## 3 The Linear Equations (4) and (5)
### Solutions of Equation (4)  
Let us obtain the solution of Equation (4) corresponding to a given initial condition \(u(0,t)\). We will see that we do not have to specify the value of \(u_{z}(0,t)\). Let us begin by calculating the Fourier transform (FT) of Equation (4) (with respect to \(t\)):  
\[iU_{z}+c_{0}U_{zz}-c_{2}\omega^{2}U+c_{3}\omega^{3}U=0 \tag{28}\]  
In this equation we have defined:  
\[U(z,\omega)=\int_{-\infty}^{+\infty}u(z,t)\ e^{i\omega t}\ dt \tag{29}\]  
The solution of Equation (28) can be obtained immediately:  
\[U(z,\omega)=a(\omega)\ exp(\lambda_{a}z)+b(\omega)\ exp(\lambda_{b}z) \tag{30}\]  
where \(\lambda_{a}\) and \(\lambda_{b}\) are the following functions of \(\omega\):  
\[\lambda_{a,b}=\frac{-i\pm\sqrt{S(\omega)}}{2c_{0}} \tag{31}\]  
where the _plus_ sign in front of the radical corresponds to \(\lambda_{a}\), the _minus_ sign corresponds to \(\lambda_{b}\), and the radicand \(S(\omega)\) is given by:  
\[S(\omega)=-1-4c_{0}\Big{(}c_{3}\omega^{3}-c_{2}\omega^{2}\Big{)} \tag{32}\]  
From Equation (30) it follows that:  
\[U_{z}(z,\omega)=\lambda_{a}a(\omega)\ exp(\lambda_{a}z)+\lambda_{b}b(\omega)\ exp( \lambda_{b}z) \tag{33}\]  
and from Equations (30) and (33) we can obtain the system:  
\[U(0,\omega)=a(\omega)\ +b(\omega) \tag{34}\]  
\[U_{z}(0,\omega)=\lambda_{a}a(\omega)+\lambda_{b}b(\omega) \tag{35}\]  
and this system gives us the values of the coefficients \(a(\omega)\) and \(b(\omega)\) in terms of the functions \(U(0,\omega)\) and \(U_{z}(0,\omega)\):  
\[a(\omega)=\frac{U_{z}(0,\omega)-\lambda_{b}U(0,\omega)}{\lambda_{a}-\lambda_{ b}} \tag{36}\]\[b(\omega)=U(0,\omega)-a(\omega) \tag{37}\]  
Now we can observe something interesting. Let us consider, for example, the particular case when \(c_{0}=1/40\), \(c_{2}=1/2\) and \(c_{3}=1/50\). In this case the radicand \(S(\omega)\) (given in (32)) has the form shown in \[U(z,\omega)=\int_{-\infty}^{+\infty}u(z,t)\ e^{i\omega t}\ dt \tag{29}\]0  
We can see in this figure that \(S(\omega)>0\) if \(\omega<\omega_{1}\) or \(\omega_{2}<\omega<\omega_{3}\), and \(S(\omega)<0\) if \(\omega_{1}<\omega<\omega_{2}\) or \(\omega>\omega_{3}\), where the frequencies \(\omega_{n}\) are the values shown in (14). Taking into account the signs of \(S(\omega)\), we can write \(\lambda_{a}\) and \(\lambda_{b}\) in the following way:  
\[\lambda_{a}=\left\{ \begin{array}{ll}\left[-i+\sqrt{S(\omega)}\right]/2c_{0},&\omega<\omega_{1}\\ \left[-i+i\sqrt{-S(\omega)}\right]/2c_{0},&\omega_{1}<\omega<\omega_{2}\\ \left[-i+\sqrt{S(\omega)}\right]/2c_{0},&\omega_{2}<\omega<\omega_{3}\\ \left[-i+i\sqrt{-S(\omega)}\right]/2c_{0}\,&\omega>\omega_{3}\end{array} \right. \tag{38}\]  
\[\lambda_{b}=\left\{ \begin{array}{ll}\left[-i-\sqrt{S(\omega)}\right]/2c_{0},&\omega<\omega_{1}\\ \left[-i-i\sqrt{-S(\omega)}\right]/2c_{0},&\omega_{1}<\omega<\omega_{2}\\ \left[-i-\sqrt{S(\omega)}\right]/2c_{0},&\omega_{2}<\omega<\omega_{3}\\ \left[-i-i\sqrt{-S(\omega)}\right]/2c_{0},&\omega>\omega_{3}\end{array} \right. \tag{39}\]  
We can see that \(\lambda_{a}\) has a positive real part when \(\omega<\omega_{1}\) and \(\omega_{2}<\omega<\omega_{3}\), and therefore for frequencies within these two intervals the first term on the right-hand-side (rhs) of Equation (30) grows exponentially in \(z\).