Stabilization of a nonlinear Euler–Bernoulli beam

Abstract

In this work, we study the vibration control of a flexible mechanical system. The dynamic of the problem is modeled as a viscoelastic nonlinear Euler–Bernoulli beam. To suppress the undesirable transversal vibrations of the beam, we adopt a control at the right boundary of the beam. This control law is simple to implement. We prove uniform stability of the system using a viscoelastic material, the multiplier method and some ideas introduced in [20]. It is shown that a large range of rates of decay of the energy can be achieved through a determined class of kernels. Unlike most of the existing classes in the market, ours are not necessarily strictly decreasing.

1 Introduction

Flexible systems exert an increasing influence in different industries and fields. For instance, we may cite flexible manipulators, flexible robot arm and marine risers for oil and gas transportation. The vibration problem of flexible systems has become a crucial topic of research. It is a widespread phenomena in engineering. The origin of these vibrations and their nature might be different. They can cause numerous harmful effects on the production process, including the damage of the equipment with significant financial consequences. There are many approaches to deal with vibration and stabilize flexible systems. Boundary control is the most practical and efficient one. In reference [5], active boundary controls to reduce vibration of an Euler–Bernoulli beam systems in one dimension are considered. In [12], nonlinear vibrations and stability issues are studied. In [4], boundary controllers are used to reduce the vibration of a coupled nonlinear flexible marine riser. Reference [11] considered an adaptive boundary control for an axially moving belt system to eliminate the vibration. In [8] using the direct method of Lyapunov, the exponential stability of a closed-loop system is proven with the help of boundary controls. Kelleche and Tatar [9] designed a nonlinear boundary control for a viscoelastic flexible system. Park et al. [14] studied the Euler–Bernoulli beam equation with memory, they proved the existence and the exponential stability of solutions for the problem

$$\begin{aligned} v _{tt}(x,t)+ v _{xxxx}(x,t)-\int _0^t k(t-s) v_{xxxx}(s)\mathrm{d}s+g(v_{t}(x,t))=0, \text {in }(0,L)\times [0,\infty ), \end{aligned}$$

under the boundary and initial conditions

$$\begin{aligned} {\left\{ \begin{array}{ll} v _{xx}(L,t)= v _{x}(0,t)= v (0,t)=0, \forall t\ge 0, \\ v _{xxx}(L,t)-\int \limits _{0}^{t}k(t-s)v _{xxx}(L,s)\mathrm{d}s=u(t), \forall t\ge 0, \\ v(x,0) =v_{0}(x), v_{t}(x,0) =v_{1}(x),\, x \in (0,L), \\ \end{array}\right. } \end{aligned}$$

where

$$\begin{aligned} {\left\{ \begin{array}{ll} u(t)=h(t)v_{t}(L,t), \\ h_{t}(t) = r v_{t}^{2} (L, t),\; h(0) = h_{0}> 0, \;t \ge 0, r > 0. \end{array}\right. } \end{aligned}$$

They supposed that the kernel k verifies

$$\begin{aligned} -c_{1}k(t)\le k_{t}(t)\le -c_{2}k(t),\; 0 < k_{tt}(t) \le c_{3}k(t) \end{aligned}$$

for some $c_{i}>0, i=1,...,3.$ Furthermore, a similar result in [15] was established under a boundary control

$$\begin{aligned} {\left\{ \begin{array}{ll} u(t)=h(t)v _{t}(L,t)+ \mu (t) sin \;t,\; t\ge 0, \\ h_{t}(t) = rv_{t}^{2} (L, t),\; h(0) = h_{0}> 0,\; t \ge 0, r > 0, \\ \mu _{t}(t)=v _{t}(L,t) sin \;t, \;\mu (0)=\mu _{0}. \end{array}\right. } \end{aligned}$$

Seghour et al. [17] investigated the following system

$$\begin{aligned} {\left\{ \begin{array}{ll} \rho v _{tt}(x,t)+EI v _{xxxx}(x,t)- EI \int _0^t k(t-s) v_{xxxx}(s)\mathrm{d}s - T v_{xx}(x,t)=0, \\ \text {in }(0,L)\times [0,\infty ), \\ v _{xx}(L,t)= v _{x}(0,t)= v (0,t)=0, \forall t\ge 0, \\ -EI v _{xxx}(L,t)+EI \int \limits _{0}^{t}k(t-s)v _{xxx}(L,s)\mathrm{d}s + T v_{x}(x,t)=u(t)- d_{s}v_{t} (L, t) \\ -M_{s} v_{tt} (L, t), \; t\ge 0, \\ v(x,0) =v_{0}(x), v_{t}(x,0) =v_{1}(x), x \in (0,L), \end{array}\right. } \end{aligned}$$

where the positives coefficients $d_{s}, M_{s}$ represents the vessel damping and the mass of the surface vessel. They showed an exponential decay result for solutions with the following conditions on the kernel:

$$\begin{aligned} 0<k^{\prime }(t)+ m k(t)\le \zeta (t), \ m\ge 0 \;t\ge 0 \end{aligned}$$

for some positive function $\zeta (t)$ and

$$\begin{aligned} u(t)=\dfrac{-K}{v_{t}(L,t)}\left\{ v _{xxx}(L,t)^{2}+ v(L,t)^{2}+\left[ \int \limits _{0}^{t}k(t-s)v _{xxx}(L,s)\mathrm{d}s\right] ^{2}\right\} , K>0, \;t\ge 0. \end{aligned}$$

Moreover, in [18], the authors considered a similar problem under $u(t)=d_{s} v_{t}(L,t)$, for kernels k verifying

$$\begin{aligned} k^{\prime }(t)\le \zeta (t), \; k(t-s) \ge \eta (t) \int _t^\infty k(t-\tau )d\tau \;t\ge 0 \end{aligned}$$

for some function $\eta (t)$. Later, the authors in [1] established uniform stability of the same problem for kernels satisfying

$$\begin{aligned} k^{\prime }(t)\le 0, \gamma (t)\zeta (t)\in L^{1}(0,\infty ). \end{aligned}$$

In [3], the authors considered the vibrating flexible beam system

$\left\{ \begin{array}{l} \rho A v _{tt}(x,t)+EI v _{xxxx}(x,t)-P_{0} v _{xx}(x,t)-\dfrac{3}{2}EA v _{xx}(x,t)v _{x}^{2}(x,t)=0, \\ \quad \text {in }(0,L)\times [0,\infty ), \\ v _{xx}(0,t)=v _{xx}(L,t)=v (0,t)=0, \forall t\ge 0 \\ -EI v _{xxx}(L,t)+P_{0} v _{x}(L,t)+\dfrac{1}{2} EA v _{x}^{3}(L,t)=-K_{1}v _{t}(L,t), \forall t\ge 0,\text { }K_{1}>0. \end{array} \right.$

They imposed a linear control force at the boundary to achieve the exponential stability of the system. Motivated by this work [3], the objective of the present paper is to consider the nonlinear viscoelastic Euler–Bernoulli beam equation

$$\begin{aligned}&\rho A v _{tt}(x,t)+EI v _{xxxx}(x,t)-P_{0}v _{xx}(x,t)-\dfrac{3}{2}EA v_{xx}(x,t)v _{x}^{2}(x,t) \nonumber \\&\quad -P_{0} \int _0^t \psi (t-s) v_{xx}(s)\mathrm{d}s =0, \, \, \text {in }(0,L)\times [0,\infty ) \end{aligned}$$

(1)

under the boundary conditions

$$\begin{aligned} {\left\{ \begin{array}{ll} v _{xx}(0,t)=v _{xx}(L,t)=v (0,t)=0, \forall t\ge 0, \\ EI v _{xxx}(L,t)=P_{0}v _{x}(L,t)+\dfrac{1}{2}EA v_{x}^{3}(L,t)-P_{0}\int \limits _{0}^{t}\psi (t-s)v _{x}(L,s)\mathrm{d}s \\ +K_{1}v _{t}(L,t), \forall t\ge 0, K_{1}>0. \end{array}\right. } \end{aligned}$$

(2)

The initial conditions are

$$\begin{aligned} v(x, 0) = v_{0}(x), \, \, v_{t}(x, 0) = v_{1}(x), x \in (0,L), \end{aligned}$$

(3)

where

EI: the flexural rigidity

$\rho A$: the mass per unit length

EA: the axial stiffness

v(x, t): the transverse displacement and

$P_{0}$: the tension force

The variance length envisaged with the tension force will be assumed to be weak compared to the overall length of the beam. We show an arbitrary decay result for problem (1)–(3) with weaker hypotheses on the relaxation function $\psi$ than the existing ones for similar problems. Namely, we do not limit ourselves to polynomially or exponentially decaying functions only. Relaxation functions that can have zero derivatives on certain subsets of $\left( 0,\infty \right)$ are considered, see [20,21,22]. We assume that the zone where the kernel is flat and is small. Consequently, a wide range of materials with various viscoelastic properties can be used in modern engineering.

The rest of our paper is arranged as follows: In Section 2, we give some useful lemmas needed for our result. The arbitrary decay of the energy result is shown in Section 3.

2 Notation and main results

We introduce the following notation

$$\begin{aligned} (\psi {\small \square }f) \left( t\right)= & {} \int \limits _{0}^{L}\int \limits _{0}^{t}\psi (t-\tau )\left[ f(t,x)-f(\tau ,x )\right] ^{2}d\tau \mathrm{d}x \\ (\psi {\small \star }f )\left( t\right)= & {} \int \limits _{0}^{L}\int \limits _{0}^{t}\psi (t-\tau )f(\tau ,x)d\tau \mathrm{d}x, \; t\ge 0. \end{aligned}$$

For the kernel $\psi$ we assume:

(H1):

$\psi : {\mathbb {R}}_{+}\rightarrow {\mathbb {R}}_{+}$ is a differentiable function satisfying

$$\begin{aligned} 0<k=\int \limits _{0}^{+\infty }\psi (s)\mathrm{d}s<1 . \end{aligned}$$

(H2):

$\psi ^{\prime }(t)\le 0$ for almost all $t\ge 0$.

(H3):

There exists a positive increasing function $\theta \left( t\right)$ such that $\dfrac{\theta ^{\prime }\left( t\right) }{\theta \left( t\right) }=u(t)$ is a decreasing function and $\int \limits _{0}^{+ \infty }\psi (s)\theta \left( s\right) \mathrm{d}s<+\infty .$

We denote

$$\begin{aligned} \begin{aligned} {\mathcal {V}}&=\left\{ v \in H^{2}\left( 0,L\right) | v (0)=0\right\} , \\ {\mathcal {H}}&=\left\{ v \in {\mathcal {V}} \cap H^{4}\left( 0,L\right) | v _{xx}(0)=v _{xx}(L)=0\right\} \end{aligned} \end{aligned}$$

and (., .), $\left\| .\right\|$ the inner product and the norm of the space $L^{2}\left( 0,L\right)$, respectively. The existence result for our problem (1)-(3) can be proved by Faedo–Galerkin method, the reader may consult [14].

Theorem 1

Suppose that (H1)-(H3) are satisfied. If $( v_{0}, v_{1}) \in {\mathcal {H}} \times L^{2}(0,L)$, then there exists a unique solution v of problem (1)-(3), in the sense that for $T> 0$,$v \in L^{\infty }([0, T ), {\mathcal {H}}), v_{t} \in L^{\infty }([0, T ), {\mathcal {V}}), v_{tt} \in L^{2}([0, T ), L^{2}(0,L)).$ Moreover, we have $v \in C([0, T ), {\mathcal {V}}), v_{t} \in C([0, T ),L^{2}(0,L))$.

We define the (classical) energy of problem (1)–(3) by

$$\begin{aligned} E(t)= & {} \frac{1}{2}\left[ \rho A\left\| v _{t}\left( t\right) \right\| ^{2}+EI\left\| v _{xx}\left( t\right) \right\| ^{2}+ P_{0} \left\| v _{x}\left( t\right) \right\| ^{2} +\frac{EA}{4}\left\| v _{x}^{2}\left( t\right) \right\| ^{2}\right] . \end{aligned}$$

(4)

Then, the time derivative of energy is equal to

$$\begin{aligned} E^{\prime }(t)=P_{0}\int \limits _{0}^{t}\psi (t-s)\left( v _{x}(s),v _{xt}(t)\right) \mathrm{d}s-K_{1}v _{t}^{2}(L,t), \,t\ge 0. \end{aligned}$$

(5)

It is easy to see that

$$\begin{aligned} \begin{aligned} \int \limits _{0}^{t}\psi (t-s)\left( v _{x}(s),v _{xt}(t)\right) \mathrm{d}s =-\dfrac{1 }{2}\left( \psi {\small \square }v _{x}\right) ^{\prime }\left( t\right) + \dfrac{1}{2}\left( \psi ^{\prime }{\small \square }v _{x}\right) \left( t\right) \\ +\dfrac{1}{2}\dfrac{d}{dt}\left( \left\| v _{x}\left( t\right) \right\| ^{2}\int \limits _{0}^{t}\psi (s)\mathrm{d}s\right) -\dfrac{1}{2} \psi (t)\left\| v _{x}\left( t\right) \right\| ^{2},t\ge 0. \end{aligned} \end{aligned}$$

Then, we consider the modified energy

$$\begin{aligned} e(t)&=\frac{\rho A}{2}\left\| v _{t}\left( t\right) \right\| ^{2}+ \frac{EI}{2}\left\| v _{xx}\left( t\right) \right\| ^{2}+\frac{P_{0}}{2 }\left( 1-\int \limits _{0}^{t}\psi (s)\mathrm{d}s\right) \left\| v _{x}\left( t\right) \right\| ^{2} \nonumber \\&\quad +\frac{EA}{8}\left\| v _{x}^{2}\left( t\right) \right\| ^{2} + \frac{P_{0}}{2} \left( \psi {\small \square }v _{x}\right) (t), \,\, t\ge 0. \end{aligned}$$

(6)

By differentiation, we obtain

$$\begin{aligned} e^{\prime }(t)=\frac{P_{0}}{2} \left( \psi ^{\prime }{\small \square }v _{x}\right) (t) -\dfrac{P_{0}}{2}\psi (t)\left\| v _{x}\left( t\right) \right\| ^{2}-K_{1}v _{t}^{2}(L,t), \,\, t\ge 0. \end{aligned}$$

(7)

If our non-negative relaxation function satisfies $\psi ^{\prime }\le 0$, it follows that e(t) is nonincreasing and uniformly bounded above by $e(0)=E(0)$. Next, we introduce the functionals

$$\begin{aligned} \varphi _{1}\left( t\right)= & {} \rho A\zeta \int \limits _{0}^{L}x v _{t}(x,t)v _{x}(x,t)\mathrm{d}x-\rho A\int \limits _{0}^{L}v _{t}\int \limits _{0}^{t}\psi (t-s)(v (t)-v (s))\mathrm{d}s\mathrm{d}x, \\ \varphi _{2}\left( t\right)= & {} P_{0}\int \limits _{0}^{L}\int \limits _{0}^{t}K_{\theta }(t-s)v _{x}(s)^{2}\mathrm{d}s\mathrm{d}x, \\ \varphi _{3}\left( t\right)= & {} EI\int \limits _{0}^{L}\int \limits _{0}^{t}K_{\theta }(t-s)v _{xx}(s)^{2}\mathrm{d}s\mathrm{d}x+\frac{EA}{2}\int \limits _{0}^{L}\int \limits _{0}^{t}K_{\theta }(t-s) v _{x}(s)^{4}\mathrm{d}s\mathrm{d}x, \end{aligned}$$

where $\zeta$ is a positive constant to be determined later,

$$\begin{aligned} K_{\theta }(t)=\theta ^{-1}\left( t\right) \int \limits _{t}^{+\infty }\psi (s)\theta \left( s\right) \mathrm{d}s \end{aligned}$$

and $\theta \left( t\right)$ is specified below. We define the second modified functional by

$$\begin{aligned} L(t)=e(t)+\sum \limits _{i=1}^{3}\lambda _{i}\text { }\varphi _{i}(t), t\ge 0 \end{aligned}$$

(8)

for $\lambda _{i}>0$, $i=1,2,3$ to be specified later. Our first result shows that this functional is an appropriate one to consider.

Proposition 2

There exist $q_{i}>0, i=1,2$ such that

$$\begin{aligned} q_{1}\left( e(t)+\varphi _{2}(t)+\varphi _{3}(t)\right) \le L(t)\le q_{2}\left( e(t)+\varphi _{2}(t)+\varphi _{3}(t)\right) ,t\ge 0. \end{aligned}$$

(9)

Proof

It is easy to see, from the above definitions, that

$$\begin{aligned} \begin{array}{clllll} \varphi _{1}(t) &{}\le \dfrac{\rho A}{2}\zeta L\left\| v _{t}\left( t\right) \right\| ^{2}+\dfrac{\rho A}{2}\zeta L\left\| v _{x}\left( t\right) \right\| ^{2} +\dfrac{\rho A}{2}\left\| v _{t}\left( t\right) \right\| ^{2}+\dfrac{\rho A c_{p}}{P_{0}}\dfrac{P_{0}}{2}k \left( \psi {\small \square }v _{x}\right) (t) \\ &{}\le \dfrac{\rho A}{2}(1+\zeta L)\left\| v _{t}\left( t\right) \right\| ^{2}+\dfrac{\rho A}{P_{0}}\zeta L\dfrac{P_{0}}{2}\left\| v _{x}\left( t\right) \right\| ^{2}+\dfrac{\rho A c_{p}}{P_{0}}\dfrac{P_{0} }{2}k \left( \psi {\small \square }v _{x}\right) (t) \\ &{}\le c_{1}\left( \dfrac{\rho A}{2}\left\| v _{t}\left( t\right) \right\| ^{2}+\dfrac{P_{0}}{2}\left\| v _{x}\left( t\right) \right\| ^{2}+\dfrac{P_{0}}{2}k \left( \psi {\small \square }v _{x}\right) (t) \right) , \end{array} \end{aligned}$$

where $c_{1}=\max (1+\zeta L, \dfrac{\rho A\zeta L}{P_{0}}, \dfrac{\rho Ac_{p}}{P_{0}})$. With these in mind, we have

$$\begin{aligned} L(t)\le & {} \left( 1+\lambda _{1}c_{1}\right) \frac{\rho A}{2}\left\| v _{t}\left( t\right) \right\| ^{2}+\frac{EI}{2}\left\| v _{xx}\left( t\right) \right\| ^{2} +\frac{EA}{8}\left\| v _{x}^{2}\left( t\right) \right\| ^{2} \\&+\left( 1-\int \limits _{0}^{t}\psi (s)\mathrm{d}s+\lambda _{1}c_{1}\right) \frac{P_{0} }{2}\left\| v _{x}\left( t\right) \right\| ^{2} +\left( 1+\lambda _{1}c_{1}k\right) \frac{P_{0}}{2} \left( \psi {\small \square }v _{x}\right) (t) \\&+\lambda _{2}\varphi _{2}\left( t\right) +\lambda _{3}\varphi _{3}\left( t\right) \end{aligned}$$

and

$$\begin{aligned} 2L(t)\ge & {} (1-c_{1}\lambda _{1})\rho A\left\| v _{t}\left( t\right) \right\| ^{2}+\left( 1-k-\lambda _{1}c_{1}\right) P_{0}\left\| v _{x}\left( t\right) \right\| ^{2} \\&+\left( 1-\lambda _{1}c_{1}k\right) P_{0}\left( \psi {\small \square }v _{x}\right) (t) +EI\left\| v _{xx}\left( t\right) \right\| ^{2} \\&+\frac{EA}{4}\left\| v _{x}^{2}\left( t\right) \right\| ^{2}+2\lambda _{2}\varphi _{2}\left( t\right) +2\lambda _{3}\varphi _{3}\left( t\right) ,t\ge 0. \end{aligned}$$

Therefore, $q_{1}\left( e(t)+\varphi _{2}\left( t\right) +\varphi _{3}(t)\right) \le L(t)\le q_{2}\left( e(t)+\varphi _{2}\left( t\right) +\varphi _{3}(t)\right)$ for some constant $q_{i}>0$ and $\lambda _{1}$ such that

$$\begin{aligned} \lambda _{1}<\frac{1-k}{c_{1}}. \end{aligned}$$

$\square$

The next result [21] gives a better estimate for

$$\begin{aligned} \int \limits _{0}^{L}v _{x}\int \limits _{0}^{t}\psi (t-s)v _{x}(s)\mathrm{d}s\mathrm{d}x. \end{aligned}$$

Lemma 3

We have for a continuous function $\psi$ on $[0, \infty )$ and $v\in H^{1}(0,L)$

$$\begin{aligned} \begin{array}{l} \displaystyle \int \limits _{0}^{L}v _{x}\displaystyle \int \limits _{0}^{t}\psi (t-s)v _{x}(s)\mathrm{d}s\mathrm{d}x\\ =\frac{1}{2 }\left( \displaystyle \int \limits _{0}^{t}\psi (s)\mathrm{d}s\right) \left\| v _{x}\right\| ^{2}+\frac{1}{2}\displaystyle \int \limits _{0}^{t}\psi (t-s)\left\| v _{x}\left( s\right) \right\| ^{2}\mathrm{d}s-\frac{1}{2}(\psi {\small \square }v _{x})(t), t\ge 0. \end{array} \end{aligned}$$

(10)

3 Asymptotic behavior

In this section we state and show our result. To this end we require some notation. For every measurable set ${\mathcal {A}}$ $\subset {\mathbb {R}}^{+}$, we define the probability measure

$$\begin{aligned} {\widehat{\psi }}\left( {\mathcal {A}}\right) =\frac{1}{1-k}\int \limits _{\mathcal {A }}\psi (s)\mathrm{d}s \end{aligned}$$

(11)

where $k=\int _{0}^{\infty }\psi (s)\mathrm{d}s.$ The flatness set and the flatness rate of $\psi$ are $\left( \text {respectively}\right)$ defined by

$$\begin{aligned} {\mathcal {F}}_{\psi }=\left\{ s\in {\mathbb {R}}^{+}, \psi (s)>0 \text { and } \psi ^{\prime }\left( s\right) =0\right\} \end{aligned}$$

(12)

and

$$\begin{aligned} {\mathcal {R}}_{\psi }={\widehat{\psi }}\left( {\mathcal {F}}_{\psi }\right) . \end{aligned}$$

Let $t^{\star }>0$ and $\int \limits _{0}^{t^{\star }}\psi (s)\mathrm{d}s=\psi _{\star }>0$.

Theorem 4

Let us suppose that $\psi$ and $\theta$ satisfy the hypotheses (H1)–(H3) and ${\mathcal {R}}_{\psi } < \dfrac{1}{4}$. Then, there exist positive constants C and $\nu$ such that

$$\begin{aligned} e(t)\le C \theta \left( t\right) ^{-\nu },\text { }t\ge 0. \end{aligned}$$

Proof

A differentiation of $\varphi _{1}\left( t\right) ,$ with respect to t along the solution of (1)-(4), gives

$$\begin{aligned} \varphi _{1}^{\prime }\left( t\right)= & {} \zeta \rho A\int \limits _{0}^{L} \left[ x v _{tt}v _{x}+x v _{t}v _{xt}\right] \mathrm{d}x \\&-\rho A\int \limits _{0}^{L}v _{t}\left[ \int \limits _{0}^{t}\psi ^{\prime }(t-s)(v (t)-v (s))\mathrm{d}s\mathrm{d}x +v _{t}\int \limits _{0}^{t}\psi (s)\mathrm{d}s\right] \mathrm{d}x \\&-\rho A\int \limits _{0}^{L}v _{tt}\int \limits _{0}^{t}\psi (t-s)(v (t)-v (s))\mathrm{d}s\mathrm{d}x, \, t\ge 0. \end{aligned}$$

We decompose the first integral into

$$\begin{aligned} \underbrace{\rho A\int \limits _{0}^{L}\left[ x v _{tt}v _{x}+x v _{t}v _{xt} \right] \mathrm{d}x}_{I_{\zeta }}=\underbrace{\rho A\int \limits _{0}^{L}x v _{t}v _{xt}\mathrm{d}x}_{I_{1}}+\underbrace{\rho A\int \limits _{0}^{L}x v _{tt}v _{x}\mathrm{d}x} _{I_{2}} \; . \end{aligned}$$

(13)

Clearly

$$\begin{aligned} I_{1}=\rho A\int \limits _{0}^{L}x\frac{d\left( v _{t}^{2}\right) }{2\mathrm{d}x} \mathrm{d}x= \frac{\rho A}{2}L v _{t}^{2}(L,t)-\frac{\rho A}{2}\left\| v _{t}\right\| ^{2}, \text { }t\ge 0 \end{aligned}$$

(14)

and

$$\begin{aligned} I_{2}= & {} \int \limits _{0}^{L}\left( -EI v _{xxxx}(x,t)+P_{0}v _{xx}(x,t)-P_{0}\int \limits _{0}^{t}\psi (t-s)v _{xx}\left( s\right) \mathrm{d}s+\dfrac{ 1}{2}EA(v _{x}^{3})_{x}\right) x v _{x}\mathrm{d}x \\= & {} \underbrace{-EI\int \limits _{0}^{L}x v _{x} v _{xxxx}(x,t)\mathrm{d}x}_{I_{21}}+ \underbrace{P_{0}\int \limits _{0}^{L}x v _{x}v _{xx}(x,t)\mathrm{d}x}_{I_{22}} \\&-\underbrace{P_{0}\int \limits _{0}^{L}x v _{x}\int \limits _{0}^{t}\psi (t-s)v _{xx}\left( s\right) \mathrm{d}s\mathrm{d}x}_{I_{23}} +\underbrace{\dfrac{EA}{2} \int \limits _{0}^{L}x v _{x}(v _{x}^{3})_{x}\mathrm{d}x}_{I_{24}}, \text { }t\ge 0. \end{aligned}$$

Moreover,

$$\begin{aligned} I_{21}= & {} -EI~L v _{x}\left( L,t\right) v _{xxx}(L,t)+ EI\int \limits _{0}^{L}\left( x v _{xx}+v _{x}\right) v _{xxx}(x,t)\mathrm{d}x \\= & {} -EI~L v _{x}\left( L,t\right) v _{xxx}(L,t)+EI\int \limits _{0}^{L}\dfrac{x}{2 }\dfrac{d}{\mathrm{d}x}\left( v _{xx}^{2}\right) \\&+EI~v _{x}\left( x,t\right) v _{xx}\left( x,t\right) |_{0}^{L}-EI\int \limits _{0}^{L}v _{xx}^{2}\mathrm{d}x \\= & {} -EI~L v _{x}\left( L,t\right) v _{xxx}(L,t)+EI\dfrac{x}{2}v _{xx}^{2}\left( x,t\right) |_{0}^{L}-\dfrac{3}{2}EI\left\| v _{xx}\right\| ^{2} \end{aligned}$$

that is

$$\begin{aligned} I_{21}= & {} -EI~L v _{x}\left( L,t\right) v _{xxx}(L,t)-\dfrac{3}{2} EI\left\| v _{xx}\right\| ^{2}, \end{aligned}$$

(15)

$$\begin{aligned} I_{22}= & {} \dfrac{P_{0}}{2}\int \limits _{0}^{L}x \dfrac{d}{\mathrm{d}x}v _{x}^{2}(x,t)\mathrm{d}x= \frac{P_{0}}{2}L v _{x}^{2}(L,t)-\frac{P_{0}}{2}\left\| v_{x}\right\| ^{2}, \end{aligned}$$

(16)

and

$$\begin{aligned} I_{23}= & {} -P_{0}L v _{x}\left( L,t\right) \int \limits _{0}^{t}\psi (t-s)v _{x}\left( L,s\right) \mathrm{d}s + P_{0}\int \limits _{0}^{L}v _{x}\int \limits _{0}^{t}\psi (t-s)v _{x}\left( s\right) \mathrm{d}s\mathrm{d}x \\&\quad +P_{0}\int \limits _{0}^{L}x v _{xx}\int \limits _{0}^{t}\psi (t-s)v _{x}\left( s\right) \mathrm{d}s\mathrm{d}x. \end{aligned}$$

Using Young and Cauchy–Schwartz inequality we estimate the integral

$$\begin{aligned} P_{0}\int \limits _{0}^{L}x v _{xx}\int \limits _{0}^{t}\psi (t-s)v _{x}\left( s\right) \mathrm{d}s\mathrm{d}x \le \dfrac{EI}{2}\left\| v _{xx}\left( t\right) \right\| ^{2}+\dfrac{P_{0}^{2}L^{2}k}{2EI}\int \limits _{0}^{t}\psi (t-s) \left\| v _{x}\left( s\right) \right\| ^{2}\mathrm{d}s. \end{aligned}$$

Next, Lemma 2 yields

$$\begin{aligned} P_{0}\int \limits _{0}^{L}v _{x}\int \limits _{0}^{t}\psi (t-s)v _{x}(s)\mathrm{d}s\mathrm{d}x= \dfrac{P_{0}}{2}\left( \int \limits _{0}^{t}\psi (s)\mathrm{d}s\right) \left\| v _{x}\right\| ^{2} \\ +\dfrac{P_{0}}{2}\int \limits _{0}^{t}\psi (t-s)\left\| v _{x}\left( s\right) \right\| ^{2}\mathrm{d}s-\dfrac{P_{0}}{2}(\psi {\small \square }v _{x})(t), \; t\ge 0. \end{aligned}$$

Then

$$\begin{aligned} I_{23} \le -P_{0}L v _{x}\left( L,t\right) \int \limits _{0}^{t}\psi (t-s)v _{x}\left( L,s\right) \mathrm{d}s +\dfrac{EI}{2}\left\| v _{xx}\right\| ^{2}+ \dfrac{P_{0}}{2}k\left\| v _{x}\right\| ^{2} \end{aligned}$$

$$\begin{aligned} +\dfrac{P_{0}}{2}\left( \dfrac{P_{0}}{EI}kL^{2}+1\right) \int \limits _{0}^{t}\psi (t-s)\left\| v _{x}\left( s\right) \right\| ^{2}\mathrm{d}s-\frac{P_{0}}{2}(\psi {\small \square }v _{x})(t) \end{aligned}$$

(17)

and

$$\begin{aligned} I_{24}= & {} \dfrac{EA}{2}L v_{x}^{4}(L,t)-\dfrac{EA}{2}\int \limits _{0}^{L} \left( v _{x}+x v _{xx}\right) v _{x}^{3}\mathrm{d}x \\= & {} \dfrac{EA}{2}L v _{x}^{4}(L,t)-\dfrac{EA}{2}\int \limits _{0}^{L}v _{x}^{4}(x,t)\mathrm{d}x-\dfrac{EA}{8}\int \limits _{0}^{L}x\dfrac{d}{\mathrm{d}x}\left( v _{x}^{4}\right) \\= & {} \dfrac{EA}{2}L v _{x}^{4}(L,t)-\dfrac{EA}{2}\int \limits _{0}^{L}v _{x}^{4}(x,t)\mathrm{d}x -\dfrac{EA}{8}L v _{x}^{4}(L,t)+\dfrac{EA}{8} \int \limits _{0}^{L}v _{x}^{4}(x,t)\mathrm{d}x \end{aligned}$$

or

$$\begin{aligned} I_{24} =\dfrac{3EA}{8}L v _{x}^{4}(L,t)-\dfrac{3EA}{8}\left\| v _{x}^{2}\right\| ^{2}. \end{aligned}$$

(18)

Taking into account (14)–(18), we have

$$\begin{aligned} I_{\zeta }\le & {} -\zeta \dfrac{\rho A}{2}\left\| v _{t}\right\| ^{2}-\zeta EI\left\| v _{xx}\left( t\right) \right\| ^{2} -\zeta \dfrac{P_{0}}{2}(1-k)\left\| v _{x}\left( t\right) \right\| ^{2}-\zeta \dfrac{3}{8}EA\left\| v _{x}^{2}\left( t\right) \right\| ^{2} \\&+\zeta \dfrac{P_{0}}{2}\left( \dfrac{P_{0}}{EI}kL^{2}+1\right) \int \limits _{0}^{t}\psi (t-s)\left\| v _{x}\left( s\right) \right\| ^{2}\mathrm{d}s-\zeta \dfrac{P_{0}}{2}(\psi {\small \square }v _{x})(t)\mathrm{d}x \\&+\left( -EI~v _{xxx}(L,t)+\frac{P_{0}}{2}v _{x}(L,t)+\dfrac{3EA}{8}EA v _{x}^{3}(L,t)-P_{0}\int \limits _{0}^{t}\psi (t-s)v _{x}\left( L,s\right) \mathrm{d}s\right) \\&\times L\zeta v _{x}\left( L,t\right) +\zeta \dfrac{\rho A}{2}L v _{t}^{2}(L,t). \end{aligned}$$

The boundary control gives us

$$\begin{aligned}&\left( -EI~v _{xxx}(L,t)+\dfrac{P_{0}}{2}v _{x}(L,t)+\dfrac{3EA}{8}EA v _{x}^{3}(L,t)-P_{0}\int \limits _{0}^{t}\psi (t-s)v _{x}\left( L,s\right) \mathrm{d}s\right) \times \\&\quad L\zeta v _{x}\left( L,t\right) = \\&\quad \left( -EI~v _{xxx}(L,t)+P_{0}v _{x}(L,t)+\dfrac{EA}{2}EA v _{x}^{3}(L,t)-P_{0}\int \limits _{0}^{t}\psi (t-s)v _{x}\left( L,s\right) \mathrm{d}s\right) \times \\&\quad L\zeta v _{x}\left( L,t\right) -\dfrac{EA}{8}L\zeta v _{x}^{4}(L,t)-\dfrac{ P_{0}}{2}\zeta L v _{x}^{2}(L,t) \\&\quad \quad =L\zeta K_{1}v _{t}(L,t)v _{x}(L,t)-\dfrac{EA}{8}\zeta L v _{x}^{4}(L,t)- \dfrac{P_{0}}{2}\zeta L v _{x}^{2}(L,t), t\ge 0. \end{aligned}$$

Moreover, by Young’s inequality

$$\begin{aligned} \zeta LK_{1}v _{t}(L,t)v _{x}(L,t)\le \dfrac{\zeta LK_{1}}{2}v _{t}^{2}(L,t)+\dfrac{\zeta LK_{1}}{2}v _{x}^{2}(L,t) \end{aligned}$$

and, therefore,

$$\begin{aligned} I_{\zeta }\le -&\zeta \dfrac{\rho A}{2}\left\| v _{t}\right\| ^{2}-\zeta EI\left\| v _{xx}\left( t\right) \right\| ^{2} -\zeta \dfrac{P_{0}}{2}(1-k)\left\| v _{x}\left( t\right) \right\| ^{2}-\zeta \frac{3}{8}EA\left\| v _{x}^{2}\left( t\right) \right\| ^{2} \nonumber \\&\quad +\zeta \dfrac{P_{0}}{2}\left( \dfrac{P_{0}}{EI}kL^{2}+1\right) \int \limits _{0}^{t}\psi (t-s)\left\| v _{x}\left( s\right) \right\| ^{2}\mathrm{d}s-\zeta \frac{P_{0}}{2}(\psi {\small \square }v _{x})(t)\qquad \qquad \qquad \nonumber \\&\quad +\dfrac{\zeta L}{2}\left( \rho A+K_{1}\right) v _{t}^{2}(L,t)+\dfrac{ \zeta L}{2}\left( K_{1}-P_{0}\right) v _{x}^{2}(L,t)-\zeta \dfrac{EA}{8}L v _{x}^{4}(L,t). \end{aligned}$$

(19)

Notice that

$$\begin{aligned}&-\rho A\int \limits _{0}^{L} v _{t}\left[ \int \limits _{0}^{t}\psi ^{\prime }(t-s)(v (t)-v (s))\mathrm{d}s\mathrm{d}x + v _{t}\int \limits _{0}^{t}\psi (s)\mathrm{d}s\right] \mathrm{d}x \nonumber \\&\quad \le \rho A\left( \delta _{1}-\int \limits _{0}^{t}\psi (s)\mathrm{d}s\right) \left\| v _{t}\right\| ^{2}-\dfrac{c_{p}\psi (0)}{4\delta _{1}}(\psi ^{\prime } {\small \square }v _{x}). \end{aligned}$$

(20)

After substitution of $-\rho A v_{tt}$ from (1) and integrating by part, we obtain

$$\begin{aligned} \begin{array}{lllllllll} &{}\underbrace{-\rho A\displaystyle \int \limits _{0}^{L}v_{tt}\displaystyle \int \limits _{0}^{t}\psi (t-s)(v (t)-v (s))\mathrm{d}s\mathrm{d}x}_{J} \\ &{}\quad =(EI v _{xxx}\left( L,t\right) -P_{0}v _{x}\left( L,t\right) -\frac{EA}{2}v _{x}^{3}\left( L,t\right) +P_{0}\displaystyle \int \limits _{0}^{t}\psi (t-s)v _{x}\left( L,s\right) \mathrm{d}s) \\ &{}\quad \quad \times \displaystyle \int \limits _{0}^{t}\psi (t-s)(v (L,t)-v (L,s))\mathrm{d}s \\ &{}\qquad +\displaystyle \int \limits _{0}^{L}\left( -EI v _{xxx}\left( t\right) +\frac{EA}{2}v _{x}^{3}\left( t\right) +P_{0}v _{x}\left( t\right) -P_{0}\displaystyle \int \limits _{0}^{t}\psi (t-s)v _{x}(s))\mathrm{d}s\right) \\ &{}\quad \quad \times \left( \displaystyle \int \limits _{0}^{t}\psi (t-s)(v _{x}(t)-v _{x}(s))\mathrm{d}s\right) \mathrm{d}x \\ &{}\quad =K_{1}v _{t}(L,t)\displaystyle \int \limits _{0}^{t}\psi (t-s)(v (L,t)-v (L,s))\mathrm{d}s \\ &{}\qquad -\displaystyle \int \limits _{0}^{L}EI v _{xxx}\left( t\right) \left( \displaystyle \int \limits _{0}^{t}\psi (t-s)(v _{x}(t)-v _{x}(s)\mathrm{d}s\right) \mathrm{d}x \\ &{}\qquad +\frac{EA}{2}\displaystyle \int \limits _{0}^{L}v _{x}^{3}\left( t\right) \left( \displaystyle \int \limits _{0}^{t}\psi (t-s)(v _{x}(t)-v _{x}(s))\mathrm{d}s\right) \mathrm{d}x \\ &{}\qquad +P_{0}\left( 1-\displaystyle \int \limits _{0}^{t}\psi (s)\mathrm{d}s\right) \displaystyle \int \limits _{0}^{L}v _{x}(t)\left( \displaystyle \int \limits _{0}^{t}\psi (t-s)(v _{x}(t)-v _{x}(s)\mathrm{d}s\right) \mathrm{d}x \\ &{}\qquad +P_{0}\displaystyle \int \limits _{0}^{L}\left( \displaystyle \int \limits _{0}^{t}\psi (t-s)(v _{x}(t)-v _{x}(s)\mathrm{d}s\right) ^{2}\mathrm{d}x \\ &{}\quad =-K_{1}v _{t}(L,t)\displaystyle \int \limits _{0}^{t}\psi (t-s)(v (L,t)-v (L,s))\mathrm{d}s+J_{1}+J_{2}+J_{3}+J_{4},\; t\ge 0. \end{array} \end{aligned}$$

Again utilizing Young’s inequality, we get

$$\begin{aligned}&-K_{1}v _{t}(L,t)\int \limits _{0}^{t}\psi (t-s)(v (L,t)-v (L,s))\mathrm{d}s \nonumber \\&\quad =-K_{1}v _{t}(L,t)v (L,t)\int \limits _{0}^{t}\psi (s)\mathrm{d}s+K_{1}v _{t}(L,t)\int \limits _{0}^{t}\psi (t-s)v (L,s)\mathrm{d}s \nonumber \\&\quad \le \dfrac{K_{1}}{2}\left( k+1\right) v _{t}^{2}(L,t)+\dfrac{kK_{1}}{2}v ^{2}(L,t)+\dfrac{K_{1}}{2}\left( \int \limits _{0}^{t}\psi (t-s)v (L,s)\mathrm{d}s\right) ^{2},\; t\ge 0. \end{aligned}$$

(21)

For the second and the third term in (21), we have

$$\begin{aligned}{}\begin{array}{l} \qquad \dfrac{K_{1}k}{2}v ^{2}(L,t)\le \dfrac{K_{1}k}{2}L\left\| v _{x}\right\| ^{2}, \\ \qquad \dfrac{K_{1}}{2}\left( \int \limits _{0}^{t}\psi (t-s)v (L,s)\mathrm{d}s\right) ^{2}=\dfrac{K_{1}}{2}\left( \int \limits _{0}^{t}\psi (t-s)\int \limits _{0}^{L}v _{x}(x,s)dxds\right) ^{2}, \\ \mathsf {and} \\ \qquad \dfrac{K_{1}}{2}\left( \int \limits _{0}^{t}\psi (t-s)v (L,s)\mathrm{d}s\right) ^{2}\le \dfrac{K_{1}}{2}Lk\int \limits _{0}^{t}\psi (t-s)\left\| v_{x}\left( s\right) \right\| ^{2}\mathrm{d}s,\; t\ge 0. \end{array} \end{aligned}$$

Hence,

$$\begin{aligned}&-K_{1}v _{t}(L,t)\int \limits _{0}^{t}\psi (t-s)(v (L,t)-v (L,s))\mathrm{d}s \nonumber \\&\quad \le \dfrac{K_{1}}{2}\left( k+1\right) v _{t}^{2}(L,t)+\dfrac{K_{1}k}{2} L\left\| v _{x}\right\| ^{2}+\dfrac{K_{1}}{2}Lk\int \limits _{0}^{t} \psi (t-s)\left\| v _{x}\left( s\right) \right\| ^{2}\mathrm{d}s. \nonumber \\ \end{aligned}$$

(22)

For $\delta _{2}> 0$, we can write

$$\begin{aligned} J_{1}= & {} \int \limits _{0}^{L}EI v _{xx}\left( t\right) \left( \int \limits _{0}^{t}\psi (t-s)(v _{xx}(t)-v _{xx}(s))\mathrm{d}s\right) \mathrm{d}x \\\le & {} EI\left( \int \limits _{0}^{t}\psi (s)\mathrm{d}s+\delta _{2}\right) \left\| v _{xx}\left( t\right) \right\| ^{2}+\frac{EI}{4\delta _{2}}\left( \int \limits _{0}^{t}\psi (s)\mathrm{d}s\right) \int \limits _{0}^{t}\psi (t-s)\left\| v _{xx}\left( s\right) \right\| ^{2}\mathrm{d}s \end{aligned}$$

and

$$\begin{aligned} J_{2}= & {} \frac{EA}{2}\int \limits _{0}^{t}\psi (s)\mathrm{d}s\text { }\left\| v _{x}^{2}\left( t\right) \right\| ^{2}-\dfrac{EA}{2}\int \limits _{0}^{L}v _{x}^{2}\left( t\right) \int \limits _{0}^{t}\psi (t-s)v _{x}(s)v _{x}(t)\mathrm{d}s\mathrm{d}x \\= & {} \dfrac{EA}{2}\int \limits _{0}^{t}\psi (s)\mathrm{d}s\text { }\left\| v _{x}^{2}\left( t\right) \right\| ^{2}-\dfrac{EA}{2}\int \limits _{0}^{L}v _{x}^{2}\left( t\right) \int \limits _{0}^{t}\psi ^{1/2}(t-s)v _{x}(s)\psi ^{1/2}(t-s)v _{x}(t)\mathrm{d}s\mathrm{d}x \\\le & {} \dfrac{EA}{2}k\left\| v _{x}^{2}\left( t\right) \right\| ^{2}+ \dfrac{EA}{2}\left( \int \limits _{0}^{L}\left( v _{x}^{2}\left( t\right) \right) ^{2}\mathrm{d}x\right) ^{1/2}\left( \int \limits _{0}^{L}(\int \limits _{0}^{t}\psi ^{1/2}(t-s)v _{x}(s)\psi ^{1/2}(t-s)v _{x}(t)\mathrm{d}s)^{2}\mathrm{d}x\right) ^{1/2} \\\le & {} \dfrac{EA}{2}k\left\| v _{x}^{2}\left( t\right) \right\| ^{2}+ \dfrac{EA}{4}\left\| v _{x}^{2}\left( t\right) \right\| ^{2}+\dfrac{EA }{4}\int \limits _{0}^{L}(\int \limits _{0}^{t}\psi ^{1/2}(t-s)v _{x}(s)\psi ^{1/2}(t-s)v _{x}(t)\mathrm{d}s)^{2}\mathrm{d}x \\\le & {} \dfrac{EA}{2}k\left\| v _{x}^{2}\left( t\right) \right\| ^{2}+ \dfrac{EA}{4}\left\| v _{x}^{2}\left( t\right) \right\| ^{2}+\dfrac{EA }{4}\int \limits _{0}^{L}\int \limits _{0}^{t}\psi (t-s)v _{x}^{2}(s)\mathrm{d}s.\int \limits _{0}^{t}\psi (t-s)v _{x}^{2}(t)\mathrm{d}s \mathrm{d}x \\\le & {} \frac{EA}{2}k\left\| v _{x}^{2}\left( t\right) \right\| ^{2}+ \dfrac{EA}{4}\left\| v _{x}^{2}\left( t\right) \right\| ^{2}+\dfrac{EA }{4}\left( \frac{k^{2}\delta _{3}}{4}\int \limits _{0}^{t}\psi (t-s)\left\| v _{x}^{2}\left( s\right) \right\| ^{2}\mathrm{d}s+\dfrac{k}{\delta _{3}} \left\| v _{x}^{2}\left( t\right) \right\| ^{2}\right) \\\le & {} \dfrac{EA}{2}\left( k+\dfrac{k}{2\delta _{3}}+\frac{1}{2}\right) \left\| v _{x}^{2}\left( t\right) \right\| ^{2}+\dfrac{EA}{16}\delta _{3}k^{2}\int \limits _{0}^{t}\psi (t-s)\left\| v _{x}^{2}\left( s\right) \right\| ^{2}\mathrm{d}s. \end{aligned}$$

Now we proceed to estimate $J_{3}.$ For all measurable sets ${\mathcal {A}}$ and ${\mathcal {F}}$ such that $\mathcal {A= {\mathbb {R}} }^{+}\backslash {\mathcal {F}}$, we see that

$$\begin{aligned} J_{3}= & {} P_{0}\left( 1-\int \limits _{0}^{t}\psi (s)\mathrm{d}s\right) \int \limits _{0}^{L}v _{x}(t)\left( \int \limits _{\mathcal {A\cap }\left[ 0,t\right] }\psi (t-s)(v _{x}(t)-v _{x}(s))\mathrm{d}s\right) \mathrm{d}x \\&\quad +P_{0}\left( 1-\int \limits _{0}^{t}\psi (s)\mathrm{d}s\right) \int \limits _{0}^{L}v _{x}(t)\left( \int \limits _{\mathcal {F\cap }\left[ 0,t\right] }\psi (t-s)(v _{x}(t)-v _{x}(s))\mathrm{d}s\right) \mathrm{d}x, t\ge 0. \end{aligned}$$

We denote ${\mathcal {Q}}_{t}={\mathcal {Q}}\mathcal {\cap }\left[ 0,t\right]$. Using Lemma 2, we obtain for $\delta _{4}>0$

$$\begin{aligned}&P_{0}\left( 1-\int \limits _{0}^{t}\psi (s)\mathrm{d}s\right) \int \limits _{0}^{L}v _{x}(t)\left( \int \limits _{{\mathcal {A}}_{t}}\psi (t-s)(v _{x}(t)-v _{x}(s))\mathrm{d}s\right) \mathrm{d}x \\&\quad \le P_{0}\left( 1-\int \limits _{0}^{t}\psi (s)\mathrm{d}s\right) \left( \delta _{4}\left\| v _{x}\left( t\right) \right\| ^{2}+\frac{k}{4\delta _{4}} \int \limits _{0}^{L}\int \limits _{{\mathcal {A}}_{t}}\psi (t-s)(v _{x}(t)-v _{x}(s))^{2}\mathrm{d}s\mathrm{d}x\right) , \end{aligned}$$

clearly

$$\begin{aligned}&\int \limits _{0}^{L}v _{x}(t)\left( \int \limits _{{\mathcal {F}} _{t}}\psi (t-s)(v _{x}(t)-v _{x}(s))\mathrm{d}s\right) \mathrm{d}x \\&\quad =\left( \int \limits _{{\mathcal {F}}_{t}}\psi (s)\mathrm{d}s\left\| v _{x}\left( t\right) \right\| ^{2}-\int \limits _{0}^{L}v _{x}(t)\left( \int \limits _{ {\mathcal {F}}_{t}}\psi (t-s)v _{x}(s)\mathrm{d}s\right) \mathrm{d}x\right) \end{aligned}$$

and

$$\begin{aligned}&-\int \limits _{0}^{L}v _{x}(t)\left( \int \limits _{{\mathcal {F}} _{t}}\psi (t-s)v _{x}(s)\mathrm{d}s\right) \mathrm{d}x \\&\quad \le \dfrac{1}{2}\int \limits _{{\mathcal {F}}_{t}}\psi (s)\mathrm{d}s\left\| v _{x}\left( t\right) \right\| ^{2}+\dfrac{1}{2}\int \limits _{{\mathcal {F}} _{t}}\psi (t-s)\left\| v _{x}\left( s\right) \right\| ^{2}\mathrm{d}s,\; t\ge 0. \end{aligned}$$

Thus,

$$\begin{aligned}&\int \limits _{0}^{L}v _{x}(t)\left( \int \limits _{{\mathcal {F}} _{t}}\psi (t-s)(v _{x}(t)-v _{x}(s))\mathrm{d}s\right) \mathrm{d}x \\&\quad \le \frac{3}{2}k\text { }{\widehat{\psi }}\left( {\mathcal {F}}_{\psi }\right) \left\| v _{x}\left( t\right) \right\| ^{2}+\frac{1}{2}\int \limits _{ {\mathcal {F}}_{t}}\psi (t-s)\left\| v _{x}\left( s\right) \right\| ^{2}\mathrm{d}s, \end{aligned}$$

where ${\widehat{\psi }}$ is defined in (11). We end up with

$$\begin{aligned} J_{3}\le & {} P_{0}\left( 1-\psi _{\star }\right) \left( \delta _{4}+\frac{3}{2 }k\text { }{\widehat{g}}\left( {\mathcal {F}}_{g}\right) \right) \left\| v _{x}\left( t\right) \right\| ^{2} \\&+P_{0}\left( 1-\psi _{\star }\right) \frac{k}{4\delta _{4}} \int \limits _{0}^{L}\int \limits _{{\mathcal {A}}_{t}}\psi (t-s)(v _{x}(t)-v _{x}(s))^{2}\mathrm{d}s\mathrm{d}x \\&+\frac{P_{0}}{2}\left( 1-\psi _{\star }\right) \int \limits _{{\mathcal {F}} _{t}}\psi (t-s)\left\| v _{x}\left( s\right) \right\| ^{2}\mathrm{d}s,\; t\ge t_{\star }. \end{aligned}$$

For $\delta _{5}>0$, we have

$$\begin{aligned} J_{4}\le & {} P_{0}\left( 1+\dfrac{1}{_{\delta _{5}}}\right) k\int \limits _{0}^{L}\int \limits _{{\mathcal {A}}_{t}}\psi (t-s)(v _{x}(t)-v _{x}(s))^{2}\mathrm{d}s\mathrm{d}x \\&+P_{0}\left( 1+\delta _{5}\right) k\text { }{\widehat{\psi }}\left( {\mathcal {F}} _{\psi }\right) \int \limits _{0}^{L}\int \limits _{{\mathcal {F}}_{t}}\psi (t-s)(v _{x}(t)-v _{x}(s))^{2}\mathrm{d}s\mathrm{d}x,\; t\ge 0. \end{aligned}$$

Taking into account (19)-(22) and the above estimations of $J_{1},$ $J_{2},$ $J_{3}$, $J_{4}$, we obtain

$$\begin{aligned} \varphi _{1}^{\prime }\left( t\right)\le & {} \rho A\left( \delta _{1}-\psi _{\star }-\dfrac{\zeta }{2}\right) \left\| v _{t}\right\| ^{2}+ EI(k+\delta _{2}-\zeta ) \left\| v _{xx}\right\| ^{2} \nonumber \\&+P_{0}\left( -\dfrac{\zeta }{2}(1-k)+\dfrac{K_{1}kL}{2P_{0}}+\left( 1-\psi _{\star }\right) (\delta _{4}+\dfrac{3}{2}k\text { }{\widehat{\psi }} \left( {\mathcal {F}}_{\psi }\right) )\right) \left\| v _{x}\right\| ^{2} \nonumber \\&+\left( \zeta \dfrac{P_{0}}{2}\left( \dfrac{P_{0}}{EI}kL^{2}+1\right) + \dfrac{K_{1}kL}{2}\right) \int \limits _{0}^{t}\psi (t-s)\left\| v _{x}\left( s\right) \right\| ^{2}\mathrm{d}s \nonumber \\&+\dfrac{EA}{2}\left( k+\dfrac{k}{2\delta _{3}}+\dfrac{1}{2}-\dfrac{3\zeta }{4}\right) \left\| v _{x}^{2}\right\| ^{2}+\dfrac{EA\delta _{3}}{16} k^{2}\int \limits _{0}^{t}\psi (t-s)\left\| v _{x}^{2}\left( s\right) \right\| ^{2}\mathrm{d}s \nonumber \\&+\dfrac{1}{2}\left( \rho A\zeta L+K_{1}\left( k+1+\zeta L\right) \right) v _{t}^{2}(L,t) +\frac{\zeta L}{2}\left( K_{1}-P_{0}\right) v _{x}^{2}(L,t) \nonumber \\&-\zeta \dfrac{EA}{8}L\ v _{x}^{4}(L,t) -\dfrac{c_{p}\psi (0)}{4\delta _{1}} (\psi ^{\prime }{\small \square }v _{x})(t)+\dfrac{kEI}{4\delta _{2}} \int \limits _{0}^{t}\psi (t-s)\left\| v _{xx}\left( s\right) \right\| ^{2}\mathrm{d}s \nonumber \\&-\dfrac{\zeta P_{0}}{2}(\psi {\small \square }v _{x})(t) +P_{0}k\left( \dfrac{1-\psi _{\star }}{4\delta _{4}}+1+\dfrac{1}{\delta _{5}}\right) \int \limits _{0}^{L}\int \limits _{{\mathcal {A}}_{t}}\psi (t-s)(v _{x}(t)-v _{x}(s))^{2}\mathrm{d}s\mathrm{d}x\nonumber \\&+P_{0}\left( 1+\delta _{5}\right) k\text { }{\widehat{\psi }}\left( {\mathcal {F}} _{\psi }\right) \int \limits _{0}^{L}\int \limits _{{\mathcal {F}}_{t}}\psi (t-s)(v _{x}(t)-v _{x}(s))^{2}\mathrm{d}s\mathrm{d}x \nonumber \\&+\dfrac{P_{0}}{2}\left( 1-\psi _{\star }\right) \int \limits _{{\mathcal {F}} _{t}}\psi (t-s)\left\| v _{x}\left( s\right) \right\| ^{2}\mathrm{d}s, t\ge t_{\star } . \end{aligned}$$

(23)

Further, a differentiation of $\varphi _{2}\left( t\right)$ yields

$$\begin{aligned} \varphi _{2}^{\prime }\left( t\right)&=P_{0}K_{\theta }\left( 0\right) \left\| v _{x}\right\| ^{2}+P_{0}\int \limits _{0}^{t}K_{\theta }^{\prime }(t-s)\left\| v _{x}(s)\right\| ^{2}\mathrm{d}s \nonumber \\&=P_{0}K_{\theta }\left( 0\right) \left\| v _{x}\right\| ^{2}-P_{0}\int \limits _{0}^{t}\frac{\theta ^{\prime }\left( t-s\right) }{ \theta \left( t-s\right) }K_{\theta }(t-s)\left\| v _{x}(s)\right\| ^{2}\mathrm{d}s-P_{0}\int \limits _{0}^{t}\psi (t-s)\left\| v _{x}(s)\right\| ^{2}\mathrm{d}s \nonumber \\&\le P_{0}K_{\theta }\left( 0\right) \left\| v _{x}\right\| ^{2}-P_{0}u(t)\int \limits _{0}^{t}K_{\theta }(t-s)\left\| v _{x}(s)\right\| ^{2}\mathrm{d}s-P_{0}\int \limits _{0}^{t}\psi (t-s)\left\| v _{x}(s)\right\| ^{2}\mathrm{d}s, t\ge 0.\nonumber \\ \end{aligned}$$

(24)

Regarding $\varphi _{3}^{\prime }\left( t\right)$ it appears that

$$\begin{aligned} \varphi _{3}^{\prime }\left( t\right)&=EI K_{\theta }\left( 0\right) \left\| v _{xx}\right\| ^{2}+EI\int \limits _{0}^{t}K_{\theta }^{\prime }(t-s)\left\| v _{xx}(s)\right\| ^{2}\mathrm{d}s +\dfrac{EA}{2} K_{\theta }\left( 0\right) \left\| v _{x}^{2}\right\| ^{2} \\&\qquad +\dfrac{EA}{2}\int \limits _{0}^{t}K_{\theta }^{\prime }(t-s)\left\| v _{x}^{2}(s)\right\| ^{2}\mathrm{d}s,\qquad \qquad \qquad \qquad \qquad \qquad \end{aligned}$$

that is

$$\begin{aligned} \varphi _{3}^{\prime }\left( t\right)&\le EI K_{\theta }\left( 0\right) \left\| v _{xx}\right\| ^{2}+\dfrac{EA}{2}K_{\theta }\left( 0\right) \left\| v _{x}^{2}\right\| ^{2} -EI\int \limits _{0}^{t}\psi (t-s)\left\| v _{xx}(s)\right\| ^{2}\mathrm{d}s\nonumber \\&\quad -\dfrac{EA}{2}u(t)\int \limits _{0}^{t}K_{\theta }(t-s)\left\| v _{x}^{2}(s)\right\| ^{2}\mathrm{d}s-\dfrac{EA}{2}\int \limits _{0}^{t}\psi (t-s)\left\| v _{x}^{2}(s)\right\| ^{2}\mathrm{d}s \nonumber \\&\qquad \qquad \qquad \qquad -EIu(t)\int \limits _{0}^{t}K_{\theta }(t-s)\left\| v _{xx}(s)\right\| ^{2}\mathrm{d}s, \; t\ge 0. \end{aligned}$$

(25)

Collecting the estimations (7), (23)–(25), we find for $t\ge t_{\star }$

$$\begin{aligned}&L^{\prime }\left( t\right) \le P_{0}\left( \dfrac{1}{2}-\lambda _{1} \dfrac{c_{p}\psi (0)}{4\delta _{1}}\right) (\psi ^{\prime }{\small \square }v _{x})(t)+\rho A\lambda _{1}(\delta _{1}-\psi _{\star }-\dfrac{\zeta }{2} )\left\| v _{t}\right\| ^{2} \nonumber \\&\quad +P_{0}\left( -\dfrac{\zeta }{2}(1-k)\lambda _{1}+\dfrac{K_{1}k}{2P_{0}} L\lambda _{1}+\lambda _{1}\left( 1-\psi _{\star }\right) (\delta _{4}+\dfrac{3 }{2}k\text { }{\widehat{\psi }}\left( {\mathcal {F}}_{\psi }\right) )+\lambda _{2}K_{\theta }\left( 0\right) \right) \left\| v _{x}\right\| ^{2} \nonumber \\&\quad +EI \left( (k+\delta _{2}-\zeta )\lambda _{1}+\lambda _{3}K_{\theta }\left( 0\right) \right) \left\| v _{xx}\right\| ^{2} +EI\left( \dfrac{ k}{4\delta _{2}}\lambda _{1}-\lambda _{3}\right) \int \limits _{0}^{t}\psi (t-s)\left\| v _{xx}\left( s\right) \right\| ^{2}\mathrm{d}s \nonumber \\&\quad +\dfrac{EA}{2}\left( \dfrac{k^{2}\delta _{3}}{8}\lambda _{1}-\lambda _{3}\right) \int \limits _{0}^{t}\psi (t-s)\left\| v _{x}^{2}\left( s\right) \right\| ^{2}\mathrm{d}s \nonumber \\&\quad -\left( K_{1}-\left( \dfrac{K_{1}}{2}\left( k+1+\zeta L\right) +\dfrac{ \rho A\zeta L}{2}\right) \lambda _{1}\right) v _{t}^{2}(L,t) \nonumber \\&\quad +\dfrac{EA}{2}\left( \lambda _{1}(k+\dfrac{k}{2\delta _{3}}+\dfrac{1}{2}- \dfrac{3\zeta }{4})+\lambda _{3}K_{\theta }\left( 0\right) \right) \left\| v _{x}^{2}\right\| ^{2} \nonumber \\&\quad -\zeta \dfrac{EA}{8}L\lambda _{1}v _{x}^{4}(L,t)-\dfrac{\zeta L}{2}\lambda _{1}\left( P_{0}-K_{1}\right) v _{x}^{2}(L,t) -\dfrac{P_{0}}{2}\lambda _{1}\zeta (\psi {\small \square } v _{x})(t) \nonumber \\&\quad +P_{0}\left( \lambda _{1}\left( \dfrac{\zeta }{2}(\dfrac{P_{0}}{EI} kL^{2}+1)+\dfrac{kK_{1}L}{2P_{0}}\right) -\lambda _{2}\right) \int \limits _{0}^{t}\psi (t-s)\left\| v _{x}\left( s\right) \right\| ^{2}\mathrm{d}s \nonumber \\&\quad +\lambda _{1}kP_{0}\left( \dfrac{1-\psi _{\star }}{4\delta _{4}}+1+\dfrac{1 }{\delta _{5}}\right) \int \limits _{0}^{L}\int \limits _{{\mathcal {A}} _{t}}\psi (t-s)(v _{x}(t)-v _{x}(s))^{2}\mathrm{d}s\mathrm{d}x \nonumber \\&\quad +\lambda _{1}P_{0}\left( 1+\delta _{5}\right) k\text { }{\widehat{\psi }} \left( {\mathcal {F}}_{\psi }\right) \int \limits _{0}^{L}\int \limits _{{\mathcal {F}} _{t}}\psi (t-s)(v _{x}(t)-v _{x}(s))^{2}\mathrm{d}s\mathrm{d}x \nonumber \\&\quad +\lambda _{1}\dfrac{P_{0}}{2}\left( 1-\psi _{\star }\right) \int \limits _{ {\mathcal {F}}_{t}}\psi (t-s)\left\| v _{x}\left( s\right) \right\| ^{2}\mathrm{d}s-\lambda _{2}P_{0}u(t)\int \limits _{0}^{t}K_{\theta }(t-s)\left\| v _{x}(s)\right\| ^{2}\mathrm{d}s \nonumber \\&\quad -\lambda _{3}u(t)\varphi _{3}\left( t\right) . \end{aligned}$$

For $n\in {\mathbb {N}}$, we introduce the sets [16]

$$\begin{aligned} {\mathcal {A}}_{n}=\left\{ s\in {\mathbb {R}} ^{+}:n\psi ^{\prime }\left( s\right) +\psi \left( s\right) \le 0\right\} . \end{aligned}$$

Notice that

$$\begin{aligned} \underset{n}{\cup }{\mathcal {A}}_{n}= {\mathbb {R}} ^{+}\backslash \left\{ {\mathcal {F}}_{\psi }\cup {\mathcal {N}}_{\psi }\right\} , \end{aligned}$$

where ${\mathcal {N}}_{\psi }$ is the null set in which $\psi ^{\prime }$ is not defined. The complement of ${\mathcal {A}}_{n}$ in ${\mathbb {R}}^{+}$ is denoted by ${\mathcal {F}}_{n}= {\mathbb {R}} ^{+}\backslash {\mathcal {A}}_{n}$. It appears that $\underset{n\rightarrow \infty }{\lim {\widehat{\psi }}\left( {\mathcal {F}}_{n}\right) =}{\widehat{\psi }} \left( {\mathcal {F}}_{\psi }\right)$ since ${\mathcal {F}}_{n+1}\subset {\mathcal {F}} _{n}$ for all n and $\underset{n}{\cap }{\mathcal {F}}_{n}={\mathcal {F}} _{\psi }\cup {\mathcal {N}}_{\psi }$. Then, we take ${\mathcal {A}}_{n}={\mathcal {A}}$ , ${\mathcal {F}}_{n}={\mathcal {F}}$, and we select $\lambda _{1}\le \dfrac{ \delta _{1}}{c_{p}\psi (0)},$ so that

$$\begin{aligned} \frac{1}{2}-\lambda _{1}\frac{c_{p}\psi (0)}{4\delta _{1}}\ge \frac{1}{4}. \end{aligned}$$

Choosing

$$\begin{aligned} \lambda _{3}=\frac{k^{2}\delta _{3}}{8}\lambda _{1},\delta _{3}= \dfrac{2}{k},\text { }\delta _{2}=1,\text { } \end{aligned}$$

we may write

$$\begin{aligned} \begin{array}{ll} &{} L^{\prime }\left( t\right) \le \rho A\left( \delta _{1}-\psi _{\star }- \dfrac{\zeta }{2}\right) \lambda _{1}\left\| v _{t}\right\| ^{2}- \dfrac{\zeta L}{2}\lambda _{1}\left( P_{0}-K_{1}\right) v_{x}^{2}(L,t) \\ &{} -\left( K_{1}-\left( \dfrac{K_{1}}{2}\left( k+1+\zeta L\right) +\dfrac{ \rho A\zeta L}{2}\right) \lambda _{1}\right) v _{t}^{2}(L,t) -\zeta \dfrac{EA }{8}L\lambda _{1}v _{x}^{4}(L,t) \\ &{} +P_{0}\left( -\dfrac{\zeta }{2}(1-k)\lambda _{1}+\dfrac{K_{1}k}{2P_{0}} L\lambda _{1}+\lambda _{1}\left( 1-\psi _{\star }\right) (\delta _{4}+\dfrac{3 }{2}k\text { }{\widehat{\psi }}\left( {\mathcal {F}}_{\psi }\right) )+\lambda _{2}K_{\theta }\left( 0\right) \right) \left\| v _{x}\right\| ^{2} \\ &{} +EI\text { }\lambda _{1}\left( k+1-\zeta +\dfrac{k}{4}K_{\theta }\left( 0\right) \right) \left\| v _{xx}\right\| ^{2} +\lambda _{1}\dfrac{P_{0} }{2}\left( 1-\psi _{\star }\right) \int \limits _{{\mathcal {F}} _{nt}}\psi (t-s)\left\| v _{x}\left( s\right) \right\| ^{2}\mathrm{d}s \\ &{} +\dfrac{EA}{2}\lambda _{1}\left( \dfrac{k^{2}}{4}+k+\dfrac{1}{2}-\dfrac{ 3\zeta }{4}+\dfrac{k}{4}K_{\theta }\left( 0\right) \right) \left\| v _{x}^{2}\right\| ^{2} -\dfrac{P_{0}\zeta }{2}\lambda _{1}(\psi {\small \square }v _{x})(t) \\ &{} +P_{0}\left( \lambda _{1}\left( \dfrac{\zeta }{2}(\dfrac{P_{0}}{EI} kL^{2}+1)+\dfrac{kK_{1}L}{2P_{0}}\right) -\lambda _{2}\right) \int \limits _{0}^{t}\psi (t-s)\left\| v _{x}\left( s\right) \right\| ^{2}\mathrm{d}s \\ &{} +\lambda _{1}kP_{0}\left( \dfrac{ 1-\psi _{\star } }{4\delta _{4}}+1+\dfrac{ 1}{\delta _{5}}\right) \int \limits _{0}^{L}\int \limits _{{\mathcal {A}} _{nt}}\psi (t-s)(v _{x}(t)-v _{x}(s))^{2}\mathrm{d}s\mathrm{d}x \\ &{} +\lambda _{1}\left( 1+\delta _{5}\right) k\text { }{\widehat{\psi }}\left( {\mathcal {F}}_{n}\right) \int \limits _{0}^{L}\int \limits _{{\mathcal {F}} _{nt}}\psi (t-s)(v _{x}(t)-v _{x}(s))^{2}\mathrm{d}s\mathrm{d}x \\ &{} -\lambda _{2}u(t)\varphi _{2}\left( t\right) -\lambda _{3}u(t)\varphi _{3}\left( t\right) , \;t\ge t_{\star }. \qquad \qquad \qquad \qquad \end{array} \nonumber \\ \end{aligned}$$

(26)

we choose $\zeta =1+2k$ so that

$$\begin{aligned} \left\{ \begin{array}{l} \dfrac{k^{2}}{4}+k+\dfrac{1}{2}-\dfrac{3\zeta }{4}+\dfrac{k}{4}K_{\theta }\left( 0\right)<0 , \\ k+1-\zeta +\dfrac{k}{4}K_{\theta }\left( 0\right) <0. \end{array} \right. \end{aligned}$$

Also, we need $K_{\theta }\left( 0\right) \le \min \left\{ 1-k+\dfrac{1}{k}, \text { }4\right\}$ and $\delta _{1}=\dfrac{\psi _{\star }+1+2k}{2}.$ For small $\delta _{5}$ and $t^{\star }, n$ large enough, we see that if ${\widehat{\psi }}\left( {\mathcal {F}}_{n}\right) <\dfrac{1}{4}$  then

$$\begin{aligned} \left( 1+\delta _{5}\right) k\text { }{\widehat{\psi }}\left( {\mathcal {F}} _{n}\right) -\frac{1}{2}\left( 1+2k\right) <0 \end{aligned}$$

and

$$\begin{aligned} \frac{3}{2}k\left( 1-\psi _{\star }\right) \text { }{\widehat{\psi }}\left( {\mathcal {F}}_{n}\right) <\sigma \left( 1+2k\right) (\frac{1-k}{2}) \end{aligned}$$

with $\sigma =\dfrac{3k\left( 1-\psi _{\star }\right) }{4\left( 1-k\right) \left( 1+2k\right) }$. For the relation

$$\begin{aligned} \lambda _{2}K_{\theta }\left( 0\right) \le \left( 1-\sigma \right) \left( 1+2k\right) (\frac{1-k}{2})\lambda _{1}-\left( \frac{K_{1}k}{2P_{0}}L+\left( 1-\psi _{\star }\right) \delta _{4}\right) \lambda _{1} \end{aligned}$$

to hold, it suffices that

$$\begin{aligned} \lambda _{2}K_{\theta }\left( 0\right)\le & {} \left( 1-\sigma \right) \left( 1+2k\right) (\frac{1-k}{3})\lambda _{1}- \frac{K_{1}k}{2P_{0}}L \lambda _{1} \\\le & {} \frac{4+k(1+3\psi _{\star }-8k-6 K_{1}L/P_{0})}{12}\lambda _{1} \end{aligned}$$

with $\delta _{4}$ small enough. Taking $K_{1} \le min \left\{ P_{0}, \dfrac{P_{0}}{30L}\right\}$, we get

$$\begin{aligned} \lambda _{2}K_{\theta }(0) \le \dfrac{4+4k/5+3k\psi _{\star }-8k^{2}}{12}. \end{aligned}$$

This is possible if $\psi _{\star }>\dfrac{8k^{2}-4-4k/5}{3k}.$ We also need $\lambda _{1}$ so small that

$$\begin{aligned} \lambda _{1}kP_{0}\left( \frac{1-\psi _{\star }}{4\delta _{4}}+(1+\dfrac{1}{ \delta _{5}})\right) -\frac{1}{4n}<0 , \\ \lambda _{1}\left\{ [\left( 2k+1\right) L+k+1]+30L^{2}\dfrac{\rho A}{ P_{0}}\left( 1+2k\right) \right\} <2. \end{aligned}$$

As a consequence of the above consideration,

$$\begin{aligned} L^{\prime }\left( t\right)\le & {} -c_{1}\frac{\psi _{\star }}{2}\rho A\lambda _{1}\left\| v _{t}\right\| ^{2}-c_{2}P_{0}\left\| v _{x}\right\| ^{2}-c_{3}EI\left\| v _{xx}\right\| ^{2}-c_{4}EA\left\| v _{x}^{2}\right\| ^{2} \\&-c_{5}P_{0}\int \limits _{0}^{L}\left( \psi {\small \square }v _{x}\right) \mathrm{d}x-\lambda _{2}u(t)\varphi _{2}\left( t\right) -\lambda _{3}u(t)\varphi _{3}\left( t\right) ,t\ge t_{\star } \end{aligned}$$

for some positive constants $c_{i}$ $i=1,...,5$. For $\lambda _{1}$ even smaller if necessary, we get

$$\begin{aligned} L^{\prime }\left( t\right) \le -C_{1}e(t)-\lambda _{2}u(t)\varphi _{2}\left( t\right) -\lambda _{3}u(t)\varphi _{3}\left( t\right) , t\ge t_{\star } \end{aligned}$$

(27)

where $C_{1}$ is some positive constant. As u(t) is nonincreasing, we have u(t) $\le u(0)$ for all $t\ge t_{\star }$. Then (27) becomes

$$\begin{aligned} L^{\prime }\left( t\right) \le -\frac{C_{1}}{u(0)}u(t)e(t)-\lambda _{2}u(t)\varphi _{2}\left( t\right) -\lambda _{3}u(t)\varphi _{3}\left( t\right) ,t\ge t_{\star }. \end{aligned}$$

By Proposition 1, we obtain

$$\begin{aligned} L^{\prime }\left( t\right) \le -C_{2}u(t)L\left( t\right) \end{aligned}$$

(28)

for some positive constant $C_{2}$. Integrating (28) over $\left[ t_{\star },t\right]$ yields

$$\begin{aligned} L\left( t\right) \le e^{-c_{2}\int \limits _{t_{\star }}^{t}u(s)\mathrm{d}s}L\left( t_{\star }\right) , t\ge t_{\star }. \end{aligned}$$

Then using inequality (9) of Proposition 1, we find

$$\begin{aligned} q_{1}\left( e(t\right) +\varphi _{2}\left( t\right) +\varphi _{3}\left( t\right) )\le e^{-c_{2}\int \limits _{t_{\star }}^{t}u(s)\mathrm{d}s}L\left( t_{\star }\right) ,t\ge t_{\star }. \end{aligned}$$

The continuity of E(t) over the interval $\left[ 0,t_{\star }\right]$ makes it possible to deduce

$$\begin{aligned} e(t)\le \frac{C}{\theta \left( t\right) ^{\nu }},\text { }t\ge 0 \end{aligned}$$

for some positive constants C and $\nu$. $\square$