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Mathematical Methods for Wave Phenomena (Computer Science and Applied Mathematics)

Norman Bleistein

Chapter 9

INVERSE METHODS FOR REFLECTOR IMAGINGS - all with Video Answers

Educators

Chapter Questions

Problem 1

(a) Let $S$ be a spherical cap of radius $a$ centered at the origin, with polar angle range $0 \leq \theta \leq \theta_0$. Define the normal $\mathbf{n}$ for $S$ to be the unit radial vector at each point $(\cos \phi \sin \theta, \sin \phi \sin \theta, \cos \theta)$. Show that
$$
\bar{\gamma}(\mathbf{k}) \sim(2 \pi i \mu a / k) e^{-i a k a}, \quad \mu=\operatorname{sign} \hat{k}_3,
$$
when either $\pm \hat{\mathbf{k}}$ is in the aperture of the cap.
(b) Find the leading term of the asymptotic expansion of $\bar{\gamma}(\mathbf{k})$ when $\hat{\mathbf{k}}$ is not in the aperture of the cap. Confirm that it is of lower order in $k$.
(c) Let $\theta_0=\pi$ so that $S$ is closed. Let $\Gamma(\mathbf{x})$ be the characteristic function of the interior of the sphere. Find the asymptotic expansion of $\Gamma(\mathbf{k})$ and confirm (9.1.4).

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01:32

Problem 1

Consider the one-dimensional inverse problem
$$
\begin{aligned}
u^{\prime \prime}(z ; \omega)+\frac{\omega^2}{v^2(z)} u(z ; \omega) & =-\delta(z), & & \text { where prime means } \frac{d}{d z}, \\
\frac{1}{v^2(z)} & =\frac{1}{c_0^2}, & & z \leq 0, \\
\frac{1}{v^2(z)} & =\frac{1}{c_0^2}[1+\alpha(z)], & & z>0 .
\end{aligned}
$$
(a) Set
$$
u(z ; \omega)=u_1(z ; \omega)+u_{\mathrm{s}}(z ; \omega)
$$
with
$$
u_1^{\prime \prime}(z ; \omega)+\left(\omega^2 / c_0^2\right) u_1(z ; \omega)=-\delta(z) .
$$
Show that
$$
u_{\mathrm{S}}^{\prime \prime}(z ; \omega)+\left(\omega^2 / c_0^2\right) u_{\mathrm{S}}(z ; \omega)=-\left(\omega^2 / c_0^2\right) x(z)\left[u_1(z ; \omega)+u_{\mathrm{S}}(z ; \omega)\right]
$$
and that the linearized equation for $u_5(0 ; \omega)$ has the solution
$$
u_{\mathrm{s}}(0 ; \omega)=\omega^2 \int_0^{\infty} \frac{\alpha(z)}{c_0^2} u_1^2(z ; \omega) d z .
$$
(b) For $c_0=$ const, show that
$$
u_{\mathrm{s}}(0 ; \omega)=-\frac{1}{4} \int_0^{\infty} \alpha(z) e^{2 i \omega z / \epsilon_0} d z=-\frac{\bar{\alpha}(2 i \omega / c)}{4},
$$
with solution
$$
\alpha(z)=-\frac{4}{\pi c_0} \int u_{\mathrm{s}}(0 ; \omega) e^{-2 i \omega z ; c_0} d \omega=-4 U_{\mathrm{s}}\left(0, \frac{2 z}{c_0}\right),
$$
with $U_{\mathrm{s}}(0, t)$ the inverse transform to the time domain of $u_{\mathrm{s}}(0 ; \omega)$.
(c) Show that for a one-dimensional reflectivity function, defined as in this section,
$$
R \gamma_{\mathrm{B}}(z)=-\frac{2 i}{\pi c_0^2} \int \omega u_{\mathrm{S}}(0 ; \omega) e^{-2 i \omega z ; c_0} d \omega .
$$

Varsha Aggarwal
Varsha Aggarwal
Numerade Educator
01:43

Problem 2

Consider the surface $S$ defined by (9.1.16). For this special parameterization $x_1=\sigma_1$ and $x_2=\sigma_2$, show that (8.4.13) agrees with (9.1.18).

Tanishq Gupta
Tanishq Gupta
Numerade Educator
11:46

Problem 3

(a) Verify $(9.1 .36)$.
(b) Introduce the notation
$$
\cos \alpha=\hat{\mathbf{0}} \cdot \frac{\partial \mathbf{x}}{\partial \sigma_1}, \quad \sin \alpha=\hat{\mathbf{0}} \cdot \frac{\partial \mathbf{x}}{\partial \sigma_2},
$$
and show that
$$
\cos \alpha=\hat{\boldsymbol{\phi}} \cdot \frac{\partial \mathbf{x}}{\partial \sigma_2}, \quad \sin \alpha=-\hat{\boldsymbol{\phi}} \cdot \frac{\partial \mathbf{x}}{\partial \sigma_1} .
$$
(c) Calculate $\operatorname{det}[A-\lambda I]$, where $I$ is the identity matrix. Show that for $\alpha=0$, this quartic in $\lambda$ factors into a pair of quadratics, each of which has two roots of opposite sign for $r_{\mathrm{n}}$ small enough. Thus, conclude that $\operatorname{sgn} A=0$ in this case.
(d) Now verify the first line in $(9.1 .37)$ for any $\alpha$.
(e) Explain why $\operatorname{sgn} A=0$ for any $\alpha$ and $r_{\mathrm{n}}$ small enough.

Chris Trentman
Chris Trentman
Numerade Educator

Problem 4

(a) Let $I(\lambda)$ be a surface integral of the form
$$
I(\lambda)=\int_S f(\mathbf{x}) e^{i \lambda \phi(\mathbf{x})} d S, \quad \mathbf{x}=\left(x_1, x_2, x_3\right) .
$$
We have seen in this section that for $\phi(\mathbf{x})$ to have a stationary point, $\nabla \phi(\mathbf{x})$
must be normal to $S$ at that point. Suppose that $\phi(\mathbf{x})$ has no stationary points on $S$. Denote the boundary curve to $S$ by $C$ and apply Stokes's theorem
$$
\int_S(\boldsymbol{\nabla} \times \mathbf{B}) \cdot \hat{\mathbf{n}} d \boldsymbol{S}=\int_C \mathbf{B} \cdot \mathbf{d x}
$$
to the vector function
$$
\mathbf{B}=\frac{\hat{\mathbf{n}} \times \mathbf{\nabla} \phi(\mathbf{x})}{i \lambda\left[(\mathbf{\nabla} \phi(\mathbf{x}))^2-(\hat{\mathbf{n}} \cdot \mathbf{\nabla} \phi(\mathbf{x}))^2\right]} f(\mathbf{x}) e^{i \lambda \phi(\mathbf{x})}
$$
to conclude that to leading order
$$
I(\lambda) \sim \frac{1}{i \lambda} \int_c \frac{\hat{\mathbf{t}} \cdot(\mathbf{\Lambda} \times \nabla \phi(\mathbf{x}))}{(\mathbf{\nabla} \phi(\mathbf{x})) 2-(\hat{\mathbf{n}} \cdot \mathbf{\nabla} \phi(\mathbf{x}))^2} f(\mathbf{x}) e^{i \lambda \phi(\mathbf{x})} d s .
$$
In this equation, $d s$ is differential arc length around the boundary of the surface $S$, and $\hat{\mathbf{t}}=d \mathbf{x}(s) / d s$ is the unit tangent vector to the boundary curve. Note that the denominators of the integrands here are nonzero under the assumption that $\phi(\mathbf{x})$ has no stationary points on $S$.
(b) Let the boundary curve $C$ be parameterized by an arc-length parameter $s: \mathbf{x}=\mathbf{x}(s)$ on $C$. Show that the stationary points are determined by $\mathbf{\nabla} \phi(\mathbf{x}) \cdot \hat{\mathbf{t}}=0$ and that
$$
I(\lambda) \sim \frac{1}{i \lambda} \sqrt{\frac{2 \pi}{\lambda}} \sum \frac{\hat{\mathbf{t}} \cdot(\mathbf{\mathbf { t }} \times \mathbf{\nabla} \phi(\mathbf{x}))}{\sqrt{|\vec{\nabla} \phi(\mathbf{x}) \cdot \mathbf{k}|}\left[(\mathbf{\nabla} \phi(\mathbf{x}))^2-(\hat{\mathbf{h}} \cdot \mathbf{\nabla} \phi(\mathbf{x}))^2\right]} f(\mathbf{x}) e^{i \lambda \phi(\mathbf{x})+i \boldsymbol{\mu} \pi / 4} .
$$
In this equation, $\boldsymbol{\kappa}$ again represents the curvature vector at the stationary point and $\mu=\operatorname{sign} \nabla \phi(\mathbf{x}) \cdot \mathbf{\kappa}$.
(c) Apply this result to the second integral in (9.1.3) under the assumption of no interior stationary points. Show that the assumption of no stationary points assures that $\hat{\mathbf{k}} \cdot \hat{\mathbf{n}} \neq \pm 1$ and that if all of the boundary stationary points are simple,
$$
\bar{\gamma}(\mathbf{k}) \sim-\frac{1}{i k} \sqrt{\frac{2 \pi}{k}} \sum \frac{\hat{\mathbf{t}} \cdot(\hat{\mathbf{n}} \times \hat{\mathbf{k}})}{\sqrt{|\overline{\mathbf{k}} \cdot \mathbf{k}|}\left[1-(\hat{\mathbf{n}} \cdot \hat{\mathbf{k}})^2\right]} e^{-i \mathbf{k} \cdot \mathbf{x}+i \mu \pi / 4} .
$$

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00:58

Problem 5

Show that the result $(9.2 .9)$ can be rewritten as
$$
R_{\mathrm{n}} \gamma_{\mathbf{8}}(\mathbf{y}) \sim \frac{4}{\pi c^2} \operatorname{Re}\left[\int_{f_0}^{f_1} \text { if } d f \int_{\Omega} d \Omega S(\boldsymbol{\xi} ; 2 \pi f) e^{4 n i f(\mathbf{y} / c}\right] .
$$

Jack Chen
Jack Chen
Numerade Educator
03:12

Problem 6

Verify (9.2.12) and (9.2.13). [Hint: Use the stationary phase result in the first line of $(9.1 .38)$ to calculate the angular integral in $(9.2 .9)$ and then interpret the integral in frequency domain as an Fourier transform back to the "time" domain with time properly interpreted.]

Amit Srivastava
Amit Srivastava
Numerade Educator
03:40

Problem 7

(a) Define $F_\lambda(t)=\lambda^2 t \exp (-\lambda t)$. Use Watson's lemma to show that
$$
\lim _{\lambda \rightarrow \infty} F_\lambda(t)=\delta(t)
$$
That is. for any fixed but large value of $\lambda, F_\lambda(t)$ is an approximate Dirac delta function.
(b) Show that the Fourier transform of $F_\lambda(t)$ is
$$
\bar{F}(2 \pi f)=\lambda^2 \int t e^{-(\lambda-2 \pi i f)} d f=\frac{\lambda^2}{(\lambda-2 \pi i f)^2} .
$$
(c) Suppose that $f_0=0.25 \lambda / 2 \pi$ and $f_1=\lambda / 2 \pi$. Show that the modulus of $\bar{F}(2 \pi f)$ varies from 0.5 to 0.96 over the bandwidth.

Adriano Chikande
Adriano Chikande
Numerade Educator
00:17

Problem 8

Verify $(9.2 .14)$.

Matt Gibson
Matt Gibson
Numerade Educator
08:13

Problem 9

The purpose of this exercise is to derive Bojarski's original POFFIS identity for the characteristic function. Suppose that the scattering obstacle to be imaged is convex and that the reflection coefficient is equal to \pm 1 . Suppose that for each experiment from the direction $\xi$, with scattering amplitude defined in $(8.4 .12)$, the corresponding experiment from the direction $-\xi$ is also carried out.
(a) Explain why this latter experiment produces the output
$$
S(-\hat{\xi} ; \omega) \frac{2 i \omega R}{c}=\int_D \hat{\mathbf{n}} \cdot \boldsymbol{\xi} e^{2 i \omega \xi \cdot \mathbf{x} / c} d S .
$$
That is, the experiment from the opposite direction produces an integral over the part of the scatterer that was dark for the original experiment.
(b) Use the divergence theorem to show that
$$
\begin{aligned}
S(\xi ; \omega)+S^*(-\hat{\xi} ; \omega) & =-\frac{2 i \omega R}{c} \int_{\mathrm{B}} \hat{\mathbf{n}} \cdot \hat{\xi} e^{2 i \omega \xi \cdot \mathbf{x} ; c} d S \\
& =-\frac{4 \omega^2 R}{c^2} \int_{\mathbf{x} \text { in } \mathrm{B}} e^{-i \omega \xi \cdot \mathbf{x}} d V=-\frac{4 \omega^2 R}{c^2} \bar{\Gamma}\left[\frac{2 \omega \xi}{c}\right] .
\end{aligned}
$$

Khoobchandra Agrawal
Khoobchandra Agrawal
Numerade Educator

Problem 11

For the same problem as in Exercise 9.10, introduce the new independent variable $\tau$ defined by
$$
\tau=\int_0^z \frac{d z^{\prime}}{v\left(z^{\prime}\right)}
$$
and set
$$
u(z(\tau) ; \omega)=w(\tau ; \omega), \quad v(z(\tau))=c(\tau) .
$$
(a) Show that the problem for $w(\tau ; \omega)$ is
$$
\begin{gathered}
\ddot{w}(\tau ; \omega)+\omega^2 w(\tau ; \omega)-\Gamma(\tau) \dot{w}(\tau ; \omega)=-v(0) \delta(\tau), \\
\Gamma(\tau)=\frac{\dot{c}(\tau)}{c(\tau)}, \quad \text { where the overdot means } \frac{d}{d \tau} .
\end{gathered}
$$
(b) Set
$$
w(\tau ; \omega)=w_{\mathrm{l}}(\tau ; \omega)+w_{\mathrm{s}}(\tau ; \omega), \quad \bar{w}_{\mathrm{l}}(\tau ; \omega)+\omega^2 w_{\mathrm{l}}(\tau ; \omega)=-v(0) \delta(\tau) .
$$
Show that the linearized equation for $w_{\mathrm{s}}(\tau ; \omega)$ has the solution
$$
w_{\mathrm{s}}(0 ; \omega)=\frac{i v(0)}{4 \omega} \int_0^{\infty} \Gamma(\tau) e^{2 i \omega \tau} d \tau,
$$
with inversion
$$
\Gamma(\tau)=-\frac{4 i}{\pi v(0)} \int_{-\infty}^{\infty} \omega w_{\mathrm{s}}(0 ; \omega) e^{-2 i \omega \tau} d \omega .
$$
The implicit solution for $v(z)$ is then given by
$$
c(\tau)=v(0) e^{\Gamma(t)}, \quad z=\int_0^z c\left(\tau^{\prime}\right) d \tau^{\prime}, \quad v(z(\tau))=c(\tau) .
$$
The solution $\Gamma(\tau)$ has been shown to produce a more accurate reconstruction of the velocity than the solution $\alpha(z)$.

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03:02

Problem 12

In Section 8.4, it was shown that the width of the main lobe of a bandlimited delta function is $t=1 /\left[f_1+f_0\right]$. Suppose two nearby reflectors are such that the zeros of their main lobes just touch. Show that this corresponds to a layer width between the the reflectors of $c /\left[f_1+f_0\right]$ and that for the values $f_0=10 \mathrm{~Hz}, f_1=40 \mathrm{~Hz}$, and $c=3000 \mathrm{~m} / \mathrm{s}$, this yields a layer width of $60 \mathrm{~m}$. This is an example of the constraints on resolution of nearby reflectors.

Neelesh Sharma
Neelesh Sharma
Numerade Educator
00:42

Problem 13

Follow the discussion after $(9.3 .20)$ to carry out the calculations that yield $(9.3 .21)$.

Suman Saurav Thakur
Suman Saurav Thakur
Numerade Educator
06:59

Problem 14

Suppose that the backscattered data from an array of experiments are given by
$$
\begin{aligned}
& U_{\mathrm{s}}(\boldsymbol{\xi}, t)=R \frac{a \delta\left(t-2[\rho-a] / c_0\right)}{8 \pi[\rho-a] \rho}, \\
& \xi=\left(\xi_1, \xi_2, 0\right), \quad \xi^2=\xi_1^2+\xi_2^2, \quad \rho=\sqrt{H^2+\xi^2} . \\
&
\end{aligned}
$$
(a) Use the ray method to show that this is the leading order backscatter response from a spherical domain of radius $a$, centered at depth $H$, with reflection coefficient $R$.
(b) Let $R a$ remain finite while $a \rightarrow 0$. Show that in this limit, the arrivals or events at the upper surface occur at the times $t=2 \rho / c$ at the distance $\rho$ from the origin on the upper surface. Plot the surface of arrivals in a spacetime plot in which the vertical axis is $c t / 2$ and the horizontal axes are the coordinates $\xi_1$ and $\xi_2$. Confirm that the surface is one sheet of a two-sheeted hyperboloid of revolution, which has as asymptote the right circular cone with opening angle whose tangent is $2 \rho / c t$.
(c) Return to nonzero $a$ Substitute the data for $U_{\mathrm{s}}(\xi, t)$ into the right side of (9.3.19) and obtain the left side asymptotically. Proceed to carry out the calculation as follows. Use the Dirac delta function to compute the time integral. Scale the variables $\mathbf{k}$ by $k=|\omega| / c_0$, and introduce the polar coordinates $\theta$ and $\phi$ :
$$
k_1=k \cos \phi \sin \theta, \quad k_2=k \sin \phi \sin \theta, \quad k_3=k \cos \theta .
$$
Now carry out the integrations in $\theta, \phi$, and $\xi$ by the method of stationary phase. The remaining integral in $k$ will be recognized as the band-limited reflectivity function multiplied by 4 .

Khoobchandra Agrawal
Khoobchandra Agrawal
Numerade Educator

Problem 15

In (9.3.19), use the dispersion relation to rewrite that integral as an integral with respect to $\omega$.
(a) Interpret the multiplication by $t$ as a derivative with respect to $\omega$ and integrate by parts. Rewrite the result as an integral in $k_3$ to obtain
$$
\beta(\mathbf{y})=\int U_{\mathrm{s}}(\boldsymbol{\xi}, t) B(\mathbf{y}-\xi, t) d \xi_1 d \xi_2 d t .
$$
with
$$
B(\mathbf{y}, t)=\frac{32 y_3}{\pi^2} \int d k_1 d k_2 d k_3 \exp \left(2 i\left\{k_1 y_1+k_2 y_2-k_3 y_3\right\}+i \omega t\right) .
$$
In this form, the reflectivity is seen to be an operator on $U_{\mathrm{s}}$ itself rather than on $t U_{\mathrm{S}}$.
(b) Consider the representation (2.4.6) of $U(\mathbf{x}, t)$, which was shown in (2.4.13) to be the three-dimensional Green's function $G(\mathbf{x}, t)$. Rewrite the denominator of the integrand as $\left(c k \text { sign } k_3\right)^2-\omega^2$, and calculate the integral as a residue sum at the zeros of this expression to obtain
$$
\begin{aligned}
G(\mathbf{x}, t) & =G_{+}(\mathbf{x}, t)+G_{-}(\mathbf{x}, t), \\
G_{ \pm}(\mathbf{x}, t) & = \pm \frac{i c}{16 \pi^3} \int \frac{d k_1 d k_2 d k_3}{k \operatorname{sign} k_3} e^{i\left(k \cdot \mathbf{x} \mp d k t \text { sign } k_3\right)}, \quad t>0 .
\end{aligned}
$$
[Remark: The $k_3$ dependence in the phase is $\left[\left|k_3\right| x_3 \mp c k t\right] \operatorname{sign} k_3$. That is, the Green's function is decomposed by this device into a down-going wave $G_{+}$and an up-going wave $G_{-}$.]
(c) In $G_{+}$, differentiate with respect to $t$ and make the change of variable of integration from $k_3$ to $-k_3$ to obtain the result
$$
\frac{\pi}{c^2} \frac{\partial G_{+}(2 \mathbf{y}, t)}{\partial t}=\frac{2}{y_3} B(\mathbf{y}, t) .
$$

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Problem 16

(a) Write the Green's function representation of the solution to (9.3.26) under the assumption that the medium is homogeneous, with constant sound speed $c$. Set $\xi_3=0$ in that solution, and take the Fourier transform in the transverse variables, as in (9.3.11). Obtain the result
$$
w\left(k_1, k_2, 0 ; \omega\right)=\left(\omega / 2 \pi c^2 k_3\right) R_n \bar{\gamma}(\mathbf{k}) \text {, }
$$
with $k_3$ defined by $(9.3 .13)$. In this result, it should also have been necessary to identify $\left|y_3\right|=y_3$ as in this section.
(b) Show that this solution for $R_n \gamma_{\mathrm{B}}(\mathbf{y})$ agrees with (9.3.17).

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07:27

Problem 17

Rewrite the result of the preceding exercise as
$$
\beta(\mathbf{y})=\int u_{\mathrm{s}}(\mathbf{x} ; \xi ; \omega) b(\mathbf{y}-\xi) d \xi_1 d \xi_2,
$$
with
$$
b(\mathbf{y})=\frac{32 y_3}{\pi^2} \int d k_1 d k_2 d k_3 \exp \left(2 i\left\{k_1 y_1+k_2 y_2-k_3 y_3\right\}\right) .
$$
Use for $u_s(\mathbf{x} ; \xi ; \omega)$ the Kirchhoff approximate backscatter field (8.4.10), thereby obtaining a sevenfold integral in $\xi, \mathbf{k}$, and $\boldsymbol{\sigma}$ for $\beta(\mathbf{y})$. Introduce the polar transformation of Exercise 9.14. Calculate the integrals in $\theta$ and $\phi$ by stationary phase. Now calculate the integrals in $\xi_1$ and $\xi_2$ by the method of stationary phase. Finally, calculate the integrals in $\sigma_1$ and $\sigma_2$ by the method of stationary phase. In this last stationary phase, the previous conditions of stationarity make the chain rule differentiations of the phase with respect to $\boldsymbol{\xi}$ at the stationary point easier to calculate. The stationarity conditions require that $\hat{\mathbf{k}}, \mathbf{y}-\mathbf{x}$, and $\boldsymbol{\xi}-\mathbf{x}$ all coalign along the normal from the scattering surface through the output point $\mathbf{y}$. The actual output is $\beta(\mathbf{y})$ scaled by the ratio of distances along the normal from $\mathbf{x}$ to $\boldsymbol{\xi}$ and from $\mathbf{x}$ to $\mathbf{y}$. Note that this ratio is equal to unity on the scattering surface.

Uma Kumari
Uma Kumari
Numerade Educator