0387986995Basic TheoryC

positive number M is sufficiently large, the solution y = e-'(= for x < -M becomes a linear combination a(()e-'t=+b(()e't= of two solutions a-K= and eK' for x > M. The main problems are (i) the properties of a(() and b(() as functions of ( and (ii) construction of u(x) for given data {a((), b(()). Keeping this introduction in mind, let us consider a differential operator

where D = and u(x) is real-valued and belongs to C°°(-oo, +oo) such that

where (is a complex parameter. First, we construct two basic solutions which are called the Jost solutions of (VI.5.3).

Theorem VI-5-1. There exist two solutions f+(x, () and f_ (x, () of (VI.5.S) such that

(i) ft are continuous for

(VI.5.4) -oc < x < +oo, 0,

(ii) ft are analytic in ( for £ (() > 0, (iii)

Proof.

Step 1. Construction of f+: Solve the integral equation

by setting

f+(x, () = E Ym(x, ()

m-o

Step 2. Proof of (i) and (ii): To prove (i) and (ii), it suffices to prove that if

for (V1.5.4).

Inequality (VI.5.6) is true for m = 0. Assume that (VI.5.6) is true for an m; then,

m + 1 C(x)P(x)m+

Remark VI-5-2. At the last estimate, the following argument was used:

(a) r-x<r<1+rforx>O,

(b)

for x < 0.

Step 3. Proof of (iii): First, the estimate

m=1

follows from (VI.5.6). However, this estimate is not enough to prove (iii). So, let us derive another estimate for large I(I. To do this, we shall prove that

"

or(x) = f iu(r)jdr.

x

Inequality (VI.5.8) is true for m = 0. Assume that (VI.5.8) is true for an m. Then,

The proof of Theorem VI-5-1 for f+ (x, () can be completed by using (VI.5.7) and

(VI.5.9).

For f-(x,(), change x by -x to derive

(VI.5.10) -LY + u(-x)y = (2y.

The solution f+ for (VI.5.10) is the solution f- for (VI.5.3). 0

Remark VI-5-3. The solutions f± are uniquely determined by the conditions given in Theorem VI-5-1. Also,

< C(x)p(x)a(x) 1+I(I

for (VI.5.4).

Remark VI-5-4. If u(x) = 0 identically on the real line, then f f (x, () = et't'.

If u(x) = 0 for Ixl _> M for some positive constant M, then f+(x, () = e'(= for x > M, while f- (x, () = e-K= for x < -M.

VI-6. Scattering data

For real (, 1+ (x, () = f+ (x, -(), where f denotes the conjugate complex of f . Both f+ and f+ are solutions of Gy = (2y and

I f+(x, f) - e Z I < C(x)P(x)

for -oo < x < +oo and -oo < f < +oo. Let W(f,gj denotes the Vlwronskian of

{f,g}. Then,

whereas the function b(() is defined and continuous on -oo < ( < +oo. Furthermore, b(() = O(Ifl-1) as I(I ---. +oo. To simplify the situation, we introduce the

following assumption.

Since a(() = 0, there exists a constant d(() such that f_(x,() = d(()f+(x,C).

Therefore,

Observation VI-6-3. The quantities a(() and b({) satisfy the following relation:

Therefore, (VI.6.8) follows from

f- = af+ + b(af- - bf_) = a(f+ + bf-) - Ib12f_

_ aaf- - Ibl2f_ = {1a12 - Ibl2}f .

Observation VI-6-4. The formula

for %() > 0 shows that the quantities q, and r({) determine a(().

Proof

(i) f (() is analytic for $(() > -2 and (94 0,

(ii)((f(() - 1) is bounded for 3(() > -2,

(iii)f (() & 0 for Qr(() >- 0 and (54 0.

0((-1) as ( -, 00 on 3(() > 0. Observe that f(() - 1 = O((-'). From this observation, it follows without any complication that

(cf. Exercise VI-18). Now, (VI.6.9) follows from If(t) = Ia(()j and

loBja(()f2 = log(Ta(f)22/

Definition V1-6-5. The set {r((), (n,,n2,... ,nN), (cl,c2,... ,CN)} is called the scattering data associated with the potential u(x).

VI-7. Reflectionless potentials

The function is called the reflection coefficient. If this coefficient is zero, the potential u(x) becomes a function of simple form. Let us look into this situation.

Observation VI-7-1. If r(() = 0, then b(() = 0. This means that f_(x,t:) = a(t) f+(x, £) = a(t) f+(x, -t) (cf. (§VI-6) ). Therefore, using the relation

we can extend f+ for all ( as a meromorphic function in (. Furthermore, if irlj (y =

1, 2,... , N) are zeros of a(() in 3'(() > 0, then -inj (j = 1, 2,... , N) are simple poles of f+(x, () in ( and

Observation VI-7-2. Set gJ (x) = e-" cj f+(x, iii ). Then, from the fact that e_,<= f+(x, () - I as I(I - +oo, it follows that

Observation VI-7-3. If we set F(x,() = e-'t= f+(x, (), then -F" - 2i(F'+uF =

Observation VI-7-4. Let us solve (VI.7.2) for the g,(x). First, set

Write the coefficient matrix of (VI.7.4) in the form IN + C(x), where IN is the

N x N identity matrix. Since

and 77, (j = 1, 2,... , N) are distinct, the matrix IN + C(x) is invertible for -oo < x < +oo. Set L(x) = det(IN + C(x)). Then, manipulating with Cramer's rule, we

and

Thus finally, from (VI.7.3), it follows that

or

This implies that u(x) is a rational function of exponential functions and satisfies

Assumption VI-6-1 of §VI-6. To see this, the following remarks are also useful:

(a) we have the identity

N N

(b) g j (-oo) = = limo g . (x) ( j = 1, 2, ... , N) exist and

from (VI.7.2). Multiplying both sides by ge, adding them up over f, and interchanging the orders of summation, we obtain (a). To show (b), calculate the inverse of the coefficient matrix of (VI.7.2). Using Cramer's rule on (VI.7.4), it can be shown that he(x) -, 0 exponentially as x -+ oo. Also, (b) implies that e2°,, xg,(x)2 is expo- nentially small as x -oo. Therefore, (a) implies that u(x) satisfies Assumption

case when N = 1, n1 = 2, and c1 = 5. On the other hand, a\straightforward calculation using formula (VI.7.7) yields u(x) _ (5+4)2. Also, u(s) = -2sech(x)

is the reflectionless potential corresponding to the data r1= 1 and c = 2.

Example VI-7-6. If N = 2, ill = 2, rh = 3, c1 = 5, and c2 = 2, we obtain

40e42(16 + 135e2-- + 600e6x + 2160e10s + 3600e12=)

U(X)

(1 + 20e4x + 75e6. + 60ebox)2

by calculating (VI.7.7).