# Solving convex optimization problems via a second order dynamical system
with implicit Hessian damping and Tikhonov regularization

Szilárd Csaba László

###### Abstract.

This paper deals with a second order dynamical system with a Tikhonov
regularization term in connection to the minimization problem of a convex
Fréchet differentiable function. The fact that beside the asymptotically
vanishing damping we also consider an implicit Hessian driven damping in the
dynamical system under study allows us, via straightforward explicit
discretization, to obtain inertial algorithms of gradient type. We show that
the value of the objective function in a generated trajectory converges
rapidly to the global minimum of the objective function and depending the
Tikhonov regularization parameter the generated trajectory converges weakly to
a minimizer of the objective function or the generated trajectory converges
strongly to the element of minimal norm from the $\operatorname*{argmin}$ set
of the objective function. We also obtain the fast convergence of the
velocities towards zero and some integral estimates. Our analysis reveals that
the Tikhonov regularization parameter and the damping parameters are strongly
correlated, there is a setting of the parameters that separates the cases when
weak convergence of the trajectories to a minimizer and strong convergence of
the trajectories to the minimal norm minimizer can be obtained.

Corresponding Author: Szilárd Csaba László, Technical University of Cluj-
Napoca, Department of Mathematics, Str. Memorandumului nr. 28, 400114 Cluj-
Napoca, Romania, e-mail<EMAIL_ADDRESS>
This work was supported by a grant of the Ministry of Research, Innovation and
Digitization, CNCS - UEFISCDI, project number PN-III-P1-1.1-TE-2021-0138,
within PNCDI III

Key Words. convex optimization; continuous second order dynamical system;
Hessian driven damping; Tikhonov regularization; convergence rate; strong
convergence.

AMS subject classification. 34G20, 47J25, 90C25, 90C30, 65K10.

##  1\. Introduction

###  1.1. The problem formulation and a model result

In this paper we study the asymptotical behaviour of the trajectories
generated by the second order dynamical system

(DS)
$\displaystyle\begin{cases}\ddot{x}(t)+\frac{\alpha}{t^{q}}\dot{x}(t)+\nabla
g\left(x(t)+\left(\gamma+\frac{\beta}{t^{q}}\right)\dot{x}(t)\right)+\epsilon(t)x(t)=0,\\\
x(t_{0})=u_{0}\in\mathcal{H},\,\dot{x}(t_{0})=v_{0}\in\mathcal{H},\end{cases}$

in connection to the minimization problem

(P) $\displaystyle\inf_{x\in\mathcal{H}}g(x),$

where $\mathcal{H}$ is a real Hilbert space endowed with the scalar product
$\langle\cdot,\cdot\rangle$ and norm $\|\cdot\|$ and
$g:\mathcal{H}\longrightarrow\mathbb{R}$ is a convex Fréchet differentiable
function. Throughout the paper we assume that ${\nabla}g$ is Lipschitz
continuous on bounded sets and the solution set of $g$, namely
$\operatorname*{argmin}g,$ is non-empty.

The starting time in (DS) is $t_{0}>0$ and we assume that
$\alpha>0,\,q\in]0,1[$ and $\gamma>0,\,\beta\in\mathbb{R}$ or
$\gamma=0,\,\beta>0$. In case $\beta<0$ we also assume that
$t_{0}>\left|\frac{\beta}{\gamma}\right|^{\frac{1}{q}}.$ Further,
$\epsilon:[t_{0},+\infty[\longrightarrow\mathbb{R}_{+}$ is a non-increasing
function of class $C^{1}$, such that $\lim\limits_{t\to+\infty}\epsilon(t)=0.$

Note that $\epsilon(t)x(t)$ in (DS) is a Tikhonov regularization term that
allows us to obtain the strong convergence of a trajectory generated by (DS)
to the minimal norm minimizer of $g,$ see [14, 6, 26, 21]. Further, we do not
allow $q=1$ since in that case (DS) becomes the dynamical system studied in
[2]. However, we show that by considering $q\in]0,1[$ the results obtained in
[2] can be improved. Further we avoid the case $\gamma=\beta=0$ since then
(DS) becomes the dynamical system studied in [26]. Our analysis shows that the
damping parameters and the Tikhonov regularization parameter are strongly
correlated, there is a setting of the parameters which assure weak convergence
of the trajectories generated by (DS) to a minimizer of $g$ and rapid
convergence of the function values to the global minimum of $g$, meanwhile for
another constellation of the parameters involved, rapid convergence of the
function values and strong convergence of the generated trajectories to the
minimal norm minimizer is obtained. In order to visualize the facts emphasized
above, let us specify $\epsilon(t)=\frac{a}{t^{p}},\,a,p>0$ and let us denote
$\beta(t)=\gamma+\frac{\beta}{t^{q}}.$ It is easy to see that
$\lim\limits_{t\to+\infty}\epsilon(t)=0$, $\epsilon(t)$ is a non-increasing
function of class $C^{1}$ and satisfies $\epsilon(t)\geq 0$ for every $t\geq
t_{0}>0$. In this case the following result holds, see Corollary 8 and
Corollary 12 and Theorem 9.

######  Theorem 1.

Suppose that the regularization parameter has the form
$\epsilon(t)=\frac{a}{t^{p}},\,a,p>0$ and for some starting points
$u_{0},\,v_{0}\in\mathcal{H}$ let $x:[t_{0},+\infty[\to\mathcal{H}$ be the
unique global solution of (DS). Then, the following results hold.

  * $(i)$

If $q+1<p\leq 2$ and for $p=2$ one has $a\geq q(1-q),$ then the trajectory
$x(t)$ converges weakly, as $t\to+\infty$, to an element of
$\operatorname*{argmin}g$. Further the following pointwise and integral
estimates are valid.
One has $\|\dot{x}(t)\|=o\left(t^{-q}\right),$
$g\left(x(t)+\beta(t)\dot{x}(t)\right)-\min g=o\left(t^{-2q}\right)\text{ as
}t\to+\infty$ and
$\int_{t_{0}}^{+\infty}t^{q}\|\dot{x}(t)\|^{2}dt<+\infty,\,\int_{t_{0}}^{+\infty}t^{q}\left(g\left(x(t)+\beta(t)\dot{x}(t)\right)-\min
g\right)dt<+\infty.$ Concerning the gradient one has the integral estimates
$\int_{t_{0}}^{+\infty}t^{2q}\left\|{\nabla}g\left(x(t)+\beta(t)\dot{x}(t)\right)\right\|^{2}dt<+\infty,$
if $\gamma>0,\,\beta\in\mathbb{R}$ and
$\int_{t_{0}}^{+\infty}t^{2q-1}\left\|{\nabla}g\left(x(t)+\beta(t)\dot{x}(t)\right)\right\|^{2}dt<+\infty,$
if $\gamma=0,\,\beta>0.$

  * $(ii)$

If $q\leq p<q+1$ and when $q=p$ one has $a\leq\dfrac{\alpha}{2\gamma},$ then
the trajectory $x(t)$ converges strongly, as $t\to+\infty$, to $x^{*}$, the
element of minimum norm of $\operatorname*{argmin}g$. Further the following
pointwise and integral estimates are valid.
One has $\|\dot{x}(t)\|=\mathcal{O}(t^{-\frac{\min(2q,p)}{2}})$ and
$g(x(t)+\beta(t)\dot{x}(t))-\min g=\mathcal{O}(t^{-\min(2q,p)})\mbox{ as
}t\to+\infty,$ and if $p>2q$ then
$\int_{t_{0}}^{+\infty}t^{q}\|\dot{x}(t)\|^{2}dt<+\infty,\,\int_{t_{0}}^{+\infty}t^{q}\left(g\left(x(t)+\beta(t)\dot{x}(t)\right)-\min
g\right)dt<+\infty.$ Concerning the gradient one has the integral estimates
$\int_{t_{0}}^{+\infty}t^{2q-1}\|{\nabla}g(x(t)+\beta(t)\dot{x}(t))\|^{2}dt<+\infty,\mbox{
if }\gamma=0,\,\beta>0\mbox{ and }p>2q$ and
$\int_{t_{0}}^{+\infty}t^{2q}\|{\nabla}g(x(t)+\beta(t)\dot{x}(t))\|^{2}dt<+\infty$
if $\gamma>0,\,\beta\in\mathbb{R}$ and $p>\max\left(\frac{2q+1}{2},2q\right).$

  * $(iii)$

The case $p=q+1$ is critical in the sense that neither weak nor strong
convergence of the trajectories can be obtained. However, also in this case
the trajectory $x(t)$ is bounded and one has
$\|\dot{x}(t)\|=\mathcal{O}\left(t^{-q}\right),$
$g\left(x(t)+\beta(t)\dot{x}(t)\right)-\min
g=\mathcal{O}\left(t^{-2q}\right)\text{ as }t\to+\infty$

###  1.2. Motivation and related works

Starting with the seminal work of Su, Boyd and Candès [33], second order
dynamical systems have been intensively studied in the literature in
connection to the minimization problem having in its objective a (convex)
smooth function, see [3, 10, 5, 8, 9, 13, 15, 16, 17, 20, 19, 22, 23, 28, 32].
These dynamical systems lead via implicit/explicit discretizations to inertial
algorithms, hence the asymptotical behaviour of the trajectories generated by
these dynamical systems give an insight into the behaviour of the sequences
generated by the algorithms obtained via discretization see [4, 29, 30, 31,
25].

Indeed, according to [33] the dynamical system

(1)
$\ddot{x}(t)+\frac{\alpha}{t}\dot{x}(t)+{\nabla}g(x(t))=0,\,\,x(t_{0})=u_{0},\,\dot{x}(t_{0})=v_{0},\,t_{0}>0,\,u_{0},v_{0}\in\mathcal{H}$

with $\alpha=3$ is the exact limit of Nesterov’s accelerated convex gradient
method [30] and that is the reason why (1) has been intensively studied in the
literature in connection to the minimization problem
$\inf_{x\in\mathcal{H}}g(x).$ In [33] the authors proved that $g(x(t))-\min
g=\mathcal{O}\left(\frac{1}{t^{2}}\right)$ for every $\alpha\geq 3$ as
$t\longrightarrow+\infty$, however they did not show the convergence of a
generated trajectory to a minimum of the objective function $g.$

In [10], Attouch, Chbani, Peypouquet and Redont considered the case $\alpha>3$
in (1), and showed that the generated trajectory $x(t)$ converges weakly to a
minimizer of $g$ as $t\longrightarrow+\infty$ and also that $g(x(t))-\min
g=o\left(\frac{1}{t^{2}}\right)$ as $t\longrightarrow+\infty$. A perturbed
version of (1) was studied by Attouch, Peypouquet and Redont in [15], see also
[5, 9, 16, 32]. They assumed that the objective $g$ is twice continuously
differentiable and the perturbation of their system is made at the damping
term. More precisely, they studied the dynamical system with Hessian driven
damping

(2)
$\ddot{x}(t)+\frac{\alpha}{t}\dot{x}(t)+\beta{\nabla}^{2}g(x(t))\dot{x}(t)+{\nabla}g(x(t))=0,\,\,x(t_{0})=u_{0},\,\dot{x}(t_{0})=v_{0},\,t_{0}>0,\,u_{0},v_{0}\in\mathcal{H},$

where $\alpha>0$ and $\beta>0.$ In case $\alpha>3,\,\beta>0$ they showed the
weak convergence of a generated trajectory to a minimizer of $g$. Moreover,
they obtained that in this case the convergence rate of the objective function
along the trajectory, that is $g(x(t))-\min g,$ is of order
${o}\left(\frac{1}{t^{2}}\right)$.

What one can notice concerning the dynamical systems (1) and (2) is that these
systems will never give through the natural implicit/explicit discretization
Nesterov type inertial algorithms, since the gradient of $g$ is evaluated in
$x(t),$ and this term via implicit/explicit discretization will become
${\nabla}g(x_{n})$ or ${\nabla}g(x_{n+1})$ and never of the form
${\nabla}g(x_{n}+\alpha_{n}(x_{n}-x_{n-1}))$ as Nesterov’s gradient method
requires. That was the reason why in [3], (see also [28]), Alecsa, László and
Pinţa introduced the dynamical system

(3) $\ddot{x}(t)+\frac{\alpha}{t}\dot{x}(t)+\nabla
g\left(x(t)+\left(\gamma+\frac{\beta}{t}\right)\dot{x}(t)\right)=0,\,x(t_{0})=u_{0},\,\dot{x}(t_{0})=v_{0},\,t_{0}>0,\,u_{0},v_{0}\in\mathcal{H}$

where $\alpha>0,\,\beta\in\mathbb{R},\,\gamma\geq 0.$ The term $\nabla
g\left(x(t)+\left(\gamma+\frac{\beta}{t}\right)\dot{x}(t)\right)$ in (3) is
called implicit Hessian damping due to the fact that by using the Taylor
expansion for ${\nabla}g(\cdot)$ we get
${\nabla}g\left(x(t)+\left(\gamma+\frac{\beta}{t}\right)\dot{x}(t)\right)\approx{\nabla}g(x(t))+\left(\gamma+\frac{\beta}{t}\right){\nabla}^{2}g(x(t))\dot{x}(t).$
This fact also explains why the trajectories of (2) and (3) share a similar
behaviour. Indeed, the objective function value in a trajectory generated by
(3) converges in order $\mathcal{O}\left(\frac{1}{t^{2}}\right)$ to the global
minimum of the objective function and the generated trajectory converges
weakly to a minimizer of the objective function $g,$ see [3]. Further,
according to [2], the dynamical system (3) can be thought as an intermediate
system between the dynamical system (1) and the dynamical system (2), which
inherits the best properties of the latter systems. Note that explicit Euler
discretization of (3) leads to Nesterov type inertial algorithms, where the
gradient is evaluated at an extrapolated term of the form $x_{n}+\frac{\gamma
n+\beta}{n+\alpha}(x_{n}-x_{n-1}),$ see [3].

Tikhonov regularized versions of (1) and (2) were studied in [11] and [18]. In
these papers, for a constellation of the parameters the authors obtained fast
convergence for the function values in a generated trajectory and weak
convergence of the generated trajectory to a minimizer of the objective
function, meanwhile for another setting of the parameters the strong
convergence result $\liminf_{t\to+\infty}\|x(t)-x^{*}\|=0$ have been obtained,
where $x(t)$ is the generated trajectory and $x^{*}$ is the minimal norm
minimizer of the objective function. For a while it was an open question
whether one can obtain fast convergence of the function values and strong
convergence results in a generated trajectory for the same constellation of
the parameters involved. The first breakthrough has been obtained in [14]
where the authors introduced the dynamical system

(4) $\displaystyle\ddot{x}(t)+\delta\sqrt{\epsilon(t)}\dot{x}(t)+\nabla
g\left(x(t)\right)+\epsilon(t)x(t)=0,\,x(t_{0})=u_{0},\,\dot{x}(t_{0})=v_{0},$

with $\delta>0$ and succeeded to obtain both fast convergence towards the
minimal value of the objective function and the strong convergence result
$\liminf_{t\to+\infty}\|x(t)-x^{*}\|=0$. Similar results have been obtained in
[2] for the Tikhonov regularized second order dynamical system (3), where the
Tikhonov regularization parameter has the form $\epsilon(t)=\frac{a}{t^{2}}$
and $\alpha>3.$ In contrast, in this paper we assume that $0<q<1$ instead of
$q=1$ considered in [2] and beside fast convergence rates we are able to
obtain the strong convergence result $\lim_{t\to+\infty}\|x(t)-x^{*}\|=0$.
Moreover these results hold for every $\alpha>0$ and every Tikhonov
regularization parameter $\epsilon(t)=\frac{a}{t^{p}},\,q\leq p<q+1.$

The results from [14] have been improved in [6] where the authors studied the
dynamical system (4) and succeeded to obtain the strong convergence result
$\lim_{t\to+\infty}\|x(t)-x^{*}\|=0.$ These results then have been extended
and further improved in [26], where it is also shown that the best choice for
the damping parameter is not $\sqrt{\epsilon(t)}.$ We emphasize that the
results of this paper are in concordance to the results obtained in [26].
Further, the results from [18] have been improved in [7] where the authors
succeeded to obtain fast convergence rates and the strong convergence result
$\lim_{t\to+\infty}\|x(t)-x^{*}\|=0.$ Since, as we emphasized before, (DS) and
the Tikhonov regularized system with Hessian driven damping considered in [7]
are strongly related, let us compare the results from this paper and the
results from [7]. The second order dynamical system with Hessian driven
damping and Tikhonov regularization studied in [7] is

(5)
$\displaystyle\ddot{x}(t)+\delta\sqrt{\epsilon(t)}\dot{x}(t)+\beta\nabla^{2}g\left(x(t)\right)\dot{x}(t)+\nabla
g\left(x(t)\right)+\epsilon(t)x(t)=0,\,x(t_{0})=u_{0},\,\dot{x}(t_{0})=v_{0},$

where $g$ is a convex function of class $C^{2}$, $\delta,\beta>0$ and
$\epsilon(t)$ is a Tikhonov regularization parameter. According to the model
result from [7], when one specify $\epsilon(t)=\frac{1}{t^{2q}},\,0<q<1$ then
the trajectories generated by the dynamical system (5) converge strongly to
the minimal norm minimizer of $g$, $g(x(t))-\min g=\mathcal{O}(t^{-2q})\mbox{
as }t\to+\infty$ and
$\int_{t_{0}}^{+\infty}t^{3q-1}\|{\nabla}g(x(t))\|^{2}dt<+\infty.$ In
contrast, we assume that our objective function is of class $C^{1}$ and
according to Theorem 1 the trajectories generated by the dynamical system (5)
converge strongly to the minimal norm minimizer of $g$ whenever $q\leq p<q+1$,
and if $2q<p<q+1$ then $g(x(t)+\beta(t)\dot{x}(t))-\min
g=\mathcal{O}(t^{-2q})\mbox{ as }t\to+\infty,$ but we also have the pointwise
estimate for the velocity $\|\dot{x}(t)\|=\mathcal{O}(t^{-q})$. Moreover, we
have the integral estimates
$\int_{t_{0}}^{+\infty}t^{q}\|\dot{x}(t)\|^{2}dt<+\infty,\,\int_{t_{0}}^{+\infty}t^{q}\left(g\left(x(t)+\beta(t)\dot{x}(t)\right)-\min
g\right)dt<+\infty$ and
$\int_{t_{0}}^{+\infty}t^{2q}\|{\nabla}g(x(t)+\beta(t)\dot{x}(t))\|^{2}dt<+\infty,\mbox{
if }\gamma>0,\,\beta\in\mathbb{R}\mbox{ and
}p>\max\left(\frac{2q+1}{2},2q\right).$

Observe that in this paper we are able to obtain extra and improved pointwise
and integral estimates, which shows again that, though the damping parameters
and the Tikhonov regularization parameter are strongly correlated, the best
choice of the damping is not $\delta\sqrt{\epsilon(t)},\,\delta>0.$

Finally, note that explicit Euler discretization of the dynamical system (DS)
yields to inertial gradient methods. Indeed, by discretizing (DS) with the
constant stepsize $h$,
$t_{n}=nh,\,\,x_{n}=x(t_{n}),\,\epsilon(t_{n})=\epsilon_{n}$ we get

$\frac{x_{n+1}-2x_{n}+x_{n-1}}{h^{2}}+\frac{\alpha}{h^{q+1}n^{q}}(x_{n}-x_{n-1})+{\nabla}g\left(x_{n}+\left(\frac{\gamma}{h}+\frac{\beta}{h^{q+1}n^{q}}\right)(x_{n}-x_{n-1})\right)+\epsilon_{n}x_{n}=0.$

Equivalently, by denoting $h^{2q}$ with $s$, $h^{1-q}\alpha$ still with
$\alpha$, $\frac{\gamma}{h}$ still with $\gamma,$ $\frac{\beta}{h^{q+1}}$
still with $\beta$ and $h^{2q}\epsilon_{n}$ still with $\epsilon_{n}$, the
latter equation can be written as a gradient type inertial algorithm with a
Tikhonov regularization term, that is,

(6)
$x_{n+1}=x_{n}+\left(1-\frac{\alpha}{n^{q}}\right)(x_{n}-x_{n-1})-s{\nabla}g\left(x_{n}+\left(\gamma+\frac{\beta}{n^{q}}\right)(x_{n}-x_{n-1})\right)-\epsilon_{n}x_{n}.$

We emphasize that algorithm (6) has a similar form as the Nesterov type
algorithm with double Tikhonov regularization studied in [24], hence the
results from this paper opens the gate for the study of the strong convergence
of inertial gradient type algorithms with a Tikhonov regularization term.

###  1.3. The organization of the paper

The outline of the paper is the following. After a short section concerning
the existence and uniqueness of the trajectories generated by the dynamical
system (DS), in section 3 we deal with the setting of the parameters that
assure weak convergence of the generated trajectories to a minimizer of the
objective function. To this purpose we introduce a general energy functional,
which will play the role of a Lyapunov function associated to the dynamical
system (DS) and beside the weak convergence of the generated trajectories to a
minimizer of the objective function we obtain fast convergence rates for the
function values and velocity but also some integral estimates involving the
gradient of the objective function. In section 4 for another constellation of
the parameters, by constructing a powerful Lyapunov function we obtain strong
convergence of the generated trajectories to the minimal norm minimizer of the
objective function, but also fast convergence of the function values in a
generated trajectory and fast convergence of the velocity. Some integral
estimates are also provided. In section 5 we conclude our paper and we present
some possible related future researches.

##  2\. Existence and uniqueness of the trajectories generated by the
dynamical system (DS)

In what follows we show the existence and uniqueness of a classical $C^{2}$
solution $x$ of the dynamical system (DS). To this purpose we rewrite (DS) as
a first order system relevant for the Cauchy-Lipschitz-Picard theorem.

######  Theorem 2.

Let $(u_{0},v_{0})\in\mathcal{H}\times\mathcal{H}$. Then, the dynamical system
(DS) admits a unique global $C^{2}(]t_{0},+\infty[,\mathcal{H})$ solution.

######  Proof .

Indeed, by using the notation $X(t):=(x(t),\dot{x}(t))$, the dynamical system
(DS) can be put in the form

(7) $\displaystyle\begin{cases}\dot{X}(t)=F(t,X(t))\\\
X(t_{0})=(u_{0},v_{0}),\end{cases}$

where
$F:[t_{0},+\infty[\times\mathcal{H}\times\mathcal{H}\longrightarrow\mathcal{H}\times\mathcal{H}$,
$F(t,u,v)=\left(v,-\frac{\alpha}{t^{q}}v-\epsilon(t)u-\nabla
g\left(u+\beta(t)v\right)\right),$ and we denote
$\beta(t)=\gamma+\frac{\beta}{t^{q}}.$ Note that due to our assumptions
$\beta(t)>0$ on the interval $[t_{0},+\infty[.$

Our proof is inspired from [8]. Since ${\nabla}g$ is Lipschitz on bounded
sets, it is obvious that for (7) the classical Cauchy-Picard theorem can be
applied, hence, there exist a unique $C^{1}$ local solution $X.$ Consequently,
(DS) has a unique $C^{2}$ local solution. Let $x$ be a maximal solution of
(DS), defined on an interval $[t_{0},T_{\max}[,\,T_{\max}\leq+\infty.$

In order to prove that $\dot{x}$ is bounded we consider the energy functional
$W:[t_{0},+\infty[\to\mathbb{R},$

(8) $\displaystyle
W(t)=g(x(t))+\frac{1}{2}\|\dot{x}(t)\|^{2}+\frac{\epsilon(t)}{2}\|x(t)\|^{2}.$

The time derivative of (8) reads as

$\displaystyle\dot{W}(t)=\langle\nabla
g(x(t)),\dot{x}(t)\rangle+\langle\dot{x}(t),\ddot{x}(t)\rangle+\frac{\dot{\epsilon}(t)}{2}\|x(t)\|^{2}+\epsilon(t)\langle
x(t),\dot{x}(t)\rangle.$

From (DS) we have
$\ddot{x}(t)=-\frac{\alpha}{t^{q}}\dot{x}(t)-\epsilon(t)x(t)-\nabla
g(x(t)+\beta(t)\dot{x}(t))$, and we obtain

$\displaystyle\dot{W}(t)$ $\displaystyle=\langle{\nabla}g(x(t))-\nabla
g(x(t)+\beta(t)\dot{x}(t)),\dot{x}(t)\rangle+\frac{\dot{\epsilon}(t)}{2}\|x(t)\|^{2}-\frac{\alpha}{t^{q}}\|\dot{x}(t)\|^{2}.$

Now, by using the fact that ${\nabla}g$ is monotone and $\dot{\epsilon}(t)\leq
0$ we get

$\displaystyle\dot{W}(t)$
$\displaystyle=-\frac{1}{\beta(t)}\langle{\nabla}g(x(t))-\nabla
g(x(t)+\beta(t)\dot{x}(t)),x(t)-(x(t)+\beta(t)\dot{x}(t))\rangle+\frac{\dot{\epsilon}(t)}{2}\|x(t)\|^{2}-\frac{\alpha}{t^{q}}\|\dot{x}(t)\|^{2}$
$\displaystyle\leq 0.$

Consequently, $W$ is non-increasing on $[t_{0},+\infty[$, hence by the fact
that $g$ is bounded from below we get $\|\dot{x}(t)\|<+\infty$ for all $t\geq
t_{0}.$

Let $\|\dot{x}_{\infty}\|=\sup_{t\in[t_{0},T_{\max}[}\|\dot{x}(t)\|$ and
assume that $T_{\max}<+\infty.$ Since
$\|x(t)-x(t^{\prime})\|\leq\|\dot{x}_{\infty}\||t-t^{\prime}|$, we get that
$\lim_{t\to T_{\max}}x(t):=x_{\infty}\in\mathcal{H}$. By (DS) the map
$\ddot{x}$ is also bounded on the interval $[t_{0},T_{\max}[$ and under the
same argument as before $\lim_{t\to T_{\max}}\dot{x}(t):=\dot{x}_{\infty}$
exists. Applying the local existence theorem with initial data
$(x_{\infty},\dot{x}_{\infty})$, we can extend the maximal solution to a
strictly larger interval, a clear contradiction. Hence $T_{\max}=+\infty,$
which completes the proof. $\blacksquare$

##  3\. Weak convergence for the trajectories of the regularized dynamical
system (DS) and fast convergence for the function values in a generated
trajectory

In this section we show that the function values in a trajectory generated by
the dynamical system (DS) converge to the global minimum of the objective
function. Further, we obtain some pointwise and integral estimates concerning
the function values in a trajectory generated by the dynamical system (DS),
but also for the gradient and velocity. Finally, we give sufficient conditions
on the Tikhonov regularization parameter in order to show that the trajectory
$x(t)$ generated by the dynamical system (DS) converges weakly to a minimizer
of our objective function $g.$

In this section we consider the following weak assumption:

(C0) $\displaystyle\text{There exist }0<K\text{ such that
}\frac{q(1-q)}{t^{2}}\leq\epsilon(t)\leq\frac{K}{t^{q}},\mbox{ for }t\mbox{
large enough.}$

Our first general result is the following.

######  Theorem 3.

For some starting points $u_{0},\,v_{0}\in\mathcal{H}$ let
$x:[t_{0},\infty)\to\mathcal{H}$ be the unique global solution of (DS). If the
Tikhonov regularization parameter $\epsilon(t)$ satisfies (C0) and
$\int_{t_{0}}^{+\infty}\frac{\epsilon(t)}{t^{q}}dt<+\infty$ then
$\lim\limits_{t\to+\infty}g\left(x(t)+\beta(t)\dot{x}(t)\right)=\min g.$ In
addition, one has that $\lim\limits_{t\to+\infty}\|\dot{x}(t)\|=0,$ therefore
$\lim\limits_{t\to+\infty}g\left(x(t)\right)=\min g.$

######  Proof .

Consider $x^{\ast}\in\operatorname*{argmin}g$ and denote
$g^{\ast}=g(x^{\ast})=\min g$. Consider $b\in]0,\alpha[$ and
$a(t)=t^{2q},\,t\geq t_{0}.$ For simplicity we denote
$\beta(t)=\gamma+\frac{\beta}{t^{q}}$ We introduce the energy functional
$\mathcal{E}:[t_{0},\infty)\to\mathbb{R},$

(9) $\displaystyle\mathcal{E}(t)=$ $\displaystyle
a(t)\left(g(x(t)+\beta(t)\dot{x}(t))-g^{\ast}\right)+\frac{t^{2q}\epsilon(t)}{2}\|x(t)\|^{2}+\frac{1}{2}\|b(x(t)-x^{\ast})+t^{q}\dot{x}(t)\|^{2}$
$\displaystyle+\frac{b(\alpha-qt^{q-1}-b)}{2}\|x(t)-x^{\ast}\|^{2}.$

From the choice of $b$, there exists a $t_{1}\geq t_{0}$ such that the
coefficient $\alpha-qt^{q-1}-b>0$ for all $t\geq t_{1}$, hence the Lyapunov
functional $\mathcal{E}(t)\geq 0$ for all $t\geq t_{1}.$

By taking the time derivative of $\mathcal{E}$ we obtain

(10) $\displaystyle\dot{\mathcal{E}}(t)$
$\displaystyle=a^{\prime}(t)\left(g(x(t)+\beta(t)\dot{x}(t))-g^{\ast}\right)+a(t)\langle{\nabla}g(x(t)+\beta(t)\dot{x}(t)),\beta(t)\ddot{x}(t)+(\beta^{\prime}(t)+1)\dot{x}(t)\rangle$
$\displaystyle+\left(qt^{2q-1}\epsilon(t)+\frac{t^{2q}\dot{\epsilon}(t)}{2}\right)\|x(t)\|^{2}+t^{2q}\epsilon(t)\langle\dot{x}(t),x(t)\rangle+\langle(b+qt^{q-1})\dot{x}(t)+t^{q}\ddot{x}(t),b(x(t)-x^{\ast})+t^{q}\dot{x}(t)\rangle$
$\displaystyle+b(\alpha-
qt^{q-1}-b)\langle\dot{x}(t),x(t)-x^{\ast}\rangle+\frac{bq(1-q)t^{q-2}}{2}\|x(t)-x^{\ast}\|^{2}.$

From the dynamical system (DS), we have that
$\ddot{x}(t)=-\epsilon(t)x(t)-\frac{\alpha}{t^{q}}\dot{x}(t)-\nabla
g(x(t)+\beta(t)\dot{x}(t)).$ Hence,

(11) $\displaystyle
a(t)\langle{\nabla}g(x(t)+\beta(t)\dot{x}(t)),\beta(t)\ddot{x}(t)+(\beta^{\prime}(t)+1)\dot{x}(t)\rangle=$
$\displaystyle
a(t)\left\langle{\nabla}g(x(t)+\beta(t)\dot{x}(t)),-\beta(t)\epsilon(t)x(t)+\left(-\beta(t)\frac{\alpha}{t^{q}}+\beta^{\prime}(t)+1\right)\dot{x}(t))-\beta(t)\nabla
g(x(t)+\beta(t)\dot{x}(t))\right\rangle=$
$\displaystyle-\beta(t)a(t)\left\|{\nabla}g(x(t)+\beta(t)\dot{x}(t))\right\|^{2}+\left(-\beta(t)\frac{\alpha}{t^{q}}+\beta^{\prime}(t)+1\right)a(t)\langle{\nabla}g(x(t)+\beta(t)\dot{x}(t)),\dot{x}(t)\rangle$
$\displaystyle-\beta(t)\epsilon(t)a(t)\langle{\nabla}g(x(t)+\beta(t)\dot{x}(t)),x(t)\rangle.$

Further,

(12)
$\displaystyle\langle(b+qt^{q-1})\dot{x}(t)+t^{q}\ddot{x}(t),b(x(t)-x^{\ast})+t^{q}\dot{x}(t)\rangle=$
$\displaystyle\langle(b+qt^{q-1}-\alpha)\dot{x}(t)-t^{q}\epsilon(t)x(t)-t^{q}\nabla
g(x(t)+\beta(t)\dot{x}(t)),b(x(t)-x^{\ast})+t^{q}\dot{x}(t)\rangle=$
$\displaystyle
b(b+qt^{q-1}-\alpha)\langle\dot{x}(t),x(t)-x^{\ast}\rangle+(b+qt^{q-1}-\alpha)t^{q}\|\dot{x}(t)\|^{2}-bt^{q}\epsilon(t)\langle
x(t),x(t)-x^{\ast}\rangle-t^{2q}\epsilon(t)\langle\dot{x}(t),x(t)\rangle$
$\displaystyle-
bt^{q}\langle{\nabla}g(x(t)+\beta(t)\dot{x}(t)),x(t)-x^{\ast}\rangle-t^{2q}\langle\nabla
g(x(t)+\beta(t)\dot{x}(t)),\dot{x}(t)\rangle.$

Combining (10), (11) and (12) we get

(13) $\displaystyle\dot{\mathcal{E}}(t)$
$\displaystyle=a^{\prime}(t)\left(g(x(t)+\beta(t)\dot{x}(t))-g^{\ast}\right)-\beta(t)a(t)\left\|{\nabla}g(x(t)+\beta(t)\dot{x}(t))\right\|^{2}+(b+qt^{q-1}-\alpha)t^{q}\|\dot{x}(t)\|^{2}$
$\displaystyle+\left(qt^{2q-1}\epsilon(t)+\frac{t^{2q}\dot{\epsilon}(t)}{2}\right)\|x(t)\|^{2}+\left(\left(-\beta(t)\frac{\alpha}{t^{q}}+\beta^{\prime}(t)+1\right)a(t)-t^{2q}\right)\langle{\nabla}g(x(t)+\beta(t)\dot{x}(t)),\dot{x}(t)\rangle$
$\displaystyle-\beta(t)\epsilon(t)a(t)\langle{\nabla}g(x(t)+\beta(t)\dot{x}(t)),x(t)\rangle-
bt^{q}\langle{\nabla}g(x(t)+\beta(t)\dot{x}(t))+\epsilon(t)x(t),x(t)-x^{\ast}\rangle$
$\displaystyle+\frac{bq(1-q)t^{q-2}}{2}\|x(t)-x^{\ast}\|^{2}.$

Consider now the strongly convex function
$g_{t}:\mathcal{H}\longrightarrow\mathbb{R},\,g_{t}(x)=g(x)+\frac{\epsilon(t)}{2}\|x\|^{2}.$
From the gradient inequality we have
$g_{t}(y)-g_{t}(x)\geq\langle{\nabla}g_{t}(x),y-x\rangle+\frac{\epsilon(t)}{2}\|x-y\|^{2},\mbox{
for all }x,y\in\mathcal{H}.$ Take now $y=x^{\ast}$ and
$x=x(t)+\beta(t)\dot{x}(t).$ We get

$\displaystyle
g(x^{\ast})+\frac{\epsilon(t)}{2}\|x^{\ast}\|^{2}-g(x(t)+\beta(t)\dot{x}(t))-\frac{\epsilon(t)}{2}\|x(t)+\beta(t)\dot{x}(t)\|^{2}\geq$
$\displaystyle-\langle{\nabla}g(x(t)+\beta(t)\dot{x}(t))+\epsilon(t)(x(t)+\beta(t)\dot{x}(t)),x(t)+\beta(t)\dot{x}(t)-x^{\ast}\rangle+\frac{\epsilon(t)}{2}\|x(t)+\beta(t)\dot{x}(t)-x^{\ast}\|^{2}.$

Consequently,

$\displaystyle-\langle{\nabla}g(x(t)+\beta(t)\dot{x}(t))+\epsilon(t)x(t),x(t)-x^{\ast}\rangle-\beta(t)\langle{\nabla}g(x(t)+\beta(t)\dot{x}(t))+\epsilon(t)x(t),\dot{x}(t)\rangle=$
$\displaystyle-\langle{\nabla}g(x(t)+\beta(t)\dot{x}(t))+\epsilon(t)x(t),x(t)+\beta(t)\dot{x}(t)-x^{\ast}\rangle\leq
g(x^{\ast})+\frac{\epsilon(t)}{2}\|x^{\ast}\|^{2}-g(x(t)+\beta(t)\dot{x}(t))$
$\displaystyle-\frac{\epsilon(t)}{2}\|x(t)+\beta(t)\dot{x}(t)\|^{2}-\frac{\epsilon(t)}{2}\|x(t)+\beta(t)\dot{x}(t)-x^{\ast}\|^{2}+\beta(t)\epsilon(t)\langle\dot{x}(t),x(t)+\beta(t)\dot{x}(t)-x^{\ast}\rangle.$

From here we get

(14)
$\displaystyle-\langle{\nabla}g(x(t)+\beta(t)\dot{x}(t))+\epsilon(t)x(t),x(t)-x^{\ast}\rangle\leq$
$\displaystyle-(g(x(t)+\beta(t)\dot{x}(t))-g(x^{\ast}))+\frac{\epsilon(t)}{2}\|x^{\ast}\|^{2}+\beta(t)\langle{\nabla}g(x(t)+\beta(t)\dot{x}(t)),\dot{x}(t)\rangle$
$\displaystyle-\frac{\epsilon(t)}{2}\|x(t)+\beta(t)\dot{x}(t)\|^{2}-\frac{\epsilon(t)}{2}\|x(t)+\beta(t)\dot{x}(t)-x^{\ast}\|^{2}+\beta(t)\epsilon(t)\langle\dot{x}(t),2x(t)+\beta(t)\dot{x}(t)-x^{\ast}\rangle.$

Further, an easy computation shows that

$\displaystyle-\frac{\epsilon(t)}{2}\|x(t)+\beta(t)\dot{x}(t)\|^{2}-\frac{\epsilon(t)}{2}\|x(t)+\beta(t)\dot{x}(t)-x^{\ast}\|^{2}+\beta(t)\epsilon(t)\langle\dot{x}(t),2x(t)+\beta(t)\dot{x}(t)-x^{\ast}\rangle=$
$\displaystyle-\frac{\epsilon(t)}{2}\|x(t)\|^{2}-\frac{\epsilon(t)}{2}\|x(t)-x^{\ast}\|^{2}.$

Hence, (14) becomes

(15)
$\displaystyle-\langle{\nabla}g(x(t)+\beta(t)\dot{x}(t))+\epsilon(t)x(t),x(t)-x^{\ast}\rangle\leq-(g(x(t)+\beta(t)\dot{x}(t))-g(x^{\ast}))-\frac{\epsilon(t)}{2}\|x(t)\|^{2}$
$\displaystyle-\frac{\epsilon(t)}{2}\|x(t)-x^{\ast}\|^{2}+\frac{\epsilon(t)}{2}\|x^{\ast}\|^{2}+\beta(t)\langle{\nabla}g(x(t)+\beta(t)\dot{x}(t)),\dot{x}(t)\rangle.$

By multiplying (15) with $bt^{q}$ and injecting in (13) we get

(16) $\displaystyle\dot{\mathcal{E}}(t)\leq$
$\displaystyle(a^{\prime}(t)-bt^{q})\left(g(x(t)+\beta(t)\dot{x}(t))-g^{\ast}\right)-\beta(t)a(t)\left\|{\nabla}g(x(t)+\beta(t)\dot{x}(t))\right\|^{2}+(b+qt^{q-1}-\alpha)t^{q}\|\dot{x}(t)\|^{2}$
$\displaystyle+bt^{q}\frac{\epsilon(t)}{2}\|x^{\ast}\|^{2}+\left(\frac{t^{2q}\dot{\epsilon}(t)}{2}+(2qt^{q-1}-b)\frac{t^{q}\epsilon(t)}{2}\right)\|x(t)\|^{2}$
$\displaystyle-\beta(t)\epsilon(t)a(t)\langle{\nabla}g(x(t)+\beta(t)\dot{x}(t)),x(t)\rangle+\frac{bq(1-q)t^{q-2}-bt^{q}\epsilon(t)}{2}\|x(t)-x^{\ast}\|^{2}$
$\displaystyle+\left(\left(-\beta(t)\frac{\alpha}{t^{q}}+\beta^{\prime}(t)+1\right)a(t)-t^{2q}+b\beta(t)t^{q}\right)\langle{\nabla}g(x(t)+\beta(t)\dot{x}(t)),\dot{x}(t)\rangle.$

Further, using the monotonicity of ${\nabla}g$ and the fact that
${\nabla}g(x^{\ast})=0$ we get for all $t\geq t_{1}$ that

(17)
$\displaystyle-\beta(t)\epsilon(t)a(t)\langle{\nabla}g(x(t)+\beta(t)\dot{x}(t)),x(t)\rangle$
$\displaystyle=-\beta(t)\epsilon(t)a(t)\langle{\nabla}g(x(t)+\beta(t)\dot{x}(t)),x(t)+\beta(t)\dot{x}(t)-x^{\ast}\rangle$
$\displaystyle+\beta(t)\epsilon(t)a(t)\langle{\nabla}g(x(t)+\beta(t)\dot{x}(t)),\beta(t)\dot{x}(t)-x^{\ast}\rangle$
$\displaystyle\leq\beta^{2}(t)\epsilon(t)a(t)\langle{\nabla}g(x(t)+\beta(t)\dot{x}(t)),\dot{x}(t)\rangle$
$\displaystyle-\beta(t)\epsilon(t)a(t)\langle{\nabla}g(x(t)+\beta(t)\dot{x}(t)),x^{\ast}\rangle.$

Consequently, by injecting (17) in (16) and taking into account that
$a(t)=t^{2q}$, it remains to estimate

$\left(\left(-\beta(t)\frac{\alpha}{t^{q}}+\beta^{\prime}(t)+\beta^{2}(t)\epsilon(t)\right)a(t)+b\beta(t)t^{q}\right)\langle{\nabla}g(x(t)+\beta(t)\dot{x}(t)),\dot{x}(t)\rangle$

and

$-\beta(t)\epsilon(t)a(t)\langle{\nabla}g(x(t)+\beta(t)\dot{x}(t)),x^{\ast}\rangle.$

Now, since $\beta(t)>0$, for all $t\geq t_{0}$ we get that for all $t\geq
t_{1}$ and for all $P,R>0$ one has

(18)
$\displaystyle\left(\left(-\beta(t)\frac{\alpha}{t^{q}}+\beta^{\prime}(t)+\beta^{2}(t)\epsilon(t)\right)a(t)+b\beta(t)t^{q}\right)\langle{\nabla}g(x(t)+\beta(t)\dot{x}(t)),\dot{x}(t)\rangle\leq$
$\displaystyle(\beta(t)(\alpha-b)t^{q}+|\beta^{\prime}(t)|t^{2q}+\beta^{2}(t)\epsilon(t)t^{2q})\left(\frac{\|{\nabla}g(x(t)+\beta(t)\dot{x}(t))\|^{2}}{2P}+\frac{P\|\dot{x}(t)\|^{2}}{2}\right)$

and

(19)
$\displaystyle-\beta(t)\epsilon(t)a(t)\langle{\nabla}g(x(t)+\beta(t)\dot{x}(t)),x^{\ast}\rangle\leq\beta(t)\epsilon(t)t^{2q}\left(\frac{Rt^{q}\|{\nabla}g(x(t)+\beta(t)\dot{x}(t))\|^{2}}{2}+\frac{\|x^{*}\|^{2}}{2Rt^{q}}\right)$

Hence, (16) leads to

(20) $\displaystyle\dot{\mathcal{E}}(t)\leq$
$\displaystyle(2qt^{2q-1}-bt^{q})\left(g(x(t)+\beta(t)\dot{x}(t))-g^{\ast}\right)$
$\displaystyle+\left((b+qt^{q-1}-\alpha)t^{q}+\frac{P(\beta(t)(\alpha-b)t^{q}+|\beta^{\prime}(t)|t^{2q}+\beta^{2}(t)\epsilon(t)t^{2q})}{2}\right)\|\dot{x}(t)\|^{2}$
$\displaystyle+\left(-\beta(t)t^{2q}+\frac{\beta(t)(\alpha-b)t^{q}+|\beta^{\prime}(t)|t^{2q}+\beta^{2}(t)\epsilon(t)t^{2q}}{2P}+\frac{R\beta(t)\epsilon(t)t^{3q}}{2}\right)\left\|{\nabla}g(x(t)+\beta(t)\dot{x}(t))\right\|^{2}$
$\displaystyle+\left(\frac{t^{2q}\dot{\epsilon}(t)}{2}+(2qt^{q-1}-b)\frac{t^{q}\epsilon(t)}{2}\right)\|x(t)\|^{2}+\frac{bq(1-q)t^{q-2}-bt^{q}\epsilon(t)}{2}\|x(t)-x^{\ast}\|^{2}$
$\displaystyle+\left(\frac{\beta(t)}{2R}+\frac{b}{2}\right)t^{q}\epsilon(t)\|x^{\ast}\|^{2}\mbox{
for all }t\geq t_{1},\,P,R>0.$

Now, according to (C0) there exist $K>0$ and $\overline{t}_{1}\geq t_{1}$ such
that $\epsilon(t)\leq\frac{K}{t^{q}}\mbox{ for every }t\geq\overline{t}_{1}$
and taking into account that $\beta(t)=\gamma+\frac{\beta}{t^{q}}$ with
$\gamma>0,\,\beta\in\mathbb{R}$ or $\gamma=0,\,\beta>0$ and $q\in]0,1[$ we
conclude the following.

There exist $t_{1}^{\prime}\geq\overline{t}_{1}$ and $r_{0}>0$ such that

$2qt^{2q-1}-bt^{q}\leq-r_{0}t^{q}\mbox{ for all }t\geq t_{1}^{\prime}.$

One can choose $P_{0}>0$ such that for some $t_{1}^{\prime\prime}\geq
t_{1}^{\prime}$ and $r_{1}>0$ one has

$(b+qt^{q-1}-\alpha)t^{q}+\frac{P_{0}(\beta(t)(\alpha-b)t^{q}+|\beta^{\prime}(t)|t^{2q}+\beta^{2}(t)\epsilon(t)t^{2q})}{2}\leq-
r_{1}t^{q}\mbox{ for all }t\geq t_{1}^{\prime\prime}.$

If $\gamma>0,\,\beta\in\mathbb{R}$ then one can choose $R_{0}>0$ such that for
some $t_{1}^{\prime\prime\prime}\geq t_{1}^{\prime\prime}$ and $r_{2}>0$ one
has

$-\beta(t)t^{2q}+\frac{\beta(t)(\alpha-b)t^{q}+|\beta^{\prime}(t)|t^{2q}+\beta^{2}(t)\epsilon(t)t^{2q}}{2P_{0}}+\frac{R_{0}\beta(t)\epsilon(t)t^{3q}}{2}\leq-
r_{2}t^{2q},\mbox{ for all }t\geq t_{1}^{\prime\prime\prime}.$

If $\gamma=0,\beta>0$ then one can choose $R_{0}>0$ such that for some
$t_{1}^{\prime\prime\prime}\geq t_{1}^{\prime\prime}$ and $r_{2}>0$ one has

$-\beta(t)t^{2q}+\frac{\beta(t)(\alpha-b)t^{q}+|\beta^{\prime}(t)|t^{2q}+\beta^{2}(t)\epsilon(t)t^{2q}}{2P_{0}}+\frac{R_{0}\beta(t)\epsilon(t)t^{3q}}{2}\leq-
r_{2}t^{q},\mbox{ for all }t\geq t_{1}^{\prime\prime\prime}.$

Finally, since $\epsilon(t)$ is decreasing we get

$\frac{t^{2q}\dot{\epsilon}(t)}{2}+(2qt^{q-1}-b)\frac{t^{q}\epsilon(t)}{2}\leq(2qt^{q-1}-b)\frac{t^{q}\epsilon(t)}{2}\leq-\frac{r_{0}}{2}t^{q}\epsilon(t),\mbox{
for all }t\geq t_{1}^{\prime}.$

Further, using that $\epsilon(t)\geq\frac{q(1-q)}{t^{2}}$ for $t$ big enough
we obtain that there exists $t_{2}\geq t_{1}^{\prime\prime\prime}$ and
$r_{3}>0$ such that $\frac{bq(1-q)t^{q-2}-bt^{q}\epsilon(t)}{2}\leq 0$ and
$\frac{\beta(t)}{2R_{0}}+\frac{b}{2}<r_{3}$ for all $t\geq t_{2}.$

Hence, by denoting $r_{2}(t)=r_{2}t^{2q}$ when $\gamma>0,\,\beta\in\mathbb{R}$
and $r_{2}(t)=r_{2}t^{q}$ whenever $\gamma=0,\,\beta>0,$ relation (20) leads
to

(21) $\displaystyle\dot{\mathcal{E}}(t)$
$\displaystyle+r_{0}t^{q}\left(g(x(t)+\beta(t)\dot{x}(t))-g^{\ast}\right)+r_{1}t^{q}\|\dot{x}(t)\|^{2}+r_{2}(t)\left\|{\nabla}g(x(t)+\beta(t)\dot{x}(t))\right\|^{2}$
$\displaystyle+\frac{r_{0}}{2}t^{q}\epsilon(t)\|x(t)\|^{2}\leq
r_{3}t^{q}\epsilon(t)\|x^{\ast}\|^{2},\mbox{ for all }t\geq t_{2}.$

By integrating (21) on the interval $[t_{2},t]$, we obtain for every $t\geq
t_{2}$ that

(22) $\displaystyle{\mathcal{E}}(t)$
$\displaystyle+Cr_{0}\int_{t_{2}}^{t}s^{q}\left(g(x(s)+\beta(s)\dot{x}(s))-g^{\ast}\right)ds+r_{1}\int_{t_{2}}^{t}s^{q}\|\dot{x}(s)\|^{2}ds+\int_{t_{2}}^{t}r_{2}(s)\left\|{\nabla}g(x(s)+\beta(s)\dot{x}(s))\right\|^{2}ds$
$\displaystyle+\frac{r_{0}}{2}\int_{t_{2}}^{t}s^{q}\epsilon(s)\|x(s)\|^{2}ds\leq\frac{r_{3}\|x^{\ast}\|^{2}}{2}\int_{t_{2}}^{t}s^{q}\epsilon(s)ds+\mathcal{E}(t_{2}).$

Now if we assume that
$\int_{t_{0}}^{+\infty}\frac{\epsilon(s)}{s^{q}}ds<+\infty,$ then from (9) and
from (22) we get that for all $t\geq t_{2}$ one has

$g(x(t)+\beta(t)\dot{x}(t))-\min
g\leq\frac{\mathcal{E}(t_{2})}{t^{2q}}+\frac{r_{3}\|x^{\ast}\|^{2}}{2t^{2q}}\int_{t_{2}}^{t}s^{q}\epsilon(s)ds,$
$\left\|\frac{b}{t^{q}}(x(t)-x^{\ast})+\dot{x}(t)\right\|^{2}\leq\frac{2\mathcal{E}(t_{2})}{t^{2q}}+\frac{r_{3}\|x^{\ast}\|^{2}}{t^{2q}}\int_{t_{2}}^{t}s^{q}\epsilon(s)ds$

and

$\frac{b(\alpha-
qt^{q-1}-b)}{2}\left\|\frac{x(t)-x^{\ast}}{t^{q}}\right\|^{2}\leq\frac{\mathcal{E}(t_{2})}{t^{2q}}+\frac{r_{3}\|x^{\ast}\|^{2}}{2t^{2q}}\int_{t_{2}}^{t}s^{q}\epsilon(s)ds.$

Obviously, $\lim\limits_{t\to+\infty}\frac{\mathcal{E}(t_{3})}{t^{2q}}=0.$
Further, Lemma (13) applied to the functions $\varphi(s)=s^{2q}$ and
$f(s)=\frac{\epsilon(s)}{s^{q}}$ provides
$\lim\limits_{t\to+\infty}\frac{1}{t^{2q}}\int_{t_{2}}^{t}s^{2q}\frac{\epsilon(s)}{s^{q}}dt=0.$
Hence,

$\lim\limits_{t\to+\infty}g(x(t)+\beta(t)\dot{x}(t))=\min g,$
$\lim\limits_{t\to+\infty}\left\|\frac{b}{t^{q}}(x(t)-x^{\ast})+\dot{x}(t)\right\|=0$

and

$\lim\limits_{t\to+\infty}\left\|\frac{x(t)-x^{\ast}}{t^{q}}\right\|=0.$

Combining the last two relations we get
$\lim\limits_{t\to+\infty}\left\|\dot{x}(t)\right\|=0,$ and from here and the
continuity of $g$ we have

$\lim\limits_{t\to+\infty}g(x(t))=\lim\limits_{t\to+\infty}g(x(t)+\beta(t)\dot{x}(t))=\min
g.$

$\blacksquare$

######  Remark 4.

In the hypotheses of Theorem 3 we assumed that the Tikhonov regularization
parameter $\epsilon(t)$ satisfies (C0) and
$\int_{t_{0}}^{+\infty}\frac{\epsilon(t)}{t^{q}}dt<+\infty.$ Note that if
$q>\frac{1}{2}$ then (C0) implies that
$\int_{t_{0}}^{+\infty}\frac{\epsilon(t)}{t^{q}}dt<+\infty.$

Indeed, according to (C0) there exists $K>0$ and $t_{1}\geq t_{0}$ such that
$\epsilon(t)\leq\frac{K}{t^{q}}$ for all $t\geq t_{1}.$ Hence,

$\int_{t_{0}}^{+\infty}\frac{\epsilon(t)}{t^{q}}dt=\int_{t_{0}}^{t_{1}}\frac{\epsilon(t)}{t^{q}}dt+\int_{t_{1}}^{+\infty}\frac{\epsilon(t)}{t^{q}}dt\leq\int_{t_{0}}^{t_{1}}\frac{\epsilon(t)}{t^{q}}dt+\int_{t_{1}}^{+\infty}\frac{K}{t^{2q}}dt<+\infty.$

When we specify $\epsilon(t)=\frac{a}{t^{p}},\,a,p>0$ then (C0) is nothing
else that $q\leq p\leq 2$ and if $p=2$ then $a\geq q(1-q).$ In this case
$\int_{t_{0}}^{+\infty}\frac{\epsilon(t)}{t^{q}}dt<+\infty$ whenever $p>1-q.$

Hence, for $\max(q,1-q)<p\leq 2$ and if $p=2$ then $a\geq q(1-q)$ then the
conclusion of Theorem 3 remains valid.

In what follows, under some stronger conditions imposed on the Tikhonov
regularization parameter, we obtain pointwise and integral estimates for the
function values, velocity and gradient.

######  Theorem 5.

For some starting points $u_{0},\,v_{0}\in\mathcal{H}$ let
$x:[t_{0},\infty)\to\mathcal{H}$ be the unique global solution of (DS).
Suppose that the condition (C0) holds and the Tikhonov regularization
parameter satisfies $\int_{t_{0}}^{+\infty}t^{q}\epsilon(t)dt<+\infty$. Then,
the following integral and pointwise estimates are valid.

One has that $\int_{t_{0}}^{+\infty}t^{q}\|\dot{x}(t)\|^{2}dt<+\infty\text{
and
}\int_{t_{0}}^{+\infty}t^{q}\left(g\left(x(t)+\beta(t)\dot{x}(t)\right)-\min
g\right)dt<+\infty.$ Further,

$\int_{t_{0}}^{+\infty}t^{2q}\left\|{\nabla}g\left(x(t)+\beta(t)\dot{x}(t)\right)\right\|^{2}dt<+\infty,$
if $\gamma>0,\,\beta\in\mathbb{R}$ and
$\int_{t_{0}}^{+\infty}t^{q}\left\|{\nabla}g\left(x(t)+\beta(t)\dot{x}(t)\right)\right\|^{2}dt<+\infty,$
whenever $\gamma=0,\,\beta>0.$

Moreover, $\|\dot{x}(t)\|=o\left(\frac{1}{t^{q}}\right)\mbox{ as
}t\to+\infty$, $g\left(x(t)+\beta(t)\dot{x}(t)\right)-\min
g=o\left(\frac{1}{t^{2q}}\right)\text{ as }t\to+\infty$ and
$\|x(t)\|=o\left(\frac{1}{t^{q}\sqrt{\epsilon(t)}}\right)\mbox{ as
}t\to+\infty.$

######  Proof .

First we evaluate the right hand side of (22). By using the assumption
$\int_{t_{0}}^{+\infty}s^{q}\epsilon(s)ds<+\infty$ we conclude that
$\frac{r_{3}}{2}\|x^{\ast}\|^{2}\int_{t_{2}}^{+\infty}t^{q}\epsilon(t)dt+\mathcal{E}(t_{2})<+\infty$,
a hence (22) leads to

(23) $\displaystyle{\mathcal{E}}(t)$
$\displaystyle+Cr_{0}\int_{t_{2}}^{t}s^{q}\left(g(x(s)+\beta(s)\dot{x}(s))-g^{\ast}\right)ds+r_{1}\int_{t_{2}}^{t}s^{q}\|\dot{x}(s)\|^{2}ds+\int_{t_{2}}^{t}r_{2}(s)\left\|{\nabla}g(x(s)+\beta(s)\dot{x}(s))\right\|^{2}ds$
$\displaystyle+\frac{r_{0}}{2}\int_{t_{2}}^{t}s^{q}\epsilon(s)\|x(s)\|^{2}ds\leq
C_{0},$

for some $C_{0}>0.$

Now, from (23) we immediately deduce that ${\mathcal{E}}(t)<+\infty$ for all
$t\geq t_{0}.$ Hence, taking into account the definition of $\mathcal{E}$ we
obtain
$g(x(t)+\beta(t)\dot{x}(t))-g^{*}=\mathcal{O}\left(\frac{1}{t^{2q}}\right),\mbox{
as }t\to+\infty,$ $\sup_{t\geq
t_{0}}\|b(x(t)-x^{\ast})+t^{q}\dot{x}(t)\|^{2}<+\infty,$ and $\sup_{t\geq
t_{0}}\|x(t)-x^{\ast}\|^{2}<+\infty.$

The latter two relations give us at once
$\|\dot{x}(t)\|=\mathcal{O}\left(\frac{1}{t^{2q}}\right),\mbox{ as
}t\to+\infty$ and that the trajectory $x(t)$ is bounded.

Further, (23) yields the following integral estimates.

(24)
$\displaystyle\int_{t_{0}}^{+\infty}t^{q}\left(g(x(t)+\beta(t)\dot{x}(t))-g^{\ast}\right)dt<+\infty,$
(25) $\displaystyle\int_{t_{0}}^{+\infty}t^{q}\|\dot{x}(t)\|^{2}dt<+\infty,$
(26)
$\displaystyle\int_{t_{0}}^{+\infty}t^{2q}\left\|{\nabla}g(x(t)+\beta(t)\dot{x}(t))\right\|^{2}dt<+\infty,\mbox{
whenever }\ \gamma>0,\,\beta\in\mathbb{R},$ (27)
$\displaystyle\int_{t_{0}}^{+\infty}t^{q}\left\|{\nabla}g(x(t)+\beta(t)\dot{x}(t))\right\|^{2}dt<+\infty,\mbox{
whenever }\ \gamma=0,\,\beta>0,$ (28)
$\displaystyle\int_{t_{0}}^{+\infty}t^{q}\epsilon(t)\|x(t)\|^{2}dt<+\infty$

In order to show that
$g(x(t)+\beta(t)\dot{x}(t))-g^{*}=o\left(\frac{1}{t^{2q}}\right),\mbox{ as
}t\to+\infty,$ and $\|\dot{x}(t)\|=o\left(\dfrac{1}{t^{q}}\right),\mbox{ as
}t\to+\infty$ assume for now that the limit
$\lim\limits_{t\to+\infty}\|x(t)-x^{\ast}\|$ exists, which shall be proved in
Theorem 6. Then, (21) provides that

$\dot{\mathcal{E}}(t)\leq\frac{r_{3}\|x^{\ast}\|^{2}}{2}t^{q}\epsilon(t),\mbox{
for all }t\geq t_{2}.$

Obviously, by the hypotheses we have
$\dfrac{r_{3}\|x^{\ast}\|^{2}}{2}t^{q}\epsilon(t)\in L^{1}([t_{2},+\infty[)$,
hence, according to Lemma (14) there exists the limit
$\lim\limits_{t\to+\infty}\mathcal{E}(t).$

Since the limit $\lim\limits_{t\to+\infty}\|x(t)-x^{\ast}\|$ exists we get
that the limit

(29)
$\lim\limits_{t\to+\infty}t^{2q}\left(g(x(t)+\beta(t)\dot{x}(t))-g^{\ast}\right)+\frac{t^{2q}\epsilon(t)}{2}\|x(t)\|^{2}+\frac{1}{2}\|t^{q}\dot{x}(t)\|^{2}$

also exists.
By combining (24), (25) and (28) we find that

(30)
$\int_{t_{0}}^{+\infty}\frac{1}{t^{q}}\left(t^{2q}\left(g(x(t)+\beta(t)\dot{x}(t))-g^{\ast}\right)+\frac{t^{2q}\epsilon(t)}{2}\|x(t)\|^{2}+\frac{1}{2}\|t^{q}\dot{x}(t)\|^{2}\right)dt<+\infty.$

Since $q<1$, the function $t\longrightarrow\frac{1}{t^{q}}\not\in
L^{1}([t_{0},+\infty[)$, therefore (29) and (30) lead to

(31)
$\lim\limits_{t\to+\infty}t^{2q}\left(g(x(t)+\beta(t)\dot{x}(t))-g^{\ast}\right)+\frac{t^{2q}\epsilon(t)}{2}\|x(t)\|^{2}+\frac{1}{2}\|t^{q}\dot{x}(t)\|^{2}=0.$

Consequently, $g(x(t)+\beta(t)\dot{x}(t))-\min
g=o\left(\frac{1}{t^{2q}}\right)\mbox{ as }t\to+\infty,$
$\|x(t)\|=o\left(\frac{1}{t^{q}\sqrt{\epsilon(t)}}\right)\mbox{ as
}t\to+\infty,$ and $\|\dot{x}(t)\|=o\left(\frac{1}{t^{q}}\right)\mbox{ as
}t\to+\infty.$ $\blacksquare$

In what follows we show that under the hypotheses of Theorem 5, the
trajectories generated by the dynamical system (DS) converge weakly to a
minimizer of the objective function $g.$

######  Theorem 6.

For some starting points $u_{0},\,v_{0}\in\mathcal{H}$ let
$x:[t_{0},\infty)\to\mathcal{H}$ be the unique global solution of (DS).
Suppose that the condition (C0) holds and the Tikhonov regularization
parameter satisfies $\int_{t_{0}}^{+\infty}t^{q}\epsilon(t)dt<+\infty$.

Then the trajectory $x(t)$ converges weakly, as $t\to+\infty$, to an element
of $\operatorname*{argmin}g$.

######  Proof .

We will use Opial’s lemma (see Lemma 15 from Appendix). In order to prove the
existence of the weak limit of $x(t)$, we will use Lemma 16 from [26] (see
Lemma 16 from Appendix) and Lemma 6 from [26]. For
$x^{\ast}\in\operatorname*{argmin}g$ let us introduce the anchor function
$w(t)=\frac{1}{2}\|x(t)-x^{\ast}\|^{2}.$ The classical derivation chain rule
leads to

$\ddot{w}(t)+\frac{\alpha}{t^{q}}\dot{w}(t)=\left\langle\ddot{x}(t)+\frac{\alpha}{t^{q}}\dot{x}(t),x(t)-x^{\ast}\right\rangle+\|\dot{x}(t)\|^{2}=\left\langle-{\nabla}g(x(t)+\beta(t)\dot{x}(t))-\epsilon(t){x}(t),x(t)-x^{\ast}\right\rangle+\|\dot{x}(t)\|^{2}.$

We consider

$\langle-{\nabla}g(x(t)+\beta(t)\dot{x}(t)),x(t)-x^{\ast}\rangle=\langle-{\nabla}g(x(t)+\beta(t)\dot{x}(t)),x(t)+\beta(t)\dot{x}(t)-x^{\ast}\rangle+\beta(t)\langle{\nabla}g(x(t)+\beta(t)\dot{x}(t)),\dot{x}(t)\rangle.$

From the monotonicity of ${\nabla}g$ we have that

$\langle-{\nabla}g(x(t)+\beta(t)\dot{x}(t)),x(t)+\beta(t)\dot{x}(t)-x^{\ast}\rangle\leq
0,$

from which it follows that

$\langle-{\nabla}g(x(t)+\beta(t)\dot{x}(t)),x(t)-x^{\ast}\rangle\leq\beta(t)\langle{\nabla}g(x(t)+\beta(t)\dot{x}(t)),\dot{x}(t)\rangle\leq\frac{|\beta(t)|}{2}\left(\|{\nabla}g(x(t)+\beta(t)\dot{x}(t))\|^{2}+\|\dot{x}(t)\|^{2}\right).$

Using that
$|\beta(t)|\leq\gamma+\dfrac{|\beta|}{t}\leq\tilde{K}=\gamma+\dfrac{|\beta|}{t_{0}}$,
it follows that

$\langle-{\nabla}g(x(t)+\beta(t)\dot{x}(t)),x(t)-x^{\ast}\rangle\leq\frac{\tilde{K}}{2}\left(\|{\nabla}g(x(t)+\beta(t)\dot{x}(t))\|^{2}+\|\dot{x}(t)\|^{2}\right).$

By employing that

$\left\langle-\epsilon(t)x(t),x(t)-x^{\ast}\right\rangle=\frac{\epsilon(t)}{2}(\|x^{\ast}\|^{2}-\|x(t)\|^{2}-\|x(t)-x^{\ast}\|^{2}),$

and using that $\epsilon(t)\in\mathbb{R}_{+}$, it follows that

(32)
$\ddot{w}(t)+\frac{\alpha}{t^{q}}\dot{w}(t)\leq\frac{\epsilon(t)}{2}\|x^{\ast}\|^{2}+\left(1+\frac{\tilde{K}}{2}\right)\|\dot{x}(t)\|^{2}+\dfrac{|\beta(t)|}{2}\|{\nabla}g(x(t)+\beta(t)\dot{x}(t))\|^{2}.$

In what follows, let us consider
$k(t)=\frac{\epsilon(t)}{2}\|x^{\ast}\|^{2}+\left(1+\frac{\tilde{K}}{2}\right)\|\dot{x}(t)\|^{2}+\dfrac{\tilde{K}}{2}\|{\nabla}g(x(t)+\beta(t)\dot{x}(t))\|^{2}$.
Note that by the hypotheses of the theorem we have
$t^{q}\frac{\epsilon(t)}{2}\|x^{\ast}\|^{2}\in L^{1}[t_{0},+\infty[$. Further,
using the integral estimate (25) it follows that $t^{q}\|\dot{x}(t)\|^{2}\in
L^{1}[t_{0},+\infty[$, hence
$\left(1+\frac{\tilde{K}}{2}\right)t^{q}\|\dot{x}(t)\|^{2}\in
L^{1}[t_{0},+\infty[$. Moreover, in one hand for the case when
$\gamma=0,\beta>0$, the integral estimate (27) leads to $\beta
t^{q}\|{\nabla}g(x(t)+\beta(t)\dot{x}(t))\|^{2}\in L^{1}[t_{0},+\infty[$. By
taking into account that in this case $t^{q}|\beta(t)|=\beta$ we find that
$\frac{|\beta(t)|}{2}t^{q}\|{\nabla}g(x(t)+\beta(t)\dot{x}(t))\|^{2}\in
L^{1}[t_{0},+\infty[$. On the other hand, when $\gamma>0,\beta\in\mathbb{R}$,
from (26) we have that $t^{2q}\|{\nabla}g(x(t)+\beta(t)\dot{x}(t))\|^{2}\in
L^{1}[t_{0},+\infty[$, which evidently implies that
$t^{q}\|{\nabla}g(x(t)+\beta(t)\dot{x}(t))\|^{2}\in L^{1}[t_{0},+\infty[$, and
we finally obtain
$\dfrac{|\beta(t)|}{2}t^{q}\|{\nabla}g(x(t)+\beta(t)\dot{x}(t))\|^{2}\in
L^{1}[t_{0},+\infty[$.

Hence, we get that $t^{q}k(t)\in L^{1}[t_{0},+\infty[$. Now, by applying the
same reasoning as in Lemma 6 from [26], we obtain the existence of the limit
$\lim\limits_{t\to+\infty}\|x(t)-x^{\ast}\|$. Now, if
$\overline{x}\in\mathcal{H}$ is a weak sequential limit point of $x(t)$ then
there exists a sequence $(t_{n})_{n\in\mathbb{N}}\subseteq[t_{0},+\infty[$
such that $\lim_{n\to+\infty}t_{n}=+\infty$ and $x(t_{n})$ converges weakly to
$\overline{x}$ as $n\to+\infty.$ Obviously the function $g$ is weakly lower
semicontinuous, since it is convex and continuous, consequently
$g(\overline{x})\leq\liminf_{n\to+\infty}g(x(t_{n}))=\lim_{n\to+\infty}g(x(t_{n}))=g^{*}=\min
g$, which shows that $\overline{x}\in\operatorname*{argmin}g.$ According to
the Opial lemma it follows that

$w-\lim_{t\to+\infty}x(t)\in\operatorname*{argmin}g.$

$\blacksquare$

######  Remark 7.

Let us consider $\epsilon(t)=\frac{a}{t^{p}}$ where $a,\,p>0$. Then, as we
have seen before, condition (C0) is satisfied whenever $2\geq p\geq q$, and
when $p=2$, then $a\geq q(1-q)$. Now, the condition
$\int_{t_{0}}^{+\infty}t^{q}\epsilon(t)dt<+\infty$ leads to $p>1+q$, thus for
$p\in]1+q,\,2]$ the conclusions of Theorem 5 and Theorem 6 hold. Note that
this fact is in concordance with the results obtained in [26] and [21].

Consequently the following result is valid.

######  Corollary 8.

Consider $0<q<1$ and $\epsilon(t)=\frac{a}{t^{p}},\,a>0$ and assume that
$q+1<p\leq 2.$ Assume further that for $p=2$ then $a\geq q(1-q)$. For some
starting points $u_{0},\,v_{0}\in\mathcal{H}$ let
$x:[t_{0},+\infty[\to\mathcal{H}$ be the unique global solution of (DS). Then,
the trajectory $x(t)$ converges weakly, as $t\to+\infty$, to an element of
$\operatorname*{argmin}g$. Further, one has
$\|\dot{x}(t)\|=o\left(t^{-q}\right),$
$g\left(x(t)+\beta(t)\dot{x}(t)\right)-\min g=o\left(t^{-2q}\right)\text{ as
}t\to+\infty$ and
$\int_{t_{0}}^{+\infty}t^{q}\|\dot{x}(t)\|^{2}dt<+\infty,\,\int_{t_{0}}^{+\infty}t^{q}\left(g\left(x(t)+\beta(t)\dot{x}(t)\right)-\min
g\right)dt<+\infty.$ Concerning the gradient one has the integral estimates
$\int_{t_{0}}^{+\infty}t^{2q}\left\|{\nabla}g\left(x(t)+\beta(t)\dot{x}(t)\right)\right\|^{2}dt<+\infty,$
if $\gamma>0,\,\beta\in\mathbb{R}$ and
$\int_{t_{0}}^{+\infty}t^{2q-1}\left\|{\nabla}g\left(x(t)+\beta(t)\dot{x}(t)\right)\right\|^{2}dt<+\infty,$
if $\gamma=0,\,\beta>0.$

Next we show that even in case $p=q+1$ the generated trajectory $x$ is bounded
and some pointwise estimates hold.

######  Theorem 9.

Consider $0<q<1$ and let $\epsilon(t)=\frac{a}{t^{q+1}},\,a>0.$ For some
starting points $u_{0},\,v_{0}\in\mathcal{H}$ let
$x:[t_{0},+\infty[\to\mathcal{H}$ be the unique global solution of (DS). Then,
the trajectory $x(t)$ is bounded and one has
$\|\dot{x}(t)\|=\mathcal{O}\left(t^{-q}\right),$
$g\left(x(t)+\beta(t)\dot{x}(t)\right)-\min
g=\mathcal{O}\left(t^{-2q}\right)\text{ as }t\to+\infty.$

######  Proof .

We consider the energy functional defined in the proof of Theorem 3. In this
special case (20) reads as

(33) $\displaystyle\dot{\mathcal{E}}(t)\leq$
$\displaystyle(2qt^{2q-1}-bt^{q})\left(g(x(t)+\beta(t)\dot{x}(t))-g^{\ast}\right)$
$\displaystyle+\left((b+qt^{q-1}-\alpha)t^{q}+\frac{P(\beta(t)(\alpha-b)t^{q}+|\beta^{\prime}(t)|t^{2q}+a\beta^{2}(t)t^{q-1})}{2}\right)\|\dot{x}(t)\|^{2}$
$\displaystyle+\left(-\beta(t)t^{2q}+\frac{\beta(t)(\alpha-b)t^{q}+|\beta^{\prime}(t)|t^{2q}+a\beta^{2}(t)t^{q-1}}{2P}+\frac{aR\beta(t)t^{2q-1}}{2}\right)\left\|{\nabla}g(x(t)+\beta(t)\dot{x}(t))\right\|^{2}$
$\displaystyle+\left(\frac{-a(q+1)t^{q-2}}{2}+(2qt^{q-1}-b)\frac{at^{-1}}{2}\right)\|x(t)\|^{2}+\frac{bq(1-q)t^{q-2}-abt^{-1}}{2}\|x(t)-x^{\ast}\|^{2}$
$\displaystyle+\left(\frac{\beta(t)}{2R}+\frac{b}{2}\right)at^{-1}\|x^{\ast}\|^{2}\mbox{
for all }t\geq t_{1},\,P,R>0.$

Further,

(34) $\displaystyle\mathcal{E}(t)$
$\displaystyle=t^{2q}\left(g(x(t)+\beta(t)\dot{x}(t))-g^{\ast}\right)+\frac{at^{q-1}}{2}\|x(t)\|^{2}+\frac{1}{2}\|b(x(t)-x^{\ast})+t^{q}\dot{x}(t)\|^{2}+\frac{b(\alpha-
qt^{q-1}-b)}{2}\|x(t)-x^{\ast}\|^{2}$ $\displaystyle\leq
t^{2q}\left(g(x(t)+\beta(t)\dot{x}(t))-g^{\ast}\right)+\frac{at^{q-1}}{2}\|x(t)\|^{2}+t^{2q}\|\dot{x}(t)\|^{2}+\frac{b(\alpha-
qt^{q-1}+b)}{2}\|x(t)-x^{\ast}\|^{2},$

for all $t\geq t_{1}.$ Let us consider now $L>0$ which will be specified
later. Then, (33) and (34) lead to

(35) $\displaystyle\dot{\mathcal{E}}(t)+\frac{L}{t}\mathcal{E}(t)\leq$
$\displaystyle((2q+L)t^{2q-1}-bt^{q})\left(g(x(t)+\beta(t)\dot{x}(t))-g^{\ast}\right)$
$\displaystyle+\left((b+(q+L)t^{q-1}-\alpha)t^{q}+\frac{P(\beta(t)(\alpha-b)t^{q}+|\beta^{\prime}(t)|t^{2q}+a\beta^{2}(t)t^{q-1})}{2}\right)\|\dot{x}(t)\|^{2}$
$\displaystyle+\left(-\beta(t)t^{2q}+\frac{\beta(t)(\alpha-b)t^{q}+|\beta^{\prime}(t)|t^{2q}+a\beta^{2}(t)t^{q-1}}{2P}+\frac{aR\beta(t)t^{2q-1}}{2}\right)\left\|{\nabla}g(x(t)+\beta(t)\dot{x}(t))\right\|^{2}$
$\displaystyle+\left(\frac{a(L+q-1)t^{q-2}}{2}-\frac{abt^{-1}}{2}\right)\|x(t)\|^{2}$
$\displaystyle+\frac{bq(1-q-L)t^{q-2}+b((\alpha+b)L-a)t^{-1}}{2}\|x(t)-x^{\ast}\|^{2}$
$\displaystyle+\left(\frac{\beta(t)}{2R}+\frac{b}{2}\right)at^{-1}\|x^{\ast}\|^{2}\mbox{
for all }t\geq t_{1},\,P,R>0.$

Now, by taking into account that $\beta(t)=\gamma+\frac{\beta}{t^{q}}$ with
$\gamma>0,\,\beta\in\mathbb{R}$ or $\gamma=0,\,\beta>0$ and $q\in]0,1[$ we
conclude the following.

There exist $t_{1}^{\prime}\geq t_{1}$ and $r_{0}>0$ such that

$(2q+L)t^{2q-1}-bt^{q}\leq 0\mbox{ for all }t\geq t_{1}^{\prime}.$

One can choose $P_{0}>0$ such that for some $t_{1}^{\prime\prime}\geq
t_{1}^{\prime}$ one has

$(b+(q+L)t^{q-1}-\alpha)t^{q}+\frac{P_{0}(\beta(t)(\alpha-b)t^{q}+|\beta^{\prime}(t)|t^{2q}+a\beta^{2}(t)t^{q-1})}{2}\leq
0\mbox{ for all }t\geq t_{1}^{\prime\prime}.$

One can choose $R_{0}>0$ such that for some $t_{1}^{\prime\prime\prime}\geq
t_{1}^{\prime\prime}$ one has

$-\beta(t)t^{2q}+\frac{\beta(t)(\alpha-b)t^{q}+|\beta^{\prime}(t)|t^{2q}+a\beta^{2}(t)t^{q-1}}{2P_{0}}+\frac{aR_{0}\beta(t)t^{2q-1}}{2}\leq
0,\mbox{ for all }t\geq t_{1}^{\prime\prime\prime}.$

Finally, let $L<\frac{a}{\alpha+b}.$ Then there exists $t_{2}\geq
t_{1}^{\prime\prime\prime}$ such that

$\frac{a(L+q-1)t^{q-2}}{2}-\frac{abt^{-1}}{2}\leq 0,\mbox{ for all }t\geq
t_{2}$

and

$bq(1-q-L)t^{q-2}+b((\alpha+b)L-a)t^{-1}\leq 0,\mbox{ for all }t\geq t_{2}.$

Consequently, by denoting $\sup_{t\geq
t_{2}}\left(\frac{\beta(t)}{2R_{0}}+\frac{b}{2}\right)a\|x^{\ast}\|^{2}=M$,
(35) leads to

(36)
$\displaystyle\dot{\mathcal{E}}(t)+\frac{L}{t}\mathcal{E}(t)\leq\frac{M}{t}\mbox{
for all }t\geq t_{2}.$

We multiply (36) with $t^{L}$ then integrate on an interval $[t_{2},T]$ in
order to obtain

$\mathcal{E}(T)\leq\frac{M}{L}+\frac{t_{2}^{L}\mathcal{E}(t_{2})-\frac{M}{L}t_{2}^{L}}{T^{L}},$

that is $\mathcal{E}(t)$ is bounded. Taking into account the definition of
$\mathcal{E}(t)$ we obtain at once that $x(t)$ is bounded,
$g(x(t)+\beta(t)\dot{x}(t))-g^{\ast}=\mathcal{O}(t^{-2q})$ and
$\|\dot{x}(t)\|=\mathcal{O}(t^{-q})$ as $t\to+\infty.$ $\blacksquare$

##  4\. On the strong convergence of the trajectories generated by the
dynamical system (DS)

In the present section we will recall the idea of Tikhonov regularization,
which is linked to the strong convergence results to a minimizer of the
objective function of minimal norm regarding our dynamical system (DS). Let us
denote $x_{t}$ the unique solution of the strongly convex minimization problem
$\min_{x\in\mathcal{H}}g_{t}(x),$ where
$g_{t}(x)=g(x)+\frac{\epsilon(t)}{2}\|x\|^{2}$. From [12] we know that the
Tikhonov approximation curve $t\to x_{t}$ satisfies
$x^{\ast}=\lim\limits_{t\to+\infty}x_{t}$, where
$x^{\ast}=\operatorname*{argmin}\limits_{x\in\operatorname*{argmin}g}\|x\|$ is
the minimal norm element from the set $\operatorname*{argmin}g.$ Therefore,
$x^{\ast}=\operatorname*{proj}_{\operatorname*{argmin}g}0$ and
$\|x_{t}\|\leq\|x^{\ast}\|$ (see [18]). By employing the definition of the
strongly convex function $g_{t}(x)$ we get that

(37) ${\nabla}g_{t}(y)={\nabla}g(y)+\epsilon(t)y,$

and by taking into account the unique minimum denoted as $x_{t}$, it follows
that

(38) ${\nabla}g_{t}(x_{t})={\nabla}g(x_{t})+\epsilon(t)x_{t}=0.$

As claimed in Lemma 2 from [6] the derivable almost everywhere function
$t\longrightarrow x_{t}$ is endowed with the property

(39)
$\left\|\frac{d}{dt}x_{t}\right\|\leq-\frac{\dot{\epsilon}(t)}{\epsilon(t)}\|x_{t}\|\mbox{
for almost every }t\geq t_{0}.$

Because $g_{t}$ is strongly convex, the gradient inequality leads to

(40)
$g_{t}(y)-g_{t}(x)\geq\langle{\nabla}g_{t}(x),y-x\rangle+\frac{\epsilon(t)}{2}\|x-y\|^{2},\mbox{
for all }x,y\in\mathcal{H}.$

In particular

(41) $g_{t}(x)-g_{t}(x_{t})\geq\frac{\epsilon(t)}{2}\|x-x_{t}\|^{2},\mbox{ for
all }x\in\mathcal{H}.$

Let $y:[t_{0},+\infty[\to\mathcal{H},s\longrightarrow y(s)$ derivable at a
point $t\in]t_{0},+\infty[.$ Then, one has that

(42)
$\frac{d}{dt}g_{t}(y(t))=\langle{\nabla}g_{t}(y(t)),\dot{y}(t)\rangle+\frac{\dot{\epsilon}(t)}{2}\|y(t)\|^{2}.$

Finally, for all $x,y\in\mathcal{H}$, one has

(43)
$g(x)-g(y)=(g_{t}(x)-g_{t}(x_{t}))+(g_{t}(x_{t})-g_{t}(y))+\frac{\epsilon(t)}{2}(\|y\|^{2}-\|x\|^{2})\leq
g_{t}(x)-g_{t}(x_{t})+\frac{\epsilon(t)}{2}\|y\|^{2}.$

Now we are ready to prove the main result of the present section.

######  Theorem 10.

Consider $0<q<1$ and assume that the Tikhonov parametrization function
satisfies the following conditions:

(C1) $\displaystyle\text{There exists }K_{1}>0\text{ and }r\in(q,1)\text{ such
that }\epsilon(t)\geq\dfrac{K_{1}}{t^{r+q}}\mbox{ for }t\mbox{ big enough}.$
(C2) $\displaystyle\text{If }\gamma\neq 0\text{ then
}\epsilon(t)\leq\dfrac{\alpha}{2\gamma t^{q}}\mbox{ for }t\mbox{ big enough.}$
(C3) $\displaystyle\text{There exists }K_{3}>0\text{ such that
}\left(\dfrac{\dot{\epsilon}(t)}{\epsilon(t)}\right)^{2}\leq\dfrac{K_{3}}{t^{2}}\mbox{
for }t\mbox{ big enough}.$

For some starting points $u_{0},\,v_{0}\in\mathcal{H}$ let
$x:[t_{0},\infty)\to\mathcal{H}$ be the unique global solution of (DS). Then,
the following statements are true.

(i) (convergence of the trajectory) $x(t)$ converges strongly, as
$t\to+\infty$, to $x^{*}$, the element of minimum norm of
$\operatorname*{argmin}g$;

(ii) (fast convergence rates)
$g_{t}(x(t)+\beta(t)\dot{x}(t))-g_{t}(x_{t})=\mathcal{O}(t^{2r-2q-2}+\epsilon(t)t^{r-1}),$
$g(x(t)+\beta(t)\dot{x}(t))-\min g=\mathcal{O}(t^{2r-2q-2}+\epsilon(t)),$
$\|\dot{x}(t)\|=\mathcal{O}(t^{r-q-1}+\sqrt{\epsilon(t)}t^{\frac{r-1}{2}})$
and
$\|x(t)-x_{t}\|=\mathcal{O}(t^{r-1}+\sqrt{\epsilon(t)}t^{\frac{2q+r-1}{2}})\mbox{
as }t\to+\infty;$

(iii) (integral estimates) If $t^{2q-1}\epsilon(t)\in L^{1}[t_{0},+\infty[$
then
$\int_{t_{0}}^{+\infty}t^{q}\left(g_{t}(x(t)+\beta(t)\dot{x}(t))-g_{t}(x_{t})\right)dt<+\infty$
and $\int_{t_{0}}^{+\infty}t^{q}\|\dot{x}(t)\|^{2}dt<+\infty.$ Further,
$\int_{t_{0}}^{+\infty}t^{2q}\|{\nabla}g(x(t)+\beta(t)\dot{x}(t))+\epsilon(t)x(t)\|^{2}dt<+\infty,\mbox{
if }\gamma>0$ and
$\int_{t_{0}}^{+\infty}t^{2q-1}\|{\nabla}g(x(t)+\beta(t)\dot{x}(t))\|^{2}dt<+\infty,\mbox{
if }\gamma=0.$

######  Proof .

Let us define, for every $t\geq t_{0}$, the following energy functional

(44) $\displaystyle E(t)$
$\displaystyle=a(t)(g_{t}(x(t)+\beta(t)\dot{x}(t))-g_{t}(x_{t}))+\frac{1}{2}\|b(x(t)-x_{t})+c(t)\dot{x}(t)\|^{2}+\frac{d(t)}{2}\|x(t)-x_{t}\|^{2},$

where $b\in(0,\alpha)$, $c(t)=t^{q}$, $d(t)=b(\alpha-b-qt^{q-1})$ and
$a(t)=\dfrac{c^{2}(t)-bc(t)\beta(t)}{1+\beta^{\prime}(t)-\dfrac{\alpha\beta(t)}{t^{q}}+\beta^{2}(t)\epsilon(t)}.$
Obviously there exists $t_{1}\geq t_{0}$ such that $a(t)\geq 0$ and $d(t)\geq
0$ for all $t\geq t_{1}$. Observe further that $a(t)=\mathcal{O}(t^{2q})$ as
$t\to+\infty.$ Therefore, by deriving $E(t)$ we get

(45) $\displaystyle\dot{E}(t)$
$\displaystyle=a^{\prime}(t)(g_{t}(x(t)+\beta(t)\dot{x}(t))-g_{t}(x_{t}))+a(t)\left(\frac{d}{dt}g_{t}(x(t)+\beta(t)\dot{x}(t))-\frac{d}{dt}g_{t}(x_{t})\right)$
$\displaystyle+\left\langle
b(x(t)-x_{t})+c(t)\dot{x}(t),b\left(\dot{x}(t)-\frac{d}{dt}x_{t}\right)+c^{\prime}(t)\dot{x}(t)+c(t)\ddot{x}(t)\right\rangle$
$\displaystyle+\frac{d^{\prime}(t)}{2}\|x(t)-x_{t}\|^{2}+d(t)\left\langle
x(t)-x_{t},\dot{x}(t)-\frac{d}{dt}x_{t}\right\rangle\,,\text{ for all }t\geq
t_{1}.$

In one hand, from (38) and (42), we find that

(46)
$\displaystyle\frac{d}{dt}g_{t}(x_{t})=\left\langle{\nabla}g_{t}(x_{t}),\frac{d}{dt}x_{t}\right\rangle+\frac{\dot{\epsilon}(t)}{2}\|x_{t}\|^{2}=\frac{\dot{\epsilon}(t)}{2}\|x_{t}\|^{2}.$

On the other hand, using (DS) along with (42) and (37) it follows that

(47) $\displaystyle\frac{d}{dt}g_{t}(x(t)+\beta(t)\dot{x}(t))$
$\displaystyle=\frac{\dot{\epsilon}(t)}{2}\|x(t)+\beta(t)\dot{x}(t)\|^{2}+\beta(t)\epsilon(t)\left(1+\beta^{\prime}(t)-\frac{\alpha\beta(t)}{t^{q}}\right)\|\dot{x}(t)\|^{2}$
$\displaystyle-\beta(t)\epsilon^{2}(t)\|x(t)\|^{2}-\beta(t)\|{\nabla}g(x(t)+\beta(t)\dot{x}(t))\|^{2}$
$\displaystyle+\left(1+\beta^{\prime}(t)-\frac{\alpha\beta(t)}{t^{q}}-\beta^{2}(t)\epsilon(t)\right)\langle{\nabla}g(x(t)+\beta(t)\dot{x}(t)),\dot{x}(t)\rangle$
$\displaystyle+\left(\epsilon(t)\left(1+\beta^{\prime}(t)-\frac{\alpha\beta(t)}{t^{q}}\right)-\beta^{2}(t)\epsilon^{2}(t)\right)\langle
x(t),\dot{x}(t)\rangle$
$\displaystyle-2\beta(t)\epsilon(t)\langle{\nabla}g(x(t)+\beta(t)\dot{x}(t)),x(t)\rangle.$

Further, by using the dynamical system (DS), we obtain

(48) $\displaystyle\left\langle
b(x(t)-x_{t})+c(t)\dot{x}(t),b\dot{x}(t)+c^{\prime}(t)\dot{x}(t)+c(t)\ddot{x}(t)\right\rangle=c(t)\left(b+c^{\prime}(t)-\frac{\alpha
c(t)}{t^{q}}\right)\|\dot{x}(t)\|^{2}$
$\displaystyle+b\left(b+c^{\prime}(t)-\frac{\alpha c(t)}{t^{q}}\right)\langle
x(t)-x_{t},\dot{x}(t)\rangle-\epsilon(t)c(t)b\langle
x(t)-x_{t},x(t)\rangle-\epsilon(t)c^{2}(t)\langle x(t),\dot{x}(t)\rangle$
$\displaystyle-
bc(t)\langle{\nabla}g(x(t)+\beta(t)\dot{x}(t)),x(t)-x_{t}\rangle-c^{2}(t)\langle{\nabla}g(x(t)+\beta(t)\dot{x}(t)),\dot{x}(t)\rangle.$

Combining (45) with (46), (47) and (48), and taking into account that
$d(t)+b\left(b+c^{\prime}(t)-\frac{\alpha c(t)}{t^{q}}\right)=0$ and
$a(t)\left(1+\beta^{\prime}(t)-\frac{\alpha\beta(t)}{t^{q}}-\beta^{2}(t)\epsilon(t)\right)=c^{2}(t)-bc(t)\beta(t)$
it follows that

(49) $\displaystyle\dot{E}(t)$
$\displaystyle=a^{\prime}(t)\left(g_{t}(x(t)+\beta(t)\dot{x}(t))-g_{t}(x_{t})\right)+\frac{a(t)\dot{\epsilon}(t)}{2}\left(\|x(t)+\beta(t)\dot{x}(t)\|^{2}-\|x_{t}\|^{2}\right)$
$\displaystyle+\frac{d^{\prime}(t)}{2}\|x(t)-x_{t}\|^{2}-a(t)\beta(t)\|{\nabla}g(x(t)+\beta(t)\dot{x}(t))\|^{2}-a(t)\beta(t)\epsilon^{2}(t)\|x(t)\|^{2}$
$\displaystyle+\left(a(t)\beta(t)\epsilon(t)\left(1+\beta^{\prime}(t)-\frac{\alpha\beta(t)}{t^{q}}\right)+c(t)\left(b+c^{\prime}(t)-\frac{\alpha
c(t)}{t^{q}}\right)\right)\|\dot{x}(t)\|^{2}$ $\displaystyle-
bc(t)\beta(t)\epsilon(t)\langle x(t),\dot{x}(t)\rangle-b\epsilon(t)c(t)\langle
x(t)-x_{t},x(t)\rangle-
bc(t)\beta(t)\langle{\nabla}g(x(t)+\beta(t)\dot{x}(t)),\dot{x}(t)\rangle$
$\displaystyle-2\beta(t)a(t)\epsilon(t)\langle{\nabla}g(x(t)+\beta(t)\dot{x}(t)),x(t)\rangle-
bc(t)\langle{\nabla}g(x(t)+\beta(t)\dot{x}(t)),x(t)-x_{t}\rangle$
$\displaystyle-
bc(t)\left\langle\dot{x}(t),\frac{d}{dt}x_{t}\right\rangle-\left(b^{2}+d(t)\right)\left\langle
x(t)-x_{t},\frac{d}{dt}x_{t}\right\rangle\,,\text{ for all }t\geq t_{1}.$

Observe that by using (40) we get

$\displaystyle-bc(t)\beta(t)\epsilon(t)\langle
x(t),\dot{x}(t)\rangle-b\epsilon(t)c(t)\langle x(t)-x_{t},x(t)\rangle-
bc(t)\beta(t)\langle{\nabla}g(x(t)+\beta(t)\dot{x}(t)),\dot{x}(t)\rangle$
$\displaystyle-
bc(t)\langle{\nabla}g(x(t)+\beta(t)\dot{x}(t)),x(t)-x_{t}\rangle=-bc(t)\langle{\nabla}g_{t}(x(t)+\beta(t)\dot{x}(t)),x(t)+\beta(t)\dot{x}(t)-x_{t}\rangle$
$\displaystyle+b\beta(t)\epsilon(t)c(t)\langle\dot{x}(t),x(t)+\beta(t)\dot{x}(t)-x_{t}\rangle\leq
bc(t)\left(g_{t}(x_{t})-g_{t}(x(t)+\beta(t)\dot{x}(t))\right)$
$\displaystyle-\frac{bc(t)\epsilon(t)}{2}\|x(t)+\beta(t)\dot{x}(t)-x_{t}\|^{2}+b\beta(t)\epsilon(t)c(t)\langle\dot{x}(t),x(t)+\beta(t)\dot{x}(t)-x_{t}\rangle=$
$\displaystyle
bc(t)\left(g_{t}(x_{t})-g_{t}(x(t)+\beta(t)\dot{x}(t))\right)-\frac{bc(t)\epsilon(t)}{2}\|x(t)-x_{t}\|^{2}+\frac{bc(t)\epsilon(t)\beta^{2}(t)}{2}\|\dot{x}(t)\|^{2}.$

Further, we have

$\displaystyle-a(t)\beta(t)\|{\nabla}g(x(t)+\beta(t)\dot{x}(t))\|^{2}-a(t)\beta(t)\epsilon^{2}(t)\|x(t)\|^{2}-2\beta(t)a(t)\epsilon(t)\langle{\nabla}g(x(t)+\beta(t)\dot{x}(t)),x(t)\rangle=$
$\displaystyle-a(t)\beta(t)\|{\nabla}g(x(t)+\beta(t)\dot{x}(t))+\epsilon(t)x(t)\|^{2}$

Consequently, (49) becomes

(50) $\displaystyle\dot{E}(t)$
$\displaystyle=(a^{\prime}(t)-bc(t))\left(g_{t}(x(t)+\beta(t)\dot{x}(t))-g_{t}(x_{t})\right)+\frac{a(t)\dot{\epsilon}(t)}{2}\left(\|x(t)+\beta(t)\dot{x}(t)\|^{2}-\|x_{t}\|^{2}\right)$
$\displaystyle+\frac{d^{\prime}(t)-bc(t)\epsilon(t)}{2}\|x(t)-x_{t}\|^{2}-a(t)\beta(t)\|{\nabla}g(x(t)+\beta(t)\dot{x}(t))+\epsilon(t)x(t)\|^{2}$
$\displaystyle+\left(a(t)\beta(t)\epsilon(t)\left(1+\beta^{\prime}(t)-\frac{\alpha\beta(t)}{t^{q}}\right)+c(t)\left(b+c^{\prime}(t)-\frac{\alpha
c(t)}{t^{q}}\right)+\frac{bc(t)\epsilon(t)\beta^{2}(t)}{2}\right)\|\dot{x}(t)\|^{2}$
$\displaystyle-
bc(t)\left\langle\dot{x}(t),\frac{d}{dt}x_{t}\right\rangle-\left(b^{2}+d(t)\right)\left\langle
x(t)-x_{t},\frac{d}{dt}x_{t}\right\rangle\,,\text{ for all }t\geq t_{1}.$

Now we use (39) for evaluating the term involving
$\left\langle\frac{d}{dt}x_{t},\dot{x}(t)\right\rangle$ and we conclude that
for every function $n(t)>0$ one has

(51) $\displaystyle-
bc(t)\left\langle\frac{d}{dt}x_{t},\dot{x}(t)\right\rangle$
$\displaystyle\leq\frac{bc(t)}{2}\left(\frac{1}{n(t)}\left(\frac{\dot{\epsilon}(t)}{\epsilon(t)}\right)^{2}\|x_{t}\|^{2}+n(t)\|\dot{x}(t)\|^{2}\right).$

Similarly, for any function $m(t)>0$ we obtain that

(52)
$\displaystyle-\left(b^{2}+d(t)\right)\left\langle\frac{d}{dt}x_{t},x(t)-x_{t}\right\rangle$
$\displaystyle\leq\frac{b^{2}+d(t)}{2}\left(m(t)\left(\frac{\dot{\epsilon}(t)}{\epsilon(t)}\right)^{2}\left\|x_{t}\right\|^{2}+\frac{1}{m(t)}\|x(t)-x_{t}\|^{2}\right).$

Hence, by using (51) and (52) we get

(53) $\displaystyle\dot{E}(t)$
$\displaystyle\leq\left(a^{\prime}(t)-bc(t)\right)\left(g_{t}(x(t)+\beta(t)\dot{x}(t))-g_{t}(x_{t})\right)+\frac{a(t)\dot{\epsilon}(t)}{2}\|x(t)+\beta(t)\dot{x}(t)\|^{2}$
$\displaystyle+\left(\frac{d^{\prime}(t)}{2}-\frac{bc(t)\epsilon(t)}{2}+\frac{b^{2}+d(t)}{2m(t)}\right)\|x(t)-x_{t}\|^{2}-a(t)\beta(t)\|{\nabla}g(x(t)+\beta(t)\dot{x}(t))+\epsilon(t)x(t)\|^{2}$
$\displaystyle+\left(a(t)\beta(t)\epsilon(t)\left(1+\beta^{\prime}(t)-\frac{\alpha\beta(t)}{t^{q}}\right)+c(t)\left(b+c^{\prime}(t)-\frac{\alpha
c(t)}{t^{q}}\right)+\frac{bc(t)\beta^{2}(t)\epsilon(t)}{2}+\frac{bn(t)c(t)}{2}\right)\|\dot{x}(t)\|^{2}$
$\displaystyle+\left(\left(\frac{bc(t)}{2n(t)}+\frac{b^{2}+d(t)}{2}m(t)\right)\left(\frac{\dot{\epsilon}(t)}{\epsilon(t)}\right)^{2}-\frac{a(t)\dot{\epsilon}(t)}{2}\right)\|x_{t}\|^{2}\,,\text{
for all }t\geq t_{1}.$

Consider now $K>0$ which shall be defined later and $q<r<1$ defined at
condition (C1). Knowing that

$\displaystyle\frac{1}{2}\|b(x(t)-x_{t})+c(t)\dot{x}(t)\|^{2}\leq
b^{2}\|x(t)-x_{t}\|^{2}+c^{2}(t)\|\dot{x}(t)\|^{2},$

we get

$\frac{K}{t^{r}}E(t)\leq\frac{Ka(t)}{t^{r}}\left(g_{t}(x(t)+\beta(t)\dot{x}(t))-g_{t}(x_{t})\right)+\frac{K}{t^{r}}c^{2}(t)\|\dot{x}(t)\|^{2}+\frac{K}{t^{r}}\left(\frac{d(t)}{2}+b^{2}\right)\|x(t)-x_{t}\|^{2}.$

Hence, we take $m(t)=\frac{1}{M}t^{r}$ and $n(t)=Nt^{q-r}$ for some $N,M>0$
and we get

(54) $\displaystyle\dot{E}(t)$
$\displaystyle+\frac{K}{t^{r}}E(t)\leq\left(a^{\prime}(t)-bc(t)+\frac{Ka(t)}{t^{r}}\right)\left(g_{t}(x(t)+\beta(t)\dot{x}(t))-g_{t}(x_{t})\right)+\frac{a(t)\dot{\epsilon}(t)}{2}\|x(t)+\beta(t)\dot{x}(t)\|^{2}$
$\displaystyle+\left(\frac{d^{\prime}(t)}{2}-\frac{bc(t)\epsilon(t)}{2}+\frac{1}{t^{r}}\left(K\frac{d(t)+2b^{2}}{2}+M\frac{b^{2}+d(t)}{2}\right)\right)\|x(t)-x_{t}\|^{2}$
$\displaystyle-a(t)\beta(t)\|{\nabla}g(x(t)+\beta(t)\dot{x}(t))+\epsilon(t)x(t)\|^{2}$
$\displaystyle+\left(a(t)\beta(t)\epsilon(t)\left(1+\beta^{\prime}(t)-\frac{\alpha\beta(t)}{t^{q}}\right)+c(t)\left(b+c^{\prime}(t)-\frac{\alpha
c(t)}{t^{q}}\right)+\frac{bc(t)\beta^{2}(t)\epsilon(t)}{2}+\left(K+\frac{bN}{2}\right)t^{2q-r}\right)\|\dot{x}(t)\|^{2}$
$\displaystyle+\left(\left(\frac{b^{2}+d(t)}{2M}+\frac{b}{2N}\right)t^{r}\left(\frac{\dot{\epsilon}(t)}{\epsilon(t)}\right)^{2}-\frac{a(t)\dot{\epsilon}(t)}{2}\right)\|x_{t}\|^{2}\,,\text{
for all }t\geq t_{1}.$

Let us evaluate the coefficients on the right hand side of (54).

Since

$a(t)=t^{2q}+\frac{(\alpha-b)\gamma t^{q}+(\alpha-b)\beta+\beta
qt^{q-1}-\left(\gamma+\frac{\beta}{t^{q}}\right)^{2}\epsilon(t)t^{2q}}{1-\frac{\beta
q}{t^{q+1}}-\frac{\alpha\gamma}{t^{q}}-\frac{\alpha\beta}{t^{2q}}+\left(\gamma+\frac{\beta}{t^{q}}\right)^{2}\epsilon(t)},$

further $q<1$ and $r>q$ we conclude that for $t$ big enough one has

$a^{\prime}(t)-bc(t)+\frac{Ka(t)}{t^{r}}<0.$

Obviously the coefficient of $\|x(t)+\beta(t)\dot{x}(t)\|^{2}$ and
$\|{\nabla}g(x(t)+\beta(t)\dot{x}(t))+\epsilon(t)x(t)\|^{2}$ are nonpositive
for all $t\geq t_{1}.$

Concerning the coefficient of $\|x(t)-x_{t}\|^{2}$ one has

$\displaystyle\frac{d^{\prime}(t)}{2}-\frac{bc(t)\epsilon(t)}{2}+\frac{1}{t^{r}}\left(K\frac{d(t)+2b^{2}}{2}+M\frac{b^{2}+d(t)}{2}\right)=$
$\displaystyle\frac{b}{2}\left(-t^{q}\epsilon(t)+(K(\alpha+b)+M\alpha)t^{-r}-q(K+M)t^{q-r-1}\right)$
$\displaystyle+\frac{bq(1-q)}{2}t^{q-2}.$

Now, according to (C1) there exists $K_{1}>0$ such that
$\epsilon(t)\geq\frac{K_{1}}{t^{q+r}}$ for $t$ big enough, hence taking into
account that $b<\alpha$ we take $K<\frac{K_{1}}{4\alpha}$,
$M<\frac{K_{1}}{2\alpha}$ and we get that

$\frac{d^{\prime}(t)}{2}-\frac{bc(t)\epsilon(t)}{2}+\frac{1}{t^{r}}\left(K\frac{d(t)+2b^{2}}{2}+M\frac{b^{2}+d(t)}{2}\right)\leq
0$

for $t$ big enough.

Finally, by taking into account that $a(t)=t^{2q}+\mathcal{O}(t^{q})$ as
$t\to+\infty$, concerning the coefficient of $\|\dot{x}(t)\|^{2}$ we conclude
the following. If $\gamma>0$ then

$\displaystyle
a(t)\beta(t)\epsilon(t)\left(1+\beta^{\prime}(t)-\frac{\alpha\beta(t)}{t^{q}}\right)+c(t)\left(b+c^{\prime}(t)-\frac{\alpha
c(t)}{t^{q}}\right)+\frac{bc(t)\beta^{2}(t)\epsilon(t)}{2}+\left(K+\frac{bN}{2}\right)t^{2q-r}=$
$\displaystyle(b-\alpha)t^{q}+\gamma\epsilon(t)t^{2q}+\mathcal{O}(t^{q}\epsilon(t))+\mathcal{O}(t^{2q-r})\mbox{
as }t\to+\infty,$

hence, by taking account condition (C2), i.e.,
$\epsilon(t)\leq\frac{\alpha}{2\gamma t^{q}}$ and by fixing
$b<\frac{\alpha}{2}$ we obtain that

$a(t)\beta(t)\epsilon(t)\left(1+\beta^{\prime}(t)-\frac{\alpha\beta(t)}{t^{q}}\right)+c(t)\left(b+c^{\prime}(t)-\frac{\alpha
c(t)}{t^{q}}\right)+\frac{bc(t)\beta^{2}(t)\epsilon(t)}{2}+\left(K+\frac{bN}{2}\right)t^{2q-r}\leq
0$

for $t$ big enough.

In case $\gamma=0,\,\beta>0$ we have

$\displaystyle
a(t)\beta(t)\epsilon(t)\left(1+\beta^{\prime}(t)-\frac{\alpha\beta(t)}{t^{q}}\right)+c(t)\left(b+c^{\prime}(t)-\frac{\alpha
c(t)}{t^{q}}\right)+\frac{bc(t)\beta^{2}(t)\epsilon(t)}{2}+\left(K+\frac{bN}{2}\right)t^{2q-r}=$
$\displaystyle(b-\alpha)t^{q}+\mathcal{O}(t^{q}\epsilon(t))+\mathcal{O}(t^{2q-r})\mbox{
as }t\to+\infty,$

and by since $b<\frac{a}{2}\alpha$ and taking into account that $\epsilon(t)$
is nonincreasing we get that also in this case

$a(t)\beta(t)\epsilon(t)\left(1+\beta^{\prime}(t)-\frac{\alpha\beta(t)}{t^{q}}\right)+c(t)\left(b+c^{\prime}(t)-\frac{\alpha
c(t)}{t^{q}}\right)+\frac{bc(t)\beta^{2}(t)\epsilon(t)}{2}+\left(K+\frac{bN}{2}\right)t^{2q-r}\leq
0$

for $t$ big enough.

Consequently, there exists $t_{2}\geq t_{1}$ such that for all $t\geq t_{2}$
one has

(55)
$\displaystyle\dot{E}(t)+\frac{K}{t^{r}}E(t)\leq\left(\left(\frac{b^{2}+d(t)}{2M}+\frac{b}{2N}\right)t^{r}\left(\frac{\dot{\epsilon}(t)}{\epsilon(t)}\right)^{2}-\frac{a(t)\dot{\epsilon}(t)}{2}\right)\|x_{t}\|^{2}.$

Now, according to (C3) there exists $K_{3}>0$ such that
$\left(\frac{\dot{\epsilon}(t)}{\epsilon(t)}\right)^{2}\leq\frac{K_{3}}{t^{2}}$
for $t$ big enough and therefore
$0\leq-\dot{\epsilon}(t)\leq\frac{\sqrt{K_{3}}}{t}\epsilon(t)$ for $t$ big
enough.

Further, one has $\|x_{t}\|\leq\|x^{*}\|$ where $x^{*}$ is the minimal norm
element from the set $\operatorname*{argmin}g.$ Hence, by taking into account
also the form of $a(t)$ we conclude that there exists $t_{3}\geq t_{2}$ and a
constant $C_{0}>0$ such that

(56)
$\displaystyle\left(\left(\frac{b^{2}+d(t)}{2M}+\frac{b}{2N}\right)t^{r}\left(\frac{\dot{\epsilon}(t)}{\epsilon(t)}\right)^{2}-\frac{a(t)\dot{\epsilon}(t)}{2}\right)\|x_{t}\|^{2}\leq
C_{0}(t^{r-2}+t^{2q-1}\epsilon(t))\mbox{ for all }t\geq t_{3}.$

Consequently, (55) leads to

(57) $\displaystyle\dot{E}(t)+\frac{K}{t^{r}}E(t)\leq
C_{0}(t^{r-2}+t^{2q-1}\epsilon(t))\mbox{ for all }t\geq t_{3}.$

By multiplying (57) with $e^{\frac{K}{1-r}t^{1-r}}$ we obtain

(58) $\displaystyle\frac{d}{dt}\left(e^{\frac{K}{1-r}t^{1-r}}E(t)\right)\leq
C_{0}(t^{r-2}+t^{2q-1}\epsilon(t))e^{\frac{K}{1-r}t^{1-r}}\mbox{ for all
}t\geq t_{3}.$

By integrating (58) on an interval $[t_{3},T],\,T>t_{3}$ we obtain

(59) $\displaystyle e^{\frac{K}{1-r}T^{1-r}}E(T)-\mathcal{C}_{0}\leq
C_{0}\int_{t_{3}}^{T}(t^{r-2}+t^{2q-1}\epsilon(t))e^{\frac{K}{1-r}t^{1-r}}dt,$

where $\mathcal{C}_{0}$ is the constant
$e^{\frac{K}{1-r}t_{3}^{1-r}}E(t_{3}).$

In what follows we will discuss the right hand side of (59). First of all,
note that for a function $t\rightarrow t^{u}e^{ct^{v}},u\in\mathbb{R},\,c,v>0$
one has

(60)
$\frac{d}{dt}\left(t^{u}e^{ct^{v}}\right)=(ut^{u-1}+cvt^{u+v-1})e^{ct^{v}}\geq
c_{1}t^{u+v-1}e^{ct^{v}},$

for some $c_{1}>0$ and $t$ big enough. Consequently, by deploying (60) we
conclude that there exists $t_{4}\geq t_{3}$ and $c_{1},c_{2}>0$ such that for
all $t\geq t_{4}$ it holds

$t^{r-2}e^{\frac{K}{1-r}t^{1-r}}\leq
c_{1}\frac{d}{dt}\left(t^{2r-2}e^{\frac{K}{1-r}t^{1-r}}\right)$

and

$t^{2q-1}\epsilon(t)e^{\frac{K}{1-r}t^{1-r}}\leq
c_{2}\epsilon(t)\frac{d}{dt}\left(t^{2q+r-1}e^{\frac{K}{1-r}t^{1-r}}\right).$

Therefore, there is a constant $\mathcal{K}$ such that

(61)
$\int_{t_{3}}^{T}t^{r-2}e^{\frac{K}{1-r}t^{1-r}}dt\leq\mathcal{C}+c_{1}\int_{t_{4}}^{T}\frac{d}{dt}\left(t^{2r-2}e^{\frac{K}{1-r}t^{1-r}}\right)dt=c_{1}T^{2r-2}e^{\frac{K}{1-r}T^{1-r}}+\mathcal{K},$

where $\mathcal{C}=\int_{t_{3}}^{t_{4}}t^{r-2}e^{\frac{K}{1-r}t^{1-r}}dt$ and
$\mathcal{K}=\mathcal{C}-c_{1}t_{4}^{2r-2}e^{\frac{K}{1-r}t_{4}^{1-r}}.$

Similarly, by using integration by parts and also that
$-\dot{\epsilon}(t)\leq\frac{\sqrt{K_{3}}}{t}\epsilon(t)$ for all $t\geq
t_{4}$ we obtain

(62)
$\displaystyle\int_{t_{3}}^{T}\epsilon(t)t^{2q-1}e^{\frac{K}{1-r}t^{1-r}}dt$
$\displaystyle\leq\mathcal{C}_{0}+c_{2}\int_{t_{4}}^{T}\epsilon(t)\frac{d}{dt}\left(t^{2q+r-1}e^{\frac{K}{1-r}t^{1-r}}\right)dt=\mathcal{C}_{1}+c_{2}\epsilon(T)T^{2q+r-1}e^{\frac{K}{1-r}T^{1-r}}$
$\displaystyle\,+c_{2}\int_{t_{4}}^{T}-\dot{\epsilon}(t)t^{2q+r-1}e^{\frac{K}{1-r}t^{1-r}}dt$
$\displaystyle\leq\mathcal{C}_{1}+c_{2}\epsilon(T)T^{2q+r-1}e^{\frac{K}{1-r}T^{1-r}}+c_{2}\sqrt{K_{3}}\int_{t_{4}}^{T}\epsilon(t)t^{2q+r-2}e^{\frac{K}{1-r}t^{1-r}}dt,$

where
$\mathcal{C}_{0}=\int_{t_{3}}^{t_{4}}\epsilon(t)t^{2q-1}e^{\frac{K}{1-r}t^{1-r}}dt$
and
$\mathcal{C}_{1}=\mathcal{C}_{0}-c_{2}\epsilon(t_{4})t_{4}^{2q+r-1}e^{\frac{K}{1-r}t_{4}^{1-r}}.$

Now, since $r<1$ there exists $t_{5}\geq t_{4}$ such that
$c_{2}\sqrt{K_{3}}\epsilon(t)t^{2q+r-2}e^{\frac{K}{1-r}t^{1-r}}\leq\frac{1}{2}\epsilon(t)t^{2q-1}e^{\frac{K}{1-r}t^{1-r}}$
for all $t\geq t_{5}$, hence

(63) $\displaystyle
c_{2}\sqrt{K_{3}}\int_{t_{4}}^{T}\epsilon(t)t^{2q+r-2}e^{\frac{K}{1-r}t^{1-r}}dt$
$\displaystyle\leq
c_{2}\sqrt{K_{3}}\int_{t_{4}}^{t_{5}}\epsilon(t)t^{2q+r-2}e^{\frac{K}{1-r}t^{1-r}}dt+\frac{1}{2}\int_{t_{5}}^{T}\epsilon(t)t^{2q-1}e^{\frac{K}{1-r}t^{1-r}}dt$
$\displaystyle=\mathcal{C}_{2}-\mathcal{C}_{3}+\frac{1}{2}\int_{t_{3}}^{T}\epsilon(t)t^{2q-1}e^{\frac{K}{1-r}t^{1-r}}dt,$

where
$\mathcal{C}_{2}=c_{2}\sqrt{K_{3}}\int_{t_{4}}^{t_{5}}\epsilon(t)t^{2q+r-2}e^{\frac{K}{1-r}t^{1-r}}dt$
and
$\mathcal{C}_{3}=\frac{1}{2}\int_{t_{3}}^{t_{5}}\epsilon(t)t^{2q-1}e^{\frac{K}{1-r}t^{1-r}}dt.$

Combining (62) and (63) we obtain that for all $T\geq t_{5}$ it holds

(64)
$\displaystyle\int_{t_{3}}^{T}\epsilon(t)t^{2q-1}e^{\frac{K}{1-r}t^{1-r}}dt$
$\displaystyle\leq
2(\mathcal{C}_{1}+\mathcal{C}_{2}-\mathcal{C}_{3})+2c_{2}\epsilon(T)T^{2q+r-1}e^{\frac{K}{1-r}T^{1-r}}=\mathcal{C}+2c_{2}\epsilon(T)T^{2q+r-1}e^{\frac{K}{1-r}T^{1-r}}.$

Hence, for $T$ big enough and for some $C_{3}>0$, (61), (64) and (59) yield to

(65) $\displaystyle E(T)$ $\displaystyle\leq
e^{-\frac{K}{1-r}T^{1-r}}\left(\mathcal{C}_{0}+C_{0}\left(\mathcal{K}+c_{1}T^{2r-2}e^{\frac{K}{1-r}T^{1-r}}+\mathcal{C}+2c_{2}\epsilon(T)T^{2q+r-1}e^{\frac{K}{1-r}T^{1-r}}\right)\right)$
$\displaystyle=\frac{\mathcal{K}_{0}}{e^{\frac{K}{1-r}T^{1-r}}}+C_{1}T^{2r-2}+C_{2}\epsilon(T)T^{2q+r-1}\leq
C_{3}(T^{2r-2}+\epsilon(T)T^{2q+r-1}),$

where $C_{1}=C_{0}c_{1},\,C_{2}=2C_{0}c_{2}$ and
$\mathcal{K}_{0}=\mathcal{C}_{0}+C_{0}(\mathcal{K}+\mathcal{C}).$

Taking into account the form of $E(t)$ we obtain at once that

$g_{t}(x(t)+\beta(t)\dot{x}(t))-g_{t}(x_{t})=\mathcal{O}(t^{2r-2q-2}+\epsilon(t)t^{r-1})\mbox{
as }t\to+\infty,$
$\|\dot{x}(t)\|=\mathcal{O}(t^{r-q-1}+\sqrt{\epsilon(t)}t^{\frac{r-1}{2}})\mbox{
as }t\to+\infty$

and

$\|x(t)-x_{t}\|=\mathcal{O}(t^{r-1}+\sqrt{\epsilon(t)}t^{\frac{2q+r-1}{2}})\mbox{
as }t\to+\infty.$

Now from $r<1,\,q<1$ and (C1) we get
$\lim_{t\to+\infty}(t^{r-1}+\sqrt{\epsilon(t)}t^{\frac{2q+r-1}{2}})=0$,
further $\lim_{t\to+\infty}x_{t}=x^{*}$, hence the latter relation shows that
$x(t)$ converges strongly to $x^{*}$, that is

$\lim_{t\to+\infty}\|x(t)-x^{*}\|=0.$

Even more, in case $\lim_{t\to+\infty}t^{2q}\epsilon(t)>0$ a better estimate
can be obtained. Indeed, by deploying (41) we get

$g_{t}(x(t)+\beta(t)\dot{x}(t))-g_{t}(x_{t})\geq\frac{\epsilon(t)}{2}\|x(t)+\beta(t)\dot{x}(t)-x_{t}\|^{2},$

hence

$\|x(t)+\beta(t)\dot{x}(t)-x_{t}\|=\mathcal{O}\left(t^{\frac{r-1}{2}}+\frac{t^{r-q-1}}{\sqrt{\epsilon(t)}}\right)\mbox{
as }t\to+\infty.$

By using the fact that
$\|\beta(t)\dot{x}(t)\|=\mathcal{O}(\beta(t)t^{r-q-1}+\beta(t)\sqrt{\epsilon(t)}t^{\frac{r-1}{2}})\mbox{
as }t\to+\infty$ we obtain that

$|x(t)-x_{t}\|=\mathcal{O}\left(t^{\frac{r-1}{2}}+\frac{t^{r-q-1}}{\sqrt{\epsilon(t)}}\right)\mbox{
as }t\to+\infty.$

Finally, by using (43) we get

$g(x(t)+\beta(t)\dot{x}(t))-g(x^{*})\leq
g_{t}(x(t)+\beta(t)\dot{x}(t))-g_{t}(x_{t})+\frac{\epsilon(t)}{2}\|x^{*}\|^{2},$

hence

$g(x(t)+\beta(t)\dot{x}(t))-\min g=\mathcal{O}(t^{2r-2q-2}+\epsilon(t))\mbox{
as }t\to+\infty.$

Concerning the integral estimates, let us take $K=0$ in (54). Then,

(66) $\displaystyle\dot{E}(t)$
$\displaystyle\leq\left(a^{\prime}(t)-bc(t)\right)\left(g_{t}(x(t)+\beta(t)\dot{x}(t))-g_{t}(x_{t})\right)+\frac{a(t)\dot{\epsilon}(t)}{2}\|x(t)+\beta(t)\dot{x}(t)\|^{2}$
$\displaystyle+\left(\frac{d^{\prime}(t)}{2}-\frac{bc(t)\epsilon(t)}{2}+M\frac{b^{2}+d(t)}{2t^{r}}\right)\|x(t)-x_{t}\|^{2}$
$\displaystyle-a(t)\beta(t)\|{\nabla}g(x(t)+\beta(t)\dot{x}(t))+\epsilon(t)x(t)\|^{2}$
$\displaystyle+\left(a(t)\beta(t)\epsilon(t)\left(1+\beta^{\prime}(t)-\frac{\alpha\beta(t)}{t^{q}}\right)+c(t)\left(b+c^{\prime}(t)-\frac{\alpha
c(t)}{t^{q}}\right)+\frac{bc(t)\beta^{2}(t)\epsilon(t)}{2}+\frac{bN}{2}t^{2q-r}\right)\|\dot{x}(t)\|^{2}$
$\displaystyle+\left(\left(\frac{b^{2}+d(t)}{2M}+\frac{b}{2N}\right)t^{r}\left(\frac{\dot{\epsilon}(t)}{\epsilon(t)}\right)^{2}-\frac{a(t)\dot{\epsilon}(t)}{2}\right)\|x_{t}\|^{2}\,,\text{
for all }t\geq t_{1}.$

Arguing as before, we conclude that for $t$ big enough, the coefficients of
$\|x(t)+\beta(t)\dot{x}(t)\|^{2}$ and $\|x(t)-x_{t}\|^{2}$ are non-positive.
Further, there exists $A_{1}>0$ such that

$a^{\prime}(t)-bc(t)\leq-A_{1}t^{q}\mbox{ for }t\mbox{ big enough.}$

Similarly, there exists $A_{2}>0$ such that for $t$ big enough
$-a(t)\beta(t)\leq-A_{2}t^{2q},$ if $\gamma\neq 0$ and $-a(t)\beta(t)\leq-
A_{2}t^{2q-1},$ if $\gamma=0.$

Finally, if $\gamma>0$ the coefficient of $\|\dot{x}(t)\|^{2}$ is

$\displaystyle
a(t)\beta(t)\epsilon(t)\left(1+\beta^{\prime}(t)-\frac{\alpha\beta(t)}{t^{q}}\right)+c(t)\left(b+c^{\prime}(t)-\frac{\alpha
c(t)}{t^{q}}\right)+\frac{bc(t)\beta^{2}(t)\epsilon(t)}{2}+\frac{bN}{2}t^{2q-r}=$
$\displaystyle(b-\alpha)t^{q}+\gamma\epsilon(t)t^{2q}+\mathcal{O}(t^{q}\epsilon(t))+\mathcal{O}(t^{2q-r})\mbox{
as }t\to+\infty,$

hence, by taking account condition (C2), i.e.,
$\epsilon(t)\leq\frac{\alpha}{2\gamma t^{q}}$ and by fixing
$b<\frac{\alpha}{2}$ we obtain that there exists $A_{3}>0$ such that

$a(t)\beta(t)\epsilon(t)\left(1+\beta^{\prime}(t)-\frac{\alpha\beta(t)}{t^{q}}\right)+c(t)\left(b+c^{\prime}(t)-\frac{\alpha
c(t)}{t^{q}}\right)+\frac{bc(t)\beta^{2}(t)\epsilon(t)}{2}+\frac{bN}{2}t^{2q-r}\leq-
A_{3}t^{q}$

for $t$ big enough.

In case $\gamma=0,\,\beta>0$ we have

$\displaystyle
a(t)\beta(t)\epsilon(t)\left(1+\beta^{\prime}(t)-\frac{\alpha\beta(t)}{t^{q}}\right)+c(t)\left(b+c^{\prime}(t)-\frac{\alpha
c(t)}{t^{q}}\right)+\frac{bc(t)\beta^{2}(t)\epsilon(t)}{2}+\left(K+\frac{bN}{2}\right)t^{2q-r}=$
$\displaystyle(b-\alpha)t^{q}+\beta\epsilon(t)t^{2q-1}+\mathcal{O}(t^{q-1}\epsilon(t))+\mathcal{O}(t^{2q-r})\mbox{
as }t\to+\infty,$

and by fixing $b<\alpha$ and taking into account that $\epsilon(t)$ is
nonincreasing we get that also in this case that there exists $A_{3}>0$ such
that

$a(t)\beta(t)\epsilon(t)\left(1+\beta^{\prime}(t)-\frac{\alpha\beta(t)}{t^{q}}\right)+c(t)\left(b+c^{\prime}(t)-\frac{\alpha
c(t)}{t^{q}}\right)+\frac{bc(t)\beta^{2}(t)\epsilon(t)}{2}+\left(K+\frac{bN}{2}\right)t^{2q-r}\leq-
A_{3}t^{q}$

for $t$ big enough.

Further, according to (56) one has
$\left(\left(\frac{b^{2}+d(t)}{2M}+\frac{b}{2N}\right)t^{r}\left(\frac{\dot{\epsilon}(t)}{\epsilon(t)}\right)^{2}-\frac{a(t)\dot{\epsilon}(t)}{2}\right)\|x_{t}\|^{2}\leq
C_{0}(t^{r-2}+t^{2q-1}\epsilon(t))$ for $t$ big enough, hence (66) leads to
the existence of $t_{1}^{\prime}\geq t_{1}$ such that

(67)
$\displaystyle\dot{E}(t)+A_{1}t^{q}\left(g_{t}(x(t)+\beta(t)\dot{x}(t))-g_{t}(x_{t})\right)+A_{2}(t)\|{\nabla}g(x(t)+\beta(t)\dot{x}(t))+\epsilon(t)x(t)\|^{2}+A_{3}t^{q}\|\dot{x}(t)\|^{2}\leq$
$\displaystyle C_{0}(t^{r-2}+t^{2q-1}\epsilon(t))\mbox{ for all }t\geq
t_{1}^{\prime},$

where $A_{2}(t)=A_{2}t^{2q}$ if $\gamma>0$ and $A_{2}(t)=A_{2}t^{2q-1}$ if
$\gamma=0$. Now, according to the hypotheses of the theorem, we have
$t^{2q-1}\epsilon(t)\in L^{1}[t_{0},+\infty[$. Further, since $r<1$ we deduce
that

$C_{0}\int_{t_{1}^{\prime}}^{+\infty}t^{r-2}+t^{2q-1}\epsilon(t)dt<+\infty.$

Consequently, by integrating (67) on an interval $[t_{1}^{\prime},T]$ we
obtain that there exists $C_{1}>0$ such that

(68) $\displaystyle
E(T)+A_{1}\int_{t_{1}^{\prime}}^{T}t^{q}\left(g_{t}(x(t)+\beta(t)\dot{x}(t))-g_{t}(x_{t})\right)dt+\int_{t_{1}^{\prime}}^{T}A_{2}(t)\|{\nabla}g(x(t)+\beta(t)\dot{x}(t))+\epsilon(t)x(t)\|^{2}dt$
$\displaystyle+A_{3}\int_{t_{1}^{\prime}}^{T}t^{q}\|\dot{x}(t)\|^{2}dt\leq
C_{1}.$

Hence,

$\int_{t_{0}}^{+\infty}t^{q}\left(g_{t}(x(t)+\beta(t)\dot{x}(t))-g_{t}(x_{t})\right)dt<+\infty,$
$\int_{t_{0}}^{+\infty}t^{q}\|\dot{x}(t)\|^{2}dt<+\infty,$
$\int_{t_{0}}^{+\infty}t^{2q}\|{\nabla}g(x(t)+\beta(t)\dot{x}(t))+\epsilon(t)x(t)\|^{2}dt<+\infty,\mbox{
if }\gamma>0$

and

$\int_{t_{0}}^{+\infty}t^{2q-1}\|{\nabla}g(x(t)+\beta(t)\dot{x}(t))+\epsilon(t)x(t)\|^{2}dt<+\infty,\mbox{
if }\gamma=0.$

Further, since $t^{2q-1}\epsilon(t)\in L^{1}[t_{0},+\infty[$ obviously
$t^{2q-1}\epsilon^{2}(t)\in L^{1}[t_{0},+\infty[$, hence

$\int_{t_{0}}^{+\infty}t^{2q-1}\|{\nabla}g(x(t)+\beta(t)\dot{x}(t))\|^{2}dt<+\infty,\mbox{
if }\gamma=0.$

$\blacksquare$

######  Remark 11.

Natural candidates for the Tikhonov regularization parameter are
$\epsilon(t)=\frac{a}{t^{p}},\,a,p>0.$ In this case condition (C1) is nothing
else that $p<q+1.$ Indeed, for $p<q+1$ one can choose $r\in]\max(q,p-q),1[$
and $K_{1}=a$ and one has $\epsilon(t)\geq\frac{K_{1}}{t^{r+q}}.$ Note that
condition (C2), in which we assume that $\gamma>0$, holds whenever $p>q$ or
$p=q$ and $a\leq\dfrac{\alpha}{2\gamma}.$ Finally, (C3) always holds.

Consequently the following result is valid.

######  Corollary 12.

Consider $0<q<1$ and $\epsilon(t)=\frac{a}{t^{p}},\,a>0$ and $q\leq p<q+1.$
Further, when $q=p$ assume that $a\leq\dfrac{\alpha}{2\gamma}.$ Let $x$ be a
trajectory generated by the dynamical system (DS). Then the following
statements hold true.

(i) $x(t)$ converges strongly, as $t\to+\infty$, to $x^{*}$, the element of
minimum norm of $\operatorname*{argmin}g$;

(ii) $g(x(t)+\beta(t)\dot{x}(t))-\min g=\mathcal{O}(t^{-\min(2q,p)})$ and
$\|\dot{x}(t)\|=\mathcal{O}(t^{-\frac{\min(2q,p)}{2}})\mbox{ as }t\to+\infty,$
further for every $r\in]\max(q,p-q),1[$ one has
$\|x(t)-x_{t}\|=\mathcal{O}(t^{r-1}+t^{\frac{2q+r-p-1}{2}})\mbox{ as
}t\to+\infty;$

(iii) If $p>2q$ then
$\int_{t_{0}}^{+\infty}t^{q}\left(g_{t}(x(t)+\beta(t)\dot{x}(t))-g_{t}(x_{t})\right)dt<+\infty$
and $\int_{t_{0}}^{+\infty}t^{q}\|\dot{x}(t)\|^{2}dt<+\infty.$ Further,
$\int_{t_{0}}^{+\infty}t^{2q}\|{\nabla}g(x(t)+\beta(t)\dot{x}(t))+t^{-p}x(t)\|^{2}dt<+\infty,\mbox{
if }\gamma>0$ and
$\int_{t_{0}}^{+\infty}t^{2q-1}\|{\nabla}g(x(t)+\beta(t)\dot{x}(t))\|^{2}dt<+\infty,\mbox{
if }\gamma=0.$

(iv) Moreover, if $p>\max\left(\frac{2q+1}{2},2q\right),\,\gamma>0$ then
$\int_{t_{0}}^{+\infty}t^{2q}\|{\nabla}g(x(t)+\beta(t)\dot{x}(t))\|^{2}dt<+\infty.$

######  Proof .

Note that (i) and (iii) are direct applications of Theorem 10, hence let us
show (ii). According to Theorem 10 the fast convergence rates in this special
case are $g(x(t)+\beta(t)\dot{x}(t))-\min g=\mathcal{O}(t^{2r-2q-2}+t^{-p}),$
$\|\dot{x}(t)\|=\mathcal{O}(t^{r-q-1}+t^{\frac{r-p-1}{2}})$ and
$\|x(t)-x_{t}\|=\mathcal{O}(t^{r-1}+t^{\frac{2q+r-p-1}{2}})\mbox{ as
}t\to+\infty.$

Now, taking into account that $r<1$ we get $t^{2r-2q-2}+t^{-p}\leq
t^{-2q}+t^{-p}\leq 2t^{-\min(2q,p)}$ the claim follows.

Concerning (iv), observe that in this case

$t^{2q}\|{\nabla}g(x(t)+\beta(t)\dot{x}(t))\|^{2}\leq
2t^{2q}\|{\nabla}g(x(t)+\beta(t)\dot{x}(t))+t^{-p}x(t)\|^{2}+2t^{2q-2p}\|x(t)\|^{2}$

and $t^{2q-2p}\|x(t)\|^{2}\in L^{1}[t_{0},+\infty[,$ hence the claim is a
consequence of (iii). $\blacksquare$

##  5\. Conclusion, perspectives

In this paper we studied a second order dynamical system with a Tikhonov
regularization term and an implicit Hessian driven damping in connection to a
minimization problem of a smooth convex function. Natural explicit
discretization of our dynamical system leads to inertial algorithm of gradient
type with a Tikhonov regularization term, hence the results of the present
paper opens the gate for study the strong convergence properties of these
algorithms. We emphasize that the literature is very poor yet in such results,
see [24, 27]. Depending on the setting of the parameters involved we obtained
both weak and strong convergence of the generated trajectories to a minimizer
of the objective function and also fast convergence rates and integral
estimates involving the function values in a generated trajectory, velocity
and gradient. Our results not only extend but also improve the results from
the literature concerning second order dynamical systems with a Tikhonov
regularization term.

##  6\. Statements and Declarations

Availability of data and materials

In this manuscript only the datasets generated by authors were analysed.

Funding

The research leading to these results received funding from Romanian Ministry
of Research, Innovation and Digitization, CNCS-UEFISCDI, project number PN-
III-P1-1.1-TE-2021-0138, within PNCDI III .

Competing interests

The author has no relevant financial or non-financial interests to disclose.

##  7\. Appendix

######  Lemma 13.

[Lemma A.3 from [11]] Let $\delta>0$ and $f\in
L^{1}((\delta,+\infty),\mathbb{R})$ be a nonnegative and continuous function.
Let $\varphi:[\delta,+\infty)\longrightarrow[0,+\infty)$ be a nondecreasing
function such that $\lim\limits_{t\to+\infty}\varphi(t)=+\infty.$ Then it
holds

$\lim\limits_{t\to+\infty}\frac{1}{\varphi(t)}\int_{\delta}^{t}\varphi(s)f(s)ds=0.$

######  Lemma 14.

[Lemma 5.1 from [1]] Suppose that $F:[t_{0},+\infty)\longrightarrow\mathbb{R}$
is locally absolutely continuous and bounded from below and that there exists
$G\in L^{1}([t_{0},+\infty),\mathbb{R})$ such that

$\frac{d}{dt}F(t)\leq G(t)$

for almost every $t\in[t_{0},+\infty)$. Then there exists
$\lim\limits_{t\to+\infty}F(t)\in\mathbb{R}$.

The continuous version of the Opial Lemma (see [10]) is the main tool for
proving weak convergence for the generated trajectory.

######  Lemma 15.

Let $S\subseteq\mathcal{H}$ be a nonempty set and $x:[t_{0},+\infty[\to H$ a
given map such that:

$\displaystyle(i)\quad\mbox{for every }z\in S\ \mbox{the limit}\
\lim\limits_{t\longrightarrow+\infty}\|x(t)-z\|\ \mbox{exists};$
$\displaystyle(ii)\quad\text{every weak sequential limit point of }x(t)\text{
belongs to the set }S.$

Then the trajectory $x(t)$ converges weakly to an element in $S$ as
$t\to+\infty$.

The following lemma from [26] is essential for proving weak convergence.

######  Lemma 16.

(Lemma 16 from [26]) Let $t_{0}>0,\,\alpha>0,\,0<q<1$ and let
$w:[t_{0},+\infty[\to\mathbb{R}$ be a continuously differentiable function
which is bounded from below. Consider $k:[t_{0},+\infty[\to\mathbb{R}$ a
nonnegative function and assume that the function
$\int_{t}^{+\infty}\frac{dT}{Q(T)}Q(t)k(t)\in L^{1}[t_{0},+\infty[$, where
$Q(T):=\exp\left(\int_{t_{0}}^{T}\frac{\alpha}{s^{q}}ds\right).$ Assume
further, that

$\ddot{w}(t)+\frac{\alpha}{t^{q}}\dot{w}(t)\leq k(t)$

for almost every $t>t_{0}.$ Then, the positive part $[\dot{w}]_{+}$ of
$\dot{w}$ belongs to $L^{1}[t_{0},+\infty[,$ and $lim_{t\to+\infty}w(t)$
exists.

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