Boundedness condition for linear system I'm studying a book on wave propagation (which uses SDE's) and I've come across a familiar looking boundedness condition which is related to energy I'm assuming. The governing acoustic wave equations are written as
$$\begin{array}{l}{\rho \frac{\partial u^{\varepsilon}}{\partial t}+\frac{\partial p^{\varepsilon}}{\partial z}+\sigma u^{\varepsilon}=0} \\ {\frac{1}{K} \frac{\partial p^{\varepsilon}}{\partial t}+\frac{\partial u^{\varepsilon}}{\partial z}=0}\end{array}$$
where $p^\epsilon$ is the pressure, $u^ε$ is the velocity, $σ$ is the dissipation of the medium, $ρ$ is the density, and $K$ is the bulk modulus. The argument follows by taking the scaled-time Fourier transform so that the system reduces to a system of ordinary differential equations:
$$\begin{aligned} \frac{d \hat{p}^{\varepsilon}}{d z}-\frac{i \omega \overline{\rho}}{\varepsilon} \hat{u}^{\varepsilon}+\sigma\left(z, \frac{z}{\varepsilon^{2}}\right) \hat{u}^{\varepsilon} &=0 \\ \frac{d \hat{u}^{\varepsilon}}{d z}-\frac{i \omega}{\overline{K} \varepsilon}\left(1+\nu\left(\frac{z}{\varepsilon^{2}}\right)\right) \hat{p}^{\varepsilon} &=0 \end{aligned}.$$
The wave modes can then be decomposed to give right and left going modes 
$$\begin{aligned} \hat{a}^{\varepsilon}(\omega, z) &=\left(\overline{\zeta}^{1 / 2} \hat{u}^{\varepsilon}(\omega, z)+\overline{\zeta}^{-1 / 2} \hat{p}^{\varepsilon}(\omega, z)\right) e^{\frac{-i \omega z}{\overline{c} \varepsilon}} \\ b^{\varepsilon}(\omega, z) &=\left(\overline{\zeta}^{1 / 2} \hat{u}^{\varepsilon}(\omega, z)-\overline{\zeta}^{-1 / 2} \hat{p}^{\varepsilon}(\omega, z)\right) e^{\frac{i \omega z}{c \varepsilon}} \end{aligned}$$
respectively, which satisfy the linear system
\begin{equation}\label{}
\frac{d}{d z} \left[ \begin{array}{c}{\hat{a}^{\varepsilon}} \\ {\hat{b}^{\varepsilon}}\end{array}\right]=\mathbf{H}_{\omega}\left(\frac{z}{\varepsilon}, \nu^{\varepsilon}(z), \sigma^{\varepsilon}(z)\right) \left[ \begin{array}{c}{\hat{a}^{\varepsilon}} \\ {\hat{b}^{\varepsilon}}\end{array}\right]
\end{equation}
where
$$\mathbf{H}_{\omega}(z, \nu, \sigma)=\frac{i \omega}{2 \overline{c}} \nu \left[ \begin{array}{c}{1} & {-e^{-2 i \omega z / \overline{c}}} \\ {e^{2 i \omega z / \overline{c}}} & {-1}\end{array}\right]+\frac{\sigma}{2 \overline{\zeta}} \left[ \begin{array}{cc}{-1} & {-e^{-2 i \omega z / \overline{c}}} \\ {e^{2 i \omega z / \overline{c}}} & {1}\end{array}\right]$$
The next step is to assume a $2\times 2$ complex valued matrix $\mathbf{P_{\omega}^{\epsilon}}$ which converts a BVP to an IVP, which solves the system
$$\frac{d}{d z} \mathbf{P}_{\omega}^{\varepsilon}(\omega,-L, z)=\mathbf{H}_{\omega}\left(\frac{z}{\varepsilon}, \nu^{\varepsilon}(z), \sigma^{\varepsilon}(z)\right) \mathbf{P}_{\omega}^{\varepsilon}(-L, z), \quad \mathbf{P}_{\omega}^{\varepsilon}(-L, z=-L)=\mathbf{I}$$
$$\mathbf{P}_{\omega}^{\varepsilon}(-L, z)=\left[ \begin{array}{c}{\alpha_{\omega, 1}^{\varepsilon}(-L, z), \quad \alpha_{\omega, 2}^{\varepsilon}(-L, z)} \\ {\beta_{\omega, 1}^{\varepsilon}(-L, z), \quad \beta_{\omega, 2}^{\varepsilon}(-L, z)}\end{array}\right]$$
where $\left(\alpha_{\omega, 1}^{\varepsilon}, \beta_{\omega, 1}^{\varepsilon}\right)^{T}$ and $\left(\alpha_{\omega, 2}^{\varepsilon}, \beta_{\omega, 2}^{\varepsilon}\right)^{T}$ are solutions of the linear system with initial conditions $(1,0)^{T}, \quad (0,1)^{T}.$
My question is this next part. They state that:
The solution vectors$\left(\alpha_{\omega, j}^{\varepsilon}, \beta_{\omega, j}^{\varepsilon}\right)$ satisfy
$$\frac{d}{d z}\left(\left|\alpha_{\omega, j}^{\varepsilon}\right|^{2}-\left|\beta_{\omega, j}^{\varepsilon}\right|^{2}\right)=-\frac{\sigma^{\varepsilon}(z)}{\overline{\zeta}}\left|\alpha_{\omega, j}^{\varepsilon} e^{i \omega z /(\overline{c} \varepsilon)}+\beta_{\omega, j}^{\varepsilon} e^{-i \omega z /(\overline{c} \varepsilon)}\right|^{2} \leq 0$$
and thus
$$\left|\alpha_{\omega, 2}^{\varepsilon}(-L, z)\right|^{2}+1 \leq\left|\beta_{\omega, 2}^{\varepsilon}(-L, z)\right|^{2}.$$
I really struggle to see how one arrives at these last two statements. Any help would be greatly appreciated. Thanks in advance and sorry for the long question. 
 A: It is obvious that the grouping with respect to the initial conditions is irrelevant, since there is no mixing of the solutions for different initial conditions, so let me start with the 4th equation above, describing the linear system for the functions $(\hat{\alpha}, \hat{\beta})$ , written succinctly and in simplified notation $(a,b)$ in the following form:
$$\frac{da}{dz}=-\Delta a-\Delta^*e^{-2i\omega z/\bar{c}}b\\
\frac{db}{dz}=\Delta^*e^{2i\omega z/\bar{c}}a+\Delta b$$ 
where $\Delta=\frac{\sigma}{2\bar{\zeta}}-\frac{i\omega\nu}{2\bar{c}}$.
We calculate the rate of change of the magnitude squared:
$$\frac{d|a|^2}{dz}=2Re(a\frac{da}{dz})=-(\Delta+\Delta^*)|a|^2-\Delta^*e^{-2i\omega z/\bar{c}}ba^*-\Delta e^{2i\omega z/\bar{c}}b^*a$$
and similarly for $b$:
$$\frac{d|b|^2}{dz}=2Re(b\frac{db}{dz})=(\Delta+\Delta^*)|b|^2+\Delta^*e^{2i\omega z/\bar{c}}b^*a+\Delta e^{-2i\omega z/\bar{c}}ba^*$$
and upon subtraction we obtain the first statement:
$$\frac{d}{dz}(|a|^2-|b|^2)=-(\Delta+\Delta^*)(|a|^2+|b|^2)-(\Delta+\Delta^*)e^{-2i\omega z/\bar{c}}ba^*-(\Delta+\Delta^*)e^{2i\omega z/\bar{c}}b^*a=-\frac{\sigma}{\bar{\zeta}}|ae^{-i\omega z/\bar{c}}-be^{i\omega z/\bar{c}}|^2$$
This equation must be satisfied regardless of initial conditions, and hence it applies for all indices $j$ in the 1st statement. 
The second statement is clearly a consequence of the first, since the fact that $\frac{d(|a|^2-|b|^2)}{dz}\leq 0$ means that the difference in square magnitudes is a decreasing function (obviously $\sigma$ is assumed positive per physicist's convention) and therefore we obtain that for the given set of initial conditions this yields an upper bound for this quantity:
$$|\hat{\alpha}^{\epsilon}(z, \omega)|^2-|\hat{\beta}^{\epsilon}(z, \omega)|^2\leq |\hat{\alpha}^{\epsilon}(-L, \omega)|^2-|\hat{\beta}^{\epsilon}(-L, \omega)|^2=-1, ~~~~ \forall z\geq -L $$
Hope this helps!
