Interacting helical vortex filaments in the 3-dimensional Ginzburg-Landau equation
aa r X i v : . [ m a t h . A P ] J u l INTERACTING HELICAL VORTEX FILAMENTS IN THE 3-DIMENSIONALGINZBURG-LANDAU EQUATION
JUAN D ´AVILA, MANUEL DEL PINO, MARIA MEDINA AND R´EMY RODIAC
Abstract.
For each given n ≥
2, we construct a family of entire solutions u ε ( z, t ), ε >
0, with helicalsymmetry to the 3-dimensional complex-valued Ginzburg-Landau equation∆ u + (1 − | u | ) u = 0 , ( z, t ) ∈ R × R ≃ R . These solutions are 2 π/ε -periodic in t and have n helix-vortex curves, with asymptotic behavior as ε → u ε ( z, t ) ≈ n Y j =1 W (cid:0) z − ε − f j ( εt ) (cid:1) , where W ( z ) = w ( r ) e iθ , z = re iθ , is the standard degree +1 vortex solution of the planar Ginzburg-Landau equation ∆ W + (1 − | W | ) W = 0 in R and f j ( t ) = √ n − e it e i ( j − π/n p | log ε | , j = 1 , . . . , n. Existence of these solutions was previously conjectured in [16], being f ( t ) = ( f ( t ) , . . . , f n ( t )) a rotatingequilibrium point for the renormalized energy of vortex filaments there derived, W ε ( f ) := π Z π (cid:16) | log ε | n X k =1 | f ′ k ( t ) | − X j = k log | f j ( t ) − f k ( t ) | (cid:17) d t, corresponding to that of a planar logarithmic n -body problem. These solutions satisfylim | z |→ + ∞ | u ε ( z, t ) | = 1 uniformly in t and have nontrivial dependence on t , thus negatively answering the Ginzburg-Landau analogue of theGibbons conjecture for the Allen-Cahn equation, a question originally formulated by H. Brezis. This paper deals with constructing entire solutions to the complex Ginzburg-Landau equation in theEuclidean space R N ∆ u + (1 − | u | ) u = 0 in R N , (1.1)where u : R N → C is a complex-valued function and N ≥
2. It is convenient for our purposes to introducea small parameter ε > ε ∆ u + (1 − | u | ) u = 0 in R N . (1.2)When regarded in a bounded region Ω ⊂ R N , equation (1.2) corresponds to the Euler-Lagrange equationfor the functional J ε ( u ) = 12 Z Ω |∇ u | + 14 ε Z Ω (1 − | u | ) , (1.3)which for N = 2 , u corresponds to a critical point of J ε in which | u | represents the density of the super-conductive property of the sample Ω (Cooper pairs of electrons). The function u isexpected to stay away from zero except on a lower-dimensional zero set, the vortex set, corresponding to defects where superconductivity is not present.In their pioneering work [8], Bethuel-Brezis-H´elein analyzed in dimension N = 2 the behavior as ε → u ε of J ε when subject to a boundary condition g : ∂ Ω → S of degree k ≥
1. Theyestablished that away from a finite number of distinct points a , . . . , a k ∈ Ω one has (up to subsequences) u ε ( x ) ≈ e iϕ ( x ) k Y j =1 x − a j | x − a j | , where ϕ ( x ) is a real harmonic function and the k -tuple ( a , . . . , a k ) minimizes a functional of points, the renormalized energy that measures through Green’s function the mutual interaction between the pointsand the boundary. Using the results in [32, 36, 41] one gets the validity of the global approximation u ε ( x ) ≈ e iϕ ( x ) k Y j =1 W (cid:18) x − a j ε (cid:19) , (1.4)where W ( z ) is the standard degree +1 vortex solution of the equation (1.1) for N = 2, namely its uniquesolution of the form W ( z ) = e iθ w ( r ) , z = re iθ , (1.5)where w > w ′′ + w ′ r − wr + (1 − w ) w = 0 in (0 , ∞ ) ,w (0 + ) = 0 , w (+ ∞ ) = 1 , (1.6)see [10, 25]. Thus before reaching the limit, the vortex set of u ε is constituted by exactly k distinctpoints, each with local degree +1. The mechanism of vortex formation in two-dimensional Ginzburg-Landau model from the action of an external constant magnetic field has been extensively studied, see [38]and references therein. Critical points of the renormalized energy are in fact in correspondence with othercritical points of J ε in (1.3) of the form (1.4) for small ε , as it has been found in [30, 15, 3, 34, 17]. In thehigher dimensional case N ≥ ε → N = 3 defectsshould typically assume the form of curves with a winding number associated: these are called vortexfilaments . The basic degree +1 vortex line is the solution u of (1.1) for N = 3 given by u ( z, t ) = W ( z ) , ( z, t ) ∈ R × R ≃ R , with W ( z ) specified in (1.5). Its zero set is of course the t -axis, and a transversal winding number +1is associated to it. In dimension N = 3, under Neumann boundary conditions, it was found in [33] alocal minimizer with energy formally corresponding to multiple vortex lines collapsing onto a segment.Motivated by this work, in [16] an expression for the renormalized energy for the interaction of nearlyparallel “degree +1 vortex lines” collapsing onto the t -axis was derived. Considering n curves t ( f i ( t ) , t ) , ≤ i ≤ n, f = ( f , · · · , f n ) , which for simplicity we assume 2 π -periodic, we look for an approximate solution of the form u ε ( z, t ) ≈ W ε ( z, t ; f ) := e iϕ ( z,t ) n Y j =1 W (cid:18) z − f j ( t ) ε (cid:19) . (1.7) NTERACTING HELICAL GINZBURG-LANDAU FILAMENTS 3
In the cylinder Ω = C = B R (0) × (0 , π ), with ϕ harmonic matching lateral zero Neumann boundaryconditions it is found in [16] that I ε ( f ) := J ε ( W ε ( · ; f )) ≈ π × nπ | log ε | + W ε ( f ) (1.8)where W ε ( f ) := π Z π (cid:16) | log ε | n X k =1 | f ′ k ( t ) | − X j = k log | f j ( t ) − f k ( t ) | (cid:17) d t. (1.9)Equilibrium location of these curves should then correspond to an approximate critical point of thefunctional I ε and hence of W ε , which is the action associated to the n -logarithmic body problem in R .This energy also appears in related problems in fluid dynamics, see e.g. [28, 27]. If we set f ( t ) = 1 p | log ε | ˜ f ( t ) , ˜ f = ( ˜ f , · · · , ˜ f n ) , this corresponds to a 2 π -periodic solution of the ODE system − ˜ f ′′ k ( t ) = 2 X i = k ˜ f k ( t ) − ˜ f i ( t ) | ˜ f k ( t ) − ˜ f i ( t ) | . (1.10)The following n -tuple ˜ f is a standard rotating solution of system (1.10).˜ f k ( t ) = √ n − e it e i ( k − π/n , k = 1 , · · · , n. (1.11)It is shown in [16] that the functional I ε in (1.8) does have a 2 π -periodic critical point f ε ( t ) such that f ε ( t ) = f ( t ) + o (1) p | log ε | , f ( t ) := 1 p | log ε | ˜ f ( t ) (1.12)uniformly as ε →
0, and it is conjectured the existence of a solution u ε ( z, t ) to the system ε ∆ u + (1 − | u | ) u = 0 , ( z, t ) ∈ R , (1.13)which is 2 π -periodic in t and has the approximate form (1.7) for f as in (1.12). The recent work [13] hasestablished a rigorous connection, in the sense of Γ-convergence, between minimizers of the functional(1.9) and minimizers in cylinders with suitable Dirichlet boundary condition, thus providing evidencetowards the conjecture in [16]. In this paper we prove this conjecture. Theorem 1.
For every n ≥ and for ε sufficiently small there exists a solution u ε ( z, t ) of (1.13) , π -periodic in the t -variable, with the following asymptotic profile: u ε ( z, t ) = n Y k =1 W (cid:18) z − f εk ( t ) ε (cid:19) + ϕ ε ( z, t ) , where f εk ( t ) is π -periodic with the asymptotic behavior (1.12) and | ϕ ε ( z, t ) | ≤ C | log ε | . Besides we have lim | z |→ + ∞ | u ε ( z, t ) | = 1 uniformly in t. (1.14) JUAN D´AVILA, MANUEL DEL PINO, MARIA MEDINA AND R´EMY RODIAC
We are also able to construct another family of solutions to (1.13). Until now we have dealt onlywith vortex filaments of degree +1. However it is believed that, in presence of several vortex filaments ofdifferent degrees d k ∈ Z the energy governing the interaction of the filaments is W ε ( f ) := π Z π (cid:16) | log ε | n X k =1 | f ′ k ( t ) | − X j = k d j d k log | f j ( t ) − f k ( t ) | (cid:17) d t. (1.15)There exist special critical points of (1.15) analogous to (1.11) for d = − d k = +1 for k = 2 , . . . , n when n ≥
5. These critical points can be written as g ( t ) = 0 , g k ( t ) = √ n − e it e i ( k − π/ ( n − , k = 2 , . . . , n, n ≥ . (1.16)From these solutions we can obtain: Theorem 2.
For every n ≥ and for ε sufficiently small there exists a solution u ε ( z, t ) of (1.13) , π -periodic in the t -variable, with the following asymptotic profile: u ε ( z, t ) = W ( z ) n Y k =2 W (cid:18) z − g εk ( t ) ε (cid:19) + ϕ ε ( z, t ) , where g εk ( t ) is π -periodic with g εk ( t ) = g k ( t ) + o ε (1) √ | log ε | , g k defined by (1.16) and | ϕ ε ( z, t ) | ≤ C | log ε | . Besides we have lim | z |→ + ∞ | u ε ( z, t ) | = 1 uniformly in t. The proofs of both theorems give a precise answer to the existence question, with an accurate descrip-tion of the solution. They take specific advantage of the geometric setting: the configuration predictedis one of multiple helix vortex curves periodically winding around each other. The Ginzburg-Landauequation has a screw driving symmetry which we take advantage of to reduce the original problem to aplanar one. For constructions of solutions with helical vortex structures we refer also to [11, 42].We observe that in terms of the parameterless equation (1.1) for N = 3, what we find is a family ofentire solutions u ε ( z, t ), πε -periodic in t with approximate form u ε ( z, t ) ≈ n Y j =1 W (cid:0) z − ε − f εj ( εt ) (cid:1) . Equation (1.1) is the complex-valued version of the Allen-Cahn equation of phase transitions, ε ∆ u + u − u = 0 in R N , (1.17)where u : R N → R . The Allen-Cahn model describes transitions of two phases between the values − ε should lie close to a minimal hypersurface.Solutions with screw-driving symmetry, whose zero level set is precisely a helicoid have been built in [22](and extended to the fractional case in [12]). Solutions with multiple interfaces whose interactions aregoverned by mechanical systems (Toda systems) have been built in [19, 18, 21, 1]. The celebrated DeGiorgi conjecture states that, at least up to dimension N = 8 solutions of (1.17) which are monotone inone direction must have one dimensional-symmetry namely their level sets must be parallel hyperplanes, NTERACTING HELICAL GINZBURG-LANDAU FILAMENTS 5 see [24, 4, 40, 20] and references therein. A variant of this statement is the
Gibbons conjecture : an entiresolution u of (1.17) such that lim | x N |→ + ∞ | u ( x ′ , x N ) | = 1 uniformly in x ′ ∈ R N − must necessarily be a function of x N only. This fact has been proven for any N ≥
2. See [5, 7, 23]. Theanalogous question for the Ginzburg-Landau equation in R N , N ≥
3, originally formulated by H. Brezisis whether or not a solution u ( z, t ), ( z, t ) ∈ R × R N − withlim | z |→ + ∞ | u ( z, t ) | = 1 uniformly in t ∈ R N − , (1.18)must necessarily be a function of z . This turns out to be false since the solutions in Theorem 1 satisfy(1.18). We observe that solutions W n ( z ) of (1.1) with total degree n , | n | ≥ W n ( z ) = e inθ w n ( r ) , w n (0) = 0 , w n (+ ∞ ) = 1 , z = re iθ , are known to exist for each n ≥
2, see [25, 10]. The solutions in Theorem 1 have transversal total degreeequal to n ≥
2. A natural question is whether or not Brezis’ statement holds true under the additionalassumption of total transversal degree equal to ±
1. See [32, 36, 41] for the corresponding question indimension N = 2 and [39] for a conjecture on the symmetry of entire solutions of (1.1) when N = 3.We will devote the rest of this paper to the proof of Theorem 1. The proof of Theorem 2 followsthe same lines. As we have mentioned, the key observation is that the invariance under screw-drivingsymmetry allows to reduce the problem to one in the plane, for which the solution to be found has afinite number of vortices with degree 1. For simplicity we treat only the case n = 2 in the following butthe arguments can be easily adapted. We will look for solutions that are close to the approximation u d ( x, y, t ) = W (cid:18) x − d cos tε , y − d sin tε (cid:19) W (cid:18) x + d cos tε , y + d sin tε (cid:19) (1.19)when ε is small. Here d is a parameter of size 1 / p | log ε | .It can be observed that the zero set of u d has the shape of a double helix and the function ˜ u d := e it u d is screw-symmetric (see Definition 1). Thus ˜ u d can be expressed as a function of two variables, whatreduces the problem to a 2-dimensional case. We will look for screw-symmetric perturbations of ˜ u d . Ourapproach will be based on the method in [17], devised to build up solutions with isolated vortices when N = 2 using a Lyapunov-Schmidt reduction. A major difficulty that we need to overcome is the presenceof very large terms in the error of approximation. In the 2-dimensional case we immediately obtain errorsthat are of size O ( ε ) while here it is O (1 / | log ε | ). This is a major difficulty since the vortex-locationadjustment arises at essentially ε -order. This is overcome by carefully decomposing the error created bythe nonlinearity in “odd” small and “even” large Fourier modes parts. The even part has at main orderno effect in the solvability conditions needed in the linear theory we devise in Proposition 5.1. Thesesteps are rather delicate and we will carry them out in detail in what follows. As a first step we reduce our 3-dimensional problem to a related 2-dimensional one. To do so, wework with a particular type of symmetry. To define this symmetry it is convenient to use cylindricalcoordinates ( r, θ, t ) ∈ R + × R × R , and to work with functions that are 2 π periodic in the second variable. Definition 1.
We say that a function u is screw-symmetric if u ( r, θ + h, t + h ) = u ( r, θ, t ) JUAN D´AVILA, MANUEL DEL PINO, MARIA MEDINA AND R´EMY RODIAC for any h ∈ R . Notice that this condition is equivalent to u ( r, θ, t + h ) = u ( r, θ − h, t ) for any h ∈ R , and then a screw-symmetric function can be expressed as a function of two-variables. Indeed, for any ( r, θ, t ) ∈ R + × R × R , u ( r, θ, t ) = u ( r, θ − t,
0) =: U ( r, θ − t ) . Writing the standard vortex of degree one in polar coordinates ( r, θ ), i.e., W ( z ) = w ( r ) e iθ , we can seethat the approximation u d defined in (1.19) satisfies u d ( r, θ, t + h ) = e ih u d ( r, θ − h, t )for any h in R . That is, u d is not screw-symmetric but ˜ u d ( r, θ, t ) := e − it u d ( r, θ, t ) is. Hence we can write u d as u d = e it ˜ u d , with ˜ u d a screw symmetric function.This suggests to look for solutions u of (1.2) that can be written as u ( r, θ, t ) = e it ˜ u ( r, θ, t )with ˜ u screw-symmetric. Thus ˜ u ( r, θ, t ) = U ( r, θ − t ), being U : R + × R π -periodic in the second variable.Denoting U = U ( r, s ) we can see that ∂ r u = e it ∂ r U ( r, s ) , ∂ rr u = e it ∂ rr U ( r, s ) , ∂ θ u = e it ∂ s U ( r, s ) , ∂ θθ u = e it ∂ ss U ( r, s ) ,∂ t u = [2 iU − ∂ s U ] e it , ∂ tt u = [ ∂ ss U − i∂ s U − U ] e it . Recalling that the Laplacian in cylindrical coordinates is expressed by ∂ rr + r ∂ r + r ∂ θθ + ∂ tt we concludethat u is a solution of (1.2) if and only if U is a solution of ε (cid:18) ∂ rr U + 1 r ∂ r U + 1 r ∂ ss U + ∂ ss U − i∂ s U − U (cid:19) + (1 − | U | ) U = 0 in R ∗ + × R . We will also work in rescaled coordinates, that is, we define V ( r, s ) := U ( εr, s ) and we search for asolution to the equation ∂ rr V + 1 r ∂ r V + 1 r ∂ ss V + ε ( ∂ ss V − i∂ s V − V ) + (1 − | V | ) V = 0 in R ∗ + × R . (2.1)From now on we will work in the plane R , and we will use the notation z = x + ix = re is . Wedenote by ∆ the Laplace operator in 2-dimensions, meaning∆ = ∂ x x + ∂ x x = ∂ rr + 1 r ∂ r + 1 r ∂ ss . Then equation (2.1) can be written as∆ V + ε ( ∂ ss V − i∂ s V − V ) + (1 − | V | ) V = 0 in R , (2.2)and the approximate solution (1.19) in the new coordinates is given by V d ( z ) = W ( z − ˜ d ) W ( z + ˜ d ) (2.3)where ˜ d := dε = ˆ dε p | log ε | , for some new parameter ˆ d = O (1). NTERACTING HELICAL GINZBURG-LANDAU FILAMENTS 7
Let S ( v ) := ∆ v + ε ( ∂ ss v − i∂ s v − v ) + (1 − | v | ) v. The equation to be solved can be written as S ( v ) = 0 . (3.1)Recall the notation z = re is = x + ix and ∆ = ∂ x x + ∂ x x . When using the coordinates ( x , x )equation (3.1) is posed in R , while if we use polar coordinates ( r, s ) the domain for (3.1) is r > s ∈ R with periodicity.Following del Pino-Kowalczyk-Musso [17], we look for a solution to (3.1) of the form v = ηV d (1 + iψ ) + (1 − η ) V d e iψ , (3.2)where V d is the ansatz (2.3) and ψ is the new unknown. The cut-off function η in (3.2) is defined as η ( z ) = η ( | z − ˜ d | ) + η ( | z + ˜ d | ) , z ∈ C = R and η : R → [0 ,
1] is a smooth cut-off function such that η ( t ) = 1 for t ≤ η ( t ) = 0 for t ≥ . (3.3)The reason for the form of the perturbation term in (3.2) is the same as in [17]. On one hand, thenonlinear terms behave better for the norms that we consider when using the multiplicative ansatz, butnear the vortices, an additive ansatz is better since it allows the position of the vortex to be adjusted.Our objective here is to rewrite (3.1) in the form L ε ψ + R + N ( ψ ) = 0and identify the linear operator L ε , the error R and the nonlinear terms N ( ψ ).It will be convenient to write S = S + S where S ( v ) = ∆ v + (1 − | v | ) v, S ( v ) = ε ( ∂ ss v − i∂ s v − v ) . (3.4)We have S ( V d + φ ) = S ( V d ) + L ( φ ) + N ( φ ) S ( V d + φ ) = S ( V d ) + S ( φ ) , where L ( φ ) = ∆ φ + (1 − | V d | ) φ − V d φ ) V d (3.5) N ( φ ) = − V d φ ) φ − | φ | ( V d + φ ) . (3.6)Rewrite (3.2) as v = V d + φφ = iV d ψ + γ ( ψ ) γ ( ψ ) = (1 − η ) V d ( e iψ − − iψ ) JUAN D´AVILA, MANUEL DEL PINO, MARIA MEDINA AND R´EMY RODIAC
Then S ( v ) = S ( V d ) + L ( iV d ψ ) + L ( γ ( ψ )) + N ( iV d ψ + γ ( ψ )) S ( v ) = S ( V d ) + S ( iV d ψ ) + S ( γ ( ψ )) . We compute L ( iV d ψ ) = iV d (cid:20) ˜ L ψ + S ( V d ) V d ψ (cid:21) , where ˜ L ( ψ ) = ∆ ψ + 2 ∇ V d ∇ ψV d − i | V d | Im( ψ ) , and so S ( v ) = iV d " − i S ( V d ) V d + ˜ L ( ψ ) + S ( V d ) V d ψ − iV d L ( γ ( ψ )) − iV d N ( iV d ψ + γ ( ψ )) . (3.7)We note that far from the vortices we have S ( v ) = S ( V d e iψ ) = iV d e iψ (cid:20) − i S ( V d ) V d + ˜ L ( ψ ) + ˜ N ( ψ ) (cid:21) , (3.8)where ˜ N ( ψ ) = i ( ∇ ψ ) + i | V d | ( e − ψ ) − ψ )) . Similarly we compute S ( iV d ψ ) = iV d (cid:18) S ( V d ) V d ψ + ˜ L ( ψ ) (cid:19) (3.9)where ˜ L ( ψ ) = ε (cid:16) ∂ ss ψ + 2 ∂ s V d V d ∂ s ψ − i∂ s ψ (cid:17) . Far away from the vortices we have S ( V d e iψ ) = iV d e iψ (cid:20) − i S ( V d ) V d + ˜ L ( ψ ) + ε i ( ∂ s ψ ) (cid:21) . (3.10)We let ˜ η ( z ) = η ( | z − ˜ d | −
1) + η ( | z + ˜ d | − η defined in (3.3). Then we write S ( v ) = 0 as0 = ˜ ηiV d " − i S ( V d ) V d + ˜ L ( ψ ) + S ( V d ) V d ψ − iV d L ( γ ( ψ )) − iV d N ( iV d ψ + γ ( ψ )) − i S ( V d ) V d + ˜ L ( ψ ) + S ( V d ) V d ψ − iV d S ( γ ( ψ )) + (1 − ˜ η ) iV d e iψ (cid:20) − i S ( V d ) V d + ˜ L ( ψ ) + ˜ N ( ψ ) − i S ( V d ) V d + ˜ L ( ψ ) + ε i ( ∂ s ψ ) (cid:21) , that is, we use expressions (3.7), (3.9) near the vortices and (3.8), (3.10) far from them. Hence S ( v ) = 0is equivalent to NTERACTING HELICAL GINZBURG-LANDAU FILAMENTS 9 L ε ( ψ ) + R + N ( ψ ) = 0where L ε ( ψ ) := ( ˜ L + ˜ L )( ψ ) + ˜ η S ( V d ) V d ψR := − i S ( V d ) V d (3.11) N ( ψ ) := ˜ η (cid:16) η + (1 − ˜ η ) e iψ − (cid:17) S ( V d ) V d ψ − iV d ˜ η ˜ η + (1 − ˜ η ) e iψ n L ( γ ( ψ )) + S ( γ ( ψ )) + N ( iV d ψ + γ ( ψ )) o + (1 − ˜ η ) e iψ ˜ η + (1 − ˜ η ) e iψ n ˜ N ( ψ ) + ε i ( ∂ s ψ ) o . Note that explicitly L ε ( ψ ) = ∆ ψ + 2 ∇ V d ∇ ψV d − i | V d | Im( ψ ) + ε (cid:16) ∂ ss ψ + 2 ∂ s V d V d ∂ s ψ − i∂ s ψ (cid:17) + ˜ η S ( V d ) V d ψ, (3.12)and that for | z ± ˜ d | ≥ N ( ψ ) = ˜ N ( ψ ) + ε i ( ∂ s ψ ) = i ( ∇ ψ ) + i | V d | ( e − ψ ) − ψ )) + ε i ( ∂ s ψ ) . In order to analyze the equation near eachvortex, it will be useful to write it in a translated variable. Namely, we set ˜ d j := ( − j ˜ d for j = 1 , z := z − ˜ d j and the function φ j (˜ z ) through the relation φ j (˜ z ) = iW (˜ z ) ψ ( z ) , | ˜ z | < ˜ d. (3.13)That is, φ ( z ) = iV d ψ ( z ) = φ j (˜ z ) α j ( z ) , where α j ( z ) = V d ( z ) W ( z − ˜ d j ) . Hence in the translated variable the unknown (3.2) becomes, in | ˜ z | < ˜ d , v ( z ) = α j (˜ z ) (cid:18) W (˜ z ) + φ j (˜ z ) + (1 − η (˜ z )) W (˜ z ) (cid:20) e φj (˜ z ) W (˜ z ) − − φ j (˜ z ) W (˜ z ) (cid:21)(cid:19) . We set E := S ( V d ) . For φ j , ψ linked through formula (3.13) we define L εj ( φ j )(˜ z ) := iW (˜ z ) L ε ( ψ )(˜ z + ˜ d j ) = L εd ( φ )( z ) α j ( z ) + ( η − E ( z ) V d ( z ) φ j (˜ z )= L εd ( φ j α j )( z ) α j ( z ) + ( η − E ( z ) V d ( z ) φ j (˜ z ) , (3.14)with L εd defined by L εd ( φ ) := ∆ φ + ε ( ∂ ss φ − i∂ s φ − φ ) + (1 − | V d | ) φ − V d φ ) V d . (3.15) Let us also define S ( V ) := ∂ rr V + 1 r ∂ r V + 1 r ∂ ss V + ε ( ∂ ss V − i∂ s V − V ) ,S ( V ) := ∂ rr V + 1 r ∂ r V + 1 r ∂ ss V + ε ( ∂ ss V − i∂ s V ) . Notice that E ( z ) = S ( α j ( z ) W (˜ z )) + (1 − | W | | α j | ) W (˜ z ) α j ( z ) , and thus, using the equation satisfied by W , E = W S ( α j ) + (1 − | W | | α j | ) α j W + 2 ∇ α j ∇ W + 2 ε ∂ s α j ∂ s W + α j S ( W )= W S ( α j ) − ε W α j + (1 − | W | | α j | ) α j W + 2 ∇ α j ∇ W + 2 ε ∂ s α j ∂ s W + α j [ ε ( ∂ ss W − i∂ s W ) − (1 − | W | ) W ] . This allows us to conclude L εj ( φ j ) = L ( φ j ) + ε ( ∂ ss φ j − i∂ s φ j − φ j ) + 2(1 − | α j | ) Re( W φ j ) W − (cid:18) ∇ α j α j ∇ WW + 2 ε ∂ s α j α j ∂ s WW + ε ( ∂ ss W − i∂ s W ) W + 4 ε (cid:19) φ j +2 ∇ α j α j ∇ φ j + 2 ε ∂ s α j α j ∂ s φ j + ˜ η E j V jd φ j , (3.16)where V jd = V d (˜ z + d j ), E j = S ( V jd ) and L is the linear operator defined by (3.5).Let us point out that for | ˜ z | < | α j (˜ z ) | = 1 + O ε ( ε | log ε | ) , ∇ α j (˜ z ) = O ε ( ε p | log ε | ) , ∆ α j = O ε ( ε | log ε | ) . (3.17)With this in mind, we can see that the linear operator L εj is a small perturbation of L . We end this section by making use of thesymmetries of the problem. Using the notation z = x + ix = re is we remark that V d satisfies V d ( − x , x ) = V d ( x , x ) and V d ( x , − x ) = V d ( x , x ) . We also remark that these symmetries are compatible with the solution operator S , that is, if S ( V ) = 0and U ( z ) = V ( − x , x ), then S ( U ) = 0, and the same for U ( z ) = V ( x , − x ). Thus we look for a solution V satisfying V ( − x , x ) = V ( x , x ) , V ( x , − x ) = V ( x , x ) , what drives to ask ψ ( x , − x ) = − ψ ( x , x ) , ψ ( − x , x ) = − ψ ( x , x ) . (3.18) In this section we compute the error of the approximation V d defined as R = − i S ( V d ) V d in (3.11).In order to measure the size of the error of our approximation we fix 0 < α, σ <
1. We recall that˜ d j := ( − j ˜ d, (4.1) NTERACTING HELICAL GINZBURG-LANDAU FILAMENTS 11 and we define the norm k h k ∗∗ := X j =1 k V d h k C α ( ρ j < + sup ρ > ,ρ > h | Re( h ) | ρ − + ρ − + ε + | Im( h ) | ρ − σ + ρ − σ + ε σ − i + sup < | z − ˜ d | < R ε , < | z − ˜ d | < R ε [Re( h )] α,B | z | / ( z ) | z − ˜ d | − − α + | z − ˜ d | − − α + sup < | z − ˜ d | < R ε , < | z − ˜ d | < R ε [Im( h )] α,B ( z ) | z − ˜ d | − σ + | z − ˜ d | − σ , (4.2)where k f k C α ( D ) = k f k C ,α ( D ) , and where we have used the notation[ f ] α,D = sup x,y ∈ D, x = y | f ( x ) − f ( y ) || x − y | α (4.3) k f k C k,α ( D ) = k X j =0 k D j f k L ∞ ( D ) + [ D k f ] α,D . (4.4) Proposition 4.1.
Let V d given by (2.3) , and denote S ( V d ) = E = iV d R = iV d ( R + iR ) . Then k R k ∗∗ ≤ C | log ε | . Proof.
Let us write V d = W a W b where W a ( z ) := W ( z − ˜ d ) , W b ( z ) := W ( z + ˜ d ) . We want to estimate E := S ( V d ), i.e., how far our approximation is to be a solution.By symmetry it suffices to compute the error in the region ( x , x ) ∈ R + × R . We denote ρ e iθ := re is − ˜ d, ρ e iθ := re is + ˜ d. We recall that ∆( f g ) = g ∆ f + f ∆ g + 2 ∇ f ∇ g and thus with S defined in (3.4): S ( V d ) = ( W ax x + W ax x ) W b + ( W bx x + W bx x ) W a + 2( W ax W bx + W ax W bx ) + (1 − | W a W b | ) W a W b . Using the fact that ∆ W + (1 − | W | ) W = 0 in R we conclude that S ( V d ) = 2( W ax W bx + W ax W bx ) + (cid:0) − | W a W b | + | W a | − | W b | − (cid:1) W a W b . (4.5)We estimate the size of this error separately in two different regions, near the vortices and far from them.Notice first that, since we work in the half-plane R + × R , we have ρ ≥ ˜ d ≥ Cε p | log ε | for some C >
Step 1:
Estimate of S ( V d ) near one vortex, i.e., when | re is − ˜ d | < Writing W = W ( ρe iθ ) we have W x = e iθ (cid:18) w ′ ( ρ ) cos θ − i w ( ρ ) ρ sin θ (cid:19) , W x = e iθ (cid:18) w ′ ( ρ ) sin θ + i w ( ρ ) ρ cos θ (cid:19) . We define w := w ( ρ ) and w := w ( ρ ) and we obtain W ax W bx = e i ( θ + θ ) n w ′ w ′ cos θ cos θ − w w ρ ρ sin θ sin θ − i (cid:20) w ′ w ρ cos θ sin θ + w ′ w ρ cos θ sin θ (cid:21)o ,W ax W bx = e i ( θ + θ ) n w ′ w ′ sin θ sin θ − w w ρ ρ cos θ cos θ + i (cid:20) w ′ w ρ sin θ cos θ + w ′ w ρ cos θ sin θ (cid:21)o . Since w ′ ( ρ ) = ρ + O ( ρ ) when ρ → + ∞ (see Lemma 7.1) and ρ ≥ Cε √ | log ε | we can see that k W ax W bx + W ax W bx k L ∞ ( ρ < ≤ Cε p | log ε | when ε is small and for some C >
0. Using now that w ( ρ ) = 1 − ρ + O ( ρ ) when ρ → + ∞ we obtain k (1 − | W a W b | + | W a | − | W b | − W a W b k L ∞ ( ρ < ≤ Cε | log ε | , and thus k E k L ∞ ( ρ < = k S ( V ) k L ∞ ( ρ < ≤ Cε p | log ε | . (4.6)Similarly, k∇ E k L ∞ ( ρ < ≤ Cε p | log ε | . (4.7) Step 2:
Estimate of S ( V d ) far away from the vortices, i.e., when | re is − ˜ d | > E = S ( V d ) = iV d ( R + iR ) with R = 2(sin θ cos θ − cos θ sin θ ) (cid:18) w ′ ρ w − w ′ ρ w (cid:19) = 2 sin( θ − θ ) (cid:18) w ′ ρ w − w ′ ρ w (cid:19) ,R =2 cos( θ − θ ) (cid:18) − w ′ w ′ w w + 1 ρ ρ (cid:19) − (cid:0) − w w + w − w − (cid:1) . Using that ρ ≤ ρ and ρ ≥ C/ε p | log ε | , along with Lemma 7.1, we conclude | R | ≤ Cε p | log ε | ρ . (4.8)Using again Lemma 7.1 we obtain1 − w w + w − w − − w + O (cid:18) ρ (cid:19) w + O (cid:18) ρ (cid:19) + w − ≤ C ρ ρ , and hence | R | ≤ C ρ ρ ≤ C (cid:16) ε | log ε | / (cid:17) σ ρ − σ , ∀ < σ < . (4.9) NTERACTING HELICAL GINZBURG-LANDAU FILAMENTS 13
To see that the previous inequality holds we can distinguish the cases 2 < ρ < ˜ d < ρ and ˜ d < ρ < ρ .We remark that we also have |∇ R | ≤ Cε p | log ε | ρ , |∇ R | ≤ C ( ε p | log ε | ) σ ρ − σ . Step 3:
Estimates of S ( V d )We recall that w := w ( ρ ) and w := w ( ρ ). Thus V d ( r, s ) =: w w e i ( θ + θ ) with ρ = q ( r cos s − ˜ d ) + r sin s, ρ = q ( r cos s + ˜ d ) + r sin s,e iθ = ( r cos s − ˜ d ) + ir sin sρ , e iθ = ( r cos s + ˜ d ) + ir sin sρ . We have ∂ s V d = [ ∂ s ρ w ′ w + ∂ s ρ w ′ w + i∂ s ( θ + θ ) w w ] e i ( θ + θ ) ∂ ss V d = nh ∂ ss ρ w ′ w + ∂ ss ρ w ′ w + ( ∂ s ρ ) w ′′ w + ( ∂ s ρ ) w ′′ w + 2 ∂ s ρ ∂ s ρ w ′ w ′ − [ ∂ s ( θ + θ )] w w i + i h ∂ s ( θ + θ )( ∂ s ρ w ′ w + ∂ s ρ w ′ w ) + ∂ ss ( θ + θ ) w w (cid:3)o e i ( θ + θ ) , and thus ε ( ∂ ss V d − i∂ s V d − V d ) = ε n ( ∂ s ρ ) w ′′ w + ( ∂ s ρ ) w ′′ w + 2( ∂ s ρ )( ∂ s ρ ) w ′ w ′ + ∂ ss ρ w ′ w + ∂ ss ρ w ′ w − ([ ∂ s ( θ + θ )] − ∂ s ( θ + θ ) + 4) w w o e i ( θ + θ ) + iε n ∂ ss ( θ + θ ) w w + (2[ ∂ s ( θ + θ )] −
4) [ ∂ s ρ w ′ w + ∂ s ρ w ′ w ] o e i ( θ + θ ) . We also need to compute the following derivatives, ∂ s ρ = r ˜ d sin sρ = ˜ d sin θ , ∂ s ρ = − r ˜ d sin sρ = − ˜ d sin θ ,∂ ss ρ = r ˜ d cos sρ − r ˜ d sin sρ = ˜ d cos θ + ˜ d cos θ ρ ,∂ ss ρ = − r ˜ d cos sρ − r ˜ d sin sρ = − ˜ d cos θ + ˜ d cos θ ρ . Now we can check that ∂ s θ = 1 + ˜ dρ ( r cos s − ˜ d ) = 1 + ˜ d cos θ ρ , ∂ s θ = 1 − ˜ dρ ( r cos s + ˜ d ) = 1 − ˜ d cos θ ρ ∂ ss θ = − ˜ dr sin sρ ( ρ + 2 ˜ d ( r cos s − ˜ d )) = − ˜ d sin θ ρ − d sin θ cos θ ρ ,∂ ss θ = ˜ dr sin sρ ( ρ − d ( r cos s + ˜ d )) = ˜ d sin θ ρ − d sin θ cos θ ρ , and besides, [ ∂ s ( θ + θ )] − ∂ s ( θ + θ ) + 4 = ˜ d (cid:18) cos θ ρ − cos θ ρ (cid:19) ,∂ ss ( θ + θ ) = ˜ d (cid:18) sin θ ρ − sin θ ρ (cid:19) − d (cid:18) sin θ cos θ ρ + sin θ cos θ ρ (cid:19) , ∂ s ( θ + θ ) − d (cid:18) cos θ ρ − cos θ ρ (cid:19) . Hence we obtain S ( V d ) = n ˆ d | log ε | (cid:16) w ′′ w sin θ + w ′′ w sin θ − w ′ w ′ sin θ sin θ + cos θ ρ w ′ w + cos θ ρ w ′ w − (cid:18) cos θ ρ − cos θ ρ (cid:19) w w (cid:17) + ε ˆ d p | log ε | (cid:16) cos θ w ′ w − cos θ w ′ w (cid:17) + i h ε ˆ d p | log ε | (cid:18) sin θ ρ − sin θ ρ (cid:19) w w − d | log ε | (cid:18) sin θ cos θ ρ + sin θ cos θ ρ (cid:19) w w + 2 ˆ d | log ε | (cid:18) cos θ ρ − cos θ ρ (cid:19) (sin θ w ′ w − sin θ w ′ w ) io e i ( θ + θ ) . (4.10)The conclusion of the proof follows from (4.6), (4.7), (4.8), (4.9) and the following Lemma 4.1. We alsouse the symmetry of the problem. (cid:3) Lemma 4.1.
Let S ( V d ) = iV d R = iV d ( R + iR )) . In the half-plane R + × R we have that k S ( V d ) k L ∞ ( ρ < ≤ C | log ε | , k∇ S ( V d ) k L ∞ ( ρ < ≤ C | log ε | , (4.11) and for ρ > : | R | ≤ C | log ε | ρ , |∇ R | ≤ C | log ε | ρ , | R | ≤ C | log ε | ρ , |∇ R | ≤ C | log ε | ρ . (4.12) Proof.
By using Lemma 7.1 we see that k S ( V d ) k L ∞ ( ρ < ≤ C | log ε | , k∇ S ( V d ) k L ∞ ( ρ < ≤ C | log ε | . For ρ > − R = ˆ d | log ε | (cid:16) w ′′ w sin θ + cos θ ρ w ′ w − (cid:18) cos θ ρ − cos θ ρ (cid:19) (cid:17) + ˆ d | log ε | (cid:16) w ′′ w sin θ − w ′ w ′ sin θ sin θ + cos θ ρ w ′ w (cid:17) + ˆ dε p | log ε | (cos θ w ′ w − cos θ w ′ w ) . NTERACTING HELICAL GINZBURG-LANDAU FILAMENTS 15
By using Lemma 7.1 and the fact that ρ ≥ ρ > d | log ε | (cid:12)(cid:12)(cid:12) w ′′ w sin θ + cos θ ρ w ′ w − (cid:18) cos θ ρ − cos θ ρ (cid:19) (cid:12)(cid:12)(cid:12) ≤ C | log ε | ρ , and ˆ dε p | log ε | (cid:12)(cid:12)(cid:12) cos θ w ′ w − cos θ w ′ w (cid:12)(cid:12)(cid:12) ≤ Cε p | log ε | ρ . Besides by using also that ρ ≥ ˜ d ≥ ˆ d/ ( ε p | log ε | ) we observe thatˆ d | log ε | (cid:12)(cid:12)(cid:12) w ′′ w sin θ − w ′ w ′ sin θ sin θ + cos θ ρ w ′ w (cid:12)(cid:12)(cid:12) ≤ C ε p | log ε | ρ . Thus we obtain (4.11) and the third estimate in (4.12). By differentiating we can also obtain the fourthestimate.Now for ρ > R = ˆ dε p | log ε | (cid:18) sin θ ρ − sin θ ρ (cid:19) − d | log ε | (cid:18) sin θ cos θ ρ + sin θ cos θ ρ (cid:19) + 2 ˆ d | log ε | (cid:20) cos θ sin θ w ′ ρ w − cos θ sin θ w ′ ρ w (cid:21) − d | log ε | cos θ ρ (sin θ w ′ w − sin θ w ′ w ) . By using Lemma 7.1 and the fact that ρ ≥ ρ we can see that the two first estimates of (4.12) hold.Actually to prove the first estimate the only difficult term to handle isˆ dε p | log ε | (cid:18) sin θ ρ − sin θ ρ (cid:19) . In the region ( R + × R ) ∩ { ρ < C/ ( ε p | log ε | ) } we have that ε p | log ε | /ρ < C/ρ . By using this andthat ρ > ρ we find that (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ˜ dε p | log ε | w w (cid:18) sin θ ρ − sin θ ρ (cid:19)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ C | log ε |
11 + ρ (4.13)in ( R + × R ) ∩ { ρ < /ε p | log ε |} .Now we use that ρ sin θ = ρ sin θ = r sin s to obtain that (cid:18) sin θ ρ − sin θ ρ (cid:19) = sin θ ρ (cid:18) − ρ ρ (cid:19) . But ρ = ρ + 4 ˜ dr cos s = ρ + 4 ˜ dρ cos θ + ˜ d . Thus when (cid:12)(cid:12)(cid:12) dρ cos θ + ˜ d (cid:12)(cid:12)(cid:12) <
1, which is true when ρ ≥ Cε √ | log ε | for an appropriate constant C >
0, we find that ρ ρ = 1 + 4 ˆ dρ cos θ + O ˜ d ρ ! . Thus (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ˆ dε p | log ε | w w (cid:18) sin θ ρ − sin θ ρ (cid:19)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ C | log ε | ρ (4.14) in ( R + × R ) ∩ { ρ > C/ ( ε p | log ε | ) } . Combining estimates (4.13) and (4.14) and differentiating we arriveat the conclusion. (cid:3) Recall the polar coordinates ρ j , θ j about ˜ d j defined by the relation z = ρ j e iθ j + ˜ d j . We can decomposea function h satisfying h ( z ) = − h ( z ) in Fourier series in θ j as h = ∞ X k =0 h k,j (4.15) h k,j ( ρ j , θ j ) = h k,j ( ρ j ) sin( kθ j ) + ih k,j ( ρ j ) cos( kθ j ) , h k,j ( ρ j ) , h k,j ( ρ j ) ∈ R , and define h e,j = X k even h k,j , h o,j = X k odd h k,j . The definitions above can also be expressed by the following. Let R j denote the reflection across theline Re( z ) = ˜ d j . Since ˜ d j ∈ R , we have R j z = 2 ˜ d j − Re( z ) + i Im( z ) . (4.16)Then h e,j and h o,j have the symmetries h o,j ( R j z ) = h o,j ( z ) , h e,j ( R j z ) = − h e,j ( z ) , and we can define equivalently h o,j ( z ) = 12 [ h ( z ) + h ( R j z )] , h e,j ( z ) = 12 [ h ( z ) − h ( R j z )] . (4.17)It is convenient to consider a global function h o defined as follows. We introduce cut-off functions η j,R ,as η j,R ( z ) = η (cid:16) | z − ˜ d j | R (cid:17) , (4.18)where η : R → [0 ,
1] is a smooth function such that η ( t ) = 1 for t ≤ η ( t ) = 0 for t ≥
2. Given h : C → C we define R ε = α ε | log ε | , (4.19)where α > R ε ≤ ˜ d and h o = η ,R ε h o, + η ,R ε h o, (4.20) h e = h o − h. For a complex function h = h + ih we introduce the new semi-norm | h | ♯♯ = X j =1 k V d h k C ,α ( ρ j < + sup <ρ Let V d given by (2.3) and denote S ( V d ) = E = iV d R. Then we can write R = R o + R e , R o = R oα + R oβ NTERACTING HELICAL GINZBURG-LANDAU FILAMENTS 17 with R o defined analogously to (4.20) and R o ( R j z ) = R o ( z ) in B R ε ( ˜ d ) ∪ B R ε ( − ˜ d ) | R oα | ♯♯ ≤ C ε p | log ε | , k R oβ k ∗∗ ≤ Cε p | log ε | , k R e k ∗∗ + k R o k ∗∗ ≤ C | log ε | Proof. The conclusion follows using the expression of S ( V d ) given by (4.10). More precisely, we define: r o, := − i ˆ d | log ε | (cid:18) θ cos θ ρ ρ + cos θ ρ (cid:19) + n ˆ dε p | log ε | − sin θ ρ − d | log ε | sin θ cos θ ρ o ,r o, := − i ˆ d | log ε | (cid:18) θ cos θ ρ ρ + cos θ ρ (cid:19) + n ˆ dε p | log ε | + sin θ ρ − d | log ε | sin θ cos θ ρ o ,R j,oα := 12 [ r o, ( z ) + r o, ( R j z )] j = 1 , R oα := η ,R ε R ,oα + η ,R ε R ,oα , with R ε defined by (4.19). We can check that R oα and R oβ := R o − R oα satisfy the desired properties. (cid:3) In the last step of the proof of Theorem 1, when we aim at canceling the Lyapunov-Schmidt coefficient,we will need the following: Lemma 4.2. In the region B ( ˜ d, ˜ d ) we have that S ( V d ) = ˆ d | log ε | W ax x W b + ˆ dε p | log ε | W ax W b + G with Re Z B ( ˜ d, ˆ dε √ | log ε | ) W ax x W ax = 0 and Re Z B ( ˜ d, ˆ dε √ | log ε | ) GW b W ax = O ε ε p | log ε | ! . Proof. It suffices to observe that S ( V d ) = E = ˆ d w e i ( θ + θ ) | log ε | (cid:18) w ′′ sin θ + cos θ (cid:18) w ′ ρ − w ρ (cid:19) + 2 i cos θ sin θ (cid:18) w ′ ρ − w ρ (cid:19)(cid:19) + ˆ dε p | log ε | (cid:18) w ′ w cos θ − iw w sin θ ρ (cid:19) e i ( θ + θ ) + G, where G := ˆ dεe i ( θ + θ ) p | log ε | w w ′ cos θ + ˆ d e i ( θ + θ ) | log ε | (cid:0) w ′′ w sin θ + w ′ w ′ sin θ sin θ + cos θ ρ w ′ w + (cid:18) θ cos θ ρ ρ + cos θ ρ (cid:19) w w (cid:1) + ie i ( θ + θ ) ˆ dε p | log ε | sin θ ρ w w − ie i ( θ + θ ) n d | log ε | sin θ cos θ ρ w w + 2 ˆ d | log ε | cos θ ρ (sin θ w ′ w + sin θ w ′ w )+ 2 ˆ d | log ε | cos θ ρ sin θ w ′ w o . (cid:3) Given h satisfying the symmetries (3.18) and appropriate decay, our aim in this section is to solve thelinear equation L ε ( ψ ) = h + c X j =1 χ j iW ( z − ˜ d j ) ( − j W x ( z − ˜ d j ) in R Re Z B (0 , χφ j W x = 0 , with φ j ( z ) = iW ( z ) ψ ( z + ˜ d j ) ψ satisfies the symmetry (3.18) . (5.1)where χ ( z ) := η (cid:18) | z | (cid:19) , χ j ( z ) := η (cid:16) ρ j (cid:17) = η | z − ˜ d j | ! with η a smooth cut-off function such that η ( t ) = 1 if t ≤ η ( t ) = 0 if t ≥ ψ and h it suffices to use one reduced parameter c and not sixas it should be the case when working with two vortices, since the linearized operator around each hasthree elements in its kernel.In order to find estimates on the solution of (5.1) we introduce some norms, for which we use thefollowing notation. Let ˜ d j , j = 1 , ρ j , θ j )are polar coordinates around ˜ d j , that is, z = ρ j e iθ j + ˜ d j .We will define two sets of norms. The first one is the following: given α, σ ∈ (0 , 1) and ψ : C → C wedefine k ψ k ∗ = X j =1 k V d ψ k C ,α ( ρ j < + k Re( ψ ) k , ∗ + k Im( ψ ) k , ∗ where, with Re ψ = ψ , Im ψ = ψ , k ψ k , ∗ = sup ρ > ,ρ > | ψ | + sup <ρ < ε , <ρ < ε |∇ ψ | ρ − + ρ − + sup r> ε h ε | ∂ r ψ | + | ∂ s ψ | i + sup <ρ Here we have used the notation (4.3)-(4.4). We recall also that the norm for the right hand side h of(5.1) is defined by (4.2). One of the main results in this section is the following. Proposition 5.1. If h satisfies (3.18) and k h k ∗∗ < + ∞ then for ε > sufficiently small there exists aunique solution ψ = T ε ( h ) to (5.1) with k ψ k ∗ < ∞ . Furthermore, there exists a constant C > dependingonly on α, σ ∈ (0 , such that this solution satisfies k ψ k ∗ ≤ C k h k ∗∗ . The proof of Proposition 5.1 is in § ψ are too weak to enable us to solve the reduced problem. This meansthat they are too weak to justify that we can choose the parameter d such that the Lyapunov-Schmidtcoefficient c in (5.1) vanishes. In order to address this difficulty we use that the largest part of the errorand ψ have a symmetry that makes them orthogonal to the kernel. To state the extra (partial) symmetryinvolved, let us consider ψ : C → C . Recall the polar coordinates ρ j , θ j about ˜ d j defined by the relation z = ρ j e iθ j + ˜ d j . We can decompose ψ in Fourier series in θ j as in (4.15) and define ψ e,j = X k even ψ k,j , ψ o,j = X k odd ψ k,j . The intuitive idea is that ψ o,j is not orthogonal to the kernel near ˜ d j but small, while ψ e,j is large butorthogonal to the kernel near ˜ d j by symmetry. With R j defined in (4.16), we have ψ o,j ( R j z ) = ψ o,j ( z ) , ψ e,j ( R j z ) = − ψ e,j ( z ) , and we can define equivalently ψ o,j ( z ) = 12 [ ψ ( z ) + ψ ( R j z )] , ψ e,j ( z ) = 12 [ ψ ( z ) − ψ ( R j z )] . (5.2)It is convenient to consider a global function ψ o defined as follows: with R ε given by (4.19) and η j,R defined in (4.18) we set ψ o = η , R ε ψ o, + η , R ε ψ o, , (5.3)That is, ψ o represents the odd part of ψ around each vortex ˜ d j , localized with a cut-off function.The part of ψ that will be small, namely ψ o will be estimated in norms that allow for growth up toa certain distance. We do this because that part arises from terms in the error R o that are small, butdecay slowly. To capture these behaviors we define | ψ | ♯ = X j =1 | log ε | − k V d ψ k C ,α ( ρ j < + | Re( ψ ) | ♯, + | Im( ψ ) | ♯, , where | ψ | ♯, = sup <ρ Suppose that h satisfies the symmetries (3.18) and k h k ∗∗ < ∞ . Suppose furthermorethat h o defined by (4.20) is decomposed as h o = h oα + h oβ where | h oα | ♯♯ < ∞ and h oα , h oβ satisfy h ok ( R j z ) = h ok ( z ) , | z − ˜ d j | < R ε , j = 1 , k = α, β, and have support in B R ε ( ˜ d ) ∪ B R ε ( ˜ d ) . Let us write ψ = ψ e + ψ o with ψ o defined by (5.3) . Then ψ o can be decomposed as ψ o = ψ oα + ψ oβ , with each function supported in B R ε ( ˜ d ) ∪ B R ε ( ˜ d ) and satisfying | ψ oα | ♯ . | h oα | ♯♯ + ε | log ε | ( k h oα k ∗∗ + k h − h o k ∗∗ ) (5.6) k ψ oβ k ∗ . k h oβ k ∗∗ , (5.7) k ψ oα k ∗ + k ψ oβ k ∗ . k h k ∗∗ + k h oα k ∗∗ + k h oβ k ∗∗ and ψ ok ( R j z ) = ψ ok ( z ) , | z − ˜ d j | < R ε , j = 1 , , k = α, β. The proof of Proposition 5.2 is in § Here we obtain a priori estimates forsolutions to L ε ( ψ ) = h in R Re Z B (0 , χ j φ j W x = 0 , with φ j ( z ) = iW ( z ) ψ ( z + ˜ d j ) ψ satisfies the symmetry (3.18) . (5.8) Lemma 5.1. There exists a constant C > such that for all ε sufficiently small and any solution ψ of (5.8) with k ψ k ∗ < ∞ one has k ψ k ∗ ≤ C k h k ∗∗ . (5.9) Proof of Lemma 5.1 . To prove Lemma 5.1 we will use first weaker norms. For ψ : C → C we define k ψ k ∗ , = X j =1 k V d ψ k L ∞ ( ρ j < + k Re( ψ ) k , ∗ , + k Im( ψ ) k , ∗ , where k ψ k , ∗ , = sup ρ > ,ρ > | ψ | + sup <ρ < ε , <ρ < ε |∇ ψ | ρ − + ρ − + sup r> ε h ε | ∂ r ψ | + | ∂ s ψ | i k ψ k , ∗ , = sup ρ > ,ρ > | ψ | ρ − σ + ρ − σ + ε σ − + sup <ρ < ε , <ρ < ε |∇ ψ | ρ − σ + ρ − σ + sup r> ε (cid:2) ε σ − | ∂ r ψ | + ε σ − | ∂ s ψ | (cid:3) . NTERACTING HELICAL GINZBURG-LANDAU FILAMENTS 21 In the expressions above the gradient of ψ j is ( ∂ x ψ j , ∂ x ψ j ), where z = ( x , x ). Since z = x + ix = re is = ρ e iθ + ˜ d = ρ e iθ − ˜ d we have |∇ ψ j | = ( ∂ x ψ j ) + ( ∂ x ψ j ) = ( ∂ r ψ j ) + 1 r ( ∂ s ψ j ) = ( ∂ ρ ψ j ) + 1 ρ ( ∂ θ ψ j ) = ( ∂ ρ ψ j ) + 1 ρ ( ∂ θ ψ j ) . We define also the norm for the right hand side h = h + ih of (5.1): k h k ∗∗ , := X j =1 k V d ) h k L ∞ ( ρ j < + sup ρ > ,ρ > h | Re( h ) | ρ − + ρ − + ε + | Im( h ) | ρ − σ + ρ − σ + ε − σ i . We claim that there exists a constant C > ε sufficiently small and any solution of(5.8) one has k ψ k ∗ , ≤ C k h k ∗∗ , . (5.10)To prove this, we assume by contradiction that there exist ε n → ψ ( n ) , h ( n ) solutions of (5.8) suchthat k ψ ( n ) k ∗ , = 1 , k h ( n ) k ∗∗ , = o n (1) . We first work near the vortices ˜ d j , and work with the function φ ( n ) j ( z ) = iW ( z ) ψ ( n ) ( z + ˜ d j ).Since k ψ ( n ) k ∗ , = 1 from Arzela-Ascoli’s Theorem we can extract a subsequence such that ˜ φ ( n ) j → φ in C ( R ). Passing to the limit in (5.8) (we use (3.16) and (3.17)), we conclude that L ( φ ) = 0 in R , with L defined in (3.5). Moreover φ inherits the symmetry φ (¯ z ) = φ ( z ). From the estimate k ψ ( n ) k ∗ , = 1 we deduce that φ ∈ L ∞ ( R ) and that ψ = Re ( φ /iW ), ψ = Im ( φ /iW ), satisfy | ψ | + | z ||∇ ψ | ≤ C, | ψ | + |∇ ψ | ≤ C | z | − σ , | z | > . By Lemma 7.2 we deduce that φ = c W x , for some c ∈ R .On the other hand, we can pass to the limit in the orthogonality conditionRe Z B (0 , χ ¯ φ ( n ) j W x = 0 , and obtain that necessarily c = 0. Hence φ ( n ) j → C ( R ). Therefore ψ ( n ) → { ρ ≥ , ρ ≥ } . (5.11)Next we derive estimates far away from the vortices. In the following we drop the superscript n forsimplicity. In { ρ > } ∩ { ρ > } we have that ψ ( n ) = ψ solves h = ∆ ψ + 2 ∇ V d ∇ ψV d − i | V d | ψ + ε ∂ ss ψ + ε (cid:16) ∂ s V d V d − i (cid:17) ∂ s ψ which for ψ = Re( ψ ), ψ = Im( ψ ) translates into the following system h = ∆ ψ + (cid:18) ∇ w w + ∇ w w (cid:19) ∇ ψ − ∇ ( θ + θ ) ∇ ψ + ε ∂ ss ψ + 2 ε (cid:20)(cid:18) ∂ s w w + ∂ s w w (cid:19) ∂ s ψ − ∂ s ( θ + θ ) ∂ s ψ (cid:21) + 4 ε ∂ s ψ (5.12) h = ∆ ψ + (cid:18) ∇ w w + ∇ w w (cid:19) ∇ ψ + ∇ ( θ + θ ) ∇ ψ − | V d | ψ + ε ∂ ss ψ + 2 ε (cid:20)(cid:18) ∇ w w + ∇ w w (cid:19) ∂ s ψ + ∂ s ( θ + θ ) ∂ s ψ (cid:21) − ε ∂ s ψ . (5.13)We start by estimating ψ . Since ψ satisfies ψ ( x , − x ) = ψ ( x , x ) and ψ ( − x , x ) = ψ ( x , x )it is sufficient to obtain estimates for ψ in the quadrant { x > , x > } .Let R > D R = { x > , x > } ∩ { ρ > R } . By the symmetries, ψ satisfieshomogeneous Neumann boundary condition at x = 0 or x = 0.In D R we have | V d | ≥ c > c . We consider (5.13) in D R and rewriteit as ∆ ψ + ε ∂ ss ψ − | V d | ψ = p , in D R , where p = h − (cid:18) ∇ w w + ∇ w w (cid:19) ∇ ψ − ∇ ( θ + θ ) ∇ ψ − ε (cid:20)(cid:18) ∂ s w w + ∂ s w w (cid:19) ∂ s ψ + ∂ s ( θ + θ ) ∂ s ψ (cid:21) + 4 ε ∂ s ψ . We use polar coordinates ( ρ , θ ) around ˜ d and ( ρ , θ ) around − ˜ d , that is, re is = ρ e iθ + ˜ d = ρ e iθ − ˜ d. From this we get that ∂ r = 1 r (cid:16) ρ + ˜ d cos θ (cid:17) ∂ ρ − ˜ d sin θ rρ ∂ θ , ∂ s = ˜ d sin θ ∂ ρ + (cid:16) d cos θ ρ (cid:17) ∂ θ ∂ r = 1 r (cid:16) ρ − ˜ d cos θ (cid:17) ∂ ρ + ˜ d sin θ rρ ∂ θ , ∂ s = − ˜ d sin θ ∂ ρ + (cid:16) − ˜ d cos θ ρ (cid:17) ∂ θ . NTERACTING HELICAL GINZBURG-LANDAU FILAMENTS 23 With these expressions and the asymptotic behaviour stated in Lemma 7.1 we see that (cid:12)(cid:12)(cid:12)(cid:12)(cid:18) ∇ w w + ∇ w w (cid:19) ∇ ψ (cid:12)(cid:12)(cid:12)(cid:12) ≤ CR (cid:16) ρ − σ + ε − σ (cid:17) k ψ k , ∗ , (cid:12)(cid:12)(cid:12) ∇ ( θ + θ ) ∇ ψ (cid:12)(cid:12)(cid:12) ≤ C (cid:0) R − σ + ε σ (cid:1) (cid:16) ρ − σ + ε − σ (cid:17) k ψ k , ∗ , ε (cid:12)(cid:12)(cid:12)(cid:12)(cid:18) ∂ s w w + ∂ s w w (cid:19) ∂ s ψ (cid:12)(cid:12)(cid:12)(cid:12) ≤ C (cid:0) R − + ε (cid:1) (cid:16) ρ − σ + ε − σ (cid:17) k ψ k , ∗ , ε | ∂ s ( θ + θ ) ∂ s ψ | ≤ C (cid:0) R − σ + ε σ (cid:1) (cid:16) ρ − σ + ε − σ (cid:17) k ψ k , ∗ , ε | ∂ s ψ | ≤ C (cid:0) R − σ + ε σ (cid:1) (cid:16) ρ − σ + ε − σ (cid:17) k ψ k , ∗ , Since we assumed k ψ k ∗ , = 1, we get | p | ≤ C ( k h k ∗∗ , + R − σ + ε σ ) (cid:16) ρ − σ + ε − σ (cid:17) . We use a barrier of the form B = M (cid:18) ρ − σ + ε − σ (cid:19) with M = C (cid:16) k h k ∗∗ , + R − σ + ε σ + k ψ k L ∞ ( B R ( ˜ d )) (cid:17) and C > ∂ ss B = ∂ B ∂ρ ˜ d sin( θ ) + ∂ B ∂ρ ˜ d cos( θ ) (cid:16) dρ cos( θ ) (cid:17) = M ( σ − σ − 3) ˜ d sin( θ ) ρ − σ + M ( σ − 2) ˜ d cos( θ ) ρ − σ (cid:16) dρ cos( θ ) (cid:17) and so ∆ B + ε ∂ ss B − | V d | B ≤ − ˜ cM (cid:16) ρ − σ + ε − σ (cid:17) in D R for some fixed ˜ c > 0. Thanks to a comparison principle in D R (a slight variant of Lemma 7.5) andstandard elliptic estimates we get that | ψ | ≤ C (cid:16) ρ − σ + ε − σ (cid:17) ( k h k ∗∗ , + R − σ + ε σ + k ψ k L ∞ ( B R ( ˜ d )) ) , in D R . (5.14)Standard elliptic estimates imply |∇ ψ | ≤ C (cid:16) ρ − σ + ε − σ (cid:17) ( k h k ∗∗ , + R − σ + ε σ + k ψ k L ∞ ( B R ( ˜ d )) ) , in D R ∩ n ρ ≤ ε o . (5.15)For points in D R with ρ > ε we use the scaling˜ ψ (˜ r, s ) = ψ ( ε − r, s )and we get the estimate ε − | ∂ r ψ | + | ∂ s ψ | ≤ Cε − σ ( k h k ∗∗ + R − σ + ε σ + k ψ k L ∞ ( B R ( ˜ d )) ) , (5.16)for points in D R with ρ > ε . Combining (5.14), (5.15) and (5.16) we get k ψ k , ∗ , ≤ C ( k h k ∗∗ , + R − σ + ε σ + k ψ k L ∞ ( B R ( ˜ d )) ) . (5.17)We next estimate ψ . We also use the symmetries satisfied by ψ , that is, ψ ( − x , x ) = − ψ ( x , x ) , ψ ( x , − x ) = − ψ ( x , x ) , to look at the equation for ψ in the quadrant { x > x > } . Let us rewrite equation (5.12) as∆ ψ + ε ∂ ss ψ = p where p = h − (cid:18) ∇ w w + ∇ w w (cid:19) ∇ ψ + ∇ ( θ + θ ) ∇ ψ − ε (cid:20)(cid:18) ∂ s w w + ∂ s w w (cid:19) ∂ s ψ − ∂ s ( θ + θ ) ∂ s ψ (cid:21) − ε ∂ s ψ . We have in D R : (cid:12)(cid:12)(cid:12)(cid:12)(cid:18) ∇ w w + ∇ w w (cid:19) ∇ ψ (cid:12)(cid:12)(cid:12)(cid:12) ≤ CRρ k ψ k , ∗ , |∇ ( θ + θ ) ∇ ψ | ≤ CR σ − (cid:16) ρ + ε (cid:17) k∇ ψ k , ∗ , ε (cid:12)(cid:12)(cid:12)(cid:12)(cid:18) ∂ s w w + ∂ s w w (cid:19) ∂ s ψ (cid:12)(cid:12)(cid:12)(cid:12) ≤ CR (cid:16) ρ + ε (cid:17) k ψ k , ∗ , ε | ∂ s ( θ + θ ) ∂ s ψ | ≤ C (cid:16) ε − σ + R σ − (cid:17)(cid:16) ε + 1 ρ (cid:17) k ψ k , ∗ , ε | ∂ s ψ | ≤ C (cid:16) ε − σ + R σ − (cid:17) (cid:18) ε + 1 ρ (cid:19) k ψ k , ∗ , . Using that k ψ k ∗ , = 1 we get that | p | ≤ C (cid:16) k h k ∗∗ , + R σ − + ε − σ (cid:17) (cid:18) ρ + ε (cid:19) We use the comparison principle with the barrier B = M θ ( π − θ )with M = C ( k h k ∗∗ , + R σ − + ε − σ + k ψ k L ∞ ( B R ( ˜ d )) ) and C a large fixed constant. We note that ∂ ss B = − ˜ d sin θ ˜ d cos θ ρ M ( π − θ )+ (cid:16) d cos θ ρ (cid:17)h − ˜ d sin θ ρ M ( π − θ ) − (cid:16) d cos θ ρ (cid:17) M i From this we get that ∆ B + ε ∂ ss B ≤ − ˜ cM (cid:16) ρ + ε (cid:17) for some fixed ˜ c > NTERACTING HELICAL GINZBURG-LANDAU FILAMENTS 25 Thanks to a comparison principle in D R (a slight variant of Lemma 7.5) we get that | ψ | ≤ C ( k h k ∗∗ , + R σ − + ε − σ + k ψ k L ∞ ( B R ( ˜ d )) ) in D R . (5.18)Elliptic estimates and a standard scaling give ρ |∇ ψ | ≤ C ( k h k ∗∗ , + R σ − + ε − σ + k ψ k L ∞ ( B R ( ˜ d )) ) (5.19)for points in D R with 2 < ρ < ε . To estimate the gradient for points in D R with ρ > ε we use thescaling ˜ ψ ( r, s ) = ψ ( ε − ˜ r, s )and see that ε − | ∂ r ψ | + | ∂ s ψ | ≤ C ( k h k ∗∗ , + R σ − + ε − σ + k ψ k L ∞ ( B R ( ˜ d )) ) . (5.20)in this region.Combining (5.18), (5.19) and (5.20) we get that k ψ k , ∗ , ≤ C ( k h k ∗∗ , + R σ − + ε − σ + k ψ k L ∞ ( B R ( ˜ d )) ) . Then using (5.17) we conclude that k ψ k ∗ , ≤ C ( k h k ∗∗ , + R − σ + ε σ + R σ − + ε − σ + k ψ k L ∞ ( B R ( ˜ d )) + k ψ k L ∞ ( B R ( ˜ d )) ) . Using then (5.11) and k h k ∗∗ = o (1), from the previous inequality we get that k ψ k ∗ , < , if from thestart R is fixed large and we take ε > (cid:3) Proof of Proposition 5.1. We first solve the problem in bounded domains. We consider the equation L ε ( ψ ) = h + c X j =1 χ j iW ( z − ˜ d j ) ( − j W x ( z − ˜ d j ) in B M (0) ψ = 0 on ∂B M (0)Re Z B (0 , χφ j W x = 0 , with φ j ( z ) = iW ( z ) ψ ( z + ˜ d j ) , j = 1 , ,ψ satisfies the symmetry (3.18) . (5.21)with M > 10 ˜ d . We set H := n φ = iV d ψ ∈ H ( B M (0) , C ); Re Z B (0 , χ ¯ φ j W x = 0 , j = 1 , , ψ satisfies (3.18) o . We equip H with the inner product[ φ, ϕ ] := Re Z B M (0) (cid:0) ∇ φ ∇ ϕ + ε ∂ s φ∂ s ϕ (cid:1) . With this inner product H is a Hilbert space. Indeed it is a closed subspace of H ( B M (0) , C ) and [ · , · ] isan inner product on H ( B M (0) , C ) thanks to the Poincar´e inequality. In terms of φ the first equation of (5.21) can be rewritten as∆ φ + (1 − | V d | ) φ − φV d ) V d + ε ( ∂ ss φ − i∂ s φ − φ ) + ( η − EV d φ = iV d h + iV d c X j =1 ˜ η ( − j χ j ( z ) W x ( z − ˜ d j ) iW ( z − ˜ d j ) . We can express this equation in its variational form. Namely, for all ϕ ∈ H− Re Z B M (0) (cid:0) ∇ φ ∇ ϕ + ε ∂ s φ∂ s ϕ (cid:1) + ε Re Z B M (0) (cid:0) iφ∂ s ϕ − φϕ (cid:1) − Z B M (0) Re( φV d ) V d ϕ + Re Z B M (0) [( η − EV d + (1 − | V d | )] φϕ = Re Z B M (0) iV d h − c X j =1 χ j ( − j W x ( z − ˜ d j ) iW ( z − ˜ d j ) ϕ. We now denote by h k ( x ) φ, ·i the linear form on H defined by h k ( x ) φ, ϕ i := ε Re Z B M (0) (cid:0) iφ∂ s ϕ − φϕ (cid:1) − Z B M (0) Re( φV d ) V d ϕ + Re Z B M (0) [( η − EV d + (1 − | V d | )] φϕ. In the same way we denote by h s, ·i the linear form defined by h s, ϕ i := Re Z B M (0) iV d h − c X j =1 χ j ( − j W x ( z − ˜ d j ) iW ( z − ˜ d j ) ϕ. Thus, the equation can be rewritten as[ φ, ϕ ] − h k ( x ) φ, ϕ i = h s, ϕ i , ∀ ϕ ∈ H . By using the Riesz representation theorem we can find a bounded linear operator K on H and S , anelement of H depending linearly on s , such that the equation has the operational form φ − K ( φ ) = S. (5.22)Besides, thanks to the compact Sobolev injections H ( B M (0) , C ) ֒ → L ( B M (0) , C ) we know that K iscompact. We can then apply Fredholm’s alternative to deduce the existence of φ such that (5.22) holds ifthe homogeneous equation only has the trivial solution. To prove this last point we establish an a prioriestimate on c . In order to do that we use the following equivalent form of the equation in the region B ( ˜ d, ˜ d ), with the translated variable it becomes: L εj ( φ j ) = h j + cχW x in B (0 , ˜ d ) . where L εj is defined in (3.14), φ j (˜ z ) = iW (˜ z ) ψ ( z − ˜ d j ) and h j (˜ z ) = iW (˜ z ) ψ ( z − ˜ d j ) for | ˜ z | < ˜ d .We can test this equation against W x to find c = − c ∗ " Re Z B (0 , ˜ d ) h j W x − Re Z B (0 , ˜ d ) L εj ( φ j ) W x , NTERACTING HELICAL GINZBURG-LANDAU FILAMENTS 27 with c ∗ := Re R B (0 , ˜ d ) χ | W x | = Re R B (0 ,R ) χ | W x | ≃ C for some C > L εj defined in(3.14). Integrating by parts we obtainRe Z B (0 , ˜ d ) L εj ( φ j ) W x = Re Z B (0 , ˜ d ) φ j ( L εj − L )( W x ) + Re nZ ∂B (0 , ˜ d ) ∂φ j ∂ν W x − φ j ∂∂ν W x o . In the previous equality we used that L ( W x ) = 0. However, using the expansion of L εj − L o in (3.16)and the estimates (3.17), we can see that (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) Re Z B (0 , ˜ d ) φ j ( L εj − L )( W x ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = O ε ( ε p | log ε | ) k ψ k ∗ . (5.23)By using the decay of φ j , ∇ φ j and W x , ∇ W x we can also check that (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) Re nZ ∂B (0 , ˜ d ) ∂φ j ∂ν W x − φ j ∂∂ν W x o(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = O ε ( ε p | log ε | ) k ψ k ∗ . Therefore we arrive at c = − c ∗ Re Z B (0 , ˜ d )) h j W x + O ε ( ε p | log ε | ) k ψ k ∗ c ∗ . To conclude the proof we note that we can apply Lemma 5.1 to conclude that a solution of the homoge-neous equation satisfies k ψ k ∗ ≤ C k c X j =1 χ j ( z )( − j W x ( z − ˜ d j ) iW ( z − ˜ d j ) k ∗∗ ≤ Cε p | log ε |k ψ k ∗ , and thus ψ = 0. Then for any M > 10 ˜ d we obtain the existence of a solution of (5.21) satisfying k ψ M k ∗ ≤ C k h k ∗∗ , with C independent of M . Note that in the previous argument the norms k · k ∗ , k · k ∗∗ are slightlyadapted to deal with the fact that we work on bounded domains. We can extract a subsequence suchthat ψ M ⇀ ψ in H ( R ) with ψ solving (5.1). From Lemma 5.1 we deduce k ψ k ∗ ≤ C k h k ∗∗ . (cid:3) Let α ∈ (0 , , σ ∈ (0 , . Then there exists a constant C > such that for all ε sufficientlysmall and any solution ψ of (5.8) with k ψ k ∗ < ∞ one has | ψ | ♯ ≤ C ( | h | ♯♯ + ε | log ε | k h k ∗∗ ) . (5.24) Proof. We work with the weaker seminorms | ψ | ♯, = X j =1 | log ε | − k V d ψ k L ∞ ( ρ j < + | Re( ψ ) | ♯, + | Im( ψ ) | ♯, , where | | ♯, , | | ♯, are defined in (5.4), (5.5) and | h | ♯♯, = X j =1 k V d h k L ∞ ( ρ j < + sup <ρ C > ε sufficiently small and any solution of (5.8)one has | ψ | ♯, ≤ C ( | h | ♯♯, + ε | log ε | k h k ∗∗ ) . (5.25) We argue by contradiction and assume that there exist ε n → ψ ( n ) , h ( n ) solutions of (5.8) suchthat | ψ ( n ) | ♯, = 1 , ( | h ( n ) | ♯♯, + ε n | log ε n | k h ( n ) k ∗∗ ) → , (5.26)as n → ∞ .We first work near the vortices and notice that, by symmetry, it is enough to consider the vortex at + ˜ d .We work with the function φ ( n ) j ( z ) = i | log ε | − W ( z ) ψ ( n ) ( z + ˜ d j ). Since | ψ ( n ) | ♯, = 1 from Arzela-Ascoli’sTheorem we can extract a subsequence such that ˜ φ ( n ) j → φ in C ( R ). Passing to the limit in (5.8),we see that L ( φ ) = 0 in R , with L defined in (3.5). The function φ inherits the symmetry φ (¯ z ) = φ ( z ) and satisfies φ ∈ L ∞ loc .Moreover, writing φ = iW ψ , ψ = ψ + iψ , we have | ψ ( z ) | ≤ | z | , | ψ ( z ) | ≤ , | z | > . Thanks to the above estimate and Lemma 7.3 we deduce φ = c W x , for some c ∈ R . On the other hand, we can pass to the limit in the orthogonality conditionRe Z B (0 , χ ¯ φ ( n ) j W x = 0 , and obtain that necessarily c = 0. Hence φ ( n ) j → C ( R ). We can also apply the same argumentnear − ˜ d and get ψ ( n ) | log ε n | → { ρ ≥ , ρ ≥ } as ε n → D R = { R < ρ < R ε } ∩ { x > } , where R > R ε is given by (4.19).We use barriers to estimate ψ ( n )2 ( z ) in ˜ D R . By the symmetries of ψ ( n )2 we get the estimates for all2 < ρ < R ε . Let us write equation (5.13) as∆ ψ + (cid:18) ∇ w w + ∇ w w (cid:19) ∇ ψ − | V d | ψ + ε ∂ ss ψ + 2 ε (cid:18) ∂ s w w + ∂ s w w (cid:19) ∂ s ψ = ˜ p where ˜ p = h − ∇ ( θ + θ ) ∇ ψ − ε ∂ s ( θ + θ ) ∂ s ψ + 4 ε ∂ s ψ . We observe that in ˜ D R it holds (cid:12)(cid:12)(cid:12) ∇ ( θ + θ ) ∇ ψ ( n )1 (cid:12)(cid:12)(cid:12) ≤ Cρ log (cid:16) R ε ρ (cid:17) | ψ ( n )1 | ♯, ε n | ∂ s ( θ + θ ) ∂ s ψ | ≤ Cρ | log ε n | log (cid:16) R ε ρ (cid:17) | ψ ( n )1 | ♯, ε n | ∂ s ψ ( n )1 | ≤ Cρ | log ε n | log (cid:16) R ε ρ (cid:17) | ψ ( n )1 | ♯, . NTERACTING HELICAL GINZBURG-LANDAU FILAMENTS 29 Using the a priori estimate of Lemma 5.1 we find that k ψ ( n ) k ∗ ≤ C k h ( n ) k ∗∗ = o (1) ε − n | log ε n | − . (5.27)Thus writing ψ ( n ) = ψ ( n )1 + iψ ( n )2 we have | ψ ( n )1 ( z ) | ≤ o (1) ε − n | log ε n | − , | ψ ( n )2 ( z ) | + |∇ ψ ( n )2 ( z ) | ≤ o (1) ε − n | log ε n | − (cid:16) ρ − σ + 1 ρ − σ (cid:17) for 2 < | z | < ε n with o (1) → n → ∞ . We note that for | z − ˜ d j | = R ε , | ψ ( n )2 ( z ) | ≤ k ψ ( n ) k ∗ R − σε = o (1) R ε R − σε = o (1) , as n → ∞ by (5.27). We use as a barrier the function˜ B = Cρ σ ( | h ( n ) | ♯♯, + k ψ ( n )2 k L ∞ ( ρ = R εn ) ) + Cρ log (cid:16) R ε ρ (cid:17) | ψ ( n )1 | ♯, + k ψ ( n )2 k L ∞ ( ρ = R ) | log ε n | ! where C > B ≤ b n ρ − σ + 1 ρ log (cid:16) R ε ρ (cid:17) (cid:16) C | ψ ( n )1 | ♯, + b n (cid:17) (5.28)in ˜ D R where b n → n → ∞ . By the maximum principle and elliptic estimates we get | ψ ( n )2 | + |∇ ψ ( n )2 | ≤ ˜ B (5.29)in ˜ D R .Next we use barriers to estimate ψ ( n )1 in ˜ D R . By the symmetries of ψ ( n )1 we get the estimates for all2 < ρ < R ε . Let us write Equation (5.12) as∆ ψ + (cid:18) ∇ w w + ∇ w w (cid:19) ∇ ψ + ε ∂ ss ψ + 2 ε (cid:18) ∂ s w w + ∂ s w w (cid:19) ∂ s ψ = p where p = h + ∇ ( θ + θ ) ∇ ψ + 2 ε ∂ s ( θ + θ ) ∂ s ψ − ε ∂ s ψ We find that in ˜ D R the following estimates hold: (cid:12)(cid:12)(cid:12) ∇ ( θ + θ ) ∇ ψ ( n )2 (cid:12)(cid:12)(cid:12) ≤ Cρ ˜ B ε n (cid:12)(cid:12)(cid:12) ∂ s ( θ + θ ) ∂ s ψ ( n )2 (cid:12)(cid:12)(cid:12) ≤ Cρ log ε n | ˜ B ε n | ∂ s ψ ( n )2 | ≤ Cρ | log ε n | ˜ B Hence using (5.28) and (5.29) we get | p | ≤ b n ρ + 1 ρ log (cid:16) R ε ρ (cid:17) (cid:16) C | ψ ( n )1 | ♯, + b n (cid:17) for a new sequence b n → Using Lemma 7.8 for part of the right hand side and the supersolution log (cid:16) R ε ρ (cid:17) (cid:16) C | ψ ( n )1 | ♯, + b n (cid:17) weconclude that | ψ ( n )1 ( z ) | ≤ Cb n ρ log (cid:16) R ε ρ (cid:17) + log (cid:16) R ε ρ (cid:17) (cid:16) C | ψ ( n )1 | ♯, + b n (cid:17) ≤ Cρ log (cid:16) R ε ρ (cid:17)(cid:16) b n + | ψ ( n )1 | ♯, R (cid:17) . This and standard elliptic estimates yield | ψ ( n )1 | ,♯ ≤ C (cid:16) b n + | ψ ( n )1 | ♯, R (cid:17) . Choosing R > | ψ ( n )1 | ,♯ → n → ∞ . Using this and (5.28), (5.29) we obtain | ψ ( n )2 | ,♯ → n → ∞ . This contradicts the assumption (5.26) and we obtain (5.25). With this inequality and standard Schauderestimates with deduce (5.24). (cid:3) As an intermediate step to obtain Proposition 5.2 we consider the symmetry properties of the solutionconstructed in Proposition 5.1, when the right hand side has symmetries. More precisely, let us considerthe local symmetry condition h ( R j z ) = − h ( z ) , | z − ˜ d j | < R ε , j = 1 , . (5.30) Lemma 5.3. Suppose that h satisfies the symmetries (3.18) and (5.30) . We assume that k h k ∗∗ < ∞ . Then there exist ψ s , ψ ∗ such that the solution ψ to (5.1) with k ψ k ∗ < ∞ can be written as ψ = ψ s + ψ ∗ with the estimates k ψ s k ∗ + k ψ ∗ k ∗ ≤ C k h k ∗∗ | ψ ∗ | ♯ ≤ Cε | log ε | k h k ∗∗ . Moreover ( ψ s , ψ ∗ ) define linear operators of h , ψ s has its support in B R ε ( ˜ d ) ∪ B R ε ( ˜ d ) and satisfies ψ s ( R j z ) = − ψ s ( z ) , | z − ˜ d j | < R ε . (5.31) Proof. To construct the function ψ s we split the operator L ε (c.f. (3.12)) into a part L εs preserving thesymmetry (5.31) and a remainder L εr . This splitting depends on which vortex ˜ d j we are considering forthe symmetry (5.31) and thus we write L εs,j , L εr,j , j = 1 , 2. It is sufficient to consider the vortex at ˜ d . NTERACTING HELICAL GINZBURG-LANDAU FILAMENTS 31 We set L εs, ( ψ ) = ∆ ψ + 2 ∇ W a ∇ ψW a − i | W a | Im( ψ )+ ε " ˜ d ∂ ρ ρ ψ sin( θ ) + ˜ d ρ ∂ ρ θ ψ sin( θ ) cos( θ ) + ∂ θ θ ψ (cid:16) d ρ cos ( θ ) (cid:17) + ∂ ρ ψ ˜ d ρ cos ( θ ) − ∂ θ ψ ˜ d ρ sin( θ ) cos( θ ) L εr, ( ψ ) = 2 ∇ W b ∇ ψW b − i ( | V d | − | W a | ) Im( ψ )+ ε " d∂ ρ θ ψ sin( θ ) + 2 ∂ θ θ ψ ˜ dρ cos( θ ) + ∂ ρ ψ ˜ d cos( θ ) − ∂ θ ψ ˜ dρ sin( θ ) + ε (cid:16) ∂ s V d V d − i (cid:17)h ∂ ρ ψ ˜ d sin( θ ) + (cid:16) dρ cos( θ ) ∂ θ ψ (cid:17)i . We use the same cut-off functions defined in (4.18) and solve L εs ( ψ , ) = hη , R ε in R Re Z B (0 , χφ , W x = 0 , with φ , ( z ) = iW ( z ) ψ , ( z + ˜ d ) ψ , satisfies ψ , (¯ z ) = − ψ , ( z ) . This is obtained as variant of Proposition 5.1 with the same proof. Note that there is no need to projectthe right hand side, since it is automatically orthogonal to the kernel by symmetry, and note also thatthe orthogonality condition for the solution holds also by symmetry. Recall that h satisfies (5.30) and weget a solution ψ , satisfying (5.31) with the estimate k ψ , k ∗ ≤ C k h k ∗∗ . In a similar way we construct ψ , centered at the vortex ˜ d and define ψ s = η , R ε ψ , , + η , R ε ψ , . (5.32)Note that we have the estimate k ψ s k ∗ ≤ C k h k ∗∗ . Let ˜ h := h − L εs, ( η , R ε ψ , ) − L εr, ( η , R ε ψ , ) − L εs, ( η , R ε ψ , ) − L εr, ( η , R ε ψ , ) . Some lengthy but direct calculations show that k ˜ h k ∗∗ ≤ C ( k h k ∗∗ + k h ∗ k ∗∗ ) | ˜ h | ♯♯ ≤ Cε | log ε | ( k h k ∗∗ + k h ∗ k ∗∗ ) . Then we solve, using Proposition 5.1, L ε ( ˜ ψ ) = ˜ h + ˜ c X j =1 χ j iW ( z − ˜ d j ) ( − j W x ( z − ˜ d j ) in R Re Z B (0 , χ ˜ φ j W x = 0 , with ˜ φ j ( z ) = iW ( z ) ˜ ψ ( z + ˜ d j )˜ ψ satisfies the symmetry (3.18) , and obtain, using also Lemma 5.2, k ˜ ψ k ∗ ≤ C k ˜ h k ∗∗ | ˜ ψ | ♯ ≤ C ( | ˜ h | ♯♯ + ε | log ε | k ˜ h k ∗∗ ) . Finally we set ψ ∗ = ˜ ψ. (5.33)The functions ψ s , ψ ∗ defined in (5.32), (5.33) satisfy the stated properties. (cid:3) Proof of Proposition 5.2. Let us define ˜ h = h − h o so that h = ˜ h + h oα + h oβ . Let ˜ ψ , ˜ ψ α , ˜ ψ β be the solution with finite k k ∗ -norm of (5.1) with right handsides ˜ h , h oα , h oβ given by Proposition 5.1. Then ψ = ˜ ψ + ˜ ψ α + ˜ ψ β and we have the estimates k ˜ ψ k ∗ . k ˜ h k ∗ k ˜ ψ j k ∗ . k h oj k ∗ , j = α, β. We have ψ o = ˜ ψ o + ˜ ψ oα + ˜ ψ oβ . We define ψ oα = ˜ ψ o + ˜ ψ oα , ψ oβ = ˜ ψ oβ . Note that by Lemma 5.2 | ˜ ψ oα | ♯ . | ˜ ψ α | ♯ . | h oα | ♯♯ + ε | log ε | k h oα k ∗∗ . According to Lemma 5.3 we can write ˜ ψ = ψ s + ψ ∗ with ψ s , ψ ∗ satisfying the properties stated inthat lemma, from which we get | ˜ ψ o | ♯ = | ( ψ ∗ ) o | ♯ . | ψ ∗ | ♯ . ε | log ε | k ˜ h k ∗∗ . Therefore | ψ oα | ♯ . | h oα | ♯♯ + ε | log ε | ( k h oα k ∗∗ + k ˜ h k ∗∗ )and this proves (5.6).On the other hand k ψ oα k ∗ ≤ k ˜ ψ o k ∗ + k ˜ ψ oα k ∗ . k ˜ ψ k ∗ + k ˜ ψ α k ∗ . k ˜ h k ∗ + k h oα k ∗ k ψ oβ k ∗ = k ˜ ψ oβ k ∗ . k ˜ ψ β k ∗ . k h oβ k ∗ and from here (5.7) follows. (cid:3) NTERACTING HELICAL GINZBURG-LANDAU FILAMENTS 33 We consider now the nonlinear projected problem L ε ( ψ ) = R + N ( ψ ) + c X j =1 χ j ( z ) iW ( z − ˜ d j ) ( − j W x ( z − ˜ d j ) in R , Re Z R χφ j W x = 0 , with φ j ( z ) = iW ( z ) ψ ( z + ˜ d j ) , j = 1 , ,ψ satisfies (3.18) . (6.1)Using the operator T ε introduced in Proposition 5.1 we can rewrite this equation in the form of a fixedpoint problem as ψ = T ε ( R + N ( ψ )) =: G ε ( ψ ) . Proposition 6.1. There exists a constant C > depending only on < α, σ < , such that for all ε sufficiently small there exists a unique solution ψ ε of (6.1) , that satisfies k ψ ε k ∗ ≤ C | log ε | . Furthermore ψ ε is a continuous function of the parameter ˆ d := p | log ε | d | ψ oε | ♯ ≤ Cε p | log ε | , (6.2) where ψ oε is defined according to (5.3) .Proof. We let F := n ψ : ψ satisfies (3.18) , Re Z R χ j φ j W x = 0 , j = 1 , , k ψ k ∗ ≤ C | log ε | , | ψ o | ♯ ≤ Cε p | log ε | o . Endowed with the norm k · k ∗ , F is a Banach space as a closed subset of the Banach space { ψ : k ψ k ∗ < + ∞} . We will show that, for ε small enough, G ε maps F into itself. Indeed, we need to check that if k ψ k ∗ ≤ C | log ε | then k T ε ( E + N ( ψ )) k ∗ ≤ C/ | log ε | .Note first that, from Proposition 4.1, k R k ∗∗ ≤ C | log ε | . Let us now estimate the size of the nonlinear term. For ρ > ρ > i ( ∇ ψ ) + i | V d | ( e − ψ − ψ ) + iε ( ∂ s ψ ) . Let us work in the right half plane (so ρ ≤ ρ ). We start with ( ∇ ψ ) . For 3 < ρ < ε we have | ( ∇ ψ ) | ≤ |∇ ψ | ≤ k ψ k ∗ ρ . For r > ε we use ( ∇ ψ ) = ( ∂ r ψ ) + 1 r ( ∂ s ψ ) 24 JUAN D´AVILA, MANUEL DEL PINO, MARIA MEDINA AND R´EMY RODIAC and estimate | ( ∂ r ψ ) | = ( ∂ r ψ ) + ( ∂ r ψ ) ≤ ε k ψ k ∗ and 1 r | ( ∂ s ψ ) | ≤ r (cid:16) ( ∂ s ψ ) + ( ∂ s ψ ) (cid:17) ≤ r k ψ k ∗ . It follows that k i ( ∇ ψ ) k ∗∗ ≤ C k ψ k ∗ . Next we consider i | V d | ( e − ψ − ψ ). We note that the real part of this function is zero. Againwe work in the right half plane. We have, for ρ > | | V d | ( e − ψ − ψ ) | ≤ C | ψ | ≤ C ( ρ − σ + ε − σ ) k ψ k , ∗ , and hence k i | V d | ( e − ψ − ψ ) k ∗∗ ≤ C k ψ k ∗ , Finally we consider iε ( ∂ s ψ ) . We have for 3 < ρ < ε | ε ( ∂ s ψ ) | ≤ ε ˜ d | ∂ ρ ψ | + (cid:16) ε + ε ˜ dρ (cid:17) | ∂ θ ψ | ≤ C ( ε + ρ − ) k ψ k ∗ . For r > ε | ε ( ∂ s ψ ) | ≤ ε k ψ k ∗ . It follows that k iε ( ∂ s ψ ) k ∗∗ ≤ k ψ k ∗ . In { ρ ≤ } ∪ { ρ ≤ } it can be checked that | iV d N ( ψ ) | ≤ C ( | D γ | + | Dγ | + | γ | + | γ + φ || φ | + | γ + φ | (1 + | γ + φ | + | γ | ) + | E d || φ | + |∇ φ | )with γ = (1 − η ) V d (cid:0) e iψ − − iψ (cid:1) . Thus we obtain that for any j = 1 , k iV d N ( ψ ) k C α ( { ρ j < } ≤ C k ψ k ∗ + | E || φ | ≤ C | log ε | . Thus for an appropriate constant C we have that G ε : ψ T ε ( E + N ( ψ )) maps the ball { ψ ; k ψ k ∗ ≤ C | log ε | } into itself.Let us now see the precise estimates on the “odd parts” and “even parts”. From Proposition 4.2 weknow that R o defined as in (4.20), can be decomposed into R o = R oα + R oβ with | R oα | ♯♯ ≤ Cε p | log ε | k R oβ k ∗∗ ≤ Cε p | log ε | . It remains to prove that |N ( ψ ) o | ♯♯ ≤ C (cid:16) ( | ψ o | ♯ + ε | log ε | / ) k ψ e k ∗ + | ψ o | ♯ (cid:17) . (6.3) NTERACTING HELICAL GINZBURG-LANDAU FILAMENTS 35 In order to do that we recall that in the decomposition of a function f in odd and even modes we havethat, near + ˜ d the function f e is exactly π -periodic in θ whereas f o is exactly 2 π -periodic in θ . Ananalogous statement is true near − ˜ d . Now we can express the product of two functions as f g = ( f e + f o )( g e + g o ) = f e g e + f e g o + g e f o + g o f o . We see that f e g e is exactly π -periodic and hence ( f g ) o = [ f e g o + g e f o + f o g o ] o . Thus we arrive at | ( f g ) o | ≤ ( | f o || g e | + | f e || g o | + | f o || g o | ) . (6.4)To estimate N ( ψ ) we use a change of variables ( r, s ) → ( ρ j , θ j ) = ( ρ, θ ) and we observe that( ∇ ψ ) = ( ∂ r ψ ) + 1 r ( ∂ s ψ ) = ( ∂ ρ ψ ) + 1 ρ ( ∂ θ ψ ) , and ε ( ∂ s ψ ) = ε ( ∂ θ ψ ) + ε ˜ d (cid:18) sin θ∂ ρ ψ∂ θ ψ + cos θρ ( ∂ θ ψ ) (cid:19) + ε ˜ d (cid:18) sin θ ( ∂ ρ ψ ) + 4 cos θ sin θρ ∂ ρ ψ∂ θ ψ + cos θρ ( ∂ θ ψ ) (cid:19) . Thus component-wise we obtain (cid:16) ˜ N ( ψ ) (cid:17) = 2( ∂ ρ ψ )( ∂ ρ ψ ) + 2( ∂ θ ψ )( ∂ θ ψ ) (cid:18) ε + 1 ρ (cid:19) + ε ˜ d (cid:18) sin θ [ ∂ ρ ψ ∂ θ ψ + ∂ θ ψ ∂ ρ ψ ] + 2 cos θρ ∂ θ ψ ∂ θ ψ (cid:19) + ε ˜ d (cid:18) θ∂ ρ ψ ∂ ρ ψ + 4 sin θ cos θρ [ ∂ ρ ψ ∂ θ ψ + ∂ θ ψ ∂ ρ ψ ]+ 2 cos θρ ∂ θ ψ ∂ θ ψ (cid:19) , (cid:16) ˜ N ( ψ ) (cid:17) = − ( ∂ ρ ψ ) + ( ∂ ρ ψ ) − (cid:18) ε + 1 ρ (cid:19) (( ∂ θ ψ ) − ( ∂ θ ψ ) )+ ε ˜ d (cid:18) sin θ ( ∂ ρ ψ ∂ θ ψ + ∂ ρ ψ ∂ θ ψ ) + cos θρ [( ∂ θ ψ ) − ( ∂ θ ψ ) ] (cid:19) + ε ˜ d (cid:18) sin θ (( ∂ ρ ψ ) − ( ∂ ρ ψ ) ) + 4 cos θ sin θρ ( ∂ ρ ψ ∂ θ ψ + ∂ ρ ψ ∂ θ ψ )+ cos θρ [( ∂ θ ψ ) − ( ∂ θ ψ ) ] (cid:19) + | V d | (1 − e ψ − ψ ) . We define A ( ψ ) := 2( ∂ ρ ψ )( ∂ ρ ψ ) + 2( ∂ θ ψ )( ∂ θ ψ ) (cid:18) ε + 1 ρ (cid:19) , B ( ψ ) := ε ˜ d (cid:18) sin θ [ ∂ ρ ψ ∂ θ ψ + ∂ θ ψ ∂ ρ ψ ] + 2 cos θρ ∂ θ ψ ∂ θ ψ (cid:19) , C ( ψ ) := ε ˜ d (cid:18) θ∂ ρ ψ ∂ ρ ψ + 4 sin θ cos θρ [ ∂ ρ ψ ∂ θ ψ + ∂ θ ψ ∂ ρ ψ ] + 2 cos θρ ∂ θ ψ ∂ θ ψ (cid:19) . We have ( N ( ψ )) = A ( ψ ) + B ( ψ ) + C ( ψ ). Besides we can see that |B ( ψ ) | ρ − + ρ − ≤ Cε p | log ε | (cid:18) |∇ ψ | ρ − + ρ − × |∇ ψ | ρ − + ρ − (cid:19) ≤ Cε p | log ε | k ψ k ∗ for 3 < ρ < R ε . Now by using the argument that a product of two π -periodic functions is π -periodicand the products of one π -periodic function and one 2 π -periodic function is 2 π -periodic and (6.4) we findthat | [ A ( ψ ) + C ( ψ )] o | ρ − + ρ − ≤ C (cid:0) k ψ k ∗ | ψ o | ♯ + | ψ o | ♯ (cid:1) . Thus we obtained that | ( N ( ψ )) o | ρ − + ρ − ≤ C k ψ k ∗ | ψ o | ♯ + | ψ o | ♯ + Cε p | log ε | k ψ k ∗ ! for 3 < ρ < R ε . We also have that1 − e ψ − ψ = (cid:16) − e ψ e − ψ e (cid:17) + (cid:16) − e ψ o − ψ o (cid:17) e ψ e + 2 ψ o (cid:16) e ψ e − (cid:17) . We notice that 1 − e ψ e − ψ e is a π -periodic function. Thus we find that (cid:12)(cid:12) (1 − e ψ − ψ ) o (cid:12)(cid:12) ≤ C (cid:0) | ψ o || ψ e | + | ψ o | (cid:1) . By using again that a product of two π -periodic functions is π -periodic and the products of one π -periodicfunction and one 2 π -periodic function is 2 π -periodic and (6.4) we can obtain that | ( N ( ψ )) o | ≤ C k ψ k ∗ | ψ o | ♯ + | ψ o | ♯ + Cε p | log ε | k ψ k ∗ ! We proceed in the same way to estimate the other terms in N ( ψ ) when ρ < ρ < G ε is a contraction for ε small enough. Indeed, if k ψ j k ∗ ≤ C | log ε | for j = 1 , kN ( ψ ) − N ( ψ ) k ∗∗ ≤ C | log ε | k ψ − ψ k ∗ . This is mainly due to the fact that N ( ψ ) is quadratic and cubic in ψ , and in the first and second derivativesof ψ . Then we can use a − b = ( a − b )( a + b ) and a − b = ( a − b )( a + ab + b ). We finally apply theBanach fixed point theorem and we find the desired solution. (cid:3) The solution ψ ε of (6.1) previously found depends continuously on ˆ d := p | log ε | d . We want to find ˆ d such that the Lyapunov-Schmidt coefficient in (6.1) satisfies c = c ( ˆ d ) = 0. We let ϕ ε := ηiV d ψ ε + (1 − η ) V d e iψ ε and φ ε := iV d ψ ε , where η was defined in (3.3). By symmetry we work only in R + × R . From the previous section we havefound ψ ε such that iW ( z ) [ L ε ( ψ ε ) + R + N ( ψ ε )] ( z + ˜ d ) = cχW x . For R ε defined in (4.19) we set c ∗ := Re Z B (0 ,R ε ) χ | W x | = Re Z B (0 , χ | W x | , NTERACTING HELICAL GINZBURG-LANDAU FILAMENTS 37 and we remark that this quantity is of order 1. We find that cc ∗ = Re Z B (0 ,R ε ) iW ( z ) R ( z + ˜ d ) W x ( z ) + Re Z B (0 ,R ε ) iW ( z ) L ε ( ψ ε )( z + ˜ d ) W x ( z )+ Re Z B (0 ,R ε ) iW N ( ψ ε )( z + ˜ d ) W x . We recall that iW ( z ) L ε ( ψ )( z + ˜ d ) = L εj ( φ j ) for j = 1 and L j defined in (3.14). Integrating by parts wefindRe Z B (0 ,R ε ) L εj ( φ j ) W x = Re Z B (0 ,R ε ) φ j ( L εj − L )( W x ) + Re nZ ∂B (0 ,R ε ) (cid:18) ∂φ j ∂ν W x − φ j ∂W x ∂ν (cid:19)o . Proceeding like in (5.23) we conclude (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) Re Z B (0 ,R ε ) L εj ( φ j ) W x (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ Cε p | log ε |k ψ k ∗ ≤ Cε p | log ε | . Now we estimate the inner product of W x and iW ( z ) N ( ψ )( z + ˜ d ). We use the orthogonality of theFourier modes to writeRe Z B (0 ,R ε ) iW ( z ) N ( ψ )( z + ˆ d ) W x = Re Z B (0 ,R ε ) iW W x ( N ( ψ )) o = Re Z B (0 ,R ε ) iw (cid:18) w ′ cos θ − iwρ sin θ (cid:19) [ N ( ψ ) o + i N o ( ψ )]= − Z B (0 ,R ε ) (cid:18) ww ′ cos θ N ( ψ ) o − w r N ( ψ ) o sin θ (cid:19) . We use that | ( N ( ψ )) o | ≤ | ( N ( ψ )) o | ♯♯ ≤ C k ψ e k ∗ | ψ o | ♯ + | ψ o | ♯ ≤ Cε | log ε | − / | ( N ( ψ )) o | ≤ C (cid:18) | ψ o | ♯ k ψ e k ∗ ρ + | ψ o | ♯ k ψ k ∗ ρ − σ + | ψ o | ♯ | ψ o | ♯ ρ − σ (cid:19) ≤ C ε | log ε | − / ρ − σ to obtain (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) Re Z B (0 ,R ε ) iW ( z ) N ( ψ )( z + ˜ d ) W x (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ C ε p | log ε | . Now we claimRe Z B (0 ,R ε ) iW ( z ) R ( z + ˜ d ) W x = − ε p | log ε | (cid:18) a ˆ d − a ˆ d (cid:19) + o ε ( ε p | log ε | )We set B := Re Z B (0 ,R ε ) iW ( z ) R ( z + ˜ d ) W x , B := Re Z B (0 ,R ε ) iW ( z ) R ( z + ˜ d ) W x , where we recall that S ( V d ) = iV d R , S ( V d ) = iV d R and S , S are given by (3.4).From Lemma 7.1 and Lemma 4.2 we find that B = ˆ dε p | log ε | Re Z { ρ 0, and a is independent of ˆ d .On the other hand, by (4.5) we have B = Re Z { ρ 1) = O ( ε | log ε | ) and w ′ ( ρ ) = 1 /ρ + O ρ (1 /ρ ). Wecan also see thatRe Z { ρ Finally we haveRe Z { ρ The third author is supported by the European Union’s Horizon 2020 research and innovation pro-gramme under the Marie Sklodowska-Curie grant agreement N.754446 and UGR Research and KnowledgeTransfer - Found Athenea3i. R.R is supported by the F.R.S.FNRS under the Mandat d’Impulsion sci-entifique F.4523.17,“Topological singularities of Sobolev maps”. Part of this work was realized duringa visit of the fourth author at the Universidad de Chile and Pontificia Universidad Cat´olica de Chilesupported by the project REC05 Wallonia-Brussels/Chile. Appendix As stated in the introduction, the buildingblock used to construct our solutions to equation (2.1) is the standard vortex of degree one, W , in R .It satisfies ∆ W + (1 − | W | ) W = 0 in R , and can be written as W ( x , x ) = w ( r ) e iθ where x = r cos θ, x = r sin θ. Here w is the unique solution of (1.6). In this section we collect useful properties of w and of the linearizedGinzburg-Landau operator around W . Lemma 7.1. The following properties hold, w (0) = 0 , w ′ (0) > , < w ( r ) < and w ′ ( r ) > for all r > , w ( r ) = 1 − r + O ( r ) for large r , w ( r ) = αr − αr + O ( r ) for r close to for some α > , if we define T ( r ) = w ′ ( r ) − wr then T (0) = 0 and T ( r ) < in (0 , + ∞ ) , w ′ ( r ) = r + O ( r ) , w ′′ ( r ) = O ( r ) . For the proof of this lemma we refer to [25, 10].An object of special importance to construct our solution is the linearized Ginzburg-Landau operatoraround W , defined by L ( φ ) := ∆ φ + (1 − | W | ) φ − W φ ) W. This operator does have a kernel, as the following result states. Lemma 7.2. Suppose that φ ∈ L ∞ ( R ) satisfies L ( φ ) = 0 in R and the symmetry φ (¯ z ) = ¯ φ ( z ) .Assume furthermore that when we write φ = iW ψ and ψ = ψ + iψ with ψ , ψ ∈ R we have | ψ | + (1 + | z | ) |∇ ψ | ≤ C, | ψ | + |∇ ψ | ≤ C | z | , | z | > . Then φ = c W x for some real constant c .Proof. The equation L ( φ ) = 0 in B (0 , c translates into∆ ψ + 2 ∇ WW ∇ ψ − i | W | Im ψ = 0 in B (0 , c . This reads 0 = ∆ ψ + 2 w ′ w ∂ r ψ + 2 r ∂ θ ψ in B (0 , c ψ + 2 w ′ w ∂ r ψ − r ∂ θ ψ − | W | ψ in B (0 , c . We thus have, by using the decay assumption on ψ , ψ , that (cid:12)(cid:12) ∆ ψ − | W | ψ (cid:12)(cid:12) ≤ C r in B (0 , c . Since | W | ≥ C > B (0 , c we can use a barrier argument and elliptic estimates to obtain that(1 + | z | ) ( | ψ | + |∇ ψ | ) ≤ C. (7.1) NTERACTING HELICAL GINZBURG-LANDAU FILAMENTS 41 We can then use the previous estimate to obtain that | ∆ ψ | ≤ C r in B (0 , c . We use that ψ ( z = x + ix ) = 0 for x = 0, a barrier argument in the half plane and elliptic estimatesto obtain | ψ | + (1 + | z | ) |∇ ψ | ≤ C (1 + | z | ) α . (7.2)for any α ∈ (0 , | φ ( z ) | + (1 + | z | ) |∇ φ | ) ≤ C (1 + | z | ) α , | z | > . (7.3)From the fact L ( φ ) = 0 in R we know thatRe Z B R (0) φ ∆ φ + Z B R (0) (1 − | W | ) | φ | − Z B R (0) | Re( W φ ) | = 0 , for any R > 0. Then integrating by parts we get Z B R (0) |∇ φ | − Re Z ∂B R (0) φ∂ ν φ − Z B R (0) (1 − | W | ) | φ | + 2 Z B R (0) | Re( W φ ) | = 0 . Using (7.3) we find | Re (cid:0) φ∂ ν φ (cid:1) | ≤ C/ (1 + | z | α +1 ). Thus (cid:12)(cid:12)(cid:12) Re Z ∂B R (0) φ∂ ν φ (cid:12)(cid:12)(cid:12) ≤ CR α . Making R → ∞ we conclude Z R |∇ φ | − Z R (1 − | W | ) | φ | + 2 Z R | Re( W φ ) | = 0 . Thanks to the decay estimates (7.3) we also have Z R (cid:2) |∇ φ | + (1 − | W | ) | φ | + | Re( ¯ W φ ) | (cid:3) < + ∞ . We can then apply Theorem 1 in [14] to obtain that there exists c , c ∈ R such that φ = c W x + c W x .Using the symmetry assumption φ (¯ z ) = ¯ φ ( z ) we can conclude that actually φ = c W x for some c ∈ R . (cid:3) Lemma 7.3. Suppose that φ ∈ L ∞ loc ( R ) satisfies L ( φ ) = 0 in R and the symmetry φ (¯ z ) = φ ( z ) .Assume furthermore that when we write φ = iW ψ and ψ = ψ + iψ with ψ , ψ ∈ R we have | ψ | + (1 + | z | ) |∇ ψ | ≤ C (1 + | z | ) α , | ψ | + |∇ ψ | ≤ C | z | , | z | > , for some α < . Then φ = c W x for some real constant c . Proof. Here we work with the change of variables φ = e iθ ψ . Then the equation L ( φ ) = 0 becomes0 = ∆ ψ − r ψ + 2 ir ∂ θ ψ + (1 − w ) ψ − iw Im( ψ ) . Writing ψ = ψ + iψ with ψ , ψ ∈ R we get the system ψ − r ψ − r ∂ θ ψ + (1 − w ) ψ ψ − r ψ + 2 r ∂ θ ψ + (1 − w ) ψ , which holds in R \ { } with the symmetry condition ψ (¯ z ) = − ψ ( z ), ψ (¯ z ) = ψ ( z ).We decompose in Fourier modes ψ = ∞ X k =1 ψ ,k ( r ) sin( kθ ) , ψ = ∞ X k =0 ψ ,k ( r ) cos( kθ ) , and obtain 0 = ∂ rr ψ ,k + 1 r ∂ r ψ ,k − k + 1 r ψ ,k + 2 kr ψ ,k + (1 − w ) ψ ,k (7.4)0 = ∂ rr ψ ,k + 1 r ∂ r ψ ,k − k + 1 r ψ ,k + 2 kr ψ ,k + (1 − w ) ψ ,k . (7.5)In particular equation (7.4) for k = 1 can be written as ∂ rr ψ , + 1 r ∂ r ψ , − r ψ , = g where g ( r ) = − r ψ , + (cid:16) w − r (cid:17) ψ , = O ( r α − )as r → ∞ . The variation of parameters formula yields a function ψ ( r ) = − r Z r ρ Z ∞ ρ g ( s ) d s d ρ, which satisfies ∂ rr ψ + 1 r ∂ r ψ − r ψ = g for r > , | ψ ( r ) | ≤ Cr α − , | ∂ r ψ ( r ) | ≤ Cr α − , for r > . (7.6)Hence ψ , ( r ) = ψ ( r ) + α r + α r − , r > , (7.7)for some α , α ∈ R . We claim that α = 0. To prove this, we note that for k = 1 the system (7.4)-(7.5)has the explicit solution ¯ ψ = (cid:20) ¯ ψ ¯ ψ (cid:21) , ¯ ψ = w ( r ) r , ¯ ψ = − w ′ ( r )Let ψ = (cid:20) ψ , ψ , (cid:21) and define the Wronskian W ( r ) = ψ · ¯ ψ r − ψ r · ¯ ψ. NTERACTING HELICAL GINZBURG-LANDAU FILAMENTS 43 We claim that W ( r ) = cr (7.8)for some c ∈ R . To prove this note that the system (7.4)-(7.5) for ψ can be written as0 = ψ rr + 1 r ψ r + Bψ where B is the 2 × B = 2 r (cid:20) − − (cid:21) + 2 r (cid:20) − w 00 1 − w (cid:21) . Then W r = − r W + ψ T ( B − B T ) ¯ ψ = − r W, because the matrix B is symmetric, and we get (7.8). Using the decomposition (7.7), the decay (7.6) andthe explicit form of ¯ ψ we see that W ( r ) = − α r . On the other hand, from the smoothness of φ near the origin we get that ψ , ( r ), ψ , ( r ) and theirderivatives remain bounded as r → 0. Since the same is true for ¯ ψ we see that W ( r ) is bounded as r → α = 0 as claimed. This in turn implies that | ψ , | ≤ Cr α ′ , | ∂ r ψ , | ≤ Cr α ′ − , for r > , where α ′ = max( − , α − < 1. Using barriers for the ODE (7.5) we get for k = 1 | ψ , | ≤ C, for r > . and for k ≥ | ψ ,k ( r ) | ≤ Ck r , for r > . Adding these inequalities we see that | ψ ( z ) | ≤ C, | z | > , and then a standard scaling and elliptic estimates show that |∇ ψ ( z ) | ≤ C | z | , | z | > . Now we can apply Lemma 7.2 and conclude that φ = c W x for some constant c ∈ R . (cid:3) In this subsection we prove elliptic estimatesthat we needed in Section 3 to develop the linear theory. More specifically we prove estimates of solutionsto some model equations.We use the notation z = ( x , x ) = re is and throughout this section ε > ∂ x x + ∂ x x = ∂ rr + 1 r ∂ r + 1 r ∂ ss . Furthermore in the equations the following term will appear: ∂ ss u = x ∂ x x u + x ∂ x x u − x x ∂ x x u − x ∂ x u − x ∂ x u. We start with recalling the statement and the proof of the comparison principle in the half-plane for theoperator ∆ + ε ∂ ss with Dirichlet boundary condition. Lemma 7.4. Let u : R × R ∗ + → R be a bounded function which is in C ( R × R ∗ + ) ∩ C ( R × R ∗ + ) andwhich satisfies (cid:26) ∆ u + ε ∂ ss u ≥ in R × R ∗ + ,u ≤ on R × { } . (7.9) Then u ≤ in R × R ∗ + .Proof. We adapt the proof of Lemma 2.1 in [6].Let us use polar coordinates ( r, s ) ∈ (0 , + ∞ ) × (0 , π ), and let ϕ > ∂ ss in( − π , π ) associated to the eigenvalue µ > 0, i.e., (cid:26) ∂ ss ϕ + µϕ = 0 on ( − π , π ) ,ϕ ( − π ) = ϕ ( π ) = 0 . We define β := √ µ and we set g ( r, s ) := r β ϕ ( s ) in (0 , + ∞ ) × ( − π , π ) and hence ∂ rr g + 1 r ∂ r g + (cid:18) r + ε (cid:19) ∂ ss g = − µε g ≤ , + ∞ ) × ( − π , π . Consider σ := u/g in (0 , + ∞ ) × (0 , π ) (note that g > u + ε ∂ ss u ≥ σ + ε ∂ ss σ + 2 g (cid:20) ∂ r g∂ r σ + (cid:18) r + ε (cid:19) ∂ s g∂ g σ (cid:21) + ∆ g + ε ∂ ss gg σ ≥ . We note that ∆ g + ε ∂ ss gg σ ≤ u is bounded lim sup r → + ∞ σ = 0. We can thus apply themaximum principle to deduce that σ ≤ , + ∞ ) × (0 , π ). Hence u ≤ , + ∞ ) × (0 , π ). (cid:3) In the same spirit we have the following comparison principle for Neumann boundary condition. Lemma 7.5. Let u : R × R ∗ + → R be a bounded function which is in C ( R × R ∗ + ) ∩ C ( R × R ∗ + ) . Let c ≥ . We assume that u satisfies (cid:26) ∆ u + ε ∂ ss u − cu ≥ in R × R ∗ + ,∂ ν u ≤ on R × { } , then u ≤ in R × R ∗ + . For a function f : R → R and ν ∈ N ∗ , α > k f k ν,α := k (1 + | z | ν ) f k L ∞ ( R ) + sup z ∈ R | z | ν + α [ f ] z,α with [ f ] z,α := sup | h | < | f ( z + h ) − f ( z ) || h | α . Our first goal is to prove Proposition 7.1. Let f : R → R be such that f ( z ) = − f ( z ) and k f k ,α < + ∞ . Then there exists aunique bounded solution of ∆ u + ε ∂ ss u = f in R (7.10) which satisfies u ( z ) = − u ( z ) and | u ( z ) | ≤ C k f k ,α , |∇ u ( z ) | ≤ C k f k ,α | z | for all z in R , (7.11) NTERACTING HELICAL GINZBURG-LANDAU FILAMENTS 45 | ε∂ s u ( z ) | ≤ C k f k ,α | z | for | z | ≥ Cε , k D u k ,α ≤ C k f k ,α . (7.12)We first prove: Lemma 7.6. Let f : R → R be such that f ( z ) = − f ( z ) and k f k ,α < + ∞ . Let u : R → R be a boundedfunction such that u (¯ z ) = − u ( z ) and ∆ u + ε ∂ ss u = f in R . (7.13) Then there exists C > independent of u, f, ε such that (7.11) , (7.12) hold.Proof. Thanks to the symmetry u ( z ) = − u ( z ) it is sufficient to consider the problem (cid:26) ∆ u + ε ∂ ss u − f ( z ) = 0 , z ∈ R × R ∗ + ,u ( x , 0) = 0 for all x ∈ R , (7.14)which we can alternatively write as (cid:26) ∆ u + ε ∂ ss u − f = 0 , ( r, s ) ∈ (0 , + ∞ ) × (0 , π ) ,u ( r, 0) = u ( r, π ) = 0 . Let us assume | f ( z ) | ≤ 11 + | z | . We want to prove that for an absolute constant C we have | u ( z ) | ≤ C. We define v ( z ) = v ( r, s ) := s ( π − s ) . We can check that∆ v + ε ∂ ss v + 11 + r = − r − ε + 11 + r < ∀ z = re is ∈ R × R ∗ + . Hence v is a positive supersolution (and − v a subsolution) for equation (7.14) in (0 , + ∞ ) × (0 , π ) and inthis set, for any bounded solution u of (7.14) we have from Lemma 7.4 | u ( z ) | ≤ | v ( z ) | in R × R ∗ + . The decay estimates in (7.11)-(7.12) follow by Schauder estimates and a standard scaling argument. (cid:3) Lemma 7.7. If u is a bounded function that satisfies ∆ u + ε ∂ ss u = 0 in R , u (¯ z ) = − u ( z ) , then u ≡ .Proof. Suppose u R u = 1. By the strong maximumprinciple the supremum cannot be attained in R \ { } . Let z n ∈ R be a sequence such that u ( z n ) → z n → | z n | → ∞ .Case z n → ∞ . Let us write z n = R n e iσ n , where R n → ∞ and σ n ∈ (0 , π ). We express u in polar coordinates ( r, s ) and define˜ u n ( r, s ) := u ( r + R n , s ) . Up to a subsequence we have ˜ u n → ˜ u uniformly in compact sets of R , where ˜ u ≤ 1, ˜ u ( p ) = 1 for somepoint p = (1 , s ) with s ∈ [0 , π ], and ∂ rr ˜ u + ε ∂ ss ˜ u = 0 in R , with the additional condition ˜ u ( r, 0) = ˜ u ( r, π ) = 0. This contradicts the strong maximum principle.Case z n → 0. Let us write z n = R n e iσ n , where R n → σ n ∈ (0 , π ). Define ˜ u n ( ζ ) := u ( R n ζ ) . Up to a subsequence ˜ u n → ˜ u uniformly in compact sets of R , where ˜ u ≤ u = 0 in R . This is a contradiction. (cid:3) Proof of proposition 7.1. We use v := k f k ,α s ( π − s ) as a super-solution to solve the problem in largehalf-balls centred at the origin. More precisely, for any M > (cid:26) ∆ u M + ε ∂ ss u M = f in B + M (0) u M = 0 on ∂B + M (0) , where B + M (0) := { ( x , x ) ∈ R × R + ; | z | < M } . Thanks to gradient estimates (7.11) we have |∇ u M | ≤ C | v | | z | in B + M (0) , for some C > M and thus we can apply Arzela-Ascoli theorem to take the limit of u M along a suitable subsequence, obtaining a solution of (7.14). The uniqueness is proved in Lemma 7.7 andthe estimates follow from Lemma 7.6. (cid:3) Proposition 7.1 is a model for the treatment of ψ the real part of ψ in Lemma 5.1. To deal with ψ we have to use an analogous proposition: Proposition 7.2. Let g : R → R be such that g (¯ z ) = g ( z ) and k g k ,α < + ∞ Then there exists a uniquebounded v : R → R such that v (¯ z ) = v ( z ) and ∆ v + ε ∂ ss v − v = g. Furthermore there exists a constant C > such that (1 + | z | ) ( | v ( z ) | + |∇ v ( z ) | ) ≤ C k g k ,α , k D v k ,α ≤ C k g k ,α , | ε | z | ∂ s v ( z ) | ≤ C k g k ,α for | z | > /ε. Proof. The symmetry assumption allows us to work in the half-plane R × R ∗ + with homogeneous Neumanncondition on the boundary. We can then apply a barrier argument and rescaled Schauder estimates toprove the proposition. (cid:3) In the course of the linear theory for our problem, when we separate even and odd modes we need ananalogue of the following lemma.Let us consider R > R < R ε < ε − , and let Ω, Ω ′ be the regionsΩ = { z ∈ R | R < | z | < R ε } Ω ′ = n z ∈ R | R < | z | < R ε o , and recall the polar coordinates notation z = re is , r > s ∈ R . NTERACTING HELICAL GINZBURG-LANDAU FILAMENTS 47 Lemma 7.8. Let f : R → R be such that f (¯ z ) = − f ( z ) and | f ( z ) | ≤ | z | . Let u be a solution of ∆ u + ε ∂ ss u = f in Ω such that u (¯ z ) = − u ( z ) and | u ( z ) | ≤ R | log ε | , | z | = R | u ( z ) | ≤ R ε , | z | = R ε . Then there is C such that | u ( z ) | ≤ C | z | log (cid:16) R ε | z | (cid:17) , ∀ z ∈ Ω ′ . Proof. We use a Fourier series decomposition, which thanks to the symmetries we can take of the form f ( r, s ) = X k ≥ f k ( r ) sin( ks ) , u ( r, s ) = X k ≥ u k ( r ) sin( ks ) . The equations on the Fourier coefficients are u ′′ k + 1 r u ′ k − k (cid:18) r + ε (cid:19) u k = f k in ( R , R ε ) . We estimate each u k using barriers. For k = 1 we define¯ u ( r ) := r log (cid:16) R ε r (cid:17) The function ¯ u satisfies ¯ u ′′ + 1 r ¯ u ′ − (cid:18) r + ε (cid:19) ¯ u < − r , for r < R ε . Thus we can use ¯ u as a barrier for u in the interval ( R , R ε ) and deduce that | u ( r ) | ≤ r log (cid:16) R ε r (cid:17) , r ∈ ( R , R ε ) . (7.15)For k ≥ u k ( r ) = C (cid:16) rk + C | log ε | (cid:16) rR (cid:17) − k + R ε (cid:16) rR ε (cid:17) k (cid:17) where C is a large fixed constant (the last two terms in ¯ u k solve almost the homogeneous equationand are there for the boundary conditions). By the maximum principle | u k | ≤ ¯ u k in ( R , R ε ) and for r ∈ (2 R , R ε ) we get ∞ X k =2 ¯ u k ( r ) ≤ C | log ε | r + Cr ≤ Cr log (cid:16) R ε r (cid:17) . This and (7.15) imply the desired conclusion. (cid:3) References [1] O. Agudelo, M. del Pino, and J. Wei. 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Soc. , 368(4):2589–2622, 2016.(Juan D´avila) Departamento de Ingenier´ıa Matem´atica-CMM Universidad de Chile, Santiago 837- 0456, Chile E-mail address : [email protected] (Manuel del Pino) Department of Mathematical Sciences University of Bath, Bat h BA2 7AY, United Kingdom,Departamento de Ingenier´ıa Matem´atica-CMM Universidad de Chile, Santiago 837- 0456, Chile, Chile E-mail address : [email protected] (Mar´ıa Medina) Universidad de Granada, Departamento de An´alisis Matem´atico, Campus Fuentenueva, 18071Granada, Spain E-mail address : [email protected] (R´emy Rodiac) Universit´e catholique de Louvain, Institut de Recherche en Math´ematique et Physique,Chemin du Cyclotron 2 bte L7.01.02, 1348 Louvain-la-Neuve, Belgium E-mail address ::