Quadratic-Argument Approach to Nonlinear Schrodinger Equation and Coupled Ones

Quadratic-Argu men t Approac h to Nonlinear Sc hr¨ odinger Equation and Co upled Ones 1 Xiaopi ng Xu Institute of Mathematics, Academ y of Mathematics & Sy stem Scie nces Chinese Academ y of Sciences, Beijing 100190, P .R. China 2 Abstract The t w o-dimensional cub ic nonlin ear S c hr¨ odinger equation is u sed to describ e the propa- gation of an intense laser b eam through a mediu m with Kerr nonlin earit y . The coupled t w o- dimensional cubic nonlinear Schr¨ o d in ger equations are used to d escrib e interact ion of el ectro- magnetic w a ves with differen t p olarizati ons in nonlinear optics. In this pap er, we so lve the ab o v e equations b y imp osing a qu adratic condition on the related argument f unctions and using their symmetry transformations. More complete families of exact solutions of suc h t yp e are obtained. Man y known in teresting solutions, suc h soliton ones, turn out to b e s p ecial cases of our solutions. 1 In tro duction The t w o-dimensional cubic nonlinear Sch r¨ odinger equation: iψ t + c ( ψ xx + ψ y y ) + a | ψ | 2 ψ = 0 (1 . 1) is used to describ e the propagation of a n in tense laser b eam through a medium with Kerr nonlinearity , where t is the distance in the direction of pro pa gation, x a nd y are the transv erse spacial co ordinates, ψ is a complex v alued function in t, x, y standing for electric field amplitude, and a, c are nonzero r eal constants. W e refer the in tro duction of [SEG] for more systematic exp osition of the equation. Akhnediev, Eleonskii a nd Kulagin [AEK] found certain exact solutions of (1.1) whose r eal and imaginary parts are linearly dep enden t o v er the functions of t . Moreo v er, Gagnon and Win ternitz [GW] found exact 1 2000 Mathematica l Sub ject Classificatio n. Prima ry 35C05, 35Q5 5; Secondary 37K 1 0. 2 Research suppo rted by C hina NSF 10 4310 40 1 solutions of the cubic and quin tic nonlinear Sc hr¨ odinger equation f or a cylindrical ge- ometry . Mihala che and P anoin [MN] used the metho d in [AEK] to obtain new solutions whic h describe the propa gation of dar k en v elop e soliton light pulses in optical fib ers in the normal group v elo city dispersion regime. F urthermore, Saied, EI-Rahman a nd Ghonam y [SEG] used v arious similarit y v ariables to reduce the ab o v e equation to certain ordinary differen tial equations and o btain some exact solutions. Ho w ev er, man y of their solutions are equiv alent to eac h other under the action of the kno wn symmetry transformations of the ab o ve equation. The coupled t w o-dimensional cubic nonlinear Schr¨ odinger equations iψ t + c 1 ( ψ xx + ψ y y ) + ( a 1 | ψ | 2 + b 1 | ϕ | 2 ) ψ = 0 , (1 . 2) iϕ t + c 2 ( ϕ xx + ϕ y y ) + ( a 2 | ψ | 2 + b 2 | ϕ | 2 ) ϕ = 0 (1 . 3) are used to describ e in teraction of electromagnetic wa v es with differen t p olarizations in nonlinear optics, where a 1 , a 2 , b 1 , b 2 , c 1 and c 2 are real constan ts. Ra dhakrishnan and Lak- shmanan [RL1] used Painlev ´ e a nalysis to find a Hir o ta bilinearization o f the ab o v e system of partia l differen tial equations and obtained bright and dark multiple soliton soutions. They also generalized their results to the coupled nonlinear Sc hr¨ o dinger equations with higher-order effects in [R L2]. G r ´ ebert and Guillot [GG] construc etd p erio dic solutions of coupled one-dimensional nonlinear Sc hr¨ odinger equations with p erio dic boundary condi- tions in some resonance situations. Moreo ve r, Hio e and Salter [HS] found a connections b et w een Lam ´ e functions and solutions of the ab ov e coupled equations. In terms of real-v alued functions, the ab ov e equations for m systems o f nonlinear partial differen tial equations. Such systems can not b e solv ed exactly without pre-assumptions. W e o bserv e that the argumen t functions of many known solutions for these equations are quadratic in the spacial v ariables x and y , in particular, those in [SEG]. Moreov er, some of these solutions are actually equiv alen t to eac h ot her unde r the Lie p oint symmetries of their corresp onding equations. These facts mo t iv ate us to solv e the ab o v e equations in this pap er b y imp osing the quadratic condition on the related argumen t functions and using their symmetry transformations. More complete families of explicit exact solutions of this t yp e with m ultiple para meter functions are obtained. Man y kno wn interes ting solutions, suc h solito n ones, turn out to b e sp ecial cases of our solutions. V arious singular solutions and p erio dic solutions that we obtain ma y refle ct some imp ortant phy sical phenomena in practical mo dels. Our solutions can also b e used to solv e some b oundary-v alue pro blems. Belo w w e giv e more details. 2 F or con v enience, w e alw ay s assume that all the in v olved partial deriv ativ es of related functions alwa ys exist and we can c hange orders of taking partia l deriv ativ es. W e also use prime ′ to denote the deriv ativ e of an y one-v ariable function. It is known that the equation (1.1) is inv arian t under the follo wing kno wn symmetric transformations: T 1 ( ψ ) = de d 3 i ψ ( d 2 t + d 2 , d ( x cos d 1 + y sin d 1 ) , d ( − x sin d 1 + y cos d 1 )) , (1 . 4) T 2 ( ψ ) = e [2( d 1 x + d 3 y ) − ( d 2 1 + d 2 3 ) t ] i/ 4 c ψ ( t, x − d 1 t + d 2 , y − d 3 t + d 4 ) , (1 . 5) where d, d 1 , d 2 , d 3 , d 4 ∈ R with d 6 = 0. In other w ords, the ab o ve tr a nsformations transform one solution of (1.1) in to another solution. Our solutions contain all the solutions in [SEG] up to the ab o v e transformations. In particular, our solutions with elliptic functions were not give n in [SEG]. Our approa ch is quite elemen tary and accessible to la rge audiences suc h a s phys icists and engineers. F or the reader’s con v enience, we list in this pap er all the solutions of t he equation (1 .1 ) found by our metho d although some o f them are kno wn and obvious . This may help non- ma t hematicians to apply the solutions of the Sch r¨ odinger equation to their fields. In f act, applying the transformat io ns in (1.4) and (1.5) to an y o f our solutions will yield more sophisticated one. Similarly , we ha v e the following known symmetric transformatio ns of the coupled equations (1.2) and (1.3): T 1 ( ψ ) = de d 3 i ψ ( d 2 t + d 2 , d ( x cos d 1 + y sin d 1 ) , d ( − x sin d 1 + y cos d 1 )) , (1 . 6) T 1 ( ϕ ) = d e d 4 i ϕ ( d 2 t + d 2 , d ( x cos d 1 + y sin d 1 ) , d ( − x sin d 1 + y cos d 1 )); (1 . 7) T 2 ( ψ ) = e [2( d 1 x + d 3 y ) − ( d 2 1 + d 2 3 ) t ] i/ 4 c 1 ψ ( t, x − d 1 t + d 2 , y − d 3 t + d 4 ) , (1 . 8) T 2 ( ϕ ) = e [2( d 1 x + d 3 y )+( d 2 1 + d 2 3 ) t ] i/ 4 c 2 ϕ ( t, x − d 1 t + d 2 , y − d 3 t + d 4 ); (1 . 9) where d, d 1 , d 2 , d 3 , d 4 ∈ R with d 6 = 0. In addition to the ab ov e sy mmetries, we also solv e the coupled equations modulo the following symmetry: ( ψ , a 1 , b 1 , c 1 ) ↔ ( ϕ , a 2 , b 2 , c 2 ) . (1 . 10) Again for the reader’s con v enience, w e list in this pap er all the solutions of the coupled equations (1 .2) and (1.3 ) found b y our metho d a lt ho ugh some of them are kno wn and ob vious. F or conv enienc e, w e alw a ys assume that all the in v olve d partial deriv ative s of related functions alw a ys exis t and we can change orders of taking partial deriv ativ es. W e also use prime ′ to denote the deriv ativ e of an y one-v ariable function. In Sec tion 2, we solv e the Sc hr¨ odinger equ atio n (1.1). In Section 3, w e use the results in Section 2 to solve the coupled Sc hr¨ odinger equations (1.2) and (1.3). 3 2 Exact S olution s of the Sc hr¨ odinge r Equation In this section, w e will presen t o ur quadratic-argumen t a ppro ac h to the tw o- dimensional cubic nonlinear Sc hr¨ odinger equation (1.1) and find more exact solutions than [SEG] in the mo dulo sense. W rite ψ = ξ ( t, x, y ) e iφ ( t,x,y ) , (2 . 1) where ξ and φ are real functions in t, x, y . Note ψ t = ( ξ t + iξ φ t ) e iφ , ψ x = ( ξ x + iξ φ x ) e iφ , ψ y = ( ξ y + iξ φ y ) e iφ , (2 . 2) ψ xx = ( ξ xx − ξ φ 2 x + i (2 ξ x φ x + ξ φ xx )) e iφ , ψ y y = ( ξ y y − ξ φ 2 y + i (2 ξ y φ y + ξ φ y y )) e iφ . (2 . 3) So the equation (1.1) b ecomes iξ t − φ t ξ + aξ 3 + c [ ξ xx + ξ y y − ξ ( φ 2 x + φ 2 y ) + i (2 ξ x φ x + 2 ξ y φ y + ξ ( φ xx + φ y y ))] = 0 , (2 . 4) equiv alen tly , ξ t + c (2 ξ x φ x + 2 ξ y φ y + ξ ( φ xx + φ y y )) = 0 , (2 . 5) − ξ [ φ t + c ( φ 2 x + φ 2 y )] + c ( ξ xx + ξ y y ) + aξ 3 = 0 . (2 . 6) Note that it is v ery difficult to solv e the a b ov e system without pre-assumptions. W e observ e that t he functions φ in all the solutions of [SEG] are quadratic in x and y . F rom the algebraic c haracteristics of the ab o ve system of partial differen tial equations, it is most affectiv e to assume that φ is quadratic in x and y . After sorting case by case, w e only ha v e the follo wing four cases tha t lead us to exact solutions, mo dulo the t r a nsformations in (1.4) and (1.5). Case 1 . φ = β ( t ) is a function of t . According to (2.5), ξ t = 0. Moreo v er, (2.6 ) becomes − β ′ ξ + c ( ξ xx + ξ y y ) + aξ 3 = 0 . (2 . 7) So w e tak e β = bt + d, b, d ∈ R . (2 . 8) If b = 0 and ac < 0, mo dulo the transformations (1.4 ), w e ta k e d = 0 and the following solutions: ξ = 1 x r − 2 c a or r − c a ( x 2 + y 2 ) . (2 . 9) 4 Next w e assume b 6 = 0. Mo dulo the transformation (1.4), w e can tak e d = 0. Note that (tan s ) ′ ′ = 2(tan 3 s + tan s ) , (sec s ) ′ ′ = 2 sec 3 s − sec s, (2 . 10) (coth s ) ′ ′ = 2(coth 3 s − coth s ) , (csc h s ) ′ ′ = 2csc h 3 s + csc h s . (2 . 11) Denote Jacobi elliptic functions sn s = sn ( s | m ) , cn s = cn ( s | m ) , dn s = dn ( s | m ) , (2 . 12) where m is t he elliptic mo dulus (e.g., cf. [W G]). Then (sn s ) ′ ′ = 2 m 2 sn 3 s − (1 + m 2 )sn s, (2 . 13) (cn s ) ′ ′ = − 2 m 2 cn 2 s + (2 m 2 − 1)cn s, (2 . 14) (dn s ) ′ ′ = − 2dn 3 s + (2 − m 2 )dn s. (2 . 15) Moreo v er, lim m → 1 sn s = tanh s, lim m → 1 cn s = lim m → 1 dn s = sec h s. (2 . 16) Consider solutions mo dulo the transformat ion (1.4). If ac < 0, we ha v e the follo wing solutions: ξ = r − 2 c a tan x, b = 2 c ; (2 . 17) ξ = r − 2 c a sec x, b = − c ; (2 . 18) ξ = r − 2 c a coth x, b = − 2 c ; (2 . 19) ξ = r − 2 c a csc h x, b = c ; (2 . 20) ξ = m r − 2 c a sn x, b = − (1 + m 2 ) c. (2 . 21) When ac > 0, w e get the follo wing solutions: ξ = m r 2 c a cn x, b = (2 m 2 − 1) c , ( 2 . 22) ξ = r 2 c a dn x, b = (2 − m 2 ) c. (2 . 23) Theorem 2.1 . L et m ∈ R . The fol lowi n g function ar e solutions ψ of the two-dimension al cubic n online ar cubic non line ar Schr¨ odinger e quation (1.1 ) : if ac < 0 , 1 x r − 2 c a , r − c a ( x 2 + y 2 ) , e 2 cti r − 2 c a tan x, e − cti r − 2 c a sec x, (2 . 24) 5 e − 2 cti r − 2 c a coth x, e cti r − 2 c a csc h x, me − (1+ m 2 ) cti r − 2 c a sn x ; (2 . 25) when ac > 0, me (2 m 2 − 1) cti r 2 c a cn x, e (2 − m 2 ) cti r 2 c a dn x. (2 . 26) Remark 2.2 . Although the ab ov e solution are simple, w e can obtain more sophisti- cated ones b y applying the transformations (1.4) and (1.5) to them. F or instance, applying the transformation (1.4) to the fir st solution in (2.24), w e get a solution: ψ = e d 2 i x cos d 1 + y sin d 1 r − 2 c a , d 1 , d 2 ∈ R . (2 . 27) Applying the transformation (1.5) to the ab o v e solution, w e obtain another solution: ψ = e [2( d 3 x + d 4 y )+( d 2 3 + d 2 4 ) t + d 2 ] i/ 4 c ( x − d 3 t ) cos d 1 + ( y − d 4 t ) sin d 1 + d 5 r − 2 c a , d 1 , d 2 , d 3 , d 4 , d 5 ∈ R . (2 . 28) Case 2 . φ = x 2 / 4 ct + β for some function β of t . In this case, (2.5) b ecomes ξ t + x t ξ x + 1 2 t ξ = 0 . (2 . 29) Th us ξ = 1 √ t ζ ( u, y ) , u = x t , (2 . 30) for some tw o-v ar iable function ζ . No w (2.6) b ecomes (2.7). Note ξ xx = t − 5 t/ 2 ζ uu , ξ y y = t − 1 / 2 ζ y y , ξ 3 = t − 3 / 2 ζ 3 . (2 . 31) So (2.7) b ecomes − β ′ √ t ζ + c ( t − 5 t/ 2 ζ uu + t − 1 / 2 ζ y y ) + at − 3 / 2 ζ 3 = 0 , (2 . 32) whose co efficien ts of t − 3 / 2 force us to take ξ = b √ t , b ∈ R . (2 . 33) No w (2.7) b ecomes − β ′ + ab 2 t = 0 = ⇒ β = ab 2 ln t (2 . 34) 6 mo dulo the transformation in (1.5). Case 3 . φ = x 2 / 4 ct + y 2 / 4 c ( t − d ) + β for some function β of t with 0 6 = d ∈ R . In this case, (2.5) b ecomes ξ t + x t ξ x + y t − d ξ y +  1 2 t + 1 2( t − d )  ξ = 0 . (2 . 35) So w e ha v e: ξ = 1 p t ( t − d ) ζ ( u, v ) , u = x t , v = y t − d , (2 . 36) for some tw o-v ar iable function ζ . Again (2.6) b ecomes (2.7). Note ξ xx = t − 5 / 2 ( t − c ) − 1 / 2 ζ uu , ξ y y = t − 1 / 2 ( t − c ) − 5 / 2 ζ vv , ξ 3 = t − 3 / 2 ( t − c ) − 3 / 2 ζ 3 . (2 . 37) So (2.7) b ecomes − β ′ p t ( t − d ) ζ + c ( t − 5 / 2 ( t − c ) − 1 / 2 ζ uu + t − 1 / 2 ( t − c ) − 5 / 2 ζ vv ) + at − 3 / 2 ( t − c ) − 3 / 2 ζ 3 = 0 , (2 . 38) whose co efficien ts of t − 3 / 2 ( t − c ) − 3 / 2 force us to take ξ = b p t ( t − d ) , b ∈ R . (2 . 39) No w (2.7) b ecomes − β ′ + ab 2 t ( t − d ) = 0 = ⇒ β = ab 2 d ln t − d t (2 . 40) mo dulo the transformation in (1.5). Theorem 2.3 . L et b, d ∈ R with d 6 = 0 . The fol lowing function ar e solutions ψ of the two-dimensional cubic nonline ar cubic no nline ar Schr¨ odinger e quation: bt ab 2 i − 1 / 2 e x 2 i/ 4 ct , bt − ab 2 i/d − 1 / 2 ( t − d ) ab 2 i/d − 1 / 2 e x 2 i/ 4 ct + y 2 i/ 4 c ( t − d ) . (2 . 41) Remark 2.4 . Applying (1.4) to the a b ov e first solution, w e get another solution ψ = dd 1 ( d 2 1 t + d 4 ) ad 2 i − 1 / 2 exp  d 2 1 ( x cos d 2 + y sin d 2 ) 2 4 c ( d 2 1 t + d 4 ) + d 3  i, (2 . 42) for d 1 , d 2 , d 3 , d 4 ∈ R . Moreo v er, we obtain a more sophisticated solution: ψ = dd 1 ( d 2 1 t + d 4 ) ad 2 i − 1 / 2 exp d 2 1 (( x − d 5 t ) cos d 2 + ( y − d 6 t ) sin d 2 + d 7 ) 2 i 4 c ( d 2 1 t + d 4 ) × exp  2( d 5 x + d 6 y ) + ( d 2 5 + d 2 6 ) t 4 c + d 3  i (2 . 43) 7 b y applying the transformation (1.5) to (2.42), where b r ∈ R . Case 4 . φ = ( x 2 + y 2 ) / 4 ct + β for some function β of t . Under our assumption, (2.5) becomes ξ t + x t ξ x + y t ξ y + 1 t ξ = 0 . (2 . 44) Th us w e hav e: ξ = 1 t ζ ( u, v ) , u = x t , v = y t , (2 . 45) for some tw o-v ar iable function ζ . Moreo v er, ( 2 .6) b ecomes − β ′ ζ + c t 2 ( ζ uu + ζ vv ) + a t 2 ζ 3 = 0 . (2 . 46) An ob vious solution is ζ = d , β = − ad 2 t , d ∈ R . (2 . 47) If ac < 0, w e hav e the simple following solutions with β = 0: ζ = 1 ℓ 1 u + ℓ 2 v + ℓ 3 r − 2 c ( ℓ 2 1 + ℓ 2 2 ) a or r − c a (( u − ℓ 1 ) 2 + ( v − ℓ 2 ) 2 ) (2 . 48) for ℓ 1 , ℓ 2 , ℓ 3 ∈ R . Next w e assume β ′ = b t 2 = ⇒ β = − b t (2 . 49) mo dulo the transformation in (1.2), where b is a real constan t to b e determined. Supp ose ζ = ℑ (  ) ,  = ℓ 1 u + ℓ 2 v + ℓ 3 (2 . 50) for ℓ 1 , ℓ 2 , ℓ 3 ∈ R suc h that ( ℓ 1 , ℓ 2 ) 6 = (0 , 0). Then (2.46) is equiv alen t to : − b ℑ + c ( ℓ 2 1 + ℓ 2 2 ) ℑ ′ ′ + a ℑ 3 = 0 . (2 . 51) According to (2.10), (2.1 1) and (2.13)-(2.15) , we ha v e the following solutions: If ac < 0 , w e ha ve the following solutions: ℑ = r − 2 c ( ℓ 2 1 + ℓ 2 2 ) a tan  , b = 2 c ( ℓ 2 1 + ℓ 2 2 ); (2 . 52) ℑ = r − 2 c ( ℓ 2 1 + ℓ 2 2 ) a sec  , b = − c ( ℓ 2 1 + ℓ 2 2 ); (2 . 53) ℑ = r − 2 c ( ℓ 2 1 + ℓ 2 2 ) a coth  , b = − 2 c ( ℓ 2 1 + ℓ 2 2 ); (2 . 54) 8 ℑ = r − 2 c ( ℓ 2 1 + ℓ 2 2 ) a csc h  , b = c ( ℓ 2 1 + ℓ 2 2 ); (2 . 55) ℑ = m r − 2 c ( ℓ 2 1 + ℓ 2 2 ) a sn  , b = − (1 + m 2 ) c ( ℓ 2 1 + ℓ 2 2 ) . (2 . 56) When ac > 0, w e get the follo wing solutions: ℑ = m r 2 c ( ℓ 2 1 + ℓ 2 2 ) a cn  , b = (2 m 2 − 1) c ( ℓ 2 1 + ℓ 2 2 ) , (2 . 57) ℑ = r 2 c ( ℓ 2 1 + ℓ 2 2 ) a dn  , b = (2 − m 2 ) c ( ℓ 2 1 + ℓ 2 2 ) . (2 . 58) Theorem 2.5 . L et ℓ 1 , ℓ 2 , ℓ 3 , ℓ 4 , ℓ 5 , m ∈ R such that ( ℓ 1 , ℓ 2 ) 6 = (0 , 0) . The fo l l o wing f unc - tions ar e solutions ψ of the two-dimensional cubic nonline a r cubic nonline ar Schr¨ odinger e quation: if ac < 0 , 1 ℓ 1 x + ℓ 2 y + ℓ 3 t r − 2 c ( ℓ 2 1 + ℓ 2 2 ) a e ( x 2 + y 2 ) i/ 4 ct , (2 . 59) r − c a (( x − ℓ 4 t ) 2 + ( y − ℓ 5 t ) 2 ) e ( x 2 + y 2 ) i/ 4 ct , (2 . 60) r − 2 c ( ℓ 2 1 + ℓ 2 2 ) a 1 t tan ℓ 1 x + ℓ 2 y + ℓ 3 t t exp  x 2 + y 2 4 ct − 2 c ( ℓ 2 1 + ℓ 2 2 ) t  i, (2 . 61) r − 2 c ( ℓ 2 1 + ℓ 2 2 ) a 1 t sec ℓ 1 x + ℓ 2 y + ℓ 3 t t exp  x 2 + y 2 4 ct + c ( ℓ 2 1 + ℓ 2 2 ) t  i, (2 . 62) r − 2 c ( ℓ 2 1 + ℓ 2 2 ) a 1 t coth ℓ 1 x + ℓ 2 y + ℓ 3 t t exp  x 2 + y 2 4 ct + 2 c ( ℓ 2 1 + ℓ 2 2 ) t  i, (2 . 63) r − 2 c ( ℓ 2 1 + ℓ 2 2 ) a 1 t csc h ℓ 1 x + ℓ 2 y + ℓ 3 t t exp  x 2 + y 2 4 ct − c ( ℓ 2 1 + ℓ 2 2 ) t  i, (2 . 64) m r − 2 c ( ℓ 2 1 + ℓ 2 2 ) a 1 t sn ℓ 1 x + ℓ 2 y + ℓ 3 t t exp  x 2 + y 2 4 ct + c (1 + m 2 )( ℓ 2 1 + ℓ 2 2 ) t  i ; (2 . 65) when ac > 0, m r 2 c ( ℓ 2 1 + ℓ 2 2 ) a 1 t cn ℓ 1 x + ℓ 2 y + ℓ 3 t t exp  x 2 + y 2 4 ct − c (2 m 2 − 1)( ℓ 2 1 + ℓ 2 2 ) t  i, (2 . 66) r 2 c ( ℓ 2 1 + ℓ 2 2 ) a 1 t dn ℓ 1 x + ℓ 2 y + ℓ 3 t t exp  x 2 + y 2 4 ct − c (2 − m 2 )( ℓ 2 1 + ℓ 2 2 ) t  i. (2 . 6 7) 9 Remark 2.6 . Applying the transformation (1 .5) to the solution (2 .59), we get another solution: ψ = e [2( d 1 x + d 3 y )+( d 2 1 + d 2 3 ) t ] i/ 4 c q − 2 c ( ℓ 2 1 + ℓ 2 2 ) a ℓ 1 ( x − d 1 t + d 2 ) + ℓ 2 ( y − d 3 t + d 4 ) + ℓ 3 t e (( x − d 1 t + d 2 ) 2 +( y − d 3 t + d 4 ) 2 ) i/ 4 ct , (2 . 68) where d 1 , d 2 , d 3 , d 4 ∈ R . 3 Exact S olution s of the Coupl ed Equ ations In this section, w e will use our r esults in previous section to find exact solutio ns of the coupled t w o-dimensional cubic nonlinear Schr¨ odinger equations (1.2) and (1.3). W rite ψ = ξ ( t, x, y ) e iφ ( t,x,y ) , ϕ = η ( t, x, y ) e iµ ( t,x,y ) (3 . 1) where ξ , φ, η and µ are real functions in t, x, y . As the arg uments in (2.1)-(2 .6 ), the system (1.2) and (1.3) is equiv alen t to the follo wing system for real functions: ξ t + c 1 (2 ξ x φ x + 2 ξ y φ y + ξ ( φ xx + φ y y )) = 0 , (3 . 2) − ξ [ φ t + c 1 ( φ 2 x + φ 2 y )] + c 1 ( ξ xx + ξ y y ) + ( a 1 ξ 2 + b 1 η 2 ) ξ = 0 , (3 . 3) η t + c 2 (2 η x µ x + 2 η y µ y + η ( µ xx + µ y y )) = 0 , (3 . 4) − η [ µ t + c 2 ( µ 2 x + µ 2 y )] + c 2 ( η xx + η y y ) + ( a 2 ξ 2 + b 2 η 2 ) η = 0 . (3 . 5) Based on o ur exp erience in last sec tion, w e will solv e the ab ov e system according to the follo wing cases. F or the con v enience, we alwa ys assume the conditions on the constants in v olv ed in an expre ssion suc h that it make sense. F o r instance, when w e use √ d 1 − d 2 , w e naturally assume d 1 ≥ d 2 . Case 1 . ( φ, µ ) = (0 , 0) and a 1 b 2 − a 2 b 1 6 = 0. In this case, ξ t = η t = 0 b y (3.2) and (3.4). Moreov er, (3.3) and (3.5) b ecome c 1 ( ξ xx + ξ y y ) + ( a 1 ξ 2 + b 1 η 2 ) ξ = 0 , c 2 ( η xx + η y y ) + ( a 2 ξ 2 + b 2 η 2 ) η = 0 . (3 . 6 ) Assume ξ = ι 1 x , η = ι 2 x . (3 . 7) Then (3.6) is eq uiv alen t to: a 1 ι 2 1 + b 1 ι 2 2 + 2 c 1 = 0 , a 2 ι 2 + b 2 ι 2 + 2 c 2 = 0 . (3 . 8) 10 Solving the ab o v e linear algebraic equations for ι 2 1 and ι 2 2 , w e ha v e: ι 2 1 = 2( b 1 c 2 − b 2 c 1 ) a 1 b 2 − a 2 b 1 , ι 2 2 = 2( a 2 c 1 − a 1 c 2 ) a 1 b 2 − a 2 b 1 . (3 . 9) Th us w e hav e t he follo wing solution ξ = ǫ 1 x s 2( b 1 c 2 − b 2 c 1 ) a 1 b 2 − a 2 b 1 , η = ǫ 2 x s 2( a 2 c 1 − a 1 c 2 ) a 1 b 2 − a 2 b 1 (3 . 10) for ǫ 1 , ǫ 2 ∈ { 1 , − 1 } . Similarly , we hav e the solution: ξ = ǫ 1 s b 1 c 2 − b 2 c 1 ( a 1 b 2 − a 2 b 1 )( x 2 + y 2 ) , η = ǫ 2 r a 2 c 1 − a 1 c 2 ( a 1 b 2 − a 2 b 1 )( x 2 + y 2 ) . (3 . 11) Case 2 . ( φ, µ ) = ( k 1 t, k 2 t ) with k 1 , k 2 ∈ R . Again w e ha v e ξ t = η t = 0 b y (3.2) and (3.4). Moreov er, (3.3) and (3.5) b ecome − k 1 ξ + c 1 ( ξ xx + ξ y y ) + ( a 1 ξ 2 + b 1 η 2 ) ξ = 0 , − k 2 η + c 2 ( η xx + η y y ) + ( a 2 ξ 2 + b 2 η 2 ) η = 0 . (3 . 1 2) First w e assume a 1 b 2 − a 2 b 1 6 = 0 and ξ = ι 1 ℑ ( x ) , η = ι 2 ℑ ( x ) . (3 . 13) Then (3.12) b ecomes − k 1 ℑ + c 1 ℑ ′ ′ + ( a 1 ι 2 1 + b 1 ι 2 2 ) ℑ 3 = 0 , − k 2 ℑ + c 2 ℑ ′ ′ + ( a 2 ι 2 + b 2 ι 2 ) ℑ 3 = 0 . (3 . 14) According to (2.10) and (2 .11), when ℑ = tan x, sec x, coth x and csc h x , w e alw ays hav e a 1 ι 2 1 + b 1 ι 2 2 + 2 c 1 = 0 , a 2 ι 2 + b 2 ι 2 + 2 c 2 = 0 . (3 . 15) Th us for ǫ 1 , ǫ 2 ∈ { 1 , − 1 } , w e hav e the following solutions: ξ = ǫ 1 s 2( b 1 c 2 − b 2 c 1 ) a 1 b 2 − a 2 b 1 tan x, η = ǫ 2 s 2( a 2 c 1 − a 1 c 2 ) a 1 b 2 − a 2 b 1 tan x, ( k 1 , k 2 ) = 2( c 1 , c 2 ); (3 . 16) ξ = ǫ 1 s 2( b 1 c 2 − b 2 c 1 ) a 1 b 2 − a 2 b 1 sec x, η = ǫ 2 s 2( a 2 c 1 − a 1 c 2 ) a 1 b 2 − a 2 b 1 sec x, ( k 1 , k 2 ) = − ( c 1 , c 2 ); (3 . 17) ξ = ǫ 1 s 2( b 1 c 2 − b 2 c 1 ) a 1 b 2 − a 2 b 1 coth x, η = ǫ 2 s 2( a 2 c 1 − a 1 c 2 ) a 1 b 2 − a 2 b 1 coth x (3 . 18) 11 and ( k 1 , k 2 ) = − 2( c 1 , c 2 ); ξ = ǫ 1 s 2( b 1 c 2 − b 2 c 1 ) a 1 b 2 − a 2 b 1 csc h x, η = ǫ 2 s 2( a 2 c 1 − a 1 c 2 ) a 1 b 2 − a 2 b 1 csc h x, ( k 1 , k 2 ) = ( c 1 , c 2 ) . (3 . 19) Similarly , (2.1 3)-(2.15) giv e us the f ollo wing solutions: ξ = mǫ 1 s 2( b 1 c 2 − b 2 c 1 ) a 1 b 2 − a 2 b 1 sn x, η = mǫ 2 s 2( a 2 c 1 − a 1 c 2 ) a 1 b 2 − a 2 b 1 sn x (3 . 20) and ( k 1 , k 2 ) = − (1 + m 2 )( c 1 , c 2 ); ξ = mǫ 1 s 2( b 2 c 1 − b 1 c 2 ) a 1 b 2 − a 2 b 1 cn x η = mǫ 2 s 2( a 1 c 2 − a 2 c 1 ) a 1 b 2 − a 2 b 1 cn x (3 . 21) and ( k 1 , k 2 ) = (2 m 2 − 1)( c 1 , c 2 ); ξ = ǫ 1 s 2( b 2 c 1 − b 1 c 2 ) a 1 b 2 − a 2 b 1 dn x, η = ǫ 2 s 2( a 1 c 2 − a 2 c 1 ) a 1 b 2 − a 2 b 1 dn x (3 . 22) and ( k 1 , k 2 ) = (2 − m 2 )( c 1 , c 2 ) . If ( a 1 , b 1 ) = a 1 (1 , d 2 ) and ( a 2 , b 2 ) = a 2 (1 , d 2 ) with d ∈ R , w e hav e the follo wing solution of (3.12): ξ = dℓ sin x, η = ℓ cos x, ( k 1 , k 2 ) = ( a 1 ( dℓ ) 2 − c 1 , a 2 ( dℓ ) 2 − c 2 ) (3 . 23) for ℓ ∈ R . Wh en ( a 1 , b 1 ) = a 1 (1 , − d 2 ) and ( a 2 , b 2 ) = a 2 (1 , − d 2 ) with d ∈ R , w e get the solution: ξ = dℓ cosh x, η = ℓ sinh x, ( k 1 , k 2 ) = ( a 1 ( dℓ ) 2 + c 1 , a 2 ( dℓ ) 2 + c 2 ) . (3 . 24) In summary , w e ha v e the following theorem. Theorem 3.1 . L et d, ℓ, m ∈ R and let ǫ 1 , ǫ 2 ∈ { 1 , − 1 } . If a 1 b 2 − a 2 b 1 6 = 0 , we h ave the fol lowing solutions of the c ouple d two-dimensional cubic nonline ar Schr¨ oding e r e quations (1.2) and (1.3): ψ = ǫ 1 x s 2( b 1 c 2 − b 2 c 1 ) a 1 b 2 − a 2 b 1 , ϕ = ǫ 2 x s 2( a 2 c 1 − a 1 c 2 ) a 1 b 2 − a 2 b 1 ; (3 . 25) ψ = ǫ 1 s b 1 c 2 − b 2 c 1 ( a 1 b 2 − a 2 b 1 )( x 2 + y 2 ) , ϕ = ǫ 2 r a 2 c 1 − a 1 c 2 ( a 1 b 2 − a 2 b 1 )( x 2 + y 2 ) ; (3 . 26) ψ = ǫ 1 s 2( b 1 c 2 − b 2 c 1 ) a 1 b 2 − a 2 b 1 e 2 c 1 ti tan x, ϕ = ǫ 2 s 2( a 2 c 1 − a 1 c 2 ) a 1 b 2 − a 2 b 1 e 2 c 2 ti tan x ; (3 . 27) 12 ψ = ǫ 1 s 2( b 1 c 2 − b 2 c 1 ) a 1 b 2 − a 2 b 1 e − c 1 ti sec x, ϕ = ǫ 2 s 2( a 2 c 1 − a 1 c 2 ) a 1 b 2 − a 2 b 1 e − c 2 ti sec x ; (3 . 28) ψ = ǫ 1 s 2( b 1 c 2 − b 2 c 1 ) a 1 b 2 − a 2 b 1 e − 2 c 1 ti coth x, ϕ = ǫ 2 s 2( a 2 c 1 − a 1 c 2 ) a 1 b 2 − a 2 b 1 e − 2 c 2 ti coth x ; (3 . 29) ψ = ǫ 1 s 2( b 1 c 2 − b 2 c 1 ) a 1 b 2 − a 2 b 1 e c 1 ti csc h x, ϕ = ǫ 2 s 2( a 2 c 1 − a 1 c 2 ) a 1 b 2 − a 2 b 1 e c 2 ti csc h x ; (3 . 30 ) ψ = mǫ 1 s 2( b 1 c 2 − b 2 c 1 ) a 1 b 2 − a 2 b 1 e − (1+ m 2 ) c 1 ti sn x, (3 . 31) ϕ = mǫ 2 s 2( a 2 c 1 − a 1 c 2 ) a 1 b 2 − a 2 b 1 e − (1+ m 2 ) c 2 ti sn x ; (3 . 32) ψ = mǫ 1 s 2( b 2 c 1 − b 1 c 2 ) a 1 b 2 − a 2 b 1 e (2 m 2 − 1) c 1 ti cn x, (3 . 33) ϕ = mǫ 2 s 2( a 1 c 2 − a 2 c 1 ) a 1 b 2 − a 2 b 1 e (2 m 2 − 1) c 2 ti cn x ; (3 . 34) ψ = ǫ 1 s 2( b 2 c 1 − b 1 c 2 ) a 1 b 2 − a 2 b 1 e (2 − m 2 ) c 1 ti dn x, ϕ = ǫ 2 s 2( a 1 c 2 − a 2 c 1 ) a 1 b 2 − a 2 b 1 e (2 − m 2 ) c 1 ti dn x. (3 . 35) If ( a 1 , b 1 ) = a 1 (1 , d 2 ) an d ( a 2 , b 2 ) = a 2 (1 , d 2 ), ψ = dℓe ( a 1 ( dℓ ) 2 − c 1 ) ti sin x, ϕ = ℓ e ( a 2 ( dℓ ) 2 − c 2 ) ti cos x. (3 . 36) When ( a 1 , b 1 ) = a 1 (1 , − d 2 ) and ( a 2 , b 2 ) = a 2 (1 , − d 2 ), ψ = dℓe ( a 1 ( dℓ ) 2 + c 1 ) ti cosh x, η = ℓe ( a 2 ( dℓ ) 2 + c 2 ) ti sinh x. (3 . 37) Remark 3.2 . Applying the symmetric transformations (1.6)-(1.9) to the ab ov e solu- tions, w e can get more sophisticated ones. F or instance, applying T 1 in (1.6) and (1.7) to (3.25) and (3.37), w e get ψ = ǫ 1 e d 2 i x cos d 1 + y sin d 1 s 2( b 1 c 2 − b 2 c 1 ) a 1 b 2 − a 2 b 1 , ϕ = ǫ 2 e d 3 i x cos d 1 + y sin d 1 s 2( a 2 c 1 − a 1 c 2 ) a 1 b 2 − a 2 b 1 ; (3 . 38) and ψ = dd 2 ℓe [( a 1 ( dℓ ) 2 + c 1 ) d 2 2 t + d 3 ] i cosh d 2 ( x cos d 1 + y sin d 1 ) , (3 . 39) η = d 2 ℓe [( a 2 ( dℓ ) 2 + c 2 ) d 2 2 t + d 4 ] i sinh d 2 ( x cos d 1 + y sin d 1 ) . (3 . 40) 13 Applying T 2 in (1.8) and (1 .9 ) to (3.26), w e obtain: ψ = ǫ 1 e [2( d 1 x + d 3 y )+( d 2 1 + d 2 3 ) t ] i/ 4 c 1 s b 1 c 2 − b 2 c 1 ( a 1 b 2 − a 2 b 1 )(( x − d 1 t + d 2 ) 2 + ( y − d 2 t + d 4 ) 2 ) , (3 . 41) ϕ = ǫ 2 e [2( d 1 x + d 3 y )+( d 2 1 + d 2 3 ) t ] i/ 4 c 2 r a 2 c 1 − a 1 c 2 ( a 1 b 2 − a 2 b 1 )(( x − d 1 t + d 2 ) 2 + ( y − d 2 t + d 4 ) 2 ) . (3 . 42 ) Case 3 . φ = x 2 / 4 c 1 t + β 1 and µ = ( x − d ) 2 / 4 c 2 ( t − ℓ ) + β 2 or µ = y 2 / 4 c 2 ( t − ℓ ) + β 2 for some f unctions β 1 and β 2 of t and real constan ts d and ℓ . First w e assume µ = ( x − d ) 2 / 4 c 2 ( t − ℓ ) + β 2 . Then (3 .2) and (3.4) become ξ t + x t ξ x + 1 2 t ξ = 0 , η t + x − d t − ℓ η x + 1 2( t − ℓ ) η = 0 . (3 . 43) Th us ξ = 1 √ t ˆ ξ ( t − 1 x, y ) , η = 1 √ t − ℓ ˆ η (( t − ℓ ) − 1 ( x − d ) , y ) (3 . 44) for some tw o-v ar iable functions ˆ ξ and ˆ η . On the other hand, (3.3) and ( 3 .5) b ecome: − β ′ 1 ξ + c 1 ( ξ xx + ξ y y ) + ( a 1 ξ 2 + b 1 η 2 ) ξ = 0 , (3 . 45) − β ′ 2 η + c 2 ( η xx + η y y ) + ( a 2 ξ 2 + b 2 η 2 ) η = 0 . (3 . 46) As (2.31)-(2.33), the ab o v e tw o equations force us to tak e ξ = k 1 √ t , η = k 2 √ t − ℓ . (3 . 47) So (3.45) and (3.46) are implied b y the equations: β ′ 1 = a 1 k 2 1 t + b 1 k 2 2 t − ℓ , β ′ 2 = a 2 k 2 1 t + b 2 k 2 2 t − ℓ . (3 . 48) Mo dulo the transformation (1.6) and (1.7), w e tak e β 1 = a 1 k 2 1 ln t + b 1 k 2 2 ln( t − ℓ ) , β 2 = a 2 k 2 1 ln t + b 2 k 2 2 ln( t − ℓ ) . (3 . 49) Exact same approac h holds for µ = y 2 / 4 c 2 ( t − ℓ ) + β 2 . Theorem 3.3 . L et d, ℓ, k 1 , k 2 ∈ R . We have the fol lowing solutions of the c ouple d two-dimensional cubic nonline ar Schr¨ odinge r e quations (1.2) and (1. 3 ) : ψ = k 1 t a 1 k 2 1 i − 1 / 2 ( t − ℓ ) b 1 k 2 2 i e x 2 i/ 2 c 1 t , ϕ = k 2 t a 2 k 2 1 i ( t − ℓ ) b 2 k 2 2 i − 1 / 2 e ( x − d ) 2 i/ 2 c 2 ( t − ℓ ) ; (3 . 50) 14 ψ = k 1 t a 1 k 2 1 i − 1 / 2 ( t − ℓ ) b 1 k 2 2 i e x 2 i/ 2 c 1 t , ϕ = k 2 t a 2 k 2 1 i ( t − ℓ ) b 2 k 2 2 i − 1 / 2 e y 2 i/ 2 c 2 ( t − ℓ ) . (3 . 51) Case 4 . φ = x 2 / 4 c 1 t + β 1 and µ = ( x − d ) 2 / 4 c 2 ( t − d 1 ) + y 2 / 4 c 2 ( t − d 2 ) + β 2 for some functions β 1 and β 2 of t and real constan ts d , d 1 and d 2 . In this case, (3.2) a nd (3.4 ) become ξ t + x t ξ x + 1 2 t ξ = 0 , η t + x − d t − d 1 η x + y t − d 2 η y +  1 2( t − d 1 ) + 1 2( t − d 2 )  ξ = 0 . (3 . 52) Th us ξ = 1 √ t ˆ ξ ( t − 1 x, y ) , η = 1 p ( t − d 1 )( t − d 2 ) ˆ η (( t − d 1 ) − 1 ( x − d ) , ( t − d 2 ) − 1 y ) (3 . 53) for some t w o-v ariable functions ˆ ξ and ˆ η . Again (3.3) and (3.5) b ecome (3 .45) and (3.46), resp ectiv ely . Moreov er, they force us t o take ξ = k 1 √ t , η = k 2 p ( t − d 1 )( t − d 2 ) . (3 . 54) So (3.3) and ( 3 .5) are implied b y the equations: β ′ 1 = a 1 k 2 1 t + b 1 k 2 2 ( t − d 1 )( t − d 2 ) , β ′ 2 = a 2 k 2 1 t + b 2 k 2 2 ( t − d 1 )( t − d 2 ) . (3 . 55) Mo dulo the transformation (1.6) and (1.7), w e get β 1 = a 1 k 2 1 ln t + b 1 k 2 2 d 2 − d 1 ln t − d 1 t − d 2 , β 2 = a 2 k 2 1 ln t + b 2 k 2 2 d 2 − d 1 ln t − d 1 t − d 2 (3 . 56) if d 1 6 = d 2 , and β 1 = a 1 k 2 1 ln t − b 1 k 2 2 t − d 1 , , β 2 = a 2 k 2 1 ln t − b 2 k 2 2 t − d 1 (3 . 57) when d 1 = d 2 . Theorem 3.4 . L et d 1 , d 2 , k 1 , k 2 ∈ R such that d 1 6 = d 2 . We ha v e the fol lo w ing solutions of the c ouple d two-dimensio n al cubic nonline ar Schr¨ odinger e quations (1.2) an d (1.3) : ψ = k 1 t a 1 k 2 1 i − 1 / 2 ( t − d 1 ) b 1 k 2 2 ( ℓ 2 − d 1 ) − 1 i ( t − d 2 ) − b 1 k 2 2 ( ℓ 2 − d 1 ) − 1 i e x 2 i/ 4 c 1 t , (3 . 58) ϕ = k 2 t a 2 k 2 1 i ( t − d 1 ) b 1 k 2 2 ( ℓ 2 − d 1 ) − 1 i − 1 / 2 ( t − d 2 ) − b 1 k 2 2 ( ℓ 2 − d 1 ) − 1 i − 1 / 2 × exp  ( x − d ) 2 i 4 c 2 ( t − d 1 ) + y 2 i 4 c 2 ( t − d 1 )  ; (3 . 59) 15 ψ = k 1 t a 1 k 2 1 i − 1 / 2 exp  x 2 i 4 c 1 t − b 1 k 2 2 i t − d 1  , ( 3 . 60) ϕ = k 2 t a 2 k 2 1 i t − d 1 exp (( x − d ) 2 + y 2 − 4 c 2 b 2 k 2 2 ) i 4 c 2 ( t − d 1 ) . (3 . 61) Case 5 . F or ℓ 1 , ℓ 2 , ℓ, d 1 , d 2 ∈ R and functions β 1 , β 2 of t , φ = x 2 4 c 1 t + y 2 4 c 1 ( t − ℓ ) + β 1 , µ = ( x − d 1 ) 2 4 c 2 ( t − ℓ 1 ) + ( y − d 2 ) 2 4 c 1 ( t − ℓ 2 ) + β 2 . (3 . 62) As in the a b o v e case, w e get ξ = k 1 p t ( t − ℓ ) , η = k 2 p ( t − ℓ 1 )( t − ℓ 2 ) . (3 . 6 3) So (3.3) and ( 3 .5) are implied b y the equations: β ′ 1 = a 1 k 2 1 t ( t − ℓ ) + b 1 k 2 2 ( t − ℓ 1 )( t − ℓ 2 ) , β ′ 2 = a 2 k 2 1 t + b 2 k 2 2 ( t − ℓ 1 )( t − ℓ 2 ) . (3 . 64) Mo dulo the transformation (1.6) and (1.7), w e ha v e β 1 = a 1 k 2 1 ℓ ln t − ℓ t + b 1 k 2 2 ℓ 2 − ℓ 1 ln t − ℓ 1 t − ℓ 2 , β 2 = a 2 k 2 1 ℓ ln t − ℓ t + b 2 k 2 2 ℓ 2 − ℓ 1 ln t − ℓ 1 t − ℓ 2 (3 . 65) if ℓ 6 = 0 and ℓ 1 6 = ℓ 2 ; β 1 = − a 1 k 2 1 t + b 1 k 2 2 ℓ 2 − ℓ 1 ln t − ℓ 1 t − ℓ 2 , β 2 = − a 2 k 2 1 t + b 2 k 2 2 ℓ 2 − ℓ 1 ln t − ℓ 1 t − ℓ 2 (3 . 66) when ℓ = 0 and ℓ 1 6 = ℓ 2 ; β 1 = a 1 k 2 1 t − b 1 k 2 2 t − ℓ 1 , , β 2 = a 2 k 2 1 t t − b 2 k 2 2 t − ℓ 1 (3 . 67) if ℓ = 0 and ℓ 1 = ℓ 2 . Therefore, w e obtain: Theorem 3.5 . L et ℓ 1 , ℓ 2 , ℓ, d 1 , d 2 , k 1 , k 2 ∈ R such that ℓ 6 = 0 and ℓ 1 6 = ℓ 2 . We have the fol lowing solutions of the c ouple d two-dimensional cubic nonline ar Schr¨ oding e r e quations (1.2) and (1.3) : ψ = k 1 t exp  ( x 2 + y 2 − 4 c 1 a 1 k 2 1 ) i 4 c 1 t − b 1 k 2 2 i t − ℓ 1  , (3 . 68) ϕ = k 2 t − ℓ 1 exp  (( x − d 1 ) 2 + ( y − d 2 ) 2 − 4 c 2 b 2 k 2 2 ) i 4 c 2 ( t − ℓ 1 ) − a 2 k 2 1 i t  ; (3 . 69) ψ = k 1 ( t − ℓ 1 ) b 1 k 2 2 ( ℓ 2 − ℓ 1 ) − 1 i ( t − ℓ 2 ) − b 1 k 2 2 ( ℓ 2 − ℓ 1 ) − 1 i t exp ( x 2 + y 2 − 4 c 1 a 1 k 2 1 ) i 4 c 1 t , (3 . 70) 16 ϕ = k 2 ( t − ℓ 1 ) b 2 k 2 2 ( ℓ 2 − ℓ 1 ) − 1 i − 1 / 2 ( t − ℓ 2 ) − b 2 k 2 2 ( ℓ 2 − ℓ 1 ) − 1 i − 1 / 2 × exp  ( x − d 1 ) 2 i 4 c 2 ( t − ℓ 1 ) + ( y − d 2 ) 2 i 4 c 2 ( t − ℓ 2 ) − a 2 k 2 1 i t  ; (3 . 71) ψ = k 1 t − a 1 k 2 1 ℓ − 1 i − 1 / 2 ( t − ℓ ) a 1 k 2 1 ℓ − 1 i − 1 / 2 ( t − ℓ 1 ) b 1 k 2 2 ( ℓ 2 − ℓ 1 ) − 1 i × ( t − ℓ 2 ) − b 1 k 2 2 ( ℓ 2 − ℓ 1 ) − 1 i exp  x 2 i 4 c 1 t + y 2 i 4 c 1 ( t − ℓ )  , (3 . 72) ϕ = k 2 t − a 2 k 2 1 ℓ − 1 i ( t − ℓ ) a 2 k 2 1 ℓ − 1 i ( t − ℓ 1 ) b 2 k 2 2 ( ℓ 2 − ℓ 1 ) − 1 i − 1 / 2 × ( t − ℓ 2 ) − b 2 k 2 2 ( ℓ 2 − ℓ 1 ) − 1 i − 1 / 2 exp  ( x − d 1 ) 2 i 4 c 2 ( t − ℓ 1 ) + ( y − d 2 ) 2 i 4 c 2 ( t − ℓ 2 )  . (3 . 73) Case 6 . F or t w o functions β 1 , β 2 of t , φ = x 2 + y 2 4 c 1 t + β 1 , µ = x 2 + y 2 4 c 2 t + β 2 . (3 . 74) As in Case 4, (3.2) and (3.4) imply ξ = 1 t ˆ ξ ( u, v ) , η = 1 t ˆ η ( u , v ) , u = x t , v = y t . (3 . 75) Moreo v er, (3.3 ) and (3.5) b ecome − β ′ 1 ˆ ξ + c 1 t 2 ( ˆ ξ uu + ˆ ξ vv ) + 1 t 2 ( a 1 ˆ ξ 2 + b 1 ˆ η 2 ) ˆ ξ = 0 , (3 . 76) − β ′ 2 ˆ η + c 2 t 2 ( ˆ η uu + ˆ η vv ) + 1 t 2 ( a 2 ˆ ξ 2 + b 2 ˆ η 2 ) ˆ η = 0 . (3 . 77) T o solv e the ab ov e sy stem, w e assume β 1 = − k 1 t , β 2 = − k 2 t , k 1 , k 2 ∈ R . (3 . 78) Then (3.77) and ( 3.78) are equiv alen t to : − k 1 ˆ ξ + c 1 ( ˆ ξ uu + ˆ ξ vv ) + ( a 1 ˆ ξ 2 + b 1 ˆ η 2 ) ˆ ξ = 0 , (3 . 79) − k 2 ˆ η + c 2 ( ˆ η uu + ˆ η vv ) + ( a 2 ˆ ξ 2 + b 2 ˆ η 2 ) ˆ η = 0 . (3 . 80) F or ℓ 1 , ℓ 2 , ℓ 3 ∈ R , w e set  = ℓ 1 u + ℓ 2 v + ℓ 3 . (3 . 81) If ( a 1 , b 1 ) = a 1 (1 , d 2 ) and ( a 2 , b 2 ) = a 2 (1 , d 2 ) with d ∈ R , w e hav e t he following solution: ˆ ξ = dℓ sin  , ˆ η = ℓ cos  , ( k 1 , k 2 ) = ( a 1 ( dℓ ) 2 − c 1 ( ℓ 2 1 + ℓ 2 2 ) , a 2 ( dℓ ) 2 − c 2 ( ℓ 2 1 + ℓ 2 2 )) (3 . 8 2) 17 for ℓ ∈ R . Wh en ( a 1 , b 1 ) = a 1 (1 , − d 2 ) and ( a 2 , b 2 ) = a 2 (1 , − d 2 ) with d ∈ R , w e get the solution: ˆ ξ = dℓ cosh  , ˆ η = ℓ sinh  , ( k 1 , k 2 ) = ( c 1 ( ℓ 2 1 + ℓ 2 2 ) + a 1 ( dℓ ) 2 , c 2 ( ℓ 2 1 + ℓ 2 2 ) + a 2 ( dℓ ) 2 ) (3 . 83) for ℓ ∈ R . Theorem 3.6 . F or d, ℓ, ℓ 1 , ℓ 2 , ℓ 3 ∈ R , we have the fol lowing solutions of the c ouple d two-dimensional cubic nonline ar Schr¨ odinge r e quations (1.2) and (1. 3 ) : ψ = dℓ sin ℓ 1 x + ℓ 2 y + ℓ 3 t t t exp  x 2 + y 2 4 c 1 t + c 1 ( ℓ 2 1 + ℓ 2 2 ) − a 1 ( dℓ ) 2 t  i, (3 . 84) ϕ = ℓ cos ℓ 1 x + ℓ 2 y + ℓ 3 t t t exp  x 2 + y 2 4 c 2 t + c 2 ( ℓ 2 1 + ℓ 2 2 ) − a 2 ( dℓ ) 2 t  i (3 . 85) if ( a 1 , b 1 ) = a 1 (1 , d 2 ) an d ( a 2 , b 2 ) = a 2 (1 , d 2 ); ψ = dℓ cosh ℓ 1 x + ℓ 2 y + ℓ 3 t t t exp  x 2 + y 2 4 c 1 t − c 1 ( ℓ 2 1 + ℓ 2 2 ) + a 1 ( dℓ ) 2 t  i, (3 . 86) ϕ = ℓ sinh ℓ 1 x + ℓ 2 y + ℓ 3 t t t exp  x 2 + y 2 4 c 2 t − c 2 ( ℓ 2 1 + ℓ 2 2 ) + a 2 ( dℓ ) 2 t  i (3 . 87) when ( a 1 , b 1 ) = a 1 (1 , − d 2 ) an d ( a 2 , b 2 ) = a 2 (1 , − d 2 ). Finally , w e assume a 1 b 2 − a 2 b 1 6 = 0. Recall the not io n in (3.81). T aking k 1 = k 2 = 0, the w e ha v e the following solutions of the system of (3.79) and (3.8 0): ˆ ξ = 1  s 2( b 1 c 2 − b 2 c 1 )( ℓ 2 1 + ℓ 2 2 ) a 1 b 2 − a 2 b 1 , ˆ η = 1  s 2( a 2 c 1 − a 1 c 2 )( ℓ 2 1 + ℓ 2 2 ) a 1 b 2 − a 2 b 1 (3 . 88) and ˆ ξ = s b 1 c 2 − b 2 c 1 ( a 1 b 2 − a 2 b 1 )(( u − d 1 ) 2 + ( v − d 2 ) 2 ) , (3 . 89) ˆ η = r a 2 c 1 − a 1 c 2 ( a 1 b 2 − a 2 b 1 )(( u − d 1 ) 2 + ( v − d 2 ) 2 ) (3 . 90) for ℓ 1 , ℓ 2 ∈ R . In general, w e assume ˆ ξ = ι 1 ℑ (  ) , ˆ η = ι 2 ℑ (  ) (3 . 91) for some o ne-v ar iable function ℑ and k 1 , k 2 ∈ R . Then (3.79) and (3.80) implied by − k 1 ℑ + c 1 ( ℓ 2 1 + ℓ 2 2 ) ℑ ′ ′ + ( a 1 ι 2 1 + b 1 ι 2 2 ) ℑ 3 = 0 , (3 . 92) − k 2 ℑ + c 2 ( ℓ 2 1 + ℓ 2 2 ) ℑ ′ ′ + ( a 2 ι 2 1 + b 2 ι 2 2 ) ℑ 3 = 0 . (3 . 93) 18 Recall ǫ 1 , ǫ 2 ∈ { 1 , − 1 } . Again b y (2.10)-(2 .15), Cas e 1 and Case 2, w e ha v e: ι 1 = ǫ 1 s 2( b 1 c 2 − b 2 c 1 )( ℓ 2 1 + ℓ 2 2 ) a 1 b 2 − a 2 b 1 , ι 2 = ǫ 2 s 2( a 2 c 1 − a 1 c 2 )( ℓ 2 1 + ℓ 2 2 ) a 1 b 2 − a 2 b 1 (3 . 94) with ( ℑ , k 1 , k 2 ) as f ollo ws: (tan  , 2 c 1 ( ℓ 2 1 + ℓ 2 2 ) , 2 c 2 ( ℓ 2 1 + ℓ 2 2 )) , (sec  , − c 1 ( ℓ 2 1 + ℓ 2 2 ) , − c 2 ( ℓ 2 1 + ℓ 2 2 )) , (3 . 95) (coth  , − 2 c 1 ( ℓ 2 1 + ℓ 2 2 ) , − 2 c 2 ( ℓ 2 1 + ℓ 2 2 )) , (csc h  , c 1 ( ℓ 2 1 + ℓ 2 2 ) , c 2 ( ℓ 2 1 + ℓ 2 2 )); (3 . 96) ι 1 = mǫ 1 s 2( b 1 c 2 − b 2 c 1 )( ℓ 2 1 + ℓ 2 2 ) a 1 b 2 − a 2 b 1 , ι 2 = mǫ 2 s 2( a 2 c 1 − a 1 c 2 )( ℓ 2 1 + ℓ 2 2 ) a 1 b 2 − a 2 b 1 , , (3 . 97) ℑ = sn  and ( k 1 , k 2 ) = − (1 + m 2 )( ℓ 2 1 + ℓ 2 2 )( c 1 , c 2 ); ι 1 = mǫ 1 s 2( b 2 c 1 − b 1 c 2 )( ℓ 2 1 + ℓ 2 2 ) a 1 b 2 − a 2 b 1 , ι 2 = mǫ 2 s 2( a 1 c 2 − a 2 c 1 )( ℓ 2 1 + ℓ 2 2 ) a 1 b 2 − a 2 b 1 , (3 . 98) ℑ = cn  and ( k 1 , k 2 ) = (2 m 2 − 1)( ℓ 2 1 + ℓ 2 2 )( c 1 , c 2 ); ι 1 = ǫ 1 s 2( b 2 c 1 − b 1 c 2 )( ℓ 2 1 + ℓ 2 2 ) a 1 b 2 − a 2 b 1 , ι 2 = ǫ 2 s 2( a 1 c 2 − a 2 c 1 )( ℓ 2 1 + ℓ 2 2 ) a 1 b 2 − a 2 b 1 , (3 . 99) ℑ = dn  and ( k 1 , k 2 ) = (2 − m 2 )( ℓ 2 1 + ℓ 2 2 )( c 1 , c 2 ) . Theorem 3.7 . L et d 1 , d 2 , ℓ 1 , ℓ 2 , ℓ 3 ∈ R and let ǫ 1 , ǫ 2 ∈ { 1 , − 1 } . We hav e the fol low i n g solutions of the c ouple d two-dimensio n al cubic nonline ar Schr¨ odinger e quations (1.2) an d (1.3) : ψ = ǫ 1 e ( x 2 + y 2 ) i/ 4 c 1 t ℓ 1 x + ℓ 2 y + ℓ 3 t s 2( b 1 c 2 − b 2 c 1 )( ℓ 2 1 + ℓ 2 2 ) a 1 b 2 − a 2 b 1 , (3 . 100) ϕ = ǫ 2 e ( x 2 + y 2 ) i/ 4 c 2 t ℓ 1 x + ℓ 2 y + ℓ 3 t s 2( a 2 c 1 − a 1 c 2 )( ℓ 2 1 + ℓ 2 2 ) a 1 b 2 − a 2 b 1 ; (3 . 101) ψ = ǫ 1 e ( x 2 + y 2 ) i/ 4 c 1 t s b 1 c 2 − b 2 c 1 ( a 1 b 2 − a 2 b 1 )(( x − d 1 t ) 2 + ( y − d 2 t ) 2 ) , (3 . 102) ϕ = ǫ 2 e ( x 2 + y 2 ) i/ 4 c 2 t r a 2 c 1 − a 1 c 2 ( a 1 b 2 − a 2 b 1 )(( x − d 1 t ) 2 + ( y − d 2 t ) 2 ) ; ( 3 . 103) ψ = ǫ 1 s 2( b 1 c 2 − b 2 c 1 )( ℓ 2 1 + ℓ 2 2 ) a 1 b 2 − a 2 b 1 1 t tan ℓ 1 x + ℓ 2 y + ℓ 3 t t × exp  x 2 + y 2 4 c 1 t − 2 c 1 ( ℓ 2 1 + ℓ 2 2 ) t  i, (3 . 104) 19 ϕ = ǫ 2 s 2( a 2 c 1 − a 1 c 2 )( ℓ 2 1 + ℓ 2 2 ) a 1 b 2 − a 2 b 1 1 t tan ℓ 1 x + ℓ 2 y + ℓ 3 t t × exp  x 2 + y 2 4 c 2 t − 2 c 2 ( ℓ 2 1 + ℓ 2 2 ) t  i ; (3 . 105 ) ψ = ǫ 1 s 2( b 1 c 2 − b 2 c 1 )( ℓ 2 1 + ℓ 2 2 ) a 1 b 2 − a 2 b 1 1 t sec ℓ 1 x + ℓ 2 y + ℓ 3 t t × exp  x 2 + y 2 4 c 1 t + c 1 ( ℓ 2 1 + ℓ 2 2 ) t  i, (3 . 106) ϕ = ǫ 2 s 2( a 2 c 1 − a 1 c 2 )( ℓ 2 1 + ℓ 2 2 ) a 1 b 2 − a 2 b 1 1 t sec ℓ 1 x + ℓ 2 y + ℓ 3 t t × exp  x 2 + y 2 4 c 2 t + c 2 ( ℓ 2 1 + ℓ 2 2 ) t  i ; (3 . 107) ψ = ǫ 1 s 2( b 1 c 2 − b 2 c 1 )( ℓ 2 1 + ℓ 2 2 ) a 1 b 2 − a 2 b 1 1 t coth ℓ 1 x + ℓ 2 y + ℓ 3 t t × exp  x 2 + y 2 4 c 1 t + 2 c 1 ( ℓ 2 1 + ℓ 2 2 ) t  i, (3 . 108 ) ϕ = ǫ 2 s 2( a 2 c 1 − a 1 c 2 )( ℓ 2 1 + ℓ 2 2 ) a 1 b 2 − a 2 b 1 1 t coth ℓ 1 x + ℓ 2 y + ℓ 3 t t × exp  x 2 + y 2 4 c 2 t + 2 c 2 ( ℓ 2 1 + ℓ 2 2 ) t  i ; (3 . 109 ) ψ = ǫ 1 s 2( b 1 c 2 − b 2 c 1 )( ℓ 2 1 + ℓ 2 2 ) a 1 b 2 − a 2 b 1 1 t csc h ℓ 1 x + ℓ 2 y + ℓ 3 t t × exp  x 2 + y 2 4 c 1 t − c 1 ( ℓ 2 1 + ℓ 2 2 ) t  i, (3 . 110) ϕ = ǫ 2 s 2( a 2 c 1 − a 1 c 2 )( ℓ 2 1 + ℓ 2 2 ) a 1 b 2 − a 2 b 1 1 t csc h ℓ 1 x + ℓ 2 y + ℓ 3 t t × exp  x 2 + y 2 4 c 2 t − c 2 ( ℓ 2 1 + ℓ 2 2 ) t  i ; (3 . 111) ψ = mǫ 1 s 2( b 1 c 2 − b 2 c 1 )( ℓ 2 1 + ℓ 2 2 ) a 1 b 2 − a 2 b 1 1 t sn ℓ 1 x + ℓ 2 y + ℓ 3 t t × exp  x 2 + y 2 4 c 1 t + c 1 (1 + m 2 )( ℓ 2 1 + ℓ 2 2 ) t  i, (3 . 112) 20 ϕ = mǫ 2 s 2( a 2 c 1 − a 1 c 2 )( ℓ 2 1 + ℓ 2 2 ) a 1 b 2 − a 2 b 1 1 t sn ℓ 1 x + ℓ 2 y + ℓ 3 t t × exp  x 2 + y 2 4 c 2 t + c 2 (1 + m 2 )( ℓ 2 1 + ℓ 2 2 ) t  i ; (3 . 113) ψ = mǫ 1 s 2( b 2 c 1 − b 1 c 2 )( ℓ 2 1 + ℓ 2 2 ) a 1 b 2 − a 2 b 1 1 t cn ℓ 1 x + ℓ 2 y + ℓ 3 t t × exp  x 2 + y 2 4 c 1 t − c 1 (2 m 2 − 1)( ℓ 2 1 + ℓ 2 2 ) t  i, (3 . 114) ϕ = mǫ 2 s 2( a 1 c 2 − a 2 c 1 )( ℓ 2 1 + ℓ 2 2 ) a 1 b 2 − a 2 b 1 1 t cn ℓ 1 x + ℓ 2 y + ℓ 3 t t × exp  x 2 + y 2 4 c 2 t − c 2 (2 m 2 − 1)( ℓ 2 1 + ℓ 2 2 ) t  i ; (3 . 115) ψ = ǫ 1 s 2( b 2 c 1 − b 1 c 2 )( ℓ 2 1 + ℓ 2 2 ) a 1 b 2 − a 2 b 1 1 t dn ℓ 1 x + ℓ 2 y + ℓ 3 t t × exp  x 2 + y 2 4 c 1 t + c 1 ( m 2 − 2)( ℓ 2 1 + ℓ 2 2 ) t  i, (3 . 116) ϕ = ǫ 2 s 2( a 1 c 2 − a 2 c 1 )( ℓ 2 1 + ℓ 2 2 ) a 1 b 2 − a 2 b 1 1 t dn ℓ 1 x + ℓ 2 y + ℓ 3 t t × exp  x 2 + y 2 4 c 2 t + c 2 ( m 2 − 2)( ℓ 2 1 + ℓ 2 2 ) t  i. (3 . 117) References [AEK] N. Akhmediev, V. Eleonskii and N. Kulagin, First-order exact solutions of the nonlinear Sc hr¨ o dinger equation, T e or et. Mat. Fiz. 72 (1987), 183-196. [GG] B. Gr ´ eb ert and J. G uillot, Periodic solutions of coupled nonlinear Sc hr¨ odinger equations in nonlinear optics: the resonan t case, Appl. Math. L e tt. 9 (1996 ), 65-68. [GW] L. Gagnon a nd P . Winternitz, Exact solutions of the cubic and quin tic nonlinear Sc hr¨ odinger equation for a cylindrical g eometry , Phys. R ev. A 22 (1989 ), 29 6 [HS] F. Hio e and T. Salter, Sp ecial set and solutions of coupled nonlinear Sc hr¨ o dinger equations, J. Phys. A: Math. Gen. 35 (200 2), no . 42 , 8913- 8928. 21 [I] N. H. Ibragimov, Lie Gr o up A nalysis of Differ ential Equations , V olume 2, CRC Hand- b o o k, CR C Press, 1995. [MP] D. Mihalac he and N. P anoin, Exact solutions of nonlinear Schr¨ odinger equation for p ositiv e group v elo cit y dispersion, J. Math. Phys. 33 (1992 ) , no. 6, 2323-2328. [RL1] R. Radhakrishnan and M. La kshmanan, Brigh t and dark soliton solutions t o cou- pled nonlinear Sc hr¨ odinger equations, J. Phys. A : Math. Gen. 28 (1995 ), no. 9, 2683-26 92. [RL2] R. Radhakrishnan and M. Lakshmanan, Exact soliton solutions to coupled non- linear Sc hr¨ odinger equations with higher-order effects, Phys. R ev. E(3) 54 (1995), no. 3, 294 9-2955. [SEG] E. Saied, R. EI-Rahman and M. Ghonam y , On the exact solution of (2+1 )- dimensional cubic nonlinear Sc hr¨ odinger (NLS) eq uation” , J. Phy. A: Math. Gen. 36 (2003), 6751-6770. [W G] Z. W ang and D. G uo, Sp e cial functions , W orld Scien tific, Singap ore, 1998. 22