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# Class Note for MATH 488 at UA

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Solitons in Mathematics and Physics MATH 488 588 Instructor VE Zakharov May 25 2007 Lecture 1 Simple waves in Hydro dynamics Let us consider the system of the Euler equation for the compressible uid 3p 3 EW 0 31 31 3p 7 EUAP 0 11 1613 A 77 9 MP We assume that uid is bamtmpz39c and presume that it depends only on density P Pp Note that 6P 67p 02p c 7 sound velocity Let us study a special class of solutions of the system 11 when velocity is de ned by density vmm Now density satis es the equation a a ltESampp70 12 1 pltmltpgtgt S p is still unknown To nd it we should study second equation which takes form 31 3p 3p 3p 7 7 7 A 7 0 13 6 at Hogh may lt gt Equations 12 and 13 Inust coincide Hence 3 71 vi ap P 2 lt6 7W 72 14 Now Si 1 i Cp 15 For the special case of polytropic gas 1 39Y P icgpo Y P0 Then 1 2 71100 7 2 16 vpv2150ltpp01711 Suppose that the density variation is small PP05P 8So516p 1 5060 Sli io 17 2 p0 For small deviations from mean density p0 equation 12 reads 66S866670 18 at P 0 1 P h P i This is the Hopf equation Coef cient 51 changes sign if y lt 71 Note7 that 3 7 S 00 405P P0 One can obtain the same results by another way Let us try to nd a function of two variables A Ap7 1 obeying the equation 3A 3A 7 Si 0 19 at 6x Equation 19 can be rewritten as follows a m 695Eax taking time derivative from 11 one gets 6A 1671 Slt6A WU371 3p 31 3A 3p 3p 3A 7 3A 3p 3A 31 Da tpa ad Ug l Aa E75 110 Coef cients before 3 a must vanish Hence we obtain 317 31 Ag SirFp P lE Compatibility condition for systern111 gives S 712 Ap 02 There are two solutions Si 1 i c Aifp 112 p C p C 39 fP idP A1 1 i idP PO p Po p Thus we have following equations dAi dAi 7 7 0 113 at i 695 Equations 113 present another form of the initial system 11 Func tions Ai are called Riernann7s invariants Suppose that A 0 Then p 1 Edp Po p in accordance with 14 This solution is called sirnple wave Lecture 2 Dispersion is included Suppose that the ow is potential 1 Vltlgt In any dimensions one can rewrite the continuity equation as follows dwpVltlgt 0 21 The Euler equation can be replaced by the Bernulli equation 31gt 1 7 7 ltigt 2 0 22 6t2V wP P 1 6Pp w p i dp lt gt m p 6 Let us consider energy of the eld HpW FedMW om eltpgt pwltpgtdp Po and calculate its functional derivatives by p7 ltIgt 6H 7 1 2 a Vq 7MP 6H E i 7dwpVltlgt 24 One can see that equations 22 can be written as follows 2 LH 6t 7 6ltIgt 6ltIgt 7 6H 25 E 67 Thus I7 p are canonically conjugated pair of variables H is Harniltonian Let us us generalize the system 21 22 including into consideration dependence of phase velocity on wave number To do this we add to the Hamiltonian an additional quadratic terrn HaHH1 22 L H1 aLc VpV dF ngLZ 1 de Vp2dF 26 0 Here L is a characteristic length7 047677 are dimensionless constants Now equations 25 read 3p 2 2 a dwpVltIgt iaLcAp poL A ltIgt YLZCZ P0 27 31gt 1 a Vltlgt2 wp aLcAlt1gt Ap For small perturbations of density one can put mmz a P0 Thereafter we replace co a c Linearization of equations 27 leads to system 6 6p p0Altlgt iaLcA p p0L2A2lt1gt 3t 2 2 28 31gt 02 yL c 7 76 aLcAltIgt A6 at P0 p P0 p Let us use Fourier harrnonics eilwtlk7 eiithrikr One gets 7212 aLck26p p0k21 L2k2ltlgt 2 29 7212 7 aLck2ltlgt 71 7L2k26p 0 Compatibility condition for system 29 leads to following dispersion re lation w 02k21 6L2k21 wk 7 aZLZkZ 210 102 02k21 8L2k2 67L2k22 211 8 yio 2 212 In the limit 13L2 a 0 one can put approximately 102 2 02k21 8L2k2 1 213 wclkl1eL2k2sclkl1 8L2k2 Suppose now that one direction is preferred 13 p131 lpl gtgtlk11 2 2 1 lkl xp k 2p77 2 19 Finally 1 1 2 3 wicp 8Lp 214 To include effects of dispersion in equation for weak sirnple wave one has to change equation 18 by more complicated equation 3 a a 1 1 a 2 a 7 grip C Swima p Ea Ai p 7 icL grip i 0 215 7 Here 6 1fx ffxdz Lecture 3 Soliton appears lfe gt 07w gt 0 this is a case of positive dispersion In the opposite case 8 lt 0 the dispersion is negative In the most physical situations 51 gt 0 Thereafter we will assume that this condition is satis ed Let us differentiate 215 by z and divide by c One has 6 6 81 6 1 8 2 64 i aE6pi16pamp6p Ai pi EL piO 31 We generalized equation 215 a bit7 assuming that in the advective terrn c p velocity 0 could have negative sign see 17 Now we introduce dimensionless variable S 716p 6U c and rescale time and spatial coordinates x L z 71 a L 16171 1 t 7L El 0 2 We set following dimensionless equation 3 u i1 363 A 4 70 32 3x 6t az WTMU 39 If there is only one perpendicular coordinate y7 one obtains a 6ui166u 62 64 0 33 395 at 3x 312 T 3954 39 If we study essentially nonstationary solutions7 one can go to the moving frame gtCt Then we get two equations lf 8 gt 0 3 3U 3U 33 3213 7 672 0 34 h E 6u6z T 3x3u This is KP l equation lf 8 lt 0 one has 3 3U 3U 33 3213 alta6uauw0 35 This is KP 2 equation In absence of dependence on y KP equations reduce to the Korteweg de Vries equations 1 6 1 7 6 i 0 3 6 at 395 3x3 39 3U 3U 3313 7 6 7 7 0 37 at 395 3x3 In fact equations 36 and 37 are equivalent Equation 37 goes to 36 after a simple transform u a 7u7 t a it Stationary solutions of equation 33 obey one of Boussinesq equations 10 77737u 38 717171371120 39 77737uz0 33910 3 3211 3411 32 7 7 7 7 2 3y 3x2 3x4 gdxzu 311 For small distortion M of solution uo one can put u 2 uo Sueiqy m We have following four dispersion relations for distortion 12 pz 194 312 12 p2 i p4 313 12 7192 194 314 12 7192 i p4 315 One can see that the trivial solution u 0 is stable only in framework of equation 38 In 39 instability takes place at p a 00 In 310 at p a 0 Equation 311 is pure elliptic All K13117 K13127 KdV and Boussinesq equations7 except 310 have soli tonic solutions We will study rst equation 37 We will look for solutions in form of propagating waves uux7Vt at t0 311 3 2 3311 iva3gu 70 u satis es the equation 2 3 571 3112 7 Vu const 316 11 But this constant should be zero in order to give us decaying solution This equation can be integrated once 2 673 113 71112 E 317 Equation 317 can be solved in elliptic functions We will study it latter on So far we are interested only in solutions decaying at in nity u 7 0 at z 7 ioo This condition implies E 07 V gt 0 Let V 4132 One can check that equation 1 3U 2 3 2 amp 11 72131170 318 has following general solution 2132 u 2 319 cosh k 7 x0 xO constant of integration Finally we obtain 2132 320 u cosh2 kz 7 0 7 413 This is soliton in a media with negative dispersion The solition is a bump7 propagating faster then sound7 in the right direction in the frame moving with sound velocity Equation 36 has following solitonic solution 232 u 72 321 cosh kz 7 0 413 In a medium with positive dispersion soliton is a dip7 propagating slower than sound7 in the left direction moving with sound velocity The Boussinesq equations 38 311 have solitonic solutions in a form u ux 7 Ag 12 At y 0 satis es one of four ODE 2 A2 71u 67 3u2 0 322 A2 71u 7 g 3u2 0 323 A2 1u g 3u2 0 324 A2 1u i g 3u2 0 325 Equation 322 at A2 lt 1 has bump type solitoniC solutions Equation 323 at A2 gt 1 has dip type solitoniC solutions Equation 324 has no solitoniC solutions Equation 325 has dip type solitoniC solution at any A Now we study solitoniC solutions for the KP equations We will seek them in a form uu7yimt For KP l one gets 3 7 3uZ2 71u 0 The dip type solitons exist for any A and any 117 satisfying the condition A2 7 1 gt 0 For KP 2 one obtains 32 3u22711u0 Bump type solitons exist for any A and any 117 satisfying the condition A271lt0 13 Lecture 4 Solitons for shallow water Lets us consider a layer of ideal uid of thickness h The bottom is solid7 the surface is free Gravitational acceleration g is perpendicular to the bottorn7 surface tension with coef cient 039 is included The uid is incompressible7 density is equal to unity X20 X Figure 41 Fluid dornain Surface elevation is 77 77z7t Flow of uid is potential 1Vltlgt and Vv0 due to incornpressibility7 potential satis es the Laplace equation AltIgt 0 41 14 Let us denote ly Ix7 t 42 apparently 61 7 0 43 6y lt gt Boundary conditions 427 43 de ne uniquely a solution of the Laplace equation Thus it is enough to follow the evolution of 77x7 t7 Tl7 t We will not prove following theorernThe proof is put in application Theorem 1 777 KT is a pair of canonical variables They obey equations JLH ati xi n dti 677 39 here H 7 total energy of fluid7 consisting of kinetic and potential energy H T U 45 Potential energy can be found in explicit form Ulgmn2dUx1Ugild 46 2 00 Kinetic energy is given by formula 1 7 W 2 1 T V i dyds ii KI Tnds 47 7h foo Here KI is normal derivative of potential tn Gs7sIlsds 48 Here Gs7 5 Gs 7 s 7 Green7s function for the Dirichlet Neurnann bound ary problem It cannot be expressed in an explicit form for arbitrary 77x7 t 15 However7 the Laplace equation can be solved approximately if kn a 07 k 7 characteristic wave number 2 2 AltIgt 273 27i 0 2ltigt ltIgt0ltIgt1 6650 0 49 ltlgt0 llz7t 2 2 6631 773 410 olly 0 yih0 17227q21y201y02 i gn20177020 2 01 373 62x11 1 2 Orwhiquot h Assuming that steepness of the surface is small kn a 07 one can put Cg 0 because we need only derivatives of ltlgt Thus 6qu 1 2 1 1 7 i 719 62 N 62 1 2 h W as 7 3s 2y y 6ltIgt 6qu a i why 67w ZigaiqBgl 12 h N as dxd 2y1L y 7 N 2 62x11 2 1 2 4 51 my 16 3x3 l 2 3 l 2 7i y2 My h y2 dy 7h 7i h2 411 2y2dy 7h 7i 2 11207 h 2hy2jh gyS 7h 2 2 h2n2hn2 n3h372h3 h3 2 1 h2 2h2 737713 77 n3n 3 Finally 1 N 2 T7 7 hd 2695 77 z 77 2772 1 3 3qu 2 h2 1 27 W quoth 7 d 411 nlth3h2 6 6952 z l 1 039 377 2 Uzi 2at i 7 d 29quot 2626 95 Using weakly nonlinear approximation 77 ltlt h and kn a 0 one can get an a 62x11 1 364KB Etawhlw gh w N 16112 6277 43912 7 35 ax quot This is a system of type 21722 In the linear approximation r 2 1 2 2 zwnihk lighk Il0 4mm 9Uk277 0 w 911130731121 1 5k 3 9 17 039 1 7 gt ghz 7 positive dispersion 9 039 1 2 i i i 7 lt gh 7 negative dispersion 9 We will show that system 412 at 039 0 is integrable 18 Lecture 5 Lax representation Let us consider the following overdetermined system of differential equations 2 D1 0 51 66 w 0 52 12x11 27 UKII AQ4 VWQ 53 Nil Nil Comparing cross derivatives one obtains may agar mD w 54 System 51752 must be compatible 7 what does that mean We demand that the common Cauchy problem for equations 51752 qjlt07 y0 WOW 7 00 lt z lt 00 has at least local solution in some domain on the y7 t plane for any arbitrary smooth enough function 110 of one variable x 19 Now we observe that operator T 7 if 7 17 is an ordinary dif ferential operator7 annihilating any arbitrary function of one variable Hence we can cancel KI in 54 and require 6f 6A 7 7 7 7 L A 7 0 55 at 3y 7 Cornrnutator is a second order differential operator identically equal to zero Hence7 one can cancel77 KI in 54 6f 6A 7 7 7 L A 56 at 3y 7 l 9V 9U 632 lL7Al 2 712 t 62V71262U26W 362W7463U7V67U 5 7 3x2 3x2 3s 3s 3x2 3x3 as 39 Condition 55 implies 261 7 1263 i 0 3s 3 V 6U O 58 Then 2 2 iv LV 126 i 251 3g 3x2 3x2 as 9U 9W 62W 63U 9U 7 7 7 7 47 7 V7 59 at 3y 3x2 3x3 as Let us put 6V W 37 3s q System 59 reads now 35 7 3y 7 3s 67U a7U637U6U67U7 510 at as 3x3 as 76y 39 20 It is equivalent to the KP 2 equation 6 9U 06U 63U 6U6U i 362U 511 as at as 3x3 as 7 3y 39 the only difference is the coef cient 3 which can be excluded by the trivial rescaling of y We found that equation 511 is a compatibility condition for following pair of a linear PDE NI 6qu i f w W 0 N 63E N N 6U 7 476953 07696 6U7h 37696 q1 0 512 This overdeterrnined system is the LAX li air77 for KP 2 equation Replacing 337g 7 7 does not change anything Let us consider following Lax Pair NI 6qu i 7 7 UKII 26 1 6x2 1 NJ 63KB 6K1 6K1 6U 7 47 O7 6U7 37 39 10 513 m1 6263 6326 azaxm Here U7 q are real functions A cornpatibility condition for 513 is equation 7 7 of 7 6U7 37 514 696 6 9U 9U 63 U 9U 32 U at as 3x3 as 3y which can be transformed to KP 1 by already mentioned transforrn t 7 7t7 U 7 7U Note that both equations 5117 514 have burnp type solitons If there is no dependence on y7 one can put 1 641 and system 517 52 goes to following 12 M5 645 i EA 70 515 Now q 0 and 63KB 6K1 6U AKII 746Ua 3EKI 21 we put 0 0 This is the classical77 Lax Pair for KdV equation7 which now can be written as follows 6f 7 L A 516 at lt gt For the Boussinesq equation one has the system w MI 6K1 7 LKI M ay 1 i1 63KB 6K1 6U 6K1 AKII47 6U7 37KB KI 07 517 6z3 6x 3m Mq M Here 0 i1 Aditional remarks 1 One can assume that U is complex and study the complex KdV 63U67U 6170 518 at 695 0653 39 where a 7 any complex number Suppose that U to a 204 Now equation 518 is equivalent to the system 1 2 62 Wait 6t 69 ex 653 6ltIgt 1 01gt 1 2 92p 55 5 a 695 519 This is a badly unstable system with Hamiltonian 1 31gt 2 1 3 Oz 32 2 Oz 3p 2 545 dx Pd Ew Wal 53920 2 One can assume that U7 V7 W are matrix functions In this case one gets matrix generalizations of Kle7 KP and Boussinesq equations The most simple is the matrix KdV equation 3 6U 36W 10 521 at ax tw 22 Here U 7 any N gtlt N matrix One can assume that U is symmetric UT U Homework Construct matrix generalisation of the KP equation 23 Lecture 6 From KdV to NLSE Let us study again overdeterrnined compatible pair of equations 3X 7 L 0 61 6y X 3X 7 A 62 at X where again 62 L g u 33 6 u A 47 6 7 37 6x3ltuc6x 6q 1 But now X is vector in C X 3 while u7 q are n gtlt 71 matrix functions zn on L t y7 and c Ct is a matrix function on It only One can easily check that equation 6L 6A 7 i 7 i A 63 at 3y 7 l takes the following form 3g 3U 1 a 7 73 E 0711 64 3U 1 3U 3U 3 2 3311 7 3g E ltcampampc3ampu 7U7ql 65 24 This system is a matrix version of the KP equation We study rst one dimensional case dudy 07 dqdy 0 and assume that c is a scalar Now equation 65 reads 3U 3U 3 631A 7 7 37 2 7 0 66 at 6695 695 6x3 W l q qt arbitrary matrix function on time Equation 66 is a matrix version on the KdV equation Let n 2 A w A z 1 0 w A 970 71 6397 Now 66 is equivalent to following pair of equations 6A 3A 3 2 2 33A 7 Ec 3amilw 70 68 31 31 3 631 7 ECg6aA ZA iO 69 We will show that system 687 69 generates well known Nonlinear Schrodinger equation Let k gt 0 7 some real number We choose 631 A 72 and introduce new time variable and new unknown function 17im im pe k1 kekltp 610 now system 68 69 turns to following a 2 7 16A 1 63A 6 2 alAi gt E f3aA 6ltp 32 1 3 33 7 25 Now letkaoo Weset 3 7 i 20 695M M 6ltp 32 E 2 lt36Altp 70 using simple scaling 739 a 73967 z a one can get 3 r 62 2 7 Ez M ltpwtltp 70 612 612 is the Nonlinear Schrodinger Equation NLSE Linear term wt can i i 1 4f wt dt be excluded by multiplication by factor 6 to 6 62 i ii 6126 0 613 Equation 3739 3x2 is defocusmg NLSE Equation 6 i lt 1ltpizltp 0 614 739 x is focusing NLSE Let us use a substitution X a Xe Ay in 61 In the absence of coef cient dependence on y 7 the equation turns to EXX7 whereX X1 X2 32 6X21 AX1 71sz AX1 z 62 7 6 i 7Xi AXZ XZ 615 We will assume ikac ikac X1 5157 X2 52677 k2 A7Zku 616 26 Equation 615 can be written as follows 2 2 M 11ltA 1 3x2 552 i 1 3252 ZE i 4551 i 52 z A51 617 Equation 62 in the vector case reads 5X1 53X1 2 5X1 5X2 kE463 3k 2AE6 E 64 37 7 2kg X1 32 0 3s 5X2 3ng 2 5X2 5X1 k7 4 3 k 2 7 i 6 7 676z3A6 was 64 3 7 367 2k X2 i 37 0 618 z Plugging 616 in 618 after few simple transformations one gets a 62 a a 1 oi Wag 3 A 1 3352 E 3x2 as 7 6A 351 i 6239 i 6ltp i 3 A 2 i 3 1 7 4 6A 351 619 Now one can make the limiting transition k a 00 and set the following equation 6 gllgiiloio glg M Z Wl 91161 31 3 1 33 0 620 27 Thus for the NLSE A910 6 0 p L740 71i 0 621 for defocusing NLSE it is self adjoint operator7 for focusing NLSE it is non self adjoint A 10 62 0 p a M 37 28 Lecture 7 From KP to DS multiscale expansion In this chapter we will start with scalar KP2 equation 3 6u3k26u63u36 2 3532qu 71 695 at 6x 6x3 695 612 39 Here k is some constant We will not use limiting transition k a 00 Instead we will use multiscale expansion We present solution of 71 in a form 7 1 1 7 u g Klequot 1167 82 uo qugemq quge m b 72 here 8 7 small parameter7 ltIgt 72k3t kx In 72 117 uo7 112 are functions of slow variables 739 Ezt7 87 77 6y Plugging 72 into 717 one can see that terms of order 8 are cancelled After substituting 72 to 71 we have to cancel terms proportional to eim b separately In equation for n 07 n 2 leading order terms are proportional to 82 Putting n 07 one obtains equation 62 62 1 62 29 Amplitude of second harmonic 112 can be found in explicit form 1 01gt 2 01gt 01gt 3 321 27 i 22 3k 22 22 112 4 I 7 0 74 6ltlgt 6ltIgt 7 7 72k3 7 7 k at am From 74 one obtains 1 12 112 75 In equation for rst harmonic n 1 rst nonvanishing term is proportional to 84 Collecting all these terms together we obtain the equation N 62 62 1 7 39k7 3 7k27 7 1761 1 711 0 Z6T 652W 22 taking 112 from 75 we set k73 7k2672672 1761 uxil7lxillzxiz 70 76 67 652 6772 0 41 39 System 737 76 is known as Davey Stewanson equation It is remarkable that sign before second spatial derivative 2372 in equa tion 73 and 76 are opposite Transition from KP2 to KPl equation is performed by change 33 a 131717 5972 a 759722 If we would start from KPl equation we should make this replacement rlNow the system takes form 32 32 l 32 27 7 7 777 2 k 652 6772 2652 Q l N 62 62 1 39k773 7 7 1761 1 71211 0 77 Z 67 ag anz 0 4k2l l Equation 73 76 have a remarkable special solutions Suppose that 37a3 9777 6 In other words7 a solution depends only on g 0477 30 Now equation 73 can be integrated 1 1 2 k2 042 k2 7 a2 74km a2 0 W2 1 7 1 2 1 2 uo 4kg 1 l 1 Equation 76 now takes form 617 1 as 2k2a2 m7 i 3k2 i 042 6T xiW 0 78 For any 04 it is the defocusmg Nonlinear Schrodinger equation If we start from KP 1 equation7 042 should be replaced by 70427 and equa tion 78 reads NI 6qu 1 1 m7 i 3k2 042 Tyiikta 3739 If 042 gt 27 this is focusing NLSE Now one can address the question 7 what is going on with Lax pairs in process of rnultiscale expansion Equation xiW 0 79 3X 7 L 0 3g X xiem 5717 X 0 710 Equation 710 is a scalar equation7 to turn it to the vector equation7 one can nd the wave function X in following form ik 4m My X X16770 X1eTgt ET 711 Equation 710 has to be expanded in Fourier series 00 mm X ane 2 7L1 Taking into account only major terrns one get following system of equations Q X1 k IX1 0 6y 7 712 a r a 1 iglizk anIM 0 31 Equation 3X 33X 2 6x 311 i E46u3kg 3 X70 713 453 3m 3y also can be transformed to a system of equations for X17 X4 To do this7 we again present X in form 711 and use expansion 72 We realize that rst non vanishing terms are of order 82 Collecting them together one see that equation 713 is now a system 3X1 32M 3X71 3amp1 7 6k 31 3k 77 0 67Z 652 65 ZUOX126 X1 6X2 52X71 49x1 36x11 7 Wi6zk 662 3K1E73zkuox1 67 xl 70 714 715 1 0 a 1 0 1 LZk0 i1aig xil 0 3 1 716 VV 1062 0x11632 ku07 A 6Zk0 i1aigz3xil 0 gig2ltquot g 72mg 32 Lecture 8 Hamiltonian formalism for waves in weakly nonlinear media We study a medium described by one pair of canonical variables p7 ltIgt 6p 7 6H 31gt 7 6H 6t 7 6ltIgt 6t 7 6p The Hamiltonian H is some functional on p7ltlgt Let us perform Fourier transform 81 W WMWWCM p02 WMgkmd 1 1 quot r 3 82 n W ltIgtFe lkrd77 pk W P7 e lkrd7j From now on during the next three lectures we will ommit the vector sign of k7 j and F After Fourier transform 6M 1 6t WWW5 EC 6H 6H 6ltlgtq kw 6ltlgtq kw 1 33 7 7 W 83 kw Md25 Hence 6H 7 1 6H 1qu mm 270 awe 1 Thus a k 1 6H L KHW d d at 2w 5 6ltlgtq T 1 However 1 7ikqr 27016 dr 6k 1 84 Finally 6pk 7 6H 6t 76ltIgtik But ltlgt7k Wk7 p7k Mk After Fourier transform equation 81 take from 3p 7 6H 6ltIgtk 7 6H E km 6t Wk 8395 For beginning we suppose that the Hamiltonian H is quadratic functional invariant with respect to spatial transitions The most general form of such functional is following H Akltlgtkltlgtkdk Bkpkltlgtkdk B k kltlgtkdk 0kpkpkdk 86 H is a real functional7 hence A 7k Ak m 0a Ck Ck 87 30 Elk 34 B7k B k this means 38 81 888 88 818 88 8H 782k Equations 85 now are a 64 AltIgt B1 B2p alt1gt a 7B1 B2ltlgt 7 Op 89 Assuming that p7 ltIgt 6th one obtains wk BgiA A 1107812 810 The medium is stable if AC 7 B gt 0 ln Virtue of 87 88 N k Mk Thus Ak and B2k are symmetric and antisymmetric parts of frequency Equation for wk has two solutions7 but only one of them has physical meaning To determine the sign in 8107 we have to introduce normal variables ak vkltIgtk 811 Such that a at 8wkak 812 sign in wk is still not known Plugging 811 to 812 and using equation 89 one gets a 31 uA 1 u 2B1 i 8AM we assume ik 11k Thus ak 8 181 i mp Dgt l NH 8a 8 81 2 mp 813 35 Then p 1mltaltkgt 7 wk 814 lt1 ak ak 7 1 815 1 83931 83931 1ifak 1a k To determine vk we substitute 8157 814 to the Hamiltonian and de mand that H wkakakdk 816 A relatively long calculation leads to a very simple result A 2 1 7 12 817 01 7 AE x2A obviously sign in 817 should coincide with the sign of A Finally 818 wk B2k signAk Ak Sign of wk has clear physical meaning 7 Waves with wave vector 1 have positive energy if wk gt 0 and negative energy in the opposite case For normal variable one gets 12 glmww 7 wk lt1 11 ak 1 118 819 p fisignA Comment In nonlinear media the Hamiltonian cannot be transformed to form 816 In a weakly nonlinear medium the Hamiltonian can be expanded in terms of canonical variables HH2H3H4 820 36 One can perform this expansion in normal variables Now H2 wkakakdk H3 H321 H 1 Hp g nggfszakaklakz c06k k1 k2dk dkl dkz 821 1 Hf m V11 12k2a2aklak2 00 6k 7 k1 i k2dk dkl dkz H12 The whole Hamiltonian can be presented as a sum H Z Hmm n g m V L m 1 Hmm Vklmknm mkaaWrl a knakn1 aknm 00 gtlt x6k1 kn 7 kW 7 i knmdk1dknm 37 Lecture 9 Three waves interactions In the last chapter we introduced normal variable ak 1k ukPk 09 and de ned coef cients ukwk from two conditions 1 ak satis es the equations 60k 79wa at 1919 2 Quadratic Hamiltonian H2 816 has diagonal form H2 wkaka2dk Now equation 92 can be presented as follows LH 3t 7 6a 91 92 93 Now we address following question Let H is an arbitrary Hamiltonian What condition we have to impose on u7 1 to provide validity of equation 93 From 91 one obtains 38 7 ak ukPk kq k 94 07k U7kPZ 7151 Then7 after differentiation of 93 by time 60k 7 dpk E kw t kw 6H 6H 6H 39 7 7 7 95 Z6a km depZ 6H 6H 6a 6H 6ak 45H 6H 7 7 7 7 96 6p 6a 6p 6ak 6p uk aZ u k6 Lk In the same way 6H 45H 6H 9 7 7 1 7 1 so 1 6 Substituting 967 97 to 95 and assuming that condition 93 is satis ed for my Hamiltonian H7 one gets z39 Hi 1167H11 LH7vu67Hu 6H 6127 1166 71ch k 1166 7k 98 Uk 60k Condition 98 must be valid for any Hamiltonian H It implies following condition on coef cients umpk uk1k um 7 mu 239 910 In previous chapter we found 1 A uk 2B1 i 2Avk vi ii Substituting 911 into 997 910 one can see that these conditions are satis ed 911 39 We have proved a very important theorem Change of variables 819 transforms any Hamiltonian equation 81 to equation 93 Suppose that Hamiltonian H is expanded in series in powers of aka We assume that the medium is homogeneous It means that Hamiltonian should be invariant with respect to transform ak a akeikgt where A arbitrary vector The Hamiltonian is H7mmm m7Mmmk H3 Hgv H 1 H321 8 Vklt13gtk2akaklak2 Vggflgzazazlazz5k 7 k1 7 k2dkdk1dk1912 1 H322 i Vklt1 igtk2aaklak2 Vggzgzakazlazzpd 7 k1 7 k2dkdk1dk1913 Equations 93 are following 361 1 6i Wka Vzi 220210225k k1 k2 1 Vljik2aklak26k 7 k1 7 k2 1455 aglahm k1 7 k2dk1dk2914 2 Note that in a general case the Hamiltonian H can be presented as follows HZHnm ngm 1 71m Hnm Vk1fvvlgnkn mkwmakl aknc kn aknm6k1 kn 7 kn117 kmm dk1dknm 915 Suppose now that wk 2 0 and 40 has nontrivial solutions In the simplest case7 when dimension of the space d 2 2 and w k gt 0 w one can nd a suf cient condition for solvability of equation 916 This condition is positivity of dispersion w gt 0 lndeed7 in this case7 for parallel vectors 161762 Now replacing k1 a k1 kh k2 a kg 7 ki where In7 k1 07 one can turn inequality 917 to equality Suppose we found three wave vectors 1017 1027 p3 comprising the resonant triad 101 102103 wmwwmwwmgt 9m Let the wave led ak is a combination of three quasi monochromatic wave trains ak Ak 32 Ck 919 We assume that Ak 0 if lk ipll gt 67 Bk 0 if lk 7le gt 67 Ck 0 if lk 7 pgl gt 6 Here 6 is a small parameter Functions A7B7C are concentrated near wave vectors 1017 pg7 p3 To construct approximate envelope equations we perform change of vari ables iwkt Ck 920 ake From 914 we obtain 3 1 7 i ikag zciicizWWWwk2t5k k1 k2 1 6t 7 WkEMWVWWwihim om ka 11gt3201Wmwwtm k1 i k2dk1dk2 922 41 Let in 920 k is close to 1017 wk close to wp1 In the right hand side of 921 most terms are fast oscillating In the limit of small amplitude lakl a 0 they can be neglected Only secular777 non oscillating terms are important Looking at the right hand side of equation 9217 one can see that only second term can be slowly oscillating7 if we assume k1 2 1027 kg 2 p3 or Viceversa Taking into account only these terms and returning back from ck to ak one can put approximately 6A E Mkmk z39 Vlj thklok26kk1k2dk1dk2 923 Now one can put k101 k11021 k21032 and introduce A fiewll l B 55544222 0 66M Then equation 923 reads Z 1101 74 W101Ak lgfjfiw bp m m 02612d1d2 924 Now we can put approximately M191 W101 117 here 01 lkp1 group velocity of the rst wave train Ak and v1gt2 N 2717d2U P1P21P32 171172173 U is a constant Now from 924 one obtains z m1A z 27r d2U 1026 7 241 7 2d1d2 END 42 Performing Fourier transform ar 1 2Wd2 fleimd fl WMrk iwdr and similar transforms for 37 5397 one ends up with following PDE 3a a 01Va z ubc Repeating this procedlure7 assuming k 2 p2 and k 2 1037 we get nally following system of nonlinear PDE 6 6 01Va wbc 6b a UgV Mai 925 30 a 113Vc z uab here align 1 3 align In these equations u is a complex constant u lt2wgtdm3 2m Mew By simple transform a a lulilZe wa b a lulilZe wb c a lulilZe wc one can put u 1 and nally get the system 6 6 111Va zbc 6b a 112Vb we 926 30 a 113Vc tab 43 This is so called three wave system This system admits following re duction a w b 2391 0 z w where u7 117 w are real functions Now 3U a 111Vu lm 31 a 02V1 uw 927 6 6 UgV LU m In the spatial homogenous case this system simpli es to the system of ODE7s Lu 6t 7 W 31 a uw 928 1w at 71M identical to the Euler equation for free rotation of three dimensional rigid body Spatially homogenous system 926 reads 3a a zbc 6b a we 929 g ml This system has following constant of motion 44 lal2 W 11 930 a 2 c 212 In the case of amplitudes tending to zero at boundaries or in in nity general system 926 has constants of motion lal2drlbl2dr11 931 lal2drlcl2dr12 These motion constants are known as Manley Row relations Now suppose that wk can change sign In other words7 we have both waves of positive and negative energy In such medium one could nd a triad satisfying following conditions 0 101 102103 93932 M101 M102 M103 0 For these triads the main nonlinear term is H1531 Repeating previous considerations one ends up with the system do a 01Va 72b 0 6b r a UgV 72a 0 933 30 a 113Vc 7211 The Manley Row relations now take form 45 lal2dr i wsz 11 934 lal2dr 7 lclzdr 12 System 933 has a simple solution a b c 7w 3U 7 uz at 935 l u to 7 t This equation describes spatially uniform explosion collapse 46 Lecture 10 From KPl Equation to 3wave System In this lecture we will derive the 3 wave system from the KP l equation and nd Lax pair for 3 wave system We start from the KP 1 equation 3 u 3112 3311 i 3211 amp a 695 6953 36720 101 After Fourier transform 1 Way g u107 06quot de dq p7 q are components of wave vector k 1971 and 1 Wm ane m mwdz dy Equation 101 takes form 311 3239 67k zwkuk i 27 uk1uk26k 7 k1 i k2dk1 dkz 102 Let in 102 p gt 0 As far as 11 u 103 one can transform equation 102 to following form Mp Q at wltpqgtultpqgt 3 3 7 27 p1gt 0 up1q1up2q26p 7 pl 7 p26q 7 11 7 QZ 172 gt 0 2u21q1um25 191 p26q Q1 7 912 dpl dpz dQl dqz Here 3q2 Mp Q 7193 104 Now we introduce normal variables 10 In gain In these variables 3a 7 3239 7 WltP7Q0P7Q 7 7p m gt 0 pp1p212UP1q1umq25p 7 pl 7 MM 7 Q1 7 24005 2W p2 gt 0 2u21q1um26p 191 p26q Q1 7 912 dpl dpz dql sz One can check directly that equation 105 can be written in the form 500 5H 7 2 3t 5aq H wpqlapq12dp dq 7 Pgt0 3 12 as at E 171 gt 0 101102103 uP1q1up2q2uP3q3 UPIQIupzqzupaqa X 106 172 gt 0 173 gt 0 X5001 102 P3591 12 93d101 dqi dpz dQ2 dps 1st One can easily check that in initial variables the Hamiltonian is H g 3 671uy2 7 us dx dy 107 Later on we will show that equation 101 has in nite number of motion constants They will be discussed in the further lectures So far we estab lished that KP 1 equation is a Hamiltonian system belonging to the class described in lecture 9 48 This could be starting point for derivation ofthe three wave system First of all we have to be sure that resonant triads do exist To see this we note rst that dispersion relation 104 can be pararneterized as follows 10 51 i 52 q i 12 108 W 45 i 53 As far as p gt 0amp1 gt 2 Since this tirne7 we will study only the case g gt 37 q gt 0 and 17 2 are positive Let us consider the triad 10151 52 19252 53 10351 53 Q1 512 55 12 5 9 13 512 53 109 m 7 4c 7 5 W2 7 463 7 5 m 7 46 7 53 Apparently 109 comprise the resonant triad P1102103 11QZQ3 W1w2ws Now we will derive a systern7 more general then 3 wave Letusde ne 1gt 2gtgt ngt0 M 7 e 7 5m 5 7 gm 4c 7 gtt 1010 PM PM 1 PM iqDlk Suppose that the wave eld u is a composition of quasi rnonochrornatic wave trains u 7 Zealwank 27139 1011 M 7 ji uij 6x 778y T8t Here 8 is a small parameter Let us look for e i bifuz It has oscillating and non oscillating terrns Neglecting oscillating terrns7 we set e i biju2 2 Z ezuikukj 1012 kg Hi 49 Plugging 1011 in 101 and keeping only terms of order 827 one obtains the following system 1117 511V ii i 3239 5139 i 61 Z uikukj 1013 ijV 06361L ill36377 11 3MIMI 3192 7 qj 12616 1014 up p2 711 glam Q 6 1 2 In the simplest case n 37 assuming U13 1 7amp312A7 U12 1 7amp312B7 1123 2 7 31ZC one obtains 3 wave system with interaction coef cient uwimwmiwwwimm To nd Lax pair for system 1013 we start with the rst Lax operator for KP 1 3X 32X 7 397 7 0 1015 2611 W uX we will seek its solutions in a form X ZXkeingggZHiggt 1016 k1 Xk Xklt 77777 Keeping in 1015 only non oscillating terms of order 87 one obtains the system 6 6 amp12 amp1 WMk0 ml 6quot 6 1w Repeating this procedure with the second equation one yields aXi 26M 7 12 3 i i 0 1018 3739 1 3g 2 511 ka One can check that the compatibility conditions for equations 10177 1018 is exactly equation 1010 50 System 1013 is a special class of more general system of n waves To construct the general system we start with the following overdetermined pair of linear equations 6amp1 6amp1 FE 12in 1019 6amp1 6amp1 E GE cin 1020 Here F7 G are commuting matrices F7 G 0 One can think that they are diagonal al 0 61 0 F G 0 an 0 bn Compatibility condition for system 10197 1020 is following 6 6 6 6 One can consider that diagonal elements of Q are zero Then equation 1021 read aQij 3Qm39 kij Eikj aibk 7 akbi ajbl 7 aibj akbj 7 ajbk Let I 7 some real diagonal matrix 2 1 All diagonal elements 1 i1 Suppose that Q satis es the condition Q4r IQI 1024 This condition is called reduction If n 37 completely anti symmetric tensor 6W has only one component 8 8123 Let us assume that I 17 11 gt 12 gt as A Q m i B 1025 Q m lt l i O Q23 72 i as Qij 7ZQ We go to the three wave system b a b VQ 7 a l l J W 1139 1739 bl i b W 7 7 ll 7 1739 8123 u a1 7 a2gtlta1 e a3gtlta2 7 as Assuming that We obtain the explosive system 933 Lecture 11 Dressing method Let 102 May z vz7y z z W be an analytic function Then 571033 75357 r57 alio 62 az ey w ax 6yZ fig695 in Virtue of Cauchy Riemann conditions What is 17 32 2 To answer this question we present 1 7 z 7 lirn 7 z gao zz8 3 2 1 22 8 i7 7 7 7 7 7 111 32 zz8 zz8 2282 2282 lf 8 a 07 expression 111 tends to zero everywhere except 2 0 One could guess that it is a 6 function of two variables with some coef cient We introduce polar coordinates now 3 1 7 8 7 uni 322 seO T2 82 lntegrating over the whole complex plane one gets after integrating over angles 00 00 d 27T rdr7re u 27139 0 7 5 0 u 5 53 Finally 6 1 7 i E i 7T6262 i 7T6z6y 112 In the same way for 20 0 We 6 1 2720 7T6706y7y0 113 Thereafter we will denote complex numbers by Greek letters Suppose that function x7 5 is not analytic but satis es on the A plane the following condition nonlocal 5 problem 3 5 39fxn7 71n7 7k7kdnd lt1L4gt at A a 00 this function tends to the constant X a X0 Using formula 113 one can transform 114 to integral equation 1T E7n7 X X0 g A g X077 dnd d d5 115 Singularity 1 5 i E 7 1m A75 saOlAi l28 According to P redholm alternative7 equation 115 has nontrivial unique solution if and only if homogeneous equation T 7 iW mmdw 11396 has only triVial zero solution We will assume that this condition is satis ed It means that the nonlocal 5 problem 114 normalized by zero condition in in nity X a 0 as A a 00 has only zero solution We assume now that the kernel T in 114 depends on coordinates x7 y7 t as follow Tmmwmwnmmwmfwgt up lt1gt 2M My 427 54 It means that T satis es the system of linear equations 6T 7 39A7 T0 695 77 6T 67 202 7 772T 0 118 Equation 114 can be rewritten symbolically as follows 7 X gt1lt T 119 Let us introduce following differential operations 6 D1X 7X ZAX 69 5X D 7 A2 1110 2X 6y X 5X D 7 4A3 3X at Z X 1111 They commute with operation 36 Applying D to 119 one can see that 6 77D Di T 1112 M X X Let PX PD17D27D3X is any polynomial on D17D27D3 lts coef cients are functions on A7 y7 It By induction one can realize that 6 7p P T 1113 m X X Let X0 17 A a 00 In neighborhood of in nity function X has asymptotic expansion X1 X2 X17 1114 X1 T 7E777777X77777d77d77d d5 1115 X2 T 7E7n7 Xn7 dnd d d5 55 and so on Now we construct following differential operator P1X D2D UX 6X 2 a r 2 P 7 A 7 A 1116 1X 6y Xlt6zz XuX 5X 52X x iad i QlAE tuX substituting 1114 in 1116 we see that at A a 00 0X1 1 P 27 7 mi 26 u0ltA Hence if 72397 1117 u 2 h 131an atAaoo However7 P1X is a solution of the nonlocal 5 prob1ern 1113 Hence P1X 0 and function X satis es the equation D2Dfux0 1118 One should remember that u is de ned by equation 1117 In the same way we construct operator P2XDg4D 6VD13iqx a a 62 63 771327 1239A7 47 1119 6 626 Z 6262 6263 6 6V a 2A 3Vm q X Substituting 1114 into 1119 and send A a 00 one can see that 2 PZX A 712 6z39V 12 6X13Vmq712 3s 3x2 as 56 If we choose 6X1 V 72 397 1120 u 2 h 5X2 02X1 127 7 6 q 32 Z 3x2 We achieve P2X a 0 as A a 00 Hence P2X 0 or D34Di 6uD13umqX0 1121 Let X peiiAmiAz Lizimat 357 4A 7A2 7413 D 7 l m y 1 1X h 5 69 A2 3 D 7 71 mi y74l t 2X 6y 5 69 2 3 D 7 ilAziA y74l t 3X at 5 In terms of p equations 11187 1121 read 6 77ultp0 1122 W 6W W 7 E46ua3umqyp0 This is the Lax pair for KP 2 equation Hence u satis es the KP 2 equation If we choose 6 D7 39A 1 6Z 6 D7 2 2 6y2 6 D7 4A3 3 6t2 We end up with KP 1 equation Now lt1gt z w Azy 105 In the linear approximation one can put X 1 in 1114 1115 Then 1 X1 7TO 7 777775167nw1 2n2y4l 3natd d dnd 1123 7r Now we see the origin of the pararnetrization 105 57 Lecture 12 Solitonic solutions of the KdV equation Suppose that the kernel in 5 problem is such that y dependence drops out To do this7 one can put T077 777 A7 A T7 mm A6 X 121 Function X satis es now the following 5 problem 73 5 52iAm4A3tX A7 75 So far T7 5 is an arbitrary analytic complex function of A This equation can be solved explicitly if one assumes N T7 Z wZ Q 7 mm 7 in 123 n1 here A17 7xn set of complex numbers From 122 one can see that X is analytic function in all complex plane with exception the set of points A17 7 An In this points it has simple poles Thus X is a rational function 58 N X1Z f 124 Let us denote tn Anx 4Af Lt n7t Using the Poincare formula one gets fn Tn52i quotXln 125 N f 172 7 quot 126 Xl m1nxm Now we have to assume that An Am 31 0 In other words the set of poles does not include re ected points An a 7A Equation 125 reads N fm Tan 127 mumquot 5 fn new Function X 124 is the following expansion at A a 00 X1 meZmfm Hence X1 Z fm7 X2 Z Amfm etc Suppose that N 17 A1 M f X1 1 ltuz4ugtgtT1T 1319 T fim a 4T4 1172239 f7 M 6 128 a 7 572 5y This is one solitonic solution of the complex KdV equation It depends on two complex parameters 11 T To make it a regular solution of the real KdV equation7 one has to put 11 m H is real 59 2mm T 27 x0 is real Then solution 128 reads u 7 2 2 cosh2nzimo39 This is a soliton 60 Lecture 13 N solitonc solutions of the KdV equation To construct N solitonic solutions7 one has to assume that Ak Mk7 m gt 0 and Tk z39Mg fk 1 wk Lk m 4321 Equation 127 reads N 9k M1362 7 7 Zn n M136sz N X1 1 97m 9k hkeLk m1 Absolutely central role in future consideration plays Theorem 1 Theorem 1 6 73971 A X1 lax 71 MIEELHLW Hkl m Ad6t 6km 61 131 132 133 In the world literature A is called T functionl First we prove Theorem 1 One can see that 739 A 133 is just a determinante of system 131 31 in h nTiT To nd a derivative from a determinate one should differentiate sequentially columns of the determinante and add the results Then 739m Z Tm T T m Tm 7 result of differentiation of the column number m In Tm only this column differs from columns in 739 iMfeLl MZZELz Tm 2 L iMne quot here we used Cramer7s theorem Finally 7 l i 7 hmeLm 7 E Comparing with 132 we accomplish the proof Note that 2 2 739 7T7 u 727 lIlT 72w 6962 T2 134 Let us look at the structure of T function One can see that it can be pre sented in the following form T173917392 135 Here 62 i 7 1 1 Hi 7 HD2 A 7 211w Hwij 7 7 7 W 7 7 7 2 7 2 Kiwi 2H 4mm Kl K7 4mm Kl K7 2 2 2 7393 E iltjltk 1 1 1 21 Hinj Hif k A i 1 i 1 Wk 7 Hil j 21W njnk 1 1 1 Hi k njnk 2mC To calculate AW we note rst that this expression is symmetric with respect to permutation of its indices Indeed 1 1 1 1 1 1 2m Hil j Hil k Hil j 21W njnk A 7 1 7 1 7 1 1 1 17k 7 Hil j 21W njnk 7 2m Hil j Hil k 1 1 1 1 1 1 Hi k njnk 2mC Hi k njnk 2mC In the last formula we replace rst and second rows Let us replace rst and second columns 1 21W Hil j njnk A i 1 i 1 7A lgk 7 ninj 2m Hi k 7 71k 1 1 1 WWW Hi k 21 Then AW can be presented in the form Pm Emmilumi nk2m nj2sj m Aijk Here Pijk symmetric polynomial of power 6 Apparently Ailk 0 Hence Pigk 2 i 7 Kj2l i 7 Kk2l j 7 K1027 where A is still inde nite constant To nd it7 one set Ki a 0 In this limit 1 Aijk 2 Hence A 1 Finally W WVU W Ra v1 Vvklz Al 7k Emmilumi nk2m nj2sj m 63 Let us introduce Ml 2mm 1 Ml 7 e Z 2 xi ln 2 2l i 2l i Now 7391 62 bk 7nkx 7 zk limit k 2 E 2lt i jgt Ki Ki 7392 Ki 6 m 02 In the same way T 7 Z 2 i1 i2m il Rik 7 lmy l 7 e H 136 2 H39 H39 i1lti2ltmltiz mgtklt Mquot wherek1landm1l Remark 2 Note that we can remove the restriction Kl gt 0 What we need is gt 0 and Kl H7 31 0 If we have exactly 71 poles T171Tn 2 Tn 52lt 1m ngt H m 7 7 j gtz39 137 W 5739 where 239 1 n and j 1 n One can see that two solitonic solution is de ned by four parameters K17 K27 17 2 1 2 1 2 2M 2 1 2gt 13 8 T 7 6 6 K1 Wye Here bl 7H1 7 1 7 4amp0 2 7H2 7 2 7 4H t ln Virtue of 134 following transform 739 7 16me7 a7b 7 constants 64 does not Change u Let in 138 bl 7 7007 2 7 nite Then 739 7 1 62 This is solution with parameters K272 lf bl 7 6 1 E62 2gt Factor 62 51 can be omittedl7 and one can put 2 771M52 2152 2 H1 62 132 7K2 7 i2 1 2 i2 2 7111M 2H2 K1 7 H22 In the same way if 152 7 700 151 7 nite 739 7 1 62 51 parameters H17 x1 if 152 7 OO 739 7 1 7 62 1 parameters 17 1 ln t 7 ioo H 151 is nite z 2 4H t H1H22 HPHW Now let K1 gt H2 gt 07 2 2 74mm 7 ght 2 7 700 if t 7 00 Z7ooift77oo lf 2 is nite z 2 limit 151 2 74K1K 7 H t 4mg 7 ght Now situation is opposite l7ooift77oo 177ooift7oo Summarizing the situation we see that if t 7 700 the 77fast77 solution is posed at 2 1 954 gt1z1771nm 2K1 K1 7 My 139 65 the slow solution is posed at z limit 7 2 If t 7 00 the fast solution is posed at z 2 4l t 7 1 The slow solution is posed at 1 2 x24H t7277lnM 1310 2H2 K1 7 H22 These results can be interpreted as follow Solutions interact like repelling particles They scatter elastically 77Fast77 solution chases 77slow77 one and hits it Then 77slow77 solution turns to 77fast77 one In the same way one can study time asyrnptotics of the N solitonic solu tion Let H1 gt H2 gt gt Kn gt 0 We will study asymptotic behavior of the soliton with number k bk 2 const z 2 limit 4 74Hmsi 7 amt lf t 7gt 700 pm 7 700 if m gt k 45m 7 00 if m lt k The most important are two terrns7 one is proportional to 62 139quot k 17 second one is proportional to 62 1quot39 k 1 k Keeping in consideration only these two terms and making 77cancelling77 of an insigni cant factor aew one gets HH1 7 K102 Isk1 7 my 2 HO K102 MH K102 It means that at t 7 700 the 77k soliton77 is posed at 721 k71 1 2 xgz4nit7xkz lnM 11 7 1311 2l k K1 7 K102 1ft7gt00 g5m7gt7007 ifmltk7 m oo7 ifmgtk 66 Repeating previous consideration one can nd that the 77k soliton77 is posed n 1 m H102 N 2 7 E xk 4l kt xk 2 ln m 2 lk1 The total shift for the 77lr soliton77 is 1 7 2 n 2 x77ln Kl k 7 Z ln KHLM lk1 m 7 2 1312 We obtained a remarkable result A soliton of an indemediaded velocity acquired positive shift after interaction with slower solitons and negative shift after interaction with faster solitons A total shift is algebraic sum of particle shifts This shift does not depend on details of interaction If two solitons have very close parameters K17 H2 they could not get so close to each other The minimal distance between them is proportional to 1 K1 7 H22 A 2 fl z 2H1 n 4 67 Lecture 14 Unchecked Scattering in the Schrodinger equation We start with equation d2 2 Qkkllux1 fooltkltoo 141 ux real function satisfying the condition 00 1 lt 00 142 00 k kn is eigenvalue if the solution fn of equation 141 tends to zero at m a 00 It is well known that this solution is unique lndeed7 if 117 112 are two solutions of 141 then KI17 12 const C 143 Here Kl17 112 Illmkllg PSigmql wronskian of functions 117 112 If 117 112 eigenfunctions7 they tend to zero at a 007 hence C 0 and 111712 are proportional to each other Eigenvalue kn must be pure imaginary lndeed7 if kn is complex 68 dzfn 2 7 7 144 W kin uf Frorn 144 one gets d i after integrating by z one obtains k2k2 n n Apparently fmfn 07 and eigenfunction F can be made real Let us introduce Jost functions 17lt1gt solutions of equation 1417 de ned by bound ary conditions 1 a 6m lt1 a 67 146 a 00 a 700 Jost functions satisfy certain integral equations One can present 1 in a form q Clem 6264M with additional condition 036139 6267 0 147 Hence q Zrkclezkz 7 Czeilkm 1 kzkl z kcie 7 C Ze ik ukll 148 Combining 1477 1487 one gets 1 7i m 1 i m c uklle k 02 iiuklle k 149 lntegrating equation 149 we take into account boundary conditions 69 1 cl 17 m uklle lkydy 1 00 1410 02 m uklleikydy One can introduce a new function A lle il 01 0264 From 1410 we conclude that A satis es the integral equation 00 AW 17 ultygtlt1e eZiklty gtgtAltkygtdy 1411 Z 1 The same operation can be performed with function ltlgt Now 1 1 cl Q m ult1gte kydy 1 m I 1412 02 17 imu e kydy Let us denote B 617 This function satis es the integral equation 00 1 Bz7k 1 7 u17 52 k yBk7ydy 1413 Z 12 Suppose now that k 277 7 77 gt 0 lthem Ml1 In 142 y gt z and this exponent tends to zero as y a 00 In 1413 lezww yll e z m y As far as y lt L this exponent also tends to zero 77am Hence both functions A7 B could be analytically continued to the upper plane They have these asyrnptotic expansions 1 00 A 177 d n M w My 1 1 m k oo Imkgt0 1414 B a 17 ydy Z foo 70 and 1 0 KI a 6 1 17 ltIgt a 6 1 Let k ZN Then iNnm Ilk Nn a e xaoo Nnm ltlgtlkmn a e xiaoo They present the same eigenfunction fn and can differ only on some factor Suppose that fn is designed by asymptotic fn a 6N xaioo fn a bTLENquotm OO Hence fn I lkiNn bn lk Nn In this point KI and ltIgt are proportional to each other 1415 1416 lkw Iikw and ltlgtk7z ltlgt7k7x also are solutions of equation 141 Apparently7 they are analytic in lower half plane Solutions IAl comprise a fundamental systern Then7 one can put lt10 akllik7 bk1k7x Pmpkw aim71m dim1m Apparently Note that kmatom 22k k7lt1gt7k 722k Calculating 007 ltlgt7k by the use of 1417 one nds WW2 WW2 1 71 1417 1418 1419 1420 We will call rnonodrorny rnatrix7 according to 1420 this matrix Qlw le is unirnodular Now from 14177 1419 we get ak ws ak nk ltIgt 1421 Hence ak is analytic in the upper half plane By plugging 1416 into 1421 one gets a a klt1 7 0 uydy 4 1 7 jungt619 1 00 uydy 1422 177 4k 00 The scattering amplitude rk is de ned as follows 00 rk 7 k Also we de ne dk amplitude of penetration through the potential barrier From 1420 we obtain 6 Wk 1610012 1 1423 This is the unitary condition By de nition the potential is re ec tionless if rk E 0 In this case ak can be found explicitly from the conditions lakl 1 for real k7 a7k 51k ak a 1 k a 00 ak analytic in the upper half plane lf ak has no zeros in upper half plane then ak E 1 ln Virtue of con dition a7k 51k all zeros are posed on the imaginary axis Apparently they are exact eigenvalues N ak can be presented as the product ak k 721 1424 For re ectionless potential function mm AH 1425 73 Lecture 15 Unchecked Solution of the inverse scattering problem for the Schrodinger equation The inverse scattering problem is formulated as follows Suppose that the re ection coef cient 77k7 set of eigenvalues an n 17 7 N and coef cients bn is known How to nd the potential Ux7 It is enough to reconstruct function ltz7 de ne d through conditions W em Bk m Xk 7 W W 1 k gt 0 151 Xk 1ke ikm A7k lmk gt 0 152 Indeed according to 1414 in the lower half plane X1 1 7 153 Xe 7 k 1 6 if U d U 27 154 X1 2 m y a Xl In a general case Mk is a funciton analytic in both upper and lower half plane It has simple poles on the upper imaginary half axis and a jump on the real axis Thus it can be presented in a form Mk 1Z 95 jo ea 155 kiten 74 Here 1Xnx are residues in poles and 96 7 X 7 jump on the real axis Equation T can be rewritten as follows 100120 7 meme momma or 1 73 k A 7k k WA 7k W ltgt ltgtrltgte ltgt Now we see that 7 Bmn 7 ltpmne quotw and I 1 96 77k62 A k521ka k Then aniz nei nm Finally 21mm 39 Xn fax JG Xltim n It is remarkable that 2b 2 1 1 7 Mn 7 7 realpositivenurnber aKn 156 157 158 159 1510 1511 1512 1513 To prove this staternent7 we differentiate equation 141 by k and put k Mn dZKI k dz 7 Rigk 7 Uxllk 72 5nqmn The other hand El n satis es the equation dzkllmn d 7 53511 i Uqmn 0 75 1514 1515 LP rorn 39Z7 T we obtain equation d 2 C74317 11 72m Mn 1516 s As far as gt 00 1 a 6 in klm x11 0 1517 Then 11131 xihm 2mnkllzmnzdx 1518 Now we differentiate equation 416 by k and put again k on We set pk a k1ik bk1k 1519 Let us calculate the Wronskian can 11 as x a foo obV ltpk7kllk bnltpk7ltp a 0 at z a foo 1520 We get a1ik71kioo Milk 100 0 1521 But Wronskian KI7117K11 25 It does not depend on x and can be calculated at z a 00 where 17k a 6 1 a 6 Putting together T and T we obtain 2b 1 if 00 gt 0 1522 all LOO Kllzds Hence 1 00 m 1 1mmx2dx 1523 Now we can accomplish derivation of equation solVing the lSP Equation T reads we 1 55525155 Xk1zz k 7 7 700 Wd 1524 This equation holds in the whole complex plane 76 Then N X M36725 lt17 2 X775 7 1 f W615 1525 m1 Kn Km 2m 700 45 As far as Mg7 Xn are found U 2 Z W r e mx d 1526 If r E 07 we obtain already derived nite system of linear algebraic equa tions N X Mie ZHW 7X Mg w 1527 ml in Hm Note that in the general case gn Xnenw W is eigenfunciton of equation 141 wan ianImn n n quot i 7 1528 g X 6 rtMn 1057 is eigenfunction of equation 141 with asyrnptotics 9 a 7 Z W x a foo 1529 MW in m M2 2H m 15 30 7yequotnequotx oo g rtMn Now b2 1 2d 7 12d 7 M2 1531 9 9 Clmay 9 fwzdx So we have now interpretation of coef cients Mn This is nothing but L2 norm of eigenfunction gn Xylen Let us calculate now 3 36 33 X k g in 2 7r MM my 1 gt6 7 x 677 1532 N 3 if W Z Mgeizwxommw 7 W mow7 607 1533 3k 1 Completely in accordance with results of lectures 11 12 77 Lecture 16 Unchecked Triangle representation and Marchenko equation Let us calculate following integral 1 MM g 0 Ak 71eik ydk 161 Ak is analytic in the upper half plane and Ak 7 1 a 0 at a 00 Hence Kz7y E 0 if z gt y Now we calculate the integral Kltnygteiqydy Kltnygteiqydy ltAltqgt71gtem 162 But Aqem Ilq7 Hence Ik7z 6 Kz7yeiqydy 163 Equation 163 is the triangle representation for the Jost function Iq7 Let us introduce function N 1 00 115 Z M5 M r ei 5d5 164 n1 700 78 and show that Kz7y satis es to the following equation mm Kz7zFzyd2Fxy 0 165 This is famous Marchenko equation To prove this note that 17q7x also has triangle representation 17k7s 6 Kz7 seik9ds From 1414 we get lt1gt i 7 6 1 Kx7ye ik5ds rk em Kx7eikzd2 166 a 1 1 Now we multiply equation 166 to 6 y gt z and integrate over k We get immediately my Foltz w KltzsgtFoltsygt 167 i eiikwkikydk eiikywdk39 a 27139 1 Here 00 Fs1 r ei 9ds 700 lntegral in the right hand can be calculated by residues if y gt z 1 00 B iikyiz r iNnth Nn 7 2w 7000 6 dmizZe amn i 168 b Nny n 7 2 Nny z E e iamn 1nan E Mne 1nan But 00 mm 54W Manama 169 79 Plugging 169 into 168 we obtain equation 165 note that this equa tion is correct only in the half plane y gt x If y lt L K7y 07 but Fzy oo K72Fzyd2 31 0 Thus we cannot use Fourier transform 162 to nd equation for Note that from 163 one gets k a 00 Imk gt 0 10195 em 17 mg z k Comparing with expansion 14147 one gets 10 uydy K7z u 72K7z We shall seek solution of the Marchenko equation in the following form Kz7y Z hn6 Nquoty Sk7ye ikydk 1610 foo Comparing coef cients before different y exponents one gets system of equations hnx M3673 z7 ze NquotZdz 0 1611 50 may rk Kszeikzdz 1612 Plugging 1610 into 16117 1612 one obtains 7ltNnNmgtz M2 mam hnzM2 5 qu 711435an 1613 W W t a in 211 Equation 1613 goes to T if one puts yn ihne N hn igne N 1614 5k7s 7rky7ke 1615 One can easily check that equations T and 1612 coincide 80 Lecture 17 Unchecked Equation integrated by the local 8 problem A very important class of nonlinear wave systerns7 including the Nonlinear Schrodinger and Sine Gordon equations could be integrated by the use of the local 3 problern Suppose that in the nonlocal 5 problern 1777 777 A7 5 R7 50607 i A6 i 5 171 Now we have the local 5 problern 73 ltR7 5 172 In the scalar case one has from T 3 1mm PM I 173 M and this problem can be solved in the explicit form 1 PM 77 7 l A 7 7 d d 174 nX 7T7777777 Thereafter we will study only the case where X7 R are ngtltn matrices 81 Suppose that X is normalized by the condition X a l x a 00 175 Here 77177 denotes a unit matrix Then X satis es the integral equation X 1 X7213Z7 dnd17 176 As before we will assume that this equation has unique solution It means that solution of equation T with zero boundary condition Xgt0 A OO 177 is identically zero We introduce a pair of commuting differential operators D i i A1 17 8 1X a lX X 3 D2X 67X XB 179 2 Here A17 B01 rational functions on 17 depending on 172 The com mutativity condition D17 D2 0 implies that 6A 6B 6727671M7Bl 0 1710 In the simplest case when matrix coef cients of A7 B do not depend on 17 2 equation T is simpli ed up to 1AltAgt1BltAgt1 0 1711 CommutatiVity of D17 D2 means that the system of equations D1ltp 6i ltpA 0 1712 3x1 D2 5iltpg 0 1713 3 In the simplest case of constant coef cients p can be chosen as follow p 57ltAltAgtw1BltAgtw2gt 1714 82 Suppose now that R7 517 17 2 depends also on 17 2 obeying the equations 37R Mo AR 1715 37RRBA BR 1716 The general common solution of equations T is RltA7X7172 77711730017 RM 1717 Here 13001751 an arbitrary matrix function7 independent of 17 2 To prove this statement one just has to remember that p 1 satis es the equations W7 71 7 3 71 7 meLBOx p 70 1719 this is because 3p 1 722716222271 Our further strategy is following We claim that in Virtue of equations T a solution X of the local 3 problems satis es the system 7 XAo uX 1720 37 XBltAgt vX 1721 1722 Here u7v again are rational functions on A with the singularities 77 no worse77 then those of A7B A more precise statement is following Suppose A is presented as A P 2A 1723 k Here P PTA Pn1 1 is a polynomial of degree 717 while Ak is a sum of fractions7 associated with the point A Ak Al A2 A k AiAk ltmgt 1724 83 we claim that u can be decomposed to the similar partial fractions 71o 13o 4 1725 in T 1301 is a polynomial of same degree Moreover Ian P Then 13 PM RHXH 1726 and again 11 Am A if k 1727 k AiAk iltmgt Similar statement is valid for BOOM1 This fundamental statement can be easily proven in the simplest case when A7 B are polynomials To make a proof we apply operators D12 to equation T and mention that 07 and operation D12 and are commuting Then we get a 6X 6B in 7 695112 X671 XRAQ 1728 in Virtue of T one gets 3 3X 77D AA RD R 1729 6A 12 Mm H 12 lt gt In the same way 6 77D D R 1730 6A 2X 2X Equations T T mean that DIX7 D2X satisfy the equation TI However7 they have a polynomial type asymptotic when A a 00 One has X1 X2 1 i i 1731 X A A2 1 X1 7 X77R77777d77d77 1732 1 X2 7 77X77R77777d77d77 1733 1734 84 To determine u7 11A one just has to introduce new operators L1X Doc 7 uX 1735 L2X D2X 7 MMX 1736 where u7 11A new polynomials and demand L1X a 0 x a 00 1737 LZX a 0 1738 Actually conditions de ne the coef cients of U71 in unique way We illustrate this statement in the next lecture So far we mention only that one can put X 1ltp 1739 and make sure that A satis es to equations N 7 A 1 1740 6951 ult gt lt gt N 67x2 v1 1741 1742 Compatibility conditions for T 311 31 67 7 67 um 1743 Equations T is the Lax pair for equation Let 1 74 Then lt77 X l 1744 This is the 77dressing formulae77 85 Lecture 18 Unchecked 21 In this lecture we will demonstrate how the scherne7 elaborated in lecture 207 works in real situation We will consider one ofthe simplest examples7 leading to integration of the Nonlinear Schrodinger Equation and its generalization Let AA IA B 1 One can choose U IAu7 V IA2UAw Function X satis es the equation a 7 XIA IAuX 3x1 6 7X xlAZ 1A2 UA wXor 3x2 67 7w m1 311 5X 672 wX AZlIXl Using the expansion of X X1 X2 X3 1 7 7 7 X AA2A3 86 181 182 183 184 185 186 187 one immediately nds 7u I7X1 188 71 I7X1 1 u 189 71 7 w LXZ 7uX1 7 w 1810 On the other hand 5X1 5X1 I 7 7 7 I 1811 1 7le 6961 MO 6961 1 7X1le Comparing T and T we nd 5X 77 1812 w 6961 LP rorn T T we get following equation 5X 5X 67 7 wX 7 A671 1813 Plugging into gives 6Xn 6Xn1 7 n 1814 3 wX 3x1 In particular 5X1 5X2 7 7 7 1815 a in 6 Then 611 1 11 17611 1 1816 6 7X1 7wX1 7 6 7X2 Now using equation T T we get the closed equation7 imposed on X1 5X1 3in 5X 77777 70 1817 169521 M 611X11X117611 lt gt Note that in T I is an arbitrary constant rnatrix Let us specify To go further7 one has to specify I Let it be a block diagonal matrix 7 11 0 I 0 712 1818 1l10l n 12 0l m 1819 87 Let us denote Q11 Q12 P11 P12 1820 X1 Q21 Q22 X2 P21 P22 Here 11171011 squared matrices of dimension 71 12271022 squared matrices of dimension m 11271012 n gtlt mrectangular matrices 12171021 m gtlt nrectangular matrices 0 Q12 U 7 I 2 1821 7x1 W 0 l lt gt LP rom equations T one obtains i 7 2 18 22 6 Q11 7 Q12Q21 Equations T provide that diagonal terms in T one satis ed Equations for off diagonal terms read 3 52Q12 3 5Q12 27 7 7 27 27 2 1823 6 Q12 6 6x1 Q12Q22 6x1 Q22 Q12Q21Q12 3 92Q21 3 5Q21 727 7 7 7 27 727 2 1824 6 Q21 6 6 Q21Q11 6x1 Q11 Q21Q12Q21 1825 LP rom 7 one get immediately 3Q12 52Q12 3Q21 52Q21 7277778 0727 8 0 1826 a h Q12Q21Q12 6 in Q21Q12Q21 This is the generalized Nonlinear Schrodinger system From system T one can immediately get equations 6 3qu2 32 5 3Q12 5Q21 2 7 7 1827 62Q12Q21 ax Q21 Q12 61Q21 6 asz Q21 Q12 6x1 3 3 3Q21 5Q12 27 7 7 7 7 1828 62Q21Q12 6 in1 Q21 6 One can easily check that equations T are nothing but diagonal terms in equation TI 88 To be sure that this is true7 one has to calculate matrix function X2 The off diagonal elements 10124721 could be found from T 6Q 12 2 2 1829 1012 6x1 112122 5921 72 7 7 2 1830 1021 6x1 121111 1831 To nd the diagonal elements 1011710227 we note that equation T generates following set identities I n 7 n7 17 n 1832 l7X 1l 6951 W 61l XllX In particular a lLXal 6X2 1mm 1833 1 LP rom T one gets 3 3 5 5Q11 5 a 7pm QQ121021 Q127q212q12q21q12 q127q217q117p22 21211012 1217112 2 11 3x1 3x1 3x1 3x1 3x1 am 1834 Equation T can be rewritten as follow 5X1 3X1 7 3X2 7 7 7 7 1835 3 3x1 3x1 Off diagonal elements in T coincide with equation TI By the use of 39Z7 one can see that diagonal elements in T are identical to equation TI To construct higher constants of motion7 one can differentiate equation T by 1 3 3 7 3 5Xn1 672671 7 671 wxn 6951 1836 Then use the relation axn Ta lLXian lI7Xn1l 1837 89 Plugging T into T we get a a 6Xn1 I n I n 7 n 1838 69521 7X1lX 1 7X 11 6951wx 6961 Let us denote Pm 117 X1andz ag 1839 6Xn1 Qn w n 1840 X 6X1 dizzy Pn and Qn 7 diagonal pairs of corresponding matrix functions They satisfy the equations gig 321 Thus In ff Pndzl look like set of matrix motion integrals 0 1841 If Xn presented in a form n n Q Q 1 X 1 13 12 q 1842 Q21 922 Pm 2 Q12Q L1 0 J 1843 0 QZIQE Thus Q12Q21 0 P 2 1844 1 l 0 Q21Q12 To calculate P2 we remember that gig pij To calculate Q17 we mentioned that 6 X2 Q11 0 7 1845 Q Q1 wX1 6dmg 0 Q22 The upper diagonal term Q11 is 5Q11 5Q12 51311 3 5Q12 7 7 7 1846 Q11 h Q11 h Q21 h Q1261 Q21 6x1 Q21 5Q22 5Q21 51322 3 5Q21 7 7 7 1847 Q22 h Q22 h Q12 h Q2161Q12 6x1 Q12 90 One can see that equation 0 1848 3 3x1 Coincides with equation TI Note that if 1127121 a 0 at lxll a 007 the same is correct for Q117Q22 Thus quantities 2q12q2172q21q12 are real constant of motion One the contrary7 Pn are just 77forrnal77 constants of rnotion7 because in a general case Qn does not vanish at lxll a 00 A pure algebraic way for construction of higher 77real77 constants of motion is not developed yet This is an interesting unsolved problem 91 Lecture 19 Unchecked 22 In this lecture we study a more general system integrated by the local 5 problem Let AAIA7BAJA7I7J0 191 LJ commuting matrices Now X satis es the equations aiXIAIUX 192 3x1 6 ixJAJA1X 193 3x2 LP rom 39Z7 T we obtain u em 194 v in m 195 Equations 39Z7 T can be presented as follows 6X1 7 6751 lLXllX 7 AlLXl 196 6X1 672 lJ7X1lX AlLLXl 197 As far as I7 J commute one gets 61W 7 17M l17lJ7X1le e J7lI7X1le 0 198 92 Equation T is the 77master equation 7 generating the integrable equation together with all its motion constants To get the basic equation7 we just put X1 instead of x We get the following equation 6 6 Elj ml aimlLXll 117 lJ7X1lX1l lJ7 lI7X1lX1l 0 199 One can simplify this equation7 remembering that X can be presented in a form X 1ltIgt 1910 Function 1 satis ed the equations 6K1 7 IA 1 1911 6951 u 6K1 7 JA 1 1912 6952 11 Compatibility condition for equations 7 read 6u 61 7 7 7 IA JA 0 1913 6952 6951 1 117 vl This condition reads I711 Lu 1914 6i6i 0 1915 gig 6x1 U71 7 39 Resolving equation T according to 397 we end up with equation 6 6 6752117X1l 7 67111le llI7X1l7lJ7X1ll 1916 One can easily check that equation T is exactly equivalent to equation TI Let now AA IA 1917 BA JAZ 1918 Same as before u IAu to JA2vAw Function KI satis es the equations N 7 IA uI1 3x1 61 672 JAZ m LU12 Compatibility conditions are 311 3 777 A A J A 0 a 61 w u7 1 111 as before 171J7u 31 767 17wu711 0 Lu 6w 3 3x1 H as in the previous case J I7 u 1 7 0 Q12 7 1011 1012 u i 2 w i 121 0 1021 1022 Equation T reduces to the form 3U Z7 w 7 67 it gives 1012 7 iaqn 21 3 am Now u w 2 Q12w21 7111sz Q12w22 w11Q12 7 1211011 7112sz 1211012 w21Q12 2 3QHQH quot1121022 w11Q12 1211011 7112sz 1211012 w21Q12 94 1919 1920 1921 1922 1923 1924 1925 1926 1927 1928 1929 1930 1931 Diagonal parts of equation T gives 6w 276 0 2 19 32 7 7 w 6x1 azthzthi 11 112121 similarly LU2 2q21q12 Finally 291221 if u M 1934 1 iii 2921Q12 off diagonal parts in T give equations 6q1262 32112 2 8 0 1935 6 112121Q12 5Q2152 32112 2 8 0 1936 h 121Q1ZQZ1 which are identical to equations TI So far 1 2 were arbitrarily complex variables Let us assume 6 6 6 6 67M 7 2E 67 7 a 1937 Equations T read 3112 32112 72239 6x2 8q12q21q12 0 1938 ql C9qu 72 61 6 8911912911 0 1939 The most important special case of system T is following Suppose n 17 m N then 1 01 Q12 Q1 117 121 3 1940 PN N 112in U 10ka 1941 k1 95 Equations T read 5 32 72397 7 8 0 Z 6 6262 Wk 01 Pk W 72 7 8 0 Z 6 6952 W Now assume that 16k aqk a i1 Then u is real N u Z akiqkiz k1 We call system T generalized Manakov system 96 1942 1943 1944 Lecture 20 Unchecked 23 Let us restate where we are Function X satis es the equations 30M 7 A I 01 695 W 7X2 3 257 7 A wX A2I7x I J 0 i1 0 J block 7 diagonalmatr202 at A 7 00 X1 X2 1 7 7 203 X A A2 Q11 Q12 0 Q12 u 2 204 X 121 QZZJ J Q21 0 J Q11 237111ZQZ1 122 i257IQZ1Q12 205 6X 2112121 1121 J w 7 206 32 J 1211 QQZ1Q12 off diagonal matrices 112421 satisfy the equations 0112 32112 2 7 7 7 8 0 207 Z at g 112121Q12 0121 32121 2 7 8 0 208 Z at Mg 121Q1ZQZ1 97 assuming that n 1 m N7 we choose I of the form 1 0 I Nfl Q12QIquotQn p21 3 209 N u qumio 101 Equations T read 22 ka 8uqk 0 2011 22m pm suqk 0 2012 2013 We assume that 10k Otka 0 i1 2014 Then u is real N u ZaleklZ 2015 k1 Now system T reduces to smaller systems 22 ka 8UQk 0 2016 This system has a trivial solution Qk ake W 2017 N uo Z aquklz 2018 k1 Solution is the 77condensate Let us assume 04k1 NN1N2 ak71 N11ltng2 2019 Now equation T is a generalized Manakov system7 partly focusing 1 S k S N17 partly defocusing One can study stability of the condensate By performing the transform Qk a qke 4w 2020 98 one transforms T to the form 22 7 1km 7 8W 7 110ng 0 2021 Assuming that qk Ake i m we derive system T to the form a AW o Ak 5 7 4W 7 u0Ak Akm 0 Assuming that Ak ak 6Ak one can linearize the system 7 The result of linearization is following system 2022 5 524 276A 0 2023 at k Qk Mg 345k 1 32 7 1kg 7 4qk6u 7 0 This system can be rewritten as follows 62 62 1 62 276A 7 4 6 7 776A 0 2025 at H 622 lt q u 2622 k Here 6U 2 Zaqk Ak By multiplication by 04qu and summation we obtain Closed equation for M 32 32 1 32 76A 7 4 6 7 776 0 2026 at kaxz 0 4622 Assuming that Su emt m we get 1 92 74120 Ep4 2027 Condensate is stable if uo lt 0 7 defocusing part is prevaling and unstable MO gt In the Manakov case 1 0 0 q1 qn 2 q 71 7101 0 0 1011 72P1Q1 I u 2 w 1022 2102Q1 0 71 7 n 0 0 39 39 2028 99 Qnm 7 2P1 In 7 7 21 In Let us present the solution of equation T in a form X KIIltIgt 2029 Here ltIgt is a common solution of the system 23 ltIgtIA 0 2030 00gt 2 ltIgtIA2 0 2031 ltIgt is the fundamental solution of the following system N111 2 21 agmi 8 3 L31 em 7 2M1 2032 00 2071 A2 2 1 1 f quk 10 2033 2 iAZqk 210k 102N111 210k 2 liq11 2034 11 In the condensate case equations T and T can be easily solved Now p7 qk do not depend on x Let us denote 1 Z quizk 2035 k1 LP rom T we get 9amp1 7 Ag 2 2036 h 1 v 31 a 7A1 7 Quill 2037 now we study time dependent solution Equations T transform now to the form 0x11 2671 A2 21001 2w 2033 0x11 2 420ml 7 2Apkkl1 i kaV 2039 100 LP rom T one get 6V 0E i A2 2uV i 2Auk111 0 Now remember that qk akeizliut pk bkezliut Assuming that I I Qk1 pk1621utqjl pleimut We can exclude time dependance from Indeed V V6672th and equations T take form 7073101 Azal 2w z39pimggwo 7AM 7 Qual systems T T are compatible One can put 71 ak eQmiilambdaqtgk Vb Z ak k1 k1 LP rom T we get the same k 1 algebraic equations for g Q A 12V00 q MW 2U51 LP rom T we get 12 27211 Q12i272u 112 iv A2 2U are not only eigenvalues of this system 2040 2041 2042 2043 2044 2045 2046 2047 2048 There is also degenerate solution q 7A Now 1 0 V5 07 Q are arbitrary constants7 satisfying the condition 2 gkak 0 k1 101 2049 Equation T has n 7 1 parametrized family of solutions7 so eigenvalue q 71 is n 7 1 time degenerated Totally we have n 1 linearly independent solutions One eigenvalue behavs at A a 00 as A all other behave like 7A Hence7 we can construct a fundamental solution 1 a WWW a 00 2050 as far as lt1gt 6 iz39IAZ X a 1 as A a 00 This is what we need Note that the inverse matrix lt1gt 11 1 satis es the equations 03gt 7 lt1gt IA 0 2051 695 u 06gt 2397 lt1gtI2 qu 0 2052 h and can be used for dressing It is especially interesting to study dressing around the condensate background 102 Lecture 21 Unchecked Solitonic solution of equation integrable by local Dproblems Let us formulate the following problem from linear algebra Let X be a Mgtlt M complex valued matrix x 7 a rational matrix valued function on complex plane A Suppose that all poles of x are simple and it can be decomposed into partial fractions N A X 1Z 211 n1 A 7 an Here an 7 some complex numbers Let X 1 7 inverse matrix It is a rational function again lts poles are zeros ofthe rational function detx In general case numerator ofthis function is a polynomial of degree N gtlt M This in general position inverse function X 1 has N gtlt M poles In this chapter we will study following class of rational functions N B 41 n bm 212 X Aibm 0 7e Certainly it is a very special class of rational functions First let N 1 The function is X 1 213 A761 103 B 71 7 X 717A7b 214 the condition XX l 1 is resolved if B 1 Alt1aib70 1biaAB70 215 Let us introduce A a 7 bP7 B 7a 7 Equation now reads P17P 0 17PP0 216 LP rorn equation we get P P and P P27 therefore P is the projective operator and 7b 7b X17aP X 1lt177bP 217 Function X7 X l could be found in form of products of the Blaschke factors77 P1 Pn 1 1 218 X lt Aia1 lt Aian P P P 417 17L71 1i 1 219 X Aibn A7 H Aibl This representation is of course not unique One can put factors including fraction in an arbitrary order 1 Aiak Another solution of the same problem can be constructed as follows as far as an7 bn all are different7 function Pn X114 Qn X IIHTL 2110 LP rorn the condition XX l 1 one get AnQn 0 Pan 0 2111 LP rorn the condition X lx 1 one obtains BnPn 0 QnAn 0 2112 of course conditions follow from P rorn we conclude that all residues An7 Bn are degenerate rnatricies 104 Equations read N N Bm Am 7 An 12 anbm 0 lt1n 1bnamgt BniO 2113 m1 These equations could be turned into linear systems We will do this in the simplest possible case when all rnatricies are rank one and can be presented as tensor products of two vectors ATPMom Engine1n 2114 In other words Ana naPn Bna gnuL749 We will assume that PW7 gm are known while Am 1 are unknown Let us denote Vnm 2 qum 2116 7 One can equation are equivalent N Vnmum Pn 0 2117 N A71397 39L7139L n 0 2118 2119 When An7 Bn are found N X1 2A 2120 n1 If N 1 equations are simple Now V P7q7 a number aib rail 7 p A M V V 1 S0 A a i b 2121 B 7a i 19 2122 Apparently P 7 is the projective operator We obtained previous result To nd the equation imposed on 137 Q we consider equations 5X 7 A A U 2123 6 x lt gt x lt gt 5X 7 B A V 2124 mm lt gt x lt gt Inverse matrix satisfy the equation 71 fax AAX 1X 1UA 2125 3x1 6 71 76X BAX 1X 1VA 2126 2 Then 6X 7 9X71 7 A 1U 7 AA 1 2127 Wm x x 6951 ltgtx lt gt 6X 71 6X71 71 7 B V 7 B A 2128 af x x 6962 ltgtx lt gt Neither U7 nor V have singularities at A ambn This imposes the following equation on vectors P and q 3P 3 quot P A 0 i 7 Ag 0 a n a 2129 3PT 3 Equations are satis ed if Pn Blown an 151 7220 2130 Here lt13 lt1gtlam Sn lt1gt 11bn The next question we would like to discuss here is a problem of reduction Suppose that J is a matrix with a constant elements7 satisfying the condition J2 1 and commuting with A and B J7AA 0 J7BA 0 2131 Let kernel of local D problem satisfy the condition RNA A 7JRA ZJ RAA 7112107 AJ 2132 106 Now compare equation for X105 A and X 175 6 X1lanbda7 R1XXTXAX 1A7A 41307 X17 502133 LP rom one gets JXTXAJ JPJO AJ JXTO AJ 7RAXJXT5AJ 2134 Comparing and one concludes X 1JXTXAJ 2135 If this condition is satis ed bn a Bn JAanJ 2136 if An An 69 Pn and En qn 69 1 the equation means qn J15 Mn XnJ 2137 Therefore we replace A a M Hence X1 1 7 2138 X e M T 1 I m 7X4 2139 x lt gt M lt gt 1 m 1 4 2140 X M and reduction means mi 1le J2 1 2141 Remembering that Q11 Q12 2142 X1 Q21 Q22 we choose 1 0 2 1 1 J 0 A A 17 7 diagonalmatrixn gtlt 71 2143 now T T Q11 Q12A 111 121 J J J 2144 X1 AQZ1 AQZZA Qiz 122 In other words 111 Q1117 Q12A 1L7 AQ21 Jim 122 AQ22A7 Q21 AQTQiz 2145 and equations for 1127 121 reduce to the single equation 61 2267 Qizw 8Q12Aqlzq12 0 2146 If A 1 this is pure focusing77 equation If A 711 this is pure defocusing77 equation For general diagonal matrix A equations are mixed Let Q12 12q and q is square matrix7 satisfying the equation 01 1 3211 27 at 2 6952 quTq 0 2147 thise equation allows further reductions for instance one can assume that q is either symmetric or antisymmetric matrix More generally a ilql 12A 0 2148 Different reductions in make possible to construct of systems of the NLSE For instance if q bl qll A 2149 1 16211112 7111 We have the following system 2atgamlal22lbl2ab260 2151 bt bxx lal2 lbl2 l6l2bal300 2152 ict 0 2b2 lol20b2amp0 2153 2154 108 Further reduction 1 0 reduces system to two equations 1 m Eaxx a22b2a525 0 2155 1 bt bex 21a12 b12ba2b0 2156 2157 In fact this system is equivalent to the system of two independent NLSE for 1 xizi aib 2158 Other simple system that deserves to be studied 0 a b a b c a b 0 Q1 7a 0 C qz b d b Q3 b 0 7b 2159 7b 70 0 c b a 0 7b 7a 109 Lecture 22 Unchecked Simple solitonic solution for the generalized Manakov System Let AA 21A B 71A 221 Function ltIgt satis es the equations 6ltIgt 7 PIA 0 222 695 Z 6ltIgt 397 i ltIgtIA2 0 223 2 at ltigt e iIQH W 224 s WWW 225 In future we assume that some reduction is performed so bn 12 Suppose we have only one pole a1 a g 2777 b1 EL Now assume that dimension ofspace is N17 Nnm Let 0 041 n n1 nm p0 0761739quot7 n77 n1777 nm q qo7q177qmqn17 qmm 10 20721777 Qn 177 Qnm 110 qO 06iama2t qk k67iama2t p0 067iaza2t pk Ekeiltma2tgt 1 g k g n 100 igke wm t 71 1 S k S n m V p7 60 26ia7ampmia27amp2t W67iai zia27fz2t n nm WZl klzi Z W k1 kn1 V is real function V KO ZEZWJrZ t W527iw2 t Solution is regular7 if W gt 01 According 2514 aid 2277 qi 1 1 rerneber that Q12 ql q For qk one sets 2239 Qk 7n 2 n50 k52i m 2 7l2t Qk K0l26727m2 tW52 25t 0 k6ia zia2 zi2t 226 227 228 229 2210 One can put k okllk Then 0 drops out of the equation Solution is de ned by a position of eigenvalue 1 23977 and by the set of complex parameters Ilk We showed that soliton does exist only if n nm Wgt0 leklzgt 2 m1 k1 kn1 The focusing77 part showed prevails over the defocusing Of course this solution could be obtained by pure elernentary rnethods 111

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Recurring Subscriptions: If you have canceled your recurring subscription on the day of renewal and have not downloaded any documents, you may request a refund by submitting an email to support@studysoup.com

Satisfaction Guarantee: If you’re not satisfied with your subscription, you can contact us for further help. Contact must be made within 3 business days of your subscription purchase and your refund request will be subject for review.

Please Note: Refunds can never be provided more than 30 days after the initial purchase date regardless of your activity on the site.