Existence Results for Polyharmonic Boundary Value Problems in the Unit Ball

Here we study the polyharmonic nonlinear elliptic boundary value problem on the unit ball B in Rn (n ≥ 2) (− )u + g(·,u) = 0, in B (in the sense of distributions) lim x→ξ∈∂B(u(x)/(1−|x|2)) = 0(ξ). Under appropriate conditions related to a Kato class on the nonlinearity g(x, t), we give some existence results. Our approach is based on estimates for the polyharmonic Green function on B with zero Dirichlet boundary conditions, including a 3G-theorem, which leeds to some useful properties on functions belonging to the Kato class.


Introduction
In this paper, we deal with higher-order elliptic Dirichlet problems (− ) m u + g(•,u) = 0, in B (in the sense of distributions), where B is the unit ball in R n (n ≥ 2), m is a positive integer, and θ is a nontrivial nonnegative continuous function on ∂B.
A basic result goes from Boggio in [1], where he gave an explicit formula for the Green function G m,n of (− ) m on B with Dirichlet boundary conditions (∂/∂ν) j u = 0, 2 Abstract and Applied Analysis 0 ≤ j ≤ m − 1. Namely, Boggio showed that the Green function is positive and is given by with k m,n is a positive constant and [x, y] 2 = |x − y| 2 + (1 − |x| 2 )(1 − |y| 2 ), for x, y in B.
The positivity of G m,n does not hold for the Green function of the m-polyharmonic operator in an arbitrary bounded domain (see, e.g., [2]).Only for the case m = 1, we do not have this restriction.
In [3], Grunau and Sweers established two-sided estimates for G m,n and so they derived a 3G-theorem result.This was improved in [4], where the authors obtained from Boggio's formula more fine estimates on G m,n .For instance, they gave a new form of the 3G-theorem: There exists C 0 > 0 such that for each x, y,z ∈ B, G m,n (x,z)G m,n (z, y) G m,n (x, y) where δ(x) = 1 − |x|.In the case m = 1, the Green function G Ω of an arbitrary bounded C 1,1 domain Ω satisfies (1.3).This has been proved by Kalton and Verbitsky [5] for n ≥ 3 and by Selmi [6] for n = 2.
On the other hand, the classical 3G-theorem related to G Ω (see [7,8]) has been exploited to introduce the classical Kato class of functions K n (Ω) (see [9,7]), which was widely used in the study of some nonlinear differential equations (see [10,11]).Similarly, in [4] the authors exploited the inequality (1.3)  (1.4) In this paper, we will use properties of this class to investigate two existence results for problem (1.1).Our plan is as follows.In Section 2, we collect some properties of functions belonging to K. In particular, we derive from the 3G-theorem that for each q ∈ K, we have In Section 3, we are interested in the following polyharmonic problem: where θ is a nontrivial nonnegative continuous function on ∂B and the functions ϕ and f verify the following assumptions.(H 1 ) ϕ is a nonnegative measurable function on B × (0,∞).(H 2 ) For each λ > 0, there exists q λ ∈ K + with α qλ ≤ 1/2 and such that for each x ∈ B, the map t → t(q λ (x) − ϕ(x,t)) is continuous and nondecreasing on [0,λ].(H 3 ) f is a nonnegative measurable function on B such that the function γ(x) := f (x)/(δ(x)) m−1 belongs to the class K.Under these hypotheses, we give an existence result for problem (1.6).In fact, we will prove that (1.6) has a positive continuous solution u satisfying for where c is a positive constant, V f (x) = B G m,n (x, y) f (y)dy, and the function ρ is de- with h being the continuous solution of the Dirichlet problem h = 0, on B and h |∂B = θ.
To establish this result, we will exploit the 3G-theorem to prove that if the coefficient q ∈ K + is sufficiently small and f is a positive function on B, then the equation has a positive solution on B. In [12], Grunau and Sweers gave a similar result with operators perturbed by small lower-order terms: In the case m = 1, problem (1.6) has been studied by Mâagli and Masmoudi in [13], where they gave an existence and a uniqueness result in a bounded domain Ω.
In Section 4, we fix a positive harmonic function h 0 in B, continuous in B and we put ρ 0 (x) = (1 − |x| 2 ) m−1 h 0 (x).Then we aim at proving an existence result for problem (1.1) with g satisfying the following assumptions.
(A 1 ) g is a nonnegative Borel measurable function on B × (0,∞), which is continuous with respect to the second variable.(A 2 ) g(x,t) ≤ ψ(x,t), where ψ is a positive Borel measurable function in B × (0,∞), such that the function t → ψ(x,t) is nonincreasing on (0,∞).(A 3 ) The function q defined on B by q(x) = ψ(x,ρ 0 (x))/ρ 0 (x) belongs to the class K.We will prove the following result.There exists a constant c 1 > 0 such that if θ ≥ (1 + c 1 )h 0 on ∂B, then problem (1.1) has a positive continuous solution u satisfying This result is a followup to the one of Athreya [14], who considered the following problem: where D is a simply connected bounded C 2 -domain and g(u) ≤ max (1,u −α ), for 0 < α < 1.Then he proved that there exists a constant c > 0 such that if ϕ ≥ (1 + c)h 0 on ∂D, then problem (*) has a positive continuous solution u such that u ≥ h 0 , where h 0 is a fixed positive harmonic function in D.
In order to simplify our statements, we define some convenient notations.
The function V f is called the m-potential of f and it is lower semicontinuous on B.
(vi) Let K + denote the set of nonnegative functions on the Kato class K.
(vii) For any ϕ ∈ Ꮾ(B), we put (viii) Let f and g be two positive functions on a set S. (1.15)

Properties of the Kato class K
We collect in this section some properties of functions belonging to the Kato class K, which are useful at stating our existence results.
Sonia Ben Othman et al. 5 Proposition 2.1 (see [4]).Let ϕ be a function in K. Then one has the following assertions.
corollary 2.2.Let q ∈ K + .Then one has and for each Proof.The result holds by (1.3).
For ζ in K + , we denote From (1.3) and Proposition 2.1(i), we derive the following proposition.
Proposition 2.4.For each q ∈ K + and h a nonnegative harmonic function in B, one has for which completes the proof.
In the next results, we will give a class of functions included in K and we will precise estimates of the m-potential of some functions in this class.
To prove Proposition 2.5, we use the next two lemmas.
To show the claim, we use the inequalities in Lemma 2.7 and the Hölder inequality.We distinguish three cases.

Abstract and Applied Analysis
Case 2 (n = 2m).Let p > 1, f ∈ L p (B), and λ < n/q.Then, for x ∈ B, we have which tends to zero as α → 0.
In the sequel, we put for (2.20) Remark 2.8.From Lemma 2.6, we note that for each x, y in B, we have (δ(x)δ(y)) m G m,n (x, y).This implies that there exists a constant c > 0 such that for each f ∈ Ꮾ(B), we have (2.21) In the next proposition, we will give upper estimates on the function v.

2.22)
To prove Proposition 2.9, we need the following key lemma.

First existence result
In this section, we are interested in the existence of positive solutions for problem (1.6).
To this end, we first introduce for q ∈ K + , such that α q ≤ 1/2, the potential kernel V q f := V m,n,q f as a solution for the pertubed polyharmonic equation (1.8).We put for x, y ∈ B that Then we have the following comparison result.
Lemma 3.1.Let q ∈ K + such that α q ≤ 1/2.Then on B 2 , one has Proof.Since α q ≤ 1/2, we deduce from (2.2) that On the other hand, we note that for x = y in B, Using these facts and (2.2), we obtain that Hence the result follows from (3.4) and (2.2).
In the sequel, for a given q ∈ K + such that α q ≤ 1/2, we define the operator V q on Ꮾ + (B) by (3.6) Lemma 3.2.Let q ∈ K + such that α q ≤ 1/2 and f ∈ Ꮾ + (B).Then V q f satisfies the following resolvent equation: Proof.From the expression of Ᏻ m,n , we deduce that for f ∈ Ꮾ + (B) such that V f < ∞, we have So we obtain that The second equality holds by integrating (3.4).
a solution of the perturbed polyharmonic equation (1.8).
Proof.Using the resolvent equation (3.7), we have Applying the operator (−Δ) m on both sides of the above equality, we obtain that (−Δ) m V q f = f − qV q f (in the sense of distributions).(3.11)This completes the proof.Now, we aim at proving an existence result for problem (1.6).We recall that the function ρ is defined on B by ρ(x) = (1 − |x| 2 ) m−1 h(x), where h is the continuous solution of the Dirichlet problem h = 0, on B and h |∂B = θ.
The main result of this section is the following.Proof.
Then by (H 2 ), there exists a function q := q β ∈ K + such that α q ≤ 1/2 and for each x ∈ B, the map which implies in particular that for each x ∈ B and t ∈ [0,β], Let We define the operator T on Λ by We claim that Λ is invariant under T. Indeed, using (3.13) and (3.7), we have for each u ∈ Λ that Moreover, from (3.13), (3.2), and Proposition 2.4, we deduce that for each u ∈ Λ, we have Next, we will prove that the operator T is nondecreasing on Λ.Indeed, let u,v ∈ Λ such that u ≤ v, then from (3.12), we obtain that Now, we consider the sequence (u k ) defined by u 0 = (1 − α q )(V f + ρ) and u k+1 = Tu k for k ∈ N. Since Λ is invariant under T, then u 1 = Tu 0 ≥ u 0 , and so from the monotonicity of T, we deduce that Hence from (3.12) and the dominated convergence theorem, we deduce that the sequence (u k ) converges to a function u which satisfies That is, Applying the operator (I + V (q•)) on both sides of the above equality and using (3.7), we deduce that u satisfies To prove continuity of u, we recall that V f /(δ(•)) m−1 ∈ C 0 (B).Then, there exists a function , and so we have by Proposition 2.
Example 3.5.Let α, β be two positive constants and q, γ are two functions in K + .Then, for each θ ∈ C + (∂B), the following polyharmonic problem has a positive continuous solution satisfying (1.7), provided that β is sufficiently small.

Second existence result
In this section, assuming that (A 1 )-(A 3 ) hold, we aim at proving an existence result for problem (1.1).We recall that h 0 is a fixed positive harmonic function in B, continuous in B. We put ρ 0 Proof.By (A 3 ), the function q defined on B by q(x) = ψ(x,ρ 0 (x))/ρ 0 (x) is in K + .Then we deduce by Proposition 2.4 that (4. 3) The result holds from Corollary 2.2.
In the sequel, we suppose (A 4 ) Proof of Theorem 4.1.We will use a fixed point argument.Let

.5)
Since h = θ on ∂B and h 0 is continuous in B, then we obtain by (A 4 ) that h ≥ (1 + c 1 )h 0 on B. So Λ is a well-defined nonempty closed bounded and convex set in C(B).

Sonia Ben Othman et al. 15
Next, let us prove the continuity of T in the uniform norm.Let (u k ) k be a sequence in Λ which converges uniformly to u ∈ Λ, then since g is continuous with respect to the second variable, we deduce by the dominated convergence theorem that ∀x ∈ B, Tu k (x) −→ Tu(x) as k −→ ∞.
.22) Finally, we need to verify that u is a positive continuous solution for problem(1.6).Since for eachx ∈ B, f (x) = (δ(x)) m−1 γ(x), where γ is in K + ,we deduce by Proposition 2.1 that f ∈ L 1 Loc (B) and by Proposition 2.3 we have that V f ∈ L 1 Loc (B).Then, since u ∼ V f + ρ and uϕ(•,u) ≤ uq, we deduce that either u and uϕ(•,u) are in L 1 Loc (B).Now, from (3.22) we can easily see that V (uϕ(•,u)) ∈ L 1 Loc (B).Hence u satisfies (in the sense of distributions) the elliptic differential equation