According to the difference of the initial energy, we consider three cases about the global existence and blow-up of the solutions for a class of coupled parabolic systems with logarithmic nonlinearity. The three cases are the low initial energy, critical initial energy and high initial energy, respectively. For the low initial energy and critical initial energy J(u0,v0)≤d, we prove the existence of global solutions with I(u0,v0)≥0 and blow up of solutions at finite time T<+∞ with I(u0,v0)<0, where I is Nehari functional. On the other hand, we give sufficient conditions for global existence and blow up of solutions in the case of high initial energy J(u0,v0)>d.
Citation: Qigang Deng, Fugeng Zeng, Dongxiu Wang. Global existence and blow up of solutions for a class of coupled parabolic systems with logarithmic nonlinearity[J]. Mathematical Biosciences and Engineering, 2022, 19(8): 8580-8600. doi: 10.3934/mbe.2022398
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According to the difference of the initial energy, we consider three cases about the global existence and blow-up of the solutions for a class of coupled parabolic systems with logarithmic nonlinearity. The three cases are the low initial energy, critical initial energy and high initial energy, respectively. For the low initial energy and critical initial energy J(u0,v0)≤d, we prove the existence of global solutions with I(u0,v0)≥0 and blow up of solutions at finite time T<+∞ with I(u0,v0)<0, where I is Nehari functional. On the other hand, we give sufficient conditions for global existence and blow up of solutions in the case of high initial energy J(u0,v0)>d.
In this paper, we consider the following initial-boundary value problem for a class of coupled parabolic systems with logarithmic nonlinearity.
{ut−Δu=|v|p|u|p−2ulog(|uv|),x∈Ω, t>0,vt−Δv=|u|p|v|p−2vlog(|uv|),x∈Ω, t>0,u(x,0)=u0(x),x∈Ω,v(x,0)=v0(x),x∈Ω,u(x,t)=v(x,t)=0,(x,t)∈∂Ω×(0,T], | (1.1) |
where (u0,v0)∈H10(Ω)×H10(Ω), T∈(0,+∞), Ω⊂Rn(n≥2) is a bounded domain with smooth boundary ∂Ω and p satisfies the following assumptions:
2<p<2∗:={∞,if n=2,2nn−2,if n≥3. | (1.2) |
Among the fields of mathematical physics, biosciences and engineering, problem (1.1) is one of the most important reaction-diffusion coupled systems with logarithmic nonlinearity. It can be used not only to predict the time evolution of various population density distributions, but also to describe the thermal propagation of a two-component combustible mixture [1,2,3]. In recent years, this kind of systematic research has attracted many mathematicians and has made remarkable progress [4,5,6,7,8]. In order to overcome the special difficulties brought by nonlinear terms, many new ideas and tools have been developed, which greatly enrich the theory of partial differential equations [9,10,11,12,13,14].
In the past years, many authors made efforts to the investigation of the existence and blow up of solutions for such kinds of systems. Galaktionov et al. [15,16] investigated the following semilinear reaction-diffusion system
{ut−Δu=vp,vt−Δv=uq. | (1.3) |
They proved the local and global existence of solutions for the initial boundary value problem of (1.3). Subsequently, Escobedo and Herrero [17] considered the initial boundary value problem of (1.3) for a bounded open domain on Rn with smooth boundary. They obtained global solution under the condition 0<pq≤1, meanwhile global solution and blow up in finite time depending on sufficient small or large initial value and pq>1. For more studies on problem (1.3) we refer the interested reader to [18,19,20] and references therein.
Recently, Xu et al. [21] considered the following nonlinear reaction-diffusion systems
{ut−Δu=(|u|2p+|v|p+1|u|p−1)u,x∈Ω,t>0,vt−Δv=(|v|2p+|u|p+1|v|p−1)v,x∈Ω,t>0,u(x,0)=u0(x),x∈Ω,v(x,0)=v0(x),x∈Ω,u(x,t)=v(x,t)=0,(x,t)∈∂Ω×(0,T]. | (1.4) |
When initial energy J(u0,v0)≤d, by virtue of Galerkin method [22] and concave function method [23], global existence and finite time blow-up of the solutions for the problem (1.4) were obtained. When initial energy J(u0,v0)>d, they discussed global existence, finite time blow-up of solutions and tried to find out the corresponding initial data with arbitrarily high initial energy. What's more, by using comparison principle and the ideas in [24,25], they described the structures of the initial data and gave some sufficient conditions of the initial data which ensured the finite time blow up and global existence of the solutions, respectively.
Inspired by the above works, we aim to use the Galerkin method, logarithmic inequalities [26], and concave function method to prove the global existence, decay, finite time blow-up of solutions for problem (1.1) with initial energy J(u0,v0)≤d. When high initial energy J(u0,v0)>d, by constructing two sets Φα and Ψα defined as (5.1) and (5.2), we prove that the weak solution will blow up in finite or infinite time if the initial value belongs to Ψα, while the weak solution will exist globally and tends to zero as time t→+∞ when the initial value belongs to Φα.
The organization of the remaining part of this paper is as follows. In Section 2, we introduce some preliminaries and lemmas of this paper. In Sections 3–5, we will give our main results and the corresponding proofs.
Throughout this paper, we denote by ‖u‖γ the norm of Lγ(Ω) for 1≤γ≤+∞ and by ‖u‖H10(Ω) the norm of H10(Ω). For u∈Lγ(Ω),
‖u‖γ={(∫Ω|u(x)|γdx)1γ,if1≤γ<+∞,ess supx∈Ω|u(x)|,ifγ=+∞, |
and for u∈H10(Ω),
‖u‖2H10(Ω)=‖u‖22+‖∇u‖22. |
By virtue of Poincaré inequality, we know that ‖u‖2H10(Ω) and ‖∇u‖22 are equivalent norms to each other, i.e., there exist C1 and C2 such that
C1‖∇u‖22≤‖u‖2H10(Ω)≤C2‖∇u‖22, |
which is denoted by ‖u‖2H10(Ω)≃‖∇u‖22. In addition, we denoted by (⋅,⋅) the inner product in L2(Ω) and c is an arbitrary positive number which may be different from line to line.
For (u,v)∈H10(Ω)×H10(Ω), we define the Nehari functional I and energy functional J as follows:
I(u,v)=∫Ω|∇u|2dx+∫Ω|∇v|2dx−2∫Ω|uv|p|log(|uv|)dx≃‖u‖2H10(Ω)+‖v‖2H10(Ω)−2∫Ω|uv|plog(|uv|)dx, | (2.1) |
J(u,v)=12∫Ω|∇u|2dx+12∫Ω|∇v|2dx+1p2∫Ω|uv|pdx−1p∫Ω|uv|plog(|uv|)dx≃12‖u‖2H10(Ω)+12‖v‖2H10(Ω)+1p2‖uv‖pp−1p∫Ω|uv|plog(|uv|)dx. | (2.2) |
From (2.1) and (2.2), we have
J(u,v)≃12pI(u,v)+p−12p(‖u‖2H10(Ω)+‖v‖2H10(Ω))+1p2‖uv‖pp. | (2.3) |
Let
N:={(u,v)∈H10(Ω)×H10(Ω)∖ {(0,0)}| I(u,v)=0} |
be the Nehari manifold. Furthermore, the potential well W and its corresponding set V are defined respectively by
W={(u,v)∈H10(Ω)×H10(Ω)∣I(u,v)>0,J(u,v)<d}∪{(0,0)}, |
V={(u,v)∈H10(Ω)×H10(Ω)∣I(u,v)<0,J(u,v)<d}, |
where
d:=inf(u,v)∈H10(Ω)×H10(Ω)∖{(0,0)}sups1,s2>0J(s1u,s2v)=inf(u,v)∈NJ(u,v) | (2.4) |
is the depth of the potential well W.
To consider the weak solution with high energy level, we need to introduce some new notions.
Jα={(u,v)∈H10(Ω)×H10(Ω)|J(u,v)<α}, |
Nα=N∩Jα={(u,v)∈N|p−12p(‖u‖2H10(Ω)+‖v‖2H10(Ω))+1p2‖uv‖pp<α}, |
and
λα=inf{12(‖u‖22+‖v‖22)|(u,v)∈Nα}, Λα=sup{12(‖u‖22+‖v‖22)|(u,v)∈Nα}forallα>d, |
Clearly, λα is nonincreasing, and Λα is nondecreasing with respect to α, respectively.
Now, we give the definitions of the weak solution, maximal existence time and finite time blow up of the problem (1.1) as follows.
Definition 1. (Weak solution) We say that (u,v) = (u(x,t),v(x,t))∈L∞([0,T),H10(Ω)×H10(Ω)) with (ut,vt)∈L2([0,T),L2(Ω)×L2(Ω)) is a weak solution of problem (1.1) on Ω×[0,T), if it satisfies the initial condition u(x,0)=u0(x),v(x,0)=v0(x) in H10(Ω),
(ut,w1)+(∇u,∇w1)=(|v|p|u|p−2ulog(|uv|),w1) |
and
(vt,w2)+(∇v,∇w2)=(|u|p|v|p−2vlog(|uv|),w2) |
for all w1,w2∈H10(Ω) and t∈(0,T). Moreover, for all t∈(0,T), we have
∫t0‖uτ‖22+‖vτ‖22dτ+J(u,v)≤J(u0,v0). | (2.5) |
Remark 1. For the global weak solution (u(t),v(t))=(u(x,t),v(x,t)) of problem (1.1), we define the ω-limit set of (u0,v0) by
ω(u0,v0):=⋂t≥0¯{u(s),v(s):s≥t}. |
Definition 2. (Maximal existence time) Let (u,v)=(u(x,t),v(x,t)) be a weak solution of problem (1.1). We define the maximal existence time of (u,v) as follows
(i) If (u,v) exists for all t∈[0,+∞), then T=+∞.
(ii) If there exists a t0∈(0,+∞) such that (u,v) exists for 0≤t<t0, but it does not exist at t=t0, then T=t0.
Definition 3. (Finite time blow-up) Let (u(t),v(t))=(u(x,t),v(x,t)) be a weak solution of problem (1.1). We say (u(t),v(t)) blows up in finite time if the maximal existence time T is finite and
limt→T‖u(t)‖22+‖v(t)‖22=+∞. |
Lemma 1. Let (u,v)∈H10(Ω)×H10(Ω)∖{(0,0)}, then the following hold
(i) limλ→0J(λu,λv)=0, limλ→+∞J(λu,λv)=−∞.
(ii) There exists a unique λ∗>0 such that ddλJ(λu,λv)|λ=λ∗=0.
(iii) J(λu,λv) is increasing on (0,λ∗), decreasing on (λ∗,+∞), and attains the maximum at λ=λ∗.
(iv) I(λu,λv)>0 for 0<λ<λ∗, I(λu,λv)<0 for λ∗<λ<+∞, and I(λ∗u,λ∗v)=0.
Proof. (i) By definition of J(u,v) and λ>0, we have
J(λu,λv)=λ22‖u‖2H10(Ω)+λ22‖v‖2H10(Ω)+λ2pp2‖uv‖pp−λ2pplogλ2‖uv‖pp−λ2pp∫Ω|uv|plog(|uv|)dx. |
Thus limλ→0J(λu,λv)=0, limλ→+∞J(λu,λv)=−∞.
(ii) Differentiating J(λu,λv) with respect to λ, we get
ddλJ(λu,λv)=λ‖u‖2H10(Ω)+λ‖v‖2H10(Ω)−2λ2p−1logλ2‖uv‖pp−2λ2p−1∫Ω|uv|plog(|uv|)dx=λ(‖u‖2H10(Ω)+‖v‖2H10(Ω)−2λ2p−2logλ2‖uv‖pp−2λ2p−2∫Ω|uv|plog(|uv|)dx). |
Setting g(λ)=‖u‖2H10(Ω)+‖v‖2H10(Ω)−2λ2p−2logλ2‖uv‖pp−2λ2p−2∫Ω|uv|plog(|uv|)dx, we have limλ→0g(λ)=‖u‖2H10(Ω)+‖v‖2H10(Ω)>0, limλ→+∞g(λ)=−∞, and
g′(λ)=−2(2p−2)λ2p−3logλ2‖uv‖pp−4λ2p−3‖uv‖pp−2(2p−2)λ2p−3∫Ω|uv|plog|uv|dx<0. |
Thus there exists a unique λ∗>0 such that g(λ∗)=0, i.e., ddλJ(λu,λv)|λ=λ∗=0.
(iii) It is easy to find that J(λu,λv) is strictly increasing on (0,λ∗], strictly decreasing on (λ∗,+∞) and taking the maximum at λ=λ∗.
(iv) Since
I(λu,λv)=‖λu‖2H10(Ω)+‖λv‖2H10(Ω)−2∫Ω|λuλv|plog(|λuλv|)dx=λddλJ(λu,λv), |
then the conclusion follows immediately.
Lemma 2. Assume (1.2) holds, let (u,v)∈H10(Ω)×H10(Ω) satisfy I(u,v)<0, then
I(u,v)<2p(J(u,v)−d). | (2.6) |
Proof. According to I(u,v)<0 and Lemma 1, we have (u,v)≠(0,0) and there exists a λ∗∈(0,1) such that I(λ∗u,λ∗v)=0, i.e., J(λ∗u,λ∗v)≥d. For λ>0, set
h(λ)=2pJ(λu,λv)−I(λu,λv)=(p−1)λ2(‖u‖2H10(Ω)+‖v‖2H10(Ω))+2pλ2p‖uv‖Pp, |
then
h′(λ)=2(p−1)λ(‖u‖2H10(Ω)+‖v‖2H10(Ω))+4λ2p−1‖uv‖Pp>0. |
Hence h(λ) is strictly increasing for λ>0. Together with λ∗∈(0,1), it follows that h(1)>h(λ∗). i.e.,
2pJ(u,v)−I(u,v)>2pJ(λ∗u,λ∗v)−I(λ∗u,λ∗v)=2pJ(λ∗u,λ∗v)≥2pd. |
Then (2.6) follows immediately.
Lemma 3. Assume (1.2) holds, let (u0,v0)∈H10(Ω)×H10(Ω) and (u(t),v(t))=(u(x,t),v(x,t)) be a weak solution of problem (1.1). If J(u0,v0)<d and I(u0,v0)<0, then (u(t),v(t))∈V for all 0≤t≤T, where T is the maximal existence time of (u(t),v(t)).
Proof. We will show that (u(t),v(t))∈V for 0≤t≤T. Arguing by contradiction, suppose that t0∈[0,T] be the smallest time for which (u(t0),v(t0))∉V, then by the continuity of the (u(t),v(t)), we get (u(t0),v(t0))∈∂V. Hence, it follows that
I(u(t0),v(t0))=0 | (2.7) |
or
J(u(t0),v(t0))=d. | (2.8) |
If (2.7) is true, then (u(t0),v(t0))∈N, J(u(t0),v(t0))>d, which contradicts with (2.5). While if (2.8) is true, it also contradicts with (2.5). Consequently, we have (u(t),v(t))∈V for all 0≤t≤T.
Lemma 4. Let (u, v) be a weak solution of problem (1.1). Then for all t∈[0,T),
ddt(‖u‖22+‖v‖22)=−2I(u,v). |
Proof. The proof of Lemma 4 directly follows by choosing w1=u,w2=v in Definition 1.
In this section, we prove global existence and finite time blow up of solutions for problem (1.1) with the initial energy J(u0,v0)<d.
Theorem 1. Assume (u0,v0)∈H10(Ω)×H10(Ω) and (1.2) hold. If J(u0,v0)<d and I(u0,v0)≥0, then the problem (1.1) has a global solution (u(t),v(t))∈L∞((0,∞);H10(Ω)×H10(Ω)) with (ut(t),vt(t))∈L2((0,∞);L2(Ω)×L2(Ω)) and (u(t),v(t))∈W for 0≤t<∞. Furthermore, if I(u0,v0)>0, then there exists a c>0 such that ‖u‖22+‖v‖22≤(‖u0‖22+‖v0‖22)e−2ct.
Proof. Since we know J(u0,v0)<d and I(u0,v0)≥0, then it follows that
(i) If 0<J(u0,v0)<d and I(u0,v0)≥0, then we have I(u0,v0)>0. In fact, if I(u0,v0)=0, then by the definition of d in (2.4), we have J(u0,v0)≥d, which is a contradiction.
(ii) If J(u0,v0)=0 and I(u0,v0)≥0, then we obtain (u0,v0)=(0,0). In fact, if (u0,v0)≠(0,0), then by the (2.3), we have p−12p(‖u0‖2H10(Ω)+‖v0‖2H10(Ω))+1p2‖u0v0‖pp<0, which is also a contradiction.
(iii) If J(u0,v0)<0 and I(u0,v0)≥0, then it is contradictive with (2.3).
From the discussions above, we consider the case 0<J(u0,v0)<d and I(u0,v0)>0. It is widely know that there is a basis {ωj(x)}∞j=1 of H10(Ω) such that ωj is an eigenfunction of the Laplacian operator corresponding to the eigenvalue λj and
{−Δωj=λjωj,x∈Ω,ωj=0,x∈∂Ω. |
Hence, we choose {ωj(x)}∞j=1 as the Galerkin basis for −Δ in H10(Ω). Then we construct the Galerkin approximate solution (um(x,t),vm(x,t)) of the problem (1.1),
{um(x,t)=∑mj=1gjm(t)ωj(x), m=1,2,⋯,vm(x,t)=∑mj=1hjm(t)ωj(x), m=1,2,⋯, |
which satisfy, for j=1,2,⋯,m,
(umt,ωj)+(∇um,∇ωj)=(|vm|p|um|p−2umlog(|umvm|),ωj) | (3.1) |
and
(vmt,ωj)+(∇vm,∇ωj)=(|um|p|vm|p−2vmlog(|umvm|),ωj), | (3.2) |
with initial condition um(x,0)=u0m, vm(x,0)=v0m, where u0m and v0m are chosen in span {ω1,ω2,⋅⋅⋅,ωm} so that
u0m=m∑j=1gjm(0)ωj(x)→u0inH10(Ω),asm→+∞ | (3.3) |
and
v0m=m∑j=1hjm(0)ωj(x)→v0inH10(Ω),asm→+∞. | (3.4) |
According to the standard ordinary differential equation theory, the system (3.1)–(3.4) admit a solution
(gjm(t),hjm(t))∈C1[0,T0)×C1[0,T0), |
where T0 is the minimum of the existence time of gjm(t) and hjm(t) for each m. Thus (um(x,t),vm(x,t))∈C1([0,T0);H10(Ω)×H10(Ω)).
Next, multiplying (3.1) and (3.2) by g′jm(t) and h′jm(t), respectively, summing for j from 1 to m, integrating with respect to t from 0 to t and adding these two equations, we get
∫t0‖umτ‖22+‖vmτ‖22dτ+J(um,vm)=J(u0m,v0m), 0≤t<T0. | (3.5) |
From J(u0,v0)<d and (3.3)–(3.4), we see that J(u0m,v0m)<d for sufficiently large m. Then we get from (3.5) that
∫t0‖umτ‖22+‖vmτ‖22dτ+J(um,vm)=J(u0m,v0m)<d, 0≤t<T0, | (3.6) |
for sufficiently large m.
By (3.3) and (3.4) and (u0,v0)∈W, we know that (u0m,v0m)∈W for large enough m. Next, we prove (um(x,t),v(x,t))∈W for large enough m and 0≤t<T0. If it is false, then there exists t0∈(0,T0) such that (um(x,t0),vm(x,t0))∈∂W, then I(um(t0),vm(t0))=0 and (um(t0),vm(t0))≠(0,0), or J(um(t0),vm(t0))=d.
By (3.6), J(um(t0),vm(t0))=d is not true. On the other hand, if I(um(t0),vm(t0))=0 and (um(t0),vm(t0))≠(0,0), then by the definition of d, we have J(um(t0),vm(t0))≥d, which is also contradiction with (3.6). So (um(x,t),vm(x,t))∈W for large enough m and 0≤t<T0.
From the fact (um(x,t),vm(x,t))∈W for large enough m, (3.6) and
J(um(t),vm(t))=p−12p(‖um(t)‖2H10(Ω)+‖vm(t)‖2H10(Ω))+1p2‖um(t)vm(t)‖pp+12pI(um(t),vm(t)), |
we obtain
∫t0‖umτ‖22+‖vmτ‖22dτ+p−12p(‖um(t)‖2H10(Ω)+‖vm(t)‖2H10(Ω))+1p2‖um(t)vm(t)‖pp<d, 0≤t<T0, | (3.7) |
for sufficiently large m, which gives
‖um(t)‖2H10(Ω)<2pp−1d, | (3.8) |
‖vm(t)‖2H10(Ω)<2pp−1d, | (3.9) |
∫t0‖umτ‖22+‖vmτ‖22dτ<d. | (3.10) |
By (3.10), we know that T0=+∞. Then by (3.8)–(3.10), there exist u,v with theirs subsequences of {um}+∞j=1 and {vm}+∞j=1, such that, as m→+∞,
um→uweaklystarinL∞(0,+∞;H10(Ω)), | (3.11) |
vm→vweaklystarinL∞(0,+∞;H10(Ω)), | (3.12) |
umt→utweaklyinL2(0,+∞;L2(Ω)), | (3.13) |
νmt⟶vtweaklyinL2(0,+∞;L2(Ω)). | (3.14) |
Then it follows Aubin-lions compactness theorem [27] that
um→ustronglyinC([0,+∞);L2(Ω)), |
vm→vstronglyinC([0,+∞);L2(Ω)). |
Clearly, this implies that
um→ua.e.inΩ×[0,+∞), |
vm→va.e.inΩ×[0,+∞). |
Furthermore, we get
|vm|p|um|p−2umlog(|umvm|)→|v|p|u|p−2ulog(|uv|)a.e.inΩ×[0,+∞), | (3.15) |
|um|p|vm|p−2vmlog(|umvm|)→|u|p|v|p−2vlog(|uv|)a.e.inΩ×[0,+∞). | (3.16) |
On the other hand,
∫Ω||vm|p|um|p−2umlog(|umvm|)|pp−1dx=∫Ω(|vm|p|um|p−1log(|umvm|))pp−1dx=∫{x∈Ω:|um(x)vm(x)|≤1}(|vm|p|um|p−1log(|umvm|))pp−1dx+∫{x∈Ω:|um(x)vm(x)|>1}(|vm|p|um|p−1log(|umvm|))pp−1dx≤∫Ω(e(p−1)−1|vm|)pp−1dx+∫Ω|um|(p−1+r)pp−1|vm|(p+r)pp−1dx≤e(p−1)−pp−1‖vm‖pp−1pp−1+‖um‖e2e‖vm‖σ2σ≤c‖vm‖pp−1H10(Ω)+c‖um‖eH10(Ω)‖vm‖σH10(Ω)≤c, | (3.17) |
where σ=(p+r)pp−1, e=(p−1+r)pp−1, and since |xq−1logx|≤(e(q−1))−1 for 0<x<1 while x−μlogx≤(eμ)−1 for x≥1,μ>0. Choosing a positive real number r so that 0<2e<2∗ and 0<2σ<2∗, and similar to the proof (3.17), we have
∫Ω||um|p|vm|p−2vmlog(|umvm|)|pp−1dx≤c. | (3.18) |
Hence, from (3.15)–(3.18) and Lion's Lemma(see [27], Lemma 1.3, p.12), we have
|vm|p|um|p−2umlog(|umvm|)→|v|p|u|p−2ulog(|uv|)weaklystarinL∞(0,+∞;Lpp−1(Ω)), | (3.19) |
|um|p|vm|p−2vmlog(|umvm|)→|u|p|v|p−2vlog(|uv|)weaklystarinL∞(0,+∞;Lpp−1(Ω)). | (3.20) |
In view of (3.11)–(3.14) and (3.19), (3.20), for j fixed, we can pass to the limit in (3.1) and (3.2) to get
(ut,ωj)+(∇u,∇ωj)=(|v|p|u|p−2ulog(|uv|),ωj) |
and
(vt,ωj)+(∇v,∇ωj)=(|u|p|v|p−2vlog(|uv|),ωj) |
for a.e., t∈(0,+∞). Since {ωj(x)}∞j=1 is the basis in H10(Ω), we have
(ut,w1)+(∇u,∇w1)=(|v|p|u|p−2ulog(|uv|),w1) | (3.21) |
and
(vt,w2)+(∇v,∇w2)=(|u|p|v|p−2vlog(|uv|),w2) | (3.22) |
for any w1,w2∈H10(Ω) and a.e., t∈(0,+∞).
Fixing any t∈(0,+∞) and integrating (3.21) and (3.22) from 0 to t, we get
(u,w1)+∫t0(∇u,∇w1)dτ=∫t0(|v|p|u|p−2ulog(|uv|),w1)dτ+(u(0),w1), ∀w1∈H10(Ω), | (3.23) |
and
(v,w2)+∫t0(∇v,∇w2)dτ=∫t0(|u|p|v|p−2vlog(|uv|),w2)dτ+(v(0),w2), ∀w2∈H10(Ω). | (3.24) |
Similarly, integrating (3.1) and (3.2) from 0 to t, and passing to the limit, we get
(u,w1)+∫t0(∇um,∇w1)dτ=∫t0(|vm|p|um|p−2umlog(|umvm|),w1)dτ+(u0,w1), ∀w1∈H10(Ω), | (3.25) |
and
(v,w2)+∫t0(∇vm,∇w2)dτ=∫t0(|um|p|vm|p−2vmlog(|umvm|),w2)dτ+(v0,w2), ∀w2∈H10(Ω). | (3.26) |
From (3.23)–(3.26), we get u(0)=u0 and v(0)=v0.
According to (3.3), (3.4), (3.7), (3.11)–(3.14), (3.19), (3.20) and since the norm is weakly lower semicontinuous, we know that the energy inequality (2.5) holds. Then from Definition 2, (u(t),v(t)) is a global weak solution and (u(t),v(t))∈W.
Next, we will prove the algebraic decay of the global solution u(x,t). Combining (2.3), (2.5) and (u(t),v(t))∈W, we have
p−12p(‖u‖2H10(Ω)+‖v‖2H10(Ω))+1p2‖uv‖pp≤J(u(t))≤J(u0). | (3.27) |
As I(u,v)<0, then there exists a λ∗∈(0,1) such that I(λ∗u,λ∗v)=0. Furthermore, we get
λp∗(p−12p(‖u‖2H10(Ω)+‖v‖2H10(Ω)))+1p2‖uv‖pp≥J(λ∗u(t))≥d. | (3.28) |
It follows from (3.27) and (3.28) that
λ∗≥(dJ(u0,v0))1p. | (3.29) |
Due to the I(λ∗u,λ∗v)=0, we have
I(λ∗u,λ∗v)=λ2∗(‖u‖2H10(Ω)+‖v‖2H10(Ω))−2λ2p∗∫Ω|uv|plog|uv|dx−2λ2p∗logλ2∗‖uv‖pp=(λ2∗−2λ2p∗)(‖u‖2H10(Ω)+‖v‖2H10(Ω))+2λ2p∗I(u,v)−2λ2p∗logλ2∗‖uv‖pp=0, |
i.e.,
I(u,v)≥(1−12λ2(p−1)∗)(‖u‖2H10(Ω)+‖v‖2H10(Ω)). | (3.30) |
Combining (3.29) with (3.30), we have
I(u,v)≥(1−12(dJ(u0,v0))2(1p−1))(‖u‖2H10(Ω)+‖v‖2H10(Ω)). |
By the emdedding H10(Ω)↪L2(Ω), we have
I(u,v)≥c(‖u‖2L2(Ω)+‖v‖2L2(Ω)). | (3.31) |
On the other hand, by Lemma 4, we know
12ddt(‖u‖22+‖v‖22)+I(u,v)=0, 0≤t<∞. |
Combining this equality with (3.31), we get
12ddt(‖u‖22+‖v‖22)+c(‖u‖2L2(Ω)+‖v‖2L2(Ω))≤0, 0≤t<∞. |
By Grönwall's inequality, we have
‖u‖22+‖v‖22≤(‖u0‖22+‖v0‖22)e−2ct, 0≤t<∞. |
The proof of Theorem 1 is complete.
Theorem 2. Assume (u0,v0)∈H10(Ω)×H10(Ω) and (1.2) hold. If J(u0,v0)<d and I(u0,v0)<0, then the weak solution (u(x,t),v(x,t)) of the problem (1.1) blows up in finite time, i.e., there exists a T>0 such that
limt→T∫t0‖u‖22+‖v‖22dτ=+∞. |
Proof. Step 1: Blow-up in finite time
By contradiction, we suppose that (u(t),v(t)) is global weak solution of problem (1.1), then Tmax=+∞. Let
G(t)=∫t0‖u‖22+‖v‖22dτ, |
then
G′(t)=‖u‖22+‖v‖22 |
and
G″(t)=2((u,ut)+(v,vt))=−2(‖∇u‖22+‖∇v‖22)+4∫Ω|u|p|v|plog(|uv|)dx=−2I(u,v). | (3.32) |
From (3.32) and energy inequality (2.5), it follows that
G″(t)≃2(p−1)(‖u‖2H10(Ω)+‖v‖2H10(Ω))+4p‖uv‖pp−4pJ(u,v)≥2(p−1)(‖u‖2H10(Ω)+‖v‖2H10(Ω))+4p‖uv‖pp−4pJ(u0,v0)+4p∫t0‖uτ‖22+‖vτ‖22dτ≥4p∫t0‖uτ‖22+‖vτ‖22dτ+2(p−1)cG′(t)−4pJ(u0,v0), | (3.33) |
where the constant c is from the Poincaré inequality ‖u‖22≤c‖∇u‖22.
Note that
(∫t0(uτ,u)+(vτ,v)dτ)2=(12∫t0ddτ(‖u‖22+‖v‖22)dτ)2=(12(‖u‖22+‖v‖22−‖u0‖22−‖v0‖22))2=14[(‖u‖22+‖v‖22)2−2(‖u0‖22+‖v0‖22)(‖u‖22+‖v‖22)+(‖u0‖22+‖v0‖22)2]=14[(G′(t))2−2G′(t)(‖u0‖22+‖v0‖22)+(‖u0‖22+‖v0‖22)2], |
then
G′(t)=4(∫t0(uτ,u)+(vτ,v)dτ)2+2G′(t)(‖u0‖22+‖v0‖22)−(‖u0‖22+‖v0‖22)2. | (3.34) |
Hence by (3.33) and (3.34) we know that
G(t)G″(t)−p(G′(t))2⩾4p∫t0‖uτ‖22+‖vτ‖22dτ∫t0‖u‖22+‖v‖22dτ−4p(∫t0(uτ,u)+(vτ,v)dτ)2+2(p−1)cG(t)G′(t)−4pG(t)J(u0,v0)−2pG′(t)(‖u0‖22+‖v0‖22)+p(‖u0‖22+‖v0‖22)2. |
By Schwartz inequality, we have
∫t0‖uτ‖22+‖vτ‖22dτ∫t0‖u‖22+‖v‖22dτ−(∫t0(uτ,u)+(vτ,v)dτ)2≥∫t0‖uτ‖22+‖vτ22dτ∫t0‖u‖22+‖v‖22dτ−(∫t0‖u‖2‖uτ‖2+‖v‖2‖vτ‖2dτ)2≥∫t0‖uτ‖22+‖vτ‖22dτ∫t0‖u‖22+‖v‖22dτ−(∫t0√‖uτ‖22+‖vτ‖22√‖u‖22+‖v‖22dτ)2≥0. |
It implies that
G(t)G′′(t)−p(G′(t))2≥2(p−1)cG′(t)G(t)−4pJ(u0,v0)G(t)−2pG′(t)(‖u0‖22+‖v0‖22). | (3.35) |
From Lemma 3 we have I(u(t),v(t))<0 for 0≤t<+∞. Thus from Lemma 2 one has
−2I(u(t),v(t))>4p(d−J(u(t),v(t))), 0≤t<+∞. | (3.36) |
Combing (3.36) and (2.5) we get
G″(t)=−2I(u,v)>4p(d−J(u,v))≥4p(d−J(u0,v0)):=C1>0, 0≤t<+∞ |
and
G′(t)≥C1t+G′(0)=C1t, 0≤t<+∞, |
G(t)≥12C1t2+G(0)=12C1t2, 0≤t<+∞. |
Hence for sufficiently large t, we have
(p−1)cG(t)>2p(‖u0‖22+‖v0|22)and(p−1)cG′(t)>4pJ(u0,v0). | (3.37) |
Combining (3.35) with (3.37), we obtain
G(t)G″(t)−p(G′(t))2≥((p−1)cG(t)−2p(‖u0‖22+‖v0‖22))G′(t)+((p−1)cG′(t)−4pJ(u0,v0))G(t)>0, |
for sufficiently large t. Note that
(G−(p−1)(t))″=−(p−1)Gp+1(t)(G(t)G″(t)−p(G′(t))2)<0. |
It follows that there exists a finite time T>0 such that limt→TG−(p−1)(t)=0, i.e., limt→T∫t0‖u‖22+‖v‖22dτ=+∞.
Step 2: Upper bound estimation of the blow-up time.
We next give an upper bound estimate of T. Suppose (u(t),v(t)) be a solution of problem (1.1) with initial value (u0,v0) satisfying I(u0,v0)<0 and J(u0,v0)<d. By Step 1, the maximal existence time T<∞. By Lemma 3, we get (u(t),v(t))∈V, ∀t∈[0,T), i.e., I(u(t),v(t))<0,t∈[0,T). For T1∈(0,T), we define the auxiliary functional M:[0,T1]→R which is defined by
M(t):=∫t0‖u‖22+‖v‖22dτ+(T−t)(‖u0‖22+‖v0‖22)+β(t+γ)2, | (3.38) |
with β>0 and γ>0 specified later. Through a direct calculation, we have
M′(t)=‖u(t)‖22+‖v(t)‖22−(‖u0‖22+‖v0‖22)+2β(t+γ)=2∫t0(uτ,u)+(vτ,v)dτ+2β(t+γ) | (3.39) |
and
M″(t)=2((ut,u)+(vt,v))+2β=2β−2I(u,v). |
It follows from Lemma 2 and (2.5) that
M″(t)>4p(d−J(u,v))+2β≥4p(d−J(u0,v0))+4p∫t0‖uτ‖22+‖vτ‖22dτ+2β. | (3.40) |
From (3.39) and Hölder inequality, we have
(M′(t))2=4[∫t0(uτ,u)+(vτ,v)dτ+2β(t+γ)]2≤4[∫t0‖uτ‖2‖u‖2+‖vτ‖2‖v‖2dτ+2β(t+γ)]2. | (3.41) |
According to the inequality
xz+yw≤(x2+y2)12(z2+w2)12, |
by setting x=‖uτ‖2, y=‖vτ‖2, z=‖u‖2, w=‖v‖2 in (3.41), we get
(M′(t))2≤4[∫t0(‖uτ‖22+‖vτ‖22)12(‖u‖22+‖v‖22)12dτ+2β(t+γ)]2. |
By the Hölder inequality, we get
(M′(t))2≤4[∫t0(‖uτ‖22+‖vτ‖22)12(‖u‖22+‖v‖22)12dτ+2β(t+γ)]2≤4[(∫t0‖uτ‖22+‖vτ‖22dτ)12(∫t0‖u‖22+‖v‖22dτ)12+2β(t+γ)]2≤4[∫t0‖uτ‖22+‖vτ‖22dτ+β][∫t0‖u‖22+‖v‖22dτdτ+β(t+γ)2]≤4M(t)[∫t0‖uτ‖22+‖vτ‖22dτ+β]. | (3.42) |
From (3.38), (3.40) and (3.42), we have
M(t)M″(t)−p(M′(t))2≥[4p(d−J(u0,v0))−2(2p−1)β]M(t). |
Restricting β to satisfy
0<β≤2p(d−J(u0,v0))2p−1, | (3.43) |
we have
M(t)M″(t)−p(M′(t))2≥0,t∈[0,T1]. |
Define y(t):=M1−p(t) for t∈[0,T1], then by M(t)>0,M′(t)>0 we get
y′(t)=−(p−1)M−p(t)M′(t)<0, |
y″(t)=−(p−1)M−p−1(t)(M″(t)M(t)−p(M′(t))2)<0 |
for all t∈[0,T1]. It follows from y″(t)<0 that
y(T1)−y(0)=y′(ξ)T1<y′(0)T1, ξ∈(0,T1), | (3.44) |
where
y(0)=M1−p(0)>0, y(T1)=M1−p(T1)>0, |
y′(0)=−(p−1)M−p(0)M′(0)=2(1−p)βγM−p(0)<0. |
Combining (3.44) and the above inequalities, we can deduce
T1≤y(T1)y′(0)−y(0)y′(0)<−y(0)y′(0)=M(0)2(p−1)βγ. |
Then by the definition of M(t) and above inequality we have
T1≤T(‖u0‖22+‖v0‖22)+βγ22(p−1)βγ=γ2(p−1)+‖u0‖22+‖v0‖222(p−1)βγT. |
Hence, letting T1→T, we get
T≤γ2(p−1)+‖u0‖22+‖v0‖222(p−1)βγT. | (3.45) |
For any β satisfying (3.43), let γ be large enough such that
‖u0‖22+‖v0‖222(p−1)β<γ<+∞, | (3.46) |
then (3.45) lead to
T≤γ2(p−1)(1−‖u0‖22+‖v0‖222(p−1)βγ)−1=βγ22(p−1)βγ−(‖u0‖22+‖v0‖22). |
Let
ρ(β,γ)=βγ22(p−1)βγ−(‖u0‖22+‖v0‖22), |
then
T≤min(β,γ)∈Φρ(β,γ), |
where Φ={(β,γ):β,γsatisfy(3.43)and(3.46)respectively}.
Since
ρ′β(β,γ)=−γ2(‖u0‖22+‖v0‖22)(2(p−1)βγ−(‖u0‖22+‖v0‖22))2<0, |
i.e., ρ(β,γ) is decreasing with respect to β. Then we have
min(β,γ)∈Φρ(β,γ)=ρ(2p(d−J(u0,v0))2p−1,γ):=ρ1(γ), |
where
ρ1(γ)=2p(d−J(u0,v0))γ24p(p−1)γ(d−J(u0,v0))−(2p−1)(‖u0‖22+‖v0‖22) |
and
(2p−1)(‖u0‖22+‖v0‖22)4p(p−1)(d−J(u0,v0))<γ<+∞. |
It is easy to get that ρ1(γ) achieves its minimum at
γ1=(2p−1)(‖u0‖22+‖v0‖22)2p(p−1)(d−J(u0,v0)), |
and
ρ1(γ1)=(2p−1)(‖u0‖22+‖v0‖22)2p(p−1)2(d−J(u0,v0)). |
Thus, we have
T≤(2p−1)(‖u0‖22+‖v0‖22)2p(p−1)2(d−J(u0,v0)). |
The proof of Theorem 2 is complete.
In this section, we prove global existence and blow up at finite time of solutions for problem (1.1) with the initial energy J(u0,v0)=d.
Theorem 3. Assume (u0,v0)∈H10(Ω)×H10(Ω) and (1.2) hold. If J(u0,v0)=d and I(u0,v0)≥0, then the problem (1.1) has a global solution (u(t),v(t))∈L∞(0,+∞;H10(Ω)×H10(Ω)) with (ut(t),vt(t))∈L2(0,+∞;L2(Ω)×L2(Ω)) and (u(t),v(t))∈¯W=W∪∂W for 0≤t<∞.
Proof. Since J(u0,v0)=d, then (u0,v0)≠(0,0). Let λm=1−1m, (u0m,v0m)=λm(u0,v0), m=1,2,⋅⋅⋅, and consider the following problem:
{umt−Δum=|vm|p|um|p−2umlog(|umvm|),x∈Ω, t>0,vmt−Δvm=|um|p|vm|p−2vmlog(|umvm|),x∈Ω, t>0,um(x,0)=u0m(x),x∈Ω,vm(x,0)=v0m(x),x∈Ω,um(x,t)=vm(x,t)=0,(x,t)∈∂Ω×(0,T]. | (4.1) |
By I(u0,v0)≥0 and Lemma 1, there exists a unique λ∗≥1 such that I(λ∗u0,λ∗v0)=0. Due to the λm<1<λ∗, we get I(λmu0,λmv0)>0,J(λmu0,λmv0)<J(u0,v0)=d. From Theorem 1, it follows that for each m problem (4.1) admits a global solution (um(t),vm(t))∈L∞(0,+∞;H10(Ω)×H10(Ω)) with (umt(t),vmt(t))∈L2(0,+∞;L2(Ω)×L2(Ω)) with the initial data
um(0)=u0m→u0inH10(Ω)asm→+∞. |
Furthermore, we have (um(t),vm(t))∈W for 0≤t<+∞,
(umt,w1)+(∇um,∇w1)=(|vm|p|um|p−2umlog(|umvm|),w1),∀w1∈H10(Ω), 0≤t<+∞, |
(vmt,w2)+(∇vm,∇w2)=(|um|p|vm|p−2vmlog(|umvm|),w2),∀w2∈H10(Ω), 0≤t<+∞, |
and
∫t0‖umτ‖22+‖vmτ‖22dτ+J(um,vm)≤J(u0m,v0m)<d, 0⩽t<+∞. | (4.2) |
From (4.2) and
J(um(t),vm(t))=p−12p(‖um(t)‖2H10(Ω)+‖vm(t)‖2H10(Ω))+1p2‖vm(t)‖pp+12pI(um(t),vm(t)), |
we obtain
∫t0‖umτ‖22+‖vmτ‖22dτ+p−12p(‖um(t)‖2H10(Ω)+‖vm(t)‖2H10(Ω))+1p2‖um(t)vm(t)‖pp<d,0⩽t<+∞. |
The remainder of the proof is similar to that in the proof of Theorem 1.
The proof of Theorem 3 is complete.
Theorem 4. Assume (u0,v0)∈H10(Ω)×H10(Ω) and (1.2) hold. If J(u0,v0)=d and I(u0,v0)<0, then the weak solution (u(x,t),v(x,t)) of the problem (1.1) blows up in finite time, i.e., there exists a T>0 such that
limt→T∫t0‖u‖22+‖v‖22dτ=+∞. |
Proof. By contradiction, we suppose that (u(t),v(t)) is a global weak solution of problem (1.1), then Tmax=+∞. Let
G(t)=∫t0‖u‖22+‖v‖22dτ. |
Taking into account to (3.35) still holds, combining the fact J(u0,v0)=d, we have
G(t)G″(t)−p(G′(t))2≥((p−1)cG(t)−2p(‖u0‖22+‖v0‖22))G′(t)+((p−1)cG′(t)−4pd)G(t). | (4.3) |
From continuities of J(u,v) and I(u,v) with respect to t, we know that there exists a sufficient small t1∈(0,+∞) such that J(u(t1),v(t1))>0 and I(u,v)<0 for 0<t<t1. By (ut,u)+(vt,v)=−I(u,v), we have (ut,u)+(vt,v)>0 and ‖ut‖22+‖vt‖22>0 for t∈[0,t1]. From (2.5), we have 0<J(u(t1),v(t1))≤d−∫t10(‖ut‖22+‖vt‖22)dt<d. Hence we take t=t1 as the initial time, and obtain (u(t1),v(t1))∈V. From Lemma 3 we have I(u(t),v(t))<0 for t1≤t<+∞. Thus from Lemma 2 one has
−2I(u(t),v(t))>4p(d−J(u(t),v(t))), t1≤t<+∞. | (4.4) |
Combing (4.4) and (2.5) we get
G″(t)=−2I(u,v)>4p(d−J(u,v))≥4p(d−J(u(t1),v(t1))):=C2>0, t1≤t<+∞ |
and
G′(t)≥C2(t−t1)+G′(t1)=C2(t−t1), t1≤t<+∞, |
G(t)≥12C22t2−C2t1t+G(t1), t1≤t<+∞. |
Hence for sufficiently large t, we have
(p−1)cG(t)>2p(‖u0‖22+‖v0|22)and(p−1)cG′(t)>4pd. | (4.5) |
Combining (4.3) with (4.5), we get
G(t)G″(t)−p(G′(t))2≥((P−1)cG(t)−2p(‖u0‖22+‖v0‖22))G′(t)+((p−1)cG′(t)−4pd)G(t)>0, |
for sufficiently large t. Then similar to the proof of Theorem 2, i.e., there exists a finite time T>0 such that limt→T∫t0‖u‖22+‖v‖22dτ=+∞.
The proof of Theorem 4 is complete.
In this section, we investigate the conditions that ensure the global existence or blow up of solution for problem (1.1) with the initial energy J(u0,v0)>d.
Theorem 5. For any α∈(d,+∞), the following conclusions hold.
(i) If (u0,v0)∈Φα, then the solution of the problem (1.1) exists globally and (u(t),v(t))→(0,0), as t→∞;
(ii) If (u0,v0)∈Ψα, then the solution of the problem (1.1) blows up in finite or infinite time, where
Φα=N+∩{(u(t),v(t))∈H10 (Ω)×H10(Ω)|12(‖u‖22+‖v‖22)<λα,d<J(u,v)⩽α}, | (5.1) |
Ψα=N−∩{(u(t),v(t))∈H10 (Ω)×H10(Ω)|12(‖u‖22+‖v‖22)>Λα,d<J(u,v)⩽α}, | (5.2) |
and
λα=inf{12(‖u‖22+‖v‖22)|(u,v)∈Nα},Λα=sup{12(‖u‖22+‖v‖22)|(u,v)∈Nα}forallα>d. |
Proof. (i) Assume that (u0,v0)∈Φα, then by the definition of Φα and the monotonicity property of λα, we have (u0,v0)∈N+, d<J(u0,v0)≤α and
12(‖u0‖22+‖v0‖22)<λα≤λJ(u0,v0). | (5.3) |
We claim that (u(t),v(t))∈N+. By contradiction, there exists a t0∈(0,T) such that (u(t),v(t))∈N+ for t∈[0,t0) and (u(t0),v(t0))∈N. By Lemma 4, we have
12ddt(‖u‖22+‖v‖22)=−I(u,v). | (5.4) |
From the definition of N+ and (5.4), we know that ‖u‖22+‖v‖22 is strictly decreasing on [0,t0). On the other hand, by (2.5), we know that J(u,v) is nonincreasing with respect to t. Therefore, we have
J(u,v)≤J(u0,v0)forallt∈[0,T). |
From (5.3), we get
12(‖u(t0)‖22+‖v(t0)‖22)<12(‖u0‖22+‖v0‖22)<λJ(u0,v0). | (5.5) |
By (u(t0),v(t0))∈N and (5.3), we get (u(t0),v(t0))∈NJ(u0,v0). According to the definition of λJ(u0,v0), we have
λJ(u0,v0)=inf{12(‖u‖22+‖v‖22)|(u,v)∈NJ(u0,v0)}≤12(‖u(t0)‖22+‖v(t0)‖22), |
which contradicts with (5.5) and prove the claim. Hence, we have (u(t),v(t))∈N+ for all t∈[0,T) and (u(t),v(t))∈JJ(u0,v0), i.e., (u(t),v(t))∈JJ(u0,v0)∩N+ for all t∈[0,T). From the definition of Nα, we have (‖u‖2H10(Ω)+‖v‖2H10(Ω))<2pp−2J(u0,v0), ∀t∈[0,T), so T=+∞. It indicates that (u(t),v(t)) is bounded uniformly in H10(Ω)×H10(Ω). Hence, ω-limit set is not an empty set.
Next, for any (ω,φ)∈ω(u0,v0), by the above discussions, we get
J(ω,φ)≤J(u0,v0)and12(‖ω‖22+‖φ‖22)<λJ(u0,v0). |
According to first inequality, it implies that (ω,φ)∈JJ(u0,v0). According to the second inequality and the definition of λJ(u0,v0), we know that (ω,φ)∉NJ(u0,v0). Since NJ(u0,v0)=N∩JJ(u0,v0), we obtain (ω,φ)∉N. Hence, ω(u0,v0)∩N=ϕ. As N include the nontrivial solutions of the problem (1.1), we have ω(u0,v0)=(0,0), i.e., (u(t),v(t))→(0,0), as t→∞.
(ii) If (u0,v0)∈Ψα, by the definition of Ψα, it is clear that (u0,v0)∈N− and d<J(u0,v0)≤α. Combing with the monotonicity of Λα, we get
12(‖u0‖22+‖v0‖22)>Λα≥ΛJ(u0,v0). |
We claim that (u(t),v(t))∈N− for t∈[0,T). By contradiction, if there exists a t1∈(0,T) such that (u(t),v(t))∈N− for t∈[0,t1) and (u(t1),v(t1))∈N. By Lemma 4, we have
12ddt(‖u‖22+‖v‖22)=−I(u,v). |
Then by the definition of N−, we deduce that 12(‖u‖22+‖v‖22) is strictly increasing on [0,t1). It along with (2.5) yields
12(‖u(t1)‖22+‖v(t1)‖22)>12(‖u0‖22+‖v0‖22)>ΛJ(u0,v0), J(u(t1),v(t1))≤J(u0,v0). | (5.6) |
By (u(t1),v(t1))∈N and (5.6), we get (u(t1),v(t1))∈NJ(u0,v0). Hence, it follows from the definition of ΛJ(u0,v0) that
ΛJ(u0,v0)=sup{12(‖u‖22+‖v‖22)|(u,v)∈NJ(u0,v0)}≥12(‖u(t1)‖22+‖v(t1)‖22), |
which is incompatible with (5.6), so we get (u(t),v(t))∈JJ(u0,v0)∩N− for all t∈[0,T).
Next, we assume that (u(t),v(t)) exists globally, i.e., T=+∞. For every (ω,φ)∈ω(u0,v0), by the above discussions, we get
J(ω,φ)≤J(u0,v0)and12(‖ω‖22+‖φ‖22)>ΛJ(u0,v0). |
According to first inequality, this shows (ω,φ)∈JJ(u0,v0). According to the second inequality and the definition of ΛJ(u0,v0), we know that (ω,φ)∉NJ(u0,v0). Since NJ(u0,v0)=N∩JJ(u0,v0), we obtain (ω,φ)∉N. Hence, ω(u0,v0)∩N=ϕ. However, since dist(0,N−)>0, we also have (0,0)∉ω(u0,v0). Thus, ω(u0,v0)=∅, it contraries to the assumption that (u(t),v(t)) is a global solution, then T<∞.
The proof of Theorem 5 is complete.
This research was supported by Guizhou Provincial Science and Technology Projects (No.Qiankehe foundation-ZK[2021]YIBAN317), and by the project of Guizhou Minzu University under (No.GZMU[2019]YB04), and central leading local science and technology development special foundation for Sichuan province, P.R. China under (No.2021ZYD0020).
The authors declare there is no conflict of interest.
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