Positive radial solutions for Dirichlet problems via a Harnack-type inequality

Received: 9 February 2022 Accepted: 18 August 2022 DOI: 10.1002/mma.8682 RESEARCH ARTICLE Positive radial solutions for Dirichlet problems via a Harnack-type inequality Radu Precup1,2 Jorge Rodríguez-López3 1Institute of Advanced Studies in Science and Technology STAR-UBB, Babe¸s-Bolyai University, Cluj-Napoca, Romania 2Tiberiu Popoviciu Institute of Numerical Analysis, Romanian Academy, Cluj-Napoca, Romania 3CITMAga and Departamento de Estatística, Análise Matemática e Optimización, Facultade de Matemáticas, Universidade de Santiago de Compostela, Santiago, Spain Correspondence Jorge Rodríguez-López, CITMAga and Departamento de Estatística, Análise Matemática e Optimización, Universidade de Santiago de Compostela, 15782, Facultade de Matemáticas, Campus Vida, Santiago, Spain. Email: [email protected] Communicated by: H. Yin Funding information AIE, Spain and FEDER, Grant/Award Number: PID2020-113275GB-I00; Institute of Advanced Studies in Science and Technology of Babes-Bolyai University of Cluj-Napoca, Grant/Award Number: CNFIS-FDI-2021-0061; Xunta de Galicia, Grant/Award Number: ED431C 2019/02 We deal with the existence and localization of positive radial solutions for Dirichlet problems involving 𝜙-Laplacian operators in a ball. In particular, p-Laplacian and Minkowski-curvature equations are considered. Our approach relies on fixed point index techniques, which work thanks to a Harnack-type inequality in terms of a seminorm. As a consequence of the localization result, it is also derived the existence of several (even infinitely many) positive solutions. KEYWORDS compression–expansion, Dirichlet problem, fixed point index, Harnack-type inequality, mean curvature operator, Positive radial solution MSC CLASSIFICATION 35J25, 35J60, 34B18, 35J92, 35J93 1INTRODUCTION In this paper, we deal with the existence, localization, and multiplicity of positive radial solutions to the Dirichlet problem involving 𝜙-Laplacian operators: −div (𝜓(|∇v|)∇v)=𝑓(|x|,v)in ,v=0on𝜕,(1.1) where is the unit open ball in Rn(n≥3)centered at the origin, the function 𝑓∶[0,1]×R+→R+is continuous, and 𝜓∶(−a,a)→Ris C1,such that 𝜙(s)∶= s𝜓(s)is an increasing homeomorphism between two intervals (−a,a)and (−b,b)(0<a,b≤+∞). This is an open access article under the terms of the Creative Commons Attribution License, which permits use, distribution and reproduction in any medium, provided the original work is properly cited. © 2022 The Authors. Mathematical Methods in the Applied Sciences published by John Wiley & Sons Ltd. Math Meth Appl Sci. 2022;1–14. wileyonlinelibrary.com/journal/mma 1 2PRECUP AND RODRÍGUEZ-LÓPEZ The following particular cases are of much interest due to their corresponding models arising from physics: (a) 𝜙∶R→R,𝜙 (s)=|s|p−2s,where p>1(herea=b=+∞), when the left side Lv in (1.1) is Lv =−div (|∇v|p−2∇v)(p−Laplace operator), involved in a nonlinear Darcy law for flows through porous media;1 (b) (singular homeomorphism) 𝜙∶(−a,a)→R,𝜙(s)=s √a2−s2(here 0 <a<+∞ and b=+∞), when Lv =−div (∇v √a2−|∇v|2)(Minkowski mean curvature operator), involved in the relativistic mechanics;2,3 (c) (bounded homeomorphism) 𝜙∶R→(−b,b),𝜙(s)=bs √1+s2(here a=+∞and 0 <b<+∞), when Lv =−bdiv (∇v √1+|∇v|2)(Euclidian mean curvature operator), associated to capillarity problems.4,5 Looking for radial solutions of (1.1), that is, functions of the form v(x)=u(r)with r=|x|, the Dirichlet problem (1.1) reduces to the mixed boundary value problem −(rn−1𝜙(u′))′=rn−1𝑓(r,u),u′(0)=u(1)=0.(1.2) Radial and nonradial solutions for the Dirichlet problem involving 𝜙-Laplace operators have been intensively investigated in the literature, both by means of topological and variational methods. We refer the interested reader to the papers6–14 and the references therein. Our approach here is based on fixed point index theory, namely, on compression–expansion-type homotopy arguments. The most known are those from Krasnosel'skiı's compression–expansion theorem in a conical annulus defined by using the max-norm of the space. Applications to one-dimensional 𝜙-Laplace equations are given in the papers.15,16 In the radial case considered in the present paper, the absence of a Harnack-type inequality in terms of the max-norm makes Krasnosel'skiı's theorem inoperative and forces us to use instead some other homotopy conditions and properties of the fixed point index. The first paper in radial solutions that uses the compression–expansion technique, but in a variational form and only for p-Laplacian equations, is Precup et al.17 As explained there, the difficulty in applying the compression–expansion method consists in the necessity that, for the considered differential operator, a Harnack-type inequality be available. In the present paper, such a key inequality is established for problem (1.2) with a general homeomorphism 𝜙satisfying some additional conditions. With its help, a precise localization of positive solutions is possible, allowing in a natural way to obtain multiple solutions. The results apply in particular for the p-Laplacian and the Minkowski mean curvature operator. Our basic assumptions are as follows: (H𝜙)𝜙∶(−a,a)→(−b,b)(0<a,b≤+∞) is an odd increasing homeomorphism such that 𝜆𝜙 (x)≥𝜙(𝜆x)for all 𝜆∈[0,1],x∈[0,a)(𝜙is convex on [0,a)).(1.3) (H𝑓)𝑓∶[0,1]×R+→[0,b)is continuous, with 𝑓(·,s)nonincreasing in [0,1]for every s∈R+and 𝑓(r,·) nondecreasing in R+for every r∈[0,1]. Note that the homeomorphisms related to the p-Laplacian for p≥2 and the Minkowski mean curvature operator both satisfy condition (H𝜙). Contrarily, the bounded homeomorphisms with a=+∞,for example, the one involved by the Euclidian mean curvature operator, are not convex on [0,+∞) and thus they do not satisfy our assumption (H𝜙). PRECUP AND RODRÍGUEZ-LÓPEZ 3 2A HARNACK-TYPE INEQUALITY In the space of functions u∈C1[0,1]satisfying u′(0)=u(1)=0,we consider the following norms: ‖u‖p=(∫1 0(rn−1||u′(r)||)pdr)1∕p for 1 ≤p<∞; ‖u‖∞=sup r∈[0,1] rn−1||u′(r)||for p=+∞. Let 𝜙satisfy (H𝜙) and denote h0(r)=−rn−1𝜙(u′(r)), J(u)=h′ 0(r)=− (rn−1𝜙(u′(r)))′. First, we prove a Harnack-type inequality for problem (1.2), given in terms of the norm ‖·‖p,with1≤p≤+∞. Theorem 2.1. Let u ∈C1[0,1]be such that u′(r)∈(−a,a)for all r ∈[0,1],h0∈C1[0,1]andJ (u)≥0on [0,1]. Then u′≤0on [0,1]. If in addition r1−nJ(u)is nonincreasing on (0,1],then u(r)≥(1−r)rn‖u‖p,r∈[0,1], for every 1≤p≤+∞. Proof. By assumption, h′ 0=J(u)≥0. Hence, h0is nondecreasing in [0,1].Since h0(0)=0,one has h0≥0 and so u′≤0on[0,1].Thus, h0(r)=rn−1𝜙(||u′(r)||).(2.1) Next, 𝜙−1(h0(r)) =𝜙−1(rn−1𝜙(||u′(r)||))≤𝜙−1(𝜙(||u′(r)||))=||u′(r)||,r∈[0,1].(2.2) Also, 𝜙−1(h0(1)) =||u′(1)||. Since both 𝜙−1and h0are nondecreasing, using (2.2), we have u(r)=∫1 r||u′(s)||ds ≥∫1 r 𝜙−1(h0(s)) ds ≥(1−r)𝜙−1(h0(r)) . Therefore, u(r)≥(1−r)𝜙−1(h0(r)) ,r∈[0,1]. Next, we prove the inequality Φ(r)∶= h0(r)−rnh0(1)≥0on [0,1].(2.3) One has Φ′(r)=J(u)( r)−nrn−1h0(1)=rn−1(r1−nJ(u)( r)−nh0(1)). By assumption, Ψ(r)∶= r1−nJ(u)( r)−nh0(1)is nonincreasing. As in the proof of Precup et al.,17, Theorem 2.1 since Φ(0)=Φ(1)=0,we deduce (2.3). Then 𝜙−1(h0(r)) ≥𝜙−1(rnh0(1))=𝜙−1(rn𝜙(||u′(1)||))≥rn||u′(1)||, 4PRECUP AND RODRÍGUEZ-LÓPEZ where the last inequality is based on (1.3). Hence, u(r)≥(1−r)rn||u′(1)||,r∈[0,1]. Finally, by (2.1) and using again (1.3), one has 𝜙−1(h0(r)) ≥rn−1|u′(r)|. Hence, we have ‖u‖p p∶= ∫1 0(rn−1||u′(r)||)pdr ≤∫1 0 𝜙−1(h0(r))pdr ≤𝜙−1(h0(1))p=||u′(1)||p,(1≤p<∞). Therefore, u(r)≥(1−r)rn‖u‖p(1≤p≤∞),(2.4) for all r∈[0,1]. A similar result has been established in Precup et al17 for the particular case of the p-Laplacian with p>n,forwhich 𝜙(s)=|s|p−2sand a=b=+∞. More exactly, it has been proved that u(r)≥(p−n p−1)1 p (1−r)rn p−1‖u‖1,p(r∈[0,1]),(2.5) where ‖u‖1,p=(∫1 0rn−1|u′(r)|pdr)1 p.In this case, since by using Hölder's inequality one has |u(r)|≤∫1 r||u′(s)||ds =∫1 r sn−1 p||u′(s)||s−n−1 pds ≤‖u‖1,p(∫1 0 sn−1 p−1ds)p−1 p =(p−1 p−n)p−1 p ‖u‖1,p, a Harnack-type inequality in terms of the max-norm |u|∞=maxr∈[0,1]|u(r)|can be immediately derived from (2.5), namely, u(r)≥p−n p−1(1−r)rn p−1|u|∞for r∈[0,1]. It is an open problem to obtain an analog result for more general homeomorphisms 𝜙satisfying (H𝜙). At this moment we are only able to establish such an inequality in terms of a max-seminorm on C[0,1].For example, taking p=+∞ in (2.4), we have the following Harnack-type inequality related to a seminorm on C[0,1]. Corollary 2.2. Under the assumptions of Theorem 2.1, if for a fixed subinterval [𝜂,𝜈]with 0<𝜂<𝜈<1,one can define in C [0,1]the seminorm [u]∞=maxr∈[𝜂,𝜈]|u(r)|,then u(r)≥(1−𝜈)𝜂2n−2(n−2)[u]∞for r ∈[𝜂,𝜈].(2.6) Proof. Clearly, |u(r)|≤∫1 r||u′(s)||ds =∫1 r sn−1||u′(s)||s−(n−1)ds ≤‖u‖∞∫1 r s−(n−1)ds ≤r2−n n−2‖u‖∞, which implies [u]∞≤𝜂2−n n−2‖u‖∞. This combined with (2.4) yields (2.6). PRECUP AND RODRÍGUEZ-LÓPEZ 5 In the sequel, inequality (2.6) is a key ingredient for the localization and multiplicity of positive radial solutions. We will use the main ideas in Precup17 in order to localize the solutions in terms of a norm and a seminorm. 3POSITIVE RADIAL SOLUTIONS Recall that by a (nonnegative) solution of (1.2), we mean a function u∈C1([0,1],R+)with u′(0)=u(1)=0,|u′(r)|<a for all r∈[0,1],suchthatrn−1𝜙(u′)∈C1[0,1]and (1.2) is satisfied. We will say that a nonnegative solution is positive if it is distinct from the identically zero function. Let Xbe the Banach space of continuous functions X=C[0,1]and K0its positive cone K0={u∈X∶u≥0on [0,1]}.It is not difficult to see that a nonnegative function uis a solution of problem (1.2) if and only if uis a fixed point of the operator T∶K0→K0given by T(u)( r)=∫1 r 𝜙−1(𝜏1−n∫𝜏 0 sn−1𝑓(s,u(s)) ds)d𝜏. (3.1) As proved in earlier studies,6,8 the operator Tis completely continuous. Let us now consider a subcone of K0related to the Harnack inequality (2.6), namely, K={u∈K0∶uis nonincreasing on [0,1]and min r∈[𝜂,𝜈]u(r)≥c[u]∞},(3.2) where c∶= (1−𝜈)𝜂2n−2(n−2). Lemma 3.1. The operator T maps the cone K into itself. Proof. Indeed, take u∈Kand let us show that v∶= Tu belongs to K.Since𝑓is nonnegative, v≥0, and moreover, J(v)≥0 and so v′≤0 (see Theorem 2.1), that is, vis nonincreasing on [0,1].Furthermore, by the monotonicity properties of 𝑓imposed in (H𝑓)and the fact that uis nonincreasing in [0,1], the composed function r→ 𝑓(r,u(r)) is nonincreasing in [0,1]. Hence, r1−nJ(v)=𝑓(r,u) is nonincreasing in [0,1]. Then Corollary 2.2 ensures that v(r)≥(1−𝜈)𝜂2n−2(n−2)[v]∞forr∈[𝜂,𝜈]. Therefore, v∈K,as claimed. Now, for any number 𝛼>0, consider the set U𝛼∶= {u∈K∶|u|∞<𝛼 }. The operator Tbeing completely continuous, the set T(U𝛼)is bounded, so there is a number 𝛼 ≥𝛼such that T(U𝛼)⊂ U𝛼.Define the operator  T∶U𝛼 →U𝛼 by  T(u)=T(min {𝛼 |u|∞ ,1}u). Lemma 3.2. If T(u)≠𝜆uforu∈Kwith|u|∞=𝛼and 𝜆≥1,(3.3) then the fixed point index i ( T,U𝛼,U𝛼)=1. Proof. Clearly, U𝛼 is a convex closed set and  Tis a compact map. Consider the homotopy H∶[0,1]×U𝛼 →U𝛼 given by H(𝜏,u)=𝜏 T(u). 6PRECUP AND RODRÍGUEZ-LÓPEZ By (3.3), this homotopy is admissible and so i( T,U𝛼,U𝛼)=i(H(1,·),U𝛼,U𝛼)=i(H(0,·),U𝛼,U𝛼)=1, where the last equality is due to the normalization property of the fixed point index, since 0 ∈U𝛼. Next, for a number 𝛽>0, consider the set V𝛽∶= {u∈U𝛼 ∶[u]∞<𝛽 }. It is clear that V𝛽is open in U𝛼. Lemma 3.3. Assume that there exists a function h ∈K such that |h|∞=𝛼,[h]∞>𝛽and (1−𝜆) T(u)+𝜆h≠uforu∈Kwith|u|∞≤𝛼, [u]∞=𝛽and 𝜆∈[0,1].(3.4) Then i ( T,V𝛽,U𝛼)=0. Proof. Observe that 𝜕V𝛽={u∈K∶[u]∞=𝛽, |u|∞≤𝛼}. Thus, (3.4) implies that (1−𝜆) T(u)+𝜆h≠ufor u∈𝜕V𝛽. By the homotopy property of the fixed point index, one has i( T,V𝛽,U𝛼)=i(h,V𝛽,U𝛼). Finally, i(h,V𝛽,U𝛼)=0, since h∈U𝛼 ⧵V𝛽. Remark 3.1. If the operator Tmaps U𝛼into itself, then 𝛼 =𝛼and condition (3.4) reduces to (1−𝜆)T(u)+𝜆h≠ufor u∈Kwith |u|∞≤𝛼, [u]∞=𝛽and 𝜆∈[0,1]. By using the previous fixed point index computations, we deduce the following existence result. Lemma 3.4. Under the assumptions of Lemmas 3.2 and 3.3, the operator T has a fixed point u in U𝛼⧵V𝛽,that is, problem (1.2) has a solution such that 𝛽<[u]∞and |u|∞<𝛼. Proof. One has 1=i( T,U𝛼,U𝛼)=i( T,U𝛼⧵V𝛽,U𝛼)+i( T,U𝛼∩V𝛽,U𝛼), 0=i( T,V𝛽,U𝛼)=i( T,V𝛽⧵U𝛼,U𝛼)+i( T,U𝛼∩V𝛽,U𝛼). As a result, i( T,U𝛼⧵V𝛽,U𝛼)−i( T,V𝛽⧵U𝛼,U𝛼)=1. In addition i( T,V𝛽⧵U𝛼,U𝛼)=0 since otherwise there would exist v∈V𝛽⧵U𝛼with  T(v)=v,that is, T(𝛼 |v|∞ v)=v, or equivalently, T(w)=𝜆w,where w=𝛼 |v|∞ vand 𝜆=|v|∞ 𝛼.Since |w|∞=𝛼and 𝜆>1, we arrived to a contradiction with (3.3). Therefore i( T,U𝛼⧵V𝛽,U𝛼)=1,which implies our conclusion. PRECUP AND RODRÍGUEZ-LÓPEZ 7 Now we give sufficient conditions in order to guarantee the assumptions of the previous lemmas hold. We will use the following notation. If b<+∞,denote A∶= 1 ∫1 0𝜙−1(b𝜏)d𝜏and B∶= ∫1 𝜂 𝜙−1(b𝜏)d𝜏. If b=+∞,denote A∶= 1 ∫1 0𝜙−1(𝜏)d𝜏and B∶= ∫1 𝜂 𝜙−1(𝜏)d𝜏. Theorem 3.5. Assume that n ≥3and conditions (H𝜙)and (H𝑓)are fulfilled. If there exist 𝛼,𝛽 > 0with 𝛽<AB𝛼, such that 𝜙−1(𝑓(0,𝛼)) <𝛼, (3.5) (1−𝜈)𝜙−1((𝜈−𝜂)𝜂n−1𝑓(𝜈,c𝛽))>𝛽, (3.6) then problem (1.2) has at least one solution u ∈K such that 𝛽<[u]∞and |u|∞<𝛼. Proof. We shall apply Lemma 3.4. First, we show that (3.3) holds. Indeed, for u∈Kwith |u|∞≤𝛼,bythe monotonicity assumptions on 𝑓,wehavethat 𝑓(s,u(s)) ≤𝑓(0,𝛼), and thus, from (3.5), |T(u)(r)|≤∫1 0 𝜙−1(𝜏1−n∫𝜏 0 sn−1𝑓(s,u(s)) ds)d𝜏 ≤∫1 0 𝜙−1(∫𝜏 0 𝑓(s,u(s)) ds)d𝜏≤𝜙−1(𝑓(0,𝛼)) <𝛼. Hence, |T(u)|∞<𝛼for all u∈Kwith |u|∞≤𝛼, which implies (3.3). In addition, on the basis of Remark 3.1, we can take 𝛼 =𝛼. Next, we prove that (3.4) holds for the following choice of h:ifb<+∞, h(r)=A𝛼∫1 r 𝜙−1(𝜏1−n∫𝜏 0 bnsn−1ds)d𝜏=A𝛼∫1 r 𝜙−1(b𝜏)d𝜏, and, otherwise, for b=+∞, h(r)=A𝛼∫1 r 𝜙−1(𝜏1−n∫𝜏 0 nsn−1ds)d𝜏=A𝛼∫1 r 𝜙−1(𝜏)d𝜏. Note that |h|∞=h(0)=𝛼and [h]∞=h(𝜂)=AB𝛼>𝛽. Assume that (3.4) does not hold. Then there exist u∈Kwith |u|∞≤𝛼,[u]∞=𝛽and 𝜆∈[0,1]such that (1−𝜆)T(u)+𝜆h=u. 8PRECUP AND RODRÍGUEZ-LÓPEZ In particular, since [u]∞=maxr∈[𝜂,𝜈]u(r)=𝛽,onehas 𝛽≥u(𝜂)=(1−𝜆)T(u)(𝜂)+𝜆h(𝜂) =(1−𝜆)∫1 𝜂 𝜙−1(𝜏1−n∫𝜏 0 sn−1𝑓(s,u(s)) ds)d𝜏+𝜆[h]∞ ≥(1−𝜆)∫1 𝜈 𝜙−1(𝜏1−n∫𝜈 𝜂 sn−1𝑓(s,u(s)) ds)d𝜏+𝜆𝛽. Since u∈Kwith [u]∞=𝛽,wehaveu(r)≥c𝛽for all r∈[𝜂,𝜈]. Thus, by (H𝑓), 𝛽≥(1−𝜆)∫1 𝜈 𝜙−1(𝜏1−n∫𝜈 𝜂 sn−1𝑓(𝜈,c𝛽)ds)d𝜏+𝜆𝛽 ≥(1−𝜆)(1−𝜈)𝜙−1((𝜈−𝜂)𝜂n−1𝑓(𝜈,c𝛽))+𝜆𝛽, that is, (1−𝜆)𝛽≥(1−𝜆)(1−𝜈)𝜙−1((𝜈−𝜂)𝜂n−1𝑓(𝜈,c𝛽)), which contradicts (3.6) for any 𝜆∈[0,1). Note that, in case 𝜆=1, one has the contradiction 𝛽≥u(𝜂)=h(𝜂)=[h]∞>𝛽. Finally, the conclusion follows from Lemma 3.4. Remark 3.2 (Asymptotic conditions).Existence of both positive numbers 𝛼and 𝛽satisfying inequalities (3.5) and (3.6) is guaranteed if the following asymptotic conditions at zero and infinity hold: lim sup x→0+ 𝜙−1((𝜈−𝜂)𝜂n−1𝑓(𝜈,x)) x>1 c(1−𝜈),lim inf x→+∞ 𝜙−1(𝑓(0,x)) x<1. Obviously, if 𝜙is a classical homeomorphism (a=b=+∞), conditions (3.5) and (3.6) can be rewritten as 𝑓(0,𝛼)<𝜙(𝛼),𝑓 (𝜈,c𝛽)>1 (𝜈−𝜂)𝜂n−1𝜙(𝛽 1−𝜈). Hence, if we assume in addition that 𝜙satisfies: lim sup x→0+ 𝜙(𝜏x) 𝜙(x)<+∞ for all 𝜏>0,(3.7) then the existence of both positive numbers 𝛼and 𝛽is guaranteed under suitable asymptotic conditions about 𝑓at zero and at infinity. Notethat assumption(3.7)holds in the case ofthep-Laplacian operatorandso it is commonly employedintheliterature, see for instance.8,12 Theorem 3.6. Assume that n ≥3, conditions (H𝜙)and (H𝑓)are fulfilled, and 𝜙is a classical homeomorphim. If (3.7) and lim sup x→0+ 𝑓(𝜈,x) 𝜙(x)=+∞,lim inf x→+∞ 𝑓(0,x) 𝜙(x)<1 (3.8) hold, then problem (1.2) has at least one positive solution. Proof. By (3.7), with 𝜏=1∕(c(1−𝜈)),thereexistsL>0sothat L>lim sup x→0+ 𝜙(x∕(c(1−𝜈))) 𝜙(x), PRECUP AND RODRÍGUEZ-LÓPEZ 9 and thus, there exists 𝜌>0suchthat L𝜙(x)≥𝜙(x c(1−𝜈))for all x∈(0,𝜌). Now, by (3.8), there exists r>0 (we may suppose r<𝜌)suchthat 𝑓(𝜈,r)>1 (𝜈−𝜂)𝜂n−1L𝜙(r), which implies that 𝑓(𝜈,r)>1 (𝜈−𝜂)𝜂n−1𝜙(r c(1−𝜈)). Finally, taking 𝛽=r∕c, condition (3.6) is obtained. On the other hand, condition lim inf x→+∞ 𝑓(0,x) 𝜙(x)<1 clearly implies the existence of a positive number 𝛼satisfying (3.5) and such that 𝛽<AB𝛼. Therefore, Theorem 3.5 ensures the existence of at least one positive solution for problem (1.2). Corollary 3.7. Assume that n ≥3,p≥2,and(H𝑓)holds. If lim sup x→0+ 𝑓(𝜈,x) xp−1=+∞and lim inf x→+∞ 𝑓(0,x) xp−1<1,(3.9) then problem −div (|∇v|p−2∇v)=𝑓(|x|,v)in ,v=0on 𝜕,(3.10) has at least one positive radial solution. Proof. It suffices to show that problem (1.2) has at least one positive solution with 𝜙(x)=|x|p−2x,p≥2. Since 𝜙is a classical homeomorphism which satisfies (H𝜙)and (3.7), the conclusion follows from Theorem 3.6. We show the applicability of our theory with an example involving radial solutions of p-Laplacian equations. Example 3.8. Consider the function 𝑓given by 𝑓(s,x)=𝑓(x)=xq+√x, with 0 ≤q<p−1andp≥2, which clearly satisfies condition (H𝑓). It is immediate to check that lim x→0+ 𝑓(x) xp−1= lim x→0+ xq+√x xp−1=+∞and lim x→+∞ 𝑓(x) xp−1= lim x→+∞ xq+√x xp−1=0. Therefore, problem (3.10) associated to this function 𝑓has at least one positive radial solution, as a consequence of Corollary 3.7. Finally, we highlight that due to the asymptotic behavior of 𝑓at zero and at infinity, this problem falls outside the scope of the results in earlier studies.12,17 On the other hand, it is worth to mention that in the case of a singular homeomorphism 𝜙(i.e., with a<+∞,b=+∞), condition (3.5) is trivially satisfied for 𝛼large enough. Hence, in that case, we only need to ensure the existence of the number 𝛽in order to obtain positive solutions for problem (1.2). Let us assume in the rest of this section that 𝜙is singular. We present an existence result inspired by those in Bereanu et al.8