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Rogue waves under influence of Raman delay

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Abstract

Rogue waves can appear in optical fibers and other optical systems as well as in natural events like water waves. Their mathematical description is based on partial differential equations that have solutions that are localized both in time and space. One example is the “Peregrine” solution of the nonlinear Schrödinger equation (NLSE). When higher-order terms in the equation are involved, the solution becomes distorted, but its main features remain localized in space and time. Although exact solutions are not obtained in all cases, approximations that describe the solutions with reasonable accuracy do exist. Here, we consider approximate rogue wave solutions of the NLSE with an optically relevant Raman delay term.

© 2018 Optical Society of America

1. INTRODUCTION

Rogue waves were first found in optics as high-amplitude pulses in supercontinuum generation [1]. However, these are not the only type of rogue waves in optics. They can also be found in the output of laser radiation [26] and in other types of optical cavities [710]. They can be influenced by Brillouin scattering [11] or by the Raman effect [12]. Indeed, there is a multiplicity of forms of rogue waves [1317], and this can make it difficult to classify them at present. Mathematically, the simplest type can be represented in the form of the “Peregrine” solution of the nonlinear Schrödinger equation (NLSE) [18]. This is a solution that is localized both in time and in space. This fundamental solution can be extended to cover more complicated cases [1922]. Rogue waves involving partial nonlinearity have been studied [23], and effects in media with decreasing dispersion or diffraction have been considered in [24]. Akhmediev breathers and rogue waves can also appear in parity-time-symmetric coupled waveguides [25]. However, presenting rogue wave solutions in exact form is not always possible.

Light propagation in optical fibers is subject to various higher-order effects [26,27], such as third-order dispersion, ψttt; fourth-order dispersion, ψtttt; a “quintic” adjustment to the Kerr nonlinearity, |ψ|4ψ; and the Raman downshift effect, τRψ(|ψ|2)t. Similarly, water wave propagation in “deep” water can be influenced by assorted physical phenomena [28], such as, wind, viscosity, and bottom friction, introducing terms like ψttt, |ψ|2ψt, and |ψ|1σψ. These higher-order effects can potentially influence the dynamics of rogue waves [29] in ways that differ from their effect on solitons. Such impacts can significantly change the detected shapes. In this work, we consider the effects of Raman delay and other disturbances on the shape of rogue waves, and, where possible, compare the results with predictions from other possible approaches.

2. BASIC EQUATIONS

We start with the extended NLSE that describes propagation of ultrashort pulses in optical fibers:

iψx+12ψtt+|ψ|2ψ=R,
where R indicates the additional terms in the form
R=τRψ(|ψ|2)tν|ψ|4ψν2n|ψ|2nψ+iγ3ψtttd4ψttttd6ψttttttis(ψ|ψ|2)t,
where n is an integer, n=3,4,. Here ν represents a nonlinearity which is of higher order than a Kerr-law response for the system. To start with, we restrict ourselves to these examples of higher-order terms that are commonly studied in the literature. Other types of terms can be considered using the same technique as we discuss here.

Equation (1) in general is not integrable and cannot be solved exactly. Therefore, we have a choice of finding approximate solutions. Among the techniques that allows us to do that we can list the Lagrangian approach and the “method of moments.” The Lagrangian technique was introduced into soliton theory by Anderson [30]. It can be used to understand the effects of various disturbances on a known system. The method of moments [31] can also be used for reducing the dimensionality of soliton dynamical systems. The Lagrangian approach is based on minimization of “action.” It can give results which are similar to those obtained by the method of moments [32]. The two methods were compared in regard to the soliton solutions of the cubic–quintic complex Ginzburg–Landau equation in [33]. They have also been used to study self-focusing and self-defocusing two-dimensional beams in dissipative media [34].

It is less obvious that the Lagrangian approach can be applied to rogue wave solutions which are localized both in space and time, in contrast to soliton solutions that are localized only in either time or space. In this work, we show that the technique can be applied to rogue waves when the exact solutions may not be available. In particular, it can be applied to nonintegrable equations such as higher-order extensions of the NLSE.

The use of the variational integral and Euler equations has been explained in [35], where the Lagrange density is introduced and its relation to the Euler–Lagrange equations for the Schrödinger equation and other conservative equations is also given. We employ the Lagrangian

L=Lddt,
where Ld is the Lagrangian density, and we have
Ld=i2(ψ*ψxψψ*x)+12|ψt|212|ψ|4,
for field ψ(x,t). When dealing with solutions on a constant background, it is common practice to subtract any constant part of Ld.

If we apply the Euler–Lagrange equations to Eq. (3), we obtain

Ldψ*+xLdψx*+tLdψt*2t2(Ldψtt*)=R,
where a subscript x or t means derivative with respect to that variable, the star * indicates complex conjugate, and R indicates the added higher-order terms.

For a trial solution containing several parameters c(j),j=1,2, the standard variational approach can be modified to allow for dissipative terms [32,33,36,37]. In this way, we obtain a separate equation

ddx(Lcxj)Lc(j)=2Re(Rψ*c(j)dt)J[cj;R]
for each parameter c(j). The set of Eqs. (4) comprise the reduced dynamical system for the rogue wave parameters. In other words, the wave profiles and, more generally, the wave evolution are described by the variation of these parameters.

3. SIMPLE EXAMPLES

A. Peregrine Rogue Wave

In order to illustrate the application of the above technique to solutions which are localized both in time and space, we first apply it to the simplest case, namely, the lowest-order rational solution of the NLSE. Thus, we first consider the case R=0, that is, the basic NLSE,

iψx+12ψtt+|ψ|2ψ=0,
and illustrate the idea using its known exact solution. Namely, we write the solution in the form of the trial function
ψ=[41+2ixc(x)+4t21]exp[ix]
with the intention of finding the unknown function, c(x).

Following Eq. (3), we first write the Lagrangian,

L=2πc(x)7/2[xc(x)2c(x)14(4x2+1)c(x)+2c(x)3+10(4x2+1)2].

Equation (4) allows us to write the equation relative to the unknown function c(x):

70π(4x2+1)(c(x)+4x2+1)c(x)9/2=0,
which results in
c(x)=1+4x2,
giving
ψ=[41+2ix1+4x2+4t21]exp[ix].

Clearly, this is the well-known rogue wave solution of the NLSE [19]. In this case, the result is exact, as we deal with the integrable NLSE and we knew the form of the solution in advance. In general, the solutions are expected to be approximate.

B. “Moving” Rogue Wave

As a more complicated example, we consider a rogue wave with an internal “velocity” or “skewness.” From previous experience, we know that this occurs for rogue waves of the Hirota equation [20] and for a multiplicity of extended equations with odd higher-order terms [38]. In order to describe this type of rogue wave, we take the solution in the form

ψ(x,t)=[41+2ixDm(x,t)1]eiFm(x,t),
where
Dm(x,t)=r(x)+4[tt0(x)]2,
and
Fm(x,t)=x+a(x)[tt0(x)]θ(x),
with the function t0(x) being responsible for the lateral motion.

We now need to find L and then the four unknown functions a(x),t0(x),r(x),θ(x). We find:

Lr7/2(x)2π=r2(x)[2Xa(x)t0(x)+Xa2(x)+xr(x)2Xθ(x)]r3(x)[22a(x)t0(x)+a2(x)2θ(x)]+14Xr(x)10X2,
where, for convenience, we define X=1+4x2.

We obtain the four ordinary differential equations (ODEs), and then solve the a(x) equation by setting a(x)=t0(x). The first (i.e., θ) equation found is

r(x)[r(x)+16x]3(4x2+1)r(x)=0,
and we use its simplest solution, namely, r(x)=4x2+1.

Then the only remaining nonzero equation is

[t0(x)]2+2θ(x)=0.

Now t0(x) is a velocity, and the simplest solution is to take it as an arbitrary constant velocity, v, so t0(x)=v. Hence t0(x)=vx and θ(x)=v2x/2, so we have

ψ=[41+2ix1+4x2+4(tvx)21]ei[vt+(1v22)x].

The Galilean transformation of the NLSE is well known {e.g., see Eq. (2.6) of [39]}. The solution given by Eq. (11) agrees with that found by applying such a transform to the basic rogue wave of Eq. (6) in Section 3.A. It reduces to that lowest-order rogue wave when v=0. We can see that the Lagrangian approach works well for these simple illustrative cases. This encourages us to move up to more complicated cases, as given below.

4. MORE GENERAL FORM OF THE TRIAL FUNCTION

As a next step, we allow for nonzero functional R in the trial function. One of the consequences may be resizing of the rogue wave, say, in the x-direction. In order to take this into account, we extend the result of Eq. (10) to allow for a stretching factor, B, by taking

ψ(x,t)=[41+2iBxDm(x,t)1]eiFm(x,t),
with Dm(x,t) still given by Eq. (8) and Fm(x,t) given by Eq. (9). Now, we apply the Euler–Lagrange formalism to the more general trial function (12).

For convenience, we split Ld into three parts:

Ld=Ld1+Ld2+Ld3,
where
Ld1=i2(ψ*ψxψψ*x)+a1a(x)t0(x)+h1θ(x)u1,Ld2=12|ψt|2,
and
Ld3=12(|ψ|41).

We need to integrate these terms over an infinite range in y1tt0(x). The terms in odd powers of y1 will integrate out to zero. In the remaining part of Ld1, a part is subtracted so that its limit is zero for high |y1|. Then, taking Lj=Ldjdt, we have L=L1+L2+L3.

Taking the limit Ly1limd1 shows that it equals to

(a11)a(x)t0(x)+(h11)θ(x)u1+1,
with unknown functions a(x),t0(x),θ(x). Setting this to zero shows that the constants needed here are a1=h1=u1=1. Now, Ld2 clearly approaches zero for large t or y1, and so needs no level adjustment. For Ld3, we note that |ψ|y1lim=1, hence we subtract this background before integrating. We then find
Lr7/2(x)2π=14Xr(x)10X2+r(x)2[a2(x)X2a(x)Xt0(x)2Xθ(x)+Bxr(x)+4B4]r(x)3[a2(x)+2B2a(x)t0(x)2θ(x)],
where X=4B2x2+1.

Using Eq. (4), we define the left-hand-side terms of the equation relating to parameter c(j) as

S[c(j)]ddx(Lcxj)Lc(j),
for each parameter c(j).

So, using Eq. (13), the a term is

r(x)3/24π[ddx(Lax)La]=r(x)3/24πS[a]=(r(x)X)[a(x)t0(x)].

The θ(x) term is

r(x)5/22π[ddx(Lθx)Lθ]=r(x)5/22πS[θ]=[3Xr(x)]r(x)16B2xr(x),
while the t0(x) term is
r(x)5/22π[ddx(Lt0(x))Lt0(x)]=r(x)5/22πS[t0]=2r(x)a(x)[r(x)X]a(x){[r(x)3X]r(x)+16B2xr(x)}.

The r(x) term is

r(x)9/2π[ddx(Lrx)Lr]=r(x)9/2πS[r]=r(x){r(x){(3Xr(x))[a2(x)2a(x)t0(x)2θ(x)]+12(B1)}+70X}70X2.

Using Eq. (4), we define the influence of any functional R on parameter cj as

J[cj;R]=2Re(Rψ*c(j)dt).

Once R is given, we thus need to solve the four ODEs, S[cj]=J[cj;R], with cj={a(x),t0(x),r(x),θ(x)}.

5. MORE EXAMPLES OF APPLICATION

A. Rogue Waves of Full Hirota Equation

The “full” Hirota equation [40] is

R=iα3(ψttt+6|ψ|2ψt)Rh.

The right-hand sides of the reduced dynamical system (4) for this equation take the form

r(x)7/212πα3J[a;Rh]=[Xr(x)]×{r(x)[(a2(x)2)r(x)+4]10X},
and
r(x)9/22πα3a(x)J[r;Rh]=210X2+r(x)×{r(x)[a2(x)(r(x)3X)+6(6r(x)+3X)]210X},
while J[t0;Rh]=J[θ;Rh]=0.

Setting r(x)=1+4B2x2 reduces the dynamical system and leaves only the equation for r, namely S[r]=J[r;Rh], to solve. It is straightforward to show that it can be solved with a(x)=k, an arbitrary constant, B=6α3k+1, θ(x)=xc0, and t0=vsx, where c0=k2(c1+c2α3k) while

vs=α3(1k22)v1k.

Hence,

α3k2(2c2+v12)2α3(v16)+(2c1+1)k=0.

This equation must be valid for all k and all α3. Consequently, we obtain c1=12, v1=6, and c2=2. Then, the functions t0(x) and θ(x) in the trial function (12) become

t0(x)=x[3α3(k22)+k],θ(x)=12k2x(4α3k+1).

This outcome agrees with the known exact result (with α2=12) found in [38]. The velocity factor, t0(x), can be zero for some combinations of k and α3, thus cancelling the effect of higher-order terms in the Hirota equation.

B. Rogue Wave of Lakshmanan—Porsezian—Daniel Equation

The even higher-order Lakshmanan—Porsezian—Daniel equation [41,42] is

iψx+12ψtt+|ψ|2ψ=R[ψ(x,t)],
where
R[ψ(x,t)]=α4(ψtttt+6ψt2ψ*+4ψ|ψt|2+8ψtt|ψ|2+2ψtt*ψ2+6ψ|ψ|4)Rl.

For this equation, the right-hand sides of the dynamical system (4) are

r(x)7/216πa(x)α4J[a;Rl]=[r(x)X]{r(x)×[(a2(x)6)r(x)+12]30X}
and
r(x)13/22πα4J[r;Rl]=2772X3+r(x)×{r(x)[12a2(x)(r(x)(r(x)[r(x)3X6]+35X)35X2)+a4(x)r(x)2[3Xr(x)]2r(x)(3r(x)[r(x)3(X+4)]+70(3X+2))+28X(15X+91)]5040X2},
while J[t0;Rl]=J[θ;Rl]=0.

Again, setting r(x)=1+4B2x2 reduces the system and leaves only the r equation, S[r]=J[r;Rl], to solve. It is straightforward to show that it can be solved with a(x)=k, an arbitrary constant, B=α4(b1k2+b2)+b0, θ(x)=x{3α4[(k22)2+n2]+k2n1}, and t0=vex, where ve=k[α4(j1+j2k2)1].

Expanding S[r]=J[r;Rl], we readily find the constants, b0=1, b1=12, b2=12, j1=24, j2=4, n1=1/2, n2=6. Hence B=112α4(k21), and

t0(x)=vex=k[4α4(k26)1]x,θ(x)={3α4[(k22)26]k22}x.

This outcome also agrees with the known exact result (with α2=12) found in [38]. Again, the velocity factor, t0(x), is zero for some combinations of k and α4.

6. SOLITON UNDER INFLUENCE OF RAMAN DELAY

The previous examples are related to integrable equations in order for us to be able to compare the results of the variational approach with exact solutions. However, the technique itself is fully justified only when there is no possibility of obtaining exact solutions. Thus, below, we consider more involved cases relevant to practical situations. First, to show the principle, we consider a soliton solution under the influence of Raman delay. This has been done earlier in [43,44]. Suppose R=τRψ(|ψ|2)tRr is the only nonzero term in Eq. (2). Various approximations have been given to estimate the soliton shape for small delay, τR, in this case [4547]. The delay makes the soliton solution nonsymmetric in t, and this requires a new trial function. Here we take

ψ(x,t)=Q(x)2w(x)sech(y1w(x))exp{i[a(x)y1θ(x)]},
where Q(x) is the energy, w(x) is the pulse width, and y1=tt0(x), with t0(x) indicating the time delay offset.

The approximating procedure described in Section 1 leads to five ODEs. The θ(x) equation is

Q(x)=0,
showing that Q(x)=Q0 is a constant. The a(x) equation is
Q(x)[a(x)t0(x)]=0,
showing that a(x)=t0(x). The w(x) equation is
Q(x)6w3(x)[w(x)Q(x)2],
showing that w(x)=2/Q0=w0. Finally, the t0(x) equation is
8τR15w04t0(x)=0,
showing that
t0(x)=a(x)=4τRQ(x)15w03x=8τR15w04x,
as in [43] (where w0=1/η, t0=t¯, and a=ω). Thus,
t0(x)=4τRx215w04.

This result for time shift is in agreement with Eqs. (5b) and (13) in [44]. Plainly, t0(x) is a close analogue of vertical projectile motion under gravity, where the velocity t0(x) changes sign at x=0, but the acceleration, t0(x)=8τR15w04, does not. This as expected, since the acceleration is caused by the Raman delay and this does not change sign.

The remaining Q(x) equation is

1w02+t0(x)2+2θ(x)=0,
showing that the rate of change of phase is θ(x)=12w0212(8τRx15w04)2, as in [43] (where θ=ϕ). If τR=0, we obtain
ψ(x,t)=1w0sech(tw0)exp{ix2w02}
=Q02sech(Q0t2)e18iQ02x,
which is the NLSE soliton of width w0 with the energy Q0. Our numerical propagation studies of this soliton, as exemplified in Fig. 1, verify the above analytic results and thus give confidence in the approach. An example of the solution, showing the parabolic path, is given in Fig. 2.

 figure: Fig. 1.

Fig. 1. Plot of the relative soliton offset, t0/τR, showing the effect of Raman delay. We compare the numerically found result (blue curve) with the analytic result of Eq. (29) (red curve). Here, delay is given by τR=0.02 and width w0=0.4.

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 figure: Fig. 2.

Fig. 2. Plot of the soliton, Eq. (23), under the effect of Raman delay, using Eqs. (24)–(30). Here, delay is given by τR=0.02 and width w0=0.4. At the ends (x=±3), the t offset is about 1.8.

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7. ROGUE WAVE UNDER INFLUENCE OF RAMAN DELAY

Now, instead of a soliton, we consider a rogue wave solution which is localized in time. For this, we use the same trial function as in Eqs. (7), (8), and (9).

The ODEs will naturally differ from the previous cases. The θ(x) equation is

r(x)[r(x)+16x]=3(4x2+1)r(x).

To solve it, we set z=X=1+4x2 and r(z)=s2(z). This gives

[3z2s2(z)]s(z)=2zs(z).

We find that the solution is

s(z)=16c(F1/3+F1/31),
where
F=54c2z21+6cz327c2z21,
where c is an arbitrary real solution parameter.

Note that this remains real even when c is small. In fact, limc0s(z)=z, so limc0r(z)=z2. Also, r(x) is even in c, so we only need to use c0. This means that r(0)1 for any c. We plot the function r(x) in Fig. 3.

 figure: Fig. 3.

Fig. 3. Plot of function r(x) for various c values. The blue curves are for c<0. This solution corresponds to the expansion in Eq. (37) with c=0.04 (top blue curve) and c=0.02 (next blue curve). The green curve, namely r(x)=z2=1+4x2, is for c=0. The two red curves are plots of Eqs. (35) and (36). The upper red curve is for c=0.02. The lower red curve is for c=0.04. In each case, the expansion, Eq. (37), is superimposed on the exact result. The curves are indistinguishable.

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 figure: Fig. 4.

Fig. 4. Plot of function t0(x)/τR, from Eq. (46) for various c: c=0.04 (blue curve), c=0.4 (magenta), c=0.2 (red), and c=0.1 (green).

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Since c is generally small (c<0.1), it is convenient to expand this result. For c0:

r(z)=z2(12cz+6c2z221c3z3+80c4z4+).

That is,

r(x)=(1+4x2)[12c1+4x2+6c2(1+4x2)21c3(1+4x2)3/2+80c4(1+4x2)2+].

So, when c=0, the solution reduces to the known NLSE result, r(x)=(1+4x2). In solving the four ODEs, we take c to be arbitrary but larger than τR.

Using Eq. (4) with S[t0] from Eq. (17) and J[t0;Rr] from Eq. (19), for t0(x), we find

r(x)9/232πJ[t0;Rr]=τR{r(x)[5Xr(x)]7X2},
so the t0(x) equation is
16τR{r(x)[r(x)5X]+7X2}+r(x)2{2r(x)(a(x)[r(x)X]8xa(x))+a(x)[3Xr(x)]r(x)}=0,
where X=1+4x2. The other right-hand-side factors, J[r;Rr],J[a;Rr],J[θ;Rr], are zero.

The r(x) equation is

r(x){r(x)[r(x)3X][a2(x)2a(x)t0(x)2θ(x)]70X}+70X2=0
and the a(x) equation is
[Xr(x)][a(x)t0(x)]=0.

Now, if we take r(x)=4x2+1 to solve the last equation, then the second one cannot be solved. So, we need to take a(x)=t0(x) to solve it. This reduces Eq. (39) to

r(x){r(x)[t0(x)((r(x)3X)r(x)+16xr(x))+2r(x)(r(x)X)t0(x)+16τR]80τRX}+112τRX2=0,
and it reduces Eq. (40) to
r(x){r(x)[3Xr(x)][t0(x)2+2θ(x)]70X}+70X2=0.

We can solve the resultant ODEs, namely, Eqs. (33), (42), and (43), to various orders in τR. We take c and τR to be small and have similar order (around 103). We assume that the offset, c, is not much smaller than τR, so we cannot have τR/c1. This excludes c=0. In fact, we expect c>τR.

Using Eq. (38), we have

r(x)=X[12cX1/2+6c2X21c3X3/2+80c4X2+].

Then

a(x)=t0(x)=2τRxcX3/2[2(8x2+3)+33cX1/2+215c2X]+33τRarctan(2x).

Now, taking t0(0)=0, we obtain (see Fig. 4)

t0(x)=τR{1c(8x2+1X1/21)+33xarctan(2x)+2152c[X1/21]}.

So, for small x, we have

t0(x)τR(215c+6c+66)x2(215c+10c+88)x4+.

In the soliton case, we also found that t0(x) varied as τRx2. If c is small, say c<0.1, then

t0(x)6τRcx2(153x2+).

Hence, we obtain the following simple form for the acceleration:

a(x)τR=t0(x)τR=2c[6X5/2+66cX2+215c2X3/2].

Thus, the acceleration, t0(x), of the ridge takes the same sign as that for the soliton case found in Eq. (27) and is also an even function, but it has a more complicated form. As for the soliton case, it is proportional to τR. The estimates for the function t0(x) are shown in Fig. 5 along with the results of numerical simulations for the same set of parameters. As can be seen from this figure, numeric solution of the four ODEs give results which are reasonably close to the above approximate solutions for small τR, say τR<0.002.

 figure: Fig. 5.

Fig. 5. Plot of functions t0(x), Eq. (46), r(x), using the results of Section 6, including Eq. (38), and θ(x), using Eq. (53), together with the same functions found numerically by solving Eqs. (33)–(41). Here, c=0.04, and the delay is given by τR=0.005. The two upper curves give r(x), while the two lower positive ones give t0(x) and the two negative ones show θ(x). In each case, the exact numeric solution and the approximation for it are very close.

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The Raman effect causes a delay, so we expect deceleration. It is analogous to the deceleration found for the Raman soliton in Eq. (28). Examples of evolution are shown in Figs. 6 and 7. Acceleration here is clearly seen.

 figure: Fig. 6.

Fig. 6. Plot of the rogue wave, Eq. (7), using Eqs. (46)–(53) for delay given by τR=0.005. Here c=0.04.

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 figure: Fig. 7.

Fig. 7. Contour plot of the rogue wave, Eq. (7), using Eqs. (46)–(53) for delay given by τR=0.005. Here c=0.04 and the propagation direction, x, is the horizontal axis, while the transverse axis, t, is vertical.

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Now, t0(0)τR=2c(6+66c+215c2), and we define this to be c0. On plotting the acceleration, we see that it plainly has a Gaussian-type form. We find that it can be approximated by

t0(x)τRc0e6x2.

Integrating shows that

t0(x)τRc0π6erf(6x),
where “erf” indicates an error function. So the final expression for velocity is t0(±)/τR±c02π6. Then
t0(x)τRc02π6[xerf(6x)1e6x26x].

The comparisons in Fig. 8 show that this is a convenient and accurate approximation, giving a simple view of the ridge offset during propagation.

 figure: Fig. 8.

Fig. 8. Plot of functions t0(x)/τR (even function of x with t0(0)c0=300 here), t0(x)/τR (odd function of x) and t0(x)/τR (even function of x with t0(0)=0) from original Eqs. (46), (45), and (48), respectively, together with a comparison with the Gaussian-based approximations for them, given by Eqs. (49), (50), and (51), all for c=0.1. Here, c0=2c(6+66c+215c2).

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So, for small x,

t0(x)τRc02x2[1x2+6x45].

For c0=2c(6+66c+215c2), we find that c0 reaches its minimum of 275 when c=0.17, and 275<c0<295 for 0.1<c<0.28. For c0=2c(6+66c), we find that c0=250 when c=0.1, and then c0 decreases monotonically as c increases. Now t0(x)c02τRx2, so we need τR<c to get realistic (not too high). For typical values of c, say c=0.2, we have c0 around 250, and hence

t0(x)125τRx2
for small x.

Equations (46)–(48) are valid for all x; thus position t0(x) and acceleration t0(x) are even functions of x, while velocity t0(x) is an odd function in x. It thus corresponds to a more complicated form than the motion under gravity of Section 6, as here the analogous particle would be subject to a force proportional to 2τRc[6X5/2+66cX2+215c2X3/2]. From Eq. (49), this force varies roughly as c0τRe6x2. It approaches zero for large |x| but does not change sign.

Now, let us turn to the exponent term θ(x). Setting θ(0)=0, we have

θ(x)=35c2arcsinh(2x).

The expansion,

θ(x)=35cx(123x2+),
agrees with the results from the partial differential equation (PDE) simulations. Now
θ(x)=35cX,
where X=1+4x2. Plainly, this is almost constant for small x.

In the next section, we plot t0(x)/τR for various c on the same diagram. We also plot θ(x)/c and (r(x)14x2)/c to estimate the next term in r(x).

8. COMPARISON WITH NUMERICAL SIMULATIONS OF PDE

For x=0, we have a(x)=θ(x)=0, so ψ is real. It maximum value, ψm, occurs when t=0 and is given by ψm=4r(0)13+8c. As noted earlier, this is a function of c only, as r(0)=12c. Using the model, at fixed x, the intensity, I, is

I={r(x)+4[tt0(x)]24}2+64x2(r(x)+4[tt0(x)]2)2.

We note that r(x) is close to 1+4x2 for small c. By setting It=0, we find that there are three stationary points. The first is at t=t0(x), and it is a maximum since 2It2<0, as 4+16x2r(x). On the either side, we have t=t0(x)±1216x2+4r(x). These are minima, since 2It2>0 for each.

In the simulations of the PDE we can start with initial conditions with r(0) close to 1 and t0(0)=θ(0)=0. At each x, the value of t0 giving the maximum intensity is then labelled as t0(x). This shows the dependence of t0(x) on p. At this point, we use the intensity maximum, Im=[r(x)4]2+64x2r(x)2, to find r(x) by solving the quadratic equation. This shows the dependence of r(x) on p. At each such point, we have x and r(x) and the complex number ψ(x,t=t0(x))=[4(1+2ix)r(x)1]ei[xθ(x)]; we directly find θ(x). So we can compare the PDE results with the predicted results found by solving the four ODEs approximately. Using the numerical PDE runs, we now find how r(x), t0(x), and θ(x) depend on c and τR.

From numerics, we can plot r(x)14x2 for various values of c, and then observe if it can be expressed as cnf(x) for some integer n, that is, whether the dependence on c and x can be separated for a suitably chosen value of n. Results show that n=1 with f(x)=2(1+4x2)k works well for k=1.865, thus roughly supporting the result of Eq. (38), which had k=3/2. Here, c is arbitrarily chosen, and it determines how much r(0) differs from 1.

Similarly, we can plot t0(x)/τR for various values of c, and then observe if it can be expressed as j[c]g(x), and plot θ(x) for various values of c, and then observe if it can be expressed as ch(x). For small x, we find that t0(x)/τRx2, agreeing with the analytic form.

For small c, the numerics show that

t0(x)25τRx2(1+14.7c),
for offset c. So, when c is not too small, t0(x) is of the same order as the analytic form, Eq. (52), found earlier. An example with c=0.1 is given in Fig. 9.

 figure: Fig. 9.

Fig. 9. Plot of function t0(x)/τR, from numerics (blue curve) for c=0.1. The magenta curve is from the analytic prediction of Eq. (52). It is in fairly good agreement with the numeric curve. Thus, the reduced model has provided a reasonably good description of the rogue wave dynamics.

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9. NUMERICAL DETERMINATION OF FUNCTIONS

Suppose θ(x)=cny(x) and r(x)=4x2+1+cnr1(x) for some unknown n. Then we find, in this model, that on the “ridge” formed by the maxima at each x,

ϕ[x,t=t0(x);c]=(1+4(1+2ix)cnr1(x)+4x2+1)ei(x+cny(x)),
while NLSE result is
ϕ[x,t=0]=(1+4(1+2ix)4x2+1)ei(x).

Using numerics, we subtract one from the other. Suppose the difference Δ is

Δ=cneix[f(x)+in(x)].

Along the ridge, we measure this Δ as a complex function of x. So, by finding a few values of Δ, we determine the dependence on c, and immediately find the value of n, which turns out to be 1.

Then, expanding eix(ϕ[x,t=t0(x);c]ϕ[x,t]Δ) for small c gives the first nonzero term. Equating it to zero gives

f(x)+in(x)+4ir1(x)i2x+(12ix)(2x3i)y(x)(2x+i)2=0.

Solving this equation, we obtain r1(x), y(x) explicitly:

r1(x)=112(4x2+1)[(34x2)f(x)+8xn(x)],
and
y(x)=13[2xf(x)n(x)].

We use this procedure to produce curves for r1(x) and y(x) for various values of c. From Fig. 10 and curves for other c values, we find r1(x)2(4x2+1)1.865. This compares with 2(4x2+1)3/2 found from the analytic result, so the approximate result is reasonably accurate.

 figure: Fig. 10.

Fig. 10. Function r1(x) found from numerical solution of the PDE. Three values of c, from (blue curve) 0.004 to (green curve) 0.006 are used, and the initial value is r(0)=12c. The analytic form, r1(x)=(r(x)14x2)/c2(1+4x2)3/2, shown by the red curve is close to the numerical results.

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From Fig. 11 and curves for other c values, we find y(x)35x and hence θ(x)35cx for x<1. For small x, we find

y(x)θ(x)/(c)=35x+2(96+353)x3+.

 figure: Fig. 11.

Fig. 11. Numerical PDE solution giving θ(x)/c. Various values of c are used, and the initial value is r(0)=12c. The analytic result is plotted as the magenta line, θ(x)/c=(35/2)arcsinh(2x)35x for x small, using Eq. (53). The numerically found curve for θ(x)/c is quite close to 35x over this range, 0<x<1.

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Plainly, for small x, which is the main region of interest for the localized rogue wave, we have θ(x)35cx, which is in good agreement with the value from the expression found numerically.

10. CONCLUSIONS

In this work, we have found rogue wave solutions of higher-order extensions of the NLSE. Our approach can be useful when exact solutions of the evolution equation cannot be found. As an example, approximate solutions in the form of rogue waves have been obtained and analyzed for an optical fiber with a Raman delay term. Based on this example, the technique can be applied to other higher-order terms in the extended NLSE.

In order to confirm the applicability of the technique, we tested it on a few integrable equations and compared the results with the exact solutions. The equation with the Raman term does not have the exact solution in the form of a rogue wave. In this case, we conducted numerical simulations confirming that the approximate solutions are sufficiently accurate. The approach presented here could also be used to study other optical systems that are only described by nonintegrable equations, for example, ultrafast nonlinear quadratic systems.

Rogue waves may be difficult to detect with existing equipment in optics. However, new techniques have been developed recently that may help in such measurements [48]. Namely, time stretching analog-to-digital converters [49,50], time microscopes [51], and time lens magnifiers [52] may be able to resolve individual ultrashort pulses that could not be done with traditional pulse accumulating methods. In this sense, theory and experiment are progressing equally well, and this may allow us to witness experimental observations of Raman rogue waves soon. The first experimental observation, using a single shot technique, has already been made in [12]. Fine details of the pulse profile that we studied here could also be observed in the near future.

Funding

Australian Research Council (ARC) (DP140100265, DP150102057, DE1312345); Volkswagen Foundation.

Acknowledgment

The authors thank Wonkeun Chang for assistance.

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Figures (11)

Fig. 1.
Fig. 1. Plot of the relative soliton offset, t 0 / τ R , showing the effect of Raman delay. We compare the numerically found result (blue curve) with the analytic result of Eq. (29) (red curve). Here, delay is given by τ R = 0.02 and width w 0 = 0.4 .
Fig. 2.
Fig. 2. Plot of the soliton, Eq. (23), under the effect of Raman delay, using Eqs. (24)–(30). Here, delay is given by τ R = 0.02 and width w 0 = 0.4 . At the ends ( x = ± 3 ), the t offset is about 1.8.
Fig. 3.
Fig. 3. Plot of function r ( x ) for various c values. The blue curves are for c < 0 . This solution corresponds to the expansion in Eq. (37) with c = 0.04 (top blue curve) and c = 0.02 (next blue curve). The green curve, namely r ( x ) = z 2 = 1 + 4 x 2 , is for c = 0 . The two red curves are plots of Eqs. (35) and (36). The upper red curve is for c = 0.02 . The lower red curve is for c = 0.04 . In each case, the expansion, Eq. (37), is superimposed on the exact result. The curves are indistinguishable.
Fig. 4.
Fig. 4. Plot of function t 0 ( x ) / τ R , from Eq. (46) for various c : c = 0.04 (blue curve), c = 0.4 (magenta), c = 0.2 (red), and c = 0.1 (green).
Fig. 5.
Fig. 5. Plot of functions t 0 ( x ) , Eq. (46), r ( x ) , using the results of Section 6, including Eq. (38), and θ ( x ) , using Eq. (53), together with the same functions found numerically by solving Eqs. (33)–(41). Here, c = 0.04 , and the delay is given by τ R = 0.005 . The two upper curves give r ( x ) , while the two lower positive ones give t 0 ( x ) and the two negative ones show θ ( x ) . In each case, the exact numeric solution and the approximation for it are very close.
Fig. 6.
Fig. 6. Plot of the rogue wave, Eq. (7), using Eqs. (46)–(53) for delay given by τ R = 0.005 . Here c = 0.04 .
Fig. 7.
Fig. 7. Contour plot of the rogue wave, Eq. (7), using Eqs. (46)–(53) for delay given by τ R = 0.005 . Here c = 0.04 and the propagation direction, x , is the horizontal axis, while the transverse axis, t , is vertical.
Fig. 8.
Fig. 8. Plot of functions t 0 ( x ) / τ R (even function of x with t 0 ( 0 ) c 0 = 300 here), t 0 ( x ) / τ R (odd function of x ) and t 0 ( x ) / τ R (even function of x with t 0 ( 0 ) = 0 ) from original Eqs. (46), (45), and (48), respectively, together with a comparison with the Gaussian-based approximations for them, given by Eqs. (49), (50), and (51), all for c = 0.1 . Here, c 0 = 2 c ( 6 + 66 c + 215 c 2 ) .
Fig. 9.
Fig. 9. Plot of function t 0 ( x ) / τ R , from numerics (blue curve) for c = 0.1 . The magenta curve is from the analytic prediction of Eq. (52). It is in fairly good agreement with the numeric curve. Thus, the reduced model has provided a reasonably good description of the rogue wave dynamics.
Fig. 10.
Fig. 10. Function r 1 ( x ) found from numerical solution of the PDE. Three values of c , from (blue curve) 0.004 to (green curve) 0.006 are used, and the initial value is r ( 0 ) = 1 2 c . The analytic form, r 1 ( x ) = ( r ( x ) 1 4 x 2 ) / c 2 ( 1 + 4 x 2 ) 3 / 2 , shown by the red curve is close to the numerical results.
Fig. 11.
Fig. 11. Numerical PDE solution giving θ ( x ) / c . Various values of c are used, and the initial value is r ( 0 ) = 1 2 c . The analytic result is plotted as the magenta line, θ ( x ) / c = ( 35 / 2 ) arcsinh ( 2 x ) 35 x for x small, using Eq. (53). The numerically found curve for θ ( x ) / c is quite close to 35 x over this range, 0 < x < 1 .

Equations (87)

Equations on this page are rendered with MathJax. Learn more.

i ψ x + 1 2 ψ t t + | ψ | 2 ψ = R ,
R = τ R ψ ( | ψ | 2 ) t ν | ψ | 4 ψ ν 2 n | ψ | 2 n ψ + i γ 3 ψ t t t d 4 ψ t t t t d 6 ψ t t t t t t i s ( ψ | ψ | 2 ) t ,
L = L d d t ,
L d = i 2 ( ψ * ψ x ψ ψ * x ) + 1 2 | ψ t | 2 1 2 | ψ | 4 ,
L d ψ * + x L d ψ x * + t L d ψ t * 2 t 2 ( L d ψ t t * ) = R ,
d d x ( L c x j ) L c ( j ) = 2 Re ( R ψ * c ( j ) d t ) J [ c j ; R ]
i ψ x + 1 2 ψ t t + | ψ | 2 ψ = 0 ,
ψ = [ 4 1 + 2 i x c ( x ) + 4 t 2 1 ] exp [ i x ]
L = 2 π c ( x ) 7 / 2 [ x c ( x ) 2 c ( x ) 14 ( 4 x 2 + 1 ) c ( x ) + 2 c ( x ) 3 + 10 ( 4 x 2 + 1 ) 2 ] .
70 π ( 4 x 2 + 1 ) ( c ( x ) + 4 x 2 + 1 ) c ( x ) 9 / 2 = 0 ,
c ( x ) = 1 + 4 x 2 ,
ψ = [ 4 1 + 2 i x 1 + 4 x 2 + 4 t 2 1 ] exp [ i x ] .
ψ ( x , t ) = [ 4 1 + 2 i x D m ( x , t ) 1 ] e i F m ( x , t ) ,
D m ( x , t ) = r ( x ) + 4 [ t t 0 ( x ) ] 2 ,
F m ( x , t ) = x + a ( x ) [ t t 0 ( x ) ] θ ( x ) ,
L r 7 / 2 ( x ) 2 π = r 2 ( x ) [ 2 X a ( x ) t 0 ( x ) + X a 2 ( x ) + x r ( x ) 2 X θ ( x ) ] r 3 ( x ) [ 2 2 a ( x ) t 0 ( x ) + a 2 ( x ) 2 θ ( x ) ] + 14 X r ( x ) 10 X 2 ,
r ( x ) [ r ( x ) + 16 x ] 3 ( 4 x 2 + 1 ) r ( x ) = 0 ,
[ t 0 ( x ) ] 2 + 2 θ ( x ) = 0 .
ψ = [ 4 1 + 2 i x 1 + 4 x 2 + 4 ( t v x ) 2 1 ] e i [ v t + ( 1 v 2 2 ) x ] .
ψ ( x , t ) = [ 4 1 + 2 i B x D m ( x , t ) 1 ] e i F m ( x , t ) ,
L d = L d 1 + L d 2 + L d 3 ,
L d 1 = i 2 ( ψ * ψ x ψ ψ * x ) + a 1 a ( x ) t 0 ( x ) + h 1 θ ( x ) u 1 , L d 2 = 1 2 | ψ t | 2 ,
L d 3 = 1 2 ( | ψ | 4 1 ) .
( a 1 1 ) a ( x ) t 0 ( x ) + ( h 1 1 ) θ ( x ) u 1 + 1 ,
L r 7 / 2 ( x ) 2 π = 14 X r ( x ) 10 X 2 + r ( x ) 2 [ a 2 ( x ) X 2 a ( x ) X t 0 ( x ) 2 X θ ( x ) + B x r ( x ) + 4 B 4 ] r ( x ) 3 [ a 2 ( x ) + 2 B 2 a ( x ) t 0 ( x ) 2 θ ( x ) ] ,
S [ c ( j ) ] d d x ( L c x j ) L c ( j ) ,
r ( x ) 3 / 2 4 π [ d d x ( L a x ) L a ] = r ( x ) 3 / 2 4 π S [ a ] = ( r ( x ) X ) [ a ( x ) t 0 ( x ) ] .
r ( x ) 5 / 2 2 π [ d d x ( L θ x ) L θ ] = r ( x ) 5 / 2 2 π S [ θ ] = [ 3 X r ( x ) ] r ( x ) 16 B 2 x r ( x ) ,
r ( x ) 5 / 2 2 π [ d d x ( L t 0 ( x ) ) L t 0 ( x ) ] = r ( x ) 5 / 2 2 π S [ t 0 ] = 2 r ( x ) a ( x ) [ r ( x ) X ] a ( x ) { [ r ( x ) 3 X ] r ( x ) + 16 B 2 x r ( x ) } .
r ( x ) 9 / 2 π [ d d x ( L r x ) L r ] = r ( x ) 9 / 2 π S [ r ] = r ( x ) { r ( x ) { ( 3 X r ( x ) ) [ a 2 ( x ) 2 a ( x ) t 0 ( x ) 2 θ ( x ) ] + 12 ( B 1 ) } + 70 X } 70 X 2 .
J [ c j ; R ] = 2 Re ( R ψ * c ( j ) d t ) .
R = i α 3 ( ψ t t t + 6 | ψ | 2 ψ t ) R h .
r ( x ) 7 / 2 12 π α 3 J [ a ; R h ] = [ X r ( x ) ] × { r ( x ) [ ( a 2 ( x ) 2 ) r ( x ) + 4 ] 10 X } ,
r ( x ) 9 / 2 2 π α 3 a ( x ) J [ r ; R h ] = 210 X 2 + r ( x ) × { r ( x ) [ a 2 ( x ) ( r ( x ) 3 X ) + 6 ( 6 r ( x ) + 3 X ) ] 210 X } ,
v s = α 3 ( 1 k 2 2 ) v 1 k .
α 3 k 2 ( 2 c 2 + v 1 2 ) 2 α 3 ( v 1 6 ) + ( 2 c 1 + 1 ) k = 0 .
t 0 ( x ) = x [ 3 α 3 ( k 2 2 ) + k ] , θ ( x ) = 1 2 k 2 x ( 4 α 3 k + 1 ) .
i ψ x + 1 2 ψ t t + | ψ | 2 ψ = R [ ψ ( x , t ) ] ,
R [ ψ ( x , t ) ] = α 4 ( ψ t t t t + 6 ψ t 2 ψ * + 4 ψ | ψ t | 2 + 8 ψ t t | ψ | 2 + 2 ψ t t * ψ 2 + 6 ψ | ψ | 4 ) R l .
r ( x ) 7 / 2 16 π a ( x ) α 4 J [ a ; R l ] = [ r ( x ) X ] { r ( x ) × [ ( a 2 ( x ) 6 ) r ( x ) + 12 ] 30 X }
r ( x ) 13 / 2 2 π α 4 J [ r ; R l ] = 2772 X 3 + r ( x ) × { r ( x ) [ 12 a 2 ( x ) ( r ( x ) ( r ( x ) [ r ( x ) 3 X 6 ] + 35 X ) 35 X 2 ) + a 4 ( x ) r ( x ) 2 [ 3 X r ( x ) ] 2 r ( x ) ( 3 r ( x ) [ r ( x ) 3 ( X + 4 ) ] + 70 ( 3 X + 2 ) ) + 28 X ( 15 X + 91 ) ] 5040 X 2 } ,
t 0 ( x ) = v e x = k [ 4 α 4 ( k 2 6 ) 1 ] x , θ ( x ) = { 3 α 4 [ ( k 2 2 ) 2 6 ] k 2 2 } x .
ψ ( x , t ) = Q ( x ) 2 w ( x ) sech ( y 1 w ( x ) ) exp { i [ a ( x ) y 1 θ ( x ) ] } ,
Q ( x ) = 0 ,
Q ( x ) [ a ( x ) t 0 ( x ) ] = 0 ,
Q ( x ) 6 w 3 ( x ) [ w ( x ) Q ( x ) 2 ] ,
8 τ R 15 w 0 4 t 0 ( x ) = 0 ,
t 0 ( x ) = a ( x ) = 4 τ R Q ( x ) 15 w 0 3 x = 8 τ R 15 w 0 4 x ,
t 0 ( x ) = 4 τ R x 2 15 w 0 4 .
1 w 0 2 + t 0 ( x ) 2 + 2 θ ( x ) = 0 ,
ψ ( x , t ) = 1 w 0 sech ( t w 0 ) exp { i x 2 w 0 2 }
= Q 0 2 sech ( Q 0 t 2 ) e 1 8 i Q 0 2 x ,
r ( x ) [ r ( x ) + 16 x ] = 3 ( 4 x 2 + 1 ) r ( x ) .
[ 3 z 2 s 2 ( z ) ] s ( z ) = 2 z s ( z ) .
s ( z ) = 1 6 c ( F 1 / 3 + F 1 / 3 1 ) ,
F = 54 c 2 z 2 1 + 6 c z 3 27 c 2 z 2 1 ,
r ( z ) = z 2 ( 1 2 c z + 6 c 2 z 2 21 c 3 z 3 + 80 c 4 z 4 + ) .
r ( x ) = ( 1 + 4 x 2 ) [ 1 2 c 1 + 4 x 2 + 6 c 2 ( 1 + 4 x 2 ) 21 c 3 ( 1 + 4 x 2 ) 3 / 2 + 80 c 4 ( 1 + 4 x 2 ) 2 + ] .
r ( x ) 9 / 2 32 π J [ t 0 ; R r ] = τ R { r ( x ) [ 5 X r ( x ) ] 7 X 2 } ,
16 τ R { r ( x ) [ r ( x ) 5 X ] + 7 X 2 } + r ( x ) 2 { 2 r ( x ) ( a ( x ) [ r ( x ) X ] 8 x a ( x ) ) + a ( x ) [ 3 X r ( x ) ] r ( x ) } = 0 ,
r ( x ) { r ( x ) [ r ( x ) 3 X ] [ a 2 ( x ) 2 a ( x ) t 0 ( x ) 2 θ ( x ) ] 70 X } + 70 X 2 = 0
[ X r ( x ) ] [ a ( x ) t 0 ( x ) ] = 0 .
r ( x ) { r ( x ) [ t 0 ( x ) ( ( r ( x ) 3 X ) r ( x ) + 16 x r ( x ) ) + 2 r ( x ) ( r ( x ) X ) t 0 ( x ) + 16 τ R ] 80 τ R X } + 112 τ R X 2 = 0 ,
r ( x ) { r ( x ) [ 3 X r ( x ) ] [ t 0 ( x ) 2 + 2 θ ( x ) ] 70 X } + 70 X 2 = 0 .
r ( x ) = X [ 1 2 c X 1 / 2 + 6 c 2 X 21 c 3 X 3 / 2 + 80 c 4 X 2 + ] .
a ( x ) = t 0 ( x ) = 2 τ R x c X 3 / 2 [ 2 ( 8 x 2 + 3 ) + 33 c X 1 / 2 + 215 c 2 X ] + 33 τ R arctan ( 2 x ) .
t 0 ( x ) = τ R { 1 c ( 8 x 2 + 1 X 1 / 2 1 ) + 33 x arctan ( 2 x ) + 215 2 c [ X 1 / 2 1 ] } .
t 0 ( x ) τ R ( 215 c + 6 c + 66 ) x 2 ( 215 c + 10 c + 88 ) x 4 + .
t 0 ( x ) 6 τ R c x 2 ( 1 5 3 x 2 + ) .
a ( x ) τ R = t 0 ( x ) τ R = 2 c [ 6 X 5 / 2 + 66 c X 2 + 215 c 2 X 3 / 2 ] .
t 0 ( x ) τ R c 0 e 6 x 2 .
t 0 ( x ) τ R c 0 π 6 erf ( 6 x ) ,
t 0 ( x ) τ R c 0 2 π 6 [ x erf ( 6 x ) 1 e 6 x 2 6 x ] .
t 0 ( x ) τ R c 0 2 x 2 [ 1 x 2 + 6 x 4 5 ] .
t 0 ( x ) 125 τ R x 2
θ ( x ) = 35 c 2 arcsinh ( 2 x ) .
θ ( x ) = 35 c x ( 1 2 3 x 2 + ) ,
θ ( x ) = 35 c X ,
I = { r ( x ) + 4 [ t t 0 ( x ) ] 2 4 } 2 + 64 x 2 ( r ( x ) + 4 [ t t 0 ( x ) ] 2 ) 2 .
t 0 ( x ) 25 τ R x 2 ( 1 + 14.7 c ) ,
ϕ [ x , t = t 0 ( x ) ; c ] = ( 1 + 4 ( 1 + 2 i x ) c n r 1 ( x ) + 4 x 2 + 1 ) e i ( x + c n y ( x ) ) ,
ϕ [ x , t = 0 ] = ( 1 + 4 ( 1 + 2 i x ) 4 x 2 + 1 ) e i ( x ) .
Δ = c n e i x [ f ( x ) + i n ( x ) ] .
f ( x ) + i n ( x ) + 4 i r 1 ( x ) i 2 x + ( 1 2 i x ) ( 2 x 3 i ) y ( x ) ( 2 x + i ) 2 = 0 .
r 1 ( x ) = 1 12 ( 4 x 2 + 1 ) [ ( 3 4 x 2 ) f ( x ) + 8 x n ( x ) ] ,
y ( x ) = 1 3 [ 2 x f ( x ) n ( x ) ] .
y ( x ) θ ( x ) / ( c ) = 35 x + 2 ( 96 + 35 3 ) x 3 + .
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