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Applied Mathematics, 2010, 1, 293-300 doi:10.4236/am.2010.14038 Published Online October 2010 (http://www.SciRP.org/journal/am) Copyright © 2010 SciRes. AM Entire Large Solutions of Quasilinear Elliptic Equations of Mixed Type* Hongxia Qin1, Zuodong Yang1,2 1Institute of Mathema tics, School of Mathematical Sciences, Nanjing Normal University, Nanjing, China 2College of Zhongbei, Nanjing Normal University, Nanjing, China E-mail: zdyang_jin@263.net Received March 10, 201 0; revised August 11, 2010; accepted August 15, 2010 Abstract In this paper, the existence and nonexistence of nonnegative entire large solutions for the quasilinear elliptic equation 2 || =()()()() m divuup xfuq x gu are established, where2m, f and g are nondecreasing and vanish at the origin. The locally Ho lder continuous functions p and qare nonnegative. We extend results previously obtained for special cases of f and g . Keywords: Entire Solutions, Large Solutions, Quasilinear Elliptic Equations 1. Introduction In this paper, we consider the problem 2 ||=()()()(),( 3) (),| | mN divuup xfuq x guinRN uxas x (1) where 2m, 1 ,([0, ),[0, ))((0, ),[0, ))fg CC , the locally Ho lder continuous functions p and q are nonnegative on N R. In addition, we assume that (0) =(0)=0;()0,()0, () ()>0,>0 fg ftgt ftgtfort (2) We call nonnegative solutions of (1) entire large solutions. In fact, this problem appears in the study of non-Newtonian fluid s [1,2] and non -Newtonian filtration [3,4], such problems also arise in the study of the sub-sonic motion of a gas [5], the electric potential in some bodies [6], and Riemannian geometry [7]. Large solutions of the problem ()=(()),, |=, uxfux x u (3) where is a bounded domain in (1) N RN have been extensively studied, see [8-20]. A problem with ()=u f ue and 2=N was first considered by Bieber- bach [13] in 1916. Bieberbach showed that if is a bounded domain in 2 R such that is a 2 C sub- manifold of 2 R, then there exisSts a unique 2()uC such that u eu = in and 2 |()(()) |uxlndx is bounded on . Here )(xd denotes the distance from a point x to . Rademacher [17], using the idea of Bieberbach, extended the above result to a smooth bounded domain in 3 R. In this case the problem plays an important role, when 2=N, in the theory of Riemann surfaces of constant negative curvature and in the theory of automorphic functions, and when 3=N, according to [17], in the study of the electric potential in a glowing hollow metal body. Lazer and McKenna [6] extended the results for a bounded domain in 1)( NRN satisfying a uniform external sphere condi- tion and the non-linearity u expuxff )(=),(= , where )(xp is continuous and strictly positive on . Lazer and McKenna [12] obtained similar results when is replaced by the Monge-Ampere operator and is a smooth, strictly convex, bound ed domain. Similar results were also obtained for a uxpf )(= with 1>a. Posteraro [16], for u euf =)( and 2N, proved the estimates for the solution )(xu of the problem (1,2) and for the measure of comparing with a problem of the same type defined in a ball. In particular, when 2=N, Posteraro [16] obtained an explicit estimate of the minimum of )(xu in terms of the measure of : ()(8/| |). minux ln *Project Supported by the National Natural Science Foundation o f China(Grant No.10871060). Project Supported by the Natural Science Foundation of the Jiangsu Higher Education Institutions of China (Grant No.08KJB110005) H. X. QIN ET AL. Copyright © 2010 SciRes. AM 294 The existence, but not uniqueness, of solutions of the problem (3) with f monotone was studied by Keller [18]. For a uuf =)( with 1>a, the problem (3) is of interest in the study of the sub-sonic motion of a gas when 2=a (see [15]) and is related to a problem involving super-diffusion, particularly for 2<1 a (see [21,22]). Pohozaev [15] proved the existence, but not uniqueness, for the problem (1.2) when 2 =)(uuf . For the case where (2)/(2) ()=( >2) NN fu uN , Loewe- ner and Nirenberg [20] proved that if consists of a disjoint union of finitely compact C manifolds, each having co-dimension less than 1/2N, then there exists a unique solution of the problem (3). The uniqueness was established for a uuf =)( with 3>a, when is a 2 C-submanifold and is replaced by a more general second-order elliptic operator, by Kondrat'ev and Nikishkin [19]. Marcus and Veron [14] proved the uniqueness for a uuf =)( with 1>a, when is compact and is locally the graph of a continuous function defined on an 1)( N-dimensional space. In [23], the authors considered the problem of existence and nonexistence of positive entire large solutions of the semilinear elliptic equation =()() ,0<.upxu qxu Recently [24], which is to extend some of these results to a more general the problem =()()()(),3, ()| |. N upxfuqxguinRN uxas x Quasilinear elliptic problems with boundary blowup 2 ||=(()),, |=, m divuufu xx u (4) have been studied, see [9,25,26] and the references therein. Diaz and Letelier [10] proved the existence and uniqueness of large solu tions to the problem (4) both for 1>,=)( muuf (super-linear case) and being of the class 2 C. Lu, Yang and E.H.Twizell [25] proved the existence of Large solutions to the problem (4) both for N Rmuuf =1,>,=)( or being a boun- ded domain (super-linear case) and N Rm =1, (sub-linear case) respectively. Recently [27], which is to extend some results of [28] to the following quasilinear elliptic problem 2 || =()(), () , m divuup xfuin ux on (5) where N R, the non-negative function )()( Cxp , and the continuous functionfsatisfies (2) and the Keller-Osserman condition 1/ 10 [()]=,()=() s m F sdsFs ftdt (6) then the author also consider the nonexistence for the non-negative non-trivial entire bounded radial solution on N R of (5) when p satisfies 1/(1) ** 0||= (( ))=,( )=(). min m xt tp tdtp tpx (7) On the other hand, if f does not satisfy (6), that is <)]([ 1/ 1dssF m, we can obtain from Lemma 2.4 in [29] that 1/( 1) 1 1< () mds fs (8) which is also shown in [30]. In this paper, we will require the above integral to be infinite, that is 1/( 1) 1 1= () mds fs (9) which is a very important condition in our main results. Furthermore, motivated by the results of [24], we also admit the following condition which is opposite to (7), that is *1/(1) * 0||= (( ))<,( )=(). max m xt tp tdtp tpx (10) As far as the authors know, however, there are no results which contain the existence criteria of positive solutions to the problem (1). In this paper, we prove the existence of the positive large solutions for the problem (1). When 2=p, the related results have been obtained by A.Lair and A.Mohammed [24]. The main results of the present paper contain extension of the results in [24] and complement of the results in [10,25,26]. The plan of the paper is as follows. In Section 2, for the convenience of the reader we give some basic lemmas that will be used in proving our results. In Section 3 we state and prove the main results. Section 4 contains some consequences of the main theorems, and other results. In Section 5 we present an Append ix wher e we state and prove three lemmas needed for proofs in previous sections. 2. Preliminary In this section, we give some results that we shall use in the rest of the paper. Lemma 2.1.(Weak comparison principle)(see [25]) Let be a bounded domain in N R(2)N with smooth boundary and )(0,)(0,: is con- tinuous and non-decreasing. Let )(, 1, 21 m Wuu satisfy 2 111 2 22 2 || () || () m m uudxudx uudxudx H. X. QIN ET AL. Copyright © 2010 SciRes. AM 295 for all non-negative )( 1, 0 m W . Then the inequality onuu 21 implies that . 21 inuu Lemma 2.2. Let f verify (9), and )[0,)[0,: be continuous. Then 11 ((( )))=( )( ),>0 (0) =,(0) = 0 NN m rvrrrfvr vv (11) admits a non-negative solution v defined on )[0, , where sss m m2 |=|)( . If in addition f is nondecrea- sing and satisfies (7), then )( rv as r . Proof. First we note that (11) has a solution )(0, 1RCv for a maximal R<0 . As a consequence of (7) we claim that =R. By way of contradiction, let us suppose that <<0 R instead. Then we must have )( rv as Rr . Let 0>},<0:)({sup:=)( ttsst Then is nondecreasing, and clearly )()( tt for 0>t. Integrating Equation (11) from 0 to r yields 11 0 (())=()(()) r NN mvr rssfvsds (12) From (12) we see that 0)( rv , therefore, v is a non-decreasing function and we can obtain from (12) that (()) ()(()) mr vr rfvr N . Then we can obtain 1/(1) 1/( 1) 1/(1) 00 () () (()) m rr m m vt t dtt dt N fvt That is 1/( 1) () 1/( 1) 1/( 1)0 1() () m vrr m m t dss ds N fs Letting Rr , and recalling that )(rv , we conclude that 1/( 1)1/( 1) 1/( 1)0 1() () mRm m R dssds N fs which is an obvious contradiction. Thus, indeed v is defined on )(0,. We now show that )(rv as r . For this we will use (7) on . Integrating the equation in (11) we find 1/(1) 11 00 1/( 1)1/(1) * 0 ()=() (()) () (()) m rt NN mrm vrtss f vsdsdt fttdt N That is 1/(1) * 0 ()( ,,)(()),>0 rm vrCmNttdt r and as a consequence of (7) we conclude that )(rv as r . 3. Main Theorems In this section, we will state the first of our main results. Theorem 3.1. Under the following hypotheses 1 1 1, 0; t Ht dst fs 1/ 11/ 1 ** 0 1/ 1 * 0 2 , , mm mt Hptp t tfptdtptsps ds where is the inverse of ; 1/ 1 * 0 3m Htqtgpt dt Let f and g satisfy (2). Furth ermore, assu me that (9) and (10) hold. If p satisfies (7), then (1) admits a solution. Proof. Let v be an entire radial large solution of )(|)(|=)|(| * 2vfxpvvdivm such that =(0)v for some 1<<0 . This is possible by Lemma 2.2, since f satisfies (9) and * p satisfies (7). Thus v is a super-solution of (1). We proceed to construct a sub-solution u of (1) such that vu on N R. Then by the standard regularity argument for elliptic problems, it is a straight forward argument to prove that (1) would have a solution w such that vwu on N R. For each positive integer n, let n u be a solution of 2 ** (| |) = (||)()(||)(),0.4, ()=,0.4 , m n n div uu pxfuqxgu cminB ux vcmon B (13) where )(0,= nBBn is the ball of radius n centered at the origin. That such a solution exists is shown in Lemma 5.2 of Appendix. Then we note that each n u is a radial solution and that 1 0<, . nn n uuvonB Let ():= (),. lim N n n uxuxxR Since each n u is radial, it follows that u is radial as well. By a standard argument we can show that u is a solution of the differential equation in (1). Clearly vu on N R. So We only prove that u is nontrivial and that )(xu as || x. Recalling that n u and v are radial and that )(=)( nvnun we see that H. X. QIN ET AL. Copyright © 2010 SciRes. AM 296 11* *1/(1) 00 1/( 1)1/( 1) 11 11 ** 00 00 (0)((( )()()())) =(0) ()()(0) ()() nt NN m nnn mm nt nt NNNN n utspsfuqsgudsdt vtspsfvdsdtvtspsfudsdt for 20, mx , we can use the inequality 1)1/(1)1/( 1)(1 mm xx , then we obtain 1/( 1)1/(1) 11* 11* 00 00 1/( 1) 11 * 00 (0)()()( )() () ()(0)= mm nt nt NN NN nn n m nt NN n u tspsfudsdttsqsgudsdt tspsf udsdtv that is (1 )/( 1)1*1/(1)11/(1) * 00 0 1/( 1) 11* 00 (0)((()( ))(()( ) ) () ()(0)= ntt Nm NmNm nnn m nt NN n utspsfudsspsfudsdt tsqsg udsdtv Since )( *sp is increasing and )( *sp is decreasing, so 1/( 1)1/( 1) (1)/(1)1*1 * 00 0 1/( 1) (1)/( 1)*1/(1)11/( 1)11/( 1) * 00 0 * 1/(1)1/( * 0 () ()() () (()()(())()() (())) () ()() mm ntt Nm NN nn m nt t Nm mNmNm nn nmm tspsfudsspsfudsdt tptsfusdsptsfusds dt ptp 1/( 1) 1)(1)/( 1)1 0 1/( 1) * 1/(1)1/(1) * 00 *1/(1)1/( 1)1/( 1) 1* 0 ()(() () ()() ()(()) ()() ()() ()((())) m t NmN n m nt mm n nmm m n tts fusdsdt ptptfusdsdt Cmptpt tfutdt and 1/( 1)1/( 1) 11* * 2 00 0 () ()()()(()) mm nt n NN nn tsqsg udsdtCmtqtg utdt Therefore we get 1/( 1) *1/(1)1/(1)1/(1)* 1* 2 0 0 (0)(,)()( )()()(())(,)( )()m n n mm m nnn uCm NptpttfudtCm Ntqtgudt (14) Now, let )(t be the inverse of the increasing function defined in (9). We note that 1)( t for all 0t. Furt h ermore, we have 0.>)),(())((=)()),((=)( ttftfttft Let w be an entire large solution of |)(|= *xpw such that 0=(0)w. Set ))((:=)( xwxa . Then )(|)(| *afxpa. Since 0>(0)=>1(0))(=(0) vwa , we invoke Lemma 2.1 in [24] to conclude that )()( xaxv for all x N R. Moreover, ,)(:=)()( * 0dtttprprw r we have |))(|()( xpxv . Now, recalling that vun for all Nn we see that *** *1/( 1)1/( 1)1/( 1) * *1/(1)1/( 1)1/( 1) * *1/(1)1/( 1)1/( 1) * () (())()(())()((())), (()( )()())((( ))) (()()()( ))((( ))) (()( )()())(((()))) n mmm n mm m mm m tqtgu ttqtgvttq tgpt ptpttfut ptpttfvt ptpttfpt Take note of (9) and (10), we invoke the Lebesgue dominated convergence theorem to infer from (14) that *1/(1)1/( 1)1/( 1) 1* 0 1/( 1) * 20 (0) (,)()()()()(()) (, )()()>0. nmm m m n u CmNptpt tfudt CmN tqtgudt This show s that u is nontrivial. Now we note that 1/(1) 11* * 00 1/(1) 11* 00 ()=(0) ()()() () () (). nn m rt NN nn m rt NN n uru tspsfuqsg udsdt tsps f udsdt Recalling that vun for all n, we invoke the Lebesgue dominated convergence theorem again, on letting n 1/( 1) 11* 00 ()() (). m rt NN u rtspsfu dsdt H. X. QIN ET AL. Copyright © 2010 SciRes. AM 297 Since u is nontrivial we see that 0>) 2 (0 r u for some 0> 0 r. Thus for 0 >rr , we have dtdsspst r ufru m N t N r r m1)1/( *1 0 1 0 0 1)1/( )() 2 ()( then dtttp N r ufru m r r m m1)1/( * 0 1)1/( 0 1)1/( ))(( 1 ) 2 ()( Therefore, as a consequence of (7) we see that )(xu as || x. To show our next main result, now we set p is c- positive on (i.e., for any 0 x satisfying 0=)(0 xp , there exists a domain 0 such that 000,x, and 0>)(xp for all 0 x.) we know that p is c-positive on N R if and only if there is a sequence n of smooth bounded domains with 1 nn for each n such that n n 1= N R and p is c-positive on each . It is easy to see that if is a non-negative and locally Ho lder continuous function in N R that satisfies (10), then the following problem admits a positive solution. 2 (||) =(), ()0,|| mN div wwxx wx x R (15) In fact .)(=)( 1)1/( *1 0 1 ||dtdssstxv m N t N x is a super-solution of (15) such that 0)( xv as || x. On the other hand, 0 is a sub-solution of (15), (See [31], Lemma 3) the assertion follows. Theorem 3.2. Suppose f and g satisfy (2). If (1) has a solution, f satisfies (8) and p is c-positive in N R, (3) admits a solution. Conversely, if gf satisfies (8), (15) admits a non-negative solution with )()(=)( xqxpx and )}(),({min:=)( xqxpx is c-positive, then (1) has a solution. Proof. Let }{ n be a sequence of bounded smooth domains in N R as provided in the definition of the c-positivity of p. Suppose (1) has a solution, say v is a solution. For each n, the problem 2 (||)=( )( ), ()=, mn n divuup xfux ux x (16) has a solution( see [29]). For each positive integer n, let n u be a solution of (16). Then by Lemma 2.1 it follows that .),()()( 1nnn xxuxuxv A standard procedure (for example, see [30]) can be used to show that ,),( lim :=)( N n nxxuxu R is the desired solution of (3). For the converse, we let n u be a solution of the problem 2 (||)=( )( )()( ), ()= , mn n divuup xfuq x gux ux x (17) The existence of such a solution is demonstrated in Lemma 5.3 of Appendix. It easily follows that the sequence }{ n u is a non-in creasing sequence. Let ()= (),. lim N n n uxu xx R A standard argument shows that u is a solution of the quasilinear equation in (17). Thus we need only show that u is nontrivial and that )(xu as || x. For this we consider the following function 1/( 1) 1 ()=,>0, () m t tdst hs (18) where )()(:=)( tgtfth . Obviously, (18) is finite for all 0>t. We also notice that 0> ))(1)(( )( =)(0,< )( 1 =)( 1)1/(1)1/( mmm thm th t th t Now fix 0> , and let nnn xxuxv ),)((=)( . Note the sequence n v is nondecreasing. Moreover, a simple computation shows that 212 2 1 (||)=|() |(||) (1)|( )|( )|| |()| (()()()()) () ()()() =()() () mmm nn nnn mm nnn m nnn nn n divvvudivuu mu uu upxfuqxgu pxfu qxgupxqx hu We can also note that 0= n v on n . Let w be a solution of (15). Thus by Lemma 2.1 we see that wvn on n for all n, letting n, and then 0 we see that wu )( on N R. Thus 0))(( xu as || x, that is )(xu as || x. 4. Consequences and Related Results We can obtain some consequences of the main theorems, and other results that are of independent interest. Theorem 4.1. Let f and g be continuous, nonde- creasing functions such that gf satisfies (9), and Suppose qp is nontrivial. If there is a solution to 2 (||)() ()()(),,3 (),| | mN divuup xfuqx g uxRN uxas x (19) then qp satisfies (7). H. X. QIN ET AL. Copyright © 2010 SciRes. AM 298 Proof. Let u be a solution of (19). Let v be a solution of the initial value (11) with * )(= qp , f replaced by gf and 0 = where 0 is chosen such that (0)> 0u . Since gf satisfies (9), we recall from Lemma 2.2 that v is defined on )[0, . Then |)(|=)( xvxw is a solution of )))(())((|)((|)(=)|(| *2xwgxwfxqpwwdiv m, and hence )()()|(|2wqgwpfwwdiv m on N R. Since 0>)(rv we see that Arv )( as r , for some A<0 . Assume that < A so that Axw )( for all N xR. Since )(xu as || x, we see that for some R, we have RxxuAxw ||),()( . Thus )()( xuxw on Rx |=| and therefore by Lemma 2.1 we find that )()( xuxw on )(0,RB . But this contradicts the choice that (0)>(0) uw . So we have = A , then ||,)( xasxw . From Equation (11) we find 1/(1) 11* 0 1/(1) 11* 0 ()=() ( )()(()) ()(())()() m r NN m r NN vrrtp qt fgvtdt rfgvrtpqtdt (20) Dividing (20) through by ))(()(1)1/( rvgfm and integrating the resulting inequality on )(0,r we have 1/( 1) 0 1/( 1) 11* 00 () ()(()) ()() r m m rt NN vt dt fg vt tspqsdsdt That is () 1/( 1) 0 1/( 1)*1/(1) 0 1 ()() 1(( )()) vr m mrm dt fg t tp qtdt N Letting r and recalling that gf satisfies (9), the claim is proved. As a consequence of Theorem 3.1 and Theorem 4.1 we also obtain the following corollaries. Corolla ry 1. Suppose (2) and ( 9) hold for f. Further, let p satisfy (10). (3) admits a solution if and only if p satisfies (7). Proof. If p satisfies (7) then Theorem 3.1, with 0=)(xq shows that (3) has a solution. The converse follows from Theorem 4.1 on taking 0)( xq again. The next corollary provides sufficient conditions for the existence and nonexistence of solutions to (1) when both p and q satisfy (7). Corollary 2. Suppose f and g satisfy (2) and p and q satisfy (7). If gf satisfies (9), then (1) has no solution. On the other hand, (1) admits a solution if gf satisfies (8) and the function )}(),({min)( xqxpx is c-positive on N R. Proof. By way of contradiction,we can obtain th e first statement from Theorem 4.1. Since qp satisfies (10) and the remark noted ju st before Theorem 3.2 shows that (15) admits a solution with qpb=. Thus the second part of the corollary is an immediate consequence of Theorem 3.2. 5. Appendix In this appendix we state and prove results that have been used in the proofs of the main results of the paper. We start by proving the existence of a solution to the following Dirichlet problem on a bounded smooth domain in N R. 2 (||)=( )( )( )( ),, ()= (),. m divuup xfuq x guin uxx on (21) Lemma 5.1. Let N R be a smooth bounded domain and let f and g satisfy (2). Let )( 2C be positive. If v is a positive super-solution of (21), then the problem (21) has a solution u such that vu <0 on . Proof. Let )( min := x x . Obviously, 0> . Now we set ,))((=)( 1)1/( 0dsshtm t where )()(=)( sgsfsh for all 0s. An application of L'H o ˆpital’s Rule shows that tt )( for all <<0 t and some 0> . Without of generality we can suppose that )(<0 . Finally, let z be a solution of the Dirichlet problem .,=)( ,),()(=)|(|2 onxz inxqxpzzdiv m The n the maximu m pri ncip le sh ows th at )(<0 xz on , we define ))((:=)( xzxw for all x, we note that)()( xzxw for allx. A simple computation shows that 0> ))(1)(( )( =)(0,>)(=)( 21)1/( 1)1/( mm m thm th ttht and 2 12 2 1 (| |) =|| (||)(1)|||| ||(( )( ))=(( )())(( )( )) (()())(( )( ))( )()( )() m mm mm m div ww 'divzzm' ''z ' pxqx fzgzpxqx f wgwpxqx pxfwqxgw and )()()( xxw for x. Thus w is a sub-solution of (21) and v is a super-solution of (21) such that vw on . By the maximum principle we note that vw on . Thus by lemma 1 in [31] we H. X. QIN ET AL. Copyright © 2010 SciRes. AM 299 conclude that (21) has a solution usuch that vuw which is what we want to show. The following lemma was used in the proof of Theorem 3.1. Lemma 5.2. Let ))[0,),([0,, Cba and B be a ball in N R centered at the origin. If f and g are nondecreasing on )[0,, then given a positive constant , there exists a radial solution to the problem 2 (||)=(||)()(||)(),, ()= ,, m divuuaxfubxguin B uxonB (22) Proof. Let }{k a and }{ k b be decreasing sequences of Ho lder continuous functions which converge uni- formly on B to a and b respectively(Se e [32]). Then by Lemma 5.1, for each k there exists a nonnegative solution k u of 2 (| |) =(||)()(||)(),, ()=,. m kk kkkk k div uu axfu bxguinB uxon B Since the sequence }{ k a and }{ k b are decreasing,it is easy to show that }{ k u is increasing. Of course,it is also bounded above by . Thus it converges, and assume uuk. Since k u satisfies the integral equation 1/(1) 11 00 ()= (0) ( )(( )( )(( ))) kk m rt NN kk kk ur u tsasfusbsgusds dt the function u will satisfy the integral equation 1/( 1) 11 00 ()= (0) ()(())() (()). m rt NN ur u tsa sfu sb s g usdsdt Since =)(Ruk for eachk, whereRis the radius of the ballB, it is clear that =)(Ru . Thusuis a non- negative solution of problem (22) on Bas claimed. Finally we state and prove a lemma that was used in the proof of Theorem 3.2. Lemma 5.3. Let N R be smooth. Suppose f and g satisfy (2). Iffsatisfies (6) andp is c-positive on , then the problem 2 (||)=( )( )()( ),, ()= ,, m divuup xfuq xg ux ux x (23) has a solution. Similarly, if instead of requiring f to satisfy (6), we require only gf to satisfy (6), and require },{min:= qp to to be c-positive on , then (23) has a solution. Proof. Since p is c-positive and f satisfies (6), let v be a large solution of )()(=)|(| 2vfxpvvdivm on (see [29]). Now for each positive integer k, let k w be a solution (See Lemma 5.1) of 2 (| |) =(()())(()()),, ()= ,, m div ww pxqxfwgwx wxkx By Lemma 2.1 we see that xxvxwxw kk),()()( 1 If )( lim =)( xwxw k k, then by a standard procedure we conclude that w is a solution of ))()())(()((=)|(| 2wgwfxqxpwwdivm on such that vw . 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