Singular extremal solutions for supercritical elliptic equations in a ball

Available online at www.sciencedirect.com

ScienceDirect J. Differential Equations 265 (2018) 2842–2885 www.elsevier.com/locate/jde

Singular extremal solutions for supercritical elliptic equations in a ball Yasuhito Miyamoto a , Y¯uki Naito b,∗ a Graduate School of Mathematical Sciences, The University of Tokyo, 3-8-1 Komaba, Meguro-ku, Tokyo 153-8914,

Japan b Department of Mathematics, Ehime University, 2-5 Bunkyo, Matsuyama 790-8577, Japan

Received 10 February 2017; revised 24 April 2018 Available online 3 May 2018

Abstract We study positive singular solutions to the Dirichlet problem for the semilinear elliptic equation u + λf (u) = 0 in the unit ball on RN . We assume that f has the form f (u) = up + g(u), where p > (N + 2)/(N − 2) and g(u) is a lower order term. We first show the uniqueness of the singular solution to the problem, and then study the existence of the singular extremal solution. In particular, we show a necessary and sufficient condition for the existence of the singular extremal solution in terms of the weak eigenvalue of the linearized problem. © 2018 Elsevier Inc. All rights reserved.

MSC: 35A02; 35B32; 35J61

Keywords: Semilinear elliptic equation; Singular extremal solution; Joseph–Lundgren exponent; Uniqueness; Bifurcation diagram; Generalized eigenvalue

* Corresponding author.

E-mail addresses: [email protected] (Y. Miyamoto), [email protected] (Y. Naito). https://doi.org/10.1016/j.jde.2018.04.055 0022-0396/© 2018 Elsevier Inc. All rights reserved.

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1. Introduction and main results We study singular solutions to the semilinear elliptic Dirichlet problem ⎧ in B, ⎨ u + λf (u) = 0 u>0 in B, ⎩ u=0 on ∂B,

(1.1)

where B = {x ∈ RN : |x| < 1} with N ≥ 3 and λ is a nonnegative constant. In (1.1) we assume that f has the form f (u) = up + g(u)

for u ≥ 0,

(1.2)

where p > pS := (N + 2)/(N − 2) and g(u) is a lower order term. By the symmetry result of Gidas–Ni–Nirenberg [16], every regular positive solution u is radially symmetric and uL∞ = u(0). It is known that the set of the solutions can be parametrized by the L∞ -norm of the solution. Let (λ, u) be a solution of (1.1) and let α = u(0) = uL∞ . Then λ becomes a graph of α, i.e., λ(α) (see, e.g., [20]). We recall some results about bifurcation diagrams of supercritical elliptic equations. Joseph– Lundgren [19] studied the Dirichlet problem ⎧ in B, ⎨ u + λ(u + 1)p = 0 u>0 in B, (1.3) ⎩ u=0 on ∂B. For p > N/(N − 2), we find the explicit singular solution (λ∗0 , u∗0 ) with λ∗0

  2 2 = N −2− p−1 p−1

and u∗0 (x) = |x|−2/(p−1) − 1.

(1.4)

Define the exponent pJ L by pJ L :=

⎧ ⎨ ⎩

1+ ∞,

4

, √ N −4−2 N −1

N ≥ 11, 2 ≤ N ≤ 10,

which is called the Joseph–Lundgren exponent introduced in [19]. It was shown by [19] that, when pS < p < pJ L , λ(α) oscillates infinitely many times around λ∗0 and converges to λ∗0 as α → ∞, and that, when N ≥ 11 and p ≥ pJ L , λ(α) is strictly increasing and it converges to λ∗0 as α → ∞. The study of the problem ⎧ ⎨ u + λu + up = 0 in B, u>0 in B, (1.5) ⎩ u=0 on ∂B was initiated by Brezis–Nirenberg [2] in the critical case p = pS . Later, the supercritical case p > pS was studied by Budd–Norbury [4], Budd [5], Merle–Peletier [23], Dolbeault–Flores [14],

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and Guo–Wei [17]. Note that (1.5) is transformed into (1.1) with f (u) = u + up by changing u → λ1/(p−1) u. The singular solution of (1.5) was constructed in [4,5,23], and the bifurcation diagram was investigated by [4,14,17]. For other bifurcation diagrams of supercritical problems see [10,11,25,26]. Our purpose is to show the uniqueness of the singular solution for the problem (1.1) and to study characterizations of a singular extremal solution which is defined later. Let us introduce a collection of hypotheses of f (u) in (1.1). (f.1) f ∈ C 1 ([0, ∞)) and f (u) > 0 for u ≥ 0. (f.2) f has the form (1.2) with p > pS , where g(u) satisfies |g(u)| ≤ C0 up−δ

and

|g  (u)| ≤ C0 up−δ−1

for u ≥ u0

(1.6)

with some constants u0 > 0, δ > 0, and C0 > 0. (f.3) f (u) is convex for u ≥ 0. Let (λ(α), u(r, α)) be a solution of (1.1) with u(0, α) = α > 0. Under the assumptions (f.1) and (f.2), all the regular solutions of (1.1) can be described as the curve {(λ(α), u(r, α)); 0 < 1 (B) = {u ∈ H 1 (B) : u = α < ∞}. (See, e.g., [24, Lemma 3.4].) We use the notations Hrad 1 (B) = {u ∈ H 1 (B) : u(1) = 0}. u(r), r = |x|}, and H0,rad rad By a singular solution u of (1.1), we mean that u(r) is a classical solution of (1.1) for 0 < r ≤ 1 and satisfies u(r) → ∞ as r → 0. It was shown by [24, Proposition 1.2 and Corollary 1.3] that, if 1 (B) (f.1) and (f.2) hold, then there exists a singular solution (λ∗ , u∗ ) of (1.1) such that u∗ ∈ H0,rad and satisfies √ u∗ (r) = A( λ∗ r)−θ (1 + O(r δθ ))

as r → 0,

(1.7)

where δ > 0 is the constant in (f.2), θ=

2 p−1

 and A :=

 1/(p−1) 2 2 . N −2− p−1 p−1

(1.8)

First our result states the uniqueness of the singular solution (λ∗ , u∗ ) and the asymptotic property of u(r, α) as α → ∞. Theorem 1.1. Suppose that (f.1) and (f.2) hold. (i) There exists a unique λ∗ > 0 such that the problem (1.1) with λ = λ∗ has a singular solution u∗ . The solution u∗ is a unique singular solution of (1.1) with λ = λ∗ . Furthermore, 1 (B) and it satisfies (1.7) with (1.8). u∗ ∈ H0,rad (ii) As α → ∞, the solution (λ(α), u(r, α)) of (1.1) satisfies λ(α) → λ∗

and u(r, α) → u∗ (r)

where (λ∗ , u∗ ) is the singular solution in (i).

1 in Cloc ((0, 1]),

(1.9)

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Remark 1.1. (i) It should be mentioned that our result requires neither u ∈ H 1 (B) nor the behavior (1.7). Thus (1.1) admits exactly one unbounded positive radial solution for p > pS . Such uniqueness results have been obtained for the special cases f (u) = up and f (u) = −u + up with p > pS by Serrin–Zou [30] and Chern et al. [8], respectively. Our approach is based on the method developed by [8] together with appropriate modifications. (ii) The asymptotic property (1.9) was shown by [23] for the problem (1.5). However, since the proof by [23] relied heavily on the nonlinearity f (u) = u + up , new ideas are needed to show (1.9) for a general nonlinearity. (iii) For properties of positive singular solutions when f (u) has the critical or subcritical growth, we refer to [13]. Define λ = sup{λ(α) : α > 0}. Since f (0) > 0, we have λ(α) → 0 as α → 0. Then, by (1.9), we have λ ∈ (0, ∞) and λ ≥ λ∗ . We say that (λ∗ , u∗ ) is the singular extremal solution if λ(α) is strictly increasing for α > 0 and λ(α) ↑ λ∗ as α → ∞. Consequently, if (1.1) has the singular extremal solution, then λ∗ = λ and (1.1) has a unique classical solution for each λ ∈ (0, λ∗ ), and a unique singular solution at λ = λ∗ , and has no classical solution for λ ≥ λ∗ . Thus the existence of the singular extremal solution determines the bifurcation diagram of the problem. The existence and properties of singular extremal solutions have been studied for the problem (1.1) with general nonlinearity f in a bounded and smooth domain . Assume that ⎫ f is C 1 , nondecreasing, convex function defined on [0, ∞) satisfying ⎪ ⎬ f (0) > 0

and

f (u) ⎭ = ∞. ⎪ u→∞ u

(1.10)

lim

Then there exists a finite positive extremal value λ such that, for 0 < λ < λ, there exists a minimal classical solution uλ ∈ C 2 () ∩ C(), and there exists a weak solution u for λ = λ. The extremal solution u is obtained as the increasing limit of uλ as λ ↑ λ, and it may be either classical or singular. It was shown by Brezis–Vázquez [3] that, if (λ∗ , u∗ ) is a singular solution satisfying u∗ ∈ H01 () and



 |∇ϕ|2 − λ∗ f  (u∗ )ϕ 2 dx ≥ 0

for all ϕ ∈ C01 (),



then u∗ is the extremal solution and λ∗ = λ. By using this result, the authors in [3] showed that, for the problem (1.3), the explicit singular solution given by (1.4) is the extremal solution when N ≥ 11 and p ≥ pJ L . It is known that, if the extremal solution u is classical, then the first eigenvalue of − −λf  (u) is 0. This fact is an immediate consequence of the implicit function theorem together with the impossibility of continuing the branch beyond λ, see [9,21,29]. Whereas, for the typical example of singular extremal solutions with f (u) = eu and f (u) = (u + 1)p , it was shown by [3] that, the first eigenvalue is positive. By Cabré and Martel [7] it was shown that, even if the first eigenvalue is positive, there always exists a positive L1 weak eigenfunction with the eigenvalue 0. For further details, we refer to [1,12,15]. In this paper, we consider the problem (1.1) in the radial case with the nonlinearity f satisfying (f.1)–(f.3), and study the existence of the singular extremal solution in terms of the linearized

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problem at the singular solution. We consider the linearized eigenvalue problem at the singular solution (λ∗ , u∗ ), that is, ⎧ ⎨ φ + λ∗ f  (u∗ )φ = −μφ ⎩

in B,

1 φ ∈ H0,rad (B).

(1.11)

1 (B), and We note here that, when p = pJ L , the equation in (1.11) has no positive solution in Hrad hence, there exists no eigenvalue of the problem (1.11). (See Proposition 4.6 below.) In order to define weak eigenvalues to this problem, we consider the following problem

⎧ ⎨ φ  + N − 1 φ  + λ∗ f  (u∗ )φ = −μφ for 0 < r < 1, r ⎩ φ ∈ C 2 ((0, 1]) and φ(1) = 0.

(1.12)

Define the first weak eigenvalue μ ˜ 1 by μ˜ 1 = sup{μ ∈ R : (1.12) has a positive solution φ(r) for 0 < r < 1},

(1.13)

and μ˜ 1 = −∞ if (1.12) has no positive solution for all μ ∈ R. By Proposition 4.6 below, we see that, if pS < p < pJ L , then μ˜ 1 = −∞, and if p > pJ L , then there exists a solution (μ1 , φ1 ) of (1.11) such that φ1 (r) > 0 for 0 < r < 1, and one has μ˜ 1 = μ1 . 1 Define C0,rad (B) = {u ∈ C01 (B) : u = u(r), r = |x|}. We obtain the following. Theorem 1.2. Suppose that (f.1)–(f.3) hold. Let (λ∗ , u∗ ) be the singular solution of (1.1). Then the following (i)–(iii) are equivalent each other. (i) (λ∗ , u∗ ) is the singular extremal solution. 1 (ii) For any φ ∈ C0,rad (B),

|∇φ|2 dx ≥ λ∗

B

f  (u∗ )φ 2 dx.

(1.14)

B

(iii) μ˜ 1 ≥ 0, where μ˜ 1 is defined by (1.13). ˜ 1 = −∞ if pS < p < pJ L . Then, by (iii), the singular Remark 1.2. (i) As mentioned above, μ solution of (1.1) is not extremal in the case pS < p < pJ L . (ii) (1.14) implies that the Morse index of u∗ is zero. For the definition and the properties of the Morse index to the problem (1.1), we refer to [24, Theorems A and B]. In (ii) we can replace 1 φ ∈ C0,rad (B) by φ ∈ C01 (B). See Remark 5.1 below. (iii) When (f.3) does not hold, the set of minimal solutions is not necessarily continuous in λ and the diagram (λ, uL∞ ) may have a turning point for some λ ∈ (0, λ). For an example, see [6]. (iv) As mentioned above, in the case of a general domain , Brezis and Vázquez [3] proved the equivalence of (i) and (ii) under the assumptions (1.10) and u∗ ∈ H01 (). In Theorem 1.2, we deal with only the radial case, however, we do not need to require that f (u) is nondecreasing.

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We have the following characterization of the nonnegative first weak eigenvalue μ ˜ 1 . For the proof, see Corollary 4.2 below. Proposition 1.1. Suppose that (f.1) and (f.2) hold. Define μ˜ 1 by (1.13). Then μ˜ 1 ≥ 0 if and only if (1.12) with μ = 0 has a positive solution. As a consequence of Theorem 1.2 and Proposition 1.1, we obtain a simpler criterion. Corollary 1.1. Suppose that (f.1)–(f.3) hold. Then (1.1) has the extremal singular solution (λ∗ , u∗ ) if and only if the problem (1.12) with μ = 0 has a positive solution. For a ≥ 0, define fa (u) = f (u + a) for u ≥ 0. In order to clarify the difference between the cases μ˜ 1 > 0 and μ˜ 1 = 0 in Theorem 1.2, we consider the problem ⎧ ⎨ u + λfa (u) = 0 u>0 ⎩ u=0

in B, in B, on ∂B.

(1.15)

We see that the problem (1.15) has the following properties. Proposition 1.2. Suppose that f satisfies (f.1)–(f.3). Let N ≥ 11 and p ≥ pJ L . Then there exists a0 ≥ 0 such that (1.15) has the singular extremal solution for all a ≥ a0 . Remark 1.3. If (f.1)–(f.3) hold, then fa also satisfies (f.1)–(f.3). Then, by Remark 1.2 (i), the singular solution of (1.15) is not extremal for all a ≥ 0 in the case pS < p < pJ L . Let (λ∗a , u∗a ) be a singular solution of (1.15), and let us consider the linearized eigenvalue problem at the singular solution ⎧ ⎨ φ  + N − 1 φ  + λ∗ f  (u∗ )φ = −μφ for 0 < r < 1, a a a r ⎩ φ ∈ C 2 ((0, 1]) and φ(1) = 0.

(1.16)

Define the first weak eigenvalue μ ˜ 1 (a) by μ˜ 1 (a) = sup{μ ∈ R : (1.16) has a positive solution φ(r) for 0 < r < 1},

(1.17)

and μ˜ 1 (a) = −∞ if (1.16) has no positive solution for all μ ∈ R. By Theorem 1.2, the problem (1.15) has the singular extremal solution if and only if μ ˜ 1(a) ≥ 0. Theorem 1.3. Suppose that (f.1)–(f.3) hold. Let N ≥ 11 and p ≥ pJ L . Assume that (1.15) with a = a0 > 0 has the singular extremal solution. (i) If μ˜ 1 (a0 ) > 0 then there exists δ > 0 such that (1.15) has the singular extremal solution for a ∈ (a0 − δ, a0 ). (ii) If μ˜ 1 (a0 ) = 0 then the singular solution of (1.15) is not extremal for a < a0 .

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Remark 1.4. Let us consider (1.15) with f (u) = up . Let a > 0, and let (λ, u) be a solution of (1.15). Put v(x) = u(x)/a. Then (a p−1 λ, v) solves (1.3). Let N ≥ 11 and p ≥ pJ L . Since (1.3) has the singular extremal solution, (1.15) has the singular extremal solution for all a > 0. By Theorem 1.3, we deduce that μ˜ 1 (a) > 0 for all a > 0. In particular, μ˜ 1 (1) > 0. Thus we derive an alternative proof of the fact that μ ˜ 1 > 0 for (1.3) if N ≥ 11 and p ≥ pJ L . In the proofs of Theorems 1.1–1.3, we consider the initial value problem ⎧ ⎨ v  + N − 1 v  + f (v) = 0 for s > 0, s ⎩ v(0) = α and v  (0) = 0,

(1.18)

where α > 0 is a parameter. We denote by v(s, α) a unique solution of (1.18), and denote by s0 (α) the first zero of v(s, α). Put u(r, α) = v(s, α)

with

r=

s , s0 (α)

(1.19)

and put λ(α) = s0 (α)2 .

(1.20)

Then (λ(α), u(r, α)) is a solution of (1.1) satisfying u(0, α) = α. We can study the properties of the eigenvalue λ(α) through an analysis of the zero s0 (α) of the solutions v(s, α) of the problem (1.18). In the proof of Theorems 1.2 and 1.3, the linearized eigenvalue problem of the singular solution φ + λ∗ f  (u∗ )φ = −μφ

in B

becomes important. When f (u) = (u + 1)p (resp. f (u) = eu ), the singular solution can be written explicitly as u∗ (r) = r −2/(p−1) − 1 (resp. u∗ (r) = −2 log r). In both cases the linearized eigenvalue problem becomes φ +

C φ = −μφ |x|2

in B,

φ=0

on ∂B,

where C is a constant. Cabré–Martel [7] gave an exact expression of all weak eigenvalues and eigenfunctions in terms of the Bessel functions. However, we cannot expect the exact expression of λ∗ f (u∗ (r)) in our case. Since λ∗ f  (u∗ ) = O(r −2 ) as r → 0, we need to study the equation of the general form u +

N −1  u + V (r)u = −μu for 0 < r < R, r

(1.21)

where V ∈ C((0, R]) satisfies r 2 V (r) → c0 as r → 0 with some constant c0 ≥ 0. We construct a general theory for the eigenvalue problem of the equation (1.21), using the principal solution whose definition is given in Section 4 below.

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The paper is organized as follows. We will prove Theorem 1.1 in Sections 2 and 3. In Section 2, we show the uniqueness of the singular solution, and in Section 3, we show the convergence of the regular solutions to the singular solution. These results play crucial roles in the proof of Theorems 1.2 and 1.3. In Section 4, we consider second order linear ODEs with singular potential, and state some propositions without proof. By employing these propositions, we give the proof of Theorems 1.2 and 1.3 in Section 5. In Sections 6, 7 and 8, we give the proof of the propositions stated in Section 4. In Section 6, we study the properties of principal solutions for second order linear ODEs, and in Section 7, we investigate the asymptotic behavior of solutions to the linearized equations. Finally, in Section 8, we study the singular eigenvalue problems. 2. Uniqueness of the singular solution In this and the following sections, we assume that p > pS and that (f.1) and (f.2) hold. We consider solutions of the equation v  +

N −1  v + f (v) = 0 s

for s > 0.

(2.1)

To prove Theorem 1.1 (i), we show the following results. Proposition 2.1. (i) Let v be a singular solution of (2.1) for 0 < s ≤ s0 with some s0 > 0. Then v satisfies lim s θ v(s) = A and

s→0

lim s θ+1 v  (s) = −θ A,

s→0

(2.2)

where θ and A are constants defined by (1.8). (ii) There exists at most one positive singular solution of (2.1). The proof of Proposition 2.1 follows from the arguments in [8] with appropriate modifications. We will only give a sketch of the proof. Since v(s) → ∞ as s → 0, there exists s0 > 0 such that v  (s) ≤ 0 and v(s) ≥ v(s0 ) > 0 for 0 < s ≤ s0 . From (f.2), there exists a constant C > 0 satisfying f (v(s)) ≤ Cv(s)p

for 0 < s ≤ s0 .

(2.3)

By Ni and Serrin [28, Theorem 2.1 and Remark following Lemma 2.1], there exists a constant C > 0 such that v(s) ≤ Cs −θ

and |v  (s)| ≤ Cs −θ−1

for 0 < s ≤ s0 .

(2.4)

Integrating (2.1) on [σ, s], and letting σ → 0, we obtain −s

N−1 

s

v (s) =

t N−1 f (v(t))dt 0

Define

for 0 < s ≤ s0 .

(2.5)

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w(t) = s θ v(s)

with t = − log s.

(2.6)

Then w satisfies w  − aw  − Ap−1 w + e−pθt f (eθt w(t)) = 0 for t ≥ t0 ,

(2.7)

where a = N − 2 − 2θ > 0 and t0 = − log s0 . Lemma 2.1. Let v be a positive singular solution of (2.1). Then lim sups→0 s θ v(s) > 0. Proof. Assume to the contrary that lims→0 s θ v(s) = 0. First we will show that (s θ v(s)) ≥ 0 for 0 < s < s1

(2.8)

with some s1 ∈ (0, s0 ]. Define w(t) by (2.6). From (2.7) and (2.3), we have w  − aw  − Ap−1 w + Cw p ≥ 0 for t ≥ t0 . Since w satisfies w(t) → 0 as t → ∞, we have (e−at w  ) ≥ e−at (Ap−1 − Cw p−1 )w > 0

for t ≥ t1

with some t1 ≥ t0 . Then e−at w  (t) is increasing for t ≥ t1 . Assume that there exists t2 ≥ t1 such that w  (t2 ) > 0. Then e−at w  (t) ≥ e−at2 w  (t2 ) > 0 for t ≥ t2 . This implies that w  (t) > e−at2 w  (t2 )eat for t > t2 , and hence w  (t) → ∞ as t → ∞. This contradicts that w(t) → 0 as t → ∞. Thus we obtain w (t) ≤ 0 for all t ≥ t1 , and hence (2.8) holds. From (2.3), we have s 2 f (v(s)) ≤ Cs 2 v(s)p−1 = C(s θ v(s))p−1 → 0 as s → 0. v(s) Take ε > 0 so that ε < 1/p. Then there exists s1 ∈ (0, s0 ] such that s 2 f (v(s)) < (N − 2 − θ )εv(s)

for 0 < s < s1 .

Note here that N − 2 − θ > 0. From (2.8) we have

s

s t N−1 f (v(t))dt ≤ (N − 2 − θ )ε

0

t N−3 v(t)dt 0

s

≤ (N − 2 − θ )εs v(s)

t N−3−θ dt = εs N−2 v(s)

θ

0

for 0 < s < s1 . From (2.5) we have

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−s N−1 v  (s) =

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s t N−1 f (v(s))dt ≤ εs N−2 v(s)

for 0 < s < s1 .

0

This implies that (s ε v(s)) ≥ 0 for 0 < s < s1 , and hence s ε v(s) ≤ s1ε v(s1 ) for 0 < s < s1 . Then we obtain v(s) = O(s −ε ) as s → 0. From (2.3) and (2.5) we obtain −s N−1 v  (s) ≤ C

s

s t N−1 v(t)p dt ≤ C

0

t N−1−pε dt = Cs N−pε . 0

Thus v  (s) = O(s 1−pε ) as s → 0. Since pε < 1, we have lims→0 v(s) < ∞, which is a contradiction. Thus we obtain lim sups→0 s θ v(s) > 0. 2 From (1.2), the equation (2.7) is reduced to w  − aw  − Ap−1 w + w p + G(t, w) = 0

for t ≥ t0 ,

(2.9)

where G(t, w) = e−pθt g(eθt w). By (f.2) there exists a constant C1 > 0 such that |g(u)| ≤ C1 (up−δ + 1) and

|g  (u)| ≤ C1 (up−δ−1 + 1) for u ≥ 0.

(2.10)

Then it follows that |G(t, w)| = |e−pθt g(eθt w)| ≤ C1 (w p−δ + e−(p−δ)θt )e−θδt .

(2.11)

Define 1 E(w)(t) = w  (t)2 + (w(t)) 2

for t ≥ t0 ,

(2.12)

where

(v) = −

Ap−1 2 1 v + v p+1 2 p+1

(2.13)

for v ≥ 0. Observe that E(w) satisfies, for t > t0 , E(w) (t) = (w  (t) − Ap−1 w + w p )w  (t) = aw  (t)2 − w  (t)G(t, w(t)). Lemma 2.2. The following (i)–(iv) hold. (i) w(t), w  (t) and w  (t) are bounded on [t0 , ∞). (ii) G(t, w(t)) → 0 as t → ∞. (iii) w  ∈ L2 ([t0 , ∞)). (iv) limt→∞ E(w)(t) = μ with some μ ∈ R.

(2.14)

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Proof. From (2.4), we see that w(t) and w  (t) are bounded on [t0 , ∞). From (2.11), we obtain

∞ |G(s, w(s))|ds < ∞

(2.15)

t0

and G(t, w(t)) → 0 as t → ∞. From (2.9), w  (t) is also bounded on [t0 , ∞). Thus (i) and (ii) hold. Integrating (2.14) on [t0 , t] we have

t E(w)(t) − E(w)(t0 ) = a



t

w (s) ds − 2

t0

w  (s)G(s, w(s))ds.

(2.16)

t0

From (2.15) we have

∞

∞



|w (s)G(s, w(s))|ds ≤ C t0

|G(s, w(s))|ds < ∞

(2.17)

t0

for some constant C > 0. Let t → ∞ in (2.16). Since E(w)(t) is bounded for t ≥ t0 , we obtain

∞

w  (s)2 ds < ∞.

(2.18)

t0

Thus (iii) holds. Letting t → ∞ in (2.16) again, from (2.17) and (2.18), we see that the limit of E(w)(t) as t → ∞ exists and is finite. Thus (iv) holds. 2 Lemma 2.3. Define w(t) by (2.6). If lim supt→∞ w(t) > 0, then lim w(t) = A

t→∞

and

lim w  (t) = 0.

(2.19)

t→∞

Proof. First we show that lim w(t) = η

(2.20)

t→∞

for some η > 0. Assume to the contrary that (2.20) does not hold. Since w(t) is positive and bounded for t ≥ t0 , there exist 0 ≤ η1 < η2 such that lim inft→∞ w(t) = η1 and lim supt→∞ w(t) = η2 . Then there exist sequences tn → ∞ and sn → ∞ such that w  (tn ) = w  (sn ) = 0,

lim w(tn ) = η1 ,

n→∞

and

lim w(sn ) = η2 .

n→∞

Thus we obtain lim E(w)(tn ) = lim (w(tn )) = (η1 ) and

n→∞

n→∞

lim E(w)(sn ) = lim (w(sn )) = (η2 ).

n→∞

n→∞

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By Lemma 2.2 (iv), we have (η1 ) = (η2 ) = μ = limt→∞ E(w)(t). Note here that (v) is strictly decreasing for 0 ≤ v < A and strictly increasing for v > A. Since 0 ≤ η1 < η2 , we conclude that η 1 < A < η2

and (A) < μ = (η1 ) = (η2 ).

Hence, there exists a sequence τn → ∞ such that w(τn ) = A for n = 1, 2, . . . . Since  lim E(w)(τn ) = lim

n→∞

n→∞

 w  (τn )2 + (A) = μ, 2

we obtain w  (τn )2 = μ − (A) > 0. n→∞ 2 lim

Hence, there exists an integer n0 such that |w  (τn )|2 ≥ μ − (A) for n ≥ n0 . Since w  (t) is bounded for t ≥ t0 by Lemma 2.2 (i), there exists ρ > 0 such that |w  (t)|2 ≥

μ − (A) > 0 for τn − ρ ≤ t ≤ τn + ρ 2

with n ≥ n0 ,

∞ which implies that t0 w  (t)2 dt = ∞. This contradicts Lemma 2.2 (iii), and hence (2.20) holds. Next we will show that lim w  (t) = 0.

t→∞

(2.21)

From (2.20) and Lemma 2.2 (iv), it follows that w  (t)2 = lim (E(w)(t) − (w(t))) = μ − (η). t→∞ t→∞ 2 lim

(2.22)

Then it suffices to show that μ = (η). Since w √ (t)2 /2 ≥ 0, we have μ ≥ (η). Assume that  μ > (η). Then, from (2.22), we obtain |w (t)| > 2(μ − (η))/2 for t ≥ t1 with some t1 ≥ t0 . This implies that limt→∞ |w(t)| = ∞, which is a contradiction. Thus we obtain μ = (η), and hence (2.21) holds. Finally, we will show that η = A in (2.20). Assume to the contrary that η = A. Letting t → ∞ in (2.9), from (2.20), (2.21) and Lemma 2.2 (ii), we obtain lim w  (t) = −Ap−1 η + ηp = 0.

t→∞

Thus we obtain |w  (t)| > | − Ap−1 η + ηp |/2 for t ≥ t2 with some t2 ≥ t0 . This implies that |w  (t)| → ∞ as t → ∞, which is a contradiction. Thus we obtain η = A in (2.20). 2 Proof of Proposition 2.1. By Lemmas 2.1 and 2.3, we obtain (2.2). Thus (i) holds. The result (ii) follows from the similar argument in [8, Proposition 3.2]. See also [22, Corollary 4.3]. 2

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Let (λ∗ , u∗ ) be a singular solution of (1.1). Put v ∗ (s) = u∗ (r)

with s =

√ λ∗ r.

(2.23)

√ Then v ∗ (s) satisfies (2.1) and v ∗ ( λ∗ ) = 0. We denote by s0∗ the first zero of v ∗ (s) for s > 0. Then s0∗ =

√ λ∗ .

(2.24)

ˆ ˆ of (1.1). Put Proof of Theorem 1.1 (i). Assume that there exists anothersingular solution (λ, u) v(s) ˆ = u(r) ˆ with s = λˆ r. Then vˆ also satisfies (2.1) and v( ˆ λˆ ) = 0. Proposition 2.1 (ii) implies that v(s) ˆ ≡ v ∗ (s). Thus we obtain λˆ = λ∗ and u(r) ˆ ≡ u∗ (r), and hence λ∗ is unique, and u∗ (r) is the unique singular solution of (1.1) with λ = λ∗ . 2 3. Convergence to the singular solution In this section, we extend the domain of f in R such that f (u) > 0 and f (u) is bounded for u ≤ 0. Then we may assume that f (u) has the form f (u) = (u+ )p + g(u) for u ∈ R,

(3.1)

where u+ = max{u, 0}, g(u) > 0 and g(u) is bounded for u ≤ 0. As a typical example, for u ≤ 0, we define f (u) = f (0) and g(u) = f (0). Let v ∗ be the singular solution of (2.1), and denote by s0∗ the first zero of v ∗ (s) for s > 0. For α > 0, denote by v(s, α) the solution of (1.18), and denote by s0 (α) the first zero of v(s, α). We extend the domains of v(s, α) and v ∗ (s) to (0, s0∗ + δ) with δ > 0 such that v(s, α) and v ∗ (s) satisfy (2.1). Since f (u) > 0 for u ∈ R, v(s, α) and v ∗ (s) are decreasing in s > 0, and hence v ∗ (s) < 0 for s0∗ < s < s0∗ + δ and v(s, α) < 0 for s0 (α) < s < s0∗ + δ if s0 (α) < s0∗ + δ. To prove Theorem 1.1 (ii), we need the following proposition. 1 ((0, s ∗ + δ]) as α → ∞. In particular, Proposition 3.1. Let δ > 0. Then v(·, α) → v ∗ in Cloc 0 ∗ s0 (α) → s0 as α → ∞.

Proof of Theorem 1.1 (ii). From (1.19) and (1.20), we see that u(r, α) = v(s0 (α)r, α) and λ(α) = s0 (α)2 . From (2.23) and (2.24), we find that u∗ (r) = v ∗ (s0∗ r) and λ∗ = (s0∗ )2 . Then, by Proposition 3.1, we obtain λ(α) → λ∗ as α → ∞. Observe that |u(r, α) − u∗ (r)| = |v(s0 (α)r, α) − v ∗ (s0∗ r)| ≤ |v(s0 (α)r, α) − v ∗ (s0 (α)r)| + |v ∗ (s0 (α)r) − v ∗ (s0∗ r)|. By Proposition 3.1, we obtain u(·, α) → u∗ in Cloc ((0, 1]) as α → ∞.

2

In the remaining part of this section, we will prove Proposition 3.1. Put μ0 > 0 by μ0 = inf{ (v) − (A) : |v − A| ≥

A }, 2

(3.2)

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where is defined by (2.13). Then it follows that |v − A| <

A 2

holds if

(v) < (A) + μ0 .

(3.3)

We obtain the following result. Lemma 3.1. Let w be a solution of (2.9), and define E(w)(t) by (2.13). Then there exists T ≥ 0 such that, if E(w)(t0 ) < (A) + μ0 /2

(3.4)

for some t0 > T , then E(w)(t) < (A) + μ0

for T ≤ t ≤ t0 .

(3.5)

Proof. First we show that there exists a positive constant C2 > 0 such that E(w) (t) ≥ −C2 (w(t)p−δ + 1)2 e−2θδt

for t ≥ 0.

(3.6)

Recall that E(w) (t) satisfies (2.14). From (2.11) with t ≥ 0 and the Young inequality, we have |w  (t)G(t, w(t))| ≤ |w  (t)|C1 (w(t)p−δ + 1)e−θδt ≤ a|w  (t)|2 +

C12 (w(t)p−δ + 1)2 e−2θδt 4a

for t ≥ 0. From (2.14), we obtain (3.6) with C2 = C12 /(4a). Take T ≥ 0 so large that C2 ((3A/2)p−δ + 1)2 −2θδT μ0 < e . 2θ δ 2

(3.7)

Assume that (3.4) holds for some t0 > T . We will show that (3.5) holds. Assume to the contrary that there exists T1 ∈ [T , t0 ) such that E(w)(t) < (A) + μ0

for T1 ≤ t < t0

and E(w)(T1 ) = (A) + μ0 .

Since (w(t)) ≤ E(w)(t), we have (w(t)) < (A) + μ0 for T ≤ t < T1 . Then, by (3.3), we obtain |w(t) − A| < A/2 for T1 ≤ t < t0 . From (3.4), (3.6) and (3.7) it follows that

t0

(A) + μ0 = E(w)(T1 ) = E(w)(t0 ) −

E(w) (s)ds

T1

∞

≤ E(w)(t0 ) + C2 ((3A/2)

p−δ

+ 1)

2 T1

e−2θδs ds

< (A) + μ0 /2 + μ0 /2 = (A) + μ0 . This is a contradiction. Thus we obtain (3.5). 2

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For the solution v(s, α) of (1.18), put z(τ, α) =

v(s, α) α

with

τ = α (p−1)/2 s.

(3.8)

Then z = z(τ, α) satisfies z +

N −1  z + α −p f (αz) = 0 for τ > 0 τ

and z(0) = 1. From (3.1), the equation is written as z +

N −1  z + (z+ )p + α −p g(αz) = 0 for τ > 0. τ

(3.9)

By (f.2), there exists C1 > 0 such that |g(u)| ≤ C1 (up−δ + 1) for u ≥ 0. Then, for α ≥ 1, we obtain   |α −p g(αu)| ≤ C1 α −δ up−δ + 1

for u ≥ 0.

(3.10)

We denote by ζ (τ ) the solution of the equation ζ  +

N −1  ζ + ζp = 0 τ

for τ > 0 and ζ (0) = 1.

(3.11)

It is known by [31, Proposition 3.4] that ζ (τ ) > 0 for τ ≥ 0 and τ (p−1)/2 ζ (τ ) → A as τ → ∞. Lemma 3.2. For any τ1 > 0, we have z(·, α) → ζ in C 1 ([0, τ1 ]) as α → ∞. Proof. Let {αk } be a sequence satisfying αk → ∞ as k → ∞. We may assume that αk ≥ 1 for k = 1, 2, . . . . Put zk (τ ) = max{z(τ, αk ), 0}. From (3.9), if zk (τ ) > 0, zk satisfies zk (τ ) = −τ 1−N

τ

  −p t N−1 zk (t)p + αk g(αk zk (t)) dt.

(3.12)

0

Since vα (s) ≤ α for s ≥ 0, we have 0 ≤ zk (τ ) ≤ 1 for τ ≥ 0. From (3.10), we have     −p zk (τ )p + αk g(αk zk (τ )) ≤ 1 + 2C1 αk−δ ≤ 1 + 2C1 for τ ≥ 0. From (3.12) we obtain |zk (τ )| ≤ Cτ for τ ≥ 0 with C = (1 + 2C1 )/N . By (3.9), {zk (τ )} is bounded for τ ≥ 0. Since zk (0) = 1 for k = 1, 2, . . . , we see that {zk (τ )} and {zk (τ )} is uniformly bounded and equi-continuous on [0, τ1 ]. By the Ascoli–Arzela theorem, there exists a subsequence which converges to some function z in C 1 ([0, τ1 ]). We will show that z ≡ ζ , where ζ is a solution of (3.11). We note that zk satisfies 1 zk (τ ) = 1 − N −2

τ  0

 −p 1 − (t/τ )N−2 tαk f (αk zk (t))dt

for τ > 0.

(3.13)

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From (3.10) we see that −p

−p

αk f (αk zk (t)) = zk (t)p + αk g(αk zk (t)) → z(t)p uniformly in t ∈ [0, τ1 ] as k → ∞. Letting k → ∞ in (3.13), we obtain 1 z(τ ) = 1 − N −2

τ 

 1 − (t/τ )N−2 tz(t)p dt

for 0 < τ ≤ τ1 ,

0

which implies that z solves (3.11), and hence z ≡ ζ . Thus we obtain zk → ζ

in C 1 ([0, τ1 ])

as k → ∞.

(3.14)

For any sequence {αk } with αk → ∞, there exists a subsequence such that (3.14) holds. This implies that z(·, α) → ζ in C 1 ([0, τ1 ]) as α → ∞. 2 For the solution v(s, α) of (1.18), define w(t, α) = s 2/(p−1) v(s, α) with t = − log s. Then, from (3.8), we have w(t, α) = s 2/(p−1) v(s, α) = τ 2/(p−1) z(τ, α)

(3.15)

d d d w(t, α) = −s (s 2/(p−1) v(s, α)) = −τ (τ 2/(p−1) z(τ, α)). dt ds dτ

(3.16)

and

Lemma 3.3. Let {αn } be a sequence such that αn → ∞ as n → ∞. Then, for any ε > 0, there exist n0 ∈ N and {tn } tending to ∞ as n → ∞ such that, if n ≥ n0 , then |w(tn , αn ) − A| < ε

and

|w  (tn , αn )| < ε.

(3.17)

Proof. Put Z(τ ) = τ 2/(p−1) ζ (τ ). By [31, Proposition 3.4] we have Z(τ ) → A as τ → ∞. Furthermore, we obtain lim inf τ |Z  (τ )| = 0. τ →∞

(3.18)

In fact, assume to the contrary that lim infτ →∞ τ |Z  (τ )| > 0. Then there exist c > 0 and τ0 > 0 such that τ |Z  (τ )| ≥ c for τ ≥ τ0 . This implies that |Z(τ )| → ∞ as τ → ∞, which is a contradiction. Thus (3.18) holds. Then there exists τ1 > 0 such that |Z(τ1 ) − A| <

ε 2

and

ε τ1 |Z  (τ1 )| < . 2

(3.19)

Put Z(τ, α) = τ 2/(p−1) z(τ, α). Lemma 3.2 implies that Z(·, αn ) → Z in C 1 ([0, τ1 ]) as n → ∞. Then there exists n0 ∈ N such that, if n ≥ n0 , then |Z(τ1 , αn ) − Z(τ1 )| <

ε 2

and

ε τ1 |Z  (τ1 , αn ) − Z  (τ1 )| < . 2

(3.20)

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It follows from (3.19) and (3.20) that |Z(τ1 , αn ) − A| < ε

and

τ1 |Z  (τ1 , αn )| < ε

for n ≥ n0 .

−(p−1)/2

Put sn = τ1 αn and tn = − log sn for n ≥ n0 . Then we have sn → 0 and tn → ∞ as n → ∞. 2/(p−1) 2/(p−1) By (3.15) and (3.16) we obtain w(tn , αn ) = sn v(sn , αn ) = τ1 z(τ1 , αn ) = Z(τ1 , αn ) and     d 2/(p−1) d 2/(p−1)  w (tn , αn ) = −sn v(s, αn )) = −τ1 z(τ, αn )) (s (τ ds dτ s=sn τ =τ1 

= −τ1 Z  (τ1 , αn ). Thus (3.17) holds.

2

Lemma 3.4. Define μ0 by (3.2). Let {αn } be a sequence tending to ∞. Then there exist T0 ≥ 0, n0 ∈ N, and {tn } tending to ∞, such that, if n ≥ n0 , then tn > T0 , |w(t, αn ) − A| <

A 2

and |w  (t, αn )| ≤



2μ0

for T0 ≤ t ≤ tn .

(3.21)

Proof. Define E(w)(t) by (2.12). In Lemma 3.3, take ε > 0 such that, if |w(t) − A| < ε and |w  (t)| < ε, then E(w)(t) < (A) + μ0 /2. Then there exist n0 ∈ N and {tn } tending to ∞ such that E(w(·, αn ))(tn ) < (A) + μ0 /2 for n ≥ n0 . By Lemma 3.1, there exists T0 ≥ 0 such that, if n ≥ n0 and tn > T0 , then

(w(t, αn )) ≤ E(w(·, αn ))(t) < (A) + μ0

for T0 ≤ t ≤ tn .

Since tn → ∞, by replacing n0 , we may assume that tn > T0 for n ≥ n0 . From (3.3), the left-hand side of (3.21) holds if n ≥ n0 . Since (w) ≥ (A) for w ≥ 0, we have w  (t, αn )2 = E(w(·, αn )(t) − (w(t, αn )) < (A) + μ0 − (w(t, αn )) ≤ μ0 2 √ for T0 ≤ t ≤ tn . This implies that |w  (t, αn )| ≤ 2μ0 for T0 ≤ t ≤ tn . 2 Proof of Proposition 3.1. Let αn → ∞. By Lemma 3.4, there exist T0 ≥ 0, n0 ∈ N and tn → ∞ such that (3.21) holds for n ≥ n0 . Take {Tk } such that T0 < T1 < T2 < · · · and Tk → ∞ as k → ∞. Since tn → ∞ as n → ∞, for each k = 1, 2, . . . , there exists nk ≥ n0 such that tn ≥ Tk if n ≥ nk . Then, if n ≥ nk , we have |w(t, αn ) − A| <

A 2

and |w  (t, αn )| <



2μ0

for T0 ≤ t ≤ Tk .

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Note that w = w(t, αn ) satisfies (2.9) and G(t, w) is estimated by (2.11). Then w  (t, αn ) is also bounded on [T0 , Tk ]. Thus {w(t, αn )} and {w  (t, αn )} are uniformly bounded and equicontinuous on [T0 , Tk ]. By the Ascoli–Arzela theorem and the diagonal argument, there exist w ∈ [T0 , ∞) and a subsequence, denoted by w(t, αn ) again, such that, on any compact subinterval I of [T0 , ∞), {w(·, αn )} converges to w in C 1 (I ) as n → ∞. Observe that w(t, αn ) satisfies

e

−at



w (t, αn ) − e

−aT0

t



w (T0 , αn ) =

  e−as Ap−1 w(s, αn ) − e−pθs f (eθs w(s, αn )) ds

T0

for t > T0 . Letting n → ∞, we obtain

e

−at



w (t) − e

−aT0



t

w (T0 ) =

  e−as Ap−1 w(s) − e−pθs f (eθs w(s)) ds

T0

for t > T0 . This implies that w solves (2.7) for t > T0 . Since (3.21) holds, w satisfies |w(t) − A| ≤

A 2

for T0 < t < ∞.

(3.22)

Put v(s) = s −2/(p−1) w(t) with t = − log s, and put S0 = e−T0 . Then v solves (2.1) for 0 < s ≤ S0 and v(·, αn ) converges to v in C 1 on any compact subinterval of (0, S0 ] as n → ∞. From (3.22) we have lims→0 v(s) = ∞, which implies that v(s) ≡ v ∗ (s) by Proposition 2.1 (ii). Thus we 1 ((0, S ]) as n → ∞. Furthermore, we obtain conclude that v(·, αn ) converges to v ∗ in Cloc 0 1 v(·, αn ) → v ∗ (·) in Cloc ((0, s0∗ + δ]) as n → ∞.

(3.23)

For any sequence αn → ∞, there exists a subsequence such that (3.23) holds. This implies that 1 ((0, s ∗ + δ]) as α → ∞. 2 v(·, α) → v ∗ in Cloc 0 4. Linear ordinary differential equations with singular potentials In this section we review some results on second order linear ordinary differential equations with singular potentials. In the proof of Theorems 1.2 and 1.3 and propositions in Section 1, these results play a crucial role. We will give the proof of these results in Sections 6, 7 and 8 below. 4.1. Properties of principal solutions Let us consider the ordinary differential equation u +

N −1  u + V (r)u = 0 for 0 < r < R, r

(4.1)

where V ∈ C((0, R]). It is known that, if (4.1) has a solution u which is positive near r = 0, then there exists a unique (neglecting a constant factor) solution u0 of (4.1) such that u0 (r) satisfies u0 (r) > 0 for 0 < r ≤ r0 with some r0 ≤ R and

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r0 0

dr = ∞. r N−1 u0 (r)2

(4.2)

Furthermore, any solution u1 which is linearly independent of u0 satisfies |u1 (r)/u0 (r)| → ∞ as r → 0 and

r1 0

dr <∞ r N−1 u1 (r)2

(4.3)

if u(r) > 0 for 0 < r ≤ r1 . The above solutions u0 and u1 are said to be principal and nonprincipal, respectively. (See, e.g., Hartman [18, Chapter XI].) Let us consider another equation u +

N −1   u + V (r)u = 0 r

for 0 < r < R,

(4.4)

 ∈ C((0, R]) satisfies V (r) ≤ V (r) for 0 < r ≤ R. By [27], we obtain the following where V comparison result. The proof will be given in Section 6 below. Proposition 4.1. Assume that u0 is a principal solution of (4.1) satisfying u0 (r) > 0 for 0 < r < R. Let u be a solution of (4.4). Assume that either the following (i) or (ii) hold. (i)

u0 (R) = 0;

(ii)

u0 (R) = 0,

u(R) = 0

and

u0 (R) u (R) ≤ . u0 (R) u(R)

Then u(r) has at least one zero in (0, R), otherwise u(r) is a constant multiple of u0 (r) and (r) for 0 < r ≤ R. V (r) ≡ V By Proposition 4.1, we obtain the following. (r), and that (4.4) has a solution u satisfying u(r) > 0 for Corollary 4.1. Assume that V (r) < V 0 < r < R and u(R) = 0. Let u0 be the principal solution of (4.1). Then u0 (r) > 0 for 0 < r ≤ R. Let I ⊂ R be an interval, and let us consider the equation u +

N −1  u + V (r, μ)u = 0 for 0 < r < R, r

(4.5)

 ∈ C((0, R]) satisfying where μ ∈ I and V (r, μ) ∈ C((0, R] × I ). Assume that there exists V (r) for all (r, μ) ∈ ((0, R] × I ) and the equation (4.4) has a solution which is positive V (r, μ) ≤ V near r = 0. By the Sturm comparison theorem, for all μ ∈ I , (4.5) has a positive solution near r = 0. Hence, there exists the principal solution of (4.5) for μ ∈ I . We denote by u0 (r, μ) the principal solution of (4.5). Multiplying constants, we may assume that u0 (r, μ) is positive near r = 0 and satisfies u0 (r1 , μ)2 + u0 (r1 , μ)2 = 1 for all μ ∈ I

(4.6)

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with some r1 ∈ (0, R). We obtain the continuous dependence of the principal solution on the parameter. The proof will be given in Section 6 below. Proposition 4.2. Let u0 (r, μ) be the principal solution of (4.5) for μ ∈ I such that u0 (r, μ) is positive near r = 0 and satisfies (4.6) with some r1 ∈ (0, R). Then, for each r ∈ (0, R], u0 (r, μ) and u0 (r, μ) are continuous for μ ∈ I . Let us consider the eigenvalue problem of the form ⎧ ⎨ φ  + N − 1 φ  + V (r)φ = −μφ for 0 < r < R, r ⎩ φ ∈ C 2 ((0, R]) and φ(R) = 0,

(4.7)

where V ∈ C((0, R]) and μ ∈ R. We assume that the equation in (4.7) has a solution which is positive near r = 0. Define μ˜ 1 by μ˜ 1 = sup{μ ∈ R : (4.7) has a positive solution φ(r) for 0 < r < R},

(4.8)

and μ˜ 1 = −∞ if (4.7) has no positive solution for all μ ∈ R. The following proposition characterizes μ˜ 1 in terms of the principal solution of the equation in (4.7), that is, φ  +

N −1  φ + V (r)φ = −μφ r

for 0 < r < R.

(4.9)

The proof will be given in Section 6 below. Proposition 4.3. Assume that μ˜ 1 > −∞. Then μ = μ˜ 1 if and only if the principal solution φ0 (r, μ) of (4.9) satisfies φ0 (r, μ) > 0 for 0 < r < R and φ0 (R, μ) = 0. By Propositions 4.1 and 4.3 we obtain the following corollary, from which Proposition 1.1 follows immediately. Corollary 4.2. Define μ˜ 1 by (4.8). The following (i)–(iv) are equivalent each other. (i) μ˜ 1 ≥ 0. (ii) The principal solution of (4.9) with μ = 0 has no zero in (0, R). (iii) There exists a positive solution of (4.9) with μ = 0 for 0 < r < R. (iv) The problem (4.7) with μ = 0 has a positive solution. Proof. Let μ˜ 1 ≥ 0. Then, by Proposition 4.3, φ0 (r, μ˜ 1 ) is positive on (0, R). If μ˜ 1 = 0, it is clear that (ii) holds. If μ˜ 1 > 0, by Corollary 4.1, the principal solution of (4.9) with μ = 0 has no zero on (0, R). Thus (i) ⇒ (ii) holds. It is clear that (ii) ⇒ (iii) holds. Assume that (4.9) with μ = 0 has a positive solution φ(r) for 0 < r < R. If φ(R) = 0, then φ solves the problem (4.7) with μ = 0. Assume that φ(R) > 0. Then, by the Sturm comparison theorem, a non-trivial solution of (4.7) with μ = 0 has no zero in (0, R). Thus (iii) ⇒ (iv) holds. It is clear that (iv) ⇒ (i) holds. 2

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4.2. Eigenvalue problems for the linearized equation In this subsection, we assume that f satisfies (f.1) and (f.2). Let v ∗ be the singular solution of (2.1), and let s0∗ be the first zero of v ∗ (s). We consider the eigenvalue problem ⎧ ⎨ ψ  + N − 1 ψ  + f  (v ∗ )ψ = −μψ for 0 < s < s ∗ , 0 s ⎩ ψ ∈ C 2 ((0, s0∗ ]) and ψ(s0∗ ) = 0.

(4.10)

μ˜ 1 = sup{μ ∈ R : (4.10) has a positive solution ψ(s) for 0 < s < s0∗ },

(4.11)

Define μ˜ 1 by

and μ˜ 1 = −∞ if (4.10) has no positive solution for all μ ∈ R. First we consider the behavior of solutions of the equation in (4.10), that is, ψ  +

N −1  ψ + f  (v ∗ )ψ = −μψ s

for 0 < s < s0∗ .

(4.12)

The following results are included in Proposition 7.3 below. Proposition 4.4. Assume that (f.1) and (f.2) hold. Let μ ∈ R. If p ≥ pJ L , then (4.12) has a solution ψ which is positive near s = 0. On the other hand, if pS < p < pJ L , then any solution ψ of (4.12) oscillates near s = 0. Remark 4.1. By the proof of Proposition 7.3, we see that s 2 f  (v ∗ (s)) = pAp−1 + o(1) as s → 0. Conditions p ≥ pJ L and p < pJ L , in Proposition 4.4, come from the relation between pAp−1 and (N − 2)2 /4 which is the optimal constant of Hardy inequality. In fact, we see that pAp−1 ≤ (N − 2)2 /4 if p ≥ pJ L and pAp−1 > (N − 2)2 /4 if p < pJ L . For r > 0, define Br = {x ∈ RN : |x| < r}. Let us consider the following eigenvalue problems ⎧ ⎨ φ + f  (v ∗ )φ = −μφ ⎩

in Bs0∗ ,

1 φ ∈ H0,rad (Bs0∗ )

(4.13)

and ⎧ ⎨ φ + μf  (v ∗ )φ = 0 in Bs0∗ , ⎩

1 φ ∈ H0,rad (Bs0∗ ).

(4.14)

We show the existence of the first eigenvalue of the problems (4.13) and (4.14) if μ˜ 1 < 0. The proof will be given in Section 8 below.

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Proposition 4.5. Assume that (f.1) and (f.2) hold. Define μ˜ 1 by (4.11). Assume μ˜ 1 < 0. (i) Let p > pJ L . Then there exists a solution (μ1 , φ1 ) of the problem (4.13) such that μ1 < 0 and φ1 (s) > 0 for 0 < s < s0∗ . (ii) Let p = pJ L . Then there exists a solution (μ1 , φ1 ) of the problem (4.14) such that μ1 ∈ (0, 1) and φ1 (s) > 0 for 0 < s < s0∗ . ˜ 1 < 0 in ProposiRemark 4.2. By employing Corollary 4.2, we can replace the assumption μ tion 4.5 by the following (i) or (ii). (i) The principal solution of (4.12) with μ = 0 has at least one zero in (0, s0∗ ). (ii) Any solution of (4.12) with μ = 0 has at least one zero in (0, s0∗ ). For the eigenvalue problem (1.11) and the equation in (1.12), we obtain the following results. The proof will be given in Section 8 below. Proposition 4.6. Suppose that (f.1) and (f.2) hold. (i) Let p > pJ L . Then there exists a solution (μ1 , φ1 ) of (1.11) such that φ1 (r) > 0 for 0 < r < 1, and one has μ1 = μ˜ 1 , μ˜ 1 is defined by (1.13). (ii) Let p = pJ L . Then there exists no eigenvalue of the problem (1.11). (iii) Let pS < p < pJ L and let μ ∈ R. Then any solution φ of the equation in (1.12) oscillates near r = 0. 5. Proof of Theorems 1.2 and 1.3 5.1. Proof of Theorem 1.2 Let v ∗ be the singular solution of (2.1), and let s0∗ be the first zero of v ∗ (s). In this subsection, we will show the following result. Proposition 5.1. Assume that (f.1)–(f.3) hold. Let v(s, α) be a solution of (1.18), and let s0 (α) be the first zero of v(s, α). Then the following (i)–(iv) are equivalent each other. (i) For all α > 0, s0 (α) < s0∗ . (ii) Let 0 < α1 < α2 . Then v(s, α1 ) < v(s, α2 ) for 0 ≤ s ≤ s0 (α1 ), and hence s0 (α1 ) < s0 (α2 ). 1 (Bs0∗ ), (iii) For any ψ ∈ C0,rad

|∇ψ|2 dx ≥

Bs ∗ 0

f  (v ∗ )ψ 2 ds.

(5.1)

Bs ∗ 0

(iv) μ˜ 1 ≥ 0, where μ˜ 1 is defined by (4.11). 1 (Bs0∗ ) by ψ ∈ C01 (Bs0∗ ). See the proof of ProposiRemark 5.1. In (iii) we can replace ψ ∈ C0,rad tion 5.1 (ii) ⇒ (iii).

Let (λ∗ , u∗ ) be the singular solution of (1.1). As a consequence of Proposition 5.1, we obtain the following.

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Corollary 5.1. Assume that (f.1)–(f.3) hold. Let (λ(α), u(r, α)) be the solution of (1.1) with u(0, α) = α > 0. Then, for the problem (1.1), the following (i)–(iv) are equivalent each other. (i) For all α > 0, λ(α) < λ∗ . (ii) λ(α) is strictly increasing in α > 0. 1 (iii) For any φ ∈ C0,rad (B), (1.14) holds. (iv) μ˜ 1 ≥ 0, where μ˜ 1 is defined by (1.13). Proof. From (1.20) and (2.24), we see that λ(α) = s0 (α)2 and λ∗ = (s0∗ )2 . For a solution φ of (1.12), put ψ(s) = φ(r)

with

s=

√ λ∗ r = s0∗ r.

(5.2)

Then, from (2.23), the problem (1.12) is reduced to (4.10). It is clear that μ˜ 1 defined by (1.13) coincides with the one defined by (4.11). By the change of variable y = x/s0∗ , we see that (5.1) is equivalent to (1.14). Then (i), (iii) and (iv) of Proposition 5.1 are equivalent to (i), (iii) and (iv) of Corollary 5.1, respectively. By Proposition 5.1, (i), (iii) and (iv), in Corollary 5.1, are equivalent each other. We will show that (i) and (ii), in Corollary 5.1, are equivalent. Note that (ii) of Proposition 5.1 implies (ii) of Corollary 5.1. Then, in Corollary 5.1, (i) ⇒ (ii) holds. Conversely, assume that (ii) holds. By Proposition 3.1, we see that λ(α) → λ∗ as α → ∞. This implies that λ(α) < λ∗ for all α > 0, and hence (i) holds. 2 Theorem 1.2 follows from Corollary 5.1 immediately. To prove Proposition 5.1, we need some lemmas. Lemma 5.1. Assume that s0 (α) < s0∗ for all α > 0. Then, for all α > 0, v(s, α) < v ∗ (s)

for 0 < s ≤ s0 (α).

(5.3)

Proof. Since f (0) > 0, we have s0 (α) → 0 as α → 0 and v(·, α)L∞ ([0,s0 (α)]) → 0 as α → 0. Then, for sufficiently small α > 0, (5.3) holds. Assume to the contrary that v(s, α1 ) − v ∗ (s) has zeros in (0, s0 (α1 )) for some α1 > 0. Define the set A = {α > 0 : v(s, α) − v ∗ (s) has at least one zero in (0, s0 (α))}, and put α0 = inf A. Since (5.3) holds for sufficiently small α > 0, we have α0 ∈ (0, α1 ]. By the uniqueness of the solution to the ordinary differential equations, each zero of v(s, α) − v ∗ (s) is simple, and hence the zero depends continuously on α > 0 and does not disappear unless the zero goes out from the boundary of (0, s0 (α)). Then the set A is open, and hence α0 ∈ / A and α0 + ε ∈ A for each small ε > 0. Thus the zero of v(s, α) − v ∗ (s) disappears at α = α0 . Since v(s, α0 ) − v ∗ (s) → −∞ as s → 0, the zero must be s0 (α0 ), and hence v(s0 (α0 ), α0 ) − v ∗ (s0 (α0 )) = 0. On the other hand, since s0 (α) < s0∗ for all α > 0, we have 0 = v(s0 (α0 ), α0 ) < v ∗ (s0 (α0 )). This is a contradiction. Thus (5.3) holds for all α > 0. 2 Lemma 5.2. Assume that (f.3) holds. Let 0 < α1 < α2 . Assume that there exists s1 ∈ (0, s0 (α1 )] satisfying

Y. Miyamoto, Y. Naito / J. Differential Equations 265 (2018) 2842–2885

v(s, α1 ) < v(s, α2 )

for 0 ≤ s < s1

and

v(s1 , α1 ) = v(s1 , α2 ).

2865

(5.4)

If α3 > α2 then v(s, α3 ) − v(s, α2 ) has at least one zero in (0, s1 ]. Proof. Assume to the contrary that v(s, α3 ) − v(s, α2 ) > 0 for 0 < s ≤ s1 . Put φ1 (s) = v(s, α2 ) − v(s, α1 ) and φ2 (s) = v(s, α3 ) − v(s, α2 ). Then φi with i = 1, 2 satisfies φi (s) > 0 for 0 < s < s1 , φi (0) = 0 and φi +

N −1  φi + mi φi = 0 for 0 < s < s1 , s

(5.5)

where mi satisfies f  (v(s, αi )) ≤ mi (s) ≤ f  (v(s, αi+1 )) for 0 ≤ s ≤ s1 . Since f (u) is convex, we have m1 (s) ≤ m2 (s) for 0 ≤ s < s1 and φ1 (s1 ) < 0 and φ2 (s1 ) > 0.

φ1 (s1 ) = 0,

(5.6)

Multiplying (5.5) with i = 1 and i = 2 by φ2 and φ1 , respectively, and integrating them on [0, s1 ], we obtain

s

N−1

(φ1 φ2

s 1

 − φ1 φ2 )

s=0

s1 =−

t N−1 (m1 − m2 )φ1 φ2 dt ≥ 0. 0

On the other hand, from (5.6), we have s 1  s N−1 (φ1 φ2 − φ1 φ2 ) = s1N−1 φ1 (s1 )φ2 (s1 ) < 0, s=0

which is a contradiction. Therefore, φ2 has at least one zero in (0, s1 ].

2

Proof of Proposition 5.1. (i) ⇒ (ii). Assume that s0 (α) < s0∗ for all α > 0. Assume to the contrary that, for some α1 and α2 with 0 < α1 < α2 , there exists s1 ∈ (0, s0 (α1 )] such that (5.4) holds. By Lemma 5.1, for all α > 0 we obtain (5.3). In particular, v(s, α2 ) satisfies v(s, α2 ) < v ∗ (s) for 0 < s ≤ s1 .

(5.7)

1 ((0, s ]) as α → ∞. Take s ∈ (0, s ) such By Proposition 3.1 we have v(s, α) → v ∗ (s) in Cloc 1 2 1 ∗ that v (s2 ) > α2 . Then, by (5.7), we can take α3 > α2 such that

v(s, α3 ) > v(s, α2 ) for s2 ≤ s ≤ s1

and

v(s2 , α3 ) > α2 .

Since v(s, α) is decreasing in s > 0, we obtain v(s, α3 ) ≥ v(s2 , α3 ) > α2 ≥ v(s, α2 )

for 0 ≤ s ≤ s2 .

Thus we obtain v(s, α3 ) > v(s, α2 ) for 0 ≤ s ≤ s1 . This contradicts Lemma 5.2, and hence (ii) holds.

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(ii) ⇒ (iii). Let α > 0. First we will show that, if (ii) holds, then



f  (v(|x|, α))ψ 2 dx

|∇ψ|2 dx ≥ Bs0 (α)

(5.8)

Bs0 (α)

for all ψ ∈ C01 (Bs0 (α) ). Put

μ1 =

|∇ψ|2 − f  (v(|x|, α))ψ 2 dx.

inf

ψ∈H01 (Bs0 (α) )

Bs0 (α)

We will show that μ1 ≥ 0. Assume to the contrary that μ1 < 0. Then, by the variational characterization, there exists a solution ψ1 ∈ H01 (Bs0 (α) ) of −ψ1 − f  (v(|x|, α))ψ1 = μ1 ψ1

in Bs0 (α)

(5.9)

satisfying ψ1 > 0 in Bs0 (α) . Take αˆ > α, and put w(r) = v(r, α) ˆ − v(r, α). By (f.3), w = w(r), r = |x|, satisfies −w − f  (v(|x|, α))w ≥ 0 in Bs0 (α) .

(5.10)

1 Since (ii) holds, we have w ∈ H0,rad (Bs0 (α) ) and w > 0 in Bs0 (α) . Multiplying (5.9) by w, and integrating by parts on Bs0 (α) , we obtain



∇ψ1 · ∇wdx −

Bs0 (α)

f  (v(|x|, α))ψ1 wdx = μ1

Bs0 (α)

Bs0 (α)

ψ1 wdx +

∂ψ1 wds, ∂n

∂Bs0 (α)

where n denotes the outer unit normal on ∂Bs0 (α) and ds denote the surface element of ∂Bs0 (α) . Since μ1 < 0 and ∂ψ1 /∂n < 0 on ∂Bs0 (α) by the Hopf lemma, we obtain

∇ψ1 · ∇wdx −

Bs0 (α)

f  (v(|x|, α))ψ1 wdx < 0.

Bs0 (α)

On the other hand, multiplying (5.10) by ψ1 , and integrating by parts on Bs0 (α) , we obtain

∇ψ1 · ∇wdx −

Bs0 (α)

f  (v(|x|, α))ψ1 wdx ≥ 0.

Bs0 (α)

This is a contradiction. Thus we obtain μ1 ≥ 0, and hence (5.8) holds for all ψ ∈ H01 (Bs0 (α) ). Since C01 (Bs0 (α) ) ⊂ H01 (Bs0 (α) ), (5.8) holds for all ψ ∈ C01 (Bs0 (α) ). We will show that (iii) holds for all ψ ∈ C01 (Bs0∗ ). Let ψ ∈ C01 (Bs0∗ ). Since s0 (α) → s0∗ as α → ∞ by Proposition 3.1 (ii), there exists α0 > 0 such that supp ψ ⊂ Bs0 (α) for α ≥ α0 . Then

Y. Miyamoto, Y. Naito / J. Differential Equations 265 (2018) 2842–2885

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ψ ∈ C01 (Bs0 (α) ), and (5.8) holds true for α ≥ α0 . Letting α → ∞ in (5.8), by the monotone convergence theorem, we obtain (5.1). Thus (iii) holds for all ψ ∈ C01 (Bs0∗ ). In particular, (iii) holds 1 (Bs0∗ ). for all ψ ∈ C0,rad (iii) ⇒ (iv). Assume to the contrary that (iii) holds but μ˜ 1 < 0. In the case p > pJ L , by 1 Proposition 4.5 (i), there exists a solution (μ1 , φ1 ) ∈ R × H0,rad (Bs0∗ ) of (4.13) satisfying μ1 < 0 and φ1 > 0 in Bs0∗ . Multiplying the equation in (4.13) by φ1 , and integrating it on Bs0∗ , we have

|∇φ1 | dx − 2

Bs ∗



f (v

∗

)φ12 dx

Bs ∗

0

= μ1

φ12 dx < 0.

Bs ∗

0

0

1 1 (B ∗ ), there exists ψ ∈ C 1 ∗ Since C0,rad (Bs0∗ ) is dense in H0,rad s0 0,rad (Bs0 ) satisfying



|∇ψ| dx − 2

Bs ∗

f  (v ∗ )ψ 2 dx < 0.

(5.11)

Bs ∗

0

0

This is a contradiction. 1 (Bs0∗ ) In the case p = pJ L , by Proposition 4.5 (ii), there exists a solution (μ1 , φ1 ) ∈ R × H0,rad of (4.14) satisfying 0 < μ1 < 1 and φ1 > 0 in Bs0∗ . Multiplying the equation in (4.14) by φ1 , and integrating it on Bs0∗ , we have

|∇φ1 | dx − μ1 2

Bs ∗

f  (v ∗ )φ12 dx = 0.

Bs ∗

0

0

Since μ1 < 1, we have

|∇φ1 |2 dx −

Bs ∗

Bs ∗

0

0

f  (v ∗ )φ12 dx = −(1 − μ1 )

f  (v ∗ )φ12 dx < 0.

Bs ∗ 0

1 1 (B ∗ ), there exists ψ ∈ C 1 ∗ Since C0,rad (Bs0∗ ) is dense in H0,rad s0 0,rad (Bs0 ) satisfying (5.11). This is a contradiction. Finally, we consider the case pS < p < pJ L . Let ψ be a solution of (4.12) with μ = −1. Then, by Proposition 4.4, ψ(s) oscillates near s = 0, and hence, ψ has at least two positive zeros s0 and s1 in (0, s0∗ ). Multiplying (4.12) by s N−1 ψ and integrating it over [s0 , s1 ], we have

s1 s s0

N−1

s1    2  ∗ 2 ψ (s) − f (v (s))ψ(s) ds = − s N−1 ψ(s)2 ds < 0. s0

˜ ˜ Define ψ˜ by ψ(s) = ψ(s) if s0 ≤ s ≤ s1 and ψ(s) = 0 if s ∈ [0, s0∗ ] \ [s0 , s1 ]. Then ψ˜ ∈ 1 H0,rad (Bs0∗ ) and satisfies

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˜ 2 dx − |∇ ψ|

Bs ∗

f (v )ψ˜ 2 dx = − 

∗

s N−1 ψ(s)2 ds < 0. s0

Bs ∗

0

s1

0

1 1 (B ∗ ), there exists ψ ∈ C 1 ∗ Since C0,rad (Bs0∗ ) is dense in H0,rad s0 0,rad (Bs0 ) satisfying (5.11). This is a contradiction. (iv) ⇒ (i). Assume to the contrary that μ˜ 1 ≥ 0 and s0 (α) ≥ s0∗ for some α > 0. Then there exists s1 ≤ s0∗ such that

v(s, α) < v ∗ (s)

for 0 < s < s1

and

v(s1 , α) = v ∗ (s1 ).

Let w(s) = v ∗ (s) − v(s, α). Then w satisfies w(s) > 0 for 0 < s < s1 and w(s1 ) = 0. Furthermore, w solves w  +

N −1  w + V (s)w = 0 for 0 < s < s1 , s

(5.12)

where V (s) =

f (v ∗ (s)) − f (v(s, α)) . v ∗ (s) − v(s, α)

By the mean value theorem and (f.3), we have V (s) ≤ f  (v ∗ (s))

for 0 < s < s1 .

Since f  (u) → ∞ as u → ∞ by (f.2) and (f.3), there exists a subinterval I ⊂ [v ∗ (s1 ), ∞) such that f  (u) is strictly increasing for u ∈ I . Then we have V (s) ≡ f  (v ∗ (s)) for 0 < s < s1 . Recall that v ∗ satisfies (2.2). Since 0 < w(s) ≤ v ∗ (s) for 0 < s < s1 and p > pS , we have

σ 0

ds ≥ s N−1 w(s)2

σ

ds =∞ s N −1 v ∗ (s)2

0

for 0 < σ < s1 . Then w is a principal solution of (5.12). Let us consider the equation ψ  +

N −1  ψ + f  (v ∗ (s))ψ = 0 for 0 < s < s0∗ . s

(5.13)

By Corollary 4.2, μ˜ 1 ≥ 0 implies that (5.13) has a positive solution ψ(s) for 0 < s < s0∗ . On the other hand, by Proposition 4.1, the solution ψ has at least one zero in (0, s1 ) ⊂ (0, s0∗ ). This is a contradiction. Thus we obtain s0 (α) < s0∗ for all α > 0. 2

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5.2. Proof of Proposition 1.2 and Theorem 1.3 Let a ≥ 0. Since v ∗ (s) is strictly decreasing in s > 0, there exists sa∗ ≤ s0∗ such that v ∗ (sa∗ ) = a. Note that sa∗ is continuous for a ≥ 0 and satisfies that sa∗ is strictly decreasing in a > 0

and sa∗ → 0

as a → ∞.

(5.14)

Define va∗ (s) = v ∗ (s) − a. Then va∗ (s) satisfies (va∗ ) +

N −1 ∗  (va ) + fa (va∗ ) = 0 for 0 < s < sa∗ s

and va∗ (sa∗ ) = 0.

Let (λ∗a , u∗a ) be the singular solution of (1.15). Then we see that u∗a (r) = va∗ (s)

with r = s/sa∗

and

λ∗a = (sa∗ )2 .

(5.15)

In fact, put u˜ ∗a (r) = va∗ (s) with r = s/sa∗ and λ˜ ∗a = (sa∗ )2 . Then (λ˜ ∗a , u˜ ∗a ) is a singular solution of (1.15). By Theorem 1.1 (i), the singular solution of (1.15) is unique. Then we have λ∗a = λ˜ ∗a and u∗a ≡ u˜ ∗a , and hence (5.15) holds. For a solution φ of the problem (1.16), put ψ(s) = φ(r) with s = sa∗ r. Then, from (5.15), ψ satisfies ⎧ ⎨ ψ  + N − 1 ψ  + f  (v ∗ (s))ψ = −μψ for 0 < s < s ∗ , a a s ⎩ ψ ∈ C 2 ((0, sa∗ ]) and ψ(sa∗ ) = 0.

(5.16)

It is clear that μ˜ 1 (a) defined by (1.17) coincides with the one defined by μ˜ 1 (a) = sup{μ ∈ R : (5.16) has a positive solution on (0, sa∗ )}. Let us consider the equation ψ  +

N −1  ψ + fa (va∗ )ψ = 0 for s > 0. s

(5.17)

Corollary 1.1 implies that (1.15) has the singular extremal solution if and  only if the problem (1.16) with μ = 0 has a positive solution φ. Put ψ(s) = φ(r) with s = λ∗a r. Then ψ(s) solves (5.17) for 0 < s ≤ sa∗ . Thus we obtain the following lemma. Lemma 5.3. Assume that (f.1)–(f.3) hold. Then (1.15) has the singular extremal solution if and only if (5.17) has a positive solution ψ(s) for 0 < s < sa∗ . Remark 5.2. Note that f  (v ∗ (s)) = fa (va∗ (s)) for s > 0. Then, in Lemma 5.3, the equation (5.17) can be replaced by ψ  +

N −1  ψ + f  (v ∗ (s))ψ = 0 for s > 0. s

(5.18)

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Proof of Proposition 1.2. By Proposition 4.4, (5.18) has a solution ψ which is positive near s = 0. Take s0 > 0 such that ψ(s) > 0 for 0 < s ≤ s0 . From (5.14), there exists a0 > 0 such that sa∗0 ≤ s0 . Let a ≥ a0 . Then sa∗ ≤ sa∗0 from (5.14), and hence the solution ψ is positive on (0, sa∗ ). By Lemma 5.3, (1.15) has the singular extremal solution for all a ≥ a0 . 2 Proof of Theorem 1.3. By Proposition 4.3, the equation ψ  +

N −1  ψ + fa (va∗ )ψ = −μ˜ 1 (a)ψ s

for s > 0

has the principal solution ψ˜ 0 satisfying ψ˜ 0 (s) > 0 for 0 < s < sa∗0 and ψ˜ 0 (sa∗0 ) = 0. Let ψ0 be the principal solution of (5.17). If μ˜ 1 (a0 ) > 0, by Corollary 4.1, ψ0 (s) > 0 for 0 < s ≤ sa∗0 . Then ψ0 (s) > 0 for 0 < s ≤ s1 with some s1 > sa∗0 . Since sa∗ is continuous for a > 0, we can take δ > 0 such that sa∗ < s1 for a ∈ (a0 − δ, a0 ). Then ψ0 is positive on (0, sa∗ ) for a ∈ (a0 − δ, a0 ). By Lemma 5.3, (1.15) has the singular extremal solution for a ∈ (a0 − δ, a0 ). Thus (i) holds. If μ˜ 1 (a0 ) = 0, then ψ˜ 0 (s) ≡ ψ0 (s). Let a < a0 . Then, from (5.14), we have sa∗ > sa∗0 , and ψ0 has a zero sa∗0 ∈ (0, sa∗ ). Note that (i) and (ii) in Remark 4.2 are equivalent each other. Then any solution of (5.17) has at least one zero in (0, sa∗ ). By Lemma 5.3, the singular solution of (1.15) is not extremal for a < a0 . Thus (ii) holds. 2 6. Properties of principal solutions: Proof of Propositions 4.1, 4.2 and 4.3 We will prove Proposition 4.1 by using Proposition A.1 in [27]. Proof of Proposition 4.1. By the change of variables v1 (t) = u0 (r) and v2 (t) = u(r) with t = r 2−N , the functions vi , for i = 1, 2, satisfies vi + Wi (t)vi = 0 for t > T with T = R 2−N , where W1 (t) =

1 t −(2N−2)/(N −2) V (t −1/(N−2) ) (N − 2)2

(6.1)

. Then we have W1 (t) ≤ W2 (t) for t ≥ T . and W2 (t) is defined by (6.1) with V replaced by V From (4.2), we observe that

r0 ∞= 0

dr 1 = r N−1 u0 (r)2 N − 2

∞ t0

dt v1 (t)2

with t0 = r02−N . We see that v1 (t) > 0 for t > T , and that v1 and v2 satisfies either v1 (T ) = 0 or v1 (T ) = 0,

v2 (T ) = 0

and

v1 (T ) v2 (T ) ≥ . v1 (T ) v2 (T )

By applying Proposition A.1 in [27], v2 (t) has at least one zero in (T , ∞), otherwise v2 (t) is a constant multiple of v1 (t) and W1 ≡ W2 . Thus u(r) satisfies either u(r) has at least one zero in (r) for 0 < r ≤ R. 2 (0, R) or u(r) is a constant multiple of u0 (r) and V (r) ≡ V

Y. Miyamoto, Y. Naito / J. Differential Equations 265 (2018) 2842–2885

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To prove Proposition 4.2, we need the following lemmas. Lemma 6.1. Let u be a solution of (4.1) such that u(r) > 0 for 0 < r ≤ r0 with some r0 ∈ (0, R] and

r0

dr r N−1 u(r)2

< ∞.

(6.2)

0

Define v by

r v(r) = u(r)

ds s N−1 u(s)2

for 0 < r ≤ r0 .

(6.3)

0

Then v is a solution of (4.1) and satisfies v(r)/u(r) → 0 as r → 0 and

r0

dr = ∞. r N−1 v(r)2

(6.4)

0

Proof. By a direct calculation, we see that v satisfies (4.1). From (6.2) and (6.3), we have v(r) = u(r)

r

ds →0 s N−1 u(s)2

as r → 0.

0

  Since r N−1 (u v − uv  ) = (r N−1 u ) v − (r N−1 v  ) u = 0, we have r N−1 (u v − uv  ) = C with some constant C ∈ R. Observe that 

u(r) v(r)



=

u (r)v(r) − u(r)v  (r) C = N−1 . v(r)2 r v(r)2

Integrating the above on [r, r0 ], we have u(r0 ) u(r) − =C v(r0 ) v(r)

r0

ds s N−1 v(s)2

.

r

Letting r → 0, since u(r)/v(r) → ∞, we obtain (6.4).

2

(r) for 0 < r ≤ R. Let u1 and u2 be solutions of (4.1) and Lemma 6.2. Assume that V (r) ≤ V (4.4), respectively. Let r0 ∈ (0, R). (i) Assume that u2 (r) > 0 for 0 < r < r0 . If u1 (r0 ) = u2 (r0 ) ≥ 0 and u1 (r0 ) < u2 (r0 ), then u1 (r) > u2 (r) for 0 < r < r0 . (ii) Assume that u2 (r) > 0 for r0 < r ≤ R. If u1 (r0 ) = u2 (r0 ) ≥ 0 and u2 (r0 ) < u1 (r0 ), then u1 (r) > u2 (r) for r0 < r ≤ R.

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Proof. We will show only (i), since (ii) follows in the same way. By the assumption, there exists δ ∈ (0, R) such that u1 (r) > u2 (r) for r0 − δ < r < r0 . Assume to the contrary that there exists r1 ∈ (0, r0 ) such that u1 (r) > u2 (r) for r1 < r < r0 and u1 (r1 ) = u2 (r1 ). Put W (r) = r N−1 (u1 (r)u2 (r) − u1 (r)u2 (r)). Then W (r0 ) ≤ 0 and (r) − V (r))u1 (r)u2 (r) ≥ 0 W  (r) = (V

for r1 < r < r0 .

Integrating the above on [r1 , r0 ], we have

r0 (s) − V (s))u1 (s)u2 (s)ds ≥ 0. W (r0 ) − W (r1 ) = (V r1

On the other hand, since u1 (r1 ) = u2 (r1 ) > 0 and u1 (r1 ) > u2 (r1 ), we obtain W (r1 ) > 0, and hence W (r0 ) − W (r1 ) < 0. This is a contradiction. Thus we obtain u1 (r) > u2 (r) for 0 < r < r0 . 2 Proof of Proposition 4.2. Let uˆ 0 be the principal solution of (4.4) which is positive near r = 0. Take r0 ∈ (0, R] such that uˆ 0 (r) > 0 for 0 < r ≤ r0 . Let w(r) ˆ be a solution of (4.4) satisfying w(r ˆ 0 ) = 0 and wˆ  (r0 ) = −1. Since uˆ 0 (r) > 0 for 0 < r ≤ r0 , by the Sturm comparison theorem, w(r) ˆ > 0 for 0 < r < r0 . Since w(r) ˆ and uˆ 0 (r) are linearly independent, w(r) ˆ is nonprincipal, and hence

r 0

ds <∞ s N−1 w(s) ˆ 2

with 0 < r < r0 .

For μ ∈ I , we denote by w(r, μ) a solution of (4.5) satisfying w(r0 , μ) = 0 and w  (r0 , μ) = −2. Lemma 6.2 (i) implies that w(r, μ) > w(r) ˆ for 0 < r < r0 . Then it follows that

r 0

ds ≤ s N−1 w(s, μ)2

r 0

ds < ∞ with 0 < r < r0 . s N−1 w(s) ˆ 2

(6.5)

By the continuous dependence of solutions to (4.5) on the parameter μ ∈ I , we have w(r, μ) → w(r, μ0 )

1 in Cloc ((0, r0 ])

as μ → μ0 .

(6.6)

Put

r W (r, μ) =

ds s N−1 w(s, μ)2

,

0

and put v(r, μ) = w(r, μ)W (r, μ) for 0 < r < r0 . Then, by Lemma 6.1, v(r, μ) is the principal solution of (4.5). Let μ0 ∈ I . We will show that, for each 0 < r < r0 ,

Y. Miyamoto, Y. Naito / J. Differential Equations 265 (2018) 2842–2885

v(r, μ) → v(r, μ0 )

and v  (r, μ) → v  (r, μ0 )

as μ → μ0 .

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(6.7)

In fact, w(r, μ) satisfies (6.5) and (6.6). Then, by the dominated convergence theorem, for 0 < r < r0 we have W (r, μ) → W (r, μ0 )

as μ → μ0 .

(6.8)

Observe that 1

v  (r, μ) = w  (r, μ)W (r, μ) −

r N−1 w(r, μ)

.

Then, using (6.6) and (6.8), we obtain (6.7) for each 0 < r < r0 . We will verify that (6.7) holds for each r ∈ (0, R]. In fact, (6.7) holds at r = r˜0 ∈ (0, r0 ). By the continuous dependence of solutions on the initial values, we obtain v(r, μ) → v(r, μ0 ) in C 1 ([˜r0 , R]) as μ → μ0 . Thus we obtain (6.7) for each r ∈ (0, R]. Define v(r, ˜ μ) by v(r, μ)

v(r, ˜ μ) = 

v(r1 , μ)2 + v  (r1 , μ)2

1/2 .

(6.9)

Then v(r ˜ 1 , μ)2 + v˜  (r1 , μ)2 = 1 for all μ ∈ I . Since the principal solution is unique neglecting a constant factor, we obtain v(r, ˜ μ) ≡ u0 (r, μ) for all (r, μ) ∈ (0, R] × I . From (6.7) and (6.9) we see that, for each r ∈ (0, R], u0 (r, μ) and u0 (r, μ) are continuous for μ ∈ I . 2 Let φ0 (r, μ) be the principal solution of (4.9). Multiplying constants, we may assume that φ0 (r, μ) is positive near r = 0 and satisfies φ0 (r1 , μ)2 + φ0 (r1 , μ)2 = 1 for all μ ∈ R with some r1 ∈ (0, R). To prove Proposition 4.3, we need the following lemma. Lemma 6.3. Define μ˜ 1 by (4.8). Assume μ˜ 1 > −∞. (i) If μ < μ˜ 1 then φ0 (r, μ) > 0 for 0 < r ≤ R. (ii) If μ > μ˜ 1 then φ0 (r, μ) has at least one zero in (0, R). Proof. (i) Let μ < μ˜ 1 . Then there exists μ0 ∈ (μ, μ˜ 1 ) such that (4.7) with μ = μ0 has a positive solution. By Corollary 4.1, φ0 (r, μ) satisfies φ0 (r, μ) > 0 for 0 < r ≤ R. (ii) Let μ > μ˜ 1 . Then, by the definition of μ˜ 1 , a solution of (4.7) has at least one zero in (0, R). By the Sturm comparison theorem, φ0 (r, μ) has at least one zero in (0, R). 2 Proof of Proposition 4.3. Assume that φ0 (s, μ) > 0 for 0 < r < R and φ0 (R, μ) = 0 for some μ ∈ R. By Lemma 6.3 (i) and (ii), we have μ ≥ μ˜ 1 and μ ≤ μ˜ 1 , respectively. Thus we obtain μ = μ˜ 1 . Conversely, we will show that φ0 (r, μ˜ 1 ) > 0 for 0 < r < R

and

φ0 (R, μ˜ 1 ) = 0.

(6.10)

By Lemma 6.3 (ii), if μ > μ˜ 1 , then φ0 (r, μ) has at least one zero in (0, R). We denote by r1 (μ) the first zero of φ0 (r, μ). Then r1 (μ) ∈ (0, R). Since φ0 (s, μ) is continuous for μ ∈ R by Proposition 4.2 and the zero is simple, r1 (μ) is continuous and does not disappear unless it goes out

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from the boundary of (0, R). Corollary 4.1 implies that r1 (μ) > r1 (μ) ˆ if μ < μˆ < μ˜ 1 . Then, letting μ ↓ μ˜ 1 , we obtain r1 (μ˜ 1 ) ∈ (0, R]. Assume that r1 (μ˜ 1 ) ∈ (0, R). Then r1 (μ) ∈ (0, R) if ˜ = R, and μ < μ˜ 1 and μ is sufficiently close to μ˜ 1 . This contradicts Lemma 6.3 (i). Thus r1 (μ) hence (6.10) holds. 2 7. Asymptotic properties of solutions to linearized equations: Proof of Proposition 4.4 In this section we consider the equation ψ  +

N −1  ψ + V (r)ψ = 0 for 0 < r < R, r

(7.1)

where V ∈ C((0, R]). In (7.1) we assume that V satisfies r 2 V (r) = c0 + O(r δ0 )

as r → 0

(7.2)

with some constants c0 > 0 and δ0 > 0. We show the following result. Proposition 7.1. Let 0 < c0 ≤ (N − 2)2 /4. Then (7.1) has a unique solution ψ0 ∈ C 2 ((0, R]) satisfying lim r ν ψ0 (r) = 1

r→0

and

lim r ν+1 ψ0 (r) = −ν,

r→0

(7.3)

where ν is the smaller root of the polynomial ν 2 − (N − 2)ν + c0 = 0.

(7.4)

Furthermore, the solution ψ0 (r) is the principal solution of (7.1). Proposition 7.2. The following (i)–(iii) hold. (i) Let 0 < c0 ≤ (N − 2)2 /4. Then (7.1) has a solution ψ which is positive near r = 0. On the other hand, let c0 > (N − 2)2 /4. Then any solution ψ of (7.1) oscillates near r = 0. 1 (B ) if and only (ii) Let 0 < c0 < (N − 2)2 /4, and let ψ be a solution of (7.1). Then ψ ∈ Hrad R if ψ is the principal solution. 1 (B ). (iii) Let c0 = (N − 2)2 /4. Then any solution ψ of (7.1) satisfies ψ ∈ / Hrad R Proof of Proposition 7.1. Define h(r) by h(r) = r 2 V (r) − c0

for 0 < r ≤ R.

(7.5)

Then h(r) = O(r δ0 ) as r → 0. Put ψ(r) = r ν v(r), where ν is the smaller root of the polynomial (7.4). Then (7.1) is reduced to v  +

N − 1 − 2ν  h(r) v + 2 v = 0 for r > 0. r r

Define F : C([0, r0 ]) → C([0, r0 ]) by

(7.6)

Y. Miyamoto, Y. Naito / J. Differential Equations 265 (2018) 2842–2885

r Fv(r) = 1 −

s −N+1+2ν

0

2875

s t N−3−2ν h(t)v(t)dtds. 0

 Since N − 3 − 2ν = −1 + (N − 2)2 − 4c0 ≥ −1 and h(t) = O(t δ0 ) as t → 0, we have t N −3−2ν h(t) ∈ L1 ([0, R]). We can easily check that F is a contraction mapping on C([0, r0 ]) for sufficiently small r0 > 0. Then, by the contraction mapping theorem, there exists a unique element v0 ∈ C([0, r0 ]) satisfying v0 = Fv0 , and hence the problem (7.6) has a unique solution v0 (r) for 0 ≤ r ≤ r0 satisfying v0 (r) = 1 + o(1) as r → 0. Note that the solution v0 (r) defined on [0, r0 ] can be extended to the solution of (7.6) defined on [0, R]. Define ψ0 (r) = r −ν v0 (r). Then, ψ0 is a solution of (7.1) and satisfies ψ0 (r) = r −ν (1 + o(1)) as r → 0. Thus the left-hand side of (7.3) holds. Differentiating ψ0 (r) with respect to r, we have ψ0 (r) = −νr −ν−1 v0 (r) + r −ν v0 (r).

(7.7)

Since v0 (r) satisfies rv0 (r) = −r −N+2+2ν

r s N−3−2ν h(s)v0 (s)ds, 0

we obtain r|v0 (r)| ≤

r

s −1 |h(s)||v0 (s)|ds = O(r δ0 )

as r → 0.

0

By (7.7) we have ψ0 (r) = −νr −ν−1 (1 + o(1)) as r → 0. Thus the right-hand side of (7.3) holds. We may assume that ψ0 (r) > 0 for 0 < r ≤ r0 . Since ν ≤ −(N − 2)/2, ψ0 (r) satisfies

r0 0

dr = ∞. r N−1 ψ0 (r)2

Thus ψ0 (r) is the principal solution of (7.1).

2

In order to show Proposition 7.2, we need some lemmas. Lemma 7.1. Assume that 0 < c0 < (N − 2)2 /4. Then the following (i) and (ii) hold. (i) There exists a solution ψ1 of (7.1) satisfying lim r νˆ ψ1 (r) = 1

r→0

and

lim r νˆ +1 ψ1 (r) = −νˆ ,

r→0

(7.8)

where νˆ is the larger root of the polynomial (7.4). 1 (B ) if and only if ψ is the principal solution. (ii) Let ψ be a solution of (7.1). Then ψ ∈ Hrad R

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Proof. (i) Let ψ0 be the solution of (7.1) obtained by Proposition 7.1. From (7.3), we may assume that ψ0 (r) > 0 for 0 < r ≤ r0 with some r0 > 0. Define ψ1 (r) by

r0 ψ1 (r) = (N − 2 − 2ν)ψ0 (r) r

ds . s N−1 ψ0 (s)2

Then, by Lemma 6.1, ψ1 (r) is the nonprincipal solution of (7.1). We show that ψ1 (r) satisfies (7.8). Since νˆ = N − 2 − ν, we obtain ⎛ ⎞

r0 ψ1 (r) ψ0 (r) ⎝ N − 2 − 2ν ds ⎠. lim = lim −ν r→0 r −ˆν r→0 r r −N+2+2ν s N−1 ψ0 (s)2 r

By L’Hopital’s rule and the left-hand side of (7.3), we have N − 2 − 2ν lim r→0 r −N+2+2ν

r0

ds s N−1 ψ

r

0

(s)2

r −N+1 ψ0 (r)−2 = 1. r→0 r −N+1+2ν

= lim

(7.9)

Then r νˆ ψ1 (r) → 1 as r → 0. Differentiating ψ1 (r) with respect to r, we have ψ1 (r) = (N

− 2 − 2ν)ψ0 (r)

r0 r

1 N − 2 − 2ν ds − N−1 . s N−1 ψ0 (s)2 r ψ0 (r)

From (7.3) and (7.9), we obtain

lim

(N − 2 − 2ν)ψ0 (r) r −ˆν −1

r→0

r0

ds s N−1 ψ

r

2 0 (s)

⎛ ⎞

r0 ψ0 (r) N − 2 − 2ν ds ⎠ = −ν = lim −ν−1 ⎝ −N+2+2ν r→0 r r s N−1 ψ0 (s)2 r

and lim

1

r→0 r −ˆν −1

N − 2 − 2ν N − 2 − 2ν = lim = N − 2 − 2ν. N−1 r ψ0 (r) r→0 r ν ψ0 (r)

Hence, it follows that lim

ψ1 (r)

r→0 r −ˆν −1

= −ν − (N − 2 − 2ν) = −N + 2 + ν = −νˆ .

Thus (7.8) holds. (ii) Let ψ be a solution of (7.1). Since ψ0 (r) and ψ1 (r) are linearly independent, there exist constants c0 , c1 ∈ R such that

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ψ(r) = c1 ψ0 (r) + c2 ψ1 (r).

(7.10)

By the uniqueness of the principal solution, ψ is the principal solution if and only if ψ is the 1 (B ) multiple constant of ψ0 (r), that is c2 = 0 in (7.10). By (7.3) and (7.8), we have ψ0 ∈ Hrad R 1 (B ), respectively. Then ψ ∈ H 1 (B ) if and only if c = 0, that is, ψ is the and ψ1 ∈ / Hrad R 2 rad R principal solution. 2 Lemma 7.2. Assume that c0 = (N − 2)2 /4. Then the following (i) and (ii) hold. (i) There exists a solution ψ1 of (7.1) satisfying lim r ν (log r)−1 ψ1 (r) = −1

r→0

and

lim r ν+1 (log r)−1 ψ1 (r) = ν,

(7.11)

r→0

where ν = (N − 2)/2. 1 (B ). (ii) Any solution ψ of (7.1) satisfies ψ ∈ / Hrad R Proof. Let ψ0 be the solution of (7.1) obtained by Proposition 7.1. Then (7.3) holds with ν = (N − 2)/2. From (7.3), we may assume that ψ0 (r) > 0 for 0 < r ≤ r0 with some r0 > 0. Define ψ1 (r) by

r0 ψ1 (r) = ψ0 (r) r

ds . s N−1 ψ0 (s)2

Then, by Lemma 6.1, ψ1 (r) is the nonprincipal solution of (7.1). We show that ψ1 (r) satisfies (7.11). We see that ⎛ ⎞

r0 ψ1 (r) ψ0 (r) ⎝ 1 ds ⎠. lim = lim −ν r→0 r −ν log r r→0 r log r s N−1 ψ0 (s)2 r

By L’Hopital’s rule and the left-hand side of (7.3), we have 1 lim r→0 log r

r0 r

1 −r −N+1 ψ0 (r)−2 ds = lim = −1. r→0 s N−1 ψ0 (s)2 r −1

(7.12)

Then r ν (log r)−1 ψ1 (r) → −1 as r → 0. Differentiating ψ1 (r) with respect to r, we have ψ1 (r) = ψ0 (r)

r0

1 s N−1 ψ

r

0

(s)2

ds −

1 r N−1 ψ

0 (r)

.

From (7.3) and (7.12), we obtain ψ0 (r) lim −ν−1 r→0 r log r

r0 r

ds ds s N−1 ψ0 (s)2

ψ0 (r) = lim −ν−1 r→0 r

⎛ ⎝ 1 log r

r0 r

⎞ 1 ds ⎠ = ν s N−1 ψ0 (s)2

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and lim

1

r→0 r −ν−1

1 1 = lim = 1. r N−1 ψ0 (r) r→0 r ν ψ0 (r)

Hence, it follows that ψ1 (r) = ν. r→0 r −ν−1 log r lim

Thus (7.11) holds. 1 (B ) and ψ ∈ 1 / Hrad By (7.3) and (7.11), we have ψ0 ∈ R 1 / Hrad (BR ), respectively. Then, for any 1 solution ψ of (7.1) satisfies ψ ∈ / Hrad (BR ). 2 Proof of Proposition 7.2. Let 0 < c0 ≤ (N − 2)2 /4. Then, by Proposition 7.1, (7.1) has a solution which is positive near r = 0. Let c0 > (N − 2)2 /4. Then there exists ε > 0 such that c0 > (N − 2)2 /4 + ε 2 . By (7.2), we have 1 V (r) > 2 r



(N − 2)2 + ε2 4

 for 0 < r ≤ r0

with some r0 ≤ R. Observe that φ(r) = r −(N−2)/2 sin(ε log r) oscillates near r = 0 and satisfies φ  +

N −1  1 φ + 2 r r



 (N − 2)2 + ε 2 φ = 0 for r > 0. 4

By the Sturm comparison theorem, any solution of (7.1) oscillates near r = 0. Thus (i) holds. From Lemmas 7.1 (ii) and 7.2 (ii), we obtain (ii) and (iii), respectively. 2 As an application, we consider the equation (4.12), that is, the case V (|x|) = f  (v ∗ (|x|)) + μ in (7.1). By Proposition 7.2, we obtain the following results, which include Proposition 4.4. Proposition 7.3. Assume that (f.1) and (f.2) hold. Let μ ∈ R. (i) Let p ≥ pJ L . Then (4.12) has a solution ψ which is positive near s = 0. On the other hand, let pS < p < pJ L . Then any solution ψ of (4.12) oscillates near s = 0. 1 (B ∗ ) if and only if ψ is the (ii) Let p > pJ L , and let ψ be a solution of (4.12). Then ψ ∈ Hrad s0 principal solution. 1 (B ∗ ). / Hrad (iii) Let p = pJ L . Then any solution ψ of (4.12) satisfies ψ ∈ s0 Proof. First we will show that f  (v ∗ (s)) =

pAp−1 (1 + O(s δθ )) s2

as s → 0.

(7.13)

In fact, for a, b ≥ 0, q > 0, we have  (a + b) − a ≤ q

q

q(a + b)q−1 b,

if q ≥ 1,

qa q−1 b,

if 0 < q < 1.

(7.14)

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Recall that v ∗ (s) is given by (2.23). Then, from (1.7), we have v ∗ (s) = As −θ (1 + O(s δθ )) as s → 0. From (1.6) and (7.14), it follows that pv ∗ (s)p−1 =

pAp−1 (1 + O(s δθ )) s2

and g  (v ∗ (s)) = O(s −2+δθ )

as s → 0.

Thus (7.13) holds. Put V (|x|) = f  (v ∗ (|x|)) + μ in (7.1). Then, by (7.13), V satisfies (7.2) with c0 = pAp−1 and δ0 = θ δ. Note that pAp−1 < (N − 2)2 /4,

pAp−1 = (N − 2)2 /4,

and

pAp−1 > (N − 2)2 /4

hold when p > pJ L , p = pJ L , and p < pJ L , respectively. Then Proposition 7.3 follows from Proposition 7.2 immediately. 2 8. Singular eigenvalue problems: Proof of Propositions 4.5 and 4.6 In order to prove Proposition 4.5, we consider the following eigenvalue problems ⎧ ⎨ φ + V (|x|)φ = −μφ ⎩

in BR , (8.1)

1 φ ∈ H0,rad (BR ),

and ⎧ ⎨ φ + μV (|x|)φ = 0 ⎩

1 φ ∈ H0,rad (BR ),

in BR , (8.2)

where V ∈ C((0, R]) and μ ∈ R is a parameter. In (8.1) and (8.2) we assume that V satisfies (7.2) with some constants c0 > 0 and δ0 > 0. Proposition 8.1. Let 0 < c0 ≤ (N − 2)2 /4 in (7.2). Assume that ψ0 is the principal solution of (7.1) and has exactly m ≥ 1 zeros in (0, R). (i) Let 0 < c0 < (N − 2)2 /4. Then the eigenvalue problem (8.1) has exactly m negative eigenvalues {μj }m j =1 ⊂ R such that μ1 < μ2 < · · · < μm < 0. (ii) Assume that V (r) ≥ 0 for 0 < r ≤ R. Let 0 < c0 ≤ (N − 2)2 /4. Then the eigenvalue problem (8.2) has exactly m eigenvalues {μj }m j =1 ⊂ R such that 0 < μ1 < μ2 < · · · < μm < 1. In the both cases (i) and (ii), the corresponding eigenfunction φj (r) has exactly j − 1 zeros in (0, R) for each j ∈ {1, 2, . . . , m}. Proof of Proposition 4.5. Put V (|x|) = f  (v ∗ (|x|)) and R = s0∗ . Then the problems (4.13) and (4.14) are reduced to the problems (8.1) and (8.2), respectively. By Remark 4.2, the principal solution of (4.12) with μ = 0 has at least one zero in (0, s0∗ ), that is, the principal solution of (7.1) has at least one zero in (0, R). From (7.13), V satisfies (7.2) with c0 = pAp−1 and δ0 = θ δ. Note that pAp−1 < (N − 2)2 /4 if p > pJ L , and pAp−1 = (N − 2)2 /4 if p = pJ L . Then, Proposition 4.5 (i) and (ii) follow from Proposition 8.1 (i) and (ii), respectively. 2

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To prove Proposition 8.1, we consider the equation u +

N −1  u + V (r, μ)u = 0 for 0 < r < R, r

(8.3)

where μ ∈ R, V (r, μ) ∈ C((0, R] × R) and V (r, μ) is strictly increasing in μ ∈ R for each fixed r ∈ (0, R]. We assume that there exists μ∗ ∈ R such that (8.3) with μ = μ∗ has a solution which is positive near r = 0. By the Sturm comparison theorem, for all μ ≤ μ∗ , (8.3) has a positive solution near r = 0. Hence, there exists the principal solution of (8.3) for all μ ≤ μ∗ . We denote by u0 (r, μ) the principal solution of (8.3) for μ ≤ μ∗ . Proposition 8.2. Assume that the following (i) and (ii) hold. (i) u0 (r, μ∗ ) has exactly m ≥ 1 zeros in (0, R). (ii) There exists μ∗ < μ∗ such that u0 (r, μ∗ ) > 0 for 0 < r ≤ R. ∗ Then there exists {μj }m j =1 with μ∗ < μ1 < μ2 < · · · < μm < μ such that the principal solution u0 (r, μj ) has exactly j − 1 zeros in (0, R) and satisfies u0 (R, μj ) = 0 for each j = 1, 2, . . . , m. Furthermore, for μ ∈ (μ∗ , μ∗ ), u0 (R, μ) = 0 if and only if μ = μj with some j ∈ {1, 2, . . . , m}. Multiplying constants, we may assume that the principal solution u0 (r, μ) is positive near r = 0 and satisfies u0 (r1 , μ)2 + u0 (r1 , μ)2 = 1 for all μ ≤ μ∗ with some r1 ∈ (0, R). By Proposition 4.2, for each fixed r ∈ (0, R], u0 (r, μ) and u0 (r, μ) are continuous for μ ≤ μ∗ . We employ the Prüfer transformation. Define the functions ρ(r, μ) and θ (r, μ) by u0 (r, μ) = ρ(r, μ) sin θ (r, μ)

and r N−1 u0 (r, μ) = ρ(r, μ) cos θ (r, μ).

(8.4)

Since u0 (r, μ) > 0 near r = 0, we take 0 < θ(r, μ) < π for simplicity. Note that u0 (r, μ) and u0 (r, μ) cannot vanish at the same point r ∈ (0, R]. Then ρ(r, μ) and θ (r, μ) are given by ρ(r, μ) = (u0 (r, μ)2 + r 2N −2 u0 (r, μ)2 )1/2

and

θ (r, μ) = arctan

u0 (r, μ) , r N−1 u0 (r, μ)

(8.5)

respectively. Then, for each fixed r ∈ (0, R], ρ(r, μ) and θ (r, μ) are continuous for μ ≤ μ∗ . By a simple calculation, we have θ  (r, μ) =

1 r N−1

cos2 θ (r, μ) + V (r, μ) sin2 θ (r, μ).

(8.6)

Then θ (r, μ) is strictly increasing when sin θ (r, μ) = 0, that is, u0 (r, μ) = 0. Then it is easy to see that u0 (r, μ) has exactly k zeros in (0, R) if and only if kπ < θ (R, μ) ≤ (k + 1)π . Furthermore, θ (r, μ) satisfies the following property. Lemma 8.1. For each fixed r0 ∈ (0, R], θ (r0 , μ) is strictly increasing in μ ≤ μ∗ . Remark 8.1. Let r0 ∈ (0, R], and assume that u0 (r0 , μ) = 0 for all μ ∈ (μ1 , μ2 ). Then, by Lemma 8.1 and the right-hand side of (8.5), u0 (r0 , μ)/u0 (r0 , μ) is strictly decreasing in μ ∈ (μ1 , μ2 ).

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Proof. Let μ1 < μ2 . First we consider the case where θ (r0 , μ1 ) ≤ π . Then u0 (r, μ1 ) > 0 for 0 < r < r0 . Assume to the contrary that θ (r0 , μ2 ) ≤ θ (r0 , μ1 ). Then, from (8.5), we have either u0 (r0 , μ1 ) = 0 or u0 (r0 , μ1 ) > 0,

u0 (r0 , μ2 ) > 0,

and

r0N−1 u0 (r0 , μ1 ) r0N−1 u0 (r0 , μ2 ) ≤ . u0 (r0 , μ1 ) u0 (r0 , μ2 )

Proposition 4.1 implies that u0 (r, μ2 ) has at least one zero in (0, r0 ), and hence θ (r0 , μ2 ) > π ≥ θ (r0 , μ1 ). This is a contradiction. Then we obtain θ (r0 , μ2 ) > θ (r0 , μ1 ). Next we consider the case where θ (r0 , μ1 ) > π . In this case, u0 (r, μ1 ) has at least one zero in (0, r0 ). Let r1 be the smallest zero in (0, r0 ). Then θ (r1 , μ1 ) = π , and by the argument above, we obtain θ (r1 , μ2 ) > θ (r1 , μ1 ) if μ1 < μ2 . Observe that θ (r, μ2 ) and θ (r, μ1 ) satisfy (8.6) with μ = μ2 and μ = μ1 , respectively. By applying the differential inequality in [18, Chapter III, Corollary 4.2], we obtain θ (r0 , μ2 ) > θ (r0 , μ1 ). 2 Proof of Proposition 8.2. Define ρ(r, μ) and θ (r, μ) by (8.4). By the assumptions (i) and (ii), we have mπ < θ (R, μ∗ ) ≤ (m + 1)π and 0 < θ (R, μ∗ ) < π . Lemma 8.1 implies that θ (R, μ) is strictly increasing in μ ≤ μ∗ . Since θ (R, μ) is continuous for μ ≤ μ∗ , there exist μ∗ < μ1 < μ2 < · · · < μm < μ∗ such that θ (R, μj ) = j π for j = 1, 2, . . . , m. Thus, for μ < μ∗ , u0 (R, μ) = 0 if and only if μ = μj for some j ∈ {1, 2, . . . , m}. Furthermore, u0 (r, μj ) has exactly j − 1 zeros in (0, R). 2 In order to prove Proposition 8.1 (ii), we consider the following equation u +

N −1  u + μV (r)u = 0 for 0 < r < R, r

(8.7)

where μ ∈ (0, 1] is a parameter and V ∈ C((0, R]) satisfies (7.2) with 0 < c0 ≤ (N − 2)2 /4 and δ0 > 0. Lemma 8.2. There exists μ > 0 such that (8.7) has a positive solution u0 (r) for 0 < r ≤ R. Proof. Define h(r) by (7.5), and put u(r) = r ν v(r), where ν is the smaller root of the polynomial (7.4). Then (8.7) is reduced to v  +

N − 1 − 2ν  μh(r) v = 0 for r > 0. v + r r2

(8.8)

Define Fμ : C([0, r0 ]) → C([0, r0 ]) by

r Fμ v(r) = 1 − μ

s 0

−N+1+2ν

s t N−3−2ν h(t)v(t)dtds. 0

Then we obtain t N−3−2ν h(t) ∈ L1 ([0, R]) by the similar argument as in the proof of Proposition 7.1. Define X = {v ∈ C([0, 1]) : 1/2 ≤ v(r) ≤ 3/2 for 0 ≤ r ≤ R}. If μ > 0 is sufficiently small, then we can show that Fμ is a contraction mapping on X. Then, by the contraction mapping theorem, there exists an element v0 ∈ X satisfying v0 = Fμ v0 , and hence (8.8) has a positive

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solution v0 on [0, R]. Define u0 (r) = r −ν v0 (r). Then, u0 (r) is a positive solution of (8.7) for 0 < r ≤ R. 2 To prove Proposition 8.1 (i), we consider the following equation u +

N −1  u + (V (r) + μ)u = 0 r

for 0 < r ≤ R,

(8.9)

where μ ≤ 0 is a parameter and V ∈ C((0, R]) satisfies (7.2) with 0 < c0 ≤ (N − 2)2 /4 and δ0 > 0. By Proposition 7.2 (i), (8.9) with μ ≤ 0 has a solution u which is positive near r = 0. Then there exists the principal solution u0 (r, μ) of (8.9) for μ ≤ 0. We may assume that u0 (r, μ) is positive near r = 0. Lemma 8.3. Assume that 0 < c0 ≤ (N − 2)2 /4. Then there exists μ∗ ≤ 0 such that u0 (r, μ∗ ) > 0 for 0 < r ≤ R. Proof. If u0 (r, 0) > 0 for 0 < r ≤ R, then take μ∗ = 0. We consider the case where u0 (r, 0) has at least one zero in (0, R). Take r0 ∈ (0, R) such that u0 (r, 0) > 0 for 0 < r ≤ r0 . By Lemma 8.1 and Remark 8.1, u0 (r, μ) > 0

on (0, r0 ]

for all μ ≤ 0,

(8.10)

and u0 (r0 , 0) u0 (r0 , μ) ≤ u0 (r0 , 0) u0 (r0 , μ)

for μ ≤ 0.

Multiplying constants, we may assume that u0 (r0 , μ) = 1 for all μ ≤ 0. Then it follows that u0 (r0 , 0) ≤ u0 (r0 , μ) for μ ≤ 0.

(8.11)

Define λ = λ(M) < 0 by λ=

−(N − 2)2 −

 (N − 2)2 + 4M , 2

where M is a constant to be determined later. Put z(r) = r0−λ r λ for r > 0. Then z satisfies z +

N −1  M z − 2 z = 0 for 0 < r ≤ R r r

and z(r0 ) = 1, z (r0 ) = λ/r0 . Take M > 0 such that λ/r0 < u (r0 , 0). Then, from (8.11), we have z (r0 ) < u (r0 , μ) for μ ≤ 0. Take μ∗ < 0 such that V (r) + μ∗ < −

M r2

for r0 ≤ r ≤ R.

Then, by Lemma 6.2 (ii), we obtain u(r, μ∗ ) ≥ z(r) > 0 for r0 ≤ r ≤ R. From (8.10) we conclude that u(r, μ∗ ) > 0 for 0 < r ≤ R. 2

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Proof of Proposition 8.1. (i) We consider the equation (8.3) with V (r, μ) = V (r) + μ, and denote by u0 (r, μ) the principal solution of (8.3). Then u0 (r, 0) has exactly m zeros in (0, R). From Lemma 8.3, there exists μ∗ < 0 such that u0 (r, μ∗ ) > 0 for 0 < r ≤ R. By Proposition 8.2, there exists {μj }m j =1 with μ∗ < μ1 < μ2 < · · · < μm < 0 such that u0 (r, μj ) has exactly j − 1 zeros in (0, R) and u0 (R, μj ) = 0 for each j = 1, 2, . . . , m. Proposition 7.2 (ii) implies that 1 for j = 1, 2, . . . , m. Then (8.1) has the eigenvalues {μj }m u0 (r, μj ) ∈ H0,rad j =1 with the corresponding eigenfunctions φj (r) = u0 (r, μj ) for j = 1, 2, . . . , m. (ii) We consider the equation (8.3) with V (r, μ) = μV (r) for μ ∈ (0, 1], and denote by u0 (r, μ) the principal solution of (8.3). Then u0 (r, 1) has exactly m zeros in (0, R). From Lemma 8.2, there exists μ∗ ∈ (0, 1) such that u0 (r, μ∗ ) > 0 for 0 < r ≤ R. By Proposition 8.2, there exists {μj }m j =1 with μ∗ < μ1 < μ2 < · · · < μm < 1 such that u0 (r, μj ) has exactly j − 1 zeros in (0, R) and u0 (R, μj ) = 0 for each j = 1, 2, . . . , m. Since V satisfies (7.2), we have r 2 μV (r) = μc0 + O(r δ0 )

as r → 0.

1 Note that μc0 < (N − 2)2 /4 if μ < 1. Then, by Proposition 7.2 (ii), we obtain u0 (r, μj ) ∈ H0,rad m for each j = 1, 2, . . . , m. Then (8.2) has the eigenvalues {μj }j =1 with the corresponding eigenfunctions φj (r) = u0 (r, μj ) for j = 1, 2, . . . , m. 2

To prove Proposition 4.6, we show the following lemma. Lemma 8.4. Assume that (f.1) and (f.2) hold. Let p > pJ L . Then there exists a solution (μ1 , φ1 ) of the problem (4.13) satisfying μ1 ∈ R and φ1 (s) > 0 for 0 < s < s0∗ . Proof. Take m ≥ 0 such that the equation ψ  +

N −1  ψ + (f  (v ∗ ) + m)ψ = 0 s

(8.12)

has a solution which has at least two zeros in (0, s0∗ ]. Then, by the Sturm comparison theorem, any solution of (8.12) has at least one zero in (0, s0∗ ). By Proposition 4.5 (i) together with Remark 4.2, the problem φ + (f  (v ∗ ) + m)φ = −μφ

in Bs0∗ ,

1 φ ∈ H0,rad (Bs0∗ ),

has a solution (μ1 , φ1 ) satisfying μ1 < 0 and φ1 (s) > 0 for 0 < s < s0∗ . Then the problem (4.13) has a solution (μ1 , φ1 ) with μ1 = μ1 + m ∈ R. 2 For a solution φ of (1.12), put ψ(s) by (5.2). Then, the problem (1.12) is reduced to (4.13) and the constant μ˜ 1 defined by (1.13) coincides with the one defined by (4.11). Proof of Proposition 4.6. Let p > pJ L , and let (μ1 , φ1 ) be the solution of (4.13) obtained by Lemma 8.4. Then (μ1 , φ1 (s0∗ r)) solves (1.11). By Proposition 7.3 (ii), φ1 (s) is the principal solution of (4.12) with μ = μ1 . Proposition 4.3 implies that μ1 = μ˜ 1 . Thus (i) holds. Let p = pJ L . By (iii) of Proposition 7.3, any solution ψ of the equation in (4.12) satisfies 1 (B ∗ ) for any μ ∈ R. Thus (ii) holds. Let p < p < p . By the latter part of (i) of ψ∈ / Hrad S JL s0 Proposition 7.3, any solution ψ of (4.12) oscillates near s = 0. Thus (iii) holds. 2

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