Electronic Journal of Differential Equations, Vol. 2020 (2020), No. 34, pp. 1–10. ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu EXISTENCE OF SOLUTIONS FOR SEMILINEAR PROBLEMS ON EXTERIOR DOMAINS JOSEPH IAIA Abstract. In this article we prove the existence of an infinite number of radial solutions to ∆u+K(r)f(u) = 0 on RN such that limr→∞ u(r) = 0 with prescribed number of zeros on the exterior of the ball of radius R > 0 where f is odd with f < 0 on (0, β), f > 0 on (β,∞) with f superlinear for large u, and K(r) ∼ r−α with α > 2(N − 1). 1. Introduction In this article we study radial solutions of ∆u+K(|x|)f(u) = 0 for R < |x| <∞, (1.1) u(x) = 0 when |x| = R, lim |x|→∞ u(x) = 0, (1.2) where u : RN → R with N > 2, R > 0, f : R→ R is odd and locally Lipschitz with (H1) f ′(0) < 0, there exists β > 0 such that f(u) < 0 on (0, β), f(u) > 0 on (β,∞). (H2) f(u) = |u|p−1u+ g(u) where p > 1 and lim u→∞ |g(u)| |u|p = 0. (H3) Denoting F (u) ≡ ∫ u 0 f(t) dt we also assume that thee exists γ with 0 < β < γ such that F < 0 on (0, γ) and F > 0 on (γ,∞). (H4) Further we assume K and K ′ are continuous on [R,∞) and K(r) > 0, there exists α > 2(N − 1) such that limr→∞ rK ′/K = −α. (H5) There exist positive constants d1, d2 such that 2(N − 1) + rK ′ K < 0, d1r −α ≤ K(r) ≤ d2r−α for r ≥ R. Our main result read as follows. Theorem 1.1. Assume (H1)–(H5) and N > 2. Then for each nonnegative integer n there exists a radial solution, un, of (1.1)–(1.2) such that un has exactly n zeros on (R,∞). 2010 Mathematics Subject Classification. 34B40, 35B05. Key words and phrases. Exterior domain; superlinear; radial solution. c©2020 Texas State University. Submitted January 12, 2019. Published April 15, 2020. 1 2 J. IAIA EJDE-2020/34 The radial solutions of (1.1)–(1.2) on RN with K(r) ≡ 1 have been well-studied. These include [2, 3, 8, 9, 10]. Recently there has been an interest in studying these problems on RN\BR(0). These include [1, 5, 6, 7]. In these papers 0 < α < 2(N−1). In this paper we consider α > 2(N − 1). Here we use a scaling argument as in [9] to prove existence of solutions. A key difference between the 0 < α < 2(N − 1) case and the α > 2(N − 1) case is that the function E(r) = 1 2 u′2 K(r) + F (u) is non-increasing for 0 < α < 2(N − 1) and nondecreasing for α > 2(N − 1). For 0 < α < 2(N − 1) this allows us to obtain important estimates on the growth of solutions. For α > 2(N −1) we are unable to do this so instead we make the change of variables u(r) = u1(r2−N ) and investigate the differential equation for u1 on [0, R2−N ]. For this equation it turns out there is a function E1 = 1 2 u′21 h(t) + F (u1) that is nondecreasing and so we can apply some similar analysis as we did in the 0 < α < 2(N − 1) case. The outline of this paper is as follows: in section two we establish existence of a radial solutions of (1.1)–(1.2) with u(R) = 0 and u′(R) > 0 on [R,∞). We then make the change of variables u1(r) = u(r2−N ) and transform our problem to the compact set [0, R2−N ] with u1(R2−N ) = 0 and u′1(R2−N ) = −b∗ < 0. The rest of section two is devoted to showing that u1(r) stays positive if b∗ > 0 stays sufficiently small and that u1(r) has more and more zeros as b∗ →∞. In section 3 we prove the main theorem by choosing appropriate values of the parameter b∗, say b∗n, such that u1,n is a solution with exactly n zeros on (0, R2−N ) for each nonnegative integer n and hence converting back to the original notation we get a solution of our original equation with exactly n zeros on (R,∞) and u(r)→ 0 as r →∞. 2. Preliminaries Since we are interested in radial solutions of (1.1)–(1.2), we denote r = |x| and write u(x) = u(|x|) where u satisfies u′′ + N − 1 r u′ +K(r)f(u) = 0 for R < r <∞, (2.1) u(R) = 0, u′(R) = b > 0. (2.2) We will occasionally write u(r, b) to emphasize the dependence of the solution on b. By the standard existence-uniqueness theorem [4] there is a unique solution of (2.1)–(2.2) on [R,R+ ε) for some ε > 0. We next we consider E(r) = 1 2 u′2 K(r) + F (u). (2.3) It is straightforward using (2.1) and (H5) to show that E′(r) = − u′2 2rK [2(N − 1) + rK ′ K ] ≥ 0. (2.4) Thus E is non-decreasing. Therefore, 1 2 u′2 K(r) + F (u) = E(r) ≥ E(R) = 1 2 b2 K(R) for r ≥ R. (2.5) Next we let u(r) = u1(r2−N ) (2.6) EJDE-2020/34 SEMILINEAR PROBLEMS ON EXTERIOR DOMAINS 3 where we denote R∗ = R2−N , b∗ = bRN−1 N − 2 . (2.7) This transforms our equation (2.1)–(2.2) into u′′1(t) + h(t)f(u1(t)) = 0 for 0 < t < R1, (2.8) where u1(R∗) = 0, u′1(R∗) = −b∗ < 0, (2.9) and h(t) = 1 (N − 2)2 t 2(N−1) 2−N K(t1/(2−N)). Since (r2(N−1)K)′ < 0 (by (H5)) and t = r 1 2−N with N > 2 it follows that h′(t) > 0 for 0 < t ≤ R∗. (2.10) In addition, from (H5) we see that 0 < d1 (N − 2)2 ≤ h(t) tq ≤ d2 (N − 2)2 for 0 < t ≤ R∗ (2.11) where q = α−2(N−1) N−2 > 0 (by (H4)). Now let E1 = 1 2 u′21 h(t) + F (u1). (2.12) Then using (2.8) and (2.10) we see that E′1 = −u ′2 1 h ′ 2h2 ≤ 0. Therefore, 1 2 u′21 h(t) + F (u1) ≥ 1 2 (b∗)2 h(R∗) on (t, R∗). (2.13) Also we consider E2 = 1 2 u′21 + h(t)F (u1). (2.14) Using (2.8) this gives E′2 = h′(t)F (u1). Integrating this on (t, R∗) gives 1 2 u′21 + h(t)F (u1) + ∫ R∗ t h′(s)F (u1) ds = 1 2 (b∗)2. (2.15) It follows from (H3) that F is bounded from below so there exists F0 > 0 such that F (u1) ≥ −F0 for all u1 ∈ R. Also since h′(t) > 0 by (2.10) we see that∫ R∗ t h′(s)F (u1) ds ≥ −F0 [h(R∗)− h(t)] . (2.16) Therefore, since h(t) > 0 and h(t) is bounded on [0, R∗] by (2.11) we see from (2.15)-(2.16) that 1 2 u′21 + h(t)F (u1) ≤ 1 2 (b∗)2 + F0[h(R∗)− h(t)] ≤ 1 2 (b∗)2 + F0h(R∗). (2.17) 4 J. IAIA EJDE-2020/34 It follows from (2.17) that for fixed b∗, then u1 and u′1 are uniformly bounded on [0, R∗] and therefore the solution u1 exists on [0, R∗]. Therefore, the solution u of (2.1)–(2.2) exists on [R,∞). Lemma 2.1. If b∗ > 0 is sufficiently small, then 0 < u1 < β on (0, R∗). Proof. We first note that if u1 has a local maximum then there exists Mb∗ with u′1 < 0 on (Mb∗ , R ∗), u′1(Mb∗) = 0, and with u′′1(Mb∗) ≤ 0. Thus f(u1(Mb∗)) ≥ 0 from (2.8) and therefore u1(Mb∗) ≥ β. Thus while 0 < u1 < β we see that u1 is monotone. So suppose now that the lemma is false. Then for every b > 0 with b sufficiently small there exists an sb∗ with 0 < sb∗ < R∗ such that u1(sb∗) = β and u′1 < 0 on (sb∗ , R ∗). Now integrating (2.8) on (t, R∗) and using (2.9) gives u′1 = −b∗ + ∫ R∗ t h(s)f(u1) ds. Integrating again on (t, R∗) gives u1(t) = b∗(R∗ − t)− ∫ R∗ t ∫ R∗ s h(x)f(u1(x)) dx ds. Observe from (H1) that there exists c1 > 0 such that f(u1) ≥ −c1u1 when u1 ≥ 0. (2.18) Then using (2.18) and the fact that u1 is decreasing on (sb∗ , R ∗) we obtain u1(t) ≤ b∗(R∗ − t) + ∫ R∗ t c1d(s)u1(s) ds (2.19) where d(s) = ∫ R∗ s h(x) dx > 0. (2.20) Then we let W (t) = ∫ R∗ t d(s)u1(s) ds (2.21) and from (2.21) we observe W ′(t) = −d(t)u1(t). Next, multiplying (2.19) by d(t) we obtain −W ′ ≤ b∗(R∗ − t)d(t) + c1d(t)W. Thus −b∗(R∗ − t)d(t) ≤W ′ + c1d(t)W. Denoting D(t) = e ∫ t 0 c1d(s) ds > 0 and multiplying the previous inequality by D(t) gives −b∗(R∗ − t)d(t)D(t) ≤ (D(t)W (t)) ′ . Integrating on (t, R∗) gives D(t)W (t) ≤ b∗ ∫ R∗ t (R∗ − s)d(s)D(s) ds thus from (2.21) and the definition of D(t) we see that∫ R∗ t d(s)u1(s) ds = W (t) ≤ b∗e− ∫ t 0 c1d(s) ds ∫ R∗ t (R∗ − s)d(s)e ∫ s 0 c1d(x) dx ds. EJDE-2020/34 SEMILINEAR PROBLEMS ON EXTERIOR DOMAINS 5 Then from (2.19) we see that u1(t) ≤ b∗ ( (R∗ − t) + c1e − ∫ t 0 c1d(s) ds ∫ R∗ t (R∗ − s)d(s)e ∫ s 0 c1d(x) dx ds ) . (2.22) Since h(t) is bounded on [0, R∗], it follows from (2.20) that d(t) is bounded on [0, R∗] and thus the term in the large parentheses in (2.22) is bounded on [0, R∗]. Therefore, from (2.22) we see there exists a c2 > 0 which is independent of b∗ such that u1(t) ≤ c2b∗ on [sb∗ , R ∗]. Evaluating this at sb∗ give 0 < β ≤ c2b ∗ → 0 as b∗ → 0 which is a contradiction. Thus we see that if b∗ > 0 is sufficiently small then 0 < u1 < β on (0, R∗). � Lemma 2.2. If b∗ is sufficiently large then u1 has a local maximum, Mb∗ , and Mb∗ → R∗ as b∗ →∞. Proof. Using (2.13) we see that if F (u1) ≤ 1 4 (b∗)2 h(R∗) , then u′21 h(t) ≥ 1 2 (b∗)2 h(R∗) . (2.23) In particular, in a neighborhood of t = R∗ we have F (u1) ≤ 1 4 (b∗)2 h(R∗) since F (u1(R∗)) = 0. Also since u′1 < 0 near t = R∗ then from (2.23): −u′1 ≥ b∗ √ h(t)√ 2h(R∗) on (t, R∗) with t near R∗. Integrating this on (t, R∗) gives u1(t) ≥ b∗√ 2h(R∗) ∫ R∗ t √ h(s) ds when F (u1) ≤ 1 4 (b∗)2 h(R∗) . (2.24) Now from (H2)-(H3) it follows that there is a c3 > 0 such that F (u1) ≥ 1 2(p+1) |u1| p+1 − c3 for all u1 ∈ R. From this and (2.23)-(2.24) we see that 1 2(p+ 1) ( b∗√ 2h(R∗) ∫ R∗ t √ h(s) ds )p+1 − c3 ≤ F (u1) ≤ (b∗)2 4h(R∗) . Rewriting this gives∫ R∗ t √ h(s) ds ≤ [ 2(p+ 1) ( c3 (b∗)p+1 + 1 4h(R∗)(b∗)p−1 )] 1 p+1√ 2h(R∗). (2.25) Since p > 1, the right-hand side of (2.25) approaches 0 as b∗ → ∞. Since∫ R∗ 0 √ h(s) ds > 0 we see that F (u1(t)) cannot be bounded by 1 4 (b∗)2h(R∗) for all t ∈ [0, R∗] and for all sufficiently large b∗. Thus for sufficiently large b∗ there exists tb∗ ∈ (0, R∗) such that F (u1(tb∗)) = (b∗)2 4h(R∗) (2.26) where 0 < u1 < u1(tb∗) on (tb∗ , R ∗). Now evaluating (2.25) at t = tb∗ and noticing the right-hand side of (2.25) goes to 0 as b∗ →∞ it follows that tb∗ → R∗ as b∗ →∞. (2.27) 6 J. IAIA EJDE-2020/34 We also note that from (H2) and (H3), there is a c4 ≥ 1 such that F (u1) ≤ c4 p+1 |u1| p+1 for all u1 ∈ R. From this and (2.26) we see that c4 p+ 1 up+1 1 (tb∗) ≥ F (u1(tb∗)) = (b∗)2 4h(R∗) (2.28) and so u1(tb∗) ≥ c5(b∗) 2 p+1 where c5 = ( (p+ 1) 4h(R∗)c4 ) 1 p+1 > 0. (2.29) Suppose now that u1 does not have a local maximum for b∗ sufficiently large so that u′1 < 0 on (0, R∗) for large b∗. We then define Q(b∗) = 1 2 inf [ 12 tb∗ ,tb∗ ] h(t) f(u1) u1 . Since tb∗ → R∗ as b∗ →∞ by (2.27) it follows that the interval [ 12 tb∗ , tb∗ ] is bounded from below by a positive constant as b∗ → ∞ and so h(t) is bounded from below on [ 12 tb∗ , tb∗ ] by a positive constant for large values of b∗. In addition, since u1 is decreasing on [ 12 tb∗ , tb∗ ] then by (2.29), u1(t) ≥ u1(tb∗) ≥ c5(b∗) 2 p+1 on [ 1 2 tb∗ , tb∗ ] (2.30) and since f(u1) u1 →∞ as u1 →∞ by (H2) it follows that Q(b∗)→∞ as b∗ →∞. (2.31) We now compare the solution of (2.8), i.e., u′′1 + [ h(t) f(u1) u1 ] u1 = 0, (2.32) with the solution of v′′1 +Q(b∗)v1 = 0, (2.33) where v1(tb∗) = u1(tb∗) > 0 and v′1(tb∗) = u′1(tb∗) < 0. Since the general solution of (2.33) is v1 = c6 sin( √ Q(b∗)(t − c7)) for some constants c6 6= 0 and c7 we see that any interval of length π√ Q(b∗) has a zero of v1. And since tb∗ → R∗ as b∗ → ∞ by (2.27), it follows from (2.31) that v1 is zero somewhere on [ 12 tb∗ , tb∗ ] since π√ Q(b∗) < 1 2 tb∗ for b∗ sufficiently large. In particular, v1 must have a local maximum, mb∗ , with mb∗ ≥ 1 2 tb∗ , v ′ 1 < 0 on (mb∗ , tb∗ ], and v1 > 0 on [mb∗ , tb∗ ]. We claim now that u1 also has a local maximum on (mb∗ , tb∗ ] for b∗ sufficiently large. So suppose not then u′1 < 0 and u1 > 0 on (mb∗ , tb∗ ]. Multiplying (2.32) by v1, multiplying (2.33) by u1, and subtracting we obtain (v1u ′ 1 − u1v′1)′ + ( h(t) f(u1) u1 −Q(b∗) ) u1v1 = 0. Integrating this on [mb∗ , tb∗ ] gives − v1(mb∗)u ′ 1(mb∗) + ∫ tb∗ mb∗ ( h(t) f(u1) u1 −Q(b∗) ) u1v1 dt = 0. (2.34) We note v1(mb∗) > 0 and that both u1 and v1 are positive on [mb∗ , tb∗ ]. Since h(t) f(u1) u1 − Q(b∗) > 0 on [mb∗ , tb∗ ], it follows from (2.34) that u′1(mb∗) > 0 which contradicts that u′1 < 0 on [mb∗ , tb∗ ]. So we see that u1 must also have a local EJDE-2020/34 SEMILINEAR PROBLEMS ON EXTERIOR DOMAINS 7 maximum, Mb∗ , with Mb∗ > mb∗ and u′1 < 0 on (Mb∗ , R ∗]. This completes the first part of the proof. Next we show Mb∗ → R∗ as b∗ →∞. Integrating (2.8) on (Mb∗ , t) gives − u′1(t) = ∫ t Mb∗ h(s)f(u1) ds. (2.35) Now since f(u1) ≥ 1 2u p 1 when u1 > 0 is large (by (H2)) and since u1 is decreasing on (Mb∗ , R ∗) then when b∗ is sufficiently large and when Mb∗ < t < tb∗ then u1(t) ≥ u1(tb∗)→∞ as b∗ →∞ by (2.29) so we obtain from (2.35): −u′1(t) ≥ 1 2 up1(t) ∫ t Mb∗ h(s) ds. Dividing by up1, integrating on (Mb∗ , tb∗), and estimating gives 1 (p− 1)up−11 (tb∗) ≥ 1 2 ∫ tb∗ Mb∗ ∫ s Mb∗ h(x) dx ds. (2.36) Now the left-hand side of (2.36) goes to 0 as b∗ → ∞ by (2.30) thus we see from (2.36) that tb∗ −Mb∗ → 0 as b∗ →∞. Also from (2.27) we know that tb∗ → R∗ as b∗ →∞. Therefore, combining these two statements we see Mb∗ → R∗ as b∗ →∞. This completes the proof. � Lemma 2.3. If b∗ is sufficiently large then u1 has an arbitrarily large number of zeros on (0, R∗). Proof. From Lemma 2.2 we know u1 has a local maximum, Mb∗ , with Mb∗ → R∗ as b∗ →∞. Recalling (2.6) it follows that u(r) = u1(r2−N ) has a local maximum, Mb, and Mb → R as b→∞. (2.37) Now we let wλ(r) = λ− 2 p−1u(Mb + r λ ) where λ 2 p−1 = u(Mb). Then w′′λ + N − 1 λMb + r w′λ +K(Mb + r λ )λ −2p p−1 f(λ 2 p−1wλ) = 0, wλ(0) = 1, w′λ(0) = 0. (2.38) Since K ′(r) < 0 and F (u) ≥ −F0 for some F0 > 0 (by (H3)), we see that(1 2 w′2λ +K(Mb + r λ )λ −2(p+1) p−1 F (λ 2 p−1wλ) )′ = − ( N − 1 λMb + r ) w′2λ + λ −2(p+1) p−1 −1K ′(Mb + r λ )F (λ 2 p−1wλ) ≤ −λ −2(p+1) p−1 −1K ′(Mb + r λ )F0. Integrating this on (0, r) gives 1 2 w′2λ +K(Mb + r λ )λ −2(p+1) p−1 F (λ 2 p−1wλ) ≤ K(Mb)λ −2(p+1) p−1 F (λ 2 p−1 )− λ −2(p+1) p−1 F0 [ K(Mb + r λ )−K(Mb) ] . (2.39) 8 J. IAIA EJDE-2020/34 Since K is bounded on [R,∞) it follows that λ −2(p+1) p−1 F0 [ K(Mb + r λ )−K(Mb) ] → 0 as λ→∞. Also from (H2) and (H3) it follows that F (λ 2 p−1 ) = 1 p+1λ 2(p+1) p−1 + G(λ 2 p−1 ) where G(u) = ∫ u 0 g(s) ds and thus by (H2) and L’Hôpital’s rule |G(u) up+1 | → 0 as u → ∞. Therefore λ −2(p+1) p−1 F (λ 2 p−1 ) = 1 p+ 1 + λ −2(p+1) p−1 G(λ 2 p−1 )→ 1 p+ 1 as λ→∞. Also by (H2) and(H3) we see that λ −2(p+1) p−1 F (λ 2 p−1wλ) = 1 p+ 1 wp+1 λ + λ −2(p+1) p−1 G(λ 2 p−1wλ). Then by (2.39) for sufficiently large λ, 1 2 w′2λ +K(Mb + r λ ) 1 p+ 1 |wλ|p+1 ≤ K(R) p+ 1 + 1− λ− 2(p+1) p−1 G(λ 2 p−1wλ). (2.40) Since |G(u) up+1 | → 0 as u→∞ it follows that |G(u)| ≤ 1 2(p+1) |u| p+1 for |u| ≥ A where A is some positive constant and |G(u)| ≤ G0 for |u| ≤ A since G is continuous. Thus |G(u)| ≤ 1 2(p+1) |u| p+1 +G0 for all u and therefore from (2.40): 1 2 w′2λ +K(Mb + r λ ) |wλ|p+1 p+ 1 ≤ K(R) p+ 1 + 1 +K(Mb + r λ ) ( |wλ|p+1 2(p+ 1) + λ− 2(p+1) p−1 G0 ) . Therefore, for sufficiently large λ and since K is bounded we have 1 2 w′2λ +K(Mb + r λ ) |wλ|p+1 2(p+ 1) ≤ K(R) p+ 1 + 2. Thus we see that |wλ| and |w′λ| are uniformly bounded on [R,∞) for large λ. So by the Arzela-Ascoli theorem a there is a subsequence (still labeled wλ) such that wλ → w uniformly on compact sets. Also, since w′λ is uniformly bounded it follows that w′λ λMb+r → 0 as λ→∞. In addition, from (H2) we have K(Mb + r λ )λ −2p p−1 f(λ 2 p−1wλ) = K(Mb + r λ )[wpλ + λ −2p p−1 g(λ 2 p−1wλ)]. Since Mb → R by Lemma 2.2 then K(Mb + r λ )wpλ → K(R)wp uniformly on compact sets. And since g(u) up → 0 as u → ∞ by (H2) it follows that K(Mb + r λ )λ −2p p−1 g(λ 2 p−1wλ)→ 0 uniformly on compact sets as λ→∞. It follows then from (2.38) that |w′′λ| is uniformly bounded. Then by the Arzela-Ascoli theorem we see for some subsequence (still labeled wλ) that wλ → w and w′λ → w′ uniformly on compact sets as λ→∞ and then from (2.38) we see that w satisfies w′′ +K(R)|w|p−1w = 0, w(0) = 1, w′(0) = 0. Now it is straightforward to show that this has infinitely many zeros on [0,∞) and therefore wλ and hence u has an arbitrarily large number of zeros on (R,∞) provided b is chosen sufficiently large. Also it follows that u1 has an arbitrarily large number of zeros provided b∗ is chosen sufficiently large. This completes the proof. � EJDE-2020/34 SEMILINEAR PROBLEMS ON EXTERIOR DOMAINS 9 3. Proof of the main theorem From Lemma 2.3 we see that the set {b∗ : u1(r, b∗) has at least one zero on (0, R∗)} is nonempty. And since 0 < u1(r, b∗) < β on (0, R∗) for b∗ > 0 sufficiently small by Lemma 2.2 then we see that this set is bounded from below by a positive constant. So we let b∗0 = inf{b∗ : u1(r, b∗) has at least one zero on 0 < t < R∗} and note that b∗0 > 0. In addition, it follows by continuity with respect to initial conditions that u1(r, b∗0) ≥ 0 on (0, R∗). We claim next that u1(r, b∗0) > 0 for 0 < t < R∗. If not then there is a z with 0 < z < R∗ such that u1(z, b∗0) = 0. Since u1(r, b∗0) ≥ 0 it follows that u′1(z, b∗0) = 0. This however implies u1 ≡ 0 contradicting u′1(R∗, b∗0) = −b∗0 < 0. Thus it must be that u1(t, b∗0) > 0 for 0 < t < R∗. Also, for b∗ > b∗0 then by definition of b0 there is a zb∗ such that u1(zb∗ , b ∗ 0) = 0. It follows that zb∗ → 0 as b∗ → (b∗0)+ otherwise a subsequence of these would converge to a z0 with 0 < z0 ≤ R∗ such that u1(z0, b ∗ 0) = 0. Since b∗0 > 0 it follows that u′1(R∗, b∗0) = −b∗0 < 0 and so z0 < R∗ but then this contradicts that u1(r, b∗0) > 0 for 0 < t < R∗. Thus zb∗ → 0 as b∗ → (b∗0)+. Then 0 = u1(zb∗ , b ∗) → u1(0, b∗0) as b∗ → (b∗0)+ thus we see that u1(0, b∗0) = 0. Thus u1(t, b∗0) is a positive solution of (2.8)-(2.9). Now if we let b0 = (N−2)b∗0 RN−1 then it follows that u(r, b0) is a positive solution of (2.1)–(2.2) and limr→∞ u(r, b0) = 0. Next by Lemma 2.3 we see that the set {b∗ : u1(t, b∗) has at least two zeros on 0 < t < R∗} is nonempty and from Lemma 2.1 this set is bounded from below. And so we let b∗1 = inf{b∗ : u1(r, b∗) has at least two zeros on 0 < t < R∗}. By [7, Lemma 2.7] it follows that if b is close to b0 then u(r, b) has at most one zero on (R,∞) and consequently u1(t, b∗) has at most zero on (0, R∗) if b∗ is close to b∗0. Therefore b∗0 < b∗1. It can then be shown that u1(t, b∗1) has exactly one zero on (0, R∗) and u1(0, b∗1) = 0. So if we let b1 = (N−2)b∗1 RN−1 then u(r, b1) is a solution of (2.1)–(2.2) with limr→∞ u(r, b1) = 0 with exactly one zero on (R,∞). Similarly it can be shown that there is a solution, un, of (2.1)–(2.2) such that limr→∞ u(r, bn) = 0 and with n interior zeros on (R,∞) where n is any nonnegative integer. This completes the proof. References [1] A. Adebe, M. Chhetri, L. Sankar, R. Shivaji; Positive solutions for a class of superlinear semipositone systems on exterior domains. Boundary Value Problems, 2014:198, 2014. [2] H. Berestycki, P. L. Lions; Non-linear scalar field equations I & II, Arch. Rational Mech. 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Weissler; Radial solutions of ∆u + f(u) = 0 with prescribed numbers of zeros, Journal of Differential Equations, Volume 83, Issue 2, 368-373, 1990. [10] W. Strauss; Existence of solitary waves in higher dimensions, Comm. Math. Phys., Volume 55, 149-162, 1977. Joseph Iaia Department of Mathematics, University of North Texas, P.O. Box 311430, Denton, TX 76203-1430, USA Email address: iaia@unt.edu 1. Introduction 2. Preliminaries 3. Proof of the main theorem References