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|−sut−Δpu=|u|q−2uln|u|,x∈Ω,t>0;u(x,t)=0,x∈∂Ω,t>0;u(x,0)=u0(x),x∈Ω, | (1.1) |
where Ω⊂RN(N≥NΩ) is a bounded domain with smooth boundary ∂Ω, Δpu=div(|∇u|p−2∇u) with u0∈W1,p0(Ω), x=(x1,x2,…,xN)∈RN with |x|=√x21+x22+⋯+x2N, and the parameters satisfy
0≤s≤2,max{2NN+2,1}<p≤q<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−∇⋅(D∇u)=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 ∇⋅(D∇u) 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|−2ut−△pu=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|−sut−△u=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|p−2△u)=|u|q−2ulog|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|p−2∇u) and logarithmic nonlinearity |u|q−2uln|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(r≥1) the norm in Lr(Ω) and by (⋅,⋅) the L2(Ω) inner product. When p>1,p≠2, we use W1,p0(Ω) to denote the Sobolev space such that both u and ∇u belong to Lp(Ω) for any u∈W1,p0(Ω), denote by W−1,p′(Ω) its dual space, and by ⟨⋅,⋅⟩ the duality pairing between them. We will equip W1,p0(Ω) with the norm ‖u‖W1,p0=‖∇u‖p, 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 0∈Ω and 0∉Ω.
If 0∈Ω, then |x|−s develops a singularity. This necessitates the use of the Hardy-Sobolev inequality, which is valid for NΩ≥3, in the proofs of our main results.
On the other hand, if 0∉Ω, 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,if0∈Ω,1,if0∉Ω. |
First, for Problem (1.1), we introduce the potential energy functional
J(u)=1p‖∇u‖pp−1q∫Ω|u|qln|u|dx+1q2‖u‖qq, | (2.1) |
and the Nehari functional
I(u)=‖∇u‖pp−∫Ω|u|qln|u|dx. | (2.2) |
By a direct computation,
J(u)=1qI(u)+(1p−1q)‖∇u‖pp+1q2‖u‖qq. | (2.3) |
By I(u) and J(u), we define the potential well:
W1={u∈X0:J(u)<d},W2={u∈X0:J(u)=d},W=W1∪W2, |
W+1={u∈W1,I(u)>0},W+2={u∈W2,I(u)>0},W+=W+1∪W+2, |
W−1={u∈W1,I(u)<0},W−2={u∈W2,I(u)<0},W−=W−1∪W−2, |
and the Nehari manifold
N={u∈X0,I(u)=0}. |
The depth of the potential well is defined as
d=infu∈NJ(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,foralls≥1, |
|splns|≤(ep)−1,forall0<s<1. |
Lemma 2.2. [15,16] Assume that q<NpN−p, i.e., q<∞ for N≤p and r≤q<NpN−p for N>p and r≥1. Then for any u∈W1,p0(Ω), it holds that
‖u‖q≤CG‖∇u‖θp‖u‖1−θr, |
where θ∈(0,1) is determined by θ=(1r−1q)(1N−1p+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)<{NpN−p,ifN>p,+∞,ifN≤p. |
Then by the Sobolev inequality, we have W1,p0(Ω)↪Lq+a(Ω) for p>1 and ∀a≥0.
Lemma 2.3. Let u(t)∈X0 and p,q satisfy max{2NN+2,1}<p≤q<p(1+2N). We have the following statements:
(i) If 0<‖u‖p≤r, then I(u)≥0;
(ii) If I(u)<0, then ‖u‖p>r;
(iii) If I(u)=0, then ‖u‖p=0 or ‖u‖p≥r,
where
r=(1Bq+α∗)1q+α−p. |
Proof. (ⅰ) A direct computation yields
ln|u(x)|<|u(x)|αα,a.e.x∈Ω,∀α>0. | (2.4) |
Then, by the definition of I(u), we have
I(u)=‖∇u‖pp−∫Ω|u|qln|u|dx=‖∇u‖pp−‖u‖q+αq+α≥(1−Bq+α∗‖∇u‖q+α−pp)‖∇u‖pp, | (2.5) |
where B∗ is the imbedding constant for W1,p0(Ω)↪Lq+α(Ω). If 0<‖∇u‖p≤r, this implies that ‖∇u‖q+α−pp≤1Bq+α∗. Therefore, we gain I(u)≥0 by (2.5).
(ⅱ) From (2.5) and I(u)<0, we can see that
(1−Bq+α∗‖∇u‖q+α−pp)‖∇u‖pp<0, |
which means that
‖∇u‖p>(1Bq+α∗)1q+α−p=r. |
(ⅲ) If I(u)=0, then from (2.5) we attain
‖∇u‖p≥(1Bq+α∗)1q+α−por‖∇u‖p=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 u∈X0, then
(i) limλ→0+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(λ∗u)=0.
Lemma 2.5. [15,18] (Logarithmic Sobolev Inequality). Let q>1, μ>0, and u∈W1,q0(RN)∖{0}. Then we have
q∫RN|u(x)|qln(|u(x)|‖u‖Lq(RN))dx+Nqln(qμeNϑq)∫RN|u(x)|qdx≤μ∫RN|∇u(x)|qdx, |
where
ϑq=qN(q−1e)q−1π−q2[Γ(N2+1)Γ(Nq−1q+1)]qN. |
Remark 2.2. If u∈W1,q0(Ω)∖{0}, then by defining u(x)=0 for x∈RN∖Ω, we derive
q∫Ω|u(x)|qln(|u(x)|‖u‖Lq(Ω))dx+Nqln(qμeNϑq)∫Ω|u(x)|qdx≤μ∫Ω|∇u(x)|qdx, | (2.6) |
for any real number μ>0.
Lemma 2.6. [14,19] (Hardy-Sobolev inequality). Let RN=Rk×RN−k,2≤k≤N and x=(y,z)∈RN=Rk×RN−k. For given n,β satisfying 1<p<N, 0≤β≤p, and β<k, let m(β,N,p)=p(N−β)(N−p). Then there exists a positive constant CH depending on β,N,p, and k such that for any u∈W1,p0(RN), it holds that
∫RN|u(x)|m|y|−βdx≤CH(∫RN|∇u|pdx)N−βN−p. |
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=2NN−s+2>2, and then Lemma 2.6 becomes
∫Ω|u(x)|2|x|−sdx≤CH(∫Ω|∇u|2NN−s+2dx)N−s+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)ds≤1ωfσ(0)f(t),∀t≥0. |
Then we have
(i) f(t)≤f(0)e1−ωt, for all t≥0, whenever σ=0.
(i) f(t)≤f(0)(1+σ1+ωσt)1σ, for all t≥0, 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))2≥0, |
where θ>0. If Ψ(0)>0 and Ψ′(0)>0, then Ψ(t)→∞ as
t→t∗≤t∗=Ψ(0)θΨ′(0). |
Lemma 2.9. [24] Suppose that 0<l<r≤1 and ϵ1,ϵ2≥0 are positive constants. If nonnegative and absolutely continuous function h(t) satisfies
h′(t)+ϵ1hl(t)≤ϵ2hr(t),t≥0, |
h(0)>0,ϵ2hr−l(0)<ϵ1, |
then we have
h(t)≤[−ϵ0(1−l)t+h1−l(0)]11−l,0<t<T0, |
and
h(t)≡0,t≥T0, |
where ϵ0=ϵ1−ϵ2hr−l(0) and T0=h1−l(0)ϵ0(1−l).
Definition 2.1. (Weak Solution). A function u:=u(x,t)∈L∞(0,T;X0) with |x|−s2ut∈L2(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|p−2∇u,∇v⟩=⟨|u|q−2uln|u|,v⟩,a.e.t∈(0,T), |
for any v∈W1,p0(Ω). Moreover,
∫t0‖|x|−s2uτ‖22dτ+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
limt→T−max∫t0‖|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 u0∈X0, and p,q satisfy max{2NN+2,1}<p≤q<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|−s2ut∈L2(0,T;L2(Ω)) satisfying u(0)=u0. Moreover, u(x,t) satisfies the energy equality
∫t0‖|x|−s2ut‖22dt+J(u)=J(u0),0≤t≤T. |
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},∀n∈N+. |
We denote the solutions corresponding to ρn of Problem (1.1) as un,
{ρn(x)unt−Δpun=|un|q−2unln|un|,x∈Ω,t>0,un(x,t)=0,x∈∂Ω,t>0,un(x,0)=un0,x∈Ω. | (3.1) |
We noticed that un0∈C∞0(Ω), and then un0→u0(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)=k∑j=1aknj(t)ωj(x)fork=1,2,⋯,j=1,2,⋯,k. |
We solve the problem
⟨ρn(x)uknt,ωj⟩+⟨|∇ukn|p−2∇ukn,∇ωj⟩+=⟨|ukn|q−2uknln|ukn|,ωj⟩, | (3.2) |
and
ukn(x,0)=k∑j=1bknjωj(x)=ukn0→un0(x)inW1,p0(Ω) | (3.3) |
as k→+∞,n→+∞. Hence {aknj}kj=1 is determined by the following Cauchy problem:
{k∑j=1(∫Ωρn(x)ωj(x)ωjdx)[aknj(t)]t=Gknj(t),aknj(0)=bknj, |
where
Gknj(t)=∫Ω|k∑j=1aknj(t)ωj(x)|q−2k∑j=1aknj(t)ωj(x)ln|k∑j=1aknj(t)ωj(x)|ωjdx−∫Ω|k∑j=1aknj(t)∇ωj(x)|p−2k∑j=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, ukn∈C1([0,T],W1,p0(Ω)).
Step 2: Priori estimates
We discuss the following two cases:
Case 1: max{2NN+2,1}<p≤q and 2≤q<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|p−2∇ukn,∇ukn⟩=⟨|ukn|q−2uknln|ukn|,ukn⟩. | (3.4) |
Integrating over (0,t) on both sides of (3.4), we get,
12‖(ρn(x))12ukn(t)‖22+∫t0‖∇ukn(s)‖ppds=∫t0∫Ω|ukn(s)|qln|ukn(s)|dxds+12‖(ρn(x))12ukn(0)‖22. |
Set
Skn(t)=12‖(ρn(x))12ukn(t)‖22+∫t0‖∇ukn(s)‖ppds. | (3.5) |
Combining the above equalities, and we have
Skn(t)≤Skn(0)+∫t0∫Ω|ukn(s)|qln|ukn(s)|dxds. | (3.6) |
From Lemma 2.1, we get
∫Ω|ukn(t)|qln|ukn(t)|dx=∫Ω1={x∈Ω;|ukn(x)|≥1}|ukn(t)|qln|ukn(t)|dx+∫Ω2={x∈Ω;|ukn(x)|<1}|ukn(t)|qln|ukn(t)|dx≤(eμ)−1∫Ω1={x∈Ω;|ukn(x)|≥1}|ukn(t)|q+μdx≤(eμ)−1‖ukn(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)−1‖ukn(t)‖q+μq+μ≤(eμ)−1CG‖∇ukn(t)‖θ(q+μ)p‖ukn(t)‖(1−θ)(q+μ)2≤(eμ)−1CGε‖∇ukn(t)‖pp+(eμ)−1CGC(ε)‖ukn(t)‖p(1−θ)(q+μ)p−θ(q+μ)2, | (3.8) |
where ε∈(0,1), and θ=(12−1q+μ)(1N−1p+12)−1=(q+μ−2)Np(q+μ)(2p−2N+Np). We note that since 0<μ<p(1+2N)−q, θ(q+μ)<p holds. Let
α=p(1−θ)(q+μ)2[p−θ(q+μ)]=p(N+q+μ)−N(q+μ)p(N+2)−N(q+μ), |
then α>1 since max{1,2NN+2}<p≤q, 2≤q<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)|2dx≤C(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)−1CGε‖∇ukn(t)‖ppds+∫t0(eμ)−1CGC(ε)‖ukn(t)‖2α2ds≤Skn(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)−1CGε∫t0(Skn(t))αds. |
Therefore,
Skn(t)≤C1+C2∫t0(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+∫t0‖∇ukn(s)‖ppds≤C3,∀n,k∈N+, | (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),0≤t≤T. | (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+(1p−CGεeμq)‖∇ukn(t)‖pp+1q2‖ukn(t)‖qq−C4≤J(ukn0(t))≤C, | (3.14) |
where C4=2CGC(ε)C(Ω)eμ(C3)α, for all n,k∈N+.
Case 2: max{1,2NN+2}<p≤q<2
Combining ln|u(x)|<|u(x)|aa a.e. x∈Ω,∀a>0 and (3.5), and taking a=2−q, we obtain
Skn(t)≤Skn(0)+12−q∫t0‖ukn(s)‖22ds. |
Together with (3.9), it can become
Skn(t)≤Skn(0)+2(2−q)C(Ω)∫t0Skn(s)ds. |
Then by means of Gronwall's inequality, we have
Skn(t)≤C5, |
and
12‖(ρn(x))12ukn(t)‖22+∫t0‖∇ukn(s)‖ppds≤C5, | (3.15) |
where C5=Skn(0)e2T(2−q)C(Ω).
From (2.1), (3.12), (3.13), and (3.15), we have
∫t0‖(ρn(x))12uknt(s)‖22ds+1p‖∇ukn(t)‖pp+1q2‖ukn(t)‖qq≤C+1q∫Ω|ukn(t)|qln|ukn(t)|dx≤C+2q(2−q)C(Ω)‖(ρn(x))12ukn(t)‖22≤C+2C5q(2−q)C(Ω), | (3.16) |
for all k,n∈N+.
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
ukn→uinC(0,T;L2(Ω)), | (3.21) |
as k,n→+∞. Thus, ukn(x,0)→u(x,0) in L2(Ω). Combining (3.3) with un0→u0(x)inW1,p0(Ω), we observe that u(x,0)=u0 in W1,p0(Ω).
From (3.21), we have ukn→u a.e.(x,t)∈Ω×(0,T). This implies
|ukn|q−2uknln|ukn|→|u|q−2uln|u|a.e.(x,t)∈Ω×(0,T). |
It follows from (3.14) and the Hölder inequality that
‖|∇ukn(t)|p−2∇ukn(t)‖W−1,p,(Ω)=supφ∈W1,p0(Ω)∖{0}∫Ω|∇ukn(t)|p−2∇ukn(t)⋅φdx‖φ‖W1,p0(Ω)≤(∫Ω||∇ukn(t)|p−2∇ukn(t)|pp−1dx)p−1p(∫Ω|φ|pdx)1p‖φ‖W1,p0(Ω)≤‖∇ukn(t)‖p−1p<C. |
That means
‖|∇ukn(t)|p−2∇ukn(t)‖L∞(0,T;W−1,p,(Ω))≤C,foranypositiveintegernandk. | (3.22) |
On the other hand, there is
∫Ω||ukn|q−2uknln|ukn||pp−1dx=∫Ω1={x∈Ω;|ukn(x)|≥1}||ukn|q−2uknln|ukn||pp−1dx+∫Ω1={x∈Ω;|ukn(x)|<1}||ukn|q−2uknln|ukn||pp−1dx. |
From Lemma 2.1 and Lemma 2.2, we have
∫Ω||ukn(t)|q−2ukn(t)ln|ukn(t)||pp−1dx=∫Ω1||ukn(t)|q−2ukn(t)ln|ukn(t)||pp−1dx+∫Ω2||ukn(t)|q−2ukn(t)ln|ukn(t)||pp−1dx≤∫Ω1||ukn(t)|−μln|ukn(t)|⋅|ukn(t)|q−1+μ|pp−1dx+∫Ω2||ukn(t)|q−1ln|ukn(t)||pp−1dx≤(eμ)−pp−1‖ukn(t)‖pp−1(q−1+μ)pp−1(q−1+μ)+[e(q−1)]−pp−1|Ω|≤(eμ)−pp−1B1‖∇ukn(t)‖pp−1(q−1+μ)p+[e(q−1)]−pp−1|Ω|<C, |
where B1 is the best constant of the Sobolev embedding W1,p0(Ω)↪Lpp−1(q−1+μ)(Ω). Here we choose 0<μ≤p(1+p−1N−p)−q, q<p(1+p−1N−p), and we know that
‖|ukn(t)|q−2ukn(t)ln|ukn(t)|‖L∞(0,T;Lpp−1(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
ukn→uweaklystarinL∞(0,T;W1,p0(Ω)) | (3.24) |
(ρn(x))12uknt→|x|−s2utweaklyinL2(0,T;L2(Ω)) | (3.25) |
|∇ukn|p−2∇ukn→χweaklystarinL∞(0,T;W−1,p′(Ω)) | (3.26) |
|ukn|q−2uknln|ukn|→|u|q−2uln|u|weaklystarinL∞(0,T;Lpp−1(Ω)). | (3.27) |
Next, by the method of Browder and Minty in the theory of monotone operators, we obtain χ=|∇u|p−2∇u.
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|p−2∇u,∇ω⟩=⟨|u|q−2uln|u|,ω⟩, | (3.28) |
for all ω∈W1,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|p−2∇u1,∇v⟩=⟨|u1|q−2u1ln|u1|,v⟩, | (3.29) |
and
⟨|x|−su2t,v⟩+⟨|∇u2|p−2∇u2,∇v⟩=⟨|u2|q−2u2ln|u2|,v⟩. | (3.30) |
Let w=u1−u2 and w(0)=0, then by subtracting (3.29) and (3.30), we can derive
∫Ω|x|−swtvdx+∫Ω(|∇u1|p−2∇u1−|∇u2|p−2∇u2)∇vdx=∫Ω(|u1|q−2u1ln|u1|−|u2|q−2u2ln|u2|)vdx. |
Let v=w, and we recall the following elementary vector inequalities that are used frequently: for all a,b∈RN, we have 0≤(p−1)|a−b|2(|a|+|b|)2−p≤(|a|p−2a−|b|p−2b)⋅(a−b), if 1<p<2. So, we obtain
12ddt‖|x|−s2w‖22≤∫Ω|u1|q−2u1ln|u1|−|u2|q−2u2ln|u2|ww2dx≤∫Ωf(u1)−f(u2)ww2dx. |
Integrating it on [0,t], we obtain
‖|x|−s2w‖22≤2∫t0∫Ωf(u1)−f(u2)ww2dxdt, | (3.31) |
where F(s)=|s|q−2sln|s|. Combining with (3.9), we get
‖w‖22≤2MT∫t0‖w‖22dt. |
By the locally Lipschitz continuity of F:RN→R, 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),0≤t≤T. | (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 u0∈W+, max{2NN+2,1}<p≤q<p(1+2N), and then Problem (1.1) admits a global solution u∈L∞(0,∞;X0) with |x|−s2ut∈L2(0,∞;L2(Ω)) and u(t)∈W+ for 0≤t≤∞.
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 u0∈W+1
From (3.32), we know that
∫t0‖|x|−s2ut(s)‖22ds+J(u(t))=J(u0)<d,0≤t≤Tmax, | (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+1forall0≤t≤Tmax. | (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∗))≥infu∈NJ(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))+(1p−1q)‖∇u(t)‖pp+1q2‖u(t)‖qq<d, | (4.5) |
and
∫t0‖|x|−s2ut(s)‖22ds+(1p−1q)‖∇u(t)‖pp+1q2‖u(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),0≤t≤+∞. |
Case 2: p=q
Similar to Case 1, we can derive
∫t0‖|x|−s2ut(s)‖22ds+1p2‖u(t)‖pp<d. |
By Lemma 2.5, we have
∫Ω|u(t)|qln|u(t)|dx≤[ln‖u(t)‖Lq(Ω)−nq2ln(qμeNϑq)]∫Ω|u(t)|qdx+μq∫Ω|∇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))+2∫Ω|u(t)|pln|u(t)|dx−‖∇u(t)‖pp≤2I(u(t))+2[ln‖u(t)‖Lp(Ω)−np2ln(pμeNϑp)]∫Ω|u(t)|pdx+(2μp−1)‖∇u(t)‖pp≤2pJ(u(t))+2[1p2+ln‖u(t)‖Lp(Ω)−np2ln(pμeNϑp)]‖u(t)‖pp≤Cd. | (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 u0∈W+2
First, we choose a sequence {θm}∞m=1⊂(0,1) such that limm→∞θm=1. Then we consider the following problem
{|x|−sut−Δpu=|u|q−2uln|u|,(x,t)∈Ω×R+,u=0,(x,t)∈∂Ω×R+,u(x,0)=u0m(x),x∈Ω, | (4.8) |
where u0m=θmu0. First of all, we claim that u0m∈W+1, and then J(u0m)<d and I(u0m)>0. In fact, from u0∈W+1, limm→∞θm=1 and I(u0)>0, and we can see
I(u0m)=θpm‖∇u0‖pp−θqmln|θm|‖u0‖qq−θqm∫Ω|u0|qln|u0|dx≥θpm(‖∇u0‖pp−θq−pm∫Ω|u0|qln|u0|dx)≥θpmI(u0)>0. |
On the other hand, by direct calculations, we obtain
dd(θm)J(θmu0)=1θm(θpm‖∇u0‖pp−θqmln|θm|‖u0‖qq−θqm∫Ω|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 u0m→u0 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 u0∈W+1, then there exist positive constants c2 such that
‖∇u(t)‖2p≤‖∇u0‖2p(p−11+c2(p−2)t)1p−2,t≥0. |
Especially, if p=2, then there exist positive constants c4 such that
‖∇u(t)‖2≤‖∇u0‖2e12(1−c4t),t≥0. |
Proof. We are now in a position to prove the algebraic decay results. Thanks to u0∈W+1, we get u(t)∈W+1. From (2.3), we have
(1p−1q)‖∇u‖pp+1q2‖u‖qq≤J(u(t))≤J(u0)<d. | (4.9) |
By (4.9), through a direct calculation, we arrive at
λ0{(1p−1q)‖∇u‖pp+1q2‖u‖qq}≥J(λ∗u(t))≥d, |
where λ0=max{(λ∗)p,(λ∗)q}. Combining with (4.9), we get
λ0≥max{(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(λ∗u)=(λ∗)p‖∇u‖pp−(λ∗)q∫Ω|u|qln|u|dx−(λ∗)qln(λ∗)‖u‖qq=(λ∗)qI(u)−((λ∗)q−(λ∗)p)‖∇u(t)‖pp−(λ∗)qln(λ∗)‖u‖qq. | (4.11) |
Namely,
I(u(t))=‖u‖qqlnλ∗+[1−(λ∗)p−q]‖∇u‖pp≥c1‖∇u(t)‖pp, | (4.12) |
where c1=1−(dJ(u0))1−qp, p<q.
According to Lemma 2.6, and (2.2), we obtain
∫TtI(u)ds=∫Tt(‖∇u‖pp−∫Ω|u|qln|u|dx)ds=−12∫Ttddt‖|x|−s2u‖22ds=12‖|x|−s2u(t)‖22−12‖|x|−s2u(T)‖22≤12‖|x|−s2u(t)‖22≤12CH‖∇u(t)‖2p. | (4.13) |
By (4.12) and (4.13), we get
∫Tt‖∇u(t)‖ppds≤CHB22c1‖∇u(t)‖2p=1c2‖∇u(t)‖2p. | (4.14) |
Let T→+∞ in (4.14), and by the virtue of Lemma 2.7, it follows that
‖∇u(t)‖2p≤‖∇u0‖2p(p−11+c2(p−2)t)1p−2,t≥0. |
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 u0∈W− and p,q satisfy 1<p≤q<2. Then u(t) blows up in infinite time.
Proof. We divide the proof into 2 steps.
Step 1: u0∈W−1
We claim that u(t)∈W−1 for all t∈[0,Tmax) provided that u0∈W−1. Let u(t) be the weak solution of Problem (1.1) with u0∈W−1, which means that u0≠0 and J(u0)<d,I(u0)<0.
If J(u0)<0. By the energy equality, we arrive at
1qI(u)+(1p−1q)‖∇u‖pp+1q2‖u‖qq=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)∈W−1.
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)∈W−1 for all t∈[0,Tmax).
From Lemma 2.4(ⅳ), as I(u(t))<0, there is a λ∗<1 such that I(λ∗u)=0. Then
d≤J(λ∗u)=1qI(λ∗u)+(λ∗)p(1p−1q)‖∇u‖pp+(λ∗)qq2‖u‖qq<(1p−1q)‖∇u‖pp+1q2‖u‖qq. | (5.3) |
Then, by taking the derivative of L(t), we obtain
ddtL(t)=∫Ωu⋅ut|x|sdx=−I(u)=−qJ(u)+(qp−1)‖∇u‖pp+1q‖u‖qq≥q(d−J(u(t)))=q(d−J(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|−s2u0‖22+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
limt→T−max‖|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. x∈Ω,∀δ>0 and (2.2), and taking δ=2−q, we obtain
I(u(t))≥‖∇u(t)‖pp−12−q‖u(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{2Ls2−q}, | (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)‖pp−12−q‖u(t)‖22≥(12L−sln(12‖|x|−s2u(t)‖22)−12−q)‖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{t−t1}ln(L(t1)),t∈[t1,Tmax], |
which implies that
‖|x|−s2u(t)‖22<‖|x|−s2u(t1)‖2exp{t−t1}2,t∈[t1,Tmax]. |
That contradicts with (5.6). Therefore, Tmax=+∞ and u(t) blows up in infinite time.
Step 2: u0∈W−2
First of all, u0∈W−2 means that I(u(t))<0,J(u(t))=d, ∀t∈[0,Tmax]. We claim that u(t)∈W−2 for all t∈[0,Tmax) provided that u0∈W−2. 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|u⋅ut|x|s|dx=−I(u(t))>0,t∈[t1,Tmax], we know that ut≠0 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 u0∈W− and p,q satisfy 2<p≤q<p(2N+1). Then u(t) blows up in finite time. Moreover,
T∗≤‖|x|−s2u(0)‖22(p−2)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)‖22≥0, |
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)=∫Ωu⋅ut|x|sdx=−I(u)=−qJ(u)+(qp−1)‖∇u‖pp+1q‖u‖qq≥−qJ(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))2≥q2L′(t)K(t), | (5.13) |
which then implies
[K(t)L−q2(t)]′=L−q2−1(t)(L(t)K′(t)−q2L′(t)K(t))≥0. | (5.14) |
Therefore,
0<k=K(0)L−q2(0)≤K(t)L−q2(t)≤1qL′(t)L−q2(t)=2(2−q)p[L2−q2(t)]′. | (5.15) |
Integrating (5.15) over [0,t] for any t∈(0,T∗) and noticing that q>2, one has
kt≤2(2−q)q[L2−q2(t)−L2−q2(0)], |
or equivalently
0≤L2−q2(t)≤L2−q2(0)−(q−2)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
T∗≤2L(0)(q−2)qK(0)=‖|x|−s2u(0)‖22(q−2)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 u0∈W−. More precisely, we have the following theorem.
Theorem 5.3. Assume that 2<p≤q<p(2N+1), u0∈W−. Then the weak solution u(t) to Problem (1.1) blows up in finite time. Furthermore, if u0∈W−1, then Tmax can be estimated from above as follows:
Tmax≤βb2(q−2)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: u0∈W−1
We claim that u(t)∈W−1 for all t∈[0,Tmax) provided that u0∈W−1. 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(λ∗u)=0. Then
d≤J(λ∗u)=1qI(λ∗u)+(λ∗)p(1p−1q)‖∇u‖pp+(λ∗)qq2‖u‖qq<(1p−1q)‖∇u‖pp+1q2‖u‖qq. | (5.17) |
We show that Tmax<+∞. For any T∈[0,Tmax), define the positive function
F(t)=∫t0L(t)dt+(T−t)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)=∫t0∫Ωu⋅ut|x|sdxdt+β(t+b), | (5.19) |
F″(t)=L′(t)+β=−I(u)+β=−qJ(u)+(qp−1)‖∇u‖pp+1q‖u‖qq+β=(qp−1)‖∇u‖pp+1q‖u‖qq+β−q[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]−[∫t0∫Ωu⋅ut|x|sdxdt+β(t+b)]2≥0. | (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(T−t)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)‖22dt−2(1+θ)β}. | (5.22) |
Choosing θ=q−22 and recalling (5.17) lets us obtain
F(t)F″(t)−q2[F′(t)]2≥F(t)[q(d−J(u0))−(q−1)β]. | (5.23) |
In view of (5.18) and (5.23), we get, for any t∈(0,Tmax) and β∈(0,q(d−J(u0))q−1], that
F(t)F″(t)−(1+q−22)[F′(t)]2≥0. |
Therefore, Lemma 2.8 guarantees that F(0)>0 and F′(0)=βb>0, and then ∃T1:0<T1<2F(0)(q−2)F′(0), such that F(t)→∞,t→T1
Tmax≤βb2(q−2)βb−‖|x|−s2u(0)‖22, |
where b>max{0,‖|x|−s2u(0)‖22(q−2)β}.
Case 2: u0∈W−2
By similar arguments as those in the proof of case 1, when u0∈W−2, by continuity, we see that there exists a t2>0 such that I(u(t2))<0 and ‖|x|−s2ut‖22>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<p≤q<p(2N+1). Let u(t) be a weak solution to Problem (1.1) that blows up at T∗. Then
T∗≥L1−α(0)CL(α−1), |
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)=∫Ωu⋅ut|x|sdx=−I(u)=−‖∇u‖pp+∫Ω|u|qln|u|dx≤((eμ)−1CGε−1)‖∇u‖pp+(eμ)−1CGC(ε)‖u‖2α2, | (5.24) |
where α>1. As (eμ)−1CGε−1<0, recalling the definition of L(t), we get
ddtL(t)≤(eμ)−1CGC(ε)C(Ω)‖|x|−s2u(t)‖2α2≤CLLα(t), | (5.25) |
where CL=(eμ)−1CGC(ε)C(Ω). Integrating (5.25) over [0,t), we get
11−α[L1−α(t)−L1−α(0)]≤CLt. |
Since α>1, letting t→T∗ in the above inequality and recalling that limt→T∗L(t)=+∞, we obtain
T∗≥L1−α(0)CL(α−1). |
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|−s2u0‖22<2(2Cp)pq+δ−p[1δ|Ω|1−q+δ2(2C(Ω))q+δ2]−2q+δ−p, |
then the weak solution of Problem (1.1) becomes extinct in finite time. Furthermore, we have the following estimates
‖|x|−s2u(t)‖22≤2[(12‖|x|−s2u0‖22)1−p2−ϵ0(1−p2)t]22−p,0<t<T∗, |
and
‖|x|−s2u(t)‖22≡0,t≥T∗. |
The extinction time is
T∗=22−p⋅(12‖|x|−s2u0‖22)1−p2ϵ0, |
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 0≤s≤2<(N+2)−2Np, which implies that 1<2NN−s+2<p in Lemma 2.6, and we can see that there exists a constant Cp>0 such that
∫Ω|u(x)|2|x|sdx≤CH‖∇u‖22NN−s+2≤Cp‖∇u‖2p. | (6.2) |
Combining (2.4), (3.9), (6.1), (6.2), and H¨older's inequality, we deduce that there exists a 0<δ≤2−q such that
L′(t)+(2Cp)p2Lp2(t)≤1δ‖u‖q+δq+δ≤1δ|Ω|1−q+δ2‖u‖q+δ2≤1δ|Ω|1−q+δ2(2C(Ω))q+δ2Lq+δ2(t). | (6.3) |
Then by Lemma 2.9, we know that
ϵ1=(2Cp)p2,ϵ2=1δ|Ω|1−q+δ2(2C(Ω))q+δ2,0<l=p2<r=q+δ2≤1. |
We assume that
0<‖|x|−s2u0‖22<2(ϵ1ϵ2)2q+δ−p=2(2Cp)pq+δ−p[1δ|Ω|1−q+δ2(2C(Ω))q+δ2]−2q+δ−p, |
and then we can see that
L(t)≤[−ϵ0(1−p2)t+L1−p2(0)]22−p,0<t<T∗, |
and
L(t)≡0,t≥T∗, |
where
ϵ0=ϵ1−ϵ2Lr−l(0)=(2Cp)p2−[1δ|Ω|1−q+δ2(2C(Ω))q+δ2]‖|x|−s2u0‖q+δ−p2, |
and
T∗=22−p⋅(12‖|x|−s2u0‖22)1−p2ϵ0. |
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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