Supercritical biharmonic equations with power-type nonlinearity

Alberto Ferrero Dipartimento di Matematica Università di Milano-Bicocca

Via Cozzi 53 20125 Milano (Italy)

Hans-Christoph Grunau Fakultät für Mathematik Otto-von-Guericke-Universität Postfach 4120 39016 Magdeburg (Germany)

Paschalis Karageorgis School of Mathematics Trinity College Dublin 2 (Ireland)

Abstract

The biharmonic supercritical equation Δ^{2}u = |u|^{p−1}u, where n > 4 and p > (n + 4)/(n − 4), is studied in the whole space ℝ^{n} as well as in a modified form with λ(1 + u)^{p} as right-hand-side with an additional eigenvalue parameter λ > 0 in the unit ball, in the latter case together with Dirichlet boundary conditions. As for entire regular radial solutions we prove oscillatory behaviour around the explicitly known radial singular solution, provided p ∈ ((n + 4)/(n − 4), p_{c}), where p_{c} ∈ ((n + 4)/(n − 4), ∞] is a further critical exponent, which was introduced in a recent work by Gazzola and the second author. The third author proved already that these oscillations do not occur in the complementing case, where p ≥ p_{c}.

Concerning the Dirichlet problem we prove existence of at least one singular solution with corresponding eigenvalue parameter. Moreover, for the extremal solution in the bifurcation diagram for this nonlinear biharmonic eigenvalue problem, we prove smoothness as long as p ∈ ((n+4)/(n− 4), p_{c}).

1 Introduction and main results

In the present paper we consider qualitative properties of entire radial solutions (defined and regular in the whole space) 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}. An important role is played by the explicitly known entire solution

u_{s}(r) = K_{0}^{1/(p−1)}r^{−4/(p−1)},  (2)

where

K_{0} = \frac{4}{p − 1} (\frac{4}{p − 1} + 2) (n − 2 − \frac{4}{p − 1}) (n − 4 − \frac{4}{p − 1}) .  (3)

It was shown in [5, 8] that positive regular entire solutions to (1) exist and that asymptotically they behave like the singular solution u_{s}:

lim_{r→∞} \frac{u(r)}{u_{s}(r)} = 1.

Moreover, for n > 12 a further critical exponent p_{c} ∈ (\frac{n+4}{n−4}, ∞) was introduced being in that interval the unique solution of the following polynomial equation:

p_{c} · \frac{4}{p_{c} − 1} · (\frac{4}{p_{c} − 1} + 2) · (n − 2 − \frac{4}{p_{c} − 1}) · (n − 4 − \frac{4}{p_{c} − 1}) = \frac{n^{2}(n − 4)^{2}}{16} .  (4)

The third author [9] proved in particular that in the “supercritical case”, i.e

p ≥ p_{c}

the convergence of u to u_{s} is monotone, i.e. ∀r : u(r) < u_{s}(r). Here, we study the reverse case:

Theorem 1. Let p_{c} ∈ ((n + 4)/(n − 4), ∞) be the number, which is defined by (4) for n ≥ 13. We assume that

\frac{n + 4}{n − 4} < p < p_{c} if n ≥ 13, \frac{n + 4}{n − 4} < p < ∞ if 5 ≤ n ≤ 12.

Let r ↦ u(r) be a radial entire solution to (1). Then, as r → ∞, u(r) oscillates infinitely many times around the singular solution u_{s}(r).

We study also existence of singular solutions as well as qualitative properties of positive solutions of the corresponding Dirichlet problem

\{ \begin{matrix}Δ^{2}u = λ(1 + u)^{p} & in B, \\ u > 0 & in B, \\ u = |∇u| = 0 & on ∂B, \end{matrix}  (5)

where B ⊂ ℝ^{n} is the unit ball, λ > 0 is an eigenvalue parameter and again n ≥ 5 and p > \frac{n+4}{n−4}. In [7] (see also [2]) it was proved that there exists an extremal parameter λ^{∗} such that for λ ∈ [0, λ^{∗}) one has a minimal solution which is regular, while not even a weak solution does exist for λ > λ^{∗}. On the extremal parameter λ = λ^{∗}, an extremal solution u^{∗} ∈ H_{0}^{2}(B) ∩ L^{p}(B) exists as monotone limit of the minimal solutions. It is expected that also in the Dirichlet problem, a singular (i.e. unbounded) solution u_{σ} corresponding to a suitable singular parameter λ_{σ} exists and will play an important role as far as the shape of the bifurcation diagram for (5) is concerned. However, in [7] we had to leave open even the existence of a singular solution which will be proved in the present paper:

Theorem 2. Let n > 4 and p > (n + 4)/(n − 4). Then, there exists a parameter λ_{σ} > 0 such that for λ = λ_{σ}, problem (5) admits a radial singular solution.

Moreover, in [7] we left open whether the extremal solution u^{∗} introduced above is singular (unbounded) or regular (bounded). The corresponding question has been settled for the exponential nonlinearity by Davila, Dupaigne, Guerra and Montenegro [6] thereby developing the previous work [1]. Here, taking advantage of an idea in [6], we prove regularity of the extremal solution of the problem with power-type nonlinearity in the “subcritical” range.

Theorem 3. Let p_{c} ∈ ((n + 4)/(n − 4), ∞) be the number, which is defined by (4) for n ≥ 13. We assume that

\frac{n + 4}{n − 4} < p < p_{c} if n ≥ 13, \frac{n + 4}{n − 4} < p < ∞ if 5 ≤ n ≤ 12.

Let u^{∗} ∈ H_{0}^{2}(B)∩L^{p}(B) be the extremal radial solution of (5) corresponding to the extremal parameter λ^{∗}, which is obtained as monotone limit of the minimal regular solutions for λ ↗ λ^{∗}. Then, u^{∗} is regular.

Related results for the corresponding second order problems were obtained e.g. in [3, 4, 10, 11].

2 Entire solutions: The corresponding autonomous system

Here we study qualitative properties of entire radial solutions r ↦ u(r) to (1) and shall prove Theorem 1. We put

v(s) := e^{4s/(p−1)}u(e^{s}) (s ∈ ℝ),  u(r) = r^{−4/(p−1)}v(log r) (r > 0).  (6)

According to [8, 9], (1) is then equivalent to

(∂_{s} − \frac{4}{p − 1} + n − 4) (∂_{s} − \frac{4}{p − 1} + n − 2) (∂_{s} − \frac{4}{p − 1} − 2) (∂_{s} − \frac{4}{p − 1}) v(s) = |v(s)|^{p−1}v(s),

(7)

s ∈ ℝ. In order to write this as an autonomous system, we define

\{ \begin{matrix}w_{1}(s) & = & v(s) \\ w_{2}(s) & = & (∂_{s} − \frac{4}{p−1}) w_{1}(s) \\ w_{3}(s) & = & (∂_{s} − \frac{4}{p−1} − 2) w_{2}(s) \\ w_{4}(s) & = & (∂_{s} − \frac{4}{p−1} + n − 2) w_{3}(s). \end{matrix}  (8)

Equation (7) is equivalent to the following system:

\{ \begin{matrix}w′_{1}(s) & = & \frac{4}{p−1}w_{1} + w_{2}, \\ w′_{2}(s) & = & (\frac{4}{p−1} + 2) w_{2} + w_{3}, \\ w′_{3}(s) & = & (\frac{4}{p−1} − (n − 2)) w_{3} + w_{4}, \\ w′_{4}(s) & = & |w_{1}(s)|^{p−1}w_{1}(s) + (\frac{4}{p−1} − (n − 4)) w_{4}. \end{matrix}  (9)

In order to perform the stability analysis around the singular solution u_{s}(r) = K_{0}^{1/(p−1)}r^{−4/(p−1)}, i.e. v(s) = K_{0}^{1/(p−1)}, we have to linearize (9) around the vector

w^{(0)} := K_{0}^{1/(p−1)} (1, −\frac{4}{p − 1}, \frac{4}{p − 1} (\frac{4}{p − 1} + 2) , (n − 2 − \frac{4}{p − 1}) \frac{4}{p − 1} (\frac{4}{p − 1} + 2))

and come up with the system w′(s) = M ∘ w(s) where

M := (\begin{matrix}\frac{4}{p−1} & 1 & 0 & 0 \\ 0 & \frac{4}{p−1} + 2 & 1 & 0 \\ 0 & 0 & \frac{4}{p−1} − (n − 2) & 1 \\ pK_{0} & 0 & 0 & \frac{4}{p−1} − (n − 4) \end{matrix}) .

The corresponding characteristic polynomial is given by

P (ν) = (ν − \frac{4}{p − 1} + n − 4) (ν − \frac{4}{p − 1} + n − 2) (ν − \frac{4}{p − 1} − 2) (ν − \frac{4}{p − 1}) − pK_{0}.

According to [8], 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.

One has ν_{1}, ν_{2} ∈ ℝ and ν_{2} < 0 < ν_{1}. For any 5 ≤ n ≤ 12 we have ν_{3}, ν_{4} ∉ ℝ and Re ν_{3} = Re ν_{4} < 0. For any n ≥ 13 and p < p_{c}, ν_{3}, ν_{4} ∉ ℝ and Re ν_{3} = Re ν_{4} < 0, while ν_{3}, ν_{4} ∈ ℝ and ν_{4} ≤ ν_{3} < 0 if p ≥ p_{c}. In any case,

ν_{2} < Re ν_{3/4} < 0 < ν_{1}.

The stable manifold of w^{(0)}, where the trajectory of any w corresponding to an entire regular solution is contained in, is tangential to the span of the eigenvectors corresponding to ν_{2}, ν_{3}, ν_{4}. In [8] the following strategy to prove Theorem 1 was outlined: in the “subcritical” setting \frac{n+4}{n−4} < p < p_{c}, any such trajectory oscillates around w^{(0)} infinitely many times except those which are tangential to the eigenvector corresponding to ν_{2}. We show that the latter can not correspond to an entire regular solution.

Proposition 1. Let w( .) be a solution of (9) in the stable manifold of w^{(0)} being tangential to the eigenvector corresponding to ν_{2}. Then the corresponding solution u of (1) is singular or even not defined for all r > 0.

In order to prove this proposition we need the following crucial observation on the sign of the components of an eigenvector corresponding to ν_{2}:

Lemma 1. One eigenvector of M corresponding to ν_{2} is given by t = (t_{1}, t_{2}, t_{3}, t_{4}) with

t_{1} = 1 > 0,
t_{2} = (ν_{2} − \frac{4}{p − 1}) < 0,
t_{3} = (ν_{2} − 2 − \frac{4}{p − 1}) (ν_{2} − \frac{4}{p − 1}) > 0,
t_{4} = (ν_{2} + n − 2 − \frac{4}{p − 1}) (ν_{2} − 2 − \frac{4}{p − 1}) (ν_{2} − \frac{4}{p − 1}) < 0.

Proof. Since ν_{2} < 0 we only have to show that

0 > ν_{2} + n − 2 − \frac{4}{p − 1} = \frac{n}{2} − \frac{1}{2(p − 1)}\sqrt{N_{2} + 4\sqrt{N_{3}}}  (10)

the latter being equivalent to proving that

N_{3} > (n − 2)^{2}(p − 1)^{4}.

Indeed, by using the supercriticality assumption (n − 4)(p − 1) > 8, we have

N_{3} − (n − 2)^{2}(p − 1)^{4} = 8(n − 2)(n − 4)(p − 1)^{4} + 8(3n^{2} − 26n + 48)(p − 1)^{3}
+16(n^{2} − 18n + 52)(p − 1)^{2} − 128(n − 6)(p − 1) + 256
= 8p(p + 1)((n − 2)(p − 1) − 4)((n − 4)(p − 1) − 4) > 0.

This proves (10) and hence the lemma. □ Proof of Proposition 1. Let w( .) be a solution to (9) being tangential for s → ∞ to the eigenvector t from the previous lemma. We may assume that w( .) exists on the whole real line ℝ because otherwise, nothing is to be proved. We put z_{1}(s) = w_{1}(s) − w_{1}^{(0)} and further

z_{1}(s) = w_{1}(s) − w_{1}^{(0)} = v(s) − K_{0}^{1/(p−1)},
z_{2}(s) = w_{2}(s) − w_{2}^{(0)} = (∂_{s} − \frac{4}{p − 1}) z_{1}(s),
z_{3}(s) = w_{3}(s) − w_{3}^{(0)} = (∂_{s} − \frac{4}{p − 1} − 2) z_{2}(s),
z_{4}(s) = w_{4}(s) − w_{4}^{(0)} = (∂_{s} − \frac{4}{p − 1} + n − 2) z_{3}(s),

so that

(∂_{s} − \frac{4}{p − 1} + n − 4) z_{4}(s) = |v(s)|^{p−1}v(s) − K_{0}^{p/(p−1)} = |w_{1}(s)|^{p−1}w_{1}(s) − |w_{1}^{(0)}|^{p−1}w_{1}^{(0)}.

Writing this more systematically yields

\{ \begin{matrix}z′_{1}(s) & = & \frac{4}{p−1}z_{1}(s) + z_{2}(s), \\ z′_{2}(s) & = & (\frac{4}{p−1} + 2) z_{2}(s) + z_{3}(s), \\ z′_{3}(s) & = & (\frac{4}{p−1} − (n − 2)) z_{3}(s) + z_{4}(s), \\ z′_{4}(s) & = & |w_{1}(s)|^{p−1}w_{1}(s) − |w_{1}^{(0)}|^{p−1}w_{1}^{(0)} + (\frac{4}{p−1} − (n − 4)) z_{4}(s). \end{matrix}  (11)

According to whether z( .) approaches the origin from “above” or “below” we distinguish two cases. First case. There exists s_{0} large enough such that

z_{1}(s_{0}) > 0, z_{2}(s_{0}) < 0, z_{3}(s_{0}) > 0, z_{4}(s_{0}) < 0.  (12)

On any interval [s, s_{0}] where z_{1}( .) = w_{1}( .) − w_{1}^{(0)} ≥ 0, we must then have

(∂_{s} + (n − 4) − \frac{4}{p − 1}) z_{4}(s) = |w_{1}(s)|^{p−1}w_{1}(s) − |w_{1}^{(0)}|^{p−1}w_{1}^{(0)} ≥ 0.

This makes e^{((n−4)− \frac{4}{p−1} )s}z_{4}(s) increasing on [s, s_{0}], and so (12) implies that

e^{((n−4)− \frac{4}{p−1} )s}z_{4}(s) ≤ e^{((n−4)− \frac{4}{p−1} )s_{0}}z_{4}(s_{0}) < 0

on [s, s_{0}]. In particular, z_{4}(s) < 0 throughout the interval, and we have

(∂_{s} + (n − 2) − \frac{4}{p − 1}) z_{3}(s) = z_{4}(s) < 0.

This makes e^{((n−2)− \frac{4}{p−1} )s}z_{3}(s) decreasing on [s, s_{0}], so we similarly find that e^{((n−2)− \frac{4}{p−1} )s}z_{3}(s) ≥ e^{((n−2)− \frac{4}{p−1} )s_{0}}z_{3}(s_{0}) > 0 by (12). Since (∂_{s} − 2 − \frac{4}{p−1}) z_{2}(s) = z_{3}(s) > 0, the exact same argument leads us to

e^{(−2− \frac{4}{p−1} )s}z_{2}(s) ≤ e^{(−2− \frac{4}{p−1} )s_{0}}z_{2}(s_{0}) < 0

by (12), hence (∂_{s} − \frac{4}{p−1}) z_{1}(s) = z_{2}(s) < 0 and we finally get

e^{− \frac{4}{p−1} s}z_{1}(s) ≥ e^{− \frac{4}{p−1} s_{0}}z_{1}(s_{0}) > 0.

That is, z_{1}(s) > 0 on any interval [s, s_{0}] where z_{1}(s) ≥ 0, so it is impossible for z_{1}(s) to become 0 at some s < s_{0}. Hence ∀s ≤ s_{0} : z_{1}(s) > 0. For the original solution this means that for r ≤ r_{0}, u( .) lies above the singular solution. This means that u( .) itself is singular at r = 0.

Second case. There exists s_{0} large enough such that

z_{1}(s_{0}) < 0, z_{2}(s_{0}) > 0, z_{3}(s_{0}) < 0, z_{4}(s_{0}) > 0.  (13)

On any interval [s, s_{0}] where z_{1}( .) = w_{1}( .) − w_{1}^{(0)} ≤ 0, we must then have

(∂_{s} + (n − 4) − \frac{4}{p − 1}) z_{4}(s) = |w_{1}(s)|^{p−1}w_{1}(s) − |w_{1}^{(0)}|^{p−1}w_{1}^{(0)} ≤ 0.

This makes e^{((n−4)− \frac{4}{p−1} )s}z_{4}(s) decreasing on [s, s_{0}], and so (13) implies that

e^{((n−4)− \frac{4}{p−1} )s}z_{4}(s) ≥ e^{((n−4)− \frac{4}{p−1} )s_{0}}z_{4}(s_{0}) > 0

on [s, s_{0}]. In particular, z_{4}(s) > 0 throughout the interval, and we have

(∂_{s} + (n − 2) − \frac{4}{p − 1}) z_{3}(s) = z_{4}(s) > 0.

This makes e^{((n−2)− \frac{4}{p−1} )s}z_{3}(s) increasing on [s, s_{0}], so we similarly find that

e^{((n−2)− \frac{4}{p−1} )s}z_{3}(s) ≤ e^{((n−2)− \frac{4}{p−1} )s_{0}}z_{3}(s_{0}) < 0  (14)

by (13). Following this approach, as in the first case, we eventually get

z_{4}(s) > 0, z_{3}(s) < 0, z_{2}(s) > 0, z_{1}(s) < 0  (15)

on any interval [s, s_{0}] where z_{1}(s) ≤ 0, so it is impossible for z_{1}(s) to become 0 at some s < s_{0}. Hence ∀s ≤ s_{0} : z_{1}(s) < 0, i.e. the corresponding u( .) is always below the singular solution. In order to prove that u( .) itself is singular also in this case, we show that z_{1}(s) → −∞ for s → −∞. Since ∀s ≤ s_{0} : z_{1}(s) < 0, we have that (14) holds true for all s ≤ s_{0}. Referring to [7, Proposition 1] would already show that also v and so u cannot be bounded. However, here it is quite easy to show this directly. For some suitable constant δ_{1} > 0 one has:

∂_{s} (e^{−(2+ \frac{4}{p−1} )s}z_{2}(s)) = e^{−(2+ \frac{4}{p−1} )s}z_{3}(s) ≤ −δ_{1}e^{−ns}

because of (14), and this implies that

e^{−(2+ \frac{4}{p−1} )s}z_{2}(s) ≥ \frac{δ_{1}}{n} e^{−ns} − \frac{δ_{1}}{n} e^{−ns_{0}} + e^{−(2+ \frac{4}{p−1} )s_{0}}z_{2}(s_{0})
≥ δ_{2}e^{−ns}

for some suitable constant δ_{2} > 0. In particular,

∂_{s} (e^{− \frac{4}{p−1} s}z_{1}(s)) = e^{− \frac{4}{p−1} s}z_{2}(s) ≥ δ_{2}e^{−(n−2)s}

and this implies that

e^{− \frac{4}{p−1} s}z_{1}(s) ≤ \frac{δ_{2}}{n − 2} (e^{−(n−2)s_{0}} − e^{−(n−2)s}) + e^{− \frac{4}{p−1} s_{0}}z_{1}(s_{0})
≤ −δ_{3}e^{−(n−2)s}

for some suitable constant δ_{3} > 0. Thus, we end up with

z_{1}(s) ≤ −δ_{3}e^{−(n−2− \frac{4}{p−1} )s} → −∞ as s → −∞,  (16)

so that also in this case, the corresponding solution u of (1) becomes singular at r = 0. □

Completing the proof of Proposition 1 also yields the proof of Theorem 1.

3 The Dirichlet problem

If we put r = |x| then the equation in (5) becomes

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) = λ(1 + u)^{p}, r ∈ [0, 1]. (17)

If we put

U(x) = 1 + u(x/\sqrt[4]{λ})  for x ∈ B\sqrt[4]{λ}(0)  (18)

then U solves the equation

Δ^{2}U = U^{p}  in B\sqrt[4]{λ}(0).  (19)

Since the equation (19) is invariant under the rescaling

U_{a}(x) = aU(a^{\frac{p−1}{4}} x)

i.e. U is a solution of (19) if and only if U_{a} is a solution of (19), it is not restrictive to concentrate our attention on solutions U of the equation (19) which satisfy the condition U(0) = 1.

Next we define U_{γ} = U_{γ}(r) as the unique solution of the initial value problem

\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} (20)

We report here the following fundamental result by [8]:

Lemma 2 ([8]). Let n > 4 and p > (n + 4)/(n − 4).

(i) There exists a unique \overline{γ} < 0 such that the solution U\overline{γ} of (20) exists on the whole interval [0, ∞), it is positive everywhere, it vanishes at infinity and it satisfies U ′ ; \overline{γ}(r) < 0 for any r ∈ (0, ∞).

(ii) If γ < \overline{γ} there exist 0 < R_{1} < R_{2} < ∞ such that the solution U_{γ} of (20) satisfies U_{γ}(R_{1}) = 0, lim_{r↑R_{2}} U_{γ}(r) = −∞ and U′_{γ}(r) < 0 for any r ∈ (0, R_{2}).

(iii) If γ > \overline{γ} there exist 0 < R_{1} < R_{2} < ∞ such that the solution U_{γ} of (20) satisfies 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) = +∞.

(iv) If γ_{1} < γ_{2} < 0 then the corresponding solutions U_{γ_{1}}, U_{γ_{2}} of (20) satisfy U_{γ_{1}} < U_{γ_{2}} and U′_{γ_{1}} < U′_{γ_{2}} as long as they both exist.

Proof. See the statement of [8, Theorem 2] and related proof and also the statement of [8, Lemma 2]. □

For any γ < 0 let U_{γ} be the unique local solution of (20). Thanks to Lemma 2 (iii), for γ > \overline{γ} we may define R_{γ} as the unique value of r > 0 for which we have U′_{γ}(R_{γ}) = 0.

Lemma 3. Let n > 4, p > (n + 4)/(n − 4) and γ ∈ (\overline{γ}, 0) with \overline{γ} as in the statement of Lemma 2. Then the map γ ↦ R_{γ} is monotonically decreasing and

lim_{γ↓\overline{γ}} R_{γ} = +∞.

Proof. The fact that the map γ ↦ R_{γ} is monotonically decreasing follows immediately by Lemma 2 (iv). This shows that the function γ ↦ R_{γ} admits a limit as γ → \overline{γ}. Suppose by contradiction that

\overline{R} := lim_{γ↓\overline{γ}} R_{γ} < +∞.

Then, by Lemma 2 (i), (iv) we have for all γ ∈ (\overline{γ}, 0) that

U_{γ}(R_{γ}) > U\overline{γ}(R_{γ}) ≥ U\overline{γ}(\overline{R}) > 0.  (21)

Define for any γ ∈ (\overline{γ}, 0), r ∈ [0, 1] the function

u_{γ}(r) = \frac{U_{γ}(R_{γ}r)}{U_{γ}(R_{γ})} − 1.  (22)

Then, u_{γ} solves the Dirichlet problem

\{ \begin{matrix}Δ^{2}u_{γ} = R_{γ}^{4}U_{γ}(R_{γ})^{p−1}(1 + u_{γ})^{p} & in B, \\ u_{γ} = |∇u_{γ}| = 0 & on ∂B. \end{matrix}  (23)

Moreover, by (21) and the fact that U_{γ}(R_{γ}) ≤ U_{γ}(r) ≤ U_{γ}(0) = 1 for any r ∈ [0, R_{γ}], we have for all γ ∈ (\overline{γ}, 0), x ∈ B

0 ≤ u_{γ}(x) ≤ U\overline{γ}(\overline{R})^{−1} − 1.  (24)

This shows that the set \{u_{γ} : γ ∈ (\overline{γ}, 0)\} is bounded in L^{∞}(B) and hence by a bootstrap argument, from (23) and the fact that R_{γ}^{4}U_{γ}(R_{γ})^{p−1} ≤ λ^{∗}, we deduce that there exists a sequence γ_{k} ↓ \overline{γ} and a function \overline{u} ∈ H_{0}^{2}(B) ∩ C^{∞}(\overline{B}) such that

u_{γ_{k}} → \overline{u}  in C^{4}(\overline{B})  (25)

as k → ∞. Since the sequence U_{γ_{k}}(R_{γ_{k}}) is monotonically decreasing and bounded from below then for any r ∈ [0, \overline{R}) we have that for sufficiently large k, U_{γ_{k}}(r) = U_{γ_{k}}(R_{γ_{k}}) [u_{γ_{k}} (r/R_{γ_{k}}) + 1] is well defined and admits a finite limit as k → ∞ which will be denoted by \overline{U}(r). In fact U_{γ_{k}} → \overline{U} in C^{4}([0, R]) for any 0 < R < \overline{R} and moreover by (25) we also have that

\overline{U}(x) = [ lim_{k→∞} U_{γ_{k}}(R_{γ_{k}})] · [\overline{u} (\frac{r}{\overline{R}}) + 1] .

Since \overline{u} ∈ H_{0}^{2}(B) we also have

lim_{r↑\overline{R}} \overline{U}′(r) = 0.  (26)

On the other hand by continuous dependence on the initial conditions we also have that

lim_{k→∞} U_{γ_{k}}(r) = U\overline{γ}(r) for all r ∈ [0, \overline{R})

and hence \overline{U}(r) = U\overline{γ}(r) for any r ∈ [0, \overline{R}). This with (26) implies

lim_{r↑\overline{R}} U ′ ; \overline{γ}(r) = 0

which is absurd since U ′ ; \overline{γ}(\overline{R}) < 0. This completes the proof of the lemma. □

Lemma 4. Let n > 4 and p > (n + 4)/(n − 4) and let u be a regular solution of (5). Then

u(x) ≤ (\frac{λ^{∗}}{λ} )^{1/(p−1)} |x|^{−4/(p−1)} − 1 for all x ∈ B\\{0\}.

Proof. Let u be a regular solution of (5) for some λ > 0 and define the rescaled function

U(x) = \frac{1}{1 + u(0)} [1 + u ( \frac{x}{\sqrt[4]{λ}(1 + u(0))^{\frac{p−1}{4}}} )]  (27)

so that U satisfies

Δ^{2}U = U^{p} in B_{R}(0)  and  U(0) = 1  (28)

where we put R = \sqrt[4]{λ}(1 + u(0))^{\frac{p−1}{4}} .

Define

M = max_{r∈[0,R]} r^{4/(p−1)}U(r)

and let \overline{R} ∈ (0, R] be such that \overline{R}^{4/(p−1)}U(\overline{R}) = M. If we define

w(r) = \frac{U(\overline{R}r)}{U(\overline{R})} − 1

then w solves the problem

\{ \begin{matrix}Δ^{2}w = \overline{R}^{4}U(\overline{R})^{p−1}(1 + w)^{p} & in B \\ w = 0 & on ∂B \\ w′ ≤ 0 & on ∂B. \end{matrix}

This proves that M^{p−1} = \overline{R}^{4}U(\overline{R})^{p−1} ≤ λ^{∗} since otherwise by the super-subsolution method (see [2, Lemma 3.3] for more details) we would obtain a solution of (5) for λ = \overline{R}^{4}U(\overline{R})^{p−1} > λ^{∗}. This yields for all r ∈ [0, R] that

U(r) ≤ Mr^{−4/(p−1)} ≤ (λ^{∗})^{1/(p−1)}r^{−4/(p−1)}.  (29)

Then reversing the identity (27), by (29) we obtain

u(r) = λ^{−1/(p−1)}R^{4/(p−1)}U(Rr) − 1 ≤ (\frac{λ^{∗}}{λ} )^{1/(p−1)} r^{−4/(p−1)} − 1

which completes the proof of the lemma. □

Proof of Theorem 2. For γ ∈ (\overline{γ}, 0) consider the corresponding solution U_{γ} of the Cauchy problem (20) and the function u_{γ} introduced in (22). If we put λ_{γ} = R_{γ}^{4}U_{γ}(R_{γ})^{p−1} then by (23) we have that u_{γ} solves

\{ \begin{matrix}Δ^{2}u_{γ} = λ_{γ}(1 + u_{γ})^{p} & in B, \\ u_{γ} = |∇u_{γ}| = 0 & on ∂B. \end{matrix}  (30)

We show that λ_{γ} remains bounded away from zero for γ > \overline{γ} sufficiently close to \overline{γ}, which is defined in Lemma 2. By [8, Theorem 3] we infer that for a fixed ε ∈ (0, K_{0}^{1/(p−1)}) there exists a corresponding r_{ε} > 0 such that

U\overline{γ}(r) > (K_{0}^{1/(p−1)} − ε)r^{−4/(p−1)}  for all r > r_{ε}.  (31)

On the other hand, by Lemma 3, we deduce that there exists γ_{0} ∈ (\overline{γ}, 0) such that for any γ ∈ (\overline{γ}, γ_{0}) then R_{γ} > r_{ε}. Therefore by Lemma 2 (iv) we obtain for all γ ∈ (\overline{γ}, γ_{0})

U_{γ}(R_{γ}) > U\overline{γ}(R_{γ}) > (K_{0}^{1/(p−1)} − ε)R_{γ}^{−4/(p−1)}

and this yields

∀γ ∈ (\overline{γ}, γ_{0}) :  λ_{γ} > (K_{0}^{1/(p−1)} − ε)^{p−1} =: C.  (32)

Combining (32) and Lemma 4 we obtain for all γ ∈ (\overline{γ}, γ_{0}), x ∈ B\\{0\}

u_{γ}(x) ≤ (\frac{λ^{∗}}{C} )^{1/(p−1)} |x|^{−4/(p−1)} − 1.  (33)

Since u_{γ} solves (30), by (33) we obtain

∫_{B} |Δu_{γ}|^{2}dx = λ_{γ} ∫_{B}(1 + u_{γ})^{p}u_{γ}dx ≤ λ^{∗} ∫_{B}(1 + u_{γ})^{p+1}dx ≤ \frac{(λ^{∗})^{\frac{2p}{p−1}}}{C^{\frac{p+1}{p−1}}} ∫_{B} |x|^{− \frac{4(p+1)}{p−1}} dx < +∞

since p > (n + 4)/(n − 4). This proves that the set \{u_{γ} : γ ∈ (\overline{γ}, γ_{0})\} is bounded in H_{0}^{2}(B) and hence there exists a sequence γ_{k} ↓ \overline{γ} and a function u ∈ H_{0}^{2}(B) such that u_{γ_{k}} ⇀ u in H_{0}^{2}(B). Moreover, by (33) and applying Lebesgue’s theorem, u weakly solves (5) for a suitable λ̃ ≥ C.

It remains to prove that the function u is unbounded. For simplicity, in the rest of the proof u_{γ_{k}}, U_{γ_{k}}, R_{γ_{k}}, λ_{γ_{k}} will be denoted respectively by u_{k}, U_{k}, R_{k}, λ_{k}.

By compact embedding we have that u_{k} → u in L^{1}(B) and hence we have

lim_{r↓0} \frac{1}{|B_{r}(0)|} ∫_{B_{r}(0)} u(x)dx = lim_{r↓0} [\frac{1}{r^{n}|B|} lim_{k→∞} ∫_{B_{r}(0)} u_{k}(x)dx]

and passing to radial coordinates, by (22) and Lemma 2 (iv), we obtain

lim_{r↓0} \frac{1}{|B_{r}(0)|} ∫_{B_{r}(0)} u(x)dx = lim_{r↓0} [−1 + \frac{n}{r^{n}} lim_{k→∞} ∫_{0}^{r} \frac{U_{k}(R_{k}ρ)}{U_{k}(R_{k})}ρ^{n−1}dρ]
= lim_{r↓0} [−1 + \frac{n}{r^{n}} lim_{k→∞} \frac{1}{R_{k}^{n}U_{k}(R_{k})} ∫_{0}^{R_{k}r} U_{k}(ρ)ρ^{n−1}dρ]
≥ lim_{r↓0} [−1 + \frac{n}{r^{n}} lim_{k→∞} \frac{1}{R_{k}^{n}U_{k}(R_{k})} ∫_{0}^{R_{k}r} U\overline{γ}(ρ)ρ^{n−1}dρ] .  (34)

By (31) we have that there exist C, R_{0} > 0 such that

∀ρ ∈ (R_{0}, ∞) :  U\overline{γ}(ρ) > Cρ^{−4/(p−1)}.  (35)

Hence we have for k > \overline{k} = \overline{k}(r)

∫_{0}^{R_{k}r} U\overline{γ}(ρ)ρ^{n−1}dρ ≥ ∫_{0}^{R_{0}} U\overline{γ}(ρ)ρ^{n−1}dρ + C (n − \frac{4}{p − 1})^{−1} (R_{k}^{n− \frac{4}{p−1}} r^{n− \frac{4}{p−1}} − R_{0}^{n− \frac{4}{p−1}} ) . (36)

Since p > (n + 4)/(n − 4) > (n + 4)/n and since by (32), λ_{k} is bounded away from zero as k → ∞ then

lim_{k→∞} R_{k}^{n}U_{k}(R_{k}) = lim_{k→∞} R_{k}^{n− \frac{4}{p−1}} λk^{\frac{1}{p−1}} = +∞

and hence by (36) we obtain

lim_{k→∞} \frac{1}{R_{k}^{n}U_{k}(R_{k})} ∫_{0}^{R_{k}r} U\overline{γ}(ρ)ρ^{n−1}dρ ≥ lim inf_{k→∞} \frac{C}{(n − \frac{4}{p−1}) R_{k}^{n}U_{k}(R_{k})} (R_{k}^{n− \frac{4}{p−1}} r^{n− \frac{4}{p−1}} − R_{0}^{n− \frac{4}{p−1}} )

= lim inf_{k→∞} \frac{Cr^{n− \frac{4}{p−1}}}{(n − \frac{4}{p−1}) λ_{k}^{1/(p−1)}} ≥ \frac{Cr^{n−4/(p−1)}}{(n − \frac{4}{p−1}) (λ^{∗})^{1/(p−1)}} =: C̃r^{n−4/(p−1)}.  (37)

Inserting (37) in (34) we obtain

lim_{r↓0} \frac{1}{|B_{r}(0)|} ∫_{B_{r}(0)} u(x)dx ≥ lim_{r↓0}(−1 + nC̃r^{−4/(p−1)}) = +∞.

This proves that u ∉ L^{∞}(B). □

Proof of Theorem 3. We make use of an idea from [6]. Let u_{λ} denote the positive minimal regular solution of (5) for 0 ≤ λ < λ^{∗}. According to [7, Theorem 2], these are stable so that one has in particular:

∀φ ∈ C_{0}^{∞}(B) : ∫_{B} (Δφ(x))^{2} dx − pλ ∫_{B}(1 + u_{λ}(x))^{p−1}φ(x)^{2} dx ≥ 0.

By taking the monotone limit we obtain that

∀φ ∈ C_{0}^{∞}(B) : ∫_{B} (Δφ(x))^{2} dx − pλ^{∗} ∫_{B}(1 + u^{∗}(x))^{p−1}φ(x)^{2} dx ≥ 0.  (38)

We assume now for contradiction that u^{∗} is singular. Then, according to [7, Theorem 5] we have the following estimate from below:

u^{∗}(x) > (\frac{K_{0}}{λ^{∗}} )^{1/(p−1)} |x|^{−4/(p−1)} − 1.  (39)

Combining this with (38) yields

∀φ ∈ C_{0}^{∞}(B) : ∫_{B} (Δφ(x))^{2} dx ≥ pK_{0} ∫_{B} |x|^{−4}φ(x)^{2} dx.  (40)

However under the subcriticality assumptions made we have that pK_{0} > n^{2}(n − 4)^{2}/16. This contradicts the optimality of the constant in Hardy’s inequality

∀φ ∈ C_{0}^{∞}(B) : ∫_{B} (Δφ(x))^{2} dx ≥ \frac{n^{2}(n − 4)^{2}}{16} ∫_{B} |x|^{−4}φ(x)^{2} dx,

so that u^{∗} has indeed to be regular. □

Acknowledgement. We are grateful to Gianni Arioli and Filippo Gazzola for fruitful discussions and interesting numerical experiments.

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