The space of 4-ended solutions to the Allen-Cahn equation on the plane
aa r X i v : . [ m a t h . A P ] S e p THE SPACE OF -ENDED SOLUTIONS TO THE ALLEN-CAHNEQUATION IN THE PLANE MICHA L KOWALCZYK, YONG LIU, AND FRANK PACARD
Abstract.
We are interested in entire solutions of the Allen-Cahn equation∆ u − F ′ ( u ) = 0 which have some special structure at infinity. In this equation,the function F is an even, bistable function. The solutions we are interestedin have their zero set asymptotic to 4 half oriented affine lines at infinity and,along each of these half affine lines, the solutions are asymptotic to the onedimensional heteroclinic solution : such solutions are called 4 -ended solutions .The main result of our paper states that, for any θ ∈ (0 , π/ θ, π − θ, π + θ and 2 π − θ with the x -axis. This paper is part of a program whose aim is toclassify all 2 k -ended solutions of the Allen-Cahn equation in dimension 2, for k ≥ Introduction
In this paper, we are interested in entire solutions of the Allen-Cahn equation(1.1) ∆ u − F ′ ( u ) = 0 , in R , where the function F is a smooth, double well potential. This means that F is even, nonnegative and has only two zeros which will be chosen to be at ± F ′′ ( ± = 0 , and also that F ′ ( t ) = 0 , for all t ∈ (0 , . It is known that (1.1) has a solution whose nodal set is any given straight line.These special solutions, which will be referred to as the heteroclinic solutions , areconstructed using the heteroclinic, one dimensional solution of (1.1), namely thefunction H defined on R , solution of H ′′ − F ′ ( H ) = 0 , (1.2)which is odd and tends to − −∞ (respectively at + ∞ ).More precisely, we have the : Definition 1.1.
Given r ∈ R and e ∈ R such that | e | = 1 , the heteroclinicsolutions with end λ := r e ⊥ + R e is defined by u ( x ) := H ( x · e ⊥ − r ) , where ⊥ denotes the rotation of angle π/ in the plane. Observe that this construction extends in any dimension to produce solutionswhose level sets are hyperplanes.The famous de Giorgi conjecture asserts that (in space dimension less than orequal to 8), if u is a bounded solution of (1.1) which is monotone in one direction,then u has to be one of the above defined heteroclinic solutions. This conjecture isknown to hold when the space dimension is equal to 2 [11], in dimension 3 [1] andin dimension 4 to 8 [18] under some mild additional assumption. Counterexampleshave been constructed in all dimensions N ≥ R and which have some special structure at infinity, namely their zero set is, atinfinity, asymptotic to 4 oriented affine lines : such solutions are called 4 -endedsolutions and will be precisely defined in the next section. The main result of ourpaper states that, for any θ ∈ (0 , π/ θ, π − θ, π + θ and 2 π − θ with the x -axis.2. The space of -ended solutions In order to proceed, we need to define precisely the class of entire solutions of(1.1) we are interested in. As already mentioned, these solutions have the propertythat their nodal sets are, away from a compact, asymptotic to a finite (even) numberof half oriented affine lines, which are called the ends of the solutions. The conceptof solutions with a finite number of ends was first introduced in [5] and, for the sakeof completeness, we recall the precise definitions in the case of 4-ended solutions.An oriented affine line λ ⊂ R can be uniquely written as λ := r e ⊥ + R e , for some r ∈ R and some unit vector e ∈ S , which defines the orientation of λ . Werecall that ⊥ denotes the rotation by π/ R . Writing e = (cos θ, sin θ ), we getthe usual coordinates ( r, θ ) which allow to identify the set of oriented affine lineswith R × S .Assume that we are given 4 oriented affine lines λ , . . . , λ ⊂ R which aredefined by λ j := r j e ⊥ j + R e j , and assume that these oriented affine lines have corresponding angles θ , . . . , θ satisfying θ < θ < θ < θ < π + θ . In this case, we will say that the 4 oriented affine lines are ordered and we willdenote by Λ ord the set of 4 oriented affine lines. It is easy to check that for all R > j = 1 , . . . ,
4, there exists s j ∈ R such that :(i) The point r j e ⊥ j + s j e j belongs to the circle ∂B R .(ii) The half affine lines(2.3) λ + j := r j e ⊥ j + s j e j + R + e j , are disjoint and included in R − B R . -ENDED SOLUTIONS 3 (iii) The minimum of the distance between two distinct half affine lines λ + i and λ + j is larger than 4.The set of half affine lines λ +1 , . . . , λ +4 together with the circle ∂B R induce adecomposition of R into 5 slightly overlapping connected components R = Ω ∪ Ω ∪ . . . ∪ Ω , where Ω := B R +1 andΩ j := (cid:0) R − B R − (cid:1) ∩ (cid:8) x ∈ R : dist( x , λ + j ) < dist( x , λ + i ) + 2 , ∀ i = j (cid:9) , (2.4)for j = 1 , . . . ,
4. Here, dist( · , λ + j ) denotes the distance to λ + j . Observe that, for all j = 1 , . . . ,
4, the set Ω j contains the half affine line λ + j .Let I , I , . . . , I be a smooth partition of unity of R which is subordinate tothe above decomposition. Hence X j =0 I j ≡ , and the support of I j is included in Ω j . Without loss of generality, we can alsoassume that I ≡ ′ := B R − , and I j ≡ ′ j := (cid:0) R − B R − (cid:1) ∩ (cid:8) x ∈ R : dist( x , λ + j ) < dist( x , λ + i ) − , ∀ i = j (cid:9) , for j = 1 , . . . ,
4. Finally, without loss of generality, we can assume that k I j k C ( R ) ≤ C. With these notations at hand, we define u λ := X j =1 ( − j I j H (dist s ( · , λ j )) , (2.5)where λ := ( λ , . . . , λ ) and(2.6) dist s ( x , λ j ) := x · e ⊥ j − r j , denotes the signed distance from a point x ∈ R to λ j .Observe that, by construction, the function u λ is, away from a compact and upto a sign, asymptotic to copies of the heteroclinic solution with ends λ , . . . , λ .Let S denote the set of functions u which are defined in R and which satisfy(2.7) u − u λ ∈ W , ( R ) , for some ordered set of oriented affine lines λ , . . . , λ ⊂ R . We also define thedecomposition operator J by J : S −→ W , ( R ) × Λ ord u ( u − u λ , λ ) . The topology on S is the one for which the operator J is continuous (the targetspace being endowed with the product topology).We now have the : MICHA L KOWALCZYK, YONG LIU, AND FRANK PACARD
Definition 2.1.
The set M is defined to be the set of solutions u of (1.1) whichbelong to S . It is known that M is not empty. For example, the saddle solution constructedin [4] belongs to M , the nodal set of this solution is the union of the two lines y = ± x . Another important fact, also proven in [4] or in [12], is that up to asign and a rigid motion, the saddle solution is the unique solution whose nodal setcoincides with the union of the two lines y = ± x . The solutions constructed in [6]are also elements of M and we shall return to this point later on.Recall from [5], that a solution u of (1.1) is said to be nondegenerate if there isno w ∈ W , ( R ) − { } which is in the kernel of L := − ∆ + F ′′ ( u ) , and which decays exponentially at infinity.As far as the structure of the set of 4-ended solutions is concerned, the mainresult of [5] asserts that : Theorem 2.1. [5]
Assume that u ∈ M is nondegenerate , then, close to u , M isa -dimensional smooth manifold. Observe that, given u ∈ M , translations and rotations of u are also elements of M and this accounts for 3 of the 4 formal dimensions of M , moreover, if u ∈ M then − u ∈ M .All the 4-ended solutions constructed so far have two axis of symmetry and infact, it follows from a result of C. Gui [13] that : Theorem 2.2. [13]
Assume that u ∈ M . Then, there exists a rigid motion g suchthat ¯ u := u ◦ g is even with respect to the x -axis and the y -axis, namely ¯ u ( x, y ) = ¯ u ( − x, y ) = ¯ u ( x, − y ) . (2.8) In addition, ¯ u is a monotone function of both the x and y variables in the upperright quadrant Q x defined by Q x := { ( x, y ) ∈ R : x > y > } , and, changing the sign of ¯ u if this is necessary, we can assume that ∂ x ¯ u < and ∂ y ¯ u > , in Q x . Thanks to this result, we can define the moduli space of 4-ended solutions by :
Definition 2.2.
The set M even is defined to be the set of u ∈ S which are solutionsof (1.1), are even with respect to the x -axis and the y -axis and which tend to +1 as at infinity along the y -axis (and tend to − at infinity along the x -axis). Inparticular, ∂ x u < and ∂ y u > , in the upper right quadrant Q x . When studying M even , we restrict our attention to functions which are evenwith respect to the x -axis and the y -axis and, in this case, a solution u ∈ M even is -ENDED SOLUTIONS 5 said to be even-nondegenerate if there is no w ∈ W , ( R ) −{ } , which is symmetricwith respect to the x -axis and the y -axis, belongs to the kernel of L := − ∆ + F ′′ ( u ) , and which decays exponentially at infinity.In the equivariant case (namely solutions which are invariant under both thesymmetry with respect to the x -axis and the y -axis), Theorem 2.1 reduces to : Theorem 2.3. [5]
Assume that u ∈ M even is even-nondegenerate , then, close to u , M even is a -dimensional smooth manifold. Any solution u ∈ M even has a nodal set which is asymptotic to 4 half orientedaffine lines and, given the symmetries of u , these half oriented affine lines areimages of each other by the symmetries with respect to the x -axis and the y -axis.In particular, there is at most one of these half oriented affine line λ := r e ⊥ + R e , which is included in the upper right quadrant Q x . Writing e = (cos θ, sin θ ) where θ ∈ (0 , π/ F : M even → ( − π/ , π/ × R ,u ( θ − π/ , r ) . For example, the image by F of the saddle solution defined in [4] is precisely (0 , F of the solutions constructed in [6] correspond to parameters( θ, r ) where θ is close to ± π/ r is close to ∓∞ . Remark 2.1.
Let us observe that, if u ∈ M even , then ¯ u defined by ¯ u ( x, y ) = − u ( y − x ) , also belongs to M even and F (¯ u ) = −F ( u ) . In this paper, we are interested in the understanding of M even . To begin with,we prove that : Theorem 2.4 (Nondegeneracy) . Any u ∈ M is nondegenerate and hence any u ∈ M even is even-nondegenerate. As a consequence of this result, we find that all connected components of M even are one-dimensional smooth manifolds. Moreover, as a byproduct of the proof ofthis result, we also obtain that the image by F of any connected component of M even is a smooth immersed curve in ( − π/ , π/ × R . Thanks to Remark 2.1, wefind that the image of any connected component of M even by F is invariant underthe action of the symmetry with respect to (0 , classifying map to be the projection of F onto the firstvariable P : M even → ( − π/ , π/ ,u θ − π/ . Our second result reads :
MICHA L KOWALCZYK, YONG LIU, AND FRANK PACARD
Theorem 2.5 (Properness) . The mapping P is proper, i.e. the pre-image of acompact in ( − π/ , π/ is compact in M even (endowed with the topology inducedby the one of S ). The solutions with almost parallel ends constructed in [6] belong to one of theconnected component of M even and we also know that the saddle solution alsobelongs to a connected component of M even . In principle, it could be possiblethat M even contained many different connected components and it could also bepossible that M even contained connected components which are diffeomorphic to S . Nevertheless, we prove that : Theorem 2.6.
All connected components of M even are diffeomorphic to R , i.e.there is no closed loop in M even . Looking at the image by P of the connected component of M even which containsthe saddle solution, we conclude from the above results that : Theorem 2.7 (Surjectivity of P ) . The mapping P is onto. As a consequence, for any θ ∈ (0 , π/ u ∈ M even whosenodal set at infinity is asymptotic to the half oriented affine lines whose angles withthe x -axis are given by θ, π − θ, π + θ and 2 π − θ .Given all the evidence we have, it is tempting to conjecture that M even has onlyone connected component and that the image of M even by F is a smooth embeddedcurve. Moreover, it is very likely that P is a diffeomorphism from M even onto( − π/ , π/ P is onto.To give credit to the above conjecture, in [16], we will show that M even has onlyone connected component which contains both the saddle solution and the solutionsconstructed in [6]. The proof of this last result is rather technical and uses toolswhich are different from the one needed to prove the results in the present paperand this is the reason why, we have chosen to present it in a separate paper [16].To complete this list of results, we mention an interesting by-product of the proofof Theorem 2.4. Assume that u is a solution of (1.1) and denote by L := − ∆ + F ′′ ( u ) , the linearized operator about L . Recall that, if Ω is a bounded domain in R , thenthe index of L in Ω is given by the number of negative eigenvalues of the operator L which belong to W , (Ω). Following [10], we have the : Definition 2.3.
The function u , solution of (1.1), has finite Morse index if theindex of every bounded domain Ω ⊂ R has a uniform upper bound. And in this paper, we prove the :
Theorem 2.8 (Morse index) . Any k -ended solution of (1.1) has finite Morseindex. We will only prove this result for 4-ended solutions but the proof extends verbatim to any 2 k -ended solution.Since the Morse index of a 2 k -ended solution u is finite (equal to m ), we knowfrom [10], that there exists a finite dimensional subspace E ⊂ L ( R ), with dim E = -ENDED SOLUTIONS 7 m , which is spanned by the eigenfunctions φ , . . . , φ m of the operator L , correspond-ing to the negative eigenvalues µ , . . . , µ m of L .We now sketch the plan of our paper.In section 3, we prove that any element of M even is even-nondegenerate (we alsoprove that it is nondegenerate, even though we do not need this result). The prooffollows the line of the proof in [15] where it is proven that the saddle solution isnondegenerate. This will prove Theorem 2.4 and, thanks to this result, it will thenfollow from the Implicit Function Theorem (see Section 8 and Theorem 2.2 in [5])that any connected component of M even is 1-dimensional.In section 4, we recall two key tools which will be needed in the analysis of theproperness of the classifying map P . The first is a well known a priori estimatefor solutions of (1.1) which states that, away from its zero set, the solutions of(1.1) tend to ± balancing formula whichholds for any solution of (1.1). This balancing formula reflects the invariance of ourproblem under translations and rotations and can be understood as a consequenceof Noether’s Theorem.In section 5, we prove the properness of the classifying map P . Assume that( u n ) n ≥ is a sequence of solutions of M even such that P ( u n ) remains bounded awayfrom − π/ π/
4, further assume that ( u n ) n ≥ converges on compacts to u (thanks to elliptic estimates, this can always be achieved up to the extraction ofa subsequence). We will show that u ∈ M even and also that P ( u ) = lim n →∞ P ( u n ) . The key tool in the proof is the use of the balancing formula introduced in theprevious section which allows one to control the nodal sets of u n as n tends toinfinity. As we will see, this compactness result implies that the image by P of theconnected component of M even which contains the saddle solution, is the entireinterval ( − π/ , π/ M even . We will show that a connected component in M even cannot be compact (i.e.cannot be diffeomorphic to S ). As a consequence, this will imply that the imageof any connected components of M even by P is the entire interval ( − π/ , π/ k -ended solution of (1.1) has finite Morse index.The proof relies on an intermediate result used in section 4, in the proof of thenondegeneracy of the 4-ended solutions of (1.1).Our results are very much inspired from a similar classification result which wasobtained in a very different framework : the theory of minimal surfaces. Let usbriefly explain the analogy between our result and the corresponding result in thetheory minimal surfaces in R .In 1834, H.F. Scherk discovered an example of a singly-periodic, embedded,minimal surface in R which, in a complement of a vertical cylinder, is asymptoticto 4 half planes with angle π/ x = 0 plane and the x = 0 plane, and it is periodic, with period 2 π in the x direction. If θ ∈ (0 , π/
2) denotes the angle between the asymptotic ends of theScherk’s surface contained in { ( x , x , x ) ∈ R : x > , x > } and the x = 0 MICHA L KOWALCZYK, YONG LIU, AND FRANK PACARD plane, then for the original Scherk surface corresponds to θ = π/
4. This surface isthe so called
Scherk’s second surface and it will denoted here by S π/ .In 1988, H. Karcher [14] found a one parameter family of Scherk’s type surfaceswith 4-ends including the original example. These minimal surfaces are parame-terized by the angle θ ∈ (0 , π/
2) between one of their asymptotic planes and the x = 0 plane. The one parameter family ( S θ ) θ ∈ (0 ,π/ of these surfaces, normalizedin such a way that the period in the x direction is 2 π , is the family of Scherk singlyperiodic minimal surfaces.We note that the 4-ended elements of Scherk family are given explicitly in termsof the Weierstrass representation, or alternatively they can be represented implicitlyas the solutions ofcos θ cosh (cid:16) x cos θ (cid:17) − sin θ cosh (cid:16) x cos θ (cid:17) = cos x . More generally, Scherk’s surfaces with 2 k -ends have also been constructed by H.Karcher [14]. They have been classified by J. Perez and M. Traizet in [17]. In somesense our result can be understood as an analog of the classification result of J.Perez and M. Traizet for 4-ended Scherk’s surfaces.3. The nondegeneracy of -ended solutions In this section, we prove that any u ∈ M even is even-nondegenerate. The prooffollows essentially the proof of the nondegeneracy of the saddle solution in [15] andsubsequently this idea was used by X. Cabr´e in [3]. The main result is the : Theorem 3.1.
Assume that u ∈ M even and δ > . Further assume that ϕ ∈ e − δ (1+ | x | ) W , ( R ) is a solution of (∆ − F ′′ ( u )) ϕ = 0 , in R which is symmetric with respect to both the x -axis and the y -axis, then ϕ ≡ . As in [15], the proof of this Proposition relies on the construction of a superso-lution for the operator L , away from a compact. To explain the main idea of theproof, let us digress slightly and consider the heteroclinic solution ( x, y ) H ( x )and define L := − ∆ + F ′′ ( H ′ ) , the linearized operator about the heteroclinic solution. Clearly, the functionΨ ( x, y ) := H ′ ( x ) , is positive and is a solution of L Ψ = 0. Since any u ∈ M even is asymptotic to aheteroclinic solution, we can transplant H ′ along the ends of u to build a positivesupersolution for L := − ∆ + F ′ ( u ). More precisely, we have the : Proposition 3.1.
Under the above assumptions, there exist R > and a function Ψ > defined in R such that (∆ − F ′′ ( u )) Ψ ≤ , in R − B (0 , R ) . -ENDED SOLUTIONS 9 Proof.
The proof follows from a direct construction of the function Ψ. In the upperright quadrant Q x , the zero set of u is asymptotic to the half of an oriented affineline λ = r e ⊥ + R e , with e := (cos θ, sin θ ). Without loss of generality, we can assume that θ ∈ [ π/ , π/
2) since, if this is not the case, we just compose − u with a rotation by π/
2. The intersection of λ with the closure of the upper right quadrant, Q x will bedenoted by ¯ λ +1 := Q x ∩ λ, and its image by the symmetry with respect to the y -axis will be denoted by ¯ λ +2 ,while its image by the symmetry with respect to the x -axis will be denoted by ¯ λ +4 .Finally, the image of ¯ λ +2 by the symmetry with respect to the x -axis is equal to thethe image of ¯ λ +4 by the symmetry with respect to the y -axis and will be denotedby ¯ λ +3 . So, at infinity, the zero set of u is asymptotic to ¯ λ +1 , . . . , ¯ λ +4 . We denote by q + j the end points of ¯ λ + j , namely { q + j } := ∂ ¯ λ + j . Observe that { q + j : j = 1 , . . . , } contains at most 2 points and we denote by Γthe line segment joining these two points. Also observe that R − (¯ λ +2 ∪ Γ ∪ ¯ λ +4 )has two connected components, one of which contains ¯ λ +1 and will be denoted by U while the other, which contains ¯ λ +3 , will be denoted by U .The crucial observation is the following : If f ( x, y ) := (1 − e − µy ) H ′ ( x ) , then, using the fact that (∆ − F ′′ ( H )) H ′ = 0, we get(∆ − F ′′ ( H )) f = − µ e − µy H ′ < . We define the function h by h (( r + s ) e ⊥ + t e ) = (1 − e − µt ) H ′ ( s ) . Observe that the function h is defined in all R . Nevertheless, since we are onlyinterested in this function in U , we define a smooth cutoff function χ which isidentically equal to 1 in U at distance 1 from ∂U and which is identically equalto 0 in U := R − U . As usual, we assume that |∇ χ | ≤ C , for some C >
0, as weare entitled to do.Using χ and h , we build our supersolution in such a way that it is invariant underthe symmetry with respect to the x -axis and under the symmetry with respect tothe y -axis. We defineΨ( x, y ) := χ ( x, y ) h ( x, y ) + χ ( − x, − y ) h ( − x, − y )+ χ ( x, − y ) h ( x, − y ) + χ ( − x, y ) h ( − x, y ) . We know from the Refined Asymptotics Theorem (Theorem 2.1 in [5]) that,as t tends to infinity, ( r, s ) u (( r + s ) e ⊥ + t e ) converges exponentially fast to( s, t ) H ( s ) uniformly in s ∈ [ − ρ, ρ ]. Using this property, we see that we canchose µ > L Ψ < − µ e − µy H ′ in a tubular neighborhood of width ρ around ∂U and away from a ball of radius R large enough, centered at the origin. Observe that the choice of µ only dependson the decay of ( s, t ) u (( r + s ) e ⊥ + t e ) towards ( s, t ) H ( s ) and does notdepend on the choice of ρ . However, increasing ρ affects the minimal value of R for which (3.9) holds.Now, we choose ρ > R > L Ψ < − µ e − µy H ′ , away from B R and away from a tubular neighborhood of width ρ around ¯ λ ∪ . . . ∪ ¯ λ +4 . Here, we simply use the fact that F ′′ ( u ) converges uniformly to F ′′ ( ±
1) awayfrom the nodal set of u . This completes the proof of result. (cid:3) Observe that this construction is not specific to the case of 4-ended solutions of(1.1) and in fact a similar construction would hold for any 2 k -ended solution. Withthis Lemma at hand, we can adapt the argument in [15] where the nondegeneracyof the saddle solution is proven and we simply adapt it to the general case where u is any 4-ended solution. Proof of Theorem 3.1.
Recall that, by definition, since u ∈ M even , we have ∂ y u > y > , and ∂ x u < x > . Let ϕ be the function as in the statement of Theorem 3.1. It follows from the LinearDecomposition Lemma (Lemma 4.2 in [5]) that ϕ decays exponentially at infinity.More precisely, there exist constants α, C > | ϕ ( x ) | ≤ C e − α | x | , for all x ∈ R − B (0 , Step 1 . We assume that ϕ is not identically equal to 0 and define Z to be thezero set of ϕ . Since ϕ is assumed to be symmetric with respect to both the x -axisand the y -axis, so is the set Z . Clearly Proposition 3.1 together with the maximumprinciple, implies that R − Z has no bounded connected component included in R − B (0 , R ), since Ψ can be used as a supersolution to get a contradiction. Step 2 . We claim that any unbounded connected component of R − Z nec-essarily contains R − B (0 , R ) for some R large enough. Indeed, if this were notthe case then, using the symmetries of ϕ , one could find Ω ⊂ R , an unboundedconnected component of R − Z , which is included in one of the four half spaces { ( x, y ) ∈ R : ± x > } or { ( x, y ) ∈ R : ± y > } . For example, let us assumethat Ω ⊂ { ( x, y ) ∈ R : y > } . Following [15], we adapt the proof of the de Giorgi conjecture in dimension 2 by N.Ghoussoub and C. Gui to derive a contradiction.We define ψ := ∂ y u which is a solution of (∆ − F ′′ ( u )) ψ = 0 in R . Moreover, ψ > { ( x, y ) ∈ R : y > } and we check from direct computation that(3.10) div (cid:0) ψ ∇ h (cid:1) = 0 , -ENDED SOLUTIONS 11 where h := ϕψ . For all R ≥
1, we consider a cutoff function ζ R which is identicallyequal to 1 in B (0 , R ), identically equal to 0 outside R − B (0 , R ) and which satisfies |∇ ζ R | ≤ C/R for some constant
C > R ≥ ζ R ψ and integrate the result over Ω. We find after anintegration by parts Z Ω |∇ h | ψ ζ R d x + 2 Z Ω ψ h ζ R ∇ h ∇ ζ R d x = 0 . Observe that, in the integration by parts, some care is needed when the boundaryof Ω touches the x -axis but it is not hard to see that the integration by parts isalso legitimate in this case (we refer to [15] for details). Then, Cauchy-Schwarzinequality yields(3.11) Z Ω |∇ h | ψ ζ R d x ≤ (cid:18)Z Ω ∩ A R |∇ h | ψ ζ R d x (cid:19) / (cid:18)Z Ω ∩ A R ϕ |∇ ζ R | d x (cid:19) / , where A R := B (0 , R ) − B (0 , R ) contains the support of ∇ ζ R . Hence, Z Ω |∇ h | ψ ζ R d x ≤ (cid:18) sup A R | ϕ | (cid:19) Z Ω ∩ A R |∇ ζ R | d x . By construction of ζ R , the integral on the right hand side is bounded independentlyof R and since ϕ ∈ W , ( R ) we know that ϕ tends to 0 at infinity. Letting R tendto infinity, we conclude that Z Ω |∇ h | ψ d x = 0 , which then implies that h ≡ ϕ ≡ ϕ ≡ R by the unique continuation theorem. This is certainly a contradictionand the proof of the claim is complete. Step 3 . By the above, we know that ϕ does not change sign away from acompact and, without loss of generality, we can assume that ϕ > R − B (0 , R )for some R > ϕ onto H ′ (composed with a suitable rotation).The nodal set of u in the upper right quadrant Q x is asymptotic to an orientedhalf line which is denoted by λ and is given by λ = r e ⊥ + R e , where e = (cos θ, sin θ ). Up to a rotation by π/ θ ∈ [ π/ , π/ λ through the symmetry withrespect to the y -axis will be denoted by ¯ λ . Notice that, since u is symmetric withrespect to the y -axis, ¯ λ is also asymptotic to the zero set of u .Given the expression of λ , we define˜ u ( s, t ) := u (( r + s ) e ⊥ + t e ) and ˜ ϕ ( s, t ) := ϕ (( r + s ) e ⊥ + t e ) . We consider the function g defined by g ( t ) := Z R ˜ ϕ ( s, t ) H ′ ( s ) ds. Since ϕ > g ≥ t > ϕg ′′ ( t ) := Z R ∂ t ˜ ϕ ( s, t ) H ′ ( s ) ds = − Z R ( ∂ s − F ′′ (˜ u ( s, t ))) ˜ ϕ ( s, t ) H ′ ( s ) ds, and an integration by parts yields g ′′ ( t ) := − Z R ˜ ϕ ( s, t ) ∂ s H ′ ( s ) ds + Z R F ′′ (˜ u ( s, t ))) ˜ ϕ ( s, t ) H ′ ( s ) ds. Finally, using the equation satisfied by H ′ , we conclude that g ′′ ( t ) = Z R ( F ′′ (˜ u ( s, t )) − F ′′ ( H ( s ))) ˜ ϕ ( s, t ) H ′ ( s ) ds. Observe that ϕ tends exponentially to 0 at infinity and hence so does g . Integratingthe above equation from t to ∞ , we conclude that g ′ ( t ) = − Z ∞ t Z R ( F ′′ (˜ u ( s, z )) − F ′′ ( H ( s ))) ˜ ϕ ( s, z ) H ′ ( s ) ds dz. Recall that ¯ λ is the image of λ through the symmetry with respect to the y -axisand that we can parameterize ¯ λ by¯ λ = ¯ r ¯ e ⊥ + R ¯ e , with obvious relations between e and ¯ e . We define¯ g ( t ) := Z R ϕ ((¯ r + s ) ¯ e ⊥ + t ¯ e ) H ′ ( s ) ds. Observe that, by symmetry of both u and ϕ , we have¯ g ( t ) = g ( t ) , for all t ≥ C > β > | g ′ ( t ) | ≤ C e − βt Z + ∞ t g ( z ) dz. for all t > Z R ˜ ϕ ( s, t ) H ′ ( s ) ds = 0 , for all t > ϕ ( s, t ) ≡ s ∈ R and all t > ϕ ≡ g , the domain ofintegration contains both a half of λ and ¯ λ (this is where we use the fact that θ ∈ [ π/ , π/ u is exponentially close to the sum of the heteroclinic solution H ′ along λ and also along ¯ λ . Close to λ , we can therefore estimate | F ′′ (˜ u ( s, t )) − F ′′ ( H ( s )) | ≤ C e − βt , -ENDED SOLUTIONS 13 for some β >
0. In fact this estimate holds at any point of { ( x, y ) ∈ R : y > } which is closer to λ than to ¯ λ .We can write any point ( r + s ) e ⊥ + t e close to ¯ λ as (¯ r + ¯ s ) ¯ e ⊥ + ¯ t ¯ e . Therefore,at any such point which is closer to ¯ λ than to λ , we simply use the fact that | F ′′ ( u (( r + s ) e ⊥ + t e )) − F ′′ ( H ( s ))) H ′ ( s ) | ≤ C e − β ¯ t H ′ (¯ s ) , and we conclude that (cid:12)(cid:12)(cid:12)(cid:12)Z ∞ t Z R ( F ′′ (˜ u ( s, z )) − F ′′ ( H ( s ))) ˜ ϕ ( s, z ) H ′ ( s ) ds dz (cid:12)(cid:12)(cid:12)(cid:12) ≤ C e − βt Z ∞ t ( g ( z ) + ¯ g ( z )) dz. Finally, the estimate (3.12) follows from the fact that ¯ g = g . (cid:3) We now explain how to prove that any u ∈ M even is non-degenerate. If φ ∈ e − δ (1+ | x | ) W , ( R ) is a solution of (∆ − F ′ ( u )) φ = 0, we can decompose φ = φ e + φ o into the sum of two functions, one of which φ e being even under the action of thesymmetry with respect to the x -axis and the other one φ o being odd under theaction of the same symmetry. Since φ o vanishes on the x -axis, we can use theargument already used in Step 2 to prove that φ o ≡ φ is even underthe action of the symmetry with respect to the x -axis. Using similar argumentsone also prove that φ is even under the action of the symmetry with respect to the y -axis and the non-degeneracy follows from Theorem 3.1.Thanks to Theorem 3.1, we can apply the Implicit Function Theorem (see Sec-tion 8 and Theorem 2.2 in [5]) to show that any connected component of M is 4-dimensional and, equivalently, we conclude that any connected component of M even is 1-dimensional (the rational being that, the formal dimension of the mod-uli space of solutions of (1.1) is equal to the number of ends but, because of thesymmetries, elements of M even have only one end in the quotient space). Moreover,a consequence of this Implicit Function Theorem is that close to u ∈ M even , thespace M even can either be parameterized by the angle α or by the distance r andthis implies that the image of any connected component of M even by the mapping F is an immersed curve. 4. Two useful tools An a priori estimate. It is well known that any solution u of (1.1) whichsatisfies | u | < ± Lemma 4.1.
Given δ ∈ (0 , , there exists ρ δ > such that, for any solution of(1.1) which satisfies | u | < , we have (4.13) B ( x , ρ δ ) ⊂ R − Z ( u ) ⇒ | u − | ≤ δ in B ( x , ρ δ ) , where Z ( u ) := { x ∈ R : u ( x ) = 0 } , denotes the nodal set of the function u . This result is a simple corollary of the result below, whose proof can already befound in [2] (see Lemma 3.1-Lemma 3.3 therein) and also in [11] :
Lemma 4.2.
There exist constants
C > and α > such that, for any solutionof (1.1) which satisfies | u | < , we have (4.14) | u ( x ) − | + |∇ u ( x ) | + |∇ u ( x ) | ≤ C e − α dist( x , Z ( u )) , for all x ∈ R .Proof. Since this Lemma plays a central role in our result, we give here a completeproof for the sake of completeness.We denote by φ R the eigenfunction which is associated to the first eigenvalue of − ∆ on the ball of radius R , under 0 Dirichlet boundary conditions. We assumethat φ R is normalized so that φ R (0) = sup B (0 ,R ) φ R = 1. Recall that the associatedeigenvalue µ R satisfies µ R = µ /R .Given δ ∈ (0 , R > − F ′ ( s ) R ≥ µ s, for all s ∈ [0 , − δ ]. Assume that R > R and that B ( x , R ) ⊂ R − Z ( u ). Tosimplify the discussion, let us also assume that u > B ( x , R ).We claim that u ≥ − δ in B ( x , R ). Indeed, if this is not the case then thereexists ¯ x ∈ B ( x , R ) such that u (¯ x ) < − δ . In this case, we define ǫ > u ≥ ǫ φ R ( · − ¯ x ) , in B (¯ x , R ). Certainly ǫ ≤ − δ and there exists z ∈ B (¯ x , R ) such that u ( z ) = ǫ φ R ( z − ¯ x ) ≤ − δ. By construction of R , we can write − ǫ ∆ φ R = µ R ǫ φ R < F ′ ( ǫφ R );Since − ∆ u = − F ′ ( u ) we conclude that − ∆( ǫφ R − u ) < , at the point z and this contradicts the fact that u − ǫφ R has a local minimum at z . The proof of the claim is complete.We now fix α > α < F ′′ (1) and we choose δ ∈ (0 ,
1) close to 1 sothat F ′′ ( t ) ≥ α for all t ∈ [1 − δ, R > R , u ≥ − δ (or u < δ −
1) in B (¯ x , R ) provided B (¯ x , R ) ⊂ R − Z ( u ).Therefore, we get − ∆(1 − u ) = − F ′ ( u ) − F ′ (1) u − − u ) ≤ − α (1 − u ) , in B (¯ x , R ). A direct computation shows that( − ∆ + α ) e − α √ r ≥ , where r := | x − ¯ x | . This, together with the maximum principle, implies the expo-nential decay of 1 − u away from Z ( u ), the zero set of u . (cid:3) -ENDED SOLUTIONS 15 The balancing formulæ.
We describe the balancing formulæ for solutionsof (1.1) in the form they were introduced in [5]. Assume that u is a smooth functiondefined in R and X a vector field also defined in R . We define the vector fieldΞ( X, u ) := (cid:18) |∇ u | + F ( u ) (cid:19) X − X ( u ) ∇ u. Recall that Killing vector fields are vector fields which generate the group ofisometries of R . They are linear combinations (with constant coefficients) of theconstant vector fields ∂ x and ∂ y generating the group of translations and the vectorfield x ∂ y − y ∂ x which generates the group of rotations in R .We have the : Lemma 4.3 (Balancing formulæ) . [5] Assume that u is a solution of (1.1) andthat X is a Killing vector field. Then div Ξ( X, u ) = 0 .Proof.
To prove this formula, just multiply the equation (1.1) by X ( u ) and usesimple manipulations on partial derivatives. (cid:3) This result is nothing but an expression of the invariance of (1.1) under theaction of rigid motions. To see how useful this result will be for us, let us assumethat u ∈ M even . By definition, the nodal set of u is, in the upper right quadrant Q x of R , asymptotic to an oriented half line λ := r e ⊥ + R e , where r ∈ R and e ∈ S . We write e := (cos θ, sin θ ) and, since we assume that theoriented line λ lies in Q x , we have θ ∈ (0 , π/ R >
0, we define the plain triangle T R := (cid:8) ( x, y ) ∈ R : x > , y > x, y ) · e < R (cid:9) , The divergence theorem implies that(4.15) Z ∂T R Ξ( X, u ) · ν ds = 0 , where ν is the (outward pointing) unit normal vector field to ∂T R .We set(4.16) c := Z + ∞−∞ (cid:18)
12 ( H ′ ) + F ( H ) (cid:19) ds. Taking X := ∂ x and letting R tend to infinity, we conclude, using the fact that u is asymptotic to ± H along the half line λ , that(4.17) c cos θ = Z x =0 , y> (cid:18) | ∂ y u | + F ( u ) (cid:19) dy. Observe that we have implicitly used the fact that ∂ x u = 0 along the y -axis. Simi-larly, taking X := ∂ y and letting R tend to infinity, we conclude that(4.18) c sin θ = Z y =0 , x> (cid:18) | ∂ x u | + F ( u ) (cid:19) dx. Finally, taking X = x ∂ y − y∂ x and letting R tend to infinity, we get(4.19) c r = Z x =0 , y> (cid:18) | ∂ y u | + F ( u ) (cid:19) y dy − Z y =0 , x> (cid:18) | ∂ x u | + F ( u ) (cid:19) x dx. The key observation is that it is possible to detect both the angle θ and the pa-rameter r which characterize the half line λ just by performing some integrationover the x -axis and the y -axis. As one can guess this property will be very usefulin the compactness analysis we are going to perform now. In some sense it willbe enough to pass to the limit in the above integrals to guarantee the convergenceof the parameters characterizing the asymptotics of the zero set of the solutions of(1.1). 5. Properness
In this section, we prove a compactness result for the set of 4-ended solutions of(1.1). More precisely, we prove that, given ( u n ) n ≥ , with u n ∈ M even , a sequenceof solutions of (1.1) whose angles θ n := π/ P ( u n ) converges to some limit angle θ ∗ ∈ (0 , π/ u ∗ with angle θ ∗ . Naturally, the fact that one can extract a subsequence whichconverges uniformly (at least on compacts of R ) to a solution u ∗ of (1.1) is notsurprising since the functions u n are uniformly bounded and, by elliptic regularity,have gradient which is uniformly bounded, hence this compactness result simplyfollows from the application of Ascoli-Arzela’s Theorem. In general, it is hard tosay anything about the limit solution u ∗ . It turns out that it is possible to controlthe zero set of the limit solution u ∗ and prove that u ∗ is also a 4-ended solution. Aswe will see, a key ingredient in this analysis is provided by the balancing formulædefined in the previous section. Theorem 5.1.
Assume that we are given a sequence ( u n ) n ≥ , with u n ∈ M even ,which converge uniformly on compacts to a solution u ∗ . If ( P ( u n )) n ≥ convergesin ( − π/ , π/ , then, u ∗ ∈ M even , lim n →∞ P ( u n ) = P ( u ∗ ) , and lim n →∞ u n = u ∗ , in the topology of S . The proof of this Theorem is decomposed into many small Lemma. Assume thatwe are given a sequence ( u n ) n ≥ , with u n ∈ M even . Recall that(5.20) ∂ x u n < ∂ y u n > , in the right upper quadrant Q x := { ( x, y ) ∈ R : x > , and y > } . We denote by Z n := { ( x, y ) ∈ R : u n ( x, y ) = 0 } , the nodal set of u n . Monotonicity of u n in Q x implies that the zero set of u n iseither a graph over the x -axis or a graph over the y -axis. In particular, Z n ∩ ∂Q x contains exactly one point which we denote by p n Z n ∩ ∂Q x = { p n } . We define θ n := π/ P ( u n ) and r n to be the parameters describing the asymp-totics of Z n in the right upper quadrant Q x . In other words, Z n ∩ Q x is asymptoticto the oriented half line λ n := r n e ⊥ n + R e n , -ENDED SOLUTIONS 17 where e n := (cos θ n , sin θ n ). Finally, we assume thatlim n →∞ θ n = θ ∗ ∈ (0 , π/ . First, we prove that the point p n where the zero set of u n meets the boundaryof the upper right quadrant Q x remains bounded as n tends to infinity. Lemma 5.1.
Under the above assumptions, the sequence ( p n ) n ≥ remains bounded,and hence converges.Proof. We argue by contradiction and for example assume that, up to a subse-quence, p n = (0 , y n ) with lim n →∞ y n = + ∞ .We define w n ( x, y ) := u n ( x, y + y n ) , Standard arguments involving elliptic estimates and Ascoli-Arzela’s Theorem implythat, up to a subsequence, the sequence ( w n ) n ≥ converges uniformly on compactsto some function w which is defined on R and which is again a solution of (1.1).Since ∂ y u n > y >
0, we conclude that ∂ y w ≥ R . Moreover, w (0 ,
0) = 0 and ∂ x w (0 , y ) = 0 for all y ∈ R since u n is even withrespect to the y -axis.Observe that, since w n is not identically equal to 0, we may apply Lemma 4.1 toconclude that w is not identically equal to 0. Indeed, thanks to (5.20) we know thatthe zero set of u n is a graph over the x -axis for some function which is increasingand u n < B n := { ( x, y ) ∈ R : | y | < y n } . In particular, provided n is chosen large enough, discs of arbitrary large radii can be inserted in B n andLemma 5.20 implies that | u n − | ≤ / { ( x, y ) ∈ R : | y | < y n − ρ / } .Therefore, w is bounded, monotone increasing in y , even with respect to the y -axis and according to De Giorgi’s conjecture in dimension 2 which is proven in[11], we conclude that w ( x, y ) = H ( y ) . Now, we use (4.17) which tells us that c cos θ n = Z x =0 , y> (cid:18) | ∂ y u n | + F ( u n ) (cid:19) dy. Passing to the limit as n tends to infinity, we conclude that c cos θ ∗ = lim n →∞ Z x =0 , y> (cid:18) | ∂ y u n | + F ( u n ) (cid:19) dy. However, given y ∗ > Z x =0 , y> (cid:18) | ∂ y u n | + F ( u n ) (cid:19) dy ≥ Z x =0 , y − y n ∈ [ − y ∗ ,y ∗ ] (cid:18) | ∂ y u n | + F ( u n ) (cid:19) dy = Z x =0 , y ∈ [ − y ∗ ,y ∗ ] (cid:18) | ∂ y w n | + F ( w n ) (cid:19) dy, for all n large enough so that y n − y ∗ >
0. Passing to the limit as n tends to infinity,we conclude thatlim n →∞ Z x =0 , y> (cid:18) | ∂ y u n | + F ( u n ) (cid:19) dy ≥ Z [ − y ∗ ,y ∗ ] (cid:18) | H ′ ( s ) | + F ( H ( s )) (cid:19) ds. Hence c cos θ ∗ ≥ Z [ − y ∗ ,y ∗ ] (cid:18) | H ′ ( s ) | + F ( H ( s )) (cid:19) ds. Since y ∗ can be chosen arbitrarily large, we get c cos θ ∗ ≥ Z R (cid:18) | H ′ ( s ) | + F ( H ( s )) (cid:19) ds = c , which is clearly in contradiction with the fact that θ ∗ >
0. If p n = ( x n ,
0) with x n tending to infinity, similar arguments using (4.18) and the fact that θ ∗ < π/
2. Thiscompletes the proof of the result. (cid:3)
Thanks to the above Lemma, we know that u ∗ is not identically constant equalto 0 or ±
1. Therefore, according to Theorem 4.4 in [12] we know that the nodal setof u ∗ in the upper right quadrant Q x must be a asymptotically a straight line whichis not parallel to the x -axis nor to the y -axis. The Refined Asymptotic Theorem(Theorem 2.1 in [5]) then implies that u ∗ is a 4-ended solution. Some commentis due in the way the Refined Asymptotic Theorem (Theorem 2.1 in [5]) is used.Indeed, in the statement of this result, one start with a solution of (1.1) whichdiffers from the model heteroclinic solution sharing the same end by some W , function. Nevertheless, close inspection of the proof of Theorem 2.1 in [5] showsthat the result remains valid provided we start from a solution which is asymptoticto the model heteroclinic solution in the L ∞ sense, and this is precisely the situationin which we need the result.To proceed, we prove that we have a uniform control on the nodal set of u n awayfrom a compact set. Again, the balancing formulæ will play an important role inthe control of the nodal set of u n .To fix the ideas, we assume that the nodal set of u n in the upper right quadrant Q x is the graph of a function y = f n ( x ). Since u n is a 4-ended solution and given thenotations introduced at the beginning of this section, we know that f n is asymptoticto the affine function given by˜ f n ( x ) := tan θ n x + r n cos θ n . In particular, given δ > n ≥
0, there exists x n,δ > | f ′ n ( x ) − tan θ n | < δ, for all x ≥ x n,δ . In the next Lemma, we prove that x n,δ can be chosen to beindependent of n ≥
0. In other words, this provides a uniform control on thederivative of f n away from a compact set. Lemma 5.2.
For all δ > , there exists x δ > such that | f ′ n ( x ) − tan θ n | < δ, for all n ≥ and for all x ≥ x δ .Proof. We now argue by contradiction. Observe that the result is true if we restrictour attention to a finite number of the u n . Hence, if the result were not true, therewould exist δ ∗ > x k ) k ≥ and ( n k ) k ≥ both tending to infinity suchthat sup x ≥ x k | f ′ n k ( x ) − tan θ n k | ≥ δ ∗ , -ENDED SOLUTIONS 19 We define ¯ x k ≥ x k to be the supremum of the x > | f ′ n k ( x ) − tan θ n k | ≥ δ ∗ . Observe that ¯ x k is well defined sincelim x →∞ f ′ n k ( x ) = tan θ n k . By definition, we have(5.21) | f ′ n k (¯ x k ) − tan θ n k | = δ ∗ , and(5.22) sup x ≥ ¯ x k | f ′ n k ( x ) − tan θ n k | ≤ δ ∗ . Moreover, the sequence (¯ x k ) k ≥ tends to infinity as k tends to infinity and, if wedefine ¯ y k := f n k (¯ x k ) , we find that the sequence (¯ y k ) k ≥ also tends to infinity as k does. This latter factis just a consequence of the fact that ( u n ) n ≥ converges on compacts to u ∗ whichis a 4-ended solution and hence the zero set of u ∗ is the graph of a function whichtends to infinity at infinity.We now consider the domain D k of Q x which contains the graph of f n k for x large enough and which is bounded by the half line t (0 , ¯ y k − ¯ x k + t ) for t > , ¯ y k − ¯ x k ) to (¯ x k + ¯ y k ,
0) and the half line t (¯ x k + ¯ y k + t, t >
0. Observe that ∂D k contains the point (¯ x k , ¯ y k ). Applying the analysis ofSection 4, we conclude that Z ∂Q x Ξ( ∂ x , u n k ) ν ds = Z ∂D k Ξ( ∂ x , u n k ) · ν ds, where ν denotes the outward pointing normal vector to the sets Q x and D k . Using(4.17), we conclude that c cos θ n k = Z ∂D k Ξ( ∂ x , u n k ) · ν ds. Observe that, up to a subsequence, the sequence of functions( x, y ) u n k ( x + ¯ x k , y + ¯ y k ) , converges, uniformly on compacts, to the heteroclinic solution whose zero set isthe line passing through the origin, of slope lim k →∞ f ′ n k (¯ x k ). Moreover, thanks to(5.22) we see that there exist constants C > β >
0, independent of k ≥ | u n k ( x ) − | + |∇ u n k ( x ) | ≤ C e − β | x − ¯ x k | , for all x ∈ ∂D k , where ¯ x k := (¯ x k , ¯ y k ). This property, together with the result ofLemma 4.2 allows one to conclude thatlim k →∞ Z ∂D k Ξ( ∂ x , u n k ) · ν ds = c cos ˜ θ ∗ , where ˜ θ ∗ is defined by tan ˜ θ ∗ = lim k →∞ f ′ n k (¯ x k ) . This is clearly in contradiction with (5.21) which implies that | tan θ ∗ − tan ˜ θ ∗ | = δ ∗ .This completes the proof of the result. (cid:3) As a consequence, we have the :
Lemma 5.3.
Under the above assumptions, we have lim n →∞ P ( u n ) = P ( u ∗ ) . Since u n converges on compacts to u ∗ which is a 4-ended solution, we conclude,with the help of the previous Lemma that the distance from a point x ∈ Z n to the x -axis and the y -axis, tends to infinity as | x | tends to infinity. We now prove a morequantitative version of this assertion in the : Lemma 5.4.
There exists constants
C > and α > , such that Z n ∩ Q x ⊂ n ( x, y ) ∈ R : x > , y > , and xα − C ≤ y ≤ αx + C o . Proof.
According to Lemma 5.2, we have a uniform control on the slopes of thenodal sets of u n away from a tubular neighborhood of the x -axis and y -axis. Thismeans that these slopes are bounded away from 0 and bounded independently of n . Next, in a ball of fixed radius, u n converges uniformly to u ∗ and the result thenfollows at once. (cid:3) Recall that F ( u n ) = ( θ n − π/ , r n ) . We set ( θ ∗ − π/ , r ∗ ) := F ( u ∗ ) . Now that we have understood the behavior of the sequence ( θ n ) n ≥ , we turn tothe behavior of the sequence ( r n ) n ≥ , the other parameter which characterizes theasymptotic of the nodal set of u n . We have the : Lemma 5.5.
Under the above assumptions, lim n →∞ r n = r ∗ .Proof. Again, the proof uses the balancing formula (4.15) but this time, we will usethe vector field X = x∂ y − y∂ x . Recall that (4.19) yields c r n = Z x =0 , y> (cid:18) | ∂ y u n | + F ( u n ) (cid:19) y dy − Z y =0 , x> (cid:18) | ∂ x u n | + F ( u n ) (cid:19) x dx. The key ingredients are Lemma 5.4 and Lemma 4.2, from which we get anexponential decay of the solution u n along the coordinate axis as | x | tends to infinity,the decay being uniform in n ≥
0. Once this decay is proven one uses the fact that( u n ) n ≥ converges in C topology to u ∗ uniformly on any given ball.Using these remarks, one can pass to the limit as n tends to infinity in the aboveequality to get to get c lim n →∞ r n = Z x =0 ,y> (cid:18) | ∂ y u ∗ | + F ( u ∗ ) (cid:19) ydy − Z y =0 ,x> (cid:18) | ∂ x u ∗ | + F ( u ∗ ) (cid:19) xdx. Since the right hand side is equal to c r ∗ , the proof is complete. (cid:3) At this point, we have shown that the sequence ( u n ) n ≥ converges uniformly oncompacts to u ∗ and the ends of u n also converges to the end of u ∗ . However, this isnot quite enough since our aim is to show the convergence of ( u n ) n ≥ to u ∗ in S .Recall that Z n , the zero set of u n , is asymptotic to λ n := r n e ⊥ n + R e n , -ENDED SOLUTIONS 21 in Q x , where e n = (cos θ n , sin θ n ). We define v n in Q x by v n ( x ) := u n ( x ) − H (cid:0) x · e ⊥ n − r n (cid:1) . Then | v n | → | x | tends to infinity in Q x .In the next Lemma, we prove that this convergence is in fact uniform in n ≥ Lemma 5.6. As | x | tends to infinity, | v n ( x ) | converges to uniformly with respectto n ≥ .Proof. The proof is by contradiction. If the result were not true, there would exists ǫ >
0, a sequence ( R j ) j ≥ tending to infinity, a sequence ( x j ) j ≥ such that | x j | ≥ R j and a sequence ( n j ) j ≥ such that(5.23) | v n j ( x j ) | ≥ ǫ. Up to a subsequence we can assume that ( θ n j , r n j ) j ≥ converges to ( θ ∗ , r ∗ ).Observe that the distance from x j to λ n j is necessarily bounded since, accordingto Lemma 4.2, v n j tends to 0 away from λ n j . Let ¯ x j be the orthogonal projectionof x j onto λ n j .Making use of elliptic estimates and Aslcoli-Arzela’s Theorem, we can assume,up to a subsequence, that ( u n j ( · − ¯ x j )) j ≥ converges uniformly on compacts to asolution of (1.1) which is non trivial and which, thanks to the de Giorgi conjecture,is an heteroclinic solution ¯ u of (1.1). The end of this heteroclinic solution is theaffine line of angle ¯ θ . As in the proof of Lemma 5.4, we use the vector field X = ∂ x in the balancing formula to conclude that θ ∗ = ¯ θ .Therefore, the parameters of the end of ¯ u are given by ( θ ∗ , ¯ r ). As in the proofof Lemma 5.4, we use the vector field X = x∂ y − y∂ x in the balancing formula toconclude that r ∗ = ¯ r . This is clearly a contradiction with (5.23). (cid:3) Thanks to the Refined Asymptotics Theorem (Theorem 2.1 in [5]) we can de-compose u n = v n + u λ n , and u ∗ = v ∗ + u λ ∗ , where λ n , λ ∗ ∈ Λ ord and where v n , v ∗ ∈ e − δ (1+ | x | ) W , ( R ) for some δ >
0. Ob-serve that, a priori , the parameter δ can vary with n but close inspection of theproof of Theorem 2.1 in [5]) shows that δ > n ≥ λ n converge to a fixed end λ ∗ .This, together with the fact that ( u n ) n ≥ converges uniformly on compacts to u ∗ implies that ( u n ) n ≥ converges to u ∗ in the topology of S . This completes theproof of the properness of the classifying map P .Let M be the connected component of M even which contains the saddle solu-tion. We claim that the properness of P implies that the image by P of M is theentire interval ( − π/ , π/ P : M → ( − π/ , π/
4) is not onto. Recall that if u ∈ M even , then ¯ u defined by ¯ u ( x, y ) := − u ( y, x ) , also belongs to M even and M is also invariant under this transformation. We willwrite ¯ u = J u . The properness of P implies that M is compact and one dimensional.Hence, it must be diffeomorphic to S . Obviously J : M → M is a diffeomorphismand the saddle solution is a fixed point of J . Since M is diffeomorphic to S , theremust be at least another fixed element v ∈ M which is a fixed point of J . Then, thezero set of v is union of the two lines y = ± x . But, according to [4] or [12], a solutionof (1.1) having as zero set the two lines y = ± x is the saddle solution. This is acontradiction and the proof of the claim is complete. Note that this argument doesnot guarantee that there are no other compact connected components in M even .To prove this fact, we will need one more result which will be described in the nextsection. In any case, instead of using the argument outlined above to show that P is onto, one can use the next section of the paper.6. Any connected component of M even is not compact We have shown in section 3 that elements in M even are even-nondegenerate.According to the moduli space theory for solutions of (1.1) (see Section 8 andTheorem 2.2 in [5]), any connected component of M even is a one dimensionalmanifold and its image by F is a smooth (possibly immersed) curve in ( − π/ , π/ × R . In particular, any compact connected component M ⊂ M even would have to bediffeomorphic to S . In this section, we show that this cannot happen. Theorem 6.1.
All connected components of M even are not compact, namely, thereis no closed loop in M even .Proof. We argue by contradiction and assume that M even contains a connectedcomponent M which is diffeomorphic to S . We choose a smooth regular parame-terization of M by σ ∈ S u ( · , σ ) ∈ M, so that ∆ u ( · , σ ) − F ′ ( u ( · , σ )) = 0 , for all σ ∈ S and ∂ σ u = 0 for all σ ∈ S . Differentiation with respect to σ impliesthat ∂ σ u ∈ T u S satisfies (∆ − F ′′ ( u )) ∂ σ u = 0 . Observe that, for all x ∈ R ,0 = u ( x , π ) − u ( x ,
0) = Z π ∂ σ u ( x , σ ) dσ. Choosing x to be the origin, this implies that there exists σ ∗ ∈ S , such that ∂ σ u ((0 , , σ ∗ ) = 0. We define φ := ∂ σ u ( · , σ ∗ ) . Observe that φ = 0 and that φ is even with respect to the symmetry about boththe x -axis and the y -axis.By definition, any element u of S can be decomposed into the sum of a functionin W , ( R ) and an element of the form u λ as defined in (2.5). Moreover, because ofthe symmetries, u λ only depends on the two parameters r and θ which characterize λ . In particular, the tangent space of S at u can be decomposed as T u S = W , (cid:0) R (cid:1) ⊕ D , -ENDED SOLUTIONS 23 where D := Span { ∂ r u λ , ∂ θ u λ } . It is easy to check that ∂ θ u λ is linearly growing along the zero set of u while ∂ r u λ is bounded.Since φ (0 ,
0) = 0 and since φ is symmetric with respect to the x -axis and the y -axis, there exists Ω, a nodal domain of φ , which is included in one of the fourhalf spaces { ( x, y ) ∈ R : ± x > } or { ( x, y ) ∈ R : ± y > } . We claimthat this nodal domain can be chosen so that φ is bounded on it. Indeed, if φ ∈ W , (cid:0) R (cid:1) ⊕ Span { ∂ r u λ } , then φ is bounded and one can select any nodaldomain contained in a half space.The other case to consider is the case where φ = a ∂ θ u λ + ˜ φ where ˜ φ is bounded.Inspection of ∂ θ u λ near the end of u shows that, away from a large ball B (0 , R ),the function φ does not vanish along the zero set of u . In this case, it is enoughto select a nodal domain of φ which is unbounded and which, away from B (0 , R ),does not contain the zero set of u . It is easy to check that φ is bounded in such anodal domain.For example, let us assume that the nodal domain Ω ⊂ { ( x, y ) ∈ R : x > } .Then, one can repeat the argument of Step 2 in the proof of Theorem 3.1, with ψ = ∂ x u , to prove that(6.24) Z Ω |∇ h | ψ ζ R d x ≤ (cid:18)Z Ω ∩ A R |∇ h | ψ ζ R d x (cid:19) / (cid:18)Z Ω ∩ A R φ |∇ ζ R | d x (cid:19) / , where h := φψ . Using the fact that φ is bounded, and letting R tend to infinity, weconclude that Z Ω |∇ h | ψ d x < + ∞ . Using this information back into (6.24), and letting R tend to infinity, we concludethat Z Ω |∇ h | ψ d x = 0 , and this implies that φ ≡ φ ≡
0, which is a contradiction. (cid:3)
We observe that from the above considerations, we can give a different proofof Theorem 2.7. Indeed, we choose M to be the connected component of M even which contains the saddle solution. Using the Implicit Function Theorem (Theorem2.2 in [5]), which applies since we have proven that any element of M even is noneven-degenerate, we conclude that M is a smooth, one dimensional manifold. ByTheorem 6.1, M is necessarily non compact and the image of M by P cannot becompact either. Hence the image of M by P contains either an interval of the form( − π/ , δ ) or ( δ, π/ P of the saddle solution is 0, we concludethat ( − π/ , δ ) or ( δ, π/
4) contains 0. Moreover, the image of M by P is symmetricwith respect to 0 and hence it has to be the whole interval ( − π/ , π/ The Morse index of -ended solutions In this section, we give a proof of Theorem 2.8. The proof follows from the resultof Proposition 3.1 together with a result of [8]. For the sake of completeness, we givehere a straightforward proof which is inspired from [9] and which is independentfrom the proof of [8].Assume that u is a 2 k -ended solution. We consider the linearized operator about u L := − ∆ + F ′′ ( u ) . From Proposition 3.1, we know that there exists R > > R such that L Ψ ≤ , in R − B R .Let φ be an eigenfunction of L in B R (with 0 Dirichlet boundary conditions),which is associated to a negative eigenvalue, namely Lφ = λ φ, in B R with ψ = 0 on ∂B R and λ <
0. For all
R > R , one can use the functionΨ as a barrier to show, using the maximum principle, that there exists a constant C > R ≥ R such that(7.25) k φ k L ∞ ( B R − B R ) ≤ C k φ k L ∞ ( ∂B R ) . Now, elliptic estimates also imply that there exists a constant
C >
0, which doesnot depend on
R > R such that(7.26) k φ k L ∞ ( B R ) ≤ C k φ k L ( B R ) . Observe that, in order to obtain this inequality, we have implicitly used the factthat the negative eigenvalues of L are bounded. Using both (7.25) and (7.26), weconclude that there exists a constant C >
R > R suchthat, for all eigenfunctions of L in B R which are associated to negative eigenvalues,we have(7.27) k φ k L ∞ ( B R ) ≤ C k φ k L ( B R ) . The proof now follows the strategy of [9] to estimate the multiplicity of theeigenvalues of the Laplace-Beltrami operator starting from H¨ormander’s estimate.Let φ , . . . , φ m be an orthonormal basis (in L ( B R )) of the vector space spannedby restrictions of the eigenfunctions of L associated to negative eigenvalues. Wedefine the Bergman kernel associated to the orthogonal projection in L ( B R ) onto V . Namely, K ( x , y ) := m X j =1 φ j ( x ) φ j ( y ) . Observe that K is independent of the choice of the orthonormal basis. Also m = Z B R K ( x , x ) d x , is the dimension of V . Obviously, there exists x ∈ ¯ B R such that K ( x , x )Vol( B R ) ≥ m. -ENDED SOLUTIONS 25 We then consider the evaluation form E x ( φ ) := φ ( x ) , and choose the orthonormal basis φ , . . . , φ m such that E x ( φ j ) = 0 for j = 2 , . . . , m .Then K ( x , x ) Vol( B R ) = φ ( x ) Vol( B R ) ≥ m But, (7.27) implies that k φ k L ∞ ( B R ) ≤ C k φ k L ( B R ) = C. Therefore m ≤ C Vol( B R ) and hence the dimension of V is bounded indepen-dently of R > R .This implies that the Morse index of L is finite. References [1] L. Ambrosio and X. Cabr´e.
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Micha l Kowalczyk, Departamento de Ingenier´ıa Matem´atica and Centro de Mode-lamiento Matem´atico (UMI 2807 CNRS), Universidad de Chile, Casilla 170 Correo 3,Santiago, Chile.
E-mail address : [email protected] Y. Liu - Departamento de Ingenier´ıa Matem´atica and Centro de Modelamiento Matem´atico(UMI 2807 CNRS), Universidad de Chile, Casilla 170 Correo 3, Santiago, Chile.
E-mail address : [email protected] Frank Pacard, Centre de Math´ematiques Laurent Schwartz, ´Ecole Polytechnique,91128 Palaiseau, France et Institut Universitaire de France
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