Handbook of mathematics for engineers and scienteists part 114 doc

Solution of the Cauchy Problem by the Inverse ScatteringProblem Method 15.12.3-1.. The solution of the Cauchy problem for nonlinear equations admitting a Lax pair or another “implicit” l

Thus, two linear systems (15.12.2.4) and (15.12.2.18) are compatible if the functions f and g satisfy the cubically nonlinear system (15.12.2.17) For

g = –k ¯ f, a=2i, (15.12.2.19)

where k is a real constant and the bar over a symbol stands for its complex conjugate, both

equations (15.12.2.18) turn into one and the same nonlinear Schr¨odinger equation

if t = f xx+2k|f2|f (f ¯ f =|f|2) (15.12.2.20)

Likewise, one can use other polynomials in λ and determine the associated linear systems

generating nonlinear evolution equations

Example 1 Searching for solutions of the determining system (15.12.2.7) in the form third-order

poly-nomials in λ results in

A = a3λ3+ a2λ2+12(a3f g + a1)λ + 12a2f g–14ia3(f g x – gf x ) + a0,

B = ia3f λ2+ (ia2f–12a3f x )λ + ia1f+12ia3f2g–12a2f x– 14a3f xx,

C = ia3gλ2+ (ia2g+12a3g x )λ + ia1g+12ia3f g2+12a2g x–14ia3g xx,

(15 12 2 21 )

where a0, a1, a2, and a3 are arbitrary constants The evolution equations for f and g corresponding to

(15.12.2.21) are

f t+ 14ia3(f xxx– 6f gf x) + 12a2(f xx– 2f2g) – ia1f x– 2a0f= 0 ,

g t+14ia3(g xxx– 6f gg x) – 12a2(g xx– 2f g2) – ia1g x+ 2a0g= 0 (15.12 2 22 ) Consider two important special cases leading to interesting nonlinear equations of mathematical physics.

1◦ If

a0= a1= a2= 0 , a3= – 4i, g= 1

the second equation of (15.12.2.22) is satisfied identically, and the first equation of (15.12.2.22) translates into the Korteweg–de Vries equation

f t + f xxx– 6f f x= 0

2◦ If

a0= a1= a2= 0 , a3= – 4i, g = f ,

both equations (15.12.2.22) translate into one and the same modified Korteweg–de Vries equation

f t + f xxx 6f2f x= 0

Example 2 Now we will look for a solution of the determining system (15.12.2.7) in the form of the

simple one-term expansion in negative powers of λ:

A = a(x, t)λ–1, B = b(x, t)λ–1, C = c(x, t)λ–1 (15 12 2 23 ) This results in the relations

a x= 12i(f g) t, b= –12if t, c= 12ig t (15 12 2 24 )

The evolution equations for functions f and g corresponding to (15.12.2.24) are written as

f xt= – 4iaf,

g xt= – 4iag. (15.12 2 25 )

If we set

a= 14i cos w, b = c = 14i sin w, f = –g = –12w x, (15 12 2 26 ) then the three relations (15.12.2.24) are reduced to one and the same equation, the sine-Gordon equation

and the two equations (15.12.2.25) coincide and give a differential consequence of equation (15.12.2.27):

w xxt = w x cos w.

Thus, the linear system of equations (15.12.2.4) and (15.12.2.5), whose coefficients are defined by

formu-las (15.12.2.23) and (15.12.2.26) with D = –A, is compatible if the function w satisfies the sine-Gordon

equation (15.12.2.27).

Remark Sometimes the determining equations (15.12.2.1)–(15.12.2.2) are also called a Lax pair, by analogy with (15.12.1.2)–(15.12.1.3).

15.12.3 Solution of the Cauchy Problem by the Inverse Scattering

Problem Method

15.12.3-1 Preliminary remarks The direct and inverse scattering problems

The solution of the Cauchy problem for nonlinear equations admitting a Lax pair or another

“implicit” linearization (see Subsection 15.12.2) falls into several successive steps Two of them involve the solution of the direct and the inverse scattering problem for auxiliary linear equations Summarized below are relevant results for the linear stationary Schr¨odinger equation

ϕ 

xx + [λ – f (x)]ϕ =0 (–∞ < x < ∞). (15.12.3.1)

It is assumed that the function f (x), called the potential, vanishes as x → ∞ and the

condition  ∞

–∞(1+|x|)|f (x)|dx<∞ holds.

Direct scattering problem Consider the linear eigenvalue problem for the ordinary

differential equation (15.12.3.1)

Eigenvalues can be of two types:

λ n= –κn2, n =1, 2, , N (discrete spectrum),

λ = k2 –∞ < k < ∞ (continuous spectrum) (15.12.3.2)

It is known that if min f (x) < λ < 0, equation (15.12.3.1) has a discrete spectrum of

eigenvalues, and if f (x) <0and λ >0, it has a continuous spectrum

Let λ n= –κn2 be discrete eigenvalues numbered so that

λ1< λ2 <· · · < λ N <0,

and let ϕ n = ϕ n (x) be the associated eigenfunctions, which vanish as x → ∞ and are

square summable Eigenfunctions are defined up to a constant factor If an eigenfunction

is fixed by its asymptotic behavior for negative x,

ϕ n → exp(κ n x) as x → –∞,

then the leading asymptotic term in the expansion of ϕ n for large positive x is expressed as

ϕ n → c nexp(–κn x ) as x → ∞. (15.12.3.3)

The eigenfunction ϕ n has n –1zero and moreover

c n= (–1)n–1|c n|

For continuous spectrum, λ = k2, the behavior of the wave function ϕ at infinity is specified by a linear combination of the exponentials exp( ikx), since f →0as|x|→ ∞.

The conditions at infinity are

ϕ → a(k)e–ikx + b(k)e ikx as x → ∞, (15.12.3.4)

and equation (15.12.3.1) uniquely determines the functions a(k) and b(k); note that they

are related by the simple constraint |a|2–|b|2 = 1 The first term in the second condition

of (15.12.3.4) corresponds to a refracted wave, and the second term to a reflected wave Therefore, the quotient

r (k) = b (k)

is called the coefficient of reflection (also reflection factor or reflectance).

The set of quantities

S ={κn , c n , r(k); n =1, , N} (15.12.3.6)

appearing in relations (15.12.3.3)–(15.12.3.5) is called the scattering data It is the deter-mination of these data for a given potential f (x) that is the purpose of the direct scattering

problem The scattering data completely determine the eigenvalue spectrum of the station-ary Schr¨odinger equation (15.12.3.1)

Note the useful formulas

|a (k)|= 1–|r (k)|2 – 1

,

arg a(k) = 1

i

N



n=1

k – iκn

k + iκ n –

1

π

 ∞

–∞

ln|a (s)|

s – k ds,

(15.12.3.7)

which can be used to restore the function a(k) for a given reflection factor r(k) The integral

in the second formula of (15.12.3.7) is understood as a Cauchy principal value integral, and

iκn are zeros of the function a(k) analytic in the upper half-plane [the κn appear in the asymptotic formulas (15.12.3.3)]

Inverse scattering problem The mapping of the potentials of equation (15.12.3.1) into

the scattering data (15.12.3.6) is unique and invertible The procedure of restoring f (x) for given S is the subject matter of the inverse scattering problem Summarized below are the

most important results of studying this problem

To determine the potential, one has to solve first the Gel’fand–Levitan–Marchenko integral equation

K (x, y) + Φ(x, y) +

 ∞

x K (x, z) Φ(z, y) dz =0, (15.12.3.8) where the functionΦ(x, y) = Φ(x + y) is determined via the scattering data (15.12.3.6) as

follows:

Φ(x, y) =N

n=1

c n

ia  (iκ n)e

– κn(x+y)+ 1

2π

 ∞

–∞ r (k)e

ik(x+y) dk, a  (k) = da

dk (15.12.3.9)

The potential is expressed in terms of the solution to the linear integral equation (15.12.3.8) as

f (x) = –2 d

dx K (x, x). (15.12.3.10) Note that the direct and inverse scattering problems are usually taught in courses on quantum mechanics For more details about these problems, see, for example, the book by Novikov, Manakov, Pitaevskii, and Zakharov (1984)

15.12.3-2 Solution of the Cauchy problem by the inverse scattering problem method Figure 15.8 depicts the general scheme of solving the Cauchy problem for nonlinear equa-tions using the method of the inverse scattering problem It is assumed that either a Lax pair (15.12.1.2)–(15.12.1.3) for the given equation has been obtained at an earlier stage or a

First step: solve the direct scattering problem

Cauchy problem = nonlinear equation + initial condition

Use the stationary equation from the Lax pair

Second step: look for the dependence

of the scattering data on time

Treat the nonstationary equation from the Lax pair

Third step: solve the inverse scattering problem

Write out the Gel'fand–Levitan–Marchenko integral equation

With the solution obtained, find the solution of the Cauchy problem

Solve the integral equation

Figure 15.8 Main stages of solving the Cauchy problem for nonlinear equations using the method of the

inverse scattering problem.

representation of the equation as the compatibility condition for the overdetermined system

of linear equations (15.12.2.1)–(15.12.2.2) is known

Let us employ this scheme for the solution of the Cauchy problem for the Korteweg–de Vries equation

∂w

∂t + ∂

3w

∂x3 –6w ∂w

with the initial condition

w = f (x) at t =0 (–∞ < x < ∞). (15.12.3.12)

The function f (x) < 0is assumed to satisfy the same conditions as the potential in equa-tion (15.12.3.1)

The solution of the Cauchy problem (15.12.3.11), (15.12.3.12) falls into several succes-sive steps (see Figure 15.8) The Korteweg–de Vries equation is represented in the form

of a Lax pair (15.12.1.7) and the results for the solution of the direct and inverse scattering problems are used

First step One considers the linear eigenvalue problem —the direct scattering

prob-lem—for the auxiliary ordinary differential equation (15.12.3.1) that results from the sub-stitution of the initial profile (15.12.3.12) into the first equation of the Lax pair (15.12.1.7) for the Korteweg–de Vries equation (15.12.3.11) On solving this problem, one determines the scattering data (15.12.3.6)

Second step For t > 0, the function w = w(x, t) must appear in the first equation (15.12.1.7) instead of the initial profile f (x) In the associated nonstationary problem, the eigenvalues (15.12.3.2) are preserved, since they are independent of t (see Paragraph 15.12.1-1), but the eigenfunctions ϕ = ϕ(x, t) are changed.

For continuous eigenvalues λ > 0, the asymptotic behavior of the eigenfunctions, ac-cording to (15.12.3.4), is as follows:

ϕ → a(k, t)e–ikx + b(k, t)e ikx as x → ∞. (15.12.3.13)

Substitute the first asymptotic relation of (15.12.3.13) into the second equation of the

Lax pair (15.12.1.7) and take into account that w → 0as x → –∞ to obtain the function

p (t):

p (t) = –4ik3= const. (15.12.3.14)

In order to determine the time dependences, a = a(k, t) and b = b(k, t), let us consider the second equation of the Lax pair (15.12.1.7), where p(t) is given by (15.12.3.14) Taking into account that w →0as x → ∞, we get the linearized equation

ϕ t+4ϕ xxx–4ik3ϕ=0 (15.12.3.15) Substituting the second asymptotic relation of (15.12.3.13) into (15.12.3.15) and equating

the coefficients of the exponentials e–ikx and e ikx, we arrive at the linear ordinary differential

equations

a 

t=0, b 

t–8ik3b=0 Integrating them gives the coefficient of reflection

r (k, t) = b (t, k)

a (t, k) = r(k,0)e8ik3t (15.12.3.16)

The asymptotic behavior the eigenfunctions for discrete eigenvalues, with k = iκn, is determined in a similar way using the linearized equation (15.12.3.15)

Consequently, we obtain the time dependences of the scattering data:

S (t) ={κn , c n (t) = c n(0)e8κn3t , r(k, t) = r(k,0)e8ik3t ; n =1, , N} (15.12.3.17)

Third step Using the scattering data (15.12.3.17), we introduce the following function

by analogy with (15.12.3.9):

Φ(x, y; t) = 21

π

 ∞

–∞ r (k,0)e i[8k3t+k(x+y)] dk+

N



n=1

c n(0)

ia  (iκn)e

8 κ 3

n t–κ n(x+y) (15.12.3.18)

In this case,Φ depends on time parametrically This function plays the same role for the

first equation of the Lax pair (15.12.1.7) for t > 0 as the function (15.12.3.9) for equa-tion (15.12.3.1); recall that equaequa-tion (15.12.3.1) coincides with the first equaequa-tion of the Lax

pair (15.12.1.7), where w is substituted for by its value in the initial condition (15.12.3.12).

In order to recover the function w(x, t) from the scattering data (15.12.3.17), one has

first to solve the linear integral equation

K (x, y; t) + Φ(x, y; t) +

 ∞

x K (x, z; t)Φ(z, y; t) dz =0, (15.12.3.19) obtained from (15.12.3.8) by simple renaming This equation contains the functionΦ of (15.12.3.18) Then, one should use the formula

w (x, t) = –2 d

dx K (x, x; t), (15.12.3.20) obtained by renaming from (15.12.3.10), to determine the solution of the Cauchy problem for the Korteweg–de Vries equation (15.12.3.11)–(15.12.3.12)

15.12.3-3 N -soliton solution of the Korteweg–de Vries equation.

Let us find an exact solution of the Korteweg–de Vries equation (15.12.3.11) for the case

of nonreflecting potentials, with zero coefficient of reflection (15.12.3.5) We proceed from the linear integral equation (15.12.3.19)

Setting r(k,0) =0in (15.12.3.18), we have

Φ(x, y; t) =N

n=1

γ n e–κn(x+y)+8κ 3

n t, γ

n= c n(0)

ia  (iκn) >0 Substituting this function into (15.12.3.19), after simple rearrangements we obtain the equation

K (x, y; t) +

N



n=1

Γn (t)e–κn(x+y)+N

n=1

Γn (t)e–κn y ∞

x e

– κn z K (x, z; t) dz =0, (15.12.3.21)

where

Γn (t) = γ n e8 κ 3

The solution of the integral equation (15.12.3.21) is sought in the form of a finite sum:

K (x, y; t) =

N



n=1

e–κn y K

n (x, t). (15.12.3.23) Inserting (15.12.3.23) in (15.12.3.21) and integrating, we have

N



n=1

e–κn y K

n (x, t) +

N



n=1

Γn(t)e–κn(x+y)+N

n=1

Γn(t)e–κn yN

m=1

K m (x, t) e

–( κn κm)x

κn+κm =0.

Rewriting this equation in the form N

n=1 ψ n (x, t)e

– κn y =0and then setting ψ n (x, t) =0, we

arrive at a nonhomogeneous system of linear algebraic equations for K n (x, t):

K n (x, t) +Γn (t)

N



m=1

1

κn+κme–(κn κm)x K m (x, t) = –Γ n (t)e–κn x, n=1, , N

(15.12.3.24)

Using Cramer’s rule, we rewrite the solution to system (15.12.3.24) as the ratio of determi-nants

K n (x, t) = detA(n) (x, t)

detA(x, t) , (15.12.3.25)

whereA is the matrix with entries

A n,m (x, t) = δ nm+ Γn (t)

κn+κm e–(κn κm)x

= δ nm+ γ n

κn+κm e–(κn κm)x+8κ

3

n t,

δ nm =

1 if n = m,

0 if n≠m,

(15.12.3.26)

andA(n) (x, t) is the matrix obtained fromA by substituting the right-hand sides of

equa-tions (15.12.3.24) for the nth column Substituting (15.12.3.25) in (15.12.3.23) yields

K (x, y; t) =

N



n=1

e–κn ydetA(n) (x, t)

detA(x, t) . Further, by setting y = x, we get

K (x, x; t) = 1

detA(x, t)

N



n=1

e–κn xdetA(n) (x, t) = ∂

∂xln detA(x, t).

In view of (15.12.3.20), we now find a solution to the Korteweg–de Vries equation (15.12.3.11) for the case of nonreflecting potentials in the form

w (x, t) = –2 ∂2

∂x2 ln detA(x, t). (15.12.3.27) This solution contains2Nfree parametersκn, γ n (n =1, , N ) and is called an N -soliton

solution.

Example In the special case N =1 , the matrixA(x, t) is characterized by a single element, defined by

n = m =1 in (15.12.3.26):

A1,1(x, t) =1 + γ

2κe–2κ x+8κ

3t.

Substituting this expression into (15.12.3.27), we get the one-soliton solution of the Korteweg–de Vries equation

w(x, t) = – 2κ 2

cosh2[κ(x –4κ 2t) + ξ0], ξ0=

1

γ

This solution represents a solitary wave traveling to the right with a constant v =4κ 2 and rapidly decaying at infinity.

For the N -soliton solution (15.12.3.27), the following asymptotic relations hold:

w(x, t)≈ – 2N

n=1

κ 2

n

cosh2

κn (x – v n t) + ξ

n  as t → ∞, (15 12 3 29 )

where v n = 4κ 2

n is the speed of the nth component. From the comparison of formulas (15.12.3.28)

and (15.12.3.29) it follows that for large times, the N -soliton solution gets broken into N independent

one-soliton solutions, and one-solitons with a higher amplitude travel with a higher speed.