Review

Age-dependent trends in adverse reactions of SARS-CoV-2 vaccines: A narrative review

  • Received: 30 April 2024 Revised: 31 July 2024 Accepted: 09 August 2024 Published: 22 August 2024
  • Since the initiation of immunization against SARS-CoV-2, the adverse reactions of vaccines have remained a major contributor to vaccine hesitancy. Their occurrence has been attributed to the immune response against the administered antigen, which is affected by different factors including the age of the vaccine recipient. The trends of the adverse reactions of COVID-19 vaccines have shown a higher frequency in younger adults (18–64 years old) than in the older population (≥65 years). One of the potential reasons is the healthy and efficient immune systems of younger individuals, which provide a robust pro-inflammatory response upon antigen encounter. Their adaptive immune system has a diverse array of naïve T and B lymphocytes, which aids in antigen elimination with the development of a long-lasting immunological memory. Age-related changes in the immune system of the elderly are referred to as immunosenescence, which reduces its efficiency. The phenomenon of cellular senescence and inflammageing can lower the vaccine responses of older recipients. Thymic involution and ageing of the bone marrow can lead to decreased levels of naïve T-cells and memory B-cells. This could compromise the ability to either fight new infections or effectively respond to vaccines and develop a lasting immune memory. Along with clinical trials of senolytic drugs, lifestyle interventions are being studied to mitigate such age-related changes. Further insights into vaccine responses in the elderly population and the means to alleviate the hallmarks of immunosenescence are required.

    Citation: Laiba Noor, Syeda Momna Ishtiaq, Farhat Batool, Muhammad Imran Arshad. Age-dependent trends in adverse reactions of SARS-CoV-2 vaccines: A narrative review[J]. AIMS Allergy and Immunology, 2024, 8(3): 146-166. doi: 10.3934/Allergy.2024008

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  • Since the initiation of immunization against SARS-CoV-2, the adverse reactions of vaccines have remained a major contributor to vaccine hesitancy. Their occurrence has been attributed to the immune response against the administered antigen, which is affected by different factors including the age of the vaccine recipient. The trends of the adverse reactions of COVID-19 vaccines have shown a higher frequency in younger adults (18–64 years old) than in the older population (≥65 years). One of the potential reasons is the healthy and efficient immune systems of younger individuals, which provide a robust pro-inflammatory response upon antigen encounter. Their adaptive immune system has a diverse array of naïve T and B lymphocytes, which aids in antigen elimination with the development of a long-lasting immunological memory. Age-related changes in the immune system of the elderly are referred to as immunosenescence, which reduces its efficiency. The phenomenon of cellular senescence and inflammageing can lower the vaccine responses of older recipients. Thymic involution and ageing of the bone marrow can lead to decreased levels of naïve T-cells and memory B-cells. This could compromise the ability to either fight new infections or effectively respond to vaccines and develop a lasting immune memory. Along with clinical trials of senolytic drugs, lifestyle interventions are being studied to mitigate such age-related changes. Further insights into vaccine responses in the elderly population and the means to alleviate the hallmarks of immunosenescence are required.



    We consider the following nonlinear parabolic system with damping and diffusion:

    {ψt=(σα)ψσθx+αψxx,θt=(1β)θ+νψx+2ψθx+βθxx,(x,t)QT, (1.1)

    with the following initial and boundary conditions:

    {(ψ,θ)(x,0)=(ψ0,θ0)(x),0x1,ψ(1,t)=ψ(0,t)=ξ(t),(θ,θx)(1,t)=(θ,θx)(0,t),0tT. (1.2)

    Here σ,α,β and ν are constants with α>0, β>0, T>0, and QT=(0,1)×(0,T). The function ξ(t) is measurable in (0,T). System (1.1) was originally proposed by Hsieh in [1] as a substitution for the Rayleigh–Benard equation for the purpose of studying chaos, and we refer the reader to [1,2,3] for the physical background. It is worthy to point out that with a truncation similar to the mode truncation from the Rayleigh–Benard equations used by Lorenz in [4], system (1.1) also leads to the Lorenz system. In spite of it being a much simpler system, it is still as rich as the Lorenz system, and some different routes to chaos, including the break of the time-periodic solution and the break of the switching solution, have been discovered via numerical simulations [5].

    Neglecting the damping and diffusion terms, system (1.1) is simplified as

    (ψθ)t=(0σν2ψ)(ψθ)x.

    It has two characteristic values: λ1=ψ+ψ2σν and λ2=ψψ2σν. Obviously, the system is elliptic if ψ2σν<0, and it is hyperbolic if ψ2σν>0. In particular, it is always strictly hyperbolic if σν<0.

    The mathematical theory of system (1.1) has been extensively investigated in a great number of papers; see [6,7,8,9,10,11,12,13,14,15,16] and the references therein. However, there is currently no global result with large initial data because it is very difficult to treat the nonlinear term of system (1.1) in analysis. Motivated by this fact, we will first study the well-posedness of global large solutions of the problem given by (1.1) and (1.2) for the case in which σν<0, and then we will discuss the limit problem as α0+, as well as the problem on the estimation of the boundary layer thickness. For the case in which σν>0, we leave the investigation in the future.

    The problem of a vanishing viscosity limit is an interesting and challenging problem in many settings, such as in the boundary layer theory (cf.[17]). Indeed, the presence of a boundary layer that accompanies the vanishing viscosity has been fundamental in fluid dynamics since the seminal work by Prandtl in 1904. In this direction, there have been extensive studies with a large number of references that we will not mention here. As important work on the mathematical basis for the laminar boundary layer theory, which is related to the problem considered in this paper, Frid and Shelukhin [18] studied the boundary layer effect of the compressible or incompressible Navier–Stokes equations with cylindrical symmetry and constant coefficients as the shear viscosity μ goes to zero; they proved the existence of a boundary layer of thickness O(μq) with any q(0,1/2). It should be pointed out that, for the incompressible case, the equations are reduced to

    vt=μ(vx+vx)x,wt=μ(wxx+wxx),0<a<x<b,t>0, (1.3)

    where μ is the shear viscosity coefficient and v and w represent the angular velocity and axial velocity, respectively. Recently, this result was investigated in more general settings; see for instance, [19,20,21] and the references therein. In the present paper, we will prove a similar result to that obtained in [18]. Note that every equation in (1.3) is linear. However, equation (1.1)2, with a nonconservative term 2ψθx, is nonlinear. It is the term that leads to new difficulties in analysis, causing all previous results on system (1.1) to be limited to small-sized initial data.

    The boundary layer problem also arises in the theory of hyperbolic systems when parabolic equations with low viscosity are applied as perturbations (see [22,23,24,25,26,27,28]).

    Formally, setting α=0, we obtain the following system:

    {¯ψt=σ¯ψσ¯θx,¯θt=(1β)¯θ+ν¯ψx+2¯ψ¯θx+β¯θxx,(x,t)QT, (1.4)

    with the following initial and boundary conditions:

    {(¯ψ,¯θ)(x,0)=(ψ0,θ0)(x),0x1,(¯θ,¯θx)(1,t)=(¯θ,¯θx)(0,t),0tT. (1.5)

    Before stating our main result, we first list some notations.

    Notations: For 1p,s, kN, and Ω=(0,1), we denote by Lp=Lp(Ω) the usual Lebesgue space on Ω with the norm Lp, and Hk=Wk,2(Ω) and H10=W1,20(Ω) as the usual Sobolev spaces on Ω with the norms Hk and H1, respectively. Ck(Ω) is the space consisting of all continuous derivatives up to order k on Ω, C(¯QT) is the set of all continuous functions on ¯QT, where QT=(0,1)×(0,T) with T>0, and Lp(0,T;B) is the space of all measurable functions from (0,T) to B with the norm Lp(0,T;B), where B=Ls or Hk. We also use the notations (f1,f2,)2B=f2B+g2B+ for functions f1,f2, belonging to the function space B equipped with a norm B, and for L2(QT)=L2(0,T;L2).

    The first result of this paper can be stated as follows.

    Theorem 1.1. Let α,β,σ and ν be constants with α>0,β(0,1) and σν<0. Assume that (ψ0,θ0)H1([0,1]) and ξC1([0,T]), and that they are compatible with the boundary conditions. Then, the following holds:

    (i) For any given α>0, there exists a unique global solution (ψ,θ) for the problem given by (1.1) and (1.2) in the following sense:

    (ψ,θ)C(¯QT)L(0,T;H1),(ψt,θt,ψxx,θxx)L2(QT).

    Moreover, for some constant C>0 independent of α(0,α0] with α0>0,

    {(ψ,θ)L(QT)C,(|ψx|,α1/4ψx,ω1/2ψx,θx)L(0,T;L2)C,(ψt,θt,α3/4ψxx,α1/4θxx,ω1/2θxx)L2(QT)C, (1.6)

    where the function ω(x):[0,1][0,1] is defined by

    ω(x)={x,0x<1/2,1x,1/2x1. (1.7)

    (ii) There exists a unique global solution (¯ψ,¯θ) for the problem given by (1.4) and (1.5) in the following sense:

    {¯ψL(QT)BV(QT),(¯ψx,ω¯ψ2x)L(0,T;L1),¯ψtL2(QT),¯θC(¯QT)L(0,T;H1),¯θtL2(QT),¯θ(1,t)=¯θ(0,t),t[0,T], (1.8)

    and

    {Qt(¯ψζtσ¯ψζν¯θxζ)dxdτ=10ψ(x,t)ζ(x,t)dx10ψ0(x)ζ(x,0)dx,QtL[¯ψ,¯θ;φ]dxdτ=10¯θ(x,t)φ(x,t)dx10θ0(x)φ(x,0)dx,a.e.t(0,T),whereL[¯ψ,¯θ;φ]=¯θφt(1β)¯θφν¯ψφx+2¯ψ¯θxφβ¯θxφx (1.9)

    for any functions ζ,φC1(¯QT) with φ(1,t)=φ(0,t) for all t[0,T] such that, as α0+,

    {ψ¯ψstronglyinLp(QT)foranyp2,ψ¯ψweaklyinM(QT),ψt¯ψtweaklyinL2(QT),αψ2x0stronglyinL2(QT), (1.10)

    where M(QT) is the set of the Radon measures on QT, and

    {θ¯θstronglyinC(¯QT),θxθ0xweaklyinL(0,T;L2),θt¯θtweaklyinL2(QT). (1.11)

    (iii) For some constant C>0 independent of α(0,1),

    (ψ¯ψ,θ¯θ)L(0,T;L2)+θx¯θxL2(QT)Cα1/4. (1.12)

    Remark 1.1. The L2 convergence rate for ψ is optimal whenever a boundary layer occurs as α0+. The reason why this is optimal will be shown by an example in Section 3.3.

    We next study the boundary layer effect of the problem given by (1.1) and (1.2). Before stating the main result, we first recall the concept of BL–thickness, defined as in [18].

    Definition 1.2. A nonnegative function δ(α) is called the BL–thickness for the problem given by (1.1) and (1.2) with a vanishing α if δ(α)0 as α0, and if

    {limα0+(ψ¯ψ,θ¯θ)L(0,T;L(δ(α),1δ(α))=0,liminfα0+(ψ¯ψ,θ¯θ)L(0,T;L(0,1))>0,

    where (ψ,θ) and (¯ψ,¯θ) are the solutions for the problems given by (1.1), (1.2) and (1.4), (1.5), respectively.

    The second result of this paper is stated as follows.

    Theorem 1.3. Under the conditions of Theorem 1.1, any function δ(α) satisfying the condition that δ(α)0 and α/δ(α)0 as α0 is a BL–thickness whenever (¯ψ(1,t),¯ψ(0,t))(ξ(t),ξ(t)) in (0,T).

    Remark 1.2. Here, the function θ satisfies the spatially periodic boundary condition that has been considered in some papers, e.g., [3,29,30]. One can see from the analysis that Theorems 1.1 and 1.3 are still valid when the boundary condition of θ is the homogeneous Neumann boundary condition or homogeneous Dirichlet boundary condition.

    The proofs of the above theorems are based on the uniform estimates given by (1.6). First, based on a key observation of the structure of system (1.1), we find two identities (see Lemma 2.1). In this step, an idea is to transform (1.1)2 into an equation with a conservative term (see (2.4)). And then, from the two identities, we deduce some basic energy-type estimates (see Lemma 2.2). The condition σν<0 plays an important role here. With the uniform estimates in hand, we derive the required uniform bound of (α1/4ψxL(0,t;L2)+α3/4ψxxL2(QT)) for the study of the boundary layer effect via standard analysis (see [18,19]). See Lemma 2.3 with proof. The uniform bound of (ψ,θ)L(QT) is derived by a more delicate analysis (see Lemma 2.4). Finally, the boundary estimate ω1/2ψxL(0,T;L2)C is established (see Lemma 2.6), through which we complete the proof for the estimation of the boundary layer thickness.

    Before ending the section, let us introduce some of the previous works on system (1.1). It should be noted that most of those works focus on the case when σν>0. In this direction, Tang and Zhao [6] considered the Cauchy problem for the following system:

    {ψt=(1α)ψθx+αψxx,θt=(1α)θ+νψx+2ψθx+αθxx, (1.13)

    with the following initial condition: (ψ,θ)(x,0)=(ψ0,θ0)(x)(0,0)asx±, where 0<α<1 and 0<ν<4α(1α). They established the global existence, nonlinear stability and optimal decay rate of solutions with small-sized initial data. Their result was extended in [7,8] to the case of the initial data with different end states, i.e.,

    (ψ,θ)(x,0)=(ψ0,θ0)(x)(ψ±,θ±)asx±, (1.14)

    where ψ±,θ± are constant states with (ψ+ψ,θ+θ)(0,0). For the initial-boundary value problem on quadrants, Duan et al. [9] obtained the global existence and the Lp decay rates of solutions for the problem given by (1.13) and (1.14) with small-sized initial data. For the Dirichlet boundary value problem, Ruan and Zhu [10] proved the global existence of system (1.1) with small-sized initial data and justified the limit as β0+ under the following condition: ν=μβ for some constant μ>0. In addition, they established the existence of a boundary layer of thickness O(βδ) with any 0<δ<1/2. Following [10], some similar results on the Dirichlet–Neumann boundary value problem were obtained in [11]. For the case when σν<0, however, there are few results on system (1.1). Chen and Zhu [12] studied the problem given by (1.13) and (1.14) with 0<α<1; they proved global existence with small-sized initial data and justified the limit as α0+. In their argument, the condition σν<0 plays a key role. Ruan and Yin [13] discussed two cases of system (1.1): α=β and αβ, and they obtained some further results that include the C convergence rate as β0+.

    We also mention that one can obtain a slightly modified system by replacing the nonlinear term 2ψθx in (1.1) with (ψθ)x. Jian and Chen [31] first obtained the global existence results for the Cauchy problem. Hsiao and Jian [29] proved the unique solvability of global smooth solutions for the spatially periodic Cauchy problem by applying the Leary–Schauder fixed-point theorem. Wang [32] discussed long time asymptotic behavior of solutions for the Cauchy problem. Some other results on this system is available in [33,34,35] and the references therein.

    The rest of the paper is organized as follows. In Section 2, we will derive the uniform a priori estimates given by (1.6). The proofs of Theorem 1.1 and Theorem 1.3 will be given in Section 3 and Section 4, respectively.

    In the section, we will derive the uniform a priori estimates by (1.6), and we suppose that the solution (ψ,θ) is smooth enough on ¯QT. From now on, we denote by C a positive generic constant that is independent of α(0,α0] for α0>0.

    First of all, we observe that (˜ψ,˜θ):=(2ψ,2θ/ν) solves the following system:

    {˜ψt=(σα)˜ψσν˜θx+α˜ψxx,˜θt=(1β)˜θ+˜ψx+˜ψ˜θx+β˜θxx, (2.1)

    with the following initial and boundary conditions:

    {(˜ψ,˜θ)(x,0)=(2ψ0,2θ0/ν)(x),0x1,˜ψ(1,t)=˜ψ(0,t)=2ξ(t),(˜θ,˜θx)(1,t)=(˜θ,˜θx)(0,t),0tT.

    From the system, we obtain the following crucial identities for our analysis.

    Lemma 2.1. (˜ψ,˜θ) satisfies the following for all t[0,T]:

    ddt10(12˜ψ2+σν˜θ)dx+10[α˜ψ2x+σν(1β)˜θ]dx=(ασ)10˜ψ2dx+α[(˜ψ˜ψx)(1,t)(˜ψ˜ψx)(0,t)] (2.2)

    and

    ddt10e˜θdx+β10e˜θ˜θ2xdx+(1β)10˜θe˜θdx=0. (2.3)

    Proof. Multiplying (2.1)1 and (2.1)2 by ˜ψ and σν, respectively, adding the equations and then integrating by parts over (0,1) with respect to x, we obtain (2.2) immediately.

    We now multiply (2.1)2 by e˜θ to obtain

    (e˜θ)t=(1β)˜θe˜θ+(˜ψe˜θ)x+βe˜θ˜θxx. (2.4)

    Integrating it over (0,1) with respect to x, we get (2.3) and complete the proof.

    Lemma 2.2. Let the assumptions of Theorem 1.1 hold. Then, we have the following for any t[0,T]:

    10ψ2dx+αQtψ2xdxdτC (2.5)

    and

    |10θdx|C. (2.6)

    Proof. Integrating (2.2) and (2.3) over (0,t), respectively, we obtain

    1210˜ψ2dx+Qt[α˜ψ2x+(σα)˜ψ2]dxdτ=10(2ψ20+2σθ0)dx+αt0ξ(t)˜ψx|x=1x=0dτσν(10˜θdx+(1β)Qt˜θdxdτ), (2.7)

    and

    10e˜θdx+βQte˜θ˜θ2xdxdτ=10e2θ0νdx(1β)Qt˜θe˜θdxdτ. (2.8)

    From the inequalities ex1+x and xex1 for all xR, and given β(0,1), we derive from (2.8) that

    10˜θdxC,

    so that

    10˜θdx+(1β)Qt˜θdxdτC. (2.9)

    We now treat the second term on the right-hand side of (2.7). Integrating (2.1)1 over (0,1) with respect to x, multiplying it by ξ(t) and then integrating over (0,t), we obtain

    |αt0ξ(τ)˜ψx|x=1x=0dτ|=|ξ(t)10˜ψdx2ξ(0)10ψ0dxt0ξt(τ)10˜ψdxdτ+(σα)t0ξ(τ)10˜ψdxdτ|C+1410˜ψ2dx+CQt˜ψ2dxdτ. (2.10)

    Combining (2.9) and (2.10) with (2.7), and noticing that σν<0, we obtain (2.5) by using the Gronwall inequality.

    Combining (2.5) and (2.10) with (2.7) yields

    |10˜θdx+(1β)Qt˜θdxdτ|C. (2.11)

    Let φ=Qt˜θdxdτ. Then, (2.11) is equivalent to

    |(e(1β)tφ)t|Ce(1β)t,

    which implies that |φ|C. This, together with (2.11), gives (2.6) and completes the proof.

    Lemma 2.3. Let the assumptions of Theorem 1.1 hold. Then, for any t[0,T],

    10θ2dx+Qt(θ2x+ψ2t)dxdτC (2.12)

    and

    α10ψ2xdx+α2Qtψ2xxdxdτC. (2.13)

    Proof. Multiplying (1.1)2 by θ, integrating over (0,1), and using (2.5) and Young's inequality, we have

    12ddt10θ2dx+β10θ2xdx=(β1)10θ2dx+10(2ψθνψ)θxdxC+β410θ2xdx+Cθ2L. (2.14)

    From the embedding theorem, W1,1L, and Young's inequality, we have

    θ2LC10θ2dx+C10|θθx|dxCε10θ2dx+ε10θ2xdx,ε(0,1). (2.15)

    Plugging it into (2.14), taking ε sufficiently small and using the Gronwall inequality, we obtain

    10θ2dx+Qtθ2xdxdτC. (2.16)

    Rewrite (1.1)1 as

    ψt+αψxx=(σα)ψ+σθx.

    Taking the square on the two sides, integrating over (0,1) and using (2.5), we obtain

    αddt10ψ2xdx+10(ψ2t+α2ψ2xx)dx=10[(σα)ψ+σθx]2dx+αξtψx|x=1x=0C+C10θ2xdx+CαψxL. (2.17)

    From the embedding theorem, W1,1L, and the Hölder inequality, we have

    ψxLC10|ψx|dx+C10|ψxx|dx;

    hence, by Young's inequality,

    αψxLCε+Cα10ψ2xdx+εα210ψ2xxdx.

    Plugging it into (2.17), taking ε sufficiently small and using (2.16), we obtain the following by applying the Gronwall inequality:

    α10ψ2xdx+Qt(ψ2t+α2ψ2xx)dxdτC.

    This completes the proof of the lemma.

    For η>0, let

    Iη(s)=s0hη(τ)dτ,wherehη(s)={1,sη,sη,|s|<η,1,s<η.

    Obviously, hηC(R),IηC1(R) and

    hη(s)0a.e.sR;|hη(s)|1,limη0+Iη(s)=|s|,sR. (2.18)

    With the help of (2.18), we obtain the following estimates, which are crucial for the study of the boundary layer effect.

    Lemma 2.4. Let the assumptions of Theorem 1.1 hold. Then, for any t[0,T],

    10(|ψx|+θ2x)dx+Qtθ2tdxdτC. (2.19)

    In particular, (ψ,θ)L(QT)C.

    Proof. Let W=ψx. Differentiating (1.1)1 with respect to x, we obtain

    Wt=(σα)Wσθxx+αWxx.

    Multiplying it by hη(W) and integrating over Qt, we get

    10Iη(W)dx=10Iη(ψ0x)dxQt[(σα)W+σθxx]hη(W)dxdταQthη(W)W2xdxdτ+αt0(Wxhη(W))|x=1x=0dτ=:10Iη(ψ0x)dx+3i=1Ii. (2.20)

    By applying |hη(s)|1, we have

    I1C+CQt(|W|+|θxx|)dxdτ. (2.21)

    By using hη(s)0 for sR, we have

    I20. (2.22)

    We now estimate I3. Since θ(1,t)=θ(0,t), it follows from the mean value theorem that, for any t[0,T], there exists some xt(0,1) such that θx(xt,t)=0; hence,

    |θx(y,t)|=|yxtθxx(x,t)dx|10|θxx|dx,y[0,1].

    It follows from (1.1)1 that

    αt0|Wx(a,t)|dτC+Ct0|θx(a,τ)|dτC+CQt|θxx|dxdτ,wherea=0,1.

    Consequently, thanks to |hη(s)|1,

    I3Cαt0[|Wx(1,τ)|+|Wx(0,τ)|]dτC+CQt|θxx|dxdτ. (2.23)

    Plugging (2.21)–(2.23) into (2.20), and letting η0+, we deduce the following by noticing that limη0+Iη(s)=|s|:

    10|ψx|dxC+CQt|ψx|dxdτ+CQt|θxx|dxdτ. (2.24)

    Plugging (1.1)2 into the final term of the right-hand side on (2.24), and by using (2.5) and (2.16), we obtain the following by using the Gronwall inequality:

    10|ψx|dxC+CQt|θt|dxdτ. (2.25)

    By using the embedding theorem, (2.5) and (2.25), we have

    ψ2L(10|ψ|dx+10|ψx|dx)2C[1+P(t)], (2.26)

    where P(t)=Qtθ2tdxdτ.

    Multiplying (1.1)2 by θt and integrating over Qt, we have

    P(t)+10(β2θ2x+1β2θ2)dxC+Qt(νψx+2ψθx)θtdxdτ. (2.27)

    By applying (2.5), (2.12), (2.16), (2.26) and Young's inequality, we deduce the following:

    νQtψxθtdxdτ=νQtψ(θx)tdxdτ=ν10ψθxdx+ν10ψ0θ0xdx+νQtψtθxdxdτC+β410θ2xdx (2.28)

    and

    2Qtψθxθtdxdτ12P(t)+Ct0(10θ2x(x,τ)dx)ψ2LdτC+12P(t)+Ct0(10θ2x(x,τ)dx)P(τ)dτ. (2.29)

    Substituting (2.28) and (2.29) into (2.27), and noticing that QTθ2xdxdtC, we obtain the following by using the Gronwall inequality:

    10θ2xdx+Qtθ2tdxdτC.

    This, together with (2.25), gives

    10|ψx|dxC.

    And, the proof of the lemma is completed.

    Lemma 2.5. Let the assumptions of Theorem 1.1 hold. Then, for any t[0,T],

    α1/210ψ2xdx+α3/2Qtψ2xxdxdτC. (2.30)

    Proof. Multiplying (1.1)2 by θxx, integrating over (0,1) and using Lemmas 2.2–2.4, we have

    12ddt10θ2xdx+β10θ2xxdx=10[(1β)θνψx2ψθx]θxxdxC+C10ψ2xdx+β410θ2xxdx+Cθx2LC+C10ψ2xdx+β210θ2xxdx, (2.31)

    where we use the estimate with ε sufficiently small based on a proof similar to (2.15):

    θx2LCε10θ2xdx+ε10θ2xxdx,ε(0,1).

    It follows from (2.31) that

    10θ2xdx+Qtθ2xxdxdτC+CQtψ2xdxdτ. (2.32)

    Similarly, multiplying (1.1)1 by αψxx, and integrating over (0,1), we have

    α2ddt10ψ2xdx+α210ψ2xxdx+α(σα)10ψ2xdx=σα10ψxθxxdx+[(σα)ξ+ξt][α˜ψx]|x=1x=0+σα(θxψx)|x=1x=0Cα10ψ2xdx+Cα10θ2xxdx+Cα[1+|θx(1,t)|+|θx(0,t)|]ψxL. (2.33)

    Then, by integrating over (0,t) and using (2.32) and the Hölder inequality, we obtain

    α10ψ2xdx+α2Qtψ2xxdxdτCα+CαQtψ2xdxdτ+CαA(t)(t0ψx2Ldτ)1/2, (2.34)

    where

    A(t):=(t0[1+|θx(1,τ)|2+|θx(0,τ)|2]dτ)1/2.

    To estimate A(t), we first integrate (1.1)2 over (x,1) for x[0,1], and then we integrate the resulting equation over (0,1) to obtain

    θx(1,t)=1β101y[θt+(1β)θ2ψθx]dxdyνξ(t)+ν10ψ(x,t)dx.

    By applying Lemmas 2.3 and 2.4, we obtain

    T0θ2x(1,t)dtC. (2.35)

    Similarly, we have

    T0θ2x(0,t)dtC. (2.36)

    Therefore,

    A(t)C. (2.37)

    From the embedding theorem, W1,1L, and the Hölder inequality, we have

    ψx2LC10ψ2xdx+C10|ψxψxx|dxC10ψ2xdx+C(10ψ2xdx)1/2(10ψ2xxdx)1/2;

    therefore, by the Hölder inequality and Young's inequality, we obtain the following for any ε(0,1):

    α(t0ψx2Ldτ)1/2Cαε+CαQtψ2xdxdτ+εα2Qtψ2xxdxdτ. (2.38)

    Combining (2.37) and (2.38) with (2.34), and taking ε sufficiently small, we obtain (2.30) by using the Gronwall inequality. The proof of the lemma is completed.

    As a consequence of Lemma 2.5 and (2.32), we have that α1/2QTθ2xxdxdtC.

    Lemma 2.6. Let the assumptions of Theorem 1.1 hold. Then, for any t[0,T],

    10ψ2xω(x)dx+Qt(θ2xx+αψ2xx)ω(x)dxdτC,

    where ω is the same as that in (1.7).

    Proof. Multiplying (1.1)1 by ψxxω(x) and integrating over Qt, we have

    1210ψ2xω(x)dx+αQtψ2xxω(x)dxdτC+CQtψ2xω(x)dxdτ+3i=1Ii, (2.39)

    where

    {I1=(ασ)Qtψψxω(x)dxdτ,I2=σQtψxθxxω(x)dxdτ,I3=σQtψxθxω(x)dxdτ.

    By applying Lemmas 2.3 and 2.4, we have

    I1CQt|ψ||ψx|xdτC, (2.40)

    and, by substituting (1.1)2 into I2, we get

    I2=σβQtψxω(x)[θt+(1β)θνψx2ψθx]dxdτC+CQtψ2xω(x)dxdτ. (2.41)

    To estimate I3, we first multiply (1.1)2 by θxω(x), and then we integrate over Qt to obtain

    νQtψxθxω(x)dxdτ=Eβ2t0[2θ2x(1/2,τ)θ2x(1,τ)θ2x(0,τ)]dτ,

    where

    E=Qtθxω(x)[θt+(1β)θ2ψθx]dxdτ.

    Note that |E|C by Lemmas 2.3 and 2.4. Then, given σν<0, we have

    I3=σνE+σνβ2t0[2θ2x(1/2,t)θ2x(1,t)θ2x(0,t)]dtC+Ct0[θ2x(1,τ)+θ2x(0,τ)]dτ. (2.42)

    Combining (2.35) and (2.36) with (2.42), we obtain

    I3C. (2.43)

    Substituting (2.40), (2.41) and (2.43) into (2.39), we complete the proof of the lemma by applying the Gronwall inequality.

    In summary, all uniform a priori estimates given by (1.6) have been obtained.

    For any fixed α>0, the global existence of strong solutions for the problem given by (1.1) and (1.2) can be shown by a routine argument (see [29,36]). First, by a smooth approximation of the initial data satisfying the conditions of Theorem 1.1, we obtain a sequence of global smooth approximate solutions by combining the a priori estimates given by (1.6) with the Leray–Schauder fixed-point theorem (see [36, Theorem 3.1]). See [29] for details. And then, a global strong solution satisfying (1.6) is constructed by means of a standard compactness argument.

    The uniqueness of solutions can be proved as follows. Let (ψ2,θ2) and (ψ1,θ1) be two strong solutions, and denote (Ψ,Θ)=(ψ2ψ1,θ2θ1). Then, (Ψ,Θ) satisfies

    {Ψt=(σα)ΨσΘx+αΨxx,Θt=(1β)Θ+νΨx+2ψ2Θx+2θ1xΨ+βΘxx, (3.1)

    with the following initial and boundary conditions:

    {(Ψ,Θ)|t=0=(0,0),0x1,(Ψ,Θ,Θx)|x=0,1=(0,0,0),0tT.

    Multiplying (3.1)1 and (3.1)2 by Ψ and Θ, respectively, integrating them over [0,1] and using Young's inequality, we obtain

    12ddt10Ψ2dx+α10Ψ2xdxβ410Θ2xdx+C10Ψ2dx (3.2)

    and

    12ddt10Θ2dx+β10Θ2xdx=10[(β1)Θ2νΘxΨ+2ψ2ΘxΘ+2θ1xΨΘ]dxβ410Θ2xdx+C10(Ψ2+Θ2)dx, (3.3)

    where we use ψ2L(QT)+θ1xL(0,T;L2)C and

    Qtθ21xΘ2dxdτCt0Θ2LdτQtΘ2dxdτ+β16QtΘ2xdxdτ.

    First, by adding (3.2) and (3.3), and then using the Gronwall inequality, we obtain that 10(Ψ2+Θ2)dx=0on[0,T] so that (Ψ,Θ)=0 on ¯QT. This completes the proof of Theorem 1.1(ⅰ).

    The aim of this part is to prove (1.10) and (1.11). Given the uniform estimates given by (1.6) and Aubin–Lions lemma (see [37]), there exists a sequence {αn}n=1, tending to zero, and a pair of functions (¯ψ,¯θ) satisfying (1.8) such that, as αn0+, the unique global solution of the problem given by (1.1) and (1.2) with α=αn, still denoted by (ψ,θ), converges to (¯ψ,¯θ) in the following sense:

    {ψ¯ψstronglyinLp(QT)foranyp2,ψ¯ψweaklyinM(QT),θ¯θstronglyinC(¯QT),θxθ0xweaklyinL(0,T;L2),(ψt,θt)(¯ψt,¯θt)weaklyinL2(QT), (3.4)

    where M(QT) is the set of Radon measures on QT. From (3.4), one can directly check that (¯ψ,¯θ) is a global solution for the problem given by (1.4) and (1.5) in the sense of (1.8) and (1.9).

    We now return to the proof of the uniqueness. Let (¯ψ2,¯θ2) and (¯ψ1,¯θ1) be two solutions, and denote (M,N)=(¯ψ2¯ψ1,¯θ2¯θ1). It follows from (1.9) that

    1210M2dx+σQtM2dxdτ=νQtMNxdxdτ, (3.5)

    and

    1210N2dx+Qt[βN2x+(1β)N2]dxdτ=Qt(νMNx+2ψ2NxN+2MNθ1x)dxdτ. (3.6)

    By using Young's inequality, we immediately obtain

    1210M2dxCQtM2dxdτ+β4QtN2xdxdτ,1210N2dx+βQtN2xdxdτCQt(M2+N2)dxdτ+β4QtN2xdxdτ,

    where we use ψ2L(QT),θ1xL(0,T;L2), and the estimate with ε sufficiently small:

    QtN2θ21xdxdτT0N2L10θ21xdxdτCεQtN2dxdτ+εQtN2xdxdτ,ε(0,1).

    Then, the Gronwall inequality gives (M,N)=(0,0) a.e. in QT. Hence, the uniqueness follows.

    Thanks to the uniqueness, the convergence results of (3.4) hold for α0+.

    Finally, by using Lemma 2.3, we immediately obtain

    α2Qtψ4xdxdτCα.

    So, αψ2x0 strongly in L2(QT) as α0+. Thus, the proof of Theorem 1.1(ⅱ) is completed.

    In addition, the following local convergence results follows from (1.6) and (3.4):

    {ψ¯ψstronglyinC([ϵ,1ϵ]×[0,T]),ψxψ0xweaklyinL(0,T;L2(ϵ,1ϵ)),θxx¯θxxweaklyinL2((ϵ,1ϵ)×(0,T))

    for any ϵ(0,1/4). Consequently, the equations comprising (1.1) hold a.e in QT.

    This purpose of this part is to prove (1.12). Let (ψi,θi) be the solution of the problem given by (1.1) and (1.2) with α=αi(0,1) for i=1,2. Denote U=ψ2ψ1 and V=θ2θ1. Then it satisfies

    {Ut=σU+α2ψ2α1ψ1σVx+α2ψ2xxα1ψ1xx,Vt=(1β)V+νUx+2ψ2Vx+2Uθ1x+βVxx. (3.7)

    Multiplying (3.7)1 and (3.7)2 by U and V, respectively, integrating them over Qt and using Lemmas 2.3–2.5, we have

    1210U2dxC(α2+α1)+CQtU2dxdτ+β4QtV2xdxdτ

    and

    1210V2dx+βQtV2xdxdτCQt(U2+V2)dxdτ+β4QtV2xdxdτ,

    where we use ψ2L(QT)+θ1xL(0,T;L2)C and the estimate with ε sufficiently small:

    QtV2(θ1x)2dxdτCt0V2LdτCεQtV2dxdτ+εQtV2xdxdτ,ε(0,1).

    Then, the Gronwall inequality gives

    10(U2+V2)dx+QtV2xdxdτC(α2+α1).

    We now fix α=α2, and then we let α10+ to obtain the desired result by using (1.10) and (1.11).

    In summary, we have completed the proof of Theorem 1.1.

    We next show the optimality of the L2 convergence rate O(α1/4) for ψ, as stated in Remark 1.1. For this purpose, we consider the following example:

    (ψ0,θ0)(0,1)on[0,1]andξ(t)t3in[0,T].

    It is easy to check that (¯ψ,¯θ)=(0,e(β1)t) is the unique solution of the problem given by (1.4) and (1.5). According to Theorem 1.3 of Section 4, a boundary layer exists as α0+. To achieve our aim, it suffices to prove the following:

    liminfα0+(α1/4ψL(0,T;L2))>0. (3.8)

    Suppose, on the contrary, that there exists a subsequence {αn} satisfying αn0+ such that the solution of the problem given by (1.1) and (1.2) with α=αn, still denoted by (ψ,θ), satisfies

    sup0<t<T10ψ2dx=o(1)α1/2n, (3.9)

    where o(1) indicates a quantity that uniformly approaches to zero as αn0+. Then, by using the embedding theorem, we obtain

    ψ2LC10ψ2dx+C10|ψ||ψx|dxCα1/2n+C(10ψ2dx10ψ2xdx)1/2.

    Hence, we get that ψL(0,T;L)0asαn0+ by using (3.9) and α1/2nψx2L2C. On the other hand, it is obvious that θ¯θL(0,T;L)0asαn0+ by using (1.11). This shows that a boundary layer does not occur as αn0+, and this leads to a contradiction. Thus, (3.8) follows. This proof is complete.

    Thanks to sup0<t<T10(ψ2x+¯ψ2x)ω(x)dxC, we have

    sup0<t<T1δδψ2xdxCδ,δ(0,1/4).

    By the embedding theorem, and from Theorem 1.1(ⅲ), we obtain the following for any δ(0,1/4):

    ψ¯ψ2L(δ,1δ)C10(ψ¯ψ)2dx+C1δδ|(ψ¯ψ)(ψx¯ψx)|dxCα+C(1δδ(ψ¯ψ)2dx1δδ(ψx¯ψx)2dx)1/2Cα+C(αδ)1/2.

    Hence, for any function δ(α) satisfying that δ(α)0 and α/δ(α)0 as α0, it holds that

    ψ¯ψL(0,T;L(δ(α),1δ(α)))0asα0+.

    On the other hand, it follows from (1.11) that θ¯θL(0,T;L)0asα0+. Consequently, for any function δ(α) satisfying that δ(α)0 and α/δ(α)0 as α0, we have

    (ψ¯ψ,θ¯θ)L(0,T;L(δ(α),1δ(α)))0asα0+.

    Finally, we observe that

    liminfα0+ψ¯ψL(0,T;L)>0

    whenever (¯ψ(1,t),¯ψ(0,t))(ξ(t),ξ(t)) on [0,T]. This ends the proof of Theorem 1.3.

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

    We would like to express deep thanks to the referees for their important comments. The research was supported in part by the National Natural Science Foundation of China (grants 12071058, 11971496) and the research project of the Liaoning Education Department (grant 2020jy002).

    The authors declare that there is no conflict of interest.



    Conflict of interest



    The authors declare no competing interests that could directly or indirectly influence the submitted work.

    Author contributions



    All authors contributed to the presented study.

    Ethics approval of research



    The manuscript is a review and did not involve any human or animal participants; therefore, it did not require approval from the ethical board.
    The authors certify that they complied with the Principles of Ethical Publishing Rules.
    Informed Consent: The data reported here were derived from studies already published and quoted in the reference list. Those papers mentioned informed consent that, depending on the studies, was implied to participate, verbal or written, or a combination of these variants during the follow-up.
    Consent to publish: This work does not contain any individual data; therefore, it did not require consent for publication by any participant.

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