Up-and Downgoing waves

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We have seen a host of examples of how physical problems in stratified source-free media reduce to the form

$\begin{displaymath} \frac{d}{dz}\ \mbox{\bf X} \eq \mbox{\bf AX}\end{displaymath}$

(1)

Where $\mbox{\bf X}$ is a vector of physical variables and $\mbox{\bf A}$ is a matrix which depends on z if material properties depend upon z. An important set of new variables in the vector $\mbox{\bf V}$ is defined by multiplying the vector of physical
variables $\mbox{\bf X}$ by a square matrix $\mbox{\bf R}$

$\begin{displaymath} \mbox{\bf V} \eq \mbox{\bf RX}\end{displaymath}$

(2)

where $\mbox{\bf R}$ is the matrix of row eigenvectors of the matrix $\mbox{\bf A}$ .Inverse to $\mbox{\bf R}$ is the matrix $\mbox{\bf C}$ of column eigenvectors of $\mbox{\bf A}$ .Premultiplying (9-3-2) by $\mbox{\bf C}$ and using $\mbox{\bf CR} = \mbox{\bf I}$ we get the inverse relation to (9-3-2) which is useful to find the physical variables $\mbox{\bf X}$ from the new variables $\mbox{\bf V}$ .

$\begin{displaymath} \mbox{\bf X} \eq \mbox{\bf CV}\end{displaymath}$

(3)

Inserting (9-3-3) into (9-3-1) we obtain

	$\begin{eqnarray} (\mbox{\bf CV}_{z}) &=& \mbox{\bf ACV} \\ \mbox{\bf CV}_{z} &=& \mbox{\bf ACV} - \mbox{\bf C}_{z} \mbox{\bf V}\end{eqnarray}$	(4)
		(5)

Premultiplying by $\mbox{\bf R}$ and using $\mbox{\bf RC} = \mbox{\bf I}$ we obtain

$\begin{displaymath} \mbox{\bf V}_{z} \eq (\mbox{\bf RAC})\mbox{\bf V} - \mbox{\bf RC}_{z}\mbox{\bf V}\end{displaymath}$

(6)

Since we have supposed $\mbox{\bf R}$ and $\mbox{\bf C}$ to be row and column eigenvector matrices of $\mbox{\bf A}$ we can replace $\mbox{\bf RAC}$ by the diagonal matrix of eigenvalues $\mbox{\boldmath$\Lambda$}$ ,that is

$\begin{displaymath} \mbox{\bf V}_{z} \eq \mbox{\boldmath$\Lambda$}\mbox{\bf V} - \mbox{\bf RC}_{z}\mbox{\bf V}\end{displaymath}$

(7)

In any region of physical space where the material is homogeneous then $\mbox{\bf A}$ , hence $\mbox{\bf C}$ ,will be independent of z and (9-3-5) will reduce to

$\begin{displaymath} \frac{d}{dz}\mbox{\bf V} \eq \mbox{\boldmath$\Lambda$}\mbox{\bf V}\end{displaymath}$

(8)

But the only matrix in (9-3-6) is a diagonal matrix, and so the problem for the different variables in the vector $\mbox{\bf V}$ decouples into a separate problem for each component. In wave problems it will be seen to be appropriate to call the components of $\mbox{\bf V}$ upgoing and downgoing wave variables. These variables flow up and down in homogeneous regions without interacting with each other. Let us consider an example.

In Sec. 9-1 we deduced that the matrix first-order differential equation for the acoustic problem in a region of no sources takes the form

$\begin{displaymath} \frac{d}{dz}\left[ \begin{array} {c} P \\ W \end{array} \ri... ...right]\, \left[ \begin{array} {c} P \\ W \end{array} \right] \end{displaymath}$ (9)

where

$\begin{eqnarray} a^{2} &=& \omega \rho \vspace{.05in}\\ b^{2} &=& \frac{\omega}{K} - \frac{k^{2}_{x}}{\omega\rho}\end{eqnarray}$ (10)

(11)

The matrix of column eigenvectors $\mbox{\bf C}$ and the matrix of row eigenvectors of the matrix of (9-3-7) are readily verified to be

$\begin{eqnarray} \mbox{\bf C} &=& \left[ \begin{array} {rc} 1 & 1 \vspace{.1in}\... ... -\frac{a}{b} \vspace{.1in}\\ 1 & \frac{a}{b} \end{array} \right]\end{eqnarray}$ (12)

(13)

It is also readily verified that the vectors are normalized, namely $\mbox{\bf RC} =\mbox{\bf CR} = \mbox{\bf I}$ and that

$\begin{displaymath} \Lambda \eq \mbox{\bf RAC} \eq \left[ \begin{array} {cc} -iab & 0 \vspace{.1in}\\ 0 &+iab \end{array} \right] \end{displaymath}$

The downgoing wave variable D is associated with the iab eigenvalue and the upcoming wave variable U is associated with the -iab. We have definitions for up- and downgoing waves as

$\begin{displaymath} \left[ \begin{array} {c} U \\ D \end{array} \right] \eq \fr... ...y} \right] \left[ \begin{array} {c} P \\ W \end{array} \right]\end{displaymath}$

(1)

Of course a row eigenvector may contain an arbitrary multiplicative scaling factor if the scaling factor is divided from the corresponding column eigenvector. This means that the definition (9-3-12a) is not unique. As it happens, the present scale factors give the up- and downgoing waves the physical dimensions of P. The physical variables P and W are found from U and D by the inverse relation

$\begin{displaymath} \left[ \begin{array} {c} P \\ W \end{array} \right] \eq \le... ...\right] \; \left[ \begin{array} {c} U \\ D \end{array} \right]\end{displaymath}$ (2)

from which we see that the pressure P is the downgoing wave plus the upcoming wave and the vertical velocity is b/a times the difference. Equation (9-3-5) governing the propagation of U and D is

$\begin{displaymath} \frac{d}{dz} \left[ \begin{array} {c} U \\ D \end{array} \r... ... \right]\, \left[ \begin{array} {c} U \\ D \end{array} \right]\end{displaymath}$ (13)

In any region of space where b/a is not a function of z we are left with the simple uncoupled equations

$\begin{displaymath} \frac{d}{dz} \left[ \begin{array} {c} U \\ D \end{array} \r... ... \right]\, \left[ \begin{array} {c} U \\ C \end{array} \right]\end{displaymath}$ (14)

Strictly, to justify the definitions of U and D as up- and downgoing waves we will have to be sure that the downgoing solution takes the form

$\begin{displaymath} d(z, t) \eq e^{- i\omega t + ik_{z}z}\end{displaymath}$ (15)

where $\omega$ and k_z must agree in sign so that constant phase is maintained as both z and t increase. The opposite sign must apply to U. In other words k_z = ab must take the sign of $\omega$ .To see that this happens we take the square root of the product of (9-3-8) and (9-3-9).

$\begin{displaymath} k_{z} \eq ab \eq \omega \bigg( \frac{\rho}{K} - \frac{k^{2}_{x}}{\omega^{2}}\bigg)^{1/2}\end{displaymath}$ (16)

For vertically propagating waves we have k_x = 0 so that k_z = ab specializes to $k_{z} = \omega (\rho/K)^{1/2}$ .Substituting this value into (9-3-15), we see that the phase angle of the exponential is constant if $z/t = (K/\rho)^{1/2}$ ,making it clear that the material's intrinsic velocity is given by

$\begin{displaymath} v \eq \bigg(\frac{K}{\rho}\bigg)^{1/2}\end{displaymath}$ (17)

Reference to Figure 9-1 shows that the angle $\theta$ between the vertical and a ray is defined by

$\begin{displaymath} \sin \theta \eq \frac{k_{x} v}{\omega}\end{displaymath}$ (18)

Inserting (9-3-17) and (9-3-18) into (9-3-16) we obtain

$\begin{displaymath} k_{z} \eq ab \eq \frac{\omega}{v} \cos \theta\end{displaymath}$ (19)

9-1
Figure 1 Rays and wavefronts in a layer. The wavelength $\lambda_{x}$ seen on the x axis and the wavelength $\lambda_{z}$ seen on the z axis are both greater than the wavelength $\lambda$ seen along the ray. Clearly, $\lambda/\lambda_{x} = \sin \theta$ and $\lambda/\lambda_{z} = \cos \theta$ so the spatial frequencies $k_{x} = 2\pi/\lambda_{x}$ and $k_{z} = 2\pi/\lambda_{z}$ satisfy $k^{2}_{x} + k^{2}_{z} = (2\pi/\lambda)^{2} = \omega^{2}/v^{2}$ , which, besides being the Pythagorean theorem (since $\sin \theta = k_{x} \, v/\omega$ ), is the Fourier transform of the wave equation. Snell's law that $(\sin \theta)/v$ is the same from layer to layer is thus equivalent to saying that $k_{x}/\omega$ is the same in each layer. That the spatial frequency k_x is the same constant in each layer is essential to the satisfaction of continuity conditions at the layer interfaces.

The time function (9-3-15) is complex. To get a real time function the expression (9-3-15) must be summed or integrated to include both positive and negative frequencies. Then, as we saw in the chapters on time series analysis, we must have $D(\omega) \eq \bar{D} (-\omega)$ .

The quantity b/a will turn out to be the material's characteristic admittance Y. Taking the square root of the ratio of (9-3-9) over (9-3-8) we have

$\begin{eqnarray} Y &=& \frac{b}{a} \eq \frac{(1-v^{2}k_{x}^{2}/\omega^{2})^{1/2}... ...vspace{.05in} \\ I &=& \frac{b}{a} \eq \frac{\rho v}{\cos \theta}\end{eqnarray}$ (20)

(21)

(22)

We shall now verify that this definition of impedance is the same as the one in the previous chapter. To do this we take a careful look at the matrizant to cross a layer $\mbox{\rm exp}[\mbox{\bf A}_{1}(z_{2} - z_{1}] = \mbox{\rm exp}(\mbox{\bf A}\, \triangle z)$ .By Sylvester's theorem we have for the matrizant

$\begin{displaymath} \mbox{\rm exp}(\mbox{\bf A}\, \triangle z) = \mbox{\bf C}\le... ... 0 & e^{+ik_{z}\; \triangle z} \end{array} \right]\mbox{\bf R}\end{displaymath}$ (23)

The matrizant relates the wave variables at the top z₁ of a layer to those at the bottom z₂. Thus (9-3-23) enables us to write

$\begin{displaymath} \left[ \begin{array} {c} P \\ W \end{array} \right]_{2} \eq... ..._{1}\, \left[ \begin{array} {c} P \\ W \end{array} \right]_{1}\end{displaymath}$ (24)

Equation (9-3-24) which seems to have jumped at us from the mysteries of Sylvester's theorem actually has a simple interpretation. Starting on the right, we interpret the multiplication of $\mbox{\bf R}_{1}$ into the P and W variables as a conversion to up- and downgoing variables. Then the multiplication by $\mbox{\rm exp}\, (\mbox{\boldmath$\Lambda_{1}$}\; \triangle z)$ carries these across the layer and the multiplication by $\mbox{\bf C}_{1}$ converts back to P and W variables which are continuous crossing an interface. Multiplying (9-3-24) through by R₂ and noting (9-3-11) and (9-3-12) we have

$\begin{displaymath} \left[ \begin{array} {c} U \\ D \end{array} \right]_{2} \eq... ...ngle z)\left[ \begin{array} {c} U \\ D \end{array} \right]_{1}\end{displaymath}$ (25)

In (9-3-25) we have now defined the up- and downgoing waves just beneath the interface as we did in the previous chapter. We should now be able to recognize the matrix as having the same form. It is

$\begin{displaymath} \frac{1}{2}\left[ \begin{array} {rr} 1 & -\frac{1}{Y_{2}} \v... ...angle z} & 0 \\ 0 & e^{iab \; \triangle z} \end{array} \right]\end{displaymath}$ (26)

Defining the Z transform variable by

$\begin{displaymath} Z \eq \mbox{\rm exp} \bigg( \frac{2i\omega\; \triangle z}{v\; \cos \theta} \bigg) \end{displaymath}$

Now we recognize that the travel time across the layer is $\triangle t = \triangle z/v\, \cos \theta$ . The layer matrix (9-3-26) is

$\begin{displaymath} \frac{Y_{1} + Y_{2}}{2Y_{2}}\left[ \begin{array} {cc} 1 & ... ...\begin{array} {cc} Z^{-1/2} & \\ & Z^{1/2} \end{array} \right]\end{displaymath}$

(27)

which may be compared to the matrix of (8-2-4) namely,

$\begin{displaymath} \frac{1}{t}\, \left[ \begin{array} {cc} 1 & c \\ c & 1 \end... ...n{array} {cc} Z^{-1/2} & 0 \\ 0 & Z^{1/2} \end{array} \right] \end{displaymath}$

establishing that the definition Y = b/a has led to the familiar definition of reflection coefficient

	$\begin{eqnarray} c &=& \frac{Y_{2} - Y_{1}}{Y_{2} + Y_{1}} \eq \frac{I_{1} - I_{... ...\rho_{1}v_{1}/\cos\, \theta_{1} + \rho_{2}v_{2}/\cos\, \theta_{2}}\end{eqnarray}$
		(28)

Exercices:

Redefine the eigenvectors so that W + D + U and P = (D - U)/Y. This transformation would be useful if we wanted t = 1 + c to refer to vertical velocity normalized variables instead of pressure variables as in Chap. 8. Deduce changes to all the equations of this section.
Write the matrizant which crosses a layer in terms of a, b, and layer thickness h.

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