Phaseless inverse uniqueness of a three-dimensional scattering problem of second type

Abstract

In this paper we discuss the phaseless inverse scattering problem in mathematical physics. We measure only the intensity of scattered wave field in far field without phase information. The modulus of the scattered wave field is an analytic function in complex plane. As the parameter of certain analytic function, the traveling time of the scattered wave field is the spectral invariant that controls the behavior of the complex-valued function. Given two sets of identical point-to-point traveling times, we compare the asymptotic behaviors of scattered wave fields in complex plane. Then, we can deduce an inverse uniqueness on the index of refraction from the inverse Radon transform in each 2-dimensional cross section.

Appendix: The Nevanlinna–Levin theorem

In the real-world applications, the scattered data are measured in real-valued frequencies/wave numbers. The assumption we have adopted in Theorem 1.1 is focused on the neighborhoods centered at real axis. In [2, Appendix], we used complex analysis to relax the assumption to be $f^1(x,k,y)\equiv f^2(x,k,y)$ for all $(x,k,y)\in S\times I\times S$ in a neighborhood $I\subset 0i+\mathbb {R}\subset \mathbb {C}$, and then manage to prove $n^{1}\equiv n^{2}$. The Fourier transforms (2.5) and (2.6) behave like the exponential functions in complex analysis on many aspects. The Nevanlinna-Levin representation theorem plays a role.

Definition 5.1

Let f(z) be an entire function. Let $M_f(r):=\max _{|z|=r}|f(z)|$. An entire function of f(z) is said to be a function of finite order if there exists a positive constant k such that the inequality

$$\begin{aligned} M_f(r)<e^{r^k} \end{aligned}$$

is valid for all sufficiently large values of r. The greatest lower bound of such numbers k is called the order of the entire function f(z). By the type $\sigma$ of an entire function f(z) of order $\rho$, we mean the greatest lower bound of positive number A for which asymptotically we have

$$\begin{aligned} M_f(r)<e^{Ar^\rho }. \end{aligned}$$

That is,

$$\begin{aligned} \sigma :=\limsup _{r\rightarrow \infty }\frac{\ln M_f(r)}{r^\rho }. \end{aligned}$$

If $0<\sigma <\infty$, then we say f(z) is of normal type or mean type.

Definition 5.2

Let f(z) be an integral function of finite order $\rho$ in the angle $[\theta _1,\theta _2]$. We call the following quantity as the indicator function of the function f(z).

$$\begin{aligned} h_f(\theta ):=\lim _{r\rightarrow \infty }\frac{\ln |f(re^{i\theta })|}{r^{\rho }}, \,\theta _1\le \theta \le \theta _2. \end{aligned}$$

The type of a function is connected the maximal value of the indicator function.

Lemma 5.3

(Levin [10], p. 72) The maximum value of the indicator $h_f(\theta )$ of the function f(z) on the interval $\alpha \le \theta \le \beta$ is equal to the type $\sigma$ of this function inside the angle $\alpha \le \arg z\le \beta$.

Lemma 5.4

The analytic function $u^{j}(x,k,y)$ in ${\mathfrak{I}}k<C$, $j=1,2,$ is bounded for $|x|\gg 0$ in the lower half complex plane.

Proof

We recall the (2.7), that is,

$$\begin{aligned} u(x,k,y)=\exp \{-ik\tau (x,y)\}\left\{ A(x,y)+O\left( \frac{1}{k}\right) \right\} ,\,x\ne y \text{ as } k\rightarrow \infty , \end{aligned}$$

which is bounded for negative ${\mathfrak{I}}k$. $\square$

Now we introduce the integral representation theorem due to Nevanlinna and Levin [1, 10, 11].

Definition 5.5

Let f(z) be an analytic function in the upper half-plane. We say u(z) is a harmonic majorant of $\log |f(z)|$ if $\log |f(z)|\le u(z)$, and u(z) is a harmonic function in the upper half-plane.

From Lemma 5.4, the function f(z) is bounded in the lower half-plane. Thus, $\log |f(z)|$ is bounded by a constant which is trivially harmonic.

Theorem 5.6

Let f(z) be an analytic function in $\{{\mathfrak{I}}z>0\}$, and let the function $\log |f(z)|$ have a positive harmonic majorant in $\{{\mathfrak{I}}z>0\}$. Then

$$\begin{aligned} \log |f(z)|=\sum _{k=1}^{\infty }\log \left| \frac{z-a_{k}}{z-\overline{a}_{k}}\right| +\frac{y}{\pi }\int _{-\infty }^{\infty }\frac{\log |f(t)|}{|t-z|^{2}}dt+\sigma y, \end{aligned}$$

(5.1)

where $\{a_{k}\}$ are the zeros of f(z) in $\{{\mathfrak{I}}z>0\}$.

Proof

We refer the proof to [11, p. 104, p. 105] and [10]. We also refer more connection of the integral representation theorem between the other analytic properties of the function to Levin’s book [11, p. 115, p. 116]. $\square$

Surely, $u^j(x,k,y)$ may have finitely many poles in ${\mathfrak{I}}k\le 0$, we apply the theorem by adding a multiple of $P^{j}(k):=\prod _{n}(k-\kappa _{n}^{j})$, $j=1,2$, where $\{\kappa _{n}^{j}\}$ are possible poles.

Theorem 5.7

If $f^1(x,k,y)= f^2(x,k,y)$ for k in a neighborhood $I\subset \mathbb {R}+0i$ with (x, y) fixed, then $|u^{1}(x,k,y)|=|u^{2}(x,k,y)|$ holds in $\mathbb {C}^{-}$.

Proof

Given $|u^1(x,k,y)|^{2}= |u^2(x,k,y)|^{2}$ for all $k\in I$, the identity for the real-analytic functions extends to hold in real axis. Hence, we have

$$\begin{aligned} |u^1(x,k,y)|= |u^2(x,k,y)|,\,k\in \mathbb {R}+0i. \end{aligned}$$

(5.2)

For fixed (x, y), the type of $u^{j}(x,k,y)P^{j}(k)$, $j=1,2,$ $\sigma =\tau ^{1}(x,y)= \tau ^{2}(x,y)$ in ${\mathfrak{I}}k\le 0$ by following Definitions 5.1, 5.2, Lemmas 5.3, and 5.4. We apply Theorem 5.6 in ${\mathfrak{I}}k\le 0$ to deduce

$$\begin{aligned}&\log |u^1(x,k,y)P^{1}(k)|-\log |u^2(x,k,y)P^{2}(k)|=\sum _{n=1}^{m}\log \left| \frac{k-a_{n}}{k-\overline{a}_{n}}\right| \\&\quad -\sum _{n=1}^{m'}\log \left| \frac{k-a_{n}'}{k-\overline{a}_{n}'}\right| ,\,{\mathfrak{I}}k\le 0, \end{aligned}$$

where $\{a_{n}\}$ and $\{a_{n}'\}$ are the zeros of $u^1(x,k,y)P^{1}(k)$ and $u^2(x,k,y)P^{2}(k)$ respectively in the lower half-plane. That is,

$$\begin{aligned} \frac{|u^1(x,k,y)P^{1}(k)|}{|u^2(x,k,y)P^{2}(k)|}=\prod _{n=1}^{m}\left| \frac{k-a_{n}}{k-\overline{a}_{n}}\right| /\prod _{n=1}^{m'}\left| \frac{k-a'_{n}}{k-\overline{a}'_{n}}\right| ,\,{\mathfrak{I}}k\le 0, \end{aligned}$$

(5.3)

from which we deduce that $\frac{u^1(x,k,y)P^{1}(k)}{u^2(x,k,y)P^{2}(k)}$ is mostly a rational function in k.

Moreover, the $|u^j(x,k,y)|^{2}$, $j=1,2$, $k\in 0i+\mathbb {R}$, extends analytically to be $u^j(x,k,y)\overline{u^j(x,\overline{k},y)}$, in $\mathbb {C}$ outside the possible poles as aforementioned in Introduction. We deduce from Theorem 1.1 assumption and the Identity Theorem in complex analysis [8] that

$$\begin{aligned} u^1(x,k,y)\overline{u^1(x,\overline{k},y)}\equiv u^2(x,k,y)\overline{u^2(x,\overline{k},y)}. \end{aligned}$$

(5.4)

Comparing the location of the zeros and poles of the identity on both sides of (5.4) in $\{{\mathfrak{I}}k\le 0\}$, we deduce from (5.3) that $\{a_{n}\}=\{a_{n}'\}$, $\{\overline{a}_{n}\}=\{\overline{a}_{n}'\}$, and then

$$\begin{aligned} \frac{|u^1(x,k,y)|}{|u^2(x,k,y)|}\equiv 1,\,{\mathfrak{I}}k\le 0. \end{aligned}$$

The theorem is thus proven. $\square$