Variational approximations to the nonlinear Schrödinger equation

Volume 156, number 7,8

PHYSICS LETTERS A

1 July 1991

Variational approximations to the nonlinear Schrodinger equation Fred Cooper Theoretical Division, T-8, LosAlamos National Laboratory, MS B285, Los A/amos, NM 87545, USA

Harvey K. Shepard Physics Department, University ofNew Hampshire, Durham, NH 03824, USA

and L.M. Simmons Jr. Santa Fe Institute, 1120 Canyon Road, Santa Fe, NM 87501, USA Received 25 February 1991; accepted for publication 29 April 1991 Communicated by D.D. Hoim 2 +gJ ~‘wI ~w=0for arbitrary We study analytically the family of nonlinear Schrodinger equations (NLSE) iaw/at+ ô ~i/öx values of the nonlinearity parameter K using two variational strategies based on two different perturbation expansions of the equation — the delta expansion (DE) (which expands in the nonlinearity) and the linear delta expansion (LDE) (which expands about an optimal linear term). For simplicity we discuss solutions to the initial value problem for which ~(x, t = 0) = C5(x). For the LDE the variational calculation gives the exact analytic solution to the NLSE for arbitrary K for this initial data. The answer has different structure for K< 1, K= 1 and K> 1. For the delta expansion the variational calculation gives a sequence of approximations to the solution valid for ic< 1.

1. Introduction (1.2) Recently two new strategies have been introduced to solve by perturbation expansions both quantum field theories and nonlinear differential equations. The first of these methods, the delta expansion [1] (DE), is based on expanding in the degree of nonlinearity of the equation. For the nonlinear Schrödinger equation with arbitrary nonlinearity described by K, ~ 82w, i~- + +g~i/i~t’~”~i=O, (1.1) this technique consists of first replacing K by the parameter ô and treating ö as a small perturbation theory parameter. To improve the dependence of the answer on the coupling constant g, one further introduces two new parameters, M and b, as follows, 436

which is equivalent to (1.1) at b = ô. The delta expansion is obtained by first expanding this equation as a power series in ô, + ôt

M~

+

3x2

g

~‘

(ô ln I ~w/MI n!

~ ‘~ — —

(1.3) Assuming a power series solution for ~ti in terms of r5, ~ çii~(x,1, M)oN, oneobtains asequenceof linear equations for the ~ with known driving terms. For the exact solution, i,ii is independent of M when b = ô. However, when we expand in ~ this is no longer the case and M can be treated as a variational parameter. That is we will assume i,i’ has an expansion to order N:

0375-9601/91/$ 03.50 © 1991

—

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~ ~ti~(x,I, M)ô~,

n=O

(1.4)

and Mwill be promoted to being a function ofxand I

1 July 1991

~ + +lflWi = —mwoln(w~wo/M), 8W2 8W 1• + -~---~-+myJ =2 —m(wT~o+y4y/~)/w~ -~-

by the requirement that

—o

(1.5)

—my,1 ~ (2.3)

-

ÔM which is known as a scaling relation because it can be derived from scaling considerations [21. The linear delta expansion [3] consists of replacingeq.(1.l)by

These equations can be simplified by letting ~ = so that 90o t92 Oo =0, ‘ l_~7+ä~ .

.801

±gAyi+~(g~~/iyi~K_g2)y,=O, (1.6)

i~ +

which contains two new parameters, ô and 2. When ö=1,thisequationreducesto (l.1).Alsoatö=1 this equation is independent of the parameter 2. At any finite order in c5, however, the solution depends on 2 and this dependence can be minimized by imposing the “principle of minimal sensitivity” [31. That is at finite order N in our expansion

820

‘~7+ ,

=J

=—mØ 0ln(Ø~Ø0/M),

ôØ~ 8202 =f2= —m(OTO0+O~O1)/O~ + 2(Ø~Ø mØ1 ln (Ø~Ø0/M) lmOo ln 0/M). —

—

(2.4) The boundary conditions on these equations are that at 1=0,

N

~ ~(x,t,A)ö~

(1.7)

and A is promoted to be a function of x and imposing —0

I

by

(1.8)

—

Ø~(x,0)=0 forn?l. (2.5) For n=0 the answer can be given in terms of g(x) as follows: 4(4~tt)”2 Øo(x, t)=e

x 2. First three terms in the delta expansions In the delta expansion we start from

J~’

exp[i(x—x’)2/4/}g(x’).

For n ~ 1 the answer can be written in terms of the Green function as follows: t

8~w

(ôlnki~w/MI)~

0 (2.1) n! where m=gM’~.To solve this for yi in a delta expansion we assume that y, can be written as —

—~+mW~

çb~(x,t)=$dt’

n=O

2yi

WWo +ôy’1 +5 2 +.... (2.2) Inserting (2.2) into (2.1) we obtain a sequence of linear equations: 81

0x2

(2.6)

0

=

J

dx’ G(x—x’,t—t’)f~(x’,I’),

—=

(2.7) where 3””4 exp(ix2/4t) (4itl)~’2 G(x, 1) =O(t)e andf~are the r.h.s. of eqs. (2.4).

(2.8)

For the linear delta expansion (LDE) we also let (2.9) and make the substitution 437

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Wn=0ne~mt,

(2.10) where now m=gA and m is assumed to be real. The structure of the 0,, equations is exactly as for the DE

1 July 1991

this suggests how we can find an exact solution. If we directly substitute this as an ansatz into (1.1), we obtain for a(t) ma(t)+ C2~~Ia*aI~~a(t)

da(I)

but with a different right-hand sidef~.For the LDE we have

dt

0 (3.8a)

(4itt)”

—

For the LDE substituting into (1.6) we instead obtain

f~=g[2Ø~ (Ø~Ø~)KØ~] —

da(i) dt _ôma(t)+~~t)=0, ____________

f2=g[AO1_(K+l)(O~Oo)KOl

(47t1)”

(2.11)

3. Dirac delta function initial conditions

(3.8b) which gives the same result at ô= 1. The boundary conditions on (3.8) are that a(I=O)=l. Writing cx(I) =R(t)elO(t) (3.9a) ,

In order to show in a simple case how the variational approximations work we will consider the mitial condition yi(x,t=0)=g(x)=Có(x). (3.1) For this initial condition, the integrals in (2.7) are trivial. To zeroth order in ö this leads (from eq. (2.6)) to (3.2)

ØO(x,I)=Ce~’4exp(ix2/4I)(4itt)~”2,

we obtain ~iR dO mR + i R dt dt 2C2(4~tt)’]”R=0. (3.9b) +g[R Imposing the boundary condition at (=0 we find —

—

—

—

R(t)=l (3.10)

Jdt~I_K.

O(t)= —mt+g(C2/4m)”~

so that I0

~

4exp(ix2/4/)(47tI)_I~/2 .

0(x, t)=Ce1mte_~~ (3.3) To obtain the rest of the c5 series we realize after iteration that for both perturbation expansions the structure of the f,, for delta function initial data is f,,(x’, t’ )=Ø 0(x’, I’ )S,,(I’) (3.4) ,

where the S,~only depends on I’. Using the properties of the Green function (2.8) we find that On(X, t)(_i)00(x,

t)Xn(I)

,

(3.5)

Inserting (3.10) into (3.9) and (3.7), we obtain an exact solution to (1.1) for all K. Let us now consider the solution by the linear delta expansion. For the linear delta expansion S

2’~(4~tI’)_K]

Treating K as a continuous variable we see that we have different analytic behavior for Xi(t) in (3.6) for K< 1, K= 1, and K> 1. For K< 1, Xi is well behaved in t at t=0. For ic~ 1 we need to introduce a lower cutoff t0 so that ~

where

x~(t)=

J

(3.6)

Sn(I’) dl’.

0

Because both these perturbation expansions lead to a solution to all orders in c5 of the form çu(x, t) =ip’ 0(x, t)a(I) 438

,

(3.7)

(3.11)

-

1 =g[A—C

2”(4it)~ln(I/t

—K)]

=g[At—C 0)] 2K(4~)_K(K_ l)~ =g[At+C x(l/t’—l/t~)],

,

,

K< 1 K= 1

K>l. (3.12)

To simplify what follows we will write x~(I)=g[At+h(K,I)]

,

(3.13)

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1 July 1991

Øi=imØot[l+ln(C2/4EMI)] Thus to order owe obtain

where

.

OimI{1+o[l+l(C2/4Mt)]}

h(K,t)=_(C2/4~yJdt’t~_K.

(3.23) (3.24)

to

If we exponentiate this result we obtain Thus to order Owe have that yi=Ø y,((x,t)=e~tOo{l_iog[2t+h(K,t)]}. Determing the variational parameter via

—o

(3.14)

(3.15)

2/4itMt)]}). 0exp(imt{l+O[l+ln(C

(3.25)

Applying the scaling law (1.5) we obtain l+ln(C2/4itMt)=0.

(3.26)

Thus, we obtain A(ic,t)t=—h(ic,t)

(3.16)

,

(3.27)

and the scaling law improved solution is so that to this order ~v=Øoexp[igt(C2e/4~tt)~] (3.17)

yi(x,t)=0oe_t~~~~t).

This is the exact answer (see (3.10)) for our boundary conditions. If we continue our calculation to order (52 we find (for K< 1) yi(x, t) = e‘~‘O~{liOg[At + h (K, t) —

—O2g2[21+h(Ic, t)]2/2+...}.

(3.18)

This series in Oexponentiates and we obtain

w(x, 1) =Oo exp{i[g2t( 1

M=C2e/4itl,

—(5)

—Ogh(K, t)

]}.

(3.19) This answer is automatically independent of the variational parameter A at (5= 1 and by direct substitution can be shown to be a solution of eq. (1.6). Thus the solution to the original NLSE is

(3.28)

.

Comparing this answer to the exact solution we find that the t dependence is correct except at (5= 1. However, the coefficient of the I dependence should have a (1 —O~’instead of the e’1 obtained by the variational calculation. So already we see that the DE as improved by the scaling condition is not getting the (5= 1 case correct. Also, it is not as accurate as the LDE which was exact at this order when improved by the principle of minimal sensitivity (1.8). Going to next order and exponentiating the answer into a phase we find to order (52 W= Oo exp{imt [1 (5(ln t_ 1) —

+~d2(ln2r—2In t+2)...]} where r= 4itMt/C2.

,

(3.29)

~u(x, t) =Ø 0e ~gI~(~ ~

(3.20)

This solution obeys the t = 0 boundary condition for K< 1. At K= 1 and for #c> 1 one gets an additional infinite phase as t—~0,unless we first let t-+t~and then let t=t0—+O. Now let us turn to the delta expansion. For the DE we have instead of (3.11) for S 1, 2/4ntM). (3.21) S1(t)=—mln(C Integrating we obtain x 2)] 1(t)=—mI[1—ln(47ttMIC so that

(3.22)

Imposing the scaling relation (1.5) one finds that the coefficient of the (52 term must be zero, or lnT=l±i.

(3.30)

Inserting this into (3.29) we obtain that the order (52 scaling improved answer is 2/4Et~5 ~=Øoexp{igt(C xexp[(l±i)O}[l—(±)iO}}. (3.31) Note that to order (52 exp[(l±i)O] [l—(±i5)] =1+0+02+...,

(3.32) 439

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PHYSICS LETTERS A

so that this factor is correct to order (52~ However by looking at the series in eq. (3.29) we can obtain a much better answer than that obtained by the scaling law improvement. If we realize that the answer cannot depend on M at b=O, then m = gM’~must be compensated by a factor r~=(1 —Oln T+02ln2’r/2+...) . Factoring this out we obtain to order (52~ Y’Oo exp [igl(4itt/C2) _ö( 1 +0+02+...)]

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principle connected with the linear delta expansion (LDE) gives the exact answer for arbitrary K after just keeping the first order result of the expansion. The exact answer had the property that it changed analytic structure at K= 1, which is the integrable case. Our study of the delta expansion DE showed it to be less accurate for this problem. The (DE) solution found in eq. (3.33) converges to the exact answer when K=O< 1. In the DE we do not seem to be able to handle the O?~1 cases.

.

(3.33) Summing the remaining (5 series we obtain the answer 2) _ö( y’=Ø0exp[igl(47tt/C 1 ~5)1] (3.34) ,

References

which is the correct exact answer for 0< 1.

4. Conclusions

[1] C.M. Bender, K.A. Milton, M. Moshe, S. Pinsky and L.M. Simmons Jr., Phys. Rev. Lett. 58 (1987) 2615; Phys. Rev. D

In this paper we have shown how to obtain a sequence ofvariational approximations to initial value problems for the nonlinear Schrodinger equation with arbitrary nonlinearity K. We have foundthat for Dirac delta function initial conditions, the variational

CM. Bender, K.A. Milton, S. Pinsky and L.M. Simmons Jr., J. Math. Phys. 30(1989)1447. [2] F. Cooper and M. Moshe, Phys. Lett. B 259 (1991) 101. [3] A. Okopinska, Phys. Rev. D 35 (1987) 1835; A. Duncan and M. Moshe, Phys. Lett. B 215 (1988) 352; H.F. Jones and M. Moshe, Phys. Lett. B 234 (1990) 492.

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37 (1988) 1472;