Citation: Xiaoming Peng, Xiaoxiao Zheng, Yadong Shang. Lower bounds for the blow-up time to a nonlinear viscoelastic wave equation with strong damping[J]. AIMS Mathematics, 2018, 3(4): 514-523. doi: 10.3934/Math.2018.4.514
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In this article, we prove the non-existence of solutions to the following quasilinear elliptic problem which has degenerate coercivity in their principal part by approximation,
{−div(a(x,u,∇u))+|u|q−1u=λ,x∈Ω,u=0,x∈∂Ω, | (1) |
where
a(x,t,ξ)⋅ξ≥c|ξ|p(1+|t|)θ(p−1), | (2) |
|a(x,t,ξ)|≤c0(|ξ|p−1+b(x)), | (3) |
[a(x,t,ξ)−a(x,t,ξ′)]⋅[ξ−ξ′]>0, | (4) |
for almost every
It is well-known that[3,9], problem
{−Δu+|u|q−1u=δ0,x∈Ω,u=0,x∈∂Ω. |
In the famous work [9], Brezis proved that if
{−Δun+|un|q−1un=fn,x∈Ω,un=0,x∈∂Ω, | (5) |
with
limn→∞∫Ω∖Bϱ(0)|fn−f|=0. |
Then
{−Δu+|u|q−1u=f,x∈Ω,u=0,x∈∂Ω. |
This fact shows that
The main goal of this paper is to study the non-existence of solutions to problem (1). More precisely, consider the limit of approximating equation (9)(see Theorem 1.2 below), our main task is to understand which is the limit of solutions to (9) and what equation it satisfies. A point worth emphasizing is that, even if
In order to state the main results of this paper, we need some definitions.
Let
capr(K,Ω)=inf{‖u‖rW1,r0:u∈C∞c(Ω),u≥χK}, |
where
Let
If
Let
limn→+∞∫Ωf+nφdx=∫Ωφdλ+,limn→+∞∫Ωf−nφdx=∫Ωφdλ−, | (6) |
for every function
‖f+n‖L1(Ω)≤C,‖f−n‖L1(Ω)≤C. | (7) |
For all
Tk(s)=max{−k,min{k,s}},Gk(s)=s−Tk(s). |
Firstly we stale the existence result.
Theorem 1.1. Let
{−div(a(x,u,∇u))+|u|q−1u=g,x∈Ω,u=0,x∈∂Ω. | (8) |
if
q<N(1−θ)N−(1+θ(p−1)). |
Moreover,
u∈Mp1(Ω),|∇u|∈Mp2(Ω), |
where
p1=N(p−1)(1−θ)N−p,p2=N(p−1)(1−θ)N−(1+θ(p−1)). |
Remark 1. The previous result gives existence and uniqueness of the entropy solution
Our main results are following:
Theorem 1.2. Let
{−div(a(x,un,∇un))+|un|q−1un=fn+gn,x∈Ω,un=0,x∈∂Ω. | (9) |
Then
σ<pq(q+1+θ(p−1))(p−1), |
if
q>r(p−1)[1+θ(p−1)]r−p, | (10) |
where
limn→+∞∫Ω|un|q−1unφdx=∫Ω|u|q−1uφdx+∫Ωφdλ,∀φ∈C(Ω). | (11) |
Remark 2. The above theorem shows that there is not a solution to problem (1) can be obtained by approximation, if
Remark 3. Boccardo et.al [7] considered the non-existence result to the following problem
{−div(a(x,∇u)(1+u)γ)+u=μ,x∈Ω,u=0,x∈∂Ω, | (12) |
where
The structure of this paper is as follows: Section 2 mainly gives some lemmas which play a important role in the process of proof of the main theorem. The proof of theorem 1.1 and 1.2 are given in Section 3.
In the following,
In order to prove Theorem 1.1 and 1.2, the following basic lemmas and definitions are required.
Lemma 2.1. (see Lemma 2.1 of [22]) Let
0≤ψ+δ≤1,0≤ψ−δ≤1,∫Ω|∇ψ+δ|rdx≤δ,∫Ω|∇ψ−δ|rdx≤δ,0≤∫Ω(1−ψ+δ)dλ+≤δ,0≤∫Ω(1−ψ−δ)dλ−≤δ,0≤∫Ωψ−δdλ+≤δ,0≤∫Ωψ+δ)dλ−≤δ,ψ+δ≡1,x∈K+,ψ+δ≤1,x∈K−, | (13) |
for every
Definition 2.2. Let
∇Tk(u)=vχ{|u|≤k},a.einΩandforeveryk>0. |
Define the gradient of
Definition 2.3. Let
∫Ωa(x,u,∇u)⋅∇Tk(u−φ)dx+∫Ω|u|q−1uTk(u−φ)dx≤∫ΩgTk(u−φ)dx, |
for every
Definition 2.4. Marcinkiewicz space
|{|υ|≥k}|≤Cks, |
for any
If
Ls(Ω)⊂Ms(Ω)⊂Ls−ε(Ω). |
Lemma 2.5. Let
∫Ω|∇Tk(u)|pdx≤Ckρ, |
for some positive constant
|∇u|∈Mpss+ρ(Ω). |
Proof. Let
|{|∇u|>σ}|=|{|∇u|>σ,|u|≤k}|+|{|∇u|>σ,|u|>k}|≤|{|∇Tk(u)|>σ}|+|{|u|>k}|. | (14) |
Moreover,
|{|∇Tk(u)|>σ}|≤1σp∫Ω|∇Tk(u)|pdx≤Ckρσp. | (15) |
Since
|{|u|>k}|≤Cks. | (16) |
Combining (14)-(16), we have
|{|∇u|>σ}|≤Ckρσp+Cks≤Ckpss+ρ. |
Therefore, by Definition 2.4, we get
Lemma 2.6. Let
∫Ω|∇Tk(un)|pdx≤Ckρ, |
for any
Lemma 2.7. Let
∫{k<|u|<k+h}|∇u|pdx≤Ckθ(p−1). |
Proof. For any given
Tk,h(s)=Th(s−Tk(s))={s−ksgn(s),k≤|s|<k+h,h,|s|≥k+h,0,|s|≤k. |
Take
∫{k<|u|<k+h}(a(x,u,∇u)⋅∇u)dx+∫Ω|u|q−1uTk,h(u)dx=∫ΩgTk,h(u)dx. | (17) |
Since
∫{k<|u|<k+h}(a(x,u,∇u)⋅∇u)dx≤∫ΩgTk,h(u)dx, | (18) |
and
∫ΩgTk,h(u)dx≤h∫{|u|>k}|g|dx≤C. | (19) |
According to the assumption (2) and (17)-(19), we get,
∫{k<|u|<k+h}|∇u|pdx≤Ckθ(p−1). |
Proposition 1. Let
∫{|u|<k}|∇u|pdx≤Ckρ | (20) |
for every
|{|u|>k}|≤Ck−p1. |
Proof. For every
‖Tk(u)‖p∗≤C(N,p,θ)‖∇Tk(u)‖p≤Ckρp, |
where
{|u|≥η}={|Tk(u)≥η|}. |
Hence
|{|u|>η}|≤‖Tk(u)‖p∗p∗ηp∗≤C(kρ)p∗pη−p∗. |
Setting
|{|u|>k}|≤Ck−N(p−ρ)N−p. |
This fact shows that
Proposition 2. Assume that
|{|∇u|>h}|≤Ch−p2, |
for every
Proof. For
ψ(k,λ)=|{|∇u|p>λ,|u|>k}|. |
Using the fact that the function
ψ(0,λ)=|{|∇u|p>λ}|≤1λ∫λ0ψ(0,s)ds≤ψ(k,0)+1λ∫λ0ψ(0,s)−ψ(k,s)ds. | (21) |
By Proposition 1,
ψ(k,0)≤Ck−p1, | (22) |
where
∫∞0ψ(0,s)−ψ(k,s)ds=∫{|u|<k}|∇u|pdx≤Ckρ. | (23) |
Combining (21)-(23), we arrive at
ψ(0,λ)≤Ckρλ+Ck−p1. | (24) |
Let
|{|∇u|>h}|≤Ch−N(p−ρ)N−ρ. |
That is
In this section we prove Theorem 1.1 and 1.2 combining the results of Sections 2.
In the proofs of Theorem 1.1 and 1.2,
limδ→0+limm→+∞limn→+∞ω(n,m,δ)=0. |
If the quantity does not depend on one or more of the three parameters
limδ→0+limn→+∞ω(n,δ)=0. |
The proof of Theorem 1.1 will be divided in several steps.
Proof. (1)Uniqueness: Let
Step 1. Assume that
I:=∫Ωa(x,u1,∇u1)⋅∇Tk(u1−Thu2)dx+∫Ωa(x,u2,∇u2)⋅∇Tk(u2−Thu1)dx=−∫Ω|u1|q−1u1Tk(u1−Thu2)dx−∫Ω|u2|q−1u2Tk(u2−Thu1)dx+∫Ωg1Tk(u1−Thu2)dx+∫Ωg2Tk(u2−Thu1)dx. | (25) |
Step 2. Denote
A0={x∈Ω:|u1−u2|<k,|u1|<h,|u2|<h},A1={x∈Ω:|u1−Thu2|<k,|u2|≥h},A2={x∈Ω:|u1−Thu2|<k,|u2|<h,|u1|≥h}. |
For
∇Tk(u1−Thu2)=∇(u1−u2) |
and
∇Tk(u2−Thu1)=∇Tk(u2−u1). |
Thus, for every
∫Ωa(x,u1,∇u1)⋅∇Tk(u1−Thu2)dx+∫Ωa(x,u2,∇u2)⋅∇Tk(u2−Thu1)dx=∫A0[a(x,u1,∇u1)−a(x,u2,∇u2)]⋅∇(u1−u2)dx:=I0. | (26) |
For
∫Ωa(x,u1,∇u1)⋅∇Tk(u1−Thu2)dx=∫A1a(x,u1,∇u1)⋅∇u1dx≥0. | (27) |
For
∫Ωa(x,u1,∇u1)⋅∇Tk(u1−Thu2)dx≥−∫A2a(x,u1,∇u1)⋅∇u2dx. | (28) |
Similarly, denote
A∗1={x∈Ω:|u2−Thu1|<k,|u1|≥h},A∗2={x∈Ω:|u2−Thu1|<k,|u1|<h,|u2|≥h}. |
Then for
∫Ωa(x,u2,∇u2)⋅∇Tk(u2−Thu1)dx=∫A∗1a(x,u2,∇u2)⋅∇u2dx≥0. | (29) |
For
∫Ωa(x,u2,∇u2)⋅∇Tk(u2−Thu1)dx≥−∫A∗2a(x,u2,∇u2)⋅∇u1dx. | (30) |
Summing up (26)-(30) in the form
I1=∫A2a(x,u1,∇u1)⋅∇u2dx+∫A∗2a(x,u2,∇u2)⋅∇u1dx:=I11+I12. |
Now, we estimate
I11≤‖a(x,u1,∇u1)‖Lp′({h≤|u1|≤h+k})‖∇u2‖Lp({h−k≤|u2|≤h})≤c0(‖∇u1‖p−1Lp′({h≤|u1|≤h+k})+‖b(x)‖Lp′({|u1|≥h}))‖∇u2‖Lp({h−k≤|u2|≤h}). |
Therefore, by Lemma 2.7 and Proposition 2,
Hence, we find
∫Ωa(x,u1,∇u1)⋅∇Tk(u1−Thu2)dx+∫Ωa(x,u2,∇u2)⋅∇Tk(u2−Thu1)dx=∫A0[a(x,u1,∇u1)−a(x,u2,∇u2)]⋅∇(u1−u2)dx+ε(h). | (31) |
Step 3. Now estimate the terms on the right hand side of (25). Denote
B0={x∈Ω:|u1|<h,|u2|<h},B1={x∈Ω:|u1|≥h},B2={x∈Ω:|u2|≥h}. |
For
∫Ω|u1|q−1u1Tk(u1−Thu2)dx+∫Ω|u2|q−1u2Tk(u2−Thu1)dx=∫B0(|u1|q−1u1−|u2|q−1u2)Tk(u1−u2)dx≥0, | (32) |
and
∫Ωg1Tk(u1−Thu2)dx+∫Ωg2Tk(u2−Thu1)dx=∫B0(g1−g2)Tk(u1−u2)dx≤0. | (33) |
For
∫Ω|u1|q−1u1Tk(u1−Thu2)dx+∫Ω|u2|q−1u2Tk(u2−Thu1)dx≤k∫B1(|u1|q−1u1+|u2|q−1u2)dx:=J1, |
and
∫Ωg1Tk(u1−Thu2)dx+∫Ωg2Tk(u2−Thu1)dx≤k∫B1(|g1|+|g2|)dx:=J2. |
For
∫Ω|u1|q−1u1Tk(u1−Thu2)dx+∫Ω|u2|q−1u2Tk(u2−Thu1)dx≤k∫B2(|u1|q−1u1+|u2|q−1u2)dx:=J∗1, |
and
∫Ωg1Tk(u1−Thu2)dx+∫Ωg2Tk(u2−Thu1)dx≤k∫B2(|g1|+|g2|)dx:=J∗2. |
According to
J1+J2+J∗1+J∗2→0ash→∞. | (34) |
Step 4. Combining (25) and (31)-(34), we have
∫A0[a(x,u1,∇u1)−a(x,u2,∇u2)]⋅∇(u1−u2)dx≤ε(h), |
where
∫{|u1−u2|<k}[a(x,u1,∇u1)−a(x,u2,∇u2)]⋅∇(u1−u2)dx≤0, |
for all
(2) Existence:
Step 1. Let
F(x,u)=g(x)−β(u), |
where
Let
γn(s)=βn(s)+1n|s|p−2s. |
Then by [20], there exists
{−diva(x,un,∇un)+γn(x,un)=gn,x∈Ω,un=0,x∈∂Ω, | (35) |
holds in the sense of distributions in
By density arguments, we can take
∫{k≤|un|<k+h}a(x,un,∇un)⋅∇undx+∫{|un|>k}γnTh(un−Tk(un))dx=∫{|un|>k}gnTh(un−Tk(un))dx, | (36) |
and
∫{|un|>k}a(x,un,∇un)⋅∇undx+∫ΩγnTk(un)dx=∫ΩgnTk(un)dx. | (37) |
Combine (36) with (2) (fix the ellipticity constant
∫{k<|un|<k+h}|∇un|pdx≤hkθ(p−1)∫{|un|>k}gndx≤hkθ(p−1)‖gn‖L1(Ω)=Ckθ(p−1). | (38) |
Since
∫{|un|>k}|γn(un)|dx≤∫{|un|>k}|gn|dx≤‖gn‖L1(Ω)≤C. | (39) |
Combine (37) with
∫{|un|<k}|∇un|pdx≤Ck1+θ(p−1). | (40) |
Step 2. Convergence. Using (38) and Proposition 1, we have
Next we prove that
For
{|un−um|>t}⊂{|un|>k}∪{|um|>k}∪{|Tk(un)−Tk(um)|>t}. |
Thus
|{|un−um|>t}|≤|{|un|>k}|+|{|um|>k}|+|{|Tk(un)−Tk(um)|>t}|. |
Choosing
Tk(un)→Tk(u)inLploc(Ω)anda.einΩ. |
Then
|{|Tk(un)−Tk(um)|>t}∩BR|≤t−q∫Ω∩BR|Tk(un)−Tk(um)|qdx≤ϵ, |
for all
Now to prove that
{|∇un−∇um|>t}∩BR⊂{|un−um|≤k,|∇un|≤l,|∇um|≤l,|∇un−∇um|>t}∪{|∇un|>l}∪{|∇um|>l}∪({|un−um|>k}∩BR). |
Choose
[a(x,t,ξ)−a(x,t,ξ′)]⋅[ξ−ξ′]≥μ. |
This is a consequence of continuity and strict monotonicity of
dn=gn−γn(x,un). | (41) |
Taking
∫{|un−um|<k}[a(x,un,∇un)−a(x,um,∇um)]⋅∇(un−um)dx=∫Ω(dn−dm)Tk(un−um)dx≤Ck1+θ(p−1). |
Then
{|un−um|≤k,|∇un|≤l,|∇um|≤l,|∇un−∇um|>t}≤1μ∫{|un−um|<k}[a(x,un,∇un)−a(x,um,∇um)]⋅∇(un−um)dx≤1μCk1+θ(p−1)≤ϵ, |
if
Since
Finally, since
Step 3. In order to prove the existence of the solution completely, we still need to prove that sequence
q∈(1,N(1−θ)N−(1+θ(p−1))). |
Indeed, by Proposition 2,
a(x,un,∇un)→a(x,u,∇u). |
It follows that
a(x,u,∇u)∈MN(1−θ)N−(1+θ(p−1))⊂Lqloc(Ω), |
for all
In this subsection, we give the proof of Theorem 1.2 following some ideas in [11,22].
Proof. Step 1 (A priori estimates). Firstly, choosing
∫Ωa(x,un,∇un)⋅∇Tk(un)(1−φδ)sdx+∫Ω|un|q−1unTk(un)(1−φδ)sdx=s∫Ωa(x,un,∇un)⋅∇φδTk(un)(1−φδ)s−1dx+∫ΩgnTk(un)(1−φδ)sdx+∫Ωf+nTk(un)(1−φδ)sdx+∫Ωf−nTk(un)(1−φδ)sdx. | (42) |
By (2), we get
∫Ωa(x,un,∇un)⋅∇Tk(un)dμ≥c∫Ω|∇Tk(un)|p(1+|Tk(un)|)θ(p−1)dμ, | (43) |
here
Since
∫Ω|un|q−1unTk(un)(1−φδ)sdx≥∫{|un|≥k}|un|q−1unTk(un)dμ≥kq+1μ({|un|≥k}). | (44) |
Using (3) and the Young inequality, we find
∫Ω|a(x,un,∇un)⋅∇φδTk(un)(1−φδ)s−1|dx≤c0k∫Ω(|∇un|p−1+b(x))(|∇φ+δ|+|∇φ+δ|)(1−φδ)s−1dx≤Ck∫Ω(|∇un|(p−1)r′+|b(x)|r′)(1−φδ)(s−1)r′dx+Ck∫Ω(|∇φ+δ|r+|∇φ+δ|r)dx≤Ck(∫Ω(|∇un|(p−1)r′+|b(x)|r′)(1−φδ)(s−1)r′dx+δ). | (45) |
Combine (42)-(45), by (7) and
∫Ω|∇Tk(un)|p(1+|Tk(un)|)θ(p−1)dμ+kq+1μ({|un|≥k})≤Ck(∫Ω|∇un|(p−1)r′(1−φδ)(s−1)r′dx+δ+μ(Ω). | (46) |
For a fixed
μ({|∇un|>σ})=μ({|∇un|>σ,|un|<k})+μ({|∇un|>σ,|un|≥k})≤1σp∫Ω|∇Tk(un)|pdμ+μ({|u|>k})≤(1+k)θ(p−1)σp∫Ω|∇Tk(un)|p(1+|Tk(un)|)θ(p−1)dμ+μ({|u|>k})≤C(∫Ω|∇un|(p−1)r′(1−φδ)(s−1)r′dx+δ+μ(Ω))((1+k)1+θ(p−1)σp+1kq), |
which implies
μ|{|∇un|>σ}|≤Cσ−pqq+1+θ(p−1)(∫Ω|∇un|(p−1)r′(1−φδ)(s−1)r′dx+δ+μ|Ω|). | (47) |
Let
(p−1)r′<η<pqq+1+θ(p−1). | (48) |
Clearly, such
∫Ω|∇un|ηdμ≤C(∫Ω|∇un|(p−1)r′(1−φδ)(s−1)r′dx+δ+μ(Ω)). |
By the Holder's inequality,
∫Ω|∇un|(p−1)r′(1−φδ)(s−1)r′dx≤C(∫Ω|∇un|ηdμ)(p−1)r′η≤C(∫Ω|∇un|(p−1)r′(1−φδ)(s−1)r′dx+δ+μ|Ω|)(p−1)r′η. |
By Lemma 2.1,
∫Ω|∇un|(p−1)r′(1−φδ)(s−1)r′dx≤C(δ+μ|Ω|)≤C(δ). | (49) |
Using (46) and (49), we conclude that
∫Ω|∇Tk(un)|pdx≤Ck1+θ(p−1). | (50) |
According to Lemma 2.5, we have
By (50) and Lemma 2.6, there exists a subsequence, still denoted by
Since
a(x,un,∇un)→a(x,u,∇u)stronglyin(Ls(Ω))N, | (51) |
for every
Step 2 (Energy estimates). Let
∫{un>2m}uqn(1−ψδ)dx=ω(n,m,δ), | (52) |
and
∫{un<−2m}|un|q(1−ψδ)dx=ω(n,m,δ). | (53) |
Choose
βm(s)={sm−1,m<s≤2m,1,s>2m,0,s≤m. |
We obtain
1m∫{m<un<2m}a(x,un,∇un)⋅∇un(1−ψδ)dx(A)−∫Ωa(x,un,∇un)⋅∇ψδβm(un)dx(B)+∫Ω|un|q−1unβm(un)(1−ψδ)dx(C)=∫Ωf+nβm(un)(1−ψδ)dx(D)−∫Ωf−nβm(un)(1−ψδ)dx(E)+∫Ωgnβm(un)(1−ψδ)dx.(F) |
Since
−(B)=∫Ωa(x,u,∇u)⋅∇ψδβm(u)dx+ω(n)=ω(n,m), |
and
(C)≥∫{un>2m}uqn(1−ψδ)dx. |
By
(D)≤∫Ωf+n(1−ψδ)dx=∫Ω(1−ψ+δ)dλ+−∫Ωψ−δdλ−+ω(n)=ω(n,δ), |
and
(F)=ω(n,m). |
We get (52), the proof of (53) is identical.
Step 3 (Passing to the limit). Now we show that
∫Ωa(x,un,∇un)⋅∇Tk(un−φ)(1−ψδ)dx(A)−∫Ωa(x,un,∇un)⋅∇ψδTk(un−φ)dx(B)+∫Ω|un|q−1unTk(un−φ)(1−ψδ)dx(C)=∫Ωf+nTk(un−φ)(1−ψδ)dx(D) |
−∫Ωf−nTk(un−φ)(1−ψδ)dx(E)+∫ΩgnTk(un−φ)(1−ψδ)dx.(F) |
By (13),
(A)=∫{|un−φ|<k}a(x,un,∇un)⋅∇un(1−ψδ)dx−∫{|un−φ|<k}a(x,un,∇un)⋅∇φ(1−ψδ)dx, |
while
∫{|un−φ|<k}a(x,un,∇un)⋅∇φ(1−ψδ)dx=∫{|u−φ|<k}a(x,u,∇u)⋅∇φdx+ω(n,δ). |
The Fatou lemma implies
∫{|u−φ|<k}a(x,u,∇u)⋅∇udx≤limn→∞inf∫{|un−φ|<k}a(x,un,∇un)⋅∇undx. |
Using (13), (51), we have
−(B)=∫Ωa(x,u,∇u)⋅∇ψδTk(u−φ)dx+ω(n)=ω(n,δ). |
While
(F)=∫ΩgTk(u−φ)dx+ω(n,δ), |
and
|(D)|+|(E)|=∫Ω(f+n+f−n)Tk(un−φ)(1−ψδ)dx≤k∫Ω(f+n+f−n)(1−ψδ)dx=ω(n,δ). |
So that we only need to deal with
(C)=∫{−2m≤un≤2m}|un|q−1unTk(un−φ)(1−ψδ)dx(G)+k∫{un>2m}uqn(1−ψδ)dx+k∫{un<−2m}|un|q(1−ψδ)dx.(H) |
By (52) and (53), we get
(H)=ω(n,m,δ), |
and
(G)=∫Ω|u|q−1uTk(u−φ)(1−ψδ)dx+ω(n,m)=∫Ω|u|q−1uTk(u−φ)dx+ω(n,m,δ). |
Summing up the result of (A)-(H), we have
∫Ωa(x,u,∇u)⋅∇Tk(u−φ)dx+∫Ω|u|q−1uTk(u−φ)dx≤∫ΩgTk(u−φ)dx. |
Thus
Finally we prove (10). Choose
∫Ωa(x,un,∇un)⋅∇φdx+∫Ω|un|q−1unφdx=∫Ω(fn+gn)φdx. |
Thanks to the assumptions of
limn→+∞∫Ω|un|q−1unφdx=−∫Ωa(x,u,∇u)⋅∇φdx+∫Ωgφdx+∫Ωφdλ. | (54) |
Since the entropy solution of (8) is also a distributional solution of the same problem, for the same
∫Ωa(x,u,∇u)⋅∇φdx+∫Ω|u|q−1uφdx=∫Ωgφdx. | (55) |
Together with (54) and (55), we find
limn→+∞∫Ω|un|q−1unφdx=∫Ω|u|q−1uφdx+∫Ωφdλ. |
Thus (11) holds for every
The authors also would like to thank the anonymous referees for their valuable comments which has helped to improve the paper.
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