MULTIPLICITY OF SOLUTIONS FOR A FOURTH ORDER ... - CAPDE

2 JUAN DÁVILA, ISABEL FLORES, AND IGNACIO GUERRA(3)(4)b) u is singular at r = 0 and satisfies8(N − 2)(N − 4)lim u(r) + 4 log(r) = logr→0 λlimr→0 ru′ (r) exists.We prove this in Section 3. In [2] the authors call u a weakly singular solutionto (2) if it satisfies (4) and is singular. It turns out that this definition is naturalfor space phase analysis, after transforming the problem to a suitable first orderautonomous system. From the PDE point of view, the following definition is alsonatural. We say that u is a weak solution of (1) if⎧⎨ u ∈ H0 2 (B), e u ∈ L 1 (B) and∫∫(5)⎩ ∆u∆ϕ = λ e u ϕ ∀ϕ ∈ C0 ∞ (B).BBIt is possible to show that weakly singular solutions are also weak solutions. Sinceweak solutions in the sense (5) are nonnegative (see [2]) as a consequence of Theorem1.1 we are showing that both notions coincide for radial functions.Combining the results of [2] and [15], some of which are obtained by a computerassisted proof, we know that for all N ≥ 5, (1) has a singular solution for someλ > 0. We give a new proof of the existence, which is not computer assisted, andshow its uniqueness.Theorem 1.2. Assume N ≥ 5. There exists a unique λ S > 0 such that (1) withλ = λ S admits a radial singular solution and this singular solution is unique in theclass of radial solutions.Let λ ∗ denote the largest value of λ ≥ 0 such that (1) has a radial classicalsolution. Then λ ∗ > 0 and finite, see [2]. Many authors have studied what happensto solutions when λ = λ ∗ (for this see [2, 15]). With respect to multiplicity ofsolutions we have the following:Theorem 1.3. Assume 5 ≤ N ≤ 12. Then λ S < λ ∗ and (1) with λ = λ S admitsinfinitely many regular radial solutions. For λ ≠ λ S then (1) has a finite numberof regular radial solutions, and this number goes to infinity as λ → λ S .The fact that if 5 ≤ N ≤ 12 then λ S < λ ∗ is also a consequence of the results in[2] and [15].Theorem 1.4. If N ≥ 13 then λ S = λ ∗ and for all 0 < λ < λ ∗ (1) has a uniqueradial solution, which is regular. For λ = λ ∗ there is a unique radial solution whichis singular.LetC = { (λ, u) ∈ (0, ∞) × C 4 (B) : u is radial and solves (1) }.Following [30] we have:Theorem 1.5. Assume N ≥ 5. The set C is homeomorphic to (0, ∞) and theidentification can be done through (λ, u) ∈ C ↦→ u(0).The inverse of the above identification can be extended continuously in a suitabletopology to [0, ∞] as 0 ↦→ (0, 0) and ∞ ↦→ (λ S , u S ) where u S is the unique singularsolution of Theorem 1.2.

**MULTIPLICITY** **FOR** ∆ 2 u = λe u 3(6)For the problem with Navier boundary conditions{∆ 2 u = λe u in Bwe have similar results.u = ∆u = 0on ∂BTheorem 1.6. Assume N ≥ 5. There exists a unique λ S > 0 such that (6) withλ = λ S admits a radial singular solution and this singular solution is unique in theclass of radial solutions.Theorem 1.7. Assume 5 ≤ N ≤ 12. Then (6) with λ = λ S admits infinitely manyregular radial solutions. For λ ≠ λ S then (6) has a finite number of regular radialsolutions and the number of radial regular solutions goes to infinity as λ → λ S .We prove the multiplicity results by phase space analysis, using ideas from thework of Bamón, Flores, del Pino [3] and which were subsequently applied also in[21, 22, 17]. By a change of variables we transform the ODE version of (1) intoa reasonable first order 4 dimensional autonomous system, which has 2 stationarypoints P 1 , P 2 . Some properties of this system, such as the local character of P 1 , P 2 ,were studied by Arioli, Gazzola, Grunau, Mitidieri [2]. We review this material inSection 2. It is important in our argument to know that there exists a heteroclinicconnection from P 1 to P 2 . This connection was found by Arioli, Gazzola, Grunau in[1], in the form of an entire solution of ∆ 2 u = e u with a special decay. We explainthis in Section 4 and show that in dimensions 5 ≤ N ≤ 12 this connection nearP 2 is a spiral. In Section 5 we study some properties of the unstable manifold atP 2 , which lead to the proof of Theorems 1.2 and 1.6. The proof of the multiplicityof solutions asserted in Theorems 1.3 and 1.6 is in Section 6. Finally Section 7 isdedicated to the study of some properties of the solution set C. In particular weprove there Theorems 1.4 and 1.5.It is natural to ask whether the uniqueness result of Theorem 1.4 is true forproblem (6). Using the techniques in this work it is possible to show that if theextremal solution u ∗ of (6) is singular, then for all λ ∈ (0, λ ∗ ) there is a uniqueradial solution. We conjecture that this is the case for all N ≥ 13, and the proofcould be done using similar ideas as in [15].The counterpart of the results in this paper for the classical problem{ −∆u = λeuin B(7)u = 0 on ∂Bare well known. In dimension 1 this problem was first studied by Liouville [35],then by Bratu [9], Chandrasekhar [11] and Frank-Kamenetskii [23]. Barenblatt[25] proved that in dimension 3 for λ = 2 there are infinitely many solutions, andJoseph and Lundgren [32] completed the description of the classical solutions to (7)in all dimensions. The literature on second order problems like (7), including othernonlinearities and general domains, is very extensive, see [4, 34].In a forthcoming work [16] we will address similar multiplicity results for thebilaplacian with power-type nonlinearities.

4 JUAN DÁVILA, ISABEL FLORES, AND IGNACIO GUERRA(8)2. PreliminariesWith the change of variables v(t) = u(r), r = e t equation (1) is equivalent to(∂ t + N − 4) (∂ t − 2) (∂ t + N − 2) ∂ t v(t) = λe v+4t for all t < 0with the boundary conditions(9)and the behavior at −∞ given byLet(10)lim v(t) ∈ R, limt→−∞⎧v 1 (t) =⎪⎨v 2 (t) = ∂ t v(t) = e t u ′ (e t )v(0) = 0, v ′ (0) = 0.λ8(N − 2)(N − 4) ev(t)+4t =t→−∞ e−t v ′ (t) = 0.v 3 (t) = (∂ t − 2 + N) v 2 (t) = e 2t ∆u(e t )⎪⎩v 4 (t) = (∂ t − 2) v 3 (t) = e 3t (∆u) ′ (e t )Then (8) becomes(11)while (9) is equivalent to(12)⎧v 1 ′ = v 1 (v 2 + 4)⎪⎨ v 2 ′ = −(N − 2)v 2 + v 3λ8(N − 2)(N − 4) eu(et )+4tv 3 ′ = 2v 3 + v 4⎪⎩v 4 ′ = −(N − 4)v 4 + 8(N − 2)(N − 4)v 1v 2 (0) = 0, v 1 (0) =The only stationary points of the system (11) are{P1 = (0, 0, 0, 0)(13)λ8(N − 2)(N − 4) .P 2 = (1, −4, −4(N − 2), 8(N − 2)).Let V = (v 1 , . . . , v 4 ). From Theorem 6 in [2] we learn that u is a regular solutionof (1) if and only iflimt→−∞ V (t) = P 1while u is a weakly singular solution if and only iflim V (t) = P 2.t→−∞The linearization of (11) around the point P 1 is given by Z ′ = M 1 Z where⎡⎤4 0 0 00 −(N − 2) 1 0M 1 =⎢⎣ 0 0 2 1⎥⎦A 0 0 −(N − 4)and A = 8(N −4)(N −2). The eigenvalues of this matrix are 2, 4, −N +4, −N +2.Thus, if N ≥ 5 then P 1 is a hyperbolic point with a 2-dimensional unstable manifoldW u (P 1 ) and a 2-dimensional stable manifold W s (P 1 ).

**MULTIPLICITY** **FOR** ∆ 2 u = λe u 5The linearization of (11) around P 2 is given by Z ′ = MZ where⎡⎤0 1 0 00 −(N − 2) 1 0(14)M =⎢⎣ 0 0 2 1⎥⎦A 0 0 −(N − 4)and A = 8(N − 4)(N − 2). The eigenvalues of M are given by⎧ν 1 = 1 (4 − N + √ )M 1 (N) + M 2 (N)2⎪⎨ ν 2 = 1 (4 − N − √ )M 1 (N) + M 2 (N)(15)2ν 3 = 1 (4 − N + √ )M 1 (N) − M 2 (N)2⎪⎩ ν 4 = 1 (4 − N − √ )M 1 (N) − M 2 (N)2whereThenM 1 (N) = (N − 2) 2 + 4,ν 2 < 0 < ν 1 .M 2 (N) = 4 √ (N − 2) 2 + AIf 5 ≤ N ≤ 12 then M 1 (N) − M 2 (N) < 0 and ν 3 , ν 4 are complex conjugate withnonzero imaginary part and negative real part. If N ≥ 13 all eigenvalues are realand ν 3 , ν 4 are negative. For all N ≥ 5, P 2 is a hyperbolic stationary point witha 1-dimensional unstable manifold W u (P 2 ) and a 3-dimensional stable manifoldW s (P 2 ).Concerning the eigenvectors of M we have:Lemma 2.1. Let v (1) , . . . , v (4) be the eigenvectors of M associated to ν 1 , . . . , ν 4 .Then(16)v (k) = [1, ν k , ν k (ν k + N − 2), ν k (ν k + N − 2)(ν k − 2)]We have that v (1) , v (2) are always real, and v (3) , v (4) are complex conjugate if 5 ≤N ≤ 12. Let us write v (i) = (v (i)1 , v(i) 2 , v(i) 3 , v(i) 4 ), i = 1, . . . , 4. If N ≥ 3 then(17)and(18)v (1)1 > 0, v (1)2 > 0, v (1)3 > 0, v (1)4 > 0,v (2)1 > 0, v (2)2 < 0, v (2)3 > 0, v (2)4 < 0.Proof. That the vectors defined by (16) are eigenvector of M follows from a directcalculation. Let v (1) = (t 1 , t 2 , t 3 , t 4 ) an eigenvector for M with eigenvalue ν 1 . Thent 1 = 1 > 0, t 2 = ν 1 > 0,t 3 = (ν 1 + N − 2)ν 1 > 0, t 4 = (ν 1 − 2)(ν 1 + N − 2)ν 1 > 0.In fact, since ν 1 > 0, it is sufficient to prove that ν 1 − 2 > 0. This holds if√M1 (N) + M 2 (N) > N and this is equivalent to A = 8(N − 2)(N − 4) > 0.The proof of (18) is similar.It will be convenient to have the following result:□

6 JUAN DÁVILA, ISABEL FLORES, AND IGNACIO GUERRALemma 2.2. The system (11) is C 1 -conjugate to its linearization around the pointP 2 .Proof. We use a result of Belickiĭ [5, 6], see also the book [39, p. 25]. To applyit we need to verify that no relation of the form Re(ν i ) = Re(ν j ) + Re(ν k ) holdsfor different indices i, j, k in {1, . . . , 4} such that Re(ν j ) < 0 and Re(ν k ) > 0,where ν 1 , . . . , ν 4 are the eigenvalues of M defined in (14). This can be verified bycalculation.□3. Behavior of singular solutionsThe purpose of this section is to prove Theorem 1.1. In what follows we assumethat u ∈ C 4 (0, 1), u ≥ 0 satisfies(19)∆ 2 u = 8(N − 2)(N − 4)e u in (0, 1),where we have assumed, by using a scaling, that λ = 8(N − 2)(N − 4). Thatthe interval is (0, 1) is not relevant for the next arguments. The arguments inthis section are based on the work of Ferrero and Grunau [18] where they showthat radial singular solutions to a problem with power-type nonlinearity are weaklysingular for that problem.Define v(t) = u(e t ), w(t) = v(t) + 4t for t ≤ 0. We also let v 1 , . . . , v 4 be definedby (10). We note that w satisfies(20)w (4) + 2(N − 4)w ′′′ + (N 2 − 10N + 20)w ′′ − 2(N − 2)(N − 4)w ′= 8(N − 2)(N − 4)(e w − 1) for all t < 0.Lemma 3.1. If δ > 0 there exists a constant C depending only on δ such that if[a, b] ⊂ (−∞, 0] is such that w(t) ≥ δ for all t ∈ [a, b] then b−a ≤ C. A consequenceis that(21)lim inf w(t) ≤ 0.t→−∞Proof. We follow an idea of Mitidieri and Pohozaev [37], which has also been usedin [1, 18, 24], called the test-function method in these references.We proceed by contradiction. We may suppose, by shifting time, that w(t) ≥ δfor all t ∈ [−L, 0] for arbitrary large L > 0. Let φ ∈ C ∞ (R) be such that, 0 ≤ φ ≤ 1,φ(t) = 0 for t ≤ −3, φ(t) > 0 for t ∈ (−3, 0), φ(t) = 0 for t ≥ 0, φ(t) = 1 fort ∈ [−2, −1], and for i = 1, 2, 3, 4∫ 0−3(φ (i) ) 2dt < +∞.φLet τ > 1 and φ τ (t) = φ(t/τ) and assume that 3τ ≤ L. Let us rewrite the equation(20) in the form4∑(22)a i w (i) (t) = A(e w − 1) for t < 0i=1where A = 8(N − 2)(N − 4) and a i ∈ R. Multiplying the equation (22) by φ τ andintegrating we find4∑∫ 0∫ 0(23)a i (−1) i φ (i)τ w dt = A (e w − 1)φ τ dt.−3τ−3τi=1

**MULTIPLICITY** **FOR** ∆ 2 u = λe u 7Let ε > 0 to be fixed later on. For all t > −3τ|wφ (i)τso that from (23) we know thatA∫ 0−3τ(e w − 1)φ τ dt ≤ εK| ≤ εw 2 φ τ + C ε(φ (i)τ ) 2∫ 0−3τφ τ∫ 0w 2 φ τ dt + C ε K maxi=1,...,4−3τ(φ (i)τ ) 2dtφ τwhere K = ∑ 4i=1 |a i|. Since w(t) ≥ δ, we can select ε > 0 sufficiently small so thatA(e w − 1) − εKw 2 ≥ δ/4 for all t ∈ [−3τ, 0]. It follows thatBut∫ 0−3τδ4 τ ≤ C εK maxi=1,...,4(φ (i)τ ) 2φ τdt = τ 1−2i ∫ 0∫ 0−3−3τ(φ (i)τ ) 2dt.φ τ(φ (i) ) 2dt ≤ C i τ 1−2i .φIt follows thatδ4 τ ≤ C εK max C iτ 1−2i for all τ > 1,i=1,...,4which is not possible.Lemma 3.2.lim sup w(t) < +∞.t→−∞Proof. We follow the idea of Lemma 1 in [18]. Assume by contradiction thatlim sup t→−∞ w(t) = +∞. Since (21) also holds there is a sequence t k → −∞ suchthat w(t k ) → +∞, and for all k ≥ 1 we have t k+1 + log 2 < t k , w(t k+1 ) ≥ w(t k ),w ′ (t k ) = 0, and w ′′ (t k ) ≤ 0. It suffices to take as t k a sequence of well separatedlocal maxima of w along which it goes to +∞.Let M k = w(t k ), r k = e t kand ρ k = r k+1r k. Note that 0 < ρ k ≤ 1/2. Defineu k (r) = u(r r k ) − M k + 4 log(r k ).where u is the original solution to (19). Then(24) ∆ 2 u k = Ae M ke u ku k (1) = 0for r ∈ (0, r −1k )□u k (ρ k ) = w(t k+1 ) − M k + 4(t k − t k+1 ) ≥ 0.Moreover, since∆u k (r) = 1 r 2 [w′′ (log(r r k )) + (N − 2)w ′ (log(r r k )) − 4(N − 2)]and w ′ (t k ) = 0 and w ′′ (t k ) ≤ 0 we have∆u k (1) = w ′′ (t k ) − 4(N − 2) ≤ 0∆u k (ρ k ) = 1 ρ 2 [w ′′ (t k+1 ) − 4(N − 2)] ≤ 0k

**MULTIPLICITY** **FOR** ∆ 2 u = λe u 9in [t, t 0 ] with t ≤ t 0 ≤ 0 we find(26)v 4 (t) = e −(N−4)t (v 4 (t 0 )e (N−4)t0 − A∫ t0t)e (N−4)s v 1 (s) dsSince v 1 (t) = λ A ew(t) and w is bounded above we have that v 1 (t) is bounded ast → −∞. Hence the integral ∫ t 0−∞ e(N−4)s v 1 (s) ds exists. If∫ t0A e (N−4)s v 1 (s) ds ≠ v 4 (t 0 )e (N−4)t0−∞we deduce from (26) that |v 4 (t)| grows exponentially as t → −∞, which contradicts(25). It follows that(27)v 4 (t 0 ) = Ae −(N−4)t0 ∫ t0−∞Since v 1 is bounded we see from this formula that|v 4 (t)| ≤ C for all t ≤ 0.e (N−4)s v 1 (s) ds ∀t 0 ≤ 0.This in turn implies that v 3 is bounded as well. In fact from (11) we have(v3 (t)e −2t) ′= e −2t v 4 (t)and integrating on [t, 0], t ≤ 0 yields(28)v 3 (t) = e 2t v 3 (0) −∫ 0te 2(t−s) v 4 (s) ds.Using that v 4 is bounded it follows that v 3 is bounded as well as t → −∞.Let us prove that v 2 remains bounded as t → −∞. Arguing as for v 4 we integratethe equation(v 2 (t)e (N−2)t) ′= e (N−2)t v 3 (t)in [t, t 0 ] with t ≤ t 0 ≤ 0:(29)The integral ∫ t 0tv 2 (t) = e −(N−2)t (v 2 (t 0 )e (N−2)t0 −∫ t0t)e (N−2)s v 3 (s) ds .e (N−2)s v 3 (s) ds converges because v 3 is bounded. If∫ t0−∞e (N−4)s v 3 (s) ds ≠ v 2 (t 0 )e (N−2)t0we deduce from (29) that |v 2 (t)| grows exponentially as t → −∞, which contradicts(25). It follows that(30)v 2 (t 0 ) = e −(N−2)t0 ∫ t0−∞e (N−4)s v 3 (s) ds ∀t 0 ≤ 0.Since v 3 is bounded we see from this formula that v 2 is also bounded.The fact that w (i) are bounded as t → −∞ follows from the formulasw ′ = v 2 + 4, w ′′ = −(N − 2)v 2 + v 3w ′′′ = (N − 2) 2 v 2 − (N − 4)v 3 + v 4w (4) = Av 1 + (N − 2) 3 v 2 − ((N − 2) 2 + 2(N − 4))v 3 − 2(N − 4)v 4 .□

10 JUAN DÁVILA, ISABEL FLORES, AND IGNACIO GUERRAAs in [1] we consider the energyE(t) = 1 2 w′′ (t) 2 − 1 2 (N 2 − 10N + 20)w ′ (t) 2 + A(e w − w).A computation reveals that if t 1 ≤ t 2 then∣E(t 2 ) − E(t 1 ) = w ′ w ′′′ ∣∣t 2∣ ∫+ 2(N − 4)w ′ w ′′ ∣∣t t22− 2(N − 4) w ′′ (s) 2 dst 1 t 1 t(31)1− 2(N − 2)∫ t2t 1w ′ (s) 2 ds.Lemma 3.5. Iflim inf w(t) = −∞t→−∞then w(t) → −∞, v i (t) → 0 as t → −∞ for i = 1, 2, 3, 4 and u is a regular solution.Proof. We first show that w(t) → −∞ as t → −∞ by contradiction. Suppose thatw(t) does not approach −∞. Then one can find sequences t k , s k → −∞ such thats k > t k ,w(t k ) remains bounded, w ′ (t k ) = 0w(s k ) → −∞, w ′ (s k ) = 0.Then by (31) E(s k ) ≤ E(t k ). But E(t k ) remains bounded while E(s k ) → ∞, whichis a contradiction.Now that we know that w(t) → −∞ as t → −∞, we deduce immediately thatv 1 (t) → 0 as t → −∞. Then using formulas (27), (28) and (30) we also obtainv i (t) → 0 as t → −∞ for i = 2, 3, 4. By Theorem 6 in [2] we deduce that u is aregular solution.□Lemma 3.6. Iflim inf w(t) > −∞t→−∞then w(t) → 0, (v 1 (t), . . . , v 4 (t)) → P 2 as t → −∞ and u is a weakly singularsolution.Proof. In this case, since w is also bounded above by Lemma 3.2, we have that wis bounded. By Lemma 3.4 the derivatives of w are bounded as well and we deducethat E(t) remains bounded as t → −∞. The boundedness of E together with theboundedness of the derivatives of w and formula (31) imply that(32)∫ 0−∞(w ′ ) 2 dt < +∞,∫ 0−∞(w ′′ ) 2 dt < +∞.Then we can select a strictly decreasing sequence t k → −∞ such thatlim (t k − t k+1 ) = 0k→∞andw ′ (t k ) → 0 as k → ∞.If t ≤ s ≤ 0 we have by (32)Hence for t ∈ [t k , t k+1 ]|w ′ (t) − w ′ (s)| ≤ C|t − s| 1/2 .|w ′ (t)| ≤ |w ′ (t k )| + C(t k+1 − t k ) 1/2 .

**MULTIPLICITY** **FOR** ∆ 2 u = λe u 11This shows that w ′ (t) → 0 as t → ∞. Using then elliptic estimates we deducew (i) (t) → 0as t → −∞for i = 1, 2, 3, 4. Using the equation we also deduce that w(t) → 0 as t → −∞. Wehence obtain that (v 1 (t), . . . , v 4 (t)) → P 2 as t → −∞. Then by Theorem 6 in [2]we have u is a weakly singular solution.□Proof of Theorem 1.1. It is consequence of Lemmas 3.5 and 3.6.□4. Heteroclinic connection from P 1 to P 2Proposition 4.1. For N ≥ 5, system (11) has an heteroclinic orbit from P 1 to P 2 .Proof. We use one of the main results in [1] on the initial value problem(33)∆ 2 u = 8(N − 2)(N − 4)e ur ∈ (0, R(β))u(0) = 0, u ′ (0) = 0, ∆u(0) = β, (∆u) ′ (0) = 0where β ∈ R is a parameter and R(β) > 0 is the maximal time of existence of thesolution. Here the constant in front of e u is taken, without loss of generality, to be8(N − 2)(N − 4). In Theorem 2 of [1] the authors show that there exists a uniqueβ such that R(β) = +∞ and(34)lim u(r) + 4 log(r) = 0.r→∞In what follows we fix β is such that (34) holds. Let v(t) = u(r) where r = e t andV = (v 1 , . . . , v 4 ) be defined by (10). Then, since u is smooth at the originand (34) tells us thatIt remains only to show that(35)limt→−∞ V (t) = P 1lim v 1(t) = 1.t→∞lim V (t) = P 2.t→∞Set w(t) = v(t) + 4t. Then w satisfies the following equation(∂ t + N − 4) (∂ t − 2) (∂ t + N − 2) ∂ t w = 8(N − 2)(N − 4)(e w − 1) in R.Note that (34) is equivalent to lim t→∞ w(t) = 0. To prove (35) it suffices to showthat(36)This follows from the lemma below.limt→∞ w(i) (t) = 0 for i = 1, 2, 3.Lemma 4.2. Assume that z : [T 0 , ∞) → R exists for some T 0 and solvesz (4) (t) + K 3 z ′′′ (t) + K 2 z ′′ (t) + K 1 z ′ (t) = f(z(t)) ∀t > T 0where f ∈ C 1 (R) and K i ∈ R. Let z 0 be such that f(z 0 ) = 0 and assume thatlim t→∞ z(t) = z 0 . Then for k = 1, . . . , 4limt→∞ z(k) (t) = 0.□

12 JUAN DÁVILA, ISABEL FLORES, AND IGNACIO GUERRAFor the proof see [18, Proposition 1].The next lemma shows that a connection from P 1 to P 2 necessarily reaches P 2in an oscillatory way if 5 ≤ N ≤ 12, but the statement below holds for all N ≥ 5.Lemma 4.3. Let N ≥ 5. Assume V : (T, ∞) → R 4 , V = (v 1 , v 2 , v 4 , v 4 ) is asolution to (11) such that lim →∞ V (t) = P 2 and either(37)or(38)V ′ (t)|V ′ (t)| + v(2)→ 0 as t → +∞|v (2) |V ′ (t)|V ′ (t)| − v(2)→ 0 as t → +∞.|v (2) |Then V can not be extended to a connection from P 1 to P 2 .Proof. Assume first that (37) holds. Then by (18) we havefor all t near +∞ and(39)v ′ 1(t) < 0, v ′ 2(t) > 0, v ′ 3(t) < 0, v ′ 4(t) > 0v 1 (t) > 1, v 2 (t) < −4, v 3 (t) > −4(N − 2), v 4 (t) < 8(N − 2)for all t near +∞. We claim that(40)v 2 (t) < −4 ∀t > T.Assume by contradiction that this fails. Then from (39) we can define t 1 > T tobe the last time such that v 2 (t 1 ) = −4. Then v ′ 2(t 1 ) ≤ 0. Using the equations (11)we deduce thatv 3 (t 1 ) ≤ −4(N − 2).Then thanks to (39) we can define t 2 ≥ t 1 to be the last time such that v 3 (t 2 ) =−4(N − 2). This implies that v ′ 3(t 2 ) ≥ 0 and by the system (11)v 4 (t 2 ) ≥ −8(N − 4).Let t 3 ≥ t 2 be the last time such that v 4 (t 3 ) = −8(N − 4). Then v ′ 4(t 3 ) ≤ 0. Wededuce from (11) thatv 1 (t 3 ) ≤ 1.Let t 4 ≥ t 3 be the last time such that v 1 (t 4 ) = 1. Then v ′ 1(t 4 ) ≥ 0 and by (11)v 2 (t 4 ) ≥ −4.But v 2 (t) < −4 for all t ∈ (t 1 , ∞), which is a contradiction. This proves the claim(40) and shows that the trajectory defined by V can not come from P 1 .Assume now that (38) holds. We claim that in this case(41)v 3 (t) < −4(N − 2) for all t > T.The proof is similar as before. Note that under the assumption (38) we have theopposite inequalities in (39). If the statement (41) fails we can define the lasttime t 1 such that v 3 (t 1 ) = −4(N − 2). Then define successively t 2 ≥ t 1 such thatv 4 (t 2 ) = 8(N − 2), v 4(t ′ 2 ) ≥ 0, t 3 ≥ t 2 such that v 1 (t 3 ) = 1, v 1(t ′ 3 ) ≤ 0, t 4 ≥ t 3 suchthat v 2 (t 4 ) = −4 and v 2(t ′ 4 ) ≥ 0, which leads to v 3 (t 4 ) ≥ −4(N − 2) which yields acontradiction. This shows that the trajectory can not come from P 1 .□

**MULTIPLICITY** **FOR** ∆ 2 u = λe u 135. The unstable manifold at P 2In this section we study W u (P 2 ) and as consequence we obtain Theorems 1.2and 1.6. Let v (j) denote the eigenvectors of the linearization of (11) at P 2 withcorresponding eigenvalue ν j . Then W u (P 2 ) is one dimensional and tangent to v (1)at P 2 . Hence, if V = (v 1 , . . . , v 4 ) : (−∞, T ) → R 4 is any trajectory in W u (P 2 )there are 2 cases:The main results in this section are〈V ′ (t), v (1) 〉 < 0 for t near −∞〈V ′ (t), v (1) 〉 > 0 for t near −∞.Proposition 5.1. Suppose that V = (v 1 , . . . , v 4 ) : (−∞, T ) → R 4 is the trajectoryin W u (P 2 ) such that 〈V ′ (t), v (1) 〉 < 0 for t near −∞. Thena) v 2 (t) < −4 for all t ∈ (−∞, T ), andb) v 3 (t) < −4(N − 2) for all t ∈ (−∞, T ).Proposition 5.2. Let V = (v 1 , . . . , v 4 ) : (−∞, T ) → R 4 be the trajectory inW u (P 2 ) such that 〈V ′ (t), v (1) 〉 > 0 for t near −∞, where T is the maximal time ofexistence. Thena) v 1 (t) > 1 for all t < T .b) There exists a unique t 0 such that v 2 (t 0 ) = 0. Moreover v ′ 2(t) > 0 for all t < T .In particular the trajectory of V intersects the hyperplane {v 2 = 0} transversally.c) There exists a unique t 1 such that v 3 (t 1 ) = 0. Moreover v ′ 3(t) > 0 for all t < T .In particular the trajectory of V intersects the hyperplane {v 3 = 0} transversally.Proof of Proposition 5.1.a) The relations (17) and the hypothesis 〈V ′ (t), v (1) 〉 < 0 for t → −∞ imply thatfor t near −∞{v1 (t) < 1, v 2 (t) < −4,(42)v 3 (t) < −4(N − 2), v 4 (t) < 8(N − 2).Assume by contradiction that v 2 (t) ≥ −4 for some t < T . Thus we may definet 0 < T the first time such that v 2 (t) = −4. Then v ′ 2(t 0 ) ≥ 0. Then by (11)0 ≤ v ′ 2(t 0 ) = v 3 (t 0 ) + 4(N − 2), that is,v 3 (t 0 ) ≥ −4(N − 2).By (42) we can define t 1 ≤ t 0 as the first time such that v 3 (t) = −4(N − 2). Thenv ′ 3(t 1 ) ≥ 0 and (11) implies(43)v 4 (t 1 ) ≥ 8(N − 2).Again using (42), let t 2 ≤ t 1 be the first time that v 4 (t) = 8(N −2). Then v ′ 4(t 2 ) ≥ 0and by (11)v 1 (t 2 ) ≥ 1.Thanks to (42) we must have a first time t 3 ≤ t 2 such that v 1 (t) = 1. But thenv ′ 1(t 3 ) ≥ 0 which by (11) impliesv 2 (t 3 ) + 4 ≥ 0.

14 JUAN DÁVILA, ISABEL FLORES, AND IGNACIO GUERRAThus v 2 (t 3 ) ≥ −4. This cannot happen if t 3 < t 0 because v 2 (t) < −4 for all t < t 0 .If t 3 = t 2 = t 1 = t 0 then v ′ 1(t 0 ) = v ′ 2(t 0 ) = v ′ 3(t 0 ) = v ′ 4(t 0 ), which means V ≡ P 2 , acontradiction. This proves that v 2 (t) < −4 for all t < T .b) Let us show now that v 3 (t) < −4(N − 2) for all t < T . If not, we can definet 1 < T as the first time such that v 3 (t) = −4(N − 2). Then v 3(t ′ 1 ) ≥ 0 and we mayrepeat the same argument starting at (43) to find t 3 ≤ t 1 such that v 2 (t 3 ) ≥ −4.This is impossible and proves the result.□Proof of Proposition 5.2.t → −∞ we have(44)for t near −∞.Let us prove first that(45)By (17) and the hypothesis 〈V ′ (t), v (1) 〉 > 0 forv ′ 1(t) > 0, v ′ 2(t) > 0, v ′ 3(t) > 0, v ′ 4(t) > 0v 1 (t) > 0 ∀t < T.This is valid for t near −∞ by (44). If v 1 (t) = 0 for some t then v 1 would beconstant by the equation, which is not possible.Before proving b) and c) we will claim that (44) is valid for all t < T.First we establish that(46)v ′ 3(t) > 0 ∀t < T.To prove (46) suppose it fails. Let s 0 < T be the first time such that v ′ 3(s 0 ) = 0.Using (11) we see that0 = v ′ 3(s 0 ) = 2v 3 (s 0 ) + v 4 (s 0 ).But v 3 (s 0 ) > −4(N − 2) we deduce v 4 (s 0 ) < 8(N − 2). Let s 1 ≤ s 0 be the first timesuch that v 4 (t) = 8(N − 2). Then v ′ 4(s 1 ) ≤ 0 and hencev 1 (s 1 ) ≤ 1.Let s 2 ≤ s 1 be the first time such that v 1 (s 2 ) = 1. Then v ′ 1(s 2 ) ≤ 0 and we conclude(47)v 2 (s 2 ) ≤ −4.Let s 3 ≤ s 2 be the first time such that v 2 (s 3 ) = −4. Then v ′ 2(s 3 ) ≤ 0 and weconclude(48)v 3 (s 2 ) ≤ −4(N − 2).Now since s 2 < s 0 , we have v 3 (s 2 ) > −4(N − 2), a contradiction. This establishesour claim (46).Since (46) holds we have then v 3 (t) > −4(N − 2) for all t < T . From the secondequation in (11), we havev ′′2 = −(N − 2)v ′ 2 + v ′ 3.We claim that v 2 ′ > 0. By contradiction, if s 0 is the first time such that v 2(s ′ 0 ) = 0then using (46), we have that v 2 ′′ (s 0 ) > 0 so v 2 has a local minimum at s 0 which isnot possible, since v 2 is increasing near t = −∞. We conclude that(49)v ′ 2(t) > 0 ∀t < T.

**MULTIPLICITY** **FOR** ∆ 2 u = λe u 15Similarly differentiating the first equation in (11), and using (45), and (49), weobtain that(50)v ′ 1(t) > 0∀t < Tand again using now the fourth equation in (11), and (50), we have(51)v ′ 4(t) > 0 ∀t < T,this proves that (44) is valid for all −∞ < t < T.Now since v ′ 1(t) > 0 for all t < T and lim t→−∞ v 1 (t) = 1, part a) of the propositionfollows.Let us prove now thatsup v i (t) = +∞, for all i = 1 . . . 4.t 0 and from Gronwall’s inequality we deduce that the solution isdefined for all times, that is T = +∞. Since v 1 is increasing, v 1 → L < +∞ ast → +∞ and v ′ 1(t k ) → 0 along some sequence t k → +∞. But v 1 , v 2 are increasingand v 2 (t) > −4, v 1 (t) > 1 for all t ∈ R. Then from the equation v ′ 1 = v 1 (v 2 + 4) weobtain a contradiction. This proves that(52)v 1 (t) → ∞ as t → T.We prove similarly that v 4 (t) → +∞ as t → T . Arguing by contradiction we havev 4 → L < ∞ as as t → T . If T = +∞ the argument is the same as before: for somesequence t k → +∞, v ′ 4(t k ) → 0. Using the equation for v ′ 4 we have a contradiction.If T < +∞, the assumption that v 4 is bounded and the system (11) imply that v 3 ,v 2 and v 1 are bounded up to T , which is not possible by (52). Thus we have proved(53)v 4 (t) → ∞ as t → T.Applying the same argument, now using (53) and the equation for v ′ 3, we obtain(54)v 3 (t) → ∞ as t → T.For v 2 , we use the same procedure now with the equation for v ′ 2 and (54), anddeduce that(55)v 2 (t) → ∞ as t → T.Finally the property b) clearly follows from (55) and v 2(t) ′ > 0 for all t < T.Similarly, c) is a consequence of (54) and that v 3(t) ′ > 0 for all t < T. □Proof of Theorems 1.2 and 1.6.Any weakly singular radial solution gives rise, through the changes of variablev(t) = u(e t ), t ≤ 0, and (10), to a solution V : (−∞, 0] → R 4 of the system(11) such that the final conditions (12) hold. Since the solution is weakly singular,lim t→−∞ V (t) = P 2 . Hence V ((−∞, 0]) is contained in W u (P 2 ) and therefore thereare 2 possibilities: either 〈V ′ (t), v (1) 〉 < 0 for t near −∞ or 〈V ′ (t), v (1) 〉 > 0 fort near −∞. The first case is not possible, because Proposition 5.1 shows that Vcannot satisfy the end condition v 2 (0) = 0. Thus we are in the second case andwe can apply Proposition 5.2 b). Therefore there exists a unique t 0 > −∞ such

16 JUAN DÁVILA, ISABEL FLORES, AND IGNACIO GUERRAthat v 2 (t 0 ) = 0 since the system is autonomous by shifting time we can assumethat v 2 (0) = 0. This concludes the proof of Theorem 1.2. The proof of Theorem1.6 is similar, we look at the component v 3 instead of v 2 , since need to show thatv 3 (0) = ∆u(1) = 0. Consequently, to conclude the proof we use Proposition 5.2c).□6. Multiplicity results: proof of Theorems 1.3 and 1.7By Propositions 5.1 and 5.2 we know that W u (P 2 ) ∩ {v 2 = 0} is a single point,which we call P ∗ = (P1 ∗ , P2 ∗ , P3 ∗ , P4 ∗ ), with P1 ∗ λ=S8(N−2)(N−4) and P 2 ∗ = 0.Let E = W u (P 1 ) ∩ {v 2 = 0}. Each regular radial solution of (1) corresponds toexactly one point v = (v 1 , . . . , v 4 ) ∈ E with v 1 > 0.Throughout this section we assume that 5 ≤ N ≤ 12. Let P 1 , P 2 be the stationarypoints of the system (11) defined in (13). Then P 1 has a 2-dimensional unstablemanifold W u (P 1 ) while P 2 has a 1-dimensional unstable manifold W u (P 2 ) and a3-dimensional stable manifold W s (P 2 ).Let V 0 : R → R 4 be the heteroclinic connection from P 1 to P 2 of Proposition 4.1and ˆV 0 = V 0 (−∞, ∞). Then ˆV 0 is contained in both W u (P 1 ) and W s (P 2 ).Lemma 6.1. W u (P 1 ) and W s (P 2 ) intersect transversally on points of ˆV 0 . Moreprecisely for points Q ∈ ˆV 0 sufficiently close to P 2 there are directions in the tangentplane to W u (P 1 ) which are almost parallel to v (1) , the tangent vector to W u (P 2 ) atP 2 .Proof. Let u(r, β) the solution to (33) defined in the maximal interval [0, R(β)).Let β 0 denote the unique value of β such that R(β 0 ) = ∞ andlim u(r, β 0) + 4 log(r) exists,r→∞see [1]. From the proof of Lemma 8 of this reference it follows that for β < β 0 thefollowing estimate holds:∂u ∂u(r, β) ≤∂r ∂r (r, β 0) − β 0 − βr ∀r ≥ 0.NThen ∂u∂β (r, β 0) satisfies the linearized equation at u(·, β 0 ) and(56)∂ ∂u∂r ∂β (r, β 0) ≥ r N∀r ≥ 0.Let v(t) = u(e t , β 0 ), t ∈ R and V = (v 1 , . . . , v 4 ) be defined by (10). Define Z = ∂V∂β .Then Z satisfiesZ ′ = (M + R(t))Zwhere M is the matrix defined in (14) and⎡⎤v 2 + 4 v 1 − 1 0 00 0 0 0R(t) =⎢⎣ 0 0 0 0⎥⎦ .0 0 0 0Recall that V (t) → P 2 as t → ∞. Moreover the convergence is exponential, that isthere are C, σ > 0 such that |V (t) − P 2 | ≤ Ce −σt for all t ≥ 0. This follows fromLemma 2.2 which shows that the system (11) is C 1 -conjugate to its linearizationnear P 2 (it suffices here to show that (11) is C 0 -conjugate to its linearization near

**MULTIPLICITY** **FOR** ∆ 2 u = λe u 17P 2 , which follows from the Hartman-Grobman theorem, see Theorem 7.1 in [31] orTheorem 1.1.3 in [28] ). Recall that the eigenvalues of M are ν 1 > 0 > ν 2 and ν 3 , ν 4which have negative real part and nonzero imaginary part. Let v (i) ∈ C 4 denote aneigenvector associated to ν i . By Theorem 8.1 in [13, Chap. 3] there are solutionsϕ k toϕ ′ k = (M + R(t))ϕ k , t > 0such that lim t→∞ ϕ k (t)e −νkt = v (k) . Then4∑(57)Z = c i ϕ ii=1for some constants c 1 , . . . , c 4 ∈ C. The condition (56) and the definitions in (10)imply that for some c > 0 ∣∣∣∣ ∂v 2∂β (t, β (58)0)∣ ≥ ce2t for all t ≥ 0.If c 1 = 0 in (57), since ν 2 , ν 3 , ν 4 have negative real, we would obtain that Z(t) → 0as t → ∞, contradicting (58). Hence c 1 ≠ 0 and thereforeZ = c 1 v (1) e ν1t + o(e ν1t ) as t → ∞.Since v (1) is the tangent vector to W u (P 2 ), we have that ∂V∂βW s (P 2 ) for t large. On the other hand ∂V∂βis not tangent tois tangent to W u (P 1 ) by construction.This shows that W s (P 2 ) and W u (P 1 ) intersect transversally on points of ˆV 0 closeto P 2 . By the invertibility of the flow away from the stationary points, W s (P 2 ) andW u (P 1 ) intersect transversally on all points of ˆV 0 .□Proof of Theorem 1.3.We will write generic points in the phase space R 4 as (v 1 , v 2 , v 3 , v 4 ). Let {e j :j = 1, . . . , 4 } denote the canonical basis of R 4 .The multiplicity results asserted in Theorem 1.3 are consequence of the followingclaims:a) E contains a spiral S about the point P ∗ ,b) S is contained in a 2-dimensional C 1 surface Σ ⊆ {v 2 = 0}, andc) the plane through P ∗ parallel to e 2 , e 3 , e 4 is transversal to the tangent planeto Σ at P ∗ .More precisely, after a C 1 diffeomorphism of a neighborhood of P ∗ to a neighborhoodof the origin in R 4 , which maps P ∗ to the origin, the curve S can beparametrized by a C 1 function of the form (r(s) cos(s), r(s) sin(s), 0, 0), s ∈ [0, ∞),such that r(s) > 0 for all s ≥ 0 and r(s) → 0 as s → ∞. Moreover onecan choose this diffeomorphism such that Σ corresponds to part of the surface{x = (x 1 , x 2 , x 3 , x 4 ) ∈ R 4 : x 3 = x 4 = 0}.Assume a), b) and c) have been proved and define hyperplane H λ = {v 1 =λ8(N−2)(N−4) } where λ > 0. If λ = λ S, the transversality condition c) ensures thatH λ is transversal to Σ, and we will see that this implies that H λ ∩ E containsinfinitely many points, which means that (1) has infinitely many radial regularsolutions. Indeed, after the C 1 diffeomorphism described above we can assume thatS = {(r(s) cos(s), r(s) sin(s), 0, 0) : s ≥ 0} and Σ = {x = (x 1 , x 2 , x 3 , x 4 ) ∈ R 4 :x 3 = x 4 = 0}. The hyperplane H λ is transformed into a C 1 hypersurface containingthe origin, which is transversal to Σ. Then H λ ∩Σ is a C 1 curve through the origin.

18 JUAN DÁVILA, ISABEL FLORES, AND IGNACIO GUERRAUsing polar coordinates we then see that H λ intersects the spiral S infinitely manytimes. If λ ≠ λ S but λ is close to λ S , by the transversality c) we have that H λ ∩ Econtains a large number of points, which yields a large number of radial regularsolutions of (1).In what follows we will prove a), b) and c). Let X t denote the flow generated by(11), that is, X t (ξ) is the solution to (11) at time t with initial condition X 0 (ξ) =ξ ∈ R 4 . For fixed ξ, X t (ξ) is defined for t in a maximal open interval containing 0.Let D be the 3 dimensional disk D = {v = (v 1 , . . . , v 4 ) : v 2 = 0, |v − P ∗ | < 1 },which by Proposition 5.2 is transversal to W u (P 2 ). Let B s ⊆ W s (P 2 ) ∩ N P2 be anopen neighborhood of P 2 relative to W s (P 2 ) diffeomorphic to a 3 dimensional disk.By choosing smaller neighborhoods if necessary, we may apply the λ-lemma of Palis[38]. Let D t be the connected component of X t (D) ∩ N P2 that contains X t (P ∗ ).Then, given ε > 0 there exists some t 0 < 0, |t 0 | large, such that D t0 contains a3-dimensional C 1 manifold M that is a ε C 1 -close to B s , which means that thereis a diffeomorphism η : M → B s such that ‖i − η‖ C 1 (M) ≤ ε where i : M → R 4 isthe inclusion map.Choose some point Q ∈ ˆV 0 such that Q ∈ N P2 . By Lemma 6.1 we may choose aC 1 curve contained in W u (P 1 ), say Γ = {γ(s) : |s| < δ } with γ : (−δ, δ) → R 4a C 1 function with γ(0) = Q, γ ′ (0) not tangent to W s (P 2 ) at Q. We can assumealso that this curve is contained in N P2 . Choosing ε small we can assume that Γintersects M.We have the following properties, which we prove after we complete the proof ofTheorem 1.3.Lemma 6.2. For large t, X t (Γ) ∩ M is a single point that we call P t and thefollowing properties hold:(59)(1) The collection of the points P t for large t forms a spiral.(2) There exists a 2 dimensional C 1 manifold ˜Σ that contains P t for all t large.(3) Let Q t0 be the intersection of M with W u (P 2 ). Then the tangent plane to˜Σ at Q t0 becomes parallel to the one generated by v (3) , v (4) (the eigenvectorscorresponding to ν 3 , ν 4 ) as ε → 0.(4) Moerover, for s > 0 suitably small the time t such that X t (γ(s)) ∈ Msatisfieswhere c > 0.s = ce −ν1t + o(e −ν1t )Let ˜S denote the collection {P t : t ≥ t 1 } where t 1 is suitably large. DefineS = X −t0 ( ˜S) and Σ = X −t0 (˜Σ). Since X −t0 is a smooth diffeomorphism from Mto a neighborhood of P ∗ inside the hyperplane {v 2 = 0} we see that S is a spiralcontained in a C 1 surface Σ. The points of S belong to W u (P 1 ) because they wereobtained though the flow from points in X t (Γ).This ends the proof of part b).We now prove statement c). It is sufficient to show that inside the space {v 2 = 0}the plane generated by e 3 , e 4 is transversal to the tangent space to Σ at P ∗ . LetV = (v 1 , . . . , v 4 ) : (−∞, 0] → R 4 denote the trajectory corresponding to the weaklysingular solution, that is, lim t→−∞ V (t) = P 2 , v 2 (0) = 0. To prove our claimwe need to transport the plane generated by e 3 and e 4 back along V and thisis accomplished by solving the linearized equation around V . More precisely, let

**MULTIPLICITY** **FOR** ∆ 2 u = λe u 19Z, ˜Z : (−∞, 0] → R 4 be solutions to the linearization of (11) around V , that is,Z = (z 1 , z 2 , z 3 , z 4 ) satisfies for t < 0⎧z 1 ′ = z 1 (v 2 + 4) + v 1 z 2⎪⎨ z 2 ′ = −(N − 2)z 2 + z 3(60)z 3 ′ = 2z 3 + z 4⎪⎩z 4 ′ = −(N − 4)z 4 + 8(N − 2)(N − 4)z 1and similarly for ˜Z = (˜z 1 , ˜z 2 , ˜z 3 , ˜z 4 ). As final conditions we take Z(0) = e 3 , ˜Z(0) =e 4 .By Theorem 8.1 in [13, Chap. 3] there are solutions ϕ k : (−∞, 0] → C 4 to (60)such that(61)lim ϕ k(t)e −νkt = v (k)t→−∞where v (1) , . . . , v (4) are the eigenvectors of M. Recall that v (1) , v (2) are real, andv (3) , v (4) are complex conjugate. Thus one can assume that ϕ 1 , ϕ 2 are real ϕ 3 , ϕ 4are complex conjugate. Then4∑4∑Z(t) = c i ϕ i (t), and ˜Z(t) = ˜c i ϕ i (t)i=1for some constants c 1 , . . . , c 4 , ˜c 1 , . . . , ˜c 4 ∈ C. We note that c 1 , c 2 , ˜c 1 , ˜c 2 are real andc 3 ϕ 3 (t) + c 4 ϕ 4 (t) ∈ R, ˜c 3 ϕ 3 (t) + ˜c 4 ϕ 4 (t) ∈ R for all t ≤ 0.We claim that(62)c 2 ≠ 0 or ˜c 2 ≠ 0.i=1Assume, by contradiction, that c 2 = 0 and ˜c 2 = 0. Definef(t) = e (N−4)t (z4 (t)˜z 1 (t)v 1 (t)− z 3 (t)˜z 2 (t) + z 2 (t)˜z 3 (t) − z 1(t)˜z 4 (t)v 1 (t)), ∀t ≤ 0.A calculation using (60) shows that f is constant. Using the final conditions for Zand ˜Z we see that f(0) = 0 and hencef(t) = 0 ∀t ≤ 0.Using (61), (16) and the assumption c 2 = 0, ˜c 2 = 0 we can computewherelim f(t) = (c 3˜c 4 − ˜c 3 c 4 )Bt→−∞B = ν 3 (ν 3 + N − 2)(ν 3 − 2) − ν 3 (ν 3 + N − 2)ν 4 + ν 4 (ν 4 + N − 2)ν 3− ν 4 (ν 4 + N − 2)(ν 4 − 2)= − 1 2 M 2(N) √ M 1 (N) − M 2 (N).Thus B ∈ iR, B ≠ 0 and we conclude that (c 3˜c 4 − ˜c 3 c 4 ) = 0. This means thatthere exists a λ ∈ C such that ˜c k = λc k , k = 3, 4. Since c 3 ϕ 3 (t) + c 4 ϕ 4 (t) ∈ R,˜c 3 ϕ 3 (t) + ˜c 4 ϕ 4 (t) ∈ R for all t ≤ 0, ν 1 > 0 and we assume that c 2 = ˜c 2 = 0, wemust have λ ∈ R. Using Z(0) = e 3 and ˜Z(0) = e 4 we see that(˜c 1 − λc 1 )ϕ 1 (0) = e 4 − λe 3 .

20 JUAN DÁVILA, ISABEL FLORES, AND IGNACIO GUERRABut ϕ 1 = cV ′ , for some constant c ∈ R, since both solve (60) and both tend to 0 ast → −∞. We know that v ′ 2(0) > 0 by Proposition 5.2 and this implies ˜c 1 − λc 1 = 0,a contradiction.Finally, the condition (62) implies the assertion c). Indeed, let us recall thatΣ = X −t0 (˜Σ) where ˜Σ is defined in Lemma 6.2 and t 0 < 0, with |t 0 | large. Usingproperty 3 of that Lemma and the condition (62) we see that for |t 0 | large at leastone of the vectors Z(t 0 ) or ˜Z(t 0 ) is transversal to the tangent plane to ˜Σ at Q t0 .To finish the proof of Theorem 1.3 we still need to verify one assertion: forλ ≠ λ S (1) has a finite number of solutions. We will do this in Proposition 7.6 ofSection 7.□Proof of Lemma 6.2. By Lemma 2.2 there is a C 1 diffeomorphism R : N P2 → N 0from an open neighborhood N P2 of P 2 to an open neighborhood N 0 of 0 withR(P 2 ) = 0, det(R ′ (P 2 )) > 0, such that RX t R −1 = L t where L t is the flow generatedby M, and the formula holds in some neighborhood of the origin. Note that L t =e Mt .Thanks to the conjugation R, to prove the lemma we may assume that P 2 is atthe origin and that near the origin the flow is given by L t = e Mt . Thus W s (P 2 ) ina neighborhood of the origin is {(y 1 , . . . , y 4 ) : y 1 = 0 } and B s = {(y 1 , . . . , y 4 ) :y 1 = 0, |y| < δ} for some δ > 0. We can also assume that the heteroclinic orbit V 0near the origin in the new variables is given by(63)V 0 (t) = (0, c 2 e ν2t , c 3 Re(e ν3t ), c 4 Im(e ν3t )), t ≥ 0for some constants c 2 , c 3 , c 4 . By Lemma 4.3 the curve V 0 cannot have a directionthat becomes parallel to e 2 = (0, 1, 0, 0) as t → ∞. Since |ν 2 | > |Re(ν 3 )| by (15),c 3 ≠ 0 or c 4 ≠ 0. By choosing ε small, we can assume that the normal vector to Mnear P ∗ is almost parallel to e 1 = (1, 0, 0, 0) after the change of variables. Thus bypassing to a subset of M we may assume that M is a C 1 graph over the variables(y 2 , y 3 , y 4 ), that is, there exists a C 1 function ψ : {y ′ = (y 2 , y 3 , y 4 ) ∈ R 3 , |y ′ | 0 such thatM = {(ψ(y ′ ), y ′ ) : y ′ ∈ R 3 , |y ′ | < δ}.By Lemma 6.1 the tangent plane to W u (P 1 ) at points close to the origin containsvectors almost parallel to e 1 = (1, 0, 0, 0) and hence γ ′ 1(0) ≠ 0. Using the implicitfunction theorem we see that for large t the intersection of M and L t (Γ) occurs atpoints of the formP t = (γ 1 (s)e ν1t , γ 2 (s)e ν2t , γ 3 (s)Re(e ν3t ), γ 4 (s)Im(e ν3t ))where s = ce −ν1t + o(e −ν1t ) as t → ∞ for some c > 0. Since c 3 ≠ 0 or c 4 ≠ 0 in(63) we can define a surface(64)˜Σ = {y = (y 1 , y 2 , y 3 , y 4 ) : |y| < δ, y 1 = ψ(y 2 , y 3 , y 4 ), y 2 = g(y 3 , y 4 )}that contains the points P t , where g is smooth away from the origin and has thepropertyg(y 3 , y 4 ) = O(|(y 3 , y 4 )| β )with β = ν 2 /Re(ν 3 ). Thanks to (15) we see that β > 1. Therefore g is C 1 and ˜Σis a C 1 surface.□Proof of Theorem 1.7.

**MULTIPLICITY** **FOR** ∆ 2 u = λe u 21By Propositions 5.1 and 5.2 we know that W u (P 2 ) ∩ {v 3 = 0} is a single point,which we call ¯P ∗ = ( ¯P 1 ∗ , ¯P 2 ∗ , ¯P 3 ∗ , ¯P 4 ∗ ), with ¯P 1 ∗ λ=S8(N−2)(N−4) and ¯P 3 ∗ = 0.As in Theorem 1.3, he multiplicity results asserted in Theorem 1.7 are consequenceof the following claims:a) E := W u (P 1 ) ∩ {v 3 = 0} contains a spiral S about the point ¯P ∗ ,b) S is contained in a 2-dimensional C 1 surface Σ ⊆ {v 3 = 0}, andc) the plane through ¯P ∗ parallel to e 2 , e 3 , e 4 is transversal to the tangent planeto Σ at ¯P ∗ .The proofs are similar to the Dirichlet case, now changing v 2 = 0 for v 3 = 0. Soto prove c) it will be sufficient now to show that inside the space {v 3 = 0} the planegenerated by e 2 , e 4 is transversal to the tangent space to Σ at ¯P ∗ . We define nowZ satisfying (60) with the final condition Z(0) = e 2 , and ˜Z remains unchanged. Inthe same form we claim that (62) holds. Indeed using the same argument as beforewith Z(0) = e 2 and ˜Z(0) = e 4 , we find(˜c 1 − λc 1 )ϕ 1 (0) = e 4 − λe 2 .But we know by Proposition 5.2 that v 3(0) ′ > 0 and this implies ˜c 1 − λc 1 = 0, acontradiction. The rest of the proof is the same.□7. Structure of the solution setIn this section we study the properties of the solution setC = { (λ, u) ∈ (0, ∞) × C 4 (B) : u is radial and solves (1) }.We assume here that N ≥ 5. We will see that all regular radial solutions u of(1) are characterized by u(0) and that this value ranges from 0 to +∞. To provethe first assertion we follow the strategy of Guo and Wei [30]. For this we recall acomparison result established by McKenna and Reichel [36, Lemma 3.2].Lemma 7.1. Assume that f : R → R is differentiable and increasing. Let u, v ∈C 4 ([0, R)), R > 0 be such that∀r ∈ [0, R)∆ 2 u(r) − f(u(r)) ≥ ∆ 2 v(r) − f(v(r)),u(0) ≥ v(0), u ′ (0) ≥ v ′ (0), ∆u(0) ≥ ∆v(0), (∆u) ′ (0) ≥ (∆v) ′ (0).Then for all r ∈ [0, R)(65)Moreoveru(r) ≥ v(r), u ′ (r) ≥ v ′ (r), ∆u(r) ≥ ∆v(r), (∆u) ′ (r) ≥ (∆v) ′ (r).i) the initial point 0 can be replaced by any initial point ρ > 0 if all four initialdata are weakly ordered,ii) a strict inequality in one of the initial data at ρ ≥ 0 or in the differentialinequality on (ρ, R) implies a strict ordering of u, u ′ , ∆u, (∆u) ′ and v, v ′ ,∆v, (∆v) ′ in (65).Analogously to [30, Lemma 5.1] we have:Lemma 7.2. Suppose that u 1 , u 2 are smooth radial solutions of (1) associated toparameters λ 1 > 0, λ 2 > 0 such that u 1 (0) = u 2 (0). Then λ 1 = λ 2 and u 1 ≡ u 2 .

22 JUAN DÁVILA, ISABEL FLORES, AND IGNACIO GUERRAProof. Suppose we have smooth radial solutions u 1 , u 2 of (1) associated to parametersλ 1 > λ 2 such that u 1 (0) = u 2 (0).For j = 1, 2v j (r) = u(λ−1/4 j r), for r ∈ [0, λ 1/4j ]u 1 (0)Then v j satisfies∆ 2 v j = f(v j ) for r ∈ [0, λ 1/4j ]where f(t) = 1u 1(0) eu1(0)t .Assume that ∆v 1 (0) < ∆v 2 (0). Then by Lemma 7.1 v 1 (r) < v 2 (r) for all r ∈[0, λ 1/42 ]. In particular v 1 (λ 1/42 ) < v 2 (λ 1/42 ) = 0 which is impossible because v 1 (r) >0 for all r ∈ [0, λ 1/41 ).Assume now that ∆v 1 (0) > ∆v 2 (0). Then by Lemma 7.1 v 1 (r) > v 2 (r), v ′ 1(r) >v ′ 2(r), ∆v 1 (r) > ∆v ′ 2(r), (∆v 1 ) ′ (r) > ∆v 2 ) ′ (r) for all r ∈ [0, λ 1/42 ]. Since v 1 isdefined up to λ 1/41 , v 2 can be extended to [0, λ 1/41 ] and the previous inequalities arevalid in this interval. Evaluating at λ 1/41 we deduce that(66)0 = v ′ 1(λ 1/41 ) > v ′ 2(λ 1/41 ).Since w = ∆v 2 satisfies ∆w = f(v 2 ) > 0 it is subharmonic and hence w(r 1 ) ≤ w(r 2 )for all 0 ≤ r 1 ≤ r 2 ≤ λ 1/41 . But the Green function for the bilaplacian in the ballof radius R > 0 with Dirichlet boundary conditions G(x, y) satisfies G(x, y) ≥c(R − |x|) 2 (R − |y|) 2 for some c > 0, see [27]. This implies that ∆v 2 (λ 1/42 ) > 0 andtherefore w(r) > 0 for all r ∈ [λ 1/42 , λ 1/41 ]. Thusr N−1 v ′ 2(r) =∫ rλ 1/42t N−1 ∆v 2 (t) dt > 0 for all r ∈ (λ 1/42 , λ 1/41 ].In particular v 2(λ ′ 1/41 ) > 0 which contradicts (66).It follows that ∆v 1 (0) = ∆v 2 (0) and hence v 1 ≡ v 2 . This implies that λ 1 = λ 2and that u 1 ≡ u 2 .□Proof of Theorem 1.4. By [2, Theorem 3] there exists λ ∗ such that if 0 ≤ λ < λ ∗then (8) has a minimal smooth solution u λ and if λ > λ ∗ then (8) has no weaksolution. The limit u ∗ = lim λ↗λ ∗ u λ exists pointwise, belongs to H 2 (B) and is aweak solution to (8) in the sense (5). The functions u λ , 0 ≤ λ < λ ∗ and u ∗ areradially symmetric and radially decreasing. Now, by [15, Theorem 1.4] we knowthat u ∗ is unbounded if N ≥ 13.Fix ¯λ ∈ (0, λ ∗ ) and let v be a smooth radial solution to (1) with parameter ¯λ.Since λ ∈ (0, λ ∗ ) → u λ (0) depends continuously on λ, and since lim λ→λ ∗ u λ (0) → ∞we see that there exists some λ ∈ (0, λ ∗ ) such that v(0) = u λ (0). By Lemma 7.2we conclude that ¯λ = λ and v = u λ .By [15, Proposition 1.8] we also know that u ∗ is a weakly singular solution. ByTheorem 1.2 there is no weakly singular solution for any other value different thanλ ∗ . Moreover, for λ = λ ∗ by [15, Theorem 1.2], u ∗ is the unique weak solution of(1). □As in Section 6, we let E = W u (P 1 )∩{v 2 = 0} and recall that each regular radialsolution of (1) corresponds to exactly one point v = (v 1 , . . . , v 4 ) ∈ E with v 1 > 0.It is therefore natural to define E 0 = W u (P 1 ) ∩ {v 2 = 0, v 1 > 0}.

**MULTIPLICITY** **FOR** ∆ 2 u = λe u 23The curve of solutions C can also be parametrized by the shooting problem (33).Let u β be the solution of (33) defined in the maximal interval of existence [0, R(β)).In Theorem 2 of [1], it is shown that for problem (33), given β ∈ (β 0 , 0) there existsa unique R 0 ∈ (0, R(β)) such that u ′ β (R 0) = 0. Moreover, u ′ β (r) < 0 in (0, R 0)and u ′ β (r) > 0 in (R 0, R(β)). It is not difficult to verify that R 0 (β) defines a C 1function of β ∈ (β 0 , 0).For β ∈ (β 0 , 0) we let V β = (v 1,β , . . . , v 4,β ) : (−∞, T (β)) → R 4 be the functionobtained from v β (t) = u β (e t ) through the transformations (10), where T (β) =log(R(β)). Define also T 0 (β) = log(R 0 (β)) for β ∈ (β 0 , 0). Then V β satisfies (11)and v 2,β (T 0 (β)) = 0. Since V β (−∞, T (β)) lies in W u (P 1 ) we have V β (T 0 (β)) ∈ E.Let us define φ : (β 0 , 0) → R 4 byφ(β) = V β (T 0 (β)) for all β ∈ (β 0 , 0).It will also be convenient to introduce, for β ∈ (β 0 , 0) the function(67)U β (r) = u β (rR 0 (β)) − u β (R 0 (β)), 0 ≤ r ≤ 1.Then U β is a solution of (1) for the value of λ = 8(N − 2)(N − 4)R 0 (β) 4 e u β(R 0(β)) .Lemma 7.3. We havelim φ(β) = P ∗ ,β→β 0lim T 0 (β) = +∞ and lim U β (0) = +∞.β→β 0 β→β 0Proof. Let M, Q and Γ = {γ(s) : |s| < δ } with γ : (−δ, δ) → R 4 be asin the proof of Theorem 1.3. We choose γ ′ (0) close to the direction v (1) . LetΓ 0 = {γ(s) : 0 < s < δ }. Then, taking δ sufficiently small, we can definethe function τ : Γ 0 → R + where τ(p) is such that X τ(p) (p) ∈ M. Then τ iscontinuous, and by (59) τ(γ(s)) = 1 ν 1log(1/s) + o(log(1/s)) as s → 0, which showsthat τ(p) → +∞ as p → Q. We note that for β ∈ (β 0 , 0) and β close to β 0 thereis some time t 1 (β) such that V β (t 1 (β)) ∈ Γ 0 . As β → β 0 , V β (t 1 (β)) → Q and thenT 0 (β) → ∞.As in the proof of Lemma 6.2 one can also show that as p → Q, p ∈ Γ 0 thepoint X τ(p) (p) approaches the intersection of M with W u (P 2 ). This shows thatφ(β) → P ∗ as β → β 0 .Finally, since T 0 (β) → ∞ as β → β 0 we see from formula (67) that U β (0) → ∞as β → β 0 .□Lemma 7.4. We havelim φ(β) = 0 and lim U β(0) = 0.β→0 β→0Proof. Using the implicit function theorem there is δ > 0 such that for λ > 0 smallthere is a unique small solution u λ of (1). The map λ ↦→ u λ is C 1 into C 4 (B). Setũ λ (r) = u λ (A λ r) − u λ (0) where A λ = ( 8(N−2)(N−4) ) 1/4 . Then ũλe u λ (0)λ is the solutionof (33) with β = β(λ) = A 2 λ ∆u λ(0) by uniqueness of that initial value problem. Inparticular U β = u λ if β = A 2 λ ∆u λ(0).1Using Theorem 4 of [2] we know that u λ /λ →8N(N+2) (1 − r2 ) 2 uniformly inB as λ → 0. By elliptic estimates the convergence is also in C 4 (B). It followsthat β(λ) = O(λ 1/2 ) as λ → 0. Thus for small β < 0 the solution of the shootingproblem (33) is ũ λ with λ > 0 such that A 2 λ ∆u λ(0) = β, and this λ > 0 is uniquelydetermined. Then as β → 0, λ → 0 and U β (0) = u λ (0) → 0. Also R 0 (β) = 1/A λ →0 and φ(β) → 0 as β → 0 (since φ(β) is expressed in terms of derivatives of u λ ). □

24 JUAN DÁVILA, ISABEL FLORES, AND IGNACIO GUERRALemma 7.5. We haveand is a real analytic curve.E 0 = {φ(β) : β ∈ (β 0 , 0)}By E 0 being real analytic we mean that each point of this set as a neighborhoodin E 0 which can be parametrized by a real analytic function.Proof. By construction φ(β) ∈ E 0 for each β ∈ (β 0 , 0). To prove E 0 ⊆ {φ(β) :β ∈ (β 0 , 0)} we need to show that given any radial regular solution u of (1) thereexists β ∈ (β 0 , 0) such that u = U β . Using Lemma 7.2 it is sufficient to find β suchthat u(0) = U β (0). We have by Lemma 7.3 that U β (0) → +∞ as β → β 0 , while byLemma 7.4 that U β (0) → 0 as β → 0. Since U β (0) varies continuously with β thereis β ∈ (β 0 , 0) such that u(0) = U β (0).The unstable manifold of P 1 is a real analytic surface, since the vector field is realanalytic, in fact a polynomial, see for instance [12, p. 104]. By the implicit functiontheorem, each point in W u (P 1 )∩{v 2 = 0} where the intersection is transversal has aneighborhood in this set which can be parametrized by a real analytic function. Atpoints in the intersection of the sets W u (P 1 ) and {v 2 = 0, v 1 > 0} the transversalitycondition holds. Indeed, the points in in this set are given bu φ(β) with β ∈ (β 0 , 0).Let U β be defined by (67) and recall that is is a positive solution of (1). We recallalso that Green function in the ball with Dirichlet boundary conditions G(x, y)satisfies G(x, y) ≥ c(1 − |x|) 2 (1 − |y|) 2 for some c > 0, see [27]. So we actually haveU ′′β (1) > 0. This implies that at t = T 0(β) we have v ′ 2(t) = e 2t ∆U β (1) > 0 whichshows that the intersection is transversal. It follows that the intersection of the setsW u (P 1 ) and {v 2 = 0, v 1 > 0} is a real analytic curve.□Proof of Theorem 1.5. It is consequence of Lemmas 7.2 and 7.5.Proposition 7.6. Assume 5 ≤ N ≤ 12.number of regular radial solutions of (1).If λ ≠ λ S , then there exists a finiteProof. By Lemmas 7.3 and 7.4 we can consider P 1 and P ∗ as the endpoints of E 0 .If λ = 0 then u = 0 is the only solution of (1). Let λ ≠ 0, λ ≠ λ ∗ . By analyticityE 0 ∩ {v 1 = λ} can only accumulate at either P 1 or P ∗ . Since P ∗ is not included in{v 1 = λ} accumulation in P ∗ is not possible. Similarly, since P 1 ∉ {v 1 = λ} the setE 0 ∩ {v 1 = λ} can not acculmulate at P 1 . Thus E 0 ∩ {v 1 = λ} consists of a finitenumber of points, which correspond to regular radial solutions of (1). □Acknowledgments. J.D. was partially supported by Fondecyt 1090167, Fondapand Basal-CMM grants, I.F. was partially supported by Fondecyt 1090518 andI.G. was partially supported by Fondecyt 1090470.References[1] G. Arioli, F. Gazzola, H.-C. Grunau, Entire solutions for a semilinear fourth order ellipticproblem with exponential nonlinearity. J. Differential Equations 230 (2006), no. 2, 743–770.[2] G. Arioli, F. Gazzola, H.-C. Grunau, E. Mitidieri, A semilinear fourth order elliptic problemwith exponential nonlinearity. SIAM J. Math. Anal. 36 (2005), no. 4, 1226–1258.[3] R. Bamón, I. Flores, M. del Pino, Ground states of semilinear elliptic equations: a geometricapproach. Ann. Inst. H. Poincaré Anal. Non Linéaire 17 (2000), no. 5, 551–581.[4] J. Bebernes, D. Eberly, Mathematical problems from combustion theory. Applied MathematicalSciences, 83. Springer-Verlag, New York, 1989.□

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26 JUAN DÁVILA, ISABEL FLORES, AND IGNACIO GUERRA[33] P. Karageorgis,Stability and intersection properties of solutions to the nonlinear biharmonicequation. Preprint.[34] P.-L. Lions, On the existence of positive solutions of semilinear elliptic equations. SIAMRev. 24 (1982), no. 4, 441–467.[35] J. Liouville, Sur l’equation aux différences partielles d2 log λdudv ± λ2a 2 = 0. J. Math. Pure Appl.36 (1853), 71–72.[36] P.J. McKenna, W. Reichel, Radial solutions of singular nonlinear biharmonic equations andapplications to conformal geometry. Electron. J. Differential Equations (2003), No. 37, 13pp.[37] È. Mitidieri, S.I. Pokhozhaev, A priori estimates and the absence of solutions of nonlinearpartial differential equations and inequalities. Proc. Steklov Inst. Math. 2001, no. 3 (234),1–362.[38] J. Palis, On Morse-Smale dynamical systems. Topology 8 (1968), 385–404.[39] D. Ruelle, Elements of differentiable dynamics and bifurcation theory. Academic Press, Inc.,Boston, MA, 1989.J. Dávila: Departamento de Ingeniería Matemática and Centro de ModelamientoMatemático (UMI 2807, CNRS), Universidad de Chile. Casilla 170/3 Correo 3, Santiago,Chile.E-mail address: jdavila@dim.uchile.clI. Flores: Departamento de Matemática, Facultad de Ciencias Físicas y MatemáticasUniversidad de Concepción, Casilla 160 C, Concepción, Chile.E-mail address: iflores@udec.clI. Guerra: Departamento de Matematica y C.C., Facultad de Ciencia, Universidad deSantiago de Chile, Casilla 307, Correo 2, Santiago, Chile.E-mail address: ignacio.guerra@usach.cl