Special Issue in honor of Alan C. Lazer Electronic Journal of Differential Equations, Special Issue 01 (2021), pp. 101–114. ISSN: 1072-6691. URL: https://ejde.math.txstate.edu or https://ejde.math.unt.edu INFINITELY MANY RADIAL SOLUTIONS FOR A p-LAPLACIAN PROBLEM WITH NEGATIVE WEIGHT AT THE ORIGIN ALFONSO CASTRO, JORGE COSSIO, SIGIFREDO HERRÓN, CARLOS VÉLEZ Abstract. We prove the existence of infinitely many sign-changing radial solutions for a Dirichlet problem in a ball defined by the p-Laplacian operator perturbed by a nonlinearity of the form W (|x|)g(u), where the weight function W changes sign exactly once, W (0) < 0, W (1) > 0, and function g is p- superlinear at infinity. Standard phase plane analysis arguments do not apply here because the solutions to the corresponding initial value problem may blow up in the region where the weight function is negative. Our result extend those in [2], where W is assumed to be positive at 0 and negative at 1. 1. Introduction We study the quasilinear Dirichlet problem ∆p u+W (|x|)g(u) = 0 in B1(0) ⊂ RN , u = 0 on ∂B1(0), (1.1) where N ≥ 2, p > 1, ∆pu = div(|∇u|p−2∇u) denotes the p-Laplacian operator, and B1(0) denotes the unit ball in RN centered at the origin. We assume that g is a non-decreasing function which is also locally Lipschitz continuous and satisfies sg(s) > 0 for s 6= 0, lim |s|→∞ g(s) s|s|p−2 =∞. (1.2) We also assume that there exists C > 0 such that |g(s)| 6 C |s|p−1, for all s ∈ [−1, 1]. (1.3) Note that the inequality in (1.2) implies g(0) = 0 and G(s) := ∫ s 0 g(t)dt > 0 for all s ∈ Rr {0}. The function W is assumed to be of class C1[0, 1] and such that there exists 0 < X < 1 with W (s) < 0 for s ∈ [0, X), W (X) = 0, W ′(X) > 0, W (s) > 0 for s ∈ (X, 1]. (1.4) 2010 Mathematics Subject Classification. 35J92, 34B15, 34G20. Key words and phrases. Indefinite weight; p-Laplace operator; phase plane; radial solution; shooting method; distributional solution. c©2021 This work is licensed under a CC BY 4.0 license. Published October 6, 2021. 101 102 A. CASTRO, J. COSSIO, S. HERRÓN, C. VÉLEZ EJDE/SI/01 The solutions to( |u′|p−2 u′ )′ (r) + N − 1 r |u′(r)|p−2u′(r) +W (r)g ( u(r) ) = 0, 0 < r < 1, (1.5) subject to u′(0) = 0, u(1) = 0, (1.6) give the radial solutions to (1.1) in the sense of distributions. More exactly, if v : B1(0) → R is a radial function and the function u : [0, 1] → R defined by u( √ x2 1 + · · ·+ x2 N ) := v(x1, . . . , xN ) satisfies (1.5)-(1.6), then v is a solution to (1.1) in the sense of distributions, see Theorem 2.10 below. Because of the singularity given by the zeros of u′ the solutions to (1.1) need not be of class C2. In fact, regularity theory for quasilinear problems indicates that the distributional solutions to (1.1) may only be expected to be in the Holder space C1,µ for some µ ∈ (0, 1), see [9, 14]. Thus, from now on, solution stands for distributional solution. Our main result is the following theorem. Theorem 1.1. If p 6 N , (1.2), (1.3) and (1.4) hold, then for each odd positive integer k problem (1.1) has a radial solution with k zeros. In particular, (1.1) has infinitely many radial solutions. For p = 2, W a positive constant, and g satisfying a growth condition such as (subcritical): lim |u|→+∞ g(u)) |u|q−1u ∈ (0,∞) with q ∈ ( 1, N + 2 N − 2 ) , or (subcritical on (0,+∞)) : lim u→+∞ g(u)) |u|q−1u ∈ (0,∞) with q ∈ ( 1, N N − 2 ) , (1.7) Theorem 1.1 was proven in [4]. This result was extended to all p > 1 in [10]. In [5] the existence of infinitely many solutions with W constant and p = 2 was extended to sub-supercritical nonlinearities; that is, nonlinearities satisfying (subcritical in (0,+∞)) : lim u→+∞ g(u)) |u|q1−1u ∈ (0,∞) with q1 ∈ ( 1, N + 2 N − 2 ) , or (supercritical in (−∞, 0)) : lim u→+∞ g(u)) |u|q2−1u ∈ (0,∞) with q2 ∈ (N + 2 N − 2 ,+∞ ) . (1.8) In turn, the results in [5] were extended in [3] to all p > 1. The approach in [4, 10, 5, 3] combines the continuous dependence of solutions to (1.5) with initial value at r = 0 with phase plane analysis in order to apply the intermediate value theorem. More recently, in [2], the authors considered the case where W is positive at zero, negative at 1 and changes sign only once. In this case, the approach in [4, 10, 5, 3] fails because the solutions to (1.5) with initial value at r = 0 may blow up in (X, 1]. Such a difficulty was overcome in [2] by figuring out a subclass of initial conditions at X such that the solutions to the corresponding initial value problems do not blow up and depend continuously on those initial conditions. Our assumption W < 0 in [0, X) causes some solutions to (1.5) with initial condition at r = 0 to blow up in that interval. We bypass this difficulty by figuring out initial conditions at X for which the solutions to initial value problems are defined in [0, X], depend continuously on such initial conditions, and yield solutions to (1.1). In establishing that such solutions exist the additional assumption p ≤ N is needed due to the singularity at r = 0, see Lemma 2.7. On the other hand, since EJDE-2021/SI/01 p-LAPLACIAN PROBLEM WITH NEGATIVE WEIGHT 103 the coefficient (N − 1)/r is bounded away from zero in [X, 1], shooting from r = 1 towards X allows for g not to have growth restrictions of the type (1.7) or (1.8). For examples of applications to problems with indefinite weight the reader is referred to [11]. For recent results on quasilinear problems with weight see [1, 2, 7, 13, 16]. For related results on the existence of infinitely many radial solutions to quasilinear problems see [3, 8, 12]. This article is organized as follows: in Section 2 we show that all solutions to (2.1) below are defined in [X, 1] and figure out the initial conditions at r = X for which the solutions to (1.5) do not blow up in [0, X]. More precisely, we find initial conditions (a, ηa), a > 0, for the initial value problem (2.9) such that u ≡ ua,η(a) is defined on [0, X], positive, increasing, and u′(0) = 0 (see Lemmas 2.6 and 2.7 below). In Section 3 we prove that, for any T ∈ (X, 1), if u(r, d) is the solution to (2.1) below then limd→−∞(u2(r, d)+(u′(r, d))2) = +∞ uniformly for r ∈ [T, 1]. We also present in Section 3 the phase plane analysis of the solutions to (2.1) in [X, 1]. The arguments in Section 3 may be traced back to work by Professor Alan C. Lazer and one the authors in [6]. In Section 4 we prove the main result by connecting at X solutions to the regular initial value problem at r = 1 with those that do not blow up in [0, X]. Namely, we prove the existence of infinitely many values of d such that the solution to (2.1) satisfies u(X) = a, u′(X) = ηa, for some a > 0. Hence, by Theorem 2.10, they give infinitely many solutions to (1.1). 2. Initial value problem and preliminaries We consider the initial value problem( rN−1 ∣∣u′∣∣p−2 u′ )′ + rN−1W (r)g ( u(r) ) = 0, 0 < r < 1, u(1) = 0, u′(1) = d. (2.1) Let Φp(x) = x|x|p−2 for x ∈ R. We observe that, for each d ∈ R, a continuous function u satisfies the integral equation u(r) = − ∫ 1 r Φ−1 p ( Φp(d)t1−N + ∫ 1 t (s t )N−1 W (s)g(u(s))ds ) dt (2.2) if and only if it is a solution to (2.1). In general, for any r0 ∈ (0, 1], a ∈ R, b ∈ R, a continuous function u satisfies u(r) = a− ∫ r0 r Φ−1 p ((r0 t )N−1 Φp(b) + ∫ r0 t (s t )N−1 W (s)g(u(s))ds ) dt, (2.3) if and only if it satisfies( rN−1 ∣∣u′∣∣p−2 u′ )′ + rN−1W (r)g ( u(r) ) = 0, 0 < r < r0, u(r0) = a, u′(r0) = b. (2.4) For d0 ∈ R− {0}, using the Contraction Mapping Principle and the fact that g is a locally Lipschitzian function, we see there exists γ ∈ (0, 1] such that for each d ∈ [d0 − γ, d0 + γ], equation (2.2) has a unique solution u(·, d) in the space of continuous functions defined on [1−γ, 1]. This and the continuity of the right hand side in (2.2) on (d, u) imply the continuous dependence of u(·, d) on d. When γ = 1 such a solution is a solution to (2.1). If γ ∈ (0, 1) the solution may be extended to [1−γ1, 1] for some γ1 > γ by applying a similar argument to (2.3) with a = u(1−γ, d) and b = u′(1 − γ, d). Hence, the function u(·, d) may be extended to a maximal 104 A. CASTRO, J. COSSIO, S. HERRÓN, C. VÉLEZ EJDE/SI/01 interval which is either [0, 1] or (θ̂(d), 1] with limt→θ̂(d)+ [u2(t)+(u′(t))2] = +∞. We remark that from the results in [15], because of hypothesis (1.3), no solution to (2.4) satisfies limt→θ̂(d)+ [u2(t) + (u′(t))2] = 0 when (a, b) 6= (0, 0). For a comprehensive study of existence, uniqueness and continuous dependence, we refer the reader to [15]. In our next lemma we prove that θ̂(d) ≤ X. Since d0 ∈ R−{0} is arbitrary, this shows the existence of a unique solution to (2.1) on [X, 1] that depends continuously on d. Lemma 2.1. For each d 6= 0, the solution to (2.1) is defined in [X, 1]. Proof. Let u(r) := u(r, d) be a solution to (2.1) and E(r, d) ≡ E(r) := p− 1 p |u′(r)|p +W (r)G ( u(r) ) . (2.5) Let C1 = p(N − 1)/(p− 1). Since W ′(X) > 0, limr→X+ W ′(r)/W (r) = +∞. Thus, there exists C2 > 0 such that W ′(r)/W (r) ≥ −C2 for r ∈ (X, 1]. We also let C3 = C1/X + C2. Assuming that θ̂(d) ≥ X, there exists s ∈ (X, 1) such that E(s) > eC3E(1). (2.6) Since |x|p, x|x|p−2 and |x|p/(p−1) are differentiable functions, and (|x|p/(p−1))′ = p p− 1 |x|(2−p)/(p−1)x, the function |u′|p−2u′ is differentiable in (s, 1] (see (2.1)). Therefore, E is differen- tiable on (s, 1] and for each r ∈ (s, 1], E ′(r) = (p− 1 p ∣∣|u′(r)|p−2u′(r) ∣∣p/(p−1) )′ +W ′(r)G(u(r)) +W (r)g(u(r))u′(r) = ∣∣|u′(r)|p−2u′(r) ∣∣(2−p)/(p−1) |u′(r)|p−2u′(r) ( |u′(r)|p−2u′(r) )′ +W ′(r)G(u(r)) +W (r)g(u(r))u′(r) = |u′(r)|2−p|u′(r)|p−2u′(r) ( |u′(r)|p−2u′(r) )′ +W ′(r)G(u(r)) +W (r)g(u(r))u′(r). (2.7) This and (2.1) yield E ′(r) = u′(r) ( − N − 1 r |u′(r)|p−2u′(r)−W (r)g(u(r)) ) +W ′(r)G(u(r)) +W (r)g(u(r))u′(r) = −N − 1 r |u′(r)|p +W ′(r)G(u(r)) = −p(N − 1) (p− 1)r E(r) +G ( u(r) )[p(N − 1) (p− 1)r W (r) +W ′(r) ] ≤W ′(r)G(u(r)), (2.8) where we used (2.5). Hence, from (2.7), E ′(r) ≥ −p(N − 1) (p− 1)r E(r) +G ( u(r) ) W ′(r) EJDE-2021/SI/01 p-LAPLACIAN PROBLEM WITH NEGATIVE WEIGHT 105 ≥ −p(N − 1) (p− 1)r E(r) + [W (r)G ( u(r) ) ]W ′(r)/W (r) ≥ −p(N − 1) (p− 1)r E(r)− C2W (r)G ( u(r) ) ≥ −p(N − 1) (p− 1)r E(r)− C2E(r) = ( − C1 r − C2 ) E(r) ≥ ( − C1 X − C2 ) E(r) := −C3E(r). Integrating on [s, 1], we have eC3E(1)− eC3sE(s) ≥ 0. Since this inequality contra- dicts (2.6) we have proven that θ̂(d) ≤ X, and hence the lemma follows. � For a > 0 and b ∈ R, we consider the initial value problem ( rN−1|u′(r)|p−2u′(r) )′ + rN−1W (r)g ( u(r) ) = 0, 0 < r < X u(X) = a, u′(X) = b. (2.9) Because of our assumptions on g, the initial value problem (2.9) has a unique solution ua,b on a maximal interval (ra,b, X]. Lemma 2.2. Let r0 ∈ (ra,b, X]. If ua,b(r0) > 0 and u′a,b(r0) ≤ 0, then u′a,b < 0 in (ra,b, r0). Proof. Let u := ua,b. Let ε > 0 be such that u(r) > u(r0)/2 for all r ∈ (r0 − ε, r0]. From (2.9), for r ∈ (r0 − ε, r0] we have rN−1|u′(r)|p−2u′(r) = rN−1 0 |u′(r0)|p−2u′(r0) + ∫ r0 r sN−1W (s)g(u(s))ds < 0. (2.10) Let r̂ = inf{r ∈ (ra,b, r0) : u′ < 0 in (r, r0)}. Hence r̂ ≤ r0 − ε. Assuming that r̂ > ra,b we have u(r̂) > 0 and, by (2.10), u′(r̂) < 0. Arguing as before, there exists δ > 0 such that u′ < 0 in (r̂− δ, r̂) which contradicts the definition of r̂ and proves that u′ < 0 in (ra,b, r0). � Lemma 2.3. If 0 < b < b̃, y > max{ra,b, ra,b̃}, then ua,b(r) > ua,b̃(r) for all r ∈ [y,X). Moreover, u′a,b(r) < u′ a,b̃ (r) for all r ∈ [y,X). Proof. Let u = ua,b and v = ua,b̃. Since b < b̃ there exists ε > 0 such that u(r) > v(r) for all r ∈ (X − ε,X). Assuming that u(r) 6> v(r) for all r ∈ [y,X), due to the continuity of u and v there exists z ∈ [y,X) such that u(z) = v(z) and 106 A. CASTRO, J. COSSIO, S. HERRÓN, C. VÉLEZ EJDE/SI/01 u(r) > v(r) for all r ∈ (z,X). On the other hand, zN−1|u′(z)|p−2u′(z) = XN−1|u′(X)|p−2u′(X) + ∫ X z sN−1W (s)g(u(s))ds 6 XN−1|u′(X)|p−2u′(X) + ∫ X z sN−1W (s)g(v(s))ds < XN−1|v′(X)|p−2v′(X) + ∫ X z sN−1W (s)g(v(s))ds = zN−1|v′(z)|p−2v′(z), (2.11) contradicting that v′(z) ≤ u′(z) since v(r) < u(r) for all r ∈ (z,X). A similar argument proves the second assertion of the lemma. Thus, the lemma is proven. � Lemma 2.4. For each a > 0 there exists ε > 0 such that if b ∈ (0, ε) then there exists rb ∈ (ra,b, X] such that u′a,b(rb) = 0 and u′a,b(r) > 0 for all r ∈ [rb, X]. Proof. Let y = ra,0 and u = ua,0. By Lemma 2.2 we have u′((y + X)/2)) < 0. This and continuous dependence on initial conditions imply the existence of ε > 0 such that if b ∈ (0, ε) then u′a,b((y + X)/2) < 0. Hence, by the intermediate value theorem, for each b ∈ (0, ε) there exists rb ∈ ((y + X)/2, X) such that u′a,b(rb) = 0. By Lemma 2.2, rb is unique. Hence u′a,b(r) does not change sign in (rb, X]. Since u′a,b(X) > 0, it follows that u′a,b(r) > 0 for all r ∈ (rb, X]. This proves the lemma. � Lemma 2.5. For each a > 0 there exists b > 0 such that ua,b(r1) = 0 for some r1 ∈ (ra,b, X]. Proof. Let b > 0 be such that bp−1 > max{− inf{W (r); r ∈ (0, X]}2g(a), 2pap−1/Xp−1} and u := ua,b. Since b > 0 there exists ε > 0 such that 0 < u(s) ≤ a for all s ∈ (X − ε,X]. Let r ∈ (0, X) be such that 0 < u(s) ≤ a for s ∈ (r,X]. Hence |u′(r)|p−2u′(r) = r1−N ( XN−1bp−1 + ∫ X r sN−1W (s)g(u(s))ds ) > r1−N ( XN−1bp−1 + XN N · inf{W (r); r ∈ (0, X]}g(a) ) ≥ bp−1 + inf{W (r); r ∈ (0, X]}g(a) X N > bp−1 2 > 2p−1ap−1 Xp−1 > 0. (2.12) Therefore, u′(r) > 2a/X. Let r1 = inf{r ∈ (0, X); 0 < u(s) ≤ a for all s ∈ (r,X]} ≤ X − ε. By the Mean Value Theorem, a ≥ a− u(r1) > (X − r1) 2a X . Hence r1 > X 2 . (2.13) EJDE-2021/SI/01 p-LAPLACIAN PROBLEM WITH NEGATIVE WEIGHT 107 Since u′ > 0 on [r1, X], we have u(r1) < a. If u(r1) > 0, by the continuity of u, there exists r2 < r1 such that u(s) ∈ (0, a] for all s ∈ [r1, X] contradicting the definition of r1. Thus u(r1) = 0 and the lemma is proven. � For a > 0 we define b̂ = b̂(a) = sup{b > 0;ua,c(r) > 0 for all r ∈ (ra,c, X] and all c ∈ (0, b)} ≡ supBa. (2.14) From Lemma 2.4 and Lemma 2.5, 0 < b̂ < +∞. Lemma 2.6. For all a > 0, ra,b̂ = 0, ua,b̂(r) > 0 for all r ∈ (0, X], and ua,b̂(r) is monotonically increasing. Proof. Let z = ua,b̂. First we prove that z(r) > 0 for all r ∈ (ra,b̂, X]. Suppose there exists r1 ∈ (0, X] such that z(r1) = 0. Without loss of generality, we may assume that z(r) > 0 for all r ∈ (r1, X]. By uniqueness of solutions to initial value problems z′(r1) > 0. Hence there exists δ > 0 such that z(r) < 0 for all r ∈ (r1 − δ, r1). Let {bj}j be an increasing sequence in Ba converging to b̂. By existence of solutions to initial value problems, there exists J such that if j ≥ J then ra,bj < r1 − δ/2. By continuous dependence on initial conditions, limj→+∞ ua,bj (r1 − δ/2) = z(r1 − δ/2) < 0 contradicting that ua,bj (r) > 0 for all r ∈ (ra,bj , X]. Hence z(r) > 0 for all r ∈ (ra,b̂, X]. Now we prove that z′(r) > 0 in (ra,b̂, X]. Assuming that z is not monotonically increasing, there exists r2 ∈ (ra,b̂, X) such that z′(r) > 0 for r ∈ (r2, X] and z′(r) < 0 for all r ∈ (ra,b̂, r2). By continuous dependence on initial conditions, there exists η > 0 such that if |b̂− b| < η then ra,b < (r2 + ra,b̂)/2. Also, there exists ρ ∈ (0, η) such that if |b̂ − b| < ρ then ua,b((r2 + ra,b̂)/2) > 0 and u′a,b((r2 + ra,b̂)/2) < 0. Hence ua,b(r) > 0 for all b < b̂ + ρ and all r ∈ (ra,b, X]. Since this contradicts the definition of b̂ we conclude that z = ua,b̂ is a monotonically increasing function. Since z is monotonically increasing and positive, if ra,b̂ > 0 then z may be extended to an interval of the form (ra,b̂− ε,X] contradicting the definition of ra,b̂. This proves that ra,b̂ = 0. � Lemma 2.7. If p ≤ N , u ∈ C1(0, 1] satisfies( rN−1 ∣∣u′∣∣p−2 u′ )′ + rN−1W (r)g ( u(r) ) = 0, 0 < r < 1, (2.15) and u > 0 and bounded in (0, X), then ζ(r) := |u′(1)|p−2u′(1) + ∫ 1 r sN−1W (s)g(u(s))ds = rN−1|u′(r)|p−2u′(r) (2.16) is non-negative and non-decreasing. Moreover, lim r→0+ ζ(r) = lim r→0+ u′(r) = 0. (2.17) Proof. For s ∈ (0, X), sN−1W (s)g(u(s)) < 0. Hence ζ increases on (0, X) and limr→0+ ζ(r) exists. Assuming there is r0 ∈ (0, X) such that rN−1 0 |u′(r0)|p−2u′(r0) =: c < 0, 108 A. CASTRO, J. COSSIO, S. HERRÓN, C. VÉLEZ EJDE/SI/01 from the monotonicity of ζ, we have rN−1|u′(r)|p−2u′(r) < c for all r ∈ (0, r0). Therefore, since p ≤ N , we have −u′(r) ≥ (−c)1/(p−1)r−1 for all r ∈ (0, r0]. Inte- grating −u′ in [r, r0] we see that limr→0+ u(r) = +∞ which contradicts that u is bounded. Hence ζ(r) ≥ 0 for all r ∈ (0, X]. To prove (2.17), we assume to the contrary that there exists κ > 0 such that rN−1|u′(r)|p−2u′(r) = |u′(1)|p−2u′(1) + ∫ 1 r sN−1W (s)g(u(s))ds ≥ κ > 0, (2.18) for all r ∈ (0, 1]. Solving for u′ in (2.18), u′(s) ≥ κ1/(p−1) s(N−1)/(p−1) ≥ κ1/(p−1) s for all s ∈ (0, 1] and p ≤ N. (2.19) Therefore, u(r) = u(1)− ∫ 1 r u′(s)ds ≤ u(1)− ∫ 1 r κ1/(p−1) s ds = u(1) + κ1/(p−1)(ln r)→ −∞, as r → 0+. This contradiction proves that limr→0+ ζ(r) = 0. Finally, by (2.16) and L’Hôspital’s rule we obtain lim r→0+ (u′(r))p−1 = lim r→0+ |u′(1)|p−2u′(1) + ∫ 1 r sN−1W (s)g(u(s))ds rN−1 = lim r→0+ −rW (r)g(u(r)) N − 1 = 0, (2.20) and thus the lemma is proven. � Remark 2.8. The conclusion of Lemma 2.7 remains valid when 1 is replaced by R ∈ (0, 1), under the assumptions that u ∈ C1(0, R] satisfies the differential equation in (2.15) on (0, R], u > 0 and bounded on some interval (0, x0) ⊂ (0, R). Lemma 2.9. If η : [0,+∞)→ [0,+∞) is defined by η(a) = b̂ and η(0) = 0, then η is a continuous function. Proof. First we prove that the function η is a non-decreasing function. Let u = ua0,η(a0) and v = ua,η(a0) with a > a0 ≥ 0. Suppose that there is r1 ∈ (ra,η(a0), X) such that u(r1) = v(r1). Without loss of generality we may also assume that u(r) < v(r) for all r ∈ (r1, X]. By the uniqueness of solutions to the initial value problem (2.9) we have v′(r1) > u′(r1). On the other hand, from (2.9) it follows that rN−1 1 (|v′(r1)|p−2v′(r1)− |u′(r1)|p−2u′(r1)) = ∫ X r1 sN−1W (s)(g(v(s))− g(u(s)))ds < 0. (2.21) Therefore v′(r1) ≤ u′(r1), which contradicts v′(r1) > u′(r1). Thus v(r) > u(r) for all r ∈ (ra,η(a0), X]. This and the definition of b̂(a) prove that η(a) > η(a0). So, we have proved η is a non decreasing function. Let {an}n be an increasing sequence that converges to a0 > 0. Hence {η(an)} is an increasing sequence bounded above by η(a0). Let u = ua0,η(a0) and un = uan,η(an). Suppose that η(a0) > limn→∞ η(an). For n sufficiently large, there exists rn ∈ (0, X) such that un(rn) = u(rn), limn→+∞ rn = X, u′n(rn) < u′(rn) EJDE-2021/SI/01 p-LAPLACIAN PROBLEM WITH NEGATIVE WEIGHT 109 and u(r) < un(r) for all r ∈ [0, rn]. Therefore, from Lemma 2.7 (see also Remark 2.8), for a fixed n large enough, 0 = lim r→0+ rN−1|u′n(r)|p−2u′n(r) = lim r→0+ ( rN−1 n (u′n(rn))p−1 + ∫ rn r sN−1W (s)g(un(s))ds ) < lim r→0+ ( rN−1 n (u′(rn))p−1 + ∫ rn r sN−1W (s)g(u(s))ds ) = 0. (2.22) This contradiction proves that η is continuous to the left. Similarly it is seen that η is continuous to the right. Thus the lemma is proven. � Theorem 2.10. If w : B1(0) → R is a radial function such that w ∈ C1(B1(0) r {0})∩C(B1(0)), and U(r) = w(r, 0, . . . , 0) satisfies the assumptions of Lemma 2.7, then w is a solution to (1.1) in the sense of distributions. Proof. Let ϕ be a function of class C∞ and compact support in B1(0) and A the (N−1)-dimensional Lebesgue measure of the unit sphere in RN . Lemma 2.7 implies |U ′(1)|p−2U ′(1) = − ∫ 1 0 sN−1g(U(s))W (s)ds. (2.23) Using that ∇w(x) = (U ′(|x|)/|x|)x and (2.23), we have∫ B1(0) ‖∇w(x)‖p−2∇w(x) · ∇ϕ(x)dx = ∫ B1(0) ‖∇w(x)‖p−2∇w(x) · (∇ϕ(x) · x)x/|x|2dx = ∫ 1 0 rN−1|U ′(r)|p−2U ′(r) ∫ ‖θ‖=1 (∇ϕ(rθ) · θ)dθ dr = ∫ 1 0 rN−1|U ′(r)|p−2U ′(r) ( d dr ∫ ‖θ‖=1 ϕ(rθ)dθ ) dr = ∫ 1 0 |U ′(1)|p−2U ′(1) ( d dr ∫ ‖θ‖=1 ϕ(rθ)dθ ) dr + ∫ 1 0 (∫ 1 r sN−1W (s)g(U(s))ds )( d dr ∫ ‖θ‖=1 ϕ(rθ)dθ ) dr = −A [ |U ′(1)|p−2U ′(1) + (∫ 1 0 rN−1W (r)g(U(r))dr )] ϕ(0) + ∫ ‖θ‖=1 ∫ 1 0 rN−1W (r)g(U(r))ϕ(rθ)drdθ = ∫ B1(0) W (|x|)g(U(|x|))ϕ(x)dx, (2.24) which proves the theorem. � 3. Energy and phase plane analysis Let T ∈ (X, 1]. Since W > 0 on (X, 1], there exists W0 = W0(T ) > 0 such that W (r) ≥W0 for each r ∈ [T, 1]. 110 A. CASTRO, J. COSSIO, S. HERRÓN, C. VÉLEZ EJDE/SI/01 We recall that, from Lemma 2.1, given d < 0 there is a unique solution u(r, d) to ( rN−1 ∣∣u′∣∣p−2 u′ )′ + rN−1W (r)g ( u(r) ) = 0, X ≤ r ≤ 1, u(1) = 0, u′(1) = d. (3.1) Lemma 3.1. Let T ∈ (X, 1) and E be as defined in (2.5). Then E(r) → +∞ as |d| → +∞, uniformly for r ∈ [T, 1]. Proof. From (2.7) we have E ′(r) ≤W ′(r)G ( u(r) ) = W ′(r) W (r) W (r)G ( u(r) ) ≤ max{|W ′(s)| : s ∈ [T, 1]} min{W (s) : s ∈ [T, 1]} W (r)G ( u(r) ) ≤ C̃W (r)G ( u(r) ) ≤ C̃E(r), for all r ∈ [T, 1]. Thus, ( e−C̃rE(r) )′ ≤ 0 for every r ∈ [T, 1]. Therefore, E(r) ≥ p− 1 p eC̃(T−1)|d|p−1 for all r ∈ [T, 1]. Thus, E(r)→ +∞ as |d| → +∞, uniformly for r ∈ [T, 1]. � Since g(0) = 0, by uniqueness of solutions to the initial value problem with initial data (0, 0), we have (u(r, d), u′(r, d)) 6= (0, 0) for all r ∈ [X, 1]. Hence, there exists a continuous function φ(r, d), for r ∈ [X, 1], such that φ(1, d) = −π/2, and u(r, d) = −ρ(r, d) cosφ(r, d), u′(r, d) = ρ(r, d) sinφ(r, d), (3.2) where ρ(r, d) = √( u(r, d) )2 + ( u′(r, d) )2 . Moreover, φ(·, d) is differentiable at r ∈ [X, 1] provided u′(r) 6= 0. Differentiating the first equation in (3.2) with respect to r, for u′(r) 6= 0, u′(r) = −ρ′(r, d) cos ( φ(r, d) ) + ρ(r, d) sin ( φ(r, d) ) · φ′(r, d). (3.3) Since W is a continuous function, there exists T ∈ (X, 1) such that W (r) ≥ W (1) 2 =: m 2 , for all r ∈ [T, 1]. (3.4) Combining (3.2) and (3.1), for r ∈ [X, 1] with u′(r) 6= 0, we have φ′(r, d) = (u′(r, d))2 ρ2(r, d) + W (r)u(r)g(u(r)) (p− 1)ρ2(r, d)|u′(r)|p−2 + (N − 1)u(r)u′(r) r(p− 1)ρ2(r, d) . (3.5) Remark 3.2. (i) By Lemma 3.1, E(r, d) → +∞ as |d| → +∞ uniformly for r ∈ [T, 1], and therefore ρ(r, d)→ +∞ as |d| → +∞ uniformly for r ∈ [T, 1]. (ii) If u(R, d) = 0 with R ∈ (X, 1), then u′(R, d) 6= 0. In addition, from (3.3) and the second equation in (3.2) it follows that φ′(R, d) = 1. EJDE-2021/SI/01 p-LAPLACIAN PROBLEM WITH NEGATIVE WEIGHT 111 (iii) If u(R, d) = 0 then φ(r, d) < φ(R, d) for every r ∈ [X,R]. Suppose, by contradiction, there exists R1 ∈ [X,R] such that φ(R1, d) = φ(R, d). By the continuity of φ(·, d) we can assume φ(r, d) < φ(R, d) for all r ∈ (R1, R) (suffices choosing R1 = inf{r ∈ [X,R] : φ(r, d) < φ(R, d)}). Since φ(r, d) < φ(R1, d) for each r ∈ (R1, R), then φ′(R1, d) ≤ 0. On the other hand, φ(R1, d) = φ(R, d) = jπ + π/2, for some j ∈ Z. Thus φ′(R1, d) = 1, which is a contradiction. Let k be a positive integer. For x0 > 0, let us define m̃(x0) = min { g(x) |x|p−2x : |x| ≥ x0 } . From the p-superlinearity of g we have m̃(x0) → +∞ as x0 → +∞. For ρ > 0 and η > 0 we define ω(ρ, η) := m̃(ρ sin(η))M1(η)/(p − 1), where M1(η) := min{sinp(η), sin2(η)}. Let T be as in (3.4), let ρ0(k) := ρ0 > 0 and δ(k) := δ ∈ (0, π/4) be such that (i) 0 < δ < min { (p− 1)T 16(N − 1) , ( (1− T )(p− 1) 2 )1/(p−1)} , (ii) ω(ρ0, δ) > 2(N − 1) m(p− 1)T , (iii) m̃(ρ0/ √ 2) ≥ 2(p/2)+5k(p− 1) m , (iv) 16δ + 8π mω(ρ0, δ) ≤ 1− T 2k . (3.6) By Remark 3.2-(i), there exists d0 < 0 such that if d < d0, then ρ(r, d) ≥ ρ0 for every r ∈ [T, 1]. (3.7) Lemma 3.3. If T ≤ r ≤ 1 and φ(r, d) ∈ [−jπ/2− δ,−jπ/2 + δ] with j > 0 an odd integer, then φ′(r, d) > 1/4. Proof. From (3.5), φ′(r, d) ≥ sin2 φ+ W (r)u(r, d)g(u(r, d)) (p− 1)ρ2(r, d)|u′(r, d)|p−2 − (N − 1)| cosφ sinφ| r(p− 1) . From (1.2) and (1.4), W (r)u(r, d)g(u(r, d)) ≥ 0 for all r ∈ [X, 1]. This, the inequal- ities | sin(φ(r, d))| ≥ cos δ, | cos(φ(r, d))| ≤ sin δ ≤ δ and (3.6)-(i) give φ′(r, d) ≥ cos2 δ − (N − 1)δ (p− 1)T ≥ cos2(π/4)− 1 16 ≥ 7 16 > 1 4 . Thus, the lemma is proven. � Lemma 3.4. If T ≤ r ≤ 1 and φ(r, d) ∈ [−(j + 1)π/2 + δ,−jπ/2− δ] with j > 0 an integer, then φ′(r, d) > mω(ρ0, δ)/4. Proof. From (3.5), φ′(r, d) > W (r)u(r, d)g(u(r, d)) (p− 1)ρ2(r, d)|u′(r, d)|p−2 − (N − 1) 2r(p− 1) ≥ W (r) p− 1 g(u(r, d)) |u(r, d)|p−2u(r, d) |u(r, d)|p ρ2(r, d)|u′(r, d)|p−2 − N − 1 2(p− 1)T 112 A. CASTRO, J. COSSIO, S. HERRÓN, C. VÉLEZ EJDE/SI/01 ≥ W (r) p− 1 g(u(r, d)) |u(r, d)|p−2u(r, d) | cosφ(r, d)|p | sinφ(r, d)|p−2 − N − 1 2(p− 1)T . From | cosφ(r, d)| ≥ sin δ, | sinφ(r, d)| ≥ sin δ, and ω(ρ0, δ) > 2(N−1) m(p−1)T (see (3.6)- (ii)), it follows that φ′(r, d) > W (r) p− 1 g(u(r, d)) |u(r, d)|p−2u(r, d) M1(δ)− mω(ρ0, δ) 4 . Since |u(r, d)| = ρ(r, d)| cosφ(r, d)| ≥ ρ0 sin δ we obtain g(u(r, d))/(|u(r, d)|p−2u(r, d)) ≥ m̃(ρ0 sin δ). This and the definition of ω(ρ0, δ) yield φ′(r, d) > W (r)ω(ρ0, δ)− mω(ρ0, δ) 4 ≥ mω(ρ0, δ) 4 . (3.8) In the latter inequality we have used W (r) ≥ m/2 for any r ∈ [T, 1]. Thus, (3.8) proves the lemma. � Lemma 3.5. If T ≤ r ≤ 1 and φ(r, d) ∈ (−jπ,−jπ+ δ]∪ [−jπ− δ,−jπ) for some positive integer j, then φ′(r, d) ≥ 8k| sin(φ(r, d))|2−p. (3.9) Proof. From δ < π/4 and | cosφ(r, d)| ≥ cos δ, it follows that u2(r, d) = ρ2(r, d)(1− sin2(δ)) ≥ ρ2(r, d)/2. This, (3.6)-(i), (3.5) and (3.6)-(iii) imply that φ′(r, d) ≥ W (r)u(r, d)g(u(r, d)) (p− 1)ρp(r, d)| sin(φ(r, d))|p−2 − (N − 1)| sin(φ(r, d))| r(p− 1) ≥ W (r)u(r, d)g(u(r, d))| sin(φ(r, d))|2−p 2p/2(p− 1)|u(r, d)|p − (N − 1)| sin(φ(r, d))| T (p− 1) ≥ ( m 2 · 2p/2(p− 1) g(u(r, d)) |u(r, d)|p−2u(r, d) − N − 1 (p− 1)T | sin(φ(r, d))|p−1 ) × | sin(φ(r, d))|2−p ≥ ( m · m̃(ρ0/ √ 2) 2(p/2)+1(p− 1) − 1 16 ) | sin(φ(r, d))|2−p, (recall | sinφ| ≤ sin δ ≤ δ) ≥ m · m̃(ρ0/ √ 2) 2(p/2)+2(p− 1) | sin(φ(r, d))|2−p ≥ 8k| sin(φ(r, d))|2−p, (3.10) which completes the proof of the lemma. � Proposition 3.6. limd→−∞ φ(X, d) = −∞. Proof. Let d < d0 be as in (3.7) and k as in (3.6). Because φ(1, d) ∈ [−π/2 − δ,−π/2 + δ], from Lemma 3.3 and (3.6)-(iv) there exists r1 ∈ [1− 4δ, 1) ⊂ [1− (1− T )/(8k), 1) such that φ(r1, d) = −π/2− δ. By Lemma 3.4 and (3.6)-(iv) there exists r2 ∈ [r1 − 2π/(mω(ρ0, δ)), r1) ⊂ [r1 − (1− T )/(8k), r1) EJDE-2021/SI/01 p-LAPLACIAN PROBLEM WITH NEGATIVE WEIGHT 113 such that φ(r2, d) = −π + δ. By Lemma 3.5, if p ≥ 2, there is r3 ∈ [r2 − δ/(8k), r2) ⊂ [r2 − (1− T )/(8k), r2) such that φ(r3, d) = −π. On the other hand, if 1 < p < 2, from Lemma 3.5 φ′(r, d) ≥ 8k| sin(φ(r, d))|2−p. (3.11) We claim that if r < r2 and φ(r, d) > −π then r2 − r < (1 − T )/(8k). Indeed, let φ(r, d) = −π + θ(r, d); then 0 < θ(r, d) ≤ δ. From (3.11) we obtain θ′(r, d) > 8k(sin θ(r, d))2−p. Since sin(θ)/θ → 1 as θ → 0, we may assume, for δ sufficiently small, sin(θ(r, d))/θ(r, d) > 1/2. Thus, θ′(r, d)(θ(r, d))p−2 > 8k 22−p > 4k. Integrating on [r, r2] we get 4k(p− 1)(r2 − r) < δp−1. By using (3.6)-(i), r2 − r < δp−1 4k(p− 1) < 1− T 8k . From this the claim follows. Hence, there exists r3 ∈ [r2 − (1 − T )/(8k), r2) such that r3 ∈ [1− 3(1− T )/(8k), 1) ⊂ [1− (1− T )/(2k), 1) and φ(r3, d) = −π. Observe that 1− r3 ≤ (1− T )/(2k) and φ(r3, d)− φ(1, d) = −π/2. Repeating this argument 2k times it is shown that there is r̂ ∈ [1− 2k(1− T )/2k, 1) = [T, 1) such that φ(r̂, d) − φ(1, d) = −kπ, namely φ(r̂, d) = −(2k + 1)π/2. Since 1 > r̂ ≥ T > X, Remark 3.2-(ii) implies φ(X, d) < φ(r̂, d) = −(2k + 1)π/2. This proves the proposition. � 4. Proof of main theorem Let u(r, d) be the solution to (3.1) with d < 0 and φ(r, d) be the argument function defined by (3.2). By Proposition 3.6 (limd→−∞ φ(X, d) = −∞) and the continuous dependence of φ(X, d) on d, for each odd positive integer k there exist real numbers d̂k and d̃k such that d̃k < d̂k, φ(X, d̂k) = −kπ − π 2 , φ(X, d̃k) = −(k + 1)π. Since k is odd, u(X, d̂k) = 0 and u′(X, d̂k) > 0; also, u(X, d̃k) > 0 and u′(X, d̃k) = 0. Because η is a continuous function (see Lemma 2.9) the set {(a, η(a)); a ≥ 0} separates {(0, y); y > 0} from {(x, 0);x > 0} in {(x, y);x ≥ 0, y ≥ 0} − {(0, 0)}, there exists dk ∈ (d̃k, d̂k) such that (u(X, dk), u′(X, dk)) = (a, η(a)) for some a > 0. Hence defining Uk(r) = ua,η(a)(r) for r ∈ [0, X] and Uk(r) = u(r, dk) for r ∈ [X, 1] we have a solution to (1.1) (see Lemmas 2.6-2.7 and Theorem 2.10). Thus, the sequence {Uk(r)}k gives us infinitely many radially symmetric solutions to (1.1), which proves the theorem. Acknowledgements. The authors wish to thank the anonymous referees for their helpful comments, and editor Julio G. Dix for obtaining referee reports and accept- ing this article. Authors Sigifredo Herrón and Carlos Vélez were supported by Universidad Na- cional de Colombia Sede Medelĺın, Facultad de Ciencias. Hermes project code 48952. 114 A. CASTRO, J. COSSIO, S. HERRÓN, C. VÉLEZ EJDE/SI/01 References [1] K. Bal, P. 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Alfonso Castro Department of Mathematics, Harvey Mudd College, Claremont, CA 91711, USA Email address: castro@g.hmc.edu Jorge Cossio Escuela de Matemáticas, Universidad Nacional de Colombia Sede Medelĺın, Medelĺın, Colombia Email address: jcossio@unal.edu.co Sigifredo Herrón Escuela de Matemáticas, Universidad Nacional de Colombia Sede Medelĺın, Medelĺın, Colombia Email address: sherron@unal.edu.co Carlos Vélez Escuela de Matemáticas, Universidad Nacional de Colombia Sede Medelĺın, Medelĺın, Colombia Email address: cauvelez@unal.edu.co 1. Introduction 2. Initial value problem and preliminaries 3. Energy and phase plane analysis 4. Proof of main theorem Acknowledgements References