Open Access

Dirichlet problems for linear and semilinear sub-Laplace equations on Carnot groups

Journal of Inequalities and Applications20122012:136

DOI: 10.1186/1029-242X-2012-136

Received: 7 December 2011

Accepted: 12 June 2012

Published: 12 June 2012

Abstract

The purpose of this article, is to study the Dirichlet problems of the sub-Laplace equation Lu + f(ξ, u) = 0, where L is the sub-Laplacian on the Carnot group G and f is a smooth function. By extending the Perron method in the Euclidean space to the Carnot group and constructing barrier functions, we establish the existence and uniqueness of solutions for the linear Dirichlet problems under certain conditions on the domains. Furthermore, the solvability of semilinear Dirichlet problems is proved via the previous results and the monotone iteration scheme corresponding to the sub-Laplacian.

Mathematics Subject Classifications: 35J25, 35J70, 35J60.

Keywords

Carnot group sub-Laplace equation Dirichlet problem Perron method monotone iteration scheme

1 Introduction

In this article we consider Dirichlet problems of the type
L u + f ( ξ , u ) = 0 , in Ω , u = φ , on Ω , https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-136/MediaObjects/13660_2011_Article_255_Equ1_HTML.gif
(1.1)

where Ω is a bounded domain in a Carnot group G and L is the sub-Laplacian. Some knowledge on G and L see next section. Hörmander's theorem permits us to judge the hypoellipticity of the operator L, i.e., if Lu C then u C (see [1]).

The investigation of the boundary value problems, concerning the operators in the form of the sum of squares of vector fields fulfilling Hörmander condition, has turned into the subject of several works, see [24]. The precursory work of Bony [2] proved a maximum principle and the solvability of the Dirichlet problem in the sense of Perron-Wiener. The Wiener type regularity of boundary points for the Dirichlet problem was considered in [3]. Thanks to the previous results, Capogna et al. [4] established the solvability of the Dirichlet problem when the boundary datum belongs to L p , 1 < p, in the group of Heisenberg type.

The Perron method (see [5, 6]) and the monotone iteration scheme (see [7, 8]) are well-known constructive methods for solving linear and semilinear Dirichlet problems, respectively. Brandolini et al. [9] applied these methods to the Dirichlet problems for sub-Laplace equations on the gauge balls in the Heisenberg group which is the simplest Carnot group of step two. Let us notice that the balls possess of legible properties. However, we do not see the reseach to the problems on other domains using these methods. Concerning the construction of barrier function, Brandolini et al. [9] used the result given in [10], which holds in the setting of Heisenberg group.

Our work is motivated by [9]. We try to extend the existence of solutions for semilinear Dirichlet problems on the Heisenberg balls in [9] to general Carnot domains. To do so, the Perron method in the Carnot group is used in this article. Based on the work in [3], we construct a barrier function in a domain of the Carnot group (see Lemma 3.10) under the hypothesis of the outer sphere condition to discuss the boundary behaviour of the Perron solutions. The method to obtain a barrier function is essentially similar to the one in [9]. Then we prove the existence of solutions for linear sub-Laplace Dirichlet problems. In the discussion of semilinear Dirichlet problems, we will use monotone iteration scheme. The main difficulty we meet is that the sub-Laplacian L does not have explicit expression. To overcome it, we use the regularity of L in [1].

The article is organized as follows. In the next section, we recall some basic definitions and collect some known results on the Carnot group which will play a role in the following sections. Section 3 is devoted to the study of the Perron method for linear equations. By finding a barrier function related to the sub-Laplacian L, we prove that the Perron solutions for linear Dirichlet problems are continuous up to the boundary. The main results are Theorems 3.8, 3.11, and 3.13. In Section 4, using the results in Section 3 and the monotone iteration scheme, we provide the solutions of the semilinear Dirichlet problems in Carnot groups with some available supersolutions and subsolutions. Finally, we give an existence of solution to the sub-Laplace equation on the whole group of Heisenberg type (a specific Carnot group of step two). The main results in this section are Theorems 4.2 and 4.3.

2 Carnot groups

We will consider G = ( N , ·) as a Carnot group with a group operation · and a family of dilations, compatible with the Lie structure.

Following [11, 12], a Carnot group G of step r ≥ 1 is a simple connected nilpotent Lie group whose Lie algebra g ̃ https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-136/MediaObjects/13660_2011_Article_255_IEq1_HTML.gif admits a stratification. That is, there exist linear subspaces V1, . . ., V r of g ̃ https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-136/MediaObjects/13660_2011_Article_255_IEq1_HTML.gif such that
g ̃ = j = 1 r V j , [ V 1 , V i ] = V i + 1 for i = 1 , 2 , , r - 1 and [ V 1 , V r ] = { 0 } . https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-136/MediaObjects/13660_2011_Article_255_Equa_HTML.gif
Via the exponential map, it is possible to induce on G a family of non-isotropic dilations defined by
δ λ ( ξ ) = δ λ ( x ( 1 ) , x ( 2 ) , , x ( r ) ) = ( λ x ( 1 ) , λ 2 x ( 2 ) , , λ r x ( r ) ) . https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-136/MediaObjects/13660_2011_Article_255_Equb_HTML.gif
Here ξ = (x(1), x(2), . . ., x(r)), x ( i ) N i https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-136/MediaObjects/13660_2011_Article_255_IEq2_HTML.gif for i = 1, . . ., r and N1 + · · · + N r = N. We denote by Q = j = 1 r j N j https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-136/MediaObjects/13660_2011_Article_255_IEq3_HTML.gif the homogeneous dimension of G attached to the dilations {δ λ }λ > 0. Let m = N1 and X = {X1, . . ., X m } be the dimension and a basis of V1, respectively. Let Xu = {X1u, . . ., X m u} denote the horizontal gradient for a function u. The sub-Laplacian associated with X on G is given by
L = j = 1 m X j 2 . https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-136/MediaObjects/13660_2011_Article_255_Equc_HTML.gif
If u and v are two measurable functions on G, their convolution is defined by
u * v ( ξ ) = G u ( η ) v ( η - 1 ξ ) d G ( η ) = G u ( ξ η - 1 ) v ( η ) d G ( η ) , https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-136/MediaObjects/13660_2011_Article_255_Equd_HTML.gif

where dG(η) denotes a fixed Haar measure on G.

Let e be the identity on G. For ξ G, we denote by ξ-1 the inverse of ξ with respect to the group operation. By [1], there exists a norm function ρ ( ξ ) C 0 G \ { e } C ( G ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-136/MediaObjects/13660_2011_Article_255_IEq4_HTML.gif satisfying
  1. (1)

    ρ(ξ) ≥ 0; Moreover, ρ(ξ) = 0 if and only if ξ = e;

     
  2. (2)

    ρ(ξ) = ρ(ξ-1).

     
The open ball of radius R centered at ξ is expressed as the set:
B G ( ξ , R ) = { η G : ρ ( ξ , η ) = ρ ( ξ - 1 η ) < R } . https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-136/MediaObjects/13660_2011_Article_255_Eque_HTML.gif
Let D https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-136/MediaObjects/13660_2011_Article_255_IEq5_HTML.gif denote the space of distributions on G. The non-isotropic Sobolev space S k, p is defined by
S k , p = { f D : D α f L p ( G ) , | α | k } , https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-136/MediaObjects/13660_2011_Article_255_Equf_HTML.gif
where α = (α1, . . ., α l ) is a multi-index, D α = D α 1 D α 2 D α l https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-136/MediaObjects/13660_2011_Article_255_IEq6_HTML.gif, and D α j { X 1 , , X m } https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-136/MediaObjects/13660_2011_Article_255_IEq7_HTML.gif. In the space S k, p , we shall adopt the norm
f S k , p = sup α k D α f L p . https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-136/MediaObjects/13660_2011_Article_255_Equg_HTML.gif
For a domain Ω in G, we define S k, p (Ω, loc) as the space of distributions f such that for every ψ ( ξ ) C 0 ( Ω ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-136/MediaObjects/13660_2011_Article_255_IEq8_HTML.gif we have S k, p . Let 0 < β < , we employ the following non-isotropic Lipschitz spaces:
  1. (i)
    for 0 < β < 1,
    Γ β : = f L C 0 : sup ξ , η | f ( η ξ ) - f ( η ) | ( ρ ( ξ , e ) ) β < , https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-136/MediaObjects/13660_2011_Article_255_Equh_HTML.gif
     
  2. (ii)
    for β = 1,
    Γ 1 : = f L C 0 : sup ξ , η | f ( η ξ ) + f ( η ξ - 1 ) - 2 f ( η ) | ( ρ ( ξ , e ) ) β < , https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-136/MediaObjects/13660_2011_Article_255_Equi_HTML.gif
     
  3. (iii)
    for β = k + β' where k = 1, 2, 3, . . . and, 0 < β' ≤ 1,
    Γ β : = f L C 0 : D α f Γ β , | α | k . https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-136/MediaObjects/13660_2011_Article_255_Equj_HTML.gif
     

We refer the reader to [1] for more information on the above.

The following results are useful.

Proposition 2.1. (i) Suppose Ω G is an open set, and suppose f , g D ( Ω ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-136/MediaObjects/13660_2011_Article_255_IEq9_HTML.gif satisfy Lf = g in Ω. If g S k, p (Ω, loc) (1 < p < ∞, k ≥ 0) then f Sk+2,p(Ω, loc).

(ii) Suppose 1 < p < ∞ and β = k - Q p > 0 https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-136/MediaObjects/13660_2011_Article_255_IEq10_HTML.gif, then S k, p Γ β .

Part (i) and (ii) are contained, respectively, in Theorems 6.1 and 5.15 of [1].

3 The Perron method and barrier function for linear problem

In this section, we study the solvability of the following linear sub-Laplace Dirichlet problem
L u - λ ( ξ ) u = f , in Ω , u = φ , on Ω , https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-136/MediaObjects/13660_2011_Article_255_Equ2_HTML.gif
(3.1)

where λ ( ξ ) C ( Ω ̄ ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-136/MediaObjects/13660_2011_Article_255_IEq11_HTML.gif satisfies λ(ξ) > 0.

Definition 3.1. A bounded open set Ω G is said to satisfy the outer sphere condition at ξ0 Ω, if there exists a ball B G (η, r) lying in G\ Ω such that
B G ( η , r ) Ω = { ξ 0 } . https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-136/MediaObjects/13660_2011_Article_255_Equk_HTML.gif

The definition in the case of general degenerate elliptic operator can be seen in [3]. Notice that in the H-type group case, every bounded convex subset accords with the condition of the outer sphere. In particular, the gauge balls in H-type group are convex domains (see [4]). From Theorem 2.12 in [13] and Theorem 5.2 in [2] respectively, one has the following two lemmas.

Lemma 3.2. (Maximum principle) Let Ω be a connected open set in a Carnot group G. If u C2(Ω) satisfies
L u - λ ( ξ ) u 0 i n Ω , https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-136/MediaObjects/13660_2011_Article_255_Equl_HTML.gif

then u cannot achieve a nonnegative maximum at an interior point unless u ≡ constant in Ω.

Lemma 3.3. Let Ω be a bounded domain in G. Then there exists a family of open subsets, denoted by F = { ω : ω ̄ Ω } https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-136/MediaObjects/13660_2011_Article_255_IEq12_HTML.gif, which is a base for the topology of Ω for which the Dirichlet problem
L u - λ ( ξ ) u = f , i n ω , u = φ , o n ω https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-136/MediaObjects/13660_2011_Article_255_Equ3_HTML.gif
(3.2)

has a unique distributional solution u C ( ω ̄ ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-136/MediaObjects/13660_2011_Article_255_IEq13_HTML.gif for any ω F https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-136/MediaObjects/13660_2011_Article_255_IEq14_HTML.gif, f C ( ω ̄ ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-136/MediaObjects/13660_2011_Article_255_IEq15_HTML.gif and φ C(∂ω). Furthermore, if f C (ω), then u C (ω).

We give notions of subsolution and supersolution for the Dirichlet problem (3.1).

Definition 3.4. Let φ C(∂Ω), f C ( Ω ̄ ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-136/MediaObjects/13660_2011_Article_255_IEq16_HTML.gif. A function u C ( Ω ̄ ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-136/MediaObjects/13660_2011_Article_255_IEq17_HTML.gif is called a subsolution of (3.1) if it fits the following properties:
  1. (i)

    uφ on ∂Ω;

     
  2. (ii)

    for every ω F https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-136/MediaObjects/13660_2011_Article_255_IEq14_HTML.gif and for every h C 2 ( ω ) C ( ω ̄ ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-136/MediaObjects/13660_2011_Article_255_IEq18_HTML.gif such that Lh - λ(ξ)h = f and uh onω, we also have uh in ω.

     

The definition of supersolution is analogous.

Lemma 3.5. Assume that u is a subsolution of (3.1) and v is a supersolution of (3.1), then either u < v in Ω or u ≡ v.

Proof. Suppose that at some point η Ω we have u(η) ≥ v(η). Set M = sup ξ Ω ( u - v ) ( ξ ) 0 https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-136/MediaObjects/13660_2011_Article_255_IEq19_HTML.gif. Take ξ0 Ω such that (u - v)(ξ0) = M, and we can know that u - v ≡ M in a neighborhood of ξ0. Otherwise there exists ω F https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-136/MediaObjects/13660_2011_Article_255_IEq14_HTML.gif such that ξ0 ω but u - v is not identically equal to M on ∂ω. Letting ū https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-136/MediaObjects/13660_2011_Article_255_IEq20_HTML.gif and v ̄ https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-136/MediaObjects/13660_2011_Article_255_IEq21_HTML.gif denote the solutions of Lw - λ(ξ)w = f in ω, equal to u and v on ∂ω respectively. Since u and v are the subsolution and the supersolution respectively, we deduce from Definition 3.4 that ū u https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-136/MediaObjects/13660_2011_Article_255_IEq22_HTML.gif and v ̄ v https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-136/MediaObjects/13660_2011_Article_255_IEq23_HTML.gif in ω. One sees that
M = sup ξ Ω ( u - v ) ( ξ ) sup ξ ω ( ū - v ̄ ) ( ξ ) ( ū - v ̄ ) ( ξ 0 ) ( u - v ) ( ξ 0 ) = M , https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-136/MediaObjects/13660_2011_Article_255_Equm_HTML.gif

and hence all the equalities above hold. By Lemma 3.2 it follows that ū - v ̄ M https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-136/MediaObjects/13660_2011_Article_255_IEq24_HTML.gif in ω and hence u - v ≡ M on ∂ω, which contradicts the choice of ω.

The previous argument implies u - v ≡ M in Ω. Combining this with Definition 3.4-(i) we obtain u ≡ v in Ω. □

Let u C ( Ω ̄ ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-136/MediaObjects/13660_2011_Article_255_IEq17_HTML.gif be a subsolution of (3.1) and ω F https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-136/MediaObjects/13660_2011_Article_255_IEq14_HTML.gif. Denote by ū https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-136/MediaObjects/13660_2011_Article_255_IEq20_HTML.gif the solution of the Dirichlet problem (see Lemma 3.3)
L ū - λ ( ξ ) ū = f ( ξ ) , in ω , ū = u , on ω , https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-136/MediaObjects/13660_2011_Article_255_Equn_HTML.gif
and define in Ω the lifting of u (in ω) by
U ( ξ ) : = ū ( ξ ) , ξ ω , u ( ξ ) , ξ Ω \ ω . https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-136/MediaObjects/13660_2011_Article_255_Equ4_HTML.gif
(3.3)

Lemma 3.6. U(ξ) is a subsolution of (3.1).

Proof. Since u(ξ) is a subsolution of (3.1), it follows that U(ξ) = u(ξ) ≤ φ(ξ) on ∂Ω. Let ω F https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-136/MediaObjects/13660_2011_Article_255_IEq25_HTML.gif and h C 2 ( ω ) C ( ω ¯ ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-136/MediaObjects/13660_2011_Article_255_IEq26_HTML.gif such that Lh - λ(ξ)h = f and Uh on ∂ω'. If ωω' = ϕ, then u = Uh on ∂ω'. It leads to U = uh in ω';

Suppose now ωω' = ϕ. Since uU, we have uh on ∂ω' and then uh in ω'. In particular, uh in ω', i.e. Uh in ω'. Thus, we have ū h https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-136/MediaObjects/13660_2011_Article_255_IEq27_HTML.gif on ∂(ω' ∩ ω). As L ( ū - h ) - λ ( ξ ) ( ū - h ) = 0 https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-136/MediaObjects/13660_2011_Article_255_IEq28_HTML.gif in ω' ∩ ω and ū - h 0 https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-136/MediaObjects/13660_2011_Article_255_IEq29_HTML.gif on ∂(ω' ∩ ω), it yields by Lemma 3.2 that ū h https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-136/MediaObjects/13660_2011_Article_255_IEq27_HTML.gif in ω' ∩ ω, and therefore Uh in ω' ∩ ω. □

The following result is a trivial consequence of Definition 3.4.

Lemma 3.7. Let u1, u2, . . ., u l be subsolutions of (3.1). Then the function
v = max { u 1 , u 2 , , u l } https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-136/MediaObjects/13660_2011_Article_255_Equo_HTML.gif

is also a subsolution of (3.1).

Let S denote the set of all subsolutions of (3.1). Notice that S is not empty, since -k2 S for k large enough. The basic result via the Perron method is contained in the following theorem.

Theorem 3.8. The function u ( ξ ) : = sup v S v ( ξ ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-136/MediaObjects/13660_2011_Article_255_IEq30_HTML.gif satisfies Lu - λ(ξ)u = f in Ω.

Proof. Notice that k2, for k large enough, is a supersolution of (3.1). By Lemma 3.5, we deduce vk2 for any v S, so u is well defined. Let η be an arbitrary fixed point of Ω. By the definition of u, there exists a sequence {v n }nsuch that v n (η) → u(η). By replacing v n with max {v1, . . ., v n }, we may assume that v1v2· · ·v n · · ·. Let ω F https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-136/MediaObjects/13660_2011_Article_255_IEq14_HTML.gif be such that η ω and define V n (η) to be the lifting of v n in ω according to (3.3). From Lemma 3.2, V n is also increasing and, since V n S (see Lemma 3.6) and V n v n , it gets V n (η) → u(η). Set V ( ξ ) : = lim n V n ( ξ ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-136/MediaObjects/13660_2011_Article_255_IEq31_HTML.gif. Obviously, we have that Vu in Ω and V (η) = u(η). Noting that every V n satisfies LV n - λ(ξ)V n = f in ω, we have, by the dominated convergence theorem that the function V satisfies LV - λ(ξ)V = f in the distributional sense in ω. Since f C (ω), we have V(ξ) C (ω) in view of the hypoellipticity of the operator L - λ(ξ).

We conclude that V ≡ u in ω. In fact, suppose V(ζ) < u(ζ) for some ζ ω, then there exists a function ū S https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-136/MediaObjects/13660_2011_Article_255_IEq32_HTML.gif such that V ( ζ ) < ū ( ζ ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-136/MediaObjects/13660_2011_Article_255_IEq33_HTML.gif. Define the increasing sequence w n = max { ū , V n } https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-136/MediaObjects/13660_2011_Article_255_IEq34_HTML.gif and then the corresponding liftings W n . Set W ( ξ ) : = lim n W n ( ξ ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-136/MediaObjects/13660_2011_Article_255_IEq35_HTML.gif. Analogously to V, W satisfies LW - λ(ξ)W = f. Since V n w n W n , we obtain VW. The equalities V(η) = u(η) = W(η) and Lemma 3.2 imply that V ≡ W in Ω. This is in contradiction with V ( ζ ) < ū ( ζ ) W ( ζ ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-136/MediaObjects/13660_2011_Article_255_IEq36_HTML.gif. Consequently, V ≡ u in ω and u satisfies Lu - λ(ξ)u = f in the classical sense. The arbitrariness of η leads to the desired result. □

Definition 3.9. Let ζ ∂Ω. Then a function w ( ξ ) C ( Ω ) C ( Ω ̄ ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-136/MediaObjects/13660_2011_Article_255_IEq37_HTML.gif is called a barrier function related to the sub-Laplacian L at ζ if the following two conditions hold:
  1. (i)

    Lw(ξ) ≤ -1 in Ω;

     
  2. (ii)

    w(ξ) > 0 on Ω ̄ \ { ζ } https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-136/MediaObjects/13660_2011_Article_255_IEq38_HTML.gif, w(ζ) = 0.

     
Lemma 3.10. Let Ω G be a bounded open domain which satisfies the outer sphere condition at every point of the boundary ∂Ω. Then for every ζ ∂Ω, the Dirichlet problem
L w = - 1 , i n Ω , w ( ξ ) = ρ ( ξ , ζ ) , o n Ω https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-136/MediaObjects/13660_2011_Article_255_Equ5_HTML.gif
(3.4)

has a unique solution w C ( Ω ) C ( Ω ̄ ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-136/MediaObjects/13660_2011_Article_255_IEq39_HTML.gif fulfilling w(ξ) > 0 on Ω ̄ \ { ζ } https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-136/MediaObjects/13660_2011_Article_255_IEq38_HTML.gif and w(ζ) = 0.

Proof. From [1], let Γ(ξ) = C Q ρ(ξ, e)-(Q-2)be the fundamental solution of the sub-Laplacian L. Define the convolution
ũ : = - Γ * χ Ω , https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-136/MediaObjects/13660_2011_Article_255_Equp_HTML.gif

where χΩ denotes the indicator function. Since Γ ( ξ ) L l o c p https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-136/MediaObjects/13660_2011_Article_255_IEq40_HTML.gif for 1 p < Q Q - 2 https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-136/MediaObjects/13660_2011_Article_255_IEq41_HTML.gif, it yields ũ C ( Ω ) C ( Ω ̄ ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-136/MediaObjects/13660_2011_Article_255_IEq42_HTML.gif.

According to Corollary 10 in [3], the problem
L v = 0 , in Ω , v ( ξ ) = ρ ( ξ , ζ ) - ũ ( ξ ) , on Ω https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-136/MediaObjects/13660_2011_Article_255_Equq_HTML.gif

has a unique solution v C ( Ω ) C ( Ω ̄ ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-136/MediaObjects/13660_2011_Article_255_IEq43_HTML.gif. Since L ũ = - χ Ω https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-136/MediaObjects/13660_2011_Article_255_IEq44_HTML.gif (see Corollary 2.8 in [1]), it follows that w : = v + ũ https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-136/MediaObjects/13660_2011_Article_255_IEq45_HTML.gif is the desired solution of (3.4). □

Theorem 3.11. Let Ω be as in Lemma 3.10. Suppose φ C(∂Ω) and f C ( Ω ) C ( Ω ̄ ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-136/MediaObjects/13660_2011_Article_255_IEq46_HTML.gif. Then the Dirichlet problem (3.1) possesses a unique solution u C ( Ω ) C ( Ω ̄ ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-136/MediaObjects/13660_2011_Article_255_IEq47_HTML.gif.

Proof. Uniqueness is a direct consequence of Lemma 3.2. Theorem 3.8 provides the existence of the solution u C (Ω). To complete the proof of the theorem, it needs only to examine that u is continuous up to the boundary of Ω.

Let ζ Ω. Since φ C(∂Ω), it follows that for any ε > 0 there exists some δ > 0 such that for every ξ ∂Ω with ρ(ξ, ζ) < δ, we have
| φ ( ξ ) - φ ( ζ ) | < ε . https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-136/MediaObjects/13660_2011_Article_255_Equr_HTML.gif
Let w(ξ) be the barrier function related to L at ζ constructed in Lemma 3.10. Set M = sup ξ Ω | φ ( ξ ) | https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-136/MediaObjects/13660_2011_Article_255_IEq48_HTML.gif and choose k1 > 0 such that k1w(ξ) ≥ 2M if ρ(ξ, ζ) ≥ δ. Set k 2 = [ | φ ( ζ ) | + ε ] max ξ Ω λ ( ξ ) + sup ξ Ω | f ( ξ ) | https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-136/MediaObjects/13660_2011_Article_255_IEq49_HTML.gif, and k = max{k1, k2}. Define that w1(ξ): = φ(ζ) + ε + kw(ξ) and w2(ξ): = φ(ζ) - ε - kw(ξ). Then we see in view of Lemma 3.10,
L w 1 - λ ( ξ ) w 1 = - k - λ ( ξ ) φ ( ζ ) - λ ( ξ ) ε - k λ ( ξ ) w ( ξ ) f in Ω . https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-136/MediaObjects/13660_2011_Article_255_Equs_HTML.gif
On the one hand, w1(ξ) = φ(ζ) + ε + kw(ξ) ≥ φ(ζ) + ε > φ(ξ) when ρ(ξ, ζ) < δ; On the other hand, w1(ξ) ≥ φ(ζ) + ε + 2M > φ(ξ) when ρ(ξ, ζ) ≥ δ. Combining these with Lemma 3.2 we can conclude that w1(ξ) is a supersolution of (3.1). Analogously, w2(ξ) is a subsolution of (3.1). Hence from the choice of u and the fact that every supersolution dominates every subsolution, we have in Ω that
w 2 ( ξ ) u ( ξ ) w 1 ( ξ ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-136/MediaObjects/13660_2011_Article_255_Equt_HTML.gif
and then
| u ( ξ ) - φ ( ζ ) | ε + k w ( ξ ) . https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-136/MediaObjects/13660_2011_Article_255_Equu_HTML.gif

Since w(ξ) 0 as ξ → ζ, we obtain u(ξ) → φ(ζ) as ξ → ζ. □

Remark 3.12. Let f C ( Ω ) C ( Ω ̄ ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-136/MediaObjects/13660_2011_Article_255_IEq46_HTML.gif and u be the solution of
L u - λ ( ξ ) u = f , in Ω , u = 0 , on Ω . https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-136/MediaObjects/13660_2011_Article_255_Equ6_HTML.gif
(3.5)

Elementary calculations show that - 1 min ξ Ω λ ( ξ ) | | f | | L ( Ω ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-136/MediaObjects/13660_2011_Article_255_IEq50_HTML.gif and 1 min ξ Ω λ ( ξ ) | | f | | L ( Ω ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-136/MediaObjects/13660_2011_Article_255_IEq51_HTML.gif are a subsolution and a supersolution of (3.5) respectively. Thus, | | u | | L ( Ω ) 1 min ξ Ω λ ( ξ ) | | f | | L ( Ω ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-136/MediaObjects/13660_2011_Article_255_IEq52_HTML.gif. It provides a L estimate for the solution of (3.5).

Theorem 3.13. Set φ C(∂Ω) and f C ( Ω ̄ ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-136/MediaObjects/13660_2011_Article_255_IEq53_HTML.gif. Then there exists a unique solution u C ( Ω ̄ ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-136/MediaObjects/13660_2011_Article_255_IEq17_HTML.gif to (3.1) in the sense of distribution.

Proof. Take a sequence f n ( ξ ) C ( Ω ) C ( Ω ̄ ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-136/MediaObjects/13660_2011_Article_255_IEq54_HTML.gif, n = 1, 2, . . ., so that {f n (ξ)} converges uniformly to f in Ω. Denote by u n the corresponding solution of the Dirichlet problem
L v - λ ( ξ ) v = f n ( ξ ) , in Ω , u = φ , on Ω . https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-136/MediaObjects/13660_2011_Article_255_Equv_HTML.gif
We obtain, in view of Remark 3.12,
| | u n - u m | | L ( Ω ) 1 min ξ Ω λ ( ξ ) | | f n - f m | | L ( Ω ) . https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-136/MediaObjects/13660_2011_Article_255_Equw_HTML.gif

In conclusion, {u n } converges uniformly to a continuous function u which is the required solution. □

4 The monotone iteration scheme for semilinear equation

Let Ω be a bounded open domain in a Carnot group G. Consider Dirichlet problem (1.1), where f(ξ, u) is a smooth function of ξ and u, φ C(∂Ω). A function μ C ( Ω ̄ ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-136/MediaObjects/13660_2011_Article_255_IEq55_HTML.gif is called a supersolution of (1.1) if it satisfies
L μ + f ( ξ , μ ) 0 , in Ω , μ ( ξ ) φ ( ξ ) , on Ω . https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-136/MediaObjects/13660_2011_Article_255_Equx_HTML.gif
Analogously, a function ν C ( Ω ̄ ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-136/MediaObjects/13660_2011_Article_255_IEq56_HTML.gif is called a subsolution of (1.1) if it satisfies
L ν + f ( ξ , ν ) 0 , in Ω , ν ( ξ ) φ ( ξ ) , on Ω . https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-136/MediaObjects/13660_2011_Article_255_Equy_HTML.gif

The above inequalities are both in the sense of distribution. Here, a function T ≥ 0 means that for any positive test function ψ, we have ≥ 0. In the following we are ready to construct a smooth solution of (1.1) commencing with a subsolution and a supersolution in S1,2(Ω, loc) by the monotone iteration scheme. We first prove a maximum principle.

Lemma 4.1. Assume that u S 1 , 2 ( Ω ) C ( Ω ̄ ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-136/MediaObjects/13660_2011_Article_255_IEq57_HTML.gif satisfies
L u - λ ( ξ ) u 0 , https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-136/MediaObjects/13660_2011_Article_255_Equz_HTML.gif

where λ ( ξ ) C ( Ω ̄ ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-136/MediaObjects/13660_2011_Article_255_IEq11_HTML.gif and λ(ξ) > 0. If u ≤ 0 on ∂Ω, then sup ξ Ω u ( ξ ) 0 https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-136/MediaObjects/13660_2011_Article_255_IEq58_HTML.gif.

Proof. Suppose that the conclusion fails. Since u is continuous on Ω ̄ https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-136/MediaObjects/13660_2011_Article_255_IEq59_HTML.gif, there exists a point ξ0 Ω such that u(ξ0) > 0. Fix ε > 0 so small that u(ξ0) - ε > 0. Consequently, the function u ε : = max{u - ε, 0} is non-negative and has compact support in Ω as u ≤ 0 on ∂Ω. By the distribution meaning of solutions, we get
Ω X u X u ε d G = Ω - u ε L u d G Ω - λ ( ξ ) u u ε d G 0 . https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-136/MediaObjects/13660_2011_Article_255_Equ7_HTML.gif
(4.1)

When u ε > 0, it follows Xu ε = Xu and Xu is not identically zero. In fact, if Xu ≡ 0, then u ≡ u(ξ0) > 0 in Ω which contradicts the assumption that u ≤ 0 on ∂Ω. Consequently the left hand side of (4.1) is positive, a contradiction. This completes the proof of the lemma. □

Theorem 4.2. Let Ω be as in Lemma 3.10. Let f C(G × (a, b)) and φ C(∂Ω). Suppose that μ and ν are, respectively, a supersolution and a subsolution of (1.1) with μ, ν S 1 , 2 ( Ω , l o c ) C ( Ω ̄ ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-136/MediaObjects/13660_2011_Article_255_IEq60_HTML.gif, νμ, and a < min ν < max μ < b. Then there exists a solution u C ( Ω ) C ( Ω ̄ ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-136/MediaObjects/13660_2011_Article_255_IEq47_HTML.gif of (1.1) satisfying νuμ.

Proof. Take K > 0 such that
f u + K 2 > 0 https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-136/MediaObjects/13660_2011_Article_255_Equ8_HTML.gif
(4.2)
on Ω ̄ × [ min ν , max μ ] https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-136/MediaObjects/13660_2011_Article_255_IEq61_HTML.gif. Let v = Tu denote the unique solution in C ( Ω ̄ ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-136/MediaObjects/13660_2011_Article_255_IEq62_HTML.gif of the Dirichlet problem (see Theorem 3.11)
( L - K 2 ) v = - [ f ( ξ , u ) + K 2 u ] , in Ω , v ( ξ ) = φ ( ξ ) , on Ω . https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-136/MediaObjects/13660_2011_Article_255_Equaa_HTML.gif
We claim that the nonlinear transformation T is monotone. To establish this we set u1 < u2 and notice that
( L - K 2 ) T u 1 = - [ f ( ξ , u 1 ) + K 2 u 1 ] , ( L - K 2 ) T u 2 = - [ f ( ξ , u 2 ) + K 2 u 2 ] , https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-136/MediaObjects/13660_2011_Article_255_Equab_HTML.gif
and Tu1 = Tu2 = φ on ∂Ω. Letting w = Tu1 - Tu2, we can obtain
( L - K 2 ) w = - [ f ( ξ , u 1 ) - f ( ξ , u 2 ) + K 2 ( u 1 - u 2 ) ] https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-136/MediaObjects/13660_2011_Article_255_Equac_HTML.gif
and w = 0 on ∂Ω. As f(ξ, u) + K2u is increasing in u by (4.2), it yields (L - K2) w ≥ 0. From
L w = K 2 w - [ f ( ξ , u 1 ) - f ( ξ , u 2 ) + K 2 ( u 1 - u 2 ) ] , https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-136/MediaObjects/13660_2011_Article_255_Equad_HTML.gif

we get w S2,2(Ω, loc) by Lw L2(Ω) and Proposition 2.1-(i). It follows that w ≤ 0 in Ω by applying Lemma 4.1, therefore, Tu1Tu2 and T is monotone. We now begin the iteration scheme.

Let u1 = . As
( L - K 2 ) u 1 = - [ f ( ξ , μ ) + K 2 μ ] , https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-136/MediaObjects/13660_2011_Article_255_Equae_HTML.gif
and u1 = φ on ∂Ω, we get by a trivial calculation that
( L - K 2 ) ( u 1 - μ ) = - [ L μ + f ( ξ , μ ) ] 0 https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-136/MediaObjects/13660_2011_Article_255_Equaf_HTML.gif

and u1 - μ ≤ 0 on ∂Ω. Arguing as in the previous gives u1μ in Ω.

Define un+1= Tu n . The monotoneity of T yields
μ u 1 u 2 . https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-136/MediaObjects/13660_2011_Article_255_Equag_HTML.gif
Analogously, starting from ν, we obtain a nondecreasing sequence
ν v 1 v 2 , https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-136/MediaObjects/13660_2011_Article_255_Equah_HTML.gif
where v1 = Tν, vn+1= Tv n . Moreover, νμ implies v1 = = u1 and, therefore, v n u n for each n . Thus
ν v 1 v 2 v n u n u 2 u 1 μ , https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-136/MediaObjects/13660_2011_Article_255_Equ9_HTML.gif
(4.3)
so that the limit u = lim n u n https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-136/MediaObjects/13660_2011_Article_255_IEq63_HTML.gif is well defined in Ω ̄ https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-136/MediaObjects/13660_2011_Article_255_IEq59_HTML.gif. Recall that
( L - K 2 ) u n + 1 = - [ f ( ξ , u n ) + K 2 u n ] . https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-136/MediaObjects/13660_2011_Article_255_Equai_HTML.gif
The dominated convergence theorem shows that
L u + f ( ξ , u ) = 0 https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-136/MediaObjects/13660_2011_Article_255_Equaj_HTML.gif

in the distributional sense. According to Proposition 2.1-(i) and the fact that f(ξ, u) L p (Ω) for 1 < p < + one has u S2,p(Ω, loc). Iterating the process, we get u S k, p (Ω, loc) for k ≥ 0. Let ψ C 0 ( Ω ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-136/MediaObjects/13660_2011_Article_255_IEq64_HTML.gif. The definition in Section 2 gives ψu S k, p . Furthermore, we obtain u C (Ω) in view of Proposition 2.1-(ii). Combining this with (4.3) we have u C ( Ω ) C ( Ω ̄ ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-136/MediaObjects/13660_2011_Article_255_IEq47_HTML.gif which is the desired solution. □

We assume henceforth that G is of Heisenberg type. Such group was introduced by Kaplan [14] and has been subsequently studied by several authors, see [4, 11, 13] and the references therein.

Let G be a Carnot group of step two whose Lie algebra g ̃ = V 1 V 2 https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-136/MediaObjects/13660_2011_Article_255_IEq65_HTML.gif. Consider the map J : V2 → End(V1) defined by
J ( ξ 2 ) ξ 1 , ξ 1 = ξ 2 , [ ξ 1 , ξ 1 ] , for ξ 1 , ξ 1 V 1 and ξ 2 V 2 . https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-136/MediaObjects/13660_2011_Article_255_Equak_HTML.gif

G is said of Heisenberg type if for every ξ2 V2, with |ξ2| = 1, the map J (ξ2): V1 → V1 is orthogonal.

In the case of the Heisenberg type groups, the gauge balls coincide with the level sets of the fundamental solution (that is a radial function in this class of groups, see [14]), and the balls B G (e, R) invade G as R tends to + since the vector fields on G satisfy the Hörmander rank condition. Thus, we get the following existence theorem in the whole space G by making use of Theorem 4.2 and the result in [4] that the gauge balls in H-type group satisfy the outer sphere condition.

Theorem 4.3. Let G be a group of Heisenberg type. Let u - (ξ), u+(ξ) S1,2(G, loc) ∩ C(G) be respectively a subsolution and a supersolution of the problem
L u + f ( ξ , u ) = 0 , https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-136/MediaObjects/13660_2011_Article_255_Equ10_HTML.gif
(4.4)
where f C (G × (a, b)) and a < u - (ξ) ≤ u+(ξ) < b. Then there exists a solution u C (G) of (4.4) satisfying
u - u u + https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-136/MediaObjects/13660_2011_Article_255_Equal_HTML.gif

in G.

Proof. Let u0 = u+, set B G (e, m) be the gauge ball of radius m centered at identity e. We construct u m inductively in the following manner. Let v m be the solution of the Dirichlet problem
L v + f ( ξ , v ) = 0 , in B G ( e , m ) , v ( ξ ) = u + ( ξ ) , on B G ( e , m ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-136/MediaObjects/13660_2011_Article_255_Equam_HTML.gif

obtained by means of Theorem 4.2 using u - and um-1, respectively, as a subsolution and a supersolution.

Define
u m ( ξ ) = v m ( ξ ) , ξ B G ( e , m ) , u + ( ξ ) , ξ B G ( e , m ) . https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-136/MediaObjects/13660_2011_Article_255_Equan_HTML.gif
Obviously, u - u m um-1. We need to prove that u m is a supersolution of (4.4). To see this, take a positive test function ψ ( ξ ) C 0 ( G ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-136/MediaObjects/13660_2011_Article_255_IEq66_HTML.gif. From the divergence theorem, we obtain
B G ( e , m ) v m L ψ d G = B G ( e , m ) ψ L v m d G + B G ( e , m ) v m A ψ , n d S - B G ( e , m ) ψ A v m , n d S https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-136/MediaObjects/13660_2011_Article_255_Equao_HTML.gif
and
G \ B G ( e , m ) u + L ψ d G = G \ B G ( e , m ) ψ L u + d G + B G ( e , m ) ψ A u + , n d S - B G ( e , m ) u + A ψ , n d S . https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-136/MediaObjects/13660_2011_Article_255_Equap_HTML.gif
The above two identities give
G u m L ψ d G = B G ( e , m ) ψ L v m d G + G \ B G ( e , m ) ψ L u + d G + B G ( e , m ) ψ A ( u + - v m ) , n d S , https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-136/MediaObjects/13660_2011_Article_255_Equ11_HTML.gif
(4.5)
where n https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-136/MediaObjects/13660_2011_Article_255_IEq67_HTML.gif denotes the outerward normal to ∂B G (e, m), and A is a fixed positive semi-definite matrix (see [4, 13]). Therefore, we may restrict ourselves to the case in which A ( u + - v m ) , n https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-136/MediaObjects/13660_2011_Article_255_IEq68_HTML.gif represents the derivative of u+ - v m in an outward direction with respect to ∂B G (e, m). Moreover, since u+ - v m ≥ 0 in B G (e, m) and u+ - v m = 0 on ∂B G (e, m), it follows
ψ A ( u + - v m ) , n 0 for ξ B G ( e , m ) . https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-136/MediaObjects/13660_2011_Article_255_Equ12_HTML.gif
(4.6)
Substitution in (4.5) gives
G u m L ψ d G - B G ( e , m ) ψ f ( ξ , v m ) d G - G \ B G ( e , m ) ψ f ( ξ , u + ) d G = - G ψ f ( ξ , u m ) d G . https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-136/MediaObjects/13660_2011_Article_255_Equaq_HTML.gif

This implies that u m is a supersolution, and we can restart the monotone iteration scheme on B G (e, m+1).

In this way we obtain iteratively a sequence of supersolutions {u m } satisfying the following properties:
  1. (i)

    {u m } is nonincreasing, and u-u m u + ;

     
  2. (ii)

    Every u m satisfies Lu m + f(ξ, u m ) = 0 in B G (e, m).

     

Set u ( ξ ) = lim m u m ( ξ ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-136/MediaObjects/13660_2011_Article_255_IEq69_HTML.gif. We observe that {u m } is a sequence of solutions of (4.4) on any B G (e, k) for mk. It follows that u is a solution on B G (e, k). Arguing as in Theorem 4.2 we know u C (B G (e, k)). The arbitrariness of k implies u C (G). Therefore, it holds that u is the required solution of (4.4). □

Declarations

Acknowledgements

We would like to thank Pengcheng Niu for research assistance and the two anonymous referees for very constructive comments. Zixia Yuan thanks the Mathematical Tianyuan Youth Foundation of China (No. 11026082) for financial support.

Authors’ Affiliations

(1)
School of Mathematical Science, University of Electronic Science and Technology of China
(2)
Department of Mathematics, Henan Institute of Science and Technology

References

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© Yuan and Yuan; licensee Springer. 2012

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