L. P. Nizhnik

THE SCATTERING PROBLEM UNDER A NONSTATIONARY PERTURBATION

(Presented by Academician S. L. Sobolev, December 29, 1959)

Mathematics

The problem of scattering of plane waves for the equation

\[ \square u(x,t)+c(x,t)u(x,t)=0, \tag{1} \]

where \(x\) is a point of three-dimensional Euclidean space \(E^3\); \(\square=\Delta-\partial^2/\partial t^2\), may be posed as follows: it is required to find a solution of equation (1) of the form
\[ u=u(x,t;\omega,\mu)=e^{i\omega(\mu\cdot x-t)}+v(x,t;\omega,\mu), \]
where \(\mu\) is the unit vector of the direction of the plane wave \(e^{i\omega(\mu\cdot x-t)}\), and \(v\), as \(|x|\to\infty\), satisfies the Fock radiation conditions \(\left({}^{1}\right.\), p. 435; \(\left.{}^{2}\right)\)

\[ v=O\!\left(\frac1{|x|}\right),\qquad \frac{\partial v}{\partial t}=O\!\left(\frac1{|x|}\right),\qquad |\operatorname{grad}v|=O\!\left(\frac1{|x|}\right),\qquad \frac{\partial v}{\partial |x|}+\frac{\partial v}{\partial t} =O\!\left(\frac1{|x|}\right). \tag{2} \]

Here we shall assume that conditions (2) are fulfilled uniformly for \(t\in(-\infty,\infty)^*\).

If \(c(x,t)\) does not depend on \(t\), then equation (1), by separation of variables, reduces to the stationary Schrödinger equation \(-\Delta u+cu=k^2u\), and the Fock radiation conditions in this case become the well-known Sommerfeld radiation conditions. The scattering problem for the Schrödinger equation was solved in \(\left({}^{3}\right)\); the inverse problem was studied in \(\left({}^{4-6}\right)\).

In the present note it will be shown that, under certain conditions of smoothness and decay at infinity imposed on the function \(c(x,t)\), there exists a unique solution of the scattering problem for equation (1). In addition, the question of reconstructing the function \(c(x,t)\) from scattering data is considered.

The results presented can be carried over to the equation \([\square-m^2+c(x,t)]u=0\). In this case the extraction of outgoing waves by means of the Fock radiation conditions must be replaced by the limiting absorption principle \(\left({}^{7}\right.\), p. 499). In this case the scattering of plane waves can be interpreted as the scattering of mesons by a nonstationary force center.

\(1^\circ\). Everywhere below we shall assume that \(c(x,t)\), together with all partial derivatives up to and including the second order, are continuous functions which are majorized by the function
\[ \frac{K(t)}{1+|x|^{3+\varepsilon}}, \]
where \(\varepsilon>0\), and the function \(K(t)\) is uniformly bounded for \(t\in(-\infty,\infty)\).

By \(C(\Omega)\) we shall denote the Banach space of continuous functions on \(\Omega\) with the uniform norm. \(\square_r^{-1}\) will denote the inverse of the operator \(\square\) by means of the retarded Green function:
\[ \square_r^{-1}f(x,t) = -\frac1{4\pi}\int \frac{f(s,t-|x-s|)}{|x-s|}\,ds. \]

\[ {}^* \text{ One could have assumed that conditions (2) are fulfilled for } t=t_0-|x| \text{ uniformly in } t_0 \text{ from an arbitrary finite interval.} \]

If there exists a solution of the scattering problem \(u=e^{i\omega(\mu\cdot x-t)}+v\) for equation (1), then
\(\Box v+cv=-ce^{i\omega(\mu\cdot x-t)}\). Since \(v\) satisfies the radiation conditions, it follows that

\[ v=-\Box_r^{-1}ce^{i\omega(\mu\cdot x-t)}-\Box_r^{-1}cv . \tag{3} \]

Conversely, every sufficiently smooth solution of equation (3), together with the plane wave \(e^{i\omega(\mu\cdot x-t)}\), gives a solution of the scattering problem. For brevity, denote the operator

\[ -\Box_r^{-1}cv=Av . \]

Lemma 1. If there exists a unique solution of the equation \(w=h+Aw\) in \(C(E^4)\) for every right-hand side \(h\), then there exists a unique solution of the scattering problem.

Proof. The uniqueness of the solution of the scattering problem follows from the uniqueness of the solution of the equation \(w=Aw\) in \(C(E^4)\). Suppose there exists a solution of the equation \(w=h+Aw\) in \(C(E^4)\) for every right-hand side \(h\). This means that the operator \((I-A^*)^{-1}\) is bounded. Consider the operator \(D_m=-d^2/dt^2+m^2\). This operator has in \(C(E^4)\) a bounded inverse \(D_m^{-1}\). The operator
\[ D_mAD_m^{-1}=A-2\Box_r^{-1}c_t\frac{\partial}{\partial t}D_m^{-1}-\Box_r^{-1}c_{tt}D_m^{-1} =A+B_m, \]
where \(B_m\) is a bounded operator which, for large \(m\), can be made arbitrarily small in norm. Therefore, for large \(m\) there exists a bounded operator
\[ (I-A^*-B_m^*)^{-1}=[I-(D_mAD_m^{-1})^*]^{-1}, \]
i.e., the equation \(\varphi=f+D_mAD_m^{-1}\varphi\) has a solution in \(C(E^4)\) for every right-hand side \(f\). If we put \(D_m^{-1}\varphi=w\), then \(w=D_m^{-1}f+Aw\). Thus, for sufficiently smooth free terms \(h\) there exist smooth solutions of the equation \(w=h+Aw\). If we choose \(h=Ae^{i\omega(\mu\cdot x-t)}\), then we obtain that the solution of equation (3) exists and is a sufficiently smooth function, i.e., there exists a solution of the scattering problem.

Theorem 1. For small \(c(x,t)\) \((K(t)<\varepsilon/3)\) there exists a unique solution of the scattering problem for equation (1).

The proof follows from Lemma 1 and from the fact that, for small \(c(x,t)\), \(\|A\|<1\).

2°. In this subsection we shall assume that
\[ K(t)\leq \frac{C}{1+|t|^{1+\delta}},\qquad \delta>0. \]
By \(L_2(p(x))\) we shall denote the Hilbert space with norm
\[ \|f(x,t)\|^2=\int |f(x,t)|^2p(x)\,dx\,dt . \]

Lemma 2. If, for some \(a>0\), there exists a unique solution of the equation \(w=h+Aw\) in \(L_2((1+|x|)^{-1-a})\) for every right-hand side \(h\), then there exists a unique solution of the scattering problem for equation (1).

The proof of this lemma is completely analogous to the proof of Lemma 1.

Lemma 3. For \(\varepsilon>1/2+\frac12 a\), the operator \(A\) is completely continuous in \(L_2((1+|x|)^{-1-a})\).

Indeed, consider the operator
\[ A_\xi f(x,t)=\frac{1}{4\pi}\int \frac{e^{-\xi|x-s|}}{|x-s|} c(s,t-|x-s|)f(s,t-|x-s|)\,ds \]
for \(\xi\geq 0\), \(A_0=A\). It is easy to prove that the operator \(A_\xi\) for \(\xi\geq0\) is bounded in \(L_2((1+|x|)^{-1-a})\), and
\[ \|A_\xi-A_0\|\to0 \]
as \(\xi\to0\). The operator \(A_\xi\) for \(\xi>0\) is completely continuous in \(L_2((1+|x|)^{-1-a})\). For this it suffices to show that the operator
\[ A_\xi' f= \frac{1}{4\pi}\int \frac{e^{-\xi|x-s|}}{|x-s|} c(s,t-|x-s|)(1+|s|)^{1/2+1/2a}f(s,t-|x-s|)\,ds \]
is completely continuous in \(L_2(E^4)\). But the unitarily equivalent operator \(T_\xi=UA_\xi' U^{-1}\), where \(U\) is the Fourier transform, has the form \((T\varphi)(p)=\)

\[ =(T_{\xi}\varphi)(p,p_0)=\frac{1}{p^2-(p_0-i\xi)^2}\int b(p-q)\varphi(q)\,d_4q \]
(\(b(p)\) is the Fourier transform of \(c(x,t)\)) \((1+|x|)^{1/2+1/2a}\), and the complete continuity of the operator \(T_\xi\) is verified analogously to Lemma 2 of [8]. Consequently, \(A_\xi'\) is completely continuous, and hence so is the operator \(A_\xi\) for \(\xi>0\). Since \(\|A_\xi-A_0\|\to0\) as \(\xi\to0\), the operator \(A_0=A\) is also completely continuous.

Lemma 4. If \(\varepsilon>1/2+1/2a\), then the equation \(w=Aw\) in \(L_2((1+|x|)^{-1-a})\) has only the trivial solution \(w=0\).

Indeed, suppose there is a nontrivial solution of the equation \(w=Aw\). Denote it by \(w_0\). Consider the operator \(A\) on functions from the half-space \(t\le T\). For sufficiently negative \(T\), \(\|A\|<1\), and therefore \(w_0=0\) for such \(t\). Let \(T_0\) be the largest number for which \(w_0=0\) for \(t\le T_0\). Estimating the equality \(w_0=Aw_0\), we obtain
\[ \int_{t\le T_0+\Delta}|w_0(x,t)|^2(1+|x|)^{-1-a}\,dx\,dt\le \]
\[ \le C_1\Delta^2\int_{t\le T_0+\Delta}|w_0(x,t)|^2(1+|x|)^{-1-a}\,dx\,dt. \]

If \(\Delta\) is a sufficiently small number, then \(w_0=0\) for \(t\le T_0+\Delta\). This contradicts the choice of \(T_0\).

From the general theory of equations in Banach spaces and Lemmas 2, 3, and 4 it follows:

Theorem 2. For \(\varepsilon>1/2\) there exists a unique solution of the scattering problem for equation (1).

3°. It is easy to prove that the solution of the scattering problem has the following asymptotic form as \(|x|\to\infty\):
\[ u(x,t;\omega,\mu)=e^{i\omega(\mu\cdot x-t)}+ \frac{\rho(t-|x|,\nu;\omega,\mu)}{|x|}+o\left(\frac{1}{|x|}\right), \]
where \(\nu\) is the unit vector in the direction of the vector \(x\); \(\rho(\tau,\nu;\omega,\mu)\) is a uniformly bounded function of its arguments.

Passing to the Fourier transform of the function \(\rho\) with respect to the first argument, we obtain
\[ u(x,t;\omega,\mu)=e^{i\omega(\mu\cdot x-t)}+ \int \frac{e^{i\gamma(|x|-t)}}{|x|}F(\gamma,\nu;\omega,\mu)\,d\gamma+ o\left(\frac{1}{|x|}\right), \]
where \(F(\gamma,\nu;\omega,\mu)\), as the Fourier transform of a bounded function, is a generalized function of measure type, and therefore
\[ \int \frac{e^{i\gamma(|x|-t)}}{|x|}F(\gamma,\nu;\omega,\mu)\,d\gamma \]
represents a sum of outgoing spherical waves of various frequencies. If the function \(c(x,t)\) satisfies the conditions of item 2°, then this sum will be continuous. In the stationary case, when \(F(\gamma,\nu;\omega,\mu)=\delta(\gamma-\omega)f(\nu,\omega,\mu)\), the integral under consideration reduces to a single term
\[ \frac{e^{i\omega(|x|-t)}}{|x|}f(\nu,\omega,\mu), \]
where \(f(\nu,\omega,\mu)\) is the usual scattering amplitude. The function \(F(\gamma,\nu;\omega,\mu)\), by analogy with the stationary case, will be called the scattering amplitude.

Using the method of [4,6], let us consider the question of reconstructing the function \(c(x,t)\) from the scattering amplitude.

Theorem 3. From the scattering amplitude \(F(\gamma,\nu;\omega,\mu)\) one uniquely reconstructs the Fourier transform of the function \(c(x,t)\) outside the light cone \(p^2-p_0^2\ge0\). Namely \((\bar p^2-p_0^2\ge0)\),
\[ \tilde c(\bar p,p_0)=\frac{1}{4\pi^2}\int c(x,t)e^{i(\bar p\cdot x+p_0t)}\,dx\,dt = \frac{1}{\pi}\lim_{\substack{\omega\to\infty\\ \omega\mu-(\omega+p_0)\nu=\bar p}} F(\omega+p_0,\nu;\omega,\mu). \]

Proof. On the basis of Lemma 1 \({}^{9}\) (or Lemma 1.3 \({}^{5}\)) one can show that, for large \(\omega\),
\(v=Ae^{i\omega(\mu\cdot x-t)}+O\left(\frac{1}{\omega}\right)\). Consequently,

\[ \begin{aligned} F(\gamma,\nu;\omega,\mu) &=\frac{1}{4\pi}\int e^{-i\gamma(\nu\cdot x-t)}C(x,t)e^{i\omega(\mu\cdot x-t)}\,dx\,dt +O\left(\frac{1}{\omega}\right) \\ &=\frac{1}{4\pi}\int C(x,t)e^{i(\bar p\cdot x+p_0t)}\,dx\,dt +O\left(\frac{1}{\omega}\right), \end{aligned} \]

where \(p_0=\gamma-\omega\), \(\bar p=\omega\mu-\gamma\nu\). Passing to the limit as \(\omega\to\infty\), we obtain the required reconstruction formula.

In conclusion, the author expresses his deep gratitude to Yu. M. Berezanskii, under whose supervision this work was carried out.

Institute of Mathematics
Academy of Sciences of the Ukrainian SSR

Received
25 XII 1959

References

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6. L. D. Faddeev, Vestnik Leningrad. Univ., No. 7 (1956).
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