# Existence of positive solutions for some polyharmonic nonlinear equations in ℝn

where m is a positive integer such that n > 2m. In the case m = 1, (1.1) contains several well-known types which have been studied extensively by many authors (see for example [1â3, 8, 9, 11, 12, 14] and the references therein). Their basic tools are essentially some properties of functions belonging to the â classical Kato class Kn (Rn ) and the subclass of Green-tight functions Kn (Rn ) (some properties pertaining to these classes can be found in [1, 4, 14]). In this paper, we are concerned with the high order. Our purpose is two folded. One is â to extend the Kato class Kn (Rn ) and the subclass Kn (Rn ) to the order m â¥ 2. The second purpose is to investigate the existence of positive solutions for (1.1). The outline of the paper is as follows. The existence results are given in Sections 3, 4 and 5. In Section 2, we give the explicit formula of the Green function Gm,n (x, y) of (âÎ)m in Rn . Namely, for each x, y in Rn Gm,n (x, y) = km,n 1 , |x â y | (1.2) Hindawi Publishing Corporation Abstract and Applied Analysis Volume 2006, Article ID 76582, Pages 1â24 DOI 10.1155/AAA/2006/76582 Solutions of polyharmonic equations in Rn where km,n is a positive constant which will be precised later. The 3G-Theorem proved in [13] for the case m = 1, is also valid for every m. Indeed, for each x, y,z in Rn , we have Gm,n (x,z)Gm,n (z, y) 2â1 Gm,n (x,z) + Gm,n (z, y) . Gm,n (x, y) This 3G-Theorem will be useful to state our existence results. Next, we study the Kato class Km,n (Rn ) deï¬ned as follows. Deï¬nition 1.1. A Borel measurable function Ï in Rn (n > 2m), belongs to the Kato class Km,n (Rn ) if lim sup Î±â0 (1.3) |x â y | d y = 0. (1.4) Indeed, ï¬rst we prove some properties of functions belonging to this class similar to those established in [1, 4]. In particular, we have the following characterization Ï â Km,n Rn ââ lim sup t â0 t 0 Rn p(s,x, y) d yds = 0, (1.5) where p(t,x, y) = (1/(4Ït)n/2 )exp(â /4t), for t â (0, â) and x, y â Rn , is the density of the Gauss semi-group on Rn . â Secondly, we study a subclass of Km,n (Rn ) denoted by Km,n (Rn ) and deï¬ned by the following. â Deï¬nition 1.2. A Borel measurable function Ï belongs to the class Km,n (Rn ) and it is n ) and satisï¬es called m-Green-tight function if Ï â Km,n (R M ââ lim | y |â¥M sup |x â y | d y = 0. (1.6) â In particular, we characterize the class Km,n (Rn ) as follows. Theorem 1.3. Let Ï â + (Rn ), (n > 2m). Then the following assertions are equivalent â (1) Ï â Km,n (Rn ). + (2) The m-potential of Ï, V Ï(x) := Rn Gm,n (x, y)d y is in C0 (Rn ). This Theorem improves the result of Zhao in [14], for the case m = 1. A more ï¬ne characterization will be given in the radial case. â One can easily check that L1 (Rn ) â© Km,n (Rn ) â Km,n (Rn ). Also we show that for p > n/2m and Î» < 2m â n/ p < Î¼, we have L p Rn 1+|Â·| â â Km,n Rn , (1.7) and we precise the behaviour of the m-potential of functions in this class. H. MË agli and M. Zribi 3 a In Section 3, we are interested in the following polyharmonic problem (â )m u + uÏ(Â·,u) = 0, in Rn (in the sense of distributions) (1.8) lim u(x) = c > 0. The function Ï is required to verify the following assumptions. (H1 ) Ï is a nonnegative measurable function on Rn Ã (0, â). â (H2 ) For each Î» > 0, there exists a nonnegative function qÎ» â Km,n (Rn ) with Î±qÎ» 1/2 n , the mapping t â t(q (x) â Ï(x,t)) is continu(see (1.24)) and such that for each x â R Î» ous and nondecreasing on [0,Î»]. Under these hypotheses, we give an existence result for the problem (1.8). In fact, we will prove the following theorem. Theorem 1.4. Assume (H1 ) and (H2 ). Then the problem (1.8) has a positive continuous solution u in Rn satisfying for each x â Rn , c/2 u(x) c. To establish this result, we use a potential theory approach. In particular, we prove that â if the function q â Km,n (Rn ) is suï¬ciently small and f is a nonnegative function on Rn , then the equation (â )m u + qu = f , (1.9) has a positive solution on Rn . In [6], Grunau and Sweers gave a similar result in the unit ball of Rn , with operators perturbed by small lower order terms: (â )m u + |k |<2m ak (u)Dk u = f . (1.10) In the case m = 1, the problem (1.8) has been studied by MË agli and Masmoudi in [7, 8], a where they gave an existence and an uniqueness result in both bounded and unbounded domain Î©. In Section 4, we are concerned with the following polyharmonic problem (â )m u = f (Â·,u), in Rn (in the sense of distributions) lim u(x) = 0. (1.11) Here f is required to satisfy the following assumptions. (H3 ) f is a nonnegative measurable function on Rn Ã (0, â), continuous with respect to the second variable. (H4 ) There exist a nonnegative function p in Rn such tha < Î±0 := p(y) Rn | y| + 1 2() d y <â (1.12) â and a nonnegative function q â Km,n (Rn ) such that for x â Rn and t > 0 p(x)h(t) f (x,t) q(x)g(t), (1.13) Solutions of polyharmonic equations in Rn where h is a nonnegative nondecreasing measurable function on [0, â) satisfying m0 := 1 h(t) < h0 := liminf â t â0+ km,n Î±0 t (1.14) and g is a nonnegative measurable function locally bounded on [0, â) satisfying 0 g â := limsup t ââ g(t) 1 < M0 := t Vq â (1.15) By using a ï¬xed point argument, we will state the following existence result. Theorem 1.5. Assume (H3 ) and (H4 ). Then the problem (1.11) has a positive continuous solution u in Rn satisfying for each x â Rn , a |x| + 1 where a, b are positive constants. This result follows up the one of Dalmasso (see [5]), who studied the problem (1.11) in the unit ball B, with more restrictive conditions on the function f . Indeed, he assumed that f is nondecreasing with respect to the second variable and satisï¬es t â0+ xâB u(x) bV q(x), (1.16) lim min f (x,t) = +â, t t â+â xâB lim max f (x,t) = 0. t (1.17) He proved the existence of a positive solution and he gave also an uniqueness result for positive radial solution when f (x,t) = f (|x|,t). When m = 1, similar conditions, but more restrictive, on the nonlinearity f have been adopted by MË agli and Masmoudi in [8]. In fact in [8], the authors studied (1.11) in an a unbounded domain D of Rn , n â¥ 3, with compact nonempty boundary âD and gave an existence result as Theorem 1.5. On the other hand, Brezis and Kamin proved in [3], the existence and the uniqueness of a positive solution for the problem âÎu = Ï(x)uÎ± in Rn , (1.18) liminf u(x) = 0, with 0 < Î± < 1 and Ï is a nonnegative measurable function satisfying some appropriate conditions. We improve in this section the result of Brezis and Kamin in [3] and the one of MË agli and Masmoudi in [8]. a In Section 5, we will study the existence of solutions to the following polyharmonic problem (â )m u = f (Â·,u), in Rn (in the sense of distributions) u(x) > 0, in Rn , (1.19) under the following assumptions on the nonlinearity f . H. MË agli and M. Zribi 5 a (H5 ) f is a nonnegative measurable function on Rn Ã (0, â), continuous with respect to the second variable on (0, â). (H6 ) f (x,t) q(x,t), where q is a nonnegative measurable function on Rn Ã (0, â) such that the function t â q(x,t) is nondecreasing on (0, â). â (H7 ) There exists a constant c > 0 such that q(Â·,c) â Km,n (Rn ) and V q(Â·,c) Put câ = c â V (q(Â·,c)) â. â < c. (1.20) We give in this section the following existence result. Theorem 1.6. Assume (H5 ), (H6 ), and (H7 ). Then for each Î´ â (0,câ ], the problem (1.19) has a positive continuous solution u in Rn satisfying for each x â Rn Î´ u(x) c, lim u(x) = Î´. (1.21) If m = 1, Yin gave in [11] an existence result of the following problem u + f (x,u) = 0, u(x) > 0, in GB , (1.22) where GB = {x â Rn , |x| > B }, for some B â¥ 0. His method relies on the technique of radial super/subsolutions. Our approach is diï¬erent, in fact we will use a ï¬xed point argument. We improve the result of Yin under more general assumptions (see Remark 5.3). In order to simplify our statements, we deï¬ne some convenient notations. Notations. (i) (Rn ) denotes the set of Borel measurable functions in Rn and + (Rn ) the set of nonnegative ones. + (ii) C0 (Rn ) := {w continuous on Rn and lim w(x) = 0} and C0 (Rn ) the set of nonnegative ones. (iii) For Ï â + (Rn ), we put the m-potential of Ï on Rn by V Ï(x) := Vm,n Ï(x) = (iv) For Ï â + (Rn ), Rn Gm,n (x, y)d y = km,n Rn |x â y | (1.23) we put Gm,n (x,z)Gm,n (z, y) Ï(z) dz. Gm,n (x, y) (1.24) Î±Ï = sup x,y âRn Rn (v) Let Î» â R, we denote by Î»+ = max (Î»,0). (vi) Let f and g be two positive functions on a set S. We call f â¼ g, if there is c > 0 such that 1 g(x) f (x) cg(x) âx â S. c (1.25) Solutions of polyharmonic equations in Rn We call f g, if there is c > 0 such that f (x) cg(x) âx â S. (1.26) The following properties will be used several times: for s,t â¥ 0, we have min(s,t) = s â§ t â¼ (s + t) p â¼ s p + t p , 2. Properties of the Kato class In this section, we characterize functions belonging to the Kato class Km,n (Rn ) and the â subclass Km,n (Rn ) of m-Green-tight functions and we prove Theorem 1.3. We recall that throughout this paper, we are concerned with n > 2m. We set p(t,x, y) = (1/(4Ït)n/2 )exp(â /4t), for t â (0, â) and x, y â Rn , the density of the Gauss semi-group on Rn . By a simple computation, we obtain that the Green function of (âÎ)m in Rn , for each m â¥ 1, is given by Gm,n (x, y) = 1 (m â 1)! â st , s+t p â R+ . (1.27) p(s,x, y)ds, for x, y in Rn . (2.1) Then we have the following explicit expression Gm,n (x, y) = km,n 1 , |x â y | for x, y in Rn , (2.2) where km,n = Î(n/2 â m)/4m Ï n/2 (m â 1)!. 2.1. The class Km,n (Rn ). We will study properties of functions belonging to Km,n (Rn ). First we remark the following comparison on the classes K j,n (Rn ), for j â¥ 1. Remark 2.1. Let j,m â N such that 1 j m, then we have for each n > 2m Kn (Rn ) := K1,n (Rn ) â K j,n (Rn ) â Km,n (Rn ), where Kn (Rn ) is the classical Kato class introduced in [1]. Example 2.2. Let Ï â (Rn ). Suppose that for p > n/2m, we have sup d y < â. (2.3) |xâ y |1 (2.4) Then by the HÂ¨ lder inequality, we conclude that Ï â Km,n (Rn ). o In particular, we have that for p > n/2m, L p (Rn ) â Km,n (Rn ). To establish the characterization (1.5) of the Kato class Km,n (Rn ), we need the following lemmas. H. MË agli and M. Zribi 7 a Lemma 2.3. For each t > 0 and x, y â Rn , we have p(s,x, y)ds Gm,n (x, y). (2.5) Moreover, for |x â y | 2 t, we have that â Gm,n (x, y) p(s,x, y)ds. (2.6) Proof. Let t > 0 and x, y â Rn . Then (2.5) follows immediately from (2.1). â If we suppose further that |x â y | 2 t, then we have p(s,x, y)ds = c smân/2â1 exp â â 4s ds (2.7) c |x â y | c â¥ |x â y | = c Gm,n (x, y), |xâ y |2 /4t â n/2âmâ1 âr r n/2âmâ1 eâr dr e dr where the letter c is a positive constant which may vary from line to line. Lemma 2.4. Let Ï â Km,n (Rn ). Then for each compact L â Rn , we have sup d y < â. (2.8) x+L In particular, we have Km,n (Rn ) â L1 (Rn ). loc Proof. Let Ï â Km,n (Rn ), then by (1.4) there exists Î± > 0 such that sup 1. |x â y | 1i p B(ai ,Î±). p (2.9) Let a1 ,...,a p â L such that L â Hence for each x â Rn , we have d y x + ai â y d y x+L i=1 B(x+ai ,Î±) p Î± (2.10) i=1 B(x+ai ,Î±) pÎ± So, sup x+L ||d y < â. Solutions of polyharmonic equations in Rn Proposition 2.5. Let Ï â Km,n (Rn ). Then for each ï¬xed Î± > 0, we have sup sup 0t 1 |xâ y |â¥Î± t mâ1 p(t,x, y) d y := M(Î±) < â. (2.11) Proof. Let Ï â Km,n (Rn ), 0 < t 1. Let Î± > 0, then we have that sup t mâ1 p(t,x, y) d y exp â |xâ y |â¥Î± exp â Î±2 /8t sup t n/2âm+1 So to prove (2.11), we need to show that sup exp â (2.12) Rn Rn d y < â. (2.13) Indeed, using Lemma 2.4, we denote by c := sup d y < â. (2.14) x+B(0,1) On the other hand, since any ball B(0,k) of radius k â¥ 1 in Rn can be covered by Î±(n) := An kn balls of radius 1, where An is a constant depending only on n (see [4, page 67]), then there exist a1 ,a2 ,...,aÎ±(n) â B(0,k) such that B(0,k) â 1iÎ±(n) B ai ,1 . (2.15) Hence for each x â Rn , we have Î±(n) x+B(0,k) i=1 B(x+ai ,1) cAn kn , (2.16) which implies that for each x â Rn , exp â â Rn k =0 exp â â k2 8 k|xâ y |k+1 d y (2.17) cAn k =0 exp â k2 (k + 1)n 8 < â. Thus (2.13) holds. This ends the proof. H. MË agli and M. Zribi 9 a Proposition 2.6. Let Ï â B(Rn ). Then Ï â Km,n (Rn ) if and only if lim sup t â0 0 Rn p(s,x, y) d yds = 0. (2.18) Proof. Suppose Ï veriï¬es (2.18), then from (2.6) we have that dy |x â y | Î±2 /4 Rn p(s,x, y) dsd y, (2.19) which implies that the function Ï satisï¬es (1.4). Conversely, suppose that Ï â Km,n (Rn ). Let Îµ > 0, then by (1.4), there exists Î± > 0 such that sup Îµ. |x â y | (2.20) Thus from (2.5) and (2.11), we deduce that for each x â Rn and t 1, we have Rn p(s,x, y) d yds p(s,x, y) d yds p(s,x, y) d yds (2.21) |xâ y |â¥Î± d y + tM(Î±) |x â y | Îµ + tM(Î±). This implies (2.18) and completes the proof. â 2.2. The class Km,n (Rn ). We will characterize the subclass of m-Green-tight functions â Km,n (Rn ). In fact, we will prove Theorem 1.3 and we give in particular a more precise characterization in the radial case. â Example 2.7. Let p > n/2m. Then L p (Rn ) â© L1 (Rn ) â Km,n (Rn ). â Proof of Theorem 1.3. Let Ï â + (Rn ). First we suppose that Ï â Km,n (Rn ), then using + similar arguments as in the proof [9, Proposition 6], we obtain easily that V Ï â C0 (Rn ). + n ). Then, we aim at proving that Ï â K â (Rn ). Conversely we suppose that V Ï â C0 (R m,n So we divide the proof into two steps. Solutions of polyharmonic equations in Rn Step 1. We will prove that Ï satisï¬es (2.18). Indeed it is clear from (2.1), that for each x â Rn , we have that V Ï(x) = p(s,x, y)d yds (m â 1)! 0 Rn â 1 p(s,x, y)d yds + (m â 1)! t Rn = I1 (x) + I2 (x). 1 (2.22) From the properties of the density p(s,x, y), we deduce that x â I1 (x) and x â I2 (x) are nonnegative lower semi-continuous functions in Rn . Then using the fact that V Ï â + + C0 (Rn ), we get that the function x â I1 (x) is also in C0 (Rn ). So, for each x â Rn , the t mâ1 + n family { 0 s Rn p(s,x, y)d yds, t > 0} is decreasing in C0 (R ), which together with n, the fact that for each x â R lim t â0 0 Rn p(s,x, y)d yds = 0 (2.23) imply by Dini Lemma, that (2.18) is satisï¬ed. + Step 2. We will prove that Ï satisï¬es (1.6). Let Îµ > 0, then since V Ï â C0 (Rn ), there exists a > 0 such that for |x| â¥ a, we have that V Ï(x) Îµ. Let M â¥ 2a, then sup sup | y |â¥M |x â y | |x|â¥a Îµ+ Rn d y + sup |x â y | |x|a | y | dy | y |â¥M |x â y | | y |â¥M (2.24) Now, since V Ï(0) < â, we deduce that lim d y = 0. | y | (2.25) M ââ | y |â¥M Then (1.6) holds and this ends the proof. â For a nonnegative function Ï in Km,n (Rn ), we denote by MÏ := Ï â Rn , |Ï| Ï . (2.26) â Proposition 2.8. For a nonnegative function Ï in Km,n (Rn ), the family of functions V MÏ := V Ï, Ï â MÏ (2.27) is uniformly bounded and equicontinuous in C0 (Rn ) and consequently it is relatively compact in C0 (Rn ). â â Proof. Let Ï â Km,n (Rn ). Obviously, since each function Ï in MÏ is in Km,n (Rn ), we obtain n ) and is uniformly bounded. Next, we by Theorem 1.3 that the family V (MÏ ) â C0 (R prove the equicontinuity of functions in V (MÏ ) on Rn âª {â} by same arguments as in the proof of [9, Proposition 6]. Thus by Ascoliâs Theorem the family V (MÏ ) is relatively compact in C0 (Rn ). This ends the proof. Remark 2.9. We recall (see [12, 14]) that for m = 1 and n â¥ 3, a radial function is in â â Kn (Rn ) if and only if 0 r |Ï(r)|dr < â. Similarly, we will give in the sequel a characterization of radial functions belonging to â Km,n (Rn ). â Proposition 2.10. Let Ï be a radial function in Rn , then Ï â Km,n (Rn ) if and only if â r 2mâ1 Ï(r) dr < â. (2.28) In order to prove Proposition 2.10, we will use the following behaviour of the mpotential of radial functions on Rn . Proposition 2.11. Let Ï â + (Rn ) be a radial function on Rn , then for x â Rn , we have â V Ï(x) â¼ Proof. Let Ï â + (Rn ). r n â1 Ï(r)dr. |x| â¨ r (2.29) First, we recall the well known results for x, y â Rn , r n â1 nâ2 Ï(r)dr, Rn 0 |x| â¨ r dz cn = . |x â z|nâ2 | y â z|nâ2 |x â y |nâ4 dy = |x â y |nâ2 â (n â 2)k1,n (2.30) Rn This implies that there exists a constant c > 0 such that dy = c |x â y |nâ4 t n â1 dtdr 0 0 (t â¨ r)nâ2 |x| â¨ t â â 1 â¥c r nâ1 Ï(r) dtdr n â3 0 |x|â¨r t â r nâ1 Ï(r) c â¥ dr. n â 4 0 |x| â¨ r nâ4 r nâ1 Ï(r) n â2 â â Rn (2.31) Hence, we obtain by recurrence that â r n â1 Ï(r)dr |x| â¨ r Rn |x â y | (2.32) Solutions of polyharmonic equations in Rn On the other hand, there exists a constant c > 0 such that for each x â Rn , dy = c |x â y | c â Ï 0 Ï 0 r nâ1 Ï(r)(sinÎ¸)nâ2 (|x|2 + r 2 â 2r |x| cosÎ¸)()/2 r nâ1 Ï(r)(sinÎ¸)nâ2 dÎ¸dr (|x| â¨ r) (sinÎ¸) â Rn â dÎ¸dr (2.33) 0 Ï 0 =c (sinÎ¸)2mâ2 dÎ¸ r nâ1 Ï(r) dr . (|x| â¨ r) Thus (2.29) holds. â Proof of Proposition 2.10. Suppose that Ï is a radial function in Km,n (Rn ), then by Theorem 1.3, V Ï(0) < â and so we deduce (2.28) from (2.29). Conversely, suppose that Ï satisï¬es (2.28). Let Î± > 0 and t = |x|, then by (2.29), we have dy |x â y | t+Î± (t âÎ±)+ t+Î± (t âÎ±)+ r n â1 Ï(r) dr (t â¨ r) r 2mâ1 (2.34) Ï(r) dr. Let Ï(s) = 0 r 2mâ1 |Ï(r)|dr, for s â [0, â]. Using (2.28), we deduce that Ï is a continuous function on [0, â]. This implies that t+Î± (t âÎ±)+ r 2mâ1 Ï(r) dr = Ï(t + Î±) â Ï (t â Î±)+ , (2.35) converges to zero as Î± â 0 uniformly for t â [0, â]. So Ï veriï¬es (1.4). Next, we have by (2.29) dy | y |â¥M |x â y | â r n â1 Ï(r) dr (t â¨ r) â r 2mâ1 Ï(r) dr, (2.36) which, using (2.28), tends to zero as M â â and so Ï veriï¬es (1.6). This completes the proof. â We close this section by giving a class of functions included in Km,n (Rn ) and we precise the behaviour of the m-potential of functions in this class. We need the following lemma. Lemma 2.12. Let Î± > 0 and a,b > 0 such that a + b < n. Then dy a b | y | |x â y | Î±nâ(a+b) . (2.37) Proof. Let Î± > 0 and a, b be nonnegative real numbers such that a + b < n. Then dy | y |a |x â y |b dy + a+b ()â©(|xâ y || y |) |x â y | Î± 0 nâ(a+b) dy a+b (| y |) | y | (2.38) r nâ1â(a+b) dr . Î± Proposition 2.13. Let p > n/2m. Then for Î» < 2m â n/ p < Î¼, we have L p Rn 1+|Â·| â â Km,n Rn . (2.39) Proof. Let p > n/2m and q â¥ 1 such that 1/ p + = 1. Let a be a function in L p (Rn ) and Î» < 2m â n/ p < Î¼. First, we will prove that the function Ï(x) := a(x)/(1 + |x|)Î¼âÎ» |x|Î» satisï¬es (1.4). Let Î± > 0, then by the HÂ¨ lder inequality and Lemma 2.12, we have for o x â Rn dy a |x â y | a dy 1 + | y| (Î¼âÎ»)q | y |Î»q |x â y |()q dy qÎ»+ |x â y |()q | y | (2.40) a p Î±2mân/ pâÎ» , which converges to zero as Î± â 0. Secondly, we claim that Ï satisï¬es (1.6). To show the claim we use the HÂ¨ lder inequalo ity. Let M > 1, then we have dy a | y |â¥M |x â y | â¼ a dy | y |â¥M 1 + | y| (Î¼âÎ»)q | y |Î»q |x â y |()q dy Î¼q |x â y |()q | y |â¥M | y | A(x) (2.41) = a Solutions of polyharmonic equations in Rn Furthermore dy | y |(+Î¼)q A(x) |x|M/2 | y |â¥M sup + sup + sup + sup dy ()q (| y |â¥M)â©(|xâ y ||x|/2) |x â y | Î¼q |x|â¥M/2 |x | dy (| y |â¥M)â©(|x|/2|xâ y |2|x|) ()q |x|â¥M/2 |x | | y |Î¼q (2.42) dy |x â y |(+Î¼)q |x|â¥M/2 (| y |â¥M)â©(|xâ y |â¥2|x|) 1 M (+Î¼)qân + sup Log(3|z|/M) , ()q |z|â¥M/2 |z | which converges to zero as M â â. This ends the proof. Remark 2.14. It is obvious to see that for each Ï â km,n |x| + 1 Rn + (Rn ), we have | y| + 1 d y V Ï(x). (2.43) We precise in the following, some upper estimates on the m-potential of functions in the class L p (Rn )/(1 + | Â· |)Î¼âÎ» . Indeed, put for a nonnegative function a â L p (Rn ) and x â Rn â â Wa(x) := V â a 1+|Â·| â (x) = Rn Gm,n (x, y) a(y) 1 + | y| | y |Î» (2.44) Then we have the following. Proposition 2.15. Let p > n/2m and Î» < 2m â n/ p < Î¼. Then there exists c > 0 such that for each nonnegative function a â L p (Rn ) and x â Rn , we have the following estimates â§ 1 âª âª âª âª â¨ 1 + |x| Wa(x) c a p âª âª âª âª â© 1 + |x| p/(pâ1) Log |x| + 1 ()â§(Î¼+n/ pâ2m) , n =n p n if Î¼ + = n. p if Î¼ + (2.45) Proof. Let p > n/2m and q â¥ 1 such that 1/ p + = 1. Let a be a nonnegative function o in L p (Rn ) and Î» < 2m â n/ p < Î¼. Put Ï(x) = a(x)/(1 + |x|)Î¼âÎ» |x|Î» , then by the HÂ¨ lder inequality, we have for each x â Rn V Ï(x) a = a dy Rn (Î¼âÎ»)q |x â y |()q 1 + | y | | y |Î»q (2.46) I(x) Furthermore, (i) if |x| 1, we have by Lemma 2.12, that I(x) dy B(x,2) |x â y |()q | y |qÎ» + + dy B c (x,2) |x â y |()q | y |Î¼q dy B(x,2) dy B c (0,2) (2.47) |x â y |()q | y |qÎ» |x â y |(+Î¼)q 1, (ii) if |x| â¥ 1, we have I(x) dy (| y |1/2) |x â y |()q | y |Î»q dy (| y |â¥1/2)â©(|xâ y ||x|/2) |x â y |()q | y |Î¼q + + dy (| y |â¥1/2)â©(|x|/2|xâ y |2|x|) |x â y |()q | y |Î¼q dy ()q | y |Î¼q (| y |â¥1/2)â©(|xâ y |â¥2|x|) |x â y | |x|()q dy Î»q (| y |1/2) | y | |x|Î¼q dy ()q (|xâ y ||x|/2) |x â y | dy |x|()q (1/2| y |3|x|) | y |Î¼q â§ âªLog |x| + 1 , if Î¼ + âª âª âª âª âª âª âª â¨ 1 |x|nâÎ¼q , if Î¼ + ()q âª |x| âª âª âª âª âª âª1, âª if Î¼ + â© dy (|xâ y |>2|x|) |x â y |(+Î¼)q n =n p n n. p (2.48) By combining the above inequalities, we get the result. â Corollary 2.16. The class of functions Lâ (Rn )/(1 + | Â· |)Î¼âÎ» is included in Km,n (Rn ) if and only if Î» < 2m < Î¼. Solutions of polyharmonic equations in Rn Proof. âââ follows from Proposition 2.13. âââ Suppose that the function Ï deï¬ned on Rn by Ï(x) = 1/(1 + |x|)Î¼âÎ» |x|Î» is in â â Km,n (Rn ). Then by Proposition 2.10, we have 0 r 2mâ1 Ï(r)dr < â. This implies that Î» < 2m < Î¼. Remark 2.17. Let Î» < 2m < Î¼ and Ï(x) = 1/(1 + |x|)Î¼âÎ» |x|Î» , for x â Rn , then by simple calculus, we obtain the following behaviour on the m-potential â§ âª âª âª â¨ âª âª â© V Ï(x) â¼ âª 1 1 + |x| 1 + |x| Log |x| + 1 , if Î¼ = n if Î¼ = n. (2.49) ()â§(Î¼â2m) , 3. First existence result In this section, we aim at proving Theorem 1.4. The following lemmas are useful. â Lemma 3.1. Let Ï be a nonnegative function in Km,n (Rn ). Then we have VÏ â Î±Ï 2 V Ï â. (3.1) Proof. By (1.3) we obtain easily that Î±Ï 2 V Ï â . On the other hand, by letting | y | â â in (1.24), we deduce from Fatou Lemma that V Ï â Î±Ï . â Lemma 3.2. Let Ï be a nonnegative function in Km,n (Rn ). Then for each x â Rn , we have V ÏGm,n (Â·, y) (x) Î±Ï Gm,n (x, y). Proof. The result holds by (1.24). (3.2) â In the sequel, let q be a nonnegative function in Km,n (Rn ) such that Î±q 1/2. For f â + (Rn ), we will deï¬ne the potential kernel V f := V q m,n,q f as a solution for the perturbed polyharmonic equation (1.9). We put for x, y â Rn , m,n (x, y) = âªkâ¥0 â§ âª â¨ (â1)k V (qÂ·) Gm,n (Â·, y) (x), if x = y (3.3) if x = y. â©â, Then we have the following comparison result. â Lemma 3.3. Let q be a nonnegative function in Km,n (Rn ) such that Î±q 1/2. Then for n , we have x, y â R 1 â Î±q Gm,n (x, y) m,n (x, y) Gm,n (x, y). (3.4) Proof. Since Î±q 1/2, we deduce from (3.2), that m,n (x, y) k â¥0 Î±q Gm,n (x, y) 1 1 â Î±q (3.5) Gm,n (x, y). Furthermore, we have for x = y in Rn m,n (x, y) = Gm,n (x, y) â V m,n (Â·, y) (x), (3.6) which together with (3.2), imply that m,n (x, y) â¥ Gm,n (x, y) â Î±q Gm,n (x, y) 1 â Î±q (3.7) 1 â 2Î±q Gm,n (x, y) 1 â Î±q â¥ 0. Hence the result follows from (3.6) and (3.2). Let us deï¬ne the operator Vq on Vq f (x) = Then we obtain the following. Lemma 3.4. Let f â + (Rn ). + (Rn ) by x â Rn . (3.8) m,n (x, y) f (y)d y, Then Vq f satisï¬es the following resolvent equation (3.9) such that V f < â, (3.10) V f = Vq f + Vq (qV f ) = Vq f + V qVq f . Proof. From the expression of m,n , we deduce that for f â (â1)k V (qÂ·) V f . k â¥0 k + (Rn ) Vq f = So we obtain that Vq (qV f ) = =â (â1)k V (qÂ·) k â¥0 V (qV f ) (â1)k V (qÂ·) V f k â¥1 (3.11) = V f â Vq f . The second equality holds by integrating (3.6). Proposition 3.5. Let f â L1 (Rn ) such that V f â L1 (Rn ). Then Vq f is a solution (in loc loc the sense of distributions) of the perturbed polyharmonic equation (1.9). Solutions of polyharmonic equations in Rn Proof. Using the resolvent equation (3.9), we have Vq f = V f â V qVq f . Applying the operator (âÎ)m on both sides of the above equality, we obtain that (âÎ)m Vq f = f â qVq f This completes the proof. Now, we are ready to prove Theorem 1.4. Proof of Theorem 1.4. Let c > 0. Then by (H2 ), there exists a nonnegative function q := â qc â Km,n (Rn ), such that Î±q 1/2 and for each x â Rn , the map t ââ t q(x) â Ï(x,t) is continuous and nondecreasing on [0, c], which implies in particular that for each x â Rn and t â [0,c], 0 Ï(x,t) q(x), Let Î := u â We deï¬ne the operator T on Î by Tu(x) := c 1 â Vq (q)(x) + Vq q â Ï(Â·,u) u (x). First, we prove that Î is invariant under T. Indeed, for each u â Î, we have Tu c 1 â Vq (q)(x) + cVq (q)(x) c. (3.18) (3.17) (3.12) (in the sense of distributions). (3.13) (3.14) (3.15) Rn : 1 â Î± q c u c . (3.16) Moreover, from (3.15), (3.4) and Lemma 3.1 we deduce that for each u â Î, we have Tu â¥ c 1 â Vq (q)(x) â¥ c 1 â V (q)(x) â¥ c 1 â Î±q . (3.19) Next, we prove that the operator T is nondecreasing on Î. Indeed, let u,v â Î such that u v, then from (3.14) we obtain that Tv â Tu = Vq q â Ï(Â·,v) v â q â Ï(Â·,u) u â¥ 0. (3.20) Now, consider the sequence (uk ) deï¬ned by u0 = (1 â Î±q )c and uk+1 = Tuk , for k â N. Then since Î is invariant under T, we obtain obviously that u1 = Tu0 â¥ u0 and so from the monotonicity of T, we have u0 u1 Â· Â· Â· uk c. (3.21) So from (3.14) and the dominated convergence theorem we deduce that the sequence (uk ) converges to a function u â Î which satisï¬es u = c 1 â Vq (q)(x) + Vq q â Ï(Â·,u) u (x). That is u â Vq (qu) = c 1 â Vq (q)(x) â Vq uÏ(Â·,u) . (3.23) (3.22) Applying the operator (I + V (qÂ·)) on both sides of the above equality and using (3.9) we deduce that u satisï¬es u = c â V uÏ(Â·,u) . (3.24) Finally, we claim that u is a positive continuous solution for the Problem (1.6). To prove the claim, we use Lemma 2.4. Indeed, since u â¼ c on Rn and 0 uÏ(Â·,u) cq, (3.25) we deduce that either u and uÏ(Â·,u) are in L1 (Rn ). loc Now, from (3.24) we can easily see that V (uÏ(Â·,u)) â L1 (Rn ). Hence u satisï¬es (in loc the sense of distributions) the elliptic diï¬erential equation (âÎ)m u + uÏ(Â·,u) = f in Rn . (3.26) On the other hand, it follows from (3.25) that uÏ(Â·,u) â Mq and so by Proposition 2.8, + we obtain that V (uÏ(Â·,u)) is in C0 (Rn ). This implies by (3.24) that lim u(x) = c, which completes the proof. Remark 3.6. Let c > 0 and u be a solution of (1.8). Then we have by Theorem 1.4 that â for each x â Rn , 0 u(x) c. Let q be the nonnegative function in Km,n (Rn ) given in the proof of Theorem 1.4. Then we deduce from (3.24) and (3.25), tha c â u(x) = V uÏ(Â·,u) (x) cV (q)(x). (3.27) Example 3.7. Let p > n/2m and a be a nonnegative function in L p (Rn ). Let Î» < 2m â n/ p < Î¼ and Î±, Î² be two nonnegative constants. Put q(x) = a(x)/(1 + |x|)Î¼âÎ» |x|Î» . Then, for each c > 0, the following polyharmonic problem (â )m u + Î²uÎ±+1 q = 0, in Rn (in the sense of distributions) lim u(x) = c, (3.28) has a positive continuous solution satisfying c/2 u(x) c, provided that Î² is suï¬ciently small. Solutions of polyharmonic equations in Rn Moreover, by Remark 3.6 and Proposition 2.15, we have â§ 1 âª âª âª âª â¨ 1 + |x| 0 c â u(x) c a p âª âª âª âª â© 1 + |x| Log |x| + 1 p/(pâ1) ()â§(Î¼+n/ pâ2m) , n =n p n if Î¼ + = n. p if Î¼ + (3.29) Remark 3.8. It is interesting to compare the asymptotics (3.29) with the results of Trubek [10], for the case m = 1. 4. Second existence result In this section, we aim at proving Theorem 1.5. Proof of Theorem 1.5. Assuming (H3 ) and (H4 ), we will use the Schauder ï¬xed point theorem. From (1.14), there exists Î· > 0 such that h(t) â¥ m0 t, for each t â [0,Î·]. (4.1) On the other hand, let Î± â (g â ,M0 ), then by (1.15), there exists Ï > 0 such that for t â¥ Ï, we have g(t) Î±t. Put Î² = sup0tÏ g(t). So we deduce tha g(t) Î±t + Î², for each t â¥ 0. (4.2) By Remark 2.14, we note that there exists a constant Î±1 > 0 such that Î±1 1 + |x| V q(x). â )}. (4.3) Let a â (0,Î·) and b = max{a/Î±1 , Î²/(1 â Î± V q Î = u â C0 Rn , a 1 + |x| So we consider the closed convex set (4.4) u(x) bV q(x), âx â Rn . Obviously by (4.3) we have that the set Î is nonempty. Next we deï¬ne the operator T on Î by Tu(x) = Rn Gm,n (x, y) f y,u(y) (4.5) Let us prove that TÎ â Î. Let u â Î, then by (4.2) we have Tu(x) Rn Gm,n (x, y)q(y)g u(y) d y Gm,n (x, y)q(y) Î±u(y) + Î² d y â Rn (4.6) Î±b V q bV q(x). + Î² V q(x) Moreover, since h is nondecreasing, we deduce by (4.1) and (1.14) that Tu(x) â¥ â¥ Rn Gm,n (x, y)p(y)h u(y) d y Gm,n (x, y)p(y)h Gm,n (x, y) a 1 + | y| p(y) Rn dy (4.7) â¥ m0 a â¥ Rn m0 akm,n 1 + |x| a = . 1 + |x| 1 + | y| p(y) 1 + | y| d y Rn 2() d y On the other hand, by (1.13), we have that for each u â Î f (Â·,u) g b V q â q. (4.8) This implies by Proposition 2.8 that Tu â V (Mq ) â C0 (Rn ). So TÎ â Î. Next, we prove the continuity of T in Î. Let (uk ) be a sequence in Î, which converges uniformly to a function u â Î. Then using (4.8) and (H3 ), we deduce by Theorem 1.3 and the dominated convergence Theorem that for x â Rn , Tuk (x) ââ Tu(x) as k ââ â. (4.9) Now, since TÎ â V (Mq ), we deduce by Proposition 2.8 that TÎ is relatively compact in C0 (Rn ), which implies that Tuk â Tu â â 0 â as k â â. â (4.10) Hence T is a compact map from Î to itself. So the Schauder ï¬xed point theorem leads to the existence of u â Î such that u = V f (Â·,u) . (4.11) Finally by (4.8) and Lemma 2.4, we conclude that y â f (y,u(y)) is in L1 (Rn ), which loc together with (4.11) imply that u satisï¬es (in the sense of distributions) the elliptic differential equation (â )m u = f (Â·,u) This ends the proof. â Example 4.1. Let p be a nonnegative function in Km,n (Rn ) and 0 Î± < 1. Then the following problem in Rn . (4.12) (â )m u + p(x)uÎ± = 0, x â Rn , (4.13) lim u(x) = 0, Solutions of polyharmonic equations in Rn has a positive solution u â C0 (Rn ) satisfying for each x â Rn 1 1 + |x| 5. Third existence result In this section, we aim at proving Theorem 1.6. Proof of Theorem 1.6. Let c > 0 be the constant given by (H7 ) and câ = c â V (q(Â·,c)) â . Let Î´ â (0,câ ]. We will use the Schauder ï¬xed point theorem, so we consider the closed convex set Î = u â C Rn âª {â} : Î´ u(x) c, âx â Rn and we deï¬ne the integral operator T on Î by Tu(x) = Î´ + V f (Â·,u) (x). (5.2) (5.1) u(x) V p(x). (4.14) First, we prove that TÎ â Î. Let u â Î, then since f is a nonnegative function, we have that Tu(x) â¥ Î´, for each x â Rn . Moreover by (H6 ), we have for x â Rn , Tu(x) Î´ + V q(Â·,u) (x) câ + V q(Â·,c) (x) c. (5.3) Furthermore by (H7 ), since for all u â Î, f (Â·,u) â Mq(Â·,c) , then it follows from Proposition 2.8 that V ( f (Â·,u)) â C0 (Rn ) and more precisely TÎ is relatively compact in C(Rn âª {â}). Therefore TÎ â Î. Next, let us prove the continuity of T in Î. Let (uk ) be a sequence in Î, which converges uniformly to a function u â Î. Since f is continuous with respect to the second variable, we deduce by the dominated convergence theorem that for each x â Rn âª {â}, Tuk (x) ââ Tu(x) as k ââ â. Now, since TÎ is relatively compact in C(Rn âª {â}), then Tuk â Tu â (5.4) â 0 â as k â â. â (5.5) Finally the Schauder ï¬xed point theorem implies the existence of u â Î such that u(x) = Î´ + V f (Â·,u) (x), â x â Rn . (5.6) Using (H6 ), (H7 ) and Lemma 2.4, we deduce that the function y â f (y,u(y)) is in L1 (Rn ). So u satisï¬es (in the sense of distributions) the elliptic diï¬erential equation loc (â )m u = f (Â·,u) in Rn . (5.7) Moreover since V ( f (Â·,u)) â C0 (Rn ), then by (5.6) it follows that lim u(x) = Î´. This ends the proof. Corollary 5.1. Assume that q(x,t) = p(x)g(t), where g is a nonnegative nondecreasing â measurable function and p is a nonnegative function in Km,n (Rn ). If the function g satisï¬es either g(t) = o(t) as t â 0 or g(t) = o(t) as t â â, then the problem (1.19) has a positive solution u â C(Rn âª {â}). Example 5.2. Among the equations of form (1.1), we have the Emden-Fowler equation of order m (â )m u + p(x)uÎ± = 0, Î± > 0, x â Rn , n > 2m, (5.8) â where p â Km,n (Rn ). (i) For the sublinear (0 < Î± < 1) or the superlinear (Î± > 1) case, let c > 0 such that Vp âc Î± â1 < 1. (5.9) Then applying Theorem 1.6, we deduce that for each Î´ â (0,c(1 â cÎ±â1 V p â )), (5.8) with Î± = 1 has a continuous positive solution u in Rn with Î´ u(x) c, for all x â Rn and lim u(x) = Î´. (ii) For the linear case (Î± = 1). If V p â < 1, then applying Theorem 1.6, we deduce that for each c > 0 and Î´ â (0,c(1 â V p â )), (5.8) has a continuous positive solution u in Rn with Î´ u(x) c, for all x â Rn and lim u(x) = Î´. Remark 5.3. We improve in this section the Yinâs result in [11]. Indeed, Yin proved in particular the existence of bounded positive solutions for the Emden-Fowler equation u + p(x)uÎ± = 0, provided that the function p satisï¬es â 0 < Î± = 1, x â Rn , n â¥ 3, (5.10) smax p(x) ds < â. |x|=s (5.11) However by taking Î» > (n â 1)/2 and p(x) = p(x ,xn ) = then we have max p(x) â¥ p(0,s) = |x|=s 2 1 + xn 1+ , n â1 2 Î» i =1 x i x â Rn , (5.12) 1 1 + s2 (5.13) which implies that (5.11) is not satisï¬ed. On the other hand, we have that p â Lâ (Rn ) â© â L1 (Rn ) â Km,n (Rn ). This implies by Corollary 5.1 that the Emden-Fowler equation (5.8) has a positive solution u â C(Rn âª {â}), for each m â¥ 1. Solutions of polyharmonic equations in Rn
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