2 Soliton Perturbation Theory

Transcription

1 Revisiting Quasistationary Perturbation Theory for Equations in 1+1 Dimensions Russell L. Herman University of North Carolina at Wilmington, Wilmington, NC Abstract We revisit quasistationary perturbation theory for integrable systems and compare the solutions to those obtained through eigenfunction expansion methods. We will focus our comparisons to the perturbed nonlinear equations such as the Korteweg de Vries, the nonlinear Schrodinger and the sine-gordon equations. 1. History 2. Soliton Perturbation Theory 3. Quasistationary Perturbations 4. Direct Integration - KdV 5. Eigenfunction Expansions - KdV 6. Comparison of Results 7. Perturbed NLS 1

2 1 History 1. Ott and Sudan Karpman and Maslov 1977 IST 3. Kaup and Newell 1977, Kaup 1976 IST 4. Keener and McLaughlin 1977 Green s Function Method 5. Kodama and Ablowitz 1981 Quasistationary 6. Sachs 1984 IST 7. Menyuk 1986 Direct 8. Kivshar and Malomed 1989 Review 9. Herman 1990 Direct + IST 10. Yan and Tang 1996 Direct + IST 11. Mann 1997 Green s Function Method 12. and many others in soliton perturbation theory 2

3 2 Soliton Perturbation Theory Multiple Scales u T + N[u] = ǫr[u], 0 < ǫ 1. (1) T = t + ǫ τ Expansion about soliton solution u 0 = u 0 (z, τ), z = x vt u(x, T) = u 0 (z, τ) + ǫu 1 (z, t, τ) (2) Linearized Equation u 1t + ˆL[u 1 ] = R[u 0 ] u 0τ = F(z). (3) Eigenfunctions and Adjoint functions Eigenfunction Expansion Method ˆLφ = λφ, ˆL ψ = λ ψ. (4) Expansion in eigenfunctions u 1 (z, t) = U(t, λ)φ(z, λ) dλ + j U j (t)φ j (z). (5) U t + λu, = F(z)ψ(z, λ) dz., U(0, λ) = 0, (6) 3

4 3 Example: KdV Equation The perturbed KdV equation u T + 6uu x + u xxx = ǫr[u], (7) Leading Order:. u 0 (z) = 2η 2 sech 2 z, z = η(x ξ), and ξ t = 4η 2 First Order: where u 1t + η 3ˆLu1 = R[u 0 ] 4ηη τ φ 1 (z) 4η 3 ξ τ φ 2 (z) F(z), (8) φ 1 (z) = (1 z tanhz) sech 2 z, φ 2 (z) = sech 2 z tanh z. (9) and ˆL = d3 dz 3 + (12 sech 2 z 4) d dz 24 sech 2 z tanhz. (10) Quasistationarity assumption: u 1 (z, t) = u 1 (z) η 3ˆLu1 = F(z). (11) 4

5 3.1 Direct Integration: Kodama and Ablowitz [5] η 3ˆLu1 = F(z). (12) Set y = tanh(z), g(y) = u 1 (z) and F(y) = F(z). Rearranging and integrating, d dy ( (1 y 2 ) dg dy ) + ( y 2)g = 1 1 y 2 F(y) dy. (13) η 3 (1 y 2 ) The Method of Variation of Parameters: g(y) = A(y)P 2 3 (y) = A(y)15y(1 y 2 ) where Solve for A(y) : [ d 15y(1 y 2 ) 2 2 A dy + 2(1 4y2 ) 2 y(1 y 2 ) A(y) = Leads to the general solution [ y(1 y 2 ) u 1 (z) = η 3 F(y) 1 1 y 2 da dy ] = F(y), (14) F(y) dy. (15) η 3 (1 y 2 ) y 1 x 1 dx dw x 2 (1 x 2 ) 3 15 w(1 w2 )F(w), (16) y 1 x ] w F(s) dx dw w ds. (17) x 2 (1 x 2 ) 3 1 s 2 y=tanh z 5

6 3.2 The Damped KdV Equation u t + 6uu x + u xxx = ǫγu. (18) F(z) = 2η ( ηγ + 2η τ 2η τ z tanh z + 2η 2 ξ τ tanh z ) sech 2 z. (19) Compatibility condition: F(z) sech 2 z dz = 0 η τ = 2 ηγ. (20) 3 The forced, linearized problem Apply the general solution to obtain η 3ˆLu1 = 2 3 γη2 sech 2 z(1 4z tanhz) 4η 3 ξ τ sech 2 z tanhz. (21) u 1 (z) = γ 6η [tanhz + z(2 z tanhz) sech 2 z] ξ τ(1 z tanhz) sech 2 z +C 1 sech 2 z tanhz + C 2 ( 1 + 3(1 z tanhz) sech 2 z), (22) The general solution to ˆLv = 0: v(z) = C 1 sech 2 z tanhz + C 2 [ 1 + 3(1 z tanh z) sech 2 z] + C 3 cosh 2 z. 6

7 3.3 Eigenfunction Expansion: Yan and Tang, Phys Rev E 54, 1996 First Order u 1t + η 3 ˆLu1 = R 1 4ηη τ φ 1 (z) 4η 3 ξ τ φ 2 (z) u 0 (z) = 2η 2 sech 2 z, z = η(x ξ) (23) Linear operator and adjoint operator: ˆL = d3 dz 3 + (12sech 2 z 4) d dz 24tanhz sech 2 z (24) ˆL = d3 dz 3 + (12sech 2 z 4) d dz (25) Eigenfunctions: ˆLφ = λφ, λ = ik(k 2 + 4) ˆL ψ = λ ψ, λ = ik(k 2 + 4) (26) Continuous States φ(z, k) = ψ(z, k) = 1 [ k(k 2 + 4) + 4i(k 2 + 2) tanhz 8k tanh 2 z 8itanh 3 z ] e ikz, 2πk(k2 + 4) 1 [ k 2 4ik tanhz 4tanh 2 z ] e ikz. (27) 2π(k2 + 4) Bound (discrete) states φ 1 (z) = (1 z tanhz) sech 2 z, φ 2 (z) = tanhz sech 2 z, ψ 1 (z) = sech 2 z, ψ 2 (z) = tanhz + z sech 2 z. (28) 7

8 Perturbation Expansion Completeness Relations and Orthogonality P φ(z,k)ψ(z,k) dk + 2 φ j (z)ψ j (z ) = δ(z z ), (29) j=1 φ(z,k)ψ(z,k ) dz = δ(k k ), φ j (z)ψ l (z) dz = δ j,l, j,k = 1,2 (30) Thus, expansions are of the form F(z) = P f(k) = F(z)ψ(z,k) dz, f j = f(k)φ(z,k) dk + 2 f j φ j (z), j=1 F(z)ψ j (z) dz, j = 1,2 (31) Then, for u 1t + η 3 ˆLu1 = F u 1 (z,t) = P U(t,k)φ(z,k) dk + 2 U j (t)φ j (z) (32) j=1 U t + η 3 λ(k)u, = f(k), U(0,k) = 0, U 1t = f 1, U 1 (0) = 0, U 2t 8η 3 U 1 = f 2, U 2 (0) = 0. (33) 8

9 Quasistationary Damped KdV - Eigenfunction Expansion Problem: η 3 ˆLu1 = 2η 2 γ sech 2 z 4ηη τ φ 1 (z) 4η 3 ξ τ φ 2 (z) F(z) (34) ˆLφ = λφ, λ = ik(k 2 + 4) (35) Let Then Expand F(z). u 1 (z) = P ˆLu 1 = P F(z) = P U(k)φ(z,k) dk + U 1 φ 1 (z) + U 2 φ 2 (z) (36) U(k)[ ik(k 2 + 4)]φ(z,k) dk 8U 1 φ 2 (z) (37) f(k)φ(z,k) dk + f 1 φ 1 (z) + f 2 φ 2 (z) (38) Equating coefficients where U(k)η 3 [ ik(k 2 + 4)] = f(k) = ψ(z,k) = 0 = f 1 = 8η 3 U 1 = f 2 = F(z)ψ(z, k) dx (39) F(z) sech 2 z dz (40) F(z)[tanhz + z sech 2 z]dz (41) 1 [ k 2 4ik tanhz 4tanh 2 z ] e ikz (42) 2πk(k2 + 4) 9

10 Damped KdV Results Expansion coefficients: f(k) = 2π γη 2 k 3 sinh( πk (43) 2 ), f 1 = 8 3 γη2 4ηη τ, f 1 = 0 η τ = 2 ηγ (44) 3 f 2 = 4η 3 ξ τ. (45) Sum over the continuum states: I = P 2π U(k) = 3iηγ U(k)φ(z, k) dk, (46) 1 (k 2 + 4) sinh( πk (47) 2 ). Inserting φ(z,k) yields The result: I = iγ 3η P I = γ 6η k(k 2 + 4) + 4i(k 2 + 2) tanhz 8k tanh 2 z 8itanh 3 z k(k 2 + 4) 2 sinh( πk 2 ) e ikz dk. (48) [ ( π z2 + 3 ) ] sech 2 z tanhz + tanhz + 2z sech 2 z. (49) 2 The full solution of the quasistationary problem: u 1 = γ [ tanhz + z(2 z tanhz) sech 2 z ] + 1 6η 2 ξ τ(1 z tanhz) sech 2 z + C sech 2 z tanhz. (50) 10

11 4 Evaluation of Integral (48) Consider the evaluation of the integrals: I = iγ 3η P k(k 2 + 4) + 4i(k 2 + 2) tanhz 8k tanh 2 z 8itanh 3 z k(k 2 + 4) 2 sinh( πk 2 ) e ikz dk. (51) Then the full solution would be given by I 1 = P I 2 = P I 3 = P I 4 = P e ikz (k 2 + 4) sinh( πk dk. (52) 2 ) e ikz k(k 2 + 4) sinh( πk dk. (53) 2 ) e ikz (k 2 + 4) 2 sinh( πk dk. (54) 2 ) e ikz k(k 2 + 4) 2 sinh( πk dk. (55) 2 ) I = iγ [ I1 + 4i(I 2 2I 4 ) tanhz 8I 3 tanh 2 z 8iI 4 tanh 3 z ]. (56) 3η 11

12 Computation of I m s For z > 0, We find that I m = 2πi I 3 = i 8 Res[f m (k n );k n = 2in] + πires[f m (k);k = 0]. (57) n=1 I 1 = i 2 i 4 (4z + 1)e 2z i I 2 = z (4z + 3)e 2z 1 4 n=2 n=2 i 192 (24z2 + 24z + 2π 2 + 9)e 2z + i 4 I 4 = z (24z2 + 48z + 2π )e 2z ( 1) n n 2 1 e 2nz, (58) ( 1) n n(n 2 1) e 2nz, (59) n=2 n=2 ( 1) n (n 2 1) 2 e 2nz, (60) ( 1) n n(n 2 1) 2 e 2nz. (61) Sum infinite series using partial fraction decomposition and the summations ( 1) n e 2nz = ln(1 + e 2z ), (62) n n=1 ( 1) n e 2nz n 2 n=1 = Li 2 (e 2z ) e 2z 0 ln(1 + x) x dx. (63) For z < 0 one obtains similar results I 1 = i 2 + i 4 (4z 1)e2z i n=2 ( 1) n n 2 1 e2nz, (64) 12

13 4.1 Comparison of Quasistationary Results For the damped KdV equation, η 3 ˆLu1 = 2 3 γη2 sech 2 z(1 4z tanhz) 4η 3 ξ τ sech 2 z tanhz, (65) Direct integration method: u 1 (z) = γ 6η [tanhz + z(2 z tanhz) sech 2 z] ξ τ(1 z tanhz) sech 2 z +C 1 sech 2 z tanhz + C 2 ( 1 + 3(1 z tanhz) sech 2 z). (66) Eigenfunction expansion method: u 1 = γ [ tanhz + z(2 z tanhz) sech 2 z ] + 1 6η 2 ξ τ(1 z tanhz) sech 2 z + C sech 2 z tanhz. (67) The same compatibility condition: η τ = 2 3 ηγ. However, these solutions differ by a solution of the homogeneous equation, v(z) 1 + 3(1 z tanhz) sech 2 z. We can write v(z) = 1 + 3(1 z tanhz) sech 2 z = 1 + 3φ 1 (z). Noting 1 = P v(z) = P h(k)φ(z,k) dk + 2φ 1 (z), (68) h(k)φ(z,k) dk + φ 1 (z). (69) where h(k) = k2 4 δ(k) k 2 +4 π. Therefore, the k = 0 pole contributes to this particular solution of the homogeneous problem. 13

14 Modification of the Eigenfunction Expansion Method We first note that ˆL[1] = 24φ 2 and recall that ˆLφ 1 = 8φ 2. We seek a solution of ˆLv = 0 by considering v = 1 + βφ 1. Then 0 = ˆL[1 + βφ 1 ] = ( 24 8β)φ 2. Therefore, we recover the missing solution v(z) = 1 3φ 1 (z). They gave the solution for u 1 (z) as u 1 (z) = γ 6η [ 1 + tanhz + 3 (1 + η γ ξ τ Comparison with Kodama and Ablowitz [5] ) (1 z tanhz) sech 2 z + z(2 z tanhz) sech 2 z]. (70) Agrees with (22) up to terms proportional to sech 2 z tanhz. In fact, this u 1 (z) can be rewritten as u 1 (z) = γ 6η [tanhz + z(2 z tanhz) sech 2 z] ξ τ(1 z tanhz) sech 2 z + γ 6η ( 1 + 3(1 z tanhz) sech 2 z), (71) showing that the extra term is proportional to v(z) above. The constant C 2 = γ 6η be obtained by imposing the boundary condition u 1 (z) 0 as z. from the general solution (66) can 14

15 5 Loose Ends Secularity Conditions? In both problems we arrived at the condition F(z) sech 2 z dz = 0. (72) In the non-quasistationary perturbation theory, there is a second condition, which gives the correction to the soliton velocity Use conservation of energy. Namely, Then F(z)[tanhz + z sech 2 z + tanh 2 z]dz = 0. (73) d u 2 dx = 2ǫγ u 2 dx. (74) dt ξ τ = γ 3η. Arbitrariness in the solution in the form of sech 2 z tanhz terms. Note that to order ǫ we have u 0 (z) = 2η 2 sech 2 (η(x ξ + ǫx 0 )) 2η 2 sech 2 z 4ǫx 0 η 3 sech 2 z tanhz. (75) Integrating ξ τ = γ 3η while using η τ = 2 3 ηγ, ξ = ξ e2γτ/3. (76) Thus, we can pick C 1 = 4x 0 η 3 in order to adjust the phase of the perturbed solution at τ = 0. 15

16 5.1 Solution Plots The solution of the forced, linearized KdV equation under the quasistationary assumption: u 1 (z) = γ 6η [ 1 + tanhz + 2(1 z tanh z) sech 2 z + z(2 z tanhz) sech 2 z]. (77) The range of validity: z ǫ 1/2. For ǫ = 0.01 this would give z 10. Kodama and Ablowitz [5] 2 1 z u1(z) -2-3 Figure 1: This is a scaled plot of the first order solution u 1 (z) in equation (77). 16

17 u(z) z Figure 2: In this figure we plot u(z) = u 0 (z) + ǫu 1 (z) for ǫ = 0.01 and γ = 6η. 17

18 6 NLS Perturbation Theory????? Perturbed NLS: iu T + u xx + 2 u 2 u = iǫr[u] Leading Order Solution u 0 (z) = 2βe iθ sechz (78) z = 2β(x ξ), ξ t = 4α, θ = 2α(x ξ) + δ, δ t = 4(α 2 + β 2 ). First Order Equation iu 1t + 8iαβu 1z + 4β 2 u 1zz + 4 u 0 2 u u 0 2 ū 1 = if 1, (79) where if 1 = ir 1 [u 0 ] ie [ iθ (4iαβξ τ 2iβδ τ )φ 1 (z) 2iα τ φ 2 (z) + 2β τ ψ 1 (z) + 4β 2 ξ τ ψ 2 (z) ] (80) φ 1 (z) = sechz, φ 2 (z) = z sech z (81) ψ 1 (z) = (1 z tanhz) sechz, ψ 2 (z) = tanhz sechz (82) For u 1 (z, t) = e iθ [A 1 (z, t) + ib 1 (z, t)] A 1t + 4β 2ˆL1 B 1 = Re(R 1 [u 0 ]) [ 2β τ ψ 1 (z) + 4β 2 ξ τ ψ 2 (z) ] B 1t 4β 2ˆL2 A 1 = Im(R 1 [u 0 ]) [(4αβξ τ 2βδ τ )φ 1 (z) 2α τ φ 2 (z)] (83) 18

19 7 NLS Perturbation Theory - Eigenfunction Expansion Quasistationary Perturbation u 1 (z) = e iθ [A 1 (z) + ib 1 (z)] +4β 2ˆL1 B 1 = Re(R 1 [u 0 ]) [ 2β τ ψ 1 (z) + 4β 2 ξ τ ψ 2 (z) ] 4β 2ˆL2 A 1 = Im(R 1 [u 0 ]) [(4αβξ τ 2βδ τ )φ 1 (z) 2α τ φ 2 (z)] (84) Linear Operators ˆL n = d2 + n(n + 1) sech 2 z 1. Eigenvalue Problem - J. Yan, et. al., Phys Rev 58 dz (1998). 0 ˆL1 ˆL 2 0 ψ φ = λ ψ φ, (85) Expansions (λ = (k 2 + 1)) ψ(z, k) = A 1 = B 1 = φ(z, k) = a 1 (k)ψ(z, k) dk + b 1 (k)φ(z, k) dk + 2 j=1 2 j=1 a 1j ψ j (z), (86) b 1j φ j (z). (87) 1 2π(k2 + 1) (1 k2 2ik tanhz)e ikz, (88) 1 2π(k2 + 1) ( 1 k2 2ik tanhz + 2 tanh 2 z)e ikz. (89) 19

20 Expansion Coefficients 4β 2 λb 1 (k) = 4β 2 λa 1 (k) = Re(R 1 [u 0 ]e iθ ) φ(z, k) dz, (90) Im(R 1 [u 0 ]e iθ ) ψ(z, k) dz, (91) 2β τ = 2α τ = 8β 2 b 12 = 4β 2 ξ τ Re(R 1 [u 0 ]e iθ ) φ 1 (z) dz, (92) Im(R 1 [u 0 ]e iθ ) ψ 2 (z) dz, (93) 8β 2 a 11 = 4αβξ τ 2βδ τ Re(R 1 [u 0 ]e iθ ) φ 2 (z) dz, (94) Im(R 1 [u 0 ]e iθ ) ψ 1 (z) dz, (95) Damped NLS: iu T + u xx + 2 u 2 u = iǫu Damped NLS Example β τ = 2β, α τ = 0. a 1 (k) = 0, 4β 2 λb 1 (k) = 2πβ sech (πk/2). 20

21 8 NLS Perturbation Theory - Direct Integration Recall: Quasistationary Perturbation u 1 (z) = e iθ [A 1 (z) + ib 1 (z)] +4β 2ˆL1 B 1 = Re(R 1 [u 0 ]) [ 2β τ ψ 1 (z) + 4β 2 ξ τ ψ 2 (z) ] 4β 2ˆL2 A 1 = Im(R 1 [u 0 ]) [(4αβξ τ 2βδ τ )φ 1 (z) 2α τ φ 2 (z)] (96) where ˆL n = d2 dz 2 + n(n + 1) sech 2 z 1. Let y = tanhz, h(z) = g(y) : Then ˆL n h(z) = F(z) where L n = d (1 dy y2 ) d 1 + n(n + 1). dy 1 y 2 Since L n Pn(y) 1 = 0, use Variation of Parameters: g(y) = f(y)pn(y). 1 Then one obtains (1 y 2 )L n g(y) = F(y), (97) [ d (P 1 dy n(y) ) 2 (1 y 2 ) df ] = F(y) dy 1 y n(y) (98) 2P1 Solution A 1 = ( 2αβξ τ + βδ τ )(1 + z tanhz) sechz + C 1 tanh z sech z, B 1 = z(2β 2 ξ τ + βz) sechz + C 2 sech z. 21

22 22

23 9 Conclusion 1. History 2. Soliton Perturbation Theory 3. Quasistationary Perturbations 4. Direct Integration - KdV 5. Eigenfunction Expansions - KdV 6. Comparison of Results 7. Perturbed NLS For the damped KdV: Same results. Care needs to be taken in obtaining solutions to homogeneous problem. Furthermore, one can compare the quasistationary results to the standard non-quasistationary results. This is given in [3] as where u 1 (z, t) = iγ 3η P 1 e ik(k2 +4)η 3 t k(k 2 + 4) 2 sinh( πk 2 )p(z, k)eikz dk, (99) p(z, k) = k(k 2 + 4) + 4i(k 2 + 2) tanhz 8k tanh 2 z 8i tanh 3 z. (100) 23

24 References [1] Karpman V I and Maslov E M 1977 A Perturbation Theory for the Korteweg-deVries Equation Phys. Lett. 60A p [2] Kaup D J and Newell A C 1978 Solitons as Particles, Oscillators, and in Slowly Changing Media: A Singular Perturbation Theory Proc. Roy. Soc. Lond. A361) p [3] Yan J and Tang Y 1996 Direct Approach to the Study of Soliton Perturbations Phys. Rev. E 54 p [4] Keener J P and McLaughlin D W 1977 Solitons Under Perturbations Phys. Rev. A16 p [5] Kodama Y and Ablowitz M J 1981 Perturbations of Solitons and Solitary Waves Stud. Appl. Math. 64 p [6] Kaup D J 1976 A Perturbation Expansion for the Zakharov-Shabat Inverse Scattering Transform SIAM J. Appl. Math. 31 p [7] Kivshar Y S and Malomed B A 1989 Dynamics of Solitons in Nearly Integrable Systems Rev. Mod. Phys. 61 p [8] Herman R L 1990 A Direct Approach to Studying Soliton Perturbations J. Phys. A. 23 p [9] Mann E 1997 The Perturbed Korteweg-deVries Equation Considered Anew J. Math. Phys. 38 p [10] Herman R L 2004 Quasistationary Perturbations of the KdV Soliton J. Phys. A. 37 p [11] Ott E and Sudan R N 1969 Damping of Solitary Waves Phys. Fluids 13 p [12] Karpman V I 1979 Soliton Evolution in the Presence of Perturbation Physica Scripta 20 p [13] Karpman V I and Maslov E M 1978 Structure of Tails Produced under the Action of Perturbations of Solitons Sov. Phys. JETP 48 p [14] Lamb Jr. G L 1980 Elements of Soliton Theory (John Wiley & Sons, New York). [15] Menyuk C R 1986 Origin of Solitons in the Real World Phys. Rev. A33 p