I'm working through the paper (1) which uses the elastic wave equation and can't understand how they obtain a set of vectors. I have the elastic wave equation defined as (2.6)-(2.8):
$$
\rho\frac{\partial^2}{\partial t^2}\bigg(\begin{array}{c}
u_1\\
u_2\end{array}\bigg)
=
\frac{\partial}{\partial x_3}\bigg(\begin{array}{c}
\sigma_{13}\\
\sigma_{23}\end{array}\bigg)
$$
where
$$
\sigma_{13} = [N+(L-N)\cos^2{\psi(x_3)}]\frac{\partial u_1}{\partial x_3} + [(L-N)\sin{\psi(x_3)}\cos{\psi(x_3)}]\frac{\partial u_2}{\partial x_3},
$$
$$
\sigma_{23} = [N+(L-N)\sin^2{\psi(x_3)}]\frac{\partial u_2}{\partial x_3} + [(L-N)\sin{\psi(x_3)}\cos{\psi(x_3)}]\frac{\partial u_1}{\partial x_3}.
$$
Now if I define $\hat f(\omega) = \int f(t)e^{i\omega t}dt,$ $f(t) = \frac{1}{2\pi}\int \hat f(\omega) e^{-i\omega t}d\omega. $
My question is, how does this next part follow? They define
$$
\boldsymbol{ \hat u}(z,w)=\bigg(\begin{array}{c}
\hat u_1\\
\hat u_2\end{array}\bigg)
$$
$$
\hat{\boldsymbol{v}}(z, \omega)=\frac{1}{2 i \omega}\left[(N+L) \left( \begin{array}{cc}{1} & {0} \\ {0} & {1}\end{array}\right)+(L-N) \left( \begin{array}{cc}{\cos 2 \psi \sin 2 \psi} \\ {\sin 2 \psi \cos 2 \psi}\end{array}\right)\right] \frac{\partial}{\partial z} \left( \begin{array}{c}{\hat{u}_{1}} \\ {\hat{u}_{2}}\end{array}\right)
$$
The four-dimensional vector $\left(\hat{u}_{1}, \hat{u}_{2}, \hat{v}_{1}, \hat{v}_{2}\right)^{t}$ satisfies the linear system (3.1)-(3.3),
$$
\frac{\partial}{\partial z} \left( \begin{array}{c}\hat{\boldsymbol{u}} \\ \hat{\boldsymbol{v}}\end{array}\right) = i \omega \left( \begin{array}{cc} \mathbf{0} & \frac{1}{K} \mathbf{I}+\frac{\kappa}{K} \mathbf{J}(z) \\ \rho \mathbf{I} & \mathbf{0} \end{array}\right) \left( \begin{array}{l}\hat{\boldsymbol{u}} \\ \hat{\boldsymbol{v}}\end{array}\right)
$$
where
$$
\mathbf{0}=\left( \begin{array}{cc}{0} & {0} \\ {0} & {0}\end{array}\right), \quad \mathbf{I}=\left( \begin{array}{cc}{1} & {0} \\ {0} & {1}\end{array}\right), \quad \mathbf{J}(z)=\left( \begin{array}{cc}{\cos 2 \psi(z)} & {\sin 2 \psi(z)} \\ {\sin 2 \psi(z)} & {-\cos 2 \psi(z)}\end{array}\right)
$$
and
$$
\frac{1}{K}=\frac{N+L}{2 L N}, \quad \kappa=\frac{N-L}{N+L}
$$
How on earth does one obtain the vector $ \boldsymbol{\hat v}$? Any help/hints/tips would be greatly appreciated.
(1) Garnier, J., & Sølna, K. (2016). Apparent attenuation of shear waves propagating through a randomly stratified anisotropic medium. Stochastics and Dynamics, 16(04), 1650009.
doi:10.1142/S021949371650009X
Best Answer
At a first glance, it seems that $\hat{\boldsymbol{v}}$ represents the time-domain Fourier transform of the velocity field. Hence, $\hat{\boldsymbol{v}} = i\omega\hat{\boldsymbol{u}}$, so that the wave equation rewrites as $i\omega\rho\hat{\boldsymbol{v}} = \partial_z (\sigma_{13}, \sigma_{23})^\top$. However, one spatial derivative $\partial_z$ and $\rho$ would be missing in the expression of $\hat{\boldsymbol{v}}$ in OP.
It could be more consistent to provide another interpretation of the notations, e.g. by considering that the time-derivative $i\omega\hat{\boldsymbol{v}}$ of this vector represents the time-domain Fourier transform of the stress field $(\sigma_{13}, \sigma_{23})^\top$. What a strange definition! With this in mind, we have $\partial_z\hat{\boldsymbol{v}} = i\omega\rho \hat{\boldsymbol{u}}$, as written in Eq. (3.1). Moreover, we can also express the constitutive law in matrix-vector form, and interprete the first line of (3.1) as the inverse of the stress-strain relationships (2.7)-(2.8). Note that if we multiply the present state vector $(\hat{\boldsymbol{u}}, \hat{\boldsymbol{v}})^\top$ by $i\omega$, then we obtain the time-domain Fourier representation of the classical "velocity-stress" first order formulation.