Electronic Journal of Differential Equations, Vol. 2021 (2021), No. 68, pp. 1–17. ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu INFINITELY MANY SOLUTIONS FOR A SINGULAR SEMILINEAR PROBLEM ON EXTERIOR DOMAINS MAGEED ALI, JOSEPH IAIA Abstract. In this article we prove the existence of an infinite number of radial solutions to ∆U + K(x)f(U) = 0 on the exterior of the ball of radius R > 0 centered at the origin in RN with U = 0 on ∂BR, and lim|x|→∞ U(x) = 0 where N > 2, f(U) ∼ −1 |U|q−1U for small U 6= 0 with 0 < q < 1, and f(U) ∼ |U |p−1U for large |U | with p > 1. Also, K(x) ∼ |x|−α with α > 2(N − 1) for large |x|. 1. Introduction In this article we consider the problem ∆U +K(|x|)f(U) = 0, x ∈ RN\BR, (1.1) U = 0 on ∂(RN\BR), (1.2) U → 0 as |x| → ∞ (1.3) where U : RN → R with N > 2, BR is the ball of radius R > 0 centered at the origin in RN and K(x) > 0. We use the following assumptions: (H1) f : R \ {0} → R and f odd, locally Lipschitz, and there exists β > 0 such that f < 0 on (0, β), f > 0 on (β,∞). (H2) f(U) = −1 |U |q−1U + g1(U) for small U 6= 0, 0 < q < 1, g1 is locally Lipschitz on R, g1(0) = 0. (H3) f(U) = |U |p−1U + g2(U) for large U where p > 1 and limU→+∞ g2(U) |U |p = 0. Now let F (U) = ∫ U 0 f(s) ds. Since f is odd it follows that F is even, and from (H2) it follows that f is integrable near U = 0. Thus F is continuous and F (0) = 0. It also follows that F is bounded below by −F0 with F0 > 0. (H4) there exists γ with 0 < β < γ such that F < 0 on (0, γ), F > 0 on (γ,∞), F > −F0 on R. (H5) K and K ′ are continuous on [R,∞) with K(r) > 0, 2(N − 1) + rK′ K < 0, there exists α such that α > 2(N − 1) and limr→∞ rK′ K = −α. (H6) There exist K1 > 0 and K2 > 0 such that K1 rα ≤ K(r) ≤ K2 rα on [R,∞). 2010 Mathematics Subject Classification. 34B40, 35B05. Key words and phrases. Exterior domain; semilinear equation; radial solution. c©2021. This work is licensed under a CC BY 4.0 license. Submitted November 2, 2020. Published August 10, 2021. 1 2 M. ALI, J. IAIA EJDE-2021/68 Since we are interested in studying radial solutions of (1.1)–(1.3), we rewrite these equations with r = |x|, U(r) = U(|x|) and see that U satisfies: U ′′(r) + N − 1 r U ′(r) +K(r)f(U(r)) = 0 on (R,∞), (1.4) U(R) = 0, lim r→∞ U(r) = 0. (1.5) Since f(U) is discontinuous at U = 0 it follows that U ′′ is not continuous at any point where U = 0. However we will see that U,U ′ are continuous on [R,∞) and satisfy rN−1U ′(r) = ∫ ∞ r sN−1K(s)f(U(s)) ds. (1.6) In this article we prove the following result. Theorem 1.1. Assuming (H1)–(H6) hold and N > 2, there exist an infinite number of nontrivial radial solutions of (1.5) and (1.6). In addition, for each nonnegative integer n, there is a solution of (1.5) and (1.6) with exactly n zeros on (0, R2−N ). The existence of a positive solution of (1.1) on RN with K(r) ≡ 1 has been studied extensively [2, 3, 9, 12]. Recently the exterior domain RN\BR(0) has been studied in [6, 7, 8, 10, 11, 13]. In addition, f(U) = −|U |q−1U + |U |p−1U with (1 < q < p) was studied in [11]. f(U) = |U |q−1U + g(U) with (1 < p < q + 1) was studied in [1]. Also f(U) = −|U |−q−1U + g(U) with (0 < q < 1 < p) was studied in [12]. 2. Preliminaries We first prove the existence of a solution of (1.4) with U(R) = 0 and U ′(R) = a > 0 (2.1) on some neighborhood to the right of R. We denote this solution by Ua(r) to emphasize the dependence on the initial parameter a. To prove existence of (1.4), (2.1) we make the change of variables Ua(r) = Va(r2−N ). (2.2) Then U ′a(r) = (2−N)r1−NV ′a(r2−N ), U ′′a (r) = (2−N)(1−N)r−NV ′a(r2−N ) + (2−N)2r2(1−N)V ′′a (r2−N ). Letting t = r2−N and r = t 1 2−N in (4), (7) we obtain V ′′a (t) + h(t)f(Va(t)) = 0 on (0, R2−N ), (2.3) Va(R2−N ) = 0, V ′a(R2−N ) = −aRN−1 N − 2 < 0, (2.4) where from (H5) and (H6), h(t) = 1 (N − 2)2 t 2(N−1) 2−N K(t 1 2−N ) ∼ tα̃ (N − 2)2 , α̃ = α− 2(N − 1) N − 2 > 0 (2.5) on (0, R2−N ). Also from (H5) and (H6) it follows that there are constants h1, h2 with 0 < h1 ≤ h2 such that h′(t) > 0, h1t α̃ ≤ h(t) ≤ h2tα̃ on (0, R2−N ). (2.6) EJDE-2021/68 INFINITELY MANY SOLUTIONS 3 For the existence of a solution of (2.3) on (R2−N − ε, R2−N ) with (2.4) for some ε > 0 we proceed as follows. First, integrate (2.3) on (t, R2−N ) and use (2.4). This gives − V ′a(t) = aRN−1 N − 2 − ∫ R2−N t h(s)f(Va(s)) ds. (2.7) Integrating again over (t, R2−N ) and using (2.4) gives Va(t) = aRN−1 N − 2 (R2−N − t)− ∫ R2−N t ∫ R2−N s h(x)f(Va(x)) dx ds. (2.8) Now let W (t) = Va(t) R2−N−t so Va(t) = (R2−N − t)W (t) and W (R2−N ) ≡ lim t→(R2−N )− Va(t) R2−N − t = −V ′a(R2−N ) = aRN−1 N − 2 . Rewriting (2.8) we have W (t) = aRN−1 N − 2 − 1 R2−N − t ∫ R2−N t ∫ R2−N s h(x)f ( (R2−N − x)W (x) ) dx ds. (2.9) We now solve this equation on [R2−N − ε, R2−N ] by a fixed point method. Let a > 0, 0 < ε < 1, and let us define S = { W ∈ C[R2−N − ε, R2−N ] : W (R2−N ) = aRN−1 N − 2 , |W (t)− aRN−1 N − 2 | ≤ aRN−1 2(N − 2) on [R2−N − ε, R2−N ] } where C[R2−N − ε, R2−N ] is the set of real-valued continuous functions on [R2−N − ε, R2−N ]. Let ‖W‖ = sup x∈[R2−N−ε,R2−N ] |W (x)|. Then (S, ‖ ·‖) is a Banach space. Now let us define a map T on S by TW (R2−N ) = aRN−1 N−2 and TW (t) = aRN−1 N − 2 − 1 R2−N − t ∫ R2−N t ∫ R2−N s h(x)f ( (R2−N − x)W (x) ) dx ds (2.10) on (R2−N − ε, R2−N ). Since W (x) ∈ S and 0 < ε < 1 we have 0 < aRN−1 2(N − 2) ≤W (x) ≤ 3aRN−1 2(N − 2) on [R2−N − ε, R2−N ]. (2.11) From (H2) we see g1(x) is locally Lipschitz and g1(0) = 0 therefore it follows that |g1((R2−N − x)W (x))| ≤ L|R2−N − x‖W (x)| (2.12) where L is the Lipschitz constant for g1 on [0, 3aR N−1 2(N−2) ]. It follows from (2.11) that | −1 (R2−N − x)qW q(x) | ≤ 2q(N − 2)q(R2−N − x)−q aq(RN−1)q (2.13) 4 M. ALI, J. IAIA EJDE-2021/68 and using (2.6), (2.12), and (2.13) we see that |h(x)f((R2−N − x)W (x))| = ∣∣h(x) ( −1 (R2−N − x)qW q(x) + g1((R2−N − x)W (x)) )∣∣ ≤ h(R2−N ) [∣∣2q(N − 2)q(R2−N − x)−q aq(RN−1)q ∣∣+ L ∣∣(R2−N − x) 3aRN−1 2(N − 2) ∣∣]. (2.14) Integrating once we obtain∫ R2−N t |h(x)f ( (R2−N − x)W (x) ) | dx ≤ h(R2−N ) [C1 aq (R2−N − t)1−q + C2a(R2−N − t)2 ] (2.15) where C1 = 2q(N − 2)q (RN−1)q(1− q) , C2 = 3LRN−1 4(N − 2) . Thus from (2.15) we have∫ R2−N t |h(x)f ( (R2−N − x)W (x) ) | dx→ 0 as t→ (R2−N )−. (2.16) Next integrating (2.15) on (t, R2−N ) and dividing by (R2−N − t) we obtain 1 R2−N − t ∫ R2−N t ∫ R2−N s |h(x)f ( (R2−N − x)W (x) ) | dx ds ≤ h(R2−N ) [C3(R2−N − t)1−q aq + aC4(R2−N − t)2 ] (2.17) where C3 = C1 2−q and C4 = C2 3 . Thus from (2.17) we see that lim t→(R2−N )− 1 R2−N − t ∫ R2−N t ∫ R2−N s |h(x)f ( (R2−N − x)W (x) ) | dx ds = 0. (2.18) Now we show that T : S → S is a contraction mapping with T (W ) ∈ S for each W ∈ S if ε > 0 is sufficiently small. First, let W ∈ S and so it follows from (2.17) and (2.18) that 1 R2−N − t ∫ R2−N t ∫ R2−N s h(x)f ( (R2−N − x)W (x) ) dx ds is continuous on [R2−N − ε, R2−N ]. Then from (2.10), (2.17), and (2.18) we see that limt→(R2−N )− TW (t) = aRN−1 N−2 , |TW (t)− aRN−1 N − 2 | ≤ aRN−1 2(N − 2) on [R2−N − ε, R2−N ] and TW is continuous if ε > 0 is sufficiently small. Thus T : S → S if ε is sufficiently small. We next prove that T is a contraction mapping if ε is sufficiently small. Let W1,W2 ∈ S. Then TW1(t)− TW2(t) = − 1 R2−N − t ∫ R2−N t ∫ R2−N s h(x) [ f((R2−N − x)W1(x)) − f((R2−N − x)W2(x)) ] dx ds. (2.19) EJDE-2021/68 INFINITELY MANY SOLUTIONS 5 By (H2) we have f((R2−N − x)W (x)) = −(R2−N − x)−qW−q(x) + g1((R2−N − x)W (x)) where 0 < q < 1. Then by (2.12) and (2.13) we first estimate |f((R2−N − x)W1)− f((R2−N − x)W2)| = ∣∣ −1 (R2−N − x)q [ 1 W q 1 − 1 W q 2 ] + g1((R2−N − x)W1)− g1((R2−N − x)W2) ∣∣ ≤ 1 (R2−N − x)q | 1 W1 q − 1 W2 q |+ L(R2−N − x)|W1 −W2| (2.20) where L is again the Lipschitz constant for g1 on [0, 3aR N−1 2(N−2) ]. Next applying the mean value theorem we see that the right-hand side of (2.20) is equal to 1 (R2−N − x)q [ q W q+1 3 |W1 −W2| ] + L(R2−N − x)|W1 −W2|, where W3 is between W1 and W2. Since Wi ∈ S for i = 1, 2, 3, and |Wi− aRN−1 N−2 | ≤ aRN−1 2(N−2) then aRN−1 2(N−2) ≤ Wi ≤ 3aRN−1 2(N−2) on [R2−N − ε, R2−N ]. Therefore W3 q+1 ≥( aRN−1 2(N−2) )q+1 , and so on [R2−N − ε, R2−N ] we have |f((R2−N − x)W1)− f((R2−N − x)W2)| ≤ |W1 −W2| [ q (R2−N − x)q (2(N − 2) aRN−1 )q+1 + L(R2−N − x) ] . (2.21) Recalling from (2.5) that h(t) is positive, continuous and increasing on (0, R2−N ], with α > 2(N − 1) we see that |TW1 − TW2| ≤ h(R2−N ) R2−N − t ∫ R2−N t ∫ R2−N s |W1 −W2| [ q (R2−N − x)q (2(N − 2) aRN−1 )q+1 + L(R2−N − x) ] dx ds ≤ h(R2−N ) R2−N − t ‖W1 −W2‖ ∫ R2−N t ∫ R2−N s [ q (R2−N − x)q (2(N − 2) aRN−1 )q+1 + L(R2−N − x) ] dx ds ≤ h(R2−N )‖W1 −W2‖ [C5ε 1−q aq+1 + C6ε 2 ] = C7,ε‖W1 −W2‖. (2.22) where C5 = q (2− q)(1− q) (2(N − 2) RN−1 )q+1 , C6 = L 6 , C7,ε = h(R2−N ) [C5ε 1−q aq+1 +C6ε 2 ] . Since lim ε→0+ C7,ε = lim ε→0+ h(R2−N ) [C5ε 1−q aq+1 + C6ε 2 ] = 0, for ε sufficiently small we see that 0 < C7,ε < 1, and therefore it follows from (2.22) that T is a contraction. Then by the contraction mapping principle on S [4] we see there exists a unique solution W ∈ S to TW = W on [R2−N − ε, R2−N ] for some 6 M. ALI, J. IAIA EJDE-2021/68 ε > 0. Then Va(t) = (R2−N − t)W (t) is a solution of (2.3) and satisfies (2.4) for some ε > 0. Now define the energy of solutions to (2.3) and (2.4) as Ea(t) = 1 2 V ′2a (t) h(t) + F (Va(t)). (2.23) Differentiating Ea, using (2.3), and using that from (2.6) that h′(t) > 0, we have E′a(t) = −V ′2 a (t)h′(t) 2h2(t) ≤ 0. (2.24) Thus Ea is non-increasing where it is defined. Therefore for these t with t < R2−N we have 0 < 1 2 a2R2(N−1) (N − 2)2h(R2−N ) = Ea(R2−N ) ≤ Ea(t) = 1 2 V ′2a (t) h(t) + F (Va(t)). (2.25) Remark 2.1. It follows from (2.3) that if Va(t0) 6= 0 then V ′′a (t0) is defined and V ′′a is continuous in a neighborhood of t0. We also note if Va is a solution of (2.7) and there exists a Za ∈ (0, R2−N ] such that Va(Za) = 0, then from (2.25) we see 0 < Ea(Za) = 1 2 V ′2a (Za) h(t) and so V ′a(Za) 6= 0. We also observe that if Va(Z0) = 0 then it follows from (2.3) and (H2) that V ′′a (Z0) is undefined and that limt→Z0 + |V ′′a (t)| = ∞. Therefore due to these considerations for the rest of this paper we will seek functions Va that are continuously differentiable on [0, R2−N ] and satisfy (2.7). Lemma 2.2. Assume -(H1)-(H6) hold, N > 2, and a > 0. Let Va(t) be the solution of (2.7) on (R2−N−ε, R2−N ) whose existence we have just proved. Then Va and V ′a are defined and continuous on [0, R2−N ]. Also |V ′a(t)| ≤ aRN−1 N−2 + √ 2F0h(R2−N ) on [0, R2−N ], |Va(t)| ≤ aR N−2 + R2−N √ 2F0h(R2−N ) on [0, R2−N ], and Va(t) satisfies (2.7) on [0, R2−N ]. Proof. It follows from (2.3) that(1 2 V ′2a (t) + h(t)F (Va(t)) )′ = h′(t)F (Va(t)). (2.26) Integrating from t to R2−N and using (2.4) yields −1 2 V ′2a (t)− h(t)F (Va(t)) = −1 2 a2R2(N−1) (N − 2)2 + ∫ R2−N t h′(s)F (Va(s)) ds. Since −F0 < F by (H4) and h > 0, h′ > 0 by (2.6) then hF0 ≥ −hF thus −1 2 V ′2a (t) + h(t)F0 ≥ − 1 2 V ′2a (t)− h(t)F (Va(t)) = −1 2 a2R2(N−1) (N − 2)2 + ∫ R2−N t h′(s)F (Va(s)) ds ≥ −1 2 a2R2(N−1) (N − 2)2 − F0 ∫ R2−N t h′(s) ds = −1 2 a2R2(N−1) (N − 2)2 − F0 ( h(R2−N )− h(t) ) . EJDE-2021/68 INFINITELY MANY SOLUTIONS 7 Therefore, V ′2a (t) ≤ a2R2(N−1) (N − 2)2 + 2F0h(R2−N ). Finally since √ x+ y ≤ √ x+ √ y for x ≥ 0 and y ≥ 0 we see that |V ′a(t)| ≤ aRN−1 N − 2 + √ 2F0h(R2−N ). (2.27) Integrating on (t, R2−N ) and using (2.3), (2.4) we obtain |Va(t)| = ∣∣ ∫ R2−N t V ′a(s) ds ∣∣ ≤ ∫ R2−N t |V ′a(s)| ds ≤ ∫ R2−N t (aRN−1 N − 2 + √ 2F0h(R2−N ) ) ds = (R2−N − t) (aRN−1 N − 2 + √ 2F0h(R2−N ) ) ≤ aR N − 2 +R2−N √ 2F0h(R2−N ). (2.28) From (2.27) and (2.28) it follows that Va and V ′a are bounded where they are defined and hence Va, V ′ a exist on [0, R2−N ] and V ′a satisfies (2.7) on [0, R2−N ]. This completes the proof of Lemma 2.2. � Lemma 2.3. Assume (H1)–(H6) hold, N > 2, a > 0, and Va(t) solves (13). Then the solutions Va(t) depend continuously on the parameter a > 0 on [0, R2−N ]. Proof. First, let 0 < a1 < a2. It follows from (2.27) and (2.28) that V ′a and Va are bounded on [0, R2−N ] and these upper bounds can be chosen to be independent of a for 0 < a1 ≤ a ≤ a2. Then from (2.27) and (2.28) we have |V ′a(t)| ≤ C8a2 + C9 on [0, R2−N ] ∀a with 0 < a1 ≤ a ≤ a2 (2.29) where C8 = R2−N N−2 , C9 = √ 2F0h(R2−N ), and |Va(t)| ≤ C10a2 + C11 on [0, R2−N ] ∀a with 0 < a1 ≤ a ≤ a2 (2.30) where C10 = R N−2 and C11 = R2−NC9. Thus we see that |V ′a| and |Va| are uniformly bounded on [0, R2−N ] for all a with 0 < a1 ≤ a ≤ a2. Next, we suppose there exists a∗ > 0, and we want to show that Va → Va∗ uniformly on [0, R2−N ] as a → a∗. By way of contradiction suppose not. Then there exist aj such that aj → a∗ as j →∞, tj ∈ [0, R2−N ] and there is an ε0 > 0 such that |Vaj (tj)− Va∗(tj)| ≥ ε0 ∀j. (2.31) Since aj → a∗ as j →∞ then if j is sufficiently large we have |aj | ≤ a∗ + 1 and by (2.29), (2.30) we see that Va and V ′a are uniformly bounded and therefore equicon- tinuous on [0, R2−N ]. Then by the Arzela-Ascoli theorem there is a subsequence ajl , of Vaj such that Vajl → V ∗a uniformly on [0, R2−N ]. So as l→∞, 0← |Vajl (tjl)− Va∗(tjl)| ≥ ε0 which is impossible. Thus Va varies continuously with a on [0, R2−N ] for all a with 0 < a1 ≤ a ≤ a2. This completes the proof of Lemma 2.3. � 8 M. ALI, J. IAIA EJDE-2021/68 Lemma 2.4. Assume (H1)–(H6), N > 2, and let Va(t) be the solution of (2.7). If a is sufficiently large then Va(t) has a local maximum, Ma, and a zero, Za, with 0 < Za < Ma < R2−N . Further Va(Ma) → ∞, Ma → R2−N , Za → R2−N , and |V ′a(Za)| → ∞ as a→∞. Proof. We first show that if a is sufficiently large then there exists ta,γ > 0 such that Va(ta,γ) = γ and 0 < Va < γ on (ta,γ , R 2−N ). Suppose not. Then 0 < Va(t) < γ on (0, R2−N ) and all sufficiently large a. Since Ea is non-increasing on 0 < t < R2−N and |Va| < γ then F (Va) < 0 and from (2.25) it follows that 1 2 V ′2a (t) h(t) ≥ 1 2 V ′2a (t) h(t) + F (Va(t)) ≥ 1 2 a2R2(N−1) (N − 2)2h(R2−N ) > 0. (2.32) Thus V ′a < 0 on (t, R2−N ) and we obtain − V ′a(t) ≥ aRN−1 (N − 2) √ h(R2−N ) √ h(t). (2.33) Integrating (2.33) from t to R2−N gives Va(t) = ∫ R2−N t −V ′a(s) ds ≥ ∫ R2−N t aRN−1 (N − 2) √ h(R2−N ) √ h(s) ds. (2.34) Evaluating this expression at t = 0 we obtain γ ≥ Va(0) ≥ aRN−1 (N − 2) √ h(R2−N ) ∫ R2−N 0 √ h(s) ds. (2.35) The right-hand side approaches infinity as a goes to infinity which contradicts the assumption that the left-hand side is bounded by γ. Thus Va gets larger than γ as a → ∞ and so there exists ta,γ with 0 < ta,γ < R2−N such that Va(ta,γ) = γ and 0 < Va(t) < γ on (ta,γ , R 2−N ). In addition, evaluating (2.34) at t = ta,γ we obtain γ = Va(ta,γ) ≥ aRN−1 (N − 2) √ h(R2−N ) ∫ R2−N ta,γ √ h(s) ds. (2.36) Thus we see that ta,γ → R2−N as a→∞. (2.37) It then follows immediately that there is ta,β such that ta,γ < ta,β < R2−N and Va(ta,β) = β. Since ta,γ → R2−N as a→∞ then it follows that ta,β → R2−N as a→∞. (2.38) Next we show that if Va is decreasing for all t ∈ [ 12R 2−N , R2−N ] then we have lima→∞ Va ( 1 2R 2−N) = ∞. We suppose by the way of contradiction that Va ( 1 2R 2−N) ≤ A where A > 0 does not depend on a for a large. For 1 2R 2−N ≤ t ≤ R2−N it follows that there exists B > 0 such that F (Va) < B on [ 12R 2−N , R2−N ] and all large a. Since Ea is non-increasing, 1 2 V ′2a (t) h(t) +B ≥ 1 2 V ′2a (t) h(t) + F (Va(t)) = Ea(t) ≥ Ea(R2−N ) = 1 2 a2R2(N−1) (N − 2)2h(R2−N ) on [R 2−N 2 , R2−N ]. Rewriting the above expression we have −V ′a(t) ≥ √ a2R2(N−1) (N − 2)2h(R2−N ) − 2B √ h(t) on [R2−N 2 , R2−N]. EJDE-2021/68 INFINITELY MANY SOLUTIONS 9 Integrating this on (t, R2−N ) we obtain: Va(t) ≥ √ a2R2(N−1) (N − 2)2h(R2−N ) − 2B ∫ R2−N t √ h(s) ds. (2.39) Now evaluating (2.39) at t = R2−N 2 we have A ≥ Va (R2−N 2 ) ≥ √ a2R2(N−1) (N − 2)2h(R2−N ) − 2B ∫ R2−N R2−N 2 √ h(s) ds. (2.40) As a→∞, the right-hand side aappraoches infinity, which is a contradiction since we were assuming A is finite. Thus lim a→∞ Va (1 2 R2−N ) =∞ if Va is decreasing on [ R2−N 2 , R2−N ]. (2.41) We next show that if Va is decreasing on [R 2−N 2 , R2−N ] then Va ( 3R2−N 4 ) → ∞ as a→∞. From (2.38) we know ta,β → R2−N as a→∞ so for a sufficiently large we have R2−N 2 ≤ ta,β and Va(t) > β on [R 2−N 2 , ta,β). From (2.3) and (H3) we see that V ′′a (t) < 0 on [R 2−N 2 , ta,β) for sufficiently large a. Thus Va(t) is concave down here so we have for 0 ≤ λ ≤ 1, Va ( λ R2−N 2 + (1− λ)ta,β ) ≥ λVa (R2−N 2 ) + (1− λ)Va(ta,β) = λVa (R2−N 2 ) + (1− λ)β ≥ λVa (R2−N 2 ) . Now for t ∈ [R 2−N 2 , ta,β ] we can write t = λ R2−N 2 + (1− λ)ta,β , i.e. λ = ta,β − t ta,β − R2−N 2 and thus 0 ≤ λ ≤ 1, and we obtain Va(t) ≥ ta,β − t ta,β − R2−N 2 Va (R2−N 2 ) on [ R2−N 2 , ta,β ]. (2.42) Evaluating at t = 3R2−N 4 gives Va (3R2−N 4 ) ≥ ta,β − 3R2−N 4 ta,β − R2−N 2 Va (R2−N 2 ) . (2.43) From (2.38) we saw that ta,β → R2−N as a → ∞ thus for sufficiently large a we have ta,β− 3R2−N 4 ta,β−R 2−N 2 ≥ 1 3 and therefore (50) along with (2.41) gives Va (3R2−N 4 ) ≥ 1 3 Va (R2−N 2 ) →∞ as a→∞. (2.44) Now let us show that Va(t) has a local maximum Ma on [R 2−N 2 , R2−N ] if a is sufficiently large. Suppose not. Then Va(t) is decreasing on [R 2−N 2 , R2−N ]. 10 M. ALI, J. IAIA EJDE-2021/68 Next let Ia = min [ 12R 2−N , 34R 2−N ] h(t)f(Va(t)) Va(t) . (2.45) Since h(t) > 0 is bounded from below on [ 12R 2−N , 34R 2−N ] then there is an h0 > 0 such that h(t) > h0 on [ 12R 2−N , 3R 2−N 4 ]. Since we are assuming Va is decreasing on [ 12R 2−N , 3R 2−N 4 ] for all a > 0 sufficiently large and since by (2.44) we have Va ( 3R2−N 4 ) → ∞ as a → ∞, it therefore follows that Va → ∞ uniformly on [ 12R 2−N , 3R 2−N 4 ]. By (H3) it then follows for sufficiently large a that f(Va) Va ≥ 1 2Va p−1 and therefore Ia = min [ 12R 2−N , 34R 2−N ] h(t)f(Va) Va ≥ h0 min [ 12R 2−N , 34R 2−N ] f(Va) Va ≥ h0 2 min [ 12R 2−N , 34R 2−N ] V p−1a ≥ h0 2 Va p−1 (3R2−N 4 ) . By (2.44) the right-hand side goes to infinity, and thus we obtain lim a→∞ Ia =∞. (2.46) Now we apply the Sturm Comparison theorem [5] on [ 12R 2−N , 3R 2−N 4 ]. Consider V ′′a + [h(t)f(Va) Va ] Va = 0, (2.47) W ′′a + IaWa = 0 (2.48) where β < Va (3 4 R2−N ) = Wa (3 4 R2−N ) , (2.49) V ′a (3 4 R2−N ) = W ′a (3 4 R2−N ) < 0. (2.50) SinceW ′′a +IaWa = 0 andWa 6≡ 0, it follows thatWa = C12 sin( √ Iat)+C13 cos( √ Iat) where C12 and C13 are not both zero. It is well-known that any interval of length π√ Ia has a zero of Wa and so it follows that Wa has a local maximum M̃a ∈ [ 34R 2−N − π√ Ia , 34R 2−N ] and Wa is decreasing on [M̃a, 3 4R 2−N ]. Also for a sufficiently large then from (2.47), 3 4R 2−N − π√ Ia > 1 2R 2−N . Multiplying (2.47) by Wa, (2.48) by Va, and subtracting we obtain (WaV ′ a − VaW ′a)′ + (h(t)f(Va) Va − Ia ) VaWa = 0. (2.51) Using (2.49), (2.50) and since Wa has a local maximum M̃a then integrating (2.51) on [M̃a, 3 4R 2−N ] we obtain −Wa(M̃a)V ′a(M̃a) + ∫ 3 4R 2−N M̃a (h(t)f(Va) Va − Ia ) VaWa = 0. (2.52) EJDE-2021/68 INFINITELY MANY SOLUTIONS 11 Since Wa(M̃a) ≥Wa ( 3 4R 2−N) > β > 0 by (2.49) and (h(t)f(Va) Va − Ia ) VaWa ≥ 0 on [M̃a, 3 4R 2−N ] then ∫ 3 4R 2−N M̃a (h(t)f(Va) Va −Ia ) VaWa > 0 and so it follows that V ′a(M̃a) > 0 which is a contradiction to the assumption that V ′a(t) < 0 on [R 2−N 2 , R2−N ). Thus Va(t) must have a local maximum, Ma, with 1 2R 2−N < Ma < R2−N and Va decreasing on (Ma, R 2−N ) if a is sufficiently large. Now let us show that Va(Ma) → ∞ as a → ∞. Suppose by the way of the contradiction that there exists a constant C14 > 0 independent of a such that Va(Ma) < C14 and so Va(t) < C14 on (Ma, R 2−N ). Integrating (2.3) on (Ma, R 2−N ) and using (2.4) gives∫ R2−N Ma V ′′a (t) dt+ ∫ R2−N Ma h(t)f(Va(t)) dt = 0. Therefore aR2−N N − 2 = ∫ R2−N Ma h(t)f(Va(t)) dt = ∫ R2−N Ma h(t)(−Va−q(t)) dt+ ∫ R2−N Ma h(t)g1(Va(t)) dt ≤ ∫ R2−N Ma h(t)g1(Va(t)) dt. (2.53) Since 0 ≤ Va(t) ≤ Va(Ma) ≤ C14 and g1 is continuous, g1(Va) ≤ C15 for some constant C15 > 0 on [Ma, R 2−N ], and since h(t) ≤ h2t α̃ (by (2.4)), estimating (2.53) gives aR2−N N − 2 ≤ h2C15 1 + α̃ [ (R2−N )1+α̃ −M1+α̃ a ] ≤ h2C15 1 + α̃ (R2−N )1+α̃. (2.54) The left-hand side of (2.54) goes to +∞ as a → ∞ but the right-hand side is bounded which contradicts the assumption that 0 ≤ Va(Ma) ≤ C14. Thus Va(Ma)→∞ as a→∞. (2.55) Now let us show that lima→∞Ma = R2−N . Since V ′′a (t) ≤ 0 on (Ma, ta,β) then Va is concave down here and so we obtain Va(λMa + (1− λ)ta,β) ≥ λVa(Ma) + (1− λ)β (2.56) where 0 ≤ λ ≤ 1. Letting λ = 1/2 gives Va (Ma + ta,β 2 ) ≥ 1 2 Va(Ma) + 1 2 β = Va(Ma) + β 2 . (2.57) From (2.55) we know that Va(Ma)→∞ as a→∞ so then (2.57) implies Va (Ma + ta,β 2 ) →∞ as a→∞. (2.58) Since Va is decreasing on [Ma, Ma+ta,β 2 ] it follows that Va → ∞ uniformly on [Ma, Ma+ta,β 2 ] for sufficiently large a. Since f(Va(t)) ≥ 1 2V p a (t) for Va large by (H3), from (2.3) −V ′′a (t) ≥ f(Va(t)) ≥ 1 2h(t)V pa (t) on [Ma, Ma+ta,β 2 ]. Since Va is decreasing on (Ma, t), integrating from Ma to t where Ma ≤ t ≤ Ma+ta,β 2 we obtain −V ′a(t) = −V ′a(t) + V ′a(Ma) 12 M. ALI, J. IAIA EJDE-2021/68 = ∫ t Ma −V ′′a (s) ds ≥ 1 2 ∫ t Ma h(s)V pa (s)ds ≥ 1 2 V pa (t) ∫ t Ma h(s) ds. Therefore, −V ′a(t) V pa (t) ≥ 1 2 ∫ t Ma h(s) ds. (2.59) Integrating on (Ma, t) gives 1 (p− 1)V p−1a (t) ≥ 1 p− 1 [V 1−p a (t)− V 1−p a (Ma)] ≥ 1 2 ∫ t Ma ∫ s Ma h(x) dx ds. (2.60) Evaluating at t = Ma+ta,β 2 gives 1 (p− 1)V p−1a ( Ma+ta,β 2 ) ≥ 1 2 ∫ Ma+ta,β 2 Ma ∫ x Ma h(x) dx ds. (2.61) The left-hand side goes to zero as a → ∞ by (2.58). Since we saw in (2.38) ta,β → R2−N as a→∞ and h(s) is continuous and positive, it follows that Ma → R2−N as a→∞. (2.62) Next we show there is a Za ∈ (0,Ma) such that Va(Za) = 0, Va(t) > 0 on (Za, R 2−N ), and Za → R2−N as a → ∞. Moreover V ′a(Za) → −∞ as a → ∞. Again we do this by contradiction. Let us assume Va(t) > 0 on (0,Ma). Since Ea(t) is non-increasing then we have F (Va(Ma)) ≤ 1 2 V ′2a h(t) + F (Va(t)) for 0 ≤ t ≤Ma. (2.63) Now if Va has a positive local minimum ma, then V ′′a (ma) ≥ 0 so f(Va(ma)) ≤ 0 so 0 < Va(ma) ≤ β but also 0 < Ea(ma) = F (Va(ma)) so Va(ma) > γ ≥ β which is a contradiction. Thus V ′a > 0 on (0,Ma). Rewriting, integrating (2.63) over [Ma 2 ,Ma], using (2.5), and making a change of variables gives∫ Va(Ma) 0 ds√ F (Va(Ma))− F (s) ≥ ∫ Va(Ma) Va(Ma2 ) ds√ F (Va(Ma))− F (s) = ∫ Ma Ma 2 |V ′a(t)| dt√ F (Va(Ma))− F (Va(t)) ≥ ∫ Ma Ma 2 √ 2h(s) ds ≥ ∫ Ma Ma 2 √ 2h1s α̃/2 ds = √ 2h1(1− 1 21+ α̃ 2 ) 1 + α̃ 2 M 1+ α̃ 2 a . (2.64) EJDE-2021/68 INFINITELY MANY SOLUTIONS 13 Now we estimate the left-hand side. It follows from (H3) that f(U) ≥ 1 2U p for U sufficiently large therefore for U large enough we see that min[ 12U,U ] f ≥ 1 2p+1U p and since p > 1, it follows that lim U→∞ U min[ 12U,U ] f = 0. (2.65) We now estimate the integral on the left-hand side of (2.64) when s ∈ [0, Va(Ma) 2 ] and a is sufficiently large. We then have F (s) < F (Va(Ma) 2 ) for all s ∈ (0, Va(Ma) 2 ) and thus F (Va(Ma))− F (Va(Ma) 2 ) < F (Va(Ma))− F (s) so ∫ Va(Ma) 2 0 ds√ F (Va(Ma))− F (s) ≤ ∫ Va(Ma) 2 0 ds√ F (Va(Ma))− F (Va(Ma) 2 ) = Va(Ma) 2√ F (Va(Ma))− F (Va(Ma) 2 ) . (2.66) By the mean value theorem there is a d1 > 0 such that Va(Ma) 2 < d1 < Va(Ma) and F (Va(Ma))− F ( Va(Ma) 2 ) = f(d1)[Va(Ma)− Va(Ma) 2 = f(d1)[ Va(Ma) 2 ] ≥ [ min [ Va(Ma) 2 ,Va(Ma)] f ]Va(Ma) 2 so Va(Ma) 2√ F (Va(Ma))− F (Va(Ma) 2 ) ≤ √ Va(Ma) 2√ min [ Va(Ma) 2 ,Va(Ma)] f ≤ 1√ 2 √ Va(Ma) min [ Va(Ma) 2 ,Va(Ma)] f → 0 (2.67) as a→∞, by (2.65). Thus by (2.66) and (2.67) we see that lim a→∞ ∫ Va(Ma) 2 0 ds√ 2 √ F (Va(Ma))− F (s) = 0. (2.68) Next, we estimate the integral on the left-hand side of (2.64) for s ∈ [Va(Ma) 2 , Va(Ma)]. By the mean value theorem there is a d2 > 0 with Va(Ma) 2 < d2 < Va(Ma) such that F (Va(Ma))− F (s) = f(d2)[Va(Ma)− s] ≥ [ min [ Va(Ma) 2 ,Va(Ma)] f ] [Va(Ma)− s]. 14 M. ALI, J. IAIA EJDE-2021/68 Therefore, ∫ Va(Ma) Va(Ma) 2 ds√ F (Va(Ma))− F (s) ≤ ∫ Va(Ma) Va(Ma) 2 ds√ [min [ Va(Ma) 2 ,Va(Ma)] f ][Va(Ma)− s] = √ 2 √ Va(Ma) min [ Va(Ma) 2 ,Va(Ma)] f . (2.69) Thus by (2.65) we see that lim a→∞ ∫ Va(Ma) Va(Ma) 2 dt√ 2 √ F (Va(Ma))− F (s) = 0. (2.70) Combining (2.67) and (2.70) we have lim a→∞ ∫ Va(Ma) 0 ds√ 2 √ F (Va(Ma))− F (s) = 0. (2.71) Thus the left-hand side of (2.64) goes to 0 as a → ∞ but the right-hand side of (2.64) does not because by (2.62) we know Ma → R2−N as a → ∞ and so we get a contradiction. Thus for a sufficiently large Va(t) has a first zero, Za, with Va(Za) = 0 and Va(t) > 0 on (Za, R 2−N ). Similarly rewriting (2.63) and integrating on (Za,Ma) we obtain∫ Va(Ma) 0 ds√ 2 √ F (Va(Ma))− F (s) ≥ √ h1 (Ma 1+ α̃ 2 − Za1+ α̃ 2 1 + α̃ 2 ) . (2.72) Since the left-hand side approaches 0 as a → ∞ (by(2.71)), we see Ma 1+ α̃ 2 − Za 1+ α̃ 2 → 0 as a → ∞. Also since we know from (2.62) that Ma → R2−N as a→∞ this then implies that Za → R2−N as a→∞. Finally we show that V ′a(Za) → +∞ as a → ∞. Since Za → R2−N as a → ∞ and Ea(t) is non-increasing, since 0 < Za ≤Ma we have 0 < F (Va(Ma)) = Ea(Ma) ≤ Ea(Za) = 1 2 V ′a 2 (Za) h(Za) and so rewriting this inequality gives 2h(Za)F (Va(Ma)) ≤ V ′a 2 (Za). (2.73) As a→∞ the left-hand side appraoches∞ because lima→∞ h(Za) = h(R2−N ) > 0 and lima→∞ F (Va(Ma)) = ∞ by (2.55). Thus V ′a 2 (Za) → ∞ as a → ∞ and thus it follows that V ′a(Za) → +∞ as a → ∞. In similar way if a > 0 is sufficiently large then Va(t) has a second zero Za,2 on (0, R2−N ) with Za,2 → R2−N as a→∞ and V ′a(Za,2) → −∞. More generally Va(t) has n zeros on (0, R2−N ) if a > 0 is sufficiently large. This completes the proof. � Lemma 2.5. Let Va(t) be the solution of (2.7), (H1)–(H6) hold, and N > 2. If R is sufficiently large then Va(t) > 0 for all t ∈ (0, R2−N ) if a sufficiently small. EJDE-2021/68 INFINITELY MANY SOLUTIONS 15 Proof. To reach a contradiction, suppose there is Za ∈ (0, R2−N ) such that Va(Za) = 0 for all a sufficiently small. Then there exists 0 < Ma < R2−N such that V ′a(Ma) = 0 and V ′a(t) < 0 on (Ma, R 2−N ). Also 0 < Ea(Ma) = F (Va(Ma)) so Va(Ma) > γ. Then by Lemma 2.2 we see that |V ′a(t)| ≤ aR2−N N−2 + √ 2F0h(R2−N ), and since Va(t) is decreasing on (Ma, R 2−N ) this gives − V ′a(t) ≤ aR2−N N − 2 + √ 2F0h(R2−N ) on (Ma, R 2−N ). (2.74) Integrating from t to R2−N and using (2.4) we obtain: Va(t) ≤ (aR2−N N − 2 + √ 2F0h(R2−N ) ) (R2−N−t) ≤ (aR2−N N − 2 + √ 2F0h(R2−N ) ) R2−N . Substituting t = Ma gives γ ≤ (aR2−N N − 2 + √ 2F0h(R2−N ) ) R2−N . Taking the limit as a→ 0+ we obtain γ ≤ √ 2F0h(R2−N )R2−N = √ 2F0h2(R2−N )α̃/2R2−N . (2.75) Then using (2.6) we obtain γ ≤ √ 2F0h2R 1−α2 where α > 2(N − 1). (2.76) Thus we see that the right-hand side of (2.76) is larger than γ for R sufficiently large but since α > 2 we see the right-hand side goes to 0 as R→∞ contradicting (2.76). Thus if R is sufficiently large then 0 < Va(t) < γ if a is sufficiently small. This completes the proof. � 3. Proof of the main Theorem 1.1 Lemma 3.1. Assume N > 2 and (H1)–(H6) hold. For a > 0 Let Va(t) be the solution of (2.7). Then Va(t) has at most a finite numbers of zeros on (0, R2−N ). Proof. Suppose by way of contradiction that there are distinct zero’s Zn ∈ (0, R2−N ) such that Va(Zn) = 0. Then either there is a decreasing subsequence (still labeled Zn) or an increasing subsequence and a Z∗ ∈ [0, R2−N ] such that Zn → Z∗ as n → ∞. By continuity Va(Z∗) = 0. Also since V ′a(R2−N ) < 0 there exists ε > 0 such that Va is not zero on (R2−N − ε, R2−N ) and thus Z∗ 6= R2−N . Therefore 0 ≤ Z∗ < R2−N . Without loss of generality assume Zn is decreasing. Then there is a local maximum or local minimum Mn of Va with Zn+1 < Mn < Zn so Mn → Z∗ as n → ∞ and notice also that since Ea(t) > 0 on [0, R2−N ] by (2.25) then Ea(Mn) = F (Va(Mn)) > 0 which implies that |Va(Mn)| > γ. Now by the mean value theorem, γ < |Va(Mn)| = |Va(Mn)− Va(Zn)| = |V ′a(cn)‖Mn − Zn|, (3.1) where cn 6= 0 and Mn < cn < Zn. Since Mn → Z∗ and Zn → Z∗ it follows that |Mn − Zn| → 0 as a → ∞. Also by (2.27) we see |V ′a(cn)| < aR2−N N−2 +√ 2F0h(R2−N ) < ∞. This implies that the right-hand side of (84) goes to zero which contradicts the fact that γ > 0. Thus Va has at most a finite numbers of zeros on (0, R2−N ). This completes the proof. � 16 M. ALI, J. IAIA EJDE-2021/68 Let Sn = { a > 0 : Va(t) has exactly n zeros on (0, R2−N ) } . By Lemma 3.1 we know that Sn is nonempty for some n. Let n0 ≥ 0 be the smallest non-negative integer n such that Sn 6= ∅ (so Sn0 6= ∅ and S0, S1, S2, . . . , Sn0−1 are all empty). By Lemma 2.3 it follows that Sn0 is bounded above. Therefore the supremum of Sn0 exists, and so we let an0 = supSn0 . If in addition R is sufficiently small then S0 6= ∅ by Lemma 2.4 and so n0 = 0. Lemma 3.2. Van(t) has exactly n zeros on (0, R2−N ) and Van(0) = 0 for all n ≥ n0. Proof. Since Sn0 is the smallest value of n such that Sn 6= ∅ this implies that Van0 (t) has at least n0 zeros on (0, R2−N ). Next we show that Van0 (t) has at most n0 zeros on (0, R2−N ). By way of contradiction, suppose there exists an (n0 + 1)st zero Z∗ with Z∗ ∈ (0, R2−N ) such that Van0 (Z∗) = 0 and 0 < Z∗ < Zn0 < · · · < Z1 < R2−N and suppose without loss of generality that Van0 > 0 on (0, Z∗). Since Ea is non- increasing then 0 < Ea(Z∗) = 1 2 V ′2an0 (Z∗) h(Z∗) which implies that V ′2an0 (Z∗) > 0. Since V ′an0 > 0 on (0, Z∗) it follows that V ′an0 (Z∗) < 0. So Van0 (Z∗ − δ) > 0 for δ > 0 sufficiently small. By continuity with respect to a it follows that if a < an0 then Va also has a (n0 + 1)st zero on (0, R2−N ) which is a contradiction to the definition of an0 . Therefore we see that Van0 (t) has exactly n0 zeros on (0, R2−N ). Now we denote Zan0 as the n0 th zero of Van0 (t). Then Van0 (t) 6= 0 if 0 < t < Zan0 . So without loss of generality we assume that Van0 < 0 on (0, Zan0 ). It follows by continuity of Van0 that Van0 (0) = limt→0+ Van0 (t) ≤ 0. Thus Van0 (0) ≤ 0. Next we show that Van0 (0) = 0. So suppose not. Then Van0 < 0 on [0, Zan0 ). From the remark before Lemma 2.2 we saw that V ′an0 (Z) 6= 0 if Van0 (Z) = 0. For an0+1 > a > an0 we see that |V ′a| ≤ |an0+1|R N−1 N−2 + √ 2F0h(R2−N ) by Lemma 2.2. It follows then that Va will also have n0 zeros on (0, R2−N ) if an0+1 > a > an0 . On the other hand, if a > an0 then by the definition of an0 we see that Va has at least (n0 + 1) zeros on (0, R2−N ) which is a contradiction. Thus the assumption that Van0 (0) < 0 is false and since Van0 (0) ≤ 0 then it follows that Van0 (0) = 0. Next let Sn0+1 = {a > 0 : Va(t) has exactly n0 + 1 zeros on (0, R2−N )}. For a slightly larger than an0 than Va has at least n0 + 1 zeros on (0, R2−N ) by definition of an0 . Next we show that Va(t) has at most n0 + 1 zeros on (0, R2−N ) if a is close to an0 and a > an0 . So suppose not and suppose that Va has an (n0 + 2)nd zero on (0, R2−N ). Then Va has a local maximum or a local minimum at some Ma where 0 < Zan0+2 < Ma < Zan0+1 and for a slightly larger than an0 . Also lima→an0 Va = Van0 uniformly on (0, R2−N ) and Zan0+1 → 0, hence Ma → 0 as a→ an0 . Since 0 < Ea(Ma) = F (Va(Ma)) it follows that |Va(Ma)| > γ > β so β ≤ |Va(Ma)| → |Van0 (0)| = 0 which is false. Thus if a > an0 and a is close to an0 then Va has at most n0 + 1 zeros on (0, R2−N ) and since we showed earlier Va has at least n0 + 1 zeros on (0, R2−N ) then it follows that Sn0+1 6= ∅. By Lemma 2.2 it follows that Sn0+1 is bounded from above. EJDE-2021/68 INFINITELY MANY SOLUTIONS 17 Let an0+1 = sup Sn0+1. In a similar fashion way we can show that Van0+1 (t) has exactly n0 + 1 zeros on (0, R2−N ) and Van0+1 (0) = 0. Proceeding inductively we can show that for each n ∈ N there exists a solution Van0+n (t) of (2.7) which has exactly n0 + n zeros on (0, R2−N ) and Van0+n(0) = 0. This completes the proof of Lemma 3.2 and the proof of the main theorem. � References [1] B. Azeroual, A. 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Iaia; Existence for semilinear equations on exterior domains, Electronic Journal of the Qualitative Theory of Differential Equations, 2016 (2016) No. 108 , 1-12. [9] C.K.R.T. Jones, T. Kupper; On the infinitely many solutions of a semilinear equation, SIAM J. Math. Anal., Volume 17 (1986), 803-835. [10] J. Joshi; Existence and nonexistence of solutions of sublinear problems with prescribed num- ber of zeros on exterior domains, Electronic Journal of Differential Equations, Volume 2017 (2017) No. 133, 1-10. [11] E.K. Lee, R. Shivaji, B. Son; Positive radial solutions to classes of singular problems on the exterior of a ball, Journal of Mathematical Analysis and Applications, 434 (2016), No. 2, 1597-1611, 2016. [12] K. McLeod, W. C. Troy, F. B. Weissler; Radial solutions of ∆u + f(u) = 0 with prescribed numbers of zeros, Journal of Differential Equations, Volume 83 (1990), Issue 2, 368-373. [13] L. Sankar, S. Sasi, R. Shivaji; Semipositone problems with falling zeros on exterior domains, Journal of Mathematical Analysis and Applications, Volume 401 (2013), Issue 1, 146-153. Mageed Ali Department of Mathematics, University of Kirkuk, Kirkuk, Iraq Email address: mageedali@uokirkuk.edu.iq Joseph Iaia Department of Mathematics, University of North Texas, Denton, TX, USA Email address: iaia@unt.edu 1. Introduction 2. Preliminaries 3. Proof of the main Theorem ?? References