The smallest radius of a ball containing the support of a compactly supported potential

1. Introduction

Let \(D\subset {\mathbb R}^3\) be a bounded domain. \(q\in C(D)\) be a real-valued compactly supported potential, \(S\) is a smooth boundary of \(D\), \(A(\beta, \alpha,k)\) be its scattering amplitude, \(k>0\) be fixed, we assume \(k=1\) without loss of generality, \(\beta\) be the unit vector in the direction of scattered field, \(\alpha\) be the unit vector in the direction of the incident field. Assume that \(D\subset Q_a:=\{x: |x|\le a\}\), \(a>0\) is the minimal number such that \(q(x)=0\) for \(|x|>a\), \(\gamma\) is the unit vector such that \(a\gamma\in S\), and for any small \(\epsilon>0\) one has \(\int_{D_\eta}|q(x)|^2dx>0\), where \(D_\eta\) is the part of \(D\) in a neighborhood of the point \(a\gamma\) in the region \(a-\epsilon<|x|<a\). This makes \(c_2>0\) in formula (6) below for an arbitrary small \(\eta>0\).

The known formula for the scattering amplitude is (see, for example, [1, 2]): \[\label{e1} A(\beta, \alpha, k)=-\frac 1 {4\pi}\int_D e^{-ik\beta \cdot x} q(x)u(x,\alpha,k) dx, \tag{1}\] where \(u(x,\alpha,k)\) is the scattering solution (see, for example, [1], pp.256–259, [2], pp.359–439). Since \(k>0\) is fixed, we omit the dependence on \(k\) in what follows. The scattering amplitude corresponding to a compactly supported potential \(q\in L^2(D)\) is an analytic function of \(\beta\) and \(\alpha\) on the complex variety \(M:=\{z\in C^3, z\cdot z=1\}, z\cdot z:=\sum_{j=1}^3 z_j^2.\), see [1].

Our goal is to derive a formula for the \(a\) in terms of the scattering amplitude.

Let us formulate some facts. A proof of the first fact can be found in [1].; the second fact is well known.

A proof of the basic result of this research note, Theorem 1, is given in §2.
Fact 1. There exists \(\nu(\alpha, \theta)\in L^2(S^2)\) such that \[\label{e2} -4\pi \int_{S^2}A(\beta, \alpha)\nu(\alpha, \theta)d\alpha=\int_D e^{-i(\beta-\theta) \cdot x} q(x) dx (1+O(\frac 1 {|\theta|})), \hspace{2cm} |\theta|\to \infty, \tag{2}\] where \[\theta\in M:=\{z\in C^3, z\cdot z=1\}, \hspace{5mm} z\cdot z= \sum_{j=1}^3 z_j^2.\]

Estimate (2) is proved in [1], pp. 260–261. A numerical procedure for calculating \(\nu(\alpha,\theta)\) is given in [1], pp. 265-266.

If \(u,v\) are real-valued vectors in \({\mathbb R}^3\), and \(z=u+iv\), then \(z\cdot z=1\) if and only if \[(u,u)-(v,v)=1, \hspace{2cm} (u,v)+(v,u)=0,\] where \((u,v):=\sum_{j=1}^3 u_j v_j\), \(|v|^2=(v,v)\).

It follows from (2) that if \(\beta-\theta=\xi\), \(\beta, \theta \in M\), and \(|\theta|\to \infty\), then \[\label{e3} \lim_{|\theta|\to \infty}[-4\pi \int_{S^2}A(\beta, \alpha)\nu(\alpha, \theta)d\alpha]=Q(\xi), \tag{3}\] where \(Q(\xi):=\int_D q(x)e^{-i\xi \cdot x}dx\) is the Fourier transform of \(q\).
Fact 2. If \(q\in L^2(D)\) is compactly supported, then its Fourier transform is an entire function of exponential type.

In §2 we prove that the the smallest number \(a>0\) such that the ball \(B_a\) contains the support of \(q\) can be calculated by the formula: \[\label{e4} a=\overline{\lim}_{|v|\to \infty}\frac{||Q(u+iv)||}{|v|}. \tag{4}\]

Here \(||Q(u)||^2=(2\pi )^3\int_{{\mathbb R}^3} |q(x)|^2 dx\), \(\|Q(u+iv)\|^2=(2\pi )^3\int_{{\mathbb R}^3} |q(x)|^2 e^{2v\cdot x}dx\).

In §2 a proof of Theorem 1 is given.

2. Proof of Theorem 1

By Plansherel’s formula one gets \[\label{e5} ||Q(u+iv)||^2= (2\pi )^3\int_{D} e^{2v\cdot x}|q(x)|^2 dx\le c^2 e^{2|v|a}, \hspace {1cm} c^2:=(2\pi )^3\int_D |q(x)|^2 dx. \tag{5}\]

Let us prove an estimate from below: \[\label{e6} ||Q(u+iv)||^2\ge c^2_2\eta e^{2|v|(a-\eta)}, \hspace {1cm} c^2_2=\int_{D_\eta} |q|^2dx, \tag{6}\] where \(\eta>0\) is arbitrarily small, and \(D_\eta\) is sufficiently small, so that \(\cos (x, \gamma)>1-\eta_1\), where \(\eta_1\) is sufficiently small, \(\gamma\) is a unit vector such that \(a\gamma\in \partial {S}\), \(S\) is the boundary of \(D\). Thus, \(v\cdot x>|v||x|(1-\eta)\). From (5)–(6) one gets: \[\label{e7} c_2e^{|v|a(1-\eta)}\le ||Q(u+iv)||\le c e^{|v|a}. \tag{7}\]

Taking natural logarithm of (7), dividing by \(|v|\) and taking \(|v|\to \infty\) yields formula (4).

Theorem 1 is proved. \(\Box\)

The main novelty in this short note is formula (4). This formula differs from the known result: the Paley-Wiener theorem ([3], p.181).

How can one choose the function \(\nu(\alpha, \theta)\) in (3)?

Let us give a method for doing this. This method is outlined on p. 266 in [1].

For \(|x|>a\) the scattering solution \[u(x,\alpha)=e^{i\alpha \cdot x}+\sum_{\ell=0}^\infty A_\ell(\alpha)Y_\ell(x_0)h_\ell(r),\] where \(x_0:=\frac x{|x|}\), \(r:=|x|\), \(h_\ell(r)\) are spherical Hankel functions (see [1], p.262), \[A_\ell(\alpha)=\int_{S^2}A(\beta, \alpha)\overline{Y_\ell(\beta)}d\beta,\] and \(Y_\ell(\beta)\) are the normalized sperical harmonics (see [1], pp261–263).

It is proved in [1], p. 265-266, that \(||\rho||_{|x|\le a}\le c|\theta|^{-1}, \hspace{0.25cm} |\theta|\to \infty\). Therefore, \(\nu(\alpha, \theta)\) can be chosen as the solution to the minimization problem: \(||\rho||_{a\le|x|\le b}=min\), where minimization is taken over \(\nu(\alpha, \theta)\in L^2(S^2)\), and \(0<a<b\) are two arbitrary numbers. Practically \(b-a\) can be chosen of the order of \(1\).

The reader may generalize Theorem 1 by assuming that \(q\in L^2(D)\) rather than \(q\in C(D)\).

3. Conclusion

A formula is given for the radius of the smallest ball containing the support of a compactly supported potential \(q(x)\). It is assumed that \(q\in L^2(D)\) is a real-valued potential, and \(D\subset {\mathbb R}^3\) is a bounded domain.