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Initial boundary value problem for fractional p-Laplacian Kirchhoff type equations with logarithmic nonlinearity

  • In this paper, we study the initial boundary value problem for a class of fractional p-Laplacian Kirchhoff type diffusion equations with logarithmic nonlinearity. Under suitable assumptions, we obtain the extinction property and accurate decay estimates of solutions by virtue of the logarithmic Sobolev inequality. Moreover, we discuss the blow-up property and global boundedness of solutions.

    Citation: Peng Shi, Min Jiang, Fugeng Zeng, Yao Huang. Initial boundary value problem for fractional p-Laplacian Kirchhoff type equations with logarithmic nonlinearity[J]. Mathematical Biosciences and Engineering, 2021, 18(3): 2832-2848. doi: 10.3934/mbe.2021144

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  • In this paper, we study the initial boundary value problem for a class of fractional p-Laplacian Kirchhoff type diffusion equations with logarithmic nonlinearity. Under suitable assumptions, we obtain the extinction property and accurate decay estimates of solutions by virtue of the logarithmic Sobolev inequality. Moreover, we discuss the blow-up property and global boundedness of solutions.



    In this paper, we study the extinction and the blow-up for the following fractional p-Laplacian Kirchhoff type equations with logarithmic nonlinearity.

    {ut+M(up)(Δ)spu=λ|u|r2uln|u|β|u|q2u,in Ω×(0,T),u(0)=u0,in Ω,u=0,on Ω×(0,T), (1.1)

    where

    u=(Q|u(x)u(y)|p|xy|N+spdxdy)1p,

    Q=R2N(CΩ×CΩ), CΩ=RNΩ, ΩRN (N>2s) is a bounded domain with Lipschitz boundary, s(0,1), 1<p<2, 1<q2, r>1, λ, β>0, (Δ)sp is the fractional p-Laplacian operator and satisfies

    (Δ)spu(x)=2limγ0+RNBγ(x)|u(x)u(y)|p2(u(x)u(y))|xy|N+spdy,

    where u(x)C and u(x) has compact support in Ω, Bγ(x)RN is the ball with center x and radius γ. u0(x)L(Ω)Ws,p0(Ω) is a nonzero non-negative function, where L(Ω) and Ws,p0(Ω) are Lebesgue space and fractional Sobolev space respectively, which will be given in section 2. M() is a Kirchhoff function with the following assumptions

    (M1) 0<s<1,M(τ):=a+bθτθ1 for τR+0:=[0,+) (a>0,b0 are two constants), θ1;

    (M2) M:R+0R+0{0} is continuous and there exits m00 such M(τ)m0 for all τ0.

    It is worth pointing out that the interest in studying problems like (1.1) relies not only on mathematical purposes, but also on their significance in real models. For example, in the study of biological populations, we can use u(x,t) to represent the density of the population at x at time t, the term (Δ)spu represents the diffusion of density, μ|u|q2u represents the internal source and λ|u|r2uln|u| denotes external influencing factors. For more practical applications of problems like (1.1), please refer to the studies [1,2,3].

    Compared with integer-order equations, it is very difficult to study the problem (1.1), which contains both non-local terms (including fractional p-Laplacian operators and Kirchhoff functions) and logarithmic nonlinearity. For the fractional order theory, we refer the readers to the studies [4,5,6]. In [7,8], the authors use Sobolev space and Nehari manifold to study the existence of solutions for fractional equations. In [9,10], the solutions for fractional equations are discussed by virtue of Nehari manifold and fibrillation diagram. By using different methods from above, the properties of the solutions for such partial differential equations are considered by the method of variational principle and topological theory in the the literature [11,12,13]. Moreover, the authors prefer to use potential well theory, Galerkin approximation and Nehari manifold method to prove the existence of solutions, decay estimation and blow-up, we refer the reader to the literature [14,15,16].

    Existence, extinction and blow-up of solutions are three important topics which regard parabolic problems; in particular, the study of extinction properties has made great progress in recent years. In [17], Liu considered the following initial boundary value problem for the fractional p-Laplacian equation

    utdiv(|u|p2u)+βuq=λur,  xΩ,  t>0, (1.2)

    where 1<p<2, q1 and r,λ,β>0. By employing the differential inequality and comparison principle, they obtained the extinction and the non-extinction of weak solutions. In [18], Sarra Toualbia et al. considered the following initial boundary value problem of a nonlocal heat equations with logarithmic nonlinearity

    utdiv(|u|p2u)=|u|p2ulog|u|Ω|u|p2ulog|u|dx, xΩ, t>0, (1.3)

    where p(2,+). By using the logarithmic Sobolev inequality and potential well method, they obtained decay, blow-up and non-extinction of solutions. In [19], Xiang and Yang studied the first initial boundary value problem of the following fractional p-Kirchhoff type

    ut+M([u]ps,p)(Δ)spu=λ|u|r2uμ|u|q2u,  (x,t)Ω×(0,), (1.4)

    where M:[0,)(0,) is a continuous function, 0<s<1<p<2, 1<q2, r>1, λ,μ>0. Under suitable assumptions, they proved the extinction and non-extinction of solutions and perfected the Gagliardo-Nirenberg inequality. For more information on the extinction properties of the solution, please refer to the studies [20,21,22,23].

    Inspired by the above works, we overcome the research difficulties of logarithmic nonlinearity, p-Laplace operator and Kirchhoff coefficients in problem (1.1) based on the potential well theory, Nehari manifold and differential inequality methods, we give the extinction and the blow-up properties of solutions. In addition, we give the global boundedness of the solution by appropriate assumptions. To the best of our knowledge, it is the first result in the literature to investigate the extinction and blow-up of solutions for fractional p-Laplacian Kirchhoff type with logarithmic nonlinearity.

    In order to introduce our main results, we first give some related definitions and sets.

    Definition 1.1(Weak solution). A function u(x,t) is said to be a weak solution of problem (1.1), if (x,t)Ω×[0,T), uLp(0,T;Ws,p0(Ω))C(0,T;L2(Ω)), utL2(0,T;L2(Ω)), u(x,0)=u0(x)Ws,p0(Ω), for all vWs,p0(Ω), t(0,T), the following equation holds

    Ωutvdx+M(up)u,v=λΩv|u|r2uln|u|dxβΩv|u|q2udx,

    where

    u,v=Q|u(x)u(y)|p2(u(x)u(y))(v(x)v(y))|xy|N+spdxdy.

    Definition 1.2(Extinction of solutions). Let u(t) be a weak solution of problem (1.1). We call u(t) an extinction if there exists T>0 such that u(x,t)>0 for all t(t,T) and u(x,t)0 for all t[T,+).

    Define the following two functionals on Ws,p0(Ω)

    E(u)=1paup+1pbuθpλ1rΩ|u|rln|u|dx+λ1r2Ω|u|rdx+β1qΩ|u|qdx, (1.5)
    I(u)=aup+θbuθpλΩ|u|rln|u|dx+βuqq. (1.6)

    Let

    Z={uLp(0,T;Ws,p0(Ω))C(0,T;L2(Ω)),utL2(0,T;L2(Ω))}.

    Remark 1 Since uZ, 1<p<2, M() is a continuous function and

    Ω|u|rln|u|dx1σur+σr+σ1σCr+σr+σur+σ,

    where 0<σ<psr, then we can claim that E(u) and I(u) are well-defined in Ws,p0(Ω). Further, by arguing essentially as in [24], one can prove the that

    uΩ|u|rln|u|dx

    is continuous from Ws,p0(Ω) to R. It follows that E(u) and I(u) are continuous.

    Define some sets as follows

    W:={uWs,p0(Ω) | I(u)>0, E(u)<h}{0}, (1.7)
    V:={uWs,p0(Ω) | I(u)<0, E(u)<h}, (1.8)

    the mountain pass level

    h:=infuNE(u), (1.9)

    the Nehari manifold

    N:={uWs,p0(Ω){0} | I(u)=0}. (1.10)

    Moreover, we define

    N+:={uWs,p0(Ω) | I(u)>0}, (1.11)
    N:={uWs,p0(Ω) | I(u)<0}. (1.12)

    Let λ1 be the first eigenvalue of the problem

    (Δ)spu=λ|u|p2u  in Ω,   u|RNΩ=0, (1.13)

    and ϕ(x)>0 a.e. in Ω be the eigenfunction corresponding to the eigenvalue λ1>0, ϕ(x)L(Ω) and ϕL(Ω)1.

    First, we give some results satisfying I(u0)>0 and q=2.

    Theorem 1.1 Assume that I(u0)>0, r=p and q=2. Let m0 be as in assumption (M2), and let

    l:=2N(s+N)psp, P1:=m0λ1L(p,Ω)+ln(R), P2:=λ1m0lpp1[λ1L(p,Ω)+ln(R)](p+l1)p1,

    where L(p,Ω) and R are given in Lemma 2.1 and Lemma 2.5. Then, there exist positive constants C1, C2, T1 and T2 such that

    (ⅰ) If λ<λ1P1, then the weak solution of (1.1) vanishes in the sense of 2 as t+.

    (ⅱ) If 2N/(N+2s)<p<2 and λ<λ1P1 or 1<p2N/(N+2s) and λ<P2, then the nonnegative solutions of (1.1) vanish in finite time, and

    {u2[(u02p2+C1β)e(p2)βtC1β]12p,t[0,T1),u20,t[T1,),

    for 2N/(N+2s)<p<2, and

    {ul+1[(u02pl+1+C2β)e(p2)βtC2β]12p,t[0,T2),ul+10,t[T2,),

    for 1<p<2N/(N+2s).

    Theorem 1.2 Assume that I(u0)>0, 0<σpsr, r>p and q=2. Let m0 be as in assumption (M2), and let

    P3:=max{m0L(r,Ω)Rrp+εΦ,βΦεr(ϑ11)pr(1ϑ1)}, P4:=max{m0lppε1(p+l1)p,βεr2(1ϑ2)pr2(1ϑ2)1},

    where

    Φ=ln(R)|Ω|s1rs1Cr(1ϑ1)ps, ps=NpNsp, s1=2ϑ1+ps(1ϑ1),ϑ1=2(rp)r(2p), r2=p(l+r+σ1)l+p1,

    Cps is the best constant of embedding from Ws,p0(Ω) to Lps(Ω), L(p,Ω) and R are given in Lemma 2.1 and Lemma 2.5, and ε,ε1>0 are two constants. Then, there exist positive constants C4, C5, C6, C7, T3 and T4 such that the non-negative weak solution of problem (1.1) vanishes in finite time and

    {u2[(u02pl+1+C5C4)e(p2)C4tC5C4]12p,t[0,T3),u20,t[T3,),

    for 2N/(N+2s)p<2, λ<P3, and

    {ul+1[(u02pl+1+C6C7)e(p2)C7tC6C7]12p,t[0,T4),ul+10,t[T4,),

    for 1<p<2N/(N+2s), λ<P4.

    Secondly, we give some results satisfying I(u0)>0 and q<2.

    Theorem 1.3 Assume that I(u0)>0, p=r, l1>l1 and 1<q<2. Let m0 be as in assumption (M2), and let

    P5:=λ1m0l1pp1[λ1L(p,Ω)+ln(R)](p+l11)p1,

    where L(p,Ω) and R are given in Lemma 2.1 and Lemma 2.5. If 2N/(N+2s)<p<2 with λ<λ1P1 or 1<p2N/(N+2s) with λ<P5, then the non-negative weak solution of problem (1.1) vanishes in finite time for any non-negative initial data.

    Theorem 1.4 Assume that I(u0)>0, l1>l1, r2 and 1<q<2. Let m0 be as in assumption (M2), and let

    P6:=max{m0L(r,Ω)Rrp,βε(1ϑ4)r4p(1ϑ4)r43C(ϑ41)r4psRr4rln(R)|Ω|2r2}, P7:=1Cβε(ϑ31)r3(1ϑ3)(p+l11)2C(ϑ31)pr3p+l11ps,
    ϑ3=[(q+l11)ps(l1+1)p](p+l11)[(q+l11)ps(p+l11)p](l1+1), ϑ4=(ps2)p(psq)2, C=1eσ|Ω|1r5l1+1Cr5l1+1Rr5r3,
    r3=(q+l11)(p+l11)ϑ3(p+l11)+(q+l11)(1ϑ3), r4=qpq(1ϑ4)+ϑ4p, r5=l1+r+σ1,

    and ε2, ε3>0 are two constants. If 2N/(N+2s)<p<2 with λ<P6 or 1<p2N/(N+2s) with λ<P7, then the non-negative weak solution of problem (1.1) vanishes in finite time for any non-negative initial data.

    Finally, we discuss the global boundedness and blow up of weak solutions.

    Theorem 1.5 Let u(x,t) be the weak solution of problem (1.1).

    (ⅰ) If E(u0)<0, r=p>q and θ=1, then the weak solution u(x,t) blows up at +;

    (ⅱ) If 0<E(u0)h and I(u0)0, then the weak solution u(x,t) is globally bounded.

    The rest of the paper is organized as follows. In Section 2, we give some related spaces and lemmas. In Section 3, we give the proof process for the main results of problem (1.1).

    In order to facilitate the proof of the main results, we start this section by introducing some symbols and Lemmas that will be used throughout the paper.

    In this section, we assume that 0<s<1<p<2 and ΩRN (N>2s) is a bounded domain with Lipschitz boundary. We denote by ui (i1) the norm of Lebesgue space Li(Ω). Let Ws,p(Ω) be the linear space of Lebesgue measurable functions u from RN to R such that the restriction to Ω of any function u in Ws,p(Ω) belongs to Lp(Ω) and

    Q|u(x)u(y)|p|xy|N+spdxdy<,

    where Q=R2N(CΩ×CΩ), CΩ=RNΩ. The space Ws,p(Ω) is equipped with the norm

    uWs,p(Ω)=(upp+Q|u(x)u(y)|p|xy|N+spdxdy)1p.

    We further give a closed linear subspace

    Ws,p0(Ω)={uWs,p(Ω)|u(x)=0  a.e. in RNΩ}.

    As shown in [19], it can be concluded that

    u=(Q|u(x)u(y)|p|xy|N+spdxdy)1p

    is an equivalent norm of Ws,p0(Ω).

    Next we give the necessary Lemmas.

    Lemma 2.1 ([25]) Let uWs,p0(Ω){0}. Then

    Ω|u|pln|u|dxL(p,Ω)up+ln(u)Ω|u|pdx,

    where L(p,Ω):=|Ω|ep+1e(psp)Cpsps, Cps is the best constant of embedding from Ws,p0(Ω) to Lps(Ω).

    Lemma 2.2 ([26]) Let y(t) be a non-negative absolutely continuous function on [T0,+) satisfying

    dydt+αyk+βy0,   t0,   y(0)0,

    where α,β>0 are constants and k(0,1). Then

    {y(t)[(y1k(T0)+αβ)e(k1)β(tT0)αβ]11k,t[T0,T),y(t)0,t[T,+),

    where T=1(1k)βln(1+βαy1k(T0)).

    Lemma 2.3 ([27]) Suppose that β0, N>sp1, and 1rq(β+1)ps, then for u such that |u|βuWs,p0(Ω), we have

    uqC1ϑβ+1psuϑr|u|βu1ϑβ+1,

    with ϑ=[(β+1)psq]r[(β+1)psr]q, where Cps is the embedding constant for Ws,p0(Ω)Lps(Ω).

    Lemma 2.4 ([19]) Let 1<p< and g:RR be an increasing function. Define

    G(t)=t0g(τ)1pdτ,   tR

    then

    |ab|p2(ab)(g(a)g(b))|G(a)G(b)|p   for all a,bR.

    Lemma 2.5 Assume that (M1) holds. Let uWs,p0(Ω){0}, 0<σpsr. We have

    (ⅰ) if 0<uR, then I(u)>0;

    (ⅱ) if I(u)0, then u>R,

    where

    R=(aσλCr+σr+σ)1r+σp,

     Cr+σ is the embedding constant for Ws,p0(Ω)Lr+σ(Ω).

    Proof. Since uWs,p0(Ω){0}, and

    σln|u(x)|<|u(x)|σ    for a.e.xΩ.

    Then by the definition of I(u), we obtain

    I(u)=aup+θbuθp+βuqqλΩ|u|rln|u|dx>aup+θbuθp+βuqqλ1σur+σr+σaupλ1σur+σr+σ(aλ1σCr+σr+σur+σp)up,

    where Cr+σ is the embedding constant for Ws,p0(Ω)Lr+σ(Ω).

    We can get

    I(u)>(aλ1σCr+σr+σur+σp)up. (2.1)

    If 0<uR, then it follows from the definition of R that

    aλ1σCr+σr+σur+σp0,

    thus (ⅰ) holds.

    If I(u)0, by (1.12), we have

    aλ1σCr+σr+σur+σp<0,

    thus (ⅱ) holds.

    Lemmas 2.6 is similar to [28,Lemmas 9], so we ignore its proof.

    Lemma 2.6 ([28]) Assume that E(u0)h, then the sets N+ and N are both invariant for u(t), i.e, if u0N (resp. u0N+), then u(t)N (resp. u(t)N+) for all t[0,T).

    Lemma 2.7 ([29]) Let α be positive. Then

    tpln(t)1eαtp+α,  for all p,t>0.

    In this section, we prove that the main results of problem (1.1).

    Proof of Theorem 1.1

    (1) Taking v=u in Definition 1.1, we obtain

    12ddtΩu2dx+M(up)up=λΩ|u|pln|u|dxβΩu2dx. (3.1)

    By Lemma 2.1, Lemma 2.5 and Lemma 2.6, we obtain

    12ddtΩu2dx+(m0λL(p,Ω)λλ1ln(R))up+βΩu2dx0.

    Since λ<λ1P1, we get

    12ddtΩu2dx+βΩu2dx0.

    thus

    u(,t)22u022e2βt.

    which implies that u(,t)20 as t+.

    (2) We consider first the case 2N/(N+2s)<p<2 with λ<λ1P1. By (3.1) and Lemma 2.1, Lemma 2.5 and Lemma 2.6, we have

    12ddtu22+(m0λλ1(λ1L(p,Ω)+ln(R)))up+βu220. (3.2)

    Using Hölder's inequality and the fractional Sobolev embedding theorem, we have

    u2|Ω|12NspNpuNpNspCps|Ω|12NspNpu, (3.3)

    where Cps>0 is the embedding constant. By (3.2), (3.3) and λ<λ1P1, we obtain

    12ddtu22+C1up2+βu220, (3.4)

    where

    C1=Cpps|Ω|NspNP2(m0λλ1(λ1L(p,Ω)+ln(R)))>0. (3.5)

    Setting y(t)=u(,t)22, y(0)=u0()22, by Lemma 2.2, we obtain

    {u2[(u02p2+C1β)e(p2)βtC1β]12p,t[0,T1),u20,t[T1,),

    where

    T1=1(2p)βln(1+βC1u02p2). (3.6)

    Next, we consider the case 1<p2N/(N+2s) and λ<λ1P2. Taking ν=ul in Definition 1.1, where l=2N(s+N)psp1, by Lemma 2.1, Lemma 2.4, Lemma 2.5 and Lemma 2.6, we obtain

    1l+1ddtul+1l+1+Gup+l1pp+βul+1l+10, (3.7)

    where G=(m0lpp(p+l1)pλpL(p,Ω)p+l1λpln(R)λ1(p+l1)). By the very choice of l and the fractional Sobolev embedding theorem, we have

    up+l1pl+1=(Ωup+l1pNpNspdx)NspNpCpsup+l1p. (3.8)

    Hence,

    1l+1ddtul+1l+1+C2up+l1l+1+βul+1l+10, (3.9)

    where

    C2=CppS(m0lpp(p+l1)pλpL(p,Ω)p+l1λpln(R)λ1(p+l1)), (3.10)

    since λ<P2, then C2>0. Setting y(t)=u(,t)l+1, y(0)=u0()l+1, by Lemma 2.2, we obtain

    {ul+1[(u02pl+1+C2β)e(p2)βtC2β]12p,t[0,T2),ul+10,t[T2,),

    where

    T2=1(2p)βln(1+βC2u02pl+1). (3.11)

    The proof is completed.

    Proof of Theorem 1.2

    We consider first the case p<r<2 and 2N/(N+2s)<p<2. Taking ν=u in Definition 1.1, we have

    12ddtu22+M(up)up=Ω|u|rln|u|dxβu22. (3.12)

    By Lemma 2.1, Lemma 2.5 and Lemma 2.6, we obtain

    12ddtu22+M(up)upλL(r,Ω)ur+λln(R)urrβu22. (3.13)

    Using Hölder's inequality and the interpolation inequality, the fractional Sobolev embedding theorem and Young inequality, we can easily obtain (see [19])

    urr|Ω|s1rs1urs1|Ω|s1rs1Cr(1ϑ1)ps(εup+εr(ϑ11)pr(1ϑ1)u22), (3.14)

    where s1>r, ϑ1(0,1), ε>0 and

    s1=2ϑ1+ps(1ϑ1),  ϑ1=2(rp)r(2p).

    By (3.13) and (3.14), we have

    12ddtu22+C3up+C4u220, (3.15)

    where

    C3=m0λL(r,Ω)Rrpλεln(R)|Ω|s1rs1Cr(1ϑ1)ps, (3.16)
    C4=βλln(R)|Ω|s1rs1Cr(1ϑ1)psεr(ϑ11)pr(1ϑ1). (3.17)

    Since 2N/(N+2s)<p<2 and λ<P3, by the fractional embedding theorem and (3.3), we obtain

    12ddtu22+C5up2+C4u220, (3.18)

    and C3>0,C4>0, where

    C5=C3Cpps|Ω|NspNP2. (3.19)

    Similar to the Theorem 1.1, one can prove the that

    {u2[(u02pl+1+C5C4)e(p2)C4tC5C4]12p,t[0,T3),u20,t[T3,),

    where

    T3=1(2p)C4ln(1+C4C5u02p2). (3.20)

    When 1<p2N/(N+2s), p<r2 and λ<P4. Taking ν=ul (l=2NspNpsp1) in Definition 1.1, by Lemma 2.4 and Lemma 2.7, we obtain

    1l+1ddtul+1l+1+m0lpp(p+l1)pul+p1ppλ1eσul+r+σ1l+r+σ1βul+1l+1 (3.21)

    further, we have

    1l+1ddtul+p1pr1r1+m0lpp(p+l1)pul+p1ppλ1eσul+p1pr2r2βul+p1pr1r1 (3.22)

    where r1=p(l+1)l+p1, r2=p(l+r+σ1)l+p1. Note that, since r1<ps, by the Hölder's inequality and the fractional Sobolev embedding theorem, we have

    ul+p1ppr1|Ω|(psr1)pr1psul+p1ppps|Ω|(psr1)pr1psCppsul+p1pp. (3.23)

    Using the same discussion as above, one can conclude that

    ul+p1pr2r2|Ω|s2r2s2Cr2(1ϑ2)ps(ε1ul+p1pp+εr2(ϑ21)pr2(1ϑ2)1ul+p1pr1r1), (3.24)

    where s2>r2, ϑ2(0,1), ε1>0 and

    s2=r1ϑ2+ps(1ϑ2),  ϑ2=r1(r2p)r2(r1p).

    Combining (3.22)-(3.24) and λ<P4, we obtain

    1l+1ddtul+p1pr1r1+C6ul+p1ppr1+C7ul+p1pr1r10, (3.25)

    and C6>0, C7>0, where

    C6=(m0lpp(l+p1)pλε11eσCr2(1ϑ2)ps|Ω|s2r2s2)|Ω|(r1ps)pr1psCpps, (3.26)
    C7=βλ1eσ|Ω|s2r2s2Cr2(1ϑ2)psεr2(ϑ21)pr2(1ϑ2)1. (3.27)

    Using Lemma 2.2 and a direct calculation, we have

    {ul+1[(u02pl+1+C6C7)e(p2)C7tC6C7]12p,t[0,T4),ul+10,t[T4,),

    where

    T4=1(2p)C7ln(C7C6u02pl+1+1). (3.28)

    The proof is completed.

    Proof of Theorem 1.3

    We consider first the case 2N/(N+2s)<p<2 with λ<λ1P1. Taking v=u in Definition 1.1, we obtain

    12ddtΩu2dx+M(up)up=λΩ|u|pln|u|dxβΩuqdx. (3.29)

    Note that upλ1upp and by Lemma 2.1, Lemma 2.5 and Lemma 2.6, we have

    12ddtu22+(m0λλ1(λ1L(p,Ω)+ln(R)))up+βuqq0. (3.30)

    By (3.3) and β>0 and λ<λ1P1, we obtain

    12ddtu22+C8up20, (3.31)

    where

    C8=Cpps|Ω|2(Nsp)Np2N(m0λλ1(λ1L(p,Ω)+ln(R)))>0. (3.32)

    By a direct calculation, we obtain

    {u2[(u02p2+C8(p2)t]12p,t[0,T5),u20,t[T5,),

    where

    T5=1(2p)C8u02p2. (3.33)

    Next, we consider the case 1<p2N(N+2s) with λ<P2. Multiplying (1.1) by ul1 (l1>l1) and integrating, by Lemma 2.1, Lemma 2.4, Lemma 2.5 and Lemma 2.6, we obtain

    1l1+1ddtul1+1l1+1+G1up+l11pp+βuq+l11q+l110, (3.34)

    where G1=(m0l1pp(p+l11)pλpL(p,Ω)p+l11λpln(R)λ1(p+l11)). By Lemma 2.3, we obtain

    ul1+1C(1ϑ3)pp+l11psup+l11p(1ϑ3)pp+l11uϑ3q+l11, (3.35)

    where

    ϑ3=[(q+l11)ps(l1+1)p](p+l11)[(q+l11)ps(p+l11)p](l1+1).

    By the choice of l1, we have 0<ϑ3<1. Hence, using the Young inequality, for every r3>0 and ε2>0, we obtain

    ur3l1+1C(1ϑ3)pr3p+l11ps(ε2up+l11pp+ε(1ϑ3)r3(1ϑ3)r3(p+l11)2uϑ3r3(p+l11)(p+l11)(1ϑ3)r3q+l11). (3.36)

    We now choose

    r3=(q+l11)(p+l11)ϑ3(p+l11)+(q+l11)(1ϑ3),

    and we notice that ϑ3r3(p+l11)(p+l11)(1ϑ3)r3=q+l11. That means

    ur3l1+1C(1ϑ3)pr3p+l11ps(ε2up+l11pp+ε(1ϑ3)r3(1ϑ3)r3(p+l11)2uq+l11q+l11). (3.37)

    We choose

    ε2=[1β(m0l1pp(p+l11)pλpL(p,Ω)p+l11λpln(R)λ1(p+l11))](1ϑ3)r3(p+l11)(p+l11).

    By (3.34) and (3.37) and λ<P5, we obtain

    1l1+1ddtul1+1l1+1+C9ur3l1+10,

    where

    C9=C(ϑ31)pr3p+l11psβ[1β(m0l1pp(p+l11)pλpL(p,Ω)p+l11λpln(R)λ1(p+l11))](ϑ31)r3(p+l11),

    and C9>0, which implies that

    {ul1+1[u0l1+1r3l1+1+(r3l11)C9t]1l1+1r3,t[0,T6),ul1+10,t[T6,),

    where

    T6=1(l1+1r3)C9u0l1+1r3l1+1.

    The proof is completed.

    Proof of Theorem 1.4

    We consider first the case r<2. When 2N/(N+2)<p<2, multiplying (1.1) by u, we obtain

    12ddtΩu2dx+M(up)up=λΩ|u|rln|u|dxβΩuqdx. (3.38)

    Similar to the Theorem 1.2, we obtain

    12ddtu22+(m0λL(r,Ω)Rrp)upλln(R)|Ω|2r2ur2βuqq. (3.39)

    Taking β=0 in Lemma 2.3, we have

    u2u(1ϑ4)psuϑ4qC(1ϑ4)pSu(1ϑ4)uϑ4q, (3.40)

    where ϑ4=(ps2)p(psq)2. Then, taking into account that ϑ4(0,1), we can apply Young's inequality: for every r4>0 and ε3>0, we have

    ur42C(1ϑ4)r4ps(ε3up+ε(ϑ41)r4p(1ϑ4)r43uϑ4r4ppr4(1ϑ4)q). (3.41)

    Taking

    r4=qpq(1ϑ4)+ϑ4p,

    then ϑ4r4ppk2(1ϑ4)=q. By (3.39), (3.41), Lemma 2.5 and λ<P6, and let ε3=(m0λL(r,Ω)Rrpβ)p(1ϑ4)r4p, we obtain

    12ddtu22+C10ur20, (3.42)

    where

    C10=βε(1ϑ4)r4p(1ϑ4)r43C(ϑ41)r4psRr4rλln(R)|Ω|2r2>0,

    which implies that our result holds.

    When 1<p2N/(N+2s), we multiply (1.1) by ul1 (l1>l1), by Lemma 2.4 and Lemma 2.7, we obtain

    1l1+1ddtul1+1l1+1+m0l1pp(p+l11)pul1+p1ppλ1eσur5r5βuq+l11q+l11, (3.43)

    where r5=l1+r+σ1. Using the Holder inequality and (3.37), and we choose

    ε2=[m0l1ppβ(p+l11)p]p+l11+(1ϑ3)r3p+l11,

    then, we have

    1l1+1ddtul1+1l1+1+βε(ϑ31)r3(1ϑ3)r3(p+l11)2C(ϑ31)pr3p+l11psur3l1+1λ1eσ|Ω|1r5l1+1ur5l1+1, (3.44)

    for 0<σ<2r. After performing some simple calculations, we finally obtain

    1l1+1ddtul1+1l1+1+C11ur3l1+10, (3.45)

    where

    C11=βε(ϑ31)r3(1ϑ3)r3(p+l11)2C(ϑ31)pr3p+l11psλC,  C=1eσ|Ω|1r5l1+1Cr5l1+1Rr5r3.

    Note that λ<P7, then C11>0, which implies that

    {ul1+1[u0l1+1r3l1+1+(r3l11)C11t]1l1+1r3,t[0,T7),ul1+10,t[T7,),

    where

    T7=1(l1+1r3)C11u0l1+1r3l1+1.

    The proof is completed.

    Proof of Theorem 1.5

    (ⅰ) By the definition of E(u) and I(u), we obtain

    E(u)=1pI(u)+1pb(1θ)uθp+β(pq)qpuqq+λ1p2upp. (3.46)

    Choosing ν=ut in Definition 1.1, we have

    Ωututdx=M(up)u,ut+λΩut|u|r2uln|u|dxβΩut|u|q2udx. (3.47)

    Note that

    ddtE(u)=1pddt(aup+buθp)λΩ|u|r2uutln|u|dx+βΩ|u|q2uutdx. (3.48)

    By (3.47) and (3.48), we obtain

    ddtE(u)+Ωututdx=0,

    which implies that

    E(u)=E(u0)t0uτ22dτ. (3.49)

    Setting Γ(t)=12Ω|u(x,t)|2dx, then we have

    Γ(t)=Ωutudx=I(u). (3.50)

    By (3.46) and (3.49) and (3.50), we obtain

    Γ(t)=pE(u)+b(1θ)uθp+1qβ(pq)uqq+λ1pupp=pE(u0)+b(1θ)uθp+1qβ(pq)uqq+λ1pupp+pt0uτ22dτ.

    Since p>q, then 1qμ(pq)uqq>0, we obtain

    Γ(t)pE(u0)>0.

    By a simple calculation, we get

    u222pE(u0)t+2u022, for all t>0,

    which implies that our result holds.

    (ⅱ) Here, we only prove the case of E(u0)<h, and the proof of E(u0)=h is similar. Choosing ν=u in Definition 1.1, we have

    12ddtu22+M(up)up=Ω|u|rln|u|dxβuqq, (3.51)

    namely

    12ddtu22+I(u)=0. (3.52)

    Next, taking ν=ut in Definition 1.1 and integrating with respect to time from 0 to t, we have

    t0uτ22dx+E(u(t))=E(u0)<h,for t>0. (3.53)

    We claim that u(x,t)W for any t>0. If it is false, there exists a t0R+0{0} such that u(t0)W, which implies

    I(u(x,t0))=0 or E(u(x,t0))=h.

    From (3.53), E(u(t0))=h is not true. So u(t0)N, then by the definition of h in (1.9), we have E(u(t0))h, which also contradicts with (3.53). Hence, u(t0)W. By (3.52) and u(t)W for all t>0, we obtain

    u22u022.

    Remark 2 Compared with problem (1.4), we not only discuss the extinction of weak solutions of problem (1.1) with logarithmic nonlinearity, but also prove that the weak solutions are globally bounded and blow up at infinity.

    This research was supported by the Project for Young Talents Growth of Guizhou Provincial Department of Education under (Grant No.Ky[2017]133), and by the project of Guizhou Minzu University under (Grant No.16yjrcxm002 and No.GZMU[2019]YB04).

    The authors declare no potential conflict of interests.



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