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Research article Special Issues

A within-host model on the interaction dynamics between innate immune cells and Mycobacterium tuberculosis

  • Received: 19 November 2024 Revised: 10 February 2025 Accepted: 12 February 2025 Published: 17 February 2025
  • Tuberculosis is the leading cause of death worldwide from a single infectious agent; it has also been declared a threat to humanity by the World Health Organization. New insights indicate that the innate immune response plays a crucial role in determining the outcome of the infection. In this study, we assessed the role of macrophages in the innate immune response through a simple mathematical model. Our results confirm that macrophages provide the primary protective response against Mycobacterium tuberculosis. However, they also highlight the importance of other innate cells in the outcome of infection. Specifically, our findings suggest that, in addition to macrophage activity, the involvement of other innate immune cells is essential for eliminating or controlling bacterial progression, ultimately leading to an adaptive immune response.

    Citation: Eduardo Ibargüen-Mondragón, M. Victoria Otero-Espinar, Miller Cerón Gómez. A within-host model on the interaction dynamics between innate immune cells and Mycobacterium tuberculosis[J]. Mathematical Biosciences and Engineering, 2025, 22(3): 511-527. doi: 10.3934/mbe.2025019

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  • Tuberculosis is the leading cause of death worldwide from a single infectious agent; it has also been declared a threat to humanity by the World Health Organization. New insights indicate that the innate immune response plays a crucial role in determining the outcome of the infection. In this study, we assessed the role of macrophages in the innate immune response through a simple mathematical model. Our results confirm that macrophages provide the primary protective response against Mycobacterium tuberculosis. However, they also highlight the importance of other innate cells in the outcome of infection. Specifically, our findings suggest that, in addition to macrophage activity, the involvement of other innate immune cells is essential for eliminating or controlling bacterial progression, ultimately leading to an adaptive immune response.



    In this paper, we study the initial boundary value problem of the nonlinear viscoelastic hyperbolic problem with variable exponents:

    {utt+2u+2uttt0g(tτ)2u(τ)dτ+|ut|m(x)2ut=|u|p(x)2u,(x,t)Ω×(0,T),u(x,t)=uν(x,t)=0,(x,t)Ω×(0,T),u(x,0)=u0(x), ut(x,0)=u1(x),xΩ, (1.1)

    where ΩRn(n1) is a bounded domain in Rn with a smooth boundary Ω, ν is the unit outer normal to Ω, the exponents m(x) and p(x) are continuous functions on ¯Ω with the logarithmic module of continuity:

    x,yΩ,|xy|<1,|m(x)m(y)|+|p(x)p(y)|ω(|xy|), (1.2)

    where

    limτ0+supω(τ)ln1τ=C<. (1.3)

    In addition to this condition, the exponents satisfy the following:

    2m:=essinfxΩm(x)m(x)m+:=esssupxΩm(x)<2(n2)n4, (1.4)
    2p:=essinfxΩp(x)p(x)p+:=esssupxΩp(x)<2(n2)n4, (1.5)

    g:R+R+ is a C1 function satisfying

    g(0)>0, g(τ)0,10g(τ)dτ=l>0. (1.6)

    The equation of Problem (1.1) arises from the modeling of various physical phenomena such as the viscoelasticity and the system governing the longitudinal motion of a viscoelastic configuration obeying a nonlinear Boltzmann's model, or electro-rheological fluids, viscoelastic fluids, processes of filtration through a porous medium, and fluids with temperature-dependent viscosity and image processing which give rise to equations with nonstandard growth conditions, that is, equations with variable exponents of nonlinearities. More details on these problems can be found in previous studies [1,2,3,4,5,6].

    When m(x) and p(x) are constants, Messaoudi [7] discussed the nonlinear viscoelastic wave equation

    uttu+t0g(tτ)u(τ)dτ+|ut|m2ut=|u|p2u,

    he proved that any weak solution with negative initial energy blows up in finite time if p>m, and a global existence result for pm. The results were improved later by Messaoudi [8], where the blow-up result in finite time with positive initial energy was obtained. Moreover, Song [9] showed the finite-time blow-up of some solutions whose initial data had arbitrarily high initial energy. In the same year, Song [10] studied the initial-boundary value problem

    |ut|ρuttu+t0g(tτ)u(τ)dτ+|ut|m2ut=|u|p2u,

    and proved the nonexistence of global solutions with positive initial energy. Cavalcanti, Domingos, and Ferreira [11] were concerned with the non-linear viscoelastic equation

    |ut|ρuttuutt+t0g(tτ)u(τ)dτγut=0,

    and proved the global existence of weak solutions. Moreover, they obtained the uniform decay rates of the energy by assuming a strong damping ut acting in the domain and providing the relaxation function which decays exponentially.

    In 2017, Messaoudi [12] considered the following nonlinear wave equation with variable exponents:

    uttu+a|ut|m(x)2ut=b|u|p(x)2u,

    where a,b are positive constants. By using the FaedoGalerkin method, the existence of a unique weak solution is established under suitable assumptions on the variable exponents m(x) and p(x). Then this paper also proved the finite-time blow-up of solutions and gave a two-dimensional numerical example to illustrate the blow up result. Park [13] showed the blow up of solutions for a viscoelastic wave equation with variable exponents

    uttu+t0g(ts)u(s)ds+a|ut|m(x)2ut=b|u|p(x)2u,

    where the exponents of nonlinearity p(x) and m(x) are given functions and a,b>0 are constants. For nonincreasing positive function g, they prove the blow-up result for the solutions with positive initial energy as well as nonpositive initial energy. Alahyane [14] discussed the nonlinear viscoelastic wave equation with variable exponents

    uttu+t0g(tτ)u(τ)dτ+μut=|u|p(x)2u,

    where μ is a nonnegative constant and the exponent of nonlinearity p(x) and g are given functions. Under arbitrary positive initial energy and specific conditions on the relaxation function g, they prove a finite-time blow-up result and give some numerical applications to illustrate their theoretical results. Ouaoua and Boughamsa [15] considered the following boundary value problem:

    utt+2uu+|ut|m(x)2ut=|u|p(x)2u,

    the authors established the local existence by using the FaedoGalerkin method with positive initial energy and suitable conditions on the variable exponents m(x) and r(x). In addition, they also proved that the local solution is global and obtained the stability estimate of the solution. Ding and Zhou [16] considered a Timoshenko-type equation

    utt+2uM(||u||22)u+|ut|p(x)2ut=|u|q(x)2u,

    they prove that the solutions blow up in finite time with positive initial energy. Therefore, the existence of finite-time blow-up solutions with arbitrarily high initial energy is established, and the upper and lower bounds of the blow-up time are derived. More related references can be found in [17,18,19,20,21,22].

    Motivated by [7,13,14], we considered the existence of the solutions and their blow-up for the nonlinear damping and viscoelastic hyperbolic problem with variable exponents. Our aim in this work is to prove the existence of the weak solutions and to find sufficient conditions on m(x) and p(x) for which the blow-up takes place.

    This article consists of three sections in addition to the introduction. In Section 2, we recall the definitions and properties of Lp(x)(Ω) and the Sobolev spaces W1,p(x)(Ω). In Section 3, we prove the existence of weak solutions for Problem (1.1). In Section 4, we state and prove the blow-up result for solutions with positive initial energy as well as nonpositive initial energy.

    In this section, we review some results regarding Lebesgue and Sobolev spaces with variable exponents first. All of these results and a comprehensive study of these spaces can be found in [23]. Here (,) and , denote the inner product in space L2(Ω) and the duality pairing between H2(Ω) and H20(Ω).

    The variable exponent Lebesgue space Lp(x)(Ω) is defined by

    Lp(x)(Ω)={u(x):uismeasurableinΩ, ρp(x)(u)=Ω|u|p(x)dx<},

    this space is endowed with the norm

    up(x)=inf {λ>0:Ω|u(x)λ|p(x)dx1}.

    The variable exponent Sobolev space W1,p(x)(Ω) is defined by

    W1,p(x)(Ω)={uLp(x)(Ω) suchthatuexistsand|u|Lp(x)(Ω)},

    the corresponding norm for this space is

    u1,p(x)=up(x)+up(x),

    define W1,p(x)0(Ω) as the closure of  C0(Ω) with respect to the W1,p(x)(Ω) norm. The spaces  Lp(x)(Ω),W1,p(x)(Ω) and W1,p(x)0(Ω) are separable and reflexive Banach spaces when 1<pp+<, where p:=essinfΩp(x) and p+:=esssupΩp(x). As usual, we denote the conjugate exponent of p(x) by p(x)=p(x)/(p(x)1) and the Sobolev exponent by

    p(x)={np(x)nkp(x),if p(x)<n,,if p(x)n.

    Lemma 2.1. If p1(x), p2(x)C+(¯Ω)={hC(¯Ω):minx¯Ωh(x)>1}, p1(x)p2(x) for any xΩ, then there exists the continuous embedding Lp2(x)(Ω)Lp1(x)(Ω), whose norm does not exceed |Ω|+1.

    Lemma 2.2. Let p(x), q(x)C+(¯Ω). Assuming that q(x)<p(x), there is a compact and continuous embedding Wk,p(x)(Ω)Lq(x)(Ω).

    Lemma 2.3. (Hölder's inequality) [24] For any uLp(x)(Ω) and vLq(x)(Ω), then the following inequality holds:

    |Ωuvdx|(1p+1q)||u||p(x)||v||q(x)2||u||p(x)||v||q(x).

    Lemma 2.4. For uLp(x)(Ω), the following relations hold:

    u0(up(x)=λρp(x)(uλ)=1),
    up(x)<1(=1;>1)ρp(x)(u)<1(=1;>1),
    up(x)>1upp(x)ρp(x)(u)up+p(x),
    up(x)<1up+p(x)ρp(x)(u)upp(x).

    Next, we give the definition of the weak solution to Problem (1.1).

    Definition 2.1. A function u(x, t) is called a weak solution for Problem (1.1), if uC(0,T;H20(Ω)) C1(0,T;H20(Ω))C2(0,T;H2(Ω)) with uttL2(0,T;H20(Ω)) and u satisfies the following conditions:

    (1) For every ωH20(Ω) and for a.e.t(0,T)

    utt,ω+(u,ω)+(utt,ω)t0g(tτ)(u(τ),ω)dτ+(|ut|m(x)2ut,ω)=(|u|p(x)2u,ω),

    (2) u(x,0)=u0(x)H20(Ω),ut(x,0)=u1(x)H20(Ω).

    In this section, we prove the existence of a weak solution for Problem (1.1) by making use of the Faedo–Galerkin method and the contraction mapping principle. For a fixed T>0, we consider the space H=C(0,T;H20(Ω))C1(0,T;H20(Ω)) with the norm ||v||2H=max0tT(||vt||22+l||v||22).

    Lemma 3.1. Assume that (1.4), (1.5), and (1.6) hold, let (u0,u1)H20(Ω)×H20(Ω), for any T>0, vH, then there exists uC(0,T;H20(Ω))C1(0,T;H20(Ω))C2(0,T;H2(Ω)) with  uttL2(0,T;H20(Ω)) satisfying

    {utt+2u+2uttt0g(tτ)2u(τ)dτ+|ut|m(x)2ut=|v|p(x)2v,(x,t)Ω×(0,T),u(x,t)=uν(x,t)=0,(x,t)Ω×(0,T),u(x,0)=u0(x), ut(x,0)=u1(x),xΩ. (3.1)

    Proof. Let {ωj}j=1 be the orthogonal basis of H20(Ω), which is the standard orthogonal basis in L2(Ω) such that

    ωj=λjωj  in  Ω,ωj=0  on  Ω,

    we denote by Vk=span{ω1,ω2,,ωk} the subspace generated by the first k vectors of the basis {ωj}j=1. By normalization, we have ||ωj||2=1. For all k1, we seek k functions ck1(t),ck2(t),,ckk(t)C2[0,T] such that

    uk(x,t)=kj=1ckj(t)ωj(x),

    satisfying the following approximate problem

    {(uktt,ωi)+(uk,ωi)+(uktt,ωi)t0g(tτ)(uk,ωi)dτ+(|ukt|m(x)2ukt,ωi)=Ω|v|p(x)2vωidx,uk(0)=uk0,   ukt(0)=uk1,     i=1,2,k, (3.2)

    where

    uk0=ki=1(u0,ωi)ωiu0   in  H20(Ω),
    uk1=ki=1(u1,ωi)ωiu1   in  H20(Ω),

    thus, (3.2) generates the initial value problem for the system of second-order differential equations with respect to cki(t):

    {(1+λ2i)ckitt(t)+λ2icki(t)=Gi(ck1t(t),,ckkt(t))+gi(cki(t)),   i=1,2,,k,cki(0)=Ωu0ωidx,      ckit(0)=Ωu1ωidx,              i=1,2,,k. (3.3)

    where

    Gi(ck1t(t),,ckkt(t))=Ω|kj=1ckjt(t)ωj(x)|m(x)2kj=1ckjt(t)ωj(x)ωi(x)dx,

    and

    gi(cki(t))=λ2it0g(tτ)cki(τ)dτ+Ω|v|p(x)2vωidx,

    by Peano's Theorem, we infer that the Problem (3.3) admits a local solution cki(t)C2[0,T].

    The first estimate. Multiplying (3.2) by ckit(t) and summing with respect to i, we arrive at the relation

    ddt(12||ukt||22+12||uk||22+12||ukt||22)+Ω|ukt|m(x)dxt0g(tτ)Ωuk(τ)uktdxdτ=Ω|v|p(x)2vuktdx. (3.4)

    By simple calculation, we have

    t0g(tτ)ΩΔuk(τ)Δuktdxdτ=12ddt(guk)12(guk)12ddtt0g(τ)dτ||Δuk||22+12g(t)||uk||22, (3.5)

    where

    (φψ)=t0φ(tτ)||ψ(t)ψ(τ)||22dτ,

    inserting (3.5) into (3.4), using Hölder's inequality and Young's inequality, we obtain

    ddt[12||ukt||22+12||ukt||22+12(guk)+12(1t0g(τ)dτ)||Δuk||22]=12(guk)12g(t)||uk||22+Ω|v|p(x)2vuktdxΩ|ukt|m(x)dxΩ|v|p(x)2vuktdx|v|p(x)2v2||ukt||2η2Ω|v|2(p(x)1)dx+12η||ukt||22, (3.6)

    using the embedding H20(Ω)L2(p(x)1)(Ω) and Lemma 2.4, we easily obtain

    Ω|v|2(p(x)1)dxmax{||v||2(p1)2(p(x)1),||v||2(p+1)2(p(x)1)}Cmax{||v||2(p1)2,||v||2(p+1)2}C, (3.7)

    where C is a positive constant. We denote by C various positive constants that may be different at different occurrences.

    Combining (3.6) and (3.7), we obtain

    ddt[12||ukt||22+12||ukt||22+12(guk)+12(1t0g(τ)dτ)||Δuk||22]η2C+12η||ukt||22,

    by Gronwall's inequality, there exists a positive constant CT such that

    ||ukt||22+||ukt||22+(guk)+l||Δuk||22CT, (3.8)

    therefore, there exists a subsequence of {uk}k=1, which we still denote by {uk}k=1, such that

    uku weakly star in L(0,T;H20(Ω)),uktut weakly star in L(0,T;H20(Ω)),uku weakly in L2(0,T;H20(Ω)),uktut weakly in L2(0,T;H20(Ω)). (3.9)

    The second estimate. Multiplying (3.2) by ckitt(t) and summing with respect to i, we obtain

    ||uktt||22+||Δuktt||22+ddt(Ω1m(x)|ukt|m(x)dx)=Ωukukttdx+t0g(tτ)ΩΔuk(τ)Δukttdxdτ+Ω|v|p(x)2vukttdx. (3.10)

    Note that we have the estimates for ε>0

    |Ωukukttdx|ε||uktt||22+14ε||uk||22, (3.11)
    Ω|v|p(x)2vukttdx|v|p(x)2v2uktt2ε||uktt||22+14εΩ|v|2(p(x)1)dx, (3.12)

    and

    |t0g(tτ)Ωuk(τ)ukttdxdτ|14εΩ(t0g(tτ)uk(τ)dτ)2dx+ε||uktt||22ε||uktt||22+14εt0g(s)dst0g(tτ)Ω|uk(τ)|2dxdτε||uktt||22+(1l)g(0)4εt0||uk(τ)||22dτ, (3.13)

    similar to (3.6) and (3.7), from H20(Ω)L2(Ω), we have

    Ω|v|p(x)2vukttdxεC||uktt||22+C4ε. (3.14)

    Taking into account (3.10)(3.14), we obtain

    ||uktt||22+(12εCε)||uktt||22+ddt(Ω1m(x)|ukt|m(x)dx)14ε||uk||22+(1l)g(0)4εt0||uk(τ)||22dτ+C4ε, (3.15)

    integrating (3.15) over (0,t), we obtain

    t0||uktt||22dτ+(12εCε)t0||uktt||22dτ+Ω1m(x)|ukt|m(x)dxC4εt0(||uk||22+τ0||uk(s)||22ds)dτ+CT, (3.16)

    taking ε small enough in (3.16), for some positive constant CT, we obtain

    t0||uktt||22dτ+t0||uktt||22dτCT, (3.17)

    we observe that estimate (3.17) implies that there exists a subsequence of {uk}k=1, which we still denote by {uk}k=1, such that

    ukttutt weakly in L2(0,T;H20(Ω)). (3.18)

    In addition, from (3.9), we have

    (uktt,ωi)=ddt(ukt,ωi)ddt(ut,ωi)=(utt,ωi)  weakly  star  in  L(0,T;H2(Ω)). (3.19)

    Next, we will deal with the nonlinear term. Combining (3.9), (3.18), and Aubin–Lions theorem [25], we deduce that there exists a subsequence of {uk}k=1 such that

    uktut strongly in C(0,T;L2(Ω)), (3.20)

    then

    |ukt|m(x)2ukt|ut|m(x)2ut  a.e. (x,t)Ω×(0,T), (3.21)

    using the embedding H20(Ω)L2(m(x)1)(Ω) and Lemma 2.4, we have

    |ukt|m(x)2ukt22=Ω|ukt|2(m(x)1)dxmax{||ukt||2(m1)2,||ukt||2(m+1)2}C, (3.22)

    hence, using (3.21) and (3.22), we obtain

    |ukt|m(x)2ukt|ut|m(x)2ut  weakly  star in L(0,T;L2(Ω)). (3.23)

    Setting up k in (3.2), combining with (3.9), (3.18), (3.19), and (3.23), we obtain

    utt,ω+(u,ω)+(utt,ω)t0g(tτ)(u(τ),ω)dτ+(|ut|m(x)2ut,ω)=(|v|p(x)2v,ω).

    To handle the initial conditions. From (3.9) and Aubin–Lions theorem, we can easily get uku in C(0,T;L2(Ω)), thus uk(0)u(0) in L2(Ω), and we also have that uk(0)=uk0u0 in H20(Ω), hence u(0)=u0 in H20(Ω). Similarly, we get that ut(0)=u1.

    Uniqueness. Suppose that (3.1) has solutions u and z, then ω=uz satisfies

    {ωtt+2ω+2ωttt0g(tτ)2ω(τ)dτ+|ut|m(x)2ut|zt|m(x)2zt=0,(x,t)Ω×(0,T),ω(x,t)=ων(x,t)=0,(x,t)Ω×(0,T),ω(x,0)=0, ωt(x,0)=0,xΩ.

    Multiplying the first equation of Problem (3.1) by ωt and integrating over Ω, we have

    12ddt[||ωt||22+(1t0g(τ)dτ)||ω||22+||ωt||22+(gω)]+12g(t)||ω||22=Ω(|ut|m(x)2ut|zt|m(x)2zt)(utzt)dx+12(gω),

    from the inequality

    (|a|m(x)2a|b|m(x)2b)(ab)0, (3.24)

    for all a,bRn and a.e. xΩ, we obtain

    ||ωt||22+l||ω||22+||ωt||22=0,

    which implies that ω=0. This completes the proof.

    Theorem 3.1. Assume that (1.4) and (1.6) hold, let the initial date (u0,u1)H20(Ω)×H20(Ω), and

    2pp(x)p+2(n3)n4,

    then there exists a unique local solution of Problem (1.1).

    Proof. For any T>0, consider MT={uH:u(0)=u0,ut(0)=u1,||u||HM}. Lemma 3.1 implies that for vMT, there exists u=S(v) such that u is the unique solution to Problem 3.1. Next, we prove that for a suitable T>0, S is a contractive map satisfying S(MT)MT.

    Multiplying the first equation of the Problem (3.1) by ut and integrating it over (0,t), we obtain

    ||ut||22+||ut||22+(gu)+l||Δu||22||u1||22+||u1||22+||Δu0||22+2t0Ω|v|p(x)2vutdxdτ, (3.25)

    using Hölder's inequality and Young's inequality, we have

    |Ω|v|p(x)2vutdx|γ||ut||22+14γΩ|v|2p(x)2dxγ||ut||22+14γ[Ω|v|2p2dx+Ω|v|2p+2dx]γ||ut||22+C4γ[||v||2p22+||v||2p+22],

    thus, (3.25) becomes

    ||ut||22+||ut||22+l||Δu||22λ0+2t0Ω|v|p(x)2vutdxdτλ0+2γTsup(0,T)||ut||22+TC2γsup(0,T)[||v||2p22+||v||2p+22],

    hence, we have

    sup(0,T)||ut||22+sup(0,T)||ut||22+lsup(0,T)||Δu||22λ0+2γTsup(0,T)||ut||22+TC2γsup(0,T)[||v||2p2H+||v||2p+2H],

    where λ0=||u1||22+||u1||22+||Δu0||22, choosing γ=12T such that

    ||u||2Hλ0+T2Csup(0,T)[||v||2p2H+||v||2p+2H].

    For any vMT, by choosing M large enough so that

    ||u||2Hλ0+2T2CM2(p+1)M2,

    and T>0, sufficiently small so that

    TM2λ02CM2(p+1),

    we obtain ||u||HM, which shows that S(MT)MT.

    Let v1,v2MT,u1=S(v1),u2=S(v2),u=u1u2, then u satisfies

    {utt+2u+2uttt0g(tτ)2u(τ)dτ+|u1t|m(x)2u1t|u2t|m(x)2u2t=|v1|p(x)2v1|v2|p(x)2v2,(x,t)Ω×(0,T),u(x,t)=uν(x,t)=0,(x,t)Ω×(0,T),u(x,0)=0, ut(x,0)=0,xΩ.

    Multiplying by ut and integrating over Ω×(0,t), we obtain

    12||ut||22+12(1t0g(τ)dτ)||u||22+12||ut||22+12(gu)+t0Ω[|u1t|m(x)2u1t|u2t|m(x)2u2t](u1tu2t)dxdτt0Ω(f(v1)f(v2))utdxdτ, (3.26)

    where f(v)=|v|p(x)2v. From (1.6) and (3.24), we obtain

    12||ut||22+l2||u||22+12||ut||22+12(gu)t0Ω(f(v1)f(v2))utdxdτ. (3.27)

    Now, we evaluate

    I=Ω|(f(v1)f(v2))||ut|dx=Ω|f(ξ)||v||ut|dx,

    where v=v1v2 and ξ=αv1+(1α)v2, 0α1. Thanks to Young's inequality and Hölder's inequality, we have

    Iδ2||ut||22+12δΩ|f(ξ)|2|v|2dxδ2||ut||22+(p+1)22δΩ|ξ|2(p(x)2)|v|2dxδ2||ut||22+(p+1)22δ(Ω|v|2nn2dx)n2n[Ω|ξ|n(p(x)2)dx]2nδ2||ut||22+(p+1)22δ(Ω|v|2nn2dx)n2n[(Ω|ξ|n(p+2)dx)2n+(Ω|ξ|n(p2)dx)2n]δ2||ut||22+(p+1)2C2δ||Δv||22[||ξ||2(p+2)2+||ξ||2(p2)2]δ2||ut||22+(p+1)2C2δ||Δv||22(M2(p+2)+M2(p2)). (3.28)

    Inserting (3.28) into (3.27), choosing δ small enough, we obtain

    ||u||2H(p+1)2CTδ(M2(p2)+M2(p+2))||v||2H,

    taking T small enough so that (p+1)2CTδ(M2(p2)+M2(p+2))<1, we conclude

    ||u||2H=||S(v1)S(v2)||2H||v1v2||2H,

    thus, the contraction mapping principle ensures the existence of a weak solution to Problem (1.1). This completes the proof.

    In this section, we show that the solution to Problem (1.1) blows up in finite time when the initial energy lies in positive as well as nonpositive. For this task, we define

    E(t)=12||ut||22+12(1t0g(τ)dτ)||u||22+12||ut||22+12(gu)Ω1p(x)|u|p(x)dx, (4.1)

    by the definition of E(t), we also have

    E(t)=Ω|ut|m(x)dx+12(gu)12g(t)||u||220. (4.2)

    Now, we set

    B1=max{1,Bl12},  λ1=(B21)2p2,  E1=(121p)(B21)pp2,

    and

    H(t)=E2E(t), (4.3)

    where the constant E2(E(0),E1) will be discussed later, and B is the best constant of the Sobolev embedding H20(Ω)Lp(x)(Ω). It follows from (4.2) that

    H(t)=E(t)0, (4.4)

    and H(t) is a nondecreasing function.

    To prove Theorem 4.1, we need the following two lemmas:

    Lemma 4.1. Suppose that (1.6) holds and the exponents m(x) and p(x) satisfy condition (1.4) and (1.5). Assume further that

    E(0)<E1  and  λ1<λ(0)=B21l||u0||22,

    then there exists a constant λ2>λ1 such that

    B21l||u||22λ2,  t0. (4.5)

    Proof. Using (1.6), (4.1), Lemma 2.4, and the embedding H20(Ω)Lp(x)(Ω), we find that

    E(t)12(1t0g(τ)dτ)||u||22Ω1p(x)|u|p(x)dxl2||u||221pΩ|u|p(x)dxl2||u||221pmax{||u||pp(x),||u||p+p(x)}l2||u||221pmax{Bp||u||p2,Bp+||u||p+2}l2||u||221pmax{Bp1lp2||u||p2,Bp+1lp+2||u||p+2}12B21λ1pmax{λp2,λp+2}:=G(λ), (4.6)

    where λ:=λ(t)=B21l||u||22. Analyzing directly the properties of G(λ), we deduce that G(λ) satisfies the following properties:

    G(λ)={12B21p+2pλp+22<0,  λ>1,12B2112λp22, 0<λ<1, 
    G+(1)=12B21p+2p<0,  G(1)=12B2112<0,
    G(λ1)=0,    0<λ1<1.

    It is easily verified that G(λ) is strictly increasing for 0<λ<λ1, strictly decreasing for λ1<λ, G(λ) as λ+, and G(λ1)=E1. Since E(0)<E1, there exists a λ2>λ1 such that G(λ2)=E(0). By (4.6), we see that G(λ(0))E(0)=G(λ2), which implies λ(0)λ2 since the condition λ(0)>λ1. To prove (4.5), we suppose by contradiction that for some t0>0, λt0=B21l||u(t0)||22<λ2. The continuity of B21l||u||22 illustrates that we could choose t0 such that λ1<λt0<λ2, then we have E(0)=G(λ2)<G(λt0)E(t0). This is a contradiction. The proof is completed.

    Lemma 4.2. Let the assumption in Lemma 4.1 be satisfied. For t[0,T), we have

    0<H(0)H(t)1pρp(x)(u).

    Proof. (4.4) indicates that H(t) is nondecreasing with respect to t, thus

    H(t)H(0)=E2E(0)>0,  t[0,T).

    It follows from (1.6), (4.1), and Lemma 4.1 that

    H(t)=E2E(t)=E212||ut||2212(1t0g(τ)dτ)||u||2212(gu)12||ut||22+Ω1p(x)|u|p(x)dxE1l2||u||22+Ω1p(x)|u|p(x)dxE112B21λ2+Ω1p(x)|u|p(x)dxE112B21λ1+Ω1p(x)|u|p(x)dxΩ1p(x)|u|p(x)dx1pρp(x)(u).

    The proof is completed.

    Our blow-up result reads as follows:

    Theorem 4.3. Suppose that

    2mm(x)m+<pp(x)p+2(n3)n4,

    and

    1l=0g(τ)dτ<p21p21+12p, (4.7)

    hold, if the following conditions

    E(0)<12(121p)(11p(p2)1ll)(B21)pp2  and  λ1<λ(0)=B21l||u0||22,

    are satisfied, then there exists T<+ such that

    limtT(||ut||22+||ut||22+||u||22+||u||p+p+)=+. (4.8)

    Proof. Assume by contradiction that (4.8) does not hold true, then for T<+ and all t[0,T], we get

    ||ut||22+||ut||22+||u||22+||u||p+p+C, (4.9)

    where C is a positive constant.

    Now, we define L(t) as follows:

    L(t)=H1α(t)+ϵΩutudx+ϵΩutudx, (4.10)

    where ε>0, small enough to be chosen later, and

    0αmin{pm+p(m+1),p22p}.

    The remaining proof will be divided into two steps.

    Step 1: Estimate for L'(t). By taking the derivative of (4.10) and using (1.1), we obtain

    L(t)=(1α)Hα(t)[Ω|ut|m(x)dx12(gu)+12g(t)||u||22]+ϵ||ut||22+ϵΩuttudx+ϵ||ut||22ϵ||u||22+ϵΩt0g(tτ)u(τ)dτudxϵΩ|ut|m(x)2utudx+ϵΩ|u|p(x)dxϵΩuttudx(1α)Hα(t)Ω|ut|m(x)dx+ϵ||ut||22ϵ||u||22+ϵΩt0g(tτ)u(τ)dτudxϵΩ|ut|m(x)2utudx+ϵΩ|u|p(x)dx+ϵ||Δut||22,

    applying Hölder's inequality and Young's inequality, we have

    ϵΩt0g(tτ)Δu(τ)Δu(t)dτdx=ϵΩt0g(tτ)Δu(t)(Δu(τ)Δu(t))dτdx+ϵt0g(tτ)dτ||Δu||22ϵt0g(tτ)||Δu(τ)Δu(t)||2||Δu(t)||2dτ+ϵt0g(tτ)dτ||Δu||22ϵp(1ε1)2(gu)+ϵ(112p(1ε1))t0g(τ)dτ||Δu||22,

    where 0<ε1<p2p, then

    L(t)(1α)Hα(t)Ω|ut|m(x)dx+ϵ||ut||22ϵ||Δu||22+ϵ||Δut||22ϵΩ|ut|m(x)2utudx+ϵΩ|u|p(x)dxϵp(1ε1)2(gu)+ϵ(112p(1ε1))t0g(τ)dτ||Δu||22,

    rewriting (4.7) to (p21)l12p(1l)>0, using (4.1) and (4.3) to substitute for (gu), choosing ε1>0 sufficiently small, we obtain

    L(t)(1α)Hα(t)Ω|ut|m(x)dx+ϵp(1ε1)H(t)+(ϵ+ϵp(1ε1)2)(||ut||22+||ut||22)+ϵ{(p(1ε1)21)(1t0g(τ)dτ)12p(1ε1)t0g(τ)dτ}||u||22ϵp(1ε1)E2ϵΩ|ut|m(x)2utudx+ϵε1Ω|u|p(x)dx(1α)Hα(t)Ω|ut|m(x)dx+ϵp(1ε1)H(t)+(ϵ+ϵp(1ε1)2)(||ut||22+||ut||22)+ϵ{(p(1ε1)21)l212p(1ε1)1l2}lλ2B21ϵp(1ε1)E2ϵΩ|ut|m(x)2utudx+ϵ{(p(1ε1)21)l212p(1ε1)1l2}||u||22+ϵε1Ω|u|p(x)dx.(1α)Hα(t)Ω|ut|m(x)dx+ϵp(1ε1)H(t)+(ϵ+ϵp(1ε1)2)(||ut||22+||ut||22)+ϵ{(p(1ε1)21)l212p(1ε1)1l2}l(B21)pp2ϵp(1ε1)E2ϵΩ|ut|m(x)2utudx+ϵ{(p(1ε1)21)l212p(1ε1)1l2}||u||22+ϵε1Ω|u|p(x)dx. (4.11)

    Step 1.1: Estimate for ϵ{(p(1ε1)21)l212p(1ε1)1l2}l(B21)pp2ϵp(1ε1)E2. It follows from the condition in Theorem 3.1 that

    E(0)<12(121p)(11lp(p2)l)(B21)pp2=(p21)l212p(1l)2lp(B21)pp2<E1,

    here, we can take ε1>0 sufficiently small and choose E2(E(0),E1) sufficiently close to E(0) such that

    ϵ(p(1ε1)21)l212p(1ε1)(1l)2l(B21)pp2ϵ(1ε1)pE2ϵ(p(1ε1)21)l212p(1ε1)(1l)2l(B21)pp2ϵ(1ε1)p(p21)l212p(1l)2lp(B21)pp20. (4.12)

    Therefore, we obtain by combining (4.11) and (4.12),

    L(t)(1α)Hα(t)Ω|ut|m(x)dx+ϵp(1ε1)H(t)+(ϵ+ϵp(1ε1)2)(||ut||22+||ut||22)+ϵε1Ω|u|p(x)dx+ϵ{(p(1ε1)21)l212p(1ε1)1l2}||u||22ϵΩ|ut|m(x)2utudx. (4.13)

    Step 1.2: Estimate for ϵΩ|ut|m(x)2utudx. Applying Young's inequality with ε2>1, the embedding Lp(x)(Ω)Lm(x)(Ω), Lemma 2.4 and Lemma 4.2, we easily have

    |Ω|ut|m(x)2utudx|Ω|ut|m(x)1Hαm(x)1m(x)(t)Hαm(x)1m(x)(t)|u|dxε2Hα(t)Ω|ut|m(x)dx+1εm12Ω|u|m(x)Hα(m(x)1)(t)dxε2Hα(t)Ω|ut|m(x)dx+2Cα(mm+)1εm12Hα(m+1)(t)Ω|u|m(x)dxε2Hα(t)Ω|ut|m(x)dx+C2εm12Hα(m+1)(t)max{||u||m+p(x),||u||mp(x)}, (4.14)

    where C1=min{H(0),1}, C2=2(1+|Ω|)m+Cα(mm+)1. Next, we have

    ||u||m+p(x)max{(Ω|u|p(x)dx)m+p+,(Ω|u|p(x)dx)m+p}max{[pH(t)]m+p+m+p,1}(Ω|u|p(x)dx)m+p,

    and

    ||u||mp(x)max{[pH(t)]mp+m+p,[pH(t)]mm+p}(Ω|u|p(x)dx)m+p,

    which illustrate

    max{||u||m+p(x),||u||mp(x)}C3(Ω|u|p(x)dx)m+p,

    where C3=2min{pH(0),1}mp+m+p. Recalling 0<αpm+p(m+1) and Lemma 4.2, apparently,

    Hα(m+1)(t)max{||u||m+p(x),||u||mp(x)}C3Hα(m+1)(t)(Ω|u|p(x)dx)m+pC3Hα(m+1)+m+p1(t)Hα(m+1)+m+p1(0)H1m+p(t)Hα(m+1)+m+p1(0)(Ω|u|p(x)dx)m+pC3(1p)1m+p(Ω|u|p(x)dx)1m+pHα(m+1)+m+p1(0)(Ω|u|p(x)dx)m+pC3(1p)1m+pCα(m+1)+m+p11Ω|u|p(x)dx, (4.15)

    it follows from (4.13), (4.14), and (4.15) that

    L(t)(1αϵε2)Hα(t)Ω|ut|m(x)dx+(ϵ+ϵp(1ε1)2)(||ut||22+||ut||22)+ϵ(1ε1)pH(t)+ϵ(ε1Cα(m+1)+m+p11C2C3(1p)1m+pεm12)Ω|u|p(x)dx+ϵ{(p(1ε1)21)l212p(1ε1)1l2}||u||22,

    let us fix the constant ε2 so that

    ε1>Cα(m+1)+m+p11C2C3(1p)1m+pεm12,

    and then choose ϵ so small that 1α>ϵε1. Therefore, we obtain

    L(t)M1(H(t)+||u||22+||ut||22+||ut||22+Ω|u|p(x)dx), (4.16)

    where

    M1=ϵmin{(1+p(1ε1)2),(1ε1)p,ε1Cα(m+1)+m+p11C2C3(1p)1m+pεm12,,(p(1ε1)21)l212p(1ε1)1l2}.

    Inequalities (4.16) and Lemma 4.2 imply L(t)L(0). Therefore, for a sufficiently small ϵ, we have

    L(0)=H1α(0)+ϵΩu1u0dx+ϵΩu1u0dx>0.

    Step 2: A differential inequality for L(t). Applying Hölder's inequality, Young's inequality and the embedding Lp(x)(Ω)L2(Ω), we easily obtain

    |Ωutudx|11α(ut2u2)11α(1+|Ω|)11α||ut||11α2||u||11αp(x)(1+|Ω|)11αμ||ut||11αμ2+(1+|Ω|)11αν||u||11ανp(x), (4.17)

    where 1μ+1ν=1. Choosing μ=2(1α)>1, then ν=2(1α)2(1α)1, further, (4.17) can be rewritten as

    |Ωutudx|11α(1+|Ω|)11αμ||ut||22+(1+|Ω|)11αν||u||22(1α)1p(x), (4.18)

    recalling 0<α<p22p, we obtain

    ||u||22(1α)1p(x)max{(Ω|u|p(x)dx)2p[2(1α)1],(Ω|u|p(x)dx)2p+[2(1α)1]}{[pH(t)]2p[2(1α)1]p[2(1α)1],[pH(t)]2p+[2(1α)1]p+[2(1α)1]}Ω|u|p(x)dxC4Ω|u|p(x)dx, (4.19)

    with C4=min{pH(0),1}2p+[2(1α)1]p+[2(1α)1]. Inserting (4.19) into (4.18), we obtain

    |Ωutudx|11α(1+|Ω|)11αμ||ut||22+(1+|Ω|)11ανC4Ω|u|p(x)dx. (4.20)

    We now estimate

    |Ωutudx|11α||ut||11α2||Δu||11α2C11αC11αH(0)H(t), (4.21)

    therefore, combining (4.20) and (4.21), we obtain

    L11α(t)=(H1α(t)+ϵΩutudx+ϵΩutudx)11αM2(H(t)+||ut||22+||ut||22+||u||22+Ω|u|p(x)dx), (4.22)

    where

    M2=max{211α(211α+ϵ11αC11αH(0)), 221αϵ11α(1+|Ω|)11αμ, 221αϵ11α(1+|Ω|)11ανC4}.

    Combining (4.16) and (4.22), we arrive at

    L(t)M1M2L11α(t),t0. (4.23)

    A simple integration of (4.23) over (0,t) yields

    Lα1α(t)1Lαα1(0)M1M2α1αt,

    this shows that L(t) blows up in finite time

    TM2M11ααLαα1(0),

    furthermore, one gets from (4.22) that

    limtT(H(t)+||ut||22+||ut||22+||u||22+Ω|u|p(x)dx)=+,

    it easily follows that

    Ω|u|p(x)dx{|u|1}|u|p+dx+{|u|<1}|u|pdx||u||p+p++|Ω|,

    and using Lemma 4.2, we obtain

    limtT(||ut||22+||ut||22+||u||22+||u||p+p+)=+,

    this leads to a contradiction with (4.9). Thus, the solution to Problem (1.1) blows up in finite time.

    Ying Chu: Methodology, Wring-original draft, Writing-review editing; Bo Wen and Libo Cheng: Methodology, Writing-original draft.

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

    The authors express their heartfelt thanks to the editors and referees who have provided some important suggestions. This work was supported by Science and Technology Development Plan Project of Jilin Province, China (20240101307JC).

    The authors declare there is no conflict of interest.



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