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

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

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

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