Method Based on Linear Integral Equations

S.10.3. Method Based on Linear Integral Equations

Below we outline the approach proposed by Zakharov and Shabat (1974) based on using linear integral equations of the form

K(x, y) = F (x, y) +

K(x, z)N (x; z, y) dz, y ≥ x, (13)

where the functions

F , N , and K can depend on some additional parameters other than the specified arguments. In each specific case, the function N is explicitly expressed through F . Define an operator A x such that

f (z)N (x; z, y) dz if y ≥ x,

A x f (y) =

0 if y<x

and assume that for each chosen N , it is possible to prove that the operator I − A x is invertible and its inverse, (I − A x ) −1 , is continuous, where I is the identity operator. The following three steps represent an algorithm for finding a nonlinear equation that can then be solved by the inverse scattering method.

1 ◦ . The function

F satisfies the following two linear ordinary (or partial) differential equations:

2 ◦ . The function K is related to F by equation (13), which can be rewritten as

(I − A x ) K=F.

3 ◦ . Applying the operators L i involved in (14) to equation (15), we obtain

L i (I − A x ) K = 0,

i = 1, 2.

This equation can be rewritten in the form

(I − A x )(L i K) = R i ,

i = 1, 2,

where R i contains all nonzero terms of the commutator [L i , (I − A x )]. Moreover, (13) and (14) should be chosen so that R i could be represented in the form

R i = (I − A x )M i ( K),

i = 1, 2,

where M i ( K) is a nonlinear functional of K. But the operator I − A x is invertible, and therefore, the function K satisfies the nonlinear differential equations

(16) It follows that each solution of the linear integral equation (13) is a solution of the nonlinear

L i K−M i ( K) = 0,

i = 1, 2.

differential equation (16).

K(x, y) = F (x, y) +

K(x, z)F (z, y) dz

and write out some identities to be used in the sequel,

K(x, z)F (z, y) dz =

F (z, y)∂ n K(x, z) dz + A n ,

K(x, z)∂ ∞ n

F (z, y) dz = (−1) n

F (z, y)∂ n

z K(x, z) dz + B n ,

where the A n are defined by the recurrence relations

F (x, y)[∂ n−1 x K(x, z)] z=x , and

A 1 =− K(x, x)F (x, y),

A n =( A n−1 ) x −

B 1 =− K(x, x)F (x, y),

B 2 =− K(x, x)∂ x

F (x, y) + [∂ z K(x, z)] z=x

F (x, y), ...

Let us introduce an operator L 1 and require that

F satisfy the linear equation L 1 F ≡ (∂ 2 x 2 − ∂ y )

F (x, y) = 0.

Applying the operator L 1 to (17) and taking into account (18), (19), we obtain 2 2 Z ( ∞ ∂

x − ∂ y ) K(x, y) =

F (x, z)(∂

x − ∂ y ) K(x, z) dz − 2F (x, y)

K(x, x). dx

Using the equation F = (I − A x ) K and taking into account that the operator I − A x is invertible, we finally get

( ∂ 2 − ∂ x 2 y ) K(x, y) + u(x)K(x, y) = 0,

where the function u(x) is defined by

u(x) = 2

dx 22 K(x, x). ( )

Require that

F satisfy the linear equation

and apply the operator L 2 to (17) to obtain

3 ∂ ∞ t +( ∂ x + ∂ y ) K(x, y) = ∂ t +( ∂ x + ∂ y ) 3 K(x, z)F (z, y) dz.

A procedure similar to the above calculations for the operator L 1 yields

t 3 +( ∂ x + ∂ y ) K + 3u(∂ x + ∂ y ) K = 0.

For the characteristic y = x, equation (24) can be rewritten in terms of u = 2(d/dx)K(x, x). Differentiating (24) with respect to x and rearranging terms, we arrive at the Korteweg–de Vries equation

u t +6 uu x + u xxx = 0.

Any function F satisfying the linear equations (20), (23) and rapidly decaying as x → +∞ generates a solution of the Korteweg–de Vries equation. To this end, one should solve the linear integral equation (17) for function K and express u

through K by formula (22). Example 6. Consider the integral equation

σ Z ∞ Z K(x, y) = F (x, y) + ∞ K(x, z)F (z, u)F (u, y) dz du,

where σ= ❳ 1. Here and in what follows, the coefficients are chosen with a view to simplifying the calculations. Let the operator L 1 have the form

which implies that

F (x, y) = F

Shifting the lower limit of integration to zero, we rewrite equation (25) in the form

K(x, y) = F

2 4 0 0 K(x, x + ζ)F

2 F 2 dζ dη,

or, equivalently,

[(I − σA x ) K](x, y) = F

2 F 2 dζ dη. Introducing the function

f (y) = 4 0 0 f (ζ)F

K 2 ( x, z) = 0 K(x, x + ζ)F

2 dζ

we can rewrite equation (25) as

K(x, y) = F

2 4 0 K x, x + η)F

2 dη.

Applying the operator L 1 of (26) to equation (29), and the operator ∂ x + ∂ z to (28), and taking into account the invertibility of I − σA x , we find, after appropriate calculations, that

( ∂ x + ∂ y ) K 2 ( x, y) = −2K(x, x)K(x, y),

∂ y ) K(x, y) = − 2 K(x, x)K 2 ( x, y).

Applying the operator ∂ x + ∂ y to (27), we get

= (I −

2 σA x ) ( ∂ x + ∂ y ) K(x, y) + 2 K 2 ( x, x)K(x, y) .

Let us require that the function

F satisfy the second linear equation

Applying the operator L 2 to equation (27) and taking into account the above auxiliary relations (30)–(32), we ultimately find that [ ∂ t +( ∂ x + ∂ 3 y 2 ) ] K(x, y) = 3σK(x, x)K(x, y)∂ x K(x, x) + 3σK ( x, x)(∂ x + ∂ y ) K(x, y)

for y ≥ x. Now, by setting q(x, t) = K(x, x; t), we rewrite equation (34), for y = x, in terms of the dependent variable q to obtain the modified Korteweg–de Vries equation

q t + q xxx =6 σq 2 q x .

Thus, each solution of the equations L i F = 0, i = 1, 2, with a sufficiently fast decay rate as x → ∞ determines a solution of ❨✂❩ equation (35). Note that we have to solve the linear integral equation (25) at an intermediate step. References for Subsection S.10.3: V. E. Zakharov and A. B. Shabat (1974), M. J. Ablowitz and H. Segur (1981),

M. J. Ablowitz and P. A. Clarkson (1991).