## 2 Results
### Numerical Stability
As a consequence, (30) renders to  
\[\alpha+\frac{y_{n}}{h_{m}h_{m+1}}+\frac{\sigma^{2}y_{n}}{k_{n}k_{n +1}}-\frac{\rho\sigma y_{n}}{2h_{m+1}k_{n+1}}-\frac{\rho\sigma y_{n}}{2h_{m}k_ {n}}+\frac{y_{n}-2r}{2h_{m}}-\frac{\kappa(\eta-y_{n})}{k_{n}} \tag{31}\] \[\leq\left|\frac{\sigma^{2}y_{n}}{k_{n}(k_{n}+k_{n+1})}-\frac{\rho \sigma y_{n}}{2h_{m}k_{n}}-\frac{\kappa(\eta-y_{n})}{k_{n}}\right|+\left|\frac {\sigma^{2}y_{n}}{k_{n+1}(k_{n}+k_{n+1})}-\frac{\rho\sigma y_{n}}{2h_{m+1}k_{ n+1}}\right|\] \[\qquad+\left|\frac{y_{n}}{h_{m+1}(h_{m}+h_{m+1})}-\frac{\rho \sigma y_{n}}{2h_{m+1}k_{n+1}}\right|+\left|\frac{\rho\sigma y_{n}}{2h_{m+1}k _{n+1}}\right|+\left|\frac{\rho\sigma y_{n}}{2h_{m}k_{n}}\right|\] \[\qquad\qquad\qquad\qquad+\left|\frac{y_{n}}{h_{m}(h_{m}+h_{m+1}) }-\frac{\rho\sigma y_{n}}{2h_{m}k_{n}}+\frac{y_{n}-2r}{2h_{m}}\right|.\]  
Recall (26) and that \(\rho>0\). We have  
\[\frac{2}{\rho\sigma}k_{n},\ \frac{2}{\rho\sigma}k_{n+1}\geq h_{m}+h_{m+1}\ \text{and}\ h_{m},\ h_{m+1}\geq\frac{\rho}{\sigma}(k_{n}+k_{n+1}).\]  
The above indicates that  
\[\frac{\sigma^{2}y_{n}}{k_{n}(k_{n}+k_{n+1})} \geq \frac{\rho\sigma y_{n}}{2h_{m}k_{n}},\] \[\frac{\sigma^{2}y_{n}}{k_{n+1}(k_{n}+k_{n+1})} \geq \frac{\rho\sigma y_{n}}{2h_{m+1}k_{n+1}},\] \[\frac{y_{n}}{h_{m+1}(h_{m}+h_{m+1})} \geq \frac{\rho\sigma y_{n}}{2h_{m+1}k_{n+1}},\] \[\frac{y_{n}}{h_{m}(h_{m}+h_{m+1})} \geq \frac{\rho\sigma y_{n}}{2h_{m}k_{n}}.\]  
Furthermore, since \(y>\eta>2r\), we conclude that  
\[-\frac{\kappa(\eta-y_{n})}{k_{n}}\geq 0\ \text{and}\ \frac{y_{n}-2r}{2h_{m}}\geq 0.\]  
Therefore, the term inside each pair of absolute signs in (31) must be positive. We may remove all absolute signs in (31), and subsequently yield  
\[\leq\left|\frac{\sigma^{2}y_{n}}{k_{n}(k_{n}+k_{n+1})}-\frac{\rho \sigma y_{n}}{2h_{m}k_{n}}-\frac{\kappa(\eta-y_{n})}{k_{n}}\right|+\left|\frac {\sigma^{2}y_{n}}{k_{n+1}(k_{n}+k_{n+1})}-\frac{\rho\sigma y_{n}}{2h_{m+1}k_{ n+1}}\right|\]0  
which is what we expect. Generalizations of the discussion for cases involving \(y\leq\eta\) are straightforward. Therefore, all eigenvalues contained in \(S_{i}\) must lie on the left half of the complex plane. Case 2: Without loss of generality, we consider the case where \(x=x_{1}\) and \(\eta<y<Y\). Thus, for any complex number \(z\in S_{i}\), where \(S_{i}\) is a Gershgorin disc satisfying  
\[\leq\left|\frac{\sigma^{2}y_{n}}{k_{n}(k_{n}+k_{n+1})}-\frac{\rho \sigma y_{n}}{2h_{m}k_{n}}-\frac{\kappa(\eta-y_{n})}{k_{n}}\right|+\left|\frac {\sigma^{2}y_{n}}{k_{n+1}(k_{n}+k_{n+1})}-\frac{\rho\sigma y_{n}}{2h_{m+1}k_{ n+1}}\right|\]1 \[\leq\left|\frac{\sigma^{2}y_{n}}{k_{n}(k_{n}+k_{n+1})}-\frac{\rho \sigma y_{n}}{2h_{m}k_{n}}-\frac{\kappa(\eta-y_{n})}{k_{n}}\right|+\left|\frac {\sigma^{2}y_{n}}{k_{n+1}(k_{n}+k_{n+1})}-\frac{\rho\sigma y_{n}}{2h_{m+1}k_{ n+1}}\right|\]2 \[\leq\left|\frac{\sigma^{2}y_{n}}{k_{n}(k_{n}+k_{n+1})}-\frac{\rho \sigma y_{n}}{2h_{m}k_{n}}-\frac{\kappa(\eta-y_{n})}{k_{n}}\right|+\left|\frac {\sigma^{2}y_{n}}{k_{n+1}(k_{n}+k_{n+1})}-\frac{\rho\sigma y_{n}}{2h_{m+1}k_{ n+1}}\right|\]3 \[\leq\left|\frac{\sigma^{2}y_{n}}{k_{n}(k_{n}+k_{n+1})}-\frac{\rho \sigma y_{n}}{2h_{m}k_{n}}-\frac{\kappa(\eta-y_{n})}{k_{n}}\right|+\left|\frac {\sigma^{2}y_{n}}{k_{n+1}(k_{n}+k_{n+1})}-\frac{\rho\sigma y_{n}}{2h_{m+1}k_{ n+1}}\right|\]4  
The above indicates that \(\alpha\), the real part of \(z\), must satisfy  
\[\leq\left|\frac{\sigma^{2}y_{n}}{k_{n}(k_{n}+k_{n+1})}-\frac{\rho \sigma y_{n}}{2h_{m}k_{n}}-\frac{\kappa(\eta-y_{n})}{k_{n}}\right|+\left|\frac {\sigma^{2}y_{n}}{k_{n+1}(k_{n}+k_{n+1})}-\frac{\rho\sigma y_{n}}{2h_{m+1}k_{ n+1}}\right|\]5  
Recall Lemma 2. Since the origin cannot lie on the boundary of every Gershgorin disc, combining results from the three cases, we conclude immediately that all eigenvalues of \(A\) must be strictly in the left half of the complex plane.