq\left(\frac{T}{N_0}\right)^NJ*\breve{w}(x,0)\geq\left(\frac{T}{N_0}\right)^N c(M)\int_0^1\breve{w}(y,0)dy, \end{equation*}

    where

    \begin{equation*} c(M) = \min\limits_{x\in[-M-1,M+1]}J(x) > 0. \end{equation*}

    Thus, there exists a positive function \eta(x,t)\in C([0,\infty),[0,\infty)) such that

    \begin{equation*} u(x,t)-v(x,t)\geq\eta(|x|,t)\int_0^1[u(y,0)-v(y,0)]dy\; \text{for}\; x\in\mathbb{R},t > 0, \end{equation*}

    which implies the assumption (A) holds.

    Now, we start to prove the following lemma, which plays an important role in the proof of Theorem 4.1.

    Lemma 4.5. Suppose that assumptions (F) and (J) hold. Then there exists a small \epsilon_0 such that if for some \tau\geq0,\; \xi\in\mathbb{R},\; \delta\in(0,\frac{\sigma}{2}] , and h>0 , there holds

    \begin{equation} \phi_2(x+c_2\tau+\xi)-\delta\leq u(x,t)\leq\phi_2(x+c_2\tau+\xi+h)+\delta\; \mathit{\text{for}}\; x\in\mathbb{R}, \end{equation} (9)

    then for every t>\tau+1 , there exist \tilde{\xi}(t),\; \tilde{\delta}(t) and \tilde{h}(t) satisfying

    \begin{align*} &\tilde{\xi}(t)\in\left[\xi-\delta\frac{\gamma}{C_m},\xi+h+\delta\frac{\gamma}{C_m}\right],\\ &\tilde{\delta}(t)\leq e^{-\omega(t-\tau-1)}\left[\delta+\epsilon_0 \min\{h,1\}\right],\\ &\tilde{h}(t)\leq\tau h-\frac{\gamma\epsilon_0}{C_m}\min\{h,1\}+2\delta\frac{\gamma}{C_m} \end{align*}

    such that 9 holds with (\tau,\xi,\delta,h) replaced by (t,\tilde{\xi}(t),\tilde{\delta}(t),\tilde{h}(t)) . Here, C_m: = \max\{\delta_-,C\} and \gamma: = \max\left\{C^v,\frac{K_V}{\omega}\right\} , where the parameters are defined in Lemmas 4.2 and 4.4.

    Proof of Lemma 4.5. In view of Lemmas 4.2 and 4.4, it is easy to see that

    \begin{equation*} \phi_2(x+c_2t-\beta_-)-\delta_-e^{-\omega t}\leq u(x,t)\leq \phi_2(x+c_2t+\beta)+Ce^{-\omega t}. \end{equation*}

    Furthermore, as in the proofs of Lemmas 4.2 and 4.4,

    \begin{equation*} \phi_2\left(x+c_2t-\beta^0+C^ve^{-\omega t}-C^v\right)-\delta_-e^{-\omega t}\leq u(x,t) \end{equation*}

    and

    \begin{equation*} \phi_2\left(x+c_2t+\beta^++\frac{K_V}{\omega}-\frac{K_V}{\omega}e^{-\omega t}\right)+Ce^{-\omega t}\geq u(x,t). \end{equation*}

    Denote

    \begin{equation*} \kappa = \max\{-\beta^0-\gamma\}, \; h = \beta^++\frac{K_V}{\omega}-\kappa. \end{equation*}

    In addition, by the definition of C_m , \gamma and letting \check{u}(x,t) = u(x-\kappa,t) , we have

    \begin{equation*} \phi_2\left(x+c_2t+\gamma e^{-\omega t}\right)-C_me^{-\omega t}\leq\check{u}(x,t) \leq \phi_2\left(x+c_2t+h+\gamma e^{-\omega t}\right)+C_me^{-\omega t}. \end{equation*}

    Let \overline{h} = \min\{h,1\} and \vartheta_0 = \frac{1}{2}\min\limits_{[0,2]}\phi'_2(x) . Then

    \begin{equation*} \int_0^1\left[\phi_2(y+\overline{h})-\phi_2(y)\right]dy\geq2\vartheta_0\overline{h}. \end{equation*}

    Therefore, at least one of the following two inequalities is true

    \begin{equation*} \text{(i)}\; \int_0^1[\check{u}(y,0)-\phi_2(y)]dy\geq\vartheta_0\overline{h},\qquad\qquad \; \; \text{(ii)}\; \int_0^1[\phi_2(y+\overline{h})-\check{u}(y,0)]dy\geq\vartheta_0\overline{h}. \end{equation*}

    Next, we consider the case (i) since the case (ii) is similar. According to the assumption (A), for \zeta = \zeta(M+c_2+2) with M\gg1 such that \frac{\gamma}{C_m}\phi'_2(x)<1 for |x|\geq M , and for every x\in[-M-c_2-\gamma,M+c_2+\gamma] , there holds

    \begin{equation*} \check{u}(x,1)-\left[\phi_2\left(x+\gamma e^{-\omega}\right)-C_me^{-\omega}\right]\geq\zeta\int_0^1[\check{u}(y,0)-(\phi_2(y)-C_m)]dy \geq\zeta\vartheta_0\overline{h}. \end{equation*}

    Now define

    \begin{equation*} \epsilon_0 = \min\left\{\frac{\sigma}{2},\frac{\gamma}{2C_m},\min\limits_{x\in[-M-c_2-\gamma,M+c_2+\gamma]} \frac{\zeta\vartheta_0\gamma}{2C_m\phi'_2(x)}\right\}. \end{equation*}

    Accordingly, there exists \tilde{\theta}\in(0,1) such that

    \begin{equation*} \begin{split} \phi_2\left(x+\gamma e^{-\omega}+2\epsilon_0\overline{h}\frac{\gamma}{C_m}\right)- \phi_2\left(x+\gamma e^{-\omega}\right) = &\phi'_2\left(x+\gamma e^{-\omega}+2\epsilon_0\overline{h}\tilde{\theta}\frac{\gamma} {C_m}\right)2\epsilon_0\overline{h}\frac{\gamma}{C_m}\\ \leq&\zeta\vartheta_0\overline{h} \end{split} \end{equation*}

    for all x\in[-M-c_2-(\gamma-1),M+c_2+(\gamma-1)]. Hence,

    \begin{equation*} \check{u}(x,1)\geq\phi_2\left(x+\gamma e^{-\omega}+2\epsilon_0\overline{h}\frac{\gamma}{C_m}\right)-C_me^{-\omega} \end{equation*}

    for x\in[-M-c_2-(\gamma-1),M+c_2+(\gamma-1)] .

    For |x|\geq M+c_2+(\gamma-1) , by the choice of M , we know that

    \begin{equation*} \phi_2\left(x+\gamma e^{-\omega}\right)\geq\phi_2\left(x+\gamma e^{-\omega}+2\epsilon_0\overline{h}\frac{\gamma}{C_m}\right)-\epsilon_0\overline{h}. \end{equation*}

    Then, it follows that

    \begin{equation*} u(x,1)\geq\phi_2\left(x+\kappa+\gamma e^{-\omega}+2\epsilon_0\overline{h}\frac{\gamma}{C_m}\right)-\epsilon_0\overline{h}-C_me^{-\omega} \; \text{for}\; x\in\mathbb{R}. \end{equation*}

    Note that p: = \delta e^{-\omega}+\epsilon_0\overline{h}\leq\sigma , repeat the operation above with 1 replaced by 1+t' , which implies that

    \begin{equation*} \begin{split} u(x,1+t')\geq&\phi_2\left(x+c_2t+\kappa+\gamma e^{-\omega}+2\epsilon_0\overline{h}\frac{\gamma}{C_m}+\gamma e^{-\omega t'}\right)-pe^{-\omega t'}\\ \geq&\phi_2\left(x+c_2(t'+1)+\kappa+\epsilon_0\overline{h}\frac{\gamma}{C_m}-\frac{\delta \gamma}{C_m}\right)-(\delta+\epsilon_0\overline{h})e^{-\omega t'}. \end{split} \end{equation*}

    Thus we finish the proof by setting

    \begin{align*} &t = 1+t',\; \tilde{\xi}(t) = \frac{\gamma\epsilon_0\overline{h}}{C_m} ,\; \tilde{\delta} = (\delta+\epsilon_0\overline{h})e^{-\omega (t-1)}\\ &\tilde{h} = \left[h+\delta\frac{\gamma}{C_m}e^{-\omega t}\right]-\tilde{\xi}(t) = h-\frac{\epsilon_0\gamma}{C_m}\overline{h}+\delta\frac{\gamma}{C_m}\left[2-e^{-\omega t}\right]. \end{align*}

    Now we shall prove Theorem 4.1.

    Proof of Theorem 4.1. We shall divide the proof in three steps.

    Step 1. Following from Lemmas 4.2 and 4.4, there exist T^*,\; M^*>0 such that

    \begin{equation} \phi_2(x+c_2T^*-M^*)-C_m\leq u(x,T^*)\leq\phi_2(x+c_2T^*+M^*)+C_m\; \text{for}\; x\in\mathbb{R}. \end{equation} (10)

    Here, C_m is defined as in the proof of Lemma 4.4. Define

    \begin{equation*} \epsilon^* = \min\left\{\frac{\sigma}{2},\frac{\epsilon_0}{4}\right\}\; \text{and} \; k_0 = \epsilon_0\frac{\gamma}{C_m}-2\epsilon^*\frac{\gamma}{C_m}\geq\frac{\gamma\epsilon_0}{2C_m} > 0. \end{equation*}

    Meanwhile, fix t^*\geq2 such that

    \begin{equation*} e^{-\omega(t^*-1)}\left[1+\frac{\epsilon_0}{\epsilon^*}\right]\leq1-k^*. \end{equation*}

    Then, replace C_m with \epsilon^* and denote M^*,\; T^* by \frac{h_0}{2},\; T_0 . Assume that h_0\geq1 , otherwise, we directly go to Step 2.

    By 10, applying Lemma 4.5 with \tau = T_0 , \xi = -\frac{h_0}{2} , h = h_0 , \delta = \epsilon^* , 9 holds with \tau = T_0+t^* , some \xi\in\left[-\frac{h_0}{2}-\frac{\gamma\epsilon^*}{C_m},\frac{h_0}{2}+\frac{\gamma\epsilon^*}{C_m}\right] , \delta = \epsilon^* and h = h_0-k^* , by the definition of k^*,\; t^* , it follows that

    \begin{equation*} \hat{\delta}(T_0+t^*)\leq[\epsilon^*+\epsilon_0]e^{-\omega t^*}\leq\epsilon^*\; \text{and}\; \hat{h}(T_0+t^*)\leq h_0-\frac{\gamma}{C_m}\epsilon_0+2\epsilon^*\frac{\gamma}{C_m}\leq h_0-k^*. \end{equation*}

    Repeat the same process, it yields that 9 holds for \tau = T_0+Nt^* , \delta = \epsilon^* and h = h_0-Nk^* with N such that h_0-(N-1)t^*\geq1 . Thus, there exists T_1>T_0 such that 9 holds for \tau = T_1 , \delta = \epsilon^* and h = 1 and some \xi\in\mathbb{R} .

    Step 2. In this step, we use a mathematical induction to show that for every nonnegative integer k , 9 holds for \xi = \xi^k\in\mathbb{R} and

    \begin{equation*} \tau = T^k: = T_1+kt^*,\; \delta = \delta^k: = (1-k^*)^k\epsilon^*,\; h = h^k: = (1-k^*)^k. \end{equation*}

    It is obvious that the assertion holds for k = 0 by Step 2. Then suppose that the assertion is true for k = l\geq0 . We show that it is true for k = l+1 . In fact, as can be seen in Lemma 4.5 with \tau = T^l and t = T^{l+1} , one can obtain that 9 holds with (\tau,\xi,\delta,h) replaced by (\hat{\tau},\hat{\xi},\hat{\delta},\hat{h}) satisfying

    \begin{align*} &\hat{\xi}\in\left[\xi^l-\delta^l\frac{\gamma}{C_m},\xi^l+\delta^l\frac{\gamma}{C_m}\right],\\ &\hat{\delta}\leq\left(\delta^l+\epsilon_0h^l\right) = [1-k^*]^l\epsilon^*\left(1 +\frac{\epsilon_0}{\epsilon^*}\right) e^{-\omega(t^*-1)}\leq(1-k^*)^{l+1}\epsilon^*,\\ &\hat{h}\leq h^l-h^l\epsilon_0\frac{\gamma}{C_m}+2\delta^l\frac{\gamma}{C_m} = [1-k^*]^l\left[1-\epsilon_0\frac{\gamma}{C_m}+2\epsilon^*\frac{\gamma}{C_m}\right] = [1-k^*]^{l+1} \end{align*}

    by the definition of \epsilon^*,\; k^* and t^* . This means that 9 holds for \tau = T^{l+1} , some \xi = \xi^{l+1}\in\left[\xi^l-\delta^l\frac{\gamma}{C_m},\xi^l+\delta^l\frac{\gamma}{C_m}\right] , \delta = [1-k^*]^{l+1}\epsilon^* and h = [1-k^*]^{l+1} . Now we finish the mathematical induction.

    Step 3. So far, we have known that 9 holds for (\tau,\xi,\delta,h) = (T^k,\xi^k,\delta^k,h^k) for all k = 0,1... . Furthermore, 9 holds with \tau\in[T^k,\infty) , \delta = \delta^k , h = h^k+2\delta^k\frac{\gamma}{C_m} and \xi = \xi^k-\delta^k\frac{\gamma}{C_m},\; k = 0,1... .

    Now we define

    \begin{equation*} \delta(t) = \delta^k, \; \xi(t) = \xi^k-\delta^k\frac{\gamma}{C_m},\; h(t) = h^k+2\delta^k\frac{\gamma}{C_m} \end{equation*}

    for t\in\left[T^k,T^{k+1}\right],\; k = 0,1... . Then,

    \begin{equation*} \phi_2(x+c_2t+\xi(t))-\delta(t)\leq u(x,t)\leq\phi_2(x+c_2t+\xi(t+h(t))+\delta(t)\; \text{for}\; t\geq T_1,\; x\in\mathbb{R}. \end{equation*}

    It follows from the definition of \delta(t) and h(t) that

    \begin{align*} &\delta(t) = \delta^k = [1-k^*]^k\epsilon^*\leq\epsilon^*\exp{\left[\left(\frac{t-T_1}{t^*}-1\right)\ln(1-k^*)\right]} \; \text{for}\; t\geq T_1,\\ &h(t) = h^k+2\delta^k\frac{\gamma}{C_m}\leq\left[1+2\epsilon^*\frac{\gamma}{C_m}\right] \exp{\left[\left(\frac{t-T_1}{t^*}-1\right)\ln(1-k^*)\right]} \; \text{for}\; t\geq T_1. \end{align*}

    Moreover, since for any t\geq\tau\geq T_1 ,

    \begin{equation*} \xi(t)\in\left[\xi(\tau)-\delta(\tau)\frac{\gamma}{C_m},\xi(\tau)+h(\tau) +\delta(\tau)\frac{\gamma}{C_m}\right], \end{equation*}

    there holds

    \begin{equation*} |\xi(t)-\xi(\tau)|\leq h(\tau)+2\delta(\tau)\frac{\gamma}{C_m}, \end{equation*}

    which implies that \xi(\infty): = \lim\limits_{t\rightarrow+\infty}\xi(t) exists and

    \begin{equation*} |\xi(\infty)-\xi(\tau)|\leq h(\tau)+2\delta(\tau)\frac{\gamma}{C_m}\leq\left[1+4\epsilon^*2\delta(\tau) \frac{\gamma}{C_m}\right]e^{\left[\left(\frac{t-T_1}{t^*}-1\right)\ln(1-k^*)\right]} \; \text{for}\; t\geq T_1. \end{equation*}

    Therefore, we have that

    \begin{equation*} |u(x,t)-\phi_2(x+c_2t+\xi(\infty))|\rightarrow0\; \text{as}\; t\rightarrow+\infty. \end{equation*}

    Furthermore, the convergence is exponential. Then we complete the proof.

    We investigate the Lyapunov stability of the entire solution obtained in Theorem 1.1 in this section. That is, the aim here is to prove Theorem 1.2. The following lemma plays an important role in proving Theorem 1.2.

    Lemma 5.1. Let u(x,t) be the unique entire solution in Theorem 1.1. Then for any \varphi\in(0,\frac{1}{2}] , there exist constants T_\varphi = T_\varphi(\varphi)>1 and K_\varphi = K_\varphi(\varphi)>0 such that

    u_t(x,t)\geq K_\varphi\; \mathit{\text{for any}}\; t\geq T_\varphi\; \mathit{\text{and}}\; x\in\Omega_{\varphi}(t),

    where

    \Omega_{\varphi}(t) = \left\{x\in\mathbb{R}:\; \varphi\leq u(x,t)\leq1-\varphi\right\}.

    Proof of Lemma 5.1. It is easy to choose T_\varphi and M_\varphi such that

    \begin{equation*} \Omega_{\varphi}(t)\subset\left\{x\in\mathbb{R}:\; |x+c_2t|\leq M_\varphi\right\}\subset\{x\in\mathbb{R}:\; x\leq-1\}. \end{equation*}

    Now suppose there exist sequences t_k\in[T_\varphi,+\infty) and x^k\in\Omega_{\varphi}(t) such that

    \begin{equation*} u_t(t_k,x^k)\rightarrow0\; \text{as}\; k\rightarrow+\infty. \end{equation*}

    Here only two cases happen, t_k\rightarrow+\infty or t_k\rightarrow t_* for some t_*\in[T_\varphi,+\infty) as k\rightarrow+\infty .

    For the former case, denote

    \begin{equation*} u_k(x,t) = u(x+x^k,t+t_k). \end{equation*}

    By Proposition 2, \{u_k(x,t)\}_{k = 1}^\infty is equicontinuous in x\in\mathbb{R} and t\in\mathbb{R} . Furthermore, there exists a subsequence still denoted by \{u_k(x,t)\}_{k = 1}^\infty such that

    \begin{equation*} u_k\rightarrow u_*\; \text{as}\; k\rightarrow+\infty \end{equation*}

    for some function u_* satisfying 1 with \frac{\partial u_*(0,0)}{\partial t} = 0 and (u_*(x,t))_t\geq0 . By the comparison principle, we further have

    \begin{equation*} \frac{\partial u_*(x,t)}{\partial t}\equiv0\; \text{for}\; t\geq0. \end{equation*}

    However, this is impossible because by Theorem 4.1

    \begin{equation*} u_*(x,t) = \phi_2(x+c_2t+\beta+a)\; \text{for some}\; a\in[-M_\eta,M_\eta]. \end{equation*}

    For the second case, x^k remains bounded by the definition of \Omega_{\varphi}(t) . Therefore, we assume that x^k\rightarrow x^* as k\rightarrow+\infty and let

    \begin{equation*} u_k(x,t): = u(x+x^k,t). \end{equation*}

    Then, each u_k(x,t) is defined for all (x,t)\in\{x\in\mathbb{R}\mid x\leq-1\}\times[T_\varphi,+\infty) by the definition of \Omega_\varphi(t) . Similarly, there exists a subsequence, again denoted by \{u_k\}^{\infty}_{k = 1} , such that

    \begin{equation*} u_k\rightarrow u^*\; \text{as}\; k\rightarrow+\infty \end{equation*}

    for some function u^* satisfying 1 with \frac{\partial u^*}{\partial t}(0,t_*) = 0,\; (u^*(x,t))_t\geq0. Then by the comparison principle, we obtain \frac{\partial u^*}{\partial t}(x,t)\equiv0 for t\geq t_* , but this is impossible since by Theorem 4.1,

    \begin{equation*} u^*(x,t)-\phi_2(x+\beta+x^*+c_2t)\rightarrow0\; \text{as}\; t\rightarrow+\infty. \end{equation*}

    This ends the proof.

    Proof of Theorem 1.2. We first define a pair of sub- and super-solutions as follows.

    \begin{equation*} U^{\pm}(x,t) = u(x+a,t+t_0\pm\tilde{\delta}\varpi(1-e^{-\omega t}))\pm\varpi e^{-\omega t}, \end{equation*}

    where \omega is defined as in Section 4 and t_0\geq T_\varphi and \tilde{\delta},\; \varpi>0 are constants. Besides, we claim that U^{\pm} are super- and sub-solutions of 1. Then, by the comparison principle, we have

    \begin{equation} U^-(x,t)\leq v(x,t)\leq U^+(x,t)\; \text{for}\; x\in\mathbb{R},\; t\geq0. \end{equation} (11)

    In view of that for all |\tau|\leq\varpi with \varpi = :\varpi(\epsilon) = \frac{\epsilon}{2\tilde{M}} and \tilde{M} = 2+\max\limits_{u\in[0,1]}(f_L(x,u))_u , we have

    \begin{equation} \sup\limits_{x\in\mathbb{R},\; t\in\mathbb{R}}|u(x,t)-u(x,t+\tau)| \leq\sup\limits_{x\in\mathbb{R},\; t\in\mathbb{R}}|u_t(x,t)|\varpi\leq\frac{\epsilon}{2}. \end{equation} (12)

    It then follows from 11 and 12 that

    \begin{equation*} |v(x,t,v_0)-u(x+a,t+t_0)| < \epsilon \end{equation*}

    for x\in\mathbb{R} and t\geq0 .

    Now we prove the claim. We show U^+(x,t) is a super-solution of 1 for t\geq0 . Then it can be similarly shown that U^-(x,t) is a sub-solution of 1 for t\geq0 . Since it is easy to see that U^+(x,0)\geq v_0(x)\; \text{for}\; x\in\mathbb{R}\; \text{and}\; \delta<\frac{\varpi}{2}, we only need to show that

    \begin{equation*} \begin{split} \mathcal{L}U^+(x,t) = &U_t^+(x,t)-(J*U^+(x,t)-U^+(x,t))-f(x,U^+)\\ = &\tilde{\delta}\varpi\omega e^{-\omega t}-\varpi\omega e^{-\omega t}+f(x,u)-f(x,U^+)\\ \geq&0. \end{split} \end{equation*}

    We go further to show \mathcal{L}U^+(x,t)\geq0 in two cases.

    Case 1. For any u = u(x+a,t+\gamma+\tilde{\delta}\varpi(1-e^{-\omega t}))\in[0,\sigma]\cup[1-\sigma,1] with \sigma defined as in Section 4,

    \begin{equation*} \begin{split} \mathcal{L}U^+(x,t)\geq&-\min\left\{\frac{|f_u(x,0)|}{2},\frac{|(f_u(x,1)|}{2}\right\}\varpi e^{-\omega t}-\omega\varpi e^{-\omega t}\\ \geq&\varpi e^{-\omega t}(2\omega-\omega)\\ \geq&0. \end{split} \end{equation*}

    Case 2. For u\in[\sigma,1-\sigma] , it follows from Lemma 5.1 that there exists K_\sigma such that u_t\geq K_\sigma . Therefore,

    \begin{equation*} \begin{split} \mathcal{L}U^+(x,t)\geq&\tilde{\delta}\varpi\omega K_\sigma e^{-\omega t}-\varpi\omega e^{-\omega t}-\|f_u(x,u)\|_\infty\varpi e^{-\omega t}\\ \geq&\varpi e^{-\omega t}\left(\tilde{\delta}\omega K_\sigma-\omega-\|f_u(x,u))\|_\infty\right). \end{split} \end{equation*}

    Let \tilde{\delta} be sufficiently large such that

    \begin{equation*} \tilde{\delta}\geq\frac{\omega K_\sigma}{\omega+\|f_u(x,u)\|_\infty}. \end{equation*}

    Thus, we have \mathcal{L}U^+(x,t)\geq0 . Now, we obtain the claim. The proof of Theorem 1.2 is completed.

    Remark 1. In this paper, we have considered the existence, uniqueness, asymptotic behavior and Lyapunov stability of entire solutions of the nonlocal dispersal equation 1 under the assumption J(x) = J(-x) , x\in\mathbb{R} . It is well-known from [8,22,28,29] that the asymmetry of J has a great influence on the profile of the traveling waves and the sign of the wave speeds, which further makes the properties of the entire solution more diverse. Naturally, an interesting problem is to consider entire solutions of 1 under asymmetric conditions.

    Remark 2. The method used here can be also applied to consider the bistable lattice differential equations with f_i satisfying (F)

    \begin{equation} \dot{u}_i(t) = u_{i+1}(t)+u_{i-1}(t)-2u_i(t)+f_i(u_i(t)),\; i\in\mathbb{Z},\; t\in\mathbb{R}. \end{equation} (13)

    The existence, uniqueness, asymptotic behavior and the Lyapunov stability of entire solutions to 13 can be similarly obtained.

    While for bistable random diffusion equations with f(x,u) satisfying (F)

    \begin{equation*} u_t = u_{xx}+f(x,u),\; x\in\mathbb{R},\; t\in\mathbb{R}, \end{equation*}

    the construction of the entire solution behaving as the traveling wave pertaining to f_1 coming from infinity is similar. In fact, since regularity of the entire solution is known, the Lyapunov function method can be used to show the approach of the entire solution to the traveling wave pertaining to f_2 , one can refer to [10,11].



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