Geology Reference
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E/ t + (C g E)/ x = E{2 i + r,t (k,x,t)/ r } (4.3.20)
If the wavenumber field k(x,t) is available - and it is for linear problems without
solving for E or M, the general initial value problems for these amplitude-like
equations can be solved in closed form using the method of characteristics (see
Chapter 2). In the foregoing equations, r,t (k,x,t) and r,x (k,x,t) denote partial
derivatives of the real frequency function with all other variables including k
held fixed. The imaginary frequency i , which is not differentiated, here
denotes the effects of true dissipation. Again, we did not explicitly consider the
attenuative effects of internal viscous dissipation, since these were not germane
to our arguments. We wished to show that an damping-like effects arose from
heterogeneity variations; the effects of irreversible friction would, in any case,
produce additive exponential terms with respect to the imaginary frequency.
Earlier we determined that the real dispersion relation satisfied A r 2 = EI
k 4 - Nk 2 . Hence, upon partial differentiation with respect to x, we obtain
2 A
r,x = - N x k 2 , thus implying that
r
-
r,x (k,x,t)/k = (N x k)/(2
A
0 )
(4.3.21)
From the imaginary part of the dispersion relation, we have
2
i = 2 N x k/(2
A
0 )
(4.3.22)
Thus, we have, on combining Equations 4.3.19, 4.3.21, and 4.3.22,
M/ t + (C g M) / x= M{2 i - r,x (k,x,t)/k}
= M [ 3 kN x /(2 A 0 )] (4.3.23)
As we are considering dynamically steady wave systems oscillating with a real
frequency 0 unchanged with time, the partial derivative with respect to “t”
vanishes, and we have
C g dM/dx + M dC g /dx = M [3kN x /(2 A 0 )] (4.3.24)
Note that only ordinary derivatives with respect to space appear in the foregoing
equation. Division by MC g throughout leads to
(1/M) dM/dx + (1/C g ) dC g /dx = [3kN x /(2 A 0 )] /C g (4.3.25)
Now, we observe that a wave group trajectory is kinematically defined by
dx/dt = C g
(4.3.26)
Thus, it follows that
d(log M)/dx + d(log C g )/dx = d{log (MC g )}/dx
= [ 3 kN x /(2 A
0 )] /C g = [ 3 kN x /(2 A
0 )] dt/dx
(4.3.27)
In turn, we therefore have
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