## Appendix A
\(\varsigma(\zeta)\) is homogeneous in the bilimit with the weights \(\tau_{0}\) and \(\tau_{\infty}\), and the degrees \(d_{V_{0}}+d_{0}\) and \(d_{V_{\infty}}+d_{\infty}\). On the basis of Lemma 1 in [30], for \(c>0\), there exists:  
\[\phi(\zeta)\leq\frac{1}{c}\varsigma(\zeta) \tag{101}\]  
Then, we can obtain:  
\[\left.\frac{\partial V(\zeta)}{\partial\zeta}g(\zeta)\right|_{\Delta=0}\leq-c \Big{(}V(\zeta)^{(dv_{0}+d_{0})/dv_{0}}+V(\zeta)^{(dv_{\infty}+d_{\infty})/dv_ {\infty}}\Big{)} \tag{102}\]  
because:  
\[ \begin{array}{l}d_{0}=\alpha-1<0\\ d_{\infty}=\beta-1>0\end{array}  \tag{103}\]  
Function (102) can be reduced as:  
\[\left.\frac{\partial V(\zeta)}{\partial\zeta}\mathbf{g}(\zeta)\right|_{\Delta =0}\leq \begin{cases}-cV(\zeta)^{(dv_{\infty}+d_{\infty})/dv_{\infty}},&V( \zeta)\geq 1\\ -cV(\zeta)^{(dv_{0}+d_{0})/dv_{0}},&V(\zeta)\leq 1\end{cases}  \tag{104}\]  
According to Lemma 3, it can be deduced that:  
\[T\leq\frac{1}{c}\bigg{(}\frac{2\alpha}{1-\alpha}+\frac{2\beta}{\beta-1}\bigg{)} \tag{105}\]  
Step 2. Suppose the disturbance of (101) is not zero, according to the Lyapunov function provided in [36], the differential of \(V(\zeta)\) can be obtained as follows, along (101):  
\[\dot{V}(\zeta)=\left.\frac{\partial V(\zeta)}{\partial\zeta}g(\zeta)=\left. \frac{\partial V(\zeta)}{\partial\zeta}g(\zeta)\right|_{\Delta=0}+\frac{ \partial V(\zeta)}{\partial\zeta_{2}}\Delta \tag{106}\]  
where \(\partial V(\zeta)/\partial\zeta_{2}\) is homogeneous in the bilimit with associated triples: \(\big{(}\tau_{0},d_{V_{0}}-r_{0,2},\partial V_{0}(\zeta)/\partial\zeta_{2} \big{)}\) and \((\tau_{\infty},d_{V_{\infty}}-r_{\infty,2},\partial V_{\infty}(\zeta)/\partial \zeta_{2})\), and \(r_{0,2}=\alpha,r_{\infty,2}=\beta\). Define the functions:  
\[ \begin{array}{l}\phi_{\Delta}(\zeta)=\left|\frac{\partial V(\zeta)}{ \partial\zeta_{2}}\right|_{\Delta=0}\\ \varsigma_{\Delta}(\zeta)=V(\zeta)^{(dv_{0}-r_{\infty,2})/dv_{0}}+V(\zeta)^{(dv_{ \infty}-r_{\infty,2})/dv_{\infty}}\end{array}  \tag{107}\]  
where \(\phi_{\Delta}(\zeta)\) is homogeneous in the bilimit with associated triples: \(\big{(}\tau_{0},d_{V_{0}}-r_{0,2},|\partial V_{0}(\zeta)/\partial\zeta_{2}| \big{)}\) and \((\tau_{\infty},d_{V_{\infty}}-r_{\infty,2},|\partial V_{\infty}(\zeta)/\partial \zeta_{2}|)\). \(\varsigma_{\Delta}(\zeta)\) is homogeneous in the bilimit with associated triples: \(\big{(}\tau_{0},d_{V_{0}}-r_{0,2},V(\zeta)^{(dv_{0}-r_{0,2})/dv_{0}}\big{)}\) and \(\big{(}\tau_{\infty},d_{V_{\infty}}-r_{\infty,2},V(\zeta)^{(dv_{\infty}-r_{ \infty,2})/dv_{\infty}}\big{)}\). On the basis of Lemma 1 in [30], for \(c_{\Delta}>0\), there exists:  
\[\phi_{\Delta}(\zeta)\leq c_{\Delta}\varsigma_{\Delta}(\zeta) \tag{108}\]By substituting (A15) into (A14), one can obtain:  
\[\left|\frac{\partial V(\zeta)}{\partial\zeta_{2}}\right|\leq c_{\Delta}\Big{(}V( \zeta)^{(d_{V_{0}}-r_{0,2})/d_{V_{0}}}+V(\zeta)^{(d_{V_{\infty}}-r_{\infty,2})/d _{V_{\infty}}}\Big{)}\] (A16)  
where \(d_{V_{0}}=2\alpha,d_{V_{\infty}}=2\beta,r_{0,2}=\alpha,\) and \(r_{\infty,2}=\beta\). Then, (A16) can be simplified into:  
\[\left|\frac{\partial V(\zeta)}{\partial\zeta_{2}}\right|\leq 2c_{\Delta}V( \zeta)^{1/2}\] (A17)  
By substituting (A9) and (A17) into (A13), we can obtain:  
\[V(\zeta)\leq-c\Big{(}V(\zeta)^{(3\alpha-1)2\alpha}+V(\zeta)^{(3\beta-1)/2p} \Big{)}+2c_{\Delta}V(\zeta)^{1/2}\Big{|}\Delta\Big{|}\] (A18)  
Rewrite (A18) as follows:  
\[\dot{V}(\zeta)\leq-cV(\zeta)^{(3\beta-1)2\beta}-\Big{(}cV(\zeta)^{(3\alpha-1) 2\alpha}-2c_{\Delta}V(\zeta)^{1/2}\Big{|}\Delta\Big{|}\Big{)}\] (A19)  
\[\dot{V}(\zeta)\leq-cV(\zeta)^{(3\alpha-1)2\alpha}-\Big{(}cV(\zeta)^{(3\beta-1 )2\beta}-2c_{\Delta}V(\zeta)^{1/2}\Big{|}\Delta\Big{|}\Big{)}\] (A20)  
where (A19) represents the Lyapunov inequality if the high-order power term is dominant, and (A20) represents the Lyapunov inequality if the low-order power term is dominant. Considering:  
\[