## 4 Minimizing Input
### Characterization by Convex Optimization  
For each fixed \((t,e)\in[0,\infty)\times\mathbb{R}^{n}\), \(\omega_{1}(t,e)\) can be viewed as a constant. Hence, a control input that minimizes \(\dot{V}(t,e,\tilde{u})=\omega_{1}(t,e)+L_{g}V(t,e)\cdot\tilde{u}\) subject to \(\tilde{u}\in\tilde{U}(t,e)\) is given as an optimal solution of the following convex optimization problem:  
\[\text{Minimize}\ L_{g}V(t,e)\cdot\tilde{u},\ \text{subject to}\ \tilde{u}\in\tilde{U}(t,e),\ (t,e)\in[0,\infty)\times\mathbb{R}^{n}. \tag{36}\]  
We define the optimal value function \(\phi(t,e)\) and the optimal solution map for the problem (36) as follows:  
\[\phi(t,e) :=\min_{\tilde{u}\in\tilde{U}(t,e)}L_{\tilde{g}}V(t,e)\cdot \tilde{u} \tag{37}\] \[\Phi(t,e) :=\{\tilde{u}\in\tilde{U}(t,e)\ |\ L_{\tilde{g}}V(t,e)\cdot\tilde{u}=\varphi(t,e)\} \tag{38}\]  
**Remark 6**. : _Note that, when \(L_{g}V(t,e)=0\), \(\phi(t,e)=0\) and \(\Phi(t,e)=\tilde{U}(t,e)\) hold, i.e., every input \(\tilde{u}\in\tilde{U}(t,e)\) attains the minimum value \(L_{\tilde{g}}V(t,e)\cdot\tilde{u}=0\)._  
The following lemma guarantees the existence of an optimal solution:  
**Lemma 2** (Existence of an optimal solution). : _The optimal value function \(\phi(t,e)\) given by (37) is finite for any fixed \((t,e)\in[0,\infty)\times\mathbb{R}^{n}\)._  
Proof. : Note that \(\tilde{U}(t,e)\) is compact for any fixed \((t,e)\in[0,\infty)\times\mathbb{R}^{n}\) (see Lemma 1), and the objective function \(L_{g}V(t,e)\cdot\tilde{u}\) is continuous with respect to \(\tilde{u}\). Hence, \(L_{g}V(t,e)\cdot\tilde{u}\) takes the minimum value on \(\tilde{U}(t,e)\) by the extreme value theorem. To discuss the optimality condition, we introduce the Lagrangian function \(L:[0,\infty)\times\mathbb{R}^{n}\times\mathbb{R}^{m}\times\mathbb{R}\to \mathbb{R}\) defined as follows [37; 38; 39]:  
\[L(t,e,\tilde{u},\lambda):=L_{g}V(t,e)\cdot\tilde{u}+\lambda\tilde{G}(t,e, \tilde{u}), \tag{39}\]  
where \(\lambda\in\mathbb{R}\) is a Lagrange multiplier. Then, a necessary and sufficient condition for optimal solutions is given by the following Karush-Kuhn-Tucker (KKT) condition [37; 38; 39]:  
**Lemma 3** (KKT condition). : _Consider the optimization problem (36) for any fixed \((t,e)\in[0,\infty)\times\mathbb{R}^{n}\). Then, \(\tilde{u}_{t,e}\in\Phi(t,e)\) (i.e., \(\tilde{u}_{t,e}\) is an optimal solution) if and only if there exists a Lagrange multiplier \(\lambda_{t,e}\in\mathbb{R}\) satisfying the following conditions:_  
\[\frac{\partial L}{\partial\tilde{u}}(t,e,\tilde{u}_{t,e},\lambda_ {t,e})=L_{g}V(t,e)+\lambda_{t,e}\frac{\partial\tilde{G}}{\partial\tilde{u}}(t,e,\tilde{u}_{t,e})=0, \tag{40}\] \[\lambda_{t,e}\tilde{G}(t,e,\tilde{u}_{t,e})=0,\;\lambda_{t,e}\geq 0,\;\tilde{G}(t,e,\tilde{u}_{t,e})\leq 0. \tag{41}\]  
Proof. : See, e.g., Section 5.5.3 of [38] or Corollary 28.3.1 of [39]. As an application of Lemma 3, here, we derive the optimal solution in the case of a norm input constraint:  
**Lemma 4**. : _Let us consider the norm input constraint:_  
\[\tilde{U}=\{\tilde{u}\in\mathbb{R}^{m}\mid\|\tilde{u}\|\leq\alpha\}, \tag{42}\]  
_where \(\alpha>0\). Then, for any fixed \((t,e)\) such that \(L_{g}V(t,e)\
eq 0\), the optimal solution for the problem (36) is given by_  
\[\tilde{u}_{t,e}=-\frac{\alpha}{\|L_{g}V(t,e)\|}L_{g}V^{\top}(t,e). \tag{43}\]  
Proof. : The norm constraint (42) is represented by \(\tilde{U}=\{u\mid\tilde{G}(\tilde{u})\leq 0\}\), where \(\tilde{G}(\tilde{u})\) is given as follows (see Example 1):  
\[\tilde{G}(\tilde{u})=\|\tilde{u}\|^{2}-\alpha^{2}. \tag{44}\]  
Substituting (44) into the KKT condition (40), we have  
\[L_{g}V(t,e)+2\lambda_{t,e}\tilde{u}_{t,e}^{\top}=0\;\;\Longleftrightarrow\; \;\tilde{u}_{t,e}=-\frac{1}{2\lambda_{t,e}}L_{g}V^{\top}(t,e). \tag{45}\]  
Since \(L_{g}V(t,e)\
eq 0\) implies \(\lambda_{t,e}>0\), the condition (41) reduces to \(\tilde{G}(\tilde{u}_{t,e})=0\). By solving this equation, we can obtain  
\[\phi(t,e) :=\min_{\tilde{u}\in\tilde{U}(t,e)}L_{\tilde{g}}V(t,e)\cdot \tilde{u} \tag{37}\]0  
and this completes the proof. **Remark 7**.