Mathematische Annalen manuscript No. (will be inserted by the editor)

Filippo Gazzola · Hans-Christoph Grunau

Radial entire solutions for supercritical biharmonic equations

Received: April 28, 2005/ Revised: October 11, 2005

Abstract We prove existence and uniqueness (up to rescaling) of positive radial entire solutions of supercritical semilinear biharmonic equations. The proof is performed with a shooting method which uses the value of the second derivative at the origin as a parameter. This method also enables us to find finite time blow up solutions. Finally, we study the convergence at infinity of smooth solutions towards the explicitly known singular solution. It turns out that the convergence is different in space dimensions n ≤ 12 and n ≥ 13.

Mathematics Subject Classification (2000) 35J60 · 35J30

1 Introduction

In the present paper we investigate existence, uniqueness, asymptotic behavior and further qualitative properties of radial solutions of the supercritical biharmonic equation

Δ^{2}u = |u|^{p−1} u  in ℝ^{n},  (1)

where n ≥ 5 and p > \frac{n+4}{n−4} . Let us mention that the subcritical case p ≤ \frac{n+4}{n−4} is by now well-established, see [5, Theorems 1.3, 1.4]. There are several motivations for the study of (1). Let us try to explain them in some detail.

Financial support by the Vigoni programme of CRUI (Rome) and DAAD (Bonn) is gratefully acknowledged.

F. Gazzola Dipartimento di Matematica, Politecnico di Milano, Piazza Leonardo da Vinci, 32, 20133 Milano (Italy) E-mail: gazzola@mate.polimi.it H.-Ch. Grunau Fakultät für Mathematik, Otto-von-Guericke-Universität, Postfach 4120, 39016 Magdeburg (Germany) E-mail: hans-christoph.grunau@mathematik.uni-magdeburg.de

We first recall that the corresponding supercritical second order equation (when n ≥ 3 and p > \frac{n+2}{n−2} )

−Δu = |u|^{p−1} u  in ℝ^{n}  (2)

was intensively studied by Gidas-Spruck [7], Pohožaev [14] and in particular detail by Xuefeng Wang [18]. For the reader’s convenience, we state those of the results obtained there, being of relevance for the present paper.

Proposition 1 [18] Let n ≥ 3 and assume that p > \frac{n+2}{n−2} . Then, for any a > 0 the equation (2) admits a unique radial positive solution u = u(r) such that u(0) = a and u(r) → 0 as r → ∞. The solution u satisfies u′(r) < 0 for all r > 0 and

lim_{r→∞}r^{2/(p−1)} u(r) = L:= ( \frac{2(np − 2p − n)}{(p − 1)^{2}} ) ^{1/(p−1)} .

Moreover, if n ≤ 10 or if n ≥ 11 and

p < p^{c} := \frac{n^{2}− 8n + 4 + 8\sqrt{n − 1}}{(n − 2)(n − 10)}

then u(r) − Lr^{−2/(p−1)} changes sign infinitely many times. If n ≥ 11 and p ≥ p^{c} then u(r) < Lr^{−2/(p−1)} for all r > 0 and the solutions are strictly ordered with respect to the initial value a = u(0) .

The main concern of the mentioned paper [18] by Wang, however, are corresponding reaction-diffusion equations.

Most of the methods employed for the proof of Proposition 1 are special for second order equations and do not apply to (1). For instance, qualitative properties of solutions require a detailed analysis of a dynamical system in the corresponding phase space which is two dimensional for (2), whereas it is four dimensional for (1). And in four dimensional spaces powerful tools such as the Poincaré-Bendixson theory are no longer available. One of our purposes is to find out which of the results in [18] continue to hold and by which new methods they can be proved.

We seek solutions u of (1) which only depend on |x| so that they also solve the corresponding ordinary differential equation. Due to their homogeneity, both equations (1) and (2) are invariant under a suitable rescaling. This means that existence of a solution immediately implies the existence of infinitely many solutions, each one of them being characterized by its value at the origin. To ensure smoothness of the solution, one needs to require that u′(0) = u′′′(0) = 0 for (1) and u′(0) = 0 for (2). But contrary to (2), solutions of (1) may be determined only by fixing a priori also the value of u′′(0). In Theorem 1, which is stated in detail like the other main results in Section 2, we show that positive radial solutions u = u(|x|) of (1) exist and are unique, up to rescaling. The proof is performed with a shooting method which uses as a free parameter the “shooting concavity”, namely the initial second derivative u′′(0). Clearly, (2) has no free parameter since one has just to fix the rescaling parameter u(0).

Theorem 2 highlights a further striking difference between (1) and (2). It states that the shooting concavity u′′(0) enables to find both positive and negative finite time blow up solutions for (1). Since no free parameter is available, no such solutions exist for (2).

In Sections 3 and 6 we transform (1) into an autonomous ordinary differential equation and, by exploiting the supercriticality assumption, we construct an energy functional which has the crucial feature of being strictly decreasing on critical points of the solution. This fact, combined with several fine estimates, enables us to prove Theorem 3, namely that positive radial entire solutions of (1) behave asymptotically as |x| → ∞ like the (positive) singular solution u_{s}(x) := C|x|^{−4/(p−1)} which solves (1) in ℝ^{n}\ {0} for a suitable value of C > 0. In other words, we show that any entire positive radial solution u = u(|x|) of (1) satisfies

lim_{|x|→∞} |x|^{4/(p−1)} u(|x|) = C  (3)

for some fixed C > 0. Although this result is similar to that obtained in [18] for (2), its proof is completely different.

The following step is to find out whether the convergence in (3) occurs monotonically or with oscillations. To this end, we perform a stability analysis for the singular solution u_{s}. It turns out that for dimensions n ≥ 13 a new critical exponent p_{c} > \frac{n+4}{n−4} arises. The stable manifold behaves differently for n ≤ 12 and p > \frac{n+4}{n−4} or n ≥ 13 and \frac{n+4}{n−4} < p < p_{c} on the one hand, and for n ≥ 13 and p ≥ p_{c} on the other hand. In Section 4 we show that strong hints give the feeling that oscillations occur in the former situation. On the other hand, in Theorem 4 we prove that monotone convergence occurs in (3) whenever n ≥ 13 and p > p_{c}. In what follows, the notion “subcritical” and “supercritical” always refers to this new critical exponent p_{c}. Our results still leave open some questions, which we describe in detail in Open Problems 1–3 in Section 4.

Finally, let us mention that our results may also shed some light on related problems in the unit ball B ⊂ ℝ^{n}. For both the cases L = −Δ and L = Δ^{2}, consider the equation

Lu = λ(1 + u)^{p} in B (4)

where λ ≥ 0. We complement (4) with homogeneous Dirichlet boundary conditions (u = 0 if L = −Δ and u = |∇u| = 0 if L = Δ^{2}). When L = −Δ, it is known [11, Théorème 4] that the extremal solution u^{∗} (corresponding to the largest value of λ for which (4) admits a positive solution) is bounded for all n and p which give rise to smooth solutions of (2) oscillating around the singular solution, see [18, Proposition 3.7]. For the remaining values of n and p (when no oscillation occurs in (2), it is known [3] that u^{∗} is unbounded. When L = Δ^{2}, similar results are not known due to several serious obstructions which arise. For instance, the singular solution of (4) cannot be explicitly determined, see [1,2]. Moreover, the link with remainder terms in Hardy inequality discovered in [3] seems to fail for higher order problems [6]. Nevertheless, the results of the present paper enable us to conjecture that, when L = Δ^{2}, extremal solutions of (4) are unbounded if and only if n ≥ 13 and p ≥ p_{c}.

This paper is organized as follows. In the next section, we state our main results. In Section 3 we transform equation (1) first into an autonomous equation and subsequently into an autonomous system. In Section 4 we study the autonomous system in the “subcritical” case (n + 4)/(n − 4) < p < p_{c}. Finally, Sections 5, 6, 7 and 8 are devoted to the proofs of the results.

2 Results

An existence result, which covers the equation (1), was given first by Serrin and Zou [16]. In Section 5, we give a different proof which is perhaps simpler and more suitable for our purposes. Moreover, we show uniqueness and complement these results with some information on the qualitative behavior of the solution.

Theorem 1 Let n ≥ 5 and assume that p > \frac{n+4}{n−4} . Then, for any a > 0 the equation

Δ^{2}u = u^{p}  in ℝ^{n}  (5)

admits a unique radial positive solution u = u(r) (r = |x|) such that u(0) = a and u(r) → 0 as r → ∞. Moreover, u satisfies:

(i) u′(r) < 0 for all r > 0. (ii) Δu(r) < 0 for all r > 0. (iii) (Δu)′(r) > 0 for all r > 0.

The solutions in Theorem 1 are constructed by means of a shooting method. We keep u(0) fixed, say u(0) = 1, and look for solutions u_{γ} of the initial value problem over [0,∞):

\begin{matrix}u_{γ}^{(4)}(r) + \frac{2(n − 1)}{r} u′′′_{γ}(r) + \frac{(n − 1)(n − 3)}{r^{2}} u′′_{γ}(r) − \frac{(n − 1)(n − 3)}{r^{3}} u′_{γ}(r) \\ = |u_{γ}(r)|^{p−1} u_{γ}(r) \\ u_{γ}(0) = 1 ,  u′_{γ}(0) = u′′′_{γ}(0) = 0 ,  u′′_{γ}(0) = γ < 0 , \end{matrix}  (6)

which is the radial version of equation (1). Then, one has the following behavior with respect to the parameter γ:

Theorem 2 There exists a unique \overline{γ} < 0 such that the solution \overline{u}of (6) (for γ = \overline{γ}) exists on the whole interval [0,∞), is positive everywhere and vanishes at infinity. If γ < \overline{γ}, there exist 0 < R_{1} < R_{2} < ∞ such that u_{γ}(R_{1}) = 0 and lim_{r→R_{2}} u_{γ}(r) = −∞. If γ > \overline{γ}, there exist 0 < R_{1} < R_{2} < ∞ such that u′_{γ}(r) < 0 for r ∈ (0,R_{1}), u′_{γ}(R_{1}) = 0, u′_{γ}(r) > 0 for r ∈ (R_{1},R_{2}) and lim_{r→R_{2}} u_{γ}(r) = +∞.

Theorem 2 shows that entire radial solutions of (1) are necessarily of one sign so that, in what follows, we restrict our attention to positive solutions. It is a simple observation that a positive singular solution u_{s} of (5) is given by

u_{s}(r) = K_{0}^{1/(p−1)} r^{−4/(p−1)} ,  (7)

where

K_{0} = \frac{8}{(p − 1)^{4}} [ (n − 2)(n − 4)(p − 1)^{3}+ 2(n^{2}− 10n + 20)(p − 1)^{2}
−16(n − 4)(p − 1) + 32] .

In contrast with the second order equation (2) discussed in [18], a priori the entire solutions of (5) found in Theorem 1 may have faster decay than the singular solution, see the discussion in Section 6. However, by transforming equation (6) into an autonomous 4 × 4 system and by means of a careful analysis of a suitable energy functional and of corresponding integrability properties, we succeed in proving the following result:

Theorem 3 Let n ≥ 5 and assume that p > \frac{n+4}{n−4} . Let u = u(r) be a positive smooth radial entire solution of (5) and let u_{s} be as in (7). Then,

u(r) < ( \frac{p + 1}{2} ) ^{1/(p−1)} u_{s}(r) for all r ≥ 0  (8)

and

lim_{r→∞} \frac{u(r)}{u_{s}(r)} = 1.  (9)

We now wish to describe in which way (9) occurs. To this end, in Section 3 and 8 we perform a stability analysis of the singular solution u_{s}. It turns out that for dimensions n ≥ 13 a new critical exponent p_{c} > \frac{n+4}{n−4} becomes important:

Theorem 4 For all n ≥ 13 there exists p_{c} > \frac{n+4}{n−4} such that if p > p_{c} and if u is a

smooth positive radial entire solution of (5), then u(r) − K_{0}^{1/(p−1)} r^{−4/(p−1)} does not change sign infinitely many times.

The number p_{c} is the unique value of p > \frac{n+4}{n−4} such that

\begin{matrix}−(n − 4)(n^{3}− 4n^{2}− 128n + 256)(p − 1)^{4}+ 128(3n − 8)(n − 6)(p − 1)^{3} \\ +256(n^{2}− 18n + 52)(p − 1)^{2}− 2048(n − 6)(p − 1) + 4096 = 0. \end{matrix}

In Proposition 2 we show that n ↦ p_{c} is decreasing for n ≥ 13 and tends to 1 as n → ∞.

Theorem 4 is a partial result concerning the “supercritical” case p > p_{c}, n ≥ 13. Section 4 is devoted to the discussion of the “subcritical” case.

3 An autonomous system

In radial coordinates r = |x|, equation (5) reads

\begin{matrix}u^{(4)}(r) + \frac{2(n − 1)}{r} u′′′(r) + \frac{(n − 1)(n − 3)}{r^{2}} u′′(r) − \frac{(n − 1)(n − 3)}{r^{3}} u′(r) \\ = u^{p}(r)  r ∈ [0,∞) . \end{matrix}  (10)

Our purpose here is to transform (10) first into an autonomous equation and, subsequently, into an autonomous system. For some of the estimates which follow, it is convenient to rewrite the original assumption p > \frac{n+4}{n−4} as

(n − 4)(p − 1) > 8 .  (11)

Inspired by the proof of [18, Proposition 3.7] (see also [7,9]) we set

u(r) = r^{−4/(p−1)} v(logr) (r > 0) ,  v(t) = e^{4t/(p−1)} u(e^{t}) (t ∈ ℝ) . (12)

Tedious calculations then show that

\frac{u′(r)}{r^{3}} = r^{−4p/(p−1)} [v′(t) − \frac{4}{p − 1} v(t)] ,  (13)

\frac{u′′(r)}{r^{2}} = r^{−4p/(p−1)} [v′′(t) − \frac{p + 7}{p − 1} v′(t) + \frac{4(p + 3)}{(p − 1)^{2}} v(t)] ,

\frac{u′′′(r)}{r} = r^{−4p/(p−1)} [v′′′(t) − \frac{3(p + 3)}{p − 1} v′′(t) + \frac{2(p^{2}+ 10p + 13)}{(p − 1)^{2}} v′(t)

−\frac{8(p + 1)(p + 3)}{(p − 1)^{3}} v(t)] ,

u^{(4)}(r) = r^{−4p/(p−1)} [v^{(4)}(t) − \frac{2(3p + 5)}{p − 1} v′′′(t) + \frac{11p^{2}+ 50p + 35}{(p − 1)^{2}} v′′(t)

−\frac{2(3p^{3}+ 35p^{2}+ 65p + 25)}{(p − 1)^{3}} v′(t) + \frac{8(p + 1)(p + 3)(3p + 1)}{(p − 1)^{4}} v(t)] .

Therefore, after the change (12), equation (10) may be rewritten as

v^{(4)}(t) + K_{3}v′′′(t) + K_{2}v′′(t) + K_{1}v′(t) + K_{0}v(t) = v^{p}(t)  t ∈ ℝ , (14)

where the constants K_{i} = K_{i}(n,p) (i = 0,...,3) are given by

K_{0} = \frac{8}{(p − 1)^{4}} [ (n − 2)(n − 4)(p − 1)^{3}+ 2(n^{2}− 10n + 20)(p − 1)^{2}
−16(n − 4)(p − 1) + 32] ,
K_{1} = −\frac{2}{(p − 1)^{3}} [ (n − 2)(n − 4)(p − 1)^{3}+ 4(n^{2}− 10n + 20)(p − 1)^{2}
−48(n − 4)(p − 1) + 128] ,
K_{2} = \frac{1}{(p − 1)^{2}} [ (n^{2}− 10n + 20)(p − 1)^{2}− 24(n − 4)(p − 1) + 96] ,
K_{3} = \frac{2}{p − 1} [ (n − 4)(p − 1) − 8] .

By using (11), it is not difficult to show that K_{1} = K_{3} = 0 if p = \frac{n+4}{n−4} and that

K_{0} > 0 ,  K_{1} < 0 ,  K_{3} > 0  ∀n ≥ 5 , p > \frac{n + 4}{n − 4} .  (15)

On the other hand, the sign of K_{2} depends on n and p. We emphasize that the sign of K_{1} and K_{3} is due to assumption (11) and will be essentially exploited in the proof of Theorem 3, see also the proof of Lemma 6.

Note that (14) admits the two constant solutions v_{0} ≡ 0 and v_{s} ≡ K_{0}^{1/(p−1)} which, by (12), correspond to the following solutions of (10):

u_{0}(r) ≡ 0 , u_{s}(r) = \frac{K_{0}^{1/(p−1)}}{r^{4/(p−1)}} .

We now write (14) as a system in ℝ^{4}. By (13) we have

u′(r) = 0 ⟺ v′(t) = \frac{4}{p − 1} v(t) .

This fact suggests us to define

w_{1}(t) = v(t), w_{2}(t) = v′(t) − \frac{4}{p − 1} v(t), w_{3}(t) = v′′(t) − \frac{4}{p − 1} v′(t),

w_{4}(t) = v′′′(t) − \frac{4}{p − 1} v′′(t)

so that (14) becomes

\{ \begin{matrix}w′_{1}(t) & = & \frac{4}{p−1} w_{1}(t) + w_{2}(t) \\ w′_{2}(t) & = & w_{3}(t) \\ w′_{3}(t) & = & w_{4}(t) \\ w′_{4}(t) & = & C_{2}w_{2}(t) + C_{3}w_{3}(t) + C_{4}w_{4}(t) + w_{1}^{p}(t) , \end{matrix}  (16)

where

C_{m}= − ∑_{k=m−1}^{4} \frac{K_{k}4^{k+1−m}}{(p − 1)^{k+1−m}}  for m = 1,2,3,4  with K_{4} = 1 .  (17)

This gives first that C_{1} = 0 so that the term C_{1}w_{1}(t) does not appear in the last equation of (16). Moreover, we have the explicit formulae:

C_{2} = \frac{2}{(p − 1)^{3}} [ (n − 2)(n − 4)(p − 1)^{3}+ 2(n^{2}− 10n + 20)(p − 1)^{2}

−16(n − 4)(p − 1) + 32] = \frac{p − 1}{4} K_{0},

C_{3} = −\frac{1}{(p − 1)^{2}} [ (n^{2}− 10n + 20)(p − 1)^{2}− 16(n − 4)(p − 1) + 48],
C_{4} = −\frac{2}{p − 1} [ (n − 4)(p − 1) − 6].

System (16) has the two stationary points (corresponding to v_{0} and v_{s})

O( 0,0,0,0) and P( K_{0}^{1/(p−1)} ,−\frac{4}{p − 1} K_{0}^{1/(p−1)} ,0,0) .

Let us consider first the “regular point” O. The linearized matrix at O is

M_{O} = ( \begin{matrix}\frac{4}{p−1} & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & C_{2} & C_{3} & C_{4} \end{matrix})

and the characteristic polynomial is

λ ↦ λ^{4}+ K_{3}λ^{3}+ K_{2}λ^{2}+ K_{1}λ + K_{0} .

Then, according to MAPLE, the eigenvalues are given by

λ_{1} = 2\frac{p + 1}{p − 1} , λ_{2} = \frac{4}{p − 1} , λ_{3} = \frac{4p}{p − 1} − n, λ_{4} = 2\frac{p + 1}{p − 1} − n.

Since we assume that p > \frac{n+4}{n−4} > \frac{n}{n−4} > \frac{n+2}{n−2} , we have

λ_{1} > λ_{2} > 0 > λ_{3} > λ_{4}.

This means that O is a hyperbolic point and that both the stable and the unstable manifolds are two-dimensional. This is the same situation as in the exponential case (see [1]) and except for λ_{2} it seems as if one could perform a formal limit p → ∞.

Around the “singular point” P the linearized matrix of the system (16) is given by

M_{P} = ( \begin{matrix}\frac{4}{p−1} & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ pK_{0} & C_{2} & C_{3} & C_{4} \end{matrix}) .  (18)

The corresponding characteristic polynomial is

ν ↦ ν^{4}+ K_{3}ν^{3}+ K_{2}ν^{2}+ K_{1}ν + (1 − p)K_{0}

and the eigenvalues are given by

ν_{1} = \frac{N_{1}+ \sqrt{N_{2}+ 4\sqrt{N_{3}}}}{2(p − 1)} , ν_{2} = \frac{N_{1}− \sqrt{N_{2}+ 4\sqrt{N_{3}}}}{2(p − 1)} ,

ν_{3} = \frac{N_{1}+ \sqrt{N_{2}− 4\sqrt{N_{3}}}}{2(p − 1)} , ν_{4} = \frac{N_{1}− \sqrt{N_{2}− 4\sqrt{N_{3}}}}{2(p − 1)} ,

where

N_{1} := −(n − 4)(p − 1) + 8, N_{2} := (n^{2}− 4n + 8)(p − 1)^{2},
N_{3} := (9n − 34)(n − 2)(p − 1)^{4}+ 8(3n − 8)(n − 6)(p − 1)^{3}
+(16n^{2}− 288n + 832)(p − 1)^{2}− 128(n − 6)(p − 1) + 256.

The stability of the stationary point P is described by the following

Proposition 2 Assume that p > \frac{n+4}{n−4} .

(i) For any n ≥ 5 we have ν_{1},ν_{2} ∈ ℝ and ν_{2} < 0 < ν_{1}. (ii) For any 5 ≤ n ≤ 12 we have ν_{3},ν_{4} ∉ ℝ and Re ν_{3} =Re ν_{4} < 0. (iii) For any n ≥ 13 there exists p_{c} > \frac{n+4}{n−4} such that:

– if p < p_{c}, then ν_{3},ν_{4} ∉ ℝ and Re ν_{3} =Re ν_{4} < 0.

– if p = p_{c}, then ν_{3},ν_{4} ∈ ℝ and ν_{4} = ν_{3} < 0.

– if p > p_{c}, then ν_{3},ν_{4} ∈ ℝ and ν_{4} < ν_{3} < 0. The number p_{c} is the unique value of p > \frac{n+4}{n−4} such that

\begin{matrix}−(n − 4)(n^{3}− 4n^{2}− 128n + 256)(p − 1)^{4}+ 128(3n − 8)(n − 6)(p − 1)^{3} \\ +256(n^{2}− 18n + 52)(p − 1)^{2}− 2048(n − 6)(p − 1) + 4096 = 0. \end{matrix}

The function n ↦ p_{c} is strictly decreasing and approaches 1 as n → ∞.

Proof See Section 8. ⊓⊔

According to Proposition 2, in any case we have

ν_{1} > 0,  ν_{2} < 0,  Reν_{3} = Reν_{4} < 0.

This means that P has a three dimensional stable manifold and a one dimensional unstable manifold (as in the exponential case, see [1, Sect. 3.1]).

Remark 1 Consider the function

ϕ(x) := x^{4}+ K_{3}x^{3}+ K_{2}x^{2}+ K_{1}x .  (19)

We have ϕ(0) = 0 and ϕ′(0) = K_{1} < 0 for every n and p. Moreover, by the previous analysis around the point O, we know that the equation ϕ(x) = −K_{0} always has 4 real solutions, 2 positive and 2 negative. By these facts we deduce that the graphic of ϕ has the shape of W with two local minima (one positive, one negative) at level below −K_{0} and the unique local maximum (negative) at strictly positive level. In particular, for any −K_{0} ≤ γ ≤ 0, the equation ϕ(x) = γ has 4 real solutions. Finally, note that the level of the local maximum of ϕ coincides with (p−1)K_{0} if and only if p = p_{c}.

4 Observations on the stable manifold of P and open problems

Let u denote a smooth positive entire radial solution of (5), let v be defined according to (12) so that it solves (14), and let w(t) = (w_{1}(t),w_{2}(t),w_{3}(t),w_{4}(t)) be the vector solution of the corresponding first order system (16).

We first state a general result which holds for any entire smooth solution:

Proposition 3 We assume that u is an entire smooth positive radial solution of (5) and that w = (w_{1},w_{2},w_{3},w_{4}) is the corresponding solution of system (16). Then,

lim_{t→∞}w(t) = P.

In particular, the trajectory w is on the stable manifold of P.

Proof See Section 8. ⊓⊔

By Proposition 3 we know that w is on the stable manifold of the singular point P while Theorem 4 gives information on the non-oscillatory behavior of u around the singular solution u_{s} in the “supercritical” case. In this section, we refer to the new critical exponent p_{c} arising in Proposition 2. Here, we are interested in the (presumably) oscillatory behavior in the “subcritical” case, i.e. in what follows we assume:

n ≤ 12  or  ( n ≥ 13 and \frac{n + 4}{n − 4} < p < p_{c}) .  (20)

We study the relative position of the hyperplane

H := {w ∈ ℝ^{4} : w_{1} = K_{0}^{1/(p−1)} }

with respect to the tangential plane of the oscillatory component of the stable manifold

OS:= {sx + ty : s,t ∈ ℝ}.

Here x ± iy denotes eigenvectors of the matrix M_{P} defined in (18) corresponding to the nonreal eigenvalues ν_{3},ν_{4}.

Proposition 4 The hyperplane H and the plane P + OS intersect transversally, i.e.

P + OS ⊄ H.

Proof See Section 8. ⊓⊔

Open Problem 1 Since we have that ν_{2} < Reν_{3} = Reν_{4} < 0 we know that all trajectories of system (16) which are in the stable manifold of P are eventually tangential to OS, except the trajectory being tangential to the eigenvector corresponding to ν_{2}. By Proposition 4 we may conclude that all these trajectories have infinitely many intersections with the hyperplane H. If the trajectory w corresponding to the solution u is among these, then we would have shown:

For t → ∞, the first component v(t) = w_{1}(t) attains infinitely many times the value K_{0}^{1/(p−1)} so that for r near ∞, u(r) oscillates infinitely often around the singular solution u_{s}, provided that the subcriticality assumption (20) is satisfied.

In order to complete the proof of this statement, it “only” remains to show that at ∞, t ↦ w(t) is not tangential to an eigenvector corresponding to ν_{2}. For this it would suffice to identify the trajectories having this property and to see that they are different from w.

Open Problem 2 Our proof of Theorem 4 relies on a result by Elias [4] which no longer applies when p = p_{c}. Nevertheless, we believe that the statement of Theorem 4 also holds true in this limit situation. If one could show that for all p > p_{c} the solutions u of (10) are approaching the singular solution u_{s} from below, then the same result would presumably also follow for p = p_{c} by continuous dependence.

Open Problem 3 With the same proof of Theorem 4, one can also show that if u_{α} and u_{β} are positive radial entire solutions of (5) with shooting levels u_{α} (0) = α and u_{β}(0) = β, then (under the assumptions of Theorem 4) u_{α} − u_{β} is nonoscillatory, i.e. it has at most a finite number of zeros. A natural question arises whether all these solutions (including the singular one) are completely ordered, i.e. they have no crossing at all, and not only eventually.

5 Proof of Theorems 1 and 2

If u = u(r) is a radial positive solution of (5) such that u(0) = 1 and u(r) → 0 as r → ∞, then

u_{a}(r) := au(a^{\frac{p−1}{4}} r) (a > 0)

is a radial positive solution of (5) such that u_{a}(0) = a and u_{a}(r) → 0 as r → ∞. Therefore, Theorem 1 follows if we prove existence and uniqueness of a solution u satisfying u(0) = 1.

Existence. In order to prove existence, we apply a shooting method with initial second derivative as parameter. We remark that nu′′(0) = Δu(0) and that by l’Hospital’s rule

(Δu)′(0) = u′′′(0) + (n − 1) lim_{r→0} \frac{ru′′(r) − u′(r)}{r^{2}} = \frac{n + 1}{2} u′′′(0).

This means that the initial conditions in (6) also read as

u(0) = 1 ,  u′(0) = (Δu)′(0) = 0 ,  Δu(0) = nγ < 0 .  (21)

For all γ < 0, (10)-(21) admits a unique local smooth solution u_{γ} defined on some right neighborhood of r = 0. Let

R_{γ} = \{ \begin{matrix}+∞ & if u_{γ}(r)u′_{γ}(r) < 0 ∀r > 0 \\ min\{r > 0; u_{γ}(r)u′_{γ}(r) = 0\} & otherwise. \end{matrix}

From now on we understand that u_{γ} is continued on [0,R_{γ}). Let

I^{+} = {γ < 0; R_{γ} < ∞, u_{γ}(R_{γ}) = 0} , I^{−} = {γ < 0; R_{γ} < ∞, u′_{γ}(R_{γ}) = 0} .

We prove the following statement:

Lemma 1 Assume p > \frac{n+4}{n−4} . If both I^{+} and I^{−} are nonempty then there exists γ < 0 such that R_{γ} = ∞. Moreover, u_{γ} is defined on [0,∞) and lim_{r→∞} u_{γ}(r) = 0.

Proof Since p ≥ \frac{n+4}{n−4} , Pohožaev’s identity (see e.g. [13, Corollary 1]) tells us that for any R > 0 the problem

\{ \begin{matrix}Δ^{2}w = w^{p} & if |x| < R \\ w = |∇w| = 0 & if |x| = R \end{matrix}

admits no positive solution. In our setting, this reads

I^{+} ∩ I^{−} = ∅ .  (22)

Moreover, by continuous dependence with respect to the variable initial datum γ, we have that

I^{+} and I^{−} are open in (−∞,0) .  (23)

Combining (22)-(23) with the assumption, we infer that there exists γ ∉ I^{+} ∪ I^{−} . Then, R_{γ} = +∞ and lim_{r→∞} u_{γ}(r) exists (recall u′_{γ} < 0). Finally, this limit is necessarily 0, since u_{γ} solves (10). ⊓⊔

Remark 2 A well-known crucial difference arises when 1 < p < \frac{n+4}{n−4} . In such case, by standard critical point theory and rescaling one has that I^{+} ∩ I^{−} ≠ ∅.

Consider now the Euler equation for Sobolev minimizers (see e.g. [17]):

\{ \begin{matrix}v^{(4)}(r) + \frac{2(n − 1)}{r} v′′′(r) + \frac{(n − 1)(n − 3)}{r^{2}} v′′(r) − \frac{(n − 1)(n − 3)}{r^{3}} v′(r) \\ = v^{\frac{n+4}{n−4}} (r), r ≥ 0, \\ v(0) = 1 ,  v′(0) = (Δv)′(0) = 0 ,  Δv(0) = nδ , \end{matrix} (24)

where δ < 0 is chosen in such a way that the unique solution of (24) is given by

v(r) = \frac{[n(n^{2}− 4)(n − 4)] ^{\frac{n−4}{4}}}{(\sqrt{n(n^{2}− 4)(n − 4)} + r^{2})^{\frac{n−4}{2}}} .  (25)

This explicit solution will serve as a comparison function for the initial value problem (10)-(21). For this purpose we quote a comparison principle, which has been observed by McKenna-Reichel [10] and which will turn out to be useful also in the proof of uniqueness below:

Lemma 2 Assume that f : ℝ → ℝ is locally Lipschitzian and strictly increasing. Let u,v ∈ C^{4}([0,R)) be such that

\{ \begin{matrix}∀r ∈ [0,R) :  Δ^{2}v(r) − f(v(r)) ≥ Δ^{2}u(r) − f(u(r)), \\ v(0) ≥ u(0), v′(0) = u′(0) = 0, Δv(0) ≥ Δu(0), \\ (Δv)′(0) = (Δu)′(0) = 0. \end{matrix}  (26)

Then we have for all r ∈ [0,R):

v(r) ≥ u(r), v′(r) ≥ u′(r), Δv(r) ≥ Δu(r), (Δv)′(r) ≥ (Δu)′(r).  (27)

Moreover,

(i) the initial point 0 can be replaced by any initial point ρ > 0 if all four initial data are weakly ordered. (ii) a strict inequality in one of the initial data at ρ ≥ 0 or in the differential inequality on (ρ,R) implies a strict ordering of v,v′,Δv,Δv′and u,u′,Δu,Δu′on (ρ,R).

With the aid of this lemma we obtain

Lemma 3 Let δ < 0 be as in (24) and let v be as in (25). Let γ < δ and let u_{γ} be the local solution of (10)-(21). Then, one of the two following facts holds true:

(i) γ ∈ I^{+} . (ii) 0 < u_{γ}(r) < v(r) for all r > 0.

Proof Since 0 < v ≤ 1 we have

Δ^{2}v − v^{p} ≥ Δ^{2}v − v^{(n+4)/(n−4)} = 0 = Δ^{2}u − |u|^{p−1} u,

as long as u exists. Hence, v(r) > u(r) and 0 > v′(r) > u′(r) for these r > 0. Assume that γ ∉ I^{+} . Then it is immediate from Lemma 2 that alternative (ii) holds true. ⊓⊔

If the alternative (ii) in Lemma 3 holds true, then the corresponding solution u_{γ} satisfies the requirements of Theorem 1 and existence follows.

If the alternative (i) in Lemma 3 holds true, in view of Lemma 1, existence is proved once we show that

I^{−} ≠ ∅ .  (28)

To this end, we consider the following Dirichlet problem

\{ \begin{matrix}Δ^{2}w = λ(1 + w)^{p} & in B \\ w > 0 & in B \\ w = \frac{∂w}{∂n} = 0 & on ∂B \end{matrix}  (29)

where λ > 0 and B ⊂ ℝ^{n} is the unit ball. Arguing as in the proof of [2, Theorem 2.3] (see also [15]) and taking into account both Lemma 6 and Theorem 1 in [1] we infer that there exists \overline{λ} > 0 such that for all λ ∈ (0,\overline{λ}] problem (29) admits a radial smooth solution w_{λ} = w_{λ}(r). So, fix one such λ and put w(r) = λ^{\frac{1}{p−1}} (1+w_{λ}(r)). Then, w satisfies

Δ^{2}w = w^{p} in B
w > λ^{\frac{1}{p−1}}  in B
w = λ^{\frac{1}{p−1}}  on ∂B
\frac{∂w}{∂n} = 0  on ∂B .

Finally, the function u_{γ}(r) := α^{\frac{4}{p−1}} w(αr) with α = w^{\frac{1−p}{4}} (0) satisfies u_{γ}(0) = 1 and

Δ^{2}u_{γ} = u_{γ}^{p}  in B_{1/α}
u_{γ} > α^{\frac{4}{p−1}} λ^{\frac{1}{p−1}}  in B_{1/α}
u_{γ} = α^{\frac{4}{p−1}} λ^{\frac{1}{p−1}}  on ∂B_{1/α}
\frac{∂u_{γ}}{∂n} = 0  on ∂B_{1/α} .

Take γ = u′′_{γ}(0) < 0. Then, R_{γ} = α^{−1} and u′_{γ}(R_{γ}) = 0. This proves that γ ∈ I^{−} and, in turn, that (28) holds true. And this proves the existence of a positive radial solution of (5) satisfying u(0) = 1 and u(x) → 0 as |x| → ∞.

Qualitative behavior. Let u = u(r) be a radial solution of (5) such that u(0) = 1 and u(r) → 0 as r → ∞. Statement (iii) follows by integrating

{r^{n−1} [Δu(r)]′}′= r^{n−1} u^{p}(r)  (30)

over [0,r] for r > 0.

In order to prove (i), we assume for contradiction that there exists R_{1} > 0, the first solution of u′(R_{1}) = 0. Then, Δu(R_{1}) = u′′(R_{1}) ≥ 0. By using the just proved statement (iii) for r > R_{1} we deduce that Δu(r) > 0 for all r > R_{1} and that u′(r) > 0 for all r > R_{1}, against the assumption of u(r) vanishing at +∞. This contradiction proves (i).

Next we shall prove (ii). For contradiction, assume now that there exists R_{1} > 0, the first solution of Δu(R_{1}) = 0. Then, by (iii), we know that there exist R_{2} > R_{1} and ε > 0 such that Δu(r) ≥ ε for all r ≥ R_{2}. Multiplying by r^{n−1} this inequality yields

[r^{n−1} u′(r)]′≥ εr^{n−1} for all r ≥ R_{2} .

Integrating this last inequality over [R_{2},r] for any r > R_{2} and dividing by r^{n−1} gives

u′(r) ≥ \frac{ε}{n}r + \frac{R_{2}^{n−1} u′(R_{2})}{r^{n−1}}− \frac{εR_{2}^{n}}{nr^{n−1}} for all r ≥ R_{2} .

Letting r → ∞ we then obtain u′(r) → +∞, contradiction. Hence, also (ii) is proved.

Uniqueness. By means of the comparison principle Lemma 2, it is immediate that the family in Lemma 3 is ordered:

Lemma 4 Let γ_{1} < γ_{2} and let u_{γ_{j}} be the corresponding local solution of (10)-(21). As long as both solutions exist, we have for r > 0 that

u_{γ_{1}}(r) < u_{γ_{2}}(r).  (31)

Since we already proved existence, the following statement makes sense:

Lemma 5 Let \overline{u} denote a positive entire radially decreasing solution of (5) such that \overline{u}(0) = 1 and \overline{u}(r) → 0 as r → ∞ and let \overline{γ} = \overline{u}′′(0). For any γ < \overline{γ} let u_{γ} be the local solution of (10)-(21). Then, for r > 0, as long as u_{γ} exists:

u′_{γ}(r) < \overline{u}′(r).  (32)

Again, the proof follows directly from Lemma 2. In particular, Lemma 5 tells us that for any γ < \overline{γ}, u_{γ}(r) vanishes in finite time. This proves uniqueness and completes the proof of Theorem 1. ⊓⊔

Proof of Theorem 2. The existence of precisely one such \overline{γ} follows from the proof of Theorem 1.

The statement in the case γ > \overline{γ} follows by arguing similarly as in Theorem 4.2 in [2]. More precisely, by Lemma 4, for r > 0 we have 0 < \overline{u}(r) < u_{γ}(r) as long as the latter exists. If there exists no R_{1} > 0 such that u′_{γ}(R_{1}) = 0, then u′_{γ}(r) < 0 for all r > 0 so that u_{γ} would be a positive global solution of (6) such that u_{γ}(r) → 0 as r → ∞, against the uniqueness stated in Theorem 1. So, let R_{1} > 0 be the first solution of u′_{γ}(R_{1}) = 0. Then, Δu_{γ}(R_{1}) ≥ 0. By integrating (30) over [0,r] for r > R_{1} we deduce that Δu_{γ}(r) > 0 for all r > R_{1} and that u′_{γ}(r) > 0 for all r > R_{1}. Then, u_{γ}(r) → +∞ at some (finite or infinite) R_{2} > R_{1}.

In order to show that R_{2} < ∞ we essentially refer to a reasoning, which was performed for the critical case in [8, Lemma 2]. Let ũ(r) := u_{γ}(r) − 1, so that it solves Δ^{2}ũ= (1 + ũ) ^{p}. Since ũ(r) ↗ ∞ for r ↗ R_{2}, successive integration of the differential equation shows that for some suitable r_{0} < R_{2}, r_{0} close enough to R_{2}, one has:

ũ(r_{0}) > 0,  ũ′(r_{0}) > 0,  Δũ(r_{0}) > 0,  (Δũ)′(r_{0}) > 0.

For any value of the rescaling parameter α > 0,

u_{0,α} (r) := α ( 1 − (r/λ_{α} )^{2}) ^{−(n−4)/2} , λ_{α} = α^{−2/(n−4)} [(n + 2)n(n − 2)(n − 4)] ^{1/4},

solves

Δ^{2}u_{0,α} = u_{0,α}^{(n+4)/(n−4)} ≤ (1 + u_{0,α})^{(n+4)/(n−4)} ≤ (1 + u_{0,α})^{p} for r ∈ [0,λ_{α}).

Choosing α > 0 small enough one may achieve that λ_{α} > r_{0} and that

ũ(r_{0}) > u_{0,α} (r_{0}),  ũ′(r_{0}) > u_{0,α} (r_{0}),  Δũ(r_{0}) > Δu_{0,α} (r_{0}),
(Δũ)′(r_{0}) > (Δu_{0,α})′(r_{0}).

That means that u_{0,α} is a subsolution for ũon [r_{0},min{R_{2},λ_{α} }). Lemma 2 yields that

ũ(r) ≥ u_{0,α} (r) on [r_{0},min{R_{2},λ_{α} }).

Consequently, R_{2} ≤ λ_{α} < ∞.

The statement in the case γ < \overline{γ} is mostly a further consequence of Lemma 5. Indeed, for γ < \overline{γ} we know that necessarily u_{γ} vanishes in finite time, say at r = R_{1}. Since by (32) u′_{γ} remains negative for all r, we necessarily have u_{γ}(r) → −∞ at some R_{2} > R_{1}. By considering −u for r < R_{2} close to R_{2} and observing that −u solves the same differential equation, the first part of the present proof shows that also here R_{2} < ∞. ⊓⊔

6 Proof of Theorem 3

In order to prove (9) we proceed in three steps. We consider the corresponding global positive solution v of (14) and show first that for t → +∞, v → 0 or v → K_{0}^{1/(p−1)} or v oscillates infinitely many times near ∞. In a second step, we exclude the first alternative. Finally, we study solutions v being oscillatory at ∞. For this purpose, an energy functional is introduced, which helps to deduce suitable L^{2}- bounds on the solution v. These bounds show that the solution again and again and even faster and faster has to be in a neighbourhood of the singular point P. By local properties of the autonomous system (16), the trajectory of v is (finally) on the stable manifold of P. For these arguments it is crucial that the coefficients K_{1} and K_{3} have the “good” sign: K_{1} < 0 and K_{3} > 0.

As a first step, we prove:

Proposition 5 Let v be a global positive solution of (14) and assume that there exists L ∈ [0,+∞] such that

lim_{t→+∞} v(t) = L .

Then, L ∈ {0,K_{0}^{1/(p−1)} }.

Proof For contradiction, assume first that L is finite and L ∉ {0,K_{0}^{1/(p−1)} }. Then, v^{p}(t)−K_{0}v(t) → α := L^{p}−K_{0}L ≠ 0 and for all ε > 0 there exists T > 0 such that

α − ε ≤ v^{(4)}(t) + K_{3}v′′′(t) + K_{2}v′′(t) + K_{1}v′(t) ≤ α + ε  ∀t ≥ T .  (33)

Take ε < |α| so that α − ε and α + ε have the same sign and let

δ := sup_{t≥T}|v(t) − v(T)| < ∞.

Integrating (33) over [T,t] for any t ≥ T yields

(α − ε)(t − T) + C − |K_{1}|δ ≤ v′′′(t) + K_{3}v′′(t) + K_{2}v′(t)
≤ (α + ε)(t − T) + C + |K_{1}|δ ∀t ≥ T ,

where C = C(T) is a constant containing all the terms v(T), v′(T), v′′(T) and v′′′(T). Repeating twice more this procedure gives

\frac{α − ε}{6} (t − T)^{3}+ O(t^{2}) ≤ v′(t) ≤ \frac{α + ε}{6} (t − T)^{3}+ O(t^{2}) as t → ∞ .

This contradicts the assumption that v admits a finite limit as t → +∞.

Next, we exclude the case L = +∞. For contradiction, assume that

lim_{t→+∞} v(t) = +∞.  (34)

Then, there exists T ∈ ℝ such that

v^{(4)}(t) + K_{3}v′′′(t) + K_{2}v′′(t) + K_{1}v′(t) ≥ \frac{v^{p}(t)}{2} ∀t ≥ T .

Moreover, by integrating this inequality over [T,t] (for t ≥ T), we get

v′′′(t) + K_{3}v′′(t) + K_{2}v′(t) + K_{1}v(t) ≥ \frac{1}{2} ∫_{T}^{t}v^{p}(s)ds + C  ∀t ≥ T , (35)

where C = C(T) is a constant containing all the terms v(T), v′(T), v′′(T) and v′′′(T). From (34) and (35) we deduce that there exists T′≥ T such that α := v′′′(T′) + K_{3}v′′(T′) + K_{2}v′(T′) + K_{1}v(T′) > 0. Since, (14) is autonomous, we may assume that T′= 0. Therefore, we have

v^{(4)}(t) + K_{3}v′′′(t) + K_{2}v′′(t) + K_{1}v′(t) ≥ \frac{v^{p}(t)}{2} ∀t ≥ 0 ,  (36)
v′′′(0) + K_{3}v′′(0) + K_{2}v′(0) + K_{1}v(0) = α > 0 .  (37)

We may now apply the test function method developed by Mitidieri-Pohožaev [12]. More precisely, fix T_{1} > T > 0 and a nonnegative function ϕ ∈ C_{c}^{4}[0,∞) such that

ϕ(t) = \{ \begin{matrix}1 & for t ∈ [0,T] \\ 0 & for t ≥ T_{1} . \end{matrix}

In particular, these properties imply that ϕ(T_{1}) = ϕ′(T_{1}) = ϕ′′(T_{1}) = ϕ′′′(T_{1}) = 0. Hence, multiplying inequality (36) by ϕ(t), integrating by parts and recalling (37) yields

∫_{0}^{T_{1}}[ϕ^{(4)}(t) − K_{3}ϕ′′′(t) + K_{2}ϕ′′(t) − K_{1}ϕ′(t)]v(t)dt ≥ \frac{1}{2} ∫_{0}^{T_{1}}v^{p}(t)ϕ(t)dt + α .

(38)

We now apply Young’s inequality in the following form: for any ε > 0 there exists C(ε) > 0 such that

vϕ^{(i)} = vϕ^{1/p} \frac{ϕ^{(i)}}{ϕ^{1/p}} ≤ εv^{p}ϕ +C(ε)\frac{|ϕ^{(i)}|^{p/(p−1)}}{ϕ^{1/(p−1)}} ,  ϕ^{(i)} = \frac{d^{i}ϕ}{dt^{i}}  (i = 1,2,3,4).

Then, provided ε is chosen sufficiently small, (38) becomes

C ∑_{i=1}^{4} ∫_{0}^{T_{1}} \frac{|ϕ^{(i)}(t)|^{p/(p−1)}}{ϕ^{1/(p−1)} (t)} dt ≥ \frac{1}{4} ∫_{0}^{T}v^{p}(t)dt + α  (39)

where C = C(ε,K_{i}) > 0. We now choose ϕ(t) = ϕ_{0}(\frac{t}{T}), where ϕ_{0} ∈ C_{c}^{4}([0,∞), ϕ_{0} ≥ 0 and

ϕ_{0}(τ) = \{ \begin{matrix}1 & for τ ∈ [0,1] \\ 0 & for τ ≥ τ_{1} > 1 . \end{matrix}

As noticed in [12], there exists a function ϕ_{0} in such class satisfying moreover

∫_{0}^{τ_{1}} \frac{|ϕ_{0}^{(i)}(τ)|^{p/(p−1)}}{ϕ_{0}^{1/(p−1)} (τ)} dτ =: A_{i} < ∞ (i = 1,2,3,4).

Then, thanks to a change of variables in the integrals, (39) becomes

C ∑_{i=1}^{4} A_{i}T^{1−ip/(p−1)} ≥ \frac{1}{4} ∫_{0}^{T}v^{p}(t)dt + α ∀T > 0.

Letting T → ∞, the previous inequality contradicts (34). ⊓⊔

In order to perform the above mentioned second step, we show that a solution v of (14) vanishes at infinity only if the corresponding vector solution w= (w_{1},w_{2},w_{3},w_{4}) of the system (16) approaches the “regular point” O.

Proposition 6 Assume that v: [T_{0},∞) → (0,∞) exists for some T_{0}, solves (14) and satisfies lim_{t→∞} v(t) = 0. Then for all k ∈ ℕ, one also has:

lim_{t→∞}v^{(k)}(t) = 0.  (40)

Proof By assumption we know that for t large enough v(t) < K_{0}^{1/(p−1)} so that by the differential equation (14) eventually v^{(4)}(t) + K_{3}v′′′(t) + K_{2}v′′(t) + K_{1}v′(t) = (v^{p−1} (t) − K_{0})v(t) < 0. This shows that

t ↦ v′′′(t) + K_{3}v′′(t) + K_{2}v′(t) + K_{1}v(t)  (41)

is eventually strictly decreasing. Using the assumption once more we see that there exists

lim_{t→∞}(v′′′(t) + K_{3}v′′(t) + K_{2}v′(t))= lim_{t→∞}(v′′′(t) + K_{3}v′′(t) + K_{2}v′(t) + K_{1}v(t))

(42)

in ℝ ∪ {−∞}. We distinguish several cases and start by assuming

lim_{t→∞}(v′′′(t) + K_{3}v′′(t) + K_{2}v′(t))= lim_{t→∞}(v′′′(t) + K_{3}v′′(t) + K_{2}v′(t) + K_{1}v(t))= 0.

(A)

In this case, since (41) is strictly decreasing, one eventually has that v′′′(t) + K_{3}v′′(t) + K_{2}v′(t) + K_{1}v(t) > 0 so that by K_{1} < 0

v′′′(t) + K_{3}v′′(t) + K_{2}v′(t) > 0 for t large enough.  (43)

This shows that t ↦ v′′(t) + K_{3}v′(t) + K_{2}v(t) is eventually strictly increasing so that there exists

lim_{t→∞}(v′′(t) + K_{3}v′(t) + K_{2}v(t))= lim_{t→∞}(v′′(t) + K_{3}v′(t))∈ ℝ ∪ {+∞}.

If this limit were equal to +∞, then also +∞ = lim_{t→∞}(v′(t) + K_{3}v(t))= lim_{t→∞}v′(t), which contradicts the assumption. Hence

lim_{t→∞}(v′′(t) + K_{3}v′(t) + K_{2}v(t))= lim_{t→∞}(v′′(t) + K_{3}v′(t))∈ ℝ.  (44)

We distinguish three further subcases and start with discussing

lim_{t→∞}(v′′(t) + K_{3}v′(t))= lim_{t→∞}(v′′(t) + K_{3}v′(t) + K_{2}v(t))= 0.  (A1)

We want to show that lim_{t→∞} v′(t) exists and assume for contradiction that limsup_{t→∞} v′(t) > liminf_{t→∞} v′(t). Then we have a sequence (t_{k})_{k∈ℕ} with t_{k} → ∞ such that consecutively v′attains local maxima and local minima in t_{k} so that in particular v′′(t_{k}) = 0. By (A1) we may conclude that lim_{k→∞} v′(t_{k}) = 0. Since v′attains consecutively its local maxima and local minima in t_{k}, this would contradict limsup_{t→∞} v′(t) > liminf_{t→∞} v′(t). Hence we have proved that lim_{t→∞} v′(t) ∈ ℝ exists. Since lim_{t→∞} v(t) = 0, we get

lim_{t→∞}v′(t) = 0.  (45)

From this and assumption (A1), we directly obtain that also lim_{t→∞} v′′(t) = 0. From assumption (A) we then get that also lim_{t→∞} v′′′(t) = 0. For k ≥ 4, the differential equation (14) finally yields lim_{t→∞} v^{(k)}(t) = 0.

Next we consider the subcase

lim_{t→∞}(v′′(t) + K_{3}v′(t))= lim_{t→∞}(v′′(t) + K_{3}v′(t) + K_{2}v(t))= 2α > 0.  (A2)

In this case, one has that eventually v′′(t)+K_{3}v′(t) ≥ α. Multiplying this inequality by exp(K_{3}t) and integrating yields

v′(t) ≥ \frac{α}{K_{3}}+ o(1) near ∞.

But this is impossible in view of our assumption lim_{t→∞} v(t) = 0.

Finally we consider the subcase

lim_{t→∞}(v′′(t) + K_{3}v′(t))= lim_{t→∞}(v′′(t) + K_{3}v′(t) + K_{2}v(t))= 2α < 0.  (A3)

With precisely the same reasoning as in the previous case we come up with v′(t) ≤ \frac{α}{K_{3}} + o(1) for t → ∞ and again, we reach a contradiction.

Now we may consider the second main case

\begin{matrix}lim_{t→∞}(v′′′(t) + K_{3}v′′(t) + K_{2}v′(t)) & = & lim_{t→∞}(v′′′(t) + K_{3}v′′(t) + K_{2}v′(t) + K_{1}v(t)) \\ = & α ≠ 0. \end{matrix}

(B)

Then t ↦ v′′(t) + K_{3}v′(t) + K_{2}v(t) is monotone near ∞ and admits a limit β ∈ ℝ ∪ {±∞}. Hence, also lim_{t→∞} (v′′(t) + K_{3}v′(t))= β. If β = 0 we proceed as in Subcase (A1) and if β ≠ 0 as in Subcases (A2) and (A3). ⊓⊔

In order to exclude the possibility L = 0 in Proposition 5, for any global smooth positive solution v of (14) and any t ∈ ℝ, we define the energy function

E(t) := E_{v}(t) := \frac{1}{p + 1} v^{p+1} (t) − \frac{K_{0}}{2}v^{2}(t) − \frac{K_{2}}{2}|v′(t)|^{2}+ \frac{1}{2}|v′′(t)|^{2}.  (46)

We prove first that on consecutive extrema of v, the energy is decreasing. For the proof of the following lemma, the sign of the coefficients K_{1},K_{3} in front of the odd order derivatives in equation (14) is absolutely crucial.

Lemma 6 Assume that t_{0} < t_{1} and that v′(t_{0}) = v′(t_{1}) = 0. Then

E(t_{0}) ≥ E(t_{1}).

If v is not constant, then the inequality is strict.

Proof From the differential equation (14) we find:

E′(s) = v^{p}(s)v′(s) − K_{0}v(s)v′(s) − K_{2}v′(s)v′′(s) + v′′(s)v′′′(s)
= (v^{p}(s) − K_{0}v(s) − K_{2}v′′(s))v′(s) + v′′(s)v′′′(s)
= ( v^{(4)}(s) + K_{3}v′′′(s) + K_{1}v′(s)) v′(s) + v′′(s)v′′′(s).

Integrating by parts, this yields:

\begin{aligned}E(t_{1}) − E(t_{0}) & = & ∫_{t_{0}}^{t_{1}}E′(s)ds = −∫_{t_{0}}^{t_{1}}v′′′(s)v′′(s)ds − K_{3}∫_{t_{0}}^{t_{1}} |v′′(s)|^{2} ds \\ & & +K_{1}∫_{t_{0}}^{t_{1}} |v′(s)|^{2} ds + ∫_{t_{0}}^{t_{1}}v′′′(s)v′′(s)ds \\ & = & −K_{3}∫_{t_{0}}^{t_{1}} |v′′(s)|^{2} ds + K_{1}∫_{t_{0}}^{t_{1}} |v′(s)|^{2} ds ≤ 0, \end{aligned} (47)

since K_{3} > 0 and K_{1} < 0. If v is not a constant, the inequality is strict. ⊓⊔

Lemma 6 enables us to prove:

Lemma 7 Assume that v: ℝ → (0,∞) solves (14) and that lim_{t→−∞} v(t) = lim_{t→−∞} v′(t) =

lim_{t→−∞} v′′(t) = 0. Then it cannot happen that also lim_{t→∞} v(t) = 0.

Proof Consider the energy function E defined in (46). By assumption, we have E(−∞) = 0. Assume for contradiction that lim_{t→∞} v(t) = 0. Then, by Proposition 6 we see that also E(+∞) = 0. By Lemma 6, this shows that v is a constant, hence v(t) ≡ 0. In turn, this contradicts the assumption that v > 0. ⊓⊔

Remark 3 In terms of dynamical systems, Lemma 7 states that the regular point O does not allow for a homoclinic orbit of system (16).

We can now exclude the possibility L = 0 in Proposition 5:

Proposition 7 Let u be a smooth positive radial solution of (5) and let v be defined according to (12). Then the first alternative in Proposition 5 does not occur, i.e. it is impossible that lim_{t→∞} v(t) = 0.

Proof Since u is assumed to be smooth near 0 and since v is defined according to (12), we have that lim_{t→−∞} v(t) = lim_{t→−∞} v′(t) = lim_{t→−∞} v′′(t) = 0. If we also had lim_{t→∞} v(t) = 0, then v(t) ≡ 0 by Lemma 7. A contradiction! ⊓⊔

As before, we assume in what follows that u is a smooth positive radial solution of (5) and that v is defined according to (12) so that v solves (14). If v is eventually monotonous, then the claim of Theorem 3 follows directly from Propositions 5 and 7. So, it remains to consider solutions v, which oscillate infinitely many times near t = ∞, i.e. have an unbounded sequence of consecutive local maxima and minima. In the sequel we always restrict to this kind of solutions without explicit mention. We first prove the following inequalities:

Lemma 8

liminf_{t→∞} v(t) > 0;  (48)
∀t ∈ ℝ: 0 < v(t) < ( \frac{p+1}{2} )^{1/(p−1)} K_{0}^{1/(p−1)} ;  (49)
∀t ∈ ℝ: v′(t) < \frac{4}{p−1} ( \frac{p+1}{2} )^{1/(p−1)} K_{0}^{1/(p−1)} .  (50)

Proof Since v is defined by means of a smooth solution of (5), we have that E(−∞) = 0. Let \overline{t} be any local maximum for v. By Lemma 6 (with t_{0} = −∞ and t_{1} = \overline{t}) we immediately get (49).

Let {t_{k}}_{k∈ℕ} denote the sequence of consecutive positive critical points of v, starting with the first local maximum t_{1} in [0,∞) of v. In particular we have that v′(t_{k}) = 0 and {t_{k}}_{k∈ℕ} is a strictly increasing sequence, diverging to +∞. Since {E(t_{k})}_{k∈ℕ} is bounded from below, by Lemma 6 we see that

lim_{k→∞}E(t_{k}) =: −δ < 0

exists. Therefore, for k large enough we have

−\frac{δ}{2} ≥ \frac{1}{p + 1} v^{p+1} (t_{k}) − \frac{K_{0}}{2}v^{2}(t_{k})

which proves (48).

Finally, note that in view of (12), statement (i) of Theorem 1 becomes

v′(t) < \frac{4}{p − 1} v(t) for all t ∈ ℝ .

This inequality, combined with (49), proves (50). ⊓⊔

By (12), we see that (49) proves (8).

In the next four lemmas we prove some summability properties over ℝ of v and of its derivatives:

Lemma 9

∫_{ℝ} |v′(s)|^{2}ds + ∫_{ℝ} |v′′(s)|^{2}ds < ∞.

Proof We take the same sequence {t_{k}}_{k∈ℕ} as in the proof of Lemma 8. Since E(−∞) = 0, we obtain from (47) that for any k :

−K_{3}∫_{−∞}^{t_{k}} |v′′(s)|^{2} ds + K_{1}∫_{−∞}^{t_{k}} |v′(s)|^{2} ds
= E(t_{k}) ≥ min_{ν∈[0,∞)} ( \frac{1}{p + 1} ν^{p+1} − \frac{K_{0}}{2}ν^{2}) > −∞.

The statement follows by letting k → ∞ and using again that K_{3} > 0 and K_{1} < 0.

⊓⊔

Lemma 10

∫_{ℝ} |v′′′(s)|^{2}ds < ∞.

Proof Here the sequence {t_{k}}_{k∈ℕ} from the previous lemmas is no longer adequate. Instead, we choose a monotonically increasing diverging sequence {τ_{k}}_{k∈ℕ} of flex points of v such that v is there increasing. By Lemma 8 we may achieve:

τ_{k} > 0, τ_{k} ↗ ∞, 0 ≤ v′(τ_{k}) < \frac{4}{p − 1} ( \frac{p + 1}{2} ) ^{1/(p−1)} K_{0}^{1/(p−1)} , v′′(τ_{k}) = 0.

(51)

We multiply the differential equation (14) by v′′and integrate over (−∞,τ_{k}):

\begin{matrix}∫_{−∞}^{τ_{k}}( v^{(4)}(s) + K_{3}v′′′(s) + K_{2}v′′(s) + K_{1}v′(s) + K_{0}v(s)) v′′(s)ds \\ = ∫_{−∞}^{τ_{k}}v^{p}(s)v′′(s)ds. \end{matrix} (52)

We show that all the lower order terms remain bounded, when k → ∞:

|∫_{−∞}^{τ_{k}}v^{p}(s)v′′(s)ds|= |v^{p}(τ_{k})v′(τ_{k}) − p∫_{−∞}^{τ_{k}}v^{p−1} (s)|v′(s)|^{2}ds|≤ O(1) (53)

22  Filippo Gazzola, Hans-Christoph Grunau

by (49), (51) and Lemma 9. With the same argument, one also gets

|∫_{−∞}^{τ_{k}}v(s)v′′(s)ds|≤ O(1).  (54)

Hölder’s inequality and Lemma 9 imply

|∫_{−∞}^{τ_{k}}v′(s)v′′(s)ds|≤ O(1).  (55)

By our choice of τ_{k} (recall that v′′(τ_{k}) = 0), we obtain:

∫_{−∞}^{τ_{k}}v′′′(s)v′′(s)ds = [ \frac{1}{2}|v′′(s)|^{2}]_{−∞}^{τ_{k}} = 0.  (56)

Finally, integrating by parts and again by our choice of τ_{k}, we find: ∫_{−∞}^{τ_{k}}v^{(4)}(s)v′′(s)ds = [v′′′(s)v′′(s)]_{−∞}^{τ_{k}} − ∫_{−∞}^{τ_{k}}(v′′′(s))^{2} ds = −∫_{−∞}^{τ_{k}}(v′′′(s))^{2} ds.

(57)

Letting k → ∞, the statement follows directly from Lemma 9 and (52)–(57). ⊓⊔ Lemma 11

∫_{ℝ} |v^{(4)}(s)|^{2}ds < ∞.

Proof In view of Lemmas 8–10 we may find a sequence {s_{k}} such that

lim_{k→∞}s_{k} = ∞, v(s_{k}) = O(1), lim_{k→∞}v′(s_{k}) = lim_{k→∞}v′′(s_{k}) = lim_{k→∞}v′′′(s_{k}) = 0.

We multiply the equation (14) by v^{(4)} and integrate over (−∞,s_{k}]:

\begin{matrix}∫_{−∞}^{s_{k}} ( v^{(4)}(s))^{2} ds \\ = ∫_{−∞}^{s_{k}} (v^{p}(s) − K_{0}v(s) − K_{1}v′(s) − K_{2}v′′(s) − K_{3}v′′′(s)) v^{(4)}(s)ds. \end{matrix}  (58)

By using Lemmas 8–10 and arguing as in the previous proofs we obtain:

∫_{−∞}^{s_{k}} v^{(4)}(s)v′′′(s)ds = [ \frac{1}{2}|v′′′(s)|^{2}]_{−∞}^{s_{k}} = o(1);

∫_{−∞}^{s_{k}} v^{(4)}(s)v′′(s)ds = o(1) − ∫_{−∞}^{s_{k}} |v′′′(s)|^{2}ds = O(1);

∫_{−∞}^{s_{k}} v^{(4)}(s)v′(s)ds = o(1) − ∫_{−∞}^{s_{k}} v′′′(s)v′′(s)ds = o(1);

∫_{−∞}^{s_{k}} v^{(4)}(s)v(s)ds = o(1) − ∫_{−∞}^{s_{k}} v′′′(s)v′(s)ds = o(1) + ∫_{−∞}^{s_{k}} |v′′(s)|^{2}ds = O(1); |∫_{−∞}^{s_{k}} v^{(4)}(s)v^{p}(s)ds|= |o(1) − p∫_{−∞}^{s_{k}} v′′′(s)v^{p−1} (s)v′(s)ds|

≤ o(1) + C ( ∫_{−∞}^{s_{k}} |v′′′(s)|^{2}ds)^{1/2}( ∫_{−∞}^{s_{k}} |v′(s)|^{2}ds)^{1/2} ≤ O(1).

Inserting all these estimates into (58), the claim follows. ⊓⊔ Lemma 12

∫_{ℝ} v^{2}(s)(v^{p−1} (s) − K_{0})^{2} ds < ∞.

Proof From the differential equation (14), we conclude

( v^{(4)}(s) + K_{3}v′′′(s) + K_{2}v′′(s) + K_{1}v′(s)) ^{2} = v^{2}(s) (v^{p−1} (s) − K_{0})^{2}.

The statement follows now immediately from Lemmas 9–11. ⊓⊔

The proof of Theorem 3 will be completed by showing:

Proposition 8 We assume that u is an entire smooth positive radial solution of (5), that v is defined according to (12) and that w = (w_{1},w_{2},w_{3},w_{4}) is the corresponding solution of system (16). We assume further that v = w_{1} has an unbounded sequence of consecutive local maxima and minima near t = ∞. Then it follows that

lim_{t→∞}w(t) = P,  (59)

where P is the “singular” steady solution of system (16). In particular, lim_{t→∞}v(t) =

K_{0}^{1/(p−1)} .

Proof By Lemmas 8–12, we can find a sequence {σ_{k}}_{k∈ℕ} such that

σ_{k+1} > σ_{k}, lim_{k→∞}(σ_{k+1} − σ_{k}) = 0, lim_{k→∞}σ_{k} = ∞, lim_{k→∞}w(σ_{k}) = P.

If (59) were not true, then there would exist a subsequence {k_{ℓ}}_{ℓ∈ℕ} with the following properties: for any ε > 0 there exists ℓ_{ε} such that for all ℓ ≥ ℓ_{ε} one has that

|w(σ_{k_{ℓ}}) − P| < ε, σ_{k_{ℓ}+1} − σ_{k_{ℓ}} < ε^{2}

and moreover that there exists θ_{ℓ} ∈ (σ_{k_{ℓ}},σ_{k_{ℓ}+1}) with

|w(s) − P| < 2ε ∀s ∈ (σ_{k_{ℓ}},θ_{ℓ}) and |w(θ_{ℓ}) − P| = 2ε .

The triangle inequality shows that |w(θ_{ℓ}) − w(σ_{k_{ℓ}})| > ε, hence

\frac{1}{θ_{ℓ}− σ_{k_{ℓ}}} |w(θ_{ℓ}) − w(σ_{k_{ℓ}})| > \frac{1}{ε} .

By the mean value Theorem we conclude that

\frac{1}{ε} < \frac{1}{θ_{ℓ}− σ_{k_{ℓ}}} |∫_{σ_{k_{ℓ}}}^{θ_{ℓ}}w′(s)ds| ≤ \frac{1}{θ_{ℓ}− σ_{k_{ℓ}}} ∫_{σ_{k_{ℓ}}}^{θ_{ℓ}} |w′(s)| ds

so that there exists τ_{ℓ} ∈ [σ_{k_{ℓ}},θ_{ℓ}] with

|w′(τ_{ℓ})| > \frac{1}{ε}.

Since ε is arbitrarily small, |w(σ_{k_{ℓ}}) − P| < ε, |w(τ_{ℓ}) − P| ≤ 2ε and since w solves system (16), this is impossible for large enough ℓ. A contradiction is achieved, thereby proving (59). ⊓⊔ 7 Proof of Theorem 4

We start this section by recalling the

Definition 1 An m-th order ordinary linear differential equation is said to be nonoscillatory on an interval I ⊂ ℝ if every nontrivial solution has only a finite number of zeros in I. It is called disconjugate on I, if every nontrivial solution has at most (m − 1) zeros in I (counting multiplicities).

Let v be defined by (12) and let ϕ be as in (19). Assume that n ≥ 13 and p > p_{c}. Then, by Proposition 2, there exists ε_{0} > 0 such that the equation ϕ(x) = (p − 1)K_{0} + ε admits four real solutions for all ε ∈ (0,ε_{0}). From now on, we fix ε = ε_{0}/2 so that the equation

ψ^{(4)}(t)+K_{3}ψ′′′(t)+K_{2}ψ′′(t)+K_{1}ψ′(t)+K_{0}ψ(t)−(pK_{0}+ ε) ψ(t) = 0 t ∈ ℝ ,

is non-oscillatory in ℝ. In other words it has four linearly independent solutions of “exponential type” ψ_{i}(t) = e^{μ_{i}t} (i = 1,...,4) for some μ_{i}’s being small perturbations of the ν_{i}’s which are all real numbers. Moreover, the differential operator

L_{0} := ( \frac{d}{dt}) ^{4}+ K_{3}( \frac{d}{dt}) ^{3}+ K_{2}( \frac{d}{dt}) ^{2}+ K_{1}( \frac{d}{dt}) + K_{0}

is disconjugate, since this is the biharmonic operator, transformed by means of (12). By differentiating (14), we obtain

L_{0}ψ(t) − pv^{p−1} (t)ψ(t) = L_{0}ψ(t) + p(t)ψ(t) = 0  t ∈ ℝ ,  (60)

where ψ(t) := v′(t) and p(t) := −pv^{p−1} (t). According to Theorem 3 we know that

∃T > 0 ∀t > T : −(pK_{0}+ ε) < p(t) < 0 .

Therefore, the equation (60) is between a disconjugate and a non-oscillatory equation so that it is also non-oscillatory in view of [4, Corollary 1]. This shows that v′(t) = ψ(t) cannot change sign infinitely many times, and therefore that v(t) − K_{0}^{1/(p−1)} does not change sign infinitely many times. ⊓⊔

8 Proof of Propositions 2, 3, 4

Proof of Proposition 2. We first observe that (11) is equivalent to

N_{1} < 0  (61)

and that (11) implies

N_{2}−N_{1}^{2} = 4(n−2)(p−1)^{2}+16(n−4)(p−1)−64 > 4(n−2)(p−1)^{2}+64 > 0.

(62)

Next, we show that

N_{3} > \frac{(N_{2}− N_{1}^{2})^{2}}{16} .  (63)

Indeed, by exploiting again (11), we have:

N_{3} − \frac{(N_{2}− N_{1}^{2})^{2}}{16}
= 8(n − 2)(n − 4)(p − 1)^{4}+ 16(n^{2}− 10n + 20)(p − 1)^{3}
−128(n − 4)(p − 1)^{2}+ 256(p − 1)
> 16(n^{2}− 6n + 12)(p − 1)^{3}− 128(n − 4)(p − 1)^{2}+ 256(p − 1)
= 64(p − 1)^{3}+ 16(n − 2)(n − 4)(p − 1)^{3}− 128(n − 4)(p − 1)^{2}+ 256(p − 1)
> 64(p − 1)^{3}+ 128(n − 2)(p − 1)^{2}− 128(n − 4)(p − 1)^{2}+ 256(p − 1)
= 64(p − 1)^{3}+ 256(p − 1)^{2}+ 256(p − 1) = 64(p + 1)^{2}(p − 1) > 0.

In particular, (63) implies that N_{3} > 0. In turn, together with the fact that N_{2} > N_{1}^{2}, this shows that \sqrt{N_{2}+ 4\sqrt{N_{3}}} > |N_{1}| which proves statement (i) in Proposition 2.

In order to discuss the stability properties of the eigenvalues ν_{3} and ν_{4} we introduce the function

\begin{aligned}N_{4} := & 16N_{3}− N_{2}^{2} = −(n − 4)(n^{3}− 4n^{2}− 128n + 256)(p − 1)^{4} \\ & +128(3n − 8)(n − 6)(p − 1)^{3}+ 256(n^{2}− 18n + 52)(p − 1)^{2} \\ & −2048(n − 6)(p − 1) + 4096. \end{aligned}  (64)

For 1.939447811...< n < 12.56534446..., the first coefficient in (64) is positive, so that assuming

5 ≤ n ≤ 12,

we obtain by means of (11):

N_{4} = −(n − 4)(n^{3}− 4n^{2}− 128n + 256)(p − 1)^{4}+ 128(3n − 8)(n − 6)(p − 1)^{3}
+256(n^{2}− 18n + 52)(p − 1)^{2}− 2048(n − 6)(p − 1) + 4096
> −8(n^{3}− 4n^{2}− 128n + 256)(p − 1)^{3}+ 128(3n − 8)(n − 6)(p − 1)^{3}
+256(n^{2}− 18n + 52)(p − 1)^{2}− 2048(n − 6)(p − 1) + 4096
= 64n^{2}(p − 1)^{3}− 8(n − 4)(n^{2}− 40n + 128)(p − 1)^{3}
+256(n^{2}− 18n + 52)(p − 1)^{2}− 2048(n − 6)(p − 1) + 4096
> 64n(n − 4)(p − 1)^{3}− 64(n^{2}− 40n + 128)(p − 1)^{2}
+256(n^{2}− 18n + 52)(p − 1)^{2}− 2048(n − 6)(p − 1) + 4096
> 512n(p − 1)^{2}+ 64(n − 4)(3n − 20)(p − 1)^{2}− 2048(n − 6)(p − 1) + 4096
= 2048(p − 1)^{2}+ 192(n − 4)^{2}(p − 1)^{2}− 2048(n − 6)(p − 1) + 4096
> 2048(p − 1)^{2}+ 1536(n − 4)(p − 1) − 2048(n − 6)(p − 1) + 4096
= 2048(p − 1)^{2}− 512(n − 12)(p − 1) + 4096 > 0,

since n ≤ 12. This, together with (61), proves statement (ii) in Proposition 2.

In order to prove statement (iii), we assume that n ≥ 13 and we study N_{4} = N_{4}(n,p) as a function of p. We compute its second derivative with respect to p:

\begin{aligned}−\frac{∂^{2}N_{4}}{∂p^{2}} = & 12(n − 4)(n^{3}− 4n^{2}− 128n + 256)(p − 1)^{2} \\ & −768(3n − 8)(n − 6)(p − 1) − 512(n^{2}− 18n + 52). \end{aligned}

This is a quadratic function of p which tends to +∞ as p → +∞. Its minimum is smaller than the Sobolev exponent (n + 4)/(n − 4) if and only if

0 < (n^{3}− 4n^{2}− 128n + 256) − 4(3n − 8)(n − 6) = (n − 18)(n^{2}+ 2n + 12) + 280.

This is certainly true for n ≥ 18, while for n = 13,...,17, we have \frac{∂^{2}N_{4}}{∂p^{2}}(n, \frac{n+4}{n−4} ) <

0. Summarizing, for p > (n + 4)/(n − 4), \frac{∂^{2}N_{4}}{∂p^{2}} has at most one zero. Therefore,

for p > \frac{n+4}{n−4} , either

p ↦ N_{4}(n,p) is always concave or it is first convex and then concave. (65)

Moreover, since the first coefficient in (64) is now negative (because n ≥ 13), we have

lim_{p→∞} N_{4}(n,p) = −∞  ∀n ≥ 13.  (66)

Finally, note that

N_{4}( n, \frac{n + 4}{n − 4} ) = \frac{32768n^{2}}{(n − 4)^{3}} > 0 and \frac{∂N_{4}}{∂p}( n, \frac{n + 4}{n − 4} ) = \frac{20480n^{2}}{(n − 4)^{2}} > 0. (67)

By (65)-(66)-(67) there exists a unique p_{c} > (n + 4)/(n − 4) such that

N_{4}(n,p) > 0 for all p < p_{c} , N_{4}(n,p_{c}) = 0 , N_{4}(n,p) < 0 for all p > p_{c}.

In order to prove that n ↦ p_{c} is strictly decreasing we calculate \frac{dp_{c}}{dn} by means of implicit differentiation and note first that the previous reasoning gives

\frac{∂N_{4}}{∂p}(n,p_{c}) < 0.  (68)

We proceed by calculating

\frac{∂N_{4}}{∂n} = −(4n^{3}− 24n^{2}− 224n + 768) (p − 1)^{4}+ 256(3n − 13)(p − 1)^{3}
+512(n − 9)(p − 1)^{2}− 2048(p − 1),
\frac{∂^{2}N_{4}}{∂n^{2}} = −(12n^{2}− 48n − 224)(p − 1)^{4}+ 768(p − 1)^{3}+ 512(p − 1)^{2},
\frac{∂^{3}N_{4}}{∂n^{3}} = −24(n − 2)(p − 1)^{4};

the latter being always negative for n > 2. Keeping p > 1 fixed, we consider now n ↦ N_{4}(n,p). First we calculate n > 4 such that p = (n + 4)/(n − 4), i.e. n = 4 + \frac{8}{p−1} . Negativity of \frac{∂^{3}N_{4}}{∂n^{3}} shows that beyond n = 4 + \frac{8}{p−1} this function is either always concave or convex first and then always concave. On the mentioned particular value we have by (67) that

N_{4}( 4 + \frac{8}{p − 1} ,p) > 0

and moreover, we find that

\frac{∂N_{4}}{∂n}( 4 + \frac{8}{p − 1} ,p) = 32( 4 + \frac{8}{p − 1} ) ( 2 + \frac{8}{p − 1} ) (p − 1)^{4} > 0.

Since N_{4}(n,p_{c}) = 0, this shows that also

\frac{∂N_{4}}{∂n}(n,p_{c}) < 0.  (69)

By implicit differentiation we conclude from (68) and (69) that

\frac{dp_{c}}{dn} = − \frac{\frac{∂N_{4}}{∂n}(n,p_{c})}{\frac{∂N_{4}}{∂p}(n,p_{c})} < 0.

Finally one reads directly from the form of N_{4} that for any p_{0} > 1, N_{4}(n,p_{0}) becomes negative, provided n is chosen large enough. This shows that \frac{n+4}{n−4} < p_{c} < p_{0} for n large enough, i.e.

lim_{n→∞} p_{c} = 1.

The proof of statement (iii) in Proposition 2 is so complete. ⊓⊔

Proof of Proposition 3. This proof is an extension of the one of Proposition 8.

It is enough to consider a solution v which converges eventually monotonically to K_{0}^{1/(p−1)} . The differential equation (14) shows that v^{(4)}(t) + K_{3}v′′′(t) + K_{2}v′′(t) + K_{1}v′(t) eventually has a fixed sign. Let us now consider ṽ(t) := v(t) − K_{0}^{1/(p−1)} . Then

lim_{t→∞} ṽ(t) = 0

and ṽ^{(4)}(t) + K_{3}ṽ′′′ (t) + K_{2}ṽ′′(t) + K_{1}ṽ′(t) is also eventually of fixed sign. This shows that

lim_{t→∞}(ṽ′′′ (t) + K_{3}ṽ′′(t) + K_{2}ṽ′(t) + K_{1}ṽ(t))= lim_{t→∞}(ṽ′′′ (t) + K_{3}ṽ′′(t) + K_{2}ṽ′(t))

exists in ℝ ∪ {±∞}. Now we may proceed precisely as in Proposition 6. ⊓⊔ Proof of Proposition 4. It is enough to show that M_{P} has no eigenvectors with first component equal to 0. Assume for contradiction that associated to some eigenvalue ν, there exists (a,b,c) ≠ (0,0,0) such that

( \begin{matrix}\frac{4}{p−1} − ν & 1 & 0 & 0 \\ 0 & −ν & 1 & 0 \\ 0 & 0 & −ν & 1 \\ pK_{0} & C_{2} & C_{3} & C_{4}− ν \end{matrix} ) ( \begin{matrix}0 \\ a \\ b \\ c \end{matrix}) = ( \begin{matrix}0 \\ 0 \\ 0 \\ 0 \end{matrix} ) .

This is clearly impossible. ⊓⊔

Acknowledgements We are grateful to the referees for their very careful reading of the manuscript and for their helpful remarks.

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