On a Riemann–Hilbert problem for the Fokas–Lenells equation

Accepted Manuscript On a Riemann-Hilbert problem for the Fokas-Lenells equation

Liping Ai, Jian Xu

PII: DOI: Reference:

S0893-9659(18)30248-9 https://doi.org/10.1016/j.aml.2018.07.027 AML 5599

To appear in:

Applied Mathematics Letters

Received date : 19 May 2018 Revised date : 18 July 2018 Accepted date : 18 July 2018 Please cite this article as: L. Ai, J. Xu, On a Riemann-Hilbert problem for the Fokas-Lenells equation, Appl. Math. Lett. (2018), https://doi.org/10.1016/j.aml.2018.07.027 This is a PDF file of an unedited manuscript that has been accepted for publication. As a service to our customers we are providing this early version of the manuscript. The manuscript will undergo copyediting, typesetting, and review of the resulting proof before it is published in its final form. Please note that during the production process errors may be discovered which could affect the content, and all legal disclaimers that apply to the journal pertain.

*Manuscript Click here to view linked References

ON A RIEMANN-HILBERT PROBLEM FOR THE FOKAS-LENELLS EQUATION LIPING AI AND JIAN XU* Abstract. The solution of the initial value problem (IVP) for the Fokas-Lenells equation (FLE) was constructed in terms of the solution M (x, t, k) of a 2 × 2 matrix Riemann-Hilbert problem (RHP)

as k → ∞, and the one-soliton solution of the FLE was derived based on this Riemann-Hilbert problem, in Nonlinearity 22(2009), [1]. However, in fact, the derivative with respect to x of the solution of the FLE (ux (x, t)) was recovered from the RHP as k → ∞.

In this paper, we construct the solution of the FLE in terms of the RHP as k → 0, because the Lax pair of the FLE contains the negative order of the spectral variable k. We show that the one-soliton solution of the FLE obtained in this paper is the same as [1], but avoiding a complex integral.

1. Introduction The Fokas-Lenells equation (FLE) utx + αβ 2 u − 2iαβux − αuxx + σiαβ 2 |u|2 ux = 0,

σ = ±1,

(1.1)

where α > 0 and β are real constant, u(x, t) is a complex-valued function, while the subscripts t and x denote the partial derivativations. It is a completely integrable nonlinear partial differential equation (here ”integrable” we mean it admits a Lax pair) which has been derived as an integrable generalization of the nonlinear Schr¨odinger equation (NLSE) using bi-Hamiltonian methods [2]. In the context of nonlinear optics, the FLE models the propagation of nonlinear light pulses Date: July 18, 2018. Key words and phrases. Riemann-Hilbert problem, Fokas-Lenells equation, Initial value problem, Negative order Lax pair. 1

2

L.AI AND J.XU

in monomode optical fibers when certain higher-order nonlinear effects are taken into account [3]. The FLE is related to the NLS equation in the same way as the Camassa-Holm equation associated with the KdV equation. The soliton solutions of the FLE have been constructed via the Riemann-Hilbert method in [1]. And the initial-boundary value problem for the FLE on the half-line was studied in [4]. A simple Nbright-soliton solution was given by Lenells [5] and the N-dark soliton solution was obtained by means of B¨acklund transformation [6]. And Matsuno get the bright and dark soliton solutions for the FL equation in [7] and [8] by a direct method. The rogue wave solutions are showed in [9, 10] by Darboux transform method. In fact, the authors of [1] obtained the x-derivative of the solution ux (x, t) of the FLE with the initial value u(x, t = 0) = u0 (x), which is smooth enough and decays rapidly as |x| → ∞, in terms of the solution

of a 2 × 2 matrix-valued Riemann-Hilbert problem as k → ∞. Then we need calculate a integral of x to get the solution u(x, t). Usually, it is

not easily to calculate the integration. Hence, a nature question is that if we can construct the solution u(x, t) directly in terms of the solution of the associated RHP. In this paper, under the same initial value assumption as [1], we show that it is possible to recover the solution u(x, t) in terms of the solution of the associated RHP as k → 0, because

the Lax pair of the FLE is an negative order with respect to spectral variable k. The outline of this paper is as follows: In Section 2, we recall some

results of [1] about the Riemann-Hilbert problem for the FLE. And then, in section 3, we introducing another transformation to control the behavior of the Jost solutions of the Lax pair as k → 0. In Section 4, we derive the soliton solutions u(x, t) from the associated Rieamnn-

Hilbert problem, especially we obtain the explicit form of the onesoliton solution.

RHP FOR THE FL EQUATION

3

2. The RHP for the Fokas-Lenells equation In this section, we review some results of the RHP for the FLE in [1]. The FLE (1.1) admits Lax pair ( Φx + ik 2 σ3 Φ = kUx Φ

iαβ 2 σ3 ( k1 U 2

Φt + iη 2 σ3 Φ = [αkUx +

where 0 u

U=

v 0

!

,

1

σ3 =

0

0 −1

!

,

η=

√

(2.1)

− U 2 )]Φ.

α(k −

β ), 2k

v = σ¯ u.

According to the paper [1], we make a transformation i

Φ(x, t, k) = e 2

Rx

−∞

ux (x0 ,t)vx (x0 ,t)dx0 σ ˆ3

i

µ(x, t, k)e− 2

R∞ x

ux (x0 ,t)vx (x0 ,t)dx0 σ3 −i(k2 x+η 2 t)σ3

e

,

(2.2)

where σ ˆ3 acts on a 2 × 2 matrix A by then σ ˆ3 A = σ3 Aσ3−1 . We can obtain the differential systems satisfied by µ(x, t, k) as ( µx + ik 2 [σ3 , µ] = V1 µ, µt + iη 2 [σ3 , µ] = V2 µ,

where

Rx

(2.3)

i (kUx − ux vx σ3 ), 2

(2.4)

Then, define two solutions µ1 and µ2 of (2.3) by ( Rx 2 0 µ1 (x, t, k) = I + −∞ eik (x −x)ˆσ3 (V1 µ1 )(x0 , t, k)dx0 , Rx 2 0 µ2 (x, t, k) = I + ∞ eik (x −x)ˆσ3 (V1 µ2 )(x0 , t, k)dx0

(2.6)

i

V 1 = e− 2 V2 = e

− 2i

Rx

−∞

ux (x0 ,t)vx (x0 ,t)dx0 σ ˆ3

0 0 0ˆ 3 −∞ ux (x ,t)vx (x ,t)dx σ



 iαβ 2 1 iα 2 2 αkUx + σ3 ( U − U ) − (ux vx − β uv)σ3 . 2 k 2 (2.5)

It follows that the columns vectors of µ1 and µ2 are bounded and analytic like   (+) (−) µ1 = [µ1 ]1 , [µ1 ]2 ,

  (−) (+) µ2 = [µ2 ]1 , [µ2 ]2 ,

where [A]1 and [A]2 denote the first and second column of a matrix A , respectively, and the superscripts (+) and (−) indicate it is analytic

4

L.AI AND J.XU

for k in D+ and D− , respectively. Here, π 3π π 3π D+ = {k| arg k ∈ (0, )∪(π, )}, D− = {k| arg k ∈ ( , π)∪( , 2π)}. 2 2 2 2 From (2.6), we can deduce that the function µ(x, t, k) = µj (x, t, k), j = 1, 2 satisfies the symmetry relations ¯ = σ1 µ(x, t, k)σ1 , µ(x, t, −k) = σ3 µ(x, t, k)σ3 , µ(x, t, σ k) (2.7) ! 0 1 . where σ1 = 1 0 And we also note that these two matrix functions µ1 and µ2 are related by a matrix s(k) which is independent of (x, t) µ2 (x, t, k) = µ1 (x, t, k)e−i(k

2 x+η 2 t)ˆ σ

3

s(k),

Imk 2 = 0.

From (2.7), it suggests the following notation for s(k): ! ¯ a(k) b(k) . s(k) = ¯ a(k) σb(k)

(2.8)

(2.9)

From the relation between µ1 and µ2 (2.8) and their analytic properties, we find that a(k) has an analytic continuation to all of D+ . Hence, we define two matrix functions M+ (x, t, k) and M− (x, t, k) as    (+)  [µ1 ]1 (+)  , [µ2 ]2 , k ∈ D+  M+ (x, t, k) = a(k)   M (x, t, k) = , (2.10) (−)  (−) [µ1 ]2  , k ∈ D−  M− (x, t, k) = [µ2 ]1 , ¯ a(k)

we can get the Riemann-Hilbert problem for the Fokas-Lenells equation (1.1) as follows: ( M+ (x, t, k) = M− (x, t, k)J(x, t, k), M (x, t, k) → I,

k → ∞.

k ∈ R ∪ iR,

(2.11)

where the jump matrix J(x, t, k) is defined by (−ik2 x−iη 2 t)ˆ σ

J(x, t, k) = e with r(k) =

¯ b(k) . a(k)

3

¯ r(k) ¯ 1 − σr(k)r(k) −σr(k)

1

!

,

(2.12)

RHP FOR THE FL EQUATION

5

In [1], the solution of Fokas-Lenells equation (1.1) was expressed by ux (x, t) = 2im(x, t)e4i

Rx

−∞

|m|2 (x0 ,t)dx0

(2.13)

where m(x, t) = lim (kM (x, t, k))12

(2.14)

k→∞

with M (x, t, k) is the unique solution of the Riemann-Hilbert problem (2.11). In view of the expression of (2.13), we know that the solution obtained by the authors of [1] is ux (x, t) not u(x, t). However, we want to obtain the solution u(x, t) directly instead of integrating the function ux (x, t) with respect to x. This is can be done if we know the asymptotic behavior of the function M (x, t, k) as k → 0. Hence, in the

following section, we control the behavior of the eigenfunctions of the Lax pair (2.1). 3. Control the behavior of M (x, t, k) as k → 0 Introducing a transformation as follows, Φ(x, t, k) = Ψ(x, t, k)e−i(k then we can get the Lax pair of Ψ, ( Ψx + ik 2 [σ3 , Ψ] = hkUx Ψ,

Ψt + iη 2 [σ3 , Ψ] = αkUx +

iαβ 2 σ3 2

2 x+η 2 t)σ 3

1 U k

,

(3.1)

− U2

We seek a solution of this equation in the form Ψ(x, t, k) = D0 + kD1 + k 2 D2 + O(k 3 ),


i

(3.2)

Ψ.

k → 0,

(3.3)

where Di , i = 0, 1, 2 are independent of k. Substituting the above expansion in the (3.2) and comparing the coefficients of the same order of k, we can find D0 = I,

D1 =

0 u v 0

!

(2)

,

D2 =

D11

0

0

D22

(2)

!

,

(3.4)

6

L.AI AND J.XU (2)

(2)

where D11 and D22 which are independent of k satisfy the following differential systems ( (2) ( (2) D11t = αux v − αuvx + 2iαβuv − D11x = ux v, (2) D22x

(2) D22t

= vx u,

= αuvx − αux v − 2iαβuv +

iαβ 2 2 2 uv 2 iαβ 2 2 2 uv 2

+ uvt , + ut v. (3.5)

As (2.6) k → ∞, we can define two eigenfunctions Ψ1 (x, t, k) and

Ψ2 (x, t, k) by ( Rx 2 0 Ψ1 (x, t, k) = I + −∞ eik (x −x)ˆσ3 (kUx Ψ1 )(x0 , t, k)dx0 , Rx 2 0 Ψ2 (x, t, k) = I + ∞ eik (x −x)ˆσ3 (kUx Ψ2 )(x0 , t, k)dx0

(3.6)

Note that the eigenfunctions µj (x, t, k) and Ψj (x, t, k), j = 1, 2 being

related to the same Lax pair (2.1), must be related to each other as i

µj (x, t, k) = e− 2 c− σ3 Ψj (x, t, k)e−i(k

2 x+η 2 t)σ

3

Cj (k)ei(k

2 x+η 2 t)σ 3

i

e 2 cσ3 , (3.7)

where Z c− =

x

0

0

0

ux (x , t)vx (x , t)dx ,

c+ =

−∞

Z

∞

ux (x0 , t)vx (x0 , t)dx0 ,

x

c = c− +c+ . (3.8)

Note that c is a quantity conserved under the dynamics governed by (1.1). Computing the Cj (k) as x → ±∞, respectively, we find i

C1 (k) = e− 2 cσ3 ,

C2 (k) = I.

(3.9) (+)

(+)

Then, from the (2.9), we can find that a(k) = det ([µ1 ]1 , [µ2 ]2 ). It follows that a(k) has the asymptotic behavior as k → 0 from (3.7), i

a(k) = e− 2 cσ3 (1 + O(k 3 )). Hence, from the definition of M (x, t, k) (2.10), we find that
i i M (x, t, k) = e− 2 c− σ3 I + kU + O(k 2 ) e 2 cσ3 , k → 0.

(3.10)

(3.11)

This implies that the solution u(x, t) of the FLE can be expressed in terms of the solution M (x, t, k) of the Riemann-Hilbert problem, ((M (x, t, 0))−1 M (x, t, k))12 . k→0 k

u(x, t)e−ic = lim

(3.12)

RHP FOR THE FL EQUATION

7

4. Soliton solutions In order to obtain the soliton solutions of the FLE, we need consider the zeros of the scattering function a(k). In the following, we assume that a(k) has simple zeros {kj }N j=1 in D+ . Hence, as P.18 of [1], M (x, t, k) satisfies the residue condition for some constant Cj as follows, 2

2

Resk=kj [M (x, t, k)]1 = Cj e2i(kj x+ηj t) [M (x, t, kj )]2 , where ηj =

√ α(kj −

(4.1)

β ). 2kj

From the symmetries (2.7), we know that a(k) is an odd function. It implies that if kj is a zero of a(k), then so is −kj . So, we can write down the first column of M (x, t, k) as, ! 2N X Cp e2i(kp2 x+ηp2 t) 1 [M (x, t, kp )]2 [M (x, t, k)]1 = + k − kp 0 p=1

(4.2)

Again, using the symmetries (2.7), the above equation can be written as ¯ M22 (x, t, σ k) ¯ M12 (x, t, σ k)

!

=

1 0

!

+

2 2 2N X Cp e2i(kp x+ηp t)

p=1

k − kp

[M (x, t, kp )]2 . (4.3)

Evaluation at σ k¯j yields ! ! 2N X Cp e2i(kp2 x+ηp2 t) M22 (x, t, kj ) 1 = + ¯j − kp [M (x, t, kp )]2 , σ k M12 (x, t, kj ) 0 p=1

j = 1, . . . , 2N. (4.4)

Solving this algebraic system for M12 (x, t, kj ), M22 (x, t, kj ), j = 1, . . . , 2N , and substituting the solution into (4.2) yields an explicit expression for [M (x, t, k)]1 . The second column [M (x, t, k)]2 comes from the symmetry. This solves the Riemann-Hilbert problem. Then, using the expression for u(x, t) (3.12), we can obtain the solution u(x, t) instead of ux (x, t) in [1]. In the following, we give the details for getting the one-soliton solution of FLE.

8

L.AI AND J.XU

Let N = 1, k2 = −k1 and C1 = C2 , the system (4.4) becomes     M22 (x, t, k1 ) = 1 + C1 e2i(k12 x+η12 t) ¯ 1 − ¯ 1 M12 (x, t, k1 ) σ k1 −k1 σ k1+k1  2 2  M12 (x, t, k1 ) = C1 e2i(k1 x+η1 t) ¯ 1 + ¯ 1 M22 (x, t, k1 ). σ k1 −k1 σ k1 +k1 (4.5)

Solving this system for M22 (x, t, k1 ) and M12 (x, t, k1 ), we get M22 (x, t, k1 ) = M12 (x, t, k1 ) = Hence, we have  M (x, t, k) = 

1

2 |C |2 e 4σk1 1 1+

,

2 −k ¯ 2 )2 (k1 1 2 t) ¯ 2 x+¯ −2i(k η 1 1

¯1 e C

2σk1 2 ¯2 2 η 2 )t 2 |C |2 e2i(k1 −k1 )x+2i(η1 −¯ 1 4σk1 1 1+ 2 2 2 ¯ (k1 −k1 )

¯2 k12 −k 1

2

1

2 −k 2 ¯ 2 )x+2i(η 2 −¯ 2i(k1 1 1 η1 )t

2

2

(4.6)

.

2

2i(k1 x+η1 t) 2i(k1 x+η1 t) + C1 e k−k M12 (x, t, k1 ) − C1 e k+k M12 (x, t, k1 ) 1 1 2 x+η 2 t) 2 x+η 2 t) 2i(k1 2i(k1 1 1 C1 e M22 (x, t, k1 ) + C1 e k+k1 M22 (x, t, k1 ) k−k1

(4.7)

 ¯ M21 (x, t, σ k)  ¯ M11 (x, t, σ k)

This implies that M (x, t, 0) is a diagonal matrix with the (1, 1)−entry is 2

¯2

2

2

4σ|C1 |2 e2i(k1 −k1 )x+2i(η1 −¯η1 )t · M11 (x, t, 0) = 1− k12 − k¯12

1 1+

2 ¯2 2 η 2 )t 1 4σk12 |C1 |2 e2i(k1 −k1 )x+2i(η1 −¯ ¯2 )2 (k12 −k 1

(4.8) and a direct calculation shows that det M (x, t, 0) = 1. Then, computing the (1, 2)−entry of the matrix (M (x, t, 0)−1 )M (x, t, k), we get 2

M −1 (x, t, 0)M (x, t, k)




12

¯1 e−2i(k1 x+¯η1 t) M22 (x, t, k1 ) 1 + 2σ C =− ¯2 k 1 1+ ¯2

2

¯2

2

¯ 2 |C1 |2 e 4σ k 1

2 −k 2 ¯ 2 )x+2i(η 2 −η 2i(k1 1 1 ¯1 )t 2 2 2 ¯ (k1 −k1 )

(4.9)

If we choose k12 = ∆2 (−σ cos γ + i sin γ),

C1 = i∆(sin γ)ei

R∞

−∞

2

¯1 )t 2 4σk1 |C1 |2 e2i(k1 −k1 )x+2i(η1 −η 2 −k ¯ 2 )2 (k1 1

ux vx dx 2iΣ0 2∆2 x0 sin γ

e

e

for ∆ > 0, γ ∈ (0, π), x0 ∈ R, Σ0 ∈ R. And denoting   β2 θ(x, t) = ∆2 sin γ x − x0 + α(1 − 4∆ )t , 4   β2 2 Σ(x, t) = −αβt − σ∆ cos γ x + α(1 + 4∆4 )t + Σ0 ,

,

·k+O(k2 ),

k → 0.

RHP FOR THE FL EQUATION

9

a direct calculation shows that the solution u(x, t) given by (3.12) is u(x, t) = −

2i sin γ e−iγσ e−2iΣ(x,t) . ∆ e2θ(x,t) + e−2θ(x,t) eiγσ

(4.10)

This is coincide with the result (5.6) on P24 of [1]. It should be pointed out that the soliton solution (4.10) obtained in this paper corresponds to the solution obtained by one-fold Darboux transform with the zero seed solution in [9]. However, the most advantage of the Riemann-Hilbert method is to analyze the long-time asymptotic behavior of the solution. In fact, based on the RiemannHilbert problem obtained in [1], one of the authors get the leading order long-time asymptotic behavior by using the nonlinear steepest descent method in [11]. Acknowledgements This work was supported by National Science Foundation of China under project NO.11501365. References [1] J. Lenells and A. S. Fokas, On a novel integrable generalization of the nonlinear Schr¨odinger equation, Nonlinearity 22 (2009), 11-27. [2] A. S. Fokas, On a class of physically important integrable equations, Physica D 87(1995), 145-150. [3] J. Lenells, Exactly solvable model for nonlinear pulse propagation in optical fibers, Stud. Appl. Math.123 (2009), 215-232. [4] J. Lenells and A. S. Fokas, An integrable generalization of the nonlinear Schrodinger equation on the half-line and solitons, Inverse Problems25 (2009), 115006 (32pp). [5] J. Lenells, Dressing for a novel integrable generalization of the nonlinear Schr¨odinger equation, J. Nonlinear Sci.20 (2010), 709-722. [6] V. E. Vekslerchik, Lattice representation and dark solitons of the Fokas– Lenells equation, Nonlinearity 24 (2011), 1165. [7] Y. Matsuno, A direct method of solution for the Fokas-Lenells derivative nonlinear Schr¨odinger equation: I. Bright soliton solutions, J. Phys. A: Math. Theor. 45 (2012) 235202 (19pp). [8] Y. Matsuno, A direct method of solution for the Fokas-Lenells derivative nonlinear Schr¨odinger equation: I. Dark soliton solutions, J. Phys. A: Math. Theor. 45 (2012) 475202 (31pp).

10

L.AI AND J.XU

[9] J.He, S.Xu and K.Porsezian, Rogue waves of the FokasCLenells equation, J. Phys. Soc. Jpn. 81 (2012) 124007 (4pp). [10] S.Xu, J.He, Y.Cheng and K.Porsezian, The n-order rogue waves of FokasLenells equation, Mathematical methods in the applied sciences, 38 (2015) ,1106-1126. [11] J. Xu and E. Fan, Long-time asymptotics for the Fokas-Lenells equation with decaying initial value problem: Without solitons, Journal of differential equations 259 (2015), 1098-1148. College of Science, University of Shanghai for Science and Technology, Shanghai 200093, People’s Republic of China College of Science, University of Shanghai for Science and Technology, Shanghai 200093, People’s Republic of China E-mail address: corresponding author:

[email protected]