[proofplan]
We construct the feedback recursively in triangular error coordinates. At each stage the new error is $z_{i+1}=x_{i+1}-\alpha_i(x_{1:i})$, the Lyapunov function is enlarged by $\frac12 z_{i+1}^2$, and the next virtual control is chosen to cancel the mixed term and add quadratic damping. The nonvanishing of each $g_i$ permits the required division, while neighbourhoods are shrunk so that all compositions are defined. At the final step the same formula gives the actual feedback $u=k(x)$, and the Lyapunov direct method gives asymptotic stability.
[/proofplan]
[step:Initialize the first backstepping identity from the given Lyapunov pair]
Define $z_1:U_1\to\mathbb R$ by $z_1(x_1)=x_1$. Define
\begin{align*}
W_1:U_1\to\mathbb R
\end{align*}
by
\begin{align*}
W_1(x_1)=-\frac{dV_1}{dx_1}(x_1)\bigl(f_1(x_1)+g_1(x_1)\alpha_1(x_1)\bigr).
\end{align*}
By hypothesis, $W_1$ is positive definite on $U_1$. Define
\begin{align*}
H_1:U_1\to\mathbb R
\end{align*}
by
\begin{align*}
H_1(x_1)=\frac{dV_1}{dx_1}(x_1)g_1(x_1).
\end{align*}
Then, for the first subsystem with $x_2$ viewed as an external input,
\begin{align*}
\frac{d}{dt}V_1(x_1)=-W_1(x_1)+H_1(x_1)\bigl(x_2-\alpha_1(x_1)\bigr).
\end{align*}
Since $V_1$ is smooth and positive definite, $x_1=0$ is a local minimum of $V_1$, hence $\frac{dV_1}{dx_1}(0)=0$. Therefore $H_1(0)=0$.
[/step]
[step:Propagate the Lyapunov identity through one recursive step]
Assume that for some $i$ with $1\le i\le n-2$ there are an open neighbourhood $U_i\subseteq D_i$ of $0$, an open neighbourhood $\widetilde U_{i+1}\subseteq D_{i+1}$ of $0$ such that $x_{1:i}\in U_i$ whenever $x_{1:i+1}\in \widetilde U_{i+1}$, smooth functions $\alpha_j\in C^\infty(U_j;\mathbb R)$ for $1\le j\le i$, error functions $z_j:U_j\to\mathbb R$ for $1\le j\le i$ and $z_{i+1}:\widetilde U_{i+1}\to\mathbb R$, a smooth positive definite function $V_i:U_i\to\mathbb R$, a smooth positive definite function $W_i:U_i\to\mathbb R$, and a smooth function $H_i:U_i\to\mathbb R$ with $H_i(0)=0$, such that
\begin{align*}
z_1(x_1)=x_1
\end{align*}
and, for $2\le j\le i+1$,
\begin{align*}
z_j(x_{1:j})=x_j-\alpha_{j-1}(x_{1:j-1}).
\end{align*}
Assume also that, along the first $i$ equations with $x_{i+1}$ viewed as an external input,
\begin{align*}
\frac{d}{dt}V_i(x_{1:i})=-W_i(x_{1:i})+H_i(x_{1:i})z_{i+1}(x_{1:i+1}).
\end{align*}
Choose a constant $c_{i+1}>0$. Since $D_{i+1}$ is open and $g_{i+1}$ is nowhere zero on $D_{i+1}$, shrink to an open neighbourhood $U_{i+1}\subseteq \widetilde U_{i+1}$ of $0$ such that all functions below are defined on $U_{i+1}$.
Define
\begin{align*}
A_{i+1}:U_{i+1}\to\mathbb R
\end{align*}
by
\begin{align*}
A_{i+1}(x_{1:i+1})=f_{i+1}(x_{1:i+1})-\sum_{j=1}^{i}\frac{\partial \alpha_i}{\partial x_j}(x_{1:i})\bigl(f_j(x_{1:j})+g_j(x_{1:j})x_{j+1}\bigr).
\end{align*}
Define the next virtual control
\begin{align*}
\alpha_{i+1}:U_{i+1}\to\mathbb R
\end{align*}
by
\begin{align*}
\alpha_{i+1}(x_{1:i+1})=-\frac{A_{i+1}(x_{1:i+1})+H_i(x_{1:i})+c_{i+1}z_{i+1}(x_{1:i+1})}{g_{i+1}(x_{1:i+1})}.
\end{align*}
This function is smooth because the numerator is smooth and $g_{i+1}$ has no zeros on $U_{i+1}$.
Now set
\begin{align*}
V_{i+1}(x_{1:i+1})=V_i(x_{1:i})+\frac12 z_{i+1}(x_{1:i+1})^2.
\end{align*}
Along the first $i+1$ equations, with $x_{i+2}$ viewed as the next external input when $i+1<n$, the chain rule gives
\begin{align*}
\dot{z}_{i+1}=A_{i+1}(x_{1:i+1})+g_{i+1}(x_{1:i+1})x_{i+2}.
\end{align*}
Therefore
\begin{align*}
\frac{d}{dt}V_{i+1}=-W_i+H_i z_{i+1}+z_{i+1}\bigl(A_{i+1}+g_{i+1}x_{i+2}\bigr).
\end{align*}
Using the definition of $\alpha_{i+1}$, this becomes
\begin{align*}
\frac{d}{dt}V_{i+1}=-W_i-c_{i+1}z_{i+1}^2+g_{i+1}z_{i+1}\bigl(x_{i+2}-\alpha_{i+1}\bigr).
\end{align*}
Thus the induction continues with
\begin{align*}
W_{i+1}(x_{1:i+1})=W_i(x_{1:i})+c_{i+1}z_{i+1}(x_{1:i+1})^2
\end{align*}
and
\begin{align*}
H_{i+1}(x_{1:i+1})=g_{i+1}(x_{1:i+1})z_{i+1}(x_{1:i+1}).
\end{align*}
[guided]
The purpose of the recursive step, for $1\le i\le n-2$, is to turn the old mixed term $H_i z_{i+1}$ into a term that can be absorbed by choosing the next virtual control. The restriction $i\le n-2$ is essential: this step constructs $\alpha_{i+1}$, so the last value it may construct is $\alpha_{n-1}$; the final equation uses the actual control input $u$ and is handled separately. We first record exactly what the derivative of the new error is. After shrinking from $\widetilde U_{i+1}$ to $U_{i+1}$, the new error is the smooth map
\begin{align*}
z_{i+1}:U_{i+1}\to\mathbb R
\end{align*}
given by
\begin{align*}
z_{i+1}(x_{1:i+1})=x_{i+1}-\alpha_i(x_{1:i}).
\end{align*}
Differentiating along the subsystem vector field gives
\begin{align*}
\dot{z}_{i+1}=f_{i+1}(x_{1:i+1})+g_{i+1}(x_{1:i+1})x_{i+2}-\sum_{j=1}^{i}\frac{\partial \alpha_i}{\partial x_j}(x_{1:i})\bigl(f_j(x_{1:j})+g_j(x_{1:j})x_{j+1}\bigr).
\end{align*}
The part of this expression not involving $x_{i+2}$ is the smooth function $A_{i+1}:U_{i+1}\to\mathbb R$ defined by
\begin{align*}
A_{i+1}(x_{1:i+1})=f_{i+1}(x_{1:i+1})-\sum_{j=1}^{i}\frac{\partial \alpha_i}{\partial x_j}(x_{1:i})\bigl(f_j(x_{1:j})+g_j(x_{1:j})x_{j+1}\bigr).
\end{align*}
The neighbourhood $U_{i+1}$ has been chosen so that $x_{1:i}\in U_i$, so every occurrence of $\alpha_i$, $V_i$, $W_i$, and $H_i$ is defined. The division by $g_{i+1}$ is valid because $g_{i+1}$ is nowhere zero on $D_{i+1}$, hence nowhere zero on $U_{i+1}$.
Now enlarge the Lyapunov function by the square of the new error:
\begin{align*}
V_{i+1}(x_{1:i+1})=V_i(x_{1:i})+\frac12 z_{i+1}(x_{1:i+1})^2.
\end{align*}
Using the induction hypothesis and the chain rule,
\begin{align*}
\frac{d}{dt}V_{i+1}=-W_i+H_i z_{i+1}+z_{i+1}\bigl(A_{i+1}+g_{i+1}x_{i+2}\bigr).
\end{align*}
The only term that prevents negativity is the coefficient of $z_{i+1}$. We choose $\alpha_{i+1}$ so that, when $x_{i+2}=\alpha_{i+1}$, this coefficient becomes $-c_{i+1}z_{i+1}$. That requirement is
\begin{align*}
H_i+A_{i+1}+g_{i+1}\alpha_{i+1}=-c_{i+1}z_{i+1}.
\end{align*}
Solving for $\alpha_{i+1}$ gives the formula
\begin{align*}
\alpha_{i+1}(x_{1:i+1})=-\frac{A_{i+1}(x_{1:i+1})+H_i(x_{1:i})+c_{i+1}z_{i+1}(x_{1:i+1})}{g_{i+1}(x_{1:i+1})}.
\end{align*}
Substituting this identity into the derivative gives
\begin{align*}
\frac{d}{dt}V_{i+1}=-W_i-c_{i+1}z_{i+1}^2+g_{i+1}z_{i+1}\bigl(x_{i+2}-\alpha_{i+1}\bigr).
\end{align*}
Thus the same structural identity is recovered at level $i+1$, with the new positive term $c_{i+1}z_{i+1}^2$ added to $W_i$ and the new mixed coefficient $H_{i+1}=g_{i+1}z_{i+1}$.
[/guided]
[/step]
[step:Verify positivity and vanishing of the recursively defined objects]
For each $i$, the triangular error map
\begin{align*}
\Phi_i:U_i\to\mathbb R^i
\end{align*}
is defined by
\begin{align*}
\Phi_i(x_{1:i})=(z_1(x_1),z_2(x_{1:2}),\dots,z_i(x_{1:i})).
\end{align*}
Its Jacobian matrix with respect to the standard coordinates is lower triangular with diagonal entries equal to $1$. Let $D(\Phi_i)_0:\mathbb R^i\to\mathbb R^i$ denote the total derivative of $\Phi_i$ at $0$. The preceding Jacobian computation shows that this [linear map](/page/Linear%20Map) is invertible, so by the [inverse function theorem](/theorems/51) $\Phi_i$ is a local diffeomorphism after shrinking $U_i$ if necessary.
The recursive formula gives
\begin{align*}
V_i(x_{1:i})=V_1(x_1)+\frac12\sum_{j=2}^{i}z_j(x_{1:j})^2.
\end{align*}
If $V_i(x_{1:i})=0$, then $V_1(x_1)=0$ and $z_j(x_{1:j})=0$ for every $2\le j\le i$. Since $V_1$ is positive definite, $x_1=0$. Recursively, $z_j=0$ gives $x_j=\alpha_{j-1}(0)=0$. Thus $x_{1:i}=0$, and $V_i$ is positive definite.
Similarly,
\begin{align*}
W_i(x_{1:i})=W_1(x_1)+\sum_{j=2}^{i}c_j z_j(x_{1:j})^2.
\end{align*}
If $W_i(x_{1:i})=0$, then each non-negative summand in this formula must vanish. Hence $W_1(x_1)=0$ and $z_j(x_{1:j})=0$ for every $2\le j\le i$. Since $W_1$ is positive definite, $x_1=0$. Then $z_2=0$ gives $x_2=\alpha_1(0)=0$, and continuing inductively, $z_j=0$ gives $x_j=\alpha_{j-1}(0)=0$ for each $2\le j\le i$. Thus $x_{1:i}=0$. Conversely, $W_i(0)=0$ by the displayed formula, so $W_i$ is positive definite.
Finally, $\alpha_{i+1}(0)=0$ at each step. Indeed, $A_{i+1}(0)=0$ because $f_j(0)=0$ and $x_{j+1}=0$ at the origin, $H_i(0)=0$ by the recursive formula, and $z_{i+1}(0)=0$. Therefore the numerator in the definition of $\alpha_{i+1}(0)$ is zero.
[/step]
[step:Choose the final feedback and obtain a strict Lyapunov decrease]
Apply the recursive construction up to level $n-1$. Define
\begin{align*}
A_n:U_n\to\mathbb R
\end{align*}
by
\begin{align*}
A_n(x_{1:n})=f_n(x_{1:n})-\sum_{j=1}^{n-1}\frac{\partial \alpha_{n-1}}{\partial x_j}(x_{1:n-1})\bigl(f_j(x_{1:j})+g_j(x_{1:j})x_{j+1}\bigr).
\end{align*}
Choose $c_n>0$, shrink $U_n\subseteq D$ if needed, and define the feedback
\begin{align*}
k:U_n\to\mathbb R
\end{align*}
by
\begin{align*}
k(x_{1:n})=-\frac{A_n(x_{1:n})+H_{n-1}(x_{1:n-1})+c_n z_n(x_{1:n})}{g_n(x_{1:n})}.
\end{align*}
The function $k$ is smooth because $g_n$ is nowhere zero on $U_n$, and $k(0)=0$ by the same calculation used for the virtual controls.
For the closed-loop system $u=k(x)$, the derivative of
\begin{align*}
V_n(x_{1:n})=V_{n-1}(x_{1:n-1})+\frac12z_n(x_{1:n})^2
\end{align*}
is
\begin{align*}
\frac{d}{dt}V_n(x_{1:n})=-W_{n-1}(x_{1:n-1})-c_n z_n(x_{1:n})^2.
\end{align*}
Define
\begin{align*}
W_n:U_n\to\mathbb R
\end{align*}
by
\begin{align*}
W_n(x_{1:n})=W_{n-1}(x_{1:n-1})+c_n z_n(x_{1:n})^2.
\end{align*}
Then $W_n$ is positive definite, and
\begin{align*}
\frac{d}{dt}V_n(x_{1:n})=-W_n(x_{1:n}).
\end{align*}
Thus the origin is an equilibrium of the closed-loop system and $V_n$ is a smooth positive definite Lyapunov function with negative definite derivative.
[/step]
[step:Apply the local and global Lyapunov criteria]
We use the standard local Lyapunov criterion for asymptotic stability: if a smooth vector field $F:U\to\mathbb R^n$ on an open neighbourhood $U$ of $0$ satisfies $F(0)=0$, and if there is a smooth function $V:U\to\mathbb R$ that is positive definite and whose derivative $\nabla V(x)\cdot F(x)$ is negative definite, then $0$ is a locally asymptotically stable equilibrium of $\dot{x}=F(x)$. In the present case, $F$ is the closed-loop vector field on $U_n$, the preceding step gives $F(0)=0$, $V=V_n$ is positive definite, and
\begin{align*}
\nabla V_n(x_{1:n})\cdot F(x_{1:n})=-W_n(x_{1:n})
\end{align*}
with $W_n$ positive definite. Therefore the origin is a locally asymptotically stable equilibrium.
In the global case, take $U_i=\mathbb R^i$ at every step. Since each $g_i$ is nowhere zero on $\mathbb R^i$, every recursive formula defines a smooth global virtual control, and the final formula defines a smooth global feedback $k:\mathbb R^n\to\mathbb R$. By the additional hypothesis, the recursively constructed $V_n$ is proper on $\mathbb R^n$. We use the standard global Lyapunov criterion: if $F:\mathbb R^n\to\mathbb R^n$ is smooth, $F(0)=0$, $V:\mathbb R^n\to\mathbb R$ is smooth, positive definite, and proper, and $\nabla V(x)\cdot F(x)$ is negative definite on $\mathbb R^n$, then $0$ is a globally asymptotically stable equilibrium of $\dot{x}=F(x)$. Here $V=V_n$ has exactly these properties and
\begin{align*}
\nabla V_n(x_{1:n})\cdot F(x_{1:n})=-W_n(x_{1:n})
\end{align*}
with $W_n$ positive definite. Hence the closed-loop origin is globally asymptotically stable.
[/step]
