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Applied Mathematics, 2010, 1, 351-356 doi:10.4236/am.2010.15046 Published Online November 2010 (http://www.SciRP.org/journal/am) Copyright © 2010 SciRes. AM Existence and Non-Existence Result for Singular Quasilinear Elliptic Equations* Mingzhu Wu1, Zuodong Yang1,2 1Institute of Mathematics, School of Mathematical Science, Nanjing Normal University, Nanjing, China 2College of Zhongbei, Nanjing Normal University, Nanjing, China E-mail: zdyang_jin@263.net Received June 1, 2010; revised August 11, 2010; accepted August 14, 2010 Abstract We prove the existence of a ground state solution for the qusilinear elliptic equation 2 (| |) p div uu =(,) f xu in N R, under suitable conditions on a locally Holder continuous non-linearity ),( txf , the non-linearity may exhibit a singularity as 0t . We also prove the non-existence of radially symmetric solutions to the singular elliptic equation 2 (||)=( )[( )( )||], pq divuudxg ur uu ()>0ux in N R, () 0ux as || ,x where ()=(||)(,(0, )), N dxdxC R 3, 0.Nq 10, (0,),(0,), ([0,), loc gC rC [0,)), 0<<1 . Keywords: Quasilinear Elliptic Equations, Existence, Non-Existence, Singular 1. Introduction In this paper, we are concerned with the existence of ground state solution or positive solution for the follow- ing problem 2 (||) =(,), >0, ()=0, p divuufx ux ux ux x (1) in which N R, () 0ux when ||x . Let :[0,)[0,)f be a locally Holder continuous function which may be singular at =0t. The problem (1) appears in the study of non-Newto- nian fluids [1,2] and non-Newtonian filtration [3]. The quantity p is a characteristic of the medium. Media with >2p are called dilatant fluids and those with <2p are called pseudoplastics. If =2p, they are Newtonian fluid. When 2p, the problem becomes more complicated since certain nice properties inherent to the case =2p seem to be lose or at least difficult to verify. The main differences between =2p and 2p can be found in [4-6]. In recent years the study of ground state solutions for =2p has received a lot of interest and gets numerous existence results (see [7-12]). For p-Laplacian equa- tions, in most papers, the focus has been on separable nonlinearities like (2). 2 (||)=( )( ), >0, ()0,| | pN N divuubx g ux uxR uxas x R (2) We refer readers to the paper [13-19]. In this paper, we consider the situation of 1>p. In [20], the author extended results to the problem (1) for =2p where f is not necessarily separable. For 2p we can see [1-6]. Motivated by papers [4-6,20], we extend the results to >1p and get two theorems. But we still have many difficulties to get entire ground state solution of (1). The second purpose is to give a result for nonexistence of solution. To the best of our knowledge, there has been very less result for nonexistence of solution about singular elliptic equation. We solve an open problem in [13] for >1p, for the case when u is a radially sym- metric solution. The paper is organized as follows. In Section 2 we recall some facts and give many lemmas that will be needed in the paper. In Section 3, we give the proof of the main result of the paper. *Project Supported by the National Natural Science Foundation o f China (No.10871060); the Natural Science Foundation of Educational Department of Jiangsu Province (No.08KJB110005) M. Z. WU ET AL. Copyright © 2010 SciRes. AM 352 2. Preliminaries Firstly we list the following assumptions and results that needed below. 2 (||) =(), >0, ()0,| | pN N divWWbxx WxR Wx asx R (3) 1 []>0Bb is a locally Holder continuous function on N R. 1 11* 1 200 [](()) < t NN p Btsbsdsdt , where *()=max{ ():||=}btbx x t for >0t. 3 []B Problem (3) has a solution b W in N R when b satisfy some condition. It can be proved that condition 2 []B implies 3 []B. This follows from the observation that () 0vr as r where 1 11* 1 0 ()=(()) t NN p r vrtsb sdsdt and satisfies the following equation 2* (||) =(). p divvvb x Thus ()= (||)wxvx is a supersolution of (3). Since 0w, and zero is clearly a sub-solution it follows from standard results that (3) admits a solution b W such that 0< b Ww. The following eigenvalue problem 21 (||) =(), 0,| | ppN divc xx as x R (4) where N R is a bounded smooth domain, and (,(0,))cC for some 0< <1 . The first eigen- value of the problem (4) will be denoted by >0 . It is easily noted from the variational characterization of eigenvalues that 12 where12 are smooth bounded domains in N R. Consider :(0, )(0, )g satisfies the following conditions: 1 []Gg is 1 C. 2 ()1 [] < limsupt b gt GtW , where =() max N x bb WWx R. The nonlinearity f in problem (1) is assumed to be a real function that satisfies the following conditions: 1 [](,) F fxt is locally Holder continuous on 0, N R and continuously differentiable in the variable t . 2 [](,) ()() F fxsbxgs for all ,0, N xs R and some functions :0, N bR and :(0, )(0, )g . ][ 3 F There is a continuous function :0, N aR and some constants >0 and 1 > such that 1 ,,,0, pN fxsaxs xs R where 1 is the first eigenvalue of the problem (4) on the ball (0 , 1 )B of radius one and centered at the origin and where ()= ()cxax. Recall that the nonlinearity (,)fxt may exhibit sin- gularity as 0t . We will consider the following Dirichlet problem for a given smooth bounded domain N R. 2 (||) =(), >0, ()=0, p divWWbxx Wx Wxx (5) To establish the main theorem, from reference [2] we give the following lemmas. Lemma 1. (Weak Comparison Principle) Let be a bounded domain in 2 NNR with smooth boundary and :(0, )(0,) is continuous and nonde- creasing. Let 1, 12 ,()() p loc uu WC satisfy 2 11 2 1222 || ()||() p p uudx udxuudxu dx for all non-negative 1, () p loc W . Then the inequality 12 sup( ()())0 lim x uxu x implies that 12 .uuin Before Lemma 2, we give the following equation 2 (||)(,) = 0, p N divuufx ux R (6) Lemma 2. Suppose that (,) f xu is defined on 1N R and is locally Holder continuous (with exponent (0,1) ) in x . Suppose moreover that there exist functions 1 , N loc vw C R such that 2 2 (||)( ,)0 (||)( ,)0 p p divvvfx v divwwfx w and () ()vx wx and that (,) f xu is locally Lipschitz continuous in u on the set ,: , N x uxvx uwxR Then Equation (6) possesses an entire solution )(xu satisfying , N vx uxwxx R Lemma 3. Let b satisfy 1 []B, 3 []B and g sati- sfy both 1 []G, 2 []G. Then there is 1 N vCR such M. Z. WU ET AL. Copyright © 2010 SciRes. AM 353 that 21 (||)( )(()), >0,()0||. pp divvv bxgvx vvx asx Proof. Since g satisfies 2 []G, we define () ˆ():=sup:>, >0 gs gtst t s (7) Note that ˆ() g t is non-increasing, positive and 1 ˆ() () g tgtt . Furthermore, by 2 []G we have 1 ˆ()< b gtW for sufficiently large t. Let 2 2ˆ ()=( ),>0 t t htgsdst t (8) Then h is 1 C, non-increasing and ˆˆ ()()( ) 2 t gt ht g for all )(0, t. Since h is non- increasing, we note that 1 ()<b ht W as t for some [0,) . Now, let us set 0 1 ():=, >0 () t tdst hs (9) On using 1 ˆ()< b gtW in (8) for sufficiently large >0t, we see from (9) that ()> b W , (10) for a sufficiently large 1 . Let 1 = be the inverse function of . By direct calculation, we see that ()=( ())tht , ()>0t , for >0t and (0) = 0 . Furthermore ()=(())(()), >0.thth tt By condition 3 []B, we take a solution b W of (3) with N R. Let us now set ():= (()) b vxW x for all x. We note from (10) that ()= ()< bb vxWxW (11) A simple computation shows that v has the desired properties. Indeed, on recalling 2 (||) = p bb div WWb , we see that 2 2 11 2 (| |) =(1)()| | ()(||) p pp p b pp p bb div vv pW div WW 111 11 =(1)(( ())( ())||() () pp ppp b pp phthtW bhv bh v 111 1 ˆ ()()(). ppp p bvg vbgv v We have used (11) in the last inequality. Since (0) = 0 , it is clear that () 0vx as ||x. Since 1 , observe that the solution v constructed in Lemma 3 also satisfies 21 ˆ (||)( ). pp divvv bgv 1 10,, 0,GgC ; 201= limup gu Gu ; 31=0 limup gu Gu . Lemma 4. Assume that g satisfies 13 [][]GG. Then there exist functions 1 g , 1 g such that 1 1 1,0,,0,igg C ; 11 1 1p gs iig sg s s , >0s and 00 1 1== lim lim ss gs gs ; 1 1,iiigg are non-increasing on (0, ); 1 1()= ()=0 lim lim ss ivgsg s . Proof. 1 11 00 ==, limlim 1 1 p pp ss gsgs s s ss 1 11 ==0. limlim 1 1 p pp ss gsgs s s ss We set >0 1 =sup 1 ts p g t gs t , then we have 1,>0 1p gt g ssandts t and )(sg is non-increasing on )(0,. Moreover, 0 ==0. lim lim s s gsand gs Now we can assume 10,gC. If not, we can replace it by 1 2 2 =,>0 s s g sgtdt s s Obviously, 1,>0; 2 s gs gs gs and M. Z. WU ET AL. Copyright © 2010 SciRes. AM 354 12 2 21 2 '= 22 s s s g sgsg gtdt ss 2 21 21 )=( 0. 22 22 ss s gsggsgs g ss s 1 1 . .,((0,),(0,)).ie gC 1 >0 () ()=inf (1) p st g t gs t Observe that 1 () 0<( ),>0; (1) p gt gs s t and )( sg is non-increasing on )(0,. Moreover, 0 ()= ()=0. lim lim s s gsandgs Now we can assume 1(0, )gC. If not, we can replace it by 1 1 2 ()=() , >0 s s gs gtdts s Obviously, 1 (1)() (),>0gsg sgss and 1'()=(1)()0,> 0g sgsgss 1 1 . .,((0,),(0,)).ie gC 3. Proof of Main Theorems In this section, we prove our main results. Theorem 1. Let N R be a bounded smooth domain that contains (0 , 1 )B, the ball of radius one centered at the origin, and let f satisfies 1 [] F , 2 [] F and 3 [] F , where b satisfies 1 []B, 3 []B and g satisfies 1 []G, 2 []G. Then the problem 2 (||) =(,), ()=0, p divuufx ux ux x (12) has a positive solution u in 1,() ()CC such that u where is an eigenfunction of the eigenvalue problem (4) on with ()= ()cxax normalized such that 0< on . Here is the constant in condition 3 [] F . Proof. Let W be the solution of (5) and set 1 ()= ()vx W where and are defined as in (9) and (10) respectively. Then =0v on , and proceeding as in the proof of Lemma 3, we note that 21 (||)( )(),. pp divvvb x gvx (13) Therefore, by condition ][ 2 F, we see that 2 (||)( ,),. p divvvfx vx (14) We recall, by the above, that v also satisfies 21 ˆ (||)( )(),. pp divvvbx gvx (15) Let be a smooth bounded domain that contains (0, 1 )B the unit ball centered at the origin. Now, let be the first eigenfunction of the problem (4) with ()= ()cxax such that 0< , where is the positive constant in 3 [] F . Invoking conditions 2 [] F and 3 [] F , we get 2 11 1 (| |) =()()(,), p pp div axaxf xx (16) Moreover, since 1<0 , we also note that, 2 11 (||)( ,) ˆ ()()()(), p pp divf x bxg bxgx Therefore, we get 21 ˆ (||)()(), pp divb x gx (17) Recalling that ˆ g is non-increasing, by Lemma 1 we note, from (15) and (17), that v . Then by the elliptic regularity theory and Lemma 2, (14) and (16), we conclude that (12) has a solution u such that uv and 1,() ()uC C . Let W be as in the proof of the above Theorem. Then we note that b WW , and hence vv where v is as in Lemma 3. Then we deduce a non-singular case. Theorem 2. If f satisfies 1 [] F , (,0)=0fx, where b satisfies 1 []B, 3 []B and g satisfies 1 []G, 2 []G, then problem (1) has a solution 1, N loc uC R. Proof. For each positive integer k, let )(0,=kBBk be the ball of radius k centered at the origin. By Theorem 1, for each positive integer k we let )( 1, kkBCu be a solution of 2 (||) =(,), ()=0, p k k divuufx uxB uxx B (18) Then 0()() kk ux vx in k B, where k v are as in Theorem 1. Corresponding to the ball k B. It is easy to see that 0 is a subsolution. Recall the above, k vv on k B for all 1k, and hence 0()() k ux vx for all 2 Bx. By a standard procedure, we conclude that {} kl u has a subsequence that converges uniformly to a function in 2, ()C . By a diagonalization process it follows that {} k u has a subsequence that converges uniformly on open bounded subsets of N R to 2, N loc uC R and that u is a solution of (1). Since 0uv, it follows M. Z. WU ET AL. Copyright © 2010 SciRes. AM 355 that () 0ux as ||x. In the last part of the paper, we prove a nonexistence result for the following problem, 2 (| |) =()[()() ||],>0, ()0,| | p qN div uu dx guruuuin uxas x R(19) The result solves an open problem in [4] for >1p, 0q for the case when u is a radially symmetric solution. Before the proof, we state some conditions which we needed at the below. 1 10,, 0,GgC; 201= limup gu Gu ; 31=0 limup gu Gu . Theorem 3. Suppose 13 ()()GG are fulfilled and d is a positive radial function, r is a nonnegative radial function that is continuous on N R and satisfies ,=))(( 1 1 1 0 1 0 ddd p NN then the problem (19) has no positive radial solution that decays to zero near infinity. Proof. Suppose (19) has such a solution )(ru . Then 2 (|()|( )) =( )[(( ))(( ))|( )|], p q divu rur dr gurrurur or, equivalently, ()ur is a solution to the problem 12 1 (||)()[ (())(())] Np N ruurdr gurrur (20) Integrating (20) from 0 to r , we have 12 1 0 | |()[(())(())]. r Np N ru udguru d Hence ()<0ur ; i.e., ()ur is non-increasing. We put ln1=>0urur for all >0r. Then we have 2 12 (|()|()) 11 =()(|()|())(1)||, 1(1) p pp p p divu ru r divu ru rpu uu and ur satisfies 22 1 1 (|' |')|'|' 1(())(()) (1)| |() (1) (1) pp p pp N uu uu r g ur rur pudr uu (21) Multiplying Equation (21) by 1N r and integrating on (0,) yields 1 12 0 1 1 0 (1) |'|'| | (1) (( ))(( )) () (1) N Np p p N p p uu ud u gu ru dd u (22) If p is even, we can deduce that ddu u p uru p p p N N r1 1 1 0 1 0)|| 1)( 1)( ((0) ~ )( ~ 1 11 1 1 00 (( ))(( )) (() ) (1) rNN p p gu ru ddd u (23) We observe that ()< (0)ur u for all >0r and ()< (0)uru for all >0r. Since ()ur is positive, then (23) implies 1 11 1 1 00 (( ))(( )) (() )(0) (1) rNN p p gu ru dddu u (24) for all 0>r. Now, using Lemma 4 in (24), we have 1 11 1 1 00 1 11 1 1 00 (()(())) (( )) (() ) (( )1) rNN p rNN p p dgu dd gu ddd u 1 11 1 1 00 (( ))(( )) (( ))(0). 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