## Appendix B Proof of Theorem 3
\begin{split} s_{1}&=(1+\beta_{2}+\beta_{1})^{2}(1+ \phi^{2}+\alpha_{1}^{2}+\alpha_{2}^{2}+\varphi^{2}+\alpha_{3}^{2})\\ s_{2}&=2(1+\beta_{2}+\beta_{1})(\alpha_{1}\tau_{1 }+\alpha_{2}\tau_{2}+\phi\tau_{1}+\alpha_{2}\gamma_{1}\sigma_{1}+\varphi\tau _{2}+\alpha_{3}\tau_{3}+\alpha_{3}\gamma_{2}\sigma_{2})\\ s_{3}&=\sigma_{1}^{2}+\sigma_{2}^{2}+\tau_{1}^{2}+ \tau_{2}^{2}+2\tau_{2}\gamma_{1}\sigma_{1}+\tau_{2}^{2}+\tau_{1}^{2}\sigma_{1 }^{2}+\tau_{1}^{2}+\tau_{3}^{2}+2\tau_{3}\gamma_{2}\sigma_{2}+\tau_{2}^{2} \sigma_{2}^{2}\\ s_{4}&=(1+\beta_{2}+\beta_{1})^{2}(2+2(\alpha_{1}^{2}+ \alpha_{2}^{2}+\alpha_{3}^{2}+\phi^{2})-2(\alpha_{1}+\phi\alpha_{2}+\varphi \alpha_{3}))\\ s_{5}&=2\sigma_{1}(1-\gamma_{1}\alpha_{1}-\phi+ \gamma_{1}\alpha_{2})-2\tau_{1}(1-\alpha_{1}+\phi-\alpha_{2})-2\sigma_{2}( \varphi-\gamma_{2}\alpha_{3}-\phi+\alpha_{2}\gamma_{2})-2\tau_{2}(\phi-\alpha _{2}+\alpha_{3}-\varphi)\\ s_{6}&=2(\sigma_{1}^{2}+\gamma_{1}^{2}\sigma_{1}^{2}+ \gamma_{2}^{2}\sigma_{2}^{2}+\sigma_{2}^{2}+\tau_{1}^{2}+\tau_{2}^{2})-2( \alpha_{1}-\tau_{1}\tau_{2}+\sigma_{1}\sigma_{2}+\gamma_{1}\sigma_{1}\gamma_{ 2}\sigma_{2})\quad\text{and}\\ \theta&=\frac{6}{\mu^{2}}+\frac{s_{6}}{\mu^{2}}+ \frac{6}{(\mu^{2}+q_{a}\Lambda)^{2}}.\end{split} \]  
That completes the proof of (II). Let the integrand of the objective functional (2) be the Lagrangian of the following form  
\[ \begin{split} f(t,x,\
u)&=P_{1}L+P_{2}L_{d}+P_{3}L_{c }+P_{4}A+P_{5}A_{d}+P_{6}A_{c}+\frac{1}{2}\sum_{i=1}^{4}Q_{i}u_{i}^{2}(t)\\ &:=f_{1}(t,x)+f_{2}(t,\
u)\end{split} \]where  
\[f_{1}(t,x)=P_{1}L+P_{2}L_{d}+P_{3}L_{c}+P_{4}A+P_{5}A_{d}+P_{6}A_{c},\text{ and }f_{2}(t,\
u)=\frac{1}{2}\sum_{i=1}^{4}Q_{i}u_{i}^{2}(t).\]  
Therefore, it is sufficient to prove that \(f_{2}(t,\
u)\) is convex in \(\
u\), the control variable. We see that \(f_{2}(t,\
u)\) is a finite linear combination of the functions denoted by \(\chi_{i}=\frac{u_{i}^{2}}{2}\) and \(Q_{i}\) for \(i=1,2,3,4\) are positive coefficients. Now, we need to prove the convexity of the map \(\chi:U\to\mathbb{R}\) defined by \(\chi(u)=\frac{u^{2}}{2}\). Also, as defined for a convex function in [46], choose \(\bar{\lambda}\in[0,1]\) and \(\tilde{z},\tilde{y}\in U\). Then,  
\[\chi((1-\bar{\lambda})\tilde{z}+\bar{\lambda}\tilde{y})-((1-\bar{ \lambda})\chi(\tilde{z})+\bar{\lambda}\chi(\tilde{y})) =\frac{1}{2}((1-\bar{\lambda})\tilde{z}+\bar{\lambda}\tilde{y})^{ 2}-\frac{1}{2}\big{(}(1-\bar{\lambda})\tilde{z}^{2}+\bar{\lambda}\tilde{y}^{2} \big{)}\] \[=\frac{1}{2}[\bar{\lambda}(\bar{\lambda}-1)(\tilde{y}^{2}-2g \tilde{z}+\tilde{z}^{2})]\] \[=\frac{1}{2}\bar{\lambda}(\bar{\lambda}-1)(\tilde{y}-\tilde{z})^{2}\] \[\leq 0.\]  
Since \(\bar{\lambda}\in[0,1]\), we have that \(\chi(\bar{\lambda}\tilde{y}+(1-\bar{\lambda})\tilde{z})\leq\bar{\lambda}\chi (\tilde{y})+(1-\bar{\lambda})\chi(\tilde{z})\). Hence, the condition (III) is satisfied. Finally, to verify the final condition (IV), we see that the Lagrangian given by  
\[f(t,x,\
u) =P_{1}L+P_{2}L_{d}+P_{3}L_{c}+P_{4}A+P_{5}A_{d}+P_{6}A_{6}+\frac {1}{2}\sum_{i=1}^{4}Q_{i}u_{i}^{2}(t)\] \[\geq\frac{1}{2}\sum_{i=1}^{4}Q_{i}u_{i}^{2}(t)\] \[\geq\Psi_{1}\left(\sum_{i=1}^{4}u_{i}^{2}\right)^{\frac{\Psi_{3} }{2}}-\Psi_{2}\]  
where \(\Psi_{1}=\frac{1}{2}min\{Q_{i},i=1,2,3,4\}\), \(\Psi_{2}\geq 0\) and \(\Psi_{3}=2\).