## 3 The Linear Equations (4) and (5)
### Solutions of Equation (4)
Consequently, a graph such as the one shown in Figure 10 tells us that the radiation arrives at the observer placed at the position \(z\)  
From these figures it follows that \(k_{p}=-40\), and therefore a horizontal line placed at \(k=-40\) intersects the dispersion relation \(k(\omega)\) shown in Figure 2 at two points: \((k,\omega)=(-40,\,0)\) and \((k,\omega)=(-40,\,25)\). The first of these points does not correspond to a traveling wave (as its frequency is zero), but the second one corresponds to a linear wave with a resonant frequency \(\omega=25\), phase velocity \(k/\omega=-1.6\) and group velocity \(dk/d\omega=-12.5\) (i.e., in fact we are talking of _inverse velocities_, \(\Delta t/\Delta z\), as mentioned in the previous paragraph). We can see, therefore, that the endpoint of the radiation wavetrain that we see in Figures 9 and 10 moves to the left of the retarded time axis at the group velocity corresponding to the resonant frequency \(\omega=25\). It should be noticed that the value \(k_{p}=-40\), which we obtained from Figures 11 and 12, could also be obtained from Equations (31) and (32). If we define \(f(\omega)=c_{3}\omega^{3}-c_{2}\omega^{2}\), we can approximate \(\sqrt{S(\omega)}\cong i(1+2c_{0}f(\omega))\), and therefore Equation (31) implies that \(\lambda_{b}\cong-i/c_{0}-if(\omega)\). Consequently, when \(a(\omega)=0\), Equation (30) takes the form \(U(z,\omega)=b(\omega)exp(\lambda_{b}z)\), and from this equation it follows that \(u(z,t)\) will contain the factor \(exp(-iz/c_{0})\), thus implying that \(u(z,t)\) has a wavenumber \(k_{p}=-1/c_{0}\). As we used \(c_{0}=1/40\) to obtain Figures 11 and 12, the origin of the value \(k_{p}=-40\) is now clear. We now understand the origin of the velocity of the endpoint of the radiation wavetrain shown in Figures 9 and 10. However, we still do not know why this endpoint exists. In other words, we still do not know why the pulse stops emitting resonant radiation at a certain instant. The clue to understand why the emission of radiation ends abruptly lies in the contrast between Figures 7 and 10. Figure 7 shows no radiation at all, while in Figure 10 the radiation is quite conspicuous. The reason for this difference must be necessarily in the initial conditions used to obtain the solutions of Equation (4) which led to these two figures. These initial conditions are shown in Figure 13, and the absolute values of their Fourier transforms are shown in Figures 14 and 15. We can see that the initial condition which lead to Figure 10 is much narrower than the initial condition corresponding to Figure 7, but the graphs which really suggest the answer to the presence of radiation in Figure 10, and its absence in Figure 7, are the Fourier transforms (FT) shown in Figures 14 and 15. The FT shown in Figure 14 shows that the spectrum corresponding to the initial condition which lead to Figure 7 is essentially contained in the interval \(-5<\omega<5\), and it is practically zero for \(\omega>\omega_{3}=10\Big{(}1+\sqrt{2}\Big{)}\approx 24.14\), which is the second interval where the dispersion relation \(k(\omega)\) is well defined (remember, from Figure 2, that \(k(\omega)\) is defined in two intervals: \(\omega_{1}<\omega<\omega_{2}\) and \(\omega>\omega_{3}\)) Therefore, it is impossible for the initial condition which lead to Figure 7 to resonate with the linear waves with frequencies \(\omega>\omega_{3}\), since this initial condition has no frequency components in this interval. On the contrary, the spectrum shown in Figure 15 shows that the initial condition which leads to Figure 10 contains significant frequency components with \(\omega>\omega_{3}\). Consequently, when this initial pulse starts traveling along the fiber, it resonates with the linear wave with wavenumber \(k_{p}=-40\) (the pulse's wavenumber) and resonant frequency \(\omega=25\), and thus the emission of radiation begins.