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Research article Special Issues

Local dynamics and coexistence of predator–prey model with directional dispersal of predator

  • 2 The author is now with the Department of Mathematics, Tamkang University, 151, Yingzhuan Road, Tamsui, New Taipei City, 251301, Taiwan.
  • Received: 15 June 2020 Accepted: 23 September 2020 Published: 30 September 2020
  • In this paper, we study the effect of directional dispersal of a predator on a predator– prey model. The prey is assumed to have traits making it undetectable to the predator and difficult to chase the prey directly. Directional dispersal of the predator is described when the predator has learned the high hunting efficiency in certain areas, thereby dispersing toward these areas instead of directly chasing the prey. We investigate the stability of the semi-trivial solution and the existence of a coexistence steady-state. Moreover, we show that the predator that moves toward a high-predation area may make the predators survive under the condition the predators cannot survive when they disperse randomly. The results are obtained through eigenvalue analysis and fixed-point index theory. Finally, we present the numerical simulation and its biological interpretations based on the obtained results.

    Citation: Kwangjoong Kim, Wonhyung Choi. Local dynamics and coexistence of predator–prey model with directional dispersal of predator[J]. Mathematical Biosciences and Engineering, 2020, 17(6): 6737-6755. doi: 10.3934/mbe.2020351

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  • In this paper, we study the effect of directional dispersal of a predator on a predator– prey model. The prey is assumed to have traits making it undetectable to the predator and difficult to chase the prey directly. Directional dispersal of the predator is described when the predator has learned the high hunting efficiency in certain areas, thereby dispersing toward these areas instead of directly chasing the prey. We investigate the stability of the semi-trivial solution and the existence of a coexistence steady-state. Moreover, we show that the predator that moves toward a high-predation area may make the predators survive under the condition the predators cannot survive when they disperse randomly. The results are obtained through eigenvalue analysis and fixed-point index theory. Finally, we present the numerical simulation and its biological interpretations based on the obtained results.


    In this article, we prove the non-existence of solutions to the following quasilinear elliptic problem which has degenerate coercivity in their principal part by approximation,

    {div(a(x,u,u))+|u|q1u=λ,xΩ,u=0,xΩ, (1)

    where 1<p<N,q>1 and λ is a Radon measure. Ω is a bounded smooth subset of RN(N>2). a(x,t,ξ):Ω×R×RNRN is the Carathéodory function (i.e: a(x,t,ξ) is measure on Ω for every (t,ξ) in R×RN, and a(,t,ξ) is continuous on R×RN for almost every x in Ω), such that the following assumptions hold,

    a(x,t,ξ)ξc|ξ|p(1+|t|)θ(p1), (2)
    |a(x,t,ξ)|c0(|ξ|p1+b(x)), (3)
    [a(x,t,ξ)a(x,t,ξ)][ξξ]>0, (4)

    for almost every xΩ,tR,ξ,ξRN with ξξ, where 0θ<1, c and c0 are two positive constants, bLp(Ω) is a non-negative function, p is the conjugate Hölder exponent of p.

    It is well-known that[3,9], problem Δu+|u|q1u=δ0 has no distributional solution if qNN2. On the other hand, if q<NN2, then there exists a unique solution to

    {Δu+|u|q1u=δ0,xΩ,u=0,xΩ.

    In the famous work [9], Brezis proved that if {un} is sequence of solution to the nonlinear elliptic problem

    {Δun+|un|q1un=fn,xΩ,un=0,xΩ, (5)

    with q>NN2, and fnL(Ω) is a sequence functions such that, for any ϱ>0,

    limnΩBϱ(0)|fnf|=0.

    Then un converges to the unique solution u to the following equation

    {Δu+|u|q1u=f,xΩ,u=0,xΩ.

    This fact shows that Bϱ(0) is a removable singularity set of solution to equation (5) provided q>NN2. Orsina and Prignet[24] extended the result of [9] to more general operator div(a(x,u,u)), where a(x,u,u) satisfies (2)-(4) with θ=0. The main results of [24] shown that problem (1) with θ=0 has a solution for every given bounded measure λ if q<r(p1)rp. Some other related results see [12,6,10,8,14,23,26,27,21,19,16] and references therein.

    The main goal of this paper is to study the non-existence of solutions to problem (1). More precisely, consider the limit of approximating equation (9)(see Theorem 1.2 below), our main task is to understand which is the limit of solutions to (9) and what equation it satisfies. A point worth emphasizing is that, even if p=2, the convergence of solutions is not true if the right hand side are distributions weakly converging in W1,2(Ω), see [5] for some counterexamples.

    In order to state the main results of this paper, we need some definitions.

    Let K be a compact subset of Ω, r>1 is a real number. The r capacity of K respect to Ω is defined as

    capr(K,Ω)=inf{urW1,r0:uCc(Ω),uχK},

    where χK is the characteristic function of K.

    Let λ be a bounded measure on Ω, we say that λ is concentrated on a set E if λ(B)=λ(BE) for every Borel subset B of Ω. Thanks to the Hahn decomposition, λ can be decomposed as the difference of two nonnegative mutually singular measure, that is λ=λ+λ.

    If λ is concentrated on a set E, as a consequence of the fact that λ+ and λ are mutually singular, we have that λ+ and λ concentrated a set E+ and E respectively and E+E=.

    Let λ=λ+λ be a measure, fn=f+nfn approximations of λ in the following way:

    limn+Ωf+nφdx=Ωφdλ+,limn+Ωfnφdx=Ωφdλ, (6)

    for every function φ, which is continuous and bounded on Ω, where {f+n} and {fn} are sequences of nonnegative L(Ω) functions. We not assume that f+n and fn are the positive and negative part of fn. Observe that choosing φ1 in (6), we obtain

    f+nL1(Ω)C,fnL1(Ω)C. (7)

    For all k>0,sR, define

    Tk(s)=max{k,min{k,s}},Gk(s)=sTk(s).

    Firstly we stale the existence result.

    Theorem 1.1. Let Ω be a bounded smooth subset of RN(N>2), 1<p<N, gL1(Ω) and (2)-(4) hold. Then there exists a unique entropy solution uW1,p0(Ω) to problem

    {div(a(x,u,u))+|u|q1u=g,xΩ,u=0,xΩ. (8)

    if

    q<N(1θ)N(1+θ(p1)).

    Moreover,

    uMp1(Ω),|u|Mp2(Ω),

    where Mp1,Mp2 represents the Marcinkiewicz space with exponent

    p1=N(p1)(1θ)Np,p2=N(p1)(1θ)N(1+θ(p1)).

    Remark 1. The previous result gives existence and uniqueness of the entropy solution uW1,p0(Ω) to (8) for every 1<p<N and 0<θ<1. If θ=0, the same result for (8) can be proved by the same techniques of [2].

    Our main results are following:

    Theorem 1.2. Let 1<p<rN and λ=λ+λ be a bounded Radon measure which is concentrated on a set E with zero r capacity. Let fn=f+nfn be a sequence of L(Ω) functions which converge to λ in the sense of (6). gL1(Ω) and let gn is a sequence of L(Ω) functions which converge to g weakly in L1(Ω). Suppose unW1,p0(Ω) is the solution to problem:

    {div(a(x,un,un))+|un|q1un=fn+gn,xΩ,un=0,xΩ. (9)

    Then |un|p1 strong converges to |u|p1 in Lσ(Ω) as n for

    σ<pq(q+1+θ(p1))(p1),

    if

    q>r(p1)[1+θ(p1)]rp, (10)

    where u is unique solution of (8). Moreover,

    limn+Ω|un|q1unφdx=Ω|u|q1uφdx+Ωφdλ,φC(Ω). (11)

    Remark 2. The above theorem shows that there is not a solution to problem (1) can be obtained by approximation, if q is large enough and the measure λ is concentrated on a set with zero r capacity.

    Remark 3. Boccardo et.al [7] considered the non-existence result to the following problem

    {div(a(x,u)(1+u)γ)+u=μ,xΩ,u=0,xΩ, (12)

    where γ>1 and μ is a non-negative Radon measure, concentrated on a set E with zero harmonic capacity, a(x,ξ) satisfies (2)-(4) with θ=0, p=2 and b(x)=0. While in Theorem 1.2, λ is a bounded Radon measure concentrated on a set E with zero r capacity with p<rN, instead of p capacity. Therefore Theorem 1.2 is not a triviality extend the results of Theorem 4.1 of [7]. Furthermore, in Theorem 1.2, θ(p1)(0,p1) since θ(0,1). Note that, in problem (12), they required that γ>1. It is worth pointing out that different ranges of γ have an important impact on the behavior of solutions to problem (12), more details see [25,18,1,17,13,4].

    The structure of this paper is as follows: Section 2 mainly gives some lemmas which play a important role in the process of proof of the main theorem. The proof of theorem 1.1 and 1.2 are given in Section 3.

    In the following, C is a constant and its value may changes from line to line.

    In order to prove Theorem 1.1 and 1.2, the following basic lemmas and definitions are required.

    Lemma 2.1. (see Lemma 2.1 of [22]) Let K+ and K be two disjoin compact subsets of Ω with zero r capacity, λ=λ+λ be a measure which is concentrated on a set with zero r capacity with 1<rN, Then there exist two functions ψ+δ and ψδ in Cc(Ω), such that

    0ψ+δ1,0ψδ1,Ω|ψ+δ|rdxδ,Ω|ψδ|rdxδ,0Ω(1ψ+δ)dλ+δ,0Ω(1ψδ)dλδ,0Ωψδdλ+δ,0Ωψ+δ)dλδ,ψ+δ1,xK+,ψ+δ1,xK, (13)

    for every δ>0.

    Definition 2.2. Let u be an measurable function on Ω such that Tk(u)W1,p0(Ω) for every k>0. Then there exist a unique measurable function v:ΩRN such that

    Tk(u)=vχ{|u|k},a.einΩandforeveryk>0.

    Define the gradient of u as the function v and denote it by v=u.

    Definition 2.3. Let fL1(Ω), q>0 and (2)-(4) hold. A measurable function u is an entropy solution to problem (8), if Tk(u)W1,p0(Ω) for every k>0, |u|qL1(Ω) and

    Ωa(x,u,u)Tk(uφ)dx+Ω|u|q1uTk(uφ)dxΩgTk(uφ)dx,

    for every φW1,p0(Ω)L1(Ω).

    Definition 2.4. Marcinkiewicz space Ms(Ω)(s>0) is the space composed of all the measurable functions v that satisfy

    |{|υ|k}|Cks,

    for any k>0, where the constant C>0.

    If |Ω| is bounded and 0<ε<s1, then the following embedding relationship hold:

    Ls(Ω)Ms(Ω)Lsε(Ω).

    Lemma 2.5. Let uMs(Ω) with s>0. If there exist a constant ρ>0, such that for any k>0,

    Ω|Tk(u)|pdxCkρ,

    for some positive constant C. Then

    |u|Mpss+ρ(Ω).

    Proof. Let σ be a fixed positive real number, for every k>0,

    |{|u|>σ}|=|{|u|>σ,|u|k}|+|{|u|>σ,|u|>k}||{|Tk(u)|>σ}|+|{|u|>k}|. (14)

    Moreover,

    |{|Tk(u)|>σ}|1σpΩ|Tk(u)|pdxCkρσp. (15)

    Since uMs(Ω), by Definition 2.4, there exist a constant C such that

    |{|u|>k}|Cks. (16)

    Combining (14)-(16), we have

    |{|u|>σ}|Ckρσp+CksCkpss+ρ.

    Therefore, by Definition 2.4, we get |u|Mpss+ρ.

    Lemma 2.6. Let {un} be a sequence in W1,p0(Ω) and assume that there exist positive constants ρ and C with p>ρ, such that

    Ω|Tk(un)|pdxCkρ,

    for any k and n. Then there exists a subsequence, still denoted by {un}, which converges to a measurable function v almost everywhere in Ω.

    Lemma 2.7. Let u be an entropy solution to (8), then

    {k<|u|<k+h}|u|pdxCkθ(p1).

    Proof. For any given h and k>0,sR, define

    Tk,h(s)=Th(sTk(s))={sksgn(s),k|s|<k+h,h,|s|k+h,0,|s|k.

    Take Tk,h(u) as test function in (8), we have

    {k<|u|<k+h}(a(x,u,u)u)dx+Ω|u|q1uTk,h(u)dx=ΩgTk,h(u)dx. (17)

    Since uTk,h(u)0, we find

    {k<|u|<k+h}(a(x,u,u)u)dxΩgTk,h(u)dx, (18)

    and

    ΩgTk,h(u)dxh{|u|>k}|g|dxC. (19)

    According to the assumption (2) and (17)-(19), we get,

    {k<|u|<k+h}|u|pdxCkθ(p1).

    Proposition 1. Let uW1,p0(Ω) be an entropy solution to (8) and satisfy

    {|u|<k}|u|pdxCkρ (20)

    for every k>0 and p>ρ. Then uMp1(Ω), where p1=N(pρ)/(Np). More precisely, there exists C=C(N,p,θ)>0 such that

    |{|u|>k}|Ckp1.

    Proof. For every k>0, by the Sobolev embedding theorem and (20),

    Tk(u)pC(N,p,θ)Tk(u)pCkρp,

    where p=NpNp. For 0<ηk, we have

    {|u|η}={|Tk(u)η|}.

    Hence

    |{|u|>η}|Tk(u)ppηpC(kρ)ppηp.

    Setting η=k, we obtain

    |{|u|>k}|CkN(pρ)Np.

    This fact shows that uMp1(Ω) with p1=N(pρ)/(Np).

    Proposition 2. Assume that uW1,p0(Ω) is an entropy solution to (8), which satisfies (20) for every k. Then uMp2(Ω), where p2=N(pρ)/(Nρ), that is there exists C=C(N,p,θ)>0 such that

    |{|u|>h}|Chp2,

    for every h>0.

    Proof. For k,λ>0, set

    ψ(k,λ)=|{|u|p>λ,|u|>k}|.

    Using the fact that the function λψ(k,λ) is nonincreasing, we get, for k,λ>0,

    ψ(0,λ)=|{|u|p>λ}|1λλ0ψ(0,s)dsψ(k,0)+1λλ0ψ(0,s)ψ(k,s)ds. (21)

    By Proposition 1,

    ψ(k,0)Ckp1, (22)

    where p1=N(pρ)/(Np). Since ψ(0,s)ψ(k,s)=|{|u|p>s,|u|<k}|, thanks to (20), we have

    0ψ(0,s)ψ(k,s)ds={|u|<k}|u|pdxCkρ. (23)

    Combining (21)-(23), we arrive at

    ψ(0,λ)Ckρλ+Ckp1. (24)

    Let Ckρλ=Ckp1 and λ=hp, (24) implies that

    |{|u|>h}|ChN(pρ)Nρ.

    That is uMp2(Ω) with p2=N(pρ)/(Nρ).

    In this section we prove Theorem 1.1 and 1.2 combining the results of Sections 2.

    In the proofs of Theorem 1.1 and 1.2, ω(n,m,δ) will denote any quantity (depending on n,m and δ) such that

    limδ0+limm+limn+ω(n,m,δ)=0.

    If the quantity does not depend on one or more of the three parameters n,m and δ, we will omit the dependence from it in ω. For example, ω(n,δ) is any quantity such that

    limδ0+limn+ω(n,δ)=0.

    The proof of Theorem 1.1 will be divided in several steps.

    Proof. (1)Uniqueness: Let u1 and u2 be two entropy solutions to equation (8). The proof of the fact that u1=u2 will follow from the following four steps.

    Step 1. Assume that giL1(Ω),(i=1,2). Choosing Tk(u1Thu2) and Tk(u2Thu1) as test function in (8) respectively, we get

    I:=Ωa(x,u1,u1)Tk(u1Thu2)dx+Ωa(x,u2,u2)Tk(u2Thu1)dx=Ω|u1|q1u1Tk(u1Thu2)dxΩ|u2|q1u2Tk(u2Thu1)dx+Ωg1Tk(u1Thu2)dx+Ωg2Tk(u2Thu1)dx. (25)

    Step 2. Denote

    A0={xΩ:|u1u2|<k,|u1|<h,|u2|<h},A1={xΩ:|u1Thu2|<k,|u2|h},A2={xΩ:|u1Thu2|<k,|u2|<h,|u1|h}.

    For xA0,

    Tk(u1Thu2)=(u1u2)

    and

    Tk(u2Thu1)=Tk(u2u1).

    Thus, for every xA0,

    Ωa(x,u1,u1)Tk(u1Thu2)dx+Ωa(x,u2,u2)Tk(u2Thu1)dx=A0[a(x,u1,u1)a(x,u2,u2)](u1u2)dx:=I0. (26)

    For xA1, Tk(u1Thu2)=(u1h)=u1. By (2), we get

    Ωa(x,u1,u1)Tk(u1Thu2)dx=A1a(x,u1,u1)u1dx0. (27)

    For xA2, Tk(u1Thu2)=(u1u2). Thus

    Ωa(x,u1,u1)Tk(u1Thu2)dxA2a(x,u1,u1)u2dx. (28)

    Similarly, denote

    A1={xΩ:|u2Thu1|<k,|u1|h},A2={xΩ:|u2Thu1|<k,|u1|<h,|u2|h}.

    Then for xA1, Tk(u2Thu1)=(u2h)=u2. By (2), we get

    Ωa(x,u2,u2)Tk(u2Thu1)dx=A1a(x,u2,u2)u2dx0. (29)

    For xA2, Tk(u2Thu1)=(u2u1). Thus

    Ωa(x,u2,u2)Tk(u2Thu1)dxA2a(x,u2,u2)u1dx. (30)

    Summing up (26)-(30) in the form II0I1, where

    I1=A2a(x,u1,u1)u2dx+A2a(x,u2,u2)u1dx:=I11+I12.

    Now, we estimate I11. By the Hölder inequality and (3), we have

    I11a(x,u1,u1)Lp({h|u1|h+k})u2Lp({hk|u2|h})c0(u1p1Lp({h|u1|h+k})+b(x)Lp({|u1|h}))u2Lp({hk|u2|h}).

    Therefore, by Lemma 2.7 and Proposition 2, I110 as h for every k>0. I120 as h for every k>0 can be obtained in the same way.

    Hence, we find

    Ωa(x,u1,u1)Tk(u1Thu2)dx+Ωa(x,u2,u2)Tk(u2Thu1)dx=A0[a(x,u1,u1)a(x,u2,u2)](u1u2)dx+ε(h). (31)

    Step 3. Now estimate the terms on the right hand side of (25). Denote

    B0={xΩ:|u1|<h,|u2|<h},B1={xΩ:|u1|h},B2={xΩ:|u2|h}.

    For xB0, since Tk(u1Thu2)=Tk(u1u2) and Tk(u2Thu1)=Tk(u2u1), we arrive at

    Ω|u1|q1u1Tk(u1Thu2)dx+Ω|u2|q1u2Tk(u2Thu1)dx=B0(|u1|q1u1|u2|q1u2)Tk(u1u2)dx0, (32)

    and

    Ωg1Tk(u1Thu2)dx+Ωg2Tk(u2Thu1)dx=B0(g1g2)Tk(u1u2)dx0. (33)

    For xB1, since Tk(u2Thu1)=Tk(u2h). Then

    Ω|u1|q1u1Tk(u1Thu2)dx+Ω|u2|q1u2Tk(u2Thu1)dxkB1(|u1|q1u1+|u2|q1u2)dx:=J1,

    and

    Ωg1Tk(u1Thu2)dx+Ωg2Tk(u2Thu1)dxkB1(|g1|+|g2|)dx:=J2.

    For xB2, since Tk(u1Thu2)=Tk(u1h), we get

    Ω|u1|q1u1Tk(u1Thu2)dx+Ω|u2|q1u2Tk(u2Thu1)dxkB2(|u1|q1u1+|u2|q1u2)dx:=J1,

    and

    Ωg1Tk(u1Thu2)dx+Ωg2Tk(u2Thu1)dxkB2(|g1|+|g2|)dx:=J2.

    According to |B1|0,|B2|0 as h and |u|qL1(Ω) for fixed k>0, we get

    J1+J2+J1+J20ash. (34)

    Step 4. Combining (25) and (31)-(34), we have

    A0[a(x,u1,u1)a(x,u2,u2)](u1u2)dxε(h),

    where ε(h)0 as h. Since A0 converges to {xΩ:|u1u2|<k} by measure as h for fixed k>0, we conclude that

    {|u1u2|<k}[a(x,u1,u1)a(x,u2,u2)](u1u2)dx0,

    for all k>0. This fact, combine with (4), implies that u1=u2 a.e in Ω. Then we get u1=u2 a.e in Ω.

    (2) Existence:

    Step 1. Let

    F(x,u)=g(x)β(u),

    where β(u)=|u|q1u, which is continuous with respect to u. Then g(x)=F(x,0)L1(RN) and β is monotonous nondecreasing with respect to u with β(0)=0 and β(u)u0.

    Let gnC0, such that gn converges to g in L1(Ω), with gnL1(Ω)gL1(Ω) for every n1. Define βn(s)=Tn(β). In this way, |βn(s)||β(s)| for every sR and xΩ. Finally we take

    γn(s)=βn(s)+1n|s|p2s.

    Then by [20], there exists unW1,p0(Ω) such that

    {diva(x,un,un)+γn(x,un)=gn,xΩ,un=0,xΩ, (35)

    holds in the sense of distributions in Ω.

    By density arguments, we can take Th(unTk(un)) and Tk(un) as the test function in (35) respectively, we have

    {k|un|<k+h}a(x,un,un)undx+{|un|>k}γnTh(unTk(un))dx={|un|>k}gnTh(unTk(un))dx, (36)

    and

    {|un|>k}a(x,un,un)undx+ΩγnTk(un)dx=ΩgnTk(un)dx. (37)

    Combine (36) with (2) (fix the ellipticity constant c=1) and γnTh(unTk(un))0, we get,

    {k<|un|<k+h}|un|pdxhkθ(p1){|un|>k}gndxhkθ(p1)gnL1(Ω)=Ckθ(p1). (38)

    Since a(x,un,un)un0 by (2), we have

    {|un|>k}|γn(un)|dx{|un|>k}|gn|dxgnL1(Ω)C. (39)

    Combine (37) with γnTk(un)0, we have

    {|un|<k}|un|pdxCk1+θ(p1). (40)

    Step 2. Convergence. Using (38) and Proposition 1, we have |{|un|>k}| is bounded uniformly for every k>0. Thanks to (40), we see that {Tk(un)} is bounded in Lploc(Ω) for every k>0.

    Next we prove that unu locally in measure.

    For t,ϵ>0, we have

    {|unum|>t}{|un|>k}{|um|>k}{|Tk(un)Tk(um)|>t}.

    Thus

    |{|unum|>t}||{|un|>k}|+|{|um|>k}|+|{|Tk(un)Tk(um)|>t}|.

    Choosing k large enough such that |{|un|>k}|<ϵ and |{|um|>k}|<ϵ. Since {Tk(un)}n is bounded in Lp(Ω) and Tk(un)W1,p0(Ω) for every k>0. Assume that {Tk(un)} is a Cauchy sequence in Lq(ΩBR) for any q<pN/(Np) and any R>0,

    Tk(un)Tk(u)inLploc(Ω)anda.einΩ.

    Then

    |{|Tk(un)Tk(um)|>t}BR|tqΩBR|Tk(un)Tk(um)|qdxϵ,

    for all n,mn0(k,t,R). This show that {un} is a Cauchy sequence in BR. Hence that unu locally.

    Now to prove that un converges to some function v locally. We need to prove that {un} is a Cauchy sequence in any ball BR. Let t,ϵ>0 again, then

    {|unum|>t}BR{|unum|k,|un|l,|um|l,|unum|>t}{|un|>l}{|um|>l}({|unum|>k}BR).

    Choose l large enough such that |{|un|>l}|ϵ for all nN. If a is a continuous function independent of x, then by (4), there exists a μ>0, such that |ξ|<l,|ξ|<l and |ξξ|>t means

    [a(x,t,ξ)a(x,t,ξ)][ξξ]μ.

    This is a consequence of continuity and strict monotonicity of a. Set

    dn=gnγn(x,un). (41)

    Taking Tk(unum) as the test function of (35) and by (37), (41), we have

    {|unum|<k}[a(x,un,un)a(x,um,um)](unum)dx=Ω(dndm)Tk(unum)dxCk1+θ(p1).

    Then

    {|unum|k,|un|l,|um|l,|unum|>t}1μ{|unum|<k}[a(x,un,un)a(x,um,um)](unum)dx1μCk1+θ(p1)ϵ,

    if k is small enough such that k1+θ(p1)μϵ/C.

    Since l and k have been confirmed, if n0 large enough, we have |({|unum|>k}BR)|ϵ for n,mn0. Then we get |{|unum|>t}BR|4ϵ. This prove that un converges to some function v locally.

    Finally, since {Tk(un)}nLp(Ω) for every k>0, it converges weakly to {Tk(u)} in Lploc(Ω). We have uW1,p0(Ω) and u=v a.e in Ω.

    Step 3. In order to prove the existence of the solution completely, we still need to prove that sequence {a(x,u,u)}n is bounded in Lqloc(Ω) for all

    q(1,N(1θ)N(1+θ(p1))).

    Indeed, by Proposition 2, |un|p1MN(1θ)N(1+θ(p1))Lqloc(Ω). And by (3), we have |a(x,un,un)|Lp(Ω)Lqloc(Ω). According to the Nemitskii's theorem, unu implies that

    a(x,un,un)a(x,u,u).

    It follows that

    a(x,u,u)MN(1θ)N(1+θ(p1))Lqloc(Ω),

    for all q(1,N(1θ)N(1+θ(p1))).

    In this subsection, we give the proof of Theorem 1.2 following some ideas in [11,22].

    Proof. Step 1 (A priori estimates). Firstly, choosing Tk(un)(1φδ)s as test function in the weak formulation of (9), where s=ηηp+1 and η will be given in (48), we have

    Ωa(x,un,un)Tk(un)(1φδ)sdx+Ω|un|q1unTk(un)(1φδ)sdx=sΩa(x,un,un)φδTk(un)(1φδ)s1dx+ΩgnTk(un)(1φδ)sdx+Ωf+nTk(un)(1φδ)sdx+ΩfnTk(un)(1φδ)sdx. (42)

    By (2), we get

    Ωa(x,un,un)Tk(un)dμcΩ|Tk(un)|p(1+|Tk(un)|)θ(p1)dμ, (43)

    here dμ:=(1φδ)sdx.

    Since unTk(un)0,

    Ω|un|q1unTk(un)(1φδ)sdx{|un|k}|un|q1unTk(un)dμkq+1μ({|un|k}). (44)

    Using (3) and the Young inequality, we find

    Ω|a(x,un,un)φδTk(un)(1φδ)s1|dxc0kΩ(|un|p1+b(x))(|φ+δ|+|φ+δ|)(1φδ)s1dxCkΩ(|un|(p1)r+|b(x)|r)(1φδ)(s1)rdx+CkΩ(|φ+δ|r+|φ+δ|r)dxCk(Ω(|un|(p1)r+|b(x)|r)(1φδ)(s1)rdx+δ). (45)

    Combine (42)-(45), by (7) and {gn}L1(Ω), bLp(Ω), we have

    Ω|Tk(un)|p(1+|Tk(un)|)θ(p1)dμ+kq+1μ({|un|k})Ck(Ω|un|(p1)r(1φδ)(s1)rdx+δ+μ(Ω). (46)

    For a fixed σ0, thanks to (46), we get

    μ({|un|>σ})=μ({|un|>σ,|un|<k})+μ({|un|>σ,|un|k})1σpΩ|Tk(un)|pdμ+μ({|u|>k})(1+k)θ(p1)σpΩ|Tk(un)|p(1+|Tk(un)|)θ(p1)dμ+μ({|u|>k})C(Ω|un|(p1)r(1φδ)(s1)rdx+δ+μ(Ω))((1+k)1+θ(p1)σp+1kq),

    which implies

    μ|{|un|>σ}|Cσpqq+1+θ(p1)(Ω|un|(p1)r(1φδ)(s1)rdx+δ+μ|Ω|). (47)

    Let

    (p1)r<η<pqq+1+θ(p1). (48)

    Clearly, such η exists by (10). In view of (47)-(48), we have

    Ω|un|ηdμC(Ω|un|(p1)r(1φδ)(s1)rdx+δ+μ(Ω)).

    By the Holder's inequality,

    Ω|un|(p1)r(1φδ)(s1)rdxC(Ω|un|ηdμ)(p1)rηC(Ω|un|(p1)r(1φδ)(s1)rdx+δ+μ|Ω|)(p1)rη.

    By Lemma 2.1, 1φδ is zero both on a neighbourhood of K+ and K. Hence

    Ω|un|(p1)r(1φδ)(s1)rdxC(δ+μ|Ω|)C(δ). (49)

    Using (46) and (49), we conclude that

    Ω|Tk(un)|pdxCk1+θ(p1). (50)

    According to Lemma 2.5, we have |un|Ms(Ω), where s=pqq+1+θ(p1).

    By (50) and Lemma 2.6, there exists a subsequence, still denoted by un, which converges to a measurable function u almost everywhere in Ω. So Tk(un)Tk(u) in Ω for every k>0.

    Since Tk(un)W1,p0(Ω), by the weak lower semi-continuity of the norm, Tk(u)W1,p0(Ω) for every k>0. Thus u has an gradient u in the sense of Definition 2.2, as a consequence of the a priori estimates on un and (4), we have

    a(x,un,un)a(x,u,u)stronglyin(Ls(Ω))N, (51)

    for every s<pq(q+1+θ(p1))(p1).

    Step 2 (Energy estimates). Let ψδ=ψ+δ+ψδ, where ψ+δ and ψδ are as in Lemma 2.1. Then

    {un>2m}uqn(1ψδ)dx=ω(n,m,δ), (52)

    and

    {un<2m}|un|q(1ψδ)dx=ω(n,m,δ). (53)

    Choose βm(un)(1ψδ) as test function in the weak formulation of (9), where

    βm(s)={sm1,m<s2m,1,s>2m,0,sm.

    We obtain

    1m{m<un<2m}a(x,un,un)un(1ψδ)dx(A)Ωa(x,un,un)ψδβm(un)dx(B)+Ω|un|q1unβm(un)(1ψδ)dx(C)=Ωf+nβm(un)(1ψδ)dx(D)Ωfnβm(un)(1ψδ)dx(E)+Ωgnβm(un)(1ψδ)dx.(F)

    Since (A) and (E) are non-negative, we can get rid of them. And since βm(um) converges to βm(u) almost everywhere in Ω and in the weaktopology of L(Ω), βm(un) converges to zero in the weaktopology of L(Ω) as m, we have

    (B)=Ωa(x,u,u)ψδβm(u)dx+ω(n)=ω(n,m),

    and

    (C){un>2m}uqn(1ψδ)dx.

    By ψδ=ψ+δ+ψδ and (6),

    (D)Ωf+n(1ψδ)dx=Ω(1ψ+δ)dλ+Ωψδdλ+ω(n)=ω(n,δ),

    and

    (F)=ω(n,m).

    We get (52), the proof of (53) is identical.

    Step 3 (Passing to the limit). Now we show that u is an entropy solution to (8) with datum g. Choose Tk(unφ)(1ψδ) as test function in the weak formulation of (9), we get

    Ωa(x,un,un)Tk(unφ)(1ψδ)dx(A)Ωa(x,un,un)ψδTk(unφ)dx(B)+Ω|un|q1unTk(unφ)(1ψδ)dx(C)=Ωf+nTk(unφ)(1ψδ)dx(D)
    ΩfnTk(unφ)(1ψδ)dx(E)+ΩgnTk(unφ)(1ψδ)dx.(F)

    By (13),

    (A)={|unφ|<k}a(x,un,un)un(1ψδ)dx{|unφ|<k}a(x,un,un)φ(1ψδ)dx,

    while

    {|unφ|<k}a(x,un,un)φ(1ψδ)dx={|uφ|<k}a(x,u,u)φdx+ω(n,δ).

    The Fatou lemma implies

    {|uφ|<k}a(x,u,u)udxlimninf{|unφ|<k}a(x,un,un)undx.

    Using (13), (51), we have

    (B)=Ωa(x,u,u)ψδTk(uφ)dx+ω(n)=ω(n,δ).

    While

    (F)=ΩgTk(uφ)dx+ω(n,δ),

    and

    |(D)|+|(E)|=Ω(f+n+fn)Tk(unφ)(1ψδ)dxkΩ(f+n+fn)(1ψδ)dx=ω(n,δ).

    So that we only need to deal with (C). Let m>k+φL(Ω) be fixed,

    (C)={2mun2m}|un|q1unTk(unφ)(1ψδ)dx(G)+k{un>2m}uqn(1ψδ)dx+k{un<2m}|un|q(1ψδ)dx.(H)

    By (52) and (53), we get

    (H)=ω(n,m,δ),

    and

    (G)=Ω|u|q1uTk(uφ)(1ψδ)dx+ω(n,m)=Ω|u|q1uTk(uφ)dx+ω(n,m,δ).

    Summing up the result of (A)-(H), we have

    Ωa(x,u,u)Tk(uφ)dx+Ω|u|q1uTk(uφ)dxΩgTk(uφ)dx.

    Thus u is the entropy solution of (8).

    Finally we prove (10). Choose φCc(Ω) as test function in the weak formulation of (9), we get

    Ωa(x,un,un)φdx+Ω|un|q1unφdx=Ω(fn+gn)φdx.

    Thanks to the assumptions of fn, gn and by (51),

    limn+Ω|un|q1unφdx=Ωa(x,u,u)φdx+Ωgφdx+Ωφdλ. (54)

    Since the entropy solution of (8) is also a distributional solution of the same problem, for the same φ,

    Ωa(x,u,u)φdx+Ω|u|q1uφdx=Ωgφdx. (55)

    Together with (54) and (55), we find

    limn+Ω|un|q1unφdx=Ω|u|q1uφdx+Ωφdλ.

    Thus (11) holds for every φCc(Ω). Since |un|q1un is bounded in L1(Ω), (11) can be extended by density to the functions in Cc(Ω).

    The authors also would like to thank the anonymous referees for their valuable comments which has helped to improve the paper.



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