Electronic Journal of Differential Equations, Vol. 2021 (2021), No. 75, pp. 1–26. ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu NONEXISTENCE RESULTS FOR HYPERBOLIC TYPE INEQUALITIES INVOLVING THE GRUSHIN OPERATOR IN EXTERIOR DOMAINS MOHAMED JLELI, BESSEM SAMET Abstract. We study the hyperbolic type differential inequality utt(t, x, y)− L`u(t, x, y) ≥ |u(t, x, y)|p, (t, x, y) ∈ (0,∞)×D1 ×D2 under the boundary conditions u(t, x, y) ≥ f(x), (t, x, y) ∈ (0,∞)× ∂D1 ×D2, u(t, x, y) ≥ g(y), (t, x, y) ∈ (0,∞)×D1 × ∂D2, where p > 1, Dk = {z ∈ RNk : |z| ≥ 1}, k = 1, 2, Nk ≥ 2, f ∈ L1(∂D1), g ∈ L1(∂D2), and L`, ` ∈ R, is the Grushin operator L`u = ∆xu + |x|2`∆yu. We obtain sufficient conditions depending on p, `, N1, N2, f , and g, for which the considered problem admits no global weak solution. We discuss separately the four cases: N1 = N2 = 2; N1 = 2, N2 ≥ 3; N1 ≥ 3, N2 = 2; N1, N2 ≥ 3. 1. Introduction This article concerns the hyperbolic type differential inequality utt(t, x, y)− L`u(t, x, y) ≥ |u(t, x, y)|p, (t, x, y) ∈ (0,∞)×D1 ×D2, u(t, x, y) ≥ f(x), (t, x, y) ∈ (0,∞)× ∂D1 ×D2, u(t, x, y) ≥ g(y), (t, x, y) ∈ (0,∞)×D1 × ∂D2, (1.1) where p > 1, D1 = {x ∈ RN1 : |x| ≥ 1}, D2 = {y ∈ RN2 : |y| ≥ 1}, N1, N2 ≥ 2, f ∈ L1(∂D1), g ∈ L1(∂D2), and L`, ` ∈ R, is the Grushin operator of the form L`u = ∆xu+ |x|2`∆yu = N1∑ i=1 ∂2u ∂x2 i + |x|2` N2∑ j=1 ∂2u ∂y2 j . (1.2) Namely, our aim is to derive sufficient conditions for which problem (1.1) admits no global weak solution. Several works have been made to investigate the nonexistence of solutions for hy- perbolic type differential inequalities. In [13], among other problems, Kato studied 2010 Mathematics Subject Classification. 35B44, 35B33, 35L10. Key words and phrases. Global weak solutions; hyperbolic type inequalities; exterior domain; Grushin operator. c©2021. This work is licensed under a CC BY 4.0 license. Submitted June 25, 2021. Published September 14, 2021. 1 2 M. JLELI, B. SAMET EJDE-2021/75 the hyperbolic inequality utt −∆u ≥ |u|p, (t, x) ∈ (0,∞)× RN . (1.3) He proved that if the initial data satisfy some suitable positivity conditions, are compactly supported, and 1 < p ≤ 1 + 2 N − 1 (N ≥ 2), then no weak solution to (1.3) can exist in (0,∞)× RN . Véron and Pohozaev [23] studied the nonexistence of nontrivial global solutions to a wide class of nonlinear hyperbolic type inequalities of the form utt ≥ Lm(ϕp(u)) + |u|q, (t, x) ∈ (0,∞)× RN , (1.4) where p > 0, ϕp is a locally bounded real valued function satisfying |ϕp(r)| ≤ c|r|p for certain c > 0, and Lm(ζ) = ∑ |α|=mD α(aα(t, x)ζ) is a homogeneous differential operator of order m in which the coefficients aα are bounded measurable functions. By an appropriate choice of test functions and the dimensional analysis, it was shown that problem (1.4) admits no weak solution such that ∫ RN ut(0, x) dx ≥ 0, provided that q > max{1, p} and either 2N−m ≤ 0 or 2N−m > 0 and N(q−p) q+1 ≤ m 2 . In [10], the authors investigated the hyperbolic inequality utt −∆u ≥ |u|p + |∇u|q + f(t, x), (t, x) ∈ (0,∞)× RN , (1.5) where p, q > 1 and f ≥ 0, f 6≡ 0. Namely, they derived general criteria for the nonexistence of global solutions to (1.5). In particular, when N ≥ 3 and f depends only on the variable space, it was shown that (1.5) admits as Fujita critical exponent the real number p∗(N, q) = { 1 + 2 N−2 if q > 1 + 1 N−1 , ∞ if q < 1 + 1 N−1 . In all the above mentioned references the considered problems are posed in the whole space RN . The study of hyperbolic type differential inequalities in other infinite domains was considered by some authors. In [16], among other problems, Laptev considered the hyperbolic inequality utt −∆u ≥ |u|p, (t, x) ∈ (0,∞)×K (1.6) under the Dirichlet type boundary condition u(t, x) ≥ 0, (t, x) ∈ (0,∞)× ∂K, (1.7) where K is the cone defined by K = {(r, ω) : r > 0, ω ∈ Ω} , and Ω is a domain of SN−1, N ≥ 3. It was shown that, if 1 < p ≤ 1 + 2 s∗ + 1 , where s∗ = N − 2 2 + √(N − 2 2 )2 + λ1 EJDE-2021/75 HYPERBOLIC TYPE INEQUALITIES IN EXTERIOR DOMAINS 3 and λ1 is the first eigenvalue of the Laplace Beltrami operator ∆θ on Ω, then prob- lem (1.6) under the boundary condition (1.7) has no nontrivial global weak solution. In [12] (see also [9]), motivated by Zhang [26], the authors investigated the nonex- istence of global weak solutions for a system of inhomogeneous wave inequalities in exterior domains under three type boundary conditions: Dirichlet type, Neumann type and mixed boundary conditions. In particular, for the hyperbolic inequality utt −∆u ≥ |x|a|u|p, (t, x) ∈ (0,∞)× Ωc, u(t, x) ≥ f(x), (t, x) ∈ (0,∞)× ∂Ω, (1.8) where a > −2, Ωc denotes the complement of Ω, Ω is a bounded smooth open set in RN containing the origin, and N ≥ 3, it was shown that, if f ∈ L1(∂Ω),∫ ∂Ω f dσ > 0, and 1 < p < N + a N − 2 , then problem (1.8) admits no global weak solution. Moreover, for p > N+a N−2 , problem (1.8) admits global solutions (namely, stationary solutions) for some f > 0. For other works related to differential inequalities in exterior domains, see e.g. [11, 20, 21] and the references therein. A large amount of works have been made to study the Grushin operator L` of the form (1.2) as well as the properties of the solutions to −L`u = f (see [1, 6, 7, 8]). Capuzzo Dolcetta and Cutri [2] studied the differential inequality − L`u ≥ up, u ≥ 0, x ∈ RN1 , y ∈ RN2 . (1.9) It was shown that, if ` > 1 and 1 < p ≤ Q Q−2 , where Q = N1 + (` + 1)N2, then (1.9) admits no nontrivial solution. D’Ambrosio and Lucente [4] investigated the differential inequality L(x, y,Dx, Dy) ≥ |x|θ1 |y|θ2 |u|q, (x, y) ∈ Rd × Rk, where L is a quasi-homogeneous differential operator including as special cases Tricomi or Grushin-type operators, q > 1, θ1, θ2 ∈ R, and k, d ≥ 1. Namely, they provided necessary conditions for existence of weak solutions to the considered inequality. For other nonexistence results for differential inequalities (stationary inequalities) involving Grushin type operators, see [3, 5, 14, 15, 17, 18, 19, 22, 24, 25, 27] and the references therein. Motivated by the above mentioned contributions, our aim in this paper is to obtain sufficient conditions depending on p, `, N1, N2, f and g, for which problem (1.1) not to admits global weak solutions. The rest of the paper is organized as follows. In Section 2, we define global weak solutions to problem (1.1) and provide the main results of this paper. In Section 3, we establish some preliminary estimates that will be used in the proofs of our main results. In Section 4, we prove the main results of this paper. We discuss separately the cases: N1 = N2 = 2; N1 = 2, N2 ≥ 3; N1 ≥ 3, N2 = 2; N1, N2 ≥ 3. The symbols C or Ci denote always generic positive constants, which are in- dependent of the scaling parameter R and the solution u. Their values could be changed from one line to another. We will use the notation µ ∼ ν for two positive functions or quantities, which satisfy C1µ ≤ ν ≤ C2µ. 4 M. JLELI, B. SAMET EJDE-2021/75 2. Main results We first fix some notation that will be used throughout this paper. Let D = D1 ×D2, Ω = (0,∞)×D, Γ1 = (0,∞)× ∂D1 ×D2, Γ2 = (0,∞)×D1 × ∂D2. We denote by n1 = n1(x) the outward unit normal vector on ∂D1 relative to D1. Similarly, we denote by n2 = n2(y) the outward unit normal vector on ∂D2 relative to D2. We introduce the test function space Φ = { ϕ ∈ C2 c (Ω) : ϕ ≥ 0, ϕ|∂D1∪∂D2 = 0, ∂xϕ ∂n1 ≤ 0, ∂yϕ ∂n2 ≤ 0 } , (2.1) where C2 c (Ω) denotes the space of C2 functions compactly supported in Ω. Here, ∂xϕ ∂n1 = ∇xϕ · n1 and ∂yϕ ∂n2 = ∇yϕ · n2. Let us mention in which sense the solutions are considered. Definition 2.1. Let f ∈ L1(∂D1) and g ∈ L1(∂D2). We say that u ∈ Lploc([0,∞)×D) is a global weak solution to (1.1), if∫ Ω |u|pϕdx dy dt− ∫ Γ1 ∂xϕ ∂n1 f(x) dσx dy dt− ∫ Γ2 |x|2` ∂yϕ ∂n2 g(y) dx dσy dt ≤ ∫ Ω u ( ϕtt −∆xϕ− |x|2`∆yϕ ) dx dy dt (2.2) for every ϕ ∈ Φ. Here, dσx denotes the surface measure on ∂D1, and dσy denotes the surface measure on ∂D2. Our first main result is the following. Theorem 2.2. Let N1 = N2 = 2, f ∈ L1(∂D1), and g ∈ L1(∂D2). (I) Let ` ≤ −1. If∫ ∂D1 f(x)dσx > 0 or ∫ ∂D1 f(x)dσx = 0, ∫ ∂D2 g(y)dσy > 0, then for all p > 1, (1.1) admits no global weak solution. (II) Let ` > −1. If∫ ∂D1 f(x)dσx > 0 or ∫ ∂D2 g(y)dσy > 0, then for all p > 1, (1.1) admits no global weak solution. Remark 2.3. Let N1 = N2 = 2. From Theorem 2.2 we deduce that, if∫ ∂D1 f(x)dσx > 0, then for all ` ∈ R and p > 1, (1.1) admits no global weak solution. EJDE-2021/75 HYPERBOLIC TYPE INEQUALITIES IN EXTERIOR DOMAINS 5 Clearly, Theorem 2.2 yields nonexistence results for the corresponding stationary problem −L`u(x, y) ≥ |u(x, y)|p, (x, y) ∈ D1 ×D2, u(x, y) ≥ f(x), (x, y) ∈ ∂D1 ×D2, u(x, y) ≥ g(y), (x, y) ∈ D1 × ∂D2. (2.3) Namely, we deduce the following result. Corollary 2.4. Let N1 = N2 = 2, f ∈ L1(∂D1), and g ∈ L1(∂D2). (I) Let ` ≤ −1. If∫ ∂D1 f(x)dσx > 0 or ∫ ∂D1 f(x)dσx = 0, ∫ ∂D2 g(y)dσy > 0, then for all p > 1, (2.3) admits no weak solution. (II) Let ` > −1. If∫ ∂D1 f(x)dσx > 0 or ∫ ∂D2 g(y)dσy > 0, then for all p > 1, (2.3) admits no weak solution. Remark 2.5. Consider the differential inequality vtt −∆v ≥ vp(v ≥ 0), (t, x) ∈ (0,∞)×D1, v(t, x) ≥ f(x), (t, x) ∈ (0,∞)× ∂D1, (2.4) where N1 = 2 and p > 1. Let v be a possible solution to (2.4) and u(t, x, y) = v(t, x), (t, x, y) ∈ (0,∞)×D1 ×D2, where N2 = 2. Then for all ` ∈ R, u is a solution to (1.1) with g ≡ 0. Taking in consideration Remark 2.3, we deduce that, if ∫ ∂D1 f(x)dσx > 0, then for all p > 1, (2.4) admits no solution. Theorem 2.6. Let N1 = 2, N2 ≥ 3, f ∈ L1(∂D1), and g ∈ L1(∂D2). (I) Let ` < −1. (i) If ∫ ∂D1 f(x)dσx > 0, then for all p > 1, (1.1) admits no global weak solution. (ii) If ∫ ∂D1 f(x)dσx = 0 and ∫ ∂D2 g(y)dσy > 0, then for all 1 < p < N2 N2−2 , (1.1) admits no global weak solution. (II) Let ` = −1. (i) If ∫ ∂D1 f(x)dσx > 0, then for all p > 1, (1.1) admits no global weak solution. (ii) If ∫ ∂D1 f(x)dσx = 0 and ∫ ∂D2 g(y)dσy > 0, then for all 1 < p ≤ N2 N2−2 , (1.1) admits no global weak solution. (III) Let −1 < ` < 0. If∫ ∂D1 f(x)dσx > 0 or ∫ ∂D2 g(y)dσy > 0, then for all p > 1, (1.1) admits no global weak solution. 6 M. JLELI, B. SAMET EJDE-2021/75 (IV) Let ` ≥ 0. (i) If ∫ ∂D1 f(x)dσx > 0, then for all p > 1, (1.1) admits no global weak solution. (ii) If ∫ ∂D2 g(y)dσy > 0, then for all 1 < p < N2 N2−2 , (1.1) admits no global weak solution. Remark 2.7. Let N1 = 2 and N2 ≥ 3. By Theorem 2.6 we deduce that, if∫ ∂D1 f(x)dσx > 0, then for all ` ∈ R and p > 1, (1.1) admits no global weak solution. Remark 2.8. Let N1 = 2, N2 ≥ 3, ` ≥ 0, g ∈ L1(∂D2), and ∫ ∂D2 g(y)dσy > 0. Then by Theorem 2.6 (IV)-(ii), if 1 < p < N2 N2 − 2 , (2.5) then (1.1) admits no global weak solution for all f ∈ L1(∂D1). Moreover, for p > N2 N2−2 , we can can check easily that u(t, x, y) = A|y|−σ, (t, x, y) ∈ (0,∞)×D1 ×D2, where A > 0 is sufficiently small and 2 p−1 < σ < N2 − 2, is a (stationary) solution to (1.1) with f ≡ 0 and g ≡ A. This shows that (2.5) is sharp. Remark 2.9. As in the previous case (see Corollary 2.4), the nonexistence results given by Theorem 2.6 hold true for the stationary problem (2.3) in the case N1 = 2 and N2 ≥ 3. Remark 2.10. Consider the differential inequality vtt −∆v ≥ vp(v ≥ 0), (t, y) ∈ (0,∞)×D2, v(t, y) ≥ g(y), (t, y) ∈ (0,∞)× ∂D2, (2.6) where N2 ≥ 3. Let v be a possible solution to (2.6) and u(t, x, y) = v(t, y), (t, x, y) ∈ (0,∞)×D1 ×D2, where N1 = 2. Then u is a solution to (1.1) with f ≡ 0 and ` = 0. Taking in consideration Remark 2.8, we deduce that, if ∫ ∂D2 g(y)dσy > 0, then for all 1 < p < N2 N2−2 , (2.6) admits no solution. We find [12, Corollary 1.9] for the case of positive solutions. Theorem 2.11. Let N1 ≥ 3, N2 = 2, f ∈ L1(∂D1), and g ∈ L1(∂D2). (I) Let ` ≤ −N1 2 . If∫ ∂D1 f(x)dσx > 0 or ∫ ∂D1 f(x)dσx = 0, ∫ ∂D2 g(y)dσy > 0, then for all 1 < p < N1 N1−2 , (1.1) admits no global weak solution. (II) Let −N1 2 < ` < −1. (i) If ∫ ∂D1 f(x)dσx > 0, then for all 1 < p < N1 N1−2 , (1.1) admits no global weak solution. (ii) If ∫ ∂D2 g(y)dσy > 0, then for all 1 < p < ` `+1 , (1.1) admits no global weak solution. (III) Let ` ≥ −1. EJDE-2021/75 HYPERBOLIC TYPE INEQUALITIES IN EXTERIOR DOMAINS 7 (i) If ∫ ∂D1 f(x)dσx > 0, then for all 1 < p < N1 N1−2 , (1.1) admits no global weak solution. (ii) If ∫ ∂D2 g(y)dσy > 0, then for all p > 1, (1.1) admits no global weak solution. Remark 2.12. Let N1 ≥ 3, N2 = 2, f ∈ L1(∂D1), and ∫ ∂D1 f(x)dσx > 0. By Theorem 2.11 we deduce that for all ` ∈ R, g ∈ L1(∂D2), and 1 < p < N1 N1 − 2 , (2.7) problem (1.1) admits no global weak solution. On the other hand, for p > N1 N1−2 , we can check easily that u(t, x, y) = A|x|−σ, (t, x, y) ∈ (0,∞)×D1 ×D2, where A > 0 is sufficiently small and 2 p−1 < σ < N1 − 2, is a (stationary) solution to (1.1) with f ≡ A and g ≡ 0. This shows that (2.7) is sharp. In the special case when ∫ ∂D1 f(x)dσx > 0 and ∫ ∂D2 g(y)dσy > 0, we deduce from Theorem 2.11 the following results. Corollary 2.13. Let N1 ≥ 3, N2 = 2, f ∈ L1(∂D1), and g ∈ L1(∂D2). Suppose that ∫ ∂D1 f(x)dσx > 0 and ∫ ∂D2 g(y)dσy > 0. (I) Let ` ≤ −N1 2 . Then for all 1 < p < N1 N1−2 , (1.1) admits no global weak solution. (II) Let −N1 2 < ` < −1. Then for all 1 < p < ` `+1 , (1.1) admits no global weak solution. (III) Let ` ≥ −1. Then for all p > 1, (1.1) admits no global weak solution. Remark 2.14. The nonexistence results given by Theorem 2.11 and Corollary 2.13 hold for the stationary problem (2.3) in the case N1 ≥ 3 and N2 = 2. Theorem 2.15. Let N1, N2 ≥ 3, f ∈ L1(∂D1), and g ∈ L1(∂D2). (I) Let ` ≤ −N1 2 . (i) If ∫ ∂D1 f(x)dσx > 0, then for all 1 < p < N1 N1−2 , (1.1) admits no global weak solution. (ii) If ` < −N1 2 , ∫ ∂D1 f(x)dσx = 0, and ∫ ∂D2 g(y)dσy > 0, then for all 1 < p < min{ N1 N1−2 , N2 N2−2}, (1.1) admits no global weak solution. (iii) If ` = −N1 2 , ∫ ∂D1 f(x)dσx = 0, and ∫ ∂D2 g(y)dσy > 0, then for all 1 < p < min{ N1 N1−2 , N2 N2−2} or p = N2 N2−2 < N1 N1−2 , (1.1) admits no global weak solution. (II) Let −N1 2 < ` < −1. 8 M. JLELI, B. SAMET EJDE-2021/75 (i) If ∫ ∂D1 f(x)dσx > 0, then for all 1 < p < N1 N1−2 , (1.1) admits no global weak solution. (ii) If ∫ ∂D2 g(y)dσy > 0, then for all 1 < p < ` `+1 , (1.1) admits no global weak solution. (III) Let −1 ≤ ` < 0. (i) If ∫ ∂D1 f(x)dσx > 0, then for all 1 < p < N1 N1−2 , (1.1) admits no global weak solution. (ii) If ∫ ∂D2 g(y)dσy > 0, then for all p > 1, (1.1) admits no global weak solution. (IV) Let ` ≥ 0. (i) If ∫ ∂D1 f(x)dσx > 0, then for all 1 < p < N1 N1−2 , (1.1) admits no global weak solution. (ii) If ∫ ∂D2 g(y)dσy > 0, then for all 1 < p < N2 N2−2 , (1.1) admits no global weak solution. Remark 2.16. From Theorem 2.15, if f ∈ L1(∂D1) and ∫ ∂D1 f(x)dσx > 0, then for all ` ∈ R, g ∈ L1(∂D2), and 1 < p < N1 N1−2 , (1.1) admits no global weak solution. We can check that the above condition is sharp (see Remark 2.12). Similarly, condition (IV)-(ii) is sharp (see Remark 2.8). In the special case when ∫ ∂D1 f(x)dσx > 0 and ∫ ∂D2 g(y)dσy > 0, we deduce from Theorem 2.15 the following results. Corollary 2.17. Let N1, N2 ≥ 3, f ∈ L1(∂D1), and g ∈ L1(∂D2). Suppose that∫ ∂D1 f(x)dσx > 0 and ∫ ∂D2 g(y)dσy > 0. (I) If ` ≤ −N1 2 , then for all 1 < p < N1 N1−2 , (1.1) admits no global weak solution. (II) If −N1 2 < ` < −1, then for all 1 < p < ` `+1 , (1.1) admits no global weak solution. (III) If −1 ≤ ` < 0, then for all p > 1, (1.1) admits no global weak solution. (IV) If ` ≥ 0, then for all 1 < p < max { N1 N1−2 , N2 N2−2 } , (1.1) admits no global weak solution. Remark 2.18. The nonexistence results given by Theorem 2.15 and Corollary 2.17 hold for the stationary problem (2.3) in the case N1, N2 ≥ 3. 3. Preliminaries Let Nk ≥ 2, k = 1, 2. We introduce the following harmonic function defined in Dk = {z ∈ RNk : |z| ≥ 1}: Hk(z) = { ln |z| if Nk = 2, 1− |z|2−Nk if Nk ≥ 3. We introduce two cut-off functions η, ξ ∈ C∞([0,∞)) satisfying respectively η ≥ 0, η 6≡ 0, supp(η) ⊂ (0, 1) and 0 ≤ ξ ≤ 1, ξ|[0,1] ≡ 1, ξ|[2,∞) ≡ 0. EJDE-2021/75 HYPERBOLIC TYPE INEQUALITIES IN EXTERIOR DOMAINS 9 For sufficiently large R and λ, let a(t) = ηλ ( t R ) , t > 0, b(x) = H1(x)ξλ ( |x| Rθ ) , x ∈ D1, c(y) = H2(y)ξλ ( |y| Rσ ) , y ∈ D2, where θ, σ > 0 are constants to be chosen later. Consider ϕR(t, x, y) = a(t)b(x)c(y), (t, x, y) ∈ Ω. (3.1) Proposition 3.1. For sufficiently large R, the function ϕR belongs to the test function space Φ, where Φ is defined by (2.1). Proof. Clearly, we have ϕR ∈ C2 c (Ω), ϕR ≥ 0, ϕR|∂D1∪∂D2 = 0. On the other hand, ∇xϕR(t, x, y) = a(t)c(y)∇x ( H1(x)ξλ ( |x| Rθ )) = a(t)c(y) [ ξλ ( |x| Rθ ) ∇xH1(x) +H1(x)∇xξλ ( |x| Rθ )] . By the definition of H1, for x ∈ ∂D1, we obtain ∇xH1(x) = { x if N1 = 2, (N1 − 2)x if N1 ≥ 3. By the properties of the function ξ, for x ∈ ∂D1, we obtain (since R is sufficiently large) ξλ ( |x| Rθ ) = 1, ∣∣∇xξλ( |x| Rθ )∣∣ = 0. Hence, for (t, x, y) ∈ Γ1, we deduce that ∂xϕR ∂n1 (t, x, y) = { −a(t)c(y) if N1 = 2 −(N1 − 2)a(t)c(y) if N1 ≥ 3 ≤ 0. (3.2) Similarly, for (t, x, y) ∈ Γ2, we obtain ∂yϕR ∂n2 (t, x, y) = { −a(t)b(x) if N1 = 2 −(N2 − 2)a(t)b(x) if N1 ≥ 3. ≤ 0. (3.3) This shows that ϕR ∈ Φ. � The following estimates follow from standard calculations. Lemma 3.2. (i) Let α ∈ R and β > −1. As R→∞, we have∫ z∈R2:1<|z| −2. (ii) Let α, β ∈ R. As R→∞, we have∫ z∈R2:R<|z|<2R |z|α(ln |z|)β dz ∼ Rα+2(lnR)β . 10 M. JLELI, B. SAMET EJDE-2021/75 Lemma 3.3. Let N ≥ 3. (i) Let α ∈ R and β > −1. As R→∞, we have ∫ z∈RN :1<|z| −N. (ii) Let α, β ∈ R. As R→∞, we have∫ z∈RN :R<|z|<2R |z|α ( 1− |z|2−N )β dz ∼ Rα+N . Lemma 3.4. Let p > 1. Then (i) ∫∞ 0 a(t) dt = CR. (ii) ∫∞ 0 a −1 p−1 (t)|a′′(t)| p p−1 dt = O ( R1− 2p p−1 ) , as R→∞. Proof. (i) is immediate, so we omit its proof. On the other hand, we have |a′′(t)| ≤ CR−2ηλ−2 ( t R ) , t ∈ (0, R), which yields a −1 p−1 (t)|a′′(t)| p p−1 ≤ CR −2p p−1 ηλ− 2p P−1 ( t R ) , t ∈ (0, R). Then ∫ ∞ 0 a −1 p−1 (t)|a′′(t)| p p−1 dt ≤ CR −2p p−1 ∫ R 0 ηλ− 2p P−1 ( t R ) dt = C (∫ 1 0 ηλ− 2p P−1 (s) ds ) R1− 2p p−1 , which proves (ii). � Lemma 3.5. As R→∞, we have∫ D1 b(x) dx = { O ( R2θ lnR ) if N1 = 2, O ( RθN1 ) if N1 ≥ 3; (3.4) and ∫ D2 c(y) dy = { O ( R2σ lnR ) if N2 = 2, O ( RσN2 ) if N2 ≥ 3. (3.5) Proof. Let N1 = 2. We have∫ D1 b(x) dx = ∫ |x|>1 H1(x)ξλ ( |x| Rθ ) dx = ∫ 1<|x|<2Rθ ln |x|ξλ ( |x| Rθ ) dx ≤ ∫ 1<|x|<2Rθ ln |x| dx. Hence, by Lemma 3.2 (with α = 0 and β = 1), we obtain∫ D1 b(x) dx ≤ CR2θ lnR. EJDE-2021/75 HYPERBOLIC TYPE INEQUALITIES IN EXTERIOR DOMAINS 11 For N1 ≥ 3, we have∫ D1 b(x) dx ≤ ∫ 1<|x|<2Rθ ( 1− |x|2−N1 ) dx. Using Lemma 3.3 (with α = 0 and β = 1), we obtain∫ D1 b(x) dx ≤ CRθN1 . Therefore, (3.4) is proved. The same argument yields (3.5). � Lemma 3.6. As R→∞, we have∫ D1 b −1 p−1 |∆xb| p p−1 dx = { O ( R −2θ p−1 lnR ) if N1 = 2, O ( R −2θp p−1 +θN1 ) if N1 ≥ 3; (3.6) and ∫ D2 c −1 p−1 |∆yc| p p−1 dx = { O ( R −2σ p−1 lnR ) if N2 = 2, O ( R −2σp p−1 +σN2 ) if N2 ≥ 3. (3.7) Proof. By the properties of the function b, we have∫ D1 b −1 p−1 |∆xb| p p−1 dx = ∫ Rθ<|x|<2Rθ b −1 p−1 |∆xb| p p−1 dx. Let N1 = 2. For Rθ < |x| < 2Rθ, we obtain ∆xb = ∆x ( (ln |x|)ξλ ( |x| Rθ )) = ln |x|∆xξ λ ( |x| Rθ ) + 2∇x(ln |x|) · ∇xξλ ( |x| Rθ ) = ln |x|∆xξ λ ( |x| Rθ ) + 2R−θλ 1 |x|2 ξλ−1 ( |x| Rθ ) x · ∇xξ ( |x| Rθ ) , where · denotes the inner product in RN1 , which yields |∆xb| ≤ CR−2θ ln |x|ξλ−2 ( |x| Rθ ) + CR−θ|x|−1ξλ−1 ( |x| Rθ ) and b −1 p−1 |∆xb| p p−1 ≤ CR −2θp p−1 (ln |x|)ξλ− 2p p−1 ( |x| Rθ ) + CR −θp p−1 |x| −p p−1 (ln |x|) −1 p−1 ξλ− p p−1 ( |x| Rθ ) ≤ C ( R −2θp p−1 (ln |x|) +R −θp p−1 |x| −p p−1 (ln |x|) −1 p−1 ) . Then, by Lemma 3.2, we deduce that∫ D1 b −1 p−1 |∆xb| p p−1 dx ≤ C ( R −2θp p−1 ∫ Rθ<|x|<2Rθ ln |x| dx+R −θp p−1 ∫ Rθ<|x|<2Rθ |x| −p p−1 (ln |x|) −1 p−1 dx ) ≤ C ( R −2θp p−1 R2θ lnR+R −θp p−1Rθ ( p−2 p−1 ) (lnR) −1 p−1 ) ≤ CR −2θ p−1 lnR. 12 M. JLELI, B. SAMET EJDE-2021/75 For N1 ≥ 3 and Rθ < |x| < 2Rθ, proceeding as above, and using Lemma 3.3, we obtain b −1 p−1 |∆xb| p p−1 ≤ C ( R −2θp p−1 ( 1− |x|2−N1 ) +R −θp p−1 |x| (1−N1)p p−1 ( 1− |x|2−N1 ) −1 p−1 ) and∫ D1 b −1 p−1 |∆xb| p p−1 dx ≤ CR −2θp p−1 ∫ Rθ<|x|<2Rθ ( 1− |x|2−N1 ) dx + CR −θp p−1 ∫ Rθ<|x|<2Rθ |x| (1−N1)p p−1 ( 1− |x|2−N1 ) −1 p−1 dx ≤ C ( R −2θp p−1 +θN1 +R −θN1 p−1 ) ≤ CR −2θp p−1 +θN1 . This proves (3.6). Similar calculations yield (3.7). � The next Lemma follows immediately from Lemmas 3.2 and 3.3. Lemma 3.7. (i) Let N1 = 2. As R→∞, we have ∫ D1 |x| 2`p p−1 b(x) dx =  O(1) if p(`+ 1) < 1, O ( (lnR)2 ) if p(`+ 1) = 1, O ( R2θ( `p p−1 +1) lnR ) if p(`+ 1) > 1. (ii) Let N1 ≥ 3. As R→∞, we have ∫ D1 |x| 2`p p−1 b(x) dx =  O(1) if p(2`+N1) < N1, O(lnR) if p(2`+N1) = N1, O ( Rθ( 2`p p−1 +N1)) if p(2`+N1) > N1. Lemma 3.8. As R→∞, we have ∫ Ω ϕ −1 p−1 R |(ϕR)tt| p p−1 dy dx dt =  O ( R1− 2p p−1 +2θ+2σ(lnR)2 ) if N1 = N2 = 2, O ( R1− 2p p−1 +2θ+σN2 lnR ) if N1 = 2, N2 ≥ 3, O ( R1− 2p p−1 +θN1+2σ lnR ) if N1 ≥ 3, N2 = 2, O ( R1− 2p p−1 +θN1+2σ ) if N1, N2 ≥ 3. Proof. By (3.1), we obtain∫ Ω ϕ −1 p−1 R |(ϕR)tt| p p−1 dy dx dt = (∫ ∞ 0 a −1 p−1 (t)|a′′(t)| p p−1 dt )(∫ D1 b(x) dx )(∫ D2 c(y) dy ) . Hence, using Lemmas 3.4 and 3.5, the desired estimates follow. � Lemma 3.9. As R→∞, we have ∫ Ω ϕ −1 p−1 R |∆xϕR| p p−1 dy dx dt =  O ( R1− 2θ p−1 +2σ(lnR)2 ) if N1 = N2 = 2, O ( R1− 2θ p−1 +σN2 lnR ) if N1 = 2, N2 ≥ 3, O ( R1+2σ− 2θp p−1 +θN1 lnR ) if N1 ≥ 3, N2 = 2, O ( R1− 2θp p−1 +θN1+σN2 ) if N1, N2 ≥ 3. EJDE-2021/75 HYPERBOLIC TYPE INEQUALITIES IN EXTERIOR DOMAINS 13 Proof. By (3.1), we have∫ Ω ϕ −1 p−1 R |∆xϕR| p p−1 dy dx dt = (∫ ∞ 0 a(t) dt )(∫ D1 b −1 p−1 |∆xb| p p−1 dx )(∫ D2 c(y) dy ) . Using Lemmas 3.4, 3.5, and 3.6, the desired estimates follow. � Lemma 3.10. As R→∞, we have ∫ Ω |x| 2`p p−1 |∆yϕ| p p−1ϕ −1 p−1 dy dx dt =  O ( R1− 2σ p−1 (lnR)A(R) ) if N1 = N2 = 2, O ( R1− 2σp p−1 +σN2A(R) ) if N1 = 2, N2 ≥ 3, O ( R1− 2σp p−1 (lnR)B(R) ) if N1 ≥ 3, N2 = 2, O ( R1− 2σp p−1 +σN2B(R) ) if N1, N2 ≥ 3, where A(R) =  1 if p(`+ 1) < 1, (lnR)2 if p(`+ 1) = 1, R2θ( `p p−1 +1) lnR if p(`+ 1) > 1; B(R) =  1 if p(2`+N1) < N1, lnR if p(2`+N1) = N1, Rθ( 2`p p−1 +N1) if p(2`+N1) > N1. (3.8) Proof. By (3.1), we have∫ Ω |x| 2`p p−1 |∆yϕ| p p−1ϕ −1 p−1 dy dx dt = (∫ ∞ 0 a(t) dt )(∫ D1 |x| 2`p p−1 b(x) dx )(∫ D2 c −1 p−1 |∆yc| p p−1 dy ) . Hence, using Lemmas 3.4, 3.6, and 3.7, the desired estimates follow. � Proposition 3.11. Let f ∈ L1(∂D1) and g ∈ L1(∂D2). If u ∈ Lploc([0,∞)×D) is a global weak solution to (1.1), then for all ϕ ∈ Φ, − ∫ Γ1 ∂xϕ ∂n1 f(x) dσx dy dt− ∫ Γ2 |x|2` ∂yϕ ∂n2 g(y) dx dσy dt ≤ C (∫ Ω ϕ −1 p−1 |ϕtt| p p−1 dy dx dt+ ∫ Ω ϕ −1 p−1 |∆xϕ| p p−1 dy dx dt + ∫ Ω |x| 2`p p−1ϕ −1 p−1 |∆yϕ| p p−1 dy dx dt ) . Proof. Let u ∈ Lploc([0,∞)×D) be a global weak solution to (1.1). Then by (2.2), for all ϕ ∈ Φ, we have∫ Ω |u|pϕdy dx dt− ∫ Γ1 ∂xϕ ∂n1 f(x) dσx dy dt− ∫ Γ2 |x|2` ∂yϕ ∂n2 g(y) dx dσy dt ≤ ∫ Ω u ( ϕtt −∆xϕ− |x|2`∆yϕ ) dy dx dt ≤ ∫ Ω |u||ϕtt| dy dx dt+ ∫ Ω |u||∆xϕ| dy dx dt+ ∫ Ω |x|2`|u||∆yϕ| dy dx dt. (3.9) 14 M. JLELI, B. SAMET EJDE-2021/75 On the other hand, by Young’s inequality, we obtain∫ Ω |u||ϕtt| dy dx dt ≤ 1 3 ∫ Ω |u|pϕdy dx dt+ C ∫ Ω ϕ −1 p−1 |ϕtt| p p−1 dy dx dt. (3.10) Similarly, we have∫ Ω |u||∆xϕ| dy dx dt ≤ 1 3 ∫ Ω |u|pϕdy dx dt+ C ∫ Ω ϕ −1 p−1 |∆xϕ| p p−1 dy dx dt (3.11) and ∫ Ω |x| 2`p p−1ϕ −1 p−1 |∆yϕ| p p−1 dy dx dt ≤ 1 3 ∫ Ω |u|pϕdy dx dt+ C ∫ Ω |x| 2`p p−1ϕ −1 p−1 |∆yϕ| p p−1 dy dx dt. (3.12) The desired estimate follows from (3.9), (3.10), (3.11), and (3.12). � 4. Proofs of main results Lemma 4.1. Let N1 = N2 = 2, f ∈ L1(∂D1), and g ∈ L1(∂D2). Suppose that u ∈ Lploc([0,∞)×D) is a global weak solution to (1.1). Then, for sufficiently large R, we have R2σ lnR ∫ ∂D1 f(x)dσx +G(R) ∫ ∂D2 g(y)dσy ≤ C ( R− 2p p−1 +2θ+2σ(lnR)2 +R− 2θ p−1 +2σ(lnR)2 +R− 2σ p−1 (lnR)A(R) ) , where A(R) is given by (3.8) and G(R) =  1 if ` < −1, (lnR)2 if ` = −1, R2θ(`+1) lnR if ` > −1. (4.1) Proof. Let u ∈ Lploc([0,∞)×D) be a global weak solution to (1.1). By Propositions 3.1 and 3.11, for sufficiently large R, we have − ∫ Γ1 ∂xϕR ∂n1 f(x) dσx dy dt− ∫ Γ2 |x|2` ∂yϕR ∂n2 g(y) dx dσy dt ≤ C (∫ Ω ϕ −1 p−1 R |(ϕR)tt| p p−1 dy dx dt+ ∫ Ω ϕ −1 p−1 R |∆xϕR| p p−1 dy dx dt + ∫ Ω |x| 2`p p−1ϕ −1 p−1 R |∆yϕR| p p−1 dy dx dt ) . (4.2) On the other hand, by (3.1), (3.2), (3.3), and Lemma 3.4-(i), we obtain − ∫ Γ1 ∂xϕr ∂n1 f(x) dσx dy dt− ∫ Γ2 |x|2` ∂ϕ ∂n2 g(y) dx dσy dt = (∫ R 0 ηλ ( t R ) dt )(∫ 1<|y|<2Rσ ln |y|ξλ ( |y| Rσ ) dy )(∫ ∂D1 f(x)dσx ) + (∫ R 0 ηλ ( t R ) dt )(∫ 1<|x|<2Rθ |x|2` ln |x|ξλ ( |x| Rθ ) dx )(∫ ∂D2 g(y)dσy ) ≥ CR (∫ 1<|y|<2Rσ ln |y|ξλ ( |y| Rσ ) dy )(∫ ∂D1 f(x)dσx ) EJDE-2021/75 HYPERBOLIC TYPE INEQUALITIES IN EXTERIOR DOMAINS 15 + CR (∫ 1<|x|<2Rθ |x|2` ln |x|ξλ ( |x| Rθ ) dx )(∫ ∂D2 g(y)dσy ) . Since ∫ 1<|y| 0. By Lemma 4.1 and (3.8), for sufficiently large R, we obtain∫ ∂D1 f(x)dσx ≤ C ( R− 2p p−1 +2θ lnR+R− 2θ p−1 lnR+R− 2σ p−1−2σ ) . In particular, for θ = 1, we have∫ ∂D1 f(x)dσx ≤ C ( R− 2 p−1 lnR+R− 2σ p−1−2σ ) . Passing to the limit as R → ∞ in the above inequality, we obtain a contradiction with ∫ ∂D1 f(x)dσx > 0. This shows that (1.1) admits no global weak solution for all p > 1. Let ` ≤ −1, ∫ ∂D1 f(x)dσx = 0, and ∫ ∂D2 g(y)dσy > 0. By Lemma 4.1 and (3.8), for sufficiently large R, we obtain∫ ∂D2 g(y)dσy ≤ C ( R− 2p p−1 +2θ+2σ(lnR)2 +R− 2θ p−1 +2σ(lnR)2 +R− 2σ p−1 (lnR) ) . 16 M. JLELI, B. SAMET EJDE-2021/75 Taking θ = 1, 0 < σ < 1 p−1 , and passing to the limit as R → ∞ in the above inequality, we obtain a contradiction with ∫ ∂D2 g(y)dσy > 0. This shows that (1.1) admits no global weak solution for all p > 1. Therefore, part (I) of Theorem 2.2 is proved. Let ` > −1 and ∫ ∂D1 f(x)dσx > 0. Using Lemma 4.1 with θ = 1 and σ > 2(`+ 1), for sufficiently large R, we obtain∫ ∂D1 f(x)dσx ≤ C ( R− 2 p−1 lnR+R− 2σ p−1−2σA(R) ) . If p(`+ 1) ≤ 1, by (3.8), we have A(R) ≤ (lnR)2. Then∫ ∂D1 f(x)dσx ≤ C ( R− 2 p−1 lnR+R− 2σ p−1−2σ(lnR)2 ) . Passing to the limit as R → ∞ in the above inequality, we obtain a contradiction with ∫ ∂D1 f(x)dσx > 0. If p(` + 1) > 1, by (3.8), we have A(R) = R2( `p p−1 +1) lnR. Then ∫ ∂D1 f(x)dσx ≤ C ( R− 2 p−1 lnR+R− 2σ p−1−2σ+2 ( `p p−1 +1 ) lnR ) . (4.5) On the other hand, for σ > 2(`+ 1), we have − 2σ p− 1 − 2σ + 2 ( `p p− 1 + 1 ) < 0. Hence, Passing to the limit as R → ∞ in (4.5), we obtain a contradiction with∫ ∂D1 f(x)dσx > 0. Then, we deduce that (1.1) admits no global weak solution for all p > 1. Let ` > −1 and ∫ ∂D2 g(y)dσy > 0. Using Lemma 4.1 with θ = 1 and 0 < σ < `+ 1, for sufficiently large R, we obtain R2(`+1) lnR ∫ ∂D2 g(y)dσy ≤ C ( R− 2 p−1 +2σ(lnR)2 +R− 2σ p−1 (lnR)A(R) ) , that is, ∫ ∂D2 g(y)dσy ≤ C ( R− 2 p−1 +2σ−2(`+1) lnR+R− 2σ p−1−2(`+1)A(R) ) . If p(`+ 1) ≤ 1, by (3.8), we have A(R) ≤ (lnR)2. Then∫ ∂D2 g(y)dσy ≤ C ( R− 2 p−1 +2σ−2(`+1) lnR+R− 2σ p−1−2(`+1)(lnR)2 ) . Hence, passing to the limit as R → ∞ in the above inequality, we obtain a con- tradiction with ∫ ∂D2 g(y)dσy > 0. If p(` + 1) > 1, by (3.8), we have A(R) = R2( `p p−1 +1) lnR. Then∫ ∂D2 g(y)dσy ≤ C ( R− 2 p−1 +2σ−2(`+1) lnR+R− 2σ p−1−2(`+1)+2( `p p−1 +1) lnR ) . (4.6) EJDE-2021/75 HYPERBOLIC TYPE INEQUALITIES IN EXTERIOR DOMAINS 17 Observe that for σ > `, − 2σ p− 1 − 2(`+ 1) + 2 ( `p p− 1 + 1 ) < 0. Hence, for max{0, `} < σ < `+1, passing to the limit as R→∞ in (4.6), we obtain a contradiction with ∫ ∂D2 g(y)dσy > 0. Therefore, we deduce that (1.1) admits no global weak solution for all p > 1. Then part (II) of Theorem 2.2 is proved. � Lemma 4.2. Let N1 = 2, N2 ≥ 3, f ∈ L1(∂D1), and g ∈ L1(∂D2). Suppose that u ∈ Lploc([0,∞)×D) is a global weak solution to (1.1). Then, for sufficiently large R, RσN2 ∫ ∂D1 f(x)dσx +G(R) ∫ ∂D2 g(y)dσy ≤ C ( R− 2p p−1 +2θ+σN2 lnR+R− 2θ p−1 +σN2 lnR+R− 2σp p−1 +σN2A(R) ) , where A(R) and G(R) are given respectively by (3.8) and (4.1). Proof. Let u ∈ Lploc([0,∞)×D) be a global weak solution to (1.1). By (3.1), (3.2), (3.3), (4.3), and Lemma 3.4-(i), for sufficiently large R, we obtain − ∫ Γ1 ∂xϕr ∂n1 f(x) dσx dy dt− ∫ Γ2 |x|2` ∂ϕ ∂n2 g(y) dx dσy dt = (∫ R 0 ηλ ( t R ) dt )(∫ 1<|y|<2Rσ ( 1− |y|2−N2 ) ξλ ( |y| Rσ ) dy )(∫ ∂D1 f(x)dσx ) + (∫ R 0 ηλ ( t R ) dt )(∫ 1<|x|<2Rθ |x|2` ln |x|ξλ ( |x| Rθ ) dx )(∫ ∂D2 g(y)dσy ) ≥ CR (∫ 1<|y|<2Rσ ( 1− |y|2−N2 ) ξλ ( |y| Rσ ) dy )(∫ ∂D1 f(x)dσx ) + CRG(R) ∫ ∂D2 g(y)dσy. Since ∫ 1<|y| 0. By Lemma 4.2, (3.8), and (4.1), for sufficiently large R, we obtain RσN2 ∫ ∂D1 f(x)dσx ≤ C ( R− 2p p−1 +2θ+σN2 lnR+R− 2θ p−1 +σN2 lnR+R− 2σp p−1 +σN2 ) , that is, ∫ ∂D1 f(x)dσx ≤ C ( R− 2p p−1 +2θ lnR+R− 2θ p−1 lnR+R− 2σp p−1 ) . Taking θ = 1, we obtain∫ ∂D1 f(x)dσx ≤ C ( R− 2θ p−1 lnR+R− 2σp p−1 ) . Passing to the limit as R → ∞ in the above inequality, we obtain a contradiction with ∫ ∂D1 f(x)dσx > 0. Hence, for all p > 1, (1.1) admits no global weak solution. This proves parts (I)-(i) and (II)-(i) of Theorem 2.6. Let ` > −1 and ∫ ∂D1 f(x)dσx > 0. Using (4.1) and Lemma 4.2 with θ = 1 and σ > ` + 1 (so σN2 > 2(` + 1)), for sufficiently large R, we obtain RσN2 ∫ ∂D1 f(x)dσx ≤ C ( R− 2 p−1 +σN2 lnR+R− 2σp p−1 +σN2A(R) ) , that is, ∫ ∂D1 f(x)dσx ≤ C ( R− 2 p−1 lnR+R− 2σp p−1A(R) ) . If p(`+ 1) ≤ 1, then by (3.8), we have A(R) ≤ (lnR)2. Then∫ ∂D1 f(x)dσx ≤ C ( R− 2 p−1 lnR+R− 2σp p−1 (lnR)2 ) . Passing to the limit as R → ∞ in the above inequality, we obtain a contradic- tion with ∫ ∂D1 f(x)dσx > 0. If p(` + 1) > 1, then by (3.8), we have A(R) = R2( `p p−1 +1) lnR. Then∫ ∂D1 f(x)dσx ≤ C ( R− 2 p−1 lnR+R− 2σp p−1 +2( `p p−1 +1) ) . (4.7) Notice that for σ > `+ 1, − 2σp p− 1 + 2 ( `p p− 1 + 1 ) < 0. Hence, passing to the limit as R → ∞ in (4.7), we obtain a contradiction with∫ ∂D1 f(x)dσx > 0. Then, we deduce that for all p > 1, (1.1) admits no global weak solution. This proves part (III) when ∫ ∂D1 f(x)dσx > 0, and part (IV)-(i). Let ` < −1 and ∫ ∂D1 f(x)dσx = 0, ∫ ∂D2 g(y)dσy > 0. EJDE-2021/75 HYPERBOLIC TYPE INEQUALITIES IN EXTERIOR DOMAINS 19 In this case, by Lemma 4.2, (3.8), and (4.1), for sufficiently large R, we obtain∫ ∂D2 g(y)dσy ≤ C ( R− 2p p−1 +2θ+σN2 lnR+R− 2θ p−1 +σN2 lnR+R− 2σp p−1 +σN2 ) . In particular, for θ = 1, we have∫ ∂D2 g(y)dσy ≤ C ( R− 2 p−1 +σN2 lnR+R−σ( 2p p−1−N2) ) . Let 1 < p < N2 N2−2 . Taking 0 < σN2 < 2 p−1 , and passing to the limit as R →∞ in the above inequality, we obtain a contradiction with ∫ ∂D2 g(y)dσy > 0. Hence, for all 1 < p < N2 N2−2 , (1.1) admits no global weak solution. This proves part (I)-(ii) of Theorem 2.6. Let ` = −1 and ∫ ∂D1 f(x)dσx = 0, ∫ ∂D2 g(y)dσy > 0. In this case, by Lemma 4.2, (3.8), and (4.1), for sufficiently large R, we obtain (lnR)2 ∫ ∂D2 g(y)dσy ≤ C ( R− 2p p−1 +2θ+σN2 lnR+R− 2θ p−1 +σN2 lnR+R− 2σp p−1 +σN2 ) , that is,∫ ∂D2 g(y)dσy ≤ C ( R− 2p p−1 +2θ+σN2(lnR)−1 +R− 2θ p−1 +σN2(lnR)−1 +R− 2σp p−1 +σN2(lnR)−2 ) . In particular, for θ = 1, we obtain∫ ∂D2 g(y)dσy ≤ C ( R− 2 p−1 +σN2(lnR)−1 +Rσ(N2− 2p p−1 )(lnR)−2 ) . Let 1 < p ≤ N2 N2−2 . Taking 0 < σN2 < 2 p−1 , and passing to the limit as R →∞ in the above inequality, we obtain a contradiction with ∫ ∂D2 g(y)dσy > 0. Hence, for all 1 < p ≤ N2 N2−2 , (1.1) admits no global weak solution. This proves part (II)-(ii). Let −1 < ` < 0 and ∫ ∂D2 g(y)dσy > 0. By (4.1) and using Lemma 4.2 with θ = 1 and 0 < σN2 < 2(`+ 1), for sufficiently large R, we obtain R2(`+1) lnR ∫ ∂D2 g(y)dσy ≤ C ( R− 2 p−1 +σN2 lnR+Rσ(N2− 2p p−1 )A(R) ) , that is,∫ ∂D2 g(y)dσy ≤ C ( R− 2 p−1 +σN2−2(`+1) lnR+Rσ(N2− 2p p−1 )−2(`+1)A(R) ) . (4.8) If p(`+ 1) ≤ 1, by (3.8) we have A(R) ≤ (lnR)2. Then∫ ∂D2 g(y)dσy ≤ C ( R− 2 p−1 +σN2−2(`+1) lnR+Rσ(N2− 2p p−1 )−2(`+1)(lnR)2 ) . 20 M. JLELI, B. SAMET EJDE-2021/75 Passing to the limit as R → ∞ in the above inequality, we obtain a contradiction with ∫ ∂D2 g(y)dσy > 0. If p(` + 1) > 1, by (3.8) we have A(R) = R2( `p p−1 +1) lnR. Then ∫ ∂D2 g(y)dσy ≤ C ( R− 2 p−1 +σN2−2(`+1) lnR+Rσ(N2− 2p p−1 )−2(`+1)+2( `p p−1 +1) lnR ) . (4.9) Observe that for 0 < σN2 < −2` p−1 (so 0 < σN2 < min{2(`+ 1), −2` p−1}), we have σ ( N2 − 2p p− 1 ) − 2(`+ 1) + 2 ( `p p− 1 + 1 ) < σN2 − 2(`+ 1) + 2 ( `p p− 1 + 1 ) < 0. Hence, passing to the limit as R → ∞ in (4.9), we obtain a contradiction with∫ ∂D2 g(y)dσy > 0. Consequently, (1.1) admits no global weak solution for all p > 1. This proves part (III) in the case ∫ ∂D2 g(y)dσy > 0. Let ` ≥ 0 and ∫ ∂D2 g(y)dσy > 0. As previously, by (4.1) and using Lemma 4.2 with θ = 1 and 0 < σN2 < 2(` + 1), for sufficiently large R, we obtain (4.8). Moreover, since ` ≥ 0 and p(`+1) ≥ p > 1, by (3.8) we have A(R) = R2( `p p−1 +1) lnR, and (4.9) holds. Observe that for all 1 < p < N2 N2−2 , σ ( N2 − 2p p− 1 ) − 2(`+ 1) + 2 ( `p p− 1 + 1 ) < 0. Hence, passing to the limit as R → ∞ in (4.9), we obtain a contradiction with∫ ∂D2 g(y)dσy > 0. Consequently, for all 1 < p < N2 N2−2 , (1.1) admits no global weak solution. This proves part (IV)-(ii). The proof of Theorem 2.6 is complete. � Case N1 ≥ 3 and N2 = 2. Proceeding as in the proofs of Lemmas 4.1 and 4.2, we obtain the following estimate. Lemma 4.3. Let N1 ≥ 3, N2 = 2, f ∈ L1(∂D1), and g ∈ L1(∂D2). Suppose that u ∈ Lploc([0,∞)×D) is a global weak solution to (1.1). Then, for sufficiently large R, R2σ lnR ∫ ∂D1 f(x)dσx + G(R) ∫ ∂D2 g(y)dσy ≤ C ( R− 2p p−1 +θN1+2σ lnR+R2σ− 2θp p−1 +θN1 lnR+R− 2σp p−1 (lnR)B(R) ) , where B(R) is given by (3.8) and G(R) =  1 if ` < −N1 2 , lnR if ` = −N1 2 , Rθ(2`+N1) if ` > −N1 2 . (4.10) Proof of Theorem 2.11. Suppose that u ∈ Lploc([0,∞)×D) is a global weak solution to (1.1). Let ` ≤ −N1 2 and ∫ ∂D1 f(x)dσx > 0. EJDE-2021/75 HYPERBOLIC TYPE INEQUALITIES IN EXTERIOR DOMAINS 21 Then, by Lemma 4.3, for sufficiently large R, we obtain R2σ lnR ∫ ∂D1 f(x)dσx ≤ C ( R− 2p p−1 +θN1+2σ lnR+R2σ− 2θp p−1 +θN1 lnR+R− 2σp p−1 (lnR)B(R) ) , that is, ∫ ∂D1 f(x)dσx ≤ C ( R− 2p p−1 +θN1 +R− 2θp p−1 +θN1 +R− 2σp p−1−2σB(R) ) . In particular, for θ = 1, we obtain∫ ∂D1 f(x)dσx ≤ C ( R− 2p p−1 +N1 +R− 2σp p−1−2σB(R) ) . (4.11) Notice that in this case, p(2`+N1) ≤ 0 < N1. Hence, by (3.8) we obtain∫ ∂D1 f(x)dσx ≤ C ( R− 2p p−1 +N1 +R− 2σp p−1−2σ ) . For 1 < p < N1 N1−2 , passing to the limit as R → ∞ in the above inequality, we obtain a contradiction with ∫ ∂D1 f(x)dσx > 0. Consequently, (1.1) admits no global weak solution for all 1 < p < N1 N1−2 . This proves part (I) of Theorem 2.11 when∫ ∂D1 f(x)dσx > 0. Let ` > −N1 2 and ∫ ∂D1 f(x)dσx > 0. In this case, using Lemma 4.3 with θ = 1 and 2σ > 2` + N1, for sufficiently large R, we obtain (4.11). Let 1 < p < N1 N1−2 . If p(2`+N1) ≤ N1, by (3.8) and (4.11) we have B(R) ≤ lnR and∫ ∂D1 f(x)dσx ≤ C ( R− 2p p−1 +N1 +R− 2σp p−1−2σ lnR ) . Then, passing to the limit as R→∞ in the above inequality, we obtain a contradic- tion with ∫ ∂D1 f(x)dσx > 0. If p(2`+N1) > N1, by (3.8) we have B(R) = R 2`p p−1 +N1 . Then ∫ ∂D1 f(x)dσx ≤ C ( R− 2p p−1 +N1 +R− 2σp p−1−2σ+ 2`p p−1 +N1 ) . Taking 2σ > 2`p p−1 +N1 (so 2σ > max{2`+N1, 2`p p−1 +N1}) and passing to the limit as R →∞ in the above inequality, we obtain a contradiction with ∫ ∂D1 f(x)dσx > 0. Consequently, (1.1) admits no global weak solution for all 1 < p < N1 N1−2 . This proves parts (II)-(i) and (III)-(i). Let ` ≤ −N1 2 and ∫ ∂D1 f(x)dσx = 0, ∫ ∂D2 g(y)dσy > 0. By Lemma 4.3 and (3.8), for sufficiently large R, we obtain∫ ∂D2 g(y)dσy ≤ C ( R− 2p p−1 +θN1+2σ lnR+R2σ− 2θp p−1 +θN1 lnR+R− 2σp p−1 lnR ) . 22 M. JLELI, B. SAMET EJDE-2021/75 In particular, for θ = 1, we have∫ ∂D2 g(y)dσy ≤ C ( R− 2p p−1 +N1+2σ lnR+R− 2σp p−1 lnR ) . Hence, for 1 < p < N1 N1−2 , taking 0 < 2σ < 2p p−1 − N1 and passing to the limit as R → ∞ in the above inequality, we obtain a contradiction with ∫ ∂D1 g(y)dσy > 0. Consequently, (1.1) admits no global weak solution for all 1 < p < N1 N1−2 . This proves part (I) of Theorem 2.11 when ∫ ∂D1 f(x)dσx = 0 and ∫ ∂D2 g(y)dσy > 0. Let −N1 2 < ` < −1 and ∫ ∂D2 g(y)dσy > 0. Using Lemma 4.3 with θ = 1 and 0 < 2σ < 2` + N1, for sufficiently large R, we obtain R2`+N1 ∫ ∂D2 g(y)dσy ≤ C ( R− 2p p−1 +N1+2σ lnR+R− 2σp p−1 (lnR)B(R) ) , that is∫ ∂D2 g(y)dσy ≤ C ( R− 2p p−1 +2σ−2` lnR+R− 2σp p−1−2`−N1(lnR)B(R) ) . (4.12) Let 1 < p < ` `+1 . If p(2`+N1) ≤ N1, by (3.8) we have B(R) ≤ lnR. Then∫ ∂D2 g(y)dσy ≤ C ( R− 2p p−1 +2σ−2` lnR+R− 2σp p−1−2`−N1(lnR)2 ) . (4.13) Taking 0 < σ < ` + p p−1 (so 0 < 2σ < min{2` + N1, 2(` + p p−1 )}) and passing to the limit as R → ∞ in the above inequality, we obtain a contradiction with∫ ∂D2 g(y)dσy > 0. If p(2`+N1) > N1, by (3.8) we have B(R) = R 2`p p−1 +N1 . Then∫ ∂D2 g(y)dσy ≤ C ( R− 2p p−1 +2σ−2` lnR+R− 2σp p−1−2`+ 2`p p−1 (lnR) ) . (4.14) Taking 0 < σ < `+ p p−1 and passing to the limit as R→∞ in the above inequality, we obtain a contradiction with ∫ ∂D2 g(y)dσy > 0. Hence, we deduce that (1.1) admits no global weak solution for all 1 < p < ` `+1 . This proves part (II)-(ii) of Theorem 2.11. Let ` ≥ −1 and ∫ ∂D2 g(y)dσy > 0. As in the previous case, using Lemma 4.3 with θ = 1 and 0 < 2σ < 2` + N1, for sufficiently large R, we obtain (4.12). If p(2`+N1) ≤ N1, by (3.8) we obtain (4.13). Notice that in this case, ` + p p−1 > 0. So, taking 0 < σ < ` + p p−1 and passing to the limit as R → ∞ in (4.13), we obtain a contradiction with ∫ ∂D2 g(y)dσy > 0. If p(2` + N1) > N1, we obtain (4.14), and the same conclusion as above follows. Consequently, (1.1) admits no global weak solution for all p > 1. This proves part (III)-(ii). The proof of Theorem 2.11 is complete. � EJDE-2021/75 HYPERBOLIC TYPE INEQUALITIES IN EXTERIOR DOMAINS 23 Case N1, N2 ≥ 3. Proceeding as in the proofs of Lemmas 4.1 and 4.2, we obtain the following estimate. Lemma 4.4. Let N1, N2 ≥ 3, f ∈ L1(∂D1), and g ∈ L1(∂D2). Suppose that u ∈ Lploc([0,∞)×D) is a global weak solution to (1.1). Then, for sufficiently large R, RσN2 (∫ ∂D1 f(x)dσx ) + G(R) (∫ ∂D2 g(y)dσy ) ≤ C ( R− 2p p−1 +θN1+σN2 +R− 2θp p−1 +θN1+σN2 +R− 2σp p−1 +σN2B(R) ) , where B(R) and G(R) are given respectively by (3.8) and (4.10). Proof of Theorem 2.15. Suppose that u ∈ Lploc([0,∞)×D) is a global weak solution to (1.1). Let ` ≤ −N1 2 and ∫ ∂D1 f(x)dσx > 0. Then by Lemma 4.4, (3.8), and (4.10), for sufficiently large R, we obtain∫ ∂D1 f(x)dσx ≤ C ( R− 2p p−1 +θN1 +R− 2θp p−1 +θN1 +R− 2σp p−1 ) . In particular, for θ = 1, we have∫ ∂D1 f(x)dσx ≤ C ( R− 2p p−1 +N1 +R− 2σp p−1 ) . Hence, for 1 < p < N1 N1−2 , passing to the limit as R → ∞ in the above inequality, we obtain a contradiction with ∫ ∂D1 f(x)dσx > 0. Therefore, (1.1) admits no global weak solution for all 1 < p < N1 N1−2 . This proves part (I)-(i) of Theorem 2.15. Let ` > −N1 2 and ∫ ∂D1 f(x)dσx > 0. In this case, using Lemma 4.4 with θ = 1 and σN2 > 2` + N1, by (4.10), for sufficiently large R, we obtain∫ ∂D1 f(x)dσx ≤ C ( R− 2p p−1 +N1 +R− 2σp p−1B(R) ) . Let 1 < p < N1 N1−2 . If p(2`+N1) ≤ N1, by (3.8) we have B(R) ≤ lnR. Then∫ ∂D1 f(x)dσx ≤ C ( R− 2p p−1 +N1 +R− 2σp p−1 lnR ) . Passing to the limit as R → ∞ in the above inequality, we obtain a contradiction with ∫ ∂D1 f(x)dσx > 0. If p(2` + N1) > N1, by (3.8) we have B(R) = R 2`p p−1 +N1 . Then ∫ ∂D1 f(x)dσx ≤ C ( R− 2p p−1 +N1 +R− 2σp p−1 + 2`p p−1 +N1 ) . Taking σ > ` + N1(p−1) 2p (so σN2 > max{2` + N1, N2(` + N1(p−1) 2p )} and passing to the limit as R → ∞ in the above inequality, we obtain a contradiction with∫ ∂D1 f(x)dσx > 0. Then, we deduce that (1.1) admits no global weak solution for all 1 < p < N1 N1−2 . This proves parts (II)-(i), (III)-(i), and (IV)-(i) of Theorem 2.15. 24 M. JLELI, B. SAMET EJDE-2021/75 Let ` < −N1 2 and ∫ ∂D1 f(x)dσx = 0, ∫ ∂D2 g(y)dσy > 0. In this case, using Lemma 4.4, (3.8), and (4.10), for sufficiently large R, we obtain∫ ∂D2 g(y)dσy ≤ C ( R− 2p p−1 +θN1+σN2 +R− 2θp p−1 +θN1+σN2 +R− 2σp p−1 +σN2 ) . In particular, for θ = 1 we have∫ ∂D2 g(y)dσy ≤ C ( R− 2p p−1 +N1+σN2 +R− 2σp p−1 +σN2 ) . Hence, for 1 < p < min{ N1 N1−2 , N2 N2−2}, taking 0 < σN2 < 2p p−1 − N1 and passing to the limit as R → ∞ in the above inequality, we obtain a contradiction with∫ ∂D2 g(y)dσy > 0. Consequently, (1.1) admits no global weak solution for all 1 < p < min{ N1 N1−2 , N2 N2−2}. This proves part (I)-(ii) of Theorem 2.15. Let ` = −N1 2 and ∫ ∂D1 f(x)dσx = 0, ∫ ∂D2 g(y)dσy > 0. Using Lemma 4.4 with θ = 1, (3.8), and (4.10), for sufficiently large R, we obtain lnR ∫ ∂D2 g(y)dσy ≤ C ( R− 2p p−1 +N1+σN2 +R− 2σp p−1 +σN2 ) , that is,∫ ∂D2 g(y)dσy ≤ C ( R− 2p p−1 +N1+σN2(lnR)−1 +R− 2σp p−1 +σN2(lnR)−1 ) . Hence, for 1 < p < min{ N1 N1−2 , N2 N2−2} or p = N2 N2−2 < N1 N1−2 , taking 0 < σN2 ≤ 2p p−1 − N1 and passing to the limit as R → ∞ in the above inequality, we obtain a contradiction with ∫ ∂D2 g(y)dσy > 0. Therefore, (1.1) admits no global weak solution for all 1 < p < min{ N1 N1−2 , N2 N2−2} or p = N2 N2−2 < N1 N1−2 . This proves part (I)-(iii). Let −N1 2 < ` < −1 and ∫ ∂D2 g(y)dσy > 0. In this case, using (4.10) and Lemma 4.4 with θ = 1 and 0 < σN2 < 2` + N1, for sufficiently large R, we obtain R2`+N1 ∫ ∂D2 g(y)dσy ≤ C ( R− 2p p−1 +N1+σN2 +R− 2σp p−1 +σN2B(R) ) , that is, ∫ ∂D2 g(y)dσy ≤ C ( R− 2p p−1 +σN2−2` +R− 2σp p−1 +σN2−2`−N1B(R) ) . Let 1 < p < ` `+1 . If p(2`+N1) ≤ N1, by (3.8) we have B(R) ≤ lnR. Then∫ ∂D2 g(y)dσy ≤ C ( R− 2p p−1 +σN2−2` +R− 2σp p−1 +σN2−2`−N1 lnR ) . (4.15) EJDE-2021/75 HYPERBOLIC TYPE INEQUALITIES IN EXTERIOR DOMAINS 25 Taking 0 < σN2 < 2(`+ p p−1 ) (so 0 < σN2 < min{2`+N1, 2(`+ p p−1 )}) and passing to the limit as R → ∞ in the above inequality, we obtain a contradiction with∫ ∂D2 g(y)dσy > 0. If p(2`+N1) > N1, by (3.8) we have B(R) = R 2`p p−1 +N1 . Then∫ ∂D2 g(y)dσy ≤ C ( R− 2p p−1 +σN2−2` +R− 2σp p−1 +σN2−2`+ 2`p p−1 ) . (4.16) Similarly, taking 0 < σN2 < 2(`+ p p−1 ) (so 0 < σN2 < min{2`+N1, 2(`+ p p−1 )}) and passing to the limit as R → ∞ in the above inequality, we obtain a contradiction with ∫ ∂D2 g(y)dσy > 0. Hence, (1.1) admits no global weak solution for all 1 < p < ` `+1 . This proves part (II)-(ii). Let −1 ≤ ` < 0 and ∫ ∂D2 g(y)dσy > 0. We use Lemma 4.4 with θ = 1 and 0 < σN2 < 2`+N1. Proceeding as in the previous case, if p(2`+N1) ≤ N1, for sufficiently large R, we obtain (4.15). Notice that since ` ≥ −1, one has `+ p p−1 > 0. Hence, taking 0 < σN2 < 2(`+ p p−1 ) (so 0 < σN2 < min{2`+N1, 2(`+ p p−1 )}) and passing to the limit as R→∞ in (4.15), we obtain a contradiction with ∫ ∂D2 g(y)dσy > 0. If p(2`+N1) > N1, then for sufficiently large R, (4.16) holds. Taking 0 < σN2 < − 2` p−1 = min{− 2` p−1 , 2(` + p p−1 ), 2` + N1} and passing to the limit as R→∞ in (4.16), the same conclusion follows. Consequently, (1.1) admits no global weak solution for all p > 1. This proves part (III)-(ii). Let ` ≥ 0 and ∫ ∂D2 g(y)dσy > 0. Using (3.8), (4.10), and Lemma 4.4 with θ = 1 and 0 < σN2 < 2` + N1, for sufficiently large R, we obtain (4.16). For 1 < p < N2 N2−2 , taking 0 < σN2 < 2(` + p p−1 ) (so 0 < σN2 < min{2` + N1, 2(` + p p−1 )}) and passing to the limit as R → ∞ in (4.16), we obtain a contradiction with ∫ ∂D2 g(y)dσy > 0. Hence, (1.1) admits no global weak solution for all 1 < p < N2 N2−2 . This proves part (IV)-(ii). The proof of Theorem 2.15 is complete. � Acknowledgments. M. 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Box 2455, Riyadh 11451, Saudi Arabia Email address: bsamet@ksu.edu.sa 1. Introduction 2. Main results 3. Preliminaries 4. Proofs of main results Case N13 and N2=2 Case N1,N23 Acknowledgments References