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# Classical Mechanics and Mathem PHY 711

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December 7 2000 PHY 711 7 Notes on Hydrodynamics 77 Solitary Waves 1 Basic assumptions We assume that we have in incompressible uid p constant a velocity potential of the form ltIgtxzt The surface of the uid is described by h Lt z The uid is contained in a tank with a structureless bottom de ned by the plane 2 0 and is lled to a vertical height h at equilibrium These functions satisfy the following conditions Poisson equation 62ltIgtzzt 62ltIgtxzt 3x2 322 Zero vertical velocity at bottom of the tank 0 u 6ltIgtx0t i Bernoulli7s equation 6ltIgtxzt 1 6ltIgtxzt 2 6ltIgtzt 2 7 at 5 lt 695 62 Ma 7W 0 3 Surface equation 7M0M0M07M0 32 3m h at J zh 0 4 In this treatment we assume seek the form of surface waves traveling along the x7 direction and assume that the effective wavelength is much larger than the height of the surface h This allows us to approximate the 27 dependence of ltIgt z t by means of a Taylor series expansion 22 32 23 33 24 34 6ltIgt mmw mawwianwigimawgagmat amp 25y IEJWQW m This expansion can be simpli ed because of the bottom boundary condition 2 which ensures that all odd derivatives 3 x 0 t vanish from the Taylor expansion In addition the Poisson equation 1 allows us to convert all even derivatives with respect to z to derivatives with respect to x Therefore the expansion 5 becomes 22 32 24 34 3ylt9 7070gww7070m 6 For convenience we de ne t E ltIgt0t Using Eq 6 the Bernoulli equation 3 then becomes 645 2 6 2 7 h 7 am 0 3x2 where we have discarded some of the higher order terms Keeping all terms up to leading order in non linearity and up to fourth order derivatives in the linear terms the Bernoulli equation becomes ltIgtzzt ltIgtz0t 7 63 hlt2 6345 r at 2 am 2 9C 0 7 6t EamZ 2 76 h 6 1lt6 32 2 i gc 0 lt8 Using a similar analysis and approximation the surface de nition equation 4 becomes a 94s h364 a 7 lthltltztgtgtamp 7g 7gio lt9 We would like to solve Eqs 8 9 for a traveling wave of the form 45967 t X96 i 0t and 967 t 7796 i 0t 10 where the speed of the wave 0 will be determined Letting u E z 7 ct Eqs 8 and 9 become lt11 Mandi 1361329 0612 07 11 and Cdziu 7 d i 612721 Wu 0 12 The modi ed surface equation 11 can be integrated once with respect to u7 choosing the constant of integration to be zero and giving the new form for the surface condition hS h NW 7 3X 077 0 13 where we have abreviated derivatives with respect to u with the 77 symbol This equation7 and the modi ed Bernoulli equation 8 are now two coupled non linear equations In order to solve them7 we use7 the modi ed Bernoulli equation to approximate x u and its higher derivatives in terms of the surface function Equation 8 becomes approximately 2 2 2 g h m 1 2 9 h 9 g 2 77 7 77 77 77 77 14 X 077 2X 2CX C77 2677 20377 Using similar approximations7 we can eliminate u and its higher derivatives from the surface equation 13 9 hzg H 92 2 hgg H h 77 7 7 7 7 7 0 15 77 077 2677 20377 6Cncn 7 where some terms involving non linearity of higher than 2 or involving higher order derivatives have been discarded Collecting the leading terms7 we obtain 9h 913 H 9 9h 2 lt17gt 7 1 77 0 For the second two terms7 Fetter and Walecka argue that it is consistent to approximate gh 02 which reduces 16 to CZ 17 E 77u 7 himmu 7 3 77le 7 0 lt17 Your text shows that a solution to Eq 17 corresponding to Eq 5630 of the text with the initial condition 770 770 and 77 0 0 is the solitary wave form Cx t 77z 7 ct 770 sech2 18 h 0 c 71noh a 9 1 37h lt19 The standard77 form of the related Korteweg de Vries equation2 is given in terms of the scaled variables 5 and i in terms of the function 77i t by with 577 577 5377 7 which has a solution new g sechz 7 ml lt21 This form is related to our results in the following way 7 3 z 7 3 ct 62770 il and tilgm 22 To show how the reduced equation 17 is related to the Korteweg de Vries equation we rst take the u derivative to nd 770 7392 m 3 7 7 7 7 7 0 23 h 77 3 77 hnn 7 where we have used the relation h 770 7 i L W 7 CZ 24 Then we notice that a d a d 77 77 77 77 7 7 7 d 7 7 25 3t Cdu an as du7 so that Eq 23 can be written 770 377 I12 3377 3 377 7777777770 26 ch 3t 3 3x3 hndz Substituting the transformation 22 into this partial di erential equation yields the Korteweg de Vries equation 20 References 1 Alexander L Fetter and John Dirk Walecka Theoretical Mechanics of Particles and Con tinua McGraw Hill 1980 Chapt 10 2 Websites concerning solitons httpwwwrnahwacuksolitons httpwwwusfuni osnabrueckdekbrauersolitonshtrnl httpwwwrnathhkyoto uacjptakasakisoliton labgalleryindeX ehtrnl i Cr gt Thus we nally have the result UEM q i 91 A 9 c October 157 2007 PHY 711 7 Notes on SturmLiouville equations 7 following Fetter and Walecka Chap 7 These notes summarize some results concerning a broad class of problems related to the Sturm Liouville system which has the general form 596 MW bow E 45096 0 1 7zgt 2495 7 MW Here7 bow is an unknown function of the dependent variable x and 7z 0z and 039 are given functions A is a scalar constant which can be given In several physics examples7 z respresents a one dimensional spatial variable7 but other examples require the dependent variable to represent time We will assume that the functions are of interest in a nite range altzltb 2 The full Sturm Liouville system also involves a speci cation of boundary conditions Since the equation is second order7 there are in general two independent conditions which can be speci ed The most common boundary conditions are listed in Eqs 4011 4015 in Fetter and Walecka involving the end points a and b In addition7 it is sometimes the case that two independent conditions can be placed at a single end point initial or nal conditions Eigenfunctions and eigenvalues For special values of A enumerated with an index 717 we can de ne eigenvalues An and their corresponding eigenfunctions fnx of the Sturm Liouville system 596 AnUWWnW 0 3 We can prove as a general property of the Sturm Liouville system7 these functions are orthogonal b Ufnfmxdx 5mm 4 where b N Uxfnx2dx 5 It can be shown that for any reasonable function hz7 de ned within the interval a S x 3 b7 we can expand that function as a linear combination of the eigenfunctions M96 x Z OnanL 6 where on 7 baz hz fnz dz 7 These ideas lead to the notion that the set of eigenfunctions form a complete set in the sense of spanning the space of all functions in the interval a S x S b as summarized by the statement m 2 M 595 7 95 s Green7s functions A related equation involves a Green7s function GAz 8x 7 Ao GAzx 6Q 7 x 9 This Green7s function can be used to solve inhomogeneous Sturm Liouville problems of the form 596 MW Wt F 96 10 where now ow is the function to be determined and the forcing function is assumed to be given In this case if the Greens function is known the solution can be evaluated according to In terms of the eigenfunctions the Greens function can be expressed as x x N sz Z f W V 3 lt11 n An 7 A While in general this expansion converges in come cases a large number of terms may be needed for accurate evaluations Another representation of the Greens function for a Sturm Liouville problem can be found in terms of the two independent solutions of the homogeneous equation at a given A For simplicity we will assume that our solution satis es the simple boundary conditions a 0 and gtb 0 595 MW 9M 07 12 where 239 E a for gaa 0 and 239 E b for gbb 0 We can show that cw ad galtzltgtgbltzgtgt lt13 where zlt means the smaller of z and x and zgt means the larger of z and x where W represents the Wronskian w E 7w WW 7 dgaltgtgbltzgtl 14 dx dx This form of the Greens function implies that the solution can be determined by evalu ating the two integrals m mom 91x m91x F dx b gb Fx dx 15 Laplace transforms Laplace transforms can be used to solve initial value problems The Laplace transform of a function is de ned as zap 2 0 em ltzgtdz lt16 Assuming that is well behaved in the interval 0 S x S 00 the following properties are useful lms1J1 0 f PAW 17 and W awMW E 10 0 P2 P 18 These identities allow us to turn a differential equation for into an algebraic equation for p We then need to perform an inverse Laplace transform to nd For illustration we will consider a simple example with T 1 0x 1 A 0 The differential equation then becomes d2 96 W FWL 19 where we will take the initial conditions to be 0 0 and d 0d 0 For our example we will also take Foe w Multiplying both sides of the equation by e m and integrating 0 S x S 00 we nd F0 5 p 20 MWM In general the inverse Laplace transform involves performing a contour integral but we can use the following simple relations 00 1 1 e mdx 7 21 0 p 00 7 1 1 xe mdx 7 22 0 p 00 7 7 1 eiw e We mdx 7 23 0 p Y Noting that F F 1 1 12 07g77712 24 phm v vp p p we see that the inverse Laplace transform gives us F0 77m7 ltzgt1ee v lt25 November 16 2007 PHY 711 7 Notes on Hydrodynamics 77Solitary Waves 1 Basic assumptions We assume that we have in incompressible uid 71 constant a velocity potential of the form ltIgtxzt where Vzt 7VltIgtzzt 1 The surface of the uid is described by h Cx t 2 It is assumed that the uid is contained in a structure lake river swimming pool etc with a structureless bottom de ned by the z 0 plane and lled to an equilibrium height of z h The functions ltIgtzzt and t satisfy the following conditions The continuity equation V V 0 becomes the Laplace equation for ltIgtz zt 62ltIgtzzt 62ltIgtzzt T T 2 If we assume irrotational ow V gtlt V 0 we also have the Bernoulli equation which at the top surface 2 h Cx t takes the form 6ltIgtxzt 2 6ltIgtxzt 2 3m 32 Here we have assumed that the potential energy is due to gravity and have taken the reference potential energy at the height 2 h The boundary conditions for this system take the form of zero vertical velocity at bottom of the tank 76ltIgtxzt 1 6 2 gzih0 forzhCzt 3 6ltIgtx 0 t 0 4 62 At the surface of the uid 2 h Cxt we expect that d a vzxztph vvlt 5 This relationship can be expressed as 76ltIgtxzt 6ltIgtxzt 6xt i 6xt 32 3m h at 0 6 zh In this treatment we assume seek the form of surface waves traveling along the x7 direction and assume that the effective wavelength is much larger than the height of the surface h This allows us to approximate the 27 dependence of ltIgtz t by means of a Taylor series expansion 31gt 22 32 23 33 24 34 9572713 W957 0713 25957071 EEQQ 071 E707t g wvo 39 quot 7 This expansion can be simpli ed because ofthe bottom boundary condition 4 which ensures that all odd derivatives 3 z 0t vanish from the Taylor expansion In addition the Laplace equation 2 allows us to convert all even derivatives with respect to z to derivatives with respect to x Therefore7 the expansion 7 becomes 22 32 24 34 7 3w707t Before seeking the form of the nonlinear equations7 we rst consider the linearized version of these equations We focus on the solution at the free surface 2 h z7t The linear version of the Bernoulli equation evaluated at the free surface is ltIgtz727t ltIgtz70t 7 07t 8 i M th t m t 0 9 The linearized surface boundary condition is i 339ltIgt x7 z t 6x t 32 3t l l o 10 zh Using the Taylor7s expansion in this surface boundary 76ltIgtxzt h62lt1gtx0t 6z7t 32 3x2 3t 11 Eliminating C from the coupled Eqs 9 and 117 we nd 2 2 3 ltIgtz70t gha ltIgtz70t at 6x2 12 a wave equation with velocity 0 xgh We now return to treating the nonlinear equations For convenience we de ne Mat E ltIgt0t Using Eq 87 the Bernoulli equation 3 then becomes 645 2 6 2 7 h 7 am 06 where we have discarded some of the higher order terms Keeping all terms up to leading order in non linearity and up to fourth order derivatives in the linear terms7 the Bernoulli equation becomes w W 1 6t 2 af x 2 9C 0 13 at 36m 5 63 h263 16452 i wcio m Using a similar analysis and approximation7 the surface de nition equation 6 becomes a 645 h364 aci ahwmmmieiio 3 6954 at 7 15 We would like to solve Eqs 14 15 for a traveling wave of the form xt xx 7 ct and Cx7 t 77z 7 ct7 16 where the speed of the wave 0 will be determined Letting u E z 7 ct Eqs 14 becomes C dxu 7 012 dsxu 1 dxm 2 du 7 du3 2 du gt gnu039 17 and 15 becomes dXugt 7 h3d4gtltu CW 18 d 7 h u 77 c du nlt du 6 du4 du The modi ed surface equation 18 can be integrated once with respect to u7 choosing the constant of integration to be zero and giving the new form for the surface condition hS hnX 7gX cn07 19 where we have abreviated derivatives with respect to u with the 77 modi ed Bernoulli equation 17 which can be slightly rearranged symbol This equation7 and the I12 1 g 2 7 i i 7 0 20 X 2X 20XCn 7 are now two coupled non linear equations In order to solve them7 we make further approximations in terms of the expected magnitudes of the terms Linear terms and lower derivatives are assumed to be larger than nonlinear terms and higher order derivatives We also want to express the equations in terms of the surface function The Bernoulli Equation 20 is thus approximated by 9 hZH 1 2 9 hzgn 922 77 7 77 77 77 77 21 X 077 2X 2CX C77 2677 26377 Using similar approximations7 we can eliminate x u and its higher derivatives from the surface equation 19 9 I129 92 2 hag h 77 7 7 7 7 7 c 0 22 77 077 2677 20377 6Cnn 7 where some terms involving non linearity of higher than 2 or involving higher order derivatives have been discarded Collecting the leading terms7 we obtain 9h 913 H 9 9h 2 lt1icizgt i 1 77 0 For the second two terms7 Fetter and Walecka argue that it is consistent to approximate gh 02 which reduces 23 to by 7392 3 2 17 W 7 377 u 7 NW 7 o lt24 Your text shows that a solution to Eq 24 corresponding to Eq 5630 of the text7 with the initial condition 770 770 and 77 0 07 is the solitary wave form 20 77z 7 ct 770 sech2 lt4 gh 7 25 with 671 lt26 The standard77 form of the related Korteweg de Vries equation2 is given in terms of the scaled variables I and i in terms of the function 77i7 t by 577 577 5377 which has a solution new g sechz 7 ml lt28 This form is related to our results in the following way 7 3 s 7 3 ct B 27707 x i wig and ti l 2n0h 29 To show how the reduced equation 24 is related to the Korteweg de Vries equation7 we rst take the u derivative to nd 770 7392 m 3 7 7 7 7 7 0 30 h 77 3 77 hnn 7 where we have used the relation h 770 9 7 1 i 7 31 h CZ Then we notice that a d a d 77 77 77 77 7 7 7 d 7 7 32 3t Cdu an as du7 so that Eq 30 can be written 770 377 I12 3377 3 377 7777777770 33 ch 3t 3 3x3 hndz Substituting the transformation 29 into this partial differential equation yields the Korteweg de Vries equation 27 References 1 Alexander L Fetter and John Dirk Walecka7 Theoretical Mechanics of Particles and Con tinua7 McGraw Hill7 19807 Chapt 10 2 Websites concerning solitons httpwwwrnahwacuksolitons October 237 2007 PHY 711 Contour Integration These notes summarize some basic properties of complex functions and their integrals An analytic function fz in a certain region of the complex plane 2 is one which takes a single non in nite value and is differentiable within that region Cauchy7s theorm states that a closed contour integral of the function within that region has the value fe z 0 1 As an example7 functions composed of integer powers of z 7 fz 2 for n01i2i3 2 fall in this catogory Notice that non integral powers are generally not analytic and that n 71 is also special ln fact7 we can show that dz 7 2 g 3 C 2 m This result follows from the fact that we can deform the contour to a unit circle about the origin so that z em Then dz 2739 em 7 d 2m 4 c 2 0 e19 One result ofthis analysis is the Cauchy integral formula which states that for any analytic function fz within a region 0 i m 2m 0272 f 2 d2 5 Another result of this analysis is the Residue Theorm which states that if the complex function 92 has poles at a nite number of points 217 within a region 0 but is otherwise analytic7 the contour integral can be avaluated according to f gzd2 2m Resgp7 6 C p where the residue is given by Bess 2 313ng 7 MW lt7 mil where m denotes the order of the pole November 7 2005 PHY 711 7 Notes on Hydrodynamics 77Solitary Waves 1 Basic assumptions We assume that we have in incompressible uid p constant a velocity potential of the form ltIgtxzt where Vzt 7VltIgtzzt 1 The surface of the uid is described by h Cxt z The uid is contained in a tank with a structureless bottom de ned by the plane 2 0 and is lled to a vertical height h at equilibrium These functions satisfy the following conditions The continuity equation V V 0 becomes the Laplace equation for ltIgtz z t 62ltIgtzzt 62ltIgtzzt T f T 2 If we assume irrotational ow V gtlt V 0 we also have the Bernoulli equation in the form 6ltIgtxzt 2 6ltIgtxzt 2 3m 32 Here we have assumed that the potential energy is due to gravity and have taken the reference potential energy at the height 2 h The boundary conditions for this system take the form of zero vertical velocity at bottom of the tank M t Mr at 9941 0 3 DlH 6ltIgt 0t 0 4 62 At the surface of the uid 2 h t we expect that 7 CK i 5C Uz95727tlzhc 7 dt 7 V39VC at 5 This becomes 6ltIgtxzt 6ltIgtzt 6xt 6xt a a a 7 6t 0 6 2 c c zh In this treatment we assume seek the form of surface waves traveling along the xi direction and assume that the effective wavelength is much larger than the height of the surface h This allows us to approximate the 27 dependence of ltIgtz t by means of a Taylor series expansion 31gt 22 32 23 33 24 34 9572713 W957 0713 25957071 EEQQ 071 E707t EEQQ 071 7 This expansion can be simpli ed because ofthe bottom boundary condition 4 which ensures that all odd derivatives 3 x 0t vanish from the Taylor expansion In addition the Poisson equation 2 allows us to convert all even derivatives with respect to z to derivatives with respect to x Therefore the expansion 7 becomes 22 32 2 3x2 464 mm 1 ltIgtzzt ltIgtz0t 4 7070 Before seeking the form of the nonlinear equations we rst consider the linearized version of these equations We focus on the solution at the free surface 2 h zt The linear version of the Bernoulli equation evaluated at the free surface is 31gt ht 74L4Jxmwo t The linearized surface boundary condition is 6ltlgtzt 6zt 7 7 7 7 0 10 32 3t Using the Taylor7s expansion in this surface boundary 76ltlgtxzt N h62ltlgtx0t 7 6zt 62 N 6262 6t 39 Eliminating C from the coupled Eqs 9 and 11 we nd 11 62ltlgtz0t 7 h62ltlgt0t at 9 6x2 a wave equation with velocity 0 xgh 12 We now return to treating the nonlinear equations For convenience we de ne t E ltlgtx0t Using Eq 8 the Bernoulli equation 3 then becomes 645 2 6 2 7 h 7 am 0 3x2 where we have discarded some of the higher order terms Keeping all terms up to leading order in non linearity and up to fourth order derivatives in the linear terms the Bernoulli equation becomes 645 h 6345 1 645 2 i 777 amp 900 14 2 3 5 h0 53 5 2 M9952 9C 07 13 6t 2 af x Using a similar analysis and approximation the surface de nition equation 6 becomes a 645 h364 egg ahwmmm3mgigia am We would like to solve Eqs 14 15 for a traveling wave of the form Wat X06 0t and CLt 7796 i 007 16 where the speed of the wave 0 will be determined Letting u E z 7 ct Eqs 14 and 15 become h nltugtgtdgtCClELUgt 7 hgdiff f Edi 17 and Golgi 7 3 2110 mu 039 18 The modi ed surface equation 17 can be integrated once with respect to u choosing the constant of integration to be zero and giving the new form for the surface condition hS h 77W 7 3X 077 7 0 19 where we have abreviated derivatives with respect to u with the 77 symbol This equation and the modi ed Bernoulli equation 14 are now two coupled non linear equations In order to solve them we use the modi ed Bernoulli equation to approximate x u and its higher derivatives in terms of the surface function Equation 14 becomes approximately 2 2 2 g h m 1 2 9 h 9 g 2 77 7 77 77 77 77 20 X 077 2X 2CX C77 2677 26377 Using similar approximations we can eliminate x u and its higher derivatives from the surface equation 19 9 hzg 92 2 hag h 77 7 7 7 7 7 c 0 21 77 077 2677 20377 6Cnn 7 where some terms involving non linearity of higher than 2 or involving higher order derivatives have been discarded Collecting the leading terms we obtain 9h 913 9 9h 2 lt17Egt ig nig 1 77 0 For the second two terms Fetter and Walecka argue that it is consistent to approximate gh 02 which reduces 22 to 7y w7gww7wwtm m Your text shows that a solution to Eq 23 corresponding to Eq 5630 of the text with the initial condition 770 770 and 77 0 0 is the solitary wave form 2 t 77z 7 ct 770 sech2 z 2h6tgt 24 h 0 c 17gn0hgihlt1gihgt39 25 The standard77 form of the related Korteweg de Vries equation2 is given in terms of the scaled variables 5 and i in terms of the function 77i f by 577 577 5377 with which has a solution 2 This form is related to our results in the following way 3 z 7 3 cf 2 77 d t 7 28 6 7707 x l2hh7 an ithh nea7 w l eiml on To show how the reduced equation 23 is related to the Korteweg de Vries equation7 we rst take the u derivative to nd 770 72 3 77 7 377 7 EW 07 29 where we have used the relation h 770 9 i 7 30 h 62 Then we notice that a d a d 77 77 77 77 a i 0 and a a 31 so that Eq 29 can be written 6 h2 33 3 6 i liiiii 7 0 32 ch 3t 3 M3 hquot Substituting the transformation 28 into this partial differential equation yields the Korteweg de Vries equation 26 References 1 Alexander L Fetter and John Dirk Walecka7 Theoretical Mechanics of Particles and Con tinua7 McGraw Hill7 19807 Chapt 10 2 Websites concerning solitons httpwwwrnahwaculltsolitons7 httpwwwusfuni osnabrueckdelltbrauersolitonshtrnl7 httpwwwrnathhkyoto uacjpwtakasakisoliton labgalleryindeX ehtrnl October 137 2008 PHY 711 7 Notes on SturmLiouville equations 7 following Fetter and Walecka Chap 7 These notes summarize some results concerning a broad class of problems related to the Sturm Liouville system which has the general form 596 MW bow E 45096 0 1 7zgt 2495 7 MW Here7 bow is an unknown function of the dependent variable x and 739z7 0z and 039 are given functions A is a scalar constant which can be given In several physics examples7 z respresents a one dimensional spatial variable7 but other examples require the dependent variable to represent time We will assume that the functions are of interest in a nite range altxltb 2 The full Sturm Liouville system also involves a speci cation of boundary conditions Since the equation is second order7 there are in general two independent conditions which can be speci ed The most common boundary conditions are listed in Eqs 4011 4015 in Fetter and Walecka involving the end points a and b In addition7 it is sometimes the case that two independent conditions can be placed at a single end point initial or nal conditions Eigenfunct ions and eigenvalues For special values of A enumerated with an index 717 we can de ne eigenvalues An and their corresponding eigenfunctions fnx of the Sturm Liouville system SQ 7 Anazfnx 0 3 We can prove as a general property of the Sturm Liouville system7 these functions are orthogonal Abazfnxfmzdx 5mm 4 where b N Uxfnx2dx 5 It can be shown that for any reasonable function hz7 de ned within the interval a S x 3 b7 we can expand that function as a linear combination of the eigenfunctions M96 x Z OnanL 6 where on 7 baz hz fnz dz 7 These ideas lead to the notion that the set of eigenfunctions form a complete set in the sense of spanning the space of all functions in the interval a S x S b as summarized by the statement 0x Z M 6W 7 x 8 Green7s functions A related equation involves a Green7s function GAz 8x 7 Ao G z z 6Q 7 x 9 This Green7s function can be used to solve inhomogeneous Sturm Liouville problems of the form 596 MW W F 96 10 where now ow is the function to be determined and the forcing function is assumed to be given In this case if the Greens function is known the solution can be evaluated according to In terms of the eigenfunctions the Greens function can be expressed as fnfn Nn GAzx Zn A i A 11 While in general this expansion converges in come cases a large number of terms may be needed for accurate evaluations Another representation of the Greens function for a Sturm Liouville problem can be found in terms of the two independent solutions of the homogeneous equation at a given A For simplicity we will assume that our solution satis es the simple boundary conditions a 0 and b 0 596 MW 9M 07 12 where 239 E a for gaa 0 and 239 E b for gbb 0 We can show that 1 GAWJ W9alt9bgtv 13 where zlt means the smaller of z and x and zgt means the larger of z and x where W represents the Wronskian W 7 we W 93 7 93 gm lt14 This form of the Greens function implies that the solution can be determined by evalu ating the two integrals m mom 91x m91z Fz dz b gbx F d 15 Laplace transforms Laplace transforms can be used to solve initial value problems The Laplace transform of a function is de ned as amszwmm m Assuming that is well behaved in the interval 0 S x S 00 the following properties are useful d dmp 0 P P7 17 d 0 dy7mm am m These identities allow us to turn a differential equation for into an algebraic equation for p We then need to perform an inverse Laplace transform to nd For illustration we will consider a simple example with T 1 U 1 A 0 The differential equation then becomes and awM 10 d2 96 W FW 19 where we will take the initial conditions to be 0 0 and d 0d 0 For our example we will also take Feed Multiplying both sides of the equation by e m and integrating 0 S x S 00 we nd F0 719 19 20 4251 In general the inverse Laplace transform involves performing a contour integral but we can use the following simple relations 00 1 1 e mdx 7 21 0 19 00 ze mdz 22 m 0 p2 0 7 7 1 eiw e We mdx 7 23 0 10 Y Noting that F F 1 1 iz oyixhaa 24 p v p v 7 12 19 p we see that the inverse Laplace transform gives us F 7317 e W 7 yz 25 November 207 2006 PHY 711 7 Notes on SturmLiouville equations 7 following Fetter and Walecka Chap 7 These notes some of the results we have derived concerning a broad class of problems related to the Sturm Liouville system which has the general form so 7 we we 2 111an we 7 we we 0 1 x dx Here7 f is an unknown function of the dependent variable x and 7z 0x and 0x are given functions A is a scalar constant which can be given In several physics examples7 z respresents a one dimensional spatial variable7 but other examples require the dependent variable to represent time We will assume that the functions are of interest in a nite range altzltb 2 The full Sturm Liouville system also involves a speci cation of boundary conditions Since the equation is second order7 there are in general two independent conditions which can be speci ed The most common boundary conditions are listed in Eqs 4011 4015 in Fetter and Walecka involving the end points a and b In addition7 it is sometimes the case that two independent conditions can be placed at a single end point initial or nal conditions Eigenfunctions and eigenvalues For special values of A enumerated with an index 717 we can de ne eigenvalues An and their corresponding eigenfunctions fnx of the Sturm Liouville system 596 AnUWWnW 0 3 We can prove as a general property of the Sturm Liouville system7 these functions are orthogonal b Ufnfmxdx 5mm 4 where b N Uxfnx2dx 5 We also showed that for any reasonable function hz7 de ned within the interval a S x 3 b7 we can expand that function as a linear combination of the eigenfunctions M96 x Z OnanL 6 where on 7 baz hz fnz dz 7 These ideas lead to the notion that the set of eigenfunctions form a complete set in the sense of spanning the space of all functions in the interval a S x S b as summarized by the statement 1795 M 595 7 95 s Green7s functions A related equation involves a Green7s function GAz 8x 7 Ao GAzx 6Q 7 x 9 This Green7s function can be used to solve inhomogeneous Sturm Liouville problems of the form 596 MW 4596 F967 10 where now ow is the function to be determined and the forcing function is assumed to be given In this case if the Greens function is known the solution can be evaluated according to b m GAxx F dx 11 In terms of the eigenfunctions the Greens function can be expressed as G 12 AW MA lt gt While in general this expansion converges in come cases a large number of terms may be needed for accurate evaluations Another representation of the Greens function for a Sturm Liouville problem can be found in terms of the two independent solutions of the homogeneous equation at a given A For simplicity we will assume that our solution satis es the simple boundary conditions a 0 and gtb 0 595 MW 9M 07 13 where 239 E a for gaa 0 and 239 E b for gbb 0 We can show that 1 GAWJ W9a95lt9b95gt7 14 where zlt means the smaller of z and x and zgt means the larger of z and x where W represents the Wronskian W 2 To W W 7 W W lt15 dx dx This form of the Greens function implies that the solution can be determined by evalu ating the two integrals m 91x A gax Fx dx bgbx F d 16 Laplace transforms Laplace transforms can be used to solve initial value problems The Laplace transform of a function is de ned as zap 2 0 em ltzgtdz lt17 Assuming that is well behaved in the interval 0 S x S 00 the following properties are useful lms1J1 0 f PAW 18 and W awMW E 10 0 P2 P 19 These identities allow us to turn a differential equation for into an algebraic equation for p We then need to perform an inverse Laplace transform to nd For illustration we will consider a simple example with T 1 0x 1 A 0 The differential equation then becomes d2 96 W FWL 20 where we will take the initial conditions to be 0 0 and d 0d 0 For our example we will also take Foe w Multiplying both sides of the equation by e m and integrating 0 S x S 00 we nd F0 LAP m 21 In general the inverse Laplace transform involves performing a contour integral but we can use the following simple relations 00 1 1 e mdx 7 22 0 p 00 7 1 1 xe mdx 7 23 0 p 00 7 7 l eiw e We mdx 7 24 0 p Y Noting that F F l l 207g77712 25 p v p v v p p p we see that the inverse Laplace transform gives us F0 77m7 ltzgt1ee v lt26 November 2 2008 PHY 711 7 Notes on Hydrodynamics 77 Solitary Waves 1 Basic assumptions We assume that we have in incompressible uid p constant a velocity potential of the form ltIgtxzt where Vzzt 7VltIgtxzt 1 The surface of the uid is described by z h zt It is assumed that the uid is contained in a structure lake river swimming pool etc with a structureless bottom de ned by the z 0 plane and lled to an equilibrium height of z h The functions ltIgtzzt and Cxt satisfy the following conditions The continuity equation V V 0 becomes the Laplace equation for ltIgtxzt 62ltIgtzzt 62ltIgtxzt i 2 3x2 322 T If we assume irrotational ow V gtlt V 0 we also have the Bernoulli equation which at the top surface 2 h Cx t takes the form 6ltIgtzt 2 6ltIgtxzt 2 3m 32 Here we have assumed that the potential energy is due to gravity and have taken the reference potential energy at the height 2 h In this form of Bernoulli7s equation we have absorbed some integration constants in the choice of time dependence of the velocity potential as discussed in your text Eq 5411 The boundary conditions for this system take the form of zero vertical velocity at bottom of the tank 6ltIgtxzt Jr 6 2 gzih0 for zhCzt 3 6ltIgt0t i T 039 4 At the surface of the uid 2 h Cxt we expect that 7 GK 7 5C Uz95727tlzhc 7 dt 7 V 39 VC at 5 This relationship can be expressed as 76ltIgtxzt 6ltIgtxzt 6xt 7 6xt 62 6x 695 at 7 0 6 zh In this treatment we assume seek the form of surface waves traveling along the x7 direction and assume that the effective wavelength is much larger than the height of the surface h This allows us to approximate the 27 dependence of ltIgt z t by means of a Taylor series expansion 64 4 2 957071 EEQAOW 7 23 33 31gt 2262ltIgt 9572713 W957 071 2957071 iQQ 071 g 32 2 322 This expansion can be simpli ed because of the bottom boundary condition 4 which ensures that all odd derivatives 3 z 0t vanish from the Taylor expansion In addition the Laplace equation 2 allows us to convert all even derivatives with respect to z to derivatives with respect to x Therefore7 the expansion 7 becomes 2 2 4 4 WH WW39 Before seeking the form of the nonlinear equations7 we rst consider the linearized version of these equations We focus on the solution at the free surface 2 h x7t The linear version of the Bernoulli equation evaluated at the free surface is ltIgtz727t ltIgtz70t 7 6q 7th7 t m t 0 9 The linearized surface boundary condition is 9 a 5G0 0 m 62 z h Using the Taylor7s expansion in this surface boundary 76ltIgtxzt h62ltIgt0t 6x7t 32 3x2 3t 11 Eliminating C from the coupled Eqs 9 and 117 we nd 3239ltIgt 7 07 t 3239ltIgt z7 07 t 6t2 911 6962 7 12 a wave equation with velocity 0 xgh We now return to treating the nonlinear equations For convenience we de ne t E z 071 Using Eq 87 the Bernoulli equation 3 then becomes 645 2 6 2 i h f Oj 3s as where we have discarded some of the higher order terms Keeping all terms up to leading order in non linearity and up to fourth order derivatives in the linear terrns7 the Bernoulli equation becomes 7 hC2 6345 i at 2 am 2 9C 07 13 6t EamZ E 611263162 f f Q wlto at Using a similar analysis and approxirnation7 the surface de nition equation 6 becomes a 94s h364 aci ghmamw e iei 0 3 6264 at T 7 15 We would like to solve Eqs 14 15 for a traveling wave of the form xt xx 7 ct and Lt 77z 7 ct7 16 where the speed of the wave 0 will be determined Letting u E z 7 ct Eqs 14 becomes C du 2 du3 2 dxu 012 dsxu 1 dxm W du gnu0 lt17 and 15 becomes dXugt h3d4gtltltugt CW 18 i M nltugtgt du H i E du4 du The modi ed surface equation 18 can be integrated once with respect to u choosing the constant of integration to be zero and giving the new form for the surface condition hS h NW 7 3X 077 0 19 where we have abreviated derivatives with respect to u with the 77 modi ed Bernoulli equation 17 which can be slightly rearranged symbol This equation and the I12 1 g 2 7 i i 7 0 20 X 2X 20XCn 7 are now two coupled non linear equations In order to solve them we make further approximations in terms of the expected magnitudes of the terms Linear terms and lower derivatives are assumed to be larger than nonlinear terms and higher order derivatives We also want to express the equations in terms of the surface function The Bernoulli Equation 20 is thus approximated by 9 hzm 1 2 9 hzgn 922 77 i if 77 77 if 21 X 077 2X 2CX C77 26 20377 Using similar approximations we can eliminate u and its higher derivatives from the surface equation 19 9 I129 92 2 hag h if i 7 i 7 7 c 0 22 77 077 2677 20377 6Cnn 7 where some terms involving non linearity of higher than 2 or involving higher order derivatives have been discarded Collecting the leading terms we obtain 9h 913 9 9h 2 lt1ic7gtni 7 ig 1 77 0 23 For the second two terms Fetter and Walecka argue that it is consistent to approximate gh 02 which reduces 23 to 1 e W 7 gm 7 M1012 o lt24 Your text shows that a solution to Eq 24 corresponding to Eq 5630 of the text with the initial condition 770 770 and 77 0 0 is the solitary wave form Cx t 77z 7 ct 770 sech2 lt4 3770 25 with 7WQWHQ72 lt26 The standard77 form of the related Korteweg de Vries equation2 is given in terms of the scaled variables 5 and i in terms of the function 77i7 t by 577 577 5377 which has a solution um g sechz 7 ml lt28 This form is related to our results in the following way 3 s 7 3 ct 2 77 d t 7 29 6 7707 x l2hh7 an ithh To show how the reduced equation 24 is related to the Korteweg de Vries equation7 we rst take the u derivative to nd 770 hz m 3 7 7 7 7 7 0 30 h 77 3 77 hnn 7 where we have used the relation h 770 9 7 1 7 7 31 h CZ Then we notice that a d a d 77 77 77 77 7 7 7 d 7 7 32 3t Cdu an as du7 so that Eq 30 can be written 7amp6n7h263n73 3777 cm 3 3 039 33 Substituting the transformation 29 into this partial di erential equation yields the Korteweg de Vries equation 27 References 1 Alexander L Fetter and John Dirk Walecka7 Theoretical Mechanics of Particles and Con tinua7 McGraw Hill7 19807 Chapt 10 2 Websites concerning solitons httpwwwrnahwacuksolitons The Kortewegde Vries Equation History exact Solutions and graphical Representation by Klaus Brauer University of OsnahriickGermany1 May 2000 Travelling waves as solutions to the Kortewegde Vries equation KdV which is a nonlinear Partial Differential Equation PDE of third order have been of some interest already since 150 years The author s aim is to present an analytical exact result to the KdV equation by means of elem entary operations as well as by using Backlund transform Special interest is devoted to nonlinear superposition of several waves the so called Solitary Waves or Solitons which is performed by Backlund transform too The derivation of the exact solutions follows the presentation of Vvedenski 1992 It is fascinating that these exact solutions which involve a lot of calculations can be done with the help of the Computer Algebra System Mathematical see Wolfram 1999 not only presenting analytical expressions but in addition 3DPlots Contour Plots and 2DPlots for discrete values of time finally animating these 2DPlots and presenting these plots as well as the solution formulae at an Internet page Another aim of the author has been to apply numerical discrete methods to nonlinear PDEs With the help of the exact solutions gained here an excellent possibility is given to test the quality of the numerical methods by checking the numerical results against the exact solution 1 Historical Background A nice story about the history and the underlying physical properperties of the Korteweg de Vries equation can be found at an Internet page of the Herriot Watt University in Edinburgh Scotland see Eilbeck 1998 The following text is taken from that page Over one hundred and y years ago while conducting experiments to determine the most ef cient design for canal boats a young Scottish engineer named John Scott Russell 1808 1882 made a remarkable scienti c discovery Here his original text as he described it in Russell 1845 I was observing the motion of a boat which was rapidly drawn along a narrow channel by a pair of horses when the boat suddenly stopped not so the mass of water in the channel which it had put in motion it accumulated round the prow of the vessel in a state of violent agitation then suddenly leaving it behind rolled forward with great velocity assuming the form of a large solitary elevation a rounded smooth and well defined heap of water which continued its course along the channel apparently without change of form or diminution of speed I followed it on horseback and overtook it still rolling on at a rate of some eight or nine miles an hour 1 Author s address Klaus Brauer University of Osnabruck Department of Mathematics amp Computer Science Applied Systems Science D49069 OSNABRUCK Germany Phone 495419692598 eMail kbrauer usfmqiosnabrueckde Internet http wwwusf uniosnabrueck deNkbrauer 2 Klaus Brauer The Korteweg de Vries Equation preserving its original figure some thirty feet long and afoot to afoot and a half in height Its height gradually diminished and after a chase of one or two miles I lost it in the windings of the channel Such in the month ofAugust I 834 was my first chance interview with that singular and beautiful phenomenon which I have called the Wave of Translation This event took place on the Union Canal at Hermiston very close to the Riccarton campus of HeriotWatt University Edinburgh Throughout his life Russell remained convinced that his solitary wave the Wave of Translation was of fundamental importance but ninteenth and early twentieth century scientists thought otherwise His fame has rested on other achievements To mention some of his many and varied activities he developed the quotwave linequot system of hull construction which revolutionized ninteenth century naval architecture and was awarded the gold medal of the Royal Society of Edinburgh in 1837 He began steam carriage service between Glasgow and Paisley in 1834 and made the first experimental observation of the quotDoppler shi quot of sound frequency as a train passes He reorganized the Royal Society of Arts founded the Institution of Naval Architects and in 1849 was elected Fellow of the Royal Society of London He designed with Brunel the quotGreat Easternquot and built it he designed the Vienna Rotunda and helped to design Britain s first armoured warship the quotWarriorquot He developed a curriculum for technical education in Britain and it has recently become known that he attempted to negotiate peace during the American Civil War It was not until the mid 1960 s when applied scientists began to use modern digital computers to study nonlinear wave propagation that the soundness of Russell s early ideas began to be appreciated He viewed the solitary wave as a selfsuf cient dynamic entity a quotthingquot displaying many properties of a particle From the modern perspective it is used as a constructive element to formulate the complex dynamical behaviour of wave systems throughout science from hydrodynamics to nonlinear optics from plasmas to shockwaves from tomados to the Great Red Spot of Jupiter from the elementary particles of matter to the elementary particles of thought For a more detailed and technical account of the solitary wave see Bullough 1988 The phenomenon described by Russell can be expressed by a nonlinear Partial Diiferential Equation of the third order To remind A partial diiferential equation PDE is a mathematical equation which contains an unknown lnction of more than one variable as well as some derivatives of that lnction with respect to the diiferent independent variables In practical applications where the PDE describes a dynamic process one of the variables has the meaning of the time hence denoted by t and the other normally only up to 3 variable have the meaning of the space hence denoted by x y and z We consider the simplest case in only one space variable x So we are looking for a lnction u depending on the variables x and t ie ux t which describes the elongation of the wave at the place x at time t Using the typical short denotations as 2 3 urban 3uxt uxogt 8uxt Lamar 8 uxt moan 8 uxt 8t 8x 8x2 8x3 Klaus Brauer The Korteweg de Vries Equation 3 the problem can be formulated as l utxt 6uxt uxxt uWxt 0 This is the Kortewegde Vries Equation KdV which is nonlinear because of the product shown in the second summand and which is of third order because of the third derivative as highest in the third summand The factor 6 is just a scaling factor to make solutions easier to describe The aim here is to find general exact solutions to l ie here we have neither initial conditions nor boundary conditions The solutions to l are called Solitons or Solitary waves The Kortewegde Vries is a hyperbolic PDE in the general sense of the hyperbolicity definition From that it follows that it describes a reversible dynamical process The author s interest for analytical solutions of l stems from the fact that in applying numerical methods to nonlinear PDEs the KdV equation is well suited as a test object since having an analytical solution statements can be made on the quality of the numerical solution in comparing the numerical result to the exact result The main part of the subsequent deduction of an analytical solution to l is taken from Vvedenskii 1992 who has shown how to find an exact solution to the KdV equation and who has used the tool of Backlund transform to obtain an analytical solution and who in addition to that performs nonlinear superposition of two three and more solutions corresponding to two three or more soliton waves by using Backlund transform again 2 Exact Solution to the KdV Equation We remember that the simplest mathematical wave is a lnction of the form u x t fx c t which eg is a solutionto the simple PDE u c ux 0 where c denotes the speed ofthe wave For the well known wave equation uquot c2 u 0 the famous d Alembert solution leads to two wave fronts represented by terms fx c t and fx c t Hence we start here with a trial solution 2 x90 Zxl3t E ZOE just denoting the parameter 0 above by 3 here and the function f by 2 Substituting the trial solution 2 into 1 we are led to the Ordinary DiiTerential Equation 3 6Z 0 Integration can be done directly since 3 is a form of atotal derivative It follows from 3 4 Klaus Brauer The Korteweg de Vries Equation 4 Bz 32Z d c where 01 is the constant of integration In order to obtain a first order equation for Z a multiplication with 21 is done ie B2 322 d22 dz Cl di d d gz 61 di 2 zdz 3zzdz dzc dz B 2 1 d39 Integration on both sides with c2 as the constant of integration leads to l3 2 3 1 dz 2 32 2 Ed i CIZC2 Now it is required thatincase x gt ion we shouldhave z gt 0 gt 0 01 2 gt 0 d From these requirements it follows c1 c2 0 Remark More general solutions can be found for other choices of c1 and c2 These solutions can be represented in terms of elliptic integrals for details see Drazin 1983 With c1 C2 0 equation 5 can be written as 6 lam 22gt d i By seperation of variables we may write 0 0 The choice of 0 for the lower integration limits does not bring any loss of generality since the starting point can be transformed linearily The integration of the le hand side of 7 can be done by using a transformation 1 2 8 s gheck w Klaus Brauer The Korteweg de Vries Equation 5 The role of 3 here is played by the variable C and we obtain 9 B 2Q3l sech2w tanhzw since the relation cosh2 w sinh2 w lholds Furthermore we have dc B sinhw 10 61W cosh3w The upper integration limit of the le hand integral in 7 due to 8 is transformed to 11 w sech391 Substituting 9 10 and l 1 into 7 we get g lf 1 sinhw dw lfCOSh2W39COShW sinhw 0 SechZW39tanhw cosh3w 0 SinhW cosh3w dw 7 5 dw w With 8 the transform back to C is done and we obtain g sech 1 2 gt z gsechag Now we use 2 and we finally get 12 uxt gsechi w BU Remarks In order to have a real solutionthe quantity 3 must be a positive number As it is easily seen from 12 for B gt 0 the solitary wave moves to the right The second point is that the amplitude is proportional to the speed which is indicated by the value of Thus larger amplitude solitary waves move with a higher speed than smaller amplitude waves To perform superpositions later see 4 we consider the following If instead of 8 we select the transformation 13 s Bcschzw 6 Klaus Brauer The Korteweg de Vries Equation then in the same way as we obtained the solution 12 we will obtain another solution which is l5 2 J5 14 uxt Ecsch Tx t The solution 14 is an irregular solution to the KdV Equation It has a singularity for vanishing argument of the cosech lnction ie for the line in the x t plane with x t0 gt tx The proof that both lnctions 12 and 14 are solutions to the original KdV Equation can be easily varified with the help of M athematica as shown in the sequel PDEKdVu x t at ux t 6 ux t 6xux t 633 ux t 2 sollx t ESechw 2 2 7 2 sol2x t CschW Bt 2 2 Next these solutions sell and 3012 are inserted into PDEKdV SimplifyPDEKdVsoll x t True SimplifyPDEKdVsol2 x t True 3 Exact Solution with Biicklund transfom We are now going to construct a solution to l with the help of Backlund transform since this technique is used later fornonlinear superposition We introduce a lnction vwiththe property vx u and it follows from 1 u wi x 15 v6vxvxxv gvt3vxzv 0 Integration with respect to x yields 16 v 3vx2 VW ft Klaus Brauer The Korteweg de Vries Equation 7 where the lnction on the right hand side is the integration lnction The new lnction v 39 is introduced with vquot v jfsds The solution u to 1 will be obtained from the solution v to 16 and since v vx there is no need for a distinction between v 39 and v So we may set f E 0 without loss of generality Now our purpose is to solve the PDE 17 vt3vxzv 0 xxx A Backlund transform which leaves an equation invariant is calledAuto B icklund transform This AutoBacklund transform is given by the following set of equations N 1 NZ v vx 5 3v v 18 a v v Gv G 2vxzvx7 7 2 The parameter 3 is the so called Backlund parameter for details we refer to Vvedenskii 1992 chapter 9 We now may generate a non trivial solution by applying 18 to the trivial solution which clearly ll lls 17 With G E 0 the first two lines of 18 result into vx 3 v 2 19 v v v 2 v 2 t xx x The first formula in 19 may easily be integrated and be inserted into the second Two types of solutions ll ll 19 vxt Wtanh xZBI 20 Jon m coth x 2 t It is easy to verify that both lnctions 20 do ll ll 19 and 17 The second lnction has a singularity for vanishing argument of coth This irregular solution here and in lture is denoted 8 Klaus Brauer The Korteweg de Vries Equation with a bar Now it is easy by integrating with respect to x to deduce from 20 two solutions a regular one and an irregular one to the KdV equation 1 These solutions are uxt vxxt 3 sech ZBO 21 17xt m 3 cschZLgm z nl Again we are led to the same solution we already found with 12 and 14 The only diiTerence is the parameter If we denote the parameter 3 in 21 by 3 and compare it with the parameter 3 from 12 and 14 then the relation 6 holds We may verify the steps withMathematica PDEoriv x t atvx t 3 6xvx t12 6x3 vx t 1 PDE1v x t ax vx t B vx v12 PDE2v x t ath t vx t 69 vx t 2 6xvx t12 With DSo lve we may obtain a solution with the following delayed lnction solux t vx t SimplifyDSolvePDE1v x t vx t x t 1 solux t TanhI3 X Zfll 39 2 Typically the arbitrary lnction is denoted by C l t Insertion into PDEl is done and the attempt is made whether the Coth lnction instead of the Tanh lnction ll lls PDEl Simplify PDEl solu x t True SimplifyPDE1solu Tanh gt Coth x t True Both the lnctions ll ll 19 rst equation Now the second equation is solved by spec C1 t DSolvePDE2solu x t C1 t thl t3Cl speco specl C1 gt 0 t5 Klaus Brauer The Korteweg de Vries Equation 9 where the integration constant has been set to zero With this result we de ne the two lnctions soluspec1x t solux t at C1 t gt at specO C1 t gt specO soluspec2x t soluspec1x t Tanh gt Coth soluspec1x t 3 v3 Tang W3 tum 2 soluspec2x t M x 2 t8 xj xE V3 Coth Both these lnctions clearly ful ll the second equation of 19 and the PDE 17 SimplifyPDE2soluspec1 x t True SimplifyPDE2 soluspec2 x t True SimplifyPDEorisoluspec1 x t True SimplifyPDEorisoluspec2 x t True Finally the solution to the KdV equation is given by the rst derivative with respect to x 0x soluspec1x t xE x 7 2 t 3 2 x 0x soluspec2x t 6 Sech 43 Csch B 2 t5 2 Though the irregular solution is not acceptable in physical sense it will be used to construct higher order solutions by using the nonlinear superposition principle 10 Klaus Brauer The Korteweg de Vries Equation 4 Nonlinear superposition Let I be any solution to 17 and let QJB be a solution obtained from I by applying the Backlund transform with the Backlund parameter To remind what we did We had chosen the trivial solution I E 0 and from that we received 20 as a solution to 17 We now consider two solutions I 31 and I 32 obtained by applying the Backlund transform to I with parameters land 3 From 18 we have only the rst line is used WI 41 131 ltIltDB1z 22 15032 ltIgtx 132 gob 032 Let I 32 31 be the solution obtained from I by successive application of 18 rst line rst step with parameter 31 and second step with parameter 32 Then we have 23 QMBP l XBl 52 BI B2B1z In the same way let I31 32 be the solution by rst applying Bl and a erwards 31 Instead of 23 we then have 24 QMBP Z XBz Bl BzqB1Bzz We now demand that 25 Bza 1 3031452 LP and try to solve 23 and 24 for 1 Subtraction of the second line in 22 from the rst yields 26 15031 15032 I31 I32 1 l31 DBz39 1 l51 NBZ 2 1 Subtracting 24 from 23 end using 25 leads to 27 quail M2 B2 1 gown z39 1 W2 2 By eliminating the le hand sides in 26 and 27 we get Klaus Brauer The Korteweg de Vries Equation 31 I32 DBl Bz39 1 l31 1032 2 1 1 I32 51 l31 l32391gtl51 Bz 2 1 231 39 I32 LP 1 I l31 1032 Bl BZ 9 91 q 2 28 13051 w 052 We have to show now that the lnction 1 from 28 indeed is a solution to 17 The expression 28 is di erentiated with respect to x and to t and three times with respect to x and is inserted into 17 The result is w339 f PW21DltDW 2 I w B I w 5 B3D B 1 2 am w my 1 2 mm w my 29 I B V I 5 a B V I B 3 2l3 B m1 m 2 12B B X1 X Z 1 2 am w my 1 2 mm w My a w I I w I 12631 32 lt am as lt RG51 m mm w M32 The second derivates with respect to x in 26 are formed resulting in 30 l31 QABZ x51 QXB239DB1 Bz 21 D 1 Dl3239 1 x 1 15632 2 15 The formula 30 has to be inserted into the last summand in 29 thus eliminating the second derivatives If we do so and keep in mind that each of the lnctions I IB1 I32 are solutions to 17 then we get from 29 2 aw aw gtgtZ aw 41w 139 3 PW 1251 32Y1 Z4 3912B1Bz1 Z4 my w 052 mil w 052 31 am w XltIsgtgtzltlt1gtltisgt M w 2ltIgtgt 6 51 32 mm w my 12 Klaus Brauer The Korteweg de Vries Equation Now from 26 the expression IB1 z 2 D is calculated and inserted into the last term of 3 l A er some rearrangement the sum of the terms on the righthandside of 3 l is zero thus showing that the lnction LP is indeed a solution to 17 The main formula for constructing lrther solutions is 28 If we take I E 0 which clearly is a solution to 17 and if we take the regular and the irregular solutions 20 which ful ll 17 then these two solutions play the role of I 31 and D32 in 28 Using two regular solutions first solution of 20 with two diiTerent values of B would lead to an irregular solution I by using 28 This can be seen as follows from 20 first equation with a xed value for the parameter B 32 lim vxt lim thh fgx 2 t im xv25 ix xVZBt iamp Since the lnction v is continous there will always be one value of the argument for wich the denominator in 28 is zero leading to a singularity in 1 So we use both the regular and the irregular solution from 20 to construct a regular solution by using 28 The solutions from 20 are denoted by v31and 7 32analogous to what we did here before the bar again denotes the irregular solution The two Backlund parameters are chosen as 32 gt 31 This ensures that the denominator in 28 does not vanish for any values of the arguments Thus we get a superpositioned lnction 2B1 B2 Vl31xat 50329590 B2 gt B1 33 Pom Since the lnction i from 33 is a solution to 17 it has to be integrated with respect to x to obtain a superpositioned solution to the original KdV equation This process of supositioning a regular and an irregular solution to 17 can be iterated We have to apply 28 again to 3 solutions Instead of I in 28 we now take the regular solution from 20 with a Backlund parameter B which is di erent from 51 and 32 and name it v3x t The role of the lnction I31 in 28 is now played by the lnction I from 33 for the two Backlund parameters land 5 and thus is named as D313xt 23 31 v3xt M3133quot B1 gt3 34 p am This solution is regular a corresponding irregular solution denoted by a bar then is gained by combining two regular solutions as mentioned above 2l32 3 35 Wz xquot z Klaus Brauer The Korteweg de Vries Equation 13 The two lnctions of the denominators of 34 and 35 are again the solutions 20 Ifwe combine the two lnctions of 34 and 35 we want to get a regular solution for a 3soliton solution It is necessary then that it holds 32 3 gt 31 Together with the regularity condition in 34right we then have 32 gt 31 gt 3 as a regularity condition Finally according to 28 the new 3soliton solution is 36 I xt v3xt L DB1Bxat DB2Bxat Again 36 is a solution to 17 it has to be intgrated with respect to x to get a solution to the original KdV equation Principally by superposition a regular nsoliton can be constructed from a pair of irregular and regular n lsolutions 5 Mathematica code for one two and three solitons The solutions from 21 as well as the solutions got from 33 and 36 are coded inMathematica as follows KdVBval x t Blockvv v w superpos2 superpos3 qurr bReg Solution Clear x t vxx tt6 26Tanh4l xx 26tt 2 wxx tt 6 V2e Cot11H xx 26tt 2 3162 superpos2xx tt 61 62 vxx tt 61 wxx tt 62 2 32 n6 Irrxx tt 32 3 vxx tt 62 vxx tt 6 2 331 Regxx tt 61 6 lex tt 6 Wxx tt 61 superpos3 xx tt 6 61 32 2 61 62 Regxx tt 61 6 Irrxx tt 62 6 39 W Sort Bval If LengthBval 1 Solutionaxvx t 6 6gtvv11 If Length Bval 2 Solution ax superpos2x t 61 62 31 gtW111I 62 gtWII2JI Solution axsuperpos3x t 6 61 62 3 gtWII1I 61 gtvvr21162 gtvvr311 1 vxx tt 6 14 Klaus Brauer The Korteweg de Vries Equation The rst parameter of this block is a list If it contains one element a onesoliton solution is calculated if it contains two or three elements 2 or 3soliton solutions are calculated It is ensured inthe code that the condition 32 gt 31 gt 3 is not Violated Calls results and proofs are KdVOpu at u 6u6x u 6x3u One KdV31 x t 2 v29 ltx 2t3n l 131 E Simpli fy KdVOp One Sech 0 Two KdV31 32 x t r 3 X72 t 51 2 Ji 2 2 131 252 Ralph 7 BHCSCMWJ Bi 2 l X72t821 59215311 l2 L NECOth xEJr xE Tanh Simpli fy KdVOp Two 0 For three solitons an analytical solution was obtainedbutMathematica was not able to verifythat it truely ll lls the KdV equation Three KdVliBl I32 Ba x t W71 x zt l 2 Sech 2 l 51 252133 7 i 2 if 2 quotf x2tfa quotf x2tfa 51Csch v39 f 2 Bl x32 Sechl v2 7 r 7 i o i 2 43 Coth 2 gme xBg 2 131 Tanh 1 2th V 2 l x 2tfil BlSech v f 2 131 133 i Sech f i i uquot v i i uquot v VZ V131 Tanh 361 722t1gt2 V133 Tanh lt3 ft3 v V2 Klaus Brauer The Korteweg de Vries Equation 15 2 131 82 4 Coth JBZ if 52quot 1 m v E Tanh 931 if 51quot 1 2 13133 A E 31 Tanh l l if t l vh 33 Tanh 9153 ligand It is easy now to write 3 additional blocks in Mathematica to produce a 3D plot a contour plot and a sequence of 2D plots which can be animated The code is straightforward the results are not shown here but may be seen visiting Brauer 2000 further information is got by visiting Eilbeck 1998 These analytical solutions are very help ll when applying numerical methods since a comparision is possible see de Frutos amp SanzSerna 1997 and Schiesser 1994 A diiTerent approach in obtaining solutions to the KdV equation is shown by Varley amp Seymour 1998 References Brauer K 2000 httpwwwusfuni osnabrueclcaleNkb 139 hfml Bullough RK 1988 quotT he Wave par excellencequot the solitary progressive great wave of equilibrium of the uid an early history of the solitary wave In Solitons M Lakshmanan Ed Springer Series in Nonlinear Dynamics 150281 New York Berlin Heidelberg etc Springer DraZin PG 1983 Solitons London Cambridge University Press London Mathematical Society Lecture Note Series 85 ISSN 00760552 Eilbeck C 1998 httpwwwmahwacaldwchrisscottirussellhiml de Frutos 1 SanzSerna JM 1997 Accuracy and conservation properties in numerical integration the case of the Korteweg de Vries equation Num Math 75 421 445 Russell J S 1845 Report on Waves Report of the 14th meeting of the British Association for the Advancement of Science York September 1844 pp 311390 Plates XLVHLVII London Schiesser WE 1994 Method of lines solution of the Korteweg de Vries equation Comp ampMaths withAppls 28 147 154 Varley E Seymour BR 1998 A Simple Derivation of the N Soliton Solutions to the Korteweg de Vries Equation SIAM Journal on Appl Math 58 904911 Vvedenskii D 1992 Partial Differential Equations with Mathematica Wokingham England etc AddisonWesley Wolfram S 1999 The Mathematica Book Cambridge UK Cambridge University Press

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