Fast Continuous Dynamics Inside the Graph of Subdifferentials of Nonsmooth Convex Functions

Abstract

In a Hilbert framework, we introduce a new class of second-order dynamical systems that combine viscous and geometric damping but also a time rescaling process for nonsmooth convex minimization. A main feature of these systems is to produce trajectories that lie in the graph of the Fenchel subdifferential of the objective. Moreover, they do not incorporate any regularization or smoothing processes. This new class originates from some combination of a continuous Nesterov-like dynamic and the Minty representation of subdifferentials. These models are investigated through first-order reformulations that amount to dynamics involving three variables: two solution trajectories (including an auxiliary one) and another one associated with subgradients. We prove the weak convergence towards equilibria for the solution trajectories, as well as properties of fast convergence to zero for their velocities. Remarkable convergence rates (possibly of exponential-type) are also established for the function values. We additionally state notable properties of fast convergence to zero for the subgradients trajectory and for its velocity. Some numerical experiments are performed so as to illustrate the efficiency of our approach. The proposed models offer a new and well-adapted framework for discrete counterparts, especially for structured minimization problems. Inertial algorithms with a correction term are then suggested relative to this latter context.

Appendix

1.1 Proof of Proposition 2.1

Let us prove that (i1) $\Rightarrow$ (i2). Consider a solution $(x,\xi ) \in {\mathcal {A}}_c\times {\mathcal {A}}_c$ to (2.2). Set $\zeta ( \cdot )=\sigma ( \cdot ) \xi ( \cdot )$ and suppose that $x( \cdot )+ \zeta ( \cdot )$ is of class $C^1$ and that $\left( x( \cdot )+ \zeta ( \cdot ) \right) ^{(1)} \in {\mathcal {A}}_c$. Clearly, for $t \ge 0$, as $\dot{x}$, ${\dot{\zeta }}$ and $\zeta$ are integrable on [0, t] (since x and $\zeta$ belong to ${\mathcal {A}}_c$), we can set as a well-defined quantity

$$\begin{aligned} z(t)=\int _0^t \left( \alpha (r)\dot{x}(r)+\beta (r) {\dot{\zeta }}(r)+b(r) \zeta (r)\right) dr- q_0. \end{aligned}$$

(1.9)

Hence, $z \in {\mathcal {A}}_c$, and by differentiating (1.9) we obtain

$$\begin{aligned} {\dot{z}(t)=\alpha (t)\dot{x}(t)+\beta (t) {\dot{\zeta }}(t)+b(t) \zeta (t), \text {for a.e. } t \ge 0}. \end{aligned}$$

(1.10)

Therefore, by (2.2b) together with the above equality, we get

$$\begin{aligned} {\big (x( \cdot ) + \zeta ( \cdot ) \big )^{(2)}(t) + \dot{z}(t)=0, \text {for a.e. } t\ge 0}. \end{aligned}$$

(1.11)

Moreover, recalling that $\big (x( \cdot ) + \zeta ( \cdot ) \big )^{(1)} \in {\mathcal {A}}_c$ and $z \in {\mathcal {A}}_c$, we get

$\frac{d}{dt} \left( \big (x( \cdot ) + \zeta ( \cdot ) \big )^{(1)} + z( \cdot ) \right) (t) = \big (x( \cdot ) + \zeta ( \cdot ) \big )^{(2)}(t)+ \dot{z}(t)$, for a.e. $t \ge 0$. Hence, we straightforwardly deduce

$$\begin{aligned} {\frac{d}{dt}\left( \big (x( \cdot ) + \zeta ( \cdot ) \big )^{(1)}(t) + z( \cdot ) \right) (t)=0, \text {for a.e. } t \ge 0}. \end{aligned}$$

(1.12)

It follows immediately that

$$\begin{aligned} {\big ( x( \cdot )+ \zeta ( \cdot ) \big )^{(1)}(t) + z(t) = \big ( x( \cdot )+ \zeta ( \cdot ) \big )^{(1)}(0) + z(0) \text { for } t \ge 0}, \end{aligned}$$

(1.13)

which, by the initial condition $\big ( x( \cdot ) + \sigma ( \cdot )\xi ( \cdot )\big )^{(1)}(0) = q_0$, yields

$$\begin{aligned} {\big ( x( \cdot )+ \zeta ( \cdot ) \big )^{(1)}(t) + z(t) = 0 \,\text {for}\, t \ge 0}, \end{aligned}$$

(1.14)

which readily implies that

$$\begin{aligned} {\dot{x}(t)+ {\dot{\zeta }}(t) + z(t) = 0\,\text {for a.e}\,t \ge 0}. \end{aligned}$$

(1.15)

Multiplying (1.15) by $\beta (t)$ and adding the resulting equality to (1.10) give us

$$\begin{aligned} {\big (\beta (t)-\alpha (t)\big )\dot{x}(t) +\dot{z}(t)+\beta (t)z(t)-b(t)\zeta (t) = 0\text { for a.e }t \ge 0}. \end{aligned}$$

(1.16)

Now, since $\theta ( \cdot )$ is positive, we introduce the function $y( \cdot )$ defined for $t \ge 0$ by

$$\begin{aligned} y(t):=\frac{1}{\theta (t)}z(t)+x(t)-\frac{\omega (t)}{\theta (t)}\zeta (t). \end{aligned}$$

(1.17)

For simplification we also set $u(t)=y(t)-x(t)$. Observe from (1.17) that we equivalently have

$$\begin{aligned} {z(t)= \theta (t) u(t) +\omega (t) \zeta (t) }. \end{aligned}$$

(1.18)

Thus, for $t \ge 0$, (1.14) in light of the above equality entails

$$\begin{aligned} {\big (x( \cdot )+ \zeta ( \cdot )\big )^{(1)}(t) + \theta (t) u(t) + \omega (t) \zeta (t) =0,} \end{aligned}$$

(1.19)

that is (2.4b). We now prove (2.4c). Differentiating (1.18), while noticing that $\{x, \xi , u\} \subset {\mathcal {A}}_c$, readily implies, for a.e. $t\ge 0$,

$$\begin{aligned} {\dot{z}(t)= \dot{\theta }(t) u(t) + \theta (t) \dot{u}(t) + {\dot{\omega }} (t) \zeta (t) + \omega (t) {\dot{\zeta }} (t)}. \end{aligned}$$

(1.20)

Moreover, using the definitions of $\alpha ( \cdot )$, $\beta ( \cdot )$ and $b( \cdot )$ given by (1.3), namely $\alpha (t) =-\frac{\dot{\theta }(t) }{\theta (t)} + \kappa -\theta (t)$, $\beta (t) = -\frac{{\dot{\theta }}(t) }{\theta (t)} + \kappa +\omega (t)$ and $b(t)= \omega (t) \left( \kappa +\frac{{\dot{\omega }} (t)}{\omega (t)}-\frac{\dot{\theta }(t)}{\theta (t)} \right)$, yields

$$\begin{aligned}{} & {} {\beta (t)-\alpha (t)-\theta (t)} = {\omega (t)}, \end{aligned}$$

(1.21a)

$$\begin{aligned}{} & {} {\beta (t) \theta (t) + \dot{\theta }(t)} = {\kappa \theta (t) + \omega (t) \theta (t)}, \end{aligned}$$

(1.21b)

$$\begin{aligned}{} & {} {\beta (t) + \frac{\dot{\omega }(t)}{\omega (t)}-\frac{b(t)}{\omega (t)}} = {\omega (t)}. \end{aligned}$$

(1.21c)

Hence, by (1.16) and using (1.20), (1.18), $\dot{u}=\dot{y} - \dot{x}$ and (1.21a), successively, we get, for a.e. $t\ge 0$,

$$\begin{aligned} \begin{array}{l} 0=\big (\beta (t)-\alpha (t)\big )\dot{x}(t) +\dot{z}(t)+\beta (t) z(t)-b(t) \zeta (t) \\ \quad =\big (\beta (t)-\alpha (t)\big )\dot{x}(t) + \dot{\theta }(t) u(t) + \theta (t) \dot{u}(t) + {\dot{\omega }} (t) \zeta (t) + \omega (t) {\dot{\zeta }} (t)\\ \qquad +\beta (t) \theta (t) u(t) + \beta (t) \omega (t) \zeta (t) -b(t) \zeta (t) \\ \quad =\big ( \beta (t)-\alpha (t)-\theta (t) \big ) \dot{x}(t) + \theta (t)\dot{y}(t) + \big (\beta (t)\theta (t)+\dot{\theta }(t) \big ) u(t) \\ \qquad + \omega (t) \big ( {\dot{\zeta }}(t) + \left( \frac{\dot{\omega }(t)}{\omega (t)} + \beta (t) - \frac{b(t) }{\omega (t)} \right) \zeta (t) \big ), \end{array} \end{aligned}$$

(1.22)

namely

$$\begin{aligned}\begin{array}{l} 0= \omega (t) \dot{x}(t) + \theta (t)\dot{y}(t) + \big ( \kappa \theta (t) + \omega (t) \theta (t) \big ) u(t) + \omega (t) \big ( {\dot{\zeta }}(t) + \omega (t) \zeta (t) \big ) \\ \quad = \omega (t) \big (\dot{x}(t) + {\dot{\zeta }} (t) +\theta (t) u(t) + \omega (t)\zeta (t) \big ) + \theta (t) \big ( \dot{y}(t) + \kappa u(t) \big ). \end{array} \end{aligned}$$

In addition, by (1.19), while recalling that $\{x,\xi \} \subset {\mathcal {A}}_c$, we obtain

$$\begin{aligned} {\dot{x}(t)+ {\dot{\zeta }} (t) + \theta (t) u(t) + \omega (t) \zeta (t) =0, \text{ for } \text{ a.e. } t\ge 0}. \end{aligned}$$

(1.23)

Thus, combining (1.22) and (1.23), in light of $\theta \ne 0$, yields

$\dot{y}(t)+\kappa (u(t))=0$ for a.e. $t\ge 0$,

that is (2.4c).

Finally, regarding the initial conditions, we have $(x(0),\xi (0))= (x_0,\xi _0)$ and $\big (x( \cdot ) + \zeta ( \cdot )\big )(0)= q_0$ (according to (i1)) while (2.4b) at time $t=0$ ensures that

$\big ( x( \cdot ) + \zeta ( \cdot ) \big )^{(1)} (0) + \theta (0) \left( y(0)-x(0) \right) + \omega (0) \zeta (0) =0$.

Hence, we deduce that $y(0) =x_0-\frac{1}{\theta (0)} \left( q_0 + \sigma (0) \omega (0) \xi _0 \right)$.

Let us prove that (i2) $\Rightarrow$ (i1). Consider a solution $(x,\xi ,y) \in {\mathcal {A}}_c\times {\mathcal {A}}_c\times C^1$ to (2.4). For simplification, we set again $u( \cdot )=y( \cdot )-x( \cdot )$ and $\zeta ( \cdot )=\sigma ( \cdot ) \xi ( \cdot )$. Clearly, by (2.4b), we have, for $t \ge 0$,

$$\begin{aligned} \big ( x( \cdot ) + \zeta ( \cdot ) \big )^{(1)} (t) + \theta (t) u(t) + \omega (t) \zeta (t)=0. \end{aligned}$$

(1.24)

This, by $(x,\xi ,y) \in {\mathcal {A}}_c\times {\mathcal {A}}_c\times C^1$ and by $\{\omega ( \cdot )$, $\theta ( \cdot )$, $\sigma ( \cdot )\} \subset C^1([0, \infty ])$, entails that $x( \cdot )+\zeta ( \cdot )$ is of class $C^1$ and that $\big ( x( \cdot ) + \sigma ( \cdot )\xi ( \cdot ) \big )^{(1)} \in {\mathcal {A}}_c$. Then, differentiating (1.24) gives us, for a.e $t \ge 0$,

$$\begin{aligned} {(x( \cdot ) + \zeta ( \cdot ))^{(2)}(t) + {\dot{\theta }}(t) u(t) + \theta (t) \dot{u}(t) + \omega (t) {\dot{\zeta }}(t) + {\dot{\omega }}(t) \zeta (t) = 0}, \end{aligned}$$

(1.25)

while we know from (2.4c) that $\dot{y}(t)= -\kappa u(t)$. Consequently, we readily obtain, for a.e $t \ge 0$,

$$\begin{aligned} (x( \cdot ) + \zeta ( \cdot ))^{(2)}(t) + {\dot{\theta }}(t) u(t) + \theta (t)\big ( -\kappa u(t) - \dot{x}(t) \big ) + \omega (t) \dot{\zeta }(t) + {\dot{\omega }}(t) \zeta (t) = 0.\nonumber \\ \end{aligned}$$

(1.26)

Furthermore, for a.e. $t \ge 0$, by (1.24) we readily have

$u(t)=-\frac{1}{\theta (t)} \big ( ( x( \cdot )+ \zeta ( \cdot )) ^{(1)} + \omega (t) \zeta (t) \big )$, which, by (1.26), entails

$$\begin{aligned} \begin{array}{l} (x( \cdot ) + \zeta ( \cdot ))^{(2)}(t) \\ = \frac{{\dot{\theta }}(t)}{\theta (t)} \big (\dot{x}(t)+ {\dot{\zeta }}(t) + \omega (t) \zeta (t)\big ) - \theta (t) \left( \frac{\kappa }{\theta (t)}\big (\dot{x}(t)+ {\dot{\zeta }}(t) + \omega (t) \zeta (t) \big )-\dot{x}(t) \right) \\ \quad - \omega (t) {\dot{\zeta }}(t) - \dot{\omega }(t) \zeta (t) \\ =-\big (\kappa -\theta (t)-\frac{\dot{\theta }(t)}{\theta (t)} \big )\dot{x}(t) - \left( \kappa + \omega (t)-\frac{\dot{\theta }(t)}{\theta (t)} \right) \dot{\zeta }(t) - \omega (t) \left( \kappa + \frac{{\dot{\omega }}(t)}{\omega (t)} - \frac{{\dot{\theta }}(t)}{\theta (t)}\right) \zeta (t). \end{array} \end{aligned}$$

Hence the expressions of $\alpha ( \cdot ), \beta ( \cdot )$ and $b( \cdot )$ defined in (1.3) amounts to (2.2b). In addition, from the initial conditions in (i2), we have $~x(0)=x_0$, $~\xi (0)=\xi _0$ and $y(0) =x_0-\frac{1}{\theta (0)} \left( q_0 + \sigma (0) \omega (0) \xi _0 \right)$. This implies that $y(0) =x(0)-\frac{1}{\theta (0)} \left( q_0 + \sigma (0) \omega (0) \xi (0) \right)$, while (2.4b) at time $t=0$ yields

$\big ( x( \cdot ) + \zeta ( \cdot ) \big )^{(1)} (0) + \theta (0) \left( y(0)-x(0) \right) + \omega (0) \zeta (0) =0$.

Hence, regarding the last two equalities, substituting the former in the latter gives us $\big ( x( \cdot ) + \zeta ( \cdot )\big )^{(1)}(0) = q_0$.

1.2 Proof of Proposition 2.2

1.2.1 The Yosida Regularization

Some useful properties of the Yosida regularization are recalled through the lemma below established in [27] (see also [8, 21, 22]).

Lemma 1.1

Let $A: {\mathcal {H}}\rightrightarrows {\mathcal {H}}$ be a maximally monotone operator such that $S:=A^{-1}(0) \ne \emptyset$. Let $\gamma ,\delta > 0$ and $x, y \in {\mathcal {H}}$. Then for $z \in A^{-1}(\{0\})$, we have

$$\begin{aligned} {\Vert \gamma A_{\gamma }x-\delta A_{\delta }y \Vert \le 2\Vert x-y\Vert + 2\frac{|\gamma - \delta |}{\gamma }\Vert x-z\Vert }, \end{aligned}$$

(1.27)

$$\begin{aligned} {\Vert A_{\gamma }x- A_{\delta }y \Vert \le \left( 3\frac{| \delta -\gamma |}{\delta \gamma } \times \Vert x-z\Vert + \frac{2}{\delta }\Vert x-y\Vert \right) }. \end{aligned}$$

(1.28)

Proof

The proof of (1.27) can be found in [8]. For proving (1.28), we simply observe that $\begin{array}{l} A_{\gamma }x- A_{\delta }y= \frac{1}{ \delta } \left( \delta A_{\gamma }x- \delta A_{\delta }y\right) = \frac{1}{ \delta } \left( ( \delta -\gamma ) A_{\gamma }x + ( \gamma A_{\gamma }x -\delta A_{\delta }y\right) , \end{array}$ from which we get $\begin{array}{l} \Vert A_{\gamma }x- A_{\delta }y \Vert \le \frac{1}{ \delta } \left( | \delta -\gamma | \times \Vert A_{\gamma }x \Vert + \Vert \gamma A_{\gamma }x -\delta A_{\delta }y\Vert \right) . \end{array}$

Consequently, by $\Vert A_{\gamma }x\Vert \le \frac{1}{\gamma } \Vert x-z\Vert$ and using (1.27), we obtain (1.28) $\square$

1.2.2 Main Proof of the Proposition

The proof follows the same lines as in [30] (see, also, [1, 2]), but it is developed through the following steps (s1)- (s3) with full details:

(s1) We begin by reformulating the (possibly) existing strong global solutions to (1.4)) (that are supposed to satisfy (2.3)) by means of the Minty representation of the maximal monotone operator $\partial f$ (see [32]). Set $J_{ \sigma ( \cdot )}^{ \partial f}=\big (I+\sigma ( \cdot ) \partial f\big )^{-1}$ and $(\partial f)_{\sigma ( \cdot ) }= \frac{1}{\sigma ( \cdot )} \big (I-J_{\sigma ( \cdot ) } ^{ \partial f}\big )$, namely the resolvent and the Yosida approximation of $\partial f$ (with index $\sigma ( \cdot )$), respectively, which are well-known to be single-valued and everywhere defined. Associated with any strong global solution $(x( \cdot ),\xi ( \cdot ),y( \cdot ))$ to (1.4), we introduce the new unknown function

$$\begin{aligned} v( \cdot )=x( \cdot )+ \sigma ( \cdot ) \xi ( \cdot ). \end{aligned}$$

(1.29)

It is readily seen that $v( \cdot )$ belongs to ${\mathcal {A}}_c$ (the set of absolutely continuous functions) and that

$v(0)= x_0+ \sigma (0) \xi _0$.

Moreover, for $t \ge 0$, by $\xi (t) \in \partial f (x(t))$ we obtain $v(t) \in x(t)+ \sigma (t) \partial f(x(t))$ and $\xi (t)=\frac{1}{\sigma (t)}(v(t)-x(t))$, hence, by Minty’s representation we simply have

$$\begin{aligned} {x(t)=J_{\sigma (t)}^{ \partial f} v(t) \text{ and } \xi (t)= \frac{1}{\sigma (t)} \big (v(t)-J_{\sigma (t)}^{ \partial f}v(t)\big )=(\partial f)_{\sigma (t)} v(t)}. \end{aligned}$$

(1.30)

Differentiating (1.29), in light of (2.3b), gives us, for a.e. $t \ge 0$,

$\dot{v}(t)=\dot{x}(t)+ (\sigma ( \cdot ) \xi ( \cdot ))^{(1)}(t) = -\theta (t) (y(t)-x(t))-\omega (t) \sigma (t) \xi (t)$, hence, by (1.30), we obtain

$$\begin{aligned} \dot{v}(t)+\theta (t) \big (y(t)-J_{\sigma (t)}^{ \partial f} v(t)\big )+\omega (t) \sigma (t) (\partial f )_{\sigma (t)} v(t) =0. \end{aligned}$$

Hence, from (2.3), we deduce that $(v( \cdot ),y( \cdot ))$ are implicitly given, for a.e. $t \in [0,\infty )$, by

$$\begin{aligned}{} & {} {\dot{v}(t) + \theta (t) \left( y(t)-J_{\sigma (t)}^{ \partial f}v(t) \right) + \omega (t) \sigma (t) (\partial f)_{\sigma (t)} v(t)=0}, \end{aligned}$$

(1.31a)

$$\begin{aligned}{} & {} \dot{y}(t)+ \kappa (y(t)- J_{\sigma (t)}^{\partial f }v(t))=0, \end{aligned}$$

(1.31b)

together with $y(0)=y_0$ and $v(0)=x_0+ \sigma (0) \xi _0$.

This shows us that any strong global solution $(x( \cdot ),\xi ( \cdot ),y( \cdot ))$ to (1.4) is entirely determined (thanks to the two formulas in (1.30)) by some (strong) solution $(v( \cdot ),y( \cdot ))$ to (1.31). So, for proving existence and uniqueness of a strong global solution to (1.4), we just state (as argued below) the existence and uniqueness of a (strong) global solution $(v( \cdot ),y( \cdot ))$ to (1.31), but also the existence of a strong global solution $(x( \cdot ),\xi ( \cdot ),y( \cdot ))$ to (1.4).

(s2) Existence, uniqueness and regularity of a (strong) global solution $(v( \cdot ),y( \cdot ))$ to (1.31). First, we show that (1.31) is relevant to the Cauchy–Lipschitz theorem. Indeed, (1.31) can be expressed as

$$\begin{aligned} \dot{U}(t)=F(t,U(t)), \end{aligned}$$

(1.32)

where $U( \cdot )=(v( \cdot ),y( \cdot ))$ and $F(t,.): {\mathcal {H}}^2 \rightarrow {\mathcal {H}}^2$ is defined for any $t \ge 0$ and $({\bar{v}}, {\bar{y}}) \in {\mathcal {H}}^2$ by $F(t,({\bar{v}},{\bar{y}}))= \big ( \phi _1(t,({\bar{v}},{\bar{y}})), \phi _2(t,({\bar{v}},{\bar{y}})) \big )$, together with

$$\begin{aligned}{} & {} { \phi _1(t,({\bar{v}},{\bar{y}})) = -\theta (t) \left( \bar{y}-J_{\sigma (t)}^{\partial f }{\bar{v}} \right) -\omega (t) \sigma (t) (\partial f)_{\sigma (t)}{\bar{v}}}, \end{aligned}$$

(1.33a)

$$\begin{aligned}{} & {} { \phi _2(t,({\bar{v}},{\bar{y}})) = -\kappa \left( {\bar{y}}- J_{\sigma (t)}^{\partial f }{\bar{v}} \right) .} \end{aligned}$$

(1.33b)

In view of applying the global Cauchy–Lipschitz theorem, we establish two main properties on F(., .) through the following items (a) and (b):

(a) Given $({\bar{v}},{\bar{y}}) \in {\mathcal {H}}^2$, we prove that $F(., (\bar{v},{\bar{y}}))$ is continuous on $[0,\infty )$. Indeed, let $z \in (\partial f)^{-1}(0)$ and $(t_1,t_2) \in [0,\infty )^2$. By Lemma 1.1 with $A= \partial f$ and $\sigma ( \cdot ) \ge \sigma _0 >0$ (from (1.5b)), we obtain $\Vert (\partial f)_{\sigma (t_1) }{\bar{y}}- (\partial f)_{\sigma (t_2) }{\bar{y}} \Vert \le 3 \frac{ | \sigma (t_1) - \sigma (t_2) |}{\sigma _0^2}\Vert {\bar{y}}-z\Vert$.

Then, the continuity of $\sigma ( \cdot )$ on $[0,\infty )$ yields that the mappings $t \rightarrow (\partial f)_{\sigma (t) }{\bar{y}}$ and $t \rightarrow J_{\sigma (t)}^{\partial f}{\bar{v}}$ (given by $J_{\sigma (t)}^{\partial f}{\bar{v}}:={\bar{v}}-\sigma (t) (\partial f)_{\sigma (t)}{\bar{v}}$) are also continuous on $[0,\infty )$. So, in light of the definition of $\phi _1(.,.)$ and $\phi _2(.,.)$, together with the continuity of $\{ \theta ( \cdot ), \omega ( \cdot ) \}$, we infer that $F(.,(\bar{v},{\bar{y}}))$ is continuous on $[0,\infty )$ (as are $\phi _1(.,.)$ and $\phi _2(.,.)$).

(b) Given $t \ge 0$, we prove that F(t, .) is $\iota (t)$-Lipschitz continuous on ${\mathcal {H}}^2$, for some continuous mapping $\iota : [0,\infty ) \rightarrow [0,\infty )$. Indeed, for $(v_i,y_i)\in {\mathcal {H}}^2$ (for $i=1,2$), while noticing that $J_{\sigma (t)}^{\partial f}$ and $\frac{1}{2}\sigma (t) (\partial f)_{\sigma (t)}$ are nonexpansive on  ${\mathcal {H}}$, by (1.33) we get

$$\begin{aligned} \begin{array}{l} \Vert \phi _1(t,(v_1,y_1)) - \phi _1(t,(v_2,y_2)) \Vert \\ \le \theta (t)\Vert y_1-y_2\Vert +\theta (t)\Vert J_{\sigma (t)}^{\partial f }v_1-J_{\sigma (t)}^{\partial f }v_2\Vert + \omega (t) \Vert {\sigma (t)} (\partial f)_{\sigma (t)}v_1-{\sigma (t)} (\partial f)_{\sigma (t)}v_2\Vert \\ \le (\theta (t)+2 \omega (t))(\Vert v_1-v_2\Vert +\Vert y_1-y_2\Vert ), \end{array} \end{aligned}$$

while an easy computation gives us

$$\begin{aligned} \begin{array}{l} \Vert \phi _2(t,v_1,y_1)-\phi _2(t,v_2,y_2)\Vert \le \kappa \left( \Vert v_1-v_2\Vert + \Vert y_1-y_2\Vert \right) .\end{array} \end{aligned}$$

It follows from the previous arguments that F(t, .) satisfies

$$\begin{aligned} \Vert F(t,(v_1,y_1))-F(t,(v_2,y_2))\Vert \le \big (\theta (t)+2 \omega (t)+ \kappa \big )\Vert (v_1,y_1)-(v_2,y_2)\Vert ,\nonumber \\ \end{aligned}$$

(1.34)

hence F(t., .) is $\iota (t)$-Lipschitz continuous on ${\mathcal {H}}^2$ along with $\iota ( \cdot )=\theta ( \cdot )+2 \omega ( \cdot )+ \kappa$ which is continuous (by the continuity of $\theta ( \cdot )$ and $\omega ( \cdot )$).

Thus, for any given $(x_0,y_0,\xi _0)\in {\mathcal {H}}^3$, applying the global Cauchy–Lipschitz theorem yields existence and uniqueness of a global classical solution $(v( \cdot ),y( \cdot ))$ to (1.31) (namely, $y( \cdot )$ and $v( \cdot )$ are of class $C^1$) such that $y(0)=y_0$ and $v(0)=x_0+ \sigma (0) \xi _0$. Furthermore, the previous arguments (a) and (b) guarantee existence and uniqueness of a strong global solution $(v( \cdot ),y( \cdot ))$ to the same problem (1.31), by invoking the version of the Cauchy–Lipschitz theorem involving absolutely continuous trajectories, see for example [25, Proposition 6.2.1.], [37, Theorem 54].

(s3) Existence of a strong global solution $(x( \cdot ),\xi ( \cdot ),y( \cdot ))$ to (1.4). Let $(x_0,\xi _0,y_0) \in {\mathcal {H}}^3$ be such that $\xi _0 \in \partial f (x_0)$. Given a global classical solution $(v( \cdot ),y( \cdot ))$ to (1.31) such that $y(0)=y_0$ and $v(0)=x_0+ \sigma (0) \xi _0$, we consider the functions $x( \cdot )$ and $\xi ( \cdot )$ defined by (1.30), and we show through the following items (s3-a)–(s3-b) that $(x( \cdot ),\xi ( \cdot ),y( \cdot ))$ is a strong global solution to (1.4):

(s3-a) Let us prove the absolute continuity of $x( \cdot )$ and $\xi ( \cdot )$ on any bounded subset of $[0, \infty )$. As $v( \cdot )$ is of class $C^1$ on $[0, \infty )$, we immediately see that $v( \cdot )$ is absolutely continuous from the characterization (i1) of Definition 2.1. Hence, given $\epsilon >0$ and finitely many intervals $I_k=(a_k,b_k)$ such that $I_k \cap I_j = \emptyset$ (for $k \ne j$), by using Definition 2.1-(i3) we know for some $\eta >0$ that $\sum _k |b_k-a_k| \le \eta$ implies that $\sum _k \Vert v(b_k)-v(a_k)\Vert \le \min (\epsilon , \frac{2 \epsilon }{\sigma _0}).$ So, invoking the non-expansiveness of $J_{\sigma ( \cdot )}^{ \partial f}$ and $\frac{1}{2} \sigma ( \cdot ) (\partial f)_{\sigma ( \cdot )}$ while recalling that $\sigma ( \cdot ) \ge \sigma _0 >0$ entails that

$\sum _k \Vert J_{\sigma (t)}^{ \partial f} v(b_k)- J_{\sigma (t)}^{\partial f} v (a_k)\Vert \le \sum _k \Vert v(b_k)- v (a_k)\Vert \le \epsilon$

and that

$\sum _k \Vert (\partial f)_{\sigma (t)} v(b_k)-(\partial f)_{\sigma (t)} v(a_k)\Vert \le \sum _k \frac{2}{\sigma (t)}\Vert v(b_k)-v(a_k)\Vert \le \frac{2}{\sigma _0} \sum _k \Vert v(b_k)-v(a_k)\Vert \le \epsilon$.

Consequently, the mappings $x( \cdot )= J_{\sigma ( \cdot )}^{\partial f }v( \cdot )$ and $\xi ( \cdot ) = (\partial f)_{\sigma ( \cdot )} v( \cdot )$ also comply with characterization (i3) of Definition 2.1, which proves that $x( \cdot )$ and $\xi ( \cdot )$ are absolutely continuous on $[0, \infty )$.

(s3-b) Let us show that the triplet $(x( \cdot ),y( \cdot ),\xi ( \cdot ))$ satisfies system (2.3). Indeed, by $x( \cdot )= J_{\sigma ( \cdot )}^{\partial f }v( \cdot )$ (from (1.30)), and $\sigma ( \cdot ) \xi ( \cdot )=v( \cdot )-x( \cdot )$, we readily deduce that $\sigma ( \cdot ) \xi ( \cdot ) \in \left( \sigma ( \cdot ) \partial f \right) (x( \cdot ))$ (because $v( \cdot ) \in x( \cdot ) + {\sigma ( \cdot )}\partial f (x( \cdot ))$), which by the positivity of $\sigma ( \cdot )$ proves (2.3a). Moreover, in (1.31), substituting $v( \cdot )$, $J_{\sigma ( \cdot )}^{\partial f }v( \cdot )$ and ${\sigma ( \cdot )} (\partial f)_{\sigma ( \cdot )}v( \cdot )$, by $x( \cdot ) + \sigma ( \cdot ) \xi ( \cdot )$, $x( \cdot )$ and $\sigma ( \cdot ) \xi ( \cdot )$, respectively, gives us immediately (2.3b) and (2.3c). In addition, regarding the initial conditions we obtain $x(0)=J_{\sigma (0)}^{\partial f } v(0)=x_0$ (since $v(0)=x_0+\sigma (0) \xi _0$ and $\xi _0 \in \partial f (x_0)$), $y(0)=y_0$ and $\sigma (0) \xi (0)=v(0)-x(0)=\sigma (0) \xi _0$.

Consequently, by items (s3-a)– (s3-b), we get the existence of a strong global solution to (1.4) $\square$

1.3 Proof of Lemma 3.2

(See [30, Lemma 4.1]). Given $t\in [ 0,\infty )$ and $h\in (0,\infty )$, we have $\xi (t)\in \partial fx(t)$ and $\xi (t+h)\in \partial f x(t+h)$ hence, by monotonicity of $\partial f$, we simply have

$$\begin{aligned} \begin{array}{l} \big <\frac{1}{h} \big (\xi (t+h)-\xi (t) \big ),\frac{1}{h} \big (x(t+h)-x(t) \big )\big >\ge 0. \end{array} \end{aligned}$$

(1.35)

Clearly, assuming that $x( \cdot )$ and $\xi ( \cdot )$ are absolutely continuous on $[0,\infty )$, yields that, for a.e. $t \in [0,\infty )$, and as $h \rightarrow 0^+$, we have

$$\begin{aligned} {\Vert \frac{1}{h}\big ( x(t+h)-x(t) \big ) \!-\! \dot{x}(t)\Vert \rightarrow 0 \text { and } \Vert \frac{ 1}{h}\big (\xi (t+h)-\xi (t)\big )\!-\! {\dot{\xi }}(t)\Vert \rightarrow 0}.\qquad \end{aligned}$$

(1.36)

Thus, letting h tend to $0^+$ in (1.35), implies $\langle \dot{\xi }(t),\dot{x}(t)\rangle \ge 0$, that is the desired result $\square$

1.4 Proof of Proposition 4.3

By Remark 3.1, we know under condition (1.15) that $\theta ( \cdot )$ is well-defined and positive on $[0,\infty )$. Moreover, by $\nu ( \cdot ) \in C^2$ and $\theta ( \cdot )=\frac{\kappa \nu ( \cdot ) -\dot{\nu }( \cdot )}{\nu ( \cdot )+ e_*}$ (hence $\theta ( \cdot )=\kappa -\frac{ \dot{\nu }( \cdot )+\kappa e_*}{\nu ( \cdot ) + e_*}$), we can see that $\theta ( \cdot ) \in C^1([0,\infty ))$ and (omitting the variable t) we readily get

$$\begin{aligned} \theta ( \kappa - \theta )= & {} \frac{(\kappa \nu - {\dot{\nu }})({\dot{\nu }} + \kappa e_*)}{(\nu + e_*)^2} \text { and } \dot{\theta }=\frac{\dot{\nu }( \dot{\nu } + \kappa e_*) - (\nu + e_*) \ddot{\nu }}{(\nu + e_*)^2} \nonumber \\{} & {} (\text {hence} \frac{\dot{\theta }}{\theta } = \frac{ \dot{\nu }( {\dot{\nu }} + \kappa e_*) - (\nu + e_*) \ddot{\nu } }{(\nu + e_*)(\kappa \nu -\dot{\nu })}). \end{aligned}$$

(1.37)

Let us recall (from (1.3)) that $\alpha := \frac{\dot{\theta }}{\theta } + \kappa - \theta$. Consequently, by the previous arguments we obtain

$\alpha := \frac{1}{\theta } (\theta ( \kappa - \theta ) - {\dot{\theta }}) = \frac{1}{\kappa \nu - {\dot{\nu }}} \left( (\dot{\nu } +\kappa e_*) \frac{\kappa \nu - 2 {\dot{\nu }}}{\nu + e_*} + \ddot{\nu }\right)$.

So, as $t \rightarrow \infty$, by $\nu (t) \rightarrow +\infty$, $\dot{\nu }(t) \rightarrow l \in [0,\infty )$ and $\ddot{\nu }(t) \rightarrow 0$ (from (4.40)), we immediately obtain that $\alpha (t) \sim \frac{ l +\kappa e_*}{\nu (t) + e_*}$. We also recall (from (1.3)) that $\beta :=-\frac{\dot{\theta }}{{\theta }}+ \kappa +\omega$, hence by the definition of $\alpha$ we equivalently have $\beta =\alpha + \theta + \omega$. Moreover, as $t \rightarrow \infty$, by $\omega :=\big (\kappa -\frac{\dot{\nu }}{\nu } \big )\vartheta -\frac{\delta }{\nu +e_*}$ (from (1.5c)), by $\nu (t) \rightarrow \infty$, ${\dot{\nu }}(t) \rightarrow l$ and $\vartheta (t) \rightarrow \vartheta _\infty$, we readily deduce that $\omega (t) \rightarrow \kappa \vartheta _\infty$. Then, as $t \rightarrow \infty$, by the latter formulation of $\beta$, and remembering that (as $t\rightarrow \infty$) $\theta (t) \rightarrow \kappa$, $\alpha (t) \rightarrow 0$ and that $\omega (t) \rightarrow \kappa \vartheta _\infty$, we get $\beta (t) \rightarrow \kappa (1+\vartheta _\infty )$. Again from (1.3) we simply have

$b(t):=\omega (t) \left( \kappa +\frac{{\dot{\omega }}(t)}{ \omega (t)}- \frac{\dot{\theta }(t)}{{\theta (t)}} \right) = \omega (t) \left( \kappa - \frac{\dot{\theta }(t)}{{\theta (t)}} \right) + {\dot{\omega }}(t)$.

In addition, we obviously see from its expression that $\omega ( \cdot )$ is of class $C^1$, and we have

$$\begin{aligned} \begin{array}{l} { {\dot{\omega }}(t)= \left( \kappa -\frac{\dot{\nu }(t)}{\nu (t)} \right) {\dot{\vartheta }}(t)- \frac{\ddot{\nu }(t) \nu (t)-{\dot{\nu }}(t) {\dot{\nu }}(t)}{\nu ^2(t)} \vartheta (t) + \frac{\delta {\dot{\nu }}(t)}{(\nu (t) + e_*)^2}}. \end{array} \end{aligned}$$

(1.38)

It is also classically deduced from the convergence of $\vartheta$ and the Lipschitz continuity of ${\dot{\vartheta }}$ that ${\dot{\vartheta (t)}} \rightarrow 0$ (as $t \rightarrow \infty$). This, in light of condition (4.40) and $\lim _{t \rightarrow \infty } \vartheta (t)= \vartheta _\infty$ entails that ${\dot{\omega }}(t) \rightarrow \kappa \vartheta _\infty$ (as $t \rightarrow \infty$). Consequently, as $t \rightarrow \infty$, by the previous formulation of b together with $\frac{\dot{\theta }(t)}{{\theta }(t)} \rightarrow 0$, $\omega (t) \rightarrow \kappa \vartheta _\infty$ and ${\dot{\omega }}(t) \rightarrow 0$, we deduce that $b(t) \rightarrow \kappa ^2 \vartheta _\infty$

$\square$