Vacuum isolating, blow up threshold and asymptotic behavior of solutions for a nonlocal parabolic equation

0.11 MB·

by  Xiaoliang Li

VACUUM ISOLATING, BLOW UP THRESHOLD AND
ASYMPTOTIC BEHAVIOR OF SOLUTIONS FOR A NONLOCAL
PARABOLIC EQUATION
7
1
0
XIAOLIANGLIANDBAIYULIU
2

n
a Abstract. Inthispaper,weconsideranonlocalparabolicequationasso-
J ciatedwithinitialandDirichletboundaryconditions.Firstly,wediscuss

5 the vacuumisolatingbehaviorofsolutionswith the helpofa familyof
1
potentialwells. Thenweobtainathresholdofglobalexistenceandblow

up for solutions with critical initial energy. Furthermore, for those so-
]
P lutions satisfy J(u ) ≤ d and I(u ) , 0, we show that globalsolutions
0 0
A decay to zero exponentially as time tends to infinity and the norm of
. blow-upsolutionsincreaseexponentially.
h
t
a
m
[

1
v 1. Introduction
2
0 In this paper, we study the following initial boundary value problem of
0
nonlocalparabolicequation
4
0
1. u = ∆u+ 1 ∗|u|p |u|p−2u, x ∈ Ω,t > 0,
t |x|n−2
0  u(x,t) = 0,(cid:16) (cid:17) x ∈ ∂Ω,t > 0, (1.1)
17  u(x,0) = u0(x), x ∈ Ω,
iv: where Ω is abounded domain in Rn (n ≥ 3), 1 < p < (n + 2)/(n − 2) and
X
1 ∗|u|p = |u(y)|p dy.
r |x|n−2 Ω |x−y|n−2
a
NonlocalpRarabolictypeequationshavebeenextensivelyusedinecology,
especiallytomodelapopulationinwhichindividualcompetesforashared
rapidlyequilibratedresourceorapopulationinwhichindividualcommuni-
catedeithervisuallyorbychemicalmeans[1–4]. Also,theycanbeapplied
tothermalphysicswithnonlocalsource[5].
2010MathematicsSubjectClassification. 35K20,35K55.
Keywordsandphrases. nonlocalparabolicequation;vacuumisolating; criticalinitial
energy;asymptoticbehavior.
∗ProjectsupportedbytheNationalNaturalScienceFoundationofChinaNo.11671031,
No.11201025.
1
2 X.LIANDB.LIU
Asamodelproblemforstudyingthecompetitionbetweenthedissipative
effect ofdiffusionandtheinfluenceofan explosivesourceterm, problem
u = ∆u+|u|p−1u, x ∈ Ω,t > 0
t
u(x,t) = 0, x ∈ ∂Ω,t ≥ 0 (1.2)

 u(x,0) = u0(x), x ∈ Ω
has been extensively studied (see [6,7,9–14] and the reference therein).
For the sub-critical case 1 < p < (n + 2)/(n − 2), blow up in infinite time
does not occur. The solution will either exist globally or blow up in finite
time. Itisnaturaltoaskunderwhatconditions,willthesolutionexistforall
time; and under what conditions, will the solution become unstable to col-
lapse. Totreattheabovequestion,Sattinger[15](seealso[16])established
apowerfulmethodwhichiscalledthepotentialwellmethod. Byusingthis
method, Ikehata and Suzuki [10], Payne and Sattinger [16] described the
behaviorofsolutionsfor(1.2)whentheinitialdatahaslowenergy(smaller
than the height of potential well). Roughly speaking, they found a thresh-
old of global solutions and blow up solutions. Liu and Zhao [7], Xu [17]
generalized the above results to the critical energy level initial data. More-
over, by generalizing the potential well method, an important phenomena
called vacuumisolatinghas been found by Liu and Zhao [7], i.e., thereis a
region which does not contain any low energy solutions. Vacuum isolating
phenomena has also been observed in various kinds of evolution equations
withvariationalstructures[18–20].
As a modelproblemofnonlocal parabolicequation, (1.1) has been stud-
iedby [21,22]. Well-posednessin Lq(Ω)has been setup. Precisely,
Theorem 1.1. [Theorem 6 and 7 in [21]] Let u ∈ Lq(Ω), n−1 ≤ q < ∞,
0
q > n(p−1)(2−1). ThenthereexistsT = T(||u || ) > 0suchthatproblem
2 p max 0 q
(1.1)possessesauniqueclassicalLq−solutionin[0,T ). Moreover,either
max
T = +∞ orlim ||u(t)|| = +∞.
max t→Tmax q
There are two natural functionals on H1(Ω) associated with the problem
0
(1.1), the energy functional and the Nehari functional, defined respectively
by
1 1 |u(y)|p|u(x)|p
J(u) = |∇u|2dx− dxdy,
2 ZΩ 2p ZΩ×Ω |x−y|n−2
1
I(u(t)) = (J′(u),u) = |∇u|2dx− ∗|u|p |u|pdx,
ZΩ ZΩ |x|n−2 !
Then alongtheflowgenerated by(1.1), wehave
d
J(u(t)) = (J′(u),u) = −||u||2 ≤ 0. (1.3)
dt t t 2
NONLOCALPARABOLICEQUATION 3
TheNehari manifoldisdefined by
N := {u ∈ H1(Ω)|I(u) = 0,u , 0}. (1.4)
0
Thedepth ofthepotentialwellis
d := inf{J(u)}. (1.5)
u∈N
By using the potential well method, Liu and Ma [21] proved that for
low energy solutions (J(u ) < d) the maximum existence time is totally
0
determinedbytheNeharifunctional I(u ). Moreprecisely,if J(u ) < d and
0 0
I(u ) > 0 then the solution exists globally, if J(u ) < d and I(u ) < 0 then
0 0 0
thesolutionblowsupin finitetime.
This paper devoted to continue the study of [21]. The first result of the
present paper deals with the solution start with initial data which has low
initial energy. We found the vacuum isolating phenomenon, by using the
familyofpotentialwells[7,8].
Let δ > 0. Define
I (u) := δ||∇u||2 − v(u)|u(x)|pdx,
δ
ZΩ
N := u ∈ H1(Ω)|I (u) = 0,||∇u|| , 0 , d(δ) = inf J(u).
δ 0 δ
n o u∈Nδ
Theorem 1.2. Let e ∈ (0,d). Suppose δ ,δ are the two roots of d(δ) =
1 2
e.Then for all solutions of problem (1.1) with J(u ) ≤ e, there is a vacuum
0
region
U = N = u ∈ H1(Ω) | I (u) = 0,u , 0,δ < δ < δ ,
e δ 0 δ 1 2
δ[<δ<δ n o
1 2
suchthatthereisnoanysolutionof problem(1.1)inU .
e
Thenwestudythecriticalinitialenergycaseandobtainthethresholdjust
likethelowinitialenergy solution.
Theorem1.3. LetΩbeasmoothboundedconvexdomaininRn(n = 3or4).
Assume1 < p < n+2, such that (p−1)(2− 1) < 4 . If u ∈ C(Ω¯)∩H1(Ω)
n−2 p n−2 0 0
and J(u ) = d,I(u ) > 0, then problem (1.1) admits a global solution u(t)
0 0
for0 < t < ∞.
Theorem 1.4. Let Ω be a smooth bounded convex domain and 1 < p <
n+2 (n ≥ 3). If u ∈ C(Ω¯) ∩ H1(Ω) and J(u ) = d,I(u ) < 0, then the
n−2 0 0 0 0
solutionofproblem(1.1)blows upin finitetime.
Afterthat,forthelowinitialenergy andcriticalinitialenergy solutionof
(1.1)i.e. J(u ) ≤ d,we studytheasymptoticbehavior.
0
4 X.LIANDB.LIU
Theorem1.5. LetΩbeasmoothboundedconvexdomaininRn(n = 3or4).
Assume1 < p < n+2, such that (p−1)(2− 1) < 4 . If u ∈ C(Ω¯)∩H1(Ω)
n−2 p n−2 0 0
satisfiesJ(u ) ≤ dandI(u ) > 0,thenfortheglobalsolutionu(t)ofproblem
0 0
(1.1)decays to 0 exponentiallyas t → ∞.
Theorem 1.6. Let Ω be a smooth bounded convex domain and 1 < p <
n+2 (n ≥ 3). Ifu ∈ C(Ω¯)∩H1(Ω)satisfies J(u ) ≤ d andI(u ) < 0,thenthe
n−2 0 0 0 0
corresponding solution of problem (1.1) grows as an exponential function
in Ln2−n2(Ω)norm.
The reminder of this paper is organized as follows. In the next section,
wegivesomepreliminariesaboutthefamilyofpotentialwells,after which
we discuss the vacuum isolating of solutions for (1.1). In Section 3, we
establishthethresholdforglobalsolutionsandfinitetimeblowupsolutions
of (1.1) at the critical initial energy level. At last, the asymptotic behavior
willbediscussedin Section4.
Throughout the paper, we denote v(u) = 1 ∗ |u|p, || · || = || · || ,
|x|n−2 p Lp(Ω)
||·|| = ||·|| and denotethemaximalexistencetimeby T .
2 max
2. VacuumIsolating
In this section, we shall introduce a family of Nehari functionals I (u)
δ
in spcace H1(Ω) and givethe corresponding lemmas, which will help us to
0
demonstratethevacuumisolatingbehaviorof(1.1).
Lemma 2.1. Let 1 < p < n+2 and δ > 0. Then is a contant C > 0
n−2 n,p,Ω,δ
satisfiesthat||∇u|| ≥ C forall u ∈ H1(Ω)\{0} and I (u) ≤ 0.
n,p,Ω,δ 0 δ
Proof. Provided that I (u) ≤ 0, applying the classical Hardy-Littlewood-
δ
Sobolevinequalitywehave
|u(y)|p|u(x)|p
δ||∇u||2 ≤ v(u)|u|pdx = dxdy ≤ C||u||2p . (2.1)
ZΩ ZΩ×Ω |x−y|n−2 2np/(n+2)
Notice that 2np < 2n due to 1 < p < n+2. By using Ho¨lder inequality and
n+2 n−2 n−2
Sobolevinequalityweobtain
||u|| ≤ C ||u|| ≤ C ||∇u||. (2.2)
2np/(n+2) n,p,Ω 2n/(n−2) n,p,Ω
Combining (2.1) and (2.2), one has δ||∇u||2 ≤ C ||∇u||2p i.e. ||∇u||2 ≥
n,p,Ω
(δ/Cn,p,Ω)2p1−2 = Cn,p,Ω,δ. (cid:3)
Lemma 2.2. Let
v(u)|u|pdx
C∗ = sup Ω . (2.3)
R ||∇u||2p
u∈H1(Ω),||∇u||,0
0
NONLOCALPARABOLICEQUATION 5
Then
1 1
d(δ) = inf J(u) = δp−11 − δp−p1 C∗−p−11. (2.4)
u∈Nδ 2 2p !
Proof. At first, by the proof of Lemma 2.1, there is C > 0 such that
n,p,Ω
v(u)|u|pdx ≤ C ||∇u||2p for all u ∈ H1(Ω)\{0}, which ensures the exis-
Ω n,p,Ω 0
RtenceofC∗.
For each u ∈ H1(Ω)\{0}, there is a unique λ = λ(δ,u) so that λ(δ,u)u ∈
0
N . A simplecalculationgives
δ
1
δ||∇u||2 2p−2
λ(δ) = ,
 v(u)|u|pdx
and RΩ 
1
1 1 ||∇u||2p p−1
J(λu) = δp−11 − δp−p1 .
2 2p ! v(u)|u|pdx
Noticingthat infu∈Nδ J(u) = infu∈H01(Ω)\{0} J(λRΩ(δ,u)u)andby usingthedefini-
tionofd(δ) weconcludethat
1 1
d(δ) = inf J(u) = inf J(λ(δ,u)u) = δp−11 − δp−p1 C∗−p−11.
u∈Nδ u∈H1(Ω)\{0} 2 2p !
0
(cid:3)
Lemma 2.3. d(δ)satisfiesthefollowingproperties:
(i) d(δ) > 0for0 < δ < p;
(ii) lim d(δ) = lim d(δ) = 0;
δ→0 δ→p
(iii) d(δ) is strictly increasing on 0 < δ ≤ 1, strictly decreasing on 1 ≤
δ < p andtakes themaximumd = d(1)atδ = 1;
(iv) d(δ)iscontinuouson 0 ≤ δ ≤ p.
Proof. From (2.4), d(δ) = 1δp−11C∗p−−11 1− δ which gives (i)(ii)(iv). By a
2 p
straightforwardcalculation,wecanver(cid:16)ifyd′((cid:17)1) = 0,d′(δ) > 0for0 < δ < 1
andd′(δ) < 0 for1 < δ < p, whichshows(iii).
(cid:3)
Remark2.4. FromtheaboveLemma, weknowthatthedepthofthepoten-
tialwell is d = (21 − 21p)C∗−p−11 = maxδ∈[0,p]d(δ).
Lemma 2.5. Let 0 < e < d andδ ,δ aretwo rootsofequationd(δ) = e. If
1 2
u ∈ H1(Ω)and J(u) ≤ e,thenthesignofI (u)remainunchangedon(δ ,δ ).
0 δ 1 2
Proof. Assume I (u) change its sign on (δ ,δ ), then there exists a δ such
δ 1 2 0
that I (u) = 0, that is to say u ∈ N and hence J(u) ≥ d(δ ). By using
δ0 δ0 0
Lemma2.3,wehave J(u) ≥ d(δ ) > d(δ ) = d(δ ),whichcontradictstothe
0 1 2
choiceofδ and δ .
1 2
6 X.LIANDB.LIU
(cid:3)
Weare nowina positiontogivetheproofofTheorem1.2.
ProofofTheorem 1.2. Let u(t) (0 ≤ t < T ) be the solution of problem
max
(1.1) correspondingtou . Weonlyneed to provethatifu , 0 and J(u ) ≤
0 0 0
e, thenforall δ ∈ (δ ,δ ), u(t) < N , i.e. I (u(t)) , 0, forallt ∈ [0,T ).
1 2 δ δ max
Atfirst,itisclearthat I (u ) , 0. Sinceif I (u ) = 0,then J(u ) ≥ d(δ) >
δ 0 δ 0 0
d(δ ) = d(δ ), which contradictswiththedefinitionofδ andδ .
1 2 1 2
Supposethere is t > 0 s.t. u(t ) ∈ U . Namely,there is someδ ∈ (δ ,δ )
1 1 e 1 2
suchthatu(t ) ∈ N . Sincetheenergyfunctional J(u)isnoincreasingalong
1 δ
the flow generated by (1.1), see (1.3). Thus, we get J(u ) ≥ J(u(t )) ≥
0 1
d(δ) > J(u ), whichleads toacontradiction. (cid:3)
0
3. Thresholdforsolutionswithcritical initial energy
In thissection,wedeal withthecriticalinitialenergy solution.
ProofofTheorem 1.3. We may assume that u(t) , 0 for all t ∈ [0,T ).
max
Actually, if there is u(t) = 0, then by uniqueness, u(s) = 0 for all s ≥ t.
Hence, theconclusionistrue.
We claim that I(u(t)) > 0 for any t ∈ [0,T ). Otherwise, supposethere
max
existsat > 0 suchthat I(u(t )) = 0, and I(u(t)) > 0 for0 < t < t . Then
0 0 0
J(u(t )) ≥ d = J(u ) (3.1)
0 0
due to u(t ) ∈ N. On the other hand, since I(u(t)) > 0 for 0 < t < t and
0 0
byusingthefactthat uudx = −I(u(t)), weobtainu , 0on(0,t ),which
Ω t t 0
indicates t0 ||u||2dτ >R 0. Integrating equation (1.3) on interval (0,t), one
0 t
has R
t
J(u(t)) = J(u )− ||u||2dτ < J(u ) = d,
0 t 0
Z
0
whichcontradictsto(3.1).
So wehave
1 1
d = J(u ) ≥ J(u(t)) ≥ (1− ) |∇u|2dx,
0
2 p ZΩ
whichindicatesthat
2p
|∇u|2dx ≤ d.
ZΩ p−1
Therefore, ||u(t)|| is uniformly bounded. For those q satisfies n − 1 ≤
H1(Ω)
0
q ≤ 2n (n = 3 or 4), and n(p −1)(2− 1) < q < 2n , we have ||u(t)|| is
n−2 2 p n−2 Lq(Ω)
bounded,byusingtheSobolevinequality. ApplyingTheorem6in[21],we
knowthat T = ∞.
max
(cid:3)
NONLOCALPARABOLICEQUATION 7
Weshall proveTheorem 1.4by usingtheconcavity method[23].
ProofofTheorem 1.4. First we prove I(u(t)) < 0 for t ∈ (0,T ). Suppose
max
itisfalse,thenthereexistsat > 0s.t. I(u(t )) = 0andI(u(t)) = I (u(t)) < 0
0 0 1
for0 ≤ t < t . On theonehand wehave||∇u(t)|| ≥ C on [0,t ) by using
0 n,p,Ω 0
Lemma2.1, which impliesu(t ) , 0. Thus weobtain
0
J(u(t )) ≥ d (3.2)
0
due to the fact that u(t ) ∈ N. On the other hand, one can see u , 0
0 t
on (0,t ) since uudx = −I(u) > 0, which indicates t0 ||u||2dτ > 0.
0 Ω t 0 t
By a similar argRument as in the proof of Theorem 1.3, weRhave J(u(t )) =
0
J(u ) − t0 ||u||2dτ < d, which contradicts with (3.2). Consequently, we
0 t
0
have R
I(u(t)) < 0, ∀t ∈ [0,T ). (3.3)
max
Moreover,by Lemma2.1,thereholds
||∇u(t)|| ≥ C , ∀t ∈ [0,T ). (3.4)
n,p,Ω max
AssumeforcontradictionthatT = +∞. DenoteM(t) = 1 t||u(τ)||2dτ.
max 2 0
Then we obtain M′(t) = 1||u(t)||2 > 0 and M′′(t) = uudx =R−I(u(t)) > 0
2 Ω t
fort > 0. Chooset > 0 suchthat R
1
t1 t1
0 < d = J(u(t )) = J(u )− ||u||2dτ = d − ||u||2dτ < d.
1 1 0 t t
Z Z
0 0
Thus, we have J(u(t)) ≤ d for each t ≥ t . It follows from (3.3), Lemma
1 1
2.1 and Lemma 2.5 that I (u(t)) < 0 for δ < δ < δ ,t ≥ t , where δ ,δ are
δ 1 2 1 1 2
two roots of equation d(δ) = d . Thus, choosing any δ ∈ (1,δ ), we have
1 0 2
I (u(t)) < 0 forallt ≥ t . Taking (3.4)intoaccount, wefind
δ0 1
M′′(t) = −I(u(t)) = (δ −1)||∇u(t)||2−I (u(t)) > (δ −1)C > 0,∀t ≥ t ,
0 δ0 0 n,p,Ω 1
whichindicates M′(t) → +∞ as t → +∞ and M(t) → +∞ as t → +∞.
Nowfort > 0,weestimatethefollowing
M′′(t) = −I(u(t)) = (p−1)||∇u(t)||2 −2pJ(u(t)) (3.5)
t
≥ 2p ||u||2dτ+(p−1)λM′(t)−2pJ(u ), (3.6)
t 0
Z
0
here constant λ satisfies ||∇u||2 ≥ λ||u||2 which from Poincare´ inequality.
2
Integrating M′′(t) = uudxon (0,t)yields
Ω t
R
t
M′(t)− M′(0) = uudxdτ.
t
Z0 ZΩ
8 X.LIANDB.LIU
Hence,
t 2
(M′(t))2 = −(M′(0))2 +2M′(t)M′(0)+ uudxdτ
t
Z0 ZΩ !
1 t 2
= − ||u ||4 + M′(t)||u ||2 + uudxdτ
0 0 t
4 Z0 ZΩ !
t 2
≤ M′(t)||u ||2 + uudxdτ . (3.7)
0 t
Z0 ZΩ !
Thencombining(3.6)and (3.7), wehave
t t t 2
MM′′ − pM′2 ≥ p ||u||2dτ· ||u||2dτ− uudxdτ
t t
+(p−Z01)λMM′ −Z20pMJ(u )− pZM0 ′Z||uΩ||2 ! 
0 0
≥ (p−1)λMM′ −2pMJ(u )− pM′||u ||2, (3.8)
0 0
wherewehaveusedSchwatz’sinequality. Since M(t) → ∞and M′(t) → ∞
ast → ∞, then thereexistsat s.t.
2
p−1 p−1
λM(t) > p||u ||2, λM′(t) > 2pJ(u ), t > t .
0 0 2
2 2
Henceweobtainby (3.8)
M(t)M′′(t)− pM′(t)2 > 0, t > t .
2
Letus considerthefunction M−p+1(t). By asimplecalculationwe have
d2
M−p+1(t) = (−p+1)M−p−1(t) M(t)M′′(t)− pM′(t)2 < 0,t > t .
dt2 2
(cid:16) (cid:17)
It guarantees that nonincreasing function M−p+1(t) is concave on (t ,∞).
2
Consequently,thereexistsafinitetimeT > 0suchthatlim M−p+1(t) = 0
t→T
i.e. lim M(t) = ∞ whichcontradictstheassumptionthatT = +∞.
t→T max
Thiscompletestheproof.
(cid:3)
Weconcludethissectionby pointingoutthefollowingremark.
Remark3.1. Fromtheproofof Theorems 1.3and 1.4, we cansee that
W′ = u ∈ H1(Ω)|J(u) ≤ d,I(u) > 0 ∪{0}and
0
n o
Z′ = u ∈ H1(Ω)|J(u) ≤ d,I(u) < 0 ,
0
n o
are both invariant for solutions of problem (1.1). Moreover, the solution
haslongtimeexistenceifu ∈ W′ andthesolutionblowsupatfinitetimeif
0
u ∈ Z′.
0
NONLOCALPARABOLICEQUATION 9
4. Exponential decay, exponential growth
In this section, we shall investigate the asymptotic behavior of solutions
for problem (1.1) with J(u ) ≤ d and give the proof of Theorem 1.5 and
0
Theorem1.6.
ProofofTheorem 1.5. Weconsiderthefollowingtwocases.
Case1. J(u ) < d.
0
By using
1 1 1
J(u) = − ||∇u||2 + I(u),
2 2p! 2p
we get J(u ) > 0. Let δ ,δ (δ < δ ) be the two roots of equation d(δ) =
0 1 2 1 2
J(u ).
0
From Proposition 10 in [21], we know J(u(t)) < d,I(u(t)) > 0 for all
t > 0, provided J(u ) < d,I(u ) > 0. Since J(u(t)) ≤ J(u ) < d,I(u(t)) > 0
0 0 0
foreach t ≥ 0, weobtain I (u(t)) > 0 forδ ∈ (δ ,δ ),t ≥ 0 byusingLemma
δ 1 2
2.5. Takinganyδ ∈ (δ ,1),wehave
0 1
1 d 1 d
||u||2 +I(u) = ||u||2 +(1−δ )||∇u||2 +I (u) = 0.
2dt 2dt 0 δ0
By applyingPoincare´ inequality,weobtain
1 d
||u||2 +(1−δ )C||u||2 < 0, t ≥ 0.
0
2dt
Consequently,byusingGronwallinequalityweknowthat
||u(t)||2 ≤ ||u ||2e−2C(1−δ0)t, 0 ≤ t < ∞.
0
Case2. J(u ) = d.
0
Indeed, given J(u ) = d,I(u ) > 0, from the proof of Theorem 1.3 we
0 0
can choose any fixed t > 0 such that 0 < J(u(t )) < d,I(u(t )) > 0. Let
0 0 0
δ ,δ are tworootsofequationd(δ) = J(u(t )). Thus byasimilarargument
1 2 0
withproofofCase1, weeasily obtain
||u(t)||2 ≤ ||u(t )||2e−2C(1−δ0)t, t ≤ t < ∞.
0 0
Thereforetheresultoftheoremfollowsimmediately. (cid:3)
In orderto proveTheorem 1.6,weneed thefollowinglemma.
Lemma4.1. Letu(t)beanontrivalsolutionofproblem(1.1)whichsatisfies
J(u0) < d,||∇u0|| > α1 := C∗2−p−12. Then there exists α2 > α1 such that
||∇u(t)|| ≥ α forall0 ≤ t < T . HereC∗ isdefined by(2.3).
2 max
Proof. Firstly,bythedefinitionofC∗ as in (2.3),we estimate
1 1 1 C∗
J(u) = ||∇u||2 − v(u)|u|pdx ≥ ||∇u||2 − ||∇u||2p. (4.1)
2 2p ZΩ 2 2p
10 X.LIANDB.LIU
Denoteg(α) = 1α2 − C∗α2p. It’seasy toverifyg(α)isstrictlyincreasingon
2 2p
(0,α ), decreasingon (α ,∞)and attainsitsmaximumat α = α :
1 1 1
1 1
g(α ) = − C∗−1/(p−1) = d. (4.2)
1
2 2p!
Let α = ||∇u || > α , we obtain g(α ) = g(||∇u ||) ≤ J(u ) < d by formula
0 0 1 0 0 0
(4.1). Hence, wecan find aα ∈ (α ,α ]suchthat g(α ) = J(u ).
2 1 0 2 0
We claim that ||∇u(t)|| ≥ α for all t ≥ 0. Otherwise, by the continuity,
2
we can choose t > 0 such that α < ||∇u(t )|| < α . Thus we know that
0 1 0 2
d = g(α ) > g(||∇u(t )||) > g(α ) = J(u ), which contradicts with the fact
1 0 2 0
that J(u ) < d.
0
Theproofisnowcomplete. (cid:3)
Withthehelp oftheabovelemma,wegivetheproofofTheorem1.6.
ProofofTheorem 1.6. Letus considerthefollowingtwocases.
Case1. J(u ) < d.
0
On theonehand, since I(u ) < 0 and byusing(2.3)weobtain
0
||∇u ||2 < v(u )|u |pdx ≤ C∗||∇u ||2p,
0 0 0 0
ZΩ
whichimplies||∇u || > C∗−1/(2p−2) = α . ApplyingLemma4.1weget
0 1
||∇u(t)|| ≥ α > α , t ∈ [0,T ). (4.3)
2 1 max
For t ≥ 0, define H(t) = d − J(u(t)), L(t) = H(t)+ 1||u(t)||2. Combining
2
(1.3), (3.5)and (4.3), wehave
H(t) > 0 (4.4)
and
dJ(u(t))
L′(t) = − −I(u(t)) = ||u||2 +(p−1)||∇u||2 +2pH(t)−2pd
t
dt
≥ (p−1)||∇u||2 +2pH(t)−2pd
α2 −α2 α2
= (p−1) 2 1||∇u||2 +(p−1) 1||∇u||2 +2pH(t)−2pd
α2 α2
2 2
α2 −α2
≥ (p−1) 2 1||∇u||2 +(p−1)α2 +2pH(t)−2pd.
α2 1
2
Noticethat2pd = (p−1)C∗−1/(p−1) = (p−1)α2 followsfromformula(4.2),
1
then
α2 −α2
L′(t) ≥ (p−1) 2 1||∇u||2 +2pH(t). (4.5)
α2
2