Research article Special Issues

Global existence and blow up of solutions for a class of coupled parabolic systems with logarithmic nonlinearity


  • According to the difference of the initial energy, we consider three cases about the global existence and blow-up of the solutions for a class of coupled parabolic systems with logarithmic nonlinearity. The three cases are the low initial energy, critical initial energy and high initial energy, respectively. For the low initial energy and critical initial energy J(u0,v0)d, we prove the existence of global solutions with I(u0,v0)0 and blow up of solutions at finite time T<+ with I(u0,v0)<0, where I is Nehari functional. On the other hand, we give sufficient conditions for global existence and blow up of solutions in the case of high initial energy J(u0,v0)>d.

    Citation: Qigang Deng, Fugeng Zeng, Dongxiu Wang. Global existence and blow up of solutions for a class of coupled parabolic systems with logarithmic nonlinearity[J]. Mathematical Biosciences and Engineering, 2022, 19(8): 8580-8600. doi: 10.3934/mbe.2022398

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  • According to the difference of the initial energy, we consider three cases about the global existence and blow-up of the solutions for a class of coupled parabolic systems with logarithmic nonlinearity. The three cases are the low initial energy, critical initial energy and high initial energy, respectively. For the low initial energy and critical initial energy J(u0,v0)d, we prove the existence of global solutions with I(u0,v0)0 and blow up of solutions at finite time T<+ with I(u0,v0)<0, where I is Nehari functional. On the other hand, we give sufficient conditions for global existence and blow up of solutions in the case of high initial energy J(u0,v0)>d.



    In this paper, we consider the following initial-boundary value problem for a class of coupled parabolic systems with logarithmic nonlinearity.

    {utΔu=|v|p|u|p2ulog(|uv|),xΩ, t>0,vtΔv=|u|p|v|p2vlog(|uv|),xΩ, t>0,u(x,0)=u0(x),xΩ,v(x,0)=v0(x),xΩ,u(x,t)=v(x,t)=0,(x,t)Ω×(0,T], (1.1)

    where (u0,v0)H10(Ω)×H10(Ω), T(0,+), ΩRn(n2) is a bounded domain with smooth boundary Ω and p satisfies the following assumptions:

    2<p<2:={,if n=2,2nn2,if n3. (1.2)

    Among the fields of mathematical physics, biosciences and engineering, problem (1.1) is one of the most important reaction-diffusion coupled systems with logarithmic nonlinearity. It can be used not only to predict the time evolution of various population density distributions, but also to describe the thermal propagation of a two-component combustible mixture [1,2,3]. In recent years, this kind of systematic research has attracted many mathematicians and has made remarkable progress [4,5,6,7,8]. In order to overcome the special difficulties brought by nonlinear terms, many new ideas and tools have been developed, which greatly enrich the theory of partial differential equations [9,10,11,12,13,14].

    In the past years, many authors made efforts to the investigation of the existence and blow up of solutions for such kinds of systems. Galaktionov et al. [15,16] investigated the following semilinear reaction-diffusion system

    {utΔu=vp,vtΔv=uq. (1.3)

    They proved the local and global existence of solutions for the initial boundary value problem of (1.3). Subsequently, Escobedo and Herrero [17] considered the initial boundary value problem of (1.3) for a bounded open domain on Rn with smooth boundary. They obtained global solution under the condition 0<pq1, meanwhile global solution and blow up in finite time depending on sufficient small or large initial value and pq>1. For more studies on problem (1.3) we refer the interested reader to [18,19,20] and references therein.

    Recently, Xu et al. [21] considered the following nonlinear reaction-diffusion systems

    {utΔu=(|u|2p+|v|p+1|u|p1)u,xΩ,t>0,vtΔv=(|v|2p+|u|p+1|v|p1)v,xΩ,t>0,u(x,0)=u0(x),xΩ,v(x,0)=v0(x),xΩ,u(x,t)=v(x,t)=0,(x,t)Ω×(0,T]. (1.4)

    When initial energy J(u0,v0)d, by virtue of Galerkin method [22] and concave function method [23], global existence and finite time blow-up of the solutions for the problem (1.4) were obtained. When initial energy J(u0,v0)>d, they discussed global existence, finite time blow-up of solutions and tried to find out the corresponding initial data with arbitrarily high initial energy. What's more, by using comparison principle and the ideas in [24,25], they described the structures of the initial data and gave some sufficient conditions of the initial data which ensured the finite time blow up and global existence of the solutions, respectively.

    Inspired by the above works, we aim to use the Galerkin method, logarithmic inequalities [26], and concave function method to prove the global existence, decay, finite time blow-up of solutions for problem (1.1) with initial energy J(u0,v0)d. When high initial energy J(u0,v0)>d, by constructing two sets Φα and Ψα defined as (5.1) and (5.2), we prove that the weak solution will blow up in finite or infinite time if the initial value belongs to Ψα, while the weak solution will exist globally and tends to zero as time t+ when the initial value belongs to Φα.

    The organization of the remaining part of this paper is as follows. In Section 2, we introduce some preliminaries and lemmas of this paper. In Sections 3–5, we will give our main results and the corresponding proofs.

    Throughout this paper, we denote by uγ the norm of Lγ(Ω) for 1γ+ and by uH10(Ω) the norm of H10(Ω). For uLγ(Ω),

    uγ={(Ω|u(x)|γdx)1γ,if1γ<+,ess supxΩ|u(x)|,ifγ=+,

    and for uH10(Ω),

    u2H10(Ω)=u22+u22.

    By virtue of Poincaré inequality, we know that u2H10(Ω) and u22 are equivalent norms to each other, i.e., there exist C1 and C2 such that

    C1u22u2H10(Ω)C2u22,

    which is denoted by u2H10(Ω)u22. In addition, we denoted by (,) the inner product in L2(Ω) and c is an arbitrary positive number which may be different from line to line.

    For (u,v)H10(Ω)×H10(Ω), we define the Nehari functional I and energy functional J as follows:

    I(u,v)=Ω|u|2dx+Ω|v|2dx2Ω|uv|p|log(|uv|)dxu2H10(Ω)+v2H10(Ω)2Ω|uv|plog(|uv|)dx, (2.1)
    J(u,v)=12Ω|u|2dx+12Ω|v|2dx+1p2Ω|uv|pdx1pΩ|uv|plog(|uv|)dx12u2H10(Ω)+12v2H10(Ω)+1p2uvpp1pΩ|uv|plog(|uv|)dx. (2.2)

    From (2.1) and (2.2), we have

    J(u,v)12pI(u,v)+p12p(u2H10(Ω)+v2H10(Ω))+1p2uvpp. (2.3)

    Let

    N:={(u,v)H10(Ω)×H10(Ω) {(0,0)}| I(u,v)=0}

    be the Nehari manifold. Furthermore, the potential well W and its corresponding set V are defined respectively by

    W={(u,v)H10(Ω)×H10(Ω)I(u,v)>0,J(u,v)<d}{(0,0)},
    V={(u,v)H10(Ω)×H10(Ω)I(u,v)<0,J(u,v)<d},

    where

    d:=inf(u,v)H10(Ω)×H10(Ω){(0,0)}sups1,s2>0J(s1u,s2v)=inf(u,v)NJ(u,v) (2.4)

    is the depth of the potential well W.

    To consider the weak solution with high energy level, we need to introduce some new notions.

    Jα={(u,v)H10(Ω)×H10(Ω)|J(u,v)<α},
    Nα=NJα={(u,v)N|p12p(u2H10(Ω)+v2H10(Ω))+1p2uvpp<α},

    and

    λα=inf{12(u22+v22)|(u,v)Nα}, Λα=sup{12(u22+v22)|(u,v)Nα}forallα>d,

    Clearly, λα is nonincreasing, and Λα is nondecreasing with respect to α, respectively.

    Now, we give the definitions of the weak solution, maximal existence time and finite time blow up of the problem (1.1) as follows.

    Definition 1. (Weak solution) We say that (u,v) = (u(x,t),v(x,t))L([0,T),H10(Ω)×H10(Ω)) with (ut,vt)L2([0,T),L2(Ω)×L2(Ω)) is a weak solution of problem (1.1) on Ω×[0,T), if it satisfies the initial condition u(x,0)=u0(x),v(x,0)=v0(x) in H10(Ω),

    (ut,w1)+(u,w1)=(|v|p|u|p2ulog(|uv|),w1)

    and

    (vt,w2)+(v,w2)=(|u|p|v|p2vlog(|uv|),w2)

    for all w1,w2H10(Ω) and t(0,T). Moreover, for all t(0,T), we have

    t0uτ22+vτ22dτ+J(u,v)J(u0,v0). (2.5)

    Remark 1. For the global weak solution (u(t),v(t))=(u(x,t),v(x,t)) of problem (1.1), we define the ω-limit set of (u0,v0) by

    ω(u0,v0):=t0¯{u(s),v(s):st}.

    Definition 2. (Maximal existence time) Let (u,v)=(u(x,t),v(x,t)) be a weak solution of problem (1.1). We define the maximal existence time of (u,v) as follows

    (i) If (u,v) exists for all t[0,+), then T=+.

    (ii) If there exists a t0(0,+) such that (u,v) exists for 0t<t0, but it does not exist at t=t0, then T=t0.

    Definition 3. (Finite time blow-up) Let (u(t),v(t))=(u(x,t),v(x,t)) be a weak solution of problem (1.1). We say (u(t),v(t)) blows up in finite time if the maximal existence time T is finite and

    limtTu(t)22+v(t)22=+.

    Lemma 1. Let (u,v)H10(Ω)×H10(Ω){(0,0)}, then the following hold

    (i) limλ0J(λu,λv)=0, limλ+J(λu,λv)=.

    (ii) There exists a unique λ>0 such that ddλJ(λu,λv)|λ=λ=0.

    (iii) J(λu,λv) is increasing on (0,λ), decreasing on (λ,+), and attains the maximum at λ=λ.

    (iv) I(λu,λv)>0 for 0<λ<λ, I(λu,λv)<0 for λ<λ<+, and I(λu,λv)=0.

    Proof. (i) By definition of J(u,v) and λ>0, we have

    J(λu,λv)=λ22u2H10(Ω)+λ22v2H10(Ω)+λ2pp2uvppλ2pplogλ2uvppλ2ppΩ|uv|plog(|uv|)dx.

    Thus limλ0J(λu,λv)=0, limλ+J(λu,λv)=.

    (ii) Differentiating J(λu,λv) with respect to λ, we get

    ddλJ(λu,λv)=λu2H10(Ω)+λv2H10(Ω)2λ2p1logλ2uvpp2λ2p1Ω|uv|plog(|uv|)dx=λ(u2H10(Ω)+v2H10(Ω)2λ2p2logλ2uvpp2λ2p2Ω|uv|plog(|uv|)dx).

    Setting g(λ)=u2H10(Ω)+v2H10(Ω)2λ2p2logλ2uvpp2λ2p2Ω|uv|plog(|uv|)dx, we have limλ0g(λ)=u2H10(Ω)+v2H10(Ω)>0, limλ+g(λ)=, and

    g(λ)=2(2p2)λ2p3logλ2uvpp4λ2p3uvpp2(2p2)λ2p3Ω|uv|plog|uv|dx<0.

    Thus there exists a unique λ>0 such that g(λ)=0, i.e., ddλJ(λu,λv)|λ=λ=0.

    (iii) It is easy to find that J(λu,λv) is strictly increasing on (0,λ], strictly decreasing on (λ,+) and taking the maximum at λ=λ.

    (iv) Since

    I(λu,λv)=λu2H10(Ω)+λv2H10(Ω)2Ω|λuλv|plog(|λuλv|)dx=λddλJ(λu,λv),

    then the conclusion follows immediately.

    Lemma 2. Assume (1.2) holds, let (u,v)H10(Ω)×H10(Ω) satisfy I(u,v)<0, then

    I(u,v)<2p(J(u,v)d). (2.6)

    Proof. According to I(u,v)<0 and Lemma 1, we have (u,v)(0,0) and there exists a λ(0,1) such that I(λu,λv)=0, i.e., J(λu,λv)d. For λ>0, set

    h(λ)=2pJ(λu,λv)I(λu,λv)=(p1)λ2(u2H10(Ω)+v2H10(Ω))+2pλ2puvPp,

    then

    h(λ)=2(p1)λ(u2H10(Ω)+v2H10(Ω))+4λ2p1uvPp>0.

    Hence h(λ) is strictly increasing for λ>0. Together with λ(0,1), it follows that h(1)>h(λ). i.e.,

    2pJ(u,v)I(u,v)>2pJ(λu,λv)I(λu,λv)=2pJ(λu,λv)2pd.

    Then (2.6) follows immediately.

    Lemma 3. Assume (1.2) holds, let (u0,v0)H10(Ω)×H10(Ω) and (u(t),v(t))=(u(x,t),v(x,t)) be a weak solution of problem (1.1). If J(u0,v0)<d and I(u0,v0)<0, then (u(t),v(t))V for all 0tT, where T is the maximal existence time of (u(t),v(t)).

    Proof. We will show that (u(t),v(t))V for 0tT. Arguing by contradiction, suppose that t0[0,T] be the smallest time for which (u(t0),v(t0))V, then by the continuity of the (u(t),v(t)), we get (u(t0),v(t0))V. Hence, it follows that

    I(u(t0),v(t0))=0 (2.7)

    or

    J(u(t0),v(t0))=d. (2.8)

    If (2.7) is true, then (u(t0),v(t0))N, J(u(t0),v(t0))>d, which contradicts with (2.5). While if (2.8) is true, it also contradicts with (2.5). Consequently, we have (u(t),v(t))V for all 0tT.

    Lemma 4. Let (u, v) be a weak solution of problem (1.1). Then for all t[0,T),

    ddt(u22+v22)=2I(u,v).

    Proof. The proof of Lemma 4 directly follows by choosing w1=u,w2=v in Definition 1.

    In this section, we prove global existence and finite time blow up of solutions for problem (1.1) with the initial energy J(u0,v0)<d.

    Theorem 1. Assume (u0,v0)H10(Ω)×H10(Ω) and (1.2) hold. If J(u0,v0)<d and I(u0,v0)0, then the problem (1.1) has a global solution (u(t),v(t))L((0,);H10(Ω)×H10(Ω)) with (ut(t),vt(t))L2((0,);L2(Ω)×L2(Ω)) and (u(t),v(t))W for 0t<. Furthermore, if I(u0,v0)>0, then there exists a c>0 such that u22+v22(u022+v022)e2ct.

    Proof. Since we know J(u0,v0)<d and I(u0,v0)0, then it follows that

    (i) If 0<J(u0,v0)<d and I(u0,v0)0, then we have I(u0,v0)>0. In fact, if I(u0,v0)=0, then by the definition of d in (2.4), we have J(u0,v0)d, which is a contradiction.

    (ii) If J(u0,v0)=0 and I(u0,v0)0, then we obtain (u0,v0)=(0,0). In fact, if (u0,v0)(0,0), then by the (2.3), we have p12p(u02H10(Ω)+v02H10(Ω))+1p2u0v0pp<0, which is also a contradiction.

    (iii) If J(u0,v0)<0 and I(u0,v0)0, then it is contradictive with (2.3).

    From the discussions above, we consider the case 0<J(u0,v0)<d and I(u0,v0)>0. It is widely know that there is a basis {ωj(x)}j=1 of H10(Ω) such that ωj is an eigenfunction of the Laplacian operator corresponding to the eigenvalue λj and

    {Δωj=λjωj,xΩ,ωj=0,xΩ.

    Hence, we choose {ωj(x)}j=1 as the Galerkin basis for Δ in H10(Ω). Then we construct the Galerkin approximate solution (um(x,t),vm(x,t)) of the problem (1.1),

    {um(x,t)=mj=1gjm(t)ωj(x), m=1,2,,vm(x,t)=mj=1hjm(t)ωj(x), m=1,2,,

    which satisfy, for j=1,2,,m,

    (umt,ωj)+(um,ωj)=(|vm|p|um|p2umlog(|umvm|),ωj) (3.1)

    and

    (vmt,ωj)+(vm,ωj)=(|um|p|vm|p2vmlog(|umvm|),ωj), (3.2)

    with initial condition um(x,0)=u0m, vm(x,0)=v0m, where u0m and v0m are chosen in span {ω1,ω2,,ωm} so that

    u0m=mj=1gjm(0)ωj(x)u0inH10(Ω),asm+ (3.3)

    and

    v0m=mj=1hjm(0)ωj(x)v0inH10(Ω),asm+. (3.4)

    According to the standard ordinary differential equation theory, the system (3.1)–(3.4) admit a solution

    (gjm(t),hjm(t))C1[0,T0)×C1[0,T0),

    where T0 is the minimum of the existence time of gjm(t) and hjm(t) for each m. Thus (um(x,t),vm(x,t))C1([0,T0);H10(Ω)×H10(Ω)).

    Next, multiplying (3.1) and (3.2) by gjm(t) and hjm(t), respectively, summing for j from 1 to m, integrating with respect to t from 0 to t and adding these two equations, we get

    t0umτ22+vmτ22dτ+J(um,vm)=J(u0m,v0m), 0t<T0. (3.5)

    From J(u0,v0)<d and (3.3)–(3.4), we see that J(u0m,v0m)<d for sufficiently large m. Then we get from (3.5) that

    t0umτ22+vmτ22dτ+J(um,vm)=J(u0m,v0m)<d, 0t<T0, (3.6)

    for sufficiently large m.

    By (3.3) and (3.4) and (u0,v0)W, we know that (u0m,v0m)W for large enough m. Next, we prove (um(x,t),v(x,t))W for large enough m and 0t<T0. If it is false, then there exists t0(0,T0) such that (um(x,t0),vm(x,t0))W, then I(um(t0),vm(t0))=0 and (um(t0),vm(t0))(0,0), or J(um(t0),vm(t0))=d.

    By (3.6), J(um(t0),vm(t0))=d is not true. On the other hand, if I(um(t0),vm(t0))=0 and (um(t0),vm(t0))(0,0), then by the definition of d, we have J(um(t0),vm(t0))d, which is also contradiction with (3.6). So (um(x,t),vm(x,t))W for large enough m and 0t<T0.

    From the fact (um(x,t),vm(x,t))W for large enough m, (3.6) and

    J(um(t),vm(t))=p12p(um(t)2H10(Ω)+vm(t)2H10(Ω))+1p2um(t)vm(t)pp+12pI(um(t),vm(t)),

    we obtain

    t0umτ22+vmτ22dτ+p12p(um(t)2H10(Ω)+vm(t)2H10(Ω))+1p2um(t)vm(t)pp<d, 0t<T0, (3.7)

    for sufficiently large m, which gives

    um(t)2H10(Ω)<2pp1d, (3.8)
    vm(t)2H10(Ω)<2pp1d, (3.9)
    t0umτ22+vmτ22dτ<d. (3.10)

    By (3.10), we know that T0=+. Then by (3.8)–(3.10), there exist u,v with theirs subsequences of {um}+j=1 and {vm}+j=1, such that, as m+,

    umuweaklystarinL(0,+;H10(Ω)), (3.11)
    vmvweaklystarinL(0,+;H10(Ω)), (3.12)
    umtutweaklyinL2(0,+;L2(Ω)), (3.13)
    νmtvtweaklyinL2(0,+;L2(Ω)). (3.14)

    Then it follows Aubin-lions compactness theorem [27] that

    umustronglyinC([0,+);L2(Ω)),
    vmvstronglyinC([0,+);L2(Ω)).

    Clearly, this implies that

    umua.e.inΩ×[0,+),
    vmva.e.inΩ×[0,+).

    Furthermore, we get

    |vm|p|um|p2umlog(|umvm|)|v|p|u|p2ulog(|uv|)a.e.inΩ×[0,+), (3.15)
    |um|p|vm|p2vmlog(|umvm|)|u|p|v|p2vlog(|uv|)a.e.inΩ×[0,+). (3.16)

    On the other hand,

    Ω||vm|p|um|p2umlog(|umvm|)|pp1dx=Ω(|vm|p|um|p1log(|umvm|))pp1dx={xΩ:|um(x)vm(x)|1}(|vm|p|um|p1log(|umvm|))pp1dx+{xΩ:|um(x)vm(x)|>1}(|vm|p|um|p1log(|umvm|))pp1dxΩ(e(p1)1|vm|)pp1dx+Ω|um|(p1+r)pp1|vm|(p+r)pp1dxe(p1)pp1vmpp1pp1+ume2evmσ2σcvmpp1H10(Ω)+cumeH10(Ω)vmσH10(Ω)c, (3.17)

    where σ=(p+r)pp1, e=(p1+r)pp1, and since |xq1logx|(e(q1))1 for 0<x<1 while xμlogx(eμ)1 for x1,μ>0. Choosing a positive real number r so that 0<2e<2 and 0<2σ<2, and similar to the proof (3.17), we have

    Ω||um|p|vm|p2vmlog(|umvm|)|pp1dxc. (3.18)

    Hence, from (3.15)–(3.18) and Lion's Lemma(see [27], Lemma 1.3, p.12), we have

    |vm|p|um|p2umlog(|umvm|)|v|p|u|p2ulog(|uv|)weaklystarinL(0,+;Lpp1(Ω)), (3.19)
    |um|p|vm|p2vmlog(|umvm|)|u|p|v|p2vlog(|uv|)weaklystarinL(0,+;Lpp1(Ω)). (3.20)

    In view of (3.11)–(3.14) and (3.19), (3.20), for j fixed, we can pass to the limit in (3.1) and (3.2) to get

    (ut,ωj)+(u,ωj)=(|v|p|u|p2ulog(|uv|),ωj)

    and

    (vt,ωj)+(v,ωj)=(|u|p|v|p2vlog(|uv|),ωj)

    for a.e., t(0,+). Since {ωj(x)}j=1 is the basis in H10(Ω), we have

    (ut,w1)+(u,w1)=(|v|p|u|p2ulog(|uv|),w1) (3.21)

    and

    (vt,w2)+(v,w2)=(|u|p|v|p2vlog(|uv|),w2) (3.22)

    for any w1,w2H10(Ω) and a.e., t(0,+).

    Fixing any t(0,+) and integrating (3.21) and (3.22) from 0 to t, we get

    (u,w1)+t0(u,w1)dτ=t0(|v|p|u|p2ulog(|uv|),w1)dτ+(u(0),w1), w1H10(Ω), (3.23)

    and

    (v,w2)+t0(v,w2)dτ=t0(|u|p|v|p2vlog(|uv|),w2)dτ+(v(0),w2), w2H10(Ω). (3.24)

    Similarly, integrating (3.1) and (3.2) from 0 to t, and passing to the limit, we get

    (u,w1)+t0(um,w1)dτ=t0(|vm|p|um|p2umlog(|umvm|),w1)dτ+(u0,w1), w1H10(Ω), (3.25)

    and

    (v,w2)+t0(vm,w2)dτ=t0(|um|p|vm|p2vmlog(|umvm|),w2)dτ+(v0,w2), w2H10(Ω). (3.26)

    From (3.23)–(3.26), we get u(0)=u0 and v(0)=v0.

    According to (3.3), (3.4), (3.7), (3.11)–(3.14), (3.19), (3.20) and since the norm is weakly lower semicontinuous, we know that the energy inequality (2.5) holds. Then from Definition 2, (u(t),v(t)) is a global weak solution and (u(t),v(t))W.

    Next, we will prove the algebraic decay of the global solution u(x,t). Combining (2.3), (2.5) and (u(t),v(t))W, we have

    p12p(u2H10(Ω)+v2H10(Ω))+1p2uvppJ(u(t))J(u0). (3.27)

    As I(u,v)<0, then there exists a λ(0,1) such that I(λu,λv)=0. Furthermore, we get

    λp(p12p(u2H10(Ω)+v2H10(Ω)))+1p2uvppJ(λu(t))d. (3.28)

    It follows from (3.27) and (3.28) that

    λ(dJ(u0,v0))1p. (3.29)

    Due to the I(λu,λv)=0, we have

    I(λu,λv)=λ2(u2H10(Ω)+v2H10(Ω))2λ2pΩ|uv|plog|uv|dx2λ2plogλ2uvpp=(λ22λ2p)(u2H10(Ω)+v2H10(Ω))+2λ2pI(u,v)2λ2plogλ2uvpp=0,

    i.e.,

    I(u,v)(112λ2(p1))(u2H10(Ω)+v2H10(Ω)). (3.30)

    Combining (3.29) with (3.30), we have

    I(u,v)(112(dJ(u0,v0))2(1p1))(u2H10(Ω)+v2H10(Ω)).

    By the emdedding H10(Ω)L2(Ω), we have

    I(u,v)c(u2L2(Ω)+v2L2(Ω)). (3.31)

    On the other hand, by Lemma 4, we know

    12ddt(u22+v22)+I(u,v)=0, 0t<.

    Combining this equality with (3.31), we get

    12ddt(u22+v22)+c(u2L2(Ω)+v2L2(Ω))0, 0t<.

    By Grönwall's inequality, we have

    u22+v22(u022+v022)e2ct, 0t<.

    The proof of Theorem 1 is complete.

    Theorem 2. Assume (u0,v0)H10(Ω)×H10(Ω) and (1.2) hold. If J(u0,v0)<d and I(u0,v0)<0, then the weak solution (u(x,t),v(x,t)) of the problem (1.1) blows up in finite time, i.e., there exists a T>0 such that

    limtTt0u22+v22dτ=+.

    Proof. Step 1: Blow-up in finite time

    By contradiction, we suppose that (u(t),v(t)) is global weak solution of problem (1.1), then Tmax=+. Let

    G(t)=t0u22+v22dτ,

    then

    G(t)=u22+v22

    and

    G(t)=2((u,ut)+(v,vt))=2(u22+v22)+4Ω|u|p|v|plog(|uv|)dx=2I(u,v). (3.32)

    From (3.32) and energy inequality (2.5), it follows that

    G(t)2(p1)(u2H10(Ω)+v2H10(Ω))+4puvpp4pJ(u,v)2(p1)(u2H10(Ω)+v2H10(Ω))+4puvpp4pJ(u0,v0)+4pt0uτ22+vτ22dτ4pt0uτ22+vτ22dτ+2(p1)cG(t)4pJ(u0,v0), (3.33)

    where the constant c is from the Poincaré inequality u22cu22.

    Note that

    (t0(uτ,u)+(vτ,v)dτ)2=(12t0ddτ(u22+v22)dτ)2=(12(u22+v22u022v022))2=14[(u22+v22)22(u022+v022)(u22+v22)+(u022+v022)2]=14[(G(t))22G(t)(u022+v022)+(u022+v022)2],

    then

    G(t)=4(t0(uτ,u)+(vτ,v)dτ)2+2G(t)(u022+v022)(u022+v022)2. (3.34)

    Hence by (3.33) and (3.34) we know that

    G(t)G(t)p(G(t))24pt0uτ22+vτ22dτt0u22+v22dτ4p(t0(uτ,u)+(vτ,v)dτ)2+2(p1)cG(t)G(t)4pG(t)J(u0,v0)2pG(t)(u022+v022)+p(u022+v022)2.

    By Schwartz inequality, we have

    t0uτ22+vτ22dτt0u22+v22dτ(t0(uτ,u)+(vτ,v)dτ)2t0uτ22+vτ22dτt0u22+v22dτ(t0u2uτ2+v2vτ2dτ)2t0uτ22+vτ22dτt0u22+v22dτ(t0uτ22+vτ22u22+v22dτ)20.

    It implies that

    G(t)G(t)p(G(t))22(p1)cG(t)G(t)4pJ(u0,v0)G(t)2pG(t)(u022+v022). (3.35)

    From Lemma 3 we have I(u(t),v(t))<0 for 0t<+. Thus from Lemma 2 one has

    2I(u(t),v(t))>4p(dJ(u(t),v(t))), 0t<+. (3.36)

    Combing (3.36) and (2.5) we get

    G(t)=2I(u,v)>4p(dJ(u,v))4p(dJ(u0,v0)):=C1>0, 0t<+

    and

    G(t)C1t+G(0)=C1t, 0t<+,
    G(t)12C1t2+G(0)=12C1t2, 0t<+.

    Hence for sufficiently large t, we have

    (p1)cG(t)>2p(u022+v0|22)and(p1)cG(t)>4pJ(u0,v0). (3.37)

    Combining (3.35) with (3.37), we obtain

    G(t)G(t)p(G(t))2((p1)cG(t)2p(u022+v022))G(t)+((p1)cG(t)4pJ(u0,v0))G(t)>0,

    for sufficiently large t. Note that

    (G(p1)(t))=(p1)Gp+1(t)(G(t)G(t)p(G(t))2)<0.

    It follows that there exists a finite time T>0 such that limtTG(p1)(t)=0, i.e., limtTt0u22+v22dτ=+.

    Step 2: Upper bound estimation of the blow-up time.

    We next give an upper bound estimate of T. Suppose (u(t),v(t)) be a solution of problem (1.1) with initial value (u0,v0) satisfying I(u0,v0)<0 and J(u0,v0)<d. By Step 1, the maximal existence time T<. By Lemma 3, we get (u(t),v(t))V, t[0,T), i.e., I(u(t),v(t))<0,t[0,T). For T1(0,T), we define the auxiliary functional M:[0,T1]R which is defined by

    M(t):=t0u22+v22dτ+(Tt)(u022+v022)+β(t+γ)2, (3.38)

    with β>0 and γ>0 specified later. Through a direct calculation, we have

    M(t)=u(t)22+v(t)22(u022+v022)+2β(t+γ)=2t0(uτ,u)+(vτ,v)dτ+2β(t+γ) (3.39)

    and

    M(t)=2((ut,u)+(vt,v))+2β=2β2I(u,v).

    It follows from Lemma 2 and (2.5) that

    M(t)>4p(dJ(u,v))+2β4p(dJ(u0,v0))+4pt0uτ22+vτ22dτ+2β. (3.40)

    From (3.39) and Hölder inequality, we have

    (M(t))2=4[t0(uτ,u)+(vτ,v)dτ+2β(t+γ)]24[t0uτ2u2+vτ2v2dτ+2β(t+γ)]2. (3.41)

    According to the inequality

    xz+yw(x2+y2)12(z2+w2)12,

    by setting x=uτ2, y=vτ2, z=u2, w=v2 in (3.41), we get

    (M(t))24[t0(uτ22+vτ22)12(u22+v22)12dτ+2β(t+γ)]2.

    By the Hölder inequality, we get

    (M(t))24[t0(uτ22+vτ22)12(u22+v22)12dτ+2β(t+γ)]24[(t0uτ22+vτ22dτ)12(t0u22+v22dτ)12+2β(t+γ)]24[t0uτ22+vτ22dτ+β][t0u22+v22dτdτ+β(t+γ)2]4M(t)[t0uτ22+vτ22dτ+β]. (3.42)

    From (3.38), (3.40) and (3.42), we have

    M(t)M(t)p(M(t))2[4p(dJ(u0,v0))2(2p1)β]M(t).

    Restricting β to satisfy

    0<β2p(dJ(u0,v0))2p1, (3.43)

    we have

    M(t)M(t)p(M(t))20,t[0,T1].

    Define y(t):=M1p(t) for t[0,T1], then by M(t)>0,M(t)>0 we get

    y(t)=(p1)Mp(t)M(t)<0,
    y(t)=(p1)Mp1(t)(M(t)M(t)p(M(t))2)<0

    for all t[0,T1]. It follows from y(t)<0 that

    y(T1)y(0)=y(ξ)T1<y(0)T1, ξ(0,T1), (3.44)

    where

    y(0)=M1p(0)>0, y(T1)=M1p(T1)>0,
    y(0)=(p1)Mp(0)M(0)=2(1p)βγMp(0)<0.

    Combining (3.44) and the above inequalities, we can deduce

    T1y(T1)y(0)y(0)y(0)<y(0)y(0)=M(0)2(p1)βγ.

    Then by the definition of M(t) and above inequality we have

    T1T(u022+v022)+βγ22(p1)βγ=γ2(p1)+u022+v0222(p1)βγT.

    Hence, letting T1T, we get

    Tγ2(p1)+u022+v0222(p1)βγT. (3.45)

    For any β satisfying (3.43), let γ be large enough such that

    u022+v0222(p1)β<γ<+, (3.46)

    then (3.45) lead to

    Tγ2(p1)(1u022+v0222(p1)βγ)1=βγ22(p1)βγ(u022+v022).

    Let

    ρ(β,γ)=βγ22(p1)βγ(u022+v022),

    then

    Tmin(β,γ)Φρ(β,γ),

    where Φ={(β,γ):β,γsatisfy(3.43)and(3.46)respectively}.

    Since

    ρβ(β,γ)=γ2(u022+v022)(2(p1)βγ(u022+v022))2<0,

    i.e., ρ(β,γ) is decreasing with respect to β. Then we have

    min(β,γ)Φρ(β,γ)=ρ(2p(dJ(u0,v0))2p1,γ):=ρ1(γ),

    where

    ρ1(γ)=2p(dJ(u0,v0))γ24p(p1)γ(dJ(u0,v0))(2p1)(u022+v022)

    and

    (2p1)(u022+v022)4p(p1)(dJ(u0,v0))<γ<+.

    It is easy to get that ρ1(γ) achieves its minimum at

    γ1=(2p1)(u022+v022)2p(p1)(dJ(u0,v0)),

    and

    ρ1(γ1)=(2p1)(u022+v022)2p(p1)2(dJ(u0,v0)).

    Thus, we have

    T(2p1)(u022+v022)2p(p1)2(dJ(u0,v0)).

    The proof of Theorem 2 is complete.

    In this section, we prove global existence and blow up at finite time of solutions for problem (1.1) with the initial energy J(u0,v0)=d.

    Theorem 3. Assume (u0,v0)H10(Ω)×H10(Ω) and (1.2) hold. If J(u0,v0)=d and I(u0,v0)0, then the problem (1.1) has a global solution (u(t),v(t))L(0,+;H10(Ω)×H10(Ω)) with (ut(t),vt(t))L2(0,+;L2(Ω)×L2(Ω)) and (u(t),v(t))¯W=WW for 0t<.

    Proof. Since J(u0,v0)=d, then (u0,v0)(0,0). Let λm=11m, (u0m,v0m)=λm(u0,v0), m=1,2,, and consider the following problem:

    {umtΔum=|vm|p|um|p2umlog(|umvm|),xΩ, t>0,vmtΔvm=|um|p|vm|p2vmlog(|umvm|),xΩ, t>0,um(x,0)=u0m(x),xΩ,vm(x,0)=v0m(x),xΩ,um(x,t)=vm(x,t)=0,(x,t)Ω×(0,T]. (4.1)

    By I(u0,v0)0 and Lemma 1, there exists a unique λ1 such that I(λu0,λv0)=0. Due to the λm<1<λ, we get I(λmu0,λmv0)>0,J(λmu0,λmv0)<J(u0,v0)=d. From Theorem 1, it follows that for each m problem (4.1) admits a global solution (um(t),vm(t))L(0,+;H10(Ω)×H10(Ω)) with (umt(t),vmt(t))L2(0,+;L2(Ω)×L2(Ω)) with the initial data

    um(0)=u0mu0inH10(Ω)asm+.

    Furthermore, we have (um(t),vm(t))W for 0t<+,

    (umt,w1)+(um,w1)=(|vm|p|um|p2umlog(|umvm|),w1),w1H10(Ω), 0t<+,
    (vmt,w2)+(vm,w2)=(|um|p|vm|p2vmlog(|umvm|),w2),w2H10(Ω), 0t<+,

    and

    t0umτ22+vmτ22dτ+J(um,vm)J(u0m,v0m)<d, 0t<+. (4.2)

    From (4.2) and

    J(um(t),vm(t))=p12p(um(t)2H10(Ω)+vm(t)2H10(Ω))+1p2vm(t)pp+12pI(um(t),vm(t)),

    we obtain

    t0umτ22+vmτ22dτ+p12p(um(t)2H10(Ω)+vm(t)2H10(Ω))+1p2um(t)vm(t)pp<d,0t<+.

    The remainder of the proof is similar to that in the proof of Theorem 1.

    The proof of Theorem 3 is complete.

    Theorem 4. Assume (u0,v0)H10(Ω)×H10(Ω) and (1.2) hold. If J(u0,v0)=d and I(u0,v0)<0, then the weak solution (u(x,t),v(x,t)) of the problem (1.1) blows up in finite time, i.e., there exists a T>0 such that

    limtTt0u22+v22dτ=+.

    Proof. By contradiction, we suppose that (u(t),v(t)) is a global weak solution of problem (1.1), then Tmax=+. Let

    G(t)=t0u22+v22dτ.

    Taking into account to (3.35) still holds, combining the fact J(u0,v0)=d, we have

    G(t)G(t)p(G(t))2((p1)cG(t)2p(u022+v022))G(t)+((p1)cG(t)4pd)G(t). (4.3)

    From continuities of J(u,v) and I(u,v) with respect to t, we know that there exists a sufficient small t1(0,+) such that J(u(t1),v(t1))>0 and I(u,v)<0 for 0<t<t1. By (ut,u)+(vt,v)=I(u,v), we have (ut,u)+(vt,v)>0 and ut22+vt22>0 for t[0,t1]. From (2.5), we have 0<J(u(t1),v(t1))dt10(ut22+vt22)dt<d. Hence we take t=t1 as the initial time, and obtain (u(t1),v(t1))V. From Lemma 3 we have I(u(t),v(t))<0 for t1t<+. Thus from Lemma 2 one has

    2I(u(t),v(t))>4p(dJ(u(t),v(t))), t1t<+. (4.4)

    Combing (4.4) and (2.5) we get

    G(t)=2I(u,v)>4p(dJ(u,v))4p(dJ(u(t1),v(t1))):=C2>0, t1t<+

    and

    G(t)C2(tt1)+G(t1)=C2(tt1), t1t<+,
    G(t)12C22t2C2t1t+G(t1), t1t<+.

    Hence for sufficiently large t, we have

    (p1)cG(t)>2p(u022+v0|22)and(p1)cG(t)>4pd. (4.5)

    Combining (4.3) with (4.5), we get

    G(t)G(t)p(G(t))2((P1)cG(t)2p(u022+v022))G(t)+((p1)cG(t)4pd)G(t)>0,

    for sufficiently large t. Then similar to the proof of Theorem 2, i.e., there exists a finite time T>0 such that limtTt0u22+v22dτ=+.

    The proof of Theorem 4 is complete.

    In this section, we investigate the conditions that ensure the global existence or blow up of solution for problem (1.1) with the initial energy J(u0,v0)>d.

    Theorem 5. For any α(d,+), the following conclusions hold.

    (i) If (u0,v0)Φα, then the solution of the problem (1.1) exists globally and (u(t),v(t))(0,0), as t;

    (ii) If (u0,v0)Ψα, then the solution of the problem (1.1) blows up in finite or infinite time, where

    Φα=N+{(u(t),v(t))H10 (Ω)×H10(Ω)|12(u22+v22)<λα,d<J(u,v)α}, (5.1)
    Ψα=N{(u(t),v(t))H10 (Ω)×H10(Ω)|12(u22+v22)>Λα,d<J(u,v)α}, (5.2)

    and

    λα=inf{12(u22+v22)|(u,v)Nα},Λα=sup{12(u22+v22)|(u,v)Nα}forallα>d.

    Proof. (i) Assume that (u0,v0)Φα, then by the definition of Φα and the monotonicity property of λα, we have (u0,v0)N+, d<J(u0,v0)α and

    12(u022+v022)<λαλJ(u0,v0). (5.3)

    We claim that (u(t),v(t))N+. By contradiction, there exists a t0(0,T) such that (u(t),v(t))N+ for t[0,t0) and (u(t0),v(t0))N. By Lemma 4, we have

    12ddt(u22+v22)=I(u,v). (5.4)

    From the definition of N+ and (5.4), we know that u22+v22 is strictly decreasing on [0,t0). On the other hand, by (2.5), we know that J(u,v) is nonincreasing with respect to t. Therefore, we have

    J(u,v)J(u0,v0)forallt[0,T).

    From (5.3), we get

    12(u(t0)22+v(t0)22)<12(u022+v022)<λJ(u0,v0). (5.5)

    By (u(t0),v(t0))N and (5.3), we get (u(t0),v(t0))NJ(u0,v0). According to the definition of λJ(u0,v0), we have

    λJ(u0,v0)=inf{12(u22+v22)|(u,v)NJ(u0,v0)}12(u(t0)22+v(t0)22),

    which contradicts with (5.5) and prove the claim. Hence, we have (u(t),v(t))N+ for all t[0,T) and (u(t),v(t))JJ(u0,v0), i.e., (u(t),v(t))JJ(u0,v0)N+ for all t[0,T). From the definition of Nα, we have (u2H10(Ω)+v2H10(Ω))<2pp2J(u0,v0), t[0,T), so T=+. It indicates that (u(t),v(t)) is bounded uniformly in H10(Ω)×H10(Ω). Hence, ω-limit set is not an empty set.

    Next, for any (ω,φ)ω(u0,v0), by the above discussions, we get

    J(ω,φ)J(u0,v0)and12(ω22+φ22)<λJ(u0,v0).

    According to first inequality, it implies that (ω,φ)JJ(u0,v0). According to the second inequality and the definition of λJ(u0,v0), we know that (ω,φ)NJ(u0,v0). Since NJ(u0,v0)=NJJ(u0,v0), we obtain (ω,φ)N. Hence, ω(u0,v0)N=ϕ. As N include the nontrivial solutions of the problem (1.1), we have ω(u0,v0)=(0,0), i.e., (u(t),v(t))(0,0), as t.

    (ii) If (u0,v0)Ψα, by the definition of Ψα, it is clear that (u0,v0)N and d<J(u0,v0)α. Combing with the monotonicity of Λα, we get

    12(u022+v022)>ΛαΛJ(u0,v0).

    We claim that (u(t),v(t))N for t[0,T). By contradiction, if there exists a t1(0,T) such that (u(t),v(t))N for t[0,t1) and (u(t1),v(t1))N. By Lemma 4, we have

    12ddt(u22+v22)=I(u,v).

    Then by the definition of N, we deduce that 12(u22+v22) is strictly increasing on [0,t1). It along with (2.5) yields

    12(u(t1)22+v(t1)22)>12(u022+v022)>ΛJ(u0,v0), J(u(t1),v(t1))J(u0,v0). (5.6)

    By (u(t1),v(t1))N and (5.6), we get (u(t1),v(t1))NJ(u0,v0). Hence, it follows from the definition of ΛJ(u0,v0) that

    ΛJ(u0,v0)=sup{12(u22+v22)|(u,v)NJ(u0,v0)}12(u(t1)22+v(t1)22),

    which is incompatible with (5.6), so we get (u(t),v(t))JJ(u0,v0)N for all t[0,T).

    Next, we assume that (u(t),v(t)) exists globally, i.e., T=+. For every (ω,φ)ω(u0,v0), by the above discussions, we get

    J(ω,φ)J(u0,v0)and12(ω22+φ22)>ΛJ(u0,v0).

    According to first inequality, this shows (ω,φ)JJ(u0,v0). According to the second inequality and the definition of ΛJ(u0,v0), we know that (ω,φ)NJ(u0,v0). Since NJ(u0,v0)=NJJ(u0,v0), we obtain (ω,φ)N. Hence, ω(u0,v0)N=ϕ. However, since dist(0,N)>0, we also have (0,0)ω(u0,v0). Thus, ω(u0,v0)=, it contraries to the assumption that (u(t),v(t)) is a global solution, then T<.

    The proof of Theorem 5 is complete.

    This research was supported by Guizhou Provincial Science and Technology Projects (No.Qiankehe foundation-ZK[2021]YIBAN317), and by the project of Guizhou Minzu University under (No.GZMU[2019]YB04), and central leading local science and technology development special foundation for Sichuan province, P.R. China under (No.2021ZYD0020).

    The authors declare there is no conflict of interest.



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