## 1 Introduction
### Evans Function
\right)\quad\text{and}\quad A_{1}(x)=\left( \begin{array}[] {cc}0&0\\ -6\text{ sech}^{2}(x)&0\end{array} \right), \tag{30}\]  
respectively. We now show how the Evans function \(E(\alpha)\) can be used in order to deduce the same conclusion. It is important to note here that  
\[\lim_{x\to\pm\in}A_{1}(x)=0\]  
and that the decay is exponentially fast. For the rest of this discussion, it will be assumed that  
\[\Re(\alpha)\geqq-1.\]  
The eigenvalues of \(A_{0}(\alpha)\) are given by  
\[\beta^{\pm}(\alpha)=\pm\sqrt{1+\alpha}, \tag{31}\]  
and the associated eigenvectors are as follows:  
\[\gamma^{\pm}(\alpha)=\big{(}1,\beta^{\pm}(\alpha)\big{)}^{T}. \tag{32}\]  
One can construct solutions \(\vec{u}(\alpha,x)\) of the Equations (27) and (30), which satisfy the following limit relationship:  
\[\lim_{x\to\pm\in}\vec{u}^{\pm}(\alpha,x)e^{-\beta^{\mp}(\alpha)x}=\gamma^{\mp }(\alpha). \tag{33}\]  
It is noted that the construction implies that  
\[\lim_{x\to\pm\in}|\vec{u}(\alpha,x)|=0.\]  
The Evans function \(E(\alpha)\) is given by  
\[E(\alpha)=\det(\vec{u}^{-},\vec{u}^{+})(\alpha,x) \tag{34}\]  
and, by [PERSON]'s formula, it is independent of \(x\), namely, the Wronskian of \(\vec{u}^{-}\) or \(\vec{u}^{+}\) is given by  
\[W(U^{\pm},V^{\pm})=c,\]where \(c\) constant and  
\[\vec{u}^{\pm}=(U^{\pm},V^{\pm}).\]  
Now, the bases of the stable and unstable subspaces can be determined numerically in the following manner. We calculate the eigenvalues of \(A_{0}(\alpha)\) with negative real part and their corresponding eigenvectors. Then, by choosing a sufficiently large number \(L\), we solve the following homogeneous equation:  
\[\frac{d\vec{u}^{+}}{dx}=\big{(}A_{0}(\alpha)+A_{1}(x)\big{)}\vec{u}^{+}\]  
in \([0,L]\) starting from the right-end point with the initial condition given by  
\[\Re(\alpha)\geqq-1.\]0  
Hence, we obtain linearly independent (approximate) solutions of the differential equation. Therefore, their values at \(x=0\) give a basis for \(E^{s}\). Similarly, by solving the differential equation in \([-L,0]\), we obtain a basis of \(E^{u}\) and the determinant defining the Evans function can be computed. The Evans function for the Equation (28) is shown in Figure 1, where we find that the Evans function has two discrete eigenvalues at \(\alpha=0\) and \(\alpha=3\). Thus, the spectrum is given as mentioned above. We now consider \(a=-1\), \(b=+1\) and \(n=2\) in (19), so that  
\[\Re(\alpha)\geqq-1.\]1  
and  
\[\Re(\alpha)\geqq-1.\]2  
Consequently, \(A_{0}(\alpha)\) and \(A_{1}(x)\) become  
\[A_{0}(\alpha)=\left( \begin{array}{cc}0&1\\ \alpha+1&0\end{array} \right)\quad\text{and}\quad A_{1}(x)=\left( \begin{array}[] {cc}0&0\\ -3\;\text{sech}^{2}\left(\frac{x}{2}\right)&0\end{array} \right), \tag{36}\]  
respectively. The [PERSON] function \(E(\alpha)\) is computed numerically for the Equation (35) and is shown in Figure 2. We find from Figure 2 that the [PERSON] function \(E(\alpha)\) has three discrete eigenvalues at \(\alpha=-\frac{3}{4}\), \(\alpha=0\) and \(\alpha=\frac{5}{4}\). Since there exists a positive eigenvalue for the [PERSON] function \(E(\alpha)\) of the linearized stability problem for a pulse solution of (24), therefore, this pulse solution is unstable. It is also appropriate to compare our analytical and numerical results with the results in [1]. We can thereby see  
Figure 1: Numerical evaluation of the [PERSON] function \(E(\alpha)\) for the Equation (28). an excellent agreement between the results developed in this article and the earlier results in [1].