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Ann. I. H. Poincaré – AN ••• (••••) •••–•••
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On planar selfdual Electroweak vortices
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Sur les vortex Électrofaibles autoduaux dans le plan
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Dongho Chae a , Gabriella Tarantello b,∗
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a Department of Mathematics, Seoul National University, Seoul 151-742, South Korea b Dipartimento di Matematica, Università degli Studi di Roma “Tor Vergata”, Via della Ricerca Scientifica, 00133 Roma, Italy
Received 29 May 2002; accepted 30 January 2003
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Abstract
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By perturbation techniques, we obtain a new class of selfdual Electroweak vortices in R2 , whose asymptotic behavior we control in an “optimal” way to yield a sort of “quantization” property for the corresponding flux. Our class of vortex-solutions complements that constructed by Spruck and Yang [Comm. Math. Phys. 144 (1992) 215–234]. 2003 Published by Éditions scientifiques et médicales Elsevier SAS.
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Résumé
En utilisant des techniques de perturbation, nous obtenons une nouvelle classe de vortex Électrofaibles autoduaux dans R2 , dont nous pouvons contrôler de façon « optimale » le comportement asymptotique en donnant une propriété de « quantisation » pour le flot corréspondant. Notre classe de solutions complète celles construites par Spruck et Yang [Comm. Math. Phys. 144 (1992) 215–234]. 2003 Published by Éditions scientifiques et médicales Elsevier SAS.
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MSC: 35J45; 35J60; 37K40; 70S15
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Introduction
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We shall be concerned with planar vortex-type (self-dual) solutions for the celebrated SU (2) × U (1)Electroweak theory of Glashow, Salam and Weinberg [10]. As observed by Ambjorn and Olesen [2–4], if the physical parameters satisfy a suitable critical condition, it is possible to determine Bogomol’nyi type equations (also known as self-dual equations) to be satisfied by Electroweak-vortices when expressed in terms of the unitary gauge variables. The selfdual equations include a gauge invariant version of the Cauchy–Riemann equation, which makes it reasonable to assume that the W -boson field admits a finite number N of zeroes (counted with
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* Corresponding author.
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E-mail addresses:
[email protected] (D. Chae),
[email protected] (G. Tarantello). 0294-1449/$ – see front matter 2003 Published by Éditions scientifiques et médicales Elsevier SAS. doi:10.1016/j.anihpc.2003.01.001
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We refer to the recent monograph of Y. Yang [17] and [5] for details, and, in particular, on how to recover the full vortex configuration out of solutions of (P). We only mention that, by virtue of the vortex ansatz and accordingly to the unitary gauge variables, the vortex solution is specified by the (complex valued) massive field W , the (scalar) field ϕ and the real valued 2-vector fields P = (Pµ )µ=1,2 and Z = (Zµ )µ=1,2 . So that, u1 and u2 determine the magnitude of W and ϕ respectively, as follows:
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ϕ 2 = e u2 ,
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|W |2 = eu1 .
(0.1)
Thus, zk corresponds to a vortex point, the integer nk ∈ N to its multiplicity k = 1, . . . , m, and N = m k=1 nk is the vortex number. Furthermore, while the 2-vector fields P and Z are defined only up to gauge transformations, their gauge invariant curls, P12 = ∂1 P2 − ∂2 P1
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and Z12 = ∂1 Z2 − ∂2 Z1
are determined by the relations: g P12 = φ 2 + 2g sin θ |W |2 , 2 sin θ 0 (0.2) g Z12 = (ϕ 2 − φ02 ) + 2g cos θ |W |2 2 cos θ (see [17]). Following the numerical evidence provided by Ambjorn and Olesen, the first rigorous results concerning Electroweak-vortices have been obtained by Spruck and Yang in [13,14]. In [13], the authors aim to obtain configurations of the type described in physics literature as the “Abrikosov’s mixed states” (see [1]), so they consider the selfdual equations subject to ’t Hooft boundary conditions [9]. In turn, this amounts to solve (P) under periodic boundary conditions, and Spruck and Yang in [13] obtain necessary and sufficient conditions on the given physical parameters, in order to ensure the presence of periodic N -vortices in the theory. Their results are sharp for N = 1, 2 and recently have been improved by Bartolucci and Tarantello in [5] to a wider range of parameters. In this paper, we shall be concerned with planar vortex-type configurations, and consider (P) over R2 subject to appropriate decay assumptions at infinity. This situation has been considered in [14], where it has been observed that necessarily, the corresponding vortex-solutions must carry infinite energy. As a consequence, solutions of (P) over R2 appear in abundance, and are distinguished according to their rate of decay at infinity. In view of (0.1), also the flux of the fields P and Z is infinite, while this may not be necessarily the case for the field: (sin θ )P + (cos θ )Z, which bares informations about the gauge potential in the original variables. We shall be concerned with such “finite-flux” selfdual solutions, by solving (P) under the condition: (0.3) eu1 , eu2 ∈ L1 R2 .
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multiplicity). Such an integer N is called the vortex number, and the corresponding vanishing points of W are called the vortex points. Moreover, by means of an approach introduced by Taubes [15,16] in the context of Ginzburg–Landau vortices, it is possible to reduce the analysis of self-dual Electroweak vortices to the study of the following elliptic system: m 2 u1 2 u2 −u = 4g e + g e − 4π nk δ(z − zk ), 1 k=1 (P) 2 u u = g e 2 − φ02 + 2g 2 eu1 2 2 cos2 θ where φ0 is a given positive parameter, g is the SU (2)-coupling constant, and θ ∈ (0, π/2) is the so called “Weinberg-angle”, related with the U (1)-coupling constant g∗ via the relation, g cos θ = 2 . (g + g∗2 )1/2
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In fact, by setting: 2g 2 eu1 , σ1 = π
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e
u2
(0.4)
R2
R2
As a first fact, we show that necessarily
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σ1 + σ2 > 2(N + 1)
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σ1 + σ2
σ2 = o(1) and σ1 = 4(N + 1) + o(1).
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R2
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(0.7)
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L1 -norm
eu2
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of uniformly estimates its To this end, we provide a priori estimates (see Theorem 1.1), so that the L∞ -norm. Thus, requiring a small σ2 , implies that we must have eu2 also small in L∞ -norm. Therefore, the first equation in (P) may be viewed as a “perturbation” of the following singular Liouville equation: m 2 u −u = 4g e − 4π nk δ(z − zk ) in R2 , k=1 eu < +∞,
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whose solutions may be classified according to Liouville formula (e.g., see [12]). When all vortex points coincide, we are able to adapt a perturbation approach introduced by Chae and Imanuvilov in [7] to construct “nontopological” Chern–Simons vortices, and obtain solutions of (P), “bifurcating” out of solutions of the singular Liouville equation mentioned above. In this way, we are able to provide rather accurate pointwise estimates on our solutions to ensure (0.7), together with a sharp control on their decay rate at infinity. We refer to Theorem 2.1 and 2.2 for the precise statements. Clearly, our result furnishes a new class of solutions for (P)–(0.3) which complements those constructed in [14].
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(0.6)
Notice that, in contrast to the periodic case, this class of solutions lacks any sort of “quantization” property for the corresponding flux (0.5). A first goal of this paper is to show that, actually, this is no longer the case, if we consider solutions of (P)–(0.3) with σ2 2N . In this case, and when all vortex points coincide, we show that the value σ2 also controls from above the value σ1 . This implies that, in certain regime, (e.g., σ2 small) the lower bound 2(N + 1) in (0.6) is no longer sharp, as σ1 + σ2 is now forced to remain close to the value 4(N + 1), and any other value is ruled out. See Lemma 1.6. Thus, for σ2 small, a sort of “quantization” property is restored as σ1 “approximately” equals to 4(N + 1). Our second goal, is to show that, in fact, such situation does occur. So, in Section 2, we shall focus our attention in constructing several families of solutions of (P)–(0.3) so that
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any assigned value in (2(N + 1), +∞).
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σ2 > 2N;
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as σ1 and σ2 determine the rate of decay at infinity of u1 and u2 (see Theorem 1.1). The existence results for (P)–(0.3) derived by Spruck and Yang in [14] imply that the lower bound (0.6) is “sharp”, as they construct solutions satisfying:
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(0.5)
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in view of (0.1), (0.2) we see that the flux of the field (sin θ )P + (cos θ )Z is given by, π = (sin θ P12 + cos θ Z12 ) = (σ1 + σ2 ). g
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g2 σ2 = 2π
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1. Asymptotic behavior
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R2
N=
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m
nk ,
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g 2 φ02 c0 = 8 cos2 θ
R2
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σ1 + σ2 > 2(N + 1).
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∈ L∞ (R2 )
and for suitable C0 > 0 (depending on c0 only)
u
e 2 ∞ 2 C0 eu2 1 2 . L (R ) L (R )
In addition, if
u2 (x) + c0 |x − x0 |2 ∈ L2 R2 1+α/2 1 + |x|
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(ii)
u+ 2
as |x| → +∞
as |x| → +∞,
for some x0 ∈ R2 and α ∈ (0, 1), then u1 (x) = 2N − (σ1 + σ2 ) ln |x| + O(1), 1 σ2 u2 (x) = −c0 |x − x0 |2 + σ1 + ln |x| + O(1) 2 cos2 θ as |x| → +∞.
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u1 (x) 2N − (σ1 + σ2 ) ln |x| − C,
for suitable C > 0, u1 (x) → 2N − (σ1 + σ2 ), ln |x| and so necessarily
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∈ L∞ (R2 ),
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Theorem 1.1. Let (u1 , u2 ) be a solution pair for (P ), then
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(1.1)
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We devote this section to obtain the asymptotic behavior, as |x| → +∞, of a solution pair (u1 , u2 ) for (P ) in terms of σ1 and σ2 , and to establish some interesting relation for such values. We have:
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R2
with g, φ0 given positive constants and θ ∈ (0, π/2). Set 2g 2 g2 eu1 , σ2 = eu2 . σ1 = π 2π
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where {z1 , . . . , zm } ⊂ R2 are given points, {n1 , . . . , nm } ⊂ N are given integers, and δ(z − p) denotes the Dirac measure with pole at p ∈ R2 . Let
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Consider the problem, m 2 u1 2 u2 −u = 4g e + g e − 4π nk δ(z − zk ), 1 k=1 g 2 u2 (P ) e − φ02 + 2g 2 eu1 in R2 , u2 = 2 2 cos θ u e 1 < +∞, eu2 < +∞,
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(1.2)
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(1.3)
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(1.4)
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(1.5)
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(1.6)
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(1.7) (1.8)
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In order to derive Theorem 1.1 we are going to recall some basically well known facts about the function: 1 ln |x − y| − ln(1 + |y|) f (y) dy (1.9) w(x) = 2π
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R2
1 and β = 2π
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f (y) dy
(i) if f 0, then
w(x) β ln |x| + 1 + C
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R2
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for suitable C > 0; w(x) (ii) → β, as |x| → +∞; ln |x| (iii) if ln 1 + |x| f ∈ L1 R2 ,
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then
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w(x) = β ln |x| + O(1),
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as |x| → +∞.
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ln |x| + 1
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or
v ∈ L2 R2 , 1+α/2 1 + |x|
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(1.15)
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(1.16)
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(1.17)
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Corollary 1.1. Assume (1.10), and let v be a solution of (1.15) which satisfies (1.16) or (1.17). Then, properties (i), (ii) and (iii) of Lemma 1.2 hold for v.
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R2
and, in view of Lemma 1.2, Proof of Corollary 1.1. Notice that the function u = v − w is harmonic over satisfies (1.16) or (1.17). In either case u must be constant. Indeed, this is a well known fact in case (1.16) holds (e.g., see Lemma 4.6.1 in [17]), while it is a consequence of Proposition 1.1 in [7] in case u satisfies (1.17). ✷
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Proof of Lemma 1.1. Let us start by observing that ln |x − y| − ln 1 + |y| ln |x| + 1 , ∀x, y ∈ R2 .
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Consequently, for f 0, property (i) can be easily derived. To obtain (ii), let |x| > 1 and consider, 1 ln |x − y| − ln 1 + |y| − ln |x| f (y) dy. σ (x) = ln |x|
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(1.12)
∈ L1loc (R2 )
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(1.14)
for some α ∈ (0, 1).
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∈ L∞ R2 ,
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(1.11)
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provided it admits a “slow” growth at infinity as expressed by one of the following conditions:
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(1.13)
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v = f
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(1.10)
We notice that properties (1.11), (1.12) and (1.14) remain valid for any other solution v of the equation:
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Lemma 1.1. Let + f ln |f | + 1 ∈ L1 R2
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with f ∈ L1 (R2 ).
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(1.18)
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σ (x) = σ1 (x) + σ2 (x) + σ3 (x),
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where σ1 , σ2 and σ3 are defined by taking the integral in (1.18) over the regions D1 = {y ∈ R2 : |x − y| < 1}, D2 = {y ∈ R2 : |x − y| > 1 and |y| < ρ} and D3 = {y ∈ R2 : |x − y| > 1 and |y| > ρ} respectively, with ρ > 1 a fixed constant. We estimate,
σ1 (x) 1
ln |x − y| − ln 1 + |y| − ln |x|
f (y) dy ln |x| {|x−y|<1}
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1 ln |x| 1 ln |x|
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ln
{|x−y|<1}
{|x−y|<1}
+2 1+
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1
f (y) dy + ln(|x| + 2) + ln |x| |x − y| ln |x|
dy + |x − y|
1 ln |x|
f (y) ln f (y) + dy
{|x−y|<1}
f (y) dy
{|y|>|x|−1}
+
1 1 π + f ln |f | L1 (R2 ) + 2 1 + ln |x| ln |x|
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{|x−y|>1,|y|<ρ}
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1 ln |x|
σ3 (x)
1 ln |x|
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{|y|>ρ}
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{|x−y|>1, |y|>2|x|}
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f (y) dy
{|y|>ρ}
f (y) dy
{|y|>ρ}
as |x| → +∞, for any fixed ρ > 1. Thus, letting ρ → +∞, we obtain the desired conclusion. Now, suppose that (1.13) holds, then to establish (iii) it remains to show that
ln |x − y| − ln |x| f (y) dy
= O(1) (1.19)
R2
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ln |x − y|
f (y) +
1 + |y|
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as |x| → +∞.
f (y) dy + 3 ln 3 f 1 2 → 4 L (R ) ln |x|
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ln |x − y| + ln 1 + |y| + ln |x| f (y) dy
{|x−y|>1, ρ<|y|<2|x|}
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1 + ln |x|
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f (y) dy,
∀a ∈ R, b > 0
ln 2(ρ + 1) f L1 (R2 ) → 0,
Finally,
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1 with a = ln |x−y| and b = |f (y)|. Consequently, σ1 (x) → 0 as |x| → +∞. Furthermore, for x ∈ R2 |x| > 2ρ, we have
ln |x − y| + ln(ρ + 1) f (y) dy
σ2 (x) 1
ln |x| |x|
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{|x−y|<1}
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f (y) dy
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a
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+
ab e + b(ln b − 1) e + b(ln b) , a
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where we have used the well known inequality:
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{|y|>|x|−1}
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Thus, we must show that σ (x) → 0 as |x| → +∞. For this purpose, write
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{|x−y|<1}
{|x−y|<1}
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π +
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R2
{|x−y|>1}
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{|x−y|>1, |y|<|x|/2}
ln |x − y|
f (y) dy +
|x|
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R2
19
for suitable C2 > 0, and (1.19) follows.
✷
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Proof of Theorem 1.1. We start with the following:
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Claim 1.
1
c0 u2
e L1 . + ln sup u2 2 π R2
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{|x−y|
33
= c0 r 2 +
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{|x−y|=r}
B1 (x)
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and conclude (1.20). Claim 2. u+ 1
∈ L R2 . ∞
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(1.20)
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(1.21)
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B1 (x)
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6
13
At this point, we can use Jensen’s inequality to estimate,
1 1 1
eu2 1 , u2 (y) dy ln eu2 (y) dy ln L π π π B1 (x)
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u2 (y) dσ,
5
12
∀r > 0. Multiply both sides of (1.21) by 2r and integrate over [0, 1] to obtain: 1 c0 u2 (x) + u2 (y) dy. 2 π
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1 2πr
4
11
{|x−y|=r}
3
10
ln |x − y| + ln |x| f (y) dy
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ln 1 + |y| f (y) dy := C1 .
Note that (1.20) immediately implies (1.5). To obtain (1.20), let us use the second equation in (P ) together with Green’s representation formula, to derive that, 4c0 r 1 ln u2 (y) dσ u2 (x) dy + 2π |x − y| 2πr
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{|x−y|<1}
{|x−y|>1, |y|>|x|/2}
C2 f L1 + ln 1 + |y| f (y) dy
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R2
On the other hand,
ln |x − y| − ln |x| f (y) dy
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f (y) ln f (y) + dy +
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as |x| → +∞. As above, we are going to estimate the integral (1.19) over various regions. Firstly, note that for |x| > 1 we have:
1
f (y) dy + ln |x − y| − ln |x| f (y) dy ln ln |x| f (y) dy
|x − y|
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1
7
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(1.22)
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B1 (x)
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in B1 (x)
max u1 C
B1/4 (x)
with a suitable constant C > 0 independent of x. To proceed further, let g2 ln |x − y| − ln 1 + |y| eu2 (y) dy. w2 (x) = 2π R2
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and
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w2 (x) → σ2 , ln |x|
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Define, u0 (x) =
30
and note that
2 u1 − u0 ∈ L∞ loc R .
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Decompose:
36
u1 (x) = u0 (x) − w2 (x) + v1 (x)
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ln |x − zk |2nk ,
k=1
31 32
m
so that v1 (x) satisfies:
39
2 v1 ∈ L∞ loc R ;
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−v1 (x) = 4g 2 eu1 (x) = 4g 2
42
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and v1+ (x) ∈ L∞ R2 . ln(|x| + 1)
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as |x| → +∞.
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Hence, by Lemma 1.1, we have w2 (x) σ2 ln |x| + 1 , ∀x ∈ R2 ,
21
4 5
9
with eu2 ∈ L1 (R2 ) ∩ L∞ (R2 ). So the argument given by Brezis and Merle [6] in the proof of Theorem 2 applies word by word and yields to the estimate:
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3
8
−u1 = 4g 2 eu1 + g 2 eu2
13
2
7
and
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R2
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RO
3
To establish (1.22) note that u1 (x) → −∞ as x → zk , k = 1, . . . , m. So, we need to show that u1 is bounded above outside a large ball which contains all points zk ’s, k = 1, . . . , m. To this end note that, if x: |x| > maxk=1,...,m |zk | + 2, then + u1 e u1 (1.23)
DP
2
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D. Chae, G. Tarantello / Ann. I. H. Poincaré – AN ••• (••••) •••–•••
8
1
anihpc46
Therefore, we may use Corollary 1.1 to derive that −v1 (x) σ1 ln |x| + 1 + C
m
10 11 12 13 14 15 16 17
(1.24)
18 19 20 21
(1.25)
22 23 24 25
(1.26)
26 27 28 29
(1.27)
30 31 32 33
(1.28)
34 35 36
(1.29)
37 38 39
|x − zk |2nk e−w2 (x) ev1 ,
(1.30)
40 41
k=1
42 43 44 45 46 47
(1.31)
48
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D. Chae, G. Tarantello / Ann. I. H. Poincaré – AN ••• (••••) •••–•••
v1 (x) → −σ1 , ln |x|
3 4
and
8
14
17
R2
1 v1 (x) = σ1 ln + O(1), |x|
22 23 24 25 26 27
32 33 34
w2 (x) = σ2 ln |x| + O(1), and
38 39 40 41 42 43 44 45 46 47 48
as |x| → +∞.
as |x| → +∞,
14 15 16 17
as |x| → +∞. Moreover, from (1.29), (1.34) and (1.35) we conclude u1 (x) =
m
ln |z − zk |
2nk
− (σ1 + σ2 ) ln |x| + O(1) as |x| → ∞,
k=1
and Theorem 1.1 is established.
✷
21 22 23 24 25 26 27
29 30 31 32 33 34 35 36 37 38 39 40
Remark 1.1. We point out that, as in Lemma 1.3 of [8], it is possible to sharpen the asymptotic behavior (1.7) and (1.8) and prove that ∂u1 ∂u1 → 2N − (σ1 + σ2 ), → 0, ∂r ∂θ ∂ 1 σ2 r u2 + c0 |x − x0 |2 → σ1 + , ∂r 2 cos2 θ
19
28
(1.35)
1 σ2 2 u2 (x) = −c0 |x − x0 | + σ1 + ln |x| + O(1) 2 cos2 θ
CO
37
13
(1.33)
20
35 36
12
Finally, suppose that (1.6) holds, then we are in position to apply Corollary 1.1 to the function v2 (x) = u2 (x) + g2 u2 + 2g 2 eu1 , and conclude that c0 |x − x0 |2 with f (x) = 2 cos 2θe u2 (x) + c0 |x − x0 |2 1 σ2 σ1 + as |x| → +∞. → ln |x| 2 cos2 θ In particular, this implies that u2 admits exponential decay at infinity. Thus, R2 ln(1 + |y|)eu2(y) dy < +∞, and taking into account (1.33), we derive:
29
31
11
18
28
30
10
(1.34)
19
21
9
Thus, we can use part (iii) of Lemma 1.1 together with Corollary 1.1 to conclude that,
18
20
as |x| → +∞.
So, (1.2) and (1.3) are established. Consequently, we must have that σ1 + σ2 > 2(N + 1) and ln 1 + |y| eu1 (y) dy < +∞.
15 16
7
RO
13
6
OF
u1 (x) → 2N − (σ1 + σ2 ) ln |x|
10
12
4 5
9
11
3
In virtue of (1.25), (1.26) and (1.27), we conclude: u1 (x) 2N − (σ1 + σ2 ) ln |x| − C, as |x| → +∞
DP
8
(1.32)
TE
7
2
as |x| → +∞.
EC
6
1
RR
5
9
for suitable C > 0, and
2
UN
1
P.9 (1-21) by:violeta p. 9
41 42 43 44
∂ u2 + c0 |x − x0 |2 → 0 ∂θ
uniformly, as r = |x| → +∞, with (r, θ ) polar coordinates in
R2 .
45 46 47 48
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D. Chae, G. Tarantello / Ann. I. H. Poincaré – AN ••• (••••) •••–•••
10
1
anihpc46
In the framework of Theorem 1.1, Spruck and Yang in [14] have constructed two families of solutions satisfying the asymptotic behavior (1.7), (1.8) (with x0 = 0) and corresponding values σ1 , σ2 free to vary as follows:
3 4 5
4 5
14 15 16 17 18 19 20 21 22
OF
RO
13
2 2 2 2+α
Xα = u ∈ Lloc R : 1 + |x| u(x) dx < ∞ ,
23
R2
24 25
which defines a Hilbert space equipped with the scalar product
26 27 28 29 30 31 32 33 34
(u, v) =
1 + |x|2+α uv.
R2
Denote with · Xα the corresponding norm on Xα . Also let 2,2 2 Yα = u ∈ Wloc R : u ∈ Xα ,
35 36
40 41 42 43 44 45 46 47 48
u2Yα = u2Xα
2
u(x)
+
2 1+α/2 1 + |x|
.
L (R2 )
CO
39
2 u 2 R . ∈ L (1 + |x|)1+α/2
It defines a Hilbert space with corresponding natural scalar product and norm:
37 38
DP
12
TE
11
EC
10
RR
9
6
In particular, Spruck–Yang’s construction allows one to obtain a solution pair for (P ) with σ1 + σ2 equals to any prescribed value in the interval (2(N + 1), +∞). So, (1.4) is sharp. On the other hand, their construction always gives, σ2 > 2N , while σ1 can take values as small as wanted. For instance, according to Spruck–Yang’s result, it is possible to exhibit a family of solutions for (P ) such that σ1 = o(1), while σ2 can take any fixed value in (2(N + 1), +∞). Clearly, those solutions deny any sort of “quantization” property to the corresponding flux (0.5). The aim of this paper is to show that the situation is quite different in case we consider solutions of (P ) that admit the asymptotic behavior (1.7), (1.8) with σ2 small. We see that in this case σ1 may not be free to take any value in the interval (2(N + 1), +∞), as (1.4) would imply. In fact, in case all vortex points coincide, we see that σ1 must remain close to the value 4(N + 1) and no other value is allowed. Thus solution of (P ) with σ2 small are very interesting, as they seem to restore a sort of “quantization” property for the flux in (0.5). We are going to construct such class of solutions. To this purpose, we consider the following function spaces introduced by Chae and Imanuvilov in [7]. For given α > 0, let
We refer to [7] for the relevant properties of those spaces. In particular, from [7] we recall, 0 (R2 ). Proposition 1.1. (a) Xα 1→ L1 (R2 ) continuously, Yα 1→ Cloc (b) If α ∈ (0, 1) and u ∈ Yα is harmonic, then u is constant. (c) There exists a constant C1 > 0 such that for all u ∈ Yα
UN
8
2 3
1. for the first family: σ1 ∈ (0, 4) and σ1 + σ2 ∈ (2(N + 2), +∞); 2. for the second family: σ1 ∈ (0, 2) and 2σ1 + σ2 ∈ (2(N + 1), +∞).
6 7
1
u(x) C1 uY ln |x| + + 1 , α
∀x ∈ R2 .
7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48
ARTICLE IN PRESS S0294-1449(03)00034-9/FLA AID:46 Vol.•••(•••) ELSGMLTM(ANIHPC):m3SC v 1.165 Prn:5/09/2003; 8:23
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Notice that assumption (1.6) is motivated by considering solutions of (P ) in the space Yα × Yα . In fact, as an easy consequence of Theorem 1.1, it follows:
3 4
4
8
5
(i) U1 := u1 − u0 ∈ Yα , U2 := u2 + c0 |x − x0 |2 ∈ Yα and eu1 ∈ Xα for every α ∈ 0, 2 σ1 + σ2 − 2(N + 1)
9
11 12 13 14 15 16 17 18
(ii)
1 2 U1
+ U2 ∈ Yα and eu2 ∈ Xα , ∀α > 0.
Remark 1.2. Clearly, estimate (1.37) bears interesting consequences in case (and hence σ2 ) is small, as it forces σ1 to remain close to the value 4(N + 1). Such a situation can actually occur, as we are going to construct family of solutions (u1,ε , u2,ε ) for (P ) (when all vortex points coincide) such that, as ε → 0, eu2,ε Xα = o(1), so
25
g2 σ2,ε = 2π
R2
26 27 28
eu2,ε = o(1),
σ1,ε =
2g 2
eu1,ε = 4(N + 1) + o(1).
π R2
30
34 35 36 37 38 39 40 41 42 43 44 45 46 47 48
Consequently, setting R(x) = 4g 2 |x|2N e−w2 (x) , we see that R(x) = O |x|2N−σ2 , and, by assumption, 2N − σ2 0. Furthermore, v1 (x) 2 R(x)e dx = 4g eu1 = 2πσ1 R2
14 15 16 17 18
20 21 22
24 25
27
(1.39)
28 29 30 31 32 33 34 35 36 37 38 39
(1.40)
40 41 42 43
(1.41)
R2
|x|2N−σ2 )ev1
13
26
RR
33
CO
32
(1.38)
Proof. Without loss of generality suppose that z1 = · · · = zm = 0. Thus, by (1.30) we have: 2 2 2N −w2 (x) v1 e v1 ∈ L∞ loc R , −v1 = 4g |x| e with w2 as defined in (1.24). Since eu2 ∈ Xα , then we also have R2 ln(1 + |x|)eu2 (x) dx < +∞, and in view of Lemma 1.2, we derive w2 (x) = σ2 ln |x| + 1 + O(1).
UN
31
12
23
EC
29
11
19
eu2 Xα
23 24
DP
22
10
Lemma 1.2. Let (u1 , u2 ) be solutions of (P ) with z1 = z2 = · · · = zm and such that eu2 ∈ Xα with α > 2 and σ2 2N . For any q ∈ (2, α+4 3 ) there exists a constant Cq = C(q, c0 , α) > 0 such that
σ1 + 2σ2 − 4(N + 1) Cq σ (q−2)/q eu2 2/q . (1.37) 2 Xα
TE
21
8 9
To complement the situation analyzed by Spruck and Yang in [14], we consider solutions of (P ) with σ2 2N . In this case, the following interesting estimate holds, when all vortex points coincide.
19 20
7
(1.36)
RO
10
6
OF
7
2 3
Corollary 1.2. Under the assumptions of Theorem 1.1 we have
5 6
1
< +∞. Thus, we are in position to apply Theorem 1 (when and, from (1.40), we get R2 (1 + 2N = σ2 ), or Theorem 2 (when 2N > σ2 ) of Chen and Li [8], to conclude:
44 45 46 47 48
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D. Chae, G. Tarantello / Ann. I. H. Poincaré – AN ••• (••••) •••–•••
12
πσ1 (σ1 − 4) =
1 2
x · ∇R(x)ev1 (x) dx
1 2
R2
2 v1 v1 = 4g 2N R(x)e − x · ∇w2 (x)R(x)e
4 5
R2
6
R2
= 4πNσ1 − 2πσ2 σ1 − 4g
7 8
2
σ1 + 2σ2 − 4(N + 1) 4g πσ1
13
x · ∇w2 (x) − σ2 R(x)ev1 (x) dx.
R2
14
On the other hand, by (1.24) we have: 2
x · ∇w2 (x) − σ2 g 2π
17 18
R2
19 20
g2 2π
21 22
|y| u2 (y) g2 e dy |x − y| 2π
1 + |y|
2+α
e
25
g2 2π
u
e 2
q−2 q L∞ (R2 )
u 2/q
e 2
Xα
q−2
x · ∇w2 (x) − σ2 C1,q eu1 q1
31
L
u q2
e 2
2 Xα (R )
32
36
2. Existence result
38
40 41 42 43 44 45
48
16 17
dy
18 19
dy
|x − y|
q q−1
dy
|x − y|
q q−1
20
q−1 q
21
(1 + |y|)
2+α−q q−1
+
{|x−y|1}
22
dy (1 + |y|)
23
q−1
24
q
2+α−1 q−1
.
25 26
(i)
1
euε2,α = O(1) ∀α > 0, Xα ε
2g 2 π
27 28 29 30
(1.43)
31 32 33 34 35 36 37 38 39 40 41 42
Theorem 2.1. Let z1 = z2 = · · · = zm = 0, there exists ε0 > 0 sufficiently small such that problem (P) admits a five-parameters family of solutions (uε1,α , uε2,α ) with ε ∈ (−ε0 , ε0 ), α ∈ R4 and |α| < ε0 satisfying:
46 47
R2
15
We devote this section to the construction of a four-parameter family of solutions for (P ) in case all vortex points coincide, such that (1.38) and (1.39) hold. Without loss of generality we take all vortex points to coincide with the origin and obtain,
CO
39
2+α q −1
12 13
RR
35
37
11
(1.42)
with a suitable constant C1,q > 0 independent of x ∈ R2 . Hence, we can use (1.43) in (1.42), and by (1.41), obtain (1.37). ✷
UN
34
10
At this point, the desired estimate (1.37) follows, by means of (1.5), and the observation that, by the choice of q 2+α−q q ∈ (2, α+4 3 ) we have 1 < q−1 < 2 < q−1 . Thus,
30
33
9
14
EC
29
dy
{|x−y|<1}
26
28
q
8
eu2 (y)
|x − y|(1 + |y|)
TE
24
2+α q
(1 + |y|)
1 qu2
R2
23
R2
7
DP
16
6
x · ∇w2 (x) − σ2 R(x)ev1 .
2
RO
12
27
5
That is,
11
15
4
R2
9 10
3
OF
3
44 45 46
e R2
43
uε1,α
= 4(N + 1) + εO(1)
as ε → 0,
47 48
ARTICLE IN PRESS S0294-1449(03)00034-9/FLA AID:46 Vol.•••(•••) ELSGMLTM(ANIHPC):m3SC v 1.165 Prn:5/09/2003; 8:23
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D. Chae, G. Tarantello / Ann. I. H. Poincaré – AN ••• (••••) •••–••• 1
(ii)
13
1
1 (N + 1)! ε ε u1,α (x) = − 2(N + 2) + ε − 1 + β1,α ln |x| + O(1), 2c0 c0N+1 ε uε2,α (x) = −c0 |x|2 + 2(N + 1) + β2,α ln |x| + ln ε + O(1) as |x| → ∞,
2 3 4 5
2 3 4 5
6 7
6
ε → 0, β ε → 0 as ε → 0, |α| → 0, and O(1) denotes a quantity bounded uniformly in (ε, α). where β1,α 2,α
7
15 16 17 18 19 20
R2
R2
with z1 , . . . , zm arbitrarily given points in R2 and small ε ∈ R. For this purpose, it is convenient to introduce complex notation and identify the pair x = (x1 , x2 ) ∈ R2 with the complex number z = x + iy ∈ C. Let
21 22
fε (z) = (N + 1)
23
25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48
m
z (z − εzk ) , nk
and Fε (z) =
k=1
For any a ∈ C, define, (z)|2
8|fε , (1 + |Fε (z) + a|2 )2 and consider the radial function ρ = ρ(r) so that, ηε,a (z) =
fε (ξ ) dξ. 0
8(N + 1)2 |z|2N ρ(|z|) := ηε=0,a=0 (z) = . (1 + |z|2N+2 )2 As well known (e.g., [12]), ∀ε 0 and ∀a ∈ C we have m
− ln ηε,a = ηε,a − 4π nk δ(z − εzk ), k=1 ηε,a = 8π(N + 1). R2
We shall look for solutions of (Pε ) with a specific structure. Namely, we set u1 (z) = ln ηε,a (z) + ε w1 (z) + v1 (z) − 2 ln 2g, u2 (z) = −c0 |z| + w2 (z) + ln ε − 2 ln g + v2 (z), 2
where w2 is the radial function given by w2 (r) = ln 1 + r 2N+2 , and so, it satisfies ρ w2 = , 2
9 10 11 12 13 14 15 16 17 18 19 20 21
(2.1)
22 23 24 25 26
(2.2)
27 28 29
(2.3)
30 31 32 33 34
in R2 ,
RR
24
RO
14
DP
13
TE
12
EC
11
CO
10
8
In order to obtain Theorem 2.1, we shall establish a more general existence result concerning the problem: m 2 u1 2 u2 −u = 4g e + g e − 4π nk δ(z − εzk ), 1 k=1 g 2 u2 (Pε ) e − φ02 + 2g 2 eu1 , u = 2 2 2 cos θ eu1 < ∞, eu2 < ∞
UN
9
OF
8
35
(2.4)
36 37 38 39 40
(2.5)
41 42 43 44
(2.6)
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ARTICLE IN PRESS S0294-1449(03)00034-9/FLA AID:46 Vol.•••(•••) ELSGMLTM(ANIHPC):m3SC v 1.165 Prn:5/09/2003; 8:23
4 5 6 7 8 9 10 11 12 13 14
Lemma 2.1. Problem (2.7) admits a radial solution w1 such that w1 ∈ Yα ,
(ii)
23 24 25 26 27
(iii)
+∞ −2ρ(r)w1 (r) 0
35 36 37 38 39 40 41 42 43 44 45 46 47 48
by setting P = (P1 , P2 ) with
14
19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36
(eε(v1+w1 ) − 1) 2 − ρw1 − e−c0 |z| +w2 , + ηε,a P1 (v1 , v2 , a, ε) = v1 + e ε ε 1 ρ 2 e−c0 |z| +v2 +w2 − ηε,a eε(v1+w1 ) − , P2 (v1 , v2 , a, ε) = v2 − 2 2 cos θ 2 2 −c0 |z|2 +w2 +v2
37 38 39 40
and extended by continuity at ε = 0. Thus, P (0, 0, 0, 0) = (0, 0), and finding a solution for problem (Pε ) is now reduced to finding a small ε0 > 0 and an implicit function ε → (v1,ε , v2,ε , aε ) : (−ε0, ε0 ) → Yα 2 × C satisfying P (v1,ε , v2,ε , aε , ε) = 0, ∀ε ∈ (−ε0 , ε0 ). To this end, let a = a1 + ia2 and observe that,
∂ηε,a (z)
∂ηε,a (z)
= −4ρϕ , = −4ρϕ− , + ∂a1 a=0,ε=0 ∂a2 a=0,ε=0
13
18
From now on, the function w1 in (2.5) is chosen according to Lemma 2.1. For fixed α > 0 define the operator P : Yα 2 × C × R → Xα 2
12
17
RR
34
11
16
CO
33
10
15
r 2(N+1) e−c0 r 2(N+1) 1 + . r dr = 2(N+1) 2 2(N+1) 2c0 (1 + r ) 1+r
UN
32
9
r2
30 31
8
TE
22
29
7
(N + 1)! 1 1 − N+1 ln r + O(1), as r → +∞; w1 (r) = 2c0 c0 rdw1 1 (N + 1)! → 1 − N+1 , as r → +∞; dr 2c0 c0
21
28
6
EC
20
5
DP
(i)
19
1 ∀α ∈ 0, ; 2
3 4
as it follows by direct computations. Notice that, by continuity problem (Pε ) may be considered also at ε = 0, and for a = 0, (Pε=0 ) admits the (trivial) solution v1 = v2 = 0. For ε small we aim to construct solutions for (Pε ) which “bifurcate” from such trivial solution. To this purpose, we start by collecting some useful properties of the function w1 , which will be established in Appendix A.
17 18
2
(2.7)
as constructed in Lemma 2.1 of [7]. Consequently, relative to the unknowns (v1 , v2 ), problem (Pε ) becomes: (eε(v1 +w1 ) − 1) 2 2 −v1 = e−c0 |z| +w2 +v2 + ηε,a − ρw1 − e−c0 |z| +w2 , ε (Pε ) ε 1 ρ 2 +w +v −c |z| 0 2 2 v2 = + ηε,a eε(v1+w1 ) − , e 2 2 2 2 cos θ
15 16
1
OF
3
while w1 is the radial solution in Yα , α ∈ (0, 12 ) of the equation: 2 w + ρw + e−c0 |z| 1 + |z|2N+2 = 0 in R2 ,
RO
2
P.14 (1-21) by:violeta p. 14
D. Chae, G. Tarantello / Ann. I. H. Poincaré – AN ••• (••••) •••–•••
14
1
anihpc46
41 42 43 44 45 46 47
(2.8)
48
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D. Chae, G. Tarantello / Ann. I. H. Poincaré – AN ••• (••••) •••–••• 1
where, in polar coordinates,
2
ϕ+ (r, θ ) =
3 4 5 6
15
1
r N+1 cos(N + 1)θ , 1 + r 2N+2
ϕ− (r, θ ) =
2
r N+1 sin(N + 1)θ . 1 + r 2N+2
3 4
= (P1,(v , P2,(v ) of P at (0, 0, 0, 0) may be easily computed Therefore, the linearized operator P(v 1 ,v2 ,a) 1 ,v2 ,a) 1 ,v2 ,a) as given by,
7
P2,(v (0, 0, 0, 0)[ψ1, ψ2 , b] = ψ2 1 ,v2 ,a)
10
ψ2 − 4ρw1 ϕ+ b1 − 4ρw1 ϕ− b2 ,
+ 2ρϕ+ b1 + 2ρϕ− b2 .
Set P(v (0, 0, 0, 0) = A, we have: 1 ,v2 ,a)
12
where ϕ0 is the radial function:
18
ϕ0 (r) =
19 20 21
Namely, if α ∈ (0,
25
29 30
then
(see Lemma 2.4 in [7]) and f ϕ± = 0 Im L = f ∈ Xα : R2
31 32
1 2 ),
Ker L = Span{ϕ+ , ϕ− , ϕ0 },
26
28
r = |z|.
L = + ρ : Yα → Xα .
23
27
1+r
, 2(N+1)
(see Proposition 2.2 in [7]). Now, let
(f1 , f2 ) ∈ Xα2 ,
33 2 +w 2
36 37
ψ2 − 4ρw1 ϕ+ b1 − 4ρw1 ϕ− b2 = f1 ,
and
ψ2 + 2ρϕ+ b1 + 2ρϕ− b2 = f2 .
38
41 42 43 44 45 46 47 48
ψ2 = φ2 + 2d1 ϕ+ + 2d2ϕ− ,
CO
40
13 14 15 16 17 18 19 20 21 22 23 24 25
(2.9)
where the constants d1 , d2 will be specified later, and φ2 ∈ Yα : φ2 ϕ± = 0. R2
28
(2.10)
29 30 31
× C, b = b1 + ib2 such that
32 33
(2.11)
34 35 36
(2.12)
37 38 39
(2.13)
40 41 42 43
(2.14)
44 45 46
Therefore, recalling that ϕ± + ρϕ± = 0, (2.12) holds if and only if φ2 = 2(d1 − b1 )ρϕ+ + 2(d2 − b2 )ρϕ− + f2 .
26 27
We start by considering (2.12). Decompose
UN
39
we seek
(ψ1 , ψ2 , b) ∈ Yα2
RR
ψ1 + ρψ1 + e−c0 |z|
34 35
12
Proof. In order to prove Proposition 2.1, we recall the following result established in [7] for the operator
22
24
1 − r 2(N+1)
11
RO
17
10
DP
16
9
is onto, and
TE
15
Proposition 2.1. Let α ∈ (0, then the operator Ker A = Span (ϕ+ , 0), (ϕ− , 0), (ϕ0 , 0), (ω1 , 1) × {0},
EC
14
A : Yα2 × C → Xα2
1 2 ),
8
OF
2 +w 2
9
13
6 7
P1,(v (0, 0, 0, 0)[ψ1, ψ2 , b] = ψ1 + ρψ1 + e−c0 |z| 1 ,v2 ,a)
8
11
5
47
(2.15)
48
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16
4
6 7 8
11 12 13 14 15 16 17
20 21
24 25 26 27 28 29
6 7 8
d2 = b2 −
R2
R2
it suffices to take 1 φ2 (x) = ln |x − y|f (y) dy, 2π R2
with
−c0 |z|2 +w2 φ2 − g1 = f1 − e
3 2π(N + 1)
37
40 41 42
R2
45 46 47 48
R2
g1 ϕ− − 2b2 R2
15 16 17 18 19 20
(2.18)
R2
e
−c0 |z|2 +w2
23 24 25
(2.19)
28 29 30
R2
R2
+∞ −c r 2 e 0 1 + r 2(N+1) − 2ρ(r)w1 (r) 0
Consequently, we may choose: c0 c0 b1 = g1 ϕ+ , b2 = g1 ϕ− , π π R2
R2
26 27
31 32 33 34 35 36
2 − 2ρw1 ϕ− = 0.
−c |z|2 +w 2 − 2ρw ϕ 2 dx = π e 0 1 ±
21 22
(2.21)
37 38 39 40
Recalling (2.6), we can use part (iii) of Lemma 2.1 to obtain:
43 44
14
and φ2 given in (2.18). Next, we are going to choose b1 and b2 in order to insure that the right hand side of (2.19) satisfies to the orthogonality conditions required by (2.10). Namely, we impose: −c |z|2 +w 2 − 2ρw ϕ 2 = 0, e 0 g1 ϕ+ − 2b1 (2.20) 1 +
38 39
R2
12 13
RR
36
11
(2.17)
ϕ+ f2 ϕ+ + ϕ− f2 ϕ− ,
CO
35
R2
10
in order to obtain a solution for (2.14)–(2.15) in Yα . Thus ψ2 in (2.13) is determined by (2.17) and (2.18). Now, insert such ψ2 into (2.11) to obtain: 2 ψ1 + ρψ1 = g1 − 2 e−c0 |z| +w2 − 2ρw1 (b1 ϕ+ + b2 ϕ− )
UN
34
f2 ϕ− .
R2
31
33
9
Hence, inserting (2.17) into (2.15), and letting 3 3 f = f2 − f2 ϕ+ ρϕ+ − f2 ϕ− ρϕ− ∈ Xα , 2π(N + 1) 2π(N + 1)
30
32
3 4π(N + 1)
EC
23
5
So (2.16) requires the choise: 3 d1 = b1 − f1 ϕ+ , 4π(N + 1)
22
3 4
R2
R2
18 19
R2
By elementary calculations, we see that 2 2 ρϕ± = π(N + 1). 3
9 10
R2
2
(2.16)
OF
5
R2
1
RO
3
In order to meet the orthogonality condition (2.14), necessarily: 2 2 f2 ϕ+ + 2(d1 − b1 ) ρϕ+ = 0; f2 ϕ− + 2(d2 − b2 ) ρϕ− = 0.
DP
2
TE
1
P.16 (1-21) by:violeta p. 16
anihpc46
41
r 2(N+1) (1 + r 2(N+1) )2
r dr =
π . 2c0
42 43 44 45 46 47 48
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2 3 4 5 6 7 8 9
and obtain that the right-hand side of (2.19) belongs to the image of the operator L = + ρ when defined over Yα (see (2.10)). Thus we derive ψ1 and the desired conclusion follows. Next, to determine Ker A, take f1 = f2 = 0 in the above computations. We find immediately, that d1 = b1 = 0, d2 = b2 = 0, and φ2 = 0. Hence, by Proposition 1.1(b), φ2 must be a constant. If φ2 = 0, then g1 = 0 and we must take ψ1 ∈ ker L. If φ2 = 0, say 2 2 φ2 = 1, then g1 = −e−c0 |z| +ω2 and ψ1 must satisfy: ψ1 + ρψ1 + e−c0 |z| +ω2 = 0. Thus, ψ1 ∈ ω1 + ker L and we conclude that ker A = W × {0} with W ⊂ Yα2 given by W = Span ker L × {0}, (ω1 , 1) (2.22)
OF
1
as claimed. ✷
10
15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31
Yα 2 = Vα ⊕ W.
34
17 18 19
21 22
uε2,α = −c0 |z|2 + w2 (z) + ln ε − 2 ln g + s + v2 (ε, α).
RR
UN
L1 =
d2 1 d +ρ + 2 r dr dr
27
29 30 31
33 34
36 37 38 39 40 41 42 43 44 45
In order to establish Lemma 2.1, let us consider the operator
47
26
35
The authors wish to thank M. Nolasco for useful comments. This research is supported partially by the grant no.2000-2-10200-002-5 from the basic research program of the KOSEF Korea, and by M.U.R.S.T. project: Variational Methods and Non Linear Differential Equation, Italy.
Appendix A
25
32
Final remark. It would be interesting to know whether or not a result similar to Theorem 2.1 remains valid even when the vortex points do not coincide.
Acknowledgements
24
28
At this point, if we take z1 = z2 = · · · = zm = 0, then problem (P) and (Pε ) are one and the same, and in view of (2.2), (2.4), (2.6), Lemma 2.1(ii), and Proposition 1.1(c), we easily derive Theorem 2.1.
45
48
12
23
43
46
11
all vanishing at ε = 0 and α = 0, such that ∀ε ∈ (−ε0 , ε0 ) problem (Pε ) admits a four-parameter family of solutions (uε1,α , uε2,α ), α = (s, α0 , α1 , α2 ) ∈ R4 , |α| < ε0 , which decompose as follows: uε1,α = ln ηε,a(ε,α) + ε (1 + s)w1 + α0 ϕ0 + α1 ϕ+ + α2 ϕ− + v1 (ε, α) − 2 ln 2g,
42
44
8
16
CO
41
7
20
38
40
6
Theorem 2.2. Let α ∈ (0, 1/2) there exists ε0 > 0 sufficiently small and continuous functions v(ε, α) = (v1 (ε, α)), v2 (ε, α) : Qε0 → Vα , a(ε, α) : Qε0 → C,
36
39
5
15
By a direct application of the Implicit Function Theorem (see Theorem 2.7.5 in [11]) we may conclude:
35
37
4
14
Furthermore, for given r > 0 denote by: Qr = (ε, α) ∈ R5 : ε ∈ (−r, r) and α = (s, α0 , α1 , α2 ) ∈ R4 , |α| < r .
32 33
3
13
RO
14
Denote by Vα ⊂ Yα the space orthogonal to W defined in (2.22) with respect to the scalar product in Yα Hence, we may write:
DP
13
2
10
2.
TE
12
2
1
9
EC
11
17
46 47
(A.1)
48
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given by the restriction of the operator L = + ρ over the radial functions. Thus, we wish to solve
2
6 7 8
(A.2)
2 −e−c0 r +w2 (r)
2 −e−c0 r (1
with f (r) = = + representation for solutions of (A.2) in case 1 1 + . f ∈ C (R ) ∩ Xα and α ∈ 0, 2
r 2(N+1) ).
Chae and Imanuvilov in [7] have obtained an integral
(A.3)
13
0
14 15
with
16 17 18
φf (r) = ϕ0 (r)
−2 (1 − r)2
r 0
where φf (1) and w(r) are extended by continuity at r = 1.
21
27 28
In (r) = 0
and notice that
29 30
2 1 − t 2n te−c0 t dt,
φf (r) = − ϕ0 (r)
31 32 33 34 35 36 37 38
43 44 45 46 47 48
IN+1 (r).
k=0
r In=1 (r) =
2 1 − t 2 te−c0 t dt
0
16 17 18 19 20 21 22 23 24
(A.5)
25 26 27 28 29
(A.6)
30 31 32 33 34
(A.7)
35 36
39 40 41 42 43 44
0
1 2 2 = 1 − e−c0 r + r 2 e−c0 r − 2 2c0
13
38
r 1 1 −c0 t 2
t =r d −c0 t 2 2 e + t dt e =−
2c0 2c dt 0 t =0
12
37
Proof. We shall proceed by induction. For n = 1,
41 42
r
Lemma A.2. The following identity holds: n 2(n−k) n! r n! 1 2 1 − n − e−c0 r 1 − . In (r) = 2c0 c0 c0k (n − k)!
39 40
−2 (1 − r)2
EC
26
r
RR
25
TE
24
CO
23
11
15
In order to obtain the solution w1 as claimed in Lemma 2.1, we shall use such a representation formula with 2 f (r) = −e−c0 r (1 + r 2(N+1) ). To this purpose, for given n ∈ N, let
UN
22
10
14
1 − r 2(N+1) ϕ0 (r) = , 1 + r 2(N+1)
ϕ0 (t)tf (t) dt,
r
19 20
9
Lemma A.1 (see Lemma 2.1 in [7]). Assume (A.3), then (A.2) admits a solution w ∈ Yα given by the formula r φf (1)r φf (s) − φf (1) (A.4) w(r) = ϕ0 (r) ds + 1−r (1 − s)2
RO
12
7 8
DP
11
4
6
9 10
3
5
OF
5
1 2
L1 w = f
3 4
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18
1
anihpc46
r te 0
−c0 t 2
dt
45 46 47 48
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5 6 7
10 11 12 13 14
r In+1 (r) =
r 1 −c0 r 2 2(n+1) −c0 r 2 2n −c0 t 2 1−e +r e − 2(n + 1) t te dt = 2c0
20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38
0
r
1 2 2 2 1 − e−c0 r + r 2(n+1) e−c0 r + 2(n + 1)In (r) − 2(n + 1) te−c0 t dt 2c0 0 1 n + 1 −c0 r 2 2 2 1 − e−c0 r + r 2(n+1) e−c0 r + 2(n + 1)In (r) + = −1 . e 2c0 c0
=
42 43 44 45
and the desired identity is established.
48
14
15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36
Furthermore, for r > 2, inserting (A.7) into (A.4), we find r w1 (r) = −ϕ0 (r)
2(N+1) 2
2
1+t 1 − t 2(N+1)
37
✷
An immediate consequence of Lemma 3.2 gives 1 (N + 1)! IN+1 (r) → 1 − N+1 := γ0 as r → +∞. 2c0 c0
46 47
13
k=0
UN
41
12
Thus, if we apply the inductive assumption (A.7), substituting above, we find n! 1 2(n + 1) 2 2 In+1 (r) = 1 − e−c0 r + r 2(n+1) e−c0 r + 1− n 2c0 2c0 c0 n n + 1 −c0 r 2 n + 1 r 2(n−k) n! −c0 r 2 1− + −e e − c0 c0 c0k (n − k)! k=0 n r 2(n−k) (n + 1)! (n + 1)! 1 −c0 r 2 2(n+1) − 1 − n+1 − e 1−r = 2c0 c0 c0k+1 (n − k)! k=0 n+1 2[(n+1)−k] r (n + 1)! (n + 1)! 1 2 1 − n+1 − e−c0 r 1 − r 2(n+1) − = 2c0 (n + 1 − k)! c0k c0 k=1 n+1 2[(n+1)−k] r (n + 1)! (n + 1)! 1 −c0 r 2 1 − n+1 − e 1− = 2c0 (n + 1 − k)! c0k c0
39 40
11
DP
19
7
10
TE
18
6
9
EC
17
5
8
2 1 − t 2(n+1) te−c0 t dt
0
15 16
4
and (A.7) is established for n = 1. Now assume that (A.7) holds for n 1, and we proceed to prove it for n + 1. To this end, notice that
8 9
3
OF
4
2
RO
3
1
RR
2
1 1 −c0 r 2 −c0 r 2 2 −c0 r 2 1−e e = +r e + −1 2c0 c0 1 1 1 2 1 − − e−c0 r 1 − r 2 − , = 2c0 c0 c0
CO
1
19
IN+1 (t) dt + O(1) t
38 39 40 41
(A.8)
42 43 44 45 46 47 48
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20
r
1
= −ϕ0 (r)
2 3
2
8 9 10 11 12 13 14
1
γ0 dt + O(1) t
2 3 4 5
Consequently,
6
w1 (r) = γ0 ln r + O(1),
as r → +∞.
(A.9)
−1 −1 r dw1 1 (r) = −rϕ0 (r) ϕ0 (r) w1 (r) − ϕ0 (r) IN+1 (r) + O dr r 2(N+1) −1 4(N + 1)r 1 = − ϕ0 (r) w1 (r) + O IN+1 (r) + . 2(N+1) 2 r (1 + r )
So, taking into account (A.9), we immediately conclude that
16
r dw1 (r) → γ0 as r → +∞ (A.10) dr and, part (ii) of Lemma 2.1 follows by (3.9) and (3.10). In order to establish (iii), notice that, by direct computation, we find 1 1 8(N + 1)2 r 4N+2 ρ(r)r 2(N+1) L1 = = , 2 (1 + r 2(N+1) )2 (1 + r 2(N+1) )4 (1 + r 2(N+1) )2
24 25 26 27 28 29
while, by definition, L1 w1 = −e−c0 r (1 + r 2(N+1) ). Therefore, in view of the asymptotic behaviors (A.9) and (A.10) of w1 as r → +∞, we can use integration by parts, to obtain: 2
+∞ ρ(r) 0
31 32 33
1 = 2
34 35 36 37
1 = 2
38 39 40
1 = 2
41 42
0 +∞
0 +∞
0 +∞
0
43
45 46
d 1 1 d dr + r w1 (r) dr dr 2 (1 + r 2(N+1) )2 1
1 L1 w1 r dr = − 2(N+1) 2 2 (1 + r )
+∞ 0
2 e−c0 r r 2(N+1)
1 + r 2(N+1)
−
+∞ 0
(r)r 2(N+1)
2ρ(r)w1 (1 + r 2(N+1) )2
0 +∞
11 12 13
16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32
+∞ d 1 d 1 1 r w1 (r) dr + ρ(r)w1 (r) r dr 2(N+1) 2 2(N+1) dr dr (1 + r 2 ) (1 + r )2
Consequently,
47 48
r 2(N+1) w1 (r)r dr (1 + r 2(N+1) )2
+∞ 1 1 L1 w1 (r)r dr = 2 (1 + r 2(N+1) )2
30
44
DP
23
TE
22
10
15
EC
21
RR
20
CO
19
UN
18
9
14
15
17
7 8
On the other hand, for r → +∞,
OF
7
2
RO
6
1 + t 2(N+1) 1 − t 2(N+1)
= −γ0 ϕ0 (r) ln r + O(1).
4 5
P.20 (1-21) by:violeta p. 20
anihpc46
33 34 35 36
ρ(r)w1 (r) 0
37
1 (1 + r 2(N+1) )2
r dr
38 39 40
e−c0 r r dr. 1 + r 2(N+1) 2
41 42 43 44
+∞ 1 2 e−c0 r r dr = . r dr = 2c0 0
45
✷
46 47 48
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References
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2
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[5]
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[6] [7]
11 12 13 14 15
[8] [9] [10] [11] [12]
16 17 18
[13] [14]
19 20 21
[15] [16] [17]
OF
5
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RO
4
2
[1] [2] [3] [4]
DP
3
22 23 24
TE
25 26 27 28
EC
29 30 31 32 33
37 38 39 40 41 42 43 44 45 46 47 48
CO
36
UN
35
RR
34
3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48