Electronic Journal of Differential Equations, Vol. 2021 (2021), No. 03, pp. 1–17. ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu EXISTENCE AND NONEXISTENCE FOR SINGULAR SUBLINEAR PROBLEMS ON EXTERIOR DOMAINS MAGEED ALI, JOSEPH A. IAIA Abstract. In this article we study the existence of radial solutions of ∆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 u near 0 with 0 < q < 1, and f(u) ∼ |u|p−1u for large |u| with 0 < p < 1. Also, K(|x|) ∼ |x|−α with N + q(N − 2) < α < 2(N − 1) for large |x|. 1. Introduction In this article we study the radial solutions of: ∆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 BR is the ball of radius R > 0 centered at the origin in RN , K(x) > 0 and u : RN → R with N > 2. In addition, we suppose f : R \ {0} → R is locally Lipschitz and (H1) f is odd, there exists β > 0 such that f < 0 on (0, β), f > 0 on (β,∞). (H2) g1 : R→ R is continuous and f(u) = −1 |u|q−1u + g1(u) where 0 < q < 1 and g1(0) = 0. (H3) g2 : R→ R is continuous and f(u) = |u|p−1u+ g2(u), where 0 < p < 1 and limu→+∞ g2(u)/|u|p = 0. We 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 and from (H3) we see there exists γ with 0 < β < γ such that (H4) F < 0 on (0, γ), F > 0 on (γ,∞), and F > −F0 on R. (H5) K and K ′ are continuous on [R,∞) with K(r) > 0, 2(N − 1) + rK′ K > 0, N + q(N − 2) < α < 2(N − 1) and limr→∞ rK ′/K = −α. (H6) There exists K1 > 0 such that limr→∞ rαK(r) = K1 > 0. 2010 Mathematics Subject Classification. 34B40, 35B05. Key words and phrases. Exterior domains; singular problem; sublinear; radial solution. c©2021 Texas State University. Submitted June 11, 2020. Published January 7, 2021. 1 2 M. ALI, J. A. IAIA EJDE-2021/03 Interest in the topic for this article comes from recent papers [2, 7, 9, 10] about solutions of differential equation problems on exterior domains. In [1] we studied (1.1)–(1.3) with K(r) ∼ r−α, where f is singular at 0 and grows superlinearly at ∞, with various values of α. We proved existence of an infinite number of solutions. In this article we consider the case when f is singular at 0 and grows sublinearly at ∞. In this article we prove the following results. Theorem 1.1. Let N > 2, R > 0, 0 < p, q < 1, N + q(N − 2) < α < 2(N − 1), and suppose (H1)–(H6) hold. Then given a non-negative integer, n0, then there are solutions u0, u1, . . . , un0 of (1.1)–(1.3) where uk has exactly k zeros on (R,∞) and limr→∞ uk(r) = 0 if R is sufficiently small. Theorem 1.2. Let N > 2, R > 0, 0 < p, q < 1, N + q(N − 2) < α < 2(N − 1), and suppose (H1)–(H6) hold. Then there are no radial solutions of (1.1)–(1.3) if R > 0 is sufficiently large. 2. Preliminaries Since we are interested in studying radial solutions of (1.1)–(1.3), we assume that r = |x| = √ x21 + x22 + · · ·+ x2N , u(r) = u(|x|) where x ∈ RN and u satisfies u′′(r) + N − 1 r u′(r) +K(r)f(u(r)) = 0 on (R,∞), (2.1) u(R) = 0, lim r→∞ u(r) = 0. (2.2) To prove existence we make the change of variables u(r) = v(r2−N ). (2.3) Then u′(r) = (2−N)r1−Nv′(r2−N ), u′′(r) = (2−N)(1−N)r−Nv′(r2−N ) + (2−N)2r2(1−N)v′′(r2−N ). Letting t = r2−N and r = t 1 2−N in (2.1)–(2.2) gives v′′(t) + h(t)f(v(t)) = 0 for 0 < t < R2−N (2.4) where from (H1)–(H6), h(t) = 1 (N − 2)2 t 2(N−1) 2−N K(t 1 2−N ) ∼ t−α̃ (N − 2)2 with α̃ = 2(N − 1)− α N − 2 > 0. (2.5) Note that 2 − α̃ = α−2 N−2 > 0. Also from (H5) and (H6) it follows that there is a constant h1 > 0 with lim t→0+ tα̃h(t) = h1, h′(t) < 0 on (0, R2−N ], 0 < α̃+ q < 1. (2.6) Then there are h0 > 0 and h2 > 0 such that h0 ≤ tα̃h(t) ≤ h2 on (0, R2−N ]. (2.7) We now consider (2.4) with v(0) = 0, v′(0) = a ≥ 0 (2.8) and we try to find a ≥ 0 such that v(R2−N ) = 0. We write va to emphasize the dependence of v on a. Let a ≥ 0. We first show that there is a solution va of equation (2.4) on (0, ε) for small ε along with (2.8) and va, v′a continuous on [0, ε). EJDE-2021/03 SINGULAR SUBLINEAR PROBLEMS ON EXTERIOR DOMAINS 3 This is a bit lengthy so we postpone this proof to the Appendix. We now assume va solves (2.4) on (0, ε) and va, v′a continuous on [0, ε). Next let (0, B) ⊂ (0, R2−N ) be the maximal open interval where the solution of (2.4) exists along with (2.8). We will show B = R2−N . First, from the proof in the appendix we have that there exists ε > 0 such that 0 < ε ≤ B ≤ R2−N . Now we define the energy of solution (2.4), (2.8) as Ea(t) = 1 2 v′2a (t) h(t) + F (va(t)) for 0 < t < B. (2.9) Differentiating Ea, using (2.4) and since we know from (2.6) that h′(t) < 0, then E′a(t) = −v ′2 a (t)h′(t) 2h2(t) ≥ 0 on (0, B). (2.10) Thus Ea is nondecreasing on (0, B). Therefore, 0 = lim t→0+ Ea(t) ≤ Ea(t) = 1 2 v′2a (t) h(t) + F (va(t)) (2.11) so it follows that Ea(t) > 0 for 0 < t < B. (2.12) Next we see that (1 2 v′2a (t) + h(t)F (va(t)) )′ = h′(t)F (va(t)). (2.13) Now let us show for fixed a ≥ 0 that va and v′a are continuous on [0, R2−N ]. Lemma 2.1. Assume (H1)–(H6) hold, N > 2, and a ≥ 0. Suppose va solves (2.4). Then |va(t)| ≤ C and |v′a(t)| ≤ C for some constant C on [0, R2−N ] and va, v ′ a are continuous on [0, R2−N ]. Proof. We first assume that there is a ta,γ ∈ [0, B) such that va(ta,γ) = γ and 0 ≤ va < γ on [0, ta,γ). We know from (H4) that F (va) ≤ 0 when t ∈ [0, ta,γ ] so we have 0 < 1 2 v′2a (t) h(t) + F (va(t)) ≤ 1 2 v′2a (t) h(t) on (0, ta,γ ]. Thus v′a > 0 on [0, ta,γ ]. Also if we multiply (2.4) by vqa, use (H2), and integrate by parts on (0, t) this gives vqav ′ a − ∫ t 0 qvq−1a (s)v′2a (s) ds+ ∫ t 0 h(s)vqa(s)g1(va(s)) ds = ∫ t 0 h(s) ds. (2.14) Thus vqav ′ a + ∫ t 0 h(s)vqa(s)g1(va(s)) ds ≥ ∫ t 0 h(s) ds. (2.15) Integrating (2.15) again and using (2.7) gives vq+1 a (t) q + 1 + ∫ t 0 ∫ s 0 h(x)vqa(x)g1(va(x)) dx ds = ∫ t 0 ∫ s 0 h(x) dx ds ≥ h0t 2−α̃ (2− α̃)(1− α̃) . (2.16) 4 M. ALI, J. A. IAIA EJDE-2021/03 Let L1 be the Lipschitz constant for g1 on [0, γ] so then |g1(va)| ≤ L1va on [0, ta,γ ]. using this and since v′a > 0 on [0, ta,γ ] then:∫ t 0 ∫ s 0 h(x)vqa(x)g1(va(x)) dx ds ≤ L1 ∫ t 0 ∫ s 0 h(x)vq+1 a (x) dx ds ≤ L1v q+1 a (t) ∫ t 0 ∫ s 0 h(x) dx ds. using this in (2.16) and using (2.7) again we see that h0t 2−α̃ (2− α̃)(1− α̃) ≤ vq+1 a (t) [ 1 q + 1 + L1h2t 2−α̃ (2− α̃)(1− α̃) ] ≤ vq+1 a (t) [ 1 q + 1 + L1h2R (2−N)(2−α̃) (2− α̃)(1− α̃) ] . Therefore va(t) ≥ C1t 2−α̃ 1+q on [0, ta,γ ] (2.17) where C1 = [ h0(q + 1) (2− α̃)(1− α̃) + L1h2(q + 1)R(2−N)(2−α̃) ] 1 q+1 > 0. Evaluating (2.17) at t = ta,γ gives ta,γ ≤ ( γ C1 ) 1+q 2−α̃ . (2.18) Then from (2.17) and (2.7) we see that h(t) vqa(t) ≤ h2 C1 q t −α̃−2q 1+q on (0, ta,γ ]. Rewriting (2.4) and substituting gives v′′a(t) = h(t) vqa(t) − h(t)g1(va(t)) ≤ h2 C1 q t −α̃−2q 1+q + h2L1t −α̃γ on (0, ta,γ ]. (2.19) Integrating on (0, t) gives v′a(t) ≤ a+ C2t 1−α̃−q 1+q + C3t 1−α̃ on [0, ta,γ ] (2.20) where C2 = h2(1+q) Cq1 (1−α̃−q) , C3 = h2L1γ 1−α̃ . Integrating (2.20) on (0, t) we have va(t) ≤ at+ C4t 2−α̃ 1+q + C3 2− α̃ t2−α̃ on [0, ta,γ ] (2.21) where C4 = h2(1 + q)2 Cq1(1− α̃− q)(2− α̃) . Evaluating (2.21) at t = ta,γ and using (2.18) we obtain γ ≤ ta,γ ( a+ C4 ( γ C1 ) 1−α̃−q 2−α̃ + C3 2− α̃ ( γ C1 ) (1−α̃)(1+q) 2−α̃ ) = ta,γ(a+ C5) (2.22) where C5 = C4 ( γ C1 ) 1−α̃−q 2−α̃ + C3 2− α̃ ( γ C1 ) (1−α̃)(1+q) 2−α̃ . EJDE-2021/03 SINGULAR SUBLINEAR PROBLEMS ON EXTERIOR DOMAINS 5 From (2.22) we have 1 ta,γ ≤ a+ C5 γ . (2.23) Now from (2.20) and for t ∈ [0, ta,γ ] we obtain 0 ≤ v′a(t) ≤ a+ C2ta,γ 1−α̃−q 1+q + C3ta,γ 1−α̃ ≤ a+ C6 on [0, ta,γ ] (2.24) where C6 = C2R (2−N)(1−α̃−q) 1+q + C3R (2−N)(1−α̃). Thus |v′a| is bounded on [0, ta,γ ] if ta,γ ≤ B. Now continuing to assume ta,γ ≤ B we integrate (2.13) on (ta,γ , t), using (2.24), h′ < 0, and −F0 ≤ F (va) (by (H4)) then we obtain 1 2 v′2a (t)− h(t)F0 ≤ 1 2 v′2a (t) + h(t)F (va) = 1 2 v′2a (ta,γ) + ∫ t ta,γ h′(s)F (va(s)) ds ≤ 1 2 (a+ C6)2 − ∫ t ta,γ h′(s)F0 ds = 1 2 (a+ C6)2 − h(t)F0 + h(ta,γ)F0. using (2.23) in the above we have 1 2 v′2a (t) ≤ 1 2 (a+ C6)2 + h(ta,γ)F0 ≤ 1 2 (a+ C6)2 + h2F0 (a+ C5 γ )α̃ . (2.25) Thus it follows from (2.25) and standard inequalities that |v′a| is bounded as |v′a| ≤ a+ C7 on [0, B) (2.26) for some C7 that does not depend on a if 0 < ta,γ ≤ B. Then |va| = ∣∣∣∫ t 0 v′a ds ∣∣∣ ≤ (a+ C7)t ≤ (a+ C7)B on [0, B) (2.27) so |va| is also bounded on [0, B) if ta,γ ≤ B. On the other hand if 0 ≤ va < γ on [0, B) then a similar argument shows that (2.17) and (2.20) hold on [0, B) and so again we see that |va|, |v′a| are bounded on [0, B). Thus limt→B− va(t) = D ∈ R. Also since h(t)F (va(t)) and h′(t)F (va(t)) are continuous on [ε, B) it follows by integrating (2.13) on [ε, B) that limt→B− v ′ a(t) = D1 ∈ R. From (2.12) we know 0 < Ea(t) ≤ 1 2 D2 1 h(B) + F (D) on [0, B) so D and D1 cannot both be zero. If B < R2−N then the solution va can be extended to [0, B+ ε) for some ε > 0 by using the fact that D,D1 are not both zero for if D 6= 0 then we can just use the standard existence theorem from differential equations and if D = 0 then D1 6= 0 and we can use the contraction mapping principle as we did in the appendix which contradicts the definition of B. Thus we see B = R2−N . Also since va, v ′ a are bounded on [0, R2−N ) then we see limt→(R2−N )− va exists and limt→(R2−N )− v ′ a exists. Thus va, v ′ a are continuous on [0, R2−N ]. This completes the proof. � 6 M. ALI, J. A. IAIA EJDE-2021/03 Lemma 2.2. Let N > 2, a ≥ 0. Assume (H1)–(H6) hold, and suppose va(t) solves (2.4), (2.8). Then the solutions va(t) continuously depend on the parameter a ≥ 0 on [0, R2−N ]. Proof. Let 0 ≤ a1 < a2. Since va, v′a are continuous on [0, R2−N ] it follows from (2.26) and (2.27) that va, v′a are bounded on [0, R2−N ]. Then notice from (2.26) and (2.27) we have |v′a(t)| ≤ a2 + C7 on [0, R2−N ] ∀a with 0 ≤ a1 ≤ a ≤ a2, (2.28) |va(t)| ≤ (a2 + C7)R2−N on [0, R2−N ] ∀a with 0 ≤ a1 ≤ a ≤ a2. (2.29) Thus we see that |v′a| and |va| are uniformly bounded on [0, R2−N ] for all a with 0 ≤ a1 ≤ a ≤ a2. Next, let a∗ ≥ 0 with 0 ≤ a1 ≤ a∗ ≤ a2. We will now show that va → va∗ uniformly on [0, R2−N ] as a→ a∗. We prove this by contradiction so suppose not. Then there exist Aj with a1 ≤ Aj ≤ a2 such that Aj → a∗ as j →∞, tj ∈ [0, R2−N ] and there is an ε2 > 0 such that |vAj (tj)− va∗(tj)| ≥ ε2 ∀j. (2.30) Since Aj → a∗ as j →∞ and 0 ≤ a1 ≤ Aj ≤ a2, by (2.28), (2.29) we see that vAj and v′Aj are uniformly bounded on [0, R2−N ] and therefore the vAj are equicontin- uous on [0, R2−N ]. Then by the Arzela-Ascoli theorem there is a subsequence vAjl of vAj such that vAjl → va∗ uniformly on [0, R2−N ]. So as l→∞, 0← |vAjl (tjl)− va∗(tjl)| ≥ ε2 > 0 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. � Lemma 2.3. Let va(t) satisfy (2.4), (2.8) and assume that (H1)–(H6) hold. Then lima→∞ max[0,R2−N ] va(t) = ∞. In addition, if va(t) has a first local maximum, Ma, with 0 < Ma ≤ R2−N , then va(Ma) → ∞ as a → ∞. Further, if a is sufficiently large, then va is increasing on [0, R2−N ] and va(R2−N )→∞ as a→∞. Proof. We assume by the way of contradiction that max[0,R2−N ] va(t) ≤ C8 for some constant C8 > 0 which does not depend on a for a large. Since f(va) = − 1 |v|q−1va + g1(va) and g1(va) is continuous on [0, C8] then there is a C9 > 0 such that |g1(va)| ≤ C9 on [0, R2−N ]. Now either v′a > 0 or va has a local maximum Ma and v′a > 0 on [0,Ma). We show that va cannot have a local maximum Ma for large a. Integrating (2.4) over (0, t) and estimating gives v′a(t) = a+ ∫ t 0 h(s) 1 |v|q−1a va ds− ∫ t 0 h(s)g1(va) ds ≥ a− C9 ∫ t 0 h(s) ds. (2.31) Recalling from (2.6) that α̃+ q < 1 and q > 0 it follows that α̃ < 1. Also from (2.7) we have −h(t) ≥ −h2t−α̃. Then using this in (2.31) implies v′a(t) ≥ a− C9h2 1− α̃ t1−α̃. (2.32) Now if va has a local maximum then evaluating (2.32) at Ma gives C9h2 1− α̃ R(2−N)(1−α̃) ≥ C9h2 1− α̃ M1−α̃ a ≥ a (2.33) EJDE-2021/03 SINGULAR SUBLINEAR PROBLEMS ON EXTERIOR DOMAINS 7 but the right-hand side goes to infinity as a → ∞ while the left-hand side is fixed and thus we obtain a contradiction. Thus we see if a > 0 is sufficiently large and va is bounded above by a constant that it is independent of a then v′a > 0 on [0, R2−N ]. Next integrating (2.32) on (0, t) we obtain: C8 ≥ va(t) ≥ at− C9h2 (1− α̃)(2− α̃) t2−α̃. (2.34) Thus C8 ≥ va(R2−N ) ≥ aR2−N − C9h2 (1− α̃)(2− α̃) (R2−N )2−α̃ (2.35) therefore the right-hand side of (2.35) approaches infinity as a approaches infin- ity, but the left-hand side is bounded by C8. so we have a contradiction. Thus lima→∞max[0,R2−N ] va(t) =∞. Now we show that if va has a first local maximum, Ma, on [0, R2−N ], then lima→∞ va(Ma) = ∞. For if not we may again appeal to (2.33) as we did earlier to again get a contradiction. Thus the assumption that va(Ma) is bounded is false. Therefore if Ma ∈ [0, R2−N ] exists, then lim a→∞ va(Ma) =∞. (2.36) Next we show that v′a > 0 on [0, R2−N ] if a is sufficiently large. So suppose not. Then there exists a first local maximum, Ma, of va, with 0 < Ma ≤ R2−N . From (2.10)–(2.12) we have Ea(t) > 0 and E′a(t) ≥ 0. Thus for 0 ≤ t ≤Ma we have 1 2 v′2a (t) h(t) + F (va(t)) ≤ F (va(Ma)). (2.37) Rewriting and integrating (2.37) on (0,Ma) gives∫ Ma 0 v′a(t) dt√ 2 √ F (va(Ma))− F (va(t)) ≤ ∫ Ma 0 √ h(t) dt ≤ √ h2 ∫ R2−N 0 t−α̃/2 dt = 2 √ h2 2− α̃ (R2−N ) 1− α̃2 . (2.38) Since va(Ma) → ∞ as a → ∞ from (2.36) it follows from (H3) that F (va(Ma)) − F (s) ≤ C10va p+1(Ma) for some constant C10 > 0. Then after changing variables on the left-hand side of (2.38) and rewriting we obtain va 1−p 2 (Ma)√ 2C10 = va(Ma)√ 2 √ C10vap+1(Ma) ≤ ∫ va(Ma) 0 ds√ 2 √ F (va(Ma))− F (s) = 2 √ h2 2− α̃ (R2−N ) 1− α̃2 . (2.39) This yields a contradiction since the right-hand side of (2.39) is finite but 0 < p < 1 and by (2.36) the left-hand side of (2.39) goes to infinity as a → ∞. Thus the assumption that va has a local maximum on [0, R2−N ] if a is sufficiently large is false. Therefore if a is sufficiently large then va is increasing on [0, R2−N ] and 8 M. ALI, J. A. IAIA EJDE-2021/03 so va(R2−N ) = max[0,R2−N ] va(t). Since from the first part of the proof we know that lima→∞max[0,R2−N ] va(t) = ∞ it follows that lima→∞ va(R2−N ) = ∞. This completes the proof. � Lemma 2.4. Let va(t) satisfy (2.4), (2.8) and assume (H1)–(H6) hold. Let R > 0 be sufficiently small. Then va(t) has a local maximum, Ma, and a zero, Za, with 0 < Ma < Za < R2−N if a is sufficiently small. In addition, if R > 0 is sufficiently small then va has n zeros on [0, R2−N ]. Proof. Let us suppose instead that v′a(t) > 0 on [0, R2−N ] for all sufficiently small a and R sufficiently small. Then from (2.18) it follows that ta,γ ≤ C11 where C11 is independent of a. Thus ta,γ < R2−N if R is sufficiently small. Since va is continuous and increasing then for t > ta,γ we have γ = va(ta,γ) < va(t). Since v′a(t) > 0 and f(va) > 0 on [γ,∞) with f(va)→∞ as va →∞ by (H3) it follows that there exists C12 > 0 such that f(va) ≥ C12 > 0 on [ta,γ , R 2−N ]. Then v′′a(t) + C12h(t) ≤ v′′a(t) + h(t)f(va(t)) = 0 on [ta,γ , R 2−N ]. (2.40) Rewriting and integrating on (ta,γ , t) gives 0 < v′a(t) ≤ v′a(ta,γ)− C12 [ t1−α̃ − t1−α̃a,γ 1− α̃ ] . (2.41) From (2.6) we know 0 < α̃ < 1 and it follows from (2.26) that if 0 ≤ a ≤ a0 then |v′a(t)| ≤ a+ C7 ≤ a0 + C7. (2.42) Thus v′a(ta,γ) is bounded by a constant that is independent of a when a is sufficiently small and so it follows that the right-hand side of (2.41) becomes negative if R is sufficiently small which contradicts the assumption that v′a(t) > 0 on [0, R2−N ]. Thus if a is sufficiently small and R is sufficiently small then there is an Ma with 0 < Ma < R2−N such that v′a > 0 on (0,Ma) and v′a(Ma) = 0. Next, we want to show that va has a zero on [0, R2−N ] if a and R are sufficiently small. In order to do this we will show that va → v0 uniformly on [0, R2−N ] as a→ 0+ where v′′0 + h(t)f(v0) = 0, v0(0) = 0 = v′0(0). Then we will show v0 has a zero and since va → v0 uniformly as a → 0+ it will follow that va has a zero if a is sufficiently small and R is sufficiently small. It follows from Lemmas 2.1 and 2.2, and (2.28)–(2.29) that va, v ′ a are uniformly bounded on [0, R2−N ] for all 0 ≤ a ≤ a0 for some a0 > 0. Therefore there is a subsequence of the va, say vaj , such that vaj → v0 uniformly on [0, R2−N ] by the Arzela-Ascoli Theorem as aj → 0. Now we assume there is a ta,β with 0 < ta,β < R2−N such that va(ta,β) = β and 0 ≤ va(t) < α on [0, ta,β). It follows from (2.21) and an argument similar to (2.22) that β ≤ ta,β(a+ C5) (2.43) and as in (2.19) we have 0 ≤ v′′a ≤ h2 Cq1 t −α̃−2q 1+q + h2L1βt −α̃ ≤ C13t −α̃−2q 1+q on [0, ta,β ] (2.44) where C13 = h2 Cq1 + h2L1βR (2−N)(2−α̃)q 1+q . EJDE-2021/03 SINGULAR SUBLINEAR PROBLEMS ON EXTERIOR DOMAINS 9 Thus for 0 < x < y < ta,β and since 0 < 1−α̃−q 1+q < 1 we have 0 ≤ v′a(y)− v′a(x) = ∫ y x v′′a(t) dt ≤ C13 ∫ y x t −α̃−2q 1+q dt = C14|y 1−α̃−q 1+q − x 1−α̃−q 1+q | ≤ C14|y − x| 1−α̃−q 1+q (2.45) where C14 = 1+q 1−α̃−qC13. And since 0 < β a0+C5 ≤ ta,β from (2.43) it follows from this that the v′a are equicontinuous on [0, β a0+C5 ] for 0 ≤ a ≤ a0 and so v′aj → v′0 uniformly on [0, β a0+C5 ] by the Arzela-Ascoli Theorem. Now if 0 < va < β on [0, R2−N ] then we see (2.44) and (2.45) hold [0, R2−N ]. Next we choose t0 with 0 < t0 < β a0+C5 . Then integrating (2.13) on (t0, t) gives: 1 2 v′2aj (t) + h(t)F (vaj (t)) = 1 2 v′2aj (t0) + ∫ t t0 h′(s)F (vaj (s)) ds. (2.46) Now since vaj → v0 uniformly and since v′aj (t0) → v′0(t0) it then follows that v′aj → v′0 uniformly on [t0, R 2−N ], and so combined with the earlier fact v′aj → v′0 uniformly on [0, β a0+C5 ] we see that v′aj → v′0 uniformly on [0, R2−N ]. Now taking limits in (2.46) gives 1 2 v′20 (t) + h(t)F (v0(t)) = 1 2 v′20 (t0) + ∫ t t0 h′(s)F (v0(s)) ds on (0, R2−N ]. Letting t0 → 0+ gives 1 2 v′20 (t) + h(t)F (v0(t)) = ∫ t 0 h′(s)F (v0(s)) ds. Then from (2.4) and (H3) we see that v′′aj → v′′0 at all points where v0(t) 6= 0 and at these points we have v′′0 + h(t)f(v0) = 0, v0(0) = v′0(0) = 0. As at the beginning of the proof of this lemma it follows that v0 has a local maximum, M0, and v0(M0) > γ if R > 0 is sufficiently small. Now we assume by way of contradiction v0 > γ on [M0, R 2−N ]. Then we have f(v0) v0 > 0 on [M0, R 2−N ] so there is a C15 > 0 such that f(v0) v0 ≥ C15 > 0 when γ ≤ v0 ≤ v0(M0). Thus substituting in (2.4) and using (2.7) we obtain v′′0 (t) + h0C15 tα̃ v0(t) ≤ 0. So v′′0 < 0 while γ ≤ v0 ≤ v0(M0). Integrating v′′0 < 0 twice on (M0 + ε, t) we have v0(t) ≤ v0(M0 + ε) + v′0(M0 + ε)(t− (M0 + ε)). (2.47) Now if R is sufficiently small then R2−N will be very large and thus we may choose t sufficiently large so that the right-hand side of (2.47) becomes negative 10 M. ALI, J. A. IAIA EJDE-2021/03 contradicting that v0 ≥ γ. So there exists tγ0 > M0 such that v0(tγ0) = γ and v′0 < 0 on (M0, tγ0) if R is sufficiently small. Next while β < γ+β 2 ≤ v0 ≤ γ then f(v0) > 0 so v′′0 < 0. Integrating v′′0 < 0 twice on (tγ0 , t) gives v0(t) < γ + v′0(tγ0)(t− tγ0) with v′0(tγ0) < 0. Now again if R is sufficiently small then R2−N is very large and so we can choose t sufficiently large from which it would follow that v0(t) < γ+β 2 contradicting that v0(t) ≥ γ+β 2 . So there is a tγ1 > tγ0 such that v0(tγ1) = γ+β 2 . Now assume v0(t) > 0 on (M0, R 2−N ). Then recall that 1 2 v′20 h(t) + F (v0) > 0 and there exists C16 > 0 so −F (v0) ≥ C16v 1−q 0 for t > tγ1 . Therefore, − v′0 v 1−q 2 0 ≥ √ 2C16h0 t −α̃/2 on (tγ1 , t). Integrating on (tγ1 , t) gives 0 < v 1+q 2 0 (t) ≤ (γ + β 2 ) 1+q 2 − (1 + q) √ 2C16h0 2− α̃ [ t 2−α̃ 2 − t 2−α̃ 2 γ1 ] . (2.48) And again if R is sufficiently small then we can choose t sufficiently large so that the right-hand side of (2.48) becomes negative contradicting that v0 > 0. Thus v0 has a first positive zero, Z1, on [0, R2−N ] if R > 0 is sufficiently small. Also 0 < 1 2 v′20 h(t) + F (v0) for t > 0 so 0 < 1 2 v′20 (Z1) h(Z1) and therefore v′0(Z1) < 0. Thus v0(Z1 + ε) < 0 for ε > 0 sufficiently small. Then since va → v0 uniformly on [0, Z1 + ε] it follows that va(Z1 + ε) < 0 if a is sufficiently small and therefore if a > 0 and R are sufficiently small we see that va has a zero 0 < Z1,a < R2−N . Then as at the beginning of the proof where we showed that va has a local maximum, a similar argument shows va has a local minimum, ma, with Z1,a < ma and then va has a second zero, Z2,a, with Z2,a > ma, if a > 0 and R are sufficiently small. Continuing in this way we can find n zeros on [0, R2−N ] if R is small enough. This completes the proof. � 3. Proof of main Results Proof of Theorem 1.1. Consider the set S0 = {a > 0 : va(t) > 0 on (0, R2−N )}. If a is sufficiently large then va(t) > 0 on (0, R2−N ) by Lemma 2.3 and therefore va ∈ S0 if a is sufficiently large. Thus S0 6= ∅. Also if a and R are sufficiently small then va has a zero on (0, R2−N ) by Lemma 2.4. Thus S0 is bounded from below by a positive constant if R is sufficiently small. Now let a0 = inf S0. We now show that va0 > 0 on (0, R2−N ) and va0(R2−N ) = 0. Suppose on the contrary that there exists a zero, Za0 ∈ (0, R2−N ), and va0 > 0 on (0, Za0) with va0(Za0) = 0. Then 0 < Ea(Za0) = 1 2 v′2a0 (Za0 ) h(Za0 ) so v′a0(Za0) < 0. Thus for Za0 < t1 < R2−N and t1 close to Za0 we have va0(t1) < 0. Then for a close to a0 with a < a0 then va(t1) < 0 by continuous dependence (Lemma EJDE-2021/03 SINGULAR SUBLINEAR PROBLEMS ON EXTERIOR DOMAINS 11 2.2) but this contradicts the definition of a0. Thus va0 > 0 on (0, R2−N ) and so va0(R2−N ) ≥ 0. Next suppose that va0(R2−N ) > 0. Then va0 > 0 on (0, R2−N ] and for a close to a0 with a < a0 then va > 0 on (0, R2−N ]. But since a < a0, it follows that a /∈ S0 so va must have a zero on (0, R2−N ] which contradicts that va > 0 on (0, R2−N ]. Thus va0(R2−N ) = 0. Also since Ea non-decreasing it follows that 0 < Ea(R2−N ) = 1 2 v′2a0 (R2−N ) h(R2−N ) so v′a0(R2−N ) < 0. Next let us define S1 = {a > 0 : va(t) solves (2.4), (2.8) and has exactly one zero on (0, R2−N )}. If we choose a slightly smaller than a0 and R sufficiently small then it follows from Lemma 2.4 that va has at least one zero, Za1 , on (0, R2−N ) and Za1 is close to R2−N . Also we know v′a0(R2−N ) < 0 so if a is sufficiently close to a0 then v′a < 0 on (Za1 , R 2−N ). Thus va has at most one zero on (0, R2−N ) if a is sufficiently close to a0. Therefore S1 is nonempty. We also know from Lemma 2.4 that if R is sufficiently small then va has a second zero on (0, R2−N ). Therefore S1 is bounded from below. So let a1 = inf S1. In a similar way we can show that va1 has exactly one zero on (0, R2−N ) and va1(R2−N ) = 0. In a similar fashion we can show that if n0 is a given nonnegative integer then if R > 0 is sufficiently small then there exists a0, a1, . . . , an0 such that vak has k zeros on (0, R2−N ) and vak(R2−N ) = 0. Finally, let uk(r) = vak(r2−N ). Then uk(r) satisfies (1.1)–(1.3) and uk has k zeros on (R,∞). This completes the proof. � Proof of Theorem 1.2. Suppose there is a solution, va, of (2.4) with va(0) = va(R2−N ) = 0. This then implies that va has a local maximum, Ma, with 0 < Ma < R2−N and v′a(Ma) = 0. Since Ea is non-decreasing (by (2.10)) then for 0 < t < Ma, 0 < 1 2 v′2a h(t) + F (va(t)) = Ea(t) ≤ Ea(Ma) = F (va(Ma)). (3.1) Thus va(Ma) > γ. Rewriting and integrating (3.1) on (0,Ma) gives∫ Ma 0 v′a(t) dt√ 2 √ F (va(Ma))− F (va(t)) ≤ ∫ Ma 0 √ h2 t −α̃/2 dt = 2 √ h2 2− α̃ M 2−α̃ 2 a ≤ 2 √ h2 2− α̃ (R2−N ) 2−α̃ 2 . (3.2) Since α̃ < 1 and from (H4) we have −F (va(t)) ≤ F0 so it follows that F (va(Ma))− F (va(t)) ≤ F (va(Ma)) + F0 which we apply to (3.2) to obtain∫ Ma 0 v′a(t) dt√ 2 √ F (va(Ma))− F (va(t)) ≥ va(Ma)√ 2 √ F (va(Ma)) + F0 . (3.3) 12 M. ALI, J. A. IAIA EJDE-2021/03 Next from (H3) it follows that there is a constant F1 > 0 such that F (x) ≤ F1|x|p+1 for all x and therefore it follows from (3.2)-(3.3) and that va(Ma) > γ that γ 1−p 2 √ 2 √ F1 + F0 γp+1 ≤ v 1−p 2 a (Ma) √ 2 √ F1 + F0 vp+1 a (Ma) ≤ 2 √ h2 2− α̃ (R2−N ) 2−α̃ 2 . (3.4) Thus the right-hand side of (3.4) goes to zero if R sufficiently large but the left-hand side of (3.4) is positive and independent of R. Thus (1.1)–(1.3) has no solutions if R is sufficiently large. This completes the proof. � 4. Appendix Lemma 4.1. Let a > 0 and (H1)–(H6) hold. Then there exists a solution va of (2.4), (2.8) on (0, ε] for some ε > 0. Proof. This is similar to the proof of existence in [1] which we include here for completeness. First integrate (2.4) over (0, t) and use (2.8). This gives v′a(t) = a− ∫ t 0 h(s)f(va(s)) ds for t > 0. (4.1) Integrate again over (0, t) and using (2.8) gives va(t) = at− ∫ t 0 ∫ s 0 h(x)f(va(x)) dx ds for t > 0. (4.2) Now let W (t) = va(t) t so va(t) = tW (t) and W (0) = limt→0+ va(t) t = v′a(0) = a. Rewriting (4.2) we obtain W (t) = a− 1 t ∫ t 0 ∫ s 0 h(x)f (xW (x)) dx ds for t > 0. (4.3) We now we solve equation (4.3) on (0, ε] by a fixed point method as follows. Let us define S = { W : [0, ε]→ R with W (0) = a > 0,W ∈ C[0, ε] and |W (t)− a| ≤ a 2 on [0, ε] } (4.4) where C[0, ε] is the set of continuous functions on [0, ε] and ε > 0. Let ‖W‖ = sup x∈[0,ε] |W (x)|. Then (S, ‖ · ‖) is a Banach space. Let us define a map T on S by TW (t) = { a for t = 0 a− 1 t ∫ t 0 ∫ s 0 h(x)f (xW (x)) d ds for 0 < t ≤ ε. From (4.4) we see 0 < a 2 ≤ W (x) ≤ 3a 2 on [0, ε] so it follows that | −1 xqW q(x) | ≤ 2qx−q aq on (0, ε] and since we know from (H1)–(H2) that g1(x) is locally Lipschitz this then implies that there exists L1 > 0 such that |g1(x)| ≤ L1|x| on [0, γ]. (4.5) Now let W ∈ S and suppose 0 < ε < 2γ 3a . Then on [0, ε] we have 0 ≤ xW (x) < ε 3a 2 < 2γ 3a 3a 2 = γ. EJDE-2021/03 SINGULAR SUBLINEAR PROBLEMS ON EXTERIOR DOMAINS 13 using (H2), (2.6), and (4.5) we estimate |h(x)f(xW (x))| = ∣∣h(x) ( −1 xqW q(x) + g1(xW (x)) )∣∣ ≤ h22q aq x−(α̃+q) + 3ah2L1 2 x1−α̃. Recalling from (2.6) that α̃+ q < 1 then integrating once over [0, t] gives∫ t 0 |h(x)f (xW (x)) | dx ≤ A1 aq t1−α̃−q +A2at 2−α̃ (4.6) where A1 = h22 q (1−α̃−q) and A2 = 3h2L1 2(2−α̃) . Thus from (4.6) we have lim t→0+ ∫ t 0 |h(x)f (xW (x)) | dx = 0. (4.7) Integrating (4.6) again gives∫ t 0 ∫ s 0 |h(x)f (xW (x)) | dx ds ≤ A3t 2−α̃−q aq + aA4t 3−α̃ (4.8) where A3 = h22 q (2−α̃−q)(1−α̃−q) and A4 = 3h2L1 2(2−α̃)(3−α̃) . So we see lim t→0+ ∫ t 0 ∫ s 0 |h(x)f (xW (x)) | dx ds = 0. (4.9) We now show that T (W ) ∈ S for each W ∈ S if ε > 0 is sufficiently small so we first let W ∈ S. It follows then from (4.9) that T (W ) is continuous on [0, ε]. Thus we see limt→0+ TW (t) = a and so |TW (t) − a| ≤ a 2 on [0, ε] if ε > 0 is sufficiently small. Therefore 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 and suppose 0 < ε < 2γ 3a . Then TW1(t)− TW2(t) = −1 t ∫ t 0 ∫ s 0 h(x)[f(xW1(x))− f(xW2(x))] d ds. (4.10) By (H2) we have f(xW (x)) = −x−qW−q(x) + g1(xW (x)) where 0 < q < 1. Then as earlier before (4.5) we see that 0 ≤ xWi ≤ ε 3a2 < γ on [0, ε] for i = 1, 2 therefore using (4.5) this gives |f(xW1(x))− f(xW2(x))| = |−1 xq [ 1 W q 1 − 1 W q 2 ] + g1(xW1(x))− g1(xW2(x))| ≤ 1 xq ∣∣ 1 W1 q − 1 W2 q ∣∣+ L1x|W1 −W2|. (4.11) Next applying the mean value theorem we see that the right-hand side of (4.11) is bounded by 1 xq [ q W q+1 3 |W1 −W2| ] + L1x|W1 −W2| where W3 is between W1 and W2. Since Wi ∈ S for i = 1, 2, 3 and |Wi − a| ≤ a 2 then a 2 ≤ Wi ≤ 3a 2 on [0, ε]. Therefore it follows that W3 q+1 ≥ ( a 2 )q+1 and so we have |f(xW1(x))− f(xW2(x))| ≤ |W1 −W2| [ q xq (2 a )q+1 + L1x ] on (0, ε]. (4.12) 14 M. ALI, J. A. IAIA EJDE-2021/03 Recalling that |h(t)| ≤ h2 tα̃ and α̃+q < 1 from (2.6), and t ∈ (0, ε], then using (4.12) in (4.10) gives |TW1 − TW2| ≤ 1 t ∫ t 0 ∫ s 0 h2 xα̃ |W1 −W2| [ q xq (2 a )q+1 + L1x ] dx ds ≤ 1 t ‖W1 −W2‖ ∫ t 0 ∫ s 0 h2 xα̃ [ q xq (2 a )q+1 + L1x ] dx ds ≤ ‖W1 −W2‖ [A5ε 1−q−α̃ aq+1 +A6ε 2−α̃], where A5 = h2 q 2q+1 (2−q−α̃)(1−q−α̃) and A6 = h2L1 (3−α̃)(2−α̃) . Since lim ε→0+ [A5ε 1−q−α̃ aq+1 +A6ε 2−α̃] = 0, for ε > 0 sufficiently small we see that |TW1 − TW2| ≤ c‖W1 −W2‖, where c = A5ε 1−q−α̃ aq+1 +A6ε 2−α̃. (4.13) Thus for ε sufficiently small we see 0 < c < 1 and therefore T is a contraction mapping on S. Thus by the contraction mapping principle [5] there exists a unique solution W ∈ S to TW = W on [0, ε] for some ε > 0. And then va(t) = tW (t) is a solution of (2.4) on (0, ε] for some ε > 0. This completest the proof. � Lemma 4.2. Let a = 0 and (H1)–(H6) hold. Then there exists a solution v0 > 0 of equation (2.4) with v0(0) = v′0(0) = 0 on (0, ε] for some ε > 0. Proof. Suppose first that v0 is a solution to (2.4) on (0, ε] with v0(0) = 0, v′0(0) = 0. (4.14) Let us determine the behavior of v0(t) on (0, ε). using the fact that f(va) = −1 |va|q−1va + g1(va) where 0 < q < 1, g1(0) = 0, and g1 is continuous at 0, then integrating (2.4) on (0, t) and using v′0(0) = 0 gives: v′0(t) = − ∫ t 0 h(s)f ( v0(s) ) ds. Integrating again on (0, t) and using v0(0) = 0 gives v0(t) = − ∫ t 0 ∫ s 0 h(x)f ( v0(x) ) dx ds. (4.15) Now let v0(t) = t 2−α̃ 1+qW (t) where W (0) 6= 0. Rewriting (4.15) we have W (t) = 1 t 2−α̃ 1+q ∫ t 0 ∫ s 0 h(x) [ 1 x (2−α̃)q 1+q W q(x) − g ( x 2−α̃ 1+qW (x) )] dx ds. (4.16) EJDE-2021/03 SINGULAR SUBLINEAR PROBLEMS ON EXTERIOR DOMAINS 15 Assuming W (t) is continuous at 0, taking the limit of (4.16) and using L’Hôpital’s rule twice gives W (0) = lim t→0+ W (t) = A7 lim t→0+ tα̃h(t) [ t−α̃ t (2−α̃)q 1+q W q(t) ] − g1 ( t 2−α̃ 1+qW (t) ) h(t) t −α̃−2q 1+q = A7 [ lim t→0+ tα̃h(t) W q(t) − lim t→0+ h(t)g1 ( t 2−α̃ 1+qW (t) ) t −α̃−2q 1+q ] = A7h1 W q(0) −A7 lim t→0+ h(t)g1 ( t 2−α̃ 1+qW (t) ) t −α̃−2q 1+q , (4.17) where A7 = ( 1+q 2−α̃ )( 1+q 1−α̃−q ) . Since tα̃h(t)→ h1 > 0, by (2.6) as t→ 0+, 0 < α̃ < 1 and |g1(v)| ≤ L1|v| on [0, γ] it follows that∣∣∣h(t)g1 ( t 2−α̃ 1+qW (t) ) t −α̃−2q 1+q ∣∣∣ ≤ h2t −α̃L1t 2−α̃ 1+q t −α̃−2q 1+q |W (t)| = h2L1t 2−α̃|W (t)| → 0 as t→ 0+. Then we have W q+1(0) = h1A7 = h1(1 + q)2 (2− α̃)(1− α̃− q) , hence W (0) = [ h1(1 + q)2 (2− α̃)(1− α̃− q) ] 1 q+1 ≡ b0. (4.18) Now let W (t) = b0Y (t). Then Y (0) = 1 and (4.16) becomes Y (t) = 1 t 2−α̃ 1+q ∫ t 0 ∫ s 0 h(x) [ 1 x (2−α̃)q 1+q b0 q+1Y q(x) − g1 ( x 2−α̃ 1+q b0Y (x) ) b0 ] dx ds. (4.19) Now we attempt to solve (4.19) by using the contraction mapping principle. Let us define J = { Y ∈ C[0, ε] : Y (0) = 1 and |Y − 1| < δ where 0 < δ < 1 is sufficiently small } . (4.20) Let ‖Y ‖ = supx∈[0,ε] |Y (x)|. Then (J, ‖ · ‖) is a Banach space. Now we define T on J by TY (t) =  1 for t = 0 1 t 2−α̃ 1+q ∫ t 0 ∫ s 0 h(x) [ 1 x (2−α̃)q 1+q b0q+1Y q(x) − g1 ( x 2−α̃ 1+q b0Y (x) ) b0 ] dx ds for 0 < t ≤ ε. It is straightforward to show TY (t) is continuous and from (4.18) and L’Hôpital’s rule limt→0+ TY (t) = 1. Thus it follows that |TY (t)−1| ≤ δ on [0, ε) if ε sufficiently small, and therefore T : J → J if ε and δ are sufficiently small. 16 M. ALI, J. A. IAIA EJDE-2021/03 Since |Y − 1| < δ < 1 then 0 < 1 − δ < Y < 1 + δ and this implies that 1 Y q ≤ 1 (1−δ)q . Let us suppose Y1, Y2 ∈ J then TY1(t)− TY2(t) = 1 t 2−α̃ 1+q ∫ t 0 ∫ s 0 h(x) [ 1 x (2−α̃)q 1+q b0 q+1 ( 1 Y q1 (x) − 1 Y q2 (x) )] dx ds − 1 b0t 2−α̃ 1+q ∫ t 0 ∫ s 0 h(x) [ g1 ( x 2−α̃ 1+q b0Y1(x) ) + g1 ( x 2−α̃ 1+q b0Y2(x) )] dx ds. (4.21) For the integral in (4.21) since Y1, Y2 ∈ J , then by the mean value theorem there is a Y3 between Y1, Y2 where |Yi−1| < δ for i = 1, 2, 3 (and therefore 1−δ < Y3 < 1+δ) then | 1 Y q1 − 1 Y q2 | = q Y3 q+1 |Y1 − Y2| ≤ q (1− δ)q+1 |Y1 − Y2|. Then using (2.6) the integral in (4.21) becomes∣∣∣ 1 t 2−α̃ 1+q ∫ t 0 ∫ s 0 h(x) [ 1 x (2−α̃)q 1+q b0 q+1 ( 1 Y1 q − 1 Y2 q )] dx ds| ≤ q (1− δ)1+qb01+q |Y1 − Y2| t (2−α̃)q 1+q ∫ t 0 ∫ s 0 h(x) x (2−α̃)q 1+q dx ds ≤ h2q (1− δ)1+qb01+q |Y1 − Y2| t (2−α̃)q 1+q ∫ t 0 ∫ s 0 x −(α̃+2q) 1+q dx ds ≤ (1 + q)2h2q (2− α̃)(1− α̃− q)(1− δ)1+q |Y1 − Y2| b0 1+q t (2−α̃)(1−q) 1+q . Recalling b0 q+1 = h1A7 = h1 ( 1+q 2−α̃ )( 1+q 1−α̃−q ) we obtain the right-hand side of (4.21) is bounded by h2q h1(1− δ)1+q ε (2−α̃)(1−q) 1+q ‖Y1 − Y2‖. Since δ > 0 and 0 < q < 1 we see that for ε > 0 sufficiently small, h2q h1(1− δ)1+q ε (2−α̃)(1−q) 1+q ≤ d < 1. For the integral in (4.21) since g1 is locally Lipschitz at 0, it follows that∣∣∣g1(x 2−α̃ 1+q b0Y1(x) ) − g1 ( x 2−α̃ 1+q b0Y2(x) )∣∣∣ ≤ L1 b0 x 2−α̃ 1+q ‖Y1 − Y2‖ so substituting this into (4.21) gives∣∣∣ −1 b0t 2−α̃ 1+q ∫ t 0 ∫ s 0 h(x) [ g1 ( x 2−α̃ 1+q b0Y1(x) ) − g1 ( x 2−α̃ 1+q b0Y2(x) )] dx ds ∣∣∣ ≤ |Y1 − Y2|h2L1 t 2−α̃ 1+q ∫ t 0 ∫ s 0 x−α̃+ 2−α̃ 1+q dx ds ≤ |Y1 − Y2|h2L1A8t 2+q 1+q (4.22) where A8 = ( 1+q 1+(2+q)(1−α̃) )( 1+q (2+q)(2−α̃) ) . Since limt→0+ h2L1A8t 2+q 1+q = 0 we can choose ε small enough so that h2L1A8t 2+q 1+q < 1−d 2 and so combining (4.21) and EJDE-2021/03 SINGULAR SUBLINEAR PROBLEMS ON EXTERIOR DOMAINS 17 (4.22) we obtain |TY1(t)− TY2(t)| ≤ 1 + d 2 ‖Y1 − Y2‖ where 0 ≤ d < 1 and thus 1+d 2 < 1. Thus T is a contraction mapping, so by the contraction mapping principle [5] there is a unique solution Y ∈ J to T (Y ) = Y on [0, ε]. Then va(t) = t 2−α̃ 1+qW (t) is a solution of (2.4), (4.14) on [0, ε] for some ε > 0. This completes the proof. � References [1] M. Ali, J. Iaia; Infinitely many solutions for a singular semilinear problem on exterior domains, submitted to Electronic Journal of Differential Equations, 2020. [2] B. Azeroual, A. Zertiti; On multiplicity of radial solutions to Dirichlet problem involving the p−Laplacian on exterior domains, International Journal of Applied Mathematics, Volume 31, Number 1, 121-147, 2018. [3] H. Berestycki, P. L. Lions, Non-linear scalar field equations I, Arch. Rational Mech. Anal., Volume 82, 313-347, 1983. [4] H. Berestycki, P. L. Lions; Non-linear scalar field equations II, Arch. Rational Mech. Anal., Volume 82, 347-375, 1983. [5] M. Berger; Nonlinearity and functional analysis, Academic Free Press, New York, 1977. [6] G. Birkhoff, G. C. Rota; Ordinary Differential Equations, John Wiley and Sons, 1962. [7] J. Iaia; Existence and nonexistence of solutions for sublinear equations on exterior domains, Electronic Journal of Differential Equations, Volume 2017, No. 214, 1-13, 2017. [8] J. Iaia; Existence of solutions for sublinear equations on exterior domains, Electronic Journal of Differential Equations, Volume 2018, No. 181, 1-14, 2018. [9] 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 No. 133, 1-10, 2017. [10] 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, No. 2, 1597-1611, 2016. [11] L. Sankar, S. Sasi, R. Shivaji; Semipositone problems with falling zeros on exterior domains, Journal of Mathematical Analysis and Applications, Volume 401, Issue 1, 146-153, 2013. Mageed Ali Department of Mathematics, University of North Texas, P.O. Box 311430, Denton, TX 76203-5017, USA. University of Kirkuk, Iraq Email address: mageedali@my.unt.edu Joseph A. Iaia Department of Mathematics, University of North Texas, P.O. Box 311430, Denton, TX 76203-5017, USA Email address: iaia@unt.edu 1. Introduction 2. Preliminaries 3. Proof of main Results 4. Appendix References