Research article Special Issues

Investigating the relationship between academic burnout and educational factors among students of Guilan University of Medical Sciences

  • Received: 08 May 2019 Accepted: 07 July 2019 Published: 29 August 2019
  • Introduction: Academic burnout is among the factors that negatively affect academic performance and has recently been studied in schools and universities. Aim: The aim of this study was to determine the relationship between academic burnout and educational factors among students of Guilan University of Medical Sciences in 2015–2016. Materials and methods: This cross sectional study was conducted on 532 students who were in second or higher semester of their study in Guilan University of Medical Sciences. The instruments used in this study were the Maslach’s Student Burnout Questionnaire and the Educational Factors Questionnaire. Data were analyzed using SPSS software version 21. Results: A significant relationship was observed between academic burnout and passion in college major (P = 0.0001), failing the courses (P = 0.0001), probation record (P < 0.009), teaching factors, educational environment and educational facilities (P < 0.05). Conclusion: The results of this study indicated a significant relationship between academic burnout and a number of educational factors. As a result, appropriate educational and teaching facilities can reduce students’ academic burnout.

    Citation: Soodabeh Gholizadeh Sarcheshmeh, Fariba Asgari, Minoo Mitra Chehrzad, Ehsan Kazemnezhad Leili. Investigating the relationship between academic burnout and educational factors among students of Guilan University of Medical Sciences[J]. AIMS Medical Science, 2019, 6(3): 230-238. doi: 10.3934/medsci.2019.3.230

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  • Introduction: Academic burnout is among the factors that negatively affect academic performance and has recently been studied in schools and universities. Aim: The aim of this study was to determine the relationship between academic burnout and educational factors among students of Guilan University of Medical Sciences in 2015–2016. Materials and methods: This cross sectional study was conducted on 532 students who were in second or higher semester of their study in Guilan University of Medical Sciences. The instruments used in this study were the Maslach’s Student Burnout Questionnaire and the Educational Factors Questionnaire. Data were analyzed using SPSS software version 21. Results: A significant relationship was observed between academic burnout and passion in college major (P = 0.0001), failing the courses (P = 0.0001), probation record (P < 0.009), teaching factors, educational environment and educational facilities (P < 0.05). Conclusion: The results of this study indicated a significant relationship between academic burnout and a number of educational factors. As a result, appropriate educational and teaching facilities can reduce students’ academic burnout.


    This paper is concerned with the initial boundary value problem

    $ {utt(x,t)uxx(x,t)+μ1(t)ut(x,t)+μ2(t)ut(x,tτ(t))=0in Ω×]0,+[,u(0,t)=u(L,t)=0on ]0,+[,u(x,0)=u0(x),ut(x,0)=u1(x)on Ω,ut(x,tτ(0))=f0(x,tτ(0))in Ω×]0,τ(0)[,
    $
    (1)

    where $ \Omega = ]0,L[ $, $ 0<\tau(t) $ are a non-constant time delay, $ \mu_1(t), \mu_2(t) $ are non-constant weights and the initial data $ (u_0,u_1,f_0) $ belong to a suitable function space.

    This problem has been first proposed and studied in Nicaise and Pignotti [22] in case of constant coefficients $ \mu_1, \mu_2 $ and constant time delay. Under suitable assumptions, the authors proved the exponential stability of the solution by introducing suitable energies and by using some observability inequalities. Some instability results are also given for the case of the some assumptions is not satisfied.

    With a weight depending on time, $ \mu_1(t), \mu_2(t) $ and constant time delay, this problem was studied in [2], where the existence of solution was made by Faedo-Galerkin method and a decay rate estimate for the energy was given by using the multiplier method.

    W. Liu in [19] studied the weak viscoelastic equation with an internal time varying delay term. By introducing suitable energy and Lyapunov functionals, he establishes a general decay rate estimate for the energy under suitable assumptions.

    F. Tahamtani and A. Peyravi [29] investigated the nonlinear viscoelastic wave equation with source term. Using the Potential well theory they showed that the solutions blow up in finite time under some restrictions on initial data and for arbitrary initial energy.

    Global existence and asymptotic behavior of solutions to the viscoelastic wave equation with a constant delay term was considered by M. Remil and A. Hakem in [28].

    Global existence and asymptotic stability for a coupled viscoelastic wave equation with time-varying delay was studied in [3] by combining the energy method with the Faedo-Galerkin's procedure.

    The stabilization problem by interior damping of the wave equation with boundary or internal time-varying delay was studied in [23] by introducing suitable Lyapunov functionals.

    Energy decay of solutions for the wave equation with a time varying delay term in the weakly nonlinear internal feedbacks was considered in [11].

    For problems with delay in different contexts we cite [9,10,30,32] with references therein. In absence of delay ($ \mu_2(t) = 0 $), the problem (1) is exponentially stable provided that $ \mu_1(t) $ is constant, see, for instance [5,6,16,17,21] and reference therein.

    Time delay is the property of a physical system by which the response to an applied force is delayed in its effect, and the central question is that delays source can destabilize a system that is asymptotically stable in the absence of delays, see [7]. In fact, an arbitrarily small delay may destabilize a system that is uniformly asymptotically stable in the absence of delay unless additional control terms have been used, see for example [8,12,31]

    By energy method in [24] was studied the stabilization of the wave equation with boundary or internal distributed delay. By semigroup approach in [27] was proved the well-posedness and exponential stability for a wave equation with frictional damping and nonlocal time-delayed condition. Transmission problem with distributed delay was studied in [18] where was established the exponential stability of the solution by introducing a suitable Lyapunov functional.

    Here we consider a wave equation with non-constant delay and nonlinear weights, thus, the present paper is a generalization of the previous ones. The remaining part of this paper is organized as follows. In the section 2 we introduce some notations and prove the dissipative property of the full energy of the system. In the section 3, for an approach combining semigroup theory (see [21] and [4]) with the energy estimate method we prove the existence and uniqueness of solution. In section 4 we present the result of exponential stability.

    We will need the following hypotheses:

    (H1) $ \mu_1:\mathbb{R}_+ \rightarrow ]0,+\infty[ $ is a non-increasing function of class $ C^1(\mathbb{R}_+) $ satisfying

    $ |μ1(t)μ1(t)|M1,0<α0μ1(t),t0,
    $
    (2)

    where $ \alpha_0 $ and $ M_1 $ are constants such that $ M_1>0 $.

    (H2) $ \mu_2:\mathbb{R}_+ \rightarrow \mathbb{R} $ is a function of class $ C^1(\mathbb{R}_+) $, which is not necessarily positives or monotones, such that

    $ |μ2(t)|βμ1(t),
    $
    (3)
    $ |μ2(t)|M2μ1(t),
    $
    (4)

    for some $ 0 < \beta < \sqrt{1-d} $ and $ M_2>0 $.

    We now state a lemma needed later.

    Lemma 2.1 (Sobolev-Poincare's inequality). Let $ q $ be a number with $ 2\leq q \leq +\infty $. Then there is a constant $ c_* = c_*(]0,L[,q) $ such that

    $ \left\| \Psi \right\|_q \leq c_* \left\| \Psi_x \right\|_2, \quad \mathit{\mbox{for}}\; \Psi \in H_0^1(]0,L[). $

    Lemma 2.2 ([13][16]). Let $ E:\mathbb{R}_+ \rightarrow \mathbb{R}_+ $ be a non increasing function and assume that there are two constants $ \sigma>-1 $ and $ \omega>0 $ such that

    $ \int_{S}^{+\infty} E^{1+\sigma}(t)\,dt \leq \frac{1}{\omega}E^\sigma(0)E(S),\ 0\leq S < +\infty. $

    Then

    $ E(t)=0 tEσ(0)ω|σ|, if1<σ<0,E(t)E(0)(1+σ1+ωσt)1σ t0, ifσ>0,E(t)E(0)e1ωt t0, ifσ=0.
    $

    As in [23], we assume that

    $ τ(t)W2,+([0,T]),  for T>0
    $
    (5)

    and there exist positive constants $ \tau_0,\tau_1 $ and $ d $ satisfying

    $ 0<τ0τ(t)τ1, t>0
    $
    (6)

    and

    $ τ(t)d<1, t>0.
    $
    (7)

    We introduce the new variable

    $ z(x,ρ,t)=ut(x,tτ(t)ρ), xΩ,ρ]0,1[,t>0.
    $
    (8)

    Then

    $ \tau(t)z_t(x,\rho,t) + (1-\tau'(t)\rho)z_\rho(x,\rho,t) = 0, \ x\in \Omega,\ \rho \in ]0,1[,\ t > 0 $

    and problem (1) takes the form

    $ {utt(x,t)uxx(x,t)+μ1(t)ut(x,t)+μ2(t)z(x,1,t)=0inΩ×]0,+[,τ(t)zt(x,ρ,t)+(1τ(t)ρ)zρ(x,ρ,t)=0inΩ×]0,1[×]0,+[,u(0,t)=u(L,t)=0on]0,+[,u(x,0)=u0(x),ut(x,0)=u1(x)onΩ,z(x,ρ,0)=ut(x,τ(0)ρ)=f0(x,τ(0)ρ)inΩ×]0,1[.
    $
    (9)

    We define the energy of the solution of problem (9) by

    $ E(t)=12ut2L2(Ω)+12ux2L2(Ω)+ξ(t)τ(t)2Ω10z2(x,ρ,t)dρdx,
    $
    (10)

    where

    $ ξ(t)=ˉξμ1(t)
    $
    (11)

    is a non-increasing function of class $ C^1(\mathbb{R}_+) $ and $ \bar{\xi} $ be a positive constant such that

    $ β1d<ˉξ<2β1d.
    $
    (12)

    Our first result states that the energy is a non-increasing function.

    Lemma 2.3. Let $ (u,z) $ be a solution to the problem (9). Then, the energy functional defined by (10) satisfies

    $ E(t)μ1(t)(1ˉξ2β21d)ut2L2(Ω)μ1(t)(ˉξ(1τ(t))2β1d2)z(x,1,t)2L2(Ω)0.
    $
    (13)

    Proof. Multiplying the first equation (9) by $ u_t(x,t) $, integrating on $ \Omega $ and using integration by parts, we get

    $ 12ddt(ut2L2(Ω)+ux2L2(Ω))+μ1(t)ut2L2(Ω)+μ2(t)Ωz(x,1,t)utdx.
    $
    (14)

    Now multiplying the second equation (9) by $ \xi(t)z(x,\rho,t) $ and integrate on $ \Omega \times ]0,1[ $, to obtain

    $ \tau(t)\xi(t)\int_{\Omega} \int_0^1\!\! z_t(x,\rho,t)z(x,\rho,t)\,d\rho\,dx = -\frac{\xi(t)}{2}\! \int_{\Omega} \int_0^1\!\!\! (1- \tau'(t)\rho)\frac{\partial}{\partial \rho}(z(x,\rho,t))^2\,d\rho\,dx. $

    Consequently,

    $ ddt(ξ(t)τ(t)2Ω10z2(x,ρ,t)dρdx)=ξ(t)2Ω10(1τ(t)ρ)ρ(z(x,ρ,t))2dρdx+ξ(t)τ(t)2Ω10z2(x,ρ,t)dρdx=ξ(t)2Ω(z2(x,0,t)z2(x,1,t))dx+ξ(t)τ(t)2Ω10z2(x,1,t)dρdx+ξ(t)τ(t)2Ω10z2(x,ρ,t)dρdx.
    $
    (15)

    From (10), (14) and (15) we obtain

    $ E(t)=ξ(t)2ut2L2(Ω)ξ(t)2z(x,1,t)2L2(Ω)+ξ(t)τ(t)2z(x,1,t)2L2(Ω)+ξ(t)τ(t)2Ω10z2(x,ρ,t)dρdxμ1(t)ut2L2(Ω)μ2(t)Ωz(x,1,t)utdx.
    $
    (16)

    Due to Young's inequality, we have

    $ μ2(t)Ωz(x,1,t)utdx|μ2(t)|21dut2L2(Ω)+|μ2(t)|1d2z(x,1,t)2L2(Ω).
    $
    (17)

    Inserting (17) into (16), we obtain

    $ E(t)(μ1(t)ξ(t)2|μ2(t)|21d)ut2L2(Ω)(ξ(t)2ξ(t)τ(t)2|μ2(t)|1d2)z(x,1,t)2L2(Ω)+ξ(t)τ(t)2Ω10z2(x,ρ,t)dρdxμ1(t)(1ˉξ2β21d)ut2L2(Ω)μ1(t)(ˉξ(1τ(t))2β1d2)z(x,1,t)2L2(Ω)0.
    $

    Lemma 2.4. Let $ (u,z) $ be a solution to the problem (9). Then the energy functional defined by (10) satisfies

    $ \|u_t(x,t)\|_{L^2(\Omega)}^{2} < -\frac{1}{\sigma}E'(t), $

    where $ \sigma = a_0\left( 1-\frac{\bar{\xi}}{2}- \frac{\beta}{2\sqrt{1-d}} \right) $.

    Proof. From Lemma 2.3, we have that

    $ E(t)μ1(t)(1ˉξ2+β21d)ut2L2(Ω)+μ1(t)(ˉξ(1τ(t))2+β1d2)z(x,1,t)2L2(Ω)0
    $

    and from (H1), we obtain

    $ 0a0(1ˉξ2+β21d)ut2L2(Ω)μ1(t)(1ˉξ2+β21d)ut2L2(Ω)E(t)
    $

    and the lemma is proved.

    For the semigroup setup we $ U = (u,u_t,z)^T $ and rewrite (9) as

    $ {Ut=A(t)U,U(0)=U0=(u0,u1,f0(,,τ(0)))T,
    $
    (18)

    where the operator $ \mathcal{A}(t) $ is defined by

    $ AU=(v,uxxμ1(t)vμ2(t)z(x,1,t),1τ(t)ρτ(t)zρ(x,ρ,t))T.
    $
    (19)

    We introduce the phase space

    $ \mathcal{H} = H_0^1(\Omega)\times L^2(\Omega)\times L^2(\Omega \times ]0,1[) $

    and the domain of $ \mathcal{A} $ is defined by

    $ D(A(t))={(u,v,z)TH/v=z(,0)  in Ω},
    $
    (20)

    where

    $ H = H^2(\Omega)\cap H_0^1(\Omega)\times H_0^1(\Omega)\times L^2(\Omega; H_0^1(]0,1[)). $

    Notice that the domain of the operator $ \mathcal{A}(t) $ is independent of the time $ t $, i.e.,

    $ D(A(t))=D(A(0)),t>0.
    $
    (21)

    $ \mathcal{H} $ is a Hilbert space provided with the inner product

    $ U,ˉUH=Ωuxˉuxdx+Ωvˉvdx+ξ(t)τ(t)Ω10zˉzdρdx,
    $
    (22)

    for $ U = (u,v,z)^T $ and $ \bar{U} = (\bar{u},\bar{v},\bar{z})^T $.

    Using this time-dependent inner product and the next theorem we will get a result of existence and uniqueness.

    Theorem 3.1. Assume that

    (i) $ Y = D(\mathcal{A}(0)) $ is dense subset of $ \mathcal{H} $,

    (ii) (21) holds,

    (iii) for all $ t \in [0,T] $, $ \mathcal{A}(t) $ generates a strongly continuous semigroup on $ \mathcal{H} $ and the family $ \mathcal{A}(t) = \left\{ \mathcal{A}(t)/ t \in [0,T] \right\} $ is stable with stability constants $ C $ and $ m $ independent of $ t $ (i.e., the semigroup $ (S_t(s))_{s\geq 0} $ generated by $ \mathcal{A}(t) $ satisfies $ \| S_t(s)u \|_{\mathcal{H}} \leq Ce^{ms} \| u \|_{\mathcal{H}} $, for all $ u \in \mathcal{H} $ and $ s\geq 0 $),

    (iv) $ \partial_t \mathcal{A}(t) $ belongs to $ L_{*}^{\infty}([0,T],B(Y, \mathcal{H})) $, which is the space of equivalent classes of essentially bounded, strongly measurable functions from $ [0,T] $ into the set $ B(Y, \mathcal{H}) $ of bounded operators from $ Y $ into $ \mathcal{H} $.

    Then, problem (18) has a solution $ U \in C([0,T],Y) \cap C^1([0,T], \mathcal{H}) $ for any initial datum in $ Y $.

    Our goal is then to check the above assumptions for problem (18).

    First, we prove $ D(\mathcal{A}(0)) $ is dense in $ \mathcal{H} $.

    The proof is the same as the one Lemma $ 2.2 $ of [25], we give it for the sake of completeness.

    Let $ (f,g,h)^T $ be orthogonal to all elements of $ D\mathcal{A}(0) $, namely

    $ 0 = \langle (u,v,z)^T,(f,g,h)^T \rangle_{\mathcal{H}} = \int_{\Omega} u_x f_x\,dx + \int_{\Omega} v g\,dx + \xi(t)\tau(t) \int_{\Omega} \int_0^1 z h\,d\rho\,dx, $

    for all $ (u,v,z)^T \in D(\mathcal{A}(0)) $.

    We first take $ u = v = 0 $ and $ z \in D(\Omega \times ]0,1[) $. As $ (0,0,z)^T \in D(\mathcal{A}(0)) $, we get

    $ \int_{\Omega} \int_0^1 zh\,d\rho\,dx = 0. $

    Since $ D(\Omega \times ]0,1[) $ is dense in $ L^2(\Omega \times ]0,1[) $, we deduce that $ h = 0 $. In the same manner, by taking $ u = z = 0 $ e $ v \in D(\Omega) $ we see that $ g = 0 $.

    The above orthogonality condition is then reduced to

    $ 0 = \int_{\Omega} u_xf_x\,dx, \quad \forall (u,v,z)^T \in D(\mathcal{A}(0)). $

    By restricting ourselves to $ v = 0 $ and $ z = 0 $, we obtain

    $ 0 = \int_{\Omega} u_xf_x\,dx, \quad \forall (u,0,0)^T \in D(\mathcal{A}(0)). $

    Since $ D(\Omega) $ is dense in $ H_0^1(\Omega) $ (equipped with the inner product $ \langle \cdot, \cdot \rangle_{H_0^1(\Omega} $), we deduce that $ f = 0 $.

    We consequently

    $ D(A(0) is dense in H.
    $
    (23)

    Secondly, we notice that

    $ ΦtΦsec2τ0|ts|,t,s[0,T],
    $
    (24)

    where $ \Phi = (u,v,z)^T $ and $ c $ is a positive constant and $ \| \cdot \| $ is the norm associated the inner product (22). For all $ t,s \in [0,T] $, we have

    $ Φ2tΦ2secτ0|ts|=(1ec2τ0|ts|)(ux2L2(Ω)+v2L2(Ω))+(ξ(t)τ(t)ξ(s)τ(s)ecτ0|ts|)Ω10z2(x,ρ,t)dρdx.
    $

    It is clear that $ 1 - e^{\frac{c}{\tau_0}|t-s|} \leq 0 $. Now we will prove $ \xi(t)\tau(t) - \xi(s)\tau(s)e^{\frac{c}{\tau_0}|t-s|} \leq 0 $ for some $ c>0 $. To do this, we have

    $ \tau(t) = \tau(s) + \tau'(r)(t-s), $

    where $ r \in ]s,t[ $.

    Hence $ \xi $ is a non increasing function and $ \xi>0 $, we get

    $ \xi(t)\tau(t) \leq \xi(s)\tau(s) + \xi(s)\tau'(r)(t-s), $

    which implies

    $ \frac{\xi(t)\tau(t)}{\xi(s)\tau(s)} \leq 1 + \frac{|\tau'(r)|}{\tau(s)}|t-s|. $

    Using (5) and $ \tau' $ is bounded, we deduce that

    $ \frac{\xi(t)\tau(t)}{\xi(s)\tau(s)} \leq 1 + \frac{c}{\tau_0}|t-s| \leq e^{\frac{c}{\tau_0}|t-s|}, $

    which proves (24) and therefore $ (iii) $ follows.

    Now we calculate $ \langle \mathcal{A}(t)U,U \rangle_t $ for a fixed $ t $. Take $ U = (u,v,z)^T \in D(\mathcal{A}(t)) $. Then

    $ A(t)U,Ut=Ωvxuxdx+Ω(uxxμ1(t)vμ2(t)z(,1))vdxξ(t)Ω10(1τ(t)ρ)zρ(x,ρ)z(x,ρ)dρdx.
    $

    Integrating by parts, we obtain

    $ A(t)U,Ut=μ1(t)v2L2(Ω)μ2(t)Ωz(,1)vdxΩ10(1τ(t)ρ)ρz2(x,ρ)dρdx.
    $

    Since

    $ \left(1- \tau'(t)\rho \right)\frac{\partial}{\partial \rho} z^2(x,\rho) = \frac{\partial}{\partial \rho} \left( \left(1- \tau'(t)\rho \right) z^2(x,\rho) \right) + \tau'(t)z^2(x,\rho), $

    we have

    $ 10(1τ(t)ρ)ρz2(x,ρ)dρ=(1τ(t))z2(x,1)z2(x,0)+τ(t)10z2(x,ρ)dρ.
    $

    So we get

    $ A(t)U,Ut=μ1(t)v2L2(Ω)μ2(t)Ωz(x,1)vdx+ξ(t)2z(x,0)2L2(Ω)ξ(t)(1τ(t))2z(x,1)2L2(Ω)ξ(t)τ(t)2Ω10z2(x,ρ)dρdx.
    $

    Therefore, from (16) and (17), we deduce

    $ A(t)U,Utμ1(t)(1ˉξ2β21d)v2L2(Ω)μ1(t)(ˉξ(1τ(t))2β1d2)z(x,1,t)2L2(Ω)+ξ(t)|τ(t)|2τ(t)τ(t)Ω10z2(x,ρ)dρdx.
    $

    Then, we have

    $ A(t)U,Utμ1(t)(1ˉξ2β21d)v2L2(Ω)μ1(t)(ˉξ(1τ(t))2β1d2)z(x,1,t)2L2(Ω)+κ(t)U,Ut,
    $

    where

    $ \kappa(t) = \frac{\sqrt{1+ \tau'(t)^2}}{2\tau(t)}. $

    From the (13), we obtain

    $ A(t)U,Utκ(t)U,Ut0,
    $
    (25)

    which means that the operator $ \tilde{\mathcal{A}} = \mathcal{A}(t) - \kappa(t)I $ is dissipative.

    Moreover, $ \kappa'(t) = \frac{\tau'(t)\tau''(t)}{2\tau(t)\sqrt{1+\tau'(t)^2}} - \frac{\tau'(t)\sqrt{1+\tau'(t)^2}}{2\tau(t)^2} $ is bounded on $ [0,T] $ for all $ T>0 $ (by (5) and (12)) and we have

    $ \frac{d}{dt}\mathcal{A}(t)U = \left(0,0,\frac{\tau''(t)\tau(t)\rho-\tau'(t)(\tau'(t)\rho-1)}{\tau(t)^2}z_{\rho} \right)^T, $

    with $ \frac{\tau''(t)\tau(t)\rho-\tau'(t)(\tau'(t)\rho-1)}{\tau(t)^2} $ bounded on $ [0,T] $ by (5) and (12). Thus

    $ ddt˜A(t)L([0,T],B(D(A(0)),H)),
    $
    (26)

    the space of equivalence classes of essentially bounded, strongly measurable functions from $ [0,T] $ into $ B(D(\mathcal{A}(0)), \mathcal{H}) $.

    Now, we will show that $ \lambda I - \mathcal{A}(t) $ is surjective for fixed $ t>0 $ and $ \lambda > 0 $. For this purpose, let $ F = (f_1,f_2,f_3)^T \in \mathcal{H} $, we seek $ U = (u,v,z)^T \in D(\mathcal{A}(t)) $ solution of

    $ \left( \lambda I - \mathcal{A}(t) \right)U = F, $

    that is verifying following system of equations

    $ {λuv=f1,λvuxx+μ1(t)vμ2(t)z(,1)=f2,λz+1τ(t)ρτ(t)zρ=f3.
    $
    (27)

    Suppose that we have found $ u $ with the appropriated regularity. Then

    $ v=λuf1.
    $
    (28)

    It is clear that $ v \in H_0^1(\Omega) $. Furthermore, by (27) we can find $ z $. From (20), we have

    $ z(x,0)=v(x),  for xΩ.
    $
    (29)

    Following the same approach as in [22], we obtain, by using equation for $ z $ in (27),

    $ z(x,\rho) = v(x)e^{-\vartheta(\rho,t)} + \tau(t)e^{-\vartheta(\rho,t)} \int_{0}^{\rho} f_3(x,s)e^{\vartheta(s,t)}\,ds, $

    if $ \tau'(t) = 0 $, where $ \vartheta(\ell,t) = \lambda \ell \tau(t) $, and

    $ z(x,\rho) = v(x)e^{\zeta(\rho,t)} + e^{\zeta(\rho,t)} \int_{0}^{\rho} \frac{\tau(t)f_3(x,s)}{1-s\tau'(s)} e^{-\zeta(s,t)}\,ds, $

    otherwise, where $ \zeta(\ell,t) = \lambda \frac{\tau(t)}{\tau'(t)}\ln(1- \ell \tau'(t)) $.

    From (28), we obtain

    $ z(x,ρ)=λu(x)eϑ(ρ,t)f1(x,ρ)eϑ(ρ,t)+τ(t)eϑ(ρ,t)ρ0f3(x,s)eϑ(s,t)ds,
    $
    (30)

    if $ \tau'(t) = 0 $, and

    $ z(x,ρ)=λu(x)eζ(ρ,t)f1(x,ρ)eζ(ρ,t)+eζ(ρ,t)ρ0τ(t)f3(x,s)1sτ(s)eζ(s,t)ds,
    $
    (31)

    otherwise.

    In particular, if $ \tau'(t) = 0 $ and from (30), we have

    $ z(x,1)=λu(x)eϑ(1,t)f1(x,1)eϑ(1,t)+τ(t)eϑ(1,t)10f3(x,s)eϑ(s,t)ds,
    $
    (32)

    and if $ \tau'(t) \neq 0 $ and from (31), we have

    $ z(x,1)=λu(x)eζ(1,t)f1(x,1)eζ(1,t)+eζ(1,t)10τ(t)f3(x,s)1sτ(s)eζ(s,t)ds.
    $
    (33)

    By using (27) and (28), the function $ u $ satisfies

    $ λ2uuxx+μ1(t)v+μ2(t)z(,1)=f2+λf1.
    $
    (34)

    Solving the equation (34) is equivalent to finding $ u \in H^2(\Omega) \cap H_0^1(\Omega) $ such that

    $ Ω(λ2uη+uxηx+μ1(t)vη+μ2(t)z(,1)η)dx=Ω(f2+λf1)ηdx,
    $
    (35)

    for all $ \eta \in H_0^1(\Omega) $.

    Consequently, the equation (35) is equivalent to the problem

    $ Υ(u,η)=L(η),
    $
    (36)

    where the bilinear form

    $ \Upsilon: H_0^1(\Omega) \times H_0^1(\Omega) \rightarrow \mathbb{R} $

    and the linear form

    $ L: H_0^1(\Omega) \rightarrow \mathbb{R} $

    are defined by

    $ \Upsilon(u, \eta) = \int_{\Omega} \left( \lambda^2 u\eta + u_x \eta_x \right)\,dx + \int_{\Omega} \lambda u \left( \mu_1(t) + \mu_2(t)N_1 \right)\eta\,dx $

    and

    $ L(\eta) = \int_{\Omega} \left( \mu_1(t)f_1 \eta + \mu_2(t)N_2 \right)\eta\,dx + \int_{\Omega} (f_2 + \lambda f_1)\eta\,dx, $

    where

    $ N_1 = \left\{ eϑ(1,t),ifτ(t)=0,eζ(1,t),ifτ(t)0
    \right. $

    and

    $ N_2 = \left\{ f1(x,1)eϑ(1,t)+τ(t)eϑ(1,t)10f3(x,s)eϑ(s,t)ds,ifτ(t)=0,f1(x,1)ezeta(1,t)+ezeta(1,t)10τ(t)f3(x,s)1sτ(t)eζ(s,t)ds,ifτ(t)0.
    \right. $

    It is easy to verify that $ \Upsilon $ is continuous and coercive, and $ L $ is continuous. So applying the Lax-Milgram theorem, we deduce that for all $ \eta \in H_0^1(\Omega) $ the problem (36) admits a unique solution

    $ u \in H_0^1(\Omega). $

    Applying the classical elliptic regularity, it follows from (35) that

    $ u \in H^2(\Omega). $

    Therefore, the operator $ \lambda I - \mathcal{A}(t) $ is surjective for any $ \lambda > 0 $ and $ t>0 $. Again as $ \kappa(t)>0 $, this prove that

    $ λI˜A(t)=(λ+κ(t))IA(t) is surjective,
    $
    (37)

    for any $ \lambda >0 $ and $ t>0 $.

    Then, (24), (25) and (37) imply that the family $ \tilde{\mathcal{A}} = \left\{ \tilde{\mathcal{A}}(t)/ t \in [0,T] \right\} $ is a stable family of generators in $ \mathcal{H} $ with stability constants independent of $ t $, by Proposition $ 1.1 $ from [14]. Therefore, the assumptions $ (i)-(iv) $ of Theorem 3.1 are verified by (21), (24), (25), (26), (37) and (23), and thus, the problem

    $ {˜Ut=˜A(t)˜U,˜U(0)=U0=(u0,u1,f0(,,τ(0)))T
    $
    (38)

    has a unique solution $ \tilde{U} \in C\left( [0,+\infty[, D(\mathcal{A}(0)) \right) \cap C^1\left( [0,+\infty[, \mathcal{H} \right) $ for $ U_0 \in D(\mathcal{A}(0)) $. The requested solution of (18) is then given by

    $ U(t) = e^{\int_0^t \kappa(s)\,ds}\tilde{U}(t) $

    because

    $ Ut(t)=κ(t)et0κ(s)ds˜U(t)+et0κ(s)ds˜Ut(t)=et0κ(s)ds(κ(t)+˜A(t))˜U(t)=A(t)et0κ(s)ds˜U(t)=A(t)U(t),
    $

    which concludes the proof.

    The existence and uniqueness are obtained by the following result.

    Theorem 3.2 (Global solution). For any initial datum $ U_0 \in \mathcal{H} $ there exists a unique solution $ U $ satisfying

    $ U \in C([0,+\infty[, \mathcal{H}) $

    for problem (18).

    Moreover, if $ U_0 \in D(\mathcal{A}(0)) $, then

    $ U \in C([0,+\infty[, D(\mathcal{A}(0))) \cap C^1([0,+\infty[, \mathcal{H}). $

    Proof. A general theory for equations of type (18) has been developed using semigroup theory [14], [15] and [26]. The simplest way to prove existence and uniqueness results in to show that the triplet $ \left\{ (\mathcal{A}, \mathcal{H}, Y) \right\} $, with $ \mathcal{A} = \left\{ \mathcal{A}(t)/ t \in [0,T] \right\} $, for some fixed $ T>0 $ and $ Y = \mathcal{A}(0) $, forms a CD-systems (or constant domain system, see [14] and [15]). More precisely, the following theorem gives the existence and uniqueness results and is proved in Theorem $ 1.9 $ of [14] (see also Theorem $ 2.13 $ of [15] or [1]).

    In this section we shall investigate the asymptotic behavior of problem (1). The stability result will be obtained using Lemma 2.2.

    Theorem 4.1 (Stability Result). Let $ (u_0,u_1,f_0(\cdot,-,\tau(0))) \in H_0^1(\Omega) \times L^2(\Omega) \times L^2(\Omega \times ]0,1[) $. Assume that the hypotheses (H1), (H2) and (5)-(7) hold. Then problem (1) admits a unique solution

    $ uC([0,+[,H10(Ω))C1([0,+[,L2(Ω)),
    $
    $ zC([0,+[,L2(Ω)×]0,1[).
    $

    Proof. From now on, we denote by $ c $ various positive constants which may be different at different occurrences.

    Given $ 0 \leq S < T < \infty $ we start by multiplying the first equation of (9) by $ uE^q $ and then integrating over $ (S,T) \times \Omega $, we obtain

    $ TSEqΩu(uttuxx+μ1(t)ut+μ2(t)z(x,1,t))dxdt=0.
    $

    Notice that

    $ u_{tt}u = \left( u_t u \right)_t - u_t^2, $

    using integration by parts and the boundary conditions we know that

    $ 0=[Eq(t)Ωuutdx]TSTSqEq1(t)E(t)ΩuutdxdtTSEq(t)ut2L2(Ω)dt+TSEq(t)ux2L2(Ω)dt+TSEq(t)Ωμ1(t)uutdxdt+TSEq(t)Ωμ2(t)uz(x,1,t)dxdt.
    $
    (39)

    Similarly, we multiply the second equation of (9) by $ E^q \xi(t) e^{-2\rho\tau(t)}z(x,\rho,t) $ and then integrate over $ \Omega \times (0,1) \times (S,T) $ to see that

    $ 0=TSΩ10Eq(t)ξ(t)e2ρτ(t)z(τ(t)zt+(1ρτ(t))zρ)dρdxdt
    $
    $ =12Ω10TSEq(t)ξ(t)e2ρτ(t)tz2dtdρdx+12TSEq(t)ξ(t)Ω10e2ρτ(t)(1ρτ(t))ρz2dρdxdt.
    $

    Using integration by parts and the boundary conditions we know that

    $ 0=[ξ(t)τ(t)2Eq(t)Ω10e2ρτ(t)z2dρdx]TS12TSqEq1(t)E(t)ξ(t)τ(t)Ω10e2ρτ(t)z2dρdxdt12TSqEq(t)ξ(t)τ(t)Ω10e2ρτ(t)z2dρdxdt+12TSEq(t)ξ(t)Ω[e2ρτ(t)(1τ(t))z2(x,1,t)z2(x,0,t)]dxdt+TSEq(t)ξ(t)τ(t)Ω10e2ρτ(t)z2dρdxdt.
    $
    (40)

    Since $ \mu_1 $ is a non-increasing function of class $ C_1(\mathbb{R}) $, its derivatives is non-positive, which implies that $ \xi'(t) \leq 0 $. This result this

    $ TSqEq(t)ξ(t)τ(t)Ω10e2ρτ(t)z2dρdxdt0.
    $
    (41)

    Moreover, as

    $ 12TSEq(t)ξ(t)Ωe2ρτ(t)(1τ(t))z2(x,1,t)dxdt0,
    $
    (42)

    then, from (40), (41) and (42), we have that

    $ TSEq(t)ξ(t)τ(t)Ω10e2ρτ(t)z2dρdxdt[ξ(t)τ(t)2Eq(t)Ω10e2ρτ(t)z2dρdx]TS+12TSqEq1(t)E(t)ξ(t)τ(t)Ω10e2ρτ(t)z2dρdxdt12TSEq(t)ξ(t)Ωz2(x,0,t)dxdt.
    $
    (43)

    Using the definition of $ E $, (39) and (43), we get

    $ γ0TSEq+1dt[Eq(t)Ωuutdx]TS[ξ(t)τ(t)2Eq(t)Ω10e2ρτ(t)z2dρdx]TS+qTSEq1(t)E(t)Ωuutdxdt+qTSξ(t)τ(t)2Eq1(t)E(t)Ω10e2ρτ(t)z2dρdxdt
    $
    $ +2TSEq(t)ut2L2(Ω)dtTSEq(t)Ωμ1(t)uutdxdtTSEq(t)Ωμ2(t)uz(x,1,t)dxdt+12TSξ(t)Eq(t)e2ρτ(t)Ωz2(x,0,t)dxdt,
    $
    (44)

    where $ \gamma_0 = 2\min \{ 1, e^{-2\tau_1} \} $.

    Using the Young and Sobolev-Poincaré inequalities and Lemma 2.3, we find that

    $ [Eq(t)Ωuutdx]TSEq(S)Ωu(x,S)ut(x,S)dxEq(T)Ωu(x,T)ut(x,T)dxcEq+1(S).
    $

    Now, we known that

    $ [ξ(t)τ(t)2Eq(t)Ω10e2ρτ(t)z2dρdx]TSξ(S)τ(S)2Eq(S)Ω10e2ρτ(S)z2(x,ρ,S)dρdxcEq(S)ξ(S)τ(S)Ω10z2(x,ρ,S)dρdxcEq+1(S).
    $

    By (13), we have

    $ TSEq1(t)E(t)ΩuutdxdtcTS(E(t))Eq(t)dtcEq+1(S).
    $

    Similarly,

    $ TSEq1(t)E(t)ξ(t)τ(t)2Ω10e2ρτ(t)z2dρdxdtcEq+1(S).
    $

    From Lemma 2.4, we deduce that

    $ TSEq(t)ut2L2(Ω)dtcTSEq(t)E(t)dtcEq+1(S).
    $

    Now, we get that

    $ |TSEq(t)Ωμ1(t)uutdxdt|μ1(0)|TSEq(t)Ωuutdxdt|c(ε1)TSEq(t)Ωu2tdxdt+ε1TSEq(t)Ωu2xdxdtc(ε1)TSEq(t)(E(t))dt+ε1TSEq(t)E(t)dtc(ε1)Eq+1(S)+ε1TSEq+1(t)dt
    $
    (45)

    and from (H2) we obtain that

    $ |TSEq(t)Ωμ2(t)uz(x,1,t)dxdt|βμ1(0)|TSEq(t)Ωφz(x,1,t)dxdt|c(ε2)Eq+1(S)+ε2TSEq+1(t)dt.
    $
    (46)

    Finally,

    $ 12TSEq(t)ξ(t)Ωz2(x,0,t)dxdtˉξμ1(0)2TSEq(t)ut2L2(Ω)dtcTSEq(t)(E(t))dtcEq+1(S).
    $

    Choosing $ \varepsilon_1 $ and $ \varepsilon_2 $ small enough, we deduce from (45) and (46) that

    $ \int_S^T E^{q+1}\,dt \leq \frac{1}{\gamma}E^{q+1}(S). $

    Since $ E(S) \leq E(0) $ for $ S \geq 0 $, we have that

    $ \int_S^T E^{q+1}\,dt \leq \frac{1}{\gamma}E(0)E^{q}(S). $

    We choose $ q = 0 $, we conclude from Lemma 2.2 that

    $ E(t) \leq E(0)e^{1-\gamma t}. $

    This ends the proof of Theorem 4.1.


    Acknowledgments



    This research was conducted as student dissertation with the financial support from the vice chancellor of research in Guilan University of Medical Sciences. The author hearty appreciate the respectable authorities of Guilan University of Medical Sciences, the Deputy of Education and all the students who contributed to the study.

    Conflict of interest



    The authors declare no conflict of interest.

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