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+△2utt−∫t0g(t−τ)△2u(τ)dτ+|ut|m(x)−2ut=|u|p(x)−2u,(x,t)∈Ω×(0,T),u(x,t)=∂u∂ν(x,t)=0,(x,t)∈∂Ω×(0,T),u(x,0)=u0(x), ut(x,0)=u1(x),x∈Ω, | (1.1) |
where Ω⊂Rn(n≥1) 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,y∈Ω,|x−y|<1,|m(x)−m(y)|+|p(x)−p(y)|≤ω(|x−y|), | (1.2) |
where
limτ→0+supω(τ)ln1τ=C<∞. | (1.3) |
In addition to this condition, the exponents satisfy the following:
2≤m−:=essinfx∈Ωm(x)≤m(x)≤m+:=esssupx∈Ωm(x)<2(n−2)n−4, | (1.4) |
2≤p−:=essinfx∈Ωp(x)≤p(x)≤p+:=esssupx∈Ωp(x)<2(n−2)n−4, | (1.5) |
g:R+→R+ is a C1 function satisfying
g(0)>0, g′(τ)≤0,1−∫∞0g(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
utt−△u+∫t0g(t−τ)△u(τ)dτ+|ut|m−2ut=|u|p−2u, |
he proved that any weak solution with negative initial energy blows up in finite time if p>m, and a global existence result for p≤m. 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|ρutt−△u+∫t0g(t−τ)△u(τ)dτ+|ut|m−2ut=|u|p−2u, |
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|ρutt−△u−△utt+∫t0g(t−τ)△u(τ)dτ−γ△ut=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:
utt−△u+a|ut|m(x)−2ut=b|u|p(x)−2u, |
where a,b are positive constants. By using the Faedo−Galerkin 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
utt−△u+∫t0g(t−s)△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
utt−△u+∫t0g(t−τ)△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+△2u−△u+|ut|m(x)−2ut=|u|p(x)−2u, |
the authors established the local existence by using the Faedo−Galerkin 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+△2u−M(||∇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 H−2(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
‖u‖p(x)=inf {λ>0:∫Ω|u(x)λ|p(x)dx≤1}. |
The variable exponent Sobolev space W1,p(x)(Ω) is defined by
W1,p(x)(Ω)={u∈Lp(x)(Ω) suchthat∇uexistsand|∇u|∈Lp(x)(Ω)}, |
the corresponding norm for this space is
‖u‖1,p(x)=‖u‖p(x)+‖∇u‖p(x), |
define W1,p(x)0(Ω) as the closure of C∞0(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<p−≤p+<∞, 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)n−kp(x),if p(x)<n,∞,if p(x)≥n. |
Lemma 2.1. If p1(x), p2(x)∈C+(¯Ω)={h∈C(¯Ω):minx∈¯Ωh(x)>1}, p1(x)≤p2(x) for any x∈Ω, 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 u∈Lp(x)(Ω) and v∈Lq(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 u∈Lp(x)(Ω), the following relations hold:
u≠0⇒(‖u‖p(x)=λ⇔ρp(x)(uλ)=1), |
‖u‖p(x)<1(=1;>1)⇔ρp(x)(u)<1(=1;>1), |
‖u‖p(x)>1⇒‖u‖p−p(x)≤ρp(x)(u)≤‖u‖p+p(x), |
‖u‖p(x)<1⇒‖u‖p+p(x)≤ρp(x)(u)≤‖u‖p−p(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 u∈C(0,T;H20(Ω)) ∩C1(0,T;H20(Ω))∩C2(0,T;H−2(Ω)) with utt∈L2(0,T;H20(Ω)) and u satisfies the following conditions:
(1) For every ω∈H20(Ω) and for a.e.t∈(0,T)
⟨utt,ω⟩+(△u,△ω)+(△utt,△ω)−∫t0g(t−τ)(△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=max0≤t≤T(||△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, v∈H, then there exists u∈C(0,T;H20(Ω))∩C1(0,T;H20(Ω))∩C2(0,T;H−2(Ω)) with utt∈L2(0,T;H20(Ω)) satisfying
{utt+△2u+△2utt−∫t0g(t−τ)△2u(τ)dτ+|ut|m(x)−2ut=|v|p(x)−2v,(x,t)∈Ω×(0,T),u(x,t)=∂u∂ν(x,t)=0,(x,t)∈∂Ω×(0,T),u(x,0)=u0(x), ut(x,0)=u1(x),x∈Ω. | (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 k≥1, we seek k functions ck1(t),ck2(t),…,ckk(t)∈C2[0,T] such that
uk(x,t)=k∑j=1ckj(t)ωj(x), |
satisfying the following approximate problem
{(uktt,ωi)+(△uk,△ωi)+(△uktt,△ωi)−∫t0g(t−τ)(△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=k∑i=1(u0,ωi)ωi→u0 in H20(Ω), |
uk1=k∑i=1(u1,ωi)ωi→u1 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|k∑j=1ckjt(t)ωj(x)|m(x)−2k∑j=1ckjt(t)ωj(x)ωi(x)dx, |
and
gi(cki(t))=λ2i∫t0g(t−τ)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)dx−∫t0g(t−τ)∫Ω△uk(τ)△uktdxdτ=∫Ω|v|p(x)−2vuktdx. | (3.4) |
By simple calculation, we have
−∫t0g(t−τ)∫ΩΔuk(τ)Δuktdxdτ=12ddt(g⋄△uk)−12(g′⋄△uk)−12ddt∫t0g(τ)dτ||Δuk||22+12g(t)||△uk||22, | (3.5) |
where
(φ⋄△ψ)=∫t0φ(t−τ)||△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(g⋄△uk)+12(1−∫t0g(τ)dτ)||Δuk||22]=12(g′⋄△uk)−12g(t)||△uk||22+∫Ω|v|p(x)−2vuktdx−∫Ω|ukt|m(x)dx≤∫Ω|v|p(x)−2vuktdx≤‖|v|p(x)−2v‖2||ukt||2≤η2∫Ω|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)dx≤max{||v||2(p−−1)2(p(x)−1),||v||2(p+−1)2(p(x)−1)}≤Cmax{||△v||2(p−−1)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(g⋄△uk)+12(1−∫t0g(τ)dτ)||Δuk||22]≤η2C+12η||ukt||22, |
by Gronwall's inequality, there exists a positive constant CT such that
||ukt||22+||△ukt||22+(g⋄△uk)+l||Δuk||22≤CT, | (3.8) |
therefore, there exists a subsequence of {uk}∞k=1, which we still denote by {uk}∞k=1, such that
uk∗⇀u weakly star in L∞(0,T;H20(Ω)),ukt∗⇀ut weakly star in L∞(0,T;H20(Ω)),uk⇀u weakly in L2(0,T;H20(Ω)),ukt⇀ut 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)=−∫Ω△uk△ukttdx+∫t0g(t−τ)∫ΩΔuk(τ)Δukttdxdτ+∫Ω|v|p(x)−2vukttdx. | (3.10) |
Note that we have the estimates for ε>0
|∫Ω△uk△ukttdx|≤ε||△uktt||22+14ε||△uk||22, | (3.11) |
∫Ω|v|p(x)−2vukttdx≤‖|v|p(x)−2v‖2‖uktt‖2≤ε||uktt||22+14ε∫Ω|v|2(p(x)−1)dx, | (3.12) |
and
|∫t0g(t−τ)∫Ω△uk(τ)△ukttdxdτ|≤14ε∫Ω(∫t0g(t−τ)△uk(τ)dτ)2dx+ε||△uktt||22≤ε||△uktt||22+14ε∫t0g(s)ds∫t0g(t−τ)∫Ω|△uk(τ)|2dxdτ≤ε||△uktt||22+(1−l)g(0)4ε∫t0||△uk(τ)||22dτ, | (3.13) |
similar to (3.6) and (3.7), from H20(Ω)↪L2(Ω), we have
∫Ω|v|p(x)−2vukttdx≤εC||△uktt||22+C4ε. | (3.14) |
Taking into account (3.10)−(3.14), we obtain
||uktt||22+(1−2ε−Cε)||△uktt||22+ddt(∫Ω1m(x)|ukt|m(x)dx)≤14ε||△uk||22+(1−l)g(0)4ε∫t0||△uk(τ)||22dτ+C4ε, | (3.15) |
integrating (3.15) over (0,t), we obtain
∫t0||uktt||22dτ+(1−2ε−Cε)∫t0||△uktt||22dτ+∫Ω1m(x)|ukt|m(x)dx≤C4ε∫t0(||△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||22dτ≤CT, | (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
uktt⇀utt 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;H−2(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
ukt→ut 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)−2ukt‖22=∫Ω|ukt|2(m(x)−1)dx≤max{||△ukt||2(m−−1)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(t−τ)(△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 uk→u in C(0,T;L2(Ω)), thus uk(0)→u(0) in L2(Ω), and we also have that uk(0)=uk0→u0 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=u−z satisfies
{ωtt+△2ω+△2ωtt−∫t0g(t−τ)△2ω(τ)dτ+|ut|m(x)−2ut−|zt|m(x)−2zt=0,(x,t)∈Ω×(0,T),ω(x,t)=∂ω∂ν(x,t)=0,(x,t)∈∂Ω×(0,T),ω(x,0)=0, ωt(x,0)=0,x∈Ω. |
Multiplying the first equation of Problem (3.1) by ωt and integrating over Ω, we have
12ddt[||ωt||22+(1−∫t0g(τ)dτ)||△ω||22+||△ωt||22+(g⋄△ω)]+12g(t)||△ω||22=−∫Ω(|ut|m(x)−2ut−|zt|m(x)−2zt)(ut−zt)dx+12(g′⋄△ω), |
from the inequality
(|a|m(x)−2a−|b|m(x)−2b)(a−b)≥0, | (3.24) |
for all a,b∈Rn and a.e. x∈Ω, 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
2≤p−≤p(x)≤p+≤2(n−3)n−4, |
then there exists a unique local solution of Problem (1.1).
Proof. For any T>0, consider MT={u∈H:u(0)=u0,ut(0)=u1,||u||H≤M}. Lemma 3.1 implies that for ∀v∈MT, 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+(g⋄△u)+l||Δu||22≤||u1||22+||△u1||22+||Δu0||22+2∫t0∫Ω|v|p(x)−2vutdxdτ, | (3.25) |
using Hölder's inequality and Young's inequality, we have
|∫Ω|v|p(x)−2vutdx|≤γ||ut||22+14γ∫Ω|v|2p(x)−2dx≤γ||ut||22+14γ[∫Ω|v|2p−−2dx+∫Ω|v|2p+−2dx]≤γ||ut||22+C4γ[||△v||2p−−22+||△v||2p+−22], |
thus, (3.25) becomes
||ut||22+||△ut||22+l||Δu||22≤λ0+2∫t0∫Ω|v|p(x)−2vutdxdτ≤λ0+2γTsup(0,T)||ut||22+TC2γsup(0,T)[||△v||2p−−22+||△v||2p+−22], |
hence, we have
sup(0,T)||ut||22+sup(0,T)||△ut||22+lsup(0,T)||Δu||22≤λ0+2γTsup(0,T)||ut||22+TC2γsup(0,T)[||v||2p−−2H+||v||2p+−2H], |
where λ0=||u1||22+||△u1||22+||Δu0||22, choosing γ=12T such that
||u||2H≤λ0+T2Csup(0,T)[||v||2p−−2H+||v||2p+−2H]. |
For any v∈MT, by choosing M large enough so that
||u||2H≤λ0+2T2CM2(p+−1)≤M2, |
and T>0, sufficiently small so that
T≤√M2−λ02CM2(p+−1), |
we obtain ||u||H≤M, which shows that S(MT)⊂MT.
Let v1,v2∈MT,u1=S(v1),u2=S(v2),u=u1−u2, then u satisfies
{utt+△2u+△2utt−∫t0g(t−τ)△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)=∂u∂ν(x,t)=0,(x,t)∈∂Ω×(0,T),u(x,0)=0, ut(x,0)=0,x∈Ω. |
Multiplying by ut and integrating over Ω×(0,t), we obtain
12||ut||22+12(1−∫t0g(τ)dτ)||△u||22+12||△ut||22+12(g⋄△u)+∫t0∫Ω[|u1t|m(x)−2u1t−|u2t|m(x)−2u2t](u1t−u2t)dxdτ≤∫t0∫Ω(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(g⋄△u)≤∫t0∫Ω(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=v1−v2 and ξ=αv1+(1−α)v2, 0≤α≤1. Thanks to Young's inequality and Hölder's inequality, we have
I≤δ2||ut||22+12δ∫Ω|f′(ξ)|2|v|2dx≤δ2||ut||22+(p+−1)22δ∫Ω|ξ|2(p(x)−2)|v|2dx≤δ2||ut||22+(p+−1)22δ(∫Ω|v|2nn−2dx)n−2n[∫Ω|ξ|n(p(x)−2)dx]2n≤δ2||ut||22+(p+−1)22δ(∫Ω|v|2nn−2dx)n−2n[(∫Ω|ξ|n(p+−2)dx)2n+(∫Ω|ξ|n(p−−2)dx)2n]≤δ2||ut||22+(p+−1)2C2δ||Δv||22[||△ξ||2(p+−2)2+||△ξ||2(p−−2)2]≤δ2||ut||22+(p+−1)2C2δ||Δv||22(M2(p+−2)+M2(p−−2)). | (3.28) |
Inserting (3.28) into (3.27), choosing δ small enough, we obtain
||u||2H≤(p+−1)2CTδ(M2(p−−2)+M2(p+−2))||v||2H, |
taking T small enough so that (p+−1)2CTδ(M2(p−−2)+M2(p+−2))<1, we conclude
||u||2H=||S(v1)−S(v2)||2H≤||v1−v2||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(1−∫t0g(τ)dτ)||△u||22+12||△ut||22+12(g⋄△u)−∫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(g′⋄△u)−12g(t)||△u||22≤0. | (4.2) |
Now, we set
B1=max{1,Bl12}, λ1=(B21)−2p−−2, E1=(12−1p−)(B21)−p−p−−2, |
and
H(t)=E2−E(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 non−decreasing 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||22≥λ2, t≥0. | (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(1−∫t0g(τ)dτ)||△u||22−∫Ω1p(x)|u|p(x)dx≥l2||△u||22−1p−∫Ω|u|p(x)dx≥l2||△u||22−1p−max{||u||p−p(x),||u||p+p(x)}≥l2||△u||22−1p−max{Bp−||△u||p−2,Bp+||△u||p+2}≥l2||△u||22−1p−max{Bp−1lp−2||△u||p−2,Bp+1lp+2||△u||p+2}≥12B21λ−1p−max{λp−2,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)={12B21−p+2p−λp+−22<0, λ>1,12B21−12λp−−22, 0<λ<1, |
G′+(1)=12B21−p+2p−<0, G′−(1)=12B21−12<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)≤1p−ρp(x)(u). |
Proof. (4.4) indicates that H(t) is nondecreasing with respect to t, thus
H(t)≥H(0)=E2−E(0)>0, ∀t∈[0,T). |
It follows from (1.6), (4.1), and Lemma 4.1 that
H(t)=E2−E(t)=E2−12||ut||22−12(1−∫t0g(τ)dτ)||△u||22−12(g⋄△u)−12||△ut||22+∫Ω1p(x)|u|p(x)dx≤E1−l2||△u||22+∫Ω1p(x)|u|p(x)dx≤E1−12B21λ2+∫Ω1p(x)|u|p(x)dx≤E1−12B21λ1+∫Ω1p(x)|u|p(x)dx≤∫Ω1p(x)|u|p(x)dx≤1p−ρp(x)(u). |
The proof is completed.
Our blow-up result reads as follows:
Theorem 4.3. Suppose that
2≤m−≤m(x)≤m+<p−≤p(x)≤p+≤2(n−3)n−4, |
and
1−l=∫∞0g(τ)dτ<p−2−1p−2−1+12p−, | (4.7) |
hold, if the following conditions
E(0)<12(12−1p−)(1−1p−(p−−2)1−ll)(B21)−p−p−−2 and λ1<λ(0)=B21l||△u0||22, |
are satisfied, then there exists T∗<+∞ such that
limt→T∗−(||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)=H1−α(t)+ϵ∫Ωutudx+ϵ∫Ω△ut△udx, | (4.10) |
where ε>0, small enough to be chosen later, and
0≤α≤min{p−−m+p−(m+−1),p−−22p−}. |
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)=(1−α)H−α(t)[∫Ω|ut|m(x)dx−12(g′⋄△u)+12g(t)||△u||22]+ϵ||ut||22+ϵ∫Ω△utt△udx+ϵ||△ut||22−ϵ||△u||22+ϵ∫Ω∫t0g(t−τ)△u(τ)dτ△udx−ϵ∫Ω|ut|m(x)−2utudx+ϵ∫Ω|u|p(x)dx−ϵ∫Ω△utt△udx≥(1−α)H−α(t)∫Ω|ut|m(x)dx+ϵ||ut||22−ϵ||△u||22+ϵ∫Ω∫t0g(t−τ)△u(τ)dτ△udx−ϵ∫Ω|ut|m(x)−2utudx+ϵ∫Ω|u|p(x)dx+ϵ||Δut||22, |
applying Hölder's inequality and Young's inequality, we have
ϵ∫Ω∫t0g(t−τ)Δu(τ)Δu(t)dτdx=ϵ∫Ω∫t0g(t−τ)Δu(t)(Δu(τ)−Δu(t))dτdx+ϵ∫t0g(t−τ)dτ||Δu||22≥−ϵ∫t0g(t−τ)||Δu(τ)−Δu(t)||2||Δu(t)||2dτ+ϵ∫t0g(t−τ)dτ||Δu||22≥−ϵp−(1−ε1)2(g⋄△u)+ϵ(1−12p−(1−ε1))∫t0g(τ)dτ||Δu||22, |
where 0<ε1<p−−2p−, then
L′(t)≥(1−α)H−α(t)∫Ω|ut|m(x)dx+ϵ||ut||22−ϵ||Δu||22+ϵ||Δut||22−ϵ∫Ω|ut|m(x)−2utudx+ϵ∫Ω|u|p(x)dx−ϵp−(1−ε1)2(g⋄△u)+ϵ(1−12p−(1−ε1))∫t0g(τ)dτ||Δu||22, |
rewriting (4.7) to (p−2−1)l−12p−(1−l)>0, using (4.1) and (4.3) to substitute for (g⋄△u), choosing ε1>0 sufficiently small, we obtain
L′(t)≥(1−α)H−α(t)∫Ω|ut|m(x)dx+ϵp−(1−ε1)H(t)+(ϵ+ϵp−(1−ε1)2)(||ut||22+||△ut||22)+ϵ{(p−(1−ε1)2−1)(1−∫t0g(τ)dτ)−12p−(1−ε1)∫t0g(τ)dτ}||△u||22−ϵp−(1−ε1)E2−ϵ∫Ω|ut|m(x)−2utudx+ϵε1∫Ω|u|p(x)dx≥(1−α)H−α(t)∫Ω|ut|m(x)dx+ϵp−(1−ε1)H(t)+(ϵ+ϵp−(1−ε1)2)(||ut||22+||△ut||22)+ϵ{(p−(1−ε1)2−1)l2−12p−(1−ε1)1−l2}lλ2B21−ϵp−(1−ε1)E2−ϵ∫Ω|ut|m(x)−2utudx+ϵ{(p−(1−ε1)2−1)l2−12p−(1−ε1)1−l2}||△u||22+ϵε1∫Ω|u|p(x)dx.≥(1−α)H−α(t)∫Ω|ut|m(x)dx+ϵp−(1−ε1)H(t)+(ϵ+ϵp−(1−ε1)2)(||ut||22+||△ut||22)+ϵ{(p−(1−ε1)2−1)l2−12p−(1−ε1)1−l2}l(B21)−p−p−−2−ϵp−(1−ε1)E2−ϵ∫Ω|ut|m(x)−2utudx+ϵ{(p−(1−ε1)2−1)l2−12p−(1−ε1)1−l2}||△u||22+ϵε1∫Ω|u|p(x)dx. | (4.11) |
Step 1.1: Estimate for ϵ{(p−(1−ε1)2−1)l2−12p−(1−ε1)1−l2}l(B21)−p−p−−2−ϵp−(1−ε1)E2. It follows from the condition in Theorem 3.1 that
E(0)<12(12−1p−)(1−1−lp−(p−−2)l)(B21)−p−p−−2=(p−2−1)l2−12p−(1−l)2lp−(B21)−p−p−−2<E1, |
here, we can take ε1>0 sufficiently small and choose E2∈(E(0),E1) sufficiently close to E(0) such that
ϵ(p−(1−ε1)2−1)l2−12p−(1−ε1)(1−l)2l(B21)−p−p−−2−ϵ(1−ε1)p−E2≥ϵ(p−(1−ε1)2−1)l2−12p−(1−ε1)(1−l)2l(B21)−p−p−−2−ϵ(1−ε1)p−(p−2−1)l2−12p−(1−l)2lp−(B21)−p−p−−2≥0. | (4.12) |
Therefore, we obtain by combining (4.11) and (4.12),
L′(t)≥(1−α)H−α(t)∫Ω|ut|m(x)dx+ϵp−(1−ε1)H(t)+(ϵ+ϵp−(1−ε1)2)(||ut||22+||△ut||22)+ϵε1∫Ω|u|p(x)dx+ϵ{(p−(1−ε1)2−1)l2−12p−(1−ε1)1−l2}||△u||22−ϵ∫Ω|ut|m(x)−2utudx. | (4.13) |
Step 1.2: Estimate for −ϵ∫Ω|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)−1H−αm(x)−1m(x)(t)Hαm(x)−1m(x)(t)|u|dx≤ε2H−α(t)∫Ω|ut|m(x)dx+1εm−−12∫Ω|u|m(x)Hα(m(x)−1)(t)dx≤ε2H−α(t)∫Ω|ut|m(x)dx+2Cα(m−−m+)1εm−−12Hα(m+−1)(t)∫Ω|u|m(x)dx≤ε2H−α(t)∫Ω|ut|m(x)dx+C2εm−−12Hα(m+−1)(t)max{||u||m+p(x),||u||m−p(x)}, | (4.14) |
where C1=min{H(0),1}, C2=2(1+|Ω|)m+Cα(m−−m+)1. Next, we have
||u||m+p(x)≤max{(∫Ω|u|p(x)dx)m+p+,(∫Ω|u|p(x)dx)m+p−}≤max{[p−H(t)]m+p+−m+p−,1}(∫Ω|u|p(x)dx)m+p−, |
and
||u||m−p(x)≤max{[p−H(t)]m−p+−m+p−,[p−H(t)]m−−m+p−}(∫Ω|u|p(x)dx)m+p−, |
which illustrate
max{||u||m+p(x),||u||m−p(x)}≤C3(∫Ω|u|p(x)dx)m+p−, |
where C3=2min{p−H(0),1}m−p+−m+p−. Recalling 0<α≤p−−m+p−(m+−1) and Lemma 4.2, apparently,
Hα(m+−1)(t)max{||u||m+p(x),||u||m−p(x)}≤C3Hα(m+−1)(t)(∫Ω|u|p(x)dx)m+p−≤C3Hα(m+−1)+m+p−−1(t)Hα(m+−1)+m+p−−1(0)H1−m+p−(t)Hα(m+−1)+m+p−−1(0)(∫Ω|u|p(x)dx)m+p−≤C3(1p−)1−m+p−(∫Ω|u|p(x)dx)1−m+p−Hα(m+−1)+m+p−−1(0)(∫Ω|u|p(x)dx)m+p−≤C3(1p−)1−m+p−Cα(m+−1)+m+p−−11∫Ω|u|p(x)dx, | (4.15) |
it follows from (4.13), (4.14), and (4.15) that
L′(t)≥(1−α−ϵε2)H−α(t)∫Ω|ut|m(x)dx+(ϵ+ϵp−(1−ε1)2)(||ut||22+||△ut||22)+ϵ(1−ε1)p−H(t)+ϵ(ε1−Cα(m+−1)+m+p−−11C2C3(1p−)1−m+p−εm−−12)∫Ω|u|p(x)dx+ϵ{(p−(1−ε1)2−1)l2−12p−(1−ε1)1−l2}||△u||22, |
let us fix the constant ε2 so that
ε1>Cα(m+−1)+m+p−−11C2C3(1p−)1−m+p−εm−−12, |
and then choose ϵ so small that 1−α>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−(1−ε1)2),(1−ε1)p−,ε1−Cα(m+−1)+m+p−−11C2C3(1p−)1−m+p−εm−−12,,(p−(1−ε1)2−1)l2−12p−(1−ε1)1−l2}. |
Inequalities (4.16) and Lemma 4.2 imply L(t)≥L(0). Therefore, for a sufficiently small ϵ, we have
L(0)=H1−α(0)+ϵ∫Ωu1u0dx+ϵ∫Ω△u1△u0dx>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|11−α≤(‖ut‖2‖u‖2)11−α≤(1+|Ω|)11−α||ut||11−α2||u||11−αp(x)≤(1+|Ω|)11−αμ||ut||11−αμ2+(1+|Ω|)11−αν||u||11−ανp(x), | (4.17) |
where 1μ+1ν=1. Choosing μ=2(1−α)>1, then ν=2(1−α)2(1−α)−1, further, (4.17) can be rewritten as
|∫Ωutudx|11−α≤(1+|Ω|)11−αμ||ut||22+(1+|Ω|)11−αν||u||22(1−α)−1p(x), | (4.18) |
recalling 0<α<p−−22p−, we obtain
||u||22(1−α)−1p(x)≤max{(∫Ω|u|p(x)dx)2p−[2(1−α)−1],(∫Ω|u|p(x)dx)2p+[2(1−α)−1]}≤{[p−H(t)]2−p−[2(1−α)−1]p−[2(1−α)−1],[p−H(t)]2−p+[2(1−α)−1]p+[2(1−α)−1]}∫Ω|u|p(x)dx≤C4∫Ω|u|p(x)dx, | (4.19) |
with C4=min{p−H(0),1}2−p+[2(1−α)−1]p+[2(1−α)−1]. Inserting (4.19) into (4.18), we obtain
|∫Ωutudx|11−α≤(1+|Ω|)11−αμ||ut||22+(1+|Ω|)11−ανC4∫Ω|u|p(x)dx. | (4.20) |
We now estimate
|∫Ω△ut△udx|11−α≤||△ut||11−α2||Δu||11−α2≤C11−α∗≤C11−α∗H(0)H(t), | (4.21) |
therefore, combining (4.20) and (4.21), we obtain
L11−α(t)=(H1−α(t)+ϵ∫Ωutudx+ϵ∫Ω△ut△udx)11−α≤M2(H(t)+||ut||22+||△ut||22+||△u||22+∫Ω|u|p(x)dx), | (4.22) |
where
M2=max{211−α(211−α+ϵ11−αC11−α∗H(0)), 221−αϵ11−α(1+|Ω|)11−αμ, 221−αϵ11−α(1+|Ω|)11−ανC4}. |
Combining (4.16) and (4.22), we arrive at
L′(t)≥M1M2L11−α(t),∀t≥0. | (4.23) |
A simple integration of (4.23) over (0,t) yields
Lα1−α(t)≥1Lαα−1(0)−M1M2α1−αt, |
this shows that L(t) blows up in finite time
T∗≤M2M11−ααLαα−1(0), |
furthermore, one gets from (4.22) that
limt→T∗−(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|p−dx≤||u||p+p++|Ω|, |
and using Lemma 4.2, we obtain
limt→T∗−(||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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