## 5 Continuous-Trajectory-Tracking Controller Design
### Proof of Theorem 2  
To prove Theorem 2, we employ the following lemmas:  
**Lemma 8**. : _The function \(P(t,e)\) defined by (63) satisfies_  
\[P(t,e)<1,\ \forall(t,e)\in([0,\infty)\times W)\cap\mathcal{X}_{1}. \tag{67}\]  
Proof. : According to Definition 4, the following holds for any \((t,e)\in[0,\infty)\times W\):  
\[\omega_{1}(t,e)+L_{\tilde{g}}V(x)\cdot\bar{k}(t,e)<-Q(e). \tag{68}\]  
Moreover, since \((t,e)\in\mathcal{X}_{1}\iff L_{\tilde{g}}V(t,e)\
eq 0\), it follows from (68) that  
\[\omega_{1}(t,e)+Q(e)<-L_{\tilde{g}}V(t,e)\cdot\bar{k}(t,e)\ \Leftrightarrow\ \frac{\omega_{1}(t,e)+Q(e)}{-L_{\tilde{g}}V(t,e)\cdot\bar{k}(t,e)}<1. \tag{69}\]  
**Lemma 9**. : _The scaling function \(\mu(t,e)\) given by (62) satisfies the following conditions:_  
**(1)**: \(\mu(t,e)\in[0,1),\ \forall(t,e)\in[0,\infty)\times W\)_;_
**(2)**: _The following holds, except_ \(e=0\)_:_  
\[\lim_{L_{\tilde{g}}V\to 0}\mu(t,e)=0. \tag{70}\]  
Proof. : Since \(\mu(t,e)=0\) when \(L_{\tilde{g}}V(t,e)=0\), we consider condition (1) for \((t,e)\in([0,\infty)\times W])\cap\mathcal{X}_{1}\). In this case, we have  
\[0\leq P(t,e)+|P(t,e)|<2 \tag{71}\]  
by Lemma 8. This shows that condition (1) holds. Then, we prove condition (2). For \((t,e)\) such that \(L_{\tilde{g}}V(t,e)=0\) and \(e\
eq 0\), the following holds by condition (B2) of Definition 1:  
\[\omega_{1}(t,e)+Q(e)<0. \tag{72}\]  
This implies the existence of a neighborhood \(\Xi\subset[0,\infty)\times W\) of \((t,e)\) such that  
\[P(t,e)+|P(t,e)|=0,\ \forall(t,e)\in\Xi. \tag{73}\]  
Hence, we can see that the following holds on \(\Xi\):  
\[\lim_{L_{\tilde{g}}V\to 0}\mu(t,e)=\lim_{L_{\tilde{g}}V\to 0}\frac{C(\|L_{\tilde{g}}V\|)} {2+C(\|L_{\tilde{g}}V\|)}=0. \tag{74}\]  
**Lemma 10**. : _The state feedback controller \(\tilde{u}=\bar{k}(t,e)\) given by (64) is continuous on \([0,\infty)\times\mathbb{R}^{n}\backslash\{0\}\)._  
Proof. : According to Lemma 7, \(\bar{k}(t,e)\) is continuous on \(\mathcal{X}_{1}\). Since \(\omega_{1}(t,e)\), \(L_{\tilde{g}}V(t,e)\), and \(Q(e)\) are continuous, \(\mu(t,e)\) is also continuous on \(\mathcal{X}_{1}\). Hence, \(\bar{k}(t,e)=\mu(t,e)\bar{k}(t,e)\) is clearly continuous on \(\mathcal{X}_{1}\). Moreover, by using Lemma 9, we can prove continuity for \((t,e)\) such that \(L_{\tilde{g}}V(t,e)=0\) and \(e\
eq 0\) as  
\[\lim_{L_{\tilde{g}}V\to 0}\bar{k}(t,e)=\lim_{L_{\tilde{g}}V\to 0}\mu(t,e)\bar{k}(t,e)=0. \tag{75}\]  
**Remark 10**. : _To prove the continuity of \(\bar{K}(t,e)\) for \(e=0\), an additional assumption is required. More precisely, the following must be held uniformly in \(t\in[0,\infty)\):_  
\[\lim_{e\to 0}|P(t,e)|=0. \tag{76}\]  
**Lemma 11**. : _The state feedback controller \(\tilde{u}=\bar{k}(t,e)\) given by (64) satisfies the input constraint \(\tilde{u}\in\tilde{U}(t,e)\) for all \((t,e)\in[0,\infty)\times W\)._  
Proof. : Note that the minimizing input \(\bar{k}(t,e)\) satisfies \(\bar{k}(t,e)\in\tilde{U}(t,e)\) and \(\mu(t,e)\in[0,1)\) on \([0,\infty)\times W\) by Lemma 9. Since \(\tilde{U}(t,e)\) is a convex set containing \(\tilde{u}=0\) (see Lemma 1), we conclude that  
\[\omega_{1}(t,e)+L_{\tilde{g}}V(x)\cdot\bar{k}(t,e)<-Q(e). \tag{68}\]0  
**Lemma 12**. : _The state feedback controller (64) asymptotically stabilizes the origin \(e=0\) of the error system (25) on \(W\subset D\)._  
Proof. : We can calculate the time derivative \(\dot{V}(t,e,\bar{k}(t,e))\) as follows:  
\[\omega_{1}(t,e)+L_{\tilde{g}}V(x)\cdot\bar{k}(t,e)<-Q(e). \tag{68}\]1 \[\omega_{1}(t,e)+L_{\tilde{g}}V(x)\cdot\bar{k}(t,e)<-Q(e). \tag{68}\]2 \[\omega_{1}(t,e)+L_{\tilde{g}}V(x)\cdot\bar{k}(t,e)<-Q(e). \tag{68}\]3 \[\omega_{1}(t,e)+L_{\tilde{g}}V(x)\cdot\bar{k}(t,e)<-Q(e). \tag{68}\]4 \[\omega_{1}(t,e)+L_{\tilde{g}}V(x)\cdot\bar{k}(t,e)<-Q(e). \tag{68}\]5 \[\omega_{1}(t,e)+L_{\tilde{g}}V(x)\cdot\bar{k}(t,e)<-Q(e). \tag{68}\]6  
Here, we note that \(e\in W\backslash\{0\}\) implies \(\omega_{1}(t,e)+L_{g}V(t,e)\cdot\bar{k}(t,e)<-Q(e)\) (see Definition 4).