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solution of the problem, corresponding to an asymmetric guided wave. Key words. .... By testing the levels of the possible symmetric solutions, we infer that for λ ...
c 1999 Society for Industrial and Applied Mathematics °

SIAM J. MATH. ANAL. Vol. 30, No. 6, pp. 1391–1400

ASYMMETRIC MODES IN SYMMETRIC NONLINEAR OPTICAL WAVEGUIDES∗ † ´ L. GAMEZ ´ DAVID ARCOYA† , SILVIA CINGOLANI‡ , AND JOSE

Abstract. We study a symmetric nonlinear value problem in all R, arising in nonlinear optics from the study of propagation of electromagnetic guided waves through a layered medium with a nonlinear response. By variational arguments, we prove the existence of a positive asymmetric solution of the problem, corresponding to an asymmetric guided wave. Key words. electromagnetic guided wave, nonlinear eigenvalue problem, asymmetric positive solution AMS subject classifications. 34B15, 35Q60, 49R05 PII. S0036141098336388

1. Introduction. The propagation of electromagnetic guided waves through a medium consisting of three layers of dielectric materials has been studied in several papers; see, for instance, [1, 13, 2]. The ability of such slab geometry to support guided waves depends upon the way in which the refractive index varies across the layers. For example, in the absence of nonlinear effects, the condition for guidance requires that the refractive index in the outer layers be smaller than that in the central region, as one might guess from Snell’s law. As discussed in [1, 13], nonlinear effects can be used to obtain guidance properties. In this paper, we consider the case where the medium is stratified in three layers of homogeneous composition perpendicular to the x-axis. In such a medium we seek solutions of Maxwell’s equations corresponding to an electric field which is monochromatic, propagating in the direction z and polarized along the y-axis. A field of this kind is given by E(x, y, z, t) = u(x)cos(βz − ωt)e2 , where β > 0, ei denotes the usual basis vector and u : R → R. This ansatz leads to a second-order nonlinear eigenvalue problem on the real axis:   1 2 ω 2 2 −¨ u(x) + β u(x) = 2 n x, u (x) u(x) for x ∈ R, c 2 where c is the speed of light in the vacuum and n2 (x, s) is the dielectric function. The guidance conditions require that all fields decay to zero as |x| → +∞ and that the total electromagnetic energy for unit length in z is finite in each plane y ≡ const. This amounts to lim u(x) = lim u(x) ˙ =0

|x|→∞

|x|→∞

∗ Received by the editors March 31, 1998; accepted for publication (in revised form) December 8, 1998; published electronically October 13, 1999. http://www.siam.org/journals/sima/30-6/33638.html † Departamento de An´ alisis Matem´ atico, Universidad de Granada, 18071-Granada, Spain ([email protected], [email protected]). These authors were supported by D.G.E.S. Ministerio de Educaci´ on y Ciencia (Spain) and Acci´ on Integrada Spain-Italy HI1997-0049 and by E.E.C. contract ERBCHRXCT940494. ‡ Dipartimento di Matematica, Politecnico di Bari, 70125-Bari, Italy ([email protected]. uniba.it). This author was supported by M.U.R.S.T. (40% and 60% funds) and E.E.C. program Human Capital Mobility contract ERBCHRXXCT 9400494.

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´ L. GAMEZ ´ DAVID ARCOYA, SILVIA CINGOLANI, AND JOSE

Z R

2

u (x) dx +

Z R

u˙ 2 (x) dx < ∞.

This problem has been studied for special choices of the dielectric function n2 . In [1], the dielectric function is taken of the form  2 if |x| < d, q + c2 2 n (x, s) = if |x| > d, q2 + s where q, c ∈ R and d > 0 denotes the thickness of the internal layer. In such a case the equation can be integrated directly, and taking λ = β 2 as a bifurcation parameter, one can obtain a family of asymmetric solutions bifurcating from the branch of symmetric ones, at a certain value λ = λ0 , yielding the existence of asymmetric bound states for any λ > λ0 . (See also [5, 6] for discussions about stability.) In a recent work [2], a class of equations which are not explicitly integrable is considered and a perturbative method is applied to yield existence of asymmetric guided waves when the internal layer of the medium is thin. In this paper, our goal is to show how variational techniques can be applied to prove existence of asymmetric modes without restrictive assumptions on the thickness of the internal layer. Precisely, we study a differential equation of the type (1.1)

−¨ u(x) + (λ − c2 h(x))u(x) = b(x)|u(x)|p−2 u(x) u ∈ H 1 (R),

for x ∈ R,

where λ > 0, p > 2, and h, b are even functions such that the supports of h and 1 − b are compact. We point out that (1.1) fits into the Akhmediev setting provided that λ = β 2 , p = 4, h = χ, b = 1 − χ, with χ(x) being the characteristic function of (−d, d). Specifically, we seek positive solutions of (1.1), under the following restrictions on h, b : R → R: (H1)

h is a bounded function with h ≥ 0, h 6≡ 0, supp h ⊂ [−1, 1];

(B1)

b is a bounded function with b(x) = 1 for x 6∈ [−1, 1], b(x) ≤ 0 for x ∈ (−1, 1).

We remark that, by making a simple change of variables, the role of the interval [−1, 1] in the assumptions (H1) and (B1) can be played by any closed and bounded interval. Our variational approach consists of minimizing the Euler functional in a suitable C 1 -manifold. In order to do this, we need to check the compactness condition of Palais–Smale, which is obtained by following closely the arguments in [8] and [3]. (See also [9, 10].) By testing the levels of the possible symmetric solutions, we infer that for λ sufficiently large, the minimizer is asymmetric. Similar arguments are used in [4] in a different setting. Precisely, our result is considered in the following theorem. Theorem 1.1. Assume (H1) and (B1). If λ > c2 khk∞ , then the problem (1.1) has, at least, one asymmetric positive solution. The paper is organized as follows: section 2 is devoted to the proof of the Palais– Smale condition. The details of the minimization and the proof that minimizers are asymmetric are given in section 3.

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2. The Palais–Smale condition. In order to prove these technical results, we need only assume the following (less restrictive) conditions on h and b: (H2) (B2)

h is a bounded function with h ≥ 0, lim|x|→∞ h(x) = 0; b is a bounded function with lim sup|x|→∞ b(x) = 1.

Since h is a bounded function, the first eigenvalue λ1 (−c2 h) of the weighted eigenvalue problem associated with the Laplacian operator of weight −c2 h is given by R R |u| ˙ 2 dx − c2 R h(x)|u|2 dx 2 2 R R . −c khk∞ ≤ λ1 (−c h) = inf u6=0 u2 dx R Moreover, assume λ > −λ1 (−c2 h) ≥ 0,

(2.1)

and set q(x) := λ − c2 h(x); then it is easy to see that Z 2 kuk := |u| ˙ 2 + q(x)|u|2 dx, u ∈ H 1 (R) R

defines a norm on H 1 (R) which is equivalent to the usual norm kuk2H 1 = Indeed, we can observe that

R ¡ R

 |u| ˙ 2 + |u|2 dx.

kuk2 ≤ max{kqk∞ , 1}kuk2H 1 . Otherwise, for k small enough, we get for any u ∈ H 1 (R), Z ¡ 2  kuk2 − kkuk2H 1 = (1 − k) |u| ˙ − c2 hu2 dx Z R  ¡ + λ − k − kc2 h u2 dx (2.2) R Z   2 2 ≥ (1 − k)λ1 (−c h) + λ − k − kc khk∞ u2 dx ≥

0.

R

From (2.2), it is easy to deduce that kuk = 0 implies u = 0. In addition, k · k satisfies the other properties of a norm and thus it is a norm equivalent to k · kH 1 . Let us consider J : H 1 (R) → R the Euler functional associated with the problem (1.1), namely, Z Z 1 1 J(u) = |u| ˙ 2 + q(x)|u|2 dx − b(x)|u|p dx, u ∈ H 1 (R). 2 R p R We now prove that the Euler functional satisfies the Palais–Smale condition on a certain sublevel. Let us define R |u| ˙ 2 + λ|u|2 R (2.3) m(λ) := inf R 2/p . u∈H 1 (R)\{0} p |u| R Lemma 2.1. Assume (H2), (B2) and (2.1). Then J satisfies the Palais–Smale p/(p−2) }; that is, if un condition on the sublevel Σ := {u ∈ H 1 (R) : J(u) < p−2 2p m(λ) 1 is a sequence in H (R), such that

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(i) J(un ) → c, (ii) J 0 (un ) → 0, p/(p−2) , then un admits a convergent subsequence. with c < L ≡ p−2 2p m(λ) Proof. We follow closely the arguments in [8] and [3]. Let {un } in H 1 (R) such that (2.4)

J(un ) = c + o(1),

J 0 (un ) = o(1)

in H −1 ,

as n → ∞, and assume c < L. From (2.4), it follows that p−2 kun k2 ≤ c + |ho(1), un i|, 2p and thus {un } is bounded in H 1 (R) and, up to a subsequence, {un } has a weak limit u ∈ H 1 (R). In order to show that {un } converges to u strongly in H 1 (R), it suffices to prove that kun k → kuk

as n → ∞.

Now we observe that for any R > 0, there results Z Z ¡  ¡  kun k2 − kuk2 ≤ |u˙ n |2 + q(x)|un |2 − |u| ˙ 2 + q(x)|u|2 |x|≤R |x|≤R Z Z ¡  ¡ 2  + |u˙ n |2 + |q(x)||un |2 + |u| ˙ + |q(x)||u|2 . |x|>R

|x|>R

Therefore since the Sobolev imbedding is compact on bounded sets, it suffices to show that for any δ > 0 there exists R > 0 such that for any n ≥ R there results Z   |u˙ n |2 + |q(x)||un |2 < δ |x|≥R

and then, by the weak lower semicontinuity of the above integral, Z   |u| ˙ 2 + |q(x)||u|2 < δ. |x|≥R

By contradiction, assume that there exists δ0 such that for any R > 0 there results Z   |u˙ n |2 + |q(x)||un |2 ≥ δ0 |x|≥R

for some n = n(R) ≥ R. As a consequence, there exists a subsequence {unk } such that Z   (2.5) |u˙ nk |2 + |q(x)||unk |2 ≥ δ0 |x|≥k

for any k ∈ N. For any r > 0, let us introduce the annulus  ª Ar = x ∈ R : r ≤ |x| ≤ r + 1 .

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Claim. For any ξ > 0 and for any R > 0 there exists r > R such that Z   (2.6) |∇unk |2 + |q(x)||unk |2 < ξ Ar

for infinitely many k ∈ N. By contradiction, assume that for some ξ0 , R0 > 0 and for any integer m ≥ [R0 ] there exists ν(m) ∈ N such that Z   |u˙ nk |2 + |q(x)||unk |2 ≥ ξ0 Am

for any k ≥ ν(m). Plainly, we can assume that the sequence {ν(m)} is nondecreasing. Therefore, for any integer m ≥ [R0 ], there exists an integer ν(m) such that Z   Z   |u˙ nk |2 + |q(x)||unk |2 ≥ |u˙ nk |2 + |q(x)||unk |2 R

[R0 ]≤|x|≤m

≥ (m − [R0 ])ξ0 for any k ≥ ν(m). This contradicts Z  Z    2 2 |u˙ nk | + |q(x)||unk | ≤ max{1, kqk∞ } |u˙ nk |2 + |unk |2 R R Z   2 ≤K |u˙ nk | + q(x)|unk |2 ≤ K1 R

with K, K1 positive constants, and it proves the claim. Now, let ξ > 0 be fixed such that λ − ξ > λ2 > 0. Taking into account (H2) and (B2), there exists R(ξ) > 0 such that (2.7)

q(x) ≥ λ − ξ

for any |x| ≥ R(ξ),

(2.8)

b(x) ≤ 1 + ξ

for any |x| ≥ R(ξ).

Let r = r(ξ) > R(ξ) be as in (2.6), and let A = Ar ; up to a subsequence, there results Z   (2.9) |u˙ nk |2 + q(x)|unk |2 < ξ A

for any k ∈ N. Now let us choose any function ρ ∈ C ∞ (R, [0, 1]) such that ρ(x) = 1 for |x| ≤ r, ρ(x) = 0 for |x| ≥ r + 1, and |ρ(x)| ˙ ≤ 2 for any x ∈ R. For any k ∈ N, let vk = ρunk and wk = (1 − ρ)unk . It is not difficult to see that (2.10)

|hJ 0 (unk ), vk i − hJ 0 (vk ), vk i| ≤ C1 ξ,

(2.11)

|hJ 0 (unk ), wk i − hJ 0 (wk ), wk i| ≤ C2 ξ,

where C1 and C2 are positive constants which do not depend on r. First, we prove (2.10): 0 hJ (unk ), vk i − hJ 0 (vk ), vk i

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Z Z ¡ ¡   = w˙ k v˙ k + q(x)wk vk + b(x) |vk |p − |unk |p−1 vk R R Z Z ¡  2 2 = ρ(1 − ρ) |u˙ nk | + q(x)|unk | + ρ(1 ˙ − 2ρ)unk u˙ nk A A Z Z − ρ˙ 2 |unk |2 + b(x)ρ(ρp−1 − 1)|unk |p A ZA Z Z ¡  2 ≤2 |u˙ nk unk | + 4 |unk |2 |u˙ nk | + q(x)|unk |2 + 6 A A A Z Z ¡  p−2 2 +2kbk∞ kunk k∞ |unk | ≤ 2 |u˙ nk |2 + q(x)|unk |2 A

A

1/2 Z

Z

|u˙ nk |2 |unk |2 +6 A A Z ¡  ≤2 |u˙ nk |2 + q(x)|unk |2 A

Z

+6

A

|u˙ nk |2 + q(x)|unk |2

1/2

Z +M

1/2 Z A

A

|unk |2

|unk |2

1/2

Z +M

A

|unk |2 .

By (2.7) and (2.9), we infer Z (λ − ξ) and thus

Z A

A

Z

|unk |2 ≤

|unk |2
0 such that αv ∈ M . Indeed, Z α=

R

p

b(x)|v| dx

Proof. (i) Let f (u) = f (u) = 1; hence (3.1)

−1/p and

kvk2

I(αv) = ¡R R

R R

2/p . b(x)|v|p dx

b(x)|u|p dx, u ∈ H 1 (R). Observe that for any u ∈ M ,

hf 0 (u), ui = −p

Z R

b(x)|u|p dx = −p.

As a consequence, f 0 (u) 6= 0 if u ∈ M . This means that M is a C 1R−manifold of codimension one. Trivially, it is closed by the continuity of the map u 7→ R b(x)|u|p dx.

´ L. GAMEZ ´ DAVID ARCOYA, SILVIA CINGOLANI, AND JOSE

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(ii) Straightforward computations show this part. Lemma 3.2. Assume (H1), (B1), and (2.1). Then inf I(v) < m(λ).

v∈M

Proof. By [12, Theorem 5.5], it is known that the infimum in (2.3) is attained at a symmetric function z ∈ H 1 (R) which is a solution of the problem −¨ u(x) + λu(x) = |u(x)|p−2 u(x) lim|x|→∞ u(x) = 0.

for x ∈ R,

In addition, by [7, 11], such a function is unique and it is given by " #1/(p−2) pλ √ , x ∈ R. z(x) = 2 cosh2 ( p−2 λx) 2 For θ ∈ R, take the translate zθ = z(· + θ) and consider, by (ii) of the preceding lemma, the positive number αθ such that αθ zθ ∈ M . The proof would be concluded if we show that I(αθ zθ ) < m(λ) for some θ. In order to verify this, observe that from √ hypotheses (H1) and (B1), the existence of the positive limit of the function z(x)e λx as x → +∞ implies that for large θ, Z Z √ √ h(x)zθ2 dx ≥ C1 e−2 λθ and (1 − b(x))|zθ |p dx ≤ C2 e−p λθ R

R

for some positive constants C1 and C2 . Then, Z Z 2 2 2 (|z| ˙ + λz ) dx − c h(x)zθ2 dx R R I(αθ zθ ) = Z 2/p Z Z ≤

|z|p dx −

R

R

R

(1 − b(x))|zθ |p dx √

(|z| ˙ 2 + λz 2 ) dx − c2 C1 e−2

Z

Z

R



R

|z|p dx − C2 e−p

(|z| ˙ 2 + λz 2 ) dx

= Z R

2/p



1 − C3 e−2



2/p



1 − C4 e−p

|z|p dx



1 − C3 e−2

= m(λ) 

λθ

λθ



1 − C4 e−p

λθ

λθ

2/p

λθ

λθ

2/p . √



Since p > 2, for large values of θ, one has (1 − C3 e−2 λθ )/(1 − C4 e−p λθ )2/p < 1 and consequently, I(αθ zθ ) < m(λ). Now we are ready to prove Theorem 1.1. Proof of Theorem 1.1. The solution of (1.1) is found as a minimizer of I constrained on M (up to a Lagrangian multiplier). In order to prove the existence of such a minimizer, and taking into account Lemmas 3.1 and 3.2, it suffices to show that I satisfies the Palais–Smale condition on the sublevel Mλ := {u ∈ M : I(u) < m(λ)}, that is, if {un } is a sequence in M , such that

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(a) I(un ) → c, (b) |I 0 (un )(v)| ≤ εn kvk for any v ∈ Tun M , with εn → 0, with c < m(λ), then {un } admits a convergent subsequence. Indeed, by (b), it follows that there exists a sequence {θn } ⊂ R such that for any n ∈ N −4un + q(x)un − θn b(x)|un |p−2 un = o(1)

in H −1 .

Testing the previous equation in un gives θn = I(un ) − ho(1), un i. Therefore, setting vn (x) = I(un )1/(p−2) un , we plainly get J 0 (vn ) = I(un )1/(p−2) o(1) = o(1)

in H −1 .

Moreover there results J(vn ) =

p−2 I(un )p/(p−2) = c0 + o(1) 2p

p/(p−2) with c0 < p−2 . Therefore, by Lemma 2.1, it follows that {vn } is precom2p m(λ) pact, and thus, taking into account (a), {un } is precompact and the Palais–Smale condition holds. Therefore, if v is a minimizer of I|M , setting u = I(v)1/(p−2) v, it results that u is ¡  p/(p−2) . a solution of (1.1). Moreover J(u) = J I(v)1/(p−2) v < p−2 2p m(λ) Therefore, the proof would be concluded if we show that u is not symmetric for λ > c2 khk∞ . This is a consequence of the fact that every possible symmetric positive p/(p−2) . Indeed, since solution u of (1.1) must have a level greater than p−2 2p m(λ) b(x) ≤ 0 for any x ∈ (−1, 1), for such u,

−¨ u(x) + (λ − c2 h(x))u(x) ≤ 0

for x ∈ (−1, 1).

¨(x) ≥ (λ − c2 h(x))u(x) ≥ 0 for any x ∈ (−1, 1). Hence, since λ > c2 khk∞ , we get u Then u is convex in (−1, 1), and thus max {u(x) : x ∈ [−1, 1]} = u(−1) = u(1), with u(1) ˙ > 0, by the Hopf lemma. This means that the maximum of u in all R is attained in some point x0 ∈ R − [−1, 1]. Now, observing that supp h ⊂ [−1, 1], we obtain −¨ u(x) + λu(x) = u(x)p−1 , for ) = 0 implies that |x| > 1, which together with the fact u(±x ˙ 0 u(x) = z(|x| − x0 )

for any |x| ≥ x0 .

z + λz = z p−1 , we obtain As a consequence, since λ > c2 khk∞ , and −¨ Z  2  2p J(u) = |u| ˙ + (λ − c2 h)u2 dx p−2 ZR  2  ≥ |u| ˙ + (λ − c2 h)u2 dx R\[−x0 ,x0 ] Z = |u| ˙ 2 + λu2 dx R\[−x0 ,x0 ] Z = |z| ˙ 2 + λz 2 dx R

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#p/(p−2) 2 2 | z| ˙ + λz dx = ¡R R 2/p |z| ˙ 2 + λz 2 dx R "R #p/(p−2) |z| ˙ 2 + q(x)|z|2 dx R ≥ 2/p ¡R |z|p dx R "

R

= m(λ)p/(p−2) , which concludes the proof. REFERENCES [1] N. N. Akhmediev, Novel class of nonlinear surface waves: Asymmetric modes in a symmetric layered structure, Sov. Phys. JEPT, 56 (1982), pp. 299–303. ´ mez, Asymmetric bound states of differential equa[2] A. Ambrosetti, D. Arcoya, and J. L. Ga tions in nonlinear optics, Rend. Sem. Mat. Univ. Padova, 100 (1998), pp. 231–247. [3] S. Cingolani and M. Lazzo, Multiple semiclassical standing waves for a class of nonlinear Schr¨ odinger equations, Topol. Methods Nonlinear Anal., 10 (1997), pp. 1–13 [4] M. J. Esteban, Nonsymmetric ground states of symmetric variational problems, Comm. Pure Appl. Math., 44 (1991), pp. 259–274. [5] M. Grillakis, J. Shatah, and W. Strauss, Stability theory of solitary waves in the presence of symmetry, J. Funct. Anal., 74 (1987), pp. 160–197. [6] C. K. R. T. Jones and J. V. Moloney, Instability of standing waves in nonlinear optical waveguides, Phys. Lett. A, 117 (1986), pp. 176–184. [7] M. K. Kwong, Uniqueness of positive solutions of ∆u − u + up = 0 in Rn , Arch. Rational Mech. Anal., 105 (1989), pp. 243–266. [8] Y. Y. Li, Existence of multiple solutions of semilinear equations in Rn , Progr. Nonlinear Differential Anal., 4 (1990), pp. 134–159. [9] P. L. Lions, The concentration-compactness method in the calculus of variations: The locally compact case. Part I, Ann. Inst. H. Poincar´ e Anal. Non Lin´eaire, 1 (1984), 109–145. [10] P. L. Lions, The concentration-compactness method in the calculus of variations. Part II, Ann. Inst. H. Poincar´ e Anal. Non Lin´eaire, 1 (1984), pp. 223–283. [11] K. Mcleod and J. Serrin, Uniqueness of positive radial solution of ∆u + f (u) = 0 in Rn , Arch. Rational Mech. Anal., 99 (1987), pp. 115–145. [12] C. Stuart, Bifurcation in Lp (RN ) for a semilinear elliptic equation, Proc. London Math. Soc. (3), 57 (1988), pp. 511–541. [13] C. Stuart, Guidance properties of nonlinear planar waveguided, Arch. Rational Mech. Anal., 125 (1993), pp. 145–200.