Research article

Decay rate of the solutions to the Lord Shulman thermoelastic Timoshenko model

  • Received: 19 March 2023 Revised: 16 April 2023 Accepted: 25 April 2023 Published: 18 May 2023
  • MSC : 35B37, 35L55, 74D05, 93D15, 93D20

  • In this work, we deal with a one-dimensional Cauchy problem in Timoshenko system with thermal effect and damping term. The heat conduction is given by the theory of Lord-Shulman. We prove that the dissipation induced by the coupling of the Timoshenko system with the heat conduction of Lord-Shulman's theory alone is strong enough to stabilize the system, but with slow decay rate. To show our result, we transform our system into a first order system and, applying the energy method in the Fourier space, we establish some pointwise estimates of the Fourier image of the solution. Using those pointwise estimates, we prove the decay estimates of the solution and show that those decay estimates are very slow.

    Citation: Abdelbaki Choucha, Sofian Abuelbacher Adam Saad, Rashid Jan, Salah Boulaaras. Decay rate of the solutions to the Lord Shulman thermoelastic Timoshenko model[J]. AIMS Mathematics, 2023, 8(7): 17246-17258. doi: 10.3934/math.2023881

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  • In this work, we deal with a one-dimensional Cauchy problem in Timoshenko system with thermal effect and damping term. The heat conduction is given by the theory of Lord-Shulman. We prove that the dissipation induced by the coupling of the Timoshenko system with the heat conduction of Lord-Shulman's theory alone is strong enough to stabilize the system, but with slow decay rate. To show our result, we transform our system into a first order system and, applying the energy method in the Fourier space, we establish some pointwise estimates of the Fourier image of the solution. Using those pointwise estimates, we prove the decay estimates of the solution and show that those decay estimates are very slow.



    Lord Shulman's thermoelasticity has also garnered a lot of interest among scientists in the past few years and there is a broad amount of contributions to explaining this theory. This theory is based on the study of a set of four hyperbolic equations with heat dissipation. In this instance, the heat conduction is also equivalent and hyperbolic in contrast to the one produced for Fourier's law. For more information and explanation about this theory and other theories, see [1,2]. Green & Naghdi [3,4] created a thermoelasticity model that incorporates thermal displacement gradient and temperature gradient among the constitutive variables and proposed a heat conduction law.

    There are numerous results from the coupling of the Fourier law of heat conduction and various systems, which has been discussed by researchers. For example, the Timoshenko system has been studied in [5,6], the Bresse system combined with the Cattaneo law of heat conduction (Bresse-Cattaneo) in [7], the Bresse system (Bresse Fourier) has been studied in [8,9,10] and MGT problem in [11]. The following papers are recommended to the reader for more information [12,13].

    Initially, the basic evolution equations for the one-dimensional Timoshenko thermo-elasticity theories with microtemperature and temperature [14,15,16,17,18] provided by

    ρtt=Tx,Iρtt=Hx+G,ρT0ζt=qx,ρEt=Px+qQ. (1.1)

    Here E,H,T,q,ζ,Q,P,T0,G denote the first moment of energy, the equilibrated stress, the stress, the heat flux vector, the entropy, the mean heat flux, the first heat flux moment, the reference temperature and the equilibrated body force, K,ρ,Iρ,I&E represent the shear modulus, the density, the polar moment of inertia of a cross section, the moment of inertia of a cross section and Young's modulus of elasticity respectively. In order to make calculations simple, let T0=ρ=Iρ=K=1 and EI=a2>0.

    In this work, the natural counterpart to the Lord-Shulman theory's microtemperatures is taken into consideration [1,2]. In this case, we must modify the constitutive equations to take the following form.

    T=T1+T2P=k2ϖx,H=H1+H2ρζ=γ0x+γ1+β1(κθt+θ),G=G1+G2Q=(k1k3)ϖ+(kk1)θx,q=kθx+k1ϖρE=β2(κϖt+ϖ)γ2x, (1.2)

    where

    T1=G1=K(x)T2=γ0(κθt+θ)H1=EIxG2=γ1(κθt+θ)μ1t,H2=γ2(κϖt+ϖ). (1.3)

    Here, the functions and denote the elastic material displacement, ϖ is the microtemperature vector and the function θ is the temperature difference, κ>0 is the relaxation parameter. β1,β2>0, the coefficients γ1,k,γ0 denote, the coupling between the volume fraction and the temperature, the thermal conductivity, the coupling between the displacement and the temperature respectively.

    As coupling is considered the coefficients k1,k2,k3,γ2,μ1>0 and satisfy the inequalities

    k21<kk3. (1.4)

    The main objective of this study is the thermal effects, let the heat capacity β1=β>0, and for more excitement in posing the problem, we do not assume the microtemperature effect, and β2=k1=k2=k3=γ2=0.

    Now, by substituting (1.2) and (1.3) into (1.1), we obtain that:

    {tt(x)x+γ0(κθt+θ)x=0,tta2xx(x)γ1(κθt+θ)+μ1t=0,β(κθt+θ)t+γ0tx+γ1tkθxx=0, (1.5)

    where

    (x,t)R×R+,

    with initial conditions

    (,t,,t,θ,θt)(x,0)=(0,1,0,1,θ0,θ1),xR. (1.6)

    We categorise this paper as follows: We utilise our preliminary findings in the remainder of this section to help us understand our major decay conclusion. The Lyapunov functional is built and the estimate for the Fourier image is discovered in the following section utilising the energy approach in Fourier space. The final portion is devoted to the conclusion.

    This, to our knowledge, is one of the first studies to look at this issue in the Fourier space. We require the following Lemma to support our main finding.

    Lemma 1.1. For any k,α0,c>0, there exist a constant C>0 in such a way that   t0 the below stated estimate hold

    ||1||kec||αtdC(1+t)(k+n)/α,Rn. (1.7)

    This section provide the decay estimate of the Fourier image of the solution for problems (1.5) and (1.6). This approach enables us to give the decay rate of the solution in the energy space by utilising Plancherel's theorem along with some integral estimates, such as Lemma 1.1. We create the appropriate Lyapunov functionals and apply the energy method in Fourier space to this problem. We conclude by demonstrating our main finding.

    Let us introduce the new variables in order to construct the Lyapunov functional in the Fourier space

    f=x,j=t,b=ax,m=t,ζ=κθt+θ,ϖ=θx. (2.1)

    Then, the system (1.5) can also be written as

    {ftjx+m=0jtfx+γ0ζx=0btayx=0mtazxfγ1ζ+μ1m=0βζt+γ0jx+γ1mkϖx=0κϖtζx+ϖ=0, (2.2)

    where initial condition are

    (f,j,b,m,ζ,ϖ)(x,0)=(f0,j0,b0,m0,ζ0,ϖ0),xR, (2.3)

    with

    f0=(0,x0),j0=1,b0=a0,x,m0=1,ζ0=κθ1θ0,ϖ0=θ0,x.

    Hence, the problems (2.2) and (2.3) can be written as

    {Ft+AFx+LF=0,F(x,0)=F0(x), (2.4)

    with F=(f,j,b,m,ζ,ϖ)T,F0=(f0,j0,b0,m0,ζ0,ϖ0) and

    AF=(jf+γ0ζayaz1β(γ0jkϖ)1κζ),LF=(m00fγ1ζ+μ1m1β(γ1m)1κϖ).

    Utilizing the Fourier transform on (2.4), we obtain that:

    {ˆFt+iAˆF+LˆF=0,ˆF(,0)=ˆF0(), (2.5)

    where ˆF(,t)=(ˆf,ˆj,ˆb,ˆm,ˆζ,ˆϖ)T(,t). The quation (2.5)1 can be written as

    {ˆftiˆj+ˆm=0ˆjtiˆf+iγ0ˆζ=0ˆbtaiˆm=0ˆmtaiˆbˆfγ1ˆζ+μ1ˆm=0βˆζt+iγ0ˆj+γ1ˆmikˆϖ=0κˆϖtiˆζ+ˆϖ=0. (2.6)

    Lemma 2.1. Let ˆF(,t) be a solution of (2.5). Then the energy functional ˆE(,t), given by

    ˆE(,t)=12{|ˆf|2+|ˆj|2+|ˆb|2+|ˆm|2+β|ˆζ|2+kκ|ˆϖ|2}, (2.7)

    satisfies

    dˆE(,t)dt=μ1|ˆm|2k|ˆϖ|20. (2.8)

    Proof. Firstly, multiplying (2.6)1,2,3,4 by ¯ˆf,¯ˆj,¯ˆb and ¯ˆm respectively, and multiplying (2.6)5,6 by ¯ˆζ,k¯ˆϖ, adding these equalities and taking the real part, we obtain that

    12ddt[|ˆf|2+|ˆj|2+|ˆb|2+|ˆm|2+β|ˆζ|2+kκ|ˆϖ|2]+μ1|ˆm|2+k|ˆϖ|2=0, (2.9)

    we find

    dˆE(,t)dt=μ1|ˆm|2k|ˆϖ|2.

    Hence, we obtain (2.7) and (2.8) (ˆE is non-increasing function).

    Now, to obtain the main result, we need the following lemmas.

    Lemma 2.2. The functional

    D1(,t):=e{i(ˆf¯ˆj+ˆm¯ˆb)}, (2.10)

    satisfies,

    dD1(,t)dt122|ˆj|2a22|ˆb|2+c(1+2)|ˆf|2+c(1+2)|ˆm|2+c2|ˆζ|2. (2.11)

    Proof. Differentiating D1 and using (2.6), we get

    dD1(,t)dt=e{iˆft¯ˆjiˆjt¯ˆf+iˆmt¯ˆbiˆbt¯ˆm}=2|ˆj|2a2|ˆb|2+2|ˆf|2+a2|ˆm|2e{iˆm¯ˆj}+e{iˆf¯ˆb}+e{iγ1ˆζ¯ˆb}e{iμ1ˆm¯ˆb}e{γ02ˆζ¯ˆf}. (2.12)

    We evaluate the terms in the RHS of (2.12), utilizing Young's inequality, for any δ1,δ2>0 we get

    e{iˆm¯ˆj}δ12|ˆj|2+c(δ1)|ˆm|2,+e{iˆf¯ˆb}δ22|ˆb|2+c(δ2)|ˆf|2,+e{iγ1ˆζ¯ˆb}δ22|ˆb|2+c(δ2)|ˆζ|2,e{iμ1ˆm¯ˆb}δ22|ˆb|2+c(δ2)|ˆm|2,e{γ02ˆζ¯ˆf}c2|ˆζ|2+c2|ˆf|2. (2.13)

    Inserting the above estimates (2.13) into (2.12) and by letting δ1=12,δ2=a6, we get (2.11).

    Lemma 2.3. The functional

    D2(,t):=e{i(κβˆϖ¯ˆζ)}, (2.14)

    holds, for any ε1>0

    dD2(,t)dtβ22|ˆζ|2+ε14(1+2)2|ˆj|2+c2|ˆm|2+c(ε1)(1+2)2|ˆϖ|2. (2.15)

    Proof. Differentiating D2 and using (2.6), we get

    D2(,t)dt=e{i(κβˆϖt¯ˆζ+κβˆϖ¯ˆζt)}=β2|ˆζ|2+kκ2|ˆϖ|2e{iβˆϖ¯ˆζ}+e{iγ1κˆm¯ˆϖ}e{γ0κ2ˆj¯ˆϖ}. (2.16)

    Similarly, we evaluate the terms in the RHS of (2.16), by applying Young's inequality, for any ε1,δ3>0 we find

    e{iβˆϖ¯ˆζ}δ32|ˆζ|2+c(δ3)|ˆϖ|2,+e{iγ1κˆm¯ˆϖ}c2|ˆm|2+c|ˆϖ|2,e{γ0κ2ˆj¯ˆϖ}ε14(1+2)2|ˆj|2+c(ε1)(1+2)2|ˆϖ|2. (2.17)

    By substituting (2.17) into (2.16) and letting δ3=β2, we obtain (2.15).

    Next, we will discuss the the below mentioned lemmas.

    Lemma 2.4. The functional

    D3(,t):=e{ˆf¯ˆm+aˆj¯ˆb}, (2.18)

    satisfies:

    (1) For a=1. Then, for any ε2>0

    dD3(,t)dt12|ˆf|2+ε221+2|ˆb|2+c|ˆm|2+c(ε2)(1+2)|ˆζ|2. (2.19)

    (2) For a1. Then, for any ε2,ε3>0

    dD3(,t)dt12|ˆf|2+ε221+2|ˆb|2+ε321+2|ˆj|2+c(ε3)(1+2)|ˆm|2+c(ε2)(1+2)|ˆζ|2. (2.20)

    Proof. Firstly, differentiating D3 and using (2.6), we get

    dD3(,t)dt=|ˆf|2+|ˆm|2e{γ1ˆζ¯ˆf}+e{μ1ˆm¯ˆf}+e{iaγ0ˆζ¯ˆb}e{i(1a2)ˆj¯ˆm}. (2.21)

    Now we will discuss two cases:

    Case1. (a=1).

    In this case, by applying the Young's inequality to the terms on the RHS of (2.21), for any ε2,δ4>0 we get

    e{γ1ˆζ¯ˆf}δ4|ˆf|2+c(δ4)|ˆζ|2,e{μ1ˆm¯ˆf}δ4|ˆf|2+c(δ4)|ˆm|2,e{iaγ0ˆζ¯ˆb}ε221+2|ˆb|2+c(ε2)(1+2)|ˆζ|2. (2.22)

    Substituting the estimates (2.22) in (2.21) and by letting δ4=14, we find (2.19).

    Case2. (a1).

    In this case, using the Young's inequality to the last terms on the RHS of (2.21), gives for any ε3>0

    (a21)e{iˆj¯ˆm}ε321+2|ˆj|2+c(ε3)(1+2)|ˆm|2. (2.23)

    Inserting (2.23) and (2.22) in (2.21), we get (2.20). Which completes the proof of Lemma 2.4.

    At this stage, we define the Lyapunov functional for the two cases by

    H(,t):=N(1+2)2ˆE(,t)+21+2{11+2N1D1(,t)+N3D3(,t)}+N2D2(,t), (2.24)

    where N,Ni,i=1,2,3 are positive constants which will be selected later.

    Lemma 2.5. There exist μ2,μ3,μ4>0 such that the functional H(,t) stated by (2.24) satisfies

    {μ2(1+2)2ˆE(,t)H(,t)μ3(1+2)2ˆE(,t),H(,t)μ4ρ()H(,t),t>0, (2.25)

    where

    ρ()=4(1+2)4. (2.26)

    Proof. Firstly, for the case a=1, by differentiating (2.24) and using (2.8), (2.11), (2.15) and (2.19), with the fact that 21+2min{1,2} and 11+21, we find

    H(,t)4(1+2)2{[12N1ε1N2]|ˆj|2+[a2N1ε2N3]|ˆb|2}(1+2)2[μ1NcN1cN2cN3]|ˆm|2(1+2)2[kNc(ε1)N2]|ˆϖ|221+2[12N3cN1]|ˆf|22[β2N2cN1c(ε2)N3]|ˆζ|2. (2.27)

    By setting

    ε1=N14N2,ε2=aN14N3,

    we obtain

    H(,t)4(1+2)2N1{14|ˆj|2+a4|ˆb|2}(1+2)2[μ1NcN1cN2cN3]|ˆm|2(1+2)2[kNc(N1,N2)N2]|ˆϖ|221+2[12N3cN1]|ˆf|22[β2N2cN1c(N1,N3)N3]|ˆζ|2. (2.28)

    Similarly, for the case a1, by differentiating (2.24) and using (2.8), (2.11), (2.15) and (2.20), with the fact that 21+2min{1,2} and 11+21, we find

    H(,t)4(1+2)2{[12N1ε1N2ε3N3]|ˆj|2+[a2N1ε2N3]|ˆb|2}(1+2)2[μ1NcN1cN2c(ε3)N3]|ˆm|2(1+2)2[kNc(ε1)N2]|ˆϖ|221+2[12N3cN1]|ˆf|22[β2N2cN1c(ε2)N3]|ˆζ|2. (2.29)

    By setting

    ε1=N18N2,ε2=aN14N3,ε3=N18N3,

    we obtain

    H(,t)4(1+2)2N1{14|ˆj|2+a4|ˆb|2}(1+2)2[μ1NcN1cN2c(N1,N3)N3]|ˆm|2(1+2)2[kNc(N1,N2)N2]|ˆϖ|221+2[12N3cN1]|ˆf|22[β2N2cN1c(N1,N3)N3]|ˆζ|2. (2.30)

    Next, for the two cases (2.28) and (2.30), we fixed N1 and choosing N3 large enough such that

    12N3cN1>0,

    then we select N2 large enough such as

    β2N2cN1c(N1,N3)N3>0.

    Hence, for the two cases we arrive at

    H(,t)4(1+2)2(α0|ˆj|2+α1|ˆb|2)(1+2)2[μ1Nc]|ˆm|221+2α2|ˆf|22α3|ˆζ|2(1+2)2[kNc]|ˆϖ|2. (2.31)

    Additionally, we have

    |H(,t)N(1+2)2ˆE(,t)|=N12(1+2)2|D1(,t)|+N321+2|D3(,t)|+N2|D2(,t)|.

    Utilizing Young's inequality and the fact that 21+2min{1,2} and 11+21, we find

    |H(,t)N(1+2)2ˆE(,t)|c(1+2)2ˆE(,t).

    Therefore, we get

    (Nc)(1+2)2ˆE(,t)H(,t)(N+c)(1+2)2ˆE(,t). (2.32)

    Now, we pick N large enough such as

    Nc>0,μ1Nc>0,kNc>0,

    and exploiting (2.7), estimates (2.31) and (2.32), respectively, there exists a positive constant α>0, t>0 and R, such that

    μ2(1+2)2ˆE(,t)H(,t)μ3(1+2)2ˆE(,t), (2.33)

    and

    H(,t)α4(1+2)2(|ˆj|2+|ˆb|2+|ˆϖ|2+|ˆf|2+|ˆm|2+|ˆζ|2), (2.34)

    then

    H(,t)λ1ρ()ˆE(,t),t0. (2.35)

    Consequently, for some positive constant μ4=λ1μ3>0, we get

    H(,t)μ4ρ()H(,t),t0, (2.36)

    where ρ()=4(1+2)4, for some λ1,μi>0,i=2,3,4. The proof of the first result (2.25) is completed.

    The pointwise estimates of the functional ˆE(,t) is stated by the following Proposition.

    Proposition 2.1. For any t0 and R, positive constants d1>0 exist in such a way that the energy functional given by (2.7) satisfies

    ˆE(,t)d1ˆE(,0)eμ4ρ()t, (2.37)

    where ρ()=4(1+2)4.

    Proof. From (2.25)2, we have by integration over (0,t)

    H(,t)H(,0)eμ4ρ()t,t0. (2.38)

    Hence, by according of (2.25) and (2.38), we establish (2.37).

    Theorem 2.1. Let s be a nonnegative integer, and F0Hs(R)L1(R). Then, the solution F of problems (2.2) and (2.3) holds, t0 the following decay estimates

    kxF(t)2CF01(1+t)18k4+C(1+t)4k+xF02, (2.39)

    where and k are nonnegative integers such that k+s and C>0 is a positive constant.

    Proof. From (2.7), we get |ˆF(,t)|2ˆE(,t), then by utilizing the Plancherel theorem and exploiting (2.37), we achieve that

    kxF(t)22=R||2k|ˆF(,t)|2dcR||2keμ4ρ()t|ˆF(,0)|2dc||1||2keμ4ρ()t|ˆF(,0)|2dR1+c||1||2keμ4ρ()t|ˆF(,0)|2dR2. (2.40)

    Now, we estimate R1,R2, the low-frequency part ||1 and the high-frequency part ||1 respectevly. Firstly, we have ρ()1164, for ||1. Then

    R1c||1||2keμ416||4t|ˆF(,0)|2dcsup||1{|ˆF(,0)|2}||1||2keμ416||4td, (2.41)

    by utilizing Lemma 1.1, we get that

    R1csup||1{|ˆF(,0)|2}(1+t)k214cF021(1+t)k214. (2.42)

    Secondly, we have ρ()1164, for ||1. Then

    R2c||1||2keμ416||4t|ˆF(,0)|2d,t0. (2.43)

    Expoiting the inequality

    sup||1{||2ec||2t}C(1+t), (2.44)

    we get that

    R2csup||1{||2eμ416||4t}||1||2(k+)|ˆF(,0)|2dc(1+t)2k+xF(x,0)22,t0. (2.45)

    Substituting (2.42) and (2.45) into (2.40), we find (2.39).

    The investigation of the general decay estimate of solutions to a one-dimensional Lord-Shulman Timoshenko system with thermal effect and damping term is the goal of this study, we prove some optimal decay results for the L2-norm of the solution. More precisely, we prove that the decay rate of the solution is of the form (1+t)1/8.

    To prove our results, we used the energy method in the Fourier space to build some very delicate Lyapunov functionals that give the desired results. In system (1.5) the presence of the mechanical damping μ1t seems to be necessary for our treatment. It is an interesting problem to prove the same result for μ1=0.

    In the upcoming study, we will attempt to utilize the same methodology in the same systems, but with the different types of the memory and the delay terms, we believe that we will obtain similar results.

    The authors declare they have not used Artificial Intelligence (AI) tools in the creation of this article.

    Researchers would like to thank the Deanship of Scientific Research, Qassim University for funding publication of this project.

    The authors declare no conflicts of interest.



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