A

Proof of Theorem 5.7.1

The proof uses a back-stepping procedure and an inductive argument. Thus, under the assumptions (i), (ii) and (iv), a global change of coordinates for the system exists such that the system is in the strict-feedback form [140, 153, 199, 212]. By augmenting the ρ linearly independent set

$z_1=h\left(x\right),z_2=L_{f}h\left(x\right),\mathrm{\dots },z_{\rho }=L_f^{\rho -1}h\left(x\right)$

with an arbitrary n−ρ linearly independent set z_ρ+1 = ψ_ρ+1(x),…, z_n = ψ_n with ψ_i(0) = 0, 〈dψ_i, G_ρ−1〉 = 0, ρ + 1 ≤ i ≤ n. Then the state feedback

$u=\frac{1}{L_{g_2}{L}_f^{\rho -1}h\left(x\right)}\left(\upsilon -L_f^{\rho }h\left(x\right)\right)$

globally transforms the system into the form:

where z_μ = (z_ρ+1,…, z_n). Moreover, in the z-coordinates we have

${\mathcal{G}}_j=span\left\{\frac{\partial }{\partial z_{\rho -j}},\mathrm{\dots },\frac{\partial }{\partial z_{\rho }}\right\},0\le j\le \rho -1,$

so that by condition (iii) the system equations (A.1) can be represented as

Now, define

$z_2^{\star }=-z_1-\frac{1}{4}\text{λ}{z}_1\left(1+{\text{Ψ}}_1^T\left(z_1,z_{\mu }\right){\text{Ψ}}_1\left(z_{1,}{z}_{\mu }\right)\right)$

and set $z_2=z_2^{\star }(z_1,z_{\mu },\text{λ})$ in (A.2). Then consider the time derivative of the function:

$V_1=\frac{1}{2}{z}_1^2$

along the trajectories of the closed-loop system:

Now if ρ = 1, then we set $v=z_2^{\star }(z_1,z_{\mu },\text{λ})$ and integrating (A.3) from t₀ with z(t₀) = 0 to some t, we have

$V_1\left(x\left(t\right)\right)-V_1\left(0\right)\le -{\displaystyle {\int }_{t_o}^t{y}^2\left(\tau \right)d\tau +\frac{1}{\lambda }{\displaystyle {\int }_{t_0}^t{\Vert w\left(t\right)\Vert }^{2}d\tau }}$

and since V₁(0) = 0, V₁(x) ≥ 0, the above inequality implies that

${\displaystyle {\int }_{t_0}^t{y}^2\left(\tau \right)d\tau \le \frac{1}{\text{λ}}}{\displaystyle {\int }_{t_0}^t{\Vert w\left(\tau \right)\Vert }^{2}d\tau .}$

For ρ > 1, we first prove the following lemma.

Lemma A.0.1 Suppose for some index i, 1 ≤ i ≤ ρ, for the system

there exist i functions

$z_j^{\star }=z_j^{\star }\left(z_1,\mathrm{\dots },z_{j-1,}\text{λ}\right),z_j^{\star }=z^{\star }\left(0,\mathrm{\dots },0,\lambda \right)=0,2\le j\le i\le +1,$

such that the function

${\text{V}}_i=\frac{1}{2}{\displaystyle \sum _{j=1}^i{\tilde{z}}_j^2,}$

where

${\tilde{z}}_1=z_1,{\tilde{z}}_j=z_j-z_j^{\star }\left(z_1,\mathrm{\dots },z_{j-1},z_{\mu ,}\text{λ}\right),2\le j\le i,$

has time derivative

${\dot{V}}_i=-{\displaystyle \sum _{j=1}^i{\tilde{z}}_j^2+\frac{c}{\text{λ}}{\Vert w\Vert }^2}$

along the trajectories of (A.4) with $z_{i+1}=z_{i+1}^{\star }$ and for some real number c > 0.

Then, for the system

there exists a function

$z_{i+2}^{\star }\left(z_1,\mathrm{\dots },z_{i+1,}{z}_{r,}\text{λ}\right),z_{i+2}^{\star }\left(0,\mathrm{\dots },\lambda \right)=0$

such that the function

${\text{V}}_{i+1}=\frac{1}{2}{\displaystyle \sum _{j=1}^{i+1}{\tilde{z}}_j^2}$

where

${\tilde{z}}_j=z_j-z_j^{\star }\left(z_1,\mathrm{\dots },z_i,z_{\mu },\text{λ}\right),1\le j\le i+1,$

has time derivative

${\dot{V}}_i=-{\displaystyle \sum _{j=1}^{i+1}{\tilde{z}}_j^2}+\frac{c+1}{\lambda }{\Vert w\Vert }^2$

along the trajectories of (A.5) with $z_{i+2}=z_{i+2}^{\star }.$

Proof: Consider the function

$V_{i+1}=\frac{1}{2}{\displaystyle \sum _{j=1}^{i+1}{\tilde{z}}_j^2}$

with $z_{i+2}=z_{i+2}^{\star }(z_1,\mathrm{\dots },z_{i+1},z_{\mu },\lambda )$ in (A.4), and using the assumption in the lemma, we have

Let

$\begin{array}{l}{\alpha }_1\left(z_1,\mathrm{\dots },z_{i+1},z_{\mu }\right)={\tilde{z}}_i-{\displaystyle \sum _{j=1}^i\frac{\partial z_{i+1}^{\star }}{\partial z_j}{z}_{j+1}-\frac{\partial z_{i+1}^{\star }}{\partial z_{\mu }}\psi ,}\\ {\alpha }_2\left(z_1,\mathrm{\dots },z_{i+1},z_{\mu }\right)={\text{Ψ}}_{i+1}-{\displaystyle \sum _{j=1}^i\frac{\partial z_{i+1}^{\star }}{\partial z_j}{\text{Ψ}}_j-\text{Ξ}{\left(\frac{\partial z_{i+1}^{\star }}{\partial z_{\mu }}\psi \right)}^T,}\\ z_{i+2}^{\star }\left(z_1,\mathrm{\dots },z_{i+1},z_{\mu }\right)=-{\alpha }_1-{\tilde{z}}_{i+1}-\frac{1}{4}\text{λ}{\tilde{z}}_{i+1}\left(1+{\alpha }_2^T{\alpha }_2\right)\end{array}$

and subsituting in (A.6), we get

$\begin{array}{l}{\dot{V}}_{i+1}\le -{\displaystyle \sum _j^{i+1}{\tilde{z}}_j^2+{\tilde{z}}_{i+1}{\alpha }_2^{T}w-\frac{1}{4}\lambda {\tilde{z}}_{i+1}^2\left(1+{\alpha }_2^T{\alpha }_2\right)+\frac{c}{\lambda }{\Vert w\Vert }^2}\\ =-{\displaystyle \sum _{j=1}^{i+1}{\tilde{z}}_j^2-\lambda \left(\frac{1}{4}{\tilde{z}}_{i+1}^2\left(1+{\alpha }_2^T{\alpha }_2\right)-\frac{{\tilde{z}}_{i+1}}{\lambda }{\alpha }_2^{T}w+\frac{{\left({\alpha }_2^{T}w\right)}^2}{{\lambda }^2\left(1+{\alpha }_2^T{\alpha }_2\right)}\right)+}\\ \frac{{\left({\alpha }_2^{T}w\right)}^2}{\text{λ}\left(1+{\alpha }_2^T{\alpha }_2\right)}+\frac{c}{\lambda }{\Vert w\Vert }^2\\ =-{\displaystyle \sum _{j=1}^{i+1}{\tilde{z}}_j^2-\lambda {\left(\frac{1}{2}{\tilde{z}}_{i+1}\sqrt{1+{\alpha }_2^T{\alpha }_2}-\frac{{\alpha }_2^{T}w}{\text{λ}\sqrt{1+{\alpha }_2^T{\alpha }_2}}\right)}^2+}\frac{{\left({\alpha }_2^{T}w\right)}^2}{\text{λ}\left(1+{\alpha }_2^T{\alpha }_2\right)}+\\ \frac{c}{\text{λ}}{\Vert w\Vert }^2\\ \le -{\displaystyle \sum _{j=1}^{i+1}{\tilde{z}}_j^2+\frac{c+1}{\lambda }{\Vert w\Vert }^2}\end{array}$

as claimed. □

To conclude the proof of the theorem, we note that the result of the lemma holds for ρ = 1 with c = 1. We now apply the result of the lemma (ρ − 1) times to arrive at the feedback contol

$\upsilon =z_{\rho +1}^{\star }\left(z_1,\mathrm{\dots },z_{\rho },z_{\mu },\text{λ}\right)$

and the function

${\text{V}}_{\rho }=\frac{1}{2}{\displaystyle \sum _{j=1}^{\rho }{\tilde{z}}_j^2}$

has time derivative

along the trajectories of (A.4) with $z_{i+1}=z_{i+1}^{\star }(z_1,\mathrm{\dots },z_{\rho },z_{\mu },\text{λ}),1\le i\le \rho .$

Finally, integrating (A.7) from t₀ and z(t₀) = 0 to some t, we get

$\text{V}\left(x\left(t\right)\right)-\text{V}\left(x\left(t_0\right)\right)\le -{\displaystyle {\int }_{t_0}^t{y}^2\left(\tau \right)d\tau -{\displaystyle \sum _{j=2}^{\rho }{\displaystyle {\int }_{t_0}^t{\tilde{z}}_j^2\left(\tau \right)d\tau +\frac{\rho }{\lambda }{\displaystyle {\int }_{t_0}^t{\Vert w\left(\tau \right)\Vert }^{2}d\tau ,}}}}$

and since $V_{\rho }\left(0\right)=0,V_{\rho }\left(x\right)\ge 0,$ we have

${\displaystyle {\int }_{t_0}^t{y}^2\left(\tau \right)d\tau \le \frac{\rho }{\lambda }{\displaystyle {\int }_{t_0}^t{\Vert w\left(\tau \right)\Vert }^{2}d\tau ,}}$

and the L₂-gain can be made arbitrarily small since λ is arbitrary. This concludes the proof of the theorem. □