Research article Special Issues

On a singular parabolic p-Laplacian equation with logarithmic nonlinearity

  • In this paper, we considered a singular parabolic p-Laplacian equation with logarithmic nonlinearity in a bounded domain with homogeneous Dirichlet boundary conditions. We established the local solvability by the technique of cut-off combining with the method of Faedo-Galerkin approximation. Based on the potential well method and Hardy-Sobolev inequality, the global existence of solutions was derived. In addition, we obtained the results of the decay. The blow-up phenomenon of solutions with different indicator ranges was also given. Moreover, we discussed the blow-up of solutions with arbitrary initial energy and the conditions of extinction.

    Citation: Xiulan Wu, Yaxin Zhao, Xiaoxin Yang. On a singular parabolic p-Laplacian equation with logarithmic nonlinearity[J]. Communications in Analysis and Mechanics, 2024, 16(3): 528-553. doi: 10.3934/cam.2024025

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  • In this paper, we considered a singular parabolic p-Laplacian equation with logarithmic nonlinearity in a bounded domain with homogeneous Dirichlet boundary conditions. We established the local solvability by the technique of cut-off combining with the method of Faedo-Galerkin approximation. Based on the potential well method and Hardy-Sobolev inequality, the global existence of solutions was derived. In addition, we obtained the results of the decay. The blow-up phenomenon of solutions with different indicator ranges was also given. Moreover, we discussed the blow-up of solutions with arbitrary initial energy and the conditions of extinction.



    In this paper, we are concerned with the following initial-boundary problem:

    {|x|sutΔpu=|u|q2uln|u|,xΩ,t>0;u(x,t)=0,xΩ,t>0;u(x,0)=u0(x),xΩ, (1.1)

    where ΩRN(NNΩ) is a bounded domain with smooth boundary Ω, Δpu=div(|u|p2u) with u0W1,p0(Ω), x=(x1,x2,,xN)RN with |x|=x21+x22++x2N, and the parameters satisfy

    0s2,max{2NN+2,1}<pq<p(1+2N).

    As is well known, according to the law of conservation, many diffusion processes with reactions can be described by the following equation (see [1]):

    ut(Du)=f(x,t,u,u), (1.2)

    where u(x,t) stands for the mass concentration in chemical reaction processes or temperature in heat conduction, at position x in the diffusion medium and time t. The function D is called the diffusion coefficient or the thermal diffusivity, the term (Du) represents the rate of change due to diffusion, and f(x,t,u,u) is the rate of change due to reaction.

    In the past few years, many researchers had focused on Equation (1.2). For more details, one can refer to [2,3,4,5,6]. For the source f(x,t,u,u)=uq, there has already been much discussion. For example, for D=|x|2, in 2004, Tan [7] considered the existence and asymptotic estimates of global solutions as well as finite time blow-up of local solutions based on the classical Hardy inequality [8]. Han [9] considered the blow-up properties of solutions to the following non-Newton filtration equation with special a medium void:

    |x|2utpu=uq. (1.3)

    A new criterion for the solutions to blow up in finite time was established by using the Hardy inequality. Moreover, the upper and lower bounds for the blow-up time were also estimated. The results solved an open problem proposed by Liu [10] in 2016.

    When the source f(u) is a logarithmic nonlinearity, Deng and Zhou [11] investigated the following semilinear heat equation with singular potential and logarithmic nonlinearity

    |x|sutu=uln|u|, (1.4)

    under an appropriate initial-boundary value condition. They did make full use of the logarithmic Sobolev inequality in [12,13] to handle the difficulty caused by the logarithmic nonlinear term uln|u|. Taking the combination of a family of potential wells, the existence of global solutions and infinite time blow-up solutions were obtained.

    Liu and Fang [14] considered a fourth-order singular parabolic equation involving logarithmic nonlinearity and p-biharmonic operator

    |x|sut+(|u|p2u)=|u|q2ulog|u|, (1.5)

    and they established the local solvability by the technique of cut-off combining with the methods of Faedo-Galerkin approximation and multiplier. Meantime, by virtue of the family of potential wells, they used the technique of modified differential inequality and the improved logarithmic Sobolev inequality to obtain the global solvability and the infinite and finite time blow-up phenomena, and derived the upper bound of blow-up time as well as the estimate of the blow-up rate. Furthermore, the results of blow-up with arbitrary initial energy and extinction phenomena were presented.

    Motivated by these works, in this paper, we consider the Problem (1.1) with the presence of nonlinear diffusion pu:=div(|u|p2u) and logarithmic nonlinearity |u|q2uln|u|. To the best our knowledge, this is the first work in the literature that takes into account a singular parabolic p-Laplacian equation with logarithmic nonlinearity.

    The rest of this paper is organized as follows. In Section 2, we introduce some symbols and definitions. In Section 3, we prove the local existence and uniqueness theorem. In Section 4, we prove the global existence and asymptotic behavior theorems of solutions. In Section 5, the blow-up phenomena of solutions are discussed. Finally, the extinction phenomenon of the solution is given in Section 6.

    In this section, we introduce some notations and lemmas that will be used throughout the paper. In what follows, we denote by r(r1) the norm in Lr(Ω) and by (,) the L2(Ω) inner product. When p>1,p2, we use W1,p0(Ω) to denote the Sobolev space such that both u and u belong to Lp(Ω) for any uW1,p0(Ω), denote by W1,p(Ω) its dual space, and by , the duality pairing between them. We will equip W1,p0(Ω) with the norm uW1,p0=up, which is equivalent to the full one due to the Poincarés inequality. We use λ1>0 to denote the first eigenvalue of in Ω under the homogeneous Dirichlet boundary condition. We also use notation X0 to denote W1,p0(Ω){0}.

    Due to the presence of the inverse coefficient |x|s, it is worth emphasizing the difference between the two cases when 0Ω and 0Ω.

    If 0Ω, then |x|s develops a singularity. This necessitates the use of the Hardy-Sobolev inequality, which is valid for NΩ3, in the proofs of our main results.

    On the other hand, if 0Ω, then there is no singularity and (1.1) can be regarded as a slight extension. In this case, our results are valid for all N{1,2,3,}. To deal with these two cases simultaneously, we employ the notation

    NΩ={3,if0Ω,1,if0Ω.

    First, for Problem (1.1), we introduce the potential energy functional

    J(u)=1pupp1qΩ|u|qln|u|dx+1q2uqq, (2.1)

    and the Nehari functional

    I(u)=uppΩ|u|qln|u|dx. (2.2)

    By a direct computation,

    J(u)=1qI(u)+(1p1q)upp+1q2uqq. (2.3)

    By I(u) and J(u), we define the potential well:

    W1={uX0:J(u)<d},W2={uX0:J(u)=d},W=W1W2,
    W+1={uW1,I(u)>0},W+2={uW2,I(u)>0},W+=W+1W+2,
    W1={uW1,I(u)<0},W2={uW2,I(u)<0},W=W1W2,

    and the Nehari manifold

    N={uX0,I(u)=0}.

    The depth of the potential well is defined as

    d=infuNJ(u).

    The solution u(x,t) to Problem (1.1) is considered in weak sense as follows. Sometimes u(x,t) will be simply written as u(t) if no confusion arises.

    Lemma 2.1. [15] Let μ be a positive number. Then we have the following inequalities:

    splns(eμ)1sp+μ,foralls1,
    |splns|(ep)1,forall0<s<1.

    Lemma 2.2. [15,16] Assume that q<NpNp, i.e., q< for Np and rq<NpNp for N>p and r1. Then for any uW1,p0(Ω), it holds that

    uqCGuθpu1θr,

    where θ(0,1) is determined by θ=(1r1q)(1N1p+1r)1 and the constant CG>0 depends on N,p,q, and r.

    Remark 2.1. From p>2NN+2, we deduce

    p(1+2N)<{NpNp,ifN>p,+,ifNp.

    Then by the Sobolev inequality, we have W1,p0(Ω)Lq+a(Ω) for p>1 and a0.

    Lemma 2.3. Let u(t)X0 and p,q satisfy max{2NN+2,1}<pq<p(1+2N). We have the following statements:

    (i) If 0<upr, then I(u)0;

    (ii) If I(u)<0, then up>r;

    (iii) If I(u)=0, then up=0 or upr,

    where

    r=(1Bq+α)1q+αp.

    Proof. (ⅰ) A direct computation yields

    ln|u(x)|<|u(x)|αα,a.e.xΩ,α>0. (2.4)

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

    I(u)=uppΩ|u|qln|u|dx=uppuq+αq+α(1Bq+αuq+αpp)upp, (2.5)

    where B is the imbedding constant for W1,p0(Ω)Lq+α(Ω). If 0<upr, this implies that uq+αpp1Bq+α. Therefore, we gain I(u)0 by (2.5).

    (ⅱ) From (2.5) and I(u)<0, we can see that

    (1Bq+αuq+αpp)upp<0,

    which means that

    up>(1Bq+α)1q+αp=r.

    (ⅲ) If I(u)=0, then from (2.5) we attain

    up(1Bq+α)1q+αporup=0.

    The prove is complete.

    Next, in Lemma 2.4, we describe some basic properties of the fiber mapping J(λu) that can be verified directly.

    Lemma 2.4. [17] Assume that uX0, then

    (i) limλ0+J(λu)=0, limλ+J(λu)=.

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

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

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

    Lemma 2.5. [15,18] (Logarithmic Sobolev Inequality). Let q>1, μ>0, and uW1,q0(RN){0}. Then we have

    qRN|u(x)|qln(|u(x)|uLq(RN))dx+Nqln(qμeNϑq)RN|u(x)|qdxμRN|u(x)|qdx,

    where

    ϑq=qN(q1e)q1πq2[Γ(N2+1)Γ(Nq1q+1)]qN.

    Remark 2.2. If uW1,q0(Ω){0}, then by defining u(x)=0 for xRNΩ, we derive

    qΩ|u(x)|qln(|u(x)|uLq(Ω))dx+Nqln(qμeNϑq)Ω|u(x)|qdxμΩ|u(x)|qdx, (2.6)

    for any real number μ>0.

    Lemma 2.6. [14,19] (Hardy-Sobolev inequality). Let RN=Rk×RNk,2kN and x=(y,z)RN=Rk×RNk. For given n,β satisfying 1<p<N, 0βp, and β<k, let m(β,N,p)=p(Nβ)(Np). Then there exists a positive constant CH depending on β,N,p, and k such that for any uW1,p0(RN), it holds that

    RN|u(x)|m|y|βdxCH(RN|u|pdx)NβNp.

    Remark 2.3. (i) When m=p=β, this inequality is the classical Hardy inequality. (ii) If m=2,β=s in Lemma 2.4, we have p=2NNs+2>2, and then Lemma 2.6 becomes

    Ω|u(x)|2|x|sdxCH(Ω|u|2NNs+2dx)Ns+2N.

    Lemma 2.7. [15,20] Let f:R+R+ be a nonincreasing function and σ be a positive constant such that

    +tf1+σ(s)ds1ωfσ(0)f(t),t0.

    Then we have

    (i) f(t)f(0)e1ωt, for all t0, whenever σ=0.

    (i) f(t)f(0)(1+σ1+ωσt)1σ, for all t0, whenever σ>0.

    The following is the concavity lemma.

    Lemma 2.8. [21,22,23] Suppose that a positive, twice-differentiable function Ψ(t) satisfies the inequality

    Ψ(t)Ψ(t)(1+θ)(Ψ(t))20,

    where θ>0. If Ψ(0)>0 and Ψ(0)>0, then Ψ(t) as

    ttt=Ψ(0)θΨ(0).

    Lemma 2.9. [24] Suppose that 0<l<r1 and ϵ1,ϵ20 are positive constants. If nonnegative and absolutely continuous function h(t) satisfies

    h(t)+ϵ1hl(t)ϵ2hr(t),t0,
    h(0)>0,ϵ2hrl(0)<ϵ1,

    then we have

    h(t)[ϵ0(1l)t+h1l(0)]11l,0<t<T0,

    and

    h(t)0,tT0,

    where ϵ0=ϵ1ϵ2hrl(0) and T0=h1l(0)ϵ0(1l).

    Definition 2.1. (Weak Solution). A function u:=u(x,t)L(0,T;X0) with |x|s2utL2(0,T;L2(Ω)) is called a weak solution of Problem (1.1) on Ω×[0,T) if u(x,0)=u0(x) in X0 and

    |x|sut,v+|u|p2u,v=|u|q2uln|u|,v,a.e.t(0,T),

    for any vW1,p0(Ω). Moreover,

    t0|x|s2uτ22dτ+J(u(x,t))=J(u0),a.e.t(0,T).

    Definition 2.2. (Maximal Existence Time). Let u(x,t) be a weak solution of Problem (1.1), we define the maximal existence time Tmax as follows:

    Tmax=sup{T>0;u(x,t)existson[0,T]}.

    (i) If Tmax=+, we say that the solution u(t) is global;

    (ii) If Tmax<+, we say that the solution u(t) blows up in finite time and Tmax is the blow-up time.

    Definition 2.3. (Finite Time Blow-Up). Let u(x,t) be a weak solution of Problem (1.1), then u(x,t) is called the finite time blow-up if the maximal existence time Tmax<+ and

    limtTmaxt0|x|s2u(τ)22dτ=+.

    Definition 2.4. (Infinite Time Blow-Up). Let u(x,t) be a weak solution of Problem (1.1), then u(x,t) is called the infinite blow-up if

    limt+|x|s2u(t)22=+.

    In this section, we state the local existence and uniqueness of weak solutions to Problem (1.1).

    Theorem 3.1. Let u0X0, and p,q satisfy max{2NN+2,1}<pq<p(1+2N). Then there exist a T>0 and a unique weak solution u(x,t)L(0,T;X0) of Problem (1.1) with |x|s2utL2(0,T;L2(Ω)) satisfying u(0)=u0. Moreover, u(x,t) satisfies the energy equality

    t0|x|s2ut22dt+J(u)=J(u0),0tT.

    Proof. We divide the proof of Theorem 3.1 into 5 steps.

    Step 1. Approximate problem

    In order to deal with the singular potential, we introduce the cut-off function

    ρn(x)=min{|x|s,n},nN+.

    We denote the solutions corresponding to ρn of Problem (1.1) as un,

    {ρn(x)untΔpun=|un|q2unln|un|,xΩ,t>0,un(x,t)=0,xΩ,t>0,un(x,0)=un0,xΩ. (3.1)

    We noticed that un0C0(Ω), and then un0u0(x)inW1,p0(Ω). Let {ωj}j=1 be a system of basis in W1,p0(Ω) which is normalized orthogonal in L2(Ω) and construct the approximate solution

    ukn(x,t)=kj=1aknj(t)ωj(x)fork=1,2,,j=1,2,,k.

    We solve the problem

    ρn(x)uknt,ωj+|ukn|p2ukn,ωj+=|ukn|q2uknln|ukn|,ωj, (3.2)

    and

    ukn(x,0)=kj=1bknjωj(x)=ukn0un0(x)inW1,p0(Ω) (3.3)

    as k+,n+. Hence {aknj}kj=1 is determined by the following Cauchy problem:

    {kj=1(Ωρn(x)ωj(x)ωjdx)[aknj(t)]t=Gknj(t),aknj(0)=bknj,

    where

    Gknj(t)=Ω|kj=1aknj(t)ωj(x)|q2kj=1aknj(t)ωj(x)ln|kj=1aknj(t)ωj(x)|ωjdxΩ|kj=1aknj(t)ωj(x)|p2kj=1aknj(t)ωj(x)ωjdx.

    Therefore, the standard theory of ordinary differential equations yields that there exists a T>0 such that aknj(t)C1([0,T]). As a consequence, uknC1([0,T],W1,p0(Ω)).

    Step 2: Priori estimates

    We discuss the following two cases:

    Case 1: max{2NN+2,1}<pq and 2q<p(1+2N)

    Multiply (3.2) by aknj(t), sum for j=1,,k, and recall ukn(x,t) to find

    ρn(x)uknt,ukn+|ukn|p2ukn,ukn=|ukn|q2uknln|ukn|,ukn. (3.4)

    Integrating over (0,t) on both sides of (3.4), we get,

    12(ρn(x))12ukn(t)22+t0ukn(s)ppds=t0Ω|ukn(s)|qln|ukn(s)|dxds+12(ρn(x))12ukn(0)22.

    Set

    Skn(t)=12(ρn(x))12ukn(t)22+t0ukn(s)ppds. (3.5)

    Combining the above equalities, and we have

    Skn(t)Skn(0)+t0Ω|ukn(s)|qln|ukn(s)|dxds. (3.6)

    From Lemma 2.1, we get

    Ω|ukn(t)|qln|ukn(t)|dx=Ω1={xΩ;|ukn(x)|1}|ukn(t)|qln|ukn(t)|dx+Ω2={xΩ;|ukn(x)|<1}|ukn(t)|qln|ukn(t)|dx(eμ)1Ω1={xΩ;|ukn(x)|1}|ukn(t)|q+μdx(eμ)1ukn(t)q+μq+μ. (3.7)

    Then, by Lemma 2.2 and Young's inequality, (3.7) becomes

    Ω|ukn(t)|qln|ukn(t)|dx(eμ)1ukn(t)q+μq+μ(eμ)1CGukn(t)θ(q+μ)pukn(t)(1θ)(q+μ)2(eμ)1CGεukn(t)pp+(eμ)1CGC(ε)ukn(t)p(1θ)(q+μ)pθ(q+μ)2, (3.8)

    where ε(0,1), and θ=(121q+μ)(1N1p+12)1=(q+μ2)Np(q+μ)(2p2N+Np). We note that since 0<μ<p(1+2N)q, θ(q+μ)<p holds. Let

    α=p(1θ)(q+μ)2[pθ(q+μ)]=p(N+q+μ)N(q+μ)p(N+2)N(q+μ),

    then α>1 since max{1,2NN+2}<pq, 2q<p(1+2N). Besides, since Ω is a bounded domain in RN, it leads to

    Ω|ukn(t)|2dx=Ω(ρn(x))1ρn(x)|ukn(t)|2dxC(Ω)(ρn(x))12ukn(t)22, (3.9)

    where C(Ω) is related to Ω.

    Thus, from (3.5), (3.6), (3.8), and (3.9), we get

    Skn(t)Skn(0)+t0(eμ)1CGεukn(t)ppds+t0(eμ)1CGC(ε)ukn(t)2α2dsSkn(0)+(eμ)1CGεSkn(t)+(eμ)1CGC(ε)C(Ω)t0(Skn(t))αds,

    and

    Skn(t)Skn(0)1(eμ)1CGε+(eμ)1CGC(ε)C(Ω)1(eμ)1CGεt0(Skn(t))αds.

    Therefore,

    Skn(t)C1+C2t0(Skn(t))αds, (3.10)

    where 1(eμ)1CGε>0, C1=Skn(0)1(eμ)1CGε, and C2=(eμ)1CGC(ε)C(Ω)1(eμ)1CGε. From the Gronwall-Bellman-Bihari inequality, we obtain

    Skn(t)C3,

    and

    12(ρn(x))12ukn(t)22+t0ukn(s)ppdsC3,n,kN+, (3.11)

    where C3 is a constant which is dependent on T.

    Multiplying (3.2) by [aknj(t)]t, summing on j=1,2,,k, and then integrating on (0,t), we know that

    t0(ρn(x))12uknt(s)22ds+J(ukn(t))=J(ukn0),0tT. (3.12)

    By the continuity of the functional J and (3.3), there exists a constant C>0 satisfying

    J(ukn0)C,foranypositiveintegernandk. (3.13)

    Applying (2.1), (3.5), (3.8), (3.11), (3.12), and (3.13), we obtain

    t0(ρn(x))12uknt(s)22ds+(1pCGεeμq)ukn(t)pp+1q2ukn(t)qqC4J(ukn0(t))C, (3.14)

    where C4=2CGC(ε)C(Ω)eμ(C3)α, for all n,kN+.

    Case 2: max{1,2NN+2}<pq<2

    Combining ln|u(x)|<|u(x)|aa a.e. xΩ,a>0 and (3.5), and taking a=2q, we obtain

    Skn(t)Skn(0)+12qt0ukn(s)22ds.

    Together with (3.9), it can become

    Skn(t)Skn(0)+2(2q)C(Ω)t0Skn(s)ds.

    Then by means of Gronwall's inequality, we have

    Skn(t)C5,

    and

    12(ρn(x))12ukn(t)22+t0ukn(s)ppdsC5, (3.15)

    where C5=Skn(0)e2T(2q)C(Ω).

    From (2.1), (3.12), (3.13), and (3.15), we have

    t0(ρn(x))12uknt(s)22ds+1pukn(t)pp+1q2ukn(t)qqC+1qΩ|ukn(t)|qln|ukn(t)|dxC+2q(2q)C(Ω)(ρn(x))12ukn(t)22C+2C5q(2q)C(Ω), (3.16)

    for all k,nN+.

    Therefore, we can derive

    ukn(t)L(0,T;W1,p0(Ω))C,foranypositiveintegernandk, (3.17)
    ukn(t)L(0,T;Lq(Ω))C,foranypositiveintegernandk, (3.18)
    (ρn(x))12uknt(t)L2(0,T;L2(Ω))C,foranypositiveintegernandk. (3.19)

    Combining (3.9) and (3.16), we have

    uknt(t)L2(0,T;L2(Ω))C,foranypositiveintegernandk. (3.20)

    Step 3: Pass to the limit

    By (3.17), (3.19), and the Aubin-Lions-Simon Lemma (see [25], Corollary 4), we get

    uknuinC(0,T;L2(Ω)), (3.21)

    as k,n+. Thus, ukn(x,0)u(x,0) in L2(Ω). Combining (3.3) with un0u0(x)inW1,p0(Ω), we observe that u(x,0)=u0 in W1,p0(Ω).

    From (3.21), we have uknu a.e.(x,t)Ω×(0,T). This implies

    |ukn|q2uknln|ukn||u|q2uln|u|a.e.(x,t)Ω×(0,T).

    It follows from (3.14) and the Hölder inequality that

    |ukn(t)|p2ukn(t)W1,p,(Ω)=supφW1,p0(Ω){0}Ω|ukn(t)|p2ukn(t)φdxφW1,p0(Ω)(Ω||ukn(t)|p2ukn(t)|pp1dx)p1p(Ω|φ|pdx)1pφW1,p0(Ω)ukn(t)p1p<C.

    That means

    |ukn(t)|p2ukn(t)L(0,T;W1,p,(Ω))C,foranypositiveintegernandk. (3.22)

    On the other hand, there is

    Ω||ukn|q2uknln|ukn||pp1dx=Ω1={xΩ;|ukn(x)|1}||ukn|q2uknln|ukn||pp1dx+Ω1={xΩ;|ukn(x)|<1}||ukn|q2uknln|ukn||pp1dx.

    From Lemma 2.1 and Lemma 2.2, we have

    Ω||ukn(t)|q2ukn(t)ln|ukn(t)||pp1dx=Ω1||ukn(t)|q2ukn(t)ln|ukn(t)||pp1dx+Ω2||ukn(t)|q2ukn(t)ln|ukn(t)||pp1dxΩ1||ukn(t)|μln|ukn(t)||ukn(t)|q1+μ|pp1dx+Ω2||ukn(t)|q1ln|ukn(t)||pp1dx(eμ)pp1ukn(t)pp1(q1+μ)pp1(q1+μ)+[e(q1)]pp1|Ω|(eμ)pp1B1ukn(t)pp1(q1+μ)p+[e(q1)]pp1|Ω|<C,

    where B1 is the best constant of the Sobolev embedding W1,p0(Ω)Lpp1(q1+μ)(Ω). Here we choose 0<μp(1+p1Np)q, q<p(1+p1Np), and we know that

    |ukn(t)|q2ukn(t)ln|ukn(t)|L(0,T;Lpp1(Ω))C,foranypositiveintegernandk. (3.23)

    By (3.17)–(3.19), (3.22), (3.23), there exist functions u, χ and a subsequence of {ukn}n,k=1 which we still denote by {ukn}n,k=1 such that

    uknuweaklystarinL(0,T;W1,p0(Ω)) (3.24)
    (ρn(x))12uknt|x|s2utweaklyinL2(0,T;L2(Ω)) (3.25)
    |ukn|p2uknχweaklystarinL(0,T;W1,p(Ω)) (3.26)
    |ukn|q2uknln|ukn||u|q2uln|u|weaklystarinL(0,T;Lpp1(Ω)). (3.27)

    Next, by the method of Browder and Minty in the theory of monotone operators, we obtain χ=|u|p2u.

    By (3.24)–(3.27), passing to the limit in (3.2) as n,k+, it follows that u satisfies the initial condition u(0)=u0,

    |x|sut,ω+|u|p2u,ω=|u|q2uln|u|,ω, (3.28)

    for all ωW1,p0(Ω), and for a.e. t[0,T].

    Step 4. Uniqueness

    Suppose there are two solutions u1 and u2 to Problem (1.1), and we have

    |x|su1t,v+|u1|p2u1,v=|u1|q2u1ln|u1|,v, (3.29)

    and

    |x|su2t,v+|u2|p2u2,v=|u2|q2u2ln|u2|,v. (3.30)

    Let w=u1u2 and w(0)=0, then by subtracting (3.29) and (3.30), we can derive

    Ω|x|swtvdx+Ω(|u1|p2u1|u2|p2u2)vdx=Ω(|u1|q2u1ln|u1||u2|q2u2ln|u2|)vdx.

    Let v=w, and we recall the following elementary vector inequalities that are used frequently: for all a,bRN, we have 0(p1)|ab|2(|a|+|b|)2p(|a|p2a|b|p2b)(ab), if 1<p<2. So, we obtain

    12ddt|x|s2w22Ω|u1|q2u1ln|u1||u2|q2u2ln|u2|ww2dxΩf(u1)f(u2)ww2dx.

    Integrating it on [0,t], we obtain

    |x|s2w222t0Ωf(u1)f(u2)ww2dxdt, (3.31)

    where F(s)=|s|q2sln|s|. Combining with (3.9), we get

    w222MTt0w22dt.

    By the locally Lipschitz continuity of F:RNR, the uniqueness follows from Gronwall's inequality.

    Step 5: Energy equality

    We multiply (1.1) with ut and integrate over Ω×(0,t) to obtain the equality

    t0|x|s2ut(s)22ds+J(u(t))=J(u0),0tT. (3.32)

    The proof of Theorem 3.1 is complete.

    In this section, we are concerned with the existence of a global weak solution to Problem (1.1).

    Theorem 4.1. Assume that u0W+, max{2NN+2,1}<pq<p(1+2N), and then Problem (1.1) admits a global solution uL(0,;X0) with |x|s2utL2(0,;L2(Ω)) and u(t)W+ for 0t.

    Proof. Now, we prove Theorem 4.1. In order to prove the existence of global weak solutions, we consider the following two steps:

    Step 1. The initial data u0W+1

    From (3.32), we know that

    t0|x|s2ut(s)22ds+J(u(t))=J(u0)<d,0tTmax, (4.1)

    where Tmax is the maximal existence time of solution u(t). We shall prove that Tmax=+. Next, we will show that

    u(t)W+1forall0tTmax. (4.2)

    Indeed, assume that (4.2) does not hold and let t be the smallest time for which u(t)W+1. Then, by the continuity of u(t), one has u(t)W+1. Hence, it follows that

    J(u(t))=d, (4.3)

    or

    I(u(t))=0. (4.4)

    Nevertheless, it is clear that (4.3) could not occur by (4.1) while if (4.4) holds then, by the definition of d, we have

    J(u(t))infuNJ(u)=d,

    which also contradicts with (4.1). Hence, (4.2) is valid.

    Next, it is discussed in two cases.

    Case 1: p<q

    As a consequence, it follows from this fact and the definition of functional J(u(t)) that

    t0|x|s2ut(s)22ds+1qI(u(t))+(1p1q)u(t)pp+1q2u(t)qq<d, (4.5)

    and

    t0|x|s2ut(s)22ds+(1p1q)u(t)pp+1q2u(t)qq<d. (4.6)

    This estimate allows us to take Tmax=+. So, we can conclude that there is a unique global weak solution u(t)W+1 of Problem (1.1) which satisfies that

    t0|x|s2ut(s)22ds+J(u(t))=J(u0),0t+.

    Case 2: p=q

    Similar to Case 1, we can derive

    t0|x|s2ut(s)22ds+1p2u(t)pp<d.

    By Lemma 2.5, we have

    Ω|u(t)|qln|u(t)|dx[lnu(t)Lq(Ω)nq2ln(qμeNϑq)]Ω|u(t)|qdx+μqΩ|u(t)|qdx.

    From (2.2), (4.2), and the above inequality, we know that

    u(t)pp=I(u(t))+Ω|u(t)|pln|u(t)|dx=2I(u(t))+2Ω|u(t)|pln|u(t)|dxu(t)pp2I(u(t))+2[lnu(t)Lp(Ω)np2ln(pμeNϑp)]Ω|u(t)|pdx+(2μp1)u(t)pp2pJ(u(t))+2[1p2+lnu(t)Lp(Ω)np2ln(pμeNϑp)]u(t)ppCd. (4.7)

    Combining the two cases above, we know that the estimate allows us to take Tmax=+. It means that there is a unique global weak solution u(t)W+1 of Problem (1.1).

    Step 2. The initial data u0W+2

    First, we choose a sequence {θm}m=1(0,1) such that limmθm=1. Then we consider the following problem

    {|x|sutΔpu=|u|q2uln|u|,(x,t)Ω×R+,u=0,(x,t)Ω×R+,u(x,0)=u0m(x),xΩ, (4.8)

    where u0m=θmu0. First of all, we claim that u0mW+1, and then J(u0m)<d and I(u0m)>0. In fact, from u0W+1, limmθm=1 and I(u0)>0, and we can see

    I(u0m)=θpmu0ppθqmln|θm|u0qqθqmΩ|u0|qln|u0|dxθpm(u0ppθqpmΩ|u0|qln|u0|dx)θpmI(u0)>0.

    On the other hand, by direct calculations, we obtain

    dd(θm)J(θmu0)=1θm(θpmu0ppθqmln|θm|u0qqθqmΩ|u0|qln|u0|dx)=1θmI(u0m)>0,

    which implies that J(θmu0) is strictly increasing with respect to θm and

    J(u0m)=J(θmu0)<J(u0)=d.

    Since u0mu0 as m+, our result can be derived by the same processes as the proof of Step 1.

    Theorem 4.1 is complete.

    Theorem 4.2. Let u(t) be the solution of Problem (1.1) and p,q satisfy

    2<p<q<p(1+2N).

    If u0W+1, then there exist positive constants c2 such that

    u(t)2pu02p(p11+c2(p2)t)1p2,t0.

    Especially, if p=2, then there exist positive constants c4 such that

    u(t)2u02e12(1c4t),t0.

    Proof. We are now in a position to prove the algebraic decay results. Thanks to u0W+1, we get u(t)W+1. From (2.3), we have

    (1p1q)upp+1q2uqqJ(u(t))J(u0)<d. (4.9)

    By (4.9), through a direct calculation, we arrive at

    λ0{(1p1q)upp+1q2uqq}J(λu(t))d,

    where λ0=max{(λ)p,(λ)q}. Combining with (4.9), we get

    λ0max{(dJ(u0))1p,(dJ(u0))1q}>1, (4.10)

    which means that λ>1, λ(dJ(u0))1p.

    From (2.2), we have

    0=I(λu)=(λ)pupp(λ)qΩ|u|qln|u|dx(λ)qln(λ)uqq=(λ)qI(u)((λ)q(λ)p)u(t)pp(λ)qln(λ)uqq. (4.11)

    Namely,

    I(u(t))=uqqlnλ+[1(λ)pq]uppc1u(t)pp, (4.12)

    where c1=1(dJ(u0))1qp, p<q.

    According to Lemma 2.6, and (2.2), we obtain

    TtI(u)ds=Tt(uppΩ|u|qln|u|dx)ds=12Ttddt|x|s2u22ds=12|x|s2u(t)2212|x|s2u(T)2212|x|s2u(t)2212CHu(t)2p. (4.13)

    By (4.12) and (4.13), we get

    Ttu(t)ppdsCHB22c1u(t)2p=1c2u(t)2p. (4.14)

    Let T+ in (4.14), and by the virtue of Lemma 2.7, it follows that

    u(t)2pu02p(p11+c2(p2)t)1p2,t0.

    Theorem 4.2 is complete.

    In this section, we present the blow-up phenomena of the solutions to (1.1) including infinite and finite time blow-up, and give some bounders of the blow-up. For simplicity, we shall write L(t)=12|x|s2u(t)22 in the sequel.

    This subsection is devoted to infinite blow-up for Problem (1.1).

    Theorem 5.1. (Infinite Blow-Up). Let u0W and p,q satisfy 1<pq<2. Then u(t) blows up in infinite time.

    Proof. We divide the proof into 2 steps.

    Step 1: u0W1

    We claim that u(t)W1 for all t[0,Tmax) provided that u0W1. Let u(t) be the weak solution of Problem (1.1) with u0W1, which means that u00 and J(u0)<d,I(u0)<0.

    If J(u0)<0. By the energy equality, we arrive at

    1qI(u)+(1p1q)upp+1q2uqq=J(u(t))J(u0)<0<d. (5.1)

    It means that u(t)0, J(u(t))<d, and I(u(t))<0, which implies that u(t)W1.

    If 0<J(u0)<d. From the energy equality, we obtain

    0<t0|x|s2ut(s)22ds+J(u(t))=J(u0)<d, (5.2)

    which means that u(x,t)0. Next, we will show that I(u(t))<0 for all t[0,Tmax). Otherwise, by the continuity of I(u), there would exist a t(0,Tmax) such that I(u(t))<0, t[0,t) and I(u(t))=0. It means that u(t)N. Then, from the definition of d, it holds that J(u(t))d which contradicts (5.2). Then u(t)W1 for all t[0,Tmax).

    From Lemma 2.4(ⅳ), as I(u(t))<0, there is a λ<1 such that I(λu)=0. Then

    dJ(λu)=1qI(λu)+(λ)p(1p1q)upp+(λ)qq2uqq<(1p1q)upp+1q2uqq. (5.3)

    Then, by taking the derivative of L(t), we obtain

    ddtL(t)=Ωuut|x|sdx=I(u)=qJ(u)+(qp1)upp+1quqqq(dJ(u(t)))=q(dJ(u0))=C0>0,t[0,Tmax]. (5.4)

    Combining (5.4) and L(t)L(0)=t0L(t)dt, we can derive

    |x|s2u(t)22|x|s2u022+C0t>0,t[0,Tmax]. (5.5)

    Now, we prove that u(t) cannot blow up in finite time. Arguing by contradiction, we assume that u(t) blows up in finite time, which implies that

    limtTmax|x|s2u(t)22=+. (5.6)

    Meantime, by (5.4), we have

    L(t)lnL(t)L(t)=12|x|s2u(t)22ln(12|x|s2u(t)22)+I(u(t)). (5.7)

    Next, combining ln|u(x)|<|u(x)|δδ a.e. xΩ,δ>0 and (2.2), and taking δ=2q, we obtain

    I(u(t))u(t)pp12qu(t)22. (5.8)

    On the other hand, from (5.4) and (5.6), we can see that there exists a t1(0,Tmax) such that

    12|x|s2u(t)22>12|x|s2u(t1)22=exp{2Ls2q}, (5.9)

    where |x|<L. Then by combining (5.7), (5.8), and (5.9) we can derive

    L(t)lnL(t)L(t)12|x|s2u(t)22ln(12|x|s2u(t)22)+u(t)pp12qu(t)22(12Lsln(12|x|s2u(t)22)12q)u(t)22>0, (5.10)

    which means that

    L(t)lnL(t)L(t)>0.

    Through a direct calculation, we have

    ddtln(L(t))=L(t)L(t)<ln(L(t)),t[t1,Tmax]. (5.11)

    Then by virtue of Gronwall's inequality, we get

    ln(L(t))<exp{tt1}ln(L(t1)),t[t1,Tmax],

    which implies that

    |x|s2u(t)22<|x|s2u(t1)2exp{tt1}2,t[t1,Tmax].

    That contradicts with (5.6). Therefore, Tmax=+ and u(t) blows up in infinite time.

    Step 2: u0W2

    First of all, u0W2 means that I(u(t))<0,J(u(t))=d, t[0,Tmax]. We claim that u(t)W2 for all t[0,Tmax) provided that u0W2. Otherwise, by continuity, there would exist a t1[0,Tmax) such that I(u(t))<0 for t[0,t1) and I(u(t1))=0. Recalling the definition of d, it is clear that J(u(t1))d. On the other hand, from Ω|uut|x|s|dx=I(u(t))>0,t[t1,Tmax], we know that ut0 and t10|x|s2uτ(τ)22dτ>0, t1[0,Tmax]. Meanwhile, it follows from the energy equality that

    J(u(t1))=J(u0)t00|x|s2uτ(τ)22dτ<J(u0)=d,

    which contradicts with J(u(t1))d. Therefore, there exists a t2[0,Tmax] such that I(u(t2))<0 and J(u(t2))<d. If we take t2 as the initial time, then similar to Step 1, we can obtain that the weak solution u(t) of Problem (1.1) blows up in infinite time.

    Theorem 5.2. (Finite Blow-Up). Let u0W and p,q satisfy 2<pq<p(2N+1). Then u(t) blows up in finite time. Moreover,

    T|x|s2u(0)22(p2)pJ(u0).

    Proof. We shall apply the first-order differential inequality technique to show the finite time blow-up result for Problem (1.1) with negative initial energy. For this, set K(t)=J(u(t)). Then L(0)>0,K(0)>0. From Problem (1.1), it follows that

    ddtK(t)=ddtJ(u(t))=|x|s2ut(t)220,

    which means that K(t)K(0)=J(u0)>0 for all t[0,T). Recalling (2.2) and (2.3), we obtain, for any t[0,T), that

    ddtL(t)=Ωuut|x|sdx=I(u)=qJ(u)+(qp1)upp+1quqqqJ(u)=qK(t)>0. (5.12)

    Making use of the Holder inequality and Cauchy-Schwarz inequality, we arrive at

    L(t)K(t)=12(|x|s2u(t)22)(|x|s2ut(t)22)12(L(t))2q2L(t)K(t), (5.13)

    which then implies

    [K(t)Lq2(t)]=Lq21(t)(L(t)K(t)q2L(t)K(t))0. (5.14)

    Therefore,

    0<k=K(0)Lq2(0)K(t)Lq2(t)1qL(t)Lq2(t)=2(2q)p[L2q2(t)]. (5.15)

    Integrating (5.15) over [0,t] for any t(0,T) and noticing that q>2, one has

    kt2(2q)q[L2q2(t)L2q2(0)],

    or equivalently

    0L2q2(t)L2q2(0)(q2)q2kt,t(0,T]. (5.16)

    It is obvious that (5.16) cannot hold for all t>0. Therefore, T<+. Moreover, it can be inferred from (5.16) that

    T2L(0)(q2)qK(0)=|x|s2u(0)22(q2)qJ(u0).

    The proof is complete.

    For the case of J(u0)0, we obtain blow-up results when the initial energy is 'subcritical' and when the initial Nehari functional is negative which means that u0W. More precisely, we have the following theorem.

    Theorem 5.3. Assume that 2<pq<p(2N+1), u0W. Then the weak solution u(t) to Problem (1.1) blows up in finite time. Furthermore, if u0W1, then Tmax can be estimated from above as follows:

    Tmaxβb2(q2)βb|x|s2u(0)22,

    where β,b are constants that will be determined in the proof.

    Proof. We will divide the proof into two cases.

    Case 1: u0W1

    We claim that u(t)W1 for all t[0,Tmax) provided that u0W1. Otherwise, by continuity, there would exist a t0[0,Tmax) such that I(u(t))>0 for t[0,t0) and I(u(t0))=0. Recalling the definition of d, it is clear that J(u(t0))d, which contradicts with J(u(t))J(u0)<d.

    From Lemma 2.4(ⅳ), as I(u(t))<0, there is a λ<1 such that I(λu)=0. Then

    dJ(λu)=1qI(λu)+(λ)p(1p1q)upp+(λ)qq2uqq<(1p1q)upp+1q2uqq. (5.17)

    We show that Tmax<+. For any T[0,Tmax), define the positive function

    F(t)=t0L(t)dt+(Tt)L(0)+β2(t+b)2, (5.18)

    where β>0,b>0. By direct computations,

    F(t)=L(t)L(0)+β(t+b)=t0ddtL(t)dt+β(t+b)=t0Ωuut|x|sdxdt+β(t+b), (5.19)
    F(t)=L(t)+β=I(u)+β=qJ(u)+(qp1)upp+1quqq+β=(qp1)upp+1quqq+βq[J(u0)t0|x|s2ut(t)22dt]. (5.20)

    Applying the Cauchy-Schwarz inequality, Young inequality, and Holder's inequality to yields

    f(t)=[t0|x|s2u(t)22dt+β(t+b)2][t0|x|s2u(t)22+β][t0Ωuut|x|sdxdt+β(t+b)]20. (5.21)

    Therefore, by recalling (5.19) and (5.20), and noticing the nonnegativity of f(t), we arrive at

    F(t)F(t)(1+θ)[F(t)]2=F(t)F(t)+(1+θ)[f(t)[2F(t)2(Tt)L(0)][t0|x|s2ut(t)22dt+β]]F(t)F(t)2(1+θ)F(t)[t0|x|s2ut(t)22dt+β]F(t){F(t)2(1+θ)t0|x|s2ut(t)22dt2(1+θ)β}. (5.22)

    Choosing θ=q22 and recalling (5.17) lets us obtain

    F(t)F(t)q2[F(t)]2F(t)[q(dJ(u0))(q1)β]. (5.23)

    In view of (5.18) and (5.23), we get, for any t(0,Tmax) and β(0,q(dJ(u0))q1], that

    F(t)F(t)(1+q22)[F(t)]20.

    Therefore, Lemma 2.8 guarantees that F(0)>0 and F(0)=βb>0, and then T1:0<T1<2F(0)(q2)F(0), such that F(t),tT1

    Tmaxβb2(q2)βb|x|s2u(0)22,

    where b>max{0,|x|s2u(0)22(q2)β}.

    Case 2: u0W2

    By similar arguments as those in the proof of case 1, when u0W2, by continuity, we see that there exists a t2>0 such that I(u(t2))<0 and |x|s2ut22>0 for all t[0,t2). From the energy equality, we get

    J(u(t2))=J(u0)t20|x|s2uτ(τ)22dτ<J(u0)=d.

    The remainder of the proof is the same as in Case 1.

    In the following, we shall derive a lower bound for the blow-up time T.

    Theorem 5.4. Assume 2<pq<p(2N+1). Let u(t) be a weak solution to Problem (1.1) that blows up at T. Then

    TL1α(0)CL(α1),

    where CL>0,α>1 are two constants that will be determined in the proof.

    Proof. Combining (2.2) and (3.8), we have

    ddtL(t)=Ωuut|x|sdx=I(u)=upp+Ω|u|qln|u|dx((eμ)1CGε1)upp+(eμ)1CGC(ε)u2α2, (5.24)

    where α>1. As (eμ)1CGε1<0, recalling the definition of L(t), we get

    ddtL(t)(eμ)1CGC(ε)C(Ω)|x|s2u(t)2α2CLLα(t), (5.25)

    where CL=(eμ)1CGC(ε)C(Ω). Integrating (5.25) over [0,t), we get

    11α[L1α(t)L1α(0)]CLt.

    Since α>1, letting tT in the above inequality and recalling that limtTL(t)=+, we obtain

    TL1α(0)CL(α1).

    The proof is complete.

    In this section, we present the result of extinction for Problem (1.1).

    Theorem 6.1. (Extinction). Assuming 2NN+2<p<q<2 and

    0<|x|s2u022<2(2Cp)pq+δp[1δ|Ω|1q+δ2(2C(Ω))q+δ2]2q+δp,

    then the weak solution of Problem (1.1) becomes extinct in finite time. Furthermore, we have the following estimates

    |x|s2u(t)222[(12|x|s2u022)1p2ϵ0(1p2)t]22p,0<t<T,

    and

    |x|s2u(t)220,tT.

    The extinction time is

    T=22p(12|x|s2u022)1p2ϵ0,

    where ϵ0,δ,Cp are given in the following.

    Proof. Multiplying (1.1) by u(t) and integrating over Ω, we have

    L(t)+u(t)pp=Ω|u|qln|u|dx. (6.1)

    Meanwhile, by 2NN+2<p<2, we have 0s2<(N+2)2Np, which implies that 1<2NNs+2<p in Lemma 2.6, and we can see that there exists a constant Cp>0 such that

    Ω|u(x)|2|x|sdxCHu22NNs+2Cpu2p. (6.2)

    Combining (2.4), (3.9), (6.1), (6.2), and H¨older's inequality, we deduce that there exists a 0<δ2q such that

    L(t)+(2Cp)p2Lp2(t)1δuq+δq+δ1δ|Ω|1q+δ2uq+δ21δ|Ω|1q+δ2(2C(Ω))q+δ2Lq+δ2(t). (6.3)

    Then by Lemma 2.9, we know that

    ϵ1=(2Cp)p2,ϵ2=1δ|Ω|1q+δ2(2C(Ω))q+δ2,0<l=p2<r=q+δ21.

    We assume that

    0<|x|s2u022<2(ϵ1ϵ2)2q+δp=2(2Cp)pq+δp[1δ|Ω|1q+δ2(2C(Ω))q+δ2]2q+δp,

    and then we can see that

    L(t)[ϵ0(1p2)t+L1p2(0)]22p,0<t<T,

    and

    L(t)0,tT,

    where

    ϵ0=ϵ1ϵ2Lrl(0)=(2Cp)p2[1δ|Ω|1q+δ2(2C(Ω))q+δ2]|x|s2u0q+δp2,

    and

    T=22p(12|x|s2u022)1p2ϵ0.

    The proof is complete.

    Xiulan Wu: Methodology, Writing-original draft, Writing-review & editing; Yanxin Zhao and Xiaoxin Yang: Methodology, Writing-original draft.

    The authors declare they have not used Artificial Intelligence (AI) tools in the creation of this article.

    The authors express their heartfelt thanks to the editors and referees who have provided some important suggestions. This work is supported by the Natural Science Foundation of Jilin Province, Free Exploration Basic Research (No. YDZJ202201ZYTS584).

    The authors declare there is no conflict of interest.



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