Research Article Singularity Analysis for a Class of ...downloads.hindawi.com/journals/amp/2016/8292835.pdf · Research Article Singularity Analysis for a Class of Porous Medium Equation

Research ArticleSingularity Analysis for a Class of Porous Medium Equationwith Time-Dependent Coefficients

Anyin Xia12 Xianxiang Pu3 and Shan Li4

1Department of Mathematics Sichuan University Chengdu 610065 China2School of Science Xihua University Chengdu 610039 China3Sichuan Sanhe College of Professionals Luzhou 646200 China4Business School Sichuan University Chengdu 610064 China

Correspondence should be addressed to Shan Li lishanscueducn

Received 22 September 2015 Accepted 24 December 2015

Academic Editor Andrei D Mironov

Copyright copy 2016 Anyin Xia et al This is an open access article distributed under the Creative Commons Attribution Licensewhich permits unrestricted use distribution and reproduction in any medium provided the original work is properly cited

This paper concerns the singularity and global regularity for the porous medium equation with time-dependent coefficients underhomogeneous Dirichlet boundary conditions Firstly some global regularity results are established Furthermore we investigatethe blow-up solution to the boundary value problemThe upper and lower estimates to the lifespan of the singular solution are alsoobtained here

1 Introduction and Main Results

In this work we consider the following porousmedium equa-tion with time-dependent coefficients under homogeneousDirichlet boundary condition

119906119905= Δ119906119898+ 119891 (119905) 119892 (119906) (119909 119905) isin Ω times (0 119879)

119906 (119909 119905) = 0 (119909 119905) isin 120597Ω times (0 119879)

119906 (119909 0) = 1199060(119909) 119909 isin Ω

(1)

where Ω sub R119873 is a smooth bounded domain 119898 gt 1 Thecoefficient 119891(119905) is a positive bounded continuous functionwith 119891(119905) le 119896 for any 119905 gt 0 The nonlinearity 119892(119906) is assumedto satisfy 119892(119906) gt 0 (119906 gt 0) and 119892(0) = 0 The initialvalue 119906

0(119909) is nontrivial nonnegative continuous function

and vanishes on 120597ΩGlobal existence and nonexistence to the nonlinear

parabolic equation are important topic and have been inves-tigated extensively please see the surveys [1ndash4] The firstpurpose in this paper is to investigate the sufficient conditionsto the global existence of the solution to the boundary valueproblem (1) The second aim of this paper is to investigatethe solution which blows up in finite time and estimatethe lifespan of the singular solution Singularity analysis

especially to evaluate the lifespan of the singular solution isalso an interesting research topic in this field

In [5ndash7] Payne and others have considered the lineardiffusion case However the degenerate diffusion makes thepresent problem more complicated and takes more essentialdifficulties here We would like to refer some results onblow-up solutions to the degenerate parabolic equations andsystem as follows Some global existence and nonexistence ofthe classical solution to degenerate parabolic equations wereestablished in [8] and then Du and his colleagues gave thesufficient conditions to the degenerate parabolic system withnonlinear nonlocal sources in [9ndash11] with nonlinear localizedsources in [10] and with nonlinear memory terms in [12]Furthermore some properties to the singular solutions suchas blow-up set uniformblow-up profile were obtained in [13ndash15]

The local existence of classical solution to system (1) canbe obtained by the standard method in [16] Firstly we givesome results on the global existence of the classical solutionto the boundary value problem (1) as follows

Theorem 1 Suppose that there exists a positive constant 119901 lt

119898 such that 119892(119906) le 119906119901 for 119906 gt 0 then every classical solutionto problem (1) is global

Hindawi Publishing CorporationAdvances in Mathematical PhysicsVolume 2016 Article ID 8292835 5 pageshttpdxdoiorg10115520168292835

2 Advances in Mathematical Physics

Secondly we give the blow-up results in the next theorem

Theorem 2 Suppose that there exists a positive constant 119901with 119901 ge 119898 such that 119892(119906) ge 119906119901 for 119906 gt 0 and 119896flinf119891(119905) gt 0then the classical solution to problem (1) blows up in finite timeprovided that the initiate data are sufficiently large

Theorem 3 Suppose that there exists a positive constant 119901with 119901 ge 119898 such that 119892(119906) ge 119906119901 for 119906 gt 0 and 119896flinf119891(119905) gt1205821 then the classical solution to problem (1) blows up in finite

time 119879 for large data 1199060(119909) where 120582

1is the first eigenvalue to

the following problem

Δ120593 + 120582120593 = 0 119909 isin Ω

120593 = 0 119909 isin 120597Ω

(2)

with 120593 gt 0 (119909 isin Ω) and intΩ120593119889119909 = 1 Moreover there exists a

constant1198790 which depends on119898119891 119892 119906

0(119909) such that119879 le 119879

0

Furthermore we give the following estimates to themaximal blow-up time 119879

Theorem 4 Suppose that Ω be a convex domain in R3 andthere exists a constant 119901 gt 0 such that 119892(119906) le 119906

119901 119906 gt 0 Ifthe solution to problem (1) blows up in finite time 119879 then thereexists a positive constant 119879

0 which depends on 119898119891 119892 119906

0(119909)

such that

119879 ge 1198790 (3)

Remark 5 We would like to mention that the results inTheorem 4 are still valid for two-dimensional case

The remainder of this paper is organized as followsGlobal existence of the solution to problem (1) is establishedin Section 2 by constructing some global upper solutionIn Sections 3 and 4 we show that the classical solution toproblem (1) will blow up in finite time and obtain upperbound of the blow-up time In the last section with aid of adifferential inequality we will establish lower estimate to themaximal blow-up time

2 Global Solution for Problem (1)

In this section we focus on the global solution of (1) and showTheorem 1

Proof Obviously if 0 lt 119901 lt 119898 then there exists a positiveconstant 120590 such that 0 lt 119898120590 lt 1 119898120590 minus 119901120590 gt 0

Next we construct a supersolution which is bounded forany 119879 gt 0 Let 120601(119909) be the solution of the following ellipticproblem

minusΔ120601 (119909) = 1 119909 isin Ω

120601 (119909) = 0 119909 isin 120597Ω

(4)

Denote 119862 = maxΩ120601(119909) Namely 0 le 120601(119909) le 119862

We define the function 119906(119909 119905) as

119906 = (119870 (120601 (119909) + 1))120590

(5)

where 120590 gt 0 satisfy 119898120590 lt 1 and 119870 gt 0 will be fixed laterClearly 119906 is bounded for any 119905 gt 0 Thus we have

119906119905minus Δ119906119898

= minus119898120590119870119898120590(119898120590 minus 1) [1 + 120601 (119909)]

119898120590minus2 1003816100381610038161003816nabla120601 (119909)

1003816100381610038161003816

2

+ [1 + 120601 (119909)]119898120590minus1

Δ120601 (119909) ge minus119898120590119870119898120590[1

+ 120601 (119909)]119898120590minus1

Δ120601 (119909) = 119898120590119870119898120590[1 + 120601 (119909)]

119898120590minus1

ge 119898120590119870119898120590(1 + 119862)

119898120590minus1

119891 (119905) 119892 (119906) le 119896119906119901le 119896 [119870 (1 + 120601 (119909))]

119901120590

le 119896 [119870 (1 + 119862)]119901120590

(6)

Denote

1198701= [

119896

119898120590

(1 + 119862)119901120590minus119898120590+1

]

1(119898120590minus119901120590)

(7)

As119898120590 minus 119901120590 gt 0 we can choose 119870 sufficiently large that 119870 gt

1198701and

[119870 (120601 (119909) + 1)]120590

ge 1199060(119909) (8)

Now it follows from (6)ndash(8) that 119906 defined by (5) is a positivesupersolution of (1) Hence 119906 le 119906 by comparison principlewhich implies 119906 exists globally

3 Blow-Up Solution to Problem (1)

In this section we will discuss the blow-up solution of (1)under some appropriate hypotheses and showTheorem 2

Proof Our strategy here is to construct blow-up subsolutionsin some subdomain of Ω in which 119906 gt 0 Some ideas areborrowed from the work [11] by Du

Let120595(119909) be a nontrivial nonnegative continuous functionand vanish on 120597Ω Without loss of generality we assume that0 isin Ω and 120595(0) gt 0 We will construct a blow-up subsolutionto complete the proof

Set

119906 (119909 119905) =

1

(119879 minus 119905)1198971205961119898

(

|119909|

(119879 minus 119905)120583) (9)

with

120596 (119903) =

1198773

12

minus

119877

4

1199032+

1

6

1199033 119903 =

|119909|

(119879 minus 119905)120583 0 le 119903 le 119877 (10)

where 119897 120583 and119879 gt 0 are determined later Clearly 0 le 120596(119903) le119877312 and 120596(119903) is nonincreasing since 1205961015840(119903) = 119903(119903minus119877)2 le 0

Note that

supp 119906 (sdot 119905) = 119861 (0 119877 (119879 minus 119905)120583) sub 119861 (0 119877119879120583) sub Ω (11)

for sufficiently small 119879 gt 0 Obviously 119906 becomesunbounded as 119905 rarr 119879

minus at the point 119909 = 0

Advances in Mathematical Physics 3

Calculating directly we obtain

119906119905(119909 119905) minus Δ119906

119898(119909 119905)

=

1198981198971205961119898

(119903) + 1205831199031205961015840(119903) 120596(1minus119898)119898

(119903)

119898 (119879 minus 119905)119897+1

+

119877 minus 2119903

2 (119879 minus 119905)119898119897+2120583

+

(119873 minus 1) (119877 minus 119903)

2 (119879 minus 119905)119898119897+120583

le

119897 (119877312)

1119898

(119879 minus 119905)119897+1

+

119873119877 minus (119873 + 1) 119903

2 (119879 minus 119905)119898119897+2120583

(12)

noticing that 119879 gt 0 is sufficiently small

Case 1 If 0 le 119903 le 119873119877(119873 + 1) we have 120596(119903) ge (3119873 +

1)119877312(119873 + 1)

3 then

119891 (119905) 119892 (119906) ge 119891 (119905) 119906119901(119909 119905) = 119891 (119905)

1

(119879 minus 119905)119901119897120596119901119898

(119903)

ge

119896

(119879 minus 119905)119901119897(

(3119873 + 1) 1198773

12 (119873 + 1)3)

119901119898

(13)

Hence119906119905(119909 119905) minus Δ119906

119898(119909 119905) minus 119891 (119905) 119892 (119906)

le

119897 (119877312)

1119898

(119879 minus 119905)119897+1

minus

119896

(119879 minus 119905)119901119897(

(3119873 + 1) 1198773

12 (119873 + 1)3)

119901119898

(14)

Case 2 If119873119877(119873 + 1) lt 119903 le 119877 then

119906119905(119909 119905) minus Δ119906

119898(119909 119905) minus 119891 (119905) 119892 (119906)

le

119897 (119877312)

1119898

(119879 minus 119905)119897+1

+

119873119877 minus (119873 + 1) 119903

2 (119879 minus 119905)119898119897+2120583

(15)

If 119901 ge 119898 gt 1 then there exists positive constant 119897 suchthat

(119898 minus 1) 119897 gt 1 (16)

Thus we get

119901119897 ge 119898119897 gt 119897 + 1

119898119897 + 2120583 gt 119897 + 1 (forall120583 gt 0)

(17)

Hence for sufficiently small 120583 gt 0 and 119879 gt 0 (11) holds And(14)-(15) imply that

119906119905(119909 119905) minus Δ119906

119898(119909 119905) minus 119891 (119905) 119892 (119906) le 0 (18)

where (119909 119905) isin 119861(0 119877(119879 minus 119905)120583) times (0 119879)Since 120595(119909) is a nontrivial nonnegative continuous func-

tion and 120595(0) gt 0 there exist two positive constants 120588 and120576 such that 120595(119909) gt 120576 for all 119909 isin 119861(0 120588) sub Ω Choose119879 small enough to insure 119861(0 119877(119879 minus 119905)120583) sub 119861(0 120588) hence119906 le 0 on 120597119861(0 119877(119879 minus 119905)

120583) times (0 119879) From (11) it follows

that 119906(119909 0) le 119872120595(119909) for sufficient large119872 By comparisonprinciple we have119906 le 119906 provided that 119906

0(119909) ge 119872120595(119909) which

implies that the solution 119906 of (1) blows up in finite time

4 Upper Bound for Blow-Up Time

In this section we will discuss upper bound for the blow-up time under some appropriate hypotheses and showTheo-rem 3

Proof Denote

119865 (119905) = int

Ω

119906120593119889119909 (19)

where 120593 is the solution to problem (2)

Case 1 (119901 gt 119898 gt 1) Combining (1) and (19) we have

1198651015840(119905) = int

Ω

[Δ119906119898+ 119891 (119905) 119892 (119906)] 120593 119889119909

ge minus1205821int

Ω

119906119898120593119889119909 + 119896int

Ω

119906119901120593119889119909

(20)

As 119901 gt 119898 gt 1 making use of Holderrsquos and Youngrsquos inequalitywe have

int

Ω

119906119898120593119889119909

le (int

Ω

119906120593119889119909)

(119901minus119898)(119901minus1)

(int

Ω

119906119901120593119889119909)

(119898minus1)(119901minus1)

le

119901 minus 119898

119901 minus 1

int

Ω

119906120593119889119909 +

119898 minus 1

119901 minus 1

int

Ω

119906119901120593119889119909

int

Ω

119906119901120593119889119909 ge (int

Ω

119906120593119889119909)

119901

(21)

Applying (19)ndash(21) we obtain the inequality

1198651015840(119905) ge minus119860

1119865 (119905) + 119860

2119865119901(119905) (22)

where1198601= 1205821((119901minus119898)(119901minus1)) 119860

2= 119896minus120582

1((119898minus1)(119901minus1)) gt

0Choosing 119906

0(119909) sufficiently large such that

1198602119865119901minus1

(0) minus 1198601gt 0 (23)

we conclude that 119865(119905) is increasing monotonously for any 119905 gt0

Moreover according to (22) and119901 gt 1 we can obtain thatthere exists 119879

1gt 0 such that

lim119905rarr119879

minus

1

119865 (119905) = +infin

1198791le int

+infin

119865(0)

119889120591

minus1198601120591 + 119860

2120591119901lt +infin

(24)

where 119865(0) = intΩ1199060120593119889119909

Case 2 (119901 = 119898 gt 1) According to (1) and (19) we can obtain

1198651015840(119905) = int

Ω

[Δ119906119898+ 119891 (119905) 119892 (119906)] 120593 119889119909

ge (119896 minus 1205821) int

Ω

119906119901120593119889119909

(25)


Moreover applying 119901 gt 1 we have

int

Ω

119906119901120593119889119909 ge (int

Ω

119906120593119889119909)

119901

(26)

Combining (25) and (26) we obtain the inequality

1198651015840(119905) ge 119860

3119865119901(119905) (27)

where 1198603= 119896 minus 120582

1gt 0

Choosing 1199060(119909) sufficiently large yields

119865 (0) gt 0 (28)

Thus we conclude that 119865(119905) is increasing monotonously forany 119905 gt 0

Furthermore according to 119901 gt 1 and (27) we can obtainthat there exists 119879

2gt 0 such that

lim119905rarr119879

minus

2

119865 (119905) = +infin

1198792le int

+infin

119865(0)

119889120591

1198603120591119901lt +infin

(29)

Hence denote 1198790= min119879

1 1198792 which depends on

1199060(119909) 119898 119891 and 119892 andTheorem 3 is completed

5 Lower Bound for Blow-Up Time

In this section we will give the lower bound to the blow-uptime as long as blow-up occurs and showTheorem 4

Proof Firstly according to Theorem 1 we have 119901 ge 119898 underthe conditions of Theorem 4

Secondly set

Φ (119905) = int

Ω

1199062119901+119898+1

119889119909 (30)

We compute

Φ1015840(119905)

= (2119901 + 119898 + 1)int

Ω

1199062119901+119898

[Δ119906119898+ 119891 (119905) 119892 (119906)] 119889119909

= minus

(2119901 + 119898 + 1) (2119901 + 119898)

(119901 + 119898)2

int

Ω

10038161003816100381610038161003816nabla119906(119901+119898)10038161003816

100381610038161003816

2

119889119909

+ (2119901 + 119898 + 1) 119891 (119905) int

Ω

1199062119901+119898

119892 (119906) 119889119909

le minus

(2119901 + 119898 + 1) (2119901 + 119898)

(119901 + 119898)2

int

Ω

10038161003816100381610038161003816nabla119906(119901+119898)10038161003816

100381610038161003816

2

119889119909

+ (2119901 + 119898 + 1) 119891 (119905) int

Ω

1199062119901+119898+119901

119889119909

(31)

Clearly making use of Holderrsquos inequality we get

119891 (119905) int

Ω

1199062119901+119898+119901

119889119909 le 119891 (119905) (int

Ω

1199062(2119901+119898)

119889119909)

12

sdot (int

Ω

1199062119901119889119909)

12

le 119891 (119905)

sdot (int

Ω

1199066(119901+119898)

119889119909)

(2119901+119898minus1)2(4119901+5119898minus1)

sdot (int

Ω

1199062119901+119898+1

119889119909)

(119901+2119898)(4119901+5119898minus1)

(int

Ω

1199062119901119889119909)

12

(32)

Applying Sobolevrsquos inequality (see [1]) in R3

(int

Ω

1205776119889119909)

16

le 4133minus12

120587minus23

(int

Ω

1003816100381610038161003816nabla1205771003816100381610038161003816

2

119889119909)

12

(33)

we can obtain

int

Ω

1199066(119901+119898)

119889119909 le

16

271205874(int

Ω

10038161003816100381610038161003816nabla119906(119901+119898)10038161003816

100381610038161003816

2

119889119909)

3

(34)

Substituting (23) into (22) yields

119891 (119905) int

Ω

1199062119901+119898+119901

119889119909 le 119891 (119905) (

23

radic2

31205873

radic120587

sdot 120576 int

Ω

10038161003816100381610038161003816nabla119906(119901+119898)10038161003816

100381610038161003816

2

119889119909)

3(2119901+119898minus1)2(4119901+5119898minus1)

sdot 120576minus3(2119901+119898minus1)(2119901+7119898+1)

(int

Ω

1199062119901+119898+1

119889119909)

(2119901+4119898)(2119901+7119898+1)

sdot (int

Ω

1199062119901119889119909)

(4119901+5119898minus1)(2119901+7119898+1)

(2119901+7119898+1)2(4119901+5119898minus1)

le

3

radic2 (2119901 + 119898 minus 1)

1205873

radic120587 (4119901 + 5119898 minus 1)

120576 int

Ω

10038161003816100381610038161003816nabla119906(119901+119898)10038161003816

100381610038161003816

2

119889119909

+

2119901 + 7119898 + 1

2 (4119901 + 5119898 minus 1)

sdot 120576minus3(2119901+119898minus1)(2119901+7119898+1)

(int

Ω

1199062119901+119898+1

119889119909)

(2119901+4119898)(2119901+7119898+1)

sdot [1198912(119905) int

Ω

1199062119901119889119909]

(4119901+5119898minus1)(2119901+7119898+1)

le

3

radic2 (2119901 + 119898 minus 1)

1205873

radic120587 (4119901 + 5119898 minus 1)

120576 int

Ω

10038161003816100381610038161003816nabla119906(119901+119898)10038161003816

100381610038161003816

2

119889119909

+

2119901 + 7119898 + 1

2 (4119901 + 5119898 minus 1)

sdot 120576minus3(2119901+119898minus1)(2119901+7119898+1)

(int

Ω

1199062119901+119898+1

119889119909)

(2119901+4119898)(2119901+7119898+1)

sdot 119891 (119905)2(4119901+5119898minus1)(2119901+7119898+1)

sdot (int

Ω

1199062119901+119898+1

119889119909)

2119901(4119901+5119898minus1)(2119901+7119898+1)(2119901+119898+1)

sdot |Ω|1minus2119901(2119901+119898+1)

(35)

where 120576 = 120587 3radic120587(2119901+119898)(4119901+5119898minus1) 3radic2(119901+119898)2(2119901+119898minus1)


Substituting (35) into (31) we obtain

Φ1015840(119905) le 119896 (119905) (int

Ω

1199062119901+119898+1

119889119909)

1205831

+1205832

= 119896 (119905)Φ (119905)1205831

+1205832

(36)

where

119896 (119905) = (2119901 + 119898 + 1)

2119901 + 7119898 + 1

2 (4119901 + 5119898 minus 1)

sdot 120576minus3(2119901+119898minus1)(2119901+7119898+1)

|Ω|1minus2119901(2119901+119898+1)

sdot 119891 (119905)2(4119901+5119898minus1)(2119901+7119898+1)

gt 0

1205831=

2119901 + 4119898

2119901 + 7119898 + 1

gt 0

1205832=

2119901 (4119901 + 5119898 minus 1)

(2119901 + 119898 + 1) (2119901 + 7119898 + 1)

gt 0

(37)

We suppose that

lim119905rarr119879

minus

Φ (119905) = +infin (0 lt 119879 lt +infin) (38)

and then there exists 1199051gt 0 such thatΦ(119905) gt 1 (119905 gt 119905

1)

Integrating (36) from 1199051to 119879 we obtain

int

+infin

1

119889120591

1205911205831

+1205832

le int

119879

1199051

119896 (119905) 119889119905 le int

119879

0

119896 (119905) 119889119905 (39)

Let Θ(119879) = int

119879

0119896(119905)119889119905 Obviously Θ(119879) is increasing

monotonously So we can obtain

119879 ge Θminus1(120579)fl119879

0gt 0 (40)

where 120579 = int+infin1

(1198891205911205911205831

+1205832

) Θminus1 is the inverse function of Θand 119879

0depends on 119906

0(119909) 119898 119891 and 119892

Thus we complete the proof of Theorem 4

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgments

The authors would like to thank the reviewers for carefulreading and giving many helpful comments to improve thepaper This work is supported in part by NSFC Grant (no61563044) the Science and Technology Major Project ofQinghai Province Natural Science Foundation (no 2015-SF-A4-3) and SRFDP (no 20100181110036)

References

[1] C Bandle and H Brunner ldquoBlow-up in diffusion equationsrdquoJournal of Computational and Applied Mathematics vol 97 no1-2 pp 3ndash22 1998

[2] F C Li and C H Xie ldquoGlobal existence and blow-up fora nonlinear porous medium equationrdquo Applied MathematicsLetters vol 16 no 2 pp 185ndash192 2003

[3] P Quittner and P Souplet Superlinear Parabolic ProblemsBlow-up Global Existence and Steady States Birkhauser BaselSwitzerland 2007

[4] F B Weissler ldquoExistence and non-existence of global solutionsfor a semilinear heat equationrdquo Israel Journal of Mathematicsvol 38 no 1-2 pp 29ndash40 1981

[5] L E Payne and P W Schaefer ldquoLower bounds for blow-uptime in parabolic problems under Dirichlet conditionsrdquo Journalof Mathematical Analysis and Applications vol 328 no 2 pp1196ndash1205 2007

[6] L E Payne and G A Philippin ldquoBlow-up phenomena inparabolic problems with time dependent coefficients underDirichlet boundary conditionsrdquo Proceedings of the AmericanMathematical Society vol 141 no 7 pp 2309ndash2318 2013

[7] L E Payne andGA Philippin ldquoBlow-up phenomena for a classof parabolic systems with time dependent coefficientsrdquo AppliedMathematics vol 3 no 4 pp 325ndash330 2012

[8] W B Deng ldquoGlobal existence and finite time blow up fora degenerate reaction-diffusion systemrdquo Nonlinear AnalysisTheoryMethodsampApplications vol 60 no 5 pp 977ndash991 2005

[9] L L Du C L Mu and M S Fan ldquoGlobal existence and non-existence for a quasilinear degenerate parabolic system withnon-local sourcerdquo Dynamical Systems vol 20 no 4 pp 401ndash412 2005

[10] L L Du ldquoBlow-up for a degenerate reaction-diffusion systemwith nonlinear localized sourcesrdquo Journal of MathematicalAnalysis and Applications vol 324 no 1 pp 304ndash320 2006

[11] L L Du ldquoBlow-up for a degenerate reaction-diffusion systemwith nonlinear nonlocal sourcesrdquo Journal of Computational andApplied Mathematics vol 202 no 2 pp 237ndash247 2007

[12] L L Du and C L Mu ldquoGlobal existence and blow-up analysisto a degenerate reaction-diffusion system with nonlinear mem-oryrdquo Nonlinear Analysis Real World Applications vol 9 no 2pp 303ndash315 2008

[13] L L Du and Z Y Xiang ldquoA further blow-up analysis for alocalized porous medium equationrdquo Applied Mathematics andComputation vol 179 no 1 pp 200ndash208 2006

[14] L L Du and Z-A Yao ldquoLocalization of blow-up points for anonlinear nonlocal porousmediumequationrdquoCommunicationson Pure and Applied Analysis vol 6 no 1 pp 183ndash190 2007

[15] M S Fan C LMu and L L Du ldquoUniform blow-up profiles fora nonlocal degenerate parabolic systemrdquo Applied MathematicalSciences vol 1 no 1ndash4 pp 13ndash23 2007

[16] E DiBenedetto Degenerate Parabolic Equations UniversitextSpringer New York NY USA 1993

Submit your manuscripts athttpwwwhindawicom

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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of


Secondly we give the blow-up results in the next theorem

Theorem 2 Suppose that there exists a positive constant 119901with 119901 ge 119898 such that 119892(119906) ge 119906119901 for 119906 gt 0 and 119896flinf119891(119905) gt 0then the classical solution to problem (1) blows up in finite timeprovided that the initiate data are sufficiently large

Theorem 3 Suppose that there exists a positive constant 119901with 119901 ge 119898 such that 119892(119906) ge 119906119901 for 119906 gt 0 and 119896flinf119891(119905) gt1205821 then the classical solution to problem (1) blows up in finite

time 119879 for large data 1199060(119909) where 120582

1is the first eigenvalue to

the following problem

Δ120593 + 120582120593 = 0 119909 isin Ω

120593 = 0 119909 isin 120597Ω

(2)

with 120593 gt 0 (119909 isin Ω) and intΩ120593119889119909 = 1 Moreover there exists a

constant1198790 which depends on119898119891 119892 119906

0(119909) such that119879 le 119879

0

Furthermore we give the following estimates to themaximal blow-up time 119879

Theorem 4 Suppose that Ω be a convex domain in R3 andthere exists a constant 119901 gt 0 such that 119892(119906) le 119906

119901 119906 gt 0 Ifthe solution to problem (1) blows up in finite time 119879 then thereexists a positive constant 119879

0 which depends on 119898119891 119892 119906

0(119909)

such that

119879 ge 1198790 (3)

Remark 5 We would like to mention that the results inTheorem 4 are still valid for two-dimensional case

The remainder of this paper is organized as followsGlobal existence of the solution to problem (1) is establishedin Section 2 by constructing some global upper solutionIn Sections 3 and 4 we show that the classical solution toproblem (1) will blow up in finite time and obtain upperbound of the blow-up time In the last section with aid of adifferential inequality we will establish lower estimate to themaximal blow-up time

2 Global Solution for Problem (1)

In this section we focus on the global solution of (1) and showTheorem 1

Proof Obviously if 0 lt 119901 lt 119898 then there exists a positiveconstant 120590 such that 0 lt 119898120590 lt 1 119898120590 minus 119901120590 gt 0

Next we construct a supersolution which is bounded forany 119879 gt 0 Let 120601(119909) be the solution of the following ellipticproblem

minusΔ120601 (119909) = 1 119909 isin Ω

120601 (119909) = 0 119909 isin 120597Ω

(4)

Denote 119862 = maxΩ120601(119909) Namely 0 le 120601(119909) le 119862

We define the function 119906(119909 119905) as

119906 = (119870 (120601 (119909) + 1))120590

(5)

where 120590 gt 0 satisfy 119898120590 lt 1 and 119870 gt 0 will be fixed laterClearly 119906 is bounded for any 119905 gt 0 Thus we have

119906119905minus Δ119906119898

= minus119898120590119870119898120590(119898120590 minus 1) [1 + 120601 (119909)]

119898120590minus2 1003816100381610038161003816nabla120601 (119909)

1003816100381610038161003816

2

+ [1 + 120601 (119909)]119898120590minus1

Δ120601 (119909) ge minus119898120590119870119898120590[1

+ 120601 (119909)]119898120590minus1

Δ120601 (119909) = 119898120590119870119898120590[1 + 120601 (119909)]

119898120590minus1

ge 119898120590119870119898120590(1 + 119862)

119898120590minus1

119891 (119905) 119892 (119906) le 119896119906119901le 119896 [119870 (1 + 120601 (119909))]

119901120590

le 119896 [119870 (1 + 119862)]119901120590

(6)

Denote

1198701= [

119896

119898120590

(1 + 119862)119901120590minus119898120590+1

]

1(119898120590minus119901120590)

(7)

As119898120590 minus 119901120590 gt 0 we can choose 119870 sufficiently large that 119870 gt

1198701and

[119870 (120601 (119909) + 1)]120590

ge 1199060(119909) (8)

Now it follows from (6)ndash(8) that 119906 defined by (5) is a positivesupersolution of (1) Hence 119906 le 119906 by comparison principlewhich implies 119906 exists globally

3 Blow-Up Solution to Problem (1)

In this section we will discuss the blow-up solution of (1)under some appropriate hypotheses and showTheorem 2

Proof Our strategy here is to construct blow-up subsolutionsin some subdomain of Ω in which 119906 gt 0 Some ideas areborrowed from the work [11] by Du

Let120595(119909) be a nontrivial nonnegative continuous functionand vanish on 120597Ω Without loss of generality we assume that0 isin Ω and 120595(0) gt 0 We will construct a blow-up subsolutionto complete the proof

Set

119906 (119909 119905) =

1

(119879 minus 119905)1198971205961119898

(

|119909|

(119879 minus 119905)120583) (9)

with

120596 (119903) =

1198773

12

minus

119877

4

1199032+

1

6

1199033 119903 =

|119909|

(119879 minus 119905)120583 0 le 119903 le 119877 (10)

where 119897 120583 and119879 gt 0 are determined later Clearly 0 le 120596(119903) le119877312 and 120596(119903) is nonincreasing since 1205961015840(119903) = 119903(119903minus119877)2 le 0

Note that

supp 119906 (sdot 119905) = 119861 (0 119877 (119879 minus 119905)120583) sub 119861 (0 119877119879120583) sub Ω (11)

for sufficiently small 119879 gt 0 Obviously 119906 becomesunbounded as 119905 rarr 119879

minus at the point 119909 = 0



119906119905(119909 119905) minus Δ119906

119898(119909 119905)

=

1198981198971205961119898

(119903) + 1205831199031205961015840(119903) 120596(1minus119898)119898

(119903)

119898 (119879 minus 119905)119897+1

+

119877 minus 2119903

2 (119879 minus 119905)119898119897+2120583

+

(119873 minus 1) (119877 minus 119903)

2 (119879 minus 119905)119898119897+120583

le

119897 (119877312)

1119898

(119879 minus 119905)119897+1

+

119873119877 minus (119873 + 1) 119903

2 (119879 minus 119905)119898119897+2120583

(12)



1)119877312(119873 + 1)

3 then

119891 (119905) 119892 (119906) ge 119891 (119905) 119906119901(119909 119905) = 119891 (119905)

1

(119879 minus 119905)119901119897120596119901119898

(119903)

ge

119896

(119879 minus 119905)119901119897(

(3119873 + 1) 1198773

12 (119873 + 1)3)

119901119898

(13)

Hence119906119905(119909 119905) minus Δ119906

119898(119909 119905) minus 119891 (119905) 119892 (119906)

le

119897 (119877312)

1119898

(119879 minus 119905)119897+1

minus

119896

(119879 minus 119905)119901119897(

(3119873 + 1) 1198773

12 (119873 + 1)3)

119901119898

(14)

Case 2 If119873119877(119873 + 1) lt 119903 le 119877 then

119906119905(119909 119905) minus Δ119906

119898(119909 119905) minus 119891 (119905) 119892 (119906)

le

119897 (119877312)

1119898

(119879 minus 119905)119897+1

+

119873119877 minus (119873 + 1) 119903

2 (119879 minus 119905)119898119897+2120583

(15)


(119898 minus 1) 119897 gt 1 (16)

Thus we get

119901119897 ge 119898119897 gt 119897 + 1

119898119897 + 2120583 gt 119897 + 1 (forall120583 gt 0)

(17)


119906119905(119909 119905) minus Δ119906

119898(119909 119905) minus 119891 (119905) 119892 (119906) le 0 (18)





0(119909) ge 119872120595(119909) which




Proof Denote

119865 (119905) = int

Ω

119906120593119889119909 (19)



1198651015840(119905) = int

Ω

[Δ119906119898+ 119891 (119905) 119892 (119906)] 120593 119889119909

ge minus1205821int

Ω

119906119898120593119889119909 + 119896int

Ω

119906119901120593119889119909

(20)


int

Ω

119906119898120593119889119909

le (int

Ω

119906120593119889119909)

(119901minus119898)(119901minus1)

(int

Ω

119906119901120593119889119909)

(119898minus1)(119901minus1)

le

119901 minus 119898

119901 minus 1

int

Ω

119906120593119889119909 +

119898 minus 1

119901 minus 1

int

Ω

119906119901120593119889119909

int

Ω

119906119901120593119889119909 ge (int

Ω

119906120593119889119909)

119901

(21)


1198651015840(119905) ge minus119860

1119865 (119905) + 119860

2119865119901(119905) (22)


2= 119896minus120582

1((119898minus1)(119901minus1)) gt

0Choosing 119906


1198602119865119901minus1

(0) minus 1198601gt 0 (23)



1gt 0 such that

lim119905rarr119879

minus

1

119865 (119905) = +infin

1198791le int

+infin

119865(0)

119889120591

minus1198601120591 + 119860

2120591119901lt +infin

(24)

where 119865(0) = intΩ1199060120593119889119909


1198651015840(119905) = int

Ω

[Δ119906119898+ 119891 (119905) 119892 (119906)] 120593 119889119909

ge (119896 minus 1205821) int

Ω

119906119901120593119889119909

(25)



int

Ω

119906119901120593119889119909 ge (int

Ω

119906120593119889119909)

119901

(26)


1198651015840(119905) ge 119860

3119865119901(119905) (27)

where 1198603= 119896 minus 120582

1gt 0


119865 (0) gt 0 (28)



2gt 0 such that

lim119905rarr119879

minus

2

119865 (119905) = +infin

1198792le int

+infin

119865(0)

119889120591

1198603120591119901lt +infin

(29)







Secondly set

Φ (119905) = int

Ω

1199062119901+119898+1

119889119909 (30)

We compute

Φ1015840(119905)

= (2119901 + 119898 + 1)int

Ω

1199062119901+119898

[Δ119906119898+ 119891 (119905) 119892 (119906)] 119889119909

= minus

(2119901 + 119898 + 1) (2119901 + 119898)

(119901 + 119898)2

int

Ω

10038161003816100381610038161003816nabla119906(119901+119898)10038161003816

100381610038161003816

2

119889119909

+ (2119901 + 119898 + 1) 119891 (119905) int

Ω

1199062119901+119898

119892 (119906) 119889119909

le minus

(2119901 + 119898 + 1) (2119901 + 119898)

(119901 + 119898)2

int

Ω

10038161003816100381610038161003816nabla119906(119901+119898)10038161003816

100381610038161003816

2

119889119909

+ (2119901 + 119898 + 1) 119891 (119905) int

Ω

1199062119901+119898+119901

119889119909

(31)


119891 (119905) int

Ω

1199062119901+119898+119901

119889119909 le 119891 (119905) (int

Ω

1199062(2119901+119898)

119889119909)

12

sdot (int

Ω

1199062119901119889119909)

12

le 119891 (119905)

sdot (int

Ω

1199066(119901+119898)

119889119909)

(2119901+119898minus1)2(4119901+5119898minus1)

sdot (int

Ω

1199062119901+119898+1

119889119909)

(119901+2119898)(4119901+5119898minus1)

(int

Ω

1199062119901119889119909)

12

(32)


(int

Ω

1205776119889119909)

16

le 4133minus12

120587minus23

(int

Ω

1003816100381610038161003816nabla1205771003816100381610038161003816

2

119889119909)

12

(33)

we can obtain

int

Ω

1199066(119901+119898)

119889119909 le

16

271205874(int

Ω

10038161003816100381610038161003816nabla119906(119901+119898)10038161003816

100381610038161003816

2

119889119909)

3

(34)


119891 (119905) int

Ω

1199062119901+119898+119901

119889119909 le 119891 (119905) (

23

radic2

31205873

radic120587

sdot 120576 int

Ω

10038161003816100381610038161003816nabla119906(119901+119898)10038161003816

100381610038161003816

2

119889119909)

3(2119901+119898minus1)2(4119901+5119898minus1)

sdot 120576minus3(2119901+119898minus1)(2119901+7119898+1)

(int

Ω

1199062119901+119898+1

119889119909)

(2119901+4119898)(2119901+7119898+1)

sdot (int

Ω

1199062119901119889119909)

(4119901+5119898minus1)(2119901+7119898+1)

(2119901+7119898+1)2(4119901+5119898minus1)

le

3

radic2 (2119901 + 119898 minus 1)

1205873

radic120587 (4119901 + 5119898 minus 1)

120576 int

Ω

10038161003816100381610038161003816nabla119906(119901+119898)10038161003816

100381610038161003816

2

119889119909

+

2119901 + 7119898 + 1

2 (4119901 + 5119898 minus 1)

sdot 120576minus3(2119901+119898minus1)(2119901+7119898+1)

(int

Ω

1199062119901+119898+1

119889119909)

(2119901+4119898)(2119901+7119898+1)

sdot [1198912(119905) int

Ω

1199062119901119889119909]

(4119901+5119898minus1)(2119901+7119898+1)

le

3

radic2 (2119901 + 119898 minus 1)

1205873

radic120587 (4119901 + 5119898 minus 1)

120576 int

Ω

10038161003816100381610038161003816nabla119906(119901+119898)10038161003816

100381610038161003816

2

119889119909

+

2119901 + 7119898 + 1

2 (4119901 + 5119898 minus 1)

sdot 120576minus3(2119901+119898minus1)(2119901+7119898+1)

(int

Ω

1199062119901+119898+1

119889119909)

(2119901+4119898)(2119901+7119898+1)

sdot 119891 (119905)2(4119901+5119898minus1)(2119901+7119898+1)

sdot (int

Ω

1199062119901+119898+1

119889119909)

2119901(4119901+5119898minus1)(2119901+7119898+1)(2119901+119898+1)

sdot |Ω|1minus2119901(2119901+119898+1)

(35)




Φ1015840(119905) le 119896 (119905) (int

Ω

1199062119901+119898+1

119889119909)

1205831

+1205832

= 119896 (119905)Φ (119905)1205831

+1205832

(36)

where

119896 (119905) = (2119901 + 119898 + 1)

2119901 + 7119898 + 1

2 (4119901 + 5119898 minus 1)

sdot 120576minus3(2119901+119898minus1)(2119901+7119898+1)

|Ω|1minus2119901(2119901+119898+1)

sdot 119891 (119905)2(4119901+5119898minus1)(2119901+7119898+1)

gt 0

1205831=

2119901 + 4119898

2119901 + 7119898 + 1

gt 0

1205832=

2119901 (4119901 + 5119898 minus 1)

(2119901 + 119898 + 1) (2119901 + 7119898 + 1)

gt 0

(37)

We suppose that

lim119905rarr119879

minus



1)


int

+infin

1

119889120591

1205911205831

+1205832

le int

119879

1199051

119896 (119905) 119889119905 le int

119879

0

119896 (119905) 119889119905 (39)

Let Θ(119879) = int

119879



119879 ge Θminus1(120579)fl119879

0gt 0 (40)


(1198891205911205911205831

+1205832


0depends on 119906

0(119909) 119898 119891 and 119892




Acknowledgments


References
























Volume 2014




Journal of











Journal of


Function Spaces






Algebra











119906119905(119909 119905) minus Δ119906

119898(119909 119905)

=

1198981198971205961119898

(119903) + 1205831199031205961015840(119903) 120596(1minus119898)119898

(119903)

119898 (119879 minus 119905)119897+1

+

119877 minus 2119903

2 (119879 minus 119905)119898119897+2120583

+

(119873 minus 1) (119877 minus 119903)

2 (119879 minus 119905)119898119897+120583

le

119897 (119877312)

1119898

(119879 minus 119905)119897+1

+

119873119877 minus (119873 + 1) 119903

2 (119879 minus 119905)119898119897+2120583

(12)



1)119877312(119873 + 1)

3 then

119891 (119905) 119892 (119906) ge 119891 (119905) 119906119901(119909 119905) = 119891 (119905)

1

(119879 minus 119905)119901119897120596119901119898

(119903)

ge

119896

(119879 minus 119905)119901119897(

(3119873 + 1) 1198773

12 (119873 + 1)3)

119901119898

(13)

Hence119906119905(119909 119905) minus Δ119906

119898(119909 119905) minus 119891 (119905) 119892 (119906)

le

119897 (119877312)

1119898

(119879 minus 119905)119897+1

minus

119896

(119879 minus 119905)119901119897(

(3119873 + 1) 1198773

12 (119873 + 1)3)

119901119898

(14)

Case 2 If119873119877(119873 + 1) lt 119903 le 119877 then

119906119905(119909 119905) minus Δ119906

119898(119909 119905) minus 119891 (119905) 119892 (119906)

le

119897 (119877312)

1119898

(119879 minus 119905)119897+1

+

119873119877 minus (119873 + 1) 119903

2 (119879 minus 119905)119898119897+2120583

(15)


(119898 minus 1) 119897 gt 1 (16)

Thus we get

119901119897 ge 119898119897 gt 119897 + 1

119898119897 + 2120583 gt 119897 + 1 (forall120583 gt 0)

(17)


119906119905(119909 119905) minus Δ119906

119898(119909 119905) minus 119891 (119905) 119892 (119906) le 0 (18)





0(119909) ge 119872120595(119909) which




Proof Denote

119865 (119905) = int

Ω

119906120593119889119909 (19)



1198651015840(119905) = int

Ω

[Δ119906119898+ 119891 (119905) 119892 (119906)] 120593 119889119909

ge minus1205821int

Ω

119906119898120593119889119909 + 119896int

Ω

119906119901120593119889119909

(20)


int

Ω

119906119898120593119889119909

le (int

Ω

119906120593119889119909)

(119901minus119898)(119901minus1)

(int

Ω

119906119901120593119889119909)

(119898minus1)(119901minus1)

le

119901 minus 119898

119901 minus 1

int

Ω

119906120593119889119909 +

119898 minus 1

119901 minus 1

int

Ω

119906119901120593119889119909

int

Ω

119906119901120593119889119909 ge (int

Ω

119906120593119889119909)

119901

(21)


1198651015840(119905) ge minus119860

1119865 (119905) + 119860

2119865119901(119905) (22)


2= 119896minus120582

1((119898minus1)(119901minus1)) gt

0Choosing 119906


1198602119865119901minus1

(0) minus 1198601gt 0 (23)



1gt 0 such that

lim119905rarr119879

minus

1

119865 (119905) = +infin

1198791le int

+infin

119865(0)

119889120591

minus1198601120591 + 119860

2120591119901lt +infin

(24)

where 119865(0) = intΩ1199060120593119889119909


1198651015840(119905) = int

Ω

[Δ119906119898+ 119891 (119905) 119892 (119906)] 120593 119889119909

ge (119896 minus 1205821) int

Ω

119906119901120593119889119909

(25)



int

Ω

119906119901120593119889119909 ge (int

Ω

119906120593119889119909)

119901

(26)


1198651015840(119905) ge 119860

3119865119901(119905) (27)

where 1198603= 119896 minus 120582

1gt 0


119865 (0) gt 0 (28)



2gt 0 such that

lim119905rarr119879

minus

2

119865 (119905) = +infin

1198792le int

+infin

119865(0)

119889120591

1198603120591119901lt +infin

(29)







Secondly set

Φ (119905) = int

Ω

1199062119901+119898+1

119889119909 (30)

We compute

Φ1015840(119905)

= (2119901 + 119898 + 1)int

Ω

1199062119901+119898

[Δ119906119898+ 119891 (119905) 119892 (119906)] 119889119909

= minus

(2119901 + 119898 + 1) (2119901 + 119898)

(119901 + 119898)2

int

Ω

10038161003816100381610038161003816nabla119906(119901+119898)10038161003816

100381610038161003816

2

119889119909

+ (2119901 + 119898 + 1) 119891 (119905) int

Ω

1199062119901+119898

119892 (119906) 119889119909

le minus

(2119901 + 119898 + 1) (2119901 + 119898)

(119901 + 119898)2

int

Ω

10038161003816100381610038161003816nabla119906(119901+119898)10038161003816

100381610038161003816

2

119889119909

+ (2119901 + 119898 + 1) 119891 (119905) int

Ω

1199062119901+119898+119901

119889119909

(31)


119891 (119905) int

Ω

1199062119901+119898+119901

119889119909 le 119891 (119905) (int

Ω

1199062(2119901+119898)

119889119909)

12

sdot (int

Ω

1199062119901119889119909)

12

le 119891 (119905)

sdot (int

Ω

1199066(119901+119898)

119889119909)

(2119901+119898minus1)2(4119901+5119898minus1)

sdot (int

Ω

1199062119901+119898+1

119889119909)

(119901+2119898)(4119901+5119898minus1)

(int

Ω

1199062119901119889119909)

12

(32)


(int

Ω

1205776119889119909)

16

le 4133minus12

120587minus23

(int

Ω

1003816100381610038161003816nabla1205771003816100381610038161003816

2

119889119909)

12

(33)

we can obtain

int

Ω

1199066(119901+119898)

119889119909 le

16

271205874(int

Ω

10038161003816100381610038161003816nabla119906(119901+119898)10038161003816

100381610038161003816

2

119889119909)

3

(34)


119891 (119905) int

Ω

1199062119901+119898+119901

119889119909 le 119891 (119905) (

23

radic2

31205873

radic120587

sdot 120576 int

Ω

10038161003816100381610038161003816nabla119906(119901+119898)10038161003816

100381610038161003816

2

119889119909)

3(2119901+119898minus1)2(4119901+5119898minus1)

sdot 120576minus3(2119901+119898minus1)(2119901+7119898+1)

(int

Ω

1199062119901+119898+1

119889119909)

(2119901+4119898)(2119901+7119898+1)

sdot (int

Ω

1199062119901119889119909)

(4119901+5119898minus1)(2119901+7119898+1)

(2119901+7119898+1)2(4119901+5119898minus1)

le

3

radic2 (2119901 + 119898 minus 1)

1205873

radic120587 (4119901 + 5119898 minus 1)

120576 int

Ω

10038161003816100381610038161003816nabla119906(119901+119898)10038161003816

100381610038161003816

2

119889119909

+

2119901 + 7119898 + 1

2 (4119901 + 5119898 minus 1)

sdot 120576minus3(2119901+119898minus1)(2119901+7119898+1)

(int

Ω

1199062119901+119898+1

119889119909)

(2119901+4119898)(2119901+7119898+1)

sdot [1198912(119905) int

Ω

1199062119901119889119909]

(4119901+5119898minus1)(2119901+7119898+1)

le

3

radic2 (2119901 + 119898 minus 1)

1205873

radic120587 (4119901 + 5119898 minus 1)

120576 int

Ω

10038161003816100381610038161003816nabla119906(119901+119898)10038161003816

100381610038161003816

2

119889119909

+

2119901 + 7119898 + 1

2 (4119901 + 5119898 minus 1)

sdot 120576minus3(2119901+119898minus1)(2119901+7119898+1)

(int

Ω

1199062119901+119898+1

119889119909)

(2119901+4119898)(2119901+7119898+1)

sdot 119891 (119905)2(4119901+5119898minus1)(2119901+7119898+1)

sdot (int

Ω

1199062119901+119898+1

119889119909)

2119901(4119901+5119898minus1)(2119901+7119898+1)(2119901+119898+1)

sdot |Ω|1minus2119901(2119901+119898+1)

(35)




Φ1015840(119905) le 119896 (119905) (int

Ω

1199062119901+119898+1

119889119909)

1205831

+1205832

= 119896 (119905)Φ (119905)1205831

+1205832

(36)

where

119896 (119905) = (2119901 + 119898 + 1)

2119901 + 7119898 + 1

2 (4119901 + 5119898 minus 1)

sdot 120576minus3(2119901+119898minus1)(2119901+7119898+1)

|Ω|1minus2119901(2119901+119898+1)

sdot 119891 (119905)2(4119901+5119898minus1)(2119901+7119898+1)

gt 0

1205831=

2119901 + 4119898

2119901 + 7119898 + 1

gt 0

1205832=

2119901 (4119901 + 5119898 minus 1)

(2119901 + 119898 + 1) (2119901 + 7119898 + 1)

gt 0

(37)

We suppose that

lim119905rarr119879

minus



1)


int

+infin

1

119889120591

1205911205831

+1205832

le int

119879

1199051

119896 (119905) 119889119905 le int

119879

0

119896 (119905) 119889119905 (39)

Let Θ(119879) = int

119879



119879 ge Θminus1(120579)fl119879

0gt 0 (40)


(1198891205911205911205831

+1205832


0depends on 119906

0(119909) 119898 119891 and 119892




Acknowledgments


References
























Volume 2014




Journal of











Journal of


Function Spaces






Algebra











int

Ω

119906119901120593119889119909 ge (int

Ω

119906120593119889119909)

119901

(26)


1198651015840(119905) ge 119860

3119865119901(119905) (27)

where 1198603= 119896 minus 120582

1gt 0


119865 (0) gt 0 (28)



2gt 0 such that

lim119905rarr119879

minus

2

119865 (119905) = +infin

1198792le int

+infin

119865(0)

119889120591

1198603120591119901lt +infin

(29)







Secondly set

Φ (119905) = int

Ω

1199062119901+119898+1

119889119909 (30)

We compute

Φ1015840(119905)

= (2119901 + 119898 + 1)int

Ω

1199062119901+119898

[Δ119906119898+ 119891 (119905) 119892 (119906)] 119889119909

= minus

(2119901 + 119898 + 1) (2119901 + 119898)

(119901 + 119898)2

int

Ω

10038161003816100381610038161003816nabla119906(119901+119898)10038161003816

100381610038161003816

2

119889119909

+ (2119901 + 119898 + 1) 119891 (119905) int

Ω

1199062119901+119898

119892 (119906) 119889119909

le minus

(2119901 + 119898 + 1) (2119901 + 119898)

(119901 + 119898)2

int

Ω

10038161003816100381610038161003816nabla119906(119901+119898)10038161003816

100381610038161003816

2

119889119909

+ (2119901 + 119898 + 1) 119891 (119905) int

Ω

1199062119901+119898+119901

119889119909

(31)


119891 (119905) int

Ω

1199062119901+119898+119901

119889119909 le 119891 (119905) (int

Ω

1199062(2119901+119898)

119889119909)

12

sdot (int

Ω

1199062119901119889119909)

12

le 119891 (119905)

sdot (int

Ω

1199066(119901+119898)

119889119909)

(2119901+119898minus1)2(4119901+5119898minus1)

sdot (int

Ω

1199062119901+119898+1

119889119909)

(119901+2119898)(4119901+5119898minus1)

(int

Ω

1199062119901119889119909)

12

(32)


(int

Ω

1205776119889119909)

16

le 4133minus12

120587minus23

(int

Ω

1003816100381610038161003816nabla1205771003816100381610038161003816

2

119889119909)

12

(33)

we can obtain

int

Ω

1199066(119901+119898)

119889119909 le

16

271205874(int

Ω

10038161003816100381610038161003816nabla119906(119901+119898)10038161003816

100381610038161003816

2

119889119909)

3

(34)


119891 (119905) int

Ω

1199062119901+119898+119901

119889119909 le 119891 (119905) (

23

radic2

31205873

radic120587

sdot 120576 int

Ω

10038161003816100381610038161003816nabla119906(119901+119898)10038161003816

100381610038161003816

2

119889119909)

3(2119901+119898minus1)2(4119901+5119898minus1)

sdot 120576minus3(2119901+119898minus1)(2119901+7119898+1)

(int

Ω

1199062119901+119898+1

119889119909)

(2119901+4119898)(2119901+7119898+1)

sdot (int

Ω

1199062119901119889119909)

(4119901+5119898minus1)(2119901+7119898+1)

(2119901+7119898+1)2(4119901+5119898minus1)

le

3

radic2 (2119901 + 119898 minus 1)

1205873

radic120587 (4119901 + 5119898 minus 1)

120576 int

Ω

10038161003816100381610038161003816nabla119906(119901+119898)10038161003816

100381610038161003816

2

119889119909

+

2119901 + 7119898 + 1

2 (4119901 + 5119898 minus 1)

sdot 120576minus3(2119901+119898minus1)(2119901+7119898+1)

(int

Ω

1199062119901+119898+1

119889119909)

(2119901+4119898)(2119901+7119898+1)

sdot [1198912(119905) int

Ω

1199062119901119889119909]

(4119901+5119898minus1)(2119901+7119898+1)

le

3

radic2 (2119901 + 119898 minus 1)

1205873

radic120587 (4119901 + 5119898 minus 1)

120576 int

Ω

10038161003816100381610038161003816nabla119906(119901+119898)10038161003816

100381610038161003816

2

119889119909

+

2119901 + 7119898 + 1

2 (4119901 + 5119898 minus 1)

sdot 120576minus3(2119901+119898minus1)(2119901+7119898+1)

(int

Ω

1199062119901+119898+1

119889119909)

(2119901+4119898)(2119901+7119898+1)

sdot 119891 (119905)2(4119901+5119898minus1)(2119901+7119898+1)

sdot (int

Ω

1199062119901+119898+1

119889119909)

2119901(4119901+5119898minus1)(2119901+7119898+1)(2119901+119898+1)

sdot |Ω|1minus2119901(2119901+119898+1)

(35)




Φ1015840(119905) le 119896 (119905) (int

Ω

1199062119901+119898+1

119889119909)

1205831

+1205832

= 119896 (119905)Φ (119905)1205831

+1205832

(36)

where

119896 (119905) = (2119901 + 119898 + 1)

2119901 + 7119898 + 1

2 (4119901 + 5119898 minus 1)

sdot 120576minus3(2119901+119898minus1)(2119901+7119898+1)

|Ω|1minus2119901(2119901+119898+1)

sdot 119891 (119905)2(4119901+5119898minus1)(2119901+7119898+1)

gt 0

1205831=

2119901 + 4119898

2119901 + 7119898 + 1

gt 0

1205832=

2119901 (4119901 + 5119898 minus 1)

(2119901 + 119898 + 1) (2119901 + 7119898 + 1)

gt 0

(37)

We suppose that

lim119905rarr119879

minus



1)


int

+infin

1

119889120591

1205911205831

+1205832

le int

119879

1199051

119896 (119905) 119889119905 le int

119879

0

119896 (119905) 119889119905 (39)

Let Θ(119879) = int

119879



119879 ge Θminus1(120579)fl119879

0gt 0 (40)


(1198891205911205911205831

+1205832


0depends on 119906

0(119909) 119898 119891 and 119892




Acknowledgments


References
























Volume 2014




Journal of











Journal of


Function Spaces






Algebra











Φ1015840(119905) le 119896 (119905) (int

Ω

1199062119901+119898+1

119889119909)

1205831

+1205832

= 119896 (119905)Φ (119905)1205831

+1205832

(36)

where

119896 (119905) = (2119901 + 119898 + 1)

2119901 + 7119898 + 1

2 (4119901 + 5119898 minus 1)

sdot 120576minus3(2119901+119898minus1)(2119901+7119898+1)

|Ω|1minus2119901(2119901+119898+1)

sdot 119891 (119905)2(4119901+5119898minus1)(2119901+7119898+1)

gt 0

1205831=

2119901 + 4119898

2119901 + 7119898 + 1

gt 0

1205832=

2119901 (4119901 + 5119898 minus 1)

(2119901 + 119898 + 1) (2119901 + 7119898 + 1)

gt 0

(37)

We suppose that

lim119905rarr119879

minus



1)


int

+infin

1

119889120591

1205911205831

+1205832

le int

119879

1199051

119896 (119905) 119889119905 le int

119879

0

119896 (119905) 119889119905 (39)

Let Θ(119879) = int

119879



119879 ge Θminus1(120579)fl119879

0gt 0 (40)


(1198891205911205911205831

+1205832


0depends on 119906

0(119909) 119898 119891 and 119892




Acknowledgments


References
























Volume 2014




Journal of











Journal of


Function Spaces






Algebra
















Volume 2014




Journal of











Journal of


Function Spaces






Algebra









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