Stochastic finite-time partial stability, partial-state stabilization, and finite-time optimal feedback control

Abstract

In many practical applications, stability with respect to part of the system’s states is often necessary with finite-time convergence to the equilibrium state of interest. Finite-time partial stability involves dynamical systems whose part of the trajectory converges to an equilibrium state in finite time. In this paper, we address finite-time partial stability in probability and uniform finite-time partial stability in probability for nonlinear stochastic dynamical systems. Specifically, we provide Lyapunov conditions involving a Lyapunov function that is positive definite and decrescent with respect to part of the system state and satisfies a differential inequality involving fractional powers for guaranteeing finite-time partial stability in probability. In addition, we show that finite-time partial stability in probability leads to uniqueness of solutions in forward time and we establish necessary and sufficient conditions for almost sure continuity of the settling-time operator of the nonlinear stochastic dynamical system. Finally, we develop a unified framework to address the problem of optimal nonlinear analysis and feedback control design for finite-time partial stochastic stability and finite-time, partial-state stochastic stabilization. Finite-time partial stability in probability of the closed-loop nonlinear system is guaranteed by means of a Lyapunov function that is positive definite and decrescent with respect to part of the system state and can clearly be seen to be the solution to the steady-state form of the stochastic Hamilton–Jacobi–Bellman equation guaranteeing both finite-time, partial-state stability and optimality. The overall framework provides the foundation for extending stochastic optimal linear–quadratic controller synthesis to nonlinear–nonquadratic optimal finite-time, partial-state stochastic stabilization.

Appendices

Appendix 1

Proof of Proposition 3.2

(i) It follows from Definition 2.2 that

$$\begin{aligned} T( x_1(0), x_2(0)) = {\mathrm{inf}}\{ t \in {\mathbb {R}}_+ {:}\, s_1(t, x_1(0), x_2(0)) = 0 \} \end{aligned}$$

(96)

for all $( x_1(0), x_2(0)) \in {\mathcal {H}}^{{{\mathbb {R}}}^{n_1} \setminus \{ 0 \}}_{n_1} \times {\mathcal {H}}_{n_2}$. Hence, $T(s_1(t_1, x_1(0), x_2(0)), s_2(t_1, x_1(0), x_2(0))) = {\mathrm{inf}}\{ t_2 \in {\mathbb {R}}_+ {:}\, s_1(t_2, s_1(t_1, x_1(0), x_2(0)), s_2(t_1, x_1(0), x_2(0))) = 0 \}$. Now, for $0 \le t_1 \le T( x_1(0),$ $x_2(0))$, the semigroup property and (96) imply that

$$\begin{aligned}&T(s_1(t_1, x_1(0), x_2(0)) , s_2(t_1, x_1(0), x_2(0))) \\&\quad = {\mathrm{inf}}\{ t_2 \in {\mathbb {R}}_+ {:}\, s_1(t_2, s_1(t_1, x_1(0), x_2(0)), s_2(t_1, x_1(0), x_2(0))) = 0 \} \\&\quad \mathop {=}\limits ^{\text {a.s.}}{\mathrm{inf}}\{ t_2 \in {\mathbb {R}}_+ {:}\, s_1(t_1+t_2, x_1(0), x_2(0)) = 0 \} \\&\quad \mathop {=}\limits ^{\text {a.s.}} T( x_1(0), x_2(0)) -t_1. \end{aligned}$$

Alternatively, for $0 \le T( x_1(0), x_2(0)) \le t_1$, $T(s_1(t_1, x_1(0), x_2(0)), s_2(t_1, x_1(0), x_2(0)))$ $\mathop {=}\limits ^{\text {a.s.}} 0$, which proves (11).

(ii) Necessity is immediate. To prove sufficiency, suppose that $T(\cdot , \cdot )$ is jointly sample continuous at $(0, x_2)$, $x_2 \in {\mathcal {H}}_{n_2}$. Let $(x_1, x_2) \in {\mathcal {H}}_{n_1} \times {\mathcal {H}}_{n_2}$ and consider the sequences $\{ x_{1n} \}_{n = 1}^{\infty } \in {\mathcal {H}}_{n_1}$ converging pointwise to $x_1$ and $\{ x_{2n} \}_{n = 1}^{\infty } \in {\mathcal {H}}_{n_2}$ converging pointwise to $x_2$. Let $\tau ^- = \liminf _{n \rightarrow \infty } T(x_{1n}, x_{2n})$ and $\tau ^+ = \limsup _{n \rightarrow \infty } T(x_{1n}, x_{2n})$ be pointwise limits. Note that $\tau ^-$, $\tau ^+ \in {\mathcal {H}}^{{\mathbb {R}}_+}_1$ and $\tau ^- \mathop {\le }\limits ^{\text {a.s.}} \tau ^+$.

Next, let $\{ x_{1n_m} \}_{m = 0}^{\infty } \in {\mathcal {H}}_{n_1}$ be a subsequence of $\{ x_{1n} \}$ and $\{ x_{2n_m} \}_{m = 0}^{\infty } \in {\mathcal {H}}_{n_2}$ be a subsequence of $\{ x_{2n} \}$ such that $T(x_{1n_m}, x_{2n_m}) \mathop {\rightarrow }\limits ^{\text {a.s.}}\tau ^+$ as $m \rightarrow \infty$. The sequence $\{ (T(x_1, x_2), x_{1n_m},$ $x_{2n_m} ) \}_{m = 1}^{\infty }$ converges in ${\mathcal {H}}^{\overline{{\mathbb {R}}}_+}_1 \times {\mathcal {H}}_{n_1} \times {\mathcal {H}}_{n_2}$ to $(T(x_1, x_2), x_1, x_2 )$ almost surely as $m \rightarrow \infty$. Since $s_1(T(x_1, x_2) + t_1, x_1, x_2) \mathop {=}\limits ^{\text {a.s.}} 0$ for all $t_1 \ge 0$ and since all solutions to (1) and (2) are sample continuous in their initial conditions [27, Thm. 7.3.1], it follows that $s_1(T(x_1, x_2), x_{1n_m}, x_{2n_m}) \mathop {\rightarrow }\limits ^{\text {a.s.}} s_1(T(x_1, x_2), x_1, x_2)\mathop {=}\limits ^{\text {a.s.}} 0$ as $m \rightarrow \infty$. Thus, since $T(0, x_2)$ is sample continuous for all $x_2 \in {\mathcal {H}}_{n_2}$, it follows that

$$\begin{aligned} T(s_1(T(x_1, x_2), x_{1n_m}, x_{2n_m}), s_2(T(x_1, x_2), x_{1n_m}, x_{2n_m}))\mathop {\rightarrow }\limits ^{\text {a.s.}} T(0, s_2(T(x_1, x_2), x_1, x_2))\mathop {=}\limits ^{\text {a.s.}} 0.\nonumber \\ \end{aligned}$$

(97)

Now, with $t_1 = T(x_1, x_2)$, $x_1(0) = x_{1n_m}$, and $x_2(0) = x_{2n_m}$, it follows from (11) and (97) that $T(s_1(T(x_1, x_2), x_{1n_m}, x_{2n_m}), s_2(T(x_1, x_2), x_{1n_m}, x_{2n_m})) \mathop {=}\limits ^{\text {a.s.}} \max \, \{ T(x_{1n_m}, x_{2n_m})- T(x_1, x_2), 0 \}$ and $\max \, \{ T(x_{1n_m}, x_{2n_m})- T(x_1, x_2) , 0 \} \mathop {\rightarrow }\limits ^{\text {a.s.}} 0$ as $m \rightarrow \infty$. Thus, $\max \, \{ \tau ^+ - T(x_1, x_2) , 0 \}\mathop {=}\limits ^{\text {a.s.}} 0$, which implies that $\tau ^+ \mathop {\le }\limits ^{\text {a.s.}} T(x_1, x_2)$.

Finally, let $\{ x_{1n_k} \}_{k = 0}^{\infty } \in {\mathcal {H}}_{n_1}$ be a subsequence of $\{ x_{1n} \}$ and $\{ x_{2n_k} \}_{k = 0}^{\infty } \in {\mathcal {H}}_{n_2}$ be a subsequence of $\{ x_{2n} \}$ such that $T(x_{1n_k}, x_{2n_k}) \mathop {\rightarrow }\limits ^{\text {a.s.}} \tau ^-$ as $k \rightarrow \infty$. It follows from $\tau ^- \mathop {\le }\limits ^{\text {a.s.}} \tau ^+$ and $\tau ^+ \mathop {\le }\limits ^{\text {a.s.}} T(x_1, x_2)$ that $\tau ^- \in {\mathcal {H}}^{{\mathbb {R}}_+}_1$, and hence, the sequence $\{ (T(x_{1n_k}, x_{2n_k}), x_{1n_k}, x_{2n_k} ) \}_{k = 1}^{\infty }$ converges pointwise to $(\tau ^-, x_1, x_2 )$ as $k \rightarrow \infty$. Since $s_1(\cdot , \cdot , \cdot )$ is jointly sample continuous, it follows that $s_1(T(x_{1n_k},$ $x_{2n_k}), x_{1n_k}, x_{2n_k} )\mathop {\rightarrow }\limits ^{\text {a.s.}} s_1(\tau ^-, x_1, x_2 )$ as $k \rightarrow \infty$. Now, since $s_1(T(x_1, x_2) + t_1, x_1, x_2) \mathop {=}\limits ^{\text {a.s.}} 0$ for all $t_1 \ge 0$, $s_1(T(x_{1n_k}, x_{2n_k}),$ $x_{1n_k}, x_{2n_k} ) \mathop {=}\limits ^{\text {a.s.}} 0$ for each k. Hence, $s_1(\tau ^-, x_1, x_2 ) \mathop {=}\limits ^{\text {a.s.}} 0$ and, by the definition of the settling-time operator, $T(x_1, x_2) \mathop {\le }\limits ^{\text {a.s.}} \tau ^-$. Now, it follows from $\tau ^- \mathop {\le }\limits ^{\text {a.s.}} \tau ^+$, $\tau ^+ \mathop {\le }\limits ^{\text {a.s.}} T(x_1, x_2)$, and $T(x_1, x_2) \mathop {\le }\limits ^{\text {a.s.}} \tau ^-$ that $\tau ^- \mathop {=}\limits ^{\text {a.s.}} T(x_1, x_2) \mathop {=}\limits ^{\text {a.s.}} \tau ^+$, and hence, $T(x_{1n}, x_{2n}) \mathop {\rightarrow }\limits ^{\text {a.s.}} T(x_1, x_2)$ as $n \rightarrow \infty$, which proves that $T(\cdot , \cdot )$ is jointly sample continuous on ${\mathcal {H}}_{n_1} \times {\mathcal {H}}_{n_2}$. $\square$

Appendix 2

Proof of Theorem 3.1

(i) Let $x_{1} \in {\mathbb {R}}^{n_1}$, $x_{20} \in {\mathbb {R}}^{n_2}$, $\varepsilon > 0$, and $\rho > 0$, and define ${\mathcal {D}}_{\varepsilon ,\rho } \triangleq \{ x_1 \in {\mathcal {B}}_{\varepsilon }(0) {:}\, V(x_1, x_{20}) < \alpha (\varepsilon )\rho \}$. Since $V(\cdot , \cdot )$ is continuous and $V(0, x_2) = 0$, it follows that ${\mathcal {D}}_{\varepsilon ,\rho }$ is nonempty and there exists $\delta =\delta (\varepsilon ,\rho , x_{20}) > 0$ such that $V(x_1, x_{20}) < \alpha (\varepsilon )\rho$, $x_1 \in {\mathcal {B}}_{\delta }(0)$. Hence, ${\mathcal {B}}_{\delta }(0) \subseteq {\mathcal {D}}_{\varepsilon ,\rho }$. Next, it follows from (14) that $V(x_1(t), x_2(t))$ is a (positive) supermartingale [26, Lemma 5.4], and hence, for every $x_1(0) \in {\mathcal {H}}^{{\mathcal {B}}_{\delta }(0)}_{n_1} \subseteq {\mathcal {H}}^{{\mathcal {D}}_{\rho }}_{n_1}$, it follows from (13), with $\alpha (\cdot )\in {\mathcal {K}}_\infty$, and the extended version of the Markov inequality for monotonically increasing functions [38, p. 193] that

$$\begin{aligned} {{\mathbb {P}}}^{x_0}\left( \sup _{t\ge 0}\Vert x_1(t)\Vert \ge \varepsilon \right)&\le \sup _{t\ge 0}\frac{{{\mathbb {E}}}^{x_0}[\alpha (\Vert x_1(t) \Vert ) ]}{\alpha (\varepsilon )} \le \sup _{t\ge 0}\frac{{{\mathbb {E}}}^{x_0}[V(x_1(t),x_2(t))]}{\alpha (\varepsilon )}\\&\le \frac{{{\mathbb {E}}}^{x_0}[V(x_1(0),x_2(0))]}{\alpha (\varepsilon )} \le \rho , \end{aligned}$$

which proves partial Lyapunov stability in probability with respect to $x_1$.

To prove global partial asymptotic stability in probability, it follows from (14) and [39, Corollary 4.2] that $\lim _{t\rightarrow \infty } k(\Vert x_2(t) \Vert ) r(V(x_1(t), x_2(t))\mathop {=}\limits ^{\text {a.s.}} 0.$ Since $k(\Vert x_2(t)\Vert )$ is ${\mathcal {F}}_t$-submartingale, it follows that $\lim _{t\rightarrow \infty }r(V(x_1(t), x_2(t))\mathop {=}\limits ^{\text {a.s.}} 0$, which, since $r{:}\,{{\mathbb {R}}}_+\rightarrow {{\mathbb {R}}}_+$, further implies that $\lim _{t\rightarrow \infty } V(x_1(t), x_2(t))\mathop {=}\limits ^{\text {a.s.}} 0$. Now, it follows from (13) that $\lim _{t\rightarrow \infty }\alpha (\Vert x_1(t) \Vert ) \le \lim _{t\rightarrow \infty }$ $V(x_1(t),x_2(t)) \mathop {=}\limits ^{\text {a.s.}} 0,$ which implies ${{\mathbb {P}}}^{x_0}\left( \lim _{t\rightarrow \infty }\Vert x_1(t) \Vert =0\right) =1$. Hence, $\mathcal{G}$ is globally partially asymptotically stable in probability and the stochastic settling-time operator $T(x_1(0),x_2(0))\le \infty$ almost surely [40].

Next, we show that $T(x_1(0),x_2(0))$ is finite with probability one and satisfies (17), and hence, ${{\mathbb {E}}}^{x_0} \left[ T(x_1(0),x_2(0))\right]$ $< \infty$. Define $T_0\buildrel {\scriptscriptstyle \triangle } \over =T(x_1(0),x_2(0))$, $x(t)\buildrel {\scriptscriptstyle \triangle } \over =[x_1^{\mathrm {T}}(t),x_2^{\mathrm {T}}(t)]^{\mathrm {T}}$, and $\alpha (V)\buildrel {\scriptscriptstyle \triangle } \over =\int _{0}^{V}\frac{{\mathrm{d}}v}{r(v)}$, $V\in \overline{{{\mathbb {R}}}}_+$. Now, using It${\hat{\mathrm{o}}}$’s (chain rule) formula the stochastic differential of V(x(t)) along the system trajectories x(t), $t\ge 0$, is given by

$$\begin{aligned} {\mathrm{d}}V(x(t)) = {\mathcal {L}}V(x(t)){\mathrm{d}}t + \frac{\partial V}{\partial x}D(x(t)){\mathrm{d}}w(t). \end{aligned}$$

Next, using (14) it follows that

$$\begin{aligned} \int ^{T_0}_{0} k(\Vert x_2(\tau )\Vert ){\mathrm{d}}\tau&= \int ^{T_0}_{0} k(\Vert x_2(\tau )\Vert )\frac{r(V(x(\tau )))}{r(V(x(\tau )))}\mathrm{d}\tau \nonumber \\&\le \int ^{T_0}_{0} -\frac{{\mathcal {L}}V(x(\tau ))}{r(V(x(\tau )))}\mathrm{d}\tau \nonumber \\&\le \int ^{T_0}_{0} -\frac{{\mathrm{d}}V(x(t))}{r(V(x(\tau )))} + \int ^{T_0}_{0} \frac{1}{r(V(x(\tau )))}\frac{\partial V}{\partial x}D(x(\tau )){\mathrm{d}}w(\tau )\nonumber \\&= \int ^{T_0}_{0} -\frac{{\mathrm{d}}\alpha (V))}{{\mathrm{d}}V}{\mathrm{d}}V(x(t)) + \int ^{T_0}_{0}\frac{1}{r(V(x(\tau )))} \frac{\partial V}{\partial x}D(x(\tau )){\mathrm{d}}w(\tau ). \end{aligned}$$

(98)

Once again, using Ito’s (chain rule) formula it follows that

$$\begin{aligned}&{\mathrm{d}}\alpha (V(x(t)))=\left[ \frac{\partial \alpha (V(x))}{\partial x} f(x(t)) + \frac{1}{2}{\mathrm{tr}}\ D^{\mathrm {T}}(x)\frac{\partial ^2 \alpha (V(x))}{\partial x^2}D(x) \right] {\mathrm{d}}t\nonumber \\&\qquad + \frac{\partial \alpha (V(x))}{\partial x}{\mathrm{d}}w(t)\nonumber \\&\quad =\left[ \frac{{\mathrm{d}}\alpha (V))}{{\mathrm{d}}V}\frac{\partial V(x)}{\partial x} f(x(t))+ \frac{1}{2}{\mathrm{tr}}\ D^{\mathrm {T}}(x)\frac{\partial }{\partial x}\left( \frac{{\mathrm{d}}\alpha (V)}{\mathrm{d}V}\frac{\partial V(x)}{\partial x}\right) D(x) \right] {\mathrm{d}}t \nonumber \\&\qquad + \frac{{\mathrm{d}}\alpha (V))}{{\mathrm{d}}V}\frac{\partial V(x)}{\partial x}{\mathrm{d}}w(t)\nonumber \\&\quad = \frac{{\mathrm{d}}\alpha (V))}{{\mathrm{d}}V}\left[ \left( \frac{\partial V(x)}{\partial x} f(x(t))+ \frac{1}{2}{\mathrm{tr}}\ D^{\mathrm {T}}(x)\frac{\partial ^2 (V(x))}{\partial x^2}D(x)\right) {\mathrm{d}}t + \frac{\partial V(x)}{\partial x}{\mathrm{d}}w(t) \right] \nonumber \\&\qquad + \frac{1}{2}{\mathrm{tr}}\ D^{\mathrm {T}}(x)\left( \frac{\partial V(x)}{\partial x}\right) ^{\mathrm {T}}\frac{\mathrm{d^2}\alpha (V)}{{\mathrm{d}}V^2}\left( \frac{\partial V(x)}{\partial x}\right) D(x){\mathrm{d}}t \nonumber \\&\quad = \frac{{\mathrm{d}}\alpha (V))}{{\mathrm{d}}V} {\mathrm{d}}V(x(t)) + \frac{1}{2}{\mathrm{tr}}\ D^{\mathrm {T}}(x)\left( \frac{\partial V(x)}{\partial x}\right) ^{\mathrm {T}}\frac{{\mathrm{d^2}}\alpha (V)}{\mathrm{d}V^2}\left( \frac{\partial V(x)}{\partial x}\right) D(x){\mathrm{d}}t. \end{aligned}$$

(99)

Hence, it follows from (98) and (16) that

$$\begin{aligned}&\int ^{T_0}_{0} k( \Vert x_2(\tau )\Vert ){\mathrm{d}}\tau \le \int ^{T_0}_{0} -{\mathrm{d}}\alpha (V(x(\tau ))) + \int ^{T_0}_{0} \frac{1}{r(V(x(\tau )))}\frac{\partial V}{\partial x}D(x(\tau ))\mathrm{d}w(\tau ) \nonumber \\&\quad \qquad + \int ^{T_0}_{0} \frac{1}{2}{\mathrm{tr}}~ D^{\mathrm {T}}(x)\left( \frac{\partial V(x)}{\partial x}\right) ^{\mathrm {T}}\frac{{\mathrm{d^2}}\alpha (V)}{\mathrm{d}V^2}\left( \frac{\partial V(x)}{\partial x}\right) D(x){\mathrm{d}}\tau \nonumber \\&\qquad = \alpha (V(x(0))) - \alpha (V(x(T_0))) + \int ^{T_0}_{0} \frac{1}{r(V(x(\tau )))}\frac{\partial V}{\partial x}D(x(\tau ))\mathrm{d}w(\tau ) \nonumber \\&\quad \qquad - \int ^{T_0}_{0} \frac{r'(V)}{r^2(V)}\frac{1}{2}{\mathrm{tr}} \left( \frac{\partial V(x)}{\partial x}D^{\mathrm {T}}(x) \right) ^{\mathrm {T}}\left( \frac{\partial V(x)}{\partial x}D(x)\right) {\mathrm{d}}\tau \nonumber \\&\qquad \le \int _{0}^{V(x(0))}\frac{{\mathrm{d}}v}{r(v)} - \int _{0}^{V(x(T_0))}\frac{{\mathrm{d}}v}{r(v)} + \int ^{T_0}_{0}\frac{1}{r(V(x(\tau )))} \frac{\partial V}{\partial x}D(x(\tau )){\mathrm{d}}w(\tau ). \end{aligned}$$

(100)

Taking the expectation on both sides of (100) and using the fact that $x(0)\mathop {=}\limits ^{\text {a.s.}} x_0$ and $x(T_0)\mathop {=}\limits ^{\text {a.s.}} 0$ yields

$$\begin{aligned} {{\mathbb {E}}}^{x_0} \left[ \int ^{T_0}_{0} k( \Vert x_2(\tau )\Vert )\mathrm{d}\tau \right] \le \int _{0}^{V(x_0)}\frac{{\mathrm{d}}v}{r(v)}. \end{aligned}$$

(101)

Next, since $q {:}\, [0, \infty ) \rightarrow {\mathbb {R}}$ is continuously differentiable and satisfies (18), and, by assumption, the process $k(\Vert x_2(t))\Vert )$ is a positive ${\mathcal {F}}_t$-submartingale, it follows that $q(\cdot )$ is convex, monotonically increasing, and invertible. Hence, applying Jensen’s inequality [38, p. 109], Fubini’s theorem [41, p. 410], and the law of iterated expectation on the random variable $q(T(x_1(0),x_2(0)))$ yields

$$\begin{aligned} {{\mathbb {E}}}^{x_0}\left[ T(x_1(0),x_2(0))\right]= & {} q^{-1}\left( q\left( {{\mathbb {E}}}^{x_0}\left[ T(x_1(0),x_2(0))\right] \right) \right) \nonumber \\\le & {} q^{-1}\left( {{\mathbb {E}}}^{x_0}\left[ q(T(x_1(0),x_2(0)))\right] \right) \nonumber \\= & {} q^{-1}\left( {{\mathbb {E}}}^{x_0} \left[ \int ^{T_0}_{0} {{\mathbb {E}}}^{x_0}\left[ k(\Vert x_2(\tau )\Vert )\right] {{\mathrm{d}}}\tau \right] \right) \nonumber \\= & {} q^{-1}\left( {{\mathbb {E}}}^{x_0} \left[ {{\mathbb {E}}}^{x_0}\left[ \int ^{T_0}_{0} k(\Vert x_2(\tau )\Vert ){\mathrm{d}}\tau |T_0)\right] \right] \right) \nonumber \\= & {} q^{-1}\left( {{\mathbb {E}}}^{x_0} \left[ \int ^{T_0}_{0} k(\Vert x_2(\tau )\Vert ){\mathrm{d}}\tau \right] \right) \nonumber \\\le & {} q^{-1}\left( \int _{0}^{V(x_0)}\frac{{{\mathrm{d}}}v}{r(v)}\right) , \end{aligned}$$

(102)

which shows that $T(x_1(0),x_2(0))$ is finite with probability one. Moreover, it follows from the stochastic finite-time stability of ${\mathcal {G}}$ with respect to $x_1$ and Proposition 3.1 that $T(\cdot , \cdot )$ can be extended to ${\mathcal {H}}_1^{\overline{{\mathbb {R}}}_+}$ and $T(0, x_{20}) \mathop {=}\limits ^{\text {a.s.}} 0$.

(ii) Let $\rho > 0$, $x_{10} \in {\mathbb {R}}^{n_1}$, and $x_{20} \in {\mathbb {R}}^{n_2}$. Since $\alpha (\cdot )$ and $\beta (\cdot )$ are class ${\mathcal {K}}_{\infty }$ functions, it follows that, for every $\varepsilon > 0$, there exists $\delta =\delta (\varepsilon ,\rho ) > 0$ such that $\beta (\delta ) \le \alpha (\varepsilon )\rho$. Now, (14) implies that $V(x_1(t), x_2(t))$ is a (positive) supermartingale, and hence, it follows from (13) and (19) that, for all $(x_1(0), x_2(0)) \in {\mathcal {H}}^{{\mathcal {B}}_{\delta }(0)}_{n_1} \times {\mathcal {H}}_{n_2}$,

$$\begin{aligned} {{\mathbb {P}}}^{x_0}\left( \sup _{t\ge 0}\Vert x_1(t)\Vert \ge \varepsilon \right)&\le \sup _{t\ge 0}\frac{{{\mathbb {E}}}^{x_0}[\alpha (\Vert x_1(t) \Vert ) ]}{\alpha (\varepsilon )}\le \sup _{t\ge 0}\frac{{{\mathbb {E}}}^{x_0}[V(x_1(t),x_2(t))]}{\alpha (\varepsilon )}\\&\le \frac{{{\mathbb {E}}}^{x_0}[V(x_1(0),x_2(0))]}{\alpha (\varepsilon )} \le \frac{\beta (\delta )}{\alpha (\varepsilon )} \le \rho . \end{aligned}$$

Hence, for every $x_1(0) \in {\mathcal {H}}^{{\mathcal {B}}_{\delta }(0)}_{n_1}$, $x_1(t) \in {\mathcal {H}}^{{\mathcal {B}}_{\varepsilon }(0)}_{n_1}$, $t \ge 0$, which proves uniform Lyapunov stability in probability with respect to $x_1$. Stochastic finite-time partial convergence follows as in the proof of (i), implying global stochastic finite-time stability of ${\mathcal {G}}$ with respect to $x_1$ uniformly in $x_{20}$. In addition, the existence of a stochastic settling-time operator $T {:}\, {\mathcal {H}}_{n_1} \times {\mathcal {H}}_{n_2} \rightarrow {\mathcal {H}}_1^{[0, \infty )}$ that verifies (17) follows as in the proof of (i).

(iii) Global uniform stochastic finite-time stability of ${\mathcal {G}}$ with respect to $x_1$ directly follows from (ii). Now, using similar arguments as in the proof of (i), $q^{-1}(\cdot )=\frac{1}{k}(\cdot )$ directly follows from (18) with $k(\Vert x_2 \Vert ) = k \in {\mathbb {R}}_+$, $x_2 \in {\mathbb {R}}^{n_2}$. Now, the existence a stochastic settling-time operator $T {:}\, {\mathcal {H}}_{n_1} \times {\mathcal {H}}_{n_2} \rightarrow {\mathcal {H}}_1^{ [0, \infty )}$ such that (20) holds follows as in the proof of (i) and (17). Since ${{\mathbb {E}}}^{x_0}[T(x_1(0),x_2(0)])]$ is not a function of $x(t), t\ge 0$, strong stochastic finite-time convergence of ${\mathcal {G}}$ with respect to $x_1$ uniformly in $x_{20}$ is immediate. Hence, the nonlinear stochastic dynamical system ${\mathcal {G}}$ is globally strongly stochastic finite-time stable with respect to $x_1$ uniformly in $x_{20}$. $\square$