. On the other hand these terms are unbounded at
. So the expression yields Previous: PS of a single wave Up: Ext. Abst.
PS of a continuous spectrum
The same mathematical procedure as in Appendix A is followed here.
Starting from the Parseval's theorem (10), and taking into account the expression for the perturbation
(4) it yields two oscillatory functions where interference terms
can be neglected. It is easy to see that these terms go to 0 as
. On the other hand these terms are unbounded at
. So the expression yields
![\begin{displaymath} \int PS dm=\frac{2 u_a^2 \mu}{L \Delta m_0^2 (1/4+\mu^2)} \... ...}{z^2} dz + \int \frac{(1- m_{01} \mu^{-1} z)}{z^2} dz \right] \end{displaymath}](img78.gif){kind=small} |
(14) |
Now a variable change from 
to
is performed. This mathematical transformation is only related
to the oscillatory terms. So we find
![\begin{displaymath} \int PS dm = \frac{2 u_a^2 \mu}{L \Delta m_0^2 (1/4+\mu^2)}... ...m_{01}^3}{m^3} \left(1-\frac{m_{01}}{m}\right)^{-2}\right] dm \end{displaymath}](img80.gif){kind=small} |
(15) |
As the integral limits are free the integrands must be equal,
This expresion gives an approximation to the amplitudes in the high wavenumber part of the spectrum.
![\begin{displaymath} PS=\frac{2 u_a^2 \mu}{L \Delta m_0^2 (1/4+\mu^2)} \left[\fra... ...rac{m_{01}^3}{m^3} \left(1-\frac{m_{01}}{m}\right)^{-2}\right] \end{displaymath}](img81.gif){kind=small}