Reports of the Academy of Sciences of the USSR
1964. Vol. 157, No. 5

MATHEMATICAL PHYSICS

A. G. RAMM

ON CONDITIONS FOR THE ANALYTICITY OF THE SCATTERING MATRIX

(Presented by Academician V. A. Fock on 21 III 1964)

In this paper it is proved that the Jost function \(f(k)\) is an entire function of the complex variable \(k\) of growth order not exceeding one and such that \(f(k)-1\) is quadratically integrable in \(k\) if and only if the potential is finite. Thus it is clarified that the class of potentials considered by Regge \((^{1})\) is the maximal admissible one for the function \(f(k)\) to be entire of growth order not exceeding one.

Let us consider the Schrödinger equation

\[ y''+[k^2-p(x)]y=0 \tag{1} \]

under the assumption on the potential

\[ \int_0^\infty x|p(x)|\,dx<\infty . \tag{2} \]

Consider the solution of equation (1) defined by the following asymptotics for large \(x\):

\[ \lim_{x\to\infty} e^{-ikx}f(x,k)=1 . \tag{3} \]

The function \(f(k)=f(0,k)\) is called the Jost function. The basic properties of solutions of equation (1) are collected in the book \((^{2})\). The analytic properties of the function \(f(k)\), under the assumption that the potential \(p(x)\) is finite, were established in \((^{1,3})\).

Our task is to invert some of the results obtained by Regge. In this direction we shall prove the theorem:

Theorem 1. Let \(f(k)\) be an entire function of growth order not exceeding one, and let \(f(k)-1\) be quadratically integrable. Suppose that \(f(k)\) is the Jost function of some scattering problem of the type (1)—(3), \(M_n\) being normalization constants.*
Then the potential \(p(x)\) is finite.

Proof. We shall base the proof on the Wiener—Paley theorem \((^{4})\) and on the properties of transformation operators \((^{2})\). It is shown in \((^{2})\) that

\[ f(x,k)=e^{ikx}+\int_x^\infty A(x,y)e^{iky}\,dy . \tag{4} \]

From (4) we obtain for \(f(k)\) the formula

\[ f(k)=1+\int_0^\infty A(0,y)e^{iky}\,dy . \tag{5} \]

* Specially chosen; see the note to formula (13).

The function \(A(x,y)\) satisfies the equation of V. A. Marchenko \((^{5})\):

\[ A(x,y)=F(x+y)+\int_x^\infty A(x,t)F(t+y)\,dt,\qquad x\leq y, \tag{6} \]

where

\[ F(t)=\frac{1}{2\pi}\int_{-\infty}^{\infty}[-S(k)+1]e^{ikt}\,dk+\sum_{n=1}^{m} M_n^2 e^{-k_n t}, \tag{7} \]

\[ S(k)=\frac{f(-k)}{f(k)}; \tag{8} \]

the points \(k=ik_n,\ k_n>0\), are poles of the function \(f(k)\).

For us the essential relation between the potential \(p(x)\) and the function \(A(x,y)\) is:

\[ p(x)=-2\frac{d}{dx}A(x,x). \tag{9} \]

All these formulas are contained in the book \((^{2})\).

From (6), for \(x=0\), we obtain the equation

\[ A(y)=F(y)+\int_0^\infty A(t)F(t+y)\,dt,\qquad y\geq 0, \tag{10} \]

where

\[ A(y)=A(0,y). \tag{11} \]

By virtue of the Wiener—Paley theorem, from the conditions imposed on \(f(k)\) and from equation (5) it follows that \(A(y)\) is a finite function, square-integrable.

Transform (10) to the form

\[ A(y)=F(y)+\int_y^\infty A(v-y)F(v)\,dv,\qquad y\geq 0. \tag{12} \]

We shall show that from equation (12) it follows that \(F(y)\equiv 0\) for sufficiently large \(y\).

Equation (12) is a Volterra equation with a difference kernel. It can be shown (see the Appendix) that there exists a solution of equation (12) having the form*:

\[ F(y)=A(y)+\int_y^\infty \Gamma(v-y)A(v)\,dv. \tag{13} \]

Since the function \(A(y)\) is finite, \(A(y)=0\) for \(y>y_0\). From (13) it follows that for \(y>y_0\), \(F(y)=0\).

To prove the finiteness of the potential we use formula (9) and the following equation, which follows from (6):

\[ A(x,x)=F(2x)+\int_x^\infty A(x,t)E(t+x)\,dt. \tag{14} \]

From equation (14) we conclude that for \(2x>y_0\), \(A(x,x)=0\). Hence, from (9), we obtain the finiteness of \(p(x)\). Theorem 1 is proved.

Appendix. Here we shall consider the question of the solvability

* With a suitable choice of the constants \(M_n\). If the discrete spectrum is absent, then all \(M_n=0\), and the reasoning can be simplified.

equations of type (12). Let

\[ g(x)=f(x)+\int_x^\infty A(y-x)f(y)\,dy. \tag{15} \]

Let, for some \(c>0\), \(e^{cx}g(x)\subset L_2(0,\infty)\), \(e^{cx}A(x)\subset L(0,\infty)\).

Theorem 2. Under the assumptions made, there exists a unique solution of equation (15) such that \(e^{c'x} f x \subset L_2(0,\infty)\), \(0<c'\). This solution has the form

\[ f(x)=g(x)+\int_x^\infty \Gamma(y-x)g(y)\,dy, \tag{16} \]

where \(\Gamma\) is some function.

Proof. Multiply both sides of equation (15) by \(e^{bx}\), \(b>0\). We obtain the equation

\[ g_b(x)=f_b(x)+\int_x^\infty A_b(y-x)f_b(y)\,dy, \tag{17} \]

where, for example, \(g_b(x)\equiv g(x)e^{bx}\).

Introduce the functions:

\[ G(x)= \begin{cases} g_b(x), & x\ge 0,\\ 0, & x<0; \end{cases} \tag{18a} \]

\[ E(x)= \begin{cases} f_b(x), & x\ge 0,\\ f_b(x), & x<0, \end{cases} \tag{18б} \]

\[ B(x)= \begin{cases} A_b(x), & x\ge 0,\\ 0, & x<0. \end{cases} \tag{18в} \]

Equation (17) can be rewritten in the form

\[ G(x)=F(x)+\int_{-\infty}^{\infty} B(y-x)F(y)\,dy. \tag{19} \]

Take the Fourier transform of (19). We obtain\({}^{6}\)

\[ \widetilde{G}(k)=\widetilde{F}(k)+\widetilde{B}(-k)\widetilde{F}(k). \tag{20} \]

Hence

\[ \widetilde{F}(k)=\frac{\widetilde{G}(k)}{1+\widetilde{B}(-k)} =\widetilde{G}(k)-\frac{\widetilde{B}(-k)\widetilde{G}(k)}{1+\widetilde{B}(-k)}. \tag{21} \]

Choosing \(b>0\) sufficiently large, we can ensure that \(1+\widetilde{B}(-k)\ne0\) for \(-\infty<k<\infty\). Then, by a well-known theorem of Wiener\({}^{5}\),

\[ \frac{1}{1+\widetilde{B}(-k)}=\widetilde{\mathcal{L}}(-k), \]

where \(\widetilde{\mathcal{L}}(-k)\) is the Fourier transform of some function from \(L(0,-\infty)\). Thus,

\[ \widetilde{F}(k)=\widetilde{G}(k)+\widetilde{\Gamma}(-k)\widetilde{G}(k), \tag{22} \]

where \(\widetilde{\Gamma}(-k)=\widetilde{B}(-k)\widetilde{\mathcal{L}}(-k)\) is the Fourier transform of some function from \(L(0,-\infty)\). Inverting (22) and using the convolution theorem, we arrive at the formula:

\[ F(x)=G(x)+\int_{-\infty}^{\infty}\Gamma(y-x)G(y)\,dy. \tag{23} \]

But for \(x<0\), \(\Gamma(x)\equiv 0\). This follows from a theorem of Titchmarsh ((\(^{6}\), 172), if one takes into account that the function

\[ \widetilde{\Gamma}(-k)=\frac{\widetilde{B}(-k)}{1+\widetilde{B}(-k)} \]

is analytic in the half-plane \(\operatorname{Im} k \leqslant 0\), together with the functions \(\widetilde{B}(-k)\) and

\[ \frac{1}{1+\widetilde{B}(-k)}. \]

(Recall that \(\widetilde{B}(-k)\) is an entire function of the complex variable \(k\), since \(f(k)\) is, by assumption, an entire function.) Therefore (23) has the form (16). The theorem is proved.

Leningrad Institute
of Precision Mechanics and Optics

Received
4 III 1964

CITED LITERATURE

\(^{1}\) T. Regge, Sbornik: Mathematics, 7, 4, 83 (1963).
\(^{2}\) Z. S. Agranovich, V. A. Marchenko, The Inverse Problem of Scattering Theory, Kharkov, 1960.
\(^{3}\) T. Regge, Sbornik: Mathematics, 7, 4, 91 (1963).
\(^{4}\) B. Ya. Levin, Distribution of Zeros of Entire Functions, Moscow, 1956.
\(^{5}\) N. Wiener, The Fourier Integral and Certain of Its Applications, Moscow, 1963.
\(^{6}\) E. Titchmarsh, Introduction to the Theory of Fourier Integrals, Moscow, 1948.