Chapter 10 Solitons Starting in the 19th century, researchers found that certain nonlinear PDEs admit exact solutions in the form of solitary waves, known today as solitons. There’s a famous story of the Scottish engineer, John Scott Russell, who in 1834 observed a hump-shaped disturbance propagating undiminished down a canal. In 1844, he published this observation1 , writing, “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, preserving its original figure some thirty feet long and a foot to a foot 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 of August 1834, was my first chance interview with that singular and beautiful phenomenon which I have called the Wave of Translation”. Russell was so taken with this phenomenon that subsequent to his discovery he built a thirty foot wave tank in his garden to reproduce the effect, which was precipitated by an initialpsudden displacement of water. Russell found empirically that the velocity obeyed g(h + um ) , where h is the average depth of the water and um is the maximum v ≃ vertical displacement of the wave. He also found that a sufficiently large initial displacement would generate two solitons, and, remarkably, that solitons can pass through one another undisturbed. It was not until 1890 that Korteweg and deVries published a theory of shallow water waves and obtained a mathematical description of Russell’s soliton. 1 J. S. Russell, Report on Waves, 14th Meeting of the British Association for the Advancement of Science, pp. 311-390. 1 2 CHAPTER 10. SOLITONS Nonlinear PDEs which admit soliton solutions typically contain two important classes of terms which feed off each other to produce the effect: DISPERSION −⇀ ↽− NONLINEARITY The effect of dispersion is to spread out pulses, while the effect of nonlinearities is, often, to draw in the disturbances. We saw this in the case of front propagation, where dispersion led to spreading and nonlinearity to steepening. In the 1970’s it was realized that several of these nonlinear PDEs yield entire families of exact solutions, and not just isolated solitons. These families contain solutions with arbitrary numbers of solitons of varying speeds and amplitudes, and undergoing mutual collisions. The three most studied systems have been • The Korteweg-deVries equation, ut + 6uux + uxxx = 0 . (10.1) This is a generic equation for ‘long waves’ in a dispersive, energy-conserving medium, to lowest order in the nonlinearity. • The Sine-Gordon equation, φtt − φxx + sin φ = 0 . (10.2) The name is a play on the Klein-Gordon equation, φtt − φxx + φ = 0. Note that the Sine-Gordon equation is periodic under φ → φ + 2π. • The nonlinear Schrödinger equation, iψt ± ψxx + 2|ψ|2 ψ = 0 . (10.3) Here, ψ is a complex scalar field. Depending on the sign of the second term, we denote this equation as either NLS(+) or NLS(−), corresponding to the so-called focusing (+) and defocusing (−) cases. Each of these three systems supports soliton solutions, including exact N -soliton solutions, and nonlinear periodic waves. 10.1 The Korteweg-deVries Equation Let h0 denote the resting depth of water in a one-dimensional channel, and y(x, t) the vertical displacement of the water’s surface. Let L be a typical horizontal scale of the wave. When |y| ≪ h0 ≪ L, and v ≈ 0 (speed of propagation small compared with c0 ), the evolution of an x-directed wave is described by the KdV equation, yt + c0 yx + 3c0 yy + 1 c h2 y =0, 2h0 x 6 0 0 xxx (10.4) 3 10.1. THE KORTEWEG-DEVRIES EQUATION p where c0 = gh0 . For small amplitude disturbances, only the first two terms are consequential, and we have yt + c0 yx ≈ 0 , (10.5) the solution to which is y(x, t) = f (x − c0 t) , (10.6) where f (ξ) is an arbitrary shape; the disturbance propagates with velocity c0 . When the dispersion and nonlinearity are included, only a particular pulse shape can propagate in an undistorted manner; this is the soliton. It is convenient to shift to a moving frame of reference: x′ = x − c0 t hence ∂ ∂ = ∂x ∂x′ Thus, yt′ + , , t′ = t , ∂ ∂ ∂ = ′ − c0 ′ . ∂t ∂t ∂x 3c0 y yx′ + 61 c0 h20 yx′ x′ x′ = 0 . 2h0 (10.7) (10.8) (10.9) Finally, rescaling position, time, and displacement, and dropping the primes, we arrive at the KdV equation, ut + 6uux + uxxx = 0 , (10.10) which is a convenient form. 10.1.1 KdV solitons We seek a solution to the KdV equation of the form u(x, t) = u(x − V t). Then with ξ ≡ x − V t, we have ∂x = ∂ξ and ∂t = −V ∂ξ when acting on u(x, t) = u(ξ). Thus, we have −V u′ + 6uu′ + u′′′ = 0 . (10.11) −V u + 3u2 + u′′ = A , (10.12) Integrating once, we have where A is a constant. We can integrate once more, obtaining − 12 V u2 + u3 + 12 (u′ )2 = Au + B , (10.13) where now both A and B are constants. We assume that u and all its derivatives vanish in the limit ξ → ±∞, which entails A = B = 0. Thus, √ du = ± u V − 2u . dξ (10.14) u = 21 V sech2 θ , (10.15) With the substitution 4 CHAPTER 10. SOLITONS Figure 10.1: Soliton solutions to the KdV equation, with five evenly spaced V values ranging from V = 2 (blue) to V = 10 (orange). The greater the speed, the narrower the shape. √ we find dθ = ∓ 12 V dξ, hence we have the solution √ u(x, t) = 12 V sech2 2V (x − V t − ξ0 ) . (10.16) Note that the maximum amplitude of the soliton is umax = 21 V , which is proportional to its velocity V . The KdV equation imposes no limitations on V other than V ≥ 0. 10.1.2 Periodic solutions : soliton trains If we relax the condition A = B = 0, new solutions to the KdV equation arise. Define the cubic P (u) = 2u3 − V u2 − 2Au − 2B ≡ 2(u − u1 )(u − u2 )(u − u3 ) , (10.17) where ui = ui (A, B, V ). We presume that A, B, and V are such that all three roots u1,2,3 are real and nondegenerate. Without further loss of generality, we may then assume u1 < u2 < u3 . Then p du = ± −P (u) . (10.18) dξ Since P (u) < 0 for u2 < u < u3 , we conclude u(ξ) must lie within this range. Therefore, we have 1/2 Zφ Zu 2 ds dθ p ξ − ξ0 = ± p =± , (10.19) u − u −P (s) 1 − k2 sin2 θ 3 1 0 u2 where u ≡ u3 − (u3 − u2 ) sin2 φ , k2 ≡ u3 − u2 . u3 − u1 (10.20) 5 10.1. THE KORTEWEG-DEVRIES EQUATION Figure 10.2: The Jacobi elliptic functions sn(ζ, k) (solid) and cn(ζ, k) (dot-dash) versus ζ/K(k), for k = 0 (blue), k = √12 (green), and k = 0.9 (red). The solution for u(ξ) is then u(ξ) = u3 − (u3 − u2 ) sn2 (ζ, k) , where r u3 − u1 ξ − ξ0 2 and sn(ζ, k) is the Jacobi elliptic function. ζ= 10.1.3 (10.21) (10.22) Interlude: primer on elliptic functions We assume 0 ≤ k2 ≤ 1 and we define Zφ ζ(φ, k) = p 0 dθ 1 − k2 sin2 θ . (10.23) The sn and cn functions are defined by the relations sn(ζ, k) = sin φ cn(ζ, k) = cos φ . (10.24) Note that sn2 (ζ, k) + cn2 (ζ, k) = 1. One also defines the function dn(ζ, k) from the relation dn2 (ζ, k) + k2 sn2 (ζ, k) = 1 . (10.25) When φ advances by one period, we have ∆φ = 2π, and therefore ∆ζ = Z, where Z2π Z= p 0 dθ 1 − k2 sin2 θ = 4 K(k) , (10.26) 6 CHAPTER 10. SOLITONS Figure 10.3: The cubic function P (u) (left), and the soliton lattice (right) for the case u1 = −1.5, u2 = −0.5, and u3 = 2.5. where K(k) is the complete elliptic integral of the first kind. Thus, sn(ζ + Z, k) = sn(ζ, k), and similarly for the cn function. In fig. 10.2, we sketch the behavior of the elliptic functions over one quarter of a period. Note that for k = 0 we have sn(ζ, 0) = sin ζ and cn(ζ, 0) = cos ζ. 10.1.4 The soliton lattice Getting back to our solution in eqn. 10.21, we see that the solution describes a soliton lattice with a wavelength √ 8 K(k) . (10.27) λ= p u3 − u1 Note that our definition of P (u) entails V = 2(u1 + u2 + u3 ) . (10.28) There is a simple mechanical analogy which merits illumination. Suppose we define and furthermore E ≡ B. Then W (u) ≡ u3 − 12 V u2 − Au , (10.29) 1 du 2 + W (u) = E , 2 dξ (10.30) which takes the form of a one-dimensional Newtonian mechanical system, if we replace ξ → t and interpret uξ as a velocity. The potential is W (u) and the total energy is E. In terms of the polynomial P (u), we have P = 2(W − E). Accordingly, the ‘motion’ u(ξ) is 7 10.1. THE KORTEWEG-DEVRIES EQUATION Figure 10.4: The cubic function P (u) (left), and the soliton lattice (right) for the case u1 = −1.5, u2 = −1.49, and u3 = 2.50. flattest for the lowest values of u, near u = u2 , which is closest to the local maximum of the function W (u). Note that specifying umin = u2 , umax = u3 , and the velocity V specifies all the parameters. Thus, we have a three parameter family of soliton lattice solutions. 10.1.5 N-soliton solutions to KdV In 1971, Ryogo Hirota2 showed that exact N -soliton solutions to the KdV equation exist. Here we discuss the Hirota solution, following the discussion in the book by Whitham. The KdV equation may be written as ut + 3u2 + uxx x = 0 , (10.31) which is in the form of the one-dimensional continuity equation ut +jx = 0, where the current is j = 3u2 + uxx . Let us define u = px . Then our continuity equation reads ptx + jx = 0, which can be integrated to yield pt + j = C, where C is a constant. Demanding that u and its derivatives vanish at spatial infinity requires C = 0. Hence, we have pt + 3p2x + pxxx = 0 . (10.32) Now consider the nonlinear transformation p = 2 (ln F )x = 2 R. Hirota, Phys. Rev. Lett. 27, 1192 (1971). 2Fx . F (10.33) 8 CHAPTER 10. SOLITONS We then have pt = 2Fxt 2Fx Ft − F F2 , px = 2Fxx 2Fx2 − 2 F F (10.34) and pxx = pxxx 2Fxxx 6Fx Fxx 4Fx3 − + 3 F F2 F 2Fxxxx 8Fx Fxxx 6Fx x2 24Fx2 Fxx 12Fx4 = − − + − . F F2 F2 F3 F4 (10.35) When we add up the combination pt + 3p2x + pxxx = 0, we find, remarkably, that the terms with F 3 and F 4 in the denominator cancel. We are then left with 2 F Ft + Fxxx x − Fx Ft + Fxxx + 3 Fxx − Fx Fxxx = 0 . (10.36) This equation has the two-parameter family of solutions F (x, t) = 1 + eφ(x,t) . (10.37) φ(x, t) = α (x − b − α2 t) , (10.38) where with α and b constants. Note that these solutions are all annihilated by the operator ∂t +∂x3 , and also by the last term in eqn. 10.36 because of the homogeneity of the derivatives. Converting back to our original field variable u(x, t), we have that these solutions are single solitons: α2 f 2(F Fxx − Fx2 ) (10.39) = = 12 α2 sech2 ( 21 φ) . u = px = F2 (1 + f )2 The velocity for these solutions is V = α2 . If eqn. 10.36 were linear, our job would be done, and we could superpose solutions. We will meet up with such a felicitous situation when we discuss the Cole-Hopf transformation for the one-dimensional Burgers’ equation. But for KdV the situation is significantly more difficult. We will write F = 1 + F (1) + F (2) + . . . + F (N ) , (10.40) with F (1) = f1 + f2 + . . . + fN , (10.41) where fj (x, t) = eφj (x,t) φj (x, t) = αj (x − α2j t − bj ) . We may then derive a hierarchy of equations, the first two levels of which are (1) (1) Ft + Fxxx =0 x (2) (2) (1) (1) (1) Ft + Fxxx = −3 Fxx Fxx − Fx(1) Fxxx . x (10.42) (10.43) 9 10.1. THE KORTEWEG-DEVRIES EQUATION Let’s explore the case N = 2. The equation for F (2) becomes (2) (2) Ft + Fxxx x = 3 α1 α2 (α2 − α1 )2 f1 f2 , with solution F (2) = α1 − α2 α1 + α2 2 f1 f2 . (10.44) (10.45) Remarkably, this completes the hierarchy for N = 2. Thus, F = 1 + f1 + f2 + 1 + f1 = det 2√α α 1 2 α +α f2 1 α1 − α2 α1 + α2 √ 2 2 α1 α2 α1 +α2 f1 1 + f2 2 f1 f2 (10.46) . What Hirota showed, quite amazingly, is that this result generalizes to the N -soliton case, F = det Q , where Q is matrix Qmn √ 2 αm αn = δmn + f . αm + αn m (10.47) (10.48) Note that for N = 2 this agrees with Eqn. 10.46. Thus, N -soliton solutions to the KdV equation may be written in the form u(x, t) = 2 ∂2 ln det Q(x, t) . ∂x2 (10.49) Here we are free to make a similarity transformation Q → S −1 Q S for any nonsingular e where matrix S. If we take Smn = eφn /2 δmn , then Q → Q, √ 2 αm αn p −φm /2 +φn /2 e Q=e Qmn e = δmn + fm fn (10.50) αm + αn is symmetric. Consider the case N = 2. Direct, if tedious, calculations lead to the expression α −α 2 α21 f1 f22 + α22 f12 f2 α21 f1 + α22 f2 + 2 (α1 − α2 )2 f1 f2 + α1 +α2 1 2 . u=2 i2 h α1 −α2 2 1 + f1 + f2 + α +α f1 f2 1 Recall that (10.51) 2 fj (x, t) = exp αj (xj − α2j t − bj ) . (10.52) Let’s consider (x, t) values for which f1 ≃ 1 is neither large nor small, and investigate what happens in the limits f2 ≪ 1 and f2 ≫ 1. In the former case, we find u≃ 2α21 f1 (1 + f1 )2 (f2 ≪ 1) , (10.53) 10 CHAPTER 10. SOLITONS Figure 10.5: Early and late time configuration of the two soliton solution to the KdV equation. which is identical to the single soliton case of eqn. 10.39. In the opposite limit, we have u≃ 2α21 g1 (1 + g1 )2 where g1 = (f2 ≫ 1) α1 − α2 α1 + α2 2 , f1 (10.54) (10.55) But multiplication of fj by a constant C is equivalent to a translation: C fj (x, t) = fj (x + α−1 j ln C , t) ≡ fj (x − ∆xj , t) . (10.56) Thus, depending on whether f2 is large or small, the solution either acquires or does not acquire a spatial shift ∆x1 , where α1 + α2 2 . (10.57) ln ∆xj = αj α1 − α2 The function f (x, t) = exp α(x − α2 t − b) is monotonically increasing in x (assuming α > 0). Thus, if at fixed t the spatial coordinate x is such that f ≪ 1, this means that the soliton lies to the right. Conversely, if f ≫ 1 the soliton lies to the left. Suppose α1 > α2 , in which case soliton #1 is stronger (i.e. greater amplitude) and faster than soliton #2. The situation is as depicted in figs. 10.5 and 10.6. Starting at early times, the strong soliton lies to the left of the weak soliton. It moves faster, hence it eventually overtakes the weak soliton. As the strong soliton passes through the weak one, it is shifted forward, and the weak soliton is shifted backward. It hardly seems fair that the strong fast soliton gets 10.1. THE KORTEWEG-DEVRIES EQUATION 11 Figure 10.6: Left: Spacetime diagram (x versus t) for the collision of two KdV solitons. The strong, fast soliton (#1) is shifted forward and the weak slow one (#2) is shifted backward. The red and blue lines indicate the centers of the two solitons. The yellow shaded circle is the ‘interaction region’ where the solution is not simply a sum of the two single soliton waveforms. Right: t versus x showing the soliton waveforms through the collision. Here, the velocity of the slow soliton is close to zero. Note that the slow soliton is again shifted backward. From P. J. Caudrey, Phil. Trans. Roy. Soc. A 28, 1215 (2011). pushed even further ahead at the expense of the weak slow one, but sometimes life is just like that3 . 10.1.6 Bäcklund transformations For certain nonlinear PDEs, a given solution may be used as a ‘seed’ to generate an entire hierarchy of solutions. This is familiar from the case of Riccati equations, which are nonlinear and nonautonomous ODEs, but for PDEs it is even more special. The general form of the Bäcklund transformation (BT) is u1,t = P (u1 , u0 , u0,t , u0,x ) (10.58) u1,x = Q(u1 , u0 , u0,t , u0,x ) , (10.59) where u0 (x, t) is the known solution. A Bäcklund transformation for the KdV equation was first obtained in 19734 . This provided a better understanding of the Hirota hierarchy. To proceed, following the discussion in the book by Scott, we consider the earlier (1968) result of Miura5 , who showed that if v(x, t) 3 See T. Piketty, Capital in the Twenty-First Century (Belknap Press, 2014). H. D. Wahlquist and F. B. Eastabrook, Phys. Rev. Lett. 31, 1386 (1973). 5 R. M. Miura, J. Math. Phys. 9, 1202 (1968). 4 12 CHAPTER 10. SOLITONS Figure 10.7: “Wrestling’s living legend” Bob Backlund, in a 1983 match, is subjected to a devastating camel clutch by the Iron Sheikh. The American Bob Backlund has nothing to do with the Bäcklund transformations discussed in the text, which are named for the 19th century Swedish mathematician Albert Bäcklund. Note that Bob Backlund’s manager has thrown in the towel (lower right). satisfies the modified KdV (MKdV) equation6 , vt − 6v 2 vx + vxxx = 0 , (10.60) u = −(v 2 + vx ) (10.61) then solves KdV: ut + 6uux + uxxx = −(v 2 + vx )t + 6(v 2 + vx )(v 2 + vx )x − (v 2 + vx )xxx = − 2v + ∂x vt − 6v 2 vx + vxxx = 0 . (10.62) From this result, we have that if vt − 6(v 2 + λ) vx + vxxx = 0 , (10.63) u = −(v 2 + vx + λ) (10.64) then solves KdV. The MKdV equation, however, is symmetric under v → −v, hence u0 = −vx − v 2 − λ u1 = +vx − v 2 − λ (10.65) both solve KdV. Now define u0 ≡ −w0,x and u1 ≡ −w1,x . Subtracting the above two equations, we find u0 − u1 = −2vx ⇒ w0 − w1 = 2v . (10.66) 6 Note that the second term in the MKdV equation is proportional to v 2 vx , as opposed to uux which appears in KdV. 13 10.1. THE KORTEWEG-DEVRIES EQUATION Adding the equations instead gives w0,x + w1,x = 2(v 2 + λ) = 12 (w0 − w1 )2 + 2λ . (10.67) Substituting for v = 12 (w0 − w1 ) and v 2 + λ − 12 (w0,x + w1,x ) into the MKdV equation, we have 2 2 (w0 − w1 )t − 3 w0,x − w1,x + (w0 − w1 )xxx = 0 . (10.68) This last equation is a Bäcklund transformation (BT), although in a somewhat nonstandard form, since the RHS of eqn. 10.58, for example, involves only the ‘new’ solution u1 and the ‘old’ solution u0 and its first derivatives. Our equation here involves third derivatives. However, we can use eqn. 10.67 to express w1,x in terms of w0 , w0,x , and w1 . Starting from the trivial solution w0 = 0, eqn. 10.67 gives w1,x = 12 w12 + 2λ . (10.69) With the choice λ = − 14 α2 < 0, we integrate and obtain w1 (x, t) = −α tanh 1 2 αx + ϕ(t) , (10.70) were ϕ(t) is at this point arbitrary. We fix ϕ(t) by invoking eqn. 10.68, which says 2 w1,t = 3w1,x − w1,xxx = 0 . (10.71) Invoking w1,x = 21 w12 + λ and differentiating twice to obtain w1,xxx , we obtain an expression for the RHS of the above equation. The result is w1,t + α2 w1,x = 0, hence h i w1 (x, t) = −α tanh 12 α (x − α2 t − b) h i u1 (x, t) = 12 α2 sech2 21 α (x − α2 t − b) , (10.72) (10.73) which recapitulates our earlier result. Of course we would like to do better, so let’s try to insert this solution into the BT and the turn the crank and see what comes out. This is unfortunately a rather difficult procedure. It becomes tractable if we assume that successive Bäcklund transformations commute, which is the case, but which we certainly have not yet proven. That is, we assume that w12 = w21 , where λ λ λ λ w0 −−−1−→ w1 −−−2−→ w12 (10.74) w0 −−−2−→ w2 −−−1−→ w21 . Invoking this result, we find that the Bäcklund transformation gives w12 = w21 = w0 − 4(λ1 − λ2 ) . w1 − w2 (10.75) Successive applications of the BT yield Hirota’s multiple soliton solutions: λ λ λ λ w0 −−−1−→ w1 −−−2−→ w12 −−−3−→ w123 −−−4−→ · · · . (10.76) 14 10.2 CHAPTER 10. SOLITONS Sine-Gordon Model Consider transverse electromagnetic waves propagating down a superconducting transmission line, shown in fig. 10.8. The transmission line is modeled by a set of inductors, capacitors, and Josephson junctions such that for a length dx of the transmission line, the capacitance is dC = C dx, the inductance is dL = L dx, and the critical current is dI0 = I0 dx. Dividing the differential voltage drop dV and shunt current dI by dx, we obtain ∂V ∂I = −L ∂x ∂t (10.77) ∂V ∂I = −C − I0 sin φ , ∂x ∂t where φ is the difference φ = φupper − φlower in the superconducting phases. The voltage is related to the rate of change of φ through the Josephson equation, ∂φ 2eV = , ∂t ~ (10.78) 2eL ∂φ =− I . ∂x ~ (10.79) ∂ 2φ 1 1 ∂ 2φ − 2 + 2 sin φ = 0 , 2 2 c ∂t ∂x λJ (10.80) and therefore Thus, we arrive at the equation where c = (L C)−1/2 is the Swihart velocity and λJ = (~/2eLI0 )1/2 is the Josephson length. We may now rescale lengths by λJ and times by λJ /c to arrive at the sine-Gordon equation, φtt − φxx + sin φ = 0 . (10.81) This equation of motion may be derived from the Lagrangian density L = 12 φ2t − 21 φ2x − U (φ) . We then obtain φtt − φxx = − ∂U , ∂φ (10.82) (10.83) and the sine-Gordon equation follows for U (φ) = 1 − cos φ. Assuming φ(x, t) = φ(x − V t) we arrive at (1 − V 2 ) φξξ = U ′ (ξ), and integrating once we obtain 2 2 1 (10.84) 2 (1 − V ) φξ − U (φ) = E . This describes a particle of mass M = 1 − V 2 moving in the inverted potential −U (φ). Assuming V 2 < 1, we may solve for φξ : φ2ξ = 2(E + U (φ)) , 1−V2 (10.85) 15 10.2. SINE-GORDON MODEL Figure 10.8: A superconducting transmission line is described by a capacitance per unit length C, an inductance per unit length L, and a critical current per unit length I0 . Based on fig. 3.4 of A. Scott, Nonlinear Science. which requires E ≥ −Umax in order for a solution to exist. For −Umax < E < −Umin , the motion is bounded by turning points. The situation for the sine-Gordon (SG) model is sketched in fig. 10.9. For the SG model, the turning points are at φ∗± = ± cos−1 (E + 1), with −2 < E < 0. We can write r E (10.86) , δ = 2 cos−1 − . φ∗± = π ± δ 2 This class of solutions describes periodic waves. From √ 2 dξ dφ √ =p , 2 1−V E + U (φ) (10.87) we have that the spatial period λ is given by λ= p ∗ 2π−φ Z 2 (1 − V 2 ) φ∗ p dφ E + U (φ) , (10.88) where φ∗ ∈ [0, π]. If E > −Umin , then φξ is always of the same sign, and φ(ξ) is a monotonic function of ξ. If U (φ) = U (φ + 2π) is periodic, then the solution is a ‘soliton lattice’ where the spatial period of φ mod 2π is e= λ r 1−V2 2 Z2π p 0 dφ . E + U (φ) (10.89) 16 CHAPTER 10. SOLITONS Figure 10.9: The inverted potential −U (φ) = cos φ − 1 for the sine-Gordon problem. For the sine-Gordon model, with U (φ) = 1 − cos φ, one finds λ= p 8(1 − V 2 ) K r E+2 2 ! , (10.90) where K(k) is the complete elliptic integral of the first kindd7 and e= λ 10.2.1 r 8 (1 − V 2 ) K E+2 r 2 E+2 ! . (10.91) Tachyon solutions When V 2 > 1, we have 2 1 2 (V − 1) φ2ξ + U (φ) = −E . Such solutions are called tachyonic. There are again three possibilities: • E > Umin : no solution. • −Umax < E < −Umin : periodic φ(ξ) with oscillations about φ = 0. • E < −Umax : tachyon lattice with monotonic φ(ξ). It turns out that the tachyon solution is unstable. 7 See the NIST Handbook of Mathematical Functions, ch. 19 (10.92) 17 10.2. SINE-GORDON MODEL 10.2.2 Hamiltonian formulation The Hamiltonian density is H = π φt − L , (10.93) ∂L = φt ∂φt (10.94) where π= is the momentum density conjugate to the field φ. Then H(π, φ) = 12 π 2 + 12 φ2x + U (φ) . (10.95) Note that the total momentum in the field is Z∞ Z∞ Z∞ P = dx π = dx φt = −V dx φx −∞ −∞ −∞ h i = −V φ(∞) − φ(−∞) = −2πnV , (10.96) where n = φ(∞) − φ(−∞) /2π is the winding number . 10.2.3 Phonons The Hamiltonian density for the SG system is minimized when U (φ) = 0 everywhere. The ground states are then classified by an integer n ∈ Z, where φ(x, t) = 2πn for ground state n. Suppose we linearize the SG equation about one of these ground states, writing φ(x, t) = 2πn + η(x, t) , (10.97) and retaining only the first order term in η from the nonlinearity. The result is the KleinGordon (KG) equation, ηtt − ηxx + η = 0 . (10.98) This is a linear equation, whose solutions may then be superposed. Fourier transforming from (x, t) to (k, ω), we obtain the equation − ω 2 + k2 + 1 η̂(k, ω) = 0 , (10.99) which entails the dispersion relation ω = ±ω(k), where p ω(k) = 1 + k2 . (10.100) Thus, the most general solution to the (1 + 1)-dimensional KG equation is η(x, t) = Z∞ −∞ dk 2π √ √ 2 2 A(k) eikx e−i 1+k t + B(k) eikx ei 1+k t . (10.101) 18 CHAPTER 10. SOLITONS For the Helmholtz equation ηtt − ηxx = 0, the dispersion is ω(k) = |k|, and the solution may be written as η(x, t) = f (x − t) + g(x + t), which is the sum of arbitrary right-moving and left-moving components. The fact that the Helmholtz equation preserves the shape of the wave is a consequence of the absence of dispersion, i.e. the phase velocity vp (k) = ωk is the same as the group velocity vg (k) = ∂ω ∂k . This is not the case for the KG equation, obviously, since √ ∂ω k 1 + k2 ω , vg (k) = =√ , (10.102) vp (k) = = k k ∂k 1 + k2 hence vp vg = 1 for KG. 10.2.4 Mechanical realization The sine-Gordon model can be realized mechanically by a set of pendula elastically coupled. The kinetic energy T and potential energy U are given by X 2 2 1 T = 2 mℓ φ̇n n U= Xh 1 2κ n φn+1 − φn 2 + mgℓ 1 − cos φn i (10.103) . Here ℓ is the distance from the hinge to the center-of-mass of the pendulum, and κ is the torsional coupling. From the Euler-Lagrange equations we obtain mℓ2 φ̈n = −κ φn+1 + φn−1 − 2φn − mgℓ sin φn . (10.104) Let a be the horizontal spacing between the pendula. Then we can write the above equation as ≈ φ′′ n ≡ c2 }| { z}|{ z " # 2 φ − φ φ − φ 1 κa g n n n+1 n−1 φ̈n = 2 · − sin φn . − mℓ a a a ℓ (10.105) The continuum limit of these coupled ODEs yields the PDE 1 1 φtt − φxx + 2 sin φ = 0 , 2 c λ (10.106) which is the sine-Gordon equation, with λ = (κa2 /mgℓ)1/2 . 10.2.5 Kinks and antikinks Let us return to eqn. 10.84 and this time set E = −Umin . With U (φ) = 1 − cos φ, we have Umin = 0, and thus dφ 2 (10.107) sin 21 φ . = ±√ dξ 1−V2 19 10.2. SINE-GORDON MODEL Figure 10.10: Kink and antikink solutions to the sine-Gordon equation. This equation may be integrated: ±√ dξ dφ = = d ln tan 14 φ . 1 2 2 sin φ 1−V 2 Thus, the solution is φ(x, t) = 4 tan −1 x − V t − x0 exp ± √ 1−V2 (10.108) . (10.109) where ξ0 is a constant of integration. This describes either a kink (with dφ/dx > 0) or an antikink (with dφ/dx < 0) propagating with velocity V , instantaneously centered at x = x0 + V t. Unlike the KdV soliton, the amplitude of the SG soliton is independent of its velocity. The SG soliton is topological, interpolating between two symmetry-related vacuum states, namely φ = 0 and φ = 2π. Note that the width of the kink and antikink solutions decreases as V increases. This is a Lorentz contraction, and should have been expected since the SG equation possesses a Lorentz invariance under transformations x′ + vt′ x= √ 1 − v2 , t′ + vx′ t= √ . 1 − v2 (10.110) One then readily finds ∂t2 − ∂x2 = ∂t2′ − ∂x2′ . (10.111) The moving soliton solutions may then be obtained by a Lorentz transformation of the stationary solution, φ(x, t) = 4 tan−1 e±(x−x0 ) . (10.112) The field φ itself is a Lorentz scalar, and hence does not change in magnitude under a Lorentz transformation. 20 10.2.6 CHAPTER 10. SOLITONS Bäcklund transformation for the sine-Gordon system Recall D’Alembert’s method of solving the Helmholtz equation, by switching to variables ζ = 21 (x − t) τ= 1 2 (x ∂x = 21 ∂ζ + 12 ∂t + t) ∂t = − 12 ∂ζ + 1 2 ∂t (10.113) . (10.114) The D’Alembertian operator then becomes ∂t2 − ∂x2 = − ∂ζ ∂τ . (10.115) This transforms the Helmholtz equation φtt − φxx = 0 to φζτ = 0, with solutions φ(ζ, τ ) = f (ζ) + g(τ ), with f and g arbitrary functions of their arguments. As applied to the SG equation, we have φζτ = sin φ . (10.116) Suppose we have a solution φ0 to this equation. Suppose further that φ1 satisfies the pair of equations, φ1 + φ0 + φ0,ζ (10.117) φ1,ζ = 2λ sin 2 2 φ1 − φ0 φ1,τ = sin − φ0,τ . (10.118) λ 2 Thus, φ1,ζτ − φ0,ζτ φ1 + φ0 = λ cos φ1,τ + φ0,τ 2 φ1 + φ0 φ1 − φ0 = 2 cos sin 2 2 (10.119) = sin φ1 − sin φ0 . Thus, if φ0,ζτ = sin φ0 , then φ1,ζτ = sin φ1 as well, and φ1 satisfies the SG equation. Eqns. 10.117 and 10.118 constitute a Bäcklund transformation for the SG system. Let’s give the ‘Bäcklund crank’ one turn, starting with the trivial solution φ0 = 0. We then have φ1,ζ = 2λ sin 12 φ1 (10.120) φ1,τ = 2λ−1 sin 12 φ1 . (10.121) The solution to these equations is easily found by direct integration: φ(ζ, τ ) = 4 tan−1 eλζ eτ /λ . (10.122) In terms of our original independent variables (x, t), we have λζ + λ−1 τ = 1 2 λ + λ−1 x − 1 2 x − vt λ − λ−1 t = ± √ , 1 − v2 (10.123) 21 10.2. SINE-GORDON MODEL where λ2 − 1 v≡ 2 λ +1 ⇐⇒ Thus, we generate the kink/antikink solution φ1 (x, t) = 4 tan −1 1+v λ=± 1−v 1/2 x − vt exp ± √ 1 − v2 . (10.124) . (10.125) As was the case with the KdV system, successive Bäcklund transformations commute. Thus, λ λ λ λ φ0 −−−1−→ φ1 −−−2−→ φ12 (10.126) φ0 −−−2−→ φ2 −−−1−→ φ21 , with φ12 = φ21 . This allows one to eliminate the derivatives and write φ12 − φ0 λ1 + λ2 φ2 − φ1 tan = . tan 4 λ1 − λ2 4 (10.127) We can now create new solutions from individual kink pairs (KK), or kink-antikink pairs (KK). For KK, taking v1 = v2 = v yields v sinh(γx) , (10.128) φKK (x, t) = 4 tan−1 cosh(γvt) where γ is the Lorentz factor, γ=√ 1 . 1 − v2 (10.129) Note that φKK (±∞, t) = ±2π and φKK (0, t) = 0, so there is a phase increase of 2π on each of the negative and positive half-lines for x, and an overall phase change of +4π. For the KK system, if we take v1 = −v2 = v, we obtain the solution sinh(γvt) −1 φKK (x, t) = 4 tan . (10.130) v cosh(γx) In this case, analytically continuing to imaginary v with v=√ iω 1 − ω2 =⇒ yields the stationary breather solution, φB (x, t) = 4 tan −1 √ γ= p 1 − ω2 1 − ω 2 sin(ωt) √ ω cosh( 1 − ω 2 x) ! (10.131) . (10.132) The breather is a localized solution to the SG system which oscillates in time. By applying a Lorentz transformation of the spacetime coordinates, one can generate a moving breather solution as well. 22 CHAPTER 10. SOLITONS Please note, in contrast to the individual kink/antikink solutions, the solutions φKK , φKK , and φB are not functions of a single variable ξ = x−V t. Indeed, a given multisoliton solution may involve components moving at several different velocities. Therefore the total momen tum P in the field is no longer given by the simple expression P = V φ(−∞) − φ(+∞) . However, in cases where the multikink solutions evolve into well-separated solitions, as happens when the individual kink velocities are distinct, the situation simplifies, as we may consider each isolated soliton as linearly independent. We then have X P = −2π ni Vi , (10.133) i where ni = +1 for kinks and ni = −1 for antikinks. 10.3 Nonlinear Schrödinger Equation The Nonlinear Schrödinger (NLS) equation arises in several physical contexts. One is the Gross-Pitaevskii action for an interacting bosonic field, Z Z ~2 ∗ d ∗ ∂ψ ∗ ∗ ∗ S[ψ, ψ ] = dt d x iψ − ∇ψ ·∇ψ − U ψ ψ + µ ψ ψ , (10.134) ∂t 2m where ψ(x, t) is a complex scalar field. The local interaction U |ψ|2 is taken to be quartic, (10.135) U |ψ|2 = 12 g |ψ|4 . Note that 2 U |ψ| 2 − µ |ψ| = 1 2g µ 2 µ2 2 − . |ψ| − g 2g (10.136) Extremizing the action with respect to ψ ∗ yields the equation δS ~2 2 ∂ψ + ∇ ψ − U ′ ψ∗ ψ ψ + µ ψ . = 0 = i ∗ δψ ∂t 2m (10.137) Extremization with respect to ψ yields the complex conjugate equation. In d = 1, we have iψt = − ~2 ψxx + U ′ ψ ∗ ψ ψ − µ ψ . 2m (10.138) We can absorb the chemical potential by making a time-dependent gauge transformation ψ(x, t) −→ eiµt ψ(x, t) . (10.139) Further rescalings of the field and independent variables yield the generic form iψt ± ψxx + 2 |ψ|2 ψ = 0 , (10.140) where the + sign pertains for the case g < 0 (attractive interaction), and the − sign for the case g > 0 (repulsive interaction). These cases are known as focusing, or NLS(+), and defocusing, or NLS(−), respectively. 23 10.3. NONLINEAR SCHRÖDINGER EQUATION 10.3.1 Amplitude-phase representation We can decompose the complex scalar ψ into its amplitude and phase: ψ = A eiφ . (10.141) We then find ψt = At + iAφt eiφ ψx = Ax + iAφx eiφ ψxx = Axx − Aφ2x (10.142) iφ + 2iAx φx + iAφxx e . Multiplying the NLS(±) equations by e−iφ and taking real and imaginary parts, we obtain the coupled nonlinear PDEs, −Aφt ± Axx − Aφ2x + 2A3 = 0 (10.143) At ± 2Ax φx + Aφxx = 0 . Note that the second of these equations may be written in the form of a continuity equation, ρt + jx = 0 , (10.144) where ρ = A2 (10.145) j = ± 2A2 φx . 10.3.2 Phonons One class of solutions to NLS(±) is the spatially uniform case 2 ψ0 (x, t) = A0 e2iA0 t , (10.146) with A = A0 and φ = 2A20 t. Let’s linearize about these solutions, writing A(x, t) = A0 + δA(x, t) (10.147) φ(x, t) = 2A20 t + δφ(x, t) . Our coupled PDEs then yield 4A20 δA ± δAxx − A0 δφt = 0 (10.148) δAt ± A0 δφxx = 0 . Fourier transforming, we obtain 4A20 ∓ k2 iA0 ω −iω ∓A0 k2 ! δÂ(k, ω) δφ̂(k, ω) =0. (10.149) 24 CHAPTER 10. SOLITONS Setting the determinant to zero, we obtain ω 2 = ∓4A20 k2 + k4 . (10.150) For NLS(−), we see that ω 2 ≥ 0 for all k, meaning that the initial solution ψ0 (x, t) is stable. For NLS(+), however, we see that wavevectors k ∈ − 2A0 , 2A0 are unstable. This is known as the Benjamin-Feir instability. 10.3.3 Soliton solutions for NLS(+) Let’s consider moving soliton solutions for NLS(+). We try a two-parameter solution of the form A(x, t) = A(x − ut) (10.151) Axx − A φ2x + vA φx + 2A3 = 0 (10.152) φ(x, t) = φ(x − vt) . This results in the coupled ODEs A φxx + 2Ax φx − uAx = 0 . Multiplying the second equation by 2A yields 2φx − u A2 = 0 =⇒ x φx = 12 u + P , 2A2 (10.153) where P is a constant of integration. Inserting this in the first equation results in Axx + 2A3 + 14 (2uv − u2 )A + 12 (v − u)P A−1 − 41 P A−3 = 0 . (10.154) We may write this as Axx + W ′ (A) = 0 , (10.155) W (A) = 12 A4 + 81 (2uv − u2 )A2 + 12 (v − u)P ln A + 81 P A−2 (10.156) where plays the role of a potential. We can integrate eqn. 10.155 to yield 1 2 2 Ax + W (A) = E , (10.157) where E is second constant of integration. This may be analyzed as a one-dimensional mechanics problem. The simplest case to consider is P = 0, in which case W (A) = 1 2 A2 + 12 uv − 14 u2 A2 . (10.158) If u2 < 2uv, then W (A) is everywhere nonnegative and convex, with a single global minimum at A = 0, where W (0) = 0. The analog mechanics problem tells us that A will oscillate between A = 0 and A = A∗ , where W (A∗ ) = E > 0. There are no solutions with 25 10.3. NONLINEAR SCHRÖDINGER EQUATION E < 0. If u2 > 2uv, then W (A) has a double well shape8 . If E > 0 then the oscillations are still between A = 0 and A = A∗ , but if E < 0 then there are two positive solutions to W (A) = E. In this latter case, we may write F (A) ≡ 2 E − W (A) = A2 − A20 A21 − A2 , (10.159) where A0 < A1 and E = − 21 A20 A21 1 2 4u − 1 2 uv = A20 + A21 (10.160) . The amplitude oscillations are now between A = A∗0 and A = A∗1 . The solution is given in terms of Jacobi elliptic functions: i h (10.161) ψ(x, t) = A1 exp 2i u (x − vt) dn A1 (x − ut − ξ0 ) , k , where k2 = 1 − A20 . A21 The simplest case is E = 0, for which A0 = 0. We then obtain h i ψ(x, t) = A∗ exp 2i u (x − vt) sech A∗ (x − ut − ξ0 ) , (10.162) (10.163) where 4A∗ 2 = u2 − 2uv. When u = 0 we obtain the stationary breather solution, for which the entire function ψ(x, t) oscillates uniformly. 10.3.4 Dark solitons for NLS(−) The small oscillations of NLS(−) are stable, as we found in our phonon calculation. It is therefore perhaps surprising to note that this system also supports solitons. We write ψ(x, t) = √ ρ0 e2iρ0 t eiα Z(x, t) , (10.164) with α an arbitrary constant. This yields iZt = Zxx + 2ρ0 1 − |Z|2 Z . (10.165) We then write Z = X + iY , yielding Xt = Yxx + 2ρ0 1 − X 2 − Y 2 Y −Yt = Xxx + 2ρ0 1 − X 2 − Y 2 X . (10.166) 8 Although we have considered A > 0 to be an amplitude, there is nothing wrong with allowing A < 0. When A(t) crosses A = 0, the phase φ(t) jumps by ± π. 26 CHAPTER 10. SOLITONS We try Y = Y0 , a constant, and set X(x, t) = X(x − V t). Then −V Xξ = 2ρ0 Y0 1 − Y02 − X 2 0 = Xξξ + 2ρ0 1 − Y02 − X 2 X Thus, Xξ = − from which it follows that 2ρ0 Y0 1 − Y02 − X 2 V , 4ρ0 Y0 XXξ V 4ρ Y 2 8ρ2 Y 2 = − 0 0 1 − Y02 − X 2 X = 0 2 0 Xξξ . V V (10.167) (10.168) Xξξ = (10.169) Thus, in order to have a solution, we must have √ V = ±2 ρ0 Y0 . (10.170) We now write ξ = x − V t, in which case, from Eqn. 10.168, √ ρ0 dξ = ∓ dX . 1 − Y02 − X 2 (10.171) p e and integrates From this point, the derivation is elementary. One writes X = 1 − Y02 X, to obtain q 2 e X(ξ) = ∓ tanh (10.172) ρ0 1 − Y0 (ξ − ξ0 ) . We simplify the notation by writing Y0 = sin β. Then √ √ iα 2iρ0 t ∓ tanh ρ0 cos β x − V t − ξ0 cos β + i sin β , ψ(x, t) = ρ0 e e (10.173) √ where α is a constant and V = ±2 ρ0 sin β. The density ρ = |ψ|2 is then given by √ ρ(x, t) = ρ0 1 − sech2 ρ0 cos β x − V t − ξ0 . (10.174) This is called a dark soliton because the density ρ(x, t) is minimized at the center of the soliton, where ρ = ρ0 sin2 β, which is smaller than the asymptotic |x| → ∞ value of ρ(±∞, t) = ρ0 .