Electronic Journal of Differential Equations, Vol. 2024 (2024), No. 06, pp. 1–14. ISSN: 1072-6691. URL: https://ejde.math.txstate.edu, https://ejde.math.unt.edu DOI: 10.58997/ejde.2024.06 EXISTENCE OF TWO INFINITE FAMILIES OF SOLUTIONS FOR SINGULAR SUPERLINEAR EQUATIONS ON EXTERIOR DOMAINS JOSEPH IAIA Abstract. In this article we study radial solutions of ∆u + K(|x|)f(u) = 0 in the exterior of the ball of radius R > 0 in RN with N > 2 where f grows superlinearly at infinity and is singular at 0 with f(u) ∼ 1 |u|q−1u and 0 < q < 1 for small u. We assume K(|x|) ∼ |x|−α for large |x| and establish existence of two infinite families of sign-changing solutions when N + q(N − 2) < α < 2(N − 1). 1. Introduction In this article we are interested in radial solutions of ∆u+K(|x|)f(u) = 0 on RN\BR, u = 0 on ∂BR, u→ 0 as |x| → ∞, (1.1) when N > 2 and where BR is the ball of radius R > 0 centered at the origin. Assuming u(x) = u(|x|) = u(r) the above problem becomes u′′ + N − 1 r u′ +K(r)f(u) = 0 for R < r <∞, (1.2) u(R) = 0, lim r→∞ u(r) = 0. (1.3) Numerous papers have proved existence of positive solutions of these equations with various nonlinearities f(u) and for various functions K(|x|) ∼ |x|−α with α > 0. See for example [1, 4, 5, 7, 11, 12, 13]. Here we prove existence of two infinite families of solutions including sign- changing solutions for this equation. We have also proved the existence of sign- changing solutions in other recent papers [2, 3, 9, 10]. We use the following assumptions: (H1) f : R\{0} → R is odd, locally Lipschitz, and f(u) = |u|p−1u+ g(u) with p > 1 for large |u| and limu→∞ |g(u)| |u|p = 0. 2020 Mathematics Subject Classification. 34B40, 35B05. Key words and phrases. Exterior domains; singular; semilinear; radial solution. ©2024. This work is licensed under a CC BY 4.0 license. Submitted July 13, 2023. Published January 23, 2024. 1 2 J. IAIA EJDE-2024/06 (H2) There exists a locally Lipschitz g1 : R→ R such that f(u) = 1 |u|q−1u + g1(u) with 0 < q < 1 for small |u| and g1(0) = 0. (H3) f > 0 on (0,∞). Let F (u) = ∫ u 0 f(t) dt. Since f is odd then F is even. Also, since 0 < q < 1 (by (H2)) it follows that f is integrable at 0 and therefore F is continuous with F (0) = 0. Also since f > 0 on (0,∞) it follows that F (u) > 0 for u > 0. Since F (u) is even then F (u) > 0 for u 6= 0. We also assume K(r) > 0 and K ′(r) are continuous on [R,∞). In addition, we assume that (H4) there exist α1, α2 and positive K1,K2,K3 such that K1 rα1 ≤ K ≤ K2 rα2 and r|K ′| K ≤ K3 on [R,∞), (1.4) where N + q(N − 2) < α2 ≤ α1 < 2(N − 1). In this article we prove the following result. Theorem 1.1. Let N > 2 and assume (H1)–(H4). If R > 0, then there exist two infinite families u±n of solutions to (1.2)-(1.3). If R > 0 is sufficiently large then there are 2 solutions, u±n , with n interior zeros on (R,∞) for all positive integers n and there is 1 positive solution. If R > 0 is sufficiently small then there is an n0 ≥ 0 such that there are 2 solutions with n zeros on (R,∞) for all n > n0 and there is one solution with n0 zeros on (R,∞). We remark that the solutions of (1.2)-(1.3) have continuous second derivatives except at points where u(r0) = 0 because limu→0 |f(u)| = ∞. Solutions, however, do turn out to be C1[R,∞). In addition, we will see in Lemma 2.1 that if a > 0 then u(r) and u′(r) cannot both be zero at any r ∈ [R,∞). In particular, if u(z) = 0 then u′(z) 6= 0 and so by (H2) it follows that rN−1Kf(u) is integrable at z. Therefore, by a C1[R,∞) solution of (1.2)-(1.3) we mean u ∈ C1[R,∞) such that rN−1u′ + ∫ r R tN−1Kf(u) dt = RN−1u′(R) for r ≥ R, u(R) = 0, and limr→∞ u(r) = 0. 2. Preliminaries Let R > 0. We begin our analysis of (1.2)-(1.3) by first making the change of variables u(r) = v(r2−N ) = v(t) and obtaining v′′(t) + h(t)f(v(t)) = 0, where 0 < h(t) = t 2(N−1) 2−N K(t 1 2−N ) (N − 2)2 . Henceforth we denote R1 = R2−N . We now attempt to solve the initial value problem v′′a + h(t)f(va) = 0 for 0 < t < R1, (2.1) va(0) = 0, v′a(0) = a > 0 (2.2) and then try to find values of a so that va(R1) = 0. (2.3) EJDE-2024/06 SINGULAR SUPERLINEAR EQUATIONS ON EXTERIOR DOMAINS 3 Let α̃1 = 2(N − 1)− α1 N − 2 , α̃2 = 2(N − 1)− α2 N − 2 . It follows from (H4) and the definition of h that there exist positive h1, h2, h3 such that 0 < h1t −α̃1 ≤ h(t) ≤ h2t−α̃2 and t|h′| h ≤ h3, (2.4) where 0 < α̃1 ≤ α̃2 < 1− q. First we prove existence of a solution to (2.1)-(2.2) on [0, ε0] for some ε0 > 0. To do this we reformulate (2.1)-(2.2) as an appropriate integral equation. Let us suppose first that va is a solution (2.1)-(2.2). Integrating on (0, t) gives: v′a + ∫ t 0 h(x)f(va(x)) dx = a for a > 0. (2.5) Integrating on (0, t) gives va + ∫ t 0 ∫ s 0 h(x)f(va(x)) dx ds = at for a > 0. (2.6) A bit of care needs to be taken here because we first need to know that the integral in (2.5) is defined. To see this notice that if va is a solution of (2.1)-(2.2) then for sufficiently small t > 0 we have a 2 t ≤ va ≤ at. In addition, it follows from (H1) and (H2) that there is a constant f1 > 0 such that f(va) ≤ f1(v−qa + vpa) and therefore by (2.4) we have 0 < h(t)f(va) ≤ f1h2 ( t−α̃2 vqa + t−α̃2vpa ) ≤ f1h2 ( t−α̃2 (a2 )qtq + t−α̃2+pap ) = f1h2 (2q aq t−α̃2−q + t−α̃2+pap ) . (2.7) From (2.4) we have 1− α̃2 − q > 0 and 1− α̃2 + p > 0 so it follows from (2.7) that h(t)f(va) is integrable near t = 0. Thus the integral in (2.5) is defined and is a continuous function. It then follows that (2.6) is also defined. Now using (H2) we see that (2.6) is equivalent to va + ∫ t 0 ∫ s 0 h(x) ( 1 vqa(x) + g1(va) ) dx ds = at. (2.8) Next let va = tw in (2.8) which gives w = a− 1 t ∫ t 0 ∫ s 0 h(x) ( 1 xqwq(x) + g1(xw) ) dx ds. (2.9) We now define Sε = {w ∈ C[0, ε] : w(0) = a > 0, and |w − a| ≤ a 2 for all t ∈ [0, ε]}. Here C[0, ε] is the set of real-valued continuous functions on [0, ε] with the supremum norm ‖ · ‖. We define T : Sε → C[0, ε] by Tw(0) = a and Tw = a− 1 t ∫ t 0 ∫ s 0 h(x) ( 1 xqwq(x) + g1(xw) ) dx ds for t > 0. 4 J. IAIA EJDE-2024/06 As mentioned in (2.4) and (2.7) it follows that 0 < h(x) xq ≤ h2x −α̃2−q and α̃2+q < 1. Hence x−α̃2−q is integrable on (0, ε). Then it is straightforward to show T maps Sε into Sε if ε > 0 is sufficiently small. Next let L be the Lipschitz constant for the function g1 defined in (H2) and suppose w1, w2 ∈ S. Using the mean value theorem and the fact that a 2 ≤ wi ≤ a for i = 1, 2 on [0, ε] we see that |Tw1 − Tw2| ≤ 1 t ∫ t 0 ∫ s 0 ( qh2 (2 a )q+1 x−α̃2−q + Lx1−α̃2 ) |w1 − w2| dx ds ≤ ‖w1 − w2‖ ( qh2 (1− α̃2 − q)(2− α̃2 − q) (2 a )q+1 t1−α̃2−q + L (2− α̃2)(3− α̃2) t2−α̃2 ) . (2.10) Since the term in parentheses in (2.10) goes to 0 as t → 0+, it follows that there exists ε0 > 0 and a c with 0 < c < 1 so that ‖Tw1 − Tw2‖ ≤ c‖w1 − w2‖ for all wi ∈ Sε0 . Thus T is a contraction and so by the contraction mapping principle T has a unique fixed point [8]. Therefore, we obtain a unique solution of (2.6) on [0, ε0]. It then follows that the integral term in (2.6) is differentiable which implies that va is differentiable and satisfies (2.5). Next we let Ea = v′2a 2h + F (va). (2.11) Recall from the comments after (H3) that F (va) ≥ 0. Therefore from (2.1) and (2.4) it follows that |E′a| = ∣∣− h′ 2h2 v′2a ∣∣ ≤ ∣∣ th′ h ∣∣ v′2a 2th ≤ h3Ea t . (2.12) Thus ( Ea th3 )′ ≤ 0 for t > 0 and therefore integrating on (ε0/2, t) (with the ε0 in the proof of existence) gives v′2a 2h + F (va) = Ea(t) ≤ C1t h3 ≤ C1R h3 1 , where C1 = Ea(ε0/2).(ε0/2)h3 . Thus va and v′a are uniformly bounded on a largest interval of the form [ε0/2, T ] ⊂ [ε0/2, R1]. It then follows from this that va and v′a are defined and continuous on all of [0, R1]. In addition, it also follows from this that the va vary continuously with respect to a. Lemma 2.1. Assume (H1)–(H4) and let va solve (2.1)-(2.2) with a > 0. Then |va|+ |v′a| > 0 on [0, R1]. Proof. First since va(0) = 0 and v′a(0) = a > 0 it follows that va and v′a cannot both be zero at any t ∈ [0, ε] for some ε > 0. Suppose now that there is a t0 ∈ (0, R1] such that va(t0) = v′a(t0) = 0. Thus Ea(t0) = 0 and then from (2.12) it follows that (Eat h3)′ ≥ 0 on (t, t0). Integrating this on (t, t0) yields Ea ≤ 0 on (t, t0). Since Ea ≥ 0 it follows then that Ea ≡ 0 on [0, t0] and thus va = v′a = 0 on [0, t0]. This however contradicts that v′a(0) = a > 0. Thus the lemma follows. � Lemma 2.2. Assume (H1)–(H4) and let va solve (2.1)-(2.2) with a > 0. Then va only has a finite number of zeros on [0, R1]. EJDE-2024/06 SINGULAR SUPERLINEAR EQUATIONS ON EXTERIOR DOMAINS 5 Proof. First since va(0) = 0 and v′a(0) = a > 0 it follows that va > 0 on (0, ε) for some ε > 0. Now suppose va(zk) = 0 for zk ∈ [ε/2, R1] with z1 < z2 < · · · ≤ R1. Then there exists z∗ with ε/2 < z∗ ≤ R1 such that zk → z∗ ∈ [ε/2, R1] and va(z∗) = 0. In addition, it follows from Lemma 2.1 that v′a(zk) 6= 0 and thus there exist local extrema, Mk, with zk < Mk < zk+1 and v′a(Mk) = 0. Thus we see Mk → z∗ and v′a(z∗) = 0. But this along with va(z∗) = 0 contradicts Lemma 2.1. Thus va has only a finite number of zeros on [0, R1]. � Lemma 2.3. Assume (H1)–(H4) and let va solve (2.1)-(2.2). Suppose a > 0 is sufficiently small. Then va has a local maximum, M1,a, and a zero, z1,a, on (0, R1). In addition, z1,a → 0, v′a(z1,a)→ 0, and va(M1,a)→ 0 as a→ 0+. More generally, if a > 0 is sufficiently small and k ≥ 1 then va has k zeros, zi,a, and k local extrema, Mi,a, with 0 < M1,a < z1,a < M2,a < z2,a < · · · < Mk,a < zk,a on (0, R1). In addition, lima→0+ zi,a = 0, lima→0+ v ′ a(zi,a) = 0, and lima→0+ |va(Mi,a)| = 0 for 1 ≤ i ≤ k. Proof. From (2.6) we have va + ∫ t 0 ∫ s 0 h(x)f(va(x)) dx ds = at. (2.13) Suppose now that va > 0 on (0, R1). Then from (H2) and (H3) there is a constant f2 > 0 such that f(va) ≥ f2v−qa . In addition, from (2.4) we see that h(t) ≥ h1t−α̃1 and 1− α̃1 − q > 0. Substituting into (2.13) gives∫ t 0 ∫ s 0 h(x)f(va(x)) dx ds ≥ f2h1 ∫ t 0 ∫ s 0 x−α̃1v−qa (x) dx ds. (2.14) Also, it follows from (2.1) and (H3) that when va > 0 we have v′′a < 0 and so integrating this inequality twice on (0, t) gives 0 < va < at. (2.15) Substituting this into (2.14) gives f2h1 ∫ t 0 ∫ s 0 x−α̃1v−qa dx ds ≥ f2h1 aq ∫ t 0 ∫ s 0 x−α̃1−q dx ds = f2h1t 2−α̃1−q aq(1− α̃1 − q)(2− α̃1 − q) . (2.16) Substituting this expression into (2.13)-(2.14) gives 0 < va ≤ at− f2h1t 2−α̃1−q aq(1− α̃1 − q)(2− α̃1 − q) . (2.17) However, the right-hand side of (2.17) is zero when t = (aq+1(1− α̃1 − q)(2− α̃1 − q) f2h1 ) 1 2−α̃1−q and notice that this value of t is less than or equal to R1 if a > 0 is sufficiently small. Thus (2.17) yields a contradiction and therefore va has a first zero, z1,a, and 0 < z1,a < R1 if a > 0 is sufficiently small. In addition, the above argument shows that 0 < z1,a ≤ (aq+1(1− α̃1 − q)(2− α̃1 − q) f2h1 ) 1 2−α̃1−q → 0 as a→ 0+. (2.18) 6 J. IAIA EJDE-2024/06 Thus lim a→0+ z1,a = 0. (2.19) Next we examine the following identity which is straightforward to establish by differentiation and (2.1), 1 2 v′2a + h(t)F (va) + ∫ t 0 (−h′(s))F (va) ds = 1 2 a2. (2.20) Evaluating at z1,a gives 1 2 v′2a (z1,a) = 1 2 a2 + ∫ z1,a 0 h′(s)F (va) ds. (2.21) Since F (t) = ∫ t 0 f(s) ds it follows from (H1) and (H2) that there is a constant f3 > 0 such that F (va) ≤ f3(v1−qa + vp+1 a ) when va > 0. (2.22) Also from (2.4) we have t|h′| h ≤ h3 and so |h′| ≤ h2h3t−1−α̃2 . (2.23) Substituting this into the right-hand side of (2.21) and using (2.15), (2.22) gives∫ z1,a 0 h′(s)F (va) ds ≤ ∫ z1,a 0 f3h2h3t −1−α̃2(a1−qt1−q + ap+1tp+1) dt = f3h2h3 (a1−qz1−α̃2−q 1,a 1− α̃2 − q + ap+1z1−α̃2+p 1,a 1− α̃2 + p ) ≤ f3h2h3a1−qR1−α̃2−q 1 ( 1 1− α̃2 − q + ap+qRp+q1 1− α̃2 + p ) . (2.24) Thus substituting (2.22) and (2.24) into (2.21) gives 1 2 v′2a (z1,a) ≤ 1 2 a2 + f3h2h3a 1−qR1−α̃2−q 1 ( 1 1− α̃2 − q + ap+qRp+q1 1− α̃2 + p ) → 0 (2.25) as a→ 0+. Therefore, lim a→0+ v′a(z1,a) = 0. (2.26) Next since va(0) = va(z1,a) = 0 and v′a(0) = a > 0 it follows that there is a local maximum, M1,a, with 0 < M1,a < z1,a. Evaluating (2.20) at M1,a gives h(M1,a)F (va(M1,a)) = 1 2 a2 + ∫ M1,a 0 h′(t)F (va) dt. (2.27) Estimating as in (2.24)-(2.24) but now on [0,M1,a] (instead of [0, z1,a]) we again obtain∫ M1,a 0 h′(t)F (va) dt ≤ f3h2h3a1−qR1−α̃2−q 1 ( 1 1− α̃2 − q + ap+qRp+q1 1− α̃2 + p ) . (2.28) Then from (2.27)-(2.28) and (2.4) we obtain F (va(M1,a)) ≤ f3h2h3a 1−qR1−α̃2+α̃1−q 1 h1 ( 1 1− α̃2 − q + ap+qRp+q1 1− α̃2 + p ) → 0 (2.29) as a→ 0+. Therefore, lim a→0+ va(M1,a) = 0. (2.30) EJDE-2024/06 SINGULAR SUPERLINEAR EQUATIONS ON EXTERIOR DOMAINS 7 In a similar way we can show va has as many zeros as desired by choosing a > 0 sufficiently small and we can also similarly establish the analogs of (2.19), (2.26), and (2.30). This completes the proof of the lemma. � Lemma 2.4. Assume (H1)–(H4) and let va solve (2.1)-(2.2). If a > 0 is sufficiently large then va has a local maximum, M1,a, on (0, R1). Proof. Suppose not and so suppose va is increasing on (0, R1) for all sufficiently large a > 0. Then va > 0 on (0, R1) and so it follows from (2.1) that v′′a < 0 on (0, R1). We now claim that va(t0)→∞ as a→∞ for any fixed t0 with 0 < t0 ≤ R1. So suppose not. Thus suppose 0 < va ≤ C2 on (0, t0] where C2 is independent of a. Using (2.15) and (2.22) we see that F (va) ≤ f3(v1−qa + vp+1 a ) = f3v 1−q a (1 + vp+qa ) ≤ f3v1−qa (1 + Cp+q2 ) = f3C3v 1−q a (2.31) where C3 = 1 + Cp+q2 . Then using (2.15) in (2.31) we obtain F (va) ≤ f3C3v 1−q a ≤ f3C3a 1−qt1−q. (2.32) Substituting this into (2.20) and using (2.4) we then have h(t) ≤ h2t−α̃2 and |h′| ≤ h2h3t −α̃2−1. This gives h(t)F (va) + ∫ t 0 (−h′(s))F (va) ds ≤ f3h2C3 ( 1 + h3 1− α̃2 − q ) a1−qt1−α̃2−q = C4a 1−qt1−α̃2−q ≤ C4a 1−qt1−α̃2−q 0 (2.33) where C4 = f3h2C3 ( 1 + h3 1−α̃2−q ) . Therefore from (2.20) and (2.33) we see that 1 2 v′2a ≥ 1 2 a2 − C4t 1−α̃−q 0 a1−q ≥ 1 2 a2 − C4R 1−α̃−q 1 a1−q ≥ 1 8 a2 for a sufficiently large. Thus v′a ≥ a/2 for a sufficiently large, and integrating this on (0, t0) gives C2 ≥ va(t0) ≥ a 2 t0 →∞ as a→∞. Hence we obtain a contradiction. Thus it follows that if va is increasing on [0, R1] then va(t0)→∞ as a→∞ for every t0 with 0 < t0 ≤ R1. Next it follows that if va is increasing on [0, R1] then since f is superlinear (by (H1)) then h(t)f(va) va →∞ uniformly on [t0, R1] for any t0 > 0 as a→∞. Therefore assuming va is increasing on [0, R1] we see that Ia = inf [t0,R1] h(t)f(va) va →∞ as a→∞. (2.34) Next we rewrite (2.1) as v′′a + (h(t)f(va) va ) va = 0. (2.35) 8 J. IAIA EJDE-2024/06 Assuming va is increasing on [0, R1], we let y solve y′′ + Iay = 0 (2.36) with y(t0) = va(t0) and y′(t0) = v′a(t0). Thus y = va(t0) cos( √ Ia(t− t0)) + v′a(t0)√ Ia sin( √ Ia(t− t0)) and so it follows that y is 2π/ √ Ia-periodic. Thus y must have a local maximum on [t0, t0 + 2π√ Ia ]. In addition, it follows from (2.34) that [t0, t0 + 2π√ Ia ] ⊂ [t0, R1] if a is sufficiently large. We will now show that va must have a local maximum on [t0, t0 + 2π√ Ia ] ⊂ [t0, R1] if a is sufficiently large. This is essentially the Sturm Comparison Theorem [6] but we write out the details because they are brief. Let a > 0 be sufficiently large so that y has a local maximum M < R1 and that y′ > 0 on [t0,M ]. Multiplying (2.35) by y, (2.36) by va, and subtracting gives (yv′a − y′va)′ + (h(t)f(va) va − Ia ) yva = 0. (2.37) Integrating this on [t0,M ] and using y′(M) = 0, y(t0) = va(t0), and y′(t0) = v′a(t0) gives y(M)v′a(M) + ∫ M t0 (h(t)f(va) va − Ia ) yva dt = 0. (2.38) On [t0,M ] we have y > 0, va > 0. In addition, the term in parentheses in (2.38) is nonnegative. Thus we see y(M)v′a(M) ≤ 0 and therefore v′a(M) ≤ 0 since y(M) > 0. Now if v′a(M) < 0 then since v′a(t0) > 0 it follows that va has a local maximum, M1,a, with t0 < M1,a < M . On the other hand, if v′a(M) = 0 then from (2.1) it follows that v′′a(M) < 0 and therefore M is a local maximum for va and we set M1,a = M . Therefore in both cases we see that va has a local maximum, M1,a, with 0 < M1,a < R1 and v′a > 0 on [0,M1,a) if a > 0 is sufficiently large. � Lemma 2.5. Assume (H1)–(H4) and let va solve (2.1)-(2.2). Suppose a > 0 is sufficiently large so that va has a smallest local maximum M1,a with v′a > 0 on [0,M1,a) and M1,a < R1. Then lima→∞ va(M1,a) =∞ and lima→∞M1,a = 0. Proof. We first show that va(M1,a) → ∞ as a → ∞. So suppose not. Mimicking the proof of Lemma 2.4, suppose there is a C5 > 0 such that va(M1,a) ≤ C5. Then using (2.31)-(2.32) and evaluating (2.20) and (2.33) at t = M1,a gives 1 2 a2 = h(M1,a)F (va(M1,a)) + ∫ M1,a 0 (−h′(s))F (va) ds ≤ f3h2C5 ( 1 + h3 1− α̃2 − q ) a1−qt1−α̃2−q = C6a 1−qM1−α̃2−q 1,a ≤ C6a 1−qR1−α̃2−q 1 (2.39) where C6 = f3h2C5(1 + h3 1−α̃2−q ). Thus 1 2 a1+q ≤ C6R 1−α̃2−q 1 . (2.40) EJDE-2024/06 SINGULAR SUPERLINEAR EQUATIONS ON EXTERIOR DOMAINS 9 However, the left-hand side of (2.40) goes to infinity as a→∞ but the right-hand side stays finite. Hence we obtain a contradiction and therefore we must have lim a→∞ va(M1,a) =∞. (2.41) Next we show M1,a → 0 as a→∞. By (H1) it follows that f(va) ≥ f4vpa when va > 0 for some constant f4 > 0. (2.42) We integrate (2.1) on (t,M1,a) and estimate using the fact that va is increasing on (t,M1,a) to obtain: v′a = ∫ M1,a t h(s)f(va) ds ≥ f4vpa ∫ M1,a t h(s) ds. (2.43) Dividing by vpa, recalling p > 1, and integrating on ( M1,a 2 ,M1,a) gives v1−pa ( M1,a 2 ) p− 1 ≥ v1−pa ( M1,a 2 )− v1−pa (M1,a) p− 1 ≥ f3 ∫ M1,a M1,a 2 ∫ M1,a s h(s) ds. (2.44) Since v′′a < 0 it follows that va is concave and thus va(λx+(1−λ)y) ≥ λva(x)+(1− λ)va(y) for 0 ≤ λ ≤ 1. In particular, for x = va(M1,a), y = 0, and λ = 1 2 we obtain va( M1,a 2 ) ≥ va(M1,a) 2 . Then it follows from this and (2.41) that va( M1,a 2 ) → ∞ as a → ∞. Since p > 1 it follows then that the left-hand side of (2.44) goes to 0 as a→∞ and thus we must have lim a→∞ M1,a = 0. (2.45) This completes the proof. � Lemma 2.6. Assume (H1)–(H4) and let va solve (2.1)-(2.2). Suppose a > 0 is sufficiently large. Then va has a zero, z1,a, with M1,a < z1,a < R1. In addition, va > 0 and v′a < 0 on (M1,a, z1,a). Further lima→∞ z1,a = 0, lima→∞ va(M1,a) = ∞, and lima→∞ v′a(z1,a) = −∞. More generally, if a is sufficiently large and k ≥ 1 then va has k zeros, zi,a, and k local extrema, Mi,a, with 0 < M1,a < z1,a < M2,a < z2,a < · · · < Mk,a < zk,a on (0, R1). In addition, lima→∞ zi,a = 0, lima→∞ |v′a(zi,a)| =∞, and lima→∞ |va(Mi,a)| =∞ for 1 ≤ i ≤ k. Proof. It follows from Lemma 2.5 that lim a→∞ va(M1,a) =∞. (2.46) Assume now that va > 0 on (M1,a, R1). Then using (2.42) and integrating on (M1,a, t) we obtain −v′a ≥ f4vpa ∫ t M1,a h(s) ds. Dividing by vpa, integrating on (M1,a, t), and using (2.4) gives v1−pa ≥ v1−pa − v1−pa (M1,a) ≥ (p− 1)f4 ∫ t M1,a ∫ s M1,a h(x) dx ds = (p− 1)f4R −α̃1 1 2 (t−M1,a)2. (2.47) 10 J. IAIA EJDE-2024/06 Evaluating (2.47) at t = R1+M1,a 2 we see v1−pa (R1 +M1,a 2 ) ≥ (p− 1)f4R −α̃1 1 2 (R1 −M1,a 2 )2 and therefore vp−1a (R1 +M1,a 2 ) ≤ 8Rα̃1 1 (p− 1)f4(R1 −M1,a)2 . (2.48) By (2.45) we see then for large a that va (R1 +M1,a 2 ) ≤ ( 32Rα̃1−2 1 (p− 1)f4 ) 1 p−1 . (2.49) Using that v′′a < 0 when va > 0 and the mean value theorem we see there is a ca with M1,a < ca < R1+M1,a 2 such that va(M1,a)− va (R1 +M1,a 2 ) = −v′a(ca) (R1 −M1,a 2 ) ≤ −v′a (R1 +M1,a 2 )(R1 2 ) . (2.50) Since v′a > 0 on (0,M1,a) it follows from (2.41) and (2.49) that the left-hand side of (2.50) goes to infinity as a→∞. And then from (2.45) and (2.50) it follows that v′a (R1 +M1,a 2 ) → −∞ as a→∞. (2.51) Since v′′a < 0 when va > 0 it follows that v′a is decreasing when va > 0 so: v′a < v′a( R1 +M1,a 2 ) for t > R1 +M1,a 2 . Integrating this on ( R1+M1,a 2 , R1) gives va(R1) < va( R1 +M1,a 2 ) + v′a( R1 +M1,a 2 )( R1 −M1,a 2 ). (2.52) It follows from (2.49) that the first term on the right-hand side (2.52) is bounded. Then from (2.45) we have M1,a → 0 as a → ∞ and this along with (2.51) implies that the right-hand side of (2.52) becomes negative while the left-hand side stays positive. Thus we obtain a contradiction and therefore there exists z1,a with M1,a < z1,a < R1 such that va(z1,a) = 0 and va > 0 on (M1,a, z1,a). From the mean value theorem and that v′′a < 0 when va > 0 it follows that there is a da such that M1,a < da < z1,a and va(M1,a) = |va(z1,a)−va(M1,a)| = |v′a(da)||z1,a−M1,a| ≤ |v′a(da)|R1 ≤ |v′a(z1,a)|R1 and since the left-hand side goes to infinity by (2.46) it then follows from the above inequality that lim a→∞ v′a(z1,a) = −∞. (2.53) Next it follows from evaluating (2.47) at M1,a+z1,a 2 that we obtain v1−p (M1,a + z1,a 2 ) ≥ (p− 1)f4R −α̃1 1 2 (M1,a − z1,a 2 )2 . (2.54) Since v′′a < 0 when va > 0 it follows that va is concave. Then it follows from this and (2.46) that va( M1,a+z1,a 2 ) ≥ va(M1,a) 2 + va(z1,a) 2 = va(M1,a) 2 → ∞. Thus we see EJDE-2024/06 SINGULAR SUPERLINEAR EQUATIONS ON EXTERIOR DOMAINS 11 the left-hand side of (2.54) goes to 0 as a → ∞ and therefore z1,a −M1,a → 0. Since M1,a → 0 by Lemma 2.5 we see then that lim a→∞ z1,a = 0. (2.55) In a similar way we can show that va as many zeros as desired on (0, R1) by choosing a > 0 sufficiently large, and we can obtain the analogs of (2.46), (2.53), and (2.55). This completes the proof. � Lemma 2.7. Assume (H1)–(H4) and let va solve (2.1)-(2.2) with a > 0. If R1 is sufficiently small then there are values of a > 0 such that va > 0 on (0, R1). Also, if R1 is sufficiently large then va has at least one zero on (0, R1) for all a > 0. Similarly, if R1 > 0 is sufficiently large then va has at least k zeros on (0, R1) for all a > 0. Proof. We prove the second part first. It follows from (H1)–(H3) that there is a constant f5 > 0 such that f(v) v ≥ f5 for all v 6= 0. In addition, we know from (2.4) that h(t) ≥ h1t−α̃1 ≥ h1R−α̃1 1 . Thus h(t)f(va) va ≥ f5 R α̃1 1 . Next we consider w′′ + (f5h1 Rα̃1 1 ) w = 0, w(0) = 0, w′(0) = a. Thus: w = c sin (√f5h1 Rα̃1 1 x ) for some c > 0, and so w has a zero on [0, √ R α̃1 1 f5h1 π]. It follows then from the Sturm Comparison Theorem [6] that va has at least one zero on [0, R1] if √ R α̃1 1 f5h1 π < R1. That is, if R1 > ( π2 f5h1 ) 1 2−α̃1 = ( π2 f5h1 ) N−2 α1−2 . Similarly, va has at least k zeros on [0, R1] if R1 > (k2π2 f5h1 ) 1 2−α̃1 = (k2π2 f5h1 ) N−2 α1−2 . Next we show that if R1 is sufficiently small then there is a value of a > 0 such that va > 0 on (0, R1). First since f(va) > 0 for va > 0 by (H3) there is a constant f6 > 0 such that f(va) ≥ f6 > 0 for va > 0. Thus it follows from this and (2.4) that h(t)f(va) ≥ f6h1t−α̃1 . Suppose now that va has a zero, za, on (0, R1). Then there is an Ma with 0 < Ma < za such that va has a local maximum at Ma. Substituting t = Ma into (2.5) then gives f6h1M 1−α̃1 a 1− α̃1 ≤ ∫ Ma 0 f6h1t −α̃1 dt ≤ ∫ Ma 0 h(t)f(va) dt = a. It follows from this that lim a→0+ Ma = 0. (2.56) 12 J. IAIA EJDE-2024/06 Returning to (2.20) and evaluating at Ma we see that 1 2 a2 = h(Ma)F (va(Ma)) + ∫ Ma 0 (−h′(t))F (va) dt. (2.57) Then using (2.15), (2.22), and (2.4) we see that∫ Ma 0 (−h′(t))F (va) dt ≤ f3h2h3 ∫ Ma 0 t−α̃2−1(a1−qt1−q + ap+1tp+1) dt = f3h2h3a 1−q ( R1−α̃2−q 1 1− α̃2 − q + ap+qR1−α̃2+p 1 1− α̃2 + p ) . (2.58) Similarly, h(Ma)F (va(Ma)) ≤ f3h2a1−q ( R1−α̃2−q 1 + ap+qR1−α̃2+p 1 ) . (2.59) Now substituting (2.58)-(2.59) into (2.57) gives 1 2 a2 ≤ f3h2a1−q ( C7R 1−α̃2−q 1 + ap+qC8R 1−α̃2+p 1 ) , (2.60) where C7 = (1 + h3 1−α̃2−q ) and C8 = (1 + h3 1−α̃2+p ). Select a = 1 and we see (2.60) becomes 1 ≤ 2f3h2 ( C7R 1−α̃2−q 1 + C8R 1−α̃2+p 1 ) (2.61) Now if R1 is sufficiently small we see that this violates (2.61). Thus if R1 is suffi- ciently small and if a = 1 then va > 0 on (0, R1). This completes the proof. � 3. Proof of Theorem 1.1 We saw from Lemma 2.2 that va has a finite number of zeros on (0, R1) for a > 0. Thus there exists an a > 0 such that va has the least number of zeros on (0, R1) among all a > 0. We denote the number of zeros of this particular va as n0 ≥ 0. (There may be more than one choice of a such that va has n0 zeros on (0, R1) but choose one such a). Now let Sn0 = {a > 0 : va solves (2.1)-(2.2) and has exactly n0 zeros on (0, R1)}. From the above comments it follows that Sn0 is nonempty and from Lemma 2.6 it follows that Sn0 is bounded above. Next let an0 = supSn0 . We now prove that van0 has exactly n0 zeros on (0, R1) and van0 (R1) = 0. From the definition of n0 it follows that van0 has at least n0 zeros on (0, R1). Now if van0 has an (n0 + 1)st zero on (0, R1) then by continuity with respect to initial conditions then so does va for a close to an0 and a < an0 but if a < an0 then va has only n0 zeros. Thus van0 has exactly n0 zeros on (0, R1). Now suppose van0 (R1) 6= 0. Without loss of generality suppose that van0 (R1) > 0. Now if a is close to an0 and a > an0 then by continuity with respect to initial conditions and the fact that if va(z) = 0 then v′a(z) 6= 0 it follows that va(R1) > 0 and also va has n0 zeros on (0, R1). But since a > an0 then va has at least n0 + 1 zeros on (0, R1) and so we obtain a contradiction. Thus it must be the case that van0 (R1) = 0 and thus we obtain a solution of (2.1)-(2.2). Then by Lemma 2.1 it follows that v′an0 (R1) 6= 0 so let us assume without loss of generality that v′an0 (R1) < 0. In a similar way we now define Sn0+1 = {a > 0 : va solves (2.1)-(2.2) and has exactly n0 + 1 zeros on (0, R1)}. EJDE-2024/06 SINGULAR SUPERLINEAR EQUATIONS ON EXTERIOR DOMAINS 13 It follows from Lemma 2.6 that Sn0+1 is bounded from above. For a > an0 and a sufficiently close to an0 it follows again by continuity with respect to initial con- ditions that va has an (n0 + 1)st zero zn0+1 < R1 and zn0+1 is close to R1. In addition, since v′an0 (R1) < 0 it follows that v′a(zn0+1) < 0. Thus va has exactly n0 + 1 zeros on (0, R1) for a > an0 and a sufficiently close to an0 . Therefore Sn0+1 is nonempty. Similarly we define an0+1 = supSn0+1 and we can similarly show that van0+1 has exactly n0 + 1 zeros on (0, R1) and van0+1(R1) = 0. Continuing in this way we see that we can find an infinite number of solutions, van , where van has exactly n zeros on (0, R1) and van(R1) = 0 for each n ≥ n0. Thus we have found one infinite family of solutions of (2.1)-(2.2). Next we let bn0 = inf Sn0 . By the above comments Sn0 is nonempty and by definition Sn0 is bounded below. Then bn0 ≤ an0 and by a similar argument we can show that vbn0 has exactly n0 zeros on (0, R1) and van0 (R1) = 0. Now it may be the case that an0 = bn0 so there may be only one solution with n0 zeros. Next we let bn0+1 = inf Sn0+1. Then we have bn0+1 < bn0 ≤ an0 < an0+1 and we can show vbn0+1 has exactly n0 + 1 zeros on (0, R1) and vbn0+1(R1) = 0. Since bn0+1 < an0+1 it follows that we have two solutions, van0 and vbn0 , with n0 + 1 zeros on (0, R1). Continuing in this way we see that if n > n0 we can find a second infinite family of solutions of (2.1)-(2.2), vbn , where vbn has exactly n zeros on (0, R1) and vbn(R1) = 0. Finally, we let u+n (t) = van(t 1 2−N ) and u−n (t) = vbn(t 1 2−N ) for all n ≥ n0. This completes the proof of Theorem 1.1. � References [1] A. Abebe, M. Chhetri, L. Sankar, R. Shivaji; Positive solutions for a class of superlinear semipositone systems on exterior domains, Boundary Value Problems, 198, 2014. [2] M. Ali, J. Iaia; Existence and nonexistence for singular, sublinear problems on exterior do- mains, Electronic Journal of Differential Equations, Vol. 2021, No. 3, 1-17, 2021. [3] M. Ali, J. Iaia; Infinitely many solutions for a singular, semilinear problem on exterior do- mains, Electronic Journal of Differential Equations, Vol. 2021, No. 68, 1-17, 2021. [4] H. Berestycki, P. L. Lions; Non-linear scalar field equations I, Arch. Rational Mech. Anal., Volume 82, 313-347, 1983. [5] H. Berestycki, P. L. Lions, L. A. Peletier; An ODE approach to the existence of positive solutions for semilinear problems on RN , Indiana University Mathematics Journal, Volume 30, No. 1, 141-157, 1981. [6] G. Birkhoff, G. C. Rota; Ordinary Differential Equations, 4th ed., Wiley, 1991. [7] M. Chhetri, L. Sankar, R. Shivaji; Positive solutions for a class of superlinear semipositone systems on exterior domains, Boundary Value Problems, 198-207, 2014. [8] L. Evans; Partial Differential Equations, 2nd ed., American Mathematical Society, 2010. [9] J. Iaia; Existence of solutions for semilinear problems on exterior domains, Electronic Journal of Differential Equations, No. 34, 1-10, 2020. [10] J. Iaia; Existence of solutions for semilinear problems with prescribed number of zeros on exterior domains, Journal of Mathematical Analysis and Applications, 446, 591-604, 2017. [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, No. 2, 1597-1611, 2016. [12] E. Lee, L. Sankar, R. Shivaji; Positive solutions for infinite semipositone problems on exterior domains, Differential and Integral Equations, Volume 24, Number 9/10, 861-875, 2011. 14 J. IAIA EJDE-2024/06 [13] 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. Joseph Iaia Department of Mathematics, University of North Texas, Denton, TX 76203-5017, USA Email address: iaia@unt.edu 1. Introduction 2. Preliminaries 3. Proof of Theorem ?? References