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Existence of positive solutions for some polyharmonic nonlinear equations in ℝn

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 [1–3, 8, 9, 11, 12, 14] and the references therein). Their basic tools are essentially some properties of functions belonging to the ∞ 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 ∞ to extend the Kato class Kn (Rn ) and the subclass Kn (Rn ) to the order m ≥ 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 (−Δ)m in Rn . Namely, for each x, y in Rn Gm,n (x, y) = km,n 1 , |x − y | (1.2) Hindawi Publishing Corporation Abstract and Applied Analysis Volume 2006, Article ID 76582, Pages 1–24 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) 2−1 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 ) defined as follows. Definition 1.1. A Borel measurable function ϕ in Rn (n > 2m), belongs to the Kato class Km,n (Rn ) if lim sup α→0 (1.3) |x − y | d y = 0. (1.4) Indeed, first we prove some properties of functions belonging to this class similar to those established in [1, 4]. In particular, we have the following characterization ϕ ∈ Km,n Rn ⇐⇒ lim sup t →0 t 0 Rn p(s,x, y) d yds = 0, (1.5) where p(t,x, y) = (1/(4πt)n/2 )exp(− /4t), for t ∈ (0, ∞) and x, y ∈ Rn , is the density of the Gauss semi-group on Rn . ∞ Secondly, we study a subclass of Km,n (Rn ) denoted by Km,n (Rn ) and defined by the following. ∞ Definition 1.2. A Borel measurable function ϕ belongs to the class Km,n (Rn ) and it is n ) and satisfies called m-Green-tight function if ϕ ∈ Km,n (R M →∞ lim | y |≥M sup |x − y | d y = 0. (1.6) ∞ In particular, we characterize the class Km,n (Rn ) as follows. Theorem 1.3. Let ϕ ∈ + (Rn ), (n > 2m). Then the following assertions are equivalent ∞ (1) ϕ ∈ Km,n (Rn ). + (2) The m-potential of ϕ, V ϕ(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 fine characterization will be given in the radial case. ∞ One can easily check that L1 (Rn ) ∩ Km,n (Rn ) ⊂ Km,n (Rn ). Also we show that for p > n/2m and λ < 2m − n/ p < μ, we have L p Rn 1+|·| ∞ ⊂ Km,n Rn , (1.7) and we precise the behaviour of the m-potential of functions in this class. H. Mˆ agli and M. Zribi 3 a In Section 3, we are interested in the following polyharmonic problem (− )m u + uϕ(·,u) = 0, in Rn (in the sense of distributions) (1.8) lim u(x) = c > 0. The function ϕ is required to verify the following assumptions. (H1 ) ϕ is a nonnegative measurable function on Rn × (0, ∞). ∞ (H2 ) For each λ > 0, there exists a nonnegative function qλ ∈ Km,n (Rn ) with αqλ 1/2 n , the mapping t → t(q (x) − ϕ(x,t)) is continu(see (1.24)) and such that for each x ∈ R λ ous and nondecreasing on [0,λ]. 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 ∈ Rn , c/2 u(x) c. To establish this result, we use a potential theory approach. In particular, we prove that ∞ if the function q ∈ Km,n (Rn ) is sufficiently small and f is a nonnegative function on Rn , then the equation (− )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: (− )m u + |k |<2m ak (u)Dk u = f . (1.10) In the case m = 1, the problem (1.8) has been studied by Mˆ agli and Masmoudi in [7, 8], a where they gave an existence and an uniqueness result in both bounded and unbounded domain Ω. In Section 4, we are concerned with the following polyharmonic problem (− )m u = f (·,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 × (0, ∞), continuous with respect to the second variable. (H4 ) There exist a nonnegative function p in Rn such tha < α0 := p(y) Rn | y| + 1 2() d y <∞ (1.12) ∞ and a nonnegative function q ∈ Km,n (Rn ) such that for x ∈ 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, ∞) satisfying m0 := 1 h(t) < h0 := liminf ∞ t →0+ km,n α0 t (1.14) and g is a nonnegative measurable function locally bounded on [0, ∞) satisfying 0 g ∞ := limsup t →∞ g(t) 1 < M0 := t Vq ∞ (1.15) By using a fixed 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 ∈ 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 satisfies t →0+ x∈B u(x) bV q(x), (1.16) lim min f (x,t) = +∞, t t →+∞ x∈B 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 Mˆ agli and Masmoudi in [8]. In fact in [8], the authors studied (1.11) in an a unbounded domain D of Rn , n ≥ 3, with compact nonempty boundary ∂D 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 −Δu = ρ(x)uα in Rn , (1.18) liminf u(x) = 0, with 0 < α < 1 and ρ 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 Mˆ agli and Masmoudi in [8]. a In Section 5, we will study the existence of solutions to the following polyharmonic problem (− )m u = f (·,u), in Rn (in the sense of distributions) u(x) > 0, in Rn , (1.19) under the following assumptions on the nonlinearity f . H. Mˆ agli and M. Zribi 5 a (H5 ) f is a nonnegative measurable function on Rn × (0, ∞), continuous with respect to the second variable on (0, ∞). (H6 ) f (x,t) q(x,t), where q is a nonnegative measurable function on Rn × (0, ∞) such that the function t → q(x,t) is nondecreasing on (0, ∞). ∞ (H7 ) There exists a constant c > 0 such that q(·,c) ∈ Km,n (Rn ) and V q(·,c) Put c∗ = c − V (q(·,c)) ∞. ∞ < c. (1.20) We give in this section the following existence result. Theorem 1.6. Assume (H5 ), (H6 ), and (H7 ). Then for each δ ∈ (0,c∗ ], the problem (1.19) has a positive continuous solution u in Rn satisfying for each x ∈ Rn δ u(x) c, lim u(x) = δ. (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 ∈ Rn , |x| > B }, for some B ≥ 0. His method relies on the technique of radial super/subsolutions. Our approach is different, in fact we will use a fixed point argument. We improve the result of Yin under more general assumptions (see Remark 5.3). In order to simplify our statements, we define 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 ϕ ∈ + (Rn ), we put the m-potential of ϕ on Rn by V ϕ(x) := Vm,n ϕ(x) = (iv) For ϕ ∈ + (Rn ), Rn Gm,n (x, y)d y = km,n Rn |x − y | (1.23) we put Gm,n (x,z)Gm,n (z, y) ϕ(z) dz. Gm,n (x, y) (1.24) αϕ = sup x,y ∈Rn Rn (v) Let λ ∈ R, we denote by λ+ = max (λ,0). (vi) Let f and g be two positive functions on a set S. We call f ∼ g, if there is c > 0 such that 1 g(x) f (x) cg(x) ∀x ∈ 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) ∀x ∈ S. (1.26) The following properties will be used several times: for s,t ≥ 0, we have min(s,t) = s ∧ t ∼ (s + t) p ∼ 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 ∞ 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/(4πt)n/2 )exp(− /4t), for t ∈ (0, ∞) and x, y ∈ Rn , the density of the Gauss semi-group on Rn . By a simple computation, we obtain that the Green function of (−Δ)m in Rn , for each m ≥ 1, is given by Gm,n (x, y) = 1 (m − 1)! ∞ st , s+t p ∈ 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 − y | for x, y in Rn , (2.2) where km,n = Γ(n/2 − m)/4m π n/2 (m − 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 ≥ 1. Remark 2.1. Let j,m ∈ N such that 1 j m, then we have for each n > 2m Kn (Rn ) := K1,n (Rn ) ⊆ K j,n (Rn ) ⊆ Km,n (Rn ), where Kn (Rn ) is the classical Kato class introduced in [1]. Example 2.2. Let ϕ ∈ (Rn ). Suppose that for p > n/2m, we have sup d y < ∞. (2.3) |x− y |1 (2.4) Then by the H¨ lder inequality, we conclude that ϕ ∈ Km,n (Rn ). o In particular, we have that for p > n/2m, L p (Rn ) ⊂ Km,n (Rn ). To establish the characterization (1.5) of the Kato class Km,n (Rn ), we need the following lemmas. H. Mˆ agli and M. Zribi 7 a Lemma 2.3. For each t > 0 and x, y ∈ Rn , we have p(s,x, y)ds Gm,n (x, y). (2.5) Moreover, for |x − y | 2 t, we have that √ Gm,n (x, y) p(s,x, y)ds. (2.6) Proof. Let t > 0 and x, y ∈ Rn . Then (2.5) follows immediately from (2.1). √ If we suppose further that |x − y | 2 t, then we have p(s,x, y)ds = c sm−n/2−1 exp − ∞ 4s ds (2.7) c |x − y | c ≥ |x − y | = c Gm,n (x, y), |x− y |2 /4t ∞ n/2−m−1 −r r n/2−m−1 e−r dr e dr where the letter c is a positive constant which may vary from line to line. Lemma 2.4. Let ϕ ∈ Km,n (Rn ). Then for each compact L ⊂ Rn , we have sup d y < ∞. (2.8) x+L In particular, we have Km,n (Rn ) ⊂ L1 (Rn ). loc Proof. Let ϕ ∈ Km,n (Rn ), then by (1.4) there exists α > 0 such that sup 1. |x − y | 1i p B(ai ,α). p (2.9) Let a1 ,...,a p ∈ L such that L ⊆ Hence for each x ∈ Rn , we have d y x + ai − y d y x+L i=1 B(x+ai ,α) p α (2.10) i=1 B(x+ai ,α) pα So, sup x+L ||d y < ∞. Solutions of polyharmonic equations in Rn Proposition 2.5. Let ϕ ∈ Km,n (Rn ). Then for each fixed α > 0, we have sup sup 0t 1 |x− y |≥α t m−1 p(t,x, y) d y := M(α) < ∞. (2.11) Proof. Let ϕ ∈ Km,n (Rn ), 0 < t 1. Let α > 0, then we have that sup t m−1 p(t,x, y) d y exp − |x− y |≥α exp − α2 /8t sup t n/2−m+1 So to prove (2.11), we need to show that sup exp − (2.12) Rn Rn d y < ∞. (2.13) Indeed, using Lemma 2.4, we denote by c := sup d y < ∞. (2.14) x+B(0,1) On the other hand, since any ball B(0,k) of radius k ≥ 1 in Rn can be covered by α(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 ,...,aα(n) ∈ B(0,k) such that B(0,k) ⊂ 1iα(n) B ai ,1 . (2.15) Hence for each x ∈ Rn , we have α(n) x+B(0,k) i=1 B(x+ai ,1) cAn kn , (2.16) which implies that for each x ∈ Rn , exp − ∞ Rn k =0 exp − ∞ k2 8 k|x− y |k+1 d y (2.17) cAn k =0 exp − k2 (k + 1)n 8 < ∞. Thus (2.13) holds. This ends the proof. H. Mˆ agli and M. Zribi 9 a Proposition 2.6. Let ϕ ∈ B(Rn ). Then ϕ ∈ Km,n (Rn ) if and only if lim sup t →0 0 Rn p(s,x, y) d yds = 0. (2.18) Proof. Suppose ϕ verifies (2.18), then from (2.6) we have that dy |x − y | α2 /4 Rn p(s,x, y) dsd y, (2.19) which implies that the function ϕ satisfies (1.4). Conversely, suppose that ϕ ∈ Km,n (Rn ). Let ε > 0, then by (1.4), there exists α > 0 such that sup ε. |x − y | (2.20) Thus from (2.5) and (2.11), we deduce that for each x ∈ 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) |x− y |≥α d y + tM(α) |x − y | ε + tM(α). This implies (2.18) and completes the proof. ∞ 2.2. The class Km,n (Rn ). We will characterize the subclass of m-Green-tight functions ∞ Km,n (Rn ). In fact, we will prove Theorem 1.3 and we give in particular a more precise characterization in the radial case. ∞ Example 2.7. Let p > n/2m. Then L p (Rn ) ∩ L1 (Rn ) ⊂ Km,n (Rn ). ∞ Proof of Theorem 1.3. Let ϕ ∈ + (Rn ). First we suppose that ϕ ∈ Km,n (Rn ), then using + similar arguments as in the proof [9, Proposition 6], we obtain easily that V ϕ ∈ C0 (Rn ). + n ). Then, we aim at proving that ϕ ∈ K ∞ (Rn ). Conversely we suppose that V ϕ ∈ C0 (R m,n So we divide the proof into two steps. Solutions of polyharmonic equations in Rn Step 1. We will prove that ϕ satisfies (2.18). Indeed it is clear from (2.1), that for each x ∈ Rn , we have that V ϕ(x) = p(s,x, y)d yds (m − 1)! 0 Rn ∞ 1 p(s,x, y)d yds + (m − 1)! t Rn = I1 (x) + I2 (x). 1 (2.22) From the properties of the density p(s,x, y), we deduce that x → I1 (x) and x → I2 (x) are nonnegative lower semi-continuous functions in Rn . Then using the fact that V ϕ ∈ + + C0 (Rn ), we get that the function x → I1 (x) is also in C0 (Rn ). So, for each x ∈ Rn , the t m−1 + 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 ∈ R lim t →0 0 Rn p(s,x, y)d yds = 0 (2.23) imply by Dini Lemma, that (2.18) is satisfied. + Step 2. We will prove that ϕ satisfies (1.6). Let ε > 0, then since V ϕ ∈ C0 (Rn ), there exists a > 0 such that for |x| ≥ a, we have that V ϕ(x) ε. Let M ≥ 2a, then sup sup | y |≥M |x − y | |x|≥a ε+ Rn d y + sup |x − y | |x|a | y | dy | y |≥M |x − y | | y |≥M (2.24) Now, since V ϕ(0) < ∞, we deduce that lim d y = 0. | y | (2.25) M →∞ | y |≥M Then (1.6) holds and this ends the proof. ∞ For a nonnegative function ρ in Km,n (Rn ), we denote by Mρ := ϕ ∈ Rn , |ϕ| ρ . (2.26) ∞ Proposition 2.8. For a nonnegative function ρ in Km,n (Rn ), the family of functions V Mρ := V ϕ, ϕ ∈ Mρ (2.27) is uniformly bounded and equicontinuous in C0 (Rn ) and consequently it is relatively compact in C0 (Rn ). ∞ ∞ Proof. Let ρ ∈ Km,n (Rn ). Obviously, since each function ϕ in Mρ is in Km,n (Rn ), we obtain n ) and is uniformly bounded. Next, we by Theorem 1.3 that the family V (Mρ ) ⊂ C0 (R prove the equicontinuity of functions in V (Mρ ) on Rn ∪ {∞} by same arguments as in the proof of [9, Proposition 6]. Thus by Ascoli’s Theorem the family V (Mρ ) is relatively compact in C0 (Rn ). This ends the proof. Remark 2.9. We recall (see [12, 14]) that for m = 1 and n ≥ 3, a radial function is in ∞ ∞ Kn (Rn ) if and only if 0 r |ϕ(r)|dr < ∞. Similarly, we will give in the sequel a characterization of radial functions belonging to ∞ Km,n (Rn ). ∞ Proposition 2.10. Let ϕ be a radial function in Rn , then ϕ ∈ Km,n (Rn ) if and only if ∞ r 2m−1 ϕ(r) dr < ∞. (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 ϕ ∈ + (Rn ) be a radial function on Rn , then for x ∈ Rn , we have ∞ V ϕ(x) ∼ Proof. Let ϕ ∈ + (Rn ). r n −1 ϕ(r)dr. |x| ∨ r (2.29) First, we recall the well known results for x, y ∈ Rn , r n −1 n−2 ϕ(r)dr, Rn 0 |x| ∨ r dz cn = . |x − z|n−2 | y − z|n−2 |x − y |n−4 dy = |x − y |n−2 ∞ (n − 2)k1,n (2.30) Rn This implies that there exists a constant c > 0 such that dy = c |x − y |n−4 t n −1 dtdr 0 0 (t ∨ r)n−2 |x| ∨ t ∞ ∞ 1 ≥c r n−1 ϕ(r) dtdr n −3 0 |x|∨r t ∞ r n−1 ϕ(r) c ≥ dr. n − 4 0 |x| ∨ r n−4 r n−1 ϕ(r) n −2 ∞ ∞ Rn (2.31) Hence, we obtain by recurrence that ∞ r n −1 ϕ(r)dr |x| ∨ r Rn |x − y | (2.32) Solutions of polyharmonic equations in Rn On the other hand, there exists a constant c > 0 such that for each x ∈ Rn , dy = c |x − y | c ∞ π 0 π 0 r n−1 ϕ(r)(sinθ)n−2 (|x|2 + r 2 − 2r |x| cosθ)()/2 r n−1 ϕ(r)(sinθ)n−2 dθdr (|x| ∨ r) (sinθ) ∞ Rn ∞ dθdr (2.33) 0 π 0 =c (sinθ)2m−2 dθ r n−1 ϕ(r) dr . (|x| ∨ r) Thus (2.29) holds. ∞ Proof of Proposition 2.10. Suppose that ϕ is a radial function in Km,n (Rn ), then by Theorem 1.3, V ϕ(0) < ∞ and so we deduce (2.28) from (2.29). Conversely, suppose that ϕ satisfies (2.28). Let α > 0 and t = |x|, then by (2.29), we have dy |x − y | t+α (t −α)+ t+α (t −α)+ r n −1 ϕ(r) dr (t ∨ r) r 2m−1 (2.34) ϕ(r) dr. Let φ(s) = 0 r 2m−1 |ϕ(r)|dr, for s ∈ [0, ∞]. Using (2.28), we deduce that φ is a continuous function on [0, ∞]. This implies that t+α (t −α)+ r 2m−1 ϕ(r) dr = φ(t + α) − φ (t − α)+ , (2.35) converges to zero as α → 0 uniformly for t ∈ [0, ∞]. So ϕ verifies (1.4). Next, we have by (2.29) dy | y |≥M |x − y | ∞ r n −1 ϕ(r) dr (t ∨ r) ∞ r 2m−1 ϕ(r) dr, (2.36) which, using (2.28), tends to zero as M → ∞ and so ϕ verifies (1.6). This completes the proof. ∞ 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 α > 0 and a,b > 0 such that a + b < n. Then dy a b | y | |x − y | αn−(a+b) . (2.37) Proof. Let α > 0 and a, b be nonnegative real numbers such that a + b < n. Then dy | y |a |x − y |b dy + a+b ()∩(|x− y || y |) |x − y | α 0 n−(a+b) dy a+b (| y |) | y | (2.38) r n−1−(a+b) dr . α Proposition 2.13. Let p > n/2m. Then for λ < 2m − n/ p < μ, we have L p Rn 1+|·| ∞ ⊂ Km,n Rn . (2.39) Proof. Let p > n/2m and q ≥ 1 such that 1/ p + = 1. Let a be a function in L p (Rn ) and λ < 2m − n/ p < μ. First, we will prove that the function ϕ(x) := a(x)/(1 + |x|)μ−λ |x|λ satisfies (1.4). Let α > 0, then by the H¨ lder inequality and Lemma 2.12, we have for o x ∈ Rn dy a |x − y | a dy 1 + | y| (μ−λ)q | y |λq |x − y |()q dy qλ+ |x − y |()q | y | (2.40) a p α2m−n/ p−λ , which converges to zero as α → 0. Secondly, we claim that ϕ satisfies (1.6). To show the claim we use the H¨ lder inequalo ity. Let M > 1, then we have dy a | y |≥M |x − y | ∼ a dy | y |≥M 1 + | y| (μ−λ)q | y |λq |x − y |()q dy μq |x − y |()q | y |≥M | y | A(x) (2.41) = a Solutions of polyharmonic equations in Rn Furthermore dy | y |(+μ)q A(x) |x|M/2 | y |≥M sup + sup + sup + sup dy ()q (| y |≥M)∩(|x− y ||x|/2) |x − y | μq |x|≥M/2 |x | dy (| y |≥M)∩(|x|/2|x− y |2|x|) ()q |x|≥M/2 |x | | y |μq (2.42) dy |x − y |(+μ)q |x|≥M/2 (| y |≥M)∩(|x− y |≥2|x|) 1 M (+μ)q−n + sup Log(3|z|/M) , ()q |z|≥M/2 |z | which converges to zero as M → ∞. This ends the proof. Remark 2.14. It is obvious to see that for each ϕ ∈ km,n |x| + 1 Rn + (Rn ), we have | y| + 1 d y V ϕ(x). (2.43) We precise in the following, some upper estimates on the m-potential of functions in the class L p (Rn )/(1 + | · |)μ−λ . Indeed, put for a nonnegative function a ∈ L p (Rn ) and x ∈ Rn ⎛ ⎞ Wa(x) := V ⎝ a 1+|·| ⎠(x) = Rn Gm,n (x, y) a(y) 1 + | y| | y |λ (2.44) Then we have the following. Proposition 2.15. Let p > n/2m and λ < 2m − n/ p < μ. Then there exists c > 0 such that for each nonnegative function a ∈ L p (Rn ) and x ∈ Rn , we have the following estimates ⎧ 1 ⎪ ⎪ ⎪ ⎪ ⎨ 1 + |x| Wa(x) c a p ⎪ ⎪ ⎪ ⎪ ⎩ 1 + |x| p/(p−1) Log |x| + 1 ()∧(μ+n/ p−2m) , n =n p n if μ + = n. p if μ + (2.45) Proof. Let p > n/2m and q ≥ 1 such that 1/ p + = 1. Let a be a nonnegative function o in L p (Rn ) and λ < 2m − n/ p < μ. Put ϕ(x) = a(x)/(1 + |x|)μ−λ |x|λ , then by the H¨ lder inequality, we have for each x ∈ Rn V ϕ(x) a = a dy Rn (μ−λ)q |x − y |()q 1 + | y | | y |λq (2.46) I(x) Furthermore, (i) if |x| 1, we have by Lemma 2.12, that I(x) dy B(x,2) |x − y |()q | y |qλ + + dy B c (x,2) |x − y |()q | y |μq dy B(x,2) dy B c (0,2) (2.47) |x − y |()q | y |qλ |x − y |(+μ)q 1, (ii) if |x| ≥ 1, we have I(x) dy (| y |1/2) |x − y |()q | y |λq dy (| y |≥1/2)∩(|x− y ||x|/2) |x − y |()q | y |μq + + dy (| y |≥1/2)∩(|x|/2|x− y |2|x|) |x − y |()q | y |μq dy ()q | y |μq (| y |≥1/2)∩(|x− y |≥2|x|) |x − y | |x|()q dy λq (| y |1/2) | y | |x|μq dy ()q (|x− y ||x|/2) |x − y | dy |x|()q (1/2| y |3|x|) | y |μq ⎧ ⎪Log |x| + 1 , if μ + ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ 1 |x|n−μq , if μ + ()q ⎪ |x| ⎪ ⎪ ⎪ ⎪ ⎪ ⎪1, ⎪ if μ + ⎩ dy (|x− y |>2|x|) |x − y |(+μ)q n =n p n n. p (2.48) By combining the above inequalities, we get the result. ∞ Corollary 2.16. The class of functions L∞ (Rn )/(1 + | · |)μ−λ is included in Km,n (Rn ) if and only if λ < 2m < μ. Solutions of polyharmonic equations in Rn Proof. “⇐” follows from Proposition 2.13. “⇒” Suppose that the function ϕ defined on Rn by ϕ(x) = 1/(1 + |x|)μ−λ |x|λ is in ∞ ∞ Km,n (Rn ). Then by Proposition 2.10, we have 0 r 2m−1 ϕ(r)dr < ∞. This implies that λ < 2m < μ. Remark 2.17. Let λ < 2m < μ and ϕ(x) = 1/(1 + |x|)μ−λ |x|λ , for x ∈ Rn , then by simple calculus, we obtain the following behaviour on the m-potential ⎧ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ V ϕ(x) ∼ ⎪ 1 1 + |x| 1 + |x| Log |x| + 1 , if μ = n if μ = n. (2.49) ()∧(μ−2m) , 3. First existence result In this section, we aim at proving Theorem 1.4. The following lemmas are useful. ∞ Lemma 3.1. Let ϕ be a nonnegative function in Km,n (Rn ). Then we have Vϕ ∞ αϕ 2 V ϕ ∞. (3.1) Proof. By (1.3) we obtain easily that αϕ 2 V ϕ ∞ . On the other hand, by letting | y | → ∞ in (1.24), we deduce from Fatou Lemma that V ϕ ∞ αϕ . ∞ Lemma 3.2. Let ϕ be a nonnegative function in Km,n (Rn ). Then for each x ∈ Rn , we have V ϕGm,n (·, y) (x) αϕ Gm,n (x, y). Proof. The result holds by (1.24). (3.2) ∞ In the sequel, let q be a nonnegative function in Km,n (Rn ) such that αq 1/2. For f ∈ + (Rn ), we will define 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 ∈ Rn , m,n (x, y) = ⎪k≥0 ⎧ ⎪ ⎨ (−1)k V (q·) Gm,n (·, y) (x), if x = y (3.3) if x = y. ⎩∞, Then we have the following comparison result. ∞ Lemma 3.3. Let q be a nonnegative function in Km,n (Rn ) such that αq 1/2. Then for n , we have x, y ∈ R 1 − αq Gm,n (x, y) m,n (x, y) Gm,n (x, y). (3.4) Proof. Since αq 1/2, we deduce from (3.2), that m,n (x, y) k ≥0 αq Gm,n (x, y) 1 1 − αq (3.5) Gm,n (x, y). Furthermore, we have for x = y in Rn m,n (x, y) = Gm,n (x, y) − V m,n (·, y) (x), (3.6) which together with (3.2), imply that m,n (x, y) ≥ Gm,n (x, y) − αq Gm,n (x, y) 1 − αq (3.7) 1 − 2αq Gm,n (x, y) 1 − αq ≥ 0. Hence the result follows from (3.6) and (3.2). Let us define the operator Vq on Vq f (x) = Then we obtain the following. Lemma 3.4. Let f ∈ + (Rn ). + (Rn ) by x ∈ Rn . (3.8) m,n (x, y) f (y)d y, Then Vq f satisfies the following resolvent equation (3.9) such that V f < ∞, (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 ∈ (−1)k V (q·) V f . k ≥0 k + (Rn ) Vq f = So we obtain that Vq (qV f ) = =− (−1)k V (q·) k ≥0 V (qV f ) (−1)k V (q·) V f k ≥1 (3.11) = V f − Vq f . The second equality holds by integrating (3.6). Proposition 3.5. Let f ∈ L1 (Rn ) such that V f ∈ 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 − V qVq f . Applying the operator (−Δ)m on both sides of the above equality, we obtain that (−Δ)m Vq f = f − 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 := ∞ qc ∈ Km,n (Rn ), such that αq 1/2 and for each x ∈ Rn , the map t −→ t q(x) − ϕ(x,t) is continuous and nondecreasing on [0, c], which implies in particular that for each x ∈ Rn and t ∈ [0,c], 0 ϕ(x,t) q(x), Let Λ := u ∈ We define the operator T on Λ by Tu(x) := c 1 − Vq (q)(x) + Vq q − ϕ(·,u) u (x). First, we prove that Λ is invariant under T. Indeed, for each u ∈ Λ, we have Tu c 1 − 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 − α q c u c . (3.16) Moreover, from (3.15), (3.4) and Lemma 3.1 we deduce that for each u ∈ Λ, we have Tu ≥ c 1 − Vq (q)(x) ≥ c 1 − V (q)(x) ≥ c 1 − αq . (3.19) Next, we prove that the operator T is nondecreasing on Λ. Indeed, let u,v ∈ Λ such that u v, then from (3.14) we obtain that Tv − Tu = Vq q − ϕ(·,v) v − q − ϕ(·,u) u ≥ 0. (3.20) Now, consider the sequence (uk ) defined by u0 = (1 − αq )c and uk+1 = Tuk , for k ∈ N. Then since Λ is invariant under T, we obtain obviously that u1 = Tu0 ≥ u0 and so from the monotonicity of T, we have u0 u1 · · · uk c. (3.21) So from (3.14) and the dominated convergence theorem we deduce that the sequence (uk ) converges to a function u ∈ Λ which satisfies u = c 1 − Vq (q)(x) + Vq q − ϕ(·,u) u (x). That is u − Vq (qu) = c 1 − Vq (q)(x) − Vq uϕ(·,u) . (3.23) (3.22) Applying the operator (I + V (q·)) on both sides of the above equality and using (3.9) we deduce that u satisfies u = c − V uϕ(·,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 ∼ c on Rn and 0 uϕ(·,u) cq, (3.25) we deduce that either u and uϕ(·,u) are in L1 (Rn ). loc Now, from (3.24) we can easily see that V (uϕ(·,u)) ∈ L1 (Rn ). Hence u satisfies (in loc the sense of distributions) the elliptic differential equation (−Δ)m u + uϕ(·,u) = f in Rn . (3.26) On the other hand, it follows from (3.25) that uϕ(·,u) ∈ Mq and so by Proposition 2.8, + we obtain that V (uϕ(·,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 ∞ for each x ∈ 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 − u(x) = V uϕ(·,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 λ < 2m − n/ p < μ and α, β be two nonnegative constants. Put q(x) = a(x)/(1 + |x|)μ−λ |x|λ . Then, for each c > 0, the following polyharmonic problem (− )m u + βuα+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 β is sufficiently small. Solutions of polyharmonic equations in Rn Moreover, by Remark 3.6 and Proposition 2.15, we have ⎧ 1 ⎪ ⎪ ⎪ ⎪ ⎨ 1 + |x| 0 c − u(x) c a p ⎪ ⎪ ⎪ ⎪ ⎩ 1 + |x| Log |x| + 1 p/(p−1) ()∧(μ+n/ p−2m) , n =n p n if μ + = n. p if μ + (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 fixed point theorem. From (1.14), there exists η > 0 such that h(t) ≥ m0 t, for each t ∈ [0,η]. (4.1) On the other hand, let α ∈ (g ∞ ,M0 ), then by (1.15), there exists ρ > 0 such that for t ≥ ρ, we have g(t) αt. Put β = sup0tρ g(t). So we deduce tha g(t) αt + β, for each t ≥ 0. (4.2) By Remark 2.14, we note that there exists a constant α1 > 0 such that α1 1 + |x| V q(x). ∞ )}. (4.3) Let a ∈ (0,η) and b = max{a/α1 , β/(1 − α V q Λ = u ∈ C0 Rn , a 1 + |x| So we consider the closed convex set (4.4) u(x) bV q(x), ∀x ∈ Rn . Obviously by (4.3) we have that the set Λ is nonempty. Next we define the operator T on Λ by Tu(x) = Rn Gm,n (x, y) f y,u(y) (4.5) Let us prove that TΛ ⊂ Λ. Let u ∈ Λ, 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) αu(y) + β d y ∞ Rn (4.6) αb V q bV q(x). + β V q(x) Moreover, since h is nondecreasing, we deduce by (4.1) and (1.14) that Tu(x) ≥ ≥ 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) ≥ m0 a ≥ 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 ∈ Λ f (·,u) g b V q ∞ q. (4.8) This implies by Proposition 2.8 that Tu ∈ V (Mq ) ⊂ C0 (Rn ). So TΛ ⊂ Λ. Next, we prove the continuity of T in Λ. Let (uk ) be a sequence in Λ, which converges uniformly to a function u ∈ Λ. Then using (4.8) and (H3 ), we deduce by Theorem 1.3 and the dominated convergence Theorem that for x ∈ Rn , Tuk (x) −→ Tu(x) as k −→ ∞. (4.9) Now, since TΛ ⊂ V (Mq ), we deduce by Proposition 2.8 that TΛ is relatively compact in C0 (Rn ), which implies that Tuk − Tu ∞ − 0 → as k − ∞. → (4.10) Hence T is a compact map from Λ to itself. So the Schauder fixed point theorem leads to the existence of u ∈ Λ such that u = V f (·,u) . (4.11) Finally by (4.8) and Lemma 2.4, we conclude that y → f (y,u(y)) is in L1 (Rn ), which loc together with (4.11) imply that u satisfies (in the sense of distributions) the elliptic differential equation (− )m u = f (·,u) This ends the proof. ∞ Example 4.1. Let p be a nonnegative function in Km,n (Rn ) and 0 α < 1. Then the following problem in Rn . (4.12) (− )m u + p(x)uα = 0, x ∈ Rn , (4.13) lim u(x) = 0, Solutions of polyharmonic equations in Rn has a positive solution u ∈ C0 (Rn ) satisfying for each x ∈ 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 c∗ = c − V (q(·,c)) ∞ . Let δ ∈ (0,c∗ ]. We will use the Schauder fixed point theorem, so we consider the closed convex set Λ = u ∈ C Rn ∪ {∞} : δ u(x) c, ∀x ∈ Rn and we define the integral operator T on Λ by Tu(x) = δ + V f (·,u) (x). (5.2) (5.1) u(x) V p(x). (4.14) First, we prove that TΛ ⊂ Λ. Let u ∈ Λ, then since f is a nonnegative function, we have that Tu(x) ≥ δ, for each x ∈ Rn . Moreover by (H6 ), we have for x ∈ Rn , Tu(x) δ + V q(·,u) (x) c∗ + V q(·,c) (x) c. (5.3) Furthermore by (H7 ), since for all u ∈ Λ, f (·,u) ∈ Mq(·,c) , then it follows from Proposition 2.8 that V ( f (·,u)) ∈ C0 (Rn ) and more precisely TΛ is relatively compact in C(Rn ∪ {∞}). Therefore TΛ ⊂ Λ. Next, let us prove the continuity of T in Λ. Let (uk ) be a sequence in Λ, which converges uniformly to a function u ∈ Λ. Since f is continuous with respect to the second variable, we deduce by the dominated convergence theorem that for each x ∈ Rn ∪ {∞}, Tuk (x) −→ Tu(x) as k −→ ∞. Now, since TΛ is relatively compact in C(Rn ∪ {∞}), then Tuk − Tu ∞ (5.4) − 0 → as k − ∞. → (5.5) Finally the Schauder fixed point theorem implies the existence of u ∈ Λ such that u(x) = δ + V f (·,u) (x), ∀ x ∈ Rn . (5.6) Using (H6 ), (H7 ) and Lemma 2.4, we deduce that the function y → f (y,u(y)) is in L1 (Rn ). So u satisfies (in the sense of distributions) the elliptic differential equation loc (− )m u = f (·,u) in Rn . (5.7) Moreover since V ( f (·,u)) ∈ C0 (Rn ), then by (5.6) it follows that lim u(x) = δ. This ends the proof. Corollary 5.1. Assume that q(x,t) = p(x)g(t), where g is a nonnegative nondecreasing ∞ measurable function and p is a nonnegative function in Km,n (Rn ). If the function g satisfies either g(t) = o(t) as t → 0 or g(t) = o(t) as t → ∞, then the problem (1.19) has a positive solution u ∈ C(Rn ∪ {∞}). Example 5.2. Among the equations of form (1.1), we have the Emden-Fowler equation of order m (− )m u + p(x)uα = 0, α > 0, x ∈ Rn , n > 2m, (5.8) ∞ where p ∈ Km,n (Rn ). (i) For the sublinear (0 < α < 1) or the superlinear (α > 1) case, let c > 0 such that Vp ∞c α −1 < 1. (5.9) Then applying Theorem 1.6, we deduce that for each δ ∈ (0,c(1 − cα−1 V p ∞ )), (5.8) with α = 1 has a continuous positive solution u in Rn with δ u(x) c, for all x ∈ Rn and lim u(x) = δ. (ii) For the linear case (α = 1). If V p ∞ < 1, then applying Theorem 1.6, we deduce that for each c > 0 and δ ∈ (0,c(1 − V p ∞ )), (5.8) has a continuous positive solution u in Rn with δ u(x) c, for all x ∈ Rn and lim u(x) = δ. Remark 5.3. We improve in this section the Yin’s result in [11]. Indeed, Yin proved in particular the existence of bounded positive solutions for the Emden-Fowler equation u + p(x)uα = 0, provided that the function p satisfies ∞ 0 < α = 1, x ∈ Rn , n ≥ 3, (5.10) smax p(x) ds < ∞. |x|=s (5.11) However by taking λ > (n − 1)/2 and p(x) = p(x ,xn ) = then we have max p(x) ≥ p(0,s) = |x|=s 2 1 + xn 1+ , n −1 2 λ i =1 x i x ∈ Rn , (5.12) 1 1 + s2 (5.13) which implies that (5.11) is not satisfied. On the other hand, we have that p ∈ L∞ (Rn ) ∩ ∞ L1 (Rn ) ⊂ Km,n (Rn ). This implies by Corollary 5.1 that the Emden-Fowler equation (5.8) has a positive solution u ∈ C(Rn ∪ {∞}), for each m ≥ 1. Solutions of polyharmonic equations in Rn http://www.deepdyve.com/assets/images/DeepDyve-Logo-lg.png

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