A Boundary-Value Problem for Normalized Finsler infinity-Laplacian Equations with Singular nonhomogeneous terms

Nonlinear Analysis 190 (2020) 111588

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Nonlinear Analysis www.elsevier.com/locate/na

A Boundary-Value Problem for Normalized Finsler infinity-Laplacian Equations with Singular nonhomogeneous terms Tesfa Biset a , Benyam Mebrate b , Ahmed Mohammed c ,∗ a b c

Department of Mathematics, Addis Ababa University, Ethiopia Department of Mathematics, Hawassa University, Hawassa, Ethiopia Department of Mathematical Sciences, Ball State University, Muncie, IN 47306, USA

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Article history: Received 12 June 2019 Accepted 1 August 2019 Communicated by Vicentiu D Radulescu

abstract We study boundary-value problems of Finsler infinity-Laplacian equations with nonhomogeneous terms that may exhibit singularity when solutions vanish on the boundary. © 2019 Elsevier Ltd. All rights reserved.

Keywords: Singular boundary-value problem Normalized Finsler–Laplacian Viscosity solution Eigenvalue problem

1. Introduction Let f be a continuous function on Ω × (0, ∞) × Rn , where Ω ⊆ Rn is an open set. We are interested in investigating existence of solutions to the boundary-value problem { −∆N F;∞ u = λf (x, u, Du) in Ω (1.1) u = 0 on ∂Ω with parameter λ > 0. The main feature of the problem we wish to draw attention to here is the possible singularity the nonlinearity f (x, t, p) may exhibit as t → 0+ . In (1.1), ∆N F;∞ stands for the anisotropic operator n ∑ ∂ 2 u ∂F ∂F (Du) (Du), (1.2) ∆N u := F,∞ ∂ξi ∂xi ∂xj ∂ξj i,j=1 where F is a Finsler–Minkowski norm (see definition in Section 3 below). This is a singular degenerate elliptic operator, a special case of which (when F (ξ) = |ξ|) is given by n ∑ ∂u ∂ 2 u ∂u −2 N ∆∞ u = |Du| . (1.3) ∂xi ∂xi ∂xj ∂xj i,j=1 ∗

Corresponding author. E-mail addresses: [email protected] (T. Biset), [email protected] (B. Mebrate), [email protected] (A. Mohammed). https://doi.org/10.1016/j.na.2019.111588 0362-546X/© 2019 Elsevier Ltd. All rights reserved.

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T. Biset, B. Mebrate and A. Mohammed / Nonlinear Analysis 190 (2020) 111588

The operator (1.3), which can be thought of as an isotropic operator, is commonly referred to as the normalized infinity-Laplacian, and has singularity when the gradient vanishes. Numerous investigations have been carried out on many interesting qualitative properties of solutions to equations involving the operator (1.3). For more on this, we direct the reader to the papers [2–4,10,11,17,20,24,28,32–35,39,42] and the references therein. The papers [3,20,34,35] focus on Dirichlet problems associated with Problem (1.3) in which the right-hand side f (x, t, p) is independent of (t, p) and is continuous and bounded on Ω . A degenerate quasilinear operator closely related to (1.2) is n ∑ ∂ 2 u ∂F ∂F (Du) (Du). ∆F,∞ u := F (Du) ∂ξi ∂xi ∂xj ∂ξj i,j=1 2

(1.4)

This is the nonnormalized analogue of (1.3). When F (ξ) = |ξ|, this becomes ∆∞ u =

n ∑ ∂u ∂ 2 u ∂u . ∂xi ∂xi ∂xj ∂xj i,j=1

(1.5)

Nonhomogeneous equations associated with the operator (1.5) have been investigated in [45,46]. More specifically, the following Dirichlet problems ∆∞ u = f (x, u) in Ω and u = φ on ∂Ω are studied [45,46], where Ω ⊆ Rn is an open bounded subset, φ ∈ C(∂Ω ) and f (x, t) is continuous on Ω × R and bounded on Ω × I for every closed subinterval I ⊆ R. The results in [45,46] extend some of the work in the earlier paper [20]. Over the years, extensive investigations have been carried out on questions of existence, uniqueness and regularity of solutions to singular boundary-value problems for various elliptic equations whose principal parts are the Laplacian, or more general non-divergence structure elliptic operators and some other nonlinear elliptic operators. Perhaps the earliest work on singular boundary-value problems is due to Fulks and Maybee, [19]. This work was followed by the seminal work of Crandall, Rabinowitz and Tartar [12] which inspired an enormous amount of work on related topics. We refer to the papers [8,9,13–16,18,21– 23,25,26,30,31,38,41,43,44,48–51], and the references therein, for many interesting investigations into singular boundary-value problems and associated topics. Turning our attention to Problem (1.1), in this paper we are primarily concerned with existence of solutions on bounded as well as on unbounded domains in Rn when the principal operator is the degenerate elliptic operator given in (1.2). We are mainly interested in showing, for large class of nonhomogeneous terms f (x, t, p) with possible singularity at t = 0, that Problem (1.1) admits positive solutions when the parameter belongs to appropriate open subintervals of R. We point out that this problem has been considered in the papers [8,18,25,38,41,50] when the principal part of the equation is the Laplacian and under various assumptions on nonhomogeneous terms that do not depend on the gradient of the solution. Let us now describe the organization of the paper. In Section 2 we state the two main results of the paper. These concern the existence of solutions to Problem (1.1) in bounded and unbounded open subsets of Rn . In Section 3, we recall some basic properties of Finsler–Minkowski norms that will be useful in the paper. We will also introduce the notion of viscosity solution, viscosity subsolution and viscosity supersolution to the equation in (1.1). The proofs of the main theorems rely on a result on an eigenvalue problem associated with the normalized Finsler infinity-Laplacian and the relevant eigenvalue problem will be studied in Section 3. Another ingredient in the proof of the main results is the existence of a positive supersolution to Problem (1.1). This will be established in Section 4. The proof of the main results will be given in Section 5. For completeness, we have included an appendix where we state and prove some simple results that are used in the paper.

T. Biset, B. Mebrate and A. Mohammed / Nonlinear Analysis 190 (2020) 111588

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Notation: We list a few notations that we use throughout the paper. • • • • • • •

R+ = (0, ∞). R+ 0 = [0, ∞). o stands for the origin in Rn . ⟨· , ·⟩ stands for the standard Euclidean inner product in Rn . a+ := max{a, 0}. a− := max{−a, 0}. For subsets O and Ω of Rn , we write O ⊂⊂ Ω to mean O ⊆ Ω .

2. Main results In this section we state the main results of the paper. Towards this objective we first introduce some basic notions related to Eq. (1.1). Throughout the paper we will assume that the function F used in the definition of the operator (1.2) is a Finsler–Minkowski norm; in other words we will suppose that F : Rn → R+ 0 satisfies the following properties. (F-1): (Regularity) F ∈ C 2 (Rn \ {o}). (F-2): (Positive homogeneity) F is positively homogeneous of degree 1; that is F (tξ) = tF (ξ) ∀ ξ ∈ Rn , and ∀ t > 0. (F-3): (Strong Convexity) D2 (F 2 )(ξ) > 0 on Rn \ {o}. In Section 3, we will state some basic properties of Finsler–Minkowski norms that will be used in the paper. We now suspend further consideration of Finsler–Minkowski norms and we introduce our basic assumptions on the nonhomogeneous term f in Problem (1.1). Suppose that Ω ⊆ Rn is an open set, which may be unbounded, and let us consider a continuous function f : Ω × R+ × Rn → R that satisfies the following two conditions. (f-a): There is t0 ∈ R+ and continuous functions h : (0, t0 ) → R+ and a : Ω → R+ 0 with a(x) ̸≡ 0 on Ω such that f (x, t, p) ≥ a(x)h(t) for all (x, t, p) ∈ Ω × (0, t0 ) × Rn . (f-b): There are continuous functions g : R+ → R+ and b : Ω → R+ such that f (x, t, p) ≤ b(x)g(t) for all (x, t, p) ∈ Ω × R+ × Rn . Throughout this section we use the notation h0 := lim inf t→0+

h(t) . t

(2.1)

We also assume that 0 < h0 ≤ ∞. It turns out that the investigation of problem (1.1) is closely related to the requirement on the function b in Condition (f-b) imposed by the solvability of the following boundary value problem. { −∆N F;∞ u = b(x), u > 0 in Ω (b-w) u = 0, on ∂Ω .

T. Biset, B. Mebrate and A. Mohammed / Nonlinear Analysis 190 (2020) 111588

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When Ω is unbounded the boundary condition in problems (1.1) and (b-w) is interpreted to mean u(x) → 0 as |x| → ∞ and x ∈ Ω . If Ω is bounded, then the Comparison Principle, [47, Theorem 4.3] shows that any solution of −∆N F;∞ u = b in Ω with u = 0 on ∂Ω must be nonnegative in Ω . This together with [37, Corollary 4.6] and [37, Theorem 5.4], shows that Problem (b-w) admits a solution w ∈ C(Ω ) such that w > 0 in Ω provided Ω is a bounded domain and b ∈ C(Ω , R+ ). Therefore, the requirement that u > 0 in Ω in (b-w) is relevant only when Ω is unbounded. When Ω is unbounded, in addition to the above conditions, we also require the following condition on f . (f-c): For each (x, p) ∈ Ω × Rn , the function t ↦→

f (x,t,tp) t

is nonincreasing on R+ .

Our first main result establishes the existence of a solution to (1.1) in bounded domains when the parameter λ belongs to an appropriate open subinterval of R+ . We emphasize that, in this paper, by a domain in Rn we will always mean an open connected subset of n R . Theorem 2.1. Let Ω ⊆ Rn be a bounded domain and suppose f : Ω ×R+ ×Rn → R is a continuous function for which conditions (f-a), (f-b) hold with b such that Problem (b-w) admits a solution in C(Ω ). Then there is a positive constant Λ(Ω ), depending on Ω and a, and a positive constant λ∗Ω , depending on g, b and Ω , ∗ such that Problem (1.1) admits a solution u = uλ ∈ C(Ω ) for any parameter λ with Λ(Ω )h−1 0 < λ < λΩ . Our next main result establishes the existence of a solution to Problem (1.1) when Ω is an unbounded domain Rn . For this we need f to meet the additional requirement (f-c). Theorem 2.2. Let Ω ⊆ Rn be an unbounded domain and suppose f : Ω ×R+ ×Rn → R+ 0 satisfies conditions (f-a) with a(x) > 0 in Ω , and (f-b) with b such that Problem (b-w) admits a solution in C(Ω ) ∩ L∞ (Ω ). If, in addition, f satisfies (f-c) then given an open bounded subset Ω0 ⊆ Ω there is a positive constant Λ(Ω0 ), depending on Ω0 and a, and there is a positive constant λ∗Ω , depending on g, b and Ω such that Problem (1.1) admits a solution u = uλ ∈ C(Ω ) for any Λ(Ω0 )h0−1 < λ < λ∗Ω . 3. On some basic preliminary results In the following lemma we recall some basic properties of Finsler–Minkowski norms. Even though the lemma is used in our work, we do not make an explicit reference to it. Further details and proofs can be found in [5]. Lemma 3.1.

Let F : Rn → [0, ∞) be a Finsler–Minkowski norm. Then

(i) F (o) = 0 and F (ξ) > 0 for ξ ∈ Rn \ {o}. (ii) F satisfies the triangle inequality. That is F (ξ + η) ≤ F (ξ) + F (η)

∀ ξ, η ∈ Rn .

Equality holds iff η = κξ for some κ ≥ 0. + n The dual F ∗ : Rn → R+ 0 of a Finsler–Minkowski norm F : R → R0 is given by

F ∗ (p) := sup ξ̸=o

⟨p, ξ⟩ , F (ξ)

and, as has been illustrated in the papers [36,37,47], it plays an important role in the investigation of the normalized Finsler infinity-Laplacian. It is easily seen from the definition that F ∗ is positively homogeneous of degree 1. In fact it is well-known that F ∗ is actually a Finsler–Minkowski norm (see [5] for details). The following lemma will be useful later, and we refer to [1] for proofs.

T. Biset, B. Mebrate and A. Mohammed / Nonlinear Analysis 190 (2020) 111588

Lemma 3.2.

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Let F be a Finsler–Minkowski norm. Then

(i) F (DF ∗ (p)) = 1 ∀ p ∈ Rn \ {o}. (ii) F ∗ (DF (ξ)) = 1 ∀ ξ ∈ Rn \ {o}. (iii) DF (DF ∗ (p)) = F ∗1(p) p ∀ p ∈ Rn \ {o}. While a proof of the above lemma can be found in [1, Lemma 3.1], we wish to make a short comment on the proof. As part of the hypotheses used in [1, Lemma 3.1], the function F is assumed to be absolutely homogeneous in the sense that F (tx) = |t|F (x) for all (t, x) ∈ R × Rn . However, a close look at the proof of [1, Lemma 3.1] that corresponds to Lemma 3.2(i) and (ii), uses condition (F-2) at each step where homogeneity of F is used. We also note that Lemma 3.2, (iii) is a direct consequence of [1, Lemma 3.1 (ii)]. To see this we observe, according to [1, Lemma 3.1 (ii)], that F (F ∗ (x)DF ∗ (x))DF (F ∗ (x)DF ∗ (x)) = x

∀ x ∈ Rn \ {o}.

(3.1)

Due to (F-2) and Lemma 3.2(i), we have F (F ∗ (x)DF ∗ (x)) = F ∗ (x)F (DF ∗ (x)) = F ∗ (x),

∀ x ∈ Rn \ {o}.

(3.2)

Since DF is homogeneous of degree 0 we have DF (F ∗ (x)DF ∗ (x)) = DF (DF ∗ (x)),

∀ x ∈ Rn \ {o}.

(3.3)

Using (3.2) and (3.3) in (3.1) we have F ∗ (x)DF (DF ∗ (x)) = x,

that is DF (DF ∗ (x)) =

1 x, F ∗ (x)

∀ x ∈ Rn \ {o},

which is Lemma 3.2, (iii). The following, which is also a consequence of Lemma 3.2, establishes the dual relationship between F and F ∗. ⟨ξ, p⟩ F (ξ) = sup ∗ , ξ ∈ Rn . F (p) p̸=o Given x ∈ Rn and r > 0 we introduce the Finsler balls BF− (x, r) = {y ∈ Rn : F ∗ (x − y) < r}. However, we will continue to use the standard notation B(x, r) to denote the Euclidean ball of radius r > 0 and centered at x ∈ Rn . We wish to stress that it is always assumed in this paper that F is a Finsler–Minkowski norm. 3.1. Definitions We begin by introducing some useful notations. For a function ϕ that is twice differentiable at x0 ∈ Rn , let {⟨ ⟩ D2 ϕ(x0 )DF (Dϕ(x0 )) , DF (Dϕ(x0 )) if Dϕ(x0 ) ̸= o + ⟨ 2 ⟩ ∆F;∞ ϕ(x0 ) = (3.4) max{ D ϕ(x0 )e , e : F ∗ (e) = 1} if Dϕ(x0 ) = o, and ∆− F;∞ ϕ(x0 )

{⟨ =

D2 ϕ(x0 )DF (Dϕ(x0 )) , DF (Dϕ(x0 )) {⟨ ⟩ } min D2 ϕ(x0 )e , e : F ∗ (e) = 1

⟩

if Dϕ(x0 ) ̸= o if Dϕ(x0 ) = o.

(3.5)

T. Biset, B. Mebrate and A. Mohammed / Nonlinear Analysis 190 (2020) 111588

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Remark 3.3. If ϕ is twice differentiable at x0 , then + ∆− F;∞ ϕ(x0 ) ≤ ∆F;∞ ϕ(x0 ).

We are now ready to give the definition of viscosity solutions of equations involving the Finsler infinity-Laplacian. Let Ω ⊆ Rn be an open set and ζ : Ω × R+ × Rn → R be a given function. We consider the equation − ∆N F;∞ u = ζ(x, u, Du),

(x ∈ Ω ).

(3.6)

We will use the notation u ≺x0 ϕ to mean that u − ϕ has a maximum at x0 . Likewise we write u ≻x0 ϕ to mean that u − ϕ has a minimum at x0 . We use U SC(Ω ) (resp., LSC(Ω )) to denote the class of upper-semicontinuous (resp., lower-semicontinuous) functions on Ω . Definition 3.4.

Let Ω ⊆ Rn be open.

(1) u ∈ USC(Ω ) is said to be a viscosity subsolution of (3.6) in Ω if and only if u > 0 in Ω and for every (ϕ , x0 ) ∈ C 2 (Ω ) × Ω such that u ≺x0 ϕ we have −∆+ F;∞ ϕ(x0 ) ≤ ζ (x0 , u(x0 ), Dϕ(x0 )) . In this case we write −∆N F;∞ u ≤ ζ(x, u, Du) in Ω . (2) u ∈ LSC(Ω ) is said to be a viscosity supersolution of (3.6) in Ω if and only if u > 0 in Ω and for every (ϕ , x0 ) ∈ C 2 (Ω ) × Ω such that u ≻x0 ϕ we have −∆− F;∞ ϕ(x0 ) ≥ ζ(x0 , u(x0 ), Dϕ(x0 )). In this case we write −∆N F;∞ u ≥ ζ(x, u, Du) in Ω . (3) u ∈ C(Ω ) is a solution of (3.6) in Ω if and only if u is a viscosity subsolution as well as a viscosity supersolution of (3.6) in Ω . If u is a viscosity solution in Ω we write −∆N F;∞ u = ζ(x, u, Du). We now wish to make several remarks on the above notation and the definition. Remark 3.5. (1) If ζ : Ω × R × Rn → R, we drop the requirement that u > 0 in Ω in Definition 3.4. (2) Suppose u and φ are twice differentiable at x. + (i) If u ≺x φ, then ∆+ F;∞ u(x) ≤ ∆F;∞ φ(x). − (ii) If u ≻x φ, then ∆− F;∞ u(x) ≥ ∆F;∞ φ(x).

(3)

(i) If u1 , u2 are subsolutions of (3.6), then max{u1 , u2 } is a subsolution of (3.6). (ii) If u1 , u2 are supersolutions of (3.6), then min{u1 , u2 } is a supersolution of (3.6).

4. An eigenvalue problem In this section we discuss an eigenvalue problem which will be useful in the study of Problem (1.1). We use a method that is reminiscent of that used in [6,7,27,40]. We consider the following eigenvalue problem { −∆N F;∞ u = λa(x)u in Ω (4.1) u = 0 in ∂Ω .

T. Biset, B. Mebrate and A. Mohammed / Nonlinear Analysis 190 (2020) 111588

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Lemma 4.1. Let Ω ⊆ Rn be a bounded domain and a ∈ C(Ω , R+ 0 ) with a ̸≡ 0 in Ω . Then there is a positive constant Λ(Ω ) such that Problem (4.1) admits a viscosity solution ϕΩ ∈ C(Ω ) in Ω such that ϕΩ > 0 in Ω . Proof . We begin by studying the perturbed problem { −∆N F;∞ u = λa(x)u + a(x) u = 0

in Ω in ∂Ω .

(4.2)

Let us consider the set Ea (Ω ) := {λ ∈ R : Problem (4.2) has a solution v ∈ C(Ω ) with v > 0 in Ω }

(4.3)

Step 1: We show that Ea (Ω ) is non-empty and bounded from above. Let v be a solution of (4.2) with λ = 0 (see [37, Theorem 6.1]). Since −∆N F;∞ v ≥ 0 in Ω , and a(x) ̸≡ 0 in Ω , it follows that v > 0 in Ω (see [37, Corollary 4.6]). Therefore 0 ∈ Ea (Ω ) showing that Ea (Ω ) is a non-empty set. Next we show that Ea (Ω ) is bounded from above. Let λ > 0 belong to Ea (Ω ), and by definition, we take a solution u ∈ C(Ω ) of (4.2) such that u > 0 in Ω . Let x0 ∈ Ω such that a(x0 ) > 0, so that a(x) > 0 on BF− (x0 , r) ⊂⊂ Ω for some r > 0. Let θ := min a(x), and ϑ := min u(x). −

−

B F (x0 ,r)

B F (x0 ,r)

We note that θ, ϑ > 0. Let us consider the nonincreasing and concave (see Lemma A.1 in the Appendix) function κ(t) := min u, 0 ≤ t ≤ r, −

B F (x0 ,t)

where we set κ(0) = u(x0 ). Note that κ(r) = ϑ, and −

κ(F ∗ (x0 − x)) ≤ u(x)

∀ x ∈ B F (x0 , r).

Therefore − − ∗ − ∆N F;∞ u ≥ λθκ(F (x0 − x)) + θ in BF (x0 , r), and u ≥ ϑ on ∂BF (x0 , r).

(4.4)

Let us define the function ∫ t∫

s

ϕ(t) := η −

(λθκ(ϱ) + θ) dϱ ds, 0

0 ≤ t ≤ r.

0

Here, ∫

r

∫

η := ϑ +

s

(λθκ(τ ) + θ) dτ ds. 0

0

Since κ is continuous, we clearly have ϕ′′ (t) = −(λθκ(t) + θ),

ϕ(0) = η, ϕ′ (0) = 0, and ϕ(r) = ϑ.

(4.5)

Let us now consider w(x) := ϕ(F ∗ (x0 − x)). According to [37, Corollary 4.6] we see that ′′ ∗ ∗ −∆N F;∞ w = −ϕ (F (x0 − x)) = λθκ(F (x0 − x)) + θ,

Therefore we have

{

−∆N F;∞ w w

= λθκ(F ∗ (x0 − x)) + θ = ϑ

x ∈ BF− (x0 , r).

in BF− (x0 , r), on ∂BF− (x0 , r).

(4.6)

From (4.4), (4.6) and the Comparison Principle, [47, Theorem 4.3], we find that −

w ≤ u in B F (x0 , r).

(4.7)

T. Biset, B. Mebrate and A. Mohammed / Nonlinear Analysis 190 (2020) 111588

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− Now let 0 ≤ t ≤ r. We remark that since −∆N F;∞ u ≥ 0 in BF (x0 , t), [37, Corollary 4.6] shows that

min

− ∂BF (x0 ,t)

−

u ≤ u(ξ) ∀ ξ ∈ B F (x0 , t).

(4.8)

Moreover, we have on recalling (4.7), that ϕ(t) = ϕ(F ∗ (x0 − x)) = w(x) ≤ u(x) for all x ∈ ∂BF− (x0 , t). This together with (4.8), and the definition of κ allows us to conclude that ϕ(t) ≤ κ(t)

∀ 0 ≤ t ≤ r.

(4.9)

Using (4.5) in (4.9) leads to ϕ′′ (t) = −λθκ(t) − θ ≤ −λθϕ(t), 0 ≤ t ≤ r, ϕ(0) = η, ϕ′ (0) = 0 and ϕ(r) = ϑ. Multiplying both sides of ϕ′′ (s) ≤ −λθϕ(s) by ϕ′ (s) and integrating on (0, t) we find (ϕ′ (t))2 ≥ λθ(η 2 − ϕ(t)2 ), that is

√

λθ ≤ − √

ϕ′ (t) η 2 − (ϕ(t))2

,

0≤t≤r

0 < t ≤ r.

Integrating this on (0, r) we find √

1 λ≤ √ r θ

∫

η

ϑ

ds

1 √ ≤ √ 2 2 r θη η −s

∫

η

ϑ

2 ds √ = √ η−s r θ

√

2 η−ϑ ≤ √ . η r θ

This shows that Ea (Ω ) is bounded from above. Step 2: Let Λ := Λ(a, Ω ) := sup Ea (Ω ). We now claim that, with λ = Λ, Problem (4.1) admits a solution u ∈ C(Ω ) with u > 0 in Ω . To see this, let {λj } be a sequence of real numbers in Ea (Ω ) such that λj → Λ. Let wj ∈ C(Ω ) be the corresponding solutions of Problem (4.2) such that wj > 0 in Ω for all j. Let us first show that {supΩ wj } is unbounded. If this is not the case, let ∥λj a(x)wj + a(x)∥L∞ (Ω) ≤ M in Ω for some positive constant M and for all j. Let w∞ ∈ C(Ω ) such that −∆N F;∞ w∞ = M in Ω and w∞ = 0 on ∂Ω . By comparison principle, [47, Theorem 4.3], we find that 0 ≤ wj ≤ w∞ on Ω for all j. As a consequence of this and [37, Theorem 4.7], we see that {wj } is equicontinuous on compact subsets of Ω . Therefore {wj } has a subsequence that converges locally uniformly to w ∈ C(Ω ). Since 0 ≤ w ≤ w∞ in Ω , we actually see that w ∈ C(Ω ) and w = 0 on ∂Ω . Therefore, by [37, Proposition 3.7] (see also the proof of Proposition 5.1 below) we find that w solves { −∆N F;∞ w = Λa(x)w + a(x) in Ω (A.2) w = 0 on ∂Ω . Now let v := 2(w + ε), where 0 < ε < 1 is a fixed constant such that 2εΛ < 1. Then −∆N F;∞ v = 2Λa(x)w + 2a(x) = a(x)[Λv + (1 − 2εΛ)] + a(x) [ ] v = a(x) Λv + (1 − 2εΛ) + a(x) 2(w + ε) [ ] 1 − 2εΛ ≥ a(x) Λ + v + a(x) 2(1 + supΩ w) = γa(x)v + a(x),

T. Biset, B. Mebrate and A. Mohammed / Nonlinear Analysis 190 (2020) 111588

where γ := Λ +

9

1 − 2εΛ . 2(1 + supΩ w)

Note that γ > Λ, z = 0 is a subsolution and z = v is a supersolution of { −∆N F;∞ z = γa(x)z + a(x) in Ω z = 0 in ∂Ω .

(4.10)

Then, according to [37, Theorem 5.4], Problem (4.10) has a solution u with 0 ≤ u ≤ v. Since −∆N F;∞ u ≥ 0 in Ω , it follows from [37, Corollary 4.6] that u > 0 in Ω . This shows that γ ∈ Ea (Ω ) which is in contradiction with the definition of Λ. Thus we have shown that the sequence {supΩ wj } is unbounded, as asserted. We note that 1 −∆N a(x). F;∞ vj = λj a(x)vj + supΩ wj where vj := Let M > 0 be a constant such that   λj a(x)vj + 

1 wj supΩ wj

1 supΩ wj

  a(x) 

j = 1, 2, . . . .

≤ M ∀ j = 1, 2, . . . .

L∞ (Ω)

Let z∗ , z ∗ ∈ C(Ω ) be solutions of N ∗ ∗ −∆N F;∞ z∗ = −M, −∆F;∞ z = M in Ω with z∗ = z = 0 on ∂Ω .

The existence of such solutions follows from [47, Theorem 5.4] or [37, Theorem 6.1]. By the Comparison Principle we see that z∗ ≤ vj ≤ z ∗ on Ω for all j ∈ N. From [37, Theorem 4.7] we see that {vj } is equicontinuous on compact subsets of Ω and is uniformly bounded in Ω . Therefore {vj } contains a subsequence that converges locally uniformly to v ∈ C(Ω ), which according to [37, Proposition 3.7], satisfies −∆N F;∞ v = Λa(x)v in Ω and, v = 0 on ∂Ω . Recalling that vj ≥ 0 in Ω for all j, we note that v ≥ 0 in Ω . Moreover, since a(x) ̸≡ 0 in Ω , we find by [37, Corollary 4.6] that v > 0 in Ω . This concludes the proof of the lemma. □ Remark 4.2. It will be convenient to call the positive number Λ = Λ(Ω ) in Lemma 4.1 as the principal eigenvalue of Problem (4.1) corresponding to Ω , and any viscosity solution ϕ ∈ C(Ω ) that satisfies Problem (4.1) with λ = Λ(Ω ) and such that ϕ > 0 in Ω as an eigenfunction of Problem (4.1) corresponding to Λ. Note that if ϕ is an eigenfunction of Problem (4.1) corresponding to Λ, so is tϕ for any t > 0. Corollary 4.3.

Suppose O and Q are bounded domains in Rn such that O ⊆ Q. Then Λ(O) ≥ Λ(Q).

Proof . It suffices to show that Ea (Q) ⊆ Ea (O) where, for a given open bounded subset Ω ⊆ Rn , Ea (Ω ) is given as in (4.3). For this, suppose λ ∈ Ea (Q) so that Problem (4.2) has a solution v ∈ C(Q) with v > 0 in Q. Note that w∗ = 0 is a subsolution while w∗ = v is a supersolution of (4.2) on O (see Lemma A.2 in the Appendix). Therefore, Problem (4.2) admits a solution u on O such that w∗ ≤ u ≤ w∗ on O (see [37, Theorem 5.4]). In particular u ≥ 0 in O. By [37, Corollary 4.6] we conclude that u > 0 in O. Therefore λ ∈ Ea (O), as claimed. □

10

T. Biset, B. Mebrate and A. Mohammed / Nonlinear Analysis 190 (2020) 111588

5. On singular boundary-value problem Let Ω ⊆ Rn be an open bounded set, h : Ω × R+ × Rn → R a continuous function, and c ∈ C(∂Ω , R+ 0 ). The next result concerns the following Dirichlet problem. { −∆N F;∞ u = h(x, u, Du) in Ω (5.1) u(x) = c on ∂Ω . We point out that h(x, t, p) may have a singularity at t = 0, where we may for instance have h(x, t, p) = ω(x) t−γ

|p| . 1 + |p|

Moreover, we allow the boundary data c to be identically zero on the boundary ∂Ω . Proposition 5.1. Let Ω ⊆ Rn be a bounded open set, and suppose h : Ω × R+ × Rn → R is a continuous function that is bounded on O × I × Rn for any subset O ⊂⊂ Ω and any interval I ⊂⊂ R+ . Suppose v ∈ C(Ω ) is a subsolution and w ∈ C(Ω ) is a supersolution of (5.1) in Ω such that 0 < v ≤ w in Ω and v = c = w on ∂Ω . Then Problem (5.1) admits a solution u ∈ C(Ω ) such that v ≤ u ≤ w in Ω . Proof . Let us consider k : Ω × R × Rn → R defined by { h(x, t, p) k(x, t, p) := h(x, v(x), p)

for t ≥ v(x) for t < v(x).

Since v > 0 on Ω , we see that supΩ×I×Rn |k(x, t, p)| < ∞ for any compact interval I ⊆ R. Let {Ωj } be an exhaustion of Ω by subdomains of Ω . For each positive integer j, let us consider the Dirichlet problem { −∆N F;∞ u = k(x, u, Du) in Ωj (5.2) u = v on ∂Ωj . We should note that v is a subsolution and w is a supersolution of (5.2) in Ωj for each j. Since v ≤ w in Ω , we invoke [37, Theorem 5.4] to find a solution uj ∈ C(Ω j ) of (5.2) such that v ≤ uj ≤ w in Ω j . We extend uj to Ω by defining uj to be v on Ω \Ωj . In this manner, we obtain a sequence {uj } in C(Ω ) such that v ≤ uj ≤ w in Ω for all j. In particular {uj } is uniformly bounded in Ω . The uniform boundedness of {uj } and [37, Theorem 4.7] show that {uj } is locally equicontinuous in Ω . Therefore, by the Cantor Diagonalization procedure, we can find a subsequence, still denoted by {uj }, that converges locally uniformly to some u ∈ C(Ω ). Since v ≤ u ≤ w, we can extend u to Ω with u = c on ∂Ω . It remains to show that u is a viscosity solution of −∆N F;∞ u = k(x, u, Du) in Ω . This follows from a slight modification of the proof of [37, Proposition 3.7]. However, for the readers’ convenience we include the proof below. As the proof that u is a supersolution runs along similar lines, we confine ourselves to showing that u is a subsolution of −∆N F;∞ u = k(x, u, Du) in Ω . Let (x0 , φ) ∈ Ω × C 2 (Ω ) such that u ≺x0 φ. We fix r > 0 sufficiently small so that for some positive integer ℓ we have B(x0 , r) ⊂⊂ Ωℓ and, (u − φ)(x) ≤ (u − φ)(x0 ),

∀ x ∈ B(x0 , r).

Given an arbitrary but fixed ε > 0, let xj ∈ B(x0 , r) be a point of maximum of ( ) ε 2 uj (x) − φ(x) + |x − x0 | , j ≥ ℓ. 2 In particular,

ε 2 |xj − x0 | ≤ uj (xj ) − φ(xj ) − (uj (x0 ) − φ(x0 )), ∀ j = ℓ, ℓ + 1, . . . . 2

(5.3)

T. Biset, B. Mebrate and A. Mohammed / Nonlinear Analysis 190 (2020) 111588

11

By passing to a subsequence, if necessary, we assume that xj → x ˆ for some x ˆ ∈ B(x0 , r). Taking the limit in (5.3) as j → ∞ we find ε 2 |ˆ x − x0 | ≤ u(ˆ x) − φ(ˆ x) − (u(x0 ) − φ(x0 )) ≤ 0, 2 so that x ˆ = x0 . We can suppose that ℓ is large enough such that xj ∈ B(x0 , r) for j ≥ ℓ. Since for j ≥ ℓ, 2 uj is a solution of (5.2) and xj is a point of maximum of uj − (φ + ε|x − x0 | /2) on B(x0 , r), on using 2 ψε (x) := φ(x) + ε|x − x0 | /2, as a test function we find − ∆+ F;∞ ψε (xj ) ≤ k(xj , uj (xj ), Dψε (xj )),

j ≥ ℓ.

(5.4)

Recalling that uj → u uniformly on B(x0 , r) and that k : Ω × R × Rn → R is a continuous function we find, on taking the limit as j → ∞ in (5.4), and then taking ε → 0, that −∆+ F;∞ φ(x0 ) ≤ k(x0 , u(x0 ), Dφ(x0 )). In fact, we justify this as follows. Note that, for j ≥ ℓ we have ∆+ F;∞ ψε (xj ) { {⟨( 2 ) ⟩ } max D φ(xj ) + εIn e , e : F ∗ (e) = 1 , if Dφ(xj ) + ε(xj − x0 ) = o ) ⟩ = ⟨( 2 D φ(xj ) + εIn DF (Dφ(xj ) + ε(xj − x0 )) , DF (Dφ(xj ) + ε(xj − x0 )) , otherwise. In any case, we conclude using Lemma 3.2, that for each j ≥ ℓ there is ej ∈ Rn such that ( 2 ) ∗ ∆+ F;∞ ψε (xj ) ≤ max{⟨ D φ(xj ) + εIn e , e⟩ : F (e) = 1} ( 2 ) = ⟨ D φ(xj ) + εIn ej , ej ⟩ = ⟨(D2 φ(xj ))ej , ej ⟩ + ε⟨ej , ej ⟩. Let {e′j } be a subsequence of {ej } such that e′j → e′ with F ∗ (e′ ) = 1. Then for such subsequence and corresponding subsequence {x′j } we have ⟨(D2 φ(x′j ))e′j , e′j ⟩ + ε⟨e′j , e′j ⟩ → ⟨D2 φ(x0 )e′ , e′ ⟩ + ε⟨e′ , e′ ⟩ Therefore, for fixed ε we have ( 2 ) ∗ lim inf ∆+ F;∞ ψε (xj ) ≤ lim inf max{⟨ D φ(xj ) + εIn e , e⟩ : F (e) = 1} j→∞

k→∞

′ ′ ≤ ∆+ F;∞ φ(x0 ) + ε⟨e , e ⟩. 2

2

Since ⟨e′ , e′ ⟩ = |e′ | ≤ (βF ∗ (e′ )/α) = (β/α)2 we have + lim lim inf ∆+ F;∞ ψε (xj ) ≤ ∆F;∞ φ(x0 ).

ε→0 j→∞

Consequently, on recalling that k ∈ C(Ω × R × Rn ) we find −k(x0 , u(x0 ), Dφ(x0 )) ≤ lim inf lim inf −k(xj , u(xj ), Dψε (xj )) ε→0

j→∞

≤ lim inf lim inf ∆+ F;∞ ψε (xj ) ε→0

j→∞

≤ ∆+ F;∞ φ(x0 ). Thus, u is a solution of (5.2). Since u ≥ v > 0 in Ω , it follows from the definition of k that, actually, u is a solution of −∆N F;∞ u = h(x, u, Du) in Ω . □

T. Biset, B. Mebrate and A. Mohammed / Nonlinear Analysis 190 (2020) 111588

12

In the investigation of problem (1.1), existence of a positive supersolution v to Problem (1.1) for appropriate range of the parameter λ > 0 will play a crucial role. As a first step towards this goal, let us introduce the continuous and positive function G : R+ × R+ → R+ defined by { } ⎧ g(s) ⎪ ⎨t sup : t < s ≤ r , for 0 < t < r s G(t, r) := g(r) ⎪ ⎩ for t ≥ r. t r Here, g is the function in Condition (f-b). We observe that for each fixed r > 0, the function G(t, r) t + is nonincreasing on R . We consider the corresponding average function Γ : R+ × R+ → R+ : ∫ 2 t G(s, γ) Γ (t, γ) := ds. t t/2 s t ↦→

(5.5)

It is easily seen that for each r ∈ R+ , the function t ↦→ Γ (t, r) is C 1 and decreasing1 on R+ . Moreover, we should note that for 0 < t < γ we have ∫ ∫ 2 t G(s, γ) 2 t G(t, γ) G(t, γ) g(t) Γ (t, γ) = ds ≥ ds = ≥ . t t/2 s t t/2 t t t Thus, given γ > 0 we have g(t) ≤ Γ (t, γ)t,

0 < t < γ.

(5.6)

In fact, we see that { } g(s) 1 g(t) ≤ Γ (t, 2t) ≤ sup : t < s ≤ 2t , t s 2 As a consequence of this estimate the following remark holds. Remark 5.2. lim Γ (t, 2t) = 0 if and only if lim

t→∞

t→∞

t > 0.

g(t) = 0. t

We also point out that if g(t) = σ, t→∞ t lim

then

lim Γ (t, 2t) = σ.

t→∞

The next result shows the existence of a positive supersolution to Problem (1.1) in any open subset of Rn . It should be noted that we make no sign restriction on the function f and that Ω is any open subset of Rn . Theorem 5.3. Let Ω ⊆ Rn be an open set and f : Ω × R+ × Rn → R be a continuous function that satisfies (f-b) with b such that Problem (b-w) admits a solution w ∈ C(Ω ) ∩ L∞ (Ω ). There exists a positive constant λ∗Ω , depending on g, b and Ω , such that for each 0 < λ < λ∗Ω , Problem (1.1) admits a positive supersolution vΩ,λ ∈ C(Ω ). Moreover, vΩ,λ satisfies − ∆N F;∞ vΩ,λ ≥ λb(x)Γ (vΩ,λ , κ) in Ω

(5.7)

for some constant κ > 1. 1

Note that ∂ Γ (t, γ) 2 =− 2 ∂t t

∫

t

t/2

G(s, γ) 1 ds + s t

1 G(t, γ) 1 ≤− + t t t

[

[

2G(t, γ) G(t/2, γ) − t t/2

2G(t, γ) G(t/2, γ) − t t/2

]

=

1 t

[

]

G(t, γ) G(t/2, γ) − t t/2

]

.

T. Biset, B. Mebrate and A. Mohammed / Nonlinear Analysis 190 (2020) 111588

13

Proof . Let wΩ ∈ C(Ω ) ∩ L∞ (Ω ) be a positive solution on Ω of Problem (b-w). We set Θ := lim inf Γ (t, 2t), t→∞

where Γ is the function given in (5.5). Let us first suppose that Θ < ∞, and define ⎧ 1 ⎨ if 0 < Θ < ∞ λ∗Ω := 2Θ ∥wΩ ∥∞ ⎩ ∞ if Θ = 0. For 0 < λ < λ∗Ω we have

(5.8)

2λ∗Ω ∥wΩ ∥∞ > 2λ∥wΩ ∥∞ .

Therefore, there is θ := θλ > 1 such that Γ (θ, 2θ) <

1 . 2λ∥wΩ ∥∞

We consider the increasing and C 2 function ∫

t

Ψ (t) := 0

ds , Γ (s, 2θ)

t > 0.

Then ∫

2θ

Ψ (2θ) = 0

ds ≥ Γ (s, 2θ)

∫ θ

2θ

ds Γ (s, 2θ)

θ ≥ , since s ↦→ Γ (s, 2θ) is decreasing. Γ (θ, 2θ) > 2λθ∥wΩ ∥∞ . We introduce the function vΩ,λ (x) := Φ (2λθwΩ (x)) ,

x ∈ Ω,

(5.9)

where Φ be the inverse of Ψ ; that is ∫ 0

Φ(t)

ds ds = t, Γ (s, 2θ)

t > 0.

We observe that vΩ,λ (x) ≤ Φ (2λθ∥wΩ ∥∞ ) ≤ 2θ.

(5.10)

It is obvious that vΩ,λ > 0 in Ω and vΩ,λ = 0 on ∂Ω . Here vΩ,λ (x) → 0 as |x| → ∞ within Ω , in case Ω is unbounded. On recalling that wΩ is a solution of (b-w), we invoke [37, Proposition 3.9] to get ′′ 2 ′ −∆N F;∞ vΩ,λ = −Φ (2λθwΩ )F (2λθDwΩ ) + 2λθΦ (2λθwΩ )b

≥ 2λθbΓ (Φ(2λθwΩ ), 2θ) g(Φ(2λθwΩ )) , by (5.6) ≥ 2λθb Φ(2λθwΩ ) ≥ λbg(Φ(2λθwΩ )), by (5.10) = λbg(vΩ,λ ) ≥ λf (x, vΩ,λ , DvΩ,λ ), Therefore, indeed vΩ,λ is a supersolution of Problem (1.1).

by Condition (f -b).

(5.11)

T. Biset, B. Mebrate and A. Mohammed / Nonlinear Analysis 190 (2020) 111588

14

Suppose now Θ = ∞. Let

) ( ∫ t ds 1 µ := sup . t 0 Γ (s, 2t) t>1

Clearly2 0 < µ < ∞. Let λ∗Ω := Then for 0 < λ < λ∗Ω we have

(5.12)

µ . ∥wΩ ∥∞

(5.13)

µ = λ∗Ω ∥wΩ ∥∞ > λ∥wΩ ∥∞ .

Therefore, there is θ := θλ > 1 such that ∫

θ

0

ds > λθ∥wΩ ∥∞ . Γ (s, 2θ)

We now set

∫

t

Ψ (t) := 0

ds , Γ (s, 2θ)

t > 0.

Then Ψ (θ) > λθ∥wΩ ∥∞ . Let vΩ,λ (x) := Φ(λθwΩ (x)),

x ∈ Ω,

where, as before, Φ is the inverse of Ψ . A similar proof as in the above shows that vΩ,λ is a supersolution of (1.1). In particular, the calculations in (5.11) hold with 2θ replaced by θ. Finally, since θ > 1 we see from (5.11) that −∆N F;∞ vΩ,λ ≥ λb(x)Γ (vλ , κ),

x ∈ Ω,

where κ := 2θ, if Θ < ∞ and κ := θ, if Θ = ∞. □ Remark 5.4. Suppose the function g in Condition (f-b) satisfies g(t) = 0. t It follows from Remark 5.2 that Θ = 0 in (5.8), so that λ∗Ω = ∞. Therefore in this case, Theorem 5.3 is valid for any parameter 0 < λ < ∞. lim

t→∞

Remark 5.5. Let O ⊆ Ω ⊆ Rn be bounded domains and suppose that Problem (b-w) admits solutions wO and wΩ corresponding to O and Ω , respectively. Note that −∆N F;∞ wΩ = b in O (see Lemma A.2 in the Appendix) and wΩ > 0 in O. By Comparison Principle we observe that 0 < wO ≤ wΩ in O. As a consequence of this we make two observations. (i) The definitions given in (5.8) or (5.13) show that λ∗Ω ≤ λ∗O . (ii) By the definition given in (5.9) we see that vO,λ ≤ vΩ,λ in O. To establish existence of solution to Problem (1.1), we need to distinguish between bounded and unbounded domains in Rn . Let us take up the bounded domain case first. 2

To see this, note that since, for a fixed t, the function s ↦→ 1/Γ (s, 2t) is increasing, we see that there is t1 > 1 such that 1 t

t

∫ 0

ds Γ (s, 2t)

≤

1

≤ 1 for all t ≥ t1 > 1.

Γ (t, 2t)

Here we have used the assumption that Θ = ∞. Therefore

{ µ ≤ max

∫ 1, sup 1
0

t1

ds Γ (s, 2t)

} < ∞.

T. Biset, B. Mebrate and A. Mohammed / Nonlinear Analysis 190 (2020) 111588

15

5.1. On bounded domains In this subsection we give the proof of Theorem 2.1. We need the following. Proposition 5.6. Let Ω ⊆ Rn be a bounded domain and f : Ω × R+ × Rn → R be a continuous function that satisfies condition (f-a). Given λ > Λ(Ω )h−1 0 , there is ϕΩ,λ ∈ C(Ω ) and ϕΩ,λ > 0 in Ω which is a subsolution of Problem (1.1) in Ω . Proof . Let λ > Λ(Ω )h−1 0 be fixed. Let us first suppose that h0 , defined in (2.1), is finite. Then, according to (2.1), there is a δ := δ(λ) with 0 < δ < t0 such that ) ( Λ(Ω ) Λ(Ω )h−1 0 h0 t = t ∀ 0 < t ≤ δ. h(t) ≥ λ λ If, on the other hand, h0 = ∞, then clearly (2.1) shows that there is 0 < δ := δ(λ) < t0 such that h(t) ≥

Λ(Ω ) t λ

∀ 0 < t ≤ δ.

In any case, we take an eigenfunction ϕΩ,λ to Problem (4.1) corresponding to the principal eigenvalue Λ(Ω ) such that 0 < ϕΩ,λ ≤ min{1, δ(λ)} in Ω . Our choice of δ, Condition (f-a) and our assumption on the parameter λ show that λf (x, ϕΩ,λ , DϕΩ,λ ) ≥ λa(x)h(ϕΩ,λ ) ≥ Λ(Ω )a(x)ϕΩ,λ = −∆N F;∞ ϕΩ,λ ,

x ∈ Ω.

This completes the proof of the proposition. □ Remark 5.7. Suppose h0 = ∞, that is lim

t→0+

h(t) =∞ t

where h is the function in Condition (f-a). Then Proposition 5.6 holds for any parameter λ > 0. Combining Theorem 5.3 and Proposition 5.6 gives us the existence of a solution to Problem (1.1) in bounded domains. Proof of Theorem 2.1. Let λ∗Ω be the positive constant in the statement of Theorem 5.3, and let Λ(Ω )h−1 < λ < λ∗Ω . Let ϕΩ,λ be the subsolution of Problem (1.1) given by Proposition 5.6. Then we 0 have −∆N F;∞ ϕΩ,λ ≤ λf (x, ϕΩ,λ , DϕΩ,λ ) ≤ λb(x)g(ϕΩ,λ ) ≤ λb(x)Γ (ϕΩ,λ , κ)ϕΩ,λ , by (5.6), since ϕΩ,λ ≤ 1 < κ ≤ λb(x)Γ (ϕΩ,λ , κ),

since ϕΩ,λ ≤ 1 in Ω .

(5.14)

Let vΩ,λ be the supersolution of Problem (1.1) in Ω , as provided by Theorem 5.3. Note that ϕΩ,λ = 0 = vΩ,λ on ∂Ω . Since t ↦→ Γ (t, κ) is a non-increasing positive function, and vΩ,λ and ϕΩ,λ satisfy (5.7) and (5.14), respectively it follows from the Comparison Principle, [47, Theorem 4.3] that 0 < ϕΩ,λ ≤ vΩ,λ in Ω . We now invoke [37, Theorem 5.4] to conclude that Problem (1.1) has a solution uλ ∈ C(Ω ) such that ϕΩ,λ ≤ uλ ≤ vλ in Ω . □

16

T. Biset, B. Mebrate and A. Mohammed / Nonlinear Analysis 190 (2020) 111588

Remark 5.8. Suppose the function h in Condition (f-a) and the function g in Condition (f-b) satisfy the following. g(t) h(t) = ∞ and lim = 0. lim + t→∞ t t t→0 Then Theorem 2.1 holds for all parameter values 0 < λ < ∞. 5.2. On unbounded domains In this section, we study the boundary-value problem (1.1) when Ω ⊆ Rn is an unbounded domain. In what follows, we fix an exhaustion {Ωj } of Ω by nonempty bounded domains so that Ωj−1 ⊂⊂ Ωj for j = 1, 2, . . . and Ω = ∪∞ j=1 Ωj . We also suppose that Ω0 ⊆ Ω1 , where Ω0 is the bounded domain in the statement of Theorem 2.2. We start with the following proposition. + Proposition 5.9. Suppose that f : Ω × R+ × Rn → R+ 0 satisfies (f-a) with a ∈ C(Ω , R ), and let −1 λ > Λ(Ω0 )h0 . If, in addition, f satisfies (f-c), then given a positive integer k there is an eigenfunction φk := φΩk corresponding to Λ(Ωk ) such that

φk ≤ zj in Ωk for any solution zj := zΩj ,λ of (1.1) in Ωj with j > k. Proof . For each (x, t, p) ∈ Ω × R × Rn , let us set f (x, et , et p) . a(x)et

h(x, t, p) := By assumption λ > Λ(Ω0 )h−1 0 . Let us fix σ such that

Λ(Ω0 ) < σ < λh0 . By the definition of h0 , there is 0 < t∗ := t∗ (λ) ≤ t0 such that h(t) ≥ σλ−1 t for 0 < t < t∗ . Therefore by (f-a) we have f (x, t, p) σ ≥ , (x, t, p) ∈ Ω × (0, t∗ ) × Rn . a(x)t λ Consequently we also have h(x, t, p) ≥

σ , λ

(x, t, p) ∈ Ω × (−∞, log t∗ ) × Rn .

Note that, by Corollary 4.3 we have λ h(x, t, p) > Λ(Ω0 ) ≥ Λ(Ωk )

∀ (x, t, p) ∈ Ω × (−∞, log t∗ ) × Rn .

(5.15)

Let φk be an eigenfunction of Problem (4.1) corresponding to the principal eigenvalue Λ(Ωk ) on the set Ωk , and zj := zΩj ,λ be a solution of (1.1) in Ωj corresponding to λ. Since f ≥ 0, by the comparison Principle [36, Theorem 6.4] and [37, Corollary 4.6], we see that zj > 0 in Ωj . We assume that φk has been normalized so that 0 < φk ≤ t∗ /2 in Ωk . We wish to show that φk ≤ zj in Ωk for all j > k. For convenience we will drop the subscripts form φk and zj , and write O := Ωk . Let v := log(φ + τ ) and u := log z, where τ > 0 is any sufficiently small constant such that φ + τ ≤ z on ∂O. Direct computation (see [37, Proposition 3.9]) shows that each of the following holds in O, in the viscosity sense. 2 −∆N F;∞ v ≤ Λ(O)a(x) + F (Dv),

and

(5.16)

T. Biset, B. Mebrate and A. Mohammed / Nonlinear Analysis 190 (2020) 111588

1 f (x, z, Dz) + 2 F 2 (Dz) z z = λa(x) h(x, u, Du) + F 2 (Du).

17

−∆N F;∞ u = λ

(5.17)

Since u, v ∈ C(O), we note (see [37, Theorem 4.7]) that u and v are locally Lipschitz continuous in O. For (x, t, p) ∈ Ω × R × Rn , let g(x, t, p) := Λ(O) a(x) + F 2 (p), and { λa(x)h(x, t, p) + F 2 (p), t < log t∗ . h(x, t, p) := ∗ 2 λa(x)h(x, log t , p) + F (p), t ≥ log t∗ . We observe that both g(x, t, p) and h(x, t, p) are continuous in O × R × Rn . Moreover, we note that g(x, t, p) < h(x, t, p) on O × R × Rn and both are nonincreasing in t ∈ R for each (x, p) ∈ O × Rn . We also recall from (5.16) and (5.17) that both of the following hold in O. −∆N F;∞ v ≤ g(x, v, Dv)

and

− ∆N F;∞ u ≥ h(x, u, Du).

Since v ≤ u on ∂O, we invoke [36, Theorem 7.2] to conclude that v ≤ u in O. Recalling that τ is arbitrary, we obtain the desired conclusion. □ Proof of Theorem 2.2. Let λ∗ := λ∗Ω be the constant in Theorem 5.3 corresponding to the unbounded open set Ω . Let Λ(Ω0 )h−1 < λ < λ∗ , where Ω0 the bounded domain in the statement of Theorem 2.2. 0 We also recall that Ω0 ⊆ Ωj for all j = 1, 2, . . .. In view of Corollary 4.3 and Remark 5.5(i), we note that −1 ∗ ∗ Λ(Ωj )h−1 0 ≤ Λ(Ω0 )h0 < λ < λΩ ≤ λΩj for all j. Therefore, we invoke Theorem 2.1 to obtain a solution uj of Problem (1.1) in the bounded domain Ωj . Note that for all j = 1, 2, . . . we have, according to Remark 5.5 (ii), 0 < uj ≤ vΩj ,λ ≤ vΩ,λ in Ωj . We extend each uj to Ω by letting uj ≡ 0 on Ω \ Ωj . Therefore, {uj } is uniformly bounded in Ω and as a consequence of [37, Theorem 4.7], the sequence {uj } is locally uniformly Lipschitz in Ω . By the standard Cantor Diagonalization process we can extract a subsequence, which we continue to denote by {uj } that converges locally uniformly to some u ∈ C(Ω ), and 0 ≤ u ≤ vΩ,λ in Ω . Let us fix a positive integer k, and let φk be the eigenfunction corresponding to Λ(Ωk ) given in Proposition 5.9 so that 0 < φk ≤ uj in Ωk ∀ j > k. Therefore we see that 0 < φk ≤ u in Ωk . Thus we have shown that 0 < u < vΩ,λ in Ω , and by adapting the argument used in Proposition 5.1, we see that u is a viscosity solution of (1.1) in Ω . Clearly u(x) → 0 as |x| → ∞. □ Acknowledgments We would like to express our appreciation to the anonymous referees for their diligence in carefully reading the original manuscript and for making suggestions which helped make the paper more readable. This work was supported by ISP (International Science Program) of Uppsala University, Sweden. Appendix In this appendix we prove the following lemma which was used in the proof of Lemma 4.1.

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Lemma A.1. Let Ω ⊆ Rn be a bounded domain and u be a viscosity solution of −∆N F;∞ u ≥ 0 in Ω with − u = 0 on ∂Ω . If R > 0 and x0 ∈ Ω such that BF (x0 , R) ⊂⊂ Ω and ∀ 0 ≤ t ≤ R.

min u

κ(t) :=

−

B F (x0 ,t)

then κ is concave on (0, R). Proof . As noted in the proof of Lemma 4.1, we see that u > 0 in Ω . It is obvious that κ is a positive and nonincreasing function in [0, R]. Let 0 < s < ϱ < t < R and set ( ) t−r ∗ w(x) := v(F (x0 − x)), where v(r) := κ(t) + (κ(s) − κ(t)) . t−s We claim that v(ϱ) ≤ κ(ϱ). Note that this holds trivially if κ(s) = κ(t). So, in what follows we will suppose that κ(s) > κ(t). By [37, Corollary 3.10], or [47, Lemma 3.1 and Remark 3.2], we note that −∆N F;∞ w = 0 − on Rn \ {x0 }. Let O := BF− (x0 , t) \ B F (x0 , s). We observe that w(x) = v(t) = κ(t) ≤ u(x) ∀x ∈ ∂BF− (x0 , t), while we similarly have w(x) = v(s) = κ(s) ≤ u(x)

∀ x ∈ ∂BF− (x0 , s).

In other words, w ≤ u on ∂O. Therefore by comparison principle we have w ≤ u in O. Let κ(ϱ) = u(xϱ ) for − some xϱ ∈ B F (x0 , ϱ). According to [37, Corollary 4.6], we note that xϱ ∈ ∂BF− (x0 , ϱ), so that F ∗ (x0 −xϱ ) = ϱ. Therefore, we have v(ϱ) = w(xϱ ) ≤ u(xϱ ) = κ(ϱ), as claimed. Thus, we have shown that ( κ(ϱ) ≥ κ(s)

t−ϱ t−s

)

( + κ(t)

ϱ−s t−s

) ,

which is the concavity of κ. □ The following lemma can be found in [29, Proposition 2.4]. We, provide a much shorter proof, with the restrictive assumptions of [29] removed. Lemma A.2. Let Ω ⊆ Rn be an open set and suppose u is a viscosity solution of (3.6) in Ω . If O ⊆ Ω is an open subset, then u is a viscosity solution of (3.6) in O. Proof . We will show that u is a subsolution of (3.6) in O. Let (x0 , φ) ∈ O × C 2 (O) such that u ≺x0 φ. By Taylor’s Theorem we recall that φ(x) = φ(x0 ) + ⟨Dφ(x0 ), x − x0 ⟩ +

⟩ 1⟨ 2 2 D φ(x0 )(x − x0 ), x − x0 + o(|x − x0 | ) as x → x0 . 2

Given 0 < ε < 1, we consider the function ϕε (x) = φ(x0 ) + ⟨Dφ(x0 ), x − x0 ⟩ +

⟩ ε 1⟨ 2 2 D φ(x0 )(x − x0 ), x − x0 + |x − x0 | . 2 2

Obviously, ϕε ∈ C 2 (Ω ). Moreover, we see that u ≺x0 ϕε . Since u is a viscosity subsolution of (3.6) in Ω we conclude that −∆+ F,∞ ϕε (x0 ) ≤ f (x0 , u(x0 ), Dϕε (x0 )).

T. Biset, B. Mebrate and A. Mohammed / Nonlinear Analysis 190 (2020) 111588

19

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