Exact Morse index of radial solutions for semilinear elliptic equations with critical exponent on annuli

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Using Jacobi elliptic functions, we show that if (p,α)=(3,N-4)\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$(p,\alpha )=(3,N-4)$$\end{document} and N≥3\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$N\ge 3$$\end{document}, then the Morse index of a positive and negative solutions m(U1,ρ±)\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$${{\textsf{m}}}(U^{\pm }_{1,\rho })$$\end{document} is completely determined by the ratio ρ/R∈(0,1)\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\rho /R\in (0,1)$$\end{document}. Upper and lower bounds for m(Un,ρ±)\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$${{\textsf{m}}}(U^{\pm }_{n,\rho })$$\end{document}, n≥1\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$n\ge 1$$\end{document}, are also obtained when (p,α)=(3,N-4)\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$(p,\alpha )=(3,N-4)$$\end{document} and N≥3\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$N\ge 3$$\end{document}.


Introduction and main results
Let N ≥ 3, R > ρ > 0 and A ρ := {x ∈ R N ; ρ < |x| < R}. We are concerned with radial solutions of the Dirichlet problem The equation in (1.1) is called the Hénon equation arising in astrophysics. Hénon [16] studied radial solutions of (1.1) on a ball. We are particularly interested in the case p = p c := N + 2 + 2α N − 2 and α > −2.
The exponent p c becomes the critical Sobolev exponent provided that α = 0. Let B denote a unit ball in R N . First we consider the problem When p ≥ p c , (1.2) has no nontrivial radial solution due to Pohozaev's identity. On the other hand, Ni [21] proved that (1.2) has a positive radial solution provided that 1 < p < p c . This result was generalized by Bartsch-Willem [7]. They proved that, for each n ≥ 2, (1.2) has radial solutions U ± n (r ) that have n nodal domains, i.e., the set {x ∈ B; U ± n (x) = 0} has exactly n connected components. The uniqueness of U + n (r ) follows from Ni-Nussbaum [22], and U − n (r ) = −U + n (r ). Therefore, p c is a threshold between existence and nonexistence of nontrivial radial solutions of (1.2).
In this paper we study the Morse index of U ± n,ρ when ρ > 0 is small. We call the Morse index m(U ± n,ρ ) of a solution U ± n,ρ of (1.1) the maximal dimension of a subspace X ⊂ H 1 0 (A ρ ) where the quadratic form associated to the linearization operator at U ± n,ρ is negative definite. In our problem (1.1) m(U ± n,ρ ) is equal to the number of the negative eigenvalues of the problem counted with their multiplicity, i.e., m(U ± n,ρ ) = {negative eigenvalues of (1.3) counted with their multiplicity}. (1.4) We call that U ± n,ρ is nondegenerate if (1.3) does not have a zero eigenvalue. We prepare some notation to state our main results. Let [ξ ], ξ ∈ R, denote the largest integer that does not exceed ξ . Let {ν j } ∞ j=0 denote the set of the eigenvalues of the Laplace-Beltrami operator − S N −1 on the sphere S N −1 , i.e., ν j := j(N + j − 2), j = 0, 1, 2, . . . . (1.5) The multiplicity of the eigenvalue ν j is given by for N ≥ 3 and j = 0, 1, 2, . . . .
The first main result is about the exact Morse index for small ρ > 0 and the divergence of the Morse index as ρ → R.
The Morse index formula given in Theorem 1.1 (i) has the following simpler form m(U ± n,ρ ) = nM (N + 1) for small ρ > 0. Explicit formulas for N = 3, 4, 5 are given in the following: Example 1.2 Let α > −2, p = p c and = α 2 + 1. Then the following hold: diverges as a → ∞. The ratio ρ/R converges to 1 as a → ∞. This divergence result of the Morse index corresponds to Theorem 1.1 (ii). Similar phenomena were also observed in previous researches including [9,17,24,25].
When ( p, α) = (3, N − 4), the Morse index of positive and negative solutions U ± 1,ρ can be completely determined by the ratio ρ/R ∈ (0, 1). In particular, the smallness of ρ > 0 is not assumed in the following: 1,ρ and U − 1,ρ be a positive and negative radial solution of (1.1), respectively. Assume that one of the following (a) and (b) holds: Here, (1.6) and K denotes the complete elliptic integral of the first kind whose definition and basic properties are recalled in Sect. 7.1.
When α ≥ 0 is not an even integer, we see by Theorem 1.1 that m(U ± n,ρ ) = m(U ± n ) for small ρ > 0. On the other hand, when α ≥ 0 is an even integer, we see that m(U ± n,ρ ) > m(U ± n ) for small ρ > 0. Hence, we can say that the critical case p = p c is more unstable than the subcritical case p < p c when α is an even integer. where := α 2 + 1. A simple proof in the case (1.2) with n ≥ 2 can be found in [12]. By Theorem 1.1 we see that m(U ± n,ρ ) does not attain the lower bound (1.11). On the other hand, if p(< p c ) is close to p c and α = 0, then we see by Proposition 1.7 that m(U ± n ) attains the lower bound (1.11) when = 1. We can say that the critical case does not have the most stable solution, while the subcritical case has.
Let us mention technical details. In the critical case by Emden's transformation we can transform a radial part of (1.1) into the scalar field equation u −u+|u| p−1 u = 0. As explained in Sect. 2, the Morse index m(U ± n,ρ ) is equal to the number of the negative eigenvalues of the weighted eigenvalue problem and all the eigenvalue of (1.12), which is denoted byλ i, j , can be written as follows: Here,λ rad,i , i = 1, 2, . . ., is the i-th radial eigenvalue of (1.12), i.e., the i-th eigenvalue of Since ν j is explicitly given by (1.5), it becomes important to study all the negative eigenvalues λ rad,i . In the critical caseλ rad,i , i = 1, 2, . . ., are given as the eigenvalues for the linearization of the scalar field equation as explained in Sect. 4. Let := [ α 2 ] + 1. The main part of our analysis is for showing that (1.14) and that, In particular,λ rad,1 < · · · <λ rad,n < 0 <λ rad,n+1 < · · · .
The limit (1.15) will be obtained in Remark 5.9. If ρ > 0 is small, then all the negative eigenvalues areλ i, j =λ rad,i + ν j for 1 ≤ i ≤ n and 0 ≤ j ≤ , which leads to Theorem 1.1. Because of a variational characterization ofλ rad,n , each orthogonal set gives an upper bound ofλ rad,n . We can obtain a sharp upper bound in Remark 5.4. On the other hand, a lower bound ofλ rad,1 is nontrivial. In this paper we use a first Neumann eigenvalue of a linearization problem as a lower bound ofλ rad,1 . Then we use a blow-up argument to show thatλ rad,1 → − This limit implies (1.15), because of (1.14). Let us again compare our problem with a problem on a ball. It was shown in [3, Proposition 3.3] that, for a certain class of nonlinear terms including |x| α |U | p−1 U , radial eigenvalues of a wighted eigenvalue problem on a ball satisfỹ It follows from (1.14) and (1.16) that the n-th radial eigenvalue is larger than that of our problem. This causes a difference of lower bounds of Morse indices. A lower bound obtained in [3, Theorem 1.1] is smaller than that in Theorem 1.1 (i). When ( p, α) = (3, N − 4),λ rad,n can be explicitly written as where k is the solution of (1.10). This explicit eigenvalue relates m(U ± n,ρ ) to the ratio ρ/R, and hence it plays a crucial role in the proof of Theorems 1.4 and 1.5 (i). There are various results about lower bounds of the Morse index, while few results are known for upper bounds. A rather explicit upper bound of the Morse index is obtained in Theorem 1.5 (ii). An upper bound of the Morse index is in general not easy to obtain, because a lower bound ofλ rad,1 is needed. In this paper we use the following explicit lower bound ofλ rad,1 : Recently, an exact expression of all the eigenvalues for the linearization of a Neumann problem u − u + u 3 = 0 is obtained in [19]. The same method is applicable to the Dirichlet problem. However, we do not use those exact expression in this paper.
In summary, thanks to the critical exponent p c , we can perform these detailed analysis of eigenvaluesλ rad,1 , . . . ,λ rad,n+1 .
The paper consists of seven sections. In Sect. 2 we recall fundamental results about eigenvalues of (1.3) and (1.12). In Sect. 3 we use Emden's transformation and transform (1.1) into the scalar filed equation on an interval. Then, we prove Theorem 1.6. In Sect. 4 we compare the weighted eigenvalue problem (1.12) with (4.1) which is an eigenvalue problem associated with the scalar field equation. In Sect. 5 we compute the Morse index of U ± n,ρ and prove Theorem 1.1. In Sect. 6 we consider the case ( p, α) = (3, N − 4) and prove Theorems 1.4 and 1.5. Section 7 is an appendix. We recall the definition and basic properties of the complete elliptic integral K (k) and Jacobi elliptic functions sn(ξ, k), cn(ξ, k), dn(ξ, k) and sd(ξ, k).

Preliminaries
Let U ± n,ρ be solutions of (1.1) with n nodal domains. In this paper we mainly study eigenvalues of the wighted eigenvalue problem (1.12). We define the number of the negative eigenvalues of (1.12) counted with their multiplicity by m(U ± n,ρ ) = {negative eigenvalues of (1.12) counted with their multiplicity}. (2.1) The following proposition plays a crucial role in the study of the Morse index for radial solutions. It was extensively used in previous researches including [1-5, 12, 13, 15].
The eigenvalue problemL ± rad,n˜ rad = −λ rad˜ rad can be also written as (1.13). From now on, σ (L ± n ) and σ (L ± rad,n ) denote the set of the eigenvalues of (1.12) and (1.13), respectively. Let Because of Proposition 2.1, we count the number of the negative eigenvalues of (1.12) instead of (1.3). The eigenvalue problem (1.12) is easier to study, since all the eigenvalues of (1.12) can be decomposed into a radial and spherical parts.
The multiplicity of an eigenvalue of (1.12) can be calculated by the following proposition: where the summation takes all pairs (i, j) satisfying Moreover, the eigenspace of (1.12) associated toλ is spanned bỹ where˜ rad,i denotes the i-th eigenfunction of (1.13), ω j (θ ) denotes an eigenfunction of − S N −1 associated to ν j and the pair (i, j) satisfies (2.4).

Exact solutions
has an exact positive radial singular solution We use the so-called Emden transformation Then, we see in the following lemma that u(t) is a solution of the problem is a solution of (3.6).
Proof By direct calculation we have Then, Since ρ < r < R, we have that 0 < t < t ρ and that u(t) satisfies the Dirichlet boundary condition. Then, u(t) satisfies (3.6). It is clear that the converse is true. The proof is complete.
It is well known that a solution of (3.6) corresponds to an orbit of the system Since a Dirichlet boundary condition is imposed in ( We see that (3.9) has two homoclinic loops connecting (0, 0) to itself which are on v 2 − u 2 + 2|u| p+1 /( p + 1) = 0. One loop surrounds (1, 0) and the other loop surround (−1, 0). Hence two loops consist of a figure eight. Let Two loops satisfy v 2 = u 2 − 2|u| p+1 /( p + 1), i.e., the boundary of . It is obvious that there is no orbit in satisfying the boundary condition of (3.6). Therefore, a solution orbit of (3.6) is in R 2 \ , and they are periodic orbits. If (u(t), v(t)) satisfies (3.9), then (−u(t), −v(t)) also satisfies (3.9). This indicates that all times from a point on the v-axis to the next point of the v-axis are equal, and hence the length of each nodal domain of u(t) is equal to each other. Hence, if u has n nodal domains, then the length of each nodal domain is t ρ /n.
, the radial solutions of (1.1) can be written explicitly in terms of Jacobi elliptic functions Proof of Theorem 1.6 Because of Lemma 3.1, it is enough to obtain an exact solution of (3.6) with n nodal domains. Since u satisfies u −u +u 3 = 0 and u satisfies the Dirichlet boundary condition, a general solution u(t) can be written in terms of elliptic functions as follows: See [10, Chapter 7, Section 10] for details of this formula and see Sect. 7 for the definition of cn(ξ, k). Since u(0) = u(t ρ ) = 0 and u has n nodal domains, we see (3.11) where K denotes the complete elliptic integral of the first kind whose definition and basic properties are recalled in Sect. 7.1. Then, By the addition formula We return to the original variables. Then we obtain (1.9). By the first equality in (3.11) we obtain (1.10). The function Thus, (1.10) has a unique solution k ∈ 1 √ 2 , 1 .

Eigenvalue problem
Let n > 1 and let u(t) be a solution of (3.6) with n nodal domains. We consider the linearized eigenvalue problem Here f (u) = −1 + p|u| p−1 . Let μ i , i ≥ 1, denote the i-th eigenvalue and let φ i denote an eigenfunction associated with μ i . We use the same change of variables as (3.4), i.e., let t := − 1 m log r R and we define where U * (r ) is the singular solution of (3.1) given by (3.2) and φ is an eigenfunction of (4.1). Then U (r ) is a solution of (1.1) and˜ (r ) satisfies Then, the pair (λ rad,i ,˜ ) satisfies (1.12), and hence the radial part of the eigenvalue problem (1.12), which is (1.13), is equivalent to (4.1). All the eigenvalues of (1.12) can be obtained by (2.3). Since all the eigenvalues of − S N −1 are explicitly given by (1.5), it is crucial to study eigenvalues of (4.1). .

Morse index
First, we consider a positive solution of We use the change of variables where m and U * (r ) are defined in (3.5) and (3.2), respectively. Then v(t) satisfies Note that T 0 → 0 as ρ → R and T 0 → ∞ as ρ → 0. A solution of (5.1) corresponds to a solution (5.2). A solution of (5.2) corresponds to an orbit of (3.9) in the right half-plane that starts from a point (0, v (−T 0 )) and arrives (0, v (T 0 )). This orbit goes across the horizontal axis at (a, 0), and a is the maximum value of v(t) for −T 0 < t < T 0 . The orbit is in R 2 \ , where is defined by (3.10). Hence a > a 0 . We study the time of this orbit. Multiplying the equation in (5.2) with v and integrating it over [T 0 , t], we have v (t) 2 2 + F(v(t)) = F(a).
Hence, T 0 can be related to a which is max −T 0 <t<T 0 v(t). It was shown in [22] that, for each pair (ρ, R), 0 < ρ < R, (5.1) has a unique solution. Therefore, (5.2) also has a unique solution for each T 0 > 0. Since T 0 (a) corresponds to a solution of (5.2) with T 0 = T 0 (a), the uniqueness of a solution of (5.2) indicates that T 0 (a) is monotone. Since a → a 0 , the corresponding orbit converges to a solution corresponding to a homoclinic loop in C loc (R), and hence T (a) → ∞ as a → a 0 . This limit, together with the existence of a solution of (5.2) for all T 0 > 0, indicates that T 0 (a) is decreasing, and T 0 (a) → 0 as a → ∞. In summary, Hence there exists an inverse function a = a(T 0 ) for 0 < T 0 < ∞. We consider the following limit problem of (5.2): Then w can be explicitly written as In particular, w(0) = a 0 . We easily see that lim Moreover, a → a 0 as T 0 → ∞.

Proof of Theorem 1.1
Let T 0 > 0 and u 1 (t) be a positive solution of Let μ D denote the first eigenvalue of the Dirichlet problem

Lemma 5.3
Let μ D be the first Dirichlet eigenvalue of (5.10). Then the following hold: Multiplying the equation of (5.9) by u and integrating it over [T 0 , x], we have By the variational characterization we see that (5.14) We take a test function ψ := u p+1 2 1 . Using (5.13) and u 1 = u 1 − u p 1 , we have

By (5.13) we have
Thus, by (5.15) and (5.14) we have Note that the largeness of T 0 > 0 is not necessary in (5.16). We have shown that (5.11) holds.
(ii) In this case we take a test function u 1 . We show that If (5.17) holds, then by (5.14) we have and hence (5.12) holds. Hereafter, we prove (5.17). We need an apriori estimate to prove (5.17). Using Hölder's inequality, we have Using this inequality, we have We have shown that (5.17) holds. The proof of (ii) is complete.

Remark 5.4
Let v(t) be a solution of (5.2) and let a := v(0). We can obtain a sharp upper bound of μ D in the following way: We can prove that However, we do not use (5.26) in this paper.
Let u 1 be a positive solution of (5.9). Let μ N denote the first eigenvalue of the Neumann Lemma 5.5 Let μ N be the first Neumann eigenvalue of (5.27). Then, Let v(t) be a solution of (5.2). Then the first eigenvalue of (5.27) is equal to that of Hereafter, we consider (5.29). Let I T 0 := (−T 0 , T 0 ). Let μ N be the first eigenvalue of (5.29) and let φ be a first eigenfunction of (5.29). We may assume that φ > 0 in R and φ L 2 (I T 0 ) = 1.
We see that and that φ attains the infimum of (5.30). Hereafter, we define φ = 0 on R\I T 0 and χ I T 0 denotes the indicator function of I T 0 . We also define v(t) = 0 on R \ I T 0 to extend the domain of v(t). We see that Since v ∞ is bounded for large T 0 > 0, by (5.31) we see that μ N is bounded for large uniformly for large T 0 > 0, we see that, for each compact set K , there are C K 1 > 0 and a compact set K 1 such that K is in the interior set of K 1 and that φ H 1 ( is bounded uniformly for large T 0 > 0. Since φ satisfies the equation in (5.29), by Schauder estimates we see that {φ} is bounded in C 2,γ (K ). It follows from Ascoli-Arzelá theorem with a diagonal argument that there exists where w is a unique solution of (5.3). Since φ ∞ is bounded uniformly for large T 0 > 0, by Lemma 5.1 we see that |v| p−1 φ 2 is dominated by an L 1 (R)-function which is independent of T 0 > 0 large. Note that T 0 → ∞ if and only if a → a 0 . The function |v| p−1 φ 2 converges pointwise to w p−1 φ 2 * in R as T 0 → ∞. By the dominated convergence theorem we obtain (5.34). Using (5.34), Fatou's lemma and Lemma 5.3 (i), we have where w is a solution of (5.3) and we used (5.32). Because of (5.35), φ * ≡ 0 in R. Since {μ N } is bounded, there exists μ N * and a subsequence of {μ N }, which is still denoted by {μ N }, such that μ N → μ N * as T 0 → ∞. Applying Fatou's lemma to the LHS of (5.33), we see that φ * ∈ L 2 (R). Since φ 2 = 1, again by Fatou's lemma we see that φ * ∈ L 2 (R), and hence φ * ∈ H 1 (R). Since φ satisfies (5.29), φ * satisfies the problem Since φ * ≡ 0 in R, by the strong maximum principle φ * > 0 in R. Thus, φ * is a first eigenfunction. By Proposition 5.2 we see that μ N * = −( p − 1)( p + 3)/4. This indicates that (5.28) holds.
(ii) Let φ N (t) be the first eigenfunction of (5.27) defined on I 1 andφ N (t) be the first eigenfunction of (5.38) defined on I . Let ξ 0 := H 1 (φ N ), η 0 := φ N 2 L 2 (I 1 ) , where, for j = 1, 2, . . . , n, We restrict a domain of the functionφ N to I j for j = By Lemma 5.6 we see that . (5.45) Sinceφ N is a first eigenfunction of (5.38), by (5.45) we see that By Lemma 5.5 we see that for each ε > 0, there exists ρ ε > 0 such that The proof is complete.
Let {λ rad,i } ∞ i=1 denote the set of the eigenvalues of (1.13) associated with the solution u ± n .
Proof of Theorem 1.1 (i) First, we consider the case where ρ > 0 is small. When ρ > 0 is small, by Corollary 5.8 and Lemma 5.10 we havẽ Therefore, if ρ > 0 is small, then (1.12) has no zero eigenvalue, and hence (1.3) has also no zero eigenvalue. Thus, U ± n,ρ is nondegenerate for small ρ > 0. By (5.51) we see that all negative eigenvalues arẽ λ rad,i + ν j for 1 ≤ i ≤ n and 0 ≤ j ≤ .
Moreover, it follows from Proposition 2.3 that a multiplicity of each eigenvalue is M j (N ). Thus, By Proposition 2.1 we see that m(U ± n,ρ ) = m(U ± n,ρ ) = n j=0 M j (N ) for small ρ > 0. We calculate j=0 M j (N ). We use the following form of the multiplicity formula of ν j : We assume that is even. Then We obtain the same formula in the odd case. Next, we consider the case where ρ > 0 is not necessarily small. Even in this case, by Corollary 5.8 (i) we havẽ Thus, by Proposition 2.1 we have The proof of (i) is complete.
Note that the smallness of ρ > 0 is not used in (5.46). By the same argument as in (ii) we see that m(U ± n,ρ ) = m(U ± n,ρ ) → ∞ as α → ∞.

Proof of Theorem 1.4
We consider the case ( p, α) = (3, N − 4). In this section u denotes a solution of (3.6) with n nodal domains for simplicity. Let a := max 0≤t≤t ρ |u(t)|. Then, a > a 0 = √ 2 and The following lemma says that the n-th eigenvalue of (4.1) with respect to a solution u with n nodal domains can be written explicitly. are the n-th eigenvalue of (4.1) and an associated eigenfunction, respectively. In particular, where t ρ is defined by (3.7).
Proof Substituting μ n and φ n into φ + (−1 + 3u 2 + μ)φ, we have Using u = u − u 3 and (6.1), we can check that the RHS of (6.2) is equal to 0. Since u(t) 2 − 2 + a 2 > 0 for 0 ≤ t ≤ t ρ and u(t) has n nodal domains, we see that φ n (t) has n − 1 zeros on (0, t ρ ). It follows from Sturm-Liouville theory that φ n is an n-th eigenfunction, and hence μ n is the n-th eigenvalue.

Upper and lower bounds of the Morse index
Proof of Theorem 1.5 (i) It follows from Proposition 2.2 that ifλ rad,n + ν < 0, then the following are negative eigenvalues of (1.12): λ rad,i + ν j for 1 ≤ i ≤ n and 0 ≤ j ≤ .
Let > 0 be given in Theorem 1.5 (i). In a similar way to the proof of Theorem 1.4 we see thatλ rad,n + ν < 0 is equivalent to ρ R > R ,n . Thus, we have shown that (1.7) holds. Proof of Theorem 1.5 (ii) Let u(t) be a solution of (3.6) with n nodal domains. Let a := u ∞ . Then u(t) can be written explicitly as (3.12) and a = 2k 2 2k 2 − 1 > a 0 = √ 2.