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Research article Special Issues

Edge computing-based intelligent monitoring system for manhole cover


  • Unusual states of manhole covers (MCs), such as being tilted, lost or flooded, can present substantial safety hazards and risks to pedestrians and vehicles on the roadway. Most MCs are still being managed through manual regular inspections and have limited information technology integration. This leads to time-consuming and labor-intensive identification with a lower level of accuracy. In this paper, we propose an edge computing-based intelligent monitoring system for manhole covers (EC-MCIMS). Sensors detect the MC and send status and positioning information via LoRa to the edge gateway located on the nearby wisdom pole. The edge gateway utilizes a lightweight machine learning model, trained on the edge impulse (EI) platform, which can predict the state of the MC. If an abnormality is detected, the display and voice device on the wisdom pole will respectively show and broadcast messages to alert pedestrians and vehicles. Simultaneously, the information is uploaded to the cloud platform, enabling remote maintenance personnel to promptly repair and restore it. Tests were performed on the EI platform and in Dongguan townships, demonstrating that the average response time for identifying MCs is 4.81 s. Higher responsiveness and lower power consumption were obtained compared to cloud computing models. Moreover, the system utilizes a lightweight model that better reduces read-only memory (ROM) and random-access memory (RAM), while maintaining an average identification accuracy of 94%.

    Citation: Liang Yu, Zhengkuan Zhang, Yangbing Lai, Yang Zhao, Fu Mo. Edge computing-based intelligent monitoring system for manhole cover[J]. Mathematical Biosciences and Engineering, 2023, 20(10): 18792-18819. doi: 10.3934/mbe.2023833

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  • Unusual states of manhole covers (MCs), such as being tilted, lost or flooded, can present substantial safety hazards and risks to pedestrians and vehicles on the roadway. Most MCs are still being managed through manual regular inspections and have limited information technology integration. This leads to time-consuming and labor-intensive identification with a lower level of accuracy. In this paper, we propose an edge computing-based intelligent monitoring system for manhole covers (EC-MCIMS). Sensors detect the MC and send status and positioning information via LoRa to the edge gateway located on the nearby wisdom pole. The edge gateway utilizes a lightweight machine learning model, trained on the edge impulse (EI) platform, which can predict the state of the MC. If an abnormality is detected, the display and voice device on the wisdom pole will respectively show and broadcast messages to alert pedestrians and vehicles. Simultaneously, the information is uploaded to the cloud platform, enabling remote maintenance personnel to promptly repair and restore it. Tests were performed on the EI platform and in Dongguan townships, demonstrating that the average response time for identifying MCs is 4.81 s. Higher responsiveness and lower power consumption were obtained compared to cloud computing models. Moreover, the system utilizes a lightweight model that better reduces read-only memory (ROM) and random-access memory (RAM), while maintaining an average identification accuracy of 94%.



    Over the past few decades, considerable attention has been paid to the rigorous analysis of mathematical models describing tumor growth and great progress has been achieved. Most work in this direction focuses on the sphere-shaped or nearly sphere-shaped tumor models; see [1,2,3,4,5,6,7,8,9] and the references therein. Observing that the work concerning models for tumors having different geometric configurations from spheroids is less frequent, in this paper we are interested in the situation of tumor cord–a kind of tumor that grows cylindrically around the central blood vessel and receives nutrient materials (such as glucose and oxygen) from the blood vessel [10]. The model only describes the evolution of the tumor cord section perpendicular to the length direction of the blood vessel due to the cord's uniformity in that direction. Assume that the radius of the blood vessel is r0 and denote by J and Γ(t) the section of the blood vessel wall and the section of the exterior surface of the tumor cord, respectively, then J={xR2;|x|=r0}. We also denote by Ω(t) the region of the section of the tumor cord, so that Ω(t) is an annular-like bounded domain in R2 and Ω(t)=JΓ(t). The mathematical formulation of the tumor model under study is as follows:

    cσt(x,t)Δσ(x,t)+σ(x,t)=0,xΩ(t),t>0, (1.1)
    Δp(x,t)=μ[σ(ξ(tτ;x,t),tτ)˜σ],xΩ(t),t>0, (1.2)
    {dξds=p(ξ,s),tτst,ξ=x,s=t, (1.3)
    σ(x,t)=ˉσ,np(x,t)=0,xJ,t>0, (1.4)
    νσ(x,t)=0,p(x,t)=γκ(x,t),xΓ(t),t>0, (1.5)
    V(x,t)=νp(x,t),xΓ(t),t>0, (1.6)
    Γ(t)=Γ0,τt0, (1.7)
    σ(x,t)=σ0(x),xΩ0,τt0, (1.8)
    p(x,t)=p0(x),xΩ0,τt0. (1.9)

    Here, σ and p denote the nutrient concentration and pressure within the tumor, respectively, which are to be determined together with Ω(t), and c=Tdiffusion /Tgrowth is the ratio of the nutrient diffusion time scale to the tumor growth (e.g., tumor doubling) time scale; thus, it is very small and can sometimes be set to be 0 (quasi-steady state approximation). Assume that the time delay τ is reflected between the time at which a cell commences mitosis and the time at which the daughter cells are produced and ξ(s;x,t) represents the cell location at time s as cells are moving with the velocity field V, then the function ξ(s;x,t) satisfies

    {dξds=V(ξ,s),tτst,ξ|s=t=x. (1.10)

    In other words, ξ tracks the path of the cell currently located at x. (1.3) is further derived from (1.10) under the assumption of a porous medium structure for the tumor, where Darcy's law V=p holds true. Because of the presence of time delay, the tumor grows at a rate that is related to the nutrient concentration when it starts mitosis and a combination of the conservation of mass and Darcy's law yields (1.2), in which μ represents the growth intensity of the tumor and ˜σ is the nutrient concentration threshold required for tumor cell growth. Additionally, ˉσ is the nutrient concentration in the blood vessel, ˉσ>˜σ, V, κ and ν denote the normal velocity, the mean curvature and the unit outward normal field of the outer boundary Γ(t), respectively, n denotes the unit outward normal field of the fixed inner boundary J, and γ is the outer surface tension coefficient. Thus, the boundary condition σ=ˉσ on J indicates that the tumor receives constant nutrient supply from the blood vessel, νσ=0 on Γ(t) implies that the nutrient cannot pass through Γ(t), np=0 on J means that tumor cells cannot pass through the blood vessel wall, p=γκ on Γ(t) is due to the cell-to-cell adhesiveness, and V=νp on Γ(t) is the well-known Stefan condition representing that the normal velocity of the tumor cord outer boundary Γ(t) is the same with that of tumor cells adjacent to Γ(t). Finally, σ0(x), p0(x), Γ0 are given initial data and Ω(t)=Ω0 for τt0.

    Before going to our interest, we prefer to recall some relevant works. Models for the growth of the strictly cylindrical tumor cord were studied in [11,12,13]. For the model (1.1)–(1.9) without the time delay, if c=0, Zhou and Cui [14] showed that the unique radially symmetric stationary solution exists and is asymptotically stable for any sufficiently small perturbations. Meanwhile, if c>0, Wu et al. [15] proved that the stationary solution is locally asymptotically stable provided that c is small enough. On the other hand, Zhao and Hu [16] considered the multicell spheroids with time delays. For the case c=0, they analyzed the linear stability of the radially symmetric stationary solution as well as the impact of the time delay.

    Motivated by the works [14,15,16], here we aim to discuss the linear stability of stationary solutions to the problem (1.1)–(1.9) with the quasi-steady-state assumption, i.e., c=0, and investigate the effect of time delay on tumor growth. Our first main result is given below.

    Theorem 1.1. For small time delay τ, the problem (1.1)–(1.9) admits a unique radially symmetric stationary solution.

    Next, in order to deal with the linear stability of the radially symmetric stationary solution, denoted by (σ,p,Ω), where Ω={xR2: r0<r=|x|<R}, we assume that the initial conditions are perturbed as follows:

    Ω(t)={xR2:r0<r<R+ερ0(θ)},τt0,σ(r,θ,t)=σ(r)+εw0(r,θ),p(r,θ,t)=p(r)+εq0(r,θ),τt0.

    The linearized problem of (1.1)–(1.9) at (σ,p,Ω) is then obtained by substituting

    Ω(t):r0<r<R+ερ(θ,t)+O(ε2), (1.11)
    σ(r,θ,t)=σ(r)+εw(r,θ,t)+O(ε2), (1.12)
    p(r,θ,t)=p(r)+εq(r,θ,t)+O(ε2) (1.13)

    into (1.1)–(1.9) and collecting the ε-order terms. Now, we can state the second main result of this paper.

    Theorem 1.2. For small time delay τ, the radially symmetric stationary solution (σ,p,Ω) of (1.1)–(1.9) with c=0 is linearly stable, i.e.,

    max0θ2π|ρ(θ,t)|Ceδt,t>0 (1.14)

    for some positive constants C and δ.

    Remark 1.1. Compared with results of the problem modeling the growth of tumor cord without time delays in [14], the introduction of the time delay does not affect the stability of the radially symmetric stationary solution even under non-radial perturbations. However, as we shall see in Subsection 3.3, the numerical result shows that adding time delay would result in a larger stationary tumor. Moreover, the stronger the growth intensity of the tumor is, the greater the influence of time delay on the size of the stationary tumor is.

    Remark 1.2. Compared with results of the nearly sphere-shaped tumor model with time delays in [16], which state that the radially symmetric stationary solution is linearly stable for small μ in the sense that limtmax0θ2π|ρ(θ,t)(a1cosθ+b1sinθ)|=0 for some constants a1 and b1, the radially symmetric stationary solution of tumor cord with the time delay is linearly stable for any μ>0 in the normal sense.

    This paper is organized as follows. In Section 2, we give the proof of Theorem 1.1 by first transforming the free boundary problem into an equivalent problem with fixed boundary and then applying the contraction mapping principle combined with Lp estimates to this fixed boundary problem. In Section 3, we prove Theorem 1.2 and Remark 1.1 by first introducing the linearization of (1.1)–(1.9) at the radially symmetric stationary solution (σ,p,Ω), and then making a delicate analysis of the expansion in the time delay τ provided that τ is sufficiently small. A brief conclusion in Section 4 completes the paper.

    In this section, we study radially symmetric stationary solutions (σ,p,Ω) to the system (1.1)–(1.9), which satisfy

    Δrσ(r)+σ(r)=0,σ(r0)=ˉσ,σ(R)=0,r0<r<R, (2.1)
    Δrp(r)=μ[σ(ξ(τ;r,0))˜σ],p(r0)=0,p(R)=γR,r0<r<R, (2.2)
    {dξds(s;r,0)=pr(ξ(s;r,0)),τs0,ξ(s;r,0)=r,s=0, (2.3)
    Rr0[σ(ξ(τ;r,0))˜σ]rdr=0, (2.4)

    where Δr is the radial part of the Laplacian in R2.

    Before proceeding further, let us recall that the modified Bessel functions Kn(r) and In(r), standard solutions of the equation

    r2y+ry(r2+n2)y=0,r>0, (2.5)

    have the following properties:

    In+1(r)=In1(r)2nrIn(r),Kn+1(r)=Kn1(r)+2nrKn(r),n1, (2.6)
    In(r)=12[In1(r)+In+1(r)],Kn(r)=12[Kn1(r)+Kn+1(r)],n1, (2.7)
    In(r)=In1(r)nrIn(r), Kn(r)=Kn1(r)nrKn(r),n1, (2.8)
    In(r)=nrIn(r)+In+1(r), Kn(r)=nrKn(r)Kn+1(r),n0, (2.9)
    In(r)Kn+1(r)+In+1(r)Kn(r)=1r,n0 (2.10)

    and

    In(r)>0,Kn(r)<0.

    Proof of Theorem 1.1 In view of (2.5), the solution of (2.1) is clearly given by

    σ(r)=ˉσI0(r)K1(R)+I1(R)K0(r)I0(r0)K1(R)+I1(R)K0(r0). (2.11)

    Introducing the notations:

    ˆr=rr0Rr0,ˆσ(ˆr)=σ(r),ˆp(ˆr)=(Rr0)p(r),ˆξ(s,ˆr,0)=ξ(s,r,0)r0Rr0,

    (2.1)–(2.4) reduces to the following system after dropping the "^" in the above variables:

    2σr2+Rr0r(Rr0)+r0σr=(Rr0)2σ,σ(0)=ˉσ,σ(1)=0, (2.12)
    {2pr2+Rr0r(Rr0)+r0pr=μ(Rr0)3[σ(r+1(Rr0)30τpr((Rr0)ξ(s;r,0)+r0)ds)˜σ],p(0)=0,p(1)=γ(Rr0)R, (2.13)
    {dξds(s;r,0)=1(Rr0)3pr((Rr0)ξ(s;r,0)+r0),τs0,ξ(s;r,0)=r,s=0, (2.14)
    10[r(Rr0)+r0][σ(r+1(Rr0)30τpr((Rr0)ξ(s;r,0)+r0)ds)˜σ]dr=0. (2.15)

    It is clear that (2.12) can be solved explicitly. For convenience, we extend the solution of (2.12) outside [0, 1]:

    σ(r;R)={ˉσI0(r(Rr0)+r0)K1(R)+I1(R)K0(r(Rr0)+r0)I0(r0)K1(R)+I1(R)K0(r0),0r1,ˉσI0(R)K1(R)+I1(R)K0(R)I0(r0)K1(R)+I1(R)K0(r0),1<r2. (2.16)

    Assume that Rmin and Rmax are positive constants to be determined later and r0<Rmin<Rmax. For any R[Rmin,Rmax], we will prove that p is also uniquely determined by applying the contraction mapping principle.

    Noticing that 0 is a lower solution of (2.14), but there is no assurance that ξ(s;r,0)1 for τs0, we suppose ξ(s;r,0)[0,2] and take

    X={pW2,[0,2];pW2,[0,2]M},

    where M>0 is to be determined. For each pX, we first solve for ξ from (2.14) and substitute it into (2.13), then the following system

    {2ˉpr2+Rr0r(Rr0)+r0ˉpr=μ(Rr0)3[σ(r+1(Rr0)30τpr((Rr0)ξ(s;r,0)+r0)ds)˜σ],ˉp(1)=γ(Rr0)R,ˉpr(0)=0 (2.17)

    allows a unique solution ˉpW2,[0,1]. Applying the strong maximum principle combined with the Hopf lemma to (2.1) shows that σ(r;R)ˉσ. Thus, integrating (2.17), we obtain

    1r(Rr0)+r0ˉprL[0,1]μ2(Rmaxr0)2(ˉσ+˜σ), (2.18)
    ˉpL[0,1]γ(Rmaxr0)Rmin+μ4R2max(Rmaxr0)(ˉσ+˜σ), (2.19)
    2ˉpr2L[0,1]3μ2(Rmaxr0)3(ˉσ+˜σ). (2.20)

    Define the mapping

    Lp(r)={ˉp(r),0r1,ˉp(1)+ˉp(1)(r1),1<r2,pX,

    then LpW2,[0,2] and LpW2,[0,2]2ˉpW2,[0,1]. Combining (2.18)–(2.20), we find

    LpW2,[0,2]2{μ2(Rmaxr0)2(ˉσ+˜σ)+3μ2(Rmaxr0)3(ˉσ+˜σ)+γ(Rmaxr0)Rmin+μ4R2max(Rmaxr0)(ˉσ+˜σ)}M1. (2.21)

    If we choose MM1, then LpX by (2.21) and L maps X to itself.

    We now show that L is a contraction. Given p(1),p(2)X, one can first get ξ(1),ξ(2) from the following two systems:

    {dξ(1)ds(s;r,0)=1(Rr0)3p(1)r((Rr0)ξ(1)(s;r,0)+r0),τs0,ξ(1)(s;r,0)=r,s=0, (2.22)
    {dξ(2)ds(s;r,0)=1(Rr0)3p(2)r((Rr0)ξ(2)(s;r,0)+r0),τs0,ξ(2)(s;r,0)=r,s=0. (2.23)

    Integrating (2.22) and (2.23) with regard to s over the interval [τ,0] and making a subtraction yield

    |ξ(1)ξ(2)|τ(Rr0)3maxτs00r1[|p(1)r((Rr0)ξ(1)+r0)p(2)r((Rr0)ξ(1)+r0)|+|p(2)r((Rr0)ξ(1)+r0)p(2)r((Rr0)ξ(2)+r0)|]τ(Rr0)3p(1)p(2)W2,[0,2]+τM(Rr0)2maxτs00r1|ξ(1)ξ(2)|

    for all τs0 and 0r1. Consequently,

    maxτs00r1|ξ(1)ξ(2)|τ(Rr0)3τM(Rr0)p(1)p(2)W2,[0,2]. (2.24)

    Next, we substitute ξ(1),ξ(2) into (2.17) and solve for ˉp(1) and ˉp(2), respectively, then it follows from (2.17) that (ˉp(1)ˉp(2))(1)=0, r(ˉp(1)ˉp(2))(0)=0 and

    2r2(ˉp(1)ˉp(2))Rr0r(Rr0)+r0r(ˉp(1)ˉp(2))=μ(Rr0)3[σ(r+1(Rr0)30τp(1)r((Rr0)ξ(1)(s;r,0)+r0)ds)σ(r+1(Rr0)30τp(2)r((Rr0)ξ(2)(s;r,0)+r0)ds)].

    Using (2.24), we derive

    1r(Rr0)+r0r(ˉp(1)ˉp(2))L[0,1]μ2(Rr0)2σ(r+1(Rr0)30τp(1)r((Rr0)ξ(1)+r0)ds)σ(r+1(Rr0)30τp(2)r((Rr0)ξ(2)+r0)ds)L[0,1]μ2(Rr0)σrL[0,2]0τ(p(1)r((Rr0)ξ(1)+r0)p(2)r((Rr0)ξ(2)+r0))dsμτ2(Rr0)σrL[0,2](p(1)p(2)W2,[0,2]+(Rr0)p(2)W2,[0,2]maxτs00r1|ξ(1)ξ(2)|)M2τp(1)p(2)W2,[0,2]

    and similarly,

    ˉp(1)ˉp(2)L[0,1]M3τp(1)p(2)W2,[0,2],2r2(ˉp(1)ˉp(2))L[0,1]M4τp(1)p(2)W2,[0,2],

    where

    M2=μˉσRmax2r0(Rmaxr0)3(Rminr0)2Mτ,M3=μˉσR3max4r0(Rmaxr0)2(Rminr0)2Mτ,M4=3μˉσRmax2r0(Rmaxr0)4(Rminr0)2Mτ.

    Here, we employed the fact that

    σrL[0,2]=σrL[0,1]ˉσRmaxr0(Rmaxr0)2

    by (2.1) and σˉσ. Let M5=M2+M3+M4, then M5 is independent of τ and

    ˉp(1)ˉp(2)W2,[0,1]M5τp(1)p(2)W2,[0,2],

    which together with Lp(1)(1)=Lp(2)(1)=γ(Rr0)R and (Lp(1))(0)=(Lp(2))(0)=0 implies that

    Lp(1)Lp(2)W2,[0,2]2M5τp(1)p(2)W2,[0,2].

    Hence, if τ is sufficiently small such that 2M5τ<1, then we derive a contracting mapping L. The existence and uniqueness of p are therefore obtained.

    It suffices to prove that there exists a unique R[Rmin,Rmax] satisfying (2.15). Substituting (2.16) into (2.15), we find that it is equivalent to solving the following equation for R:

    G(R,τ)=10r(Rr0)+r0R+r0[σ(r+1(Rr0)30τpr((Rr0)ξ(s;r,0)+r0)ds)˜σ]dr=0.

    Clearly,

    G(R,0)=10(σ(r;R)˜σ)r(Rr0)+r0R+r0dr=10σ(r;R)r(Rr0)+r0R+r0dr˜σ2.

    Using Lemma 3.1 and Theorem 3.2 in [14] and the condition ˉσ>˜σ, we know that

    limRr0G(R,0)=ˉσ˜σ2>0,limRG(R,0)=˜σ2<0,G(R,0)R<0,

    which implies that the equation G(R,0)=0 has a unique solution, denoted by RS, and

    G(12(RS+r0),0)>0,G(32RS,0)<0.

    Since

    G(R,τ)R=G(R,0)R+2G(R,η)Rττ+O(τ2),0ητ

    when τ is sufficiently small, G(R,τ)R and G(R,0)R have the same sign. Thus, G(R,τ) is monotone decreasing in R. Using the fact that G(R,τ) is continuous in τ, we further have

    G(12(RS+r0),τ)>0,G(32RS,τ)<0.

    Hence, when τ is sufficiently small, the equation G(R,τ)=0 has a unique solution R. Taking Rmin=12(RS+r0) and Rmax=32RS, we complete the proof of the theorem.

    This section is devoted to the linear stability of the radially symmetric stationary solution (σ,p,Ω) of the problem (1.1)–(1.9) and the effect of time delay on the stability and the size of the stationary tumor. Let (σ,p,Ω(t)), given by (1.11)–(1.13), be solutions to (1.1)–(1.9), and denote by er, eθ the unit normal vectors in r, θ directions, respectively. Written in the rectangular coordinates in R2,

    er=(cosθ,sinθ)T,eθ=(sinθ,cosθ)T.

    Using the notation ξ1(s;r,θ,t), ξ2(s;r,θ,t) for the polar radius and angle of ξ(s;r,θ,t), respectively, we have

    ξ(s;r,θ,t)=ξ1(s;r,θ,t)er(ξ)=ξ1(s;r,θ,t)(cosξ2(s;r,θ,t),sinξ2(s;r,θ,t))T.

    Expand ξ1, ξ2 in ε as

    {ξ1=ξ10+εξ11+O(ε2),ξ2=ξ20+εξ21+O(ε2), (3.1)

    then we derive from (1.3) and (1.13) that

    {dξ10ds=pr(ξ10),tτst,ξ10|s=t=r; (3.2)
    {dξ11ds=2pr2(ξ10)ξ11qr(ξ10,ξ20,s),tτst,ξ11|s=t=0; (3.3)
    {dξ20ds=0,tτst,ξ20|s=t=θ; (3.4)
    {dξ21ds=1ξ210qθ(ξ10,ξ20,s),tτst,ξ21|s=t=0. (3.5)

    It is evident that ξ20θ. Noticing that the equation for ξ10 is the same as that for ξ in the radially symmetric case, ξ10 is independent of θ.

    Substituting (1.11)–(1.13) and (3.1)–(3.5) into (1.1), (1.2), (1.4)–(1.6), using the mean-curvature formula in the 2-dimensional case for the curve r=ρ(θ):

    κ=ρ2+2ρ2θρρθθ(ρ2+(ρθ)2)3/2

    and collecting the ε-order terms, we obtain the linearized system in BR×{t>0}:

    Δω(r,θ,t)=ω(r,θ,t),ω(r0,θ,t)=0,ωr(R,θ,t)+σ(R)ρ(θ,t)=0, (3.6)
    {Δq(r,θ,t)=μσr(ξ10(tτ;r,t))ξ11(tτ;r,θ,t)μw(ξ10(tτ;r,t),θ,tτ),qr(r0,θ,t)=0,q(R,θ,t)+γR2(ρ(θ,t)+2ρθ2(θ,t))=0, (3.7)
    ρ(θ,t)t=qr(R,θ,t)2pr2(R,θ,t)ρ(θ,t). (3.8)

    Due to the presence of the time delay, the linearization problem (3.6)–(3.8) cannot be solved explicitly. Assume that ω,q, ρ and ξ11 have the following Fourier expansions:

    {ω(r,θ,t)=A0(r,t)+n=1[An(r,t)cosnθ+Bn(r,t)sinnθ],q(r,θ,t)=E0(r,t)+n=1[En(r,t)cosnθ+Fn(r,t)sinnθ],ρ(θ,t)=a0(t)+n=1[an(t)cosnθ+bn(t)sinnθ],ξ11(s;r,θ,t)=e0(s;r,t)+n=1[en(s;r,t)cosnθ+fn(s;r,t)sinnθ]. (3.9)

    Substituting (3.9) into (3.6)–(3.8) yields the following system in BR×{t>0}:

    {2Anr2(r,t)+1rAnr(r,t)n2r2An(r,t)=An(r,t),An(r0,t)=0,Anr(R,t)+σ(R)an(t)=0, (3.10)
    {2Bnr2(r,t)+1rBnr(r,t)n2r2Bn(r,t)=Bn(r,t),Bn(r0,t)=0,Bnr(R,t)+σ(R)bn(t)=0, (3.11)
    {2Enr2(r,t)+1rEnr(r,t)n2r2En(r,t)=μσr(ξ10(tτ;r,t))en(tτ;r,t)μAn(ξ10(tτ;r,t),tτ),Enr(r0,t)=0,En(R,t)+γ(1n2)R2an(t)=0, (3.12)
    {2Fnr2(r,t)+1rFnr(r,t)n2r2Fn(r,t)=μσr(ξ10(tτ;r,t))fn(tτ;r,t)μBn(ξ10(tτ;r,t),tτ),Fnr(r0,t)=0,Fn(R,t)+γ(1n2)R2bn(t)=0, (3.13)
    {ens(s;r,t)=2pr2(ξ10)en(s;r,t)Enr(ξ10,s),tτst,ens=t=0, (3.14)
    {fns(s;r,t)=2pr2(ξ10)fn(s;r,t)Fnr(ξ10,s),tτst,fns=t=0, (3.15)
    dan(t)dt=2pr2(R)an(t)Enr(R,t), (3.16)
    dbn(t)dt=2pr2(R)bn(t)Fnr(R,t). (3.17)

    Since it is impossible to solve the systems (2.1)–(2.4) and (3.10)–(3.17) explicitly and the time delay τ is actually very small, in what follows, we analyze the expansion in τ for (2.1)–(2.4) and (3.10)–(3.17).

    Let

    R=R0+τR1+O(τ2),σ=σ0+τσ1+O(τ2),p=p0+τp1+O(τ2),An=A0n+τA1n+O(τ2),Bn=B0n+τB1n+O(τ2),En=E0n+τE1n+O(τ2),Fn=F0n+τF1n+O(τ2),an=a0n+τa1n+O(τ2),bn=b0n+τb1n+O(τ2).

    Substitute these expansions into (2.1)–(2.4) and (3.10)–(3.17). Since an(t) and bn(t) have the same asymptotic behavior at , we will only make an analysis of an(t). For this, we discuss the expansions of R, σ, p, An, En and an. Since the equations for the expansions of σ, p, An, En and an are the same as those in [16], here we only compute the expansions of the boundary conditions of σ, p, An and En.

    Expansions of the boundary conditions of σ:

    It follows from (2.10) and (2.11) that

    σ(r)=ˉσK1(R)I0(r)+I1(R)K0(r)I0(r0)K1(R)+I1(R)K0(r0)=ˉσK1(R0)I0(r)+I1(R0)K0(r)I0(r0)K1(R0)+I1(R0)K0(r0)+τˉσR1R0I0(r0)K0(r)K0(r0)I0(r)[I0(r0)K1(R0)+I1(R0)K0(r0)]2+O(τ2),

    which implies

    σ0(r)=ˉσK1(R0)I0(r)+I1(R0)K0(r)I0(r0)K1(R0)+I1(R0)K0(r0), (3.18)
    σ1(r)=ˉσR1R0I0(r0)K0(r)K0(r0)I0(r)[I0(r0)K1(R0)+I1(R0)K0(r0)]2. (3.19)

    By the boundary conditions in (2.1), we find

    σ0(r0)+τσ1(r0)+O(τ2)=ˉσ,σ0r(R0)+τ2σ0r2(R0)R1+τσ1r(R0)+O(τ2)=0.

    Expansions of the boundary conditions of p:

    One obtains from the boundary conditions in (2.2) that

    p0r(r0)+τp1r(r0)+O(τ2)=0,p0(R0)+τp0r(R0)R1+τp1(R0)+O(τ2)=γR0τγR1(R0)2+O(τ2).

    Expansion of (2.4):

    In view of (4.31) in [16], there holds

    0=Rr0[σ(ξ(τ;r,0))˜σ]rdr=Rr0[σ0(r)˜σ]rdr+τR0r0(σ0r(r)p0r(r)+σ1(r))rdr+O(τ2). (3.20)

    Using (3.18), we compute

    Rr0[σ0(r)˜σ]rdr=Rr0(ˉσK1(R0)I0(r)+I1(R0)K0(r)I0(r0)K1(R0)+I1(R0)K0(r0)˜σ)rdr=ˉσr0I1(R0)K1(r0)I1(r0)K1(R0)I0(r0)K1(R0)+I1(R0)K0(r0)+˜σ2[r20(R0)2]+τR1(ˉσI0(r0)K1(R0)+I1(R0)K0(r0)˜σR0)+O(τ2). (3.21)

    A combination of (3.20) and (3.21) gives

    τ[ˉσR1I0(r0)K1(R0)+I1(R0)K0(r0)˜σR0R1+R0r0(σ0r(r)p0r(r)+σ1(r))rdr]+ˉσr0I1(R0)K1(r0)I1(r0)K1(R0)I0(r0)K1(R0)+I1(R0)K0(r0)+˜σ2[r20(R0)2]+O(τ2)=0. (3.22)

    Expansions of the boundary conditions of An:

    We derive from the boundary conditions in (3.10) that

    A0n(r0,t)+τA1n(r0,t)+O(τ2)=0,0=A0nr(R0+τR1,t)+τA1nr(R0,t)+[σ0(R0+τR1)+τσ1(R0)][a0n(t)+τa1n(t)]+O(τ2)=A0nr(R0,t)+σ0(R0)a0n(t)+τ(2A0nr2(R0,t)R1+A1nr(R0,t)+σ0r(R0,t)R1a0n(t)+σ0(R0)a1n(t)+σ1(R0)a0n(t))+O(τ2).

    Expansions of the boundary conditions of En:

    Substituting the expansion of En into the boundary conditions in (3.12) yields

    E0nr(r0,t)+τE1nr(r0,t)+O(τ2)=0,0=E0n(R0+τR1,t)+τE1n(R0,t)+γ1n2(R0+τR1)2[a0n(t)+τa1n(t)]+O(τ2)=E0n(R0,t)+γ1n2(R0)2a0n(t)+τ(E0nr(R0,t)R1+E1n(R0,t)2γ1n2(R0)3R1a0n(t)+γ1n2(R0)2a1n(t))+O(τ2).

    Collecting all zeroth-order terms in \tau leads to the following system for r_0 < r < R^0_\ast :

    \begin{align} &-\frac{\partial^2 \sigma^0_\ast}{\partial r^2}-\frac{1}{r}\frac{\partial \sigma^0_\ast}{\partial r} = -\sigma^0_\ast, \qquad \sigma^0_\ast(r_0) = \bar{\sigma}, \qquad \frac{\partial \sigma^0_\ast}{\partial r}(R_\ast^0) = 0, \end{align} (3.23)
    \begin{align} &-\frac{\partial^2 p^0_\ast}{\partial r^2}-\frac{1}{r}\frac{\partial p^0_\ast}{\partial r} = \mu(\sigma^0_\ast-\tilde{\sigma}), \quad \frac{\partial p^0_\ast}{\partial r}(r_0) = 0, \quad p^0_\ast(R^0_\ast) = \frac{\gamma}{R^0_\ast}, \end{align} (3.24)
    \begin{align} &\bar{\sigma}r_0\frac{I_1(R^0_\ast)K_1(r_0)-I_1(r_0)K_1(R^0_\ast)}{I_0(r_0)K_1(R^0_\ast)+I_1(R^0_\ast)K_0(r_0)} +\frac{\widetilde{\sigma}}{2}[r_0^2-(R^0_\ast)^2] = 0, \end{align} (3.25)
    \begin{align} &\left\{\begin{array}{l} -\frac{\partial^2 A^0_n}{\partial r^2}-\frac{1}{r}\frac{\partial A^0_n}{\partial r}+\bigg(\frac{n^2}{r^2}+1\bigg)A^0_n = 0, \\ A^0_n(r_0,t) = 0,\quad\frac{\partial A^0_n}{\partial r}(R^0_\ast,t)+\sigma^0_\ast(R^0_\ast)a^0_n(t) = 0, \end{array}\right. \end{align} (3.26)
    \begin{align} &\left\{\begin{array}{l} -\frac{\partial^2 B^0_n}{\partial r^2}-\frac{1}{r}\frac{\partial B^0_n}{\partial r}+\bigg(\frac{n^2}{r^2}+1\bigg)B^0_n = 0, \\ B^0_n(r_0,t) = 0,\quad\frac{\partial B^0_n}{\partial r}(R^0_\ast,t)+\sigma^0_\ast(R^0_\ast)b^0_n(t) = 0, \end{array}\right. \end{align} (3.27)
    \begin{align} &\left\{\begin{array}{l} -\frac{\partial^2 E^0_n}{\partial r^2}-\frac{1}{r}\frac{\partial E^0_n}{\partial r}+\frac{n^2}{r^2}E^0_n = \mu A^0_n, \\ \frac{\partial E^0_n}{\partial r}(r_0,t) = 0, \quad E^0_n(R^0_\ast,t) = \gamma\frac{n^2-1}{(R_\ast^0)^2}a^0_n(t), \end{array}\right. \end{align} (3.28)
    \begin{align} &\left\{\begin{array}{l} -\frac{\partial^2 F^0_n}{\partial r^2}-\frac{1}{r}\frac{\partial F^0_n}{\partial r}+\frac{n^2}{r^2}F^0_n = \mu B^0_n, \\ \frac{\partial F^0_n}{\partial r}(r_0,t) = 0, \quad F^0_n(R^0_\ast,t) = \gamma\frac{n^2-1}{(R_\ast^0)^2}b^0_n(t), \end{array}\right. \end{align} (3.29)
    \begin{align} &\frac{\mathrm{d}a^0_n(t)}{\mathrm{d}t} = -\frac{\partial^2 p^0_\ast}{\partial r^2}(R^0_\ast)a^0_n(t)-\frac{\partial E^0_n}{\partial r}(R^0_\ast,t), \end{align} (3.30)
    \begin{align} &\frac{\mathrm{d}b^0_n(t)}{\mathrm{d}t} = -\frac{\partial^2 p^0_\ast}{\partial r^2}(R^0_\ast)b^0_n(t)-\frac{\partial F^0_n}{\partial r}(R^0_\ast,t). \end{align} (3.31)

    A direct calculation gives

    \begin{align} &\frac{\partial p^0_\ast}{\partial r}(r) = \frac{\mu\bar{\sigma}}{I_0(r_0)K_1(R^0_\ast)+I_1(R^0_\ast)K_0(r_0)}\bigg\{K_1(R^0_\ast)\left(\frac{r_0}{r}I_1(r_0)-I_1(r)\right) \\ &\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \qquad\qquad+I_1(R^0_\ast)\left(K_1(r)-\frac{r_0}{r}K_1(r_0)\right)\bigg\}+\frac{\mu\tilde{\sigma}r}{2}-\frac{\mu\tilde{\sigma}r^2_0}{2r}, \end{align} (3.32)
    \begin{align} &\frac{\partial^2 p^0_\ast}{\partial r^2}(R^0_\ast) = \frac{\mu\bar{\sigma}}{R^0_\ast[(R^0_\ast)^2-r^2_0]}\frac{2r_0R^0_\ast[I_1(R^0_\ast)K_1(r_0)-I_1(r_0)K_1(R^0_\ast)]-(R^0_\ast)^2+r^2_0}{I_0(r_0)K_1(R^0_\ast)+I_1(R^0_\ast)K_0(r_0)}, \end{align} (3.33)
    \begin{align} &A^0_n(r,t) = \frac{\bar{\sigma}a^0_n(t)h_n(r_0,R^0_\ast)}{R^0_\ast }\frac{K_n(r_0)I_n(r)-I_n(r_0)K_n(r)}{I_0(r_0)K_1(R^0_\ast)+I_1(R^0_\ast)K_0(r_0)}, \end{align} (3.34)
    \begin{align} &\frac{\partial A^0_n}{\partial r}(r,t) = \frac{-\bar{\sigma}a^0_n(t)h_n(r_0,R^0_\ast)}{R^0_\ast [I_0(r_0)K_1(R^0_\ast)+I_1(R^0_\ast)K_0(r_0)]}\frac1{h_n(r_0,r)}, \end{align} (3.35)
    \begin{align} &\frac{\partial^2 A^0_n}{\partial r^2}(R^0_\ast,t) = \frac{-\bar{\sigma}a^0_n(t)h_n(r_0,R^0_\ast)g_n(r_0,R^0_\ast)}{R^0_\ast [I_0(r_0)K_1(R^0_\ast)+I_1(R^0_\ast)K_0(r_0)]}, \end{align} (3.36)

    where

    \begin{align} &h_n(r_0,x) = \frac1{I_n(r_0)\left(\frac{n}{x}K_n(x)-K_{n+1}(x)\right) -K_n(r_0)\left(\frac{n}{x}I_n(x)+I_{n+1}(x)\right)}, \\ &g_n(r_0,x) = I_n(r_0)\left[\left(1+\frac{n(n-1)}{x^2}\right)K_n(x)+\frac{1}{x}K_{n+1}(x)\right] \\ &\qquad\qquad\qquad\qquad-K_n(r_0)\bigg[\left(1+\frac{n(n-1)}{x^2}\right)I_n(x)-\frac{1}{x}I_{n+1}(x)\bigg]. \end{align} (3.37)

    Let \eta^0_n = E^0_n+\mu A^0_n , then we find from (3.26) and (3.28) that \eta^0_n satisfies

    \begin{align*} -\frac{\partial^2 \eta^0_n}{\partial r^2}-\frac{1}{r}\frac{\partial \eta^0_n}{\partial r}+\frac{n^2}{r^2}\eta^0_n = 0, \end{align*}

    whose solution is

    \begin{align*} \eta^0_n(r,t) = C_1(t)r^n+C_2(t)r^{-n} \end{align*}

    and thus,

    \begin{equation} E^0_n(r,t) = \eta^0_n(r,t)-\mu A^0_n(r,t) = C_1(t)r^n+C_2(t)r^{-n}-\mu A^0_n(r,t), \end{equation} (3.38)

    where C_1(t) and C_2(t) are to be determined by the boundary conditions in (3.28). By (2.10), (3.34), (3.35) and (3.37), we get

    \begin{align} C_1(t) = &\frac{\mu\bar{\sigma} a^0_n(t)h_n(r_0,R^0_\ast)}{nR^0_\ast[(R^0_\ast)^{2n}+r_0^{2n}]}\frac{n(R^0_\ast)^n[I_n(R^0_\ast)K_n(r_0)-I_n(r_0)K_n(R^0_\ast)]+r^n_0}{I_0(r_0)K_1(R^0_\ast)+I_1(R^0_\ast)K_0(r_0)} \\ &+\frac{(n^2-1)\gamma a^0_n(t)}{(R^0_\ast)^{2n}+r_0^{2n}}(R^0_\ast)^{n-2}, \end{align} (3.39)
    \begin{align} C_2(t) = &\frac{\mu\bar{\sigma} a^0_n(t)h_n(r_0,R^0_\ast)r_0^n(R^0_\ast)^{n}}{nR^0_\ast[(R^0_\ast)^{2n}+r_0^{2n}]}\frac{nr_0^n[I_n(R^0_\ast)K_n(r_0)-I_n(r_0)K_n(R^0_\ast)]-(R^0_\ast)^n}{I_0(r_0)K_1(R^0_\ast)+I_1(R^0_\ast)K_0(r_0)} \\ &+\frac{(n^2-1)\gamma a^0_n(t)}{(R^0_\ast)^{2n}+r_0^{2n}}r_0^{2n}(R^0_\ast)^{n-2}. \end{align} (3.40)

    Using (3.35) and (3.38)–(3.40), we further derive

    \begin{align} \frac{\partial E^0_n}{\partial r}(r,t) = &\frac{\mu\bar{\sigma} a^0_n(t)h_n(r_0,R^0_\ast)}{R^0_\ast [I_0(r_0)K_1(R^0_\ast)+I_1(R^0_\ast)K_0(r_0)]}\bigg\{r_0^{n}\frac{r^{n-1}+(R^0_\ast)^{2n}r^{-n-1}}{(R^0_\ast)^{2n}+r_0^{2n}} \\ &\; \; \; \; \; \; \; \; \; \; +n(R^0_\ast)^n\frac{r^{n-1}-r_0^{2n}r^{-n-1}}{(R^0_\ast)^{2n}+r_0^{2n}}[I_n(R^0_\ast)K_n(r_0)-I_n(r_0)K_n(R^0_\ast)] \\ &\; \; \; \; \; \; \; \; \; \; +\frac1{h_n(r_0,r)}\bigg\}+n(n^2-1)\gamma a^0_n(t)(R^0_\ast)^{n-2}\frac{r^{n-1}-r_0^{2n}r^{-n-1}}{(R^0_\ast)^{2n}+r_0^{2n}}. \end{align} (3.41)

    Substituting (3.33) and (3.41) into (3.30) yields

    \begin{equation} \frac{\mathrm{d}a^0_n(t)}{\mathrm{d}t} = U_n(r_0,R^0_\ast)a^0_n(t), \end{equation} (3.42)

    whose solution is explicitly given by

    \begin{equation} a^0_n(t) = a^0_n(0)\exp\{U_n(r_0,R^0_\ast)t\}. \end{equation} (3.43)

    Here,

    \begin{align} U_n(r_0,R^0_\ast) = &\frac{\mu\bar{\sigma}h_n(r_0,R^0_\ast)} {I_0(r_0)K_1(R^0_\ast)+I_1(R^0_\ast)K_0(r_0)} \bigg[\frac{2r_0(I_1(r_0)K_1(R^0_\ast)-I_1(R^0_\ast)K_1(r_0))} {h_n(r_0,R^0_\ast)[(R^0_\ast)^2-r_0^2]} \\ &+\frac{n}{(R^0_\ast)^2}\frac{(R^0_\ast)^{2n}-r_0^{2n}}{(R^0_\ast)^{2n}+r_0^{2n}} (I_n(r_0)K_n(R^0_\ast)-I_n(R^0_\ast)K_n(r_0)) -\frac{2r_0^n(R^0_\ast)^{n-2}}{(R^0_\ast)^{2n}+r_0^{2n}}\bigg] \\ &-\frac{\gamma n(n^2-1) }{(R^0_\ast)^3}\frac{(R^0_\ast)^{2n}-r_0^{2n}}{(R^0_\ast)^{2n}+r_0^{2n}}. \end{align} (3.44)

    It was proven in Lemma 4.4 of [14] that U_n(r_0, R^0_\ast) < 0 for any n\geq 0 . Thus, we have the following:

    Lemma 3.1. For any n\geq 0 , there exists \delta > 0 such that |a_0^n(t)|\leq|a_0^n(0)|e^{-\delta t} for all t > 0 .

    Lemma 3.1 shows that a_n^0(t) decays to 0 exponentially at +\infty ; hence, when \tau = 0 , the radially symmetric stationary solution is asymptotically stable for all \mu > 0 .

    Recalling that R_\ast = R^0_\ast + \tau R^1_\ast + O(\tau^2) , in order to see the effect of the time delay \tau on the size of the stationary tumor, in this subsection we discuss the sign of R^1_\ast by a theoretical analysis combined with numerical simulations.

    We obtain from (3.22) that

    \begin{equation} \frac{\bar{\sigma}R^1_\ast}{I_0(r_0)K_1(R^0_\ast)+I_1(R^0_\ast)K_0(r_0)}-\tilde{\sigma}R^0_\ast R^1_\ast+\int^{R^0_\ast}_{r_0}\left(\frac{\partial \sigma^0_\ast}{\partial r}(r)\frac{\partial p^0_\ast}{\partial r}(r)+\sigma^1_\ast(r)\right)rdr = 0, \end{equation} (3.45)

    then by using (2.6)–(2.10), (3.18), (3.19), (3.25) and (3.32), one can solve (3.45) to obtain

    \begin{equation} R^1_\ast = -\frac{\mu\bar{\sigma}}{2R^0_\ast}\frac{T(r_0,R^0_\ast)}{S(r_0,R^0_\ast)}, \end{equation} (3.46)

    where

    \begin{align*} T(r_0,R^0_\ast)& = 1-r_0^2[I_0(r_0)K_1(R^0_\ast)+I_1(R^0_\ast)K_0(r_0)]^2 +r_0^2\frac{(R^0_\ast)^2-r_0^2-4}{(R^0_\ast)^2-r_0^2} [I_1(R^0_\ast)K_1(r_0)-I_1(r_0)K_1(R^0_\ast)]^2, \\ S(r_0,R^0_\ast)& = \frac{2r_0}{(R^0_\ast)^2-r_0^2}[I_1(r_0)K_1(R^0_\ast)-I_1(R^0_\ast)K_1(r_0)] [I_0(r_0)K_1(R^0_\ast) +I_1(R^0_\ast)K_0(r_0)]+\frac{1}{(R^0_\ast)^2}. \end{align*}

    Since

    U_0(r_0, R^0_\ast) = \frac{\mu\bar{\sigma}}{[I_0(r_0)K_1(R^0_\ast)+I_1(R^0_\ast)K_0(r_0)]^2}S(r_0,R^0_\ast)

    by (3.44), we know S(r_0, R^0_\ast) < 0 . Additionally, we numerically compute the function T(r_0, R^0_\ast) and find it is positive (see Figure 1). Hence, it follows from (3.46) that R^1_\ast > 0 and R^1_\ast is monotone increasing in \mu .

    Figure 1.  The graph of \min_{0.0001\le r_0 < R^0_\ast}T(r_0, R^0_\ast) with R^0_\ast\in[0.0001, 10] .

    Remark 3.1. The discussion above indicates that the presence of the time delay leads to a larger stationary tumor. Furthermore, the bigger the tumor aggressive parameter \mu is, the greater the effect of time delay on the size of the stationary tumor is.

    Now, we tackle the system consisting of all the first-order terms in \tau for r_0 < r < R^0_\ast :

    \begin{align} &-\frac{\partial^2 \sigma^1_\ast}{\partial r^2}-\frac{1}{r}\frac{\partial \sigma^1_\ast}{\partial r} = -\sigma^1_\ast, \quad \sigma^1_\ast(r_0) = 0, \quad \frac{\partial \sigma^1_\ast}{\partial r}(R^0_\ast)+\frac{\partial^2 \sigma^0_\ast}{\partial r^2}(R^0_\ast)R^1_\ast = 0, \end{align} (3.47)
    \begin{align} &\left\{\begin{array}{l} -\frac{\partial^2 p^1_\ast}{\partial r^2}-\frac{1}{r}\frac{\partial p^1_\ast}{\partial r} = \mu\frac{\partial \sigma^0_\ast}{\partial r}\frac{\partial p^0_\ast}{\partial r}+\mu\sigma^1_\ast, \\ \frac{\partial p^1_\ast}{\partial r}(r_0) = 0,\quad p_\ast^1(R_\ast^0) = -\frac{\gamma R_\ast^1}{(R_\ast^0)^2}-\frac{\partial p_\ast^0}{\partial r}(R_\ast^0)R_\ast^1, \end{array}\right. \end{align} (3.48)
    \begin{align} &\frac{\bar{\sigma}R^1_\ast}{I_0(r_0)K_1(R^0_\ast)+I_1(R^0_\ast)K_0(r_0)}-\tilde{\sigma}R^0_\ast R^1_\ast+\int^{R^0_\ast}_{r_0}\left(\frac{\partial \sigma^0_\ast}{\partial r}(r)\frac{\partial p^0_\ast}{\partial r}(r)+\sigma^1_\ast(r)\right)rdr = 0, \end{align} (3.49)
    \begin{align} &\left\{\begin{array}{l} -\frac{\partial^2 A^1_n}{\partial r^2}-\frac{1}{r}\frac{\partial A^1_n}{\partial r}+\left(\frac{n^2}{r^2}+1\right)A^1_n = 0, \\ A^1_n(r_0,t) = 0, \quad\frac{\partial^2 A^0_n}{\partial r^2}(R^0_\ast,t)R^1_\ast+\frac{\partial A^1_n}{\partial r}(R^0_\ast,t)+\sigma^0_\ast(R^0_\ast)a^1_n(t)+\sigma^1_\ast(R^0_\ast)a^0_n(t) = 0, \end{array}\right. \end{align} (3.50)
    \begin{align} &\left\{\begin{array}{l} -\frac{\partial^2 B^1_n}{\partial r^2}-\frac{1}{r}\frac{\partial B^1_n}{\partial r}+\left(\frac{n^2}{r^2}+1\right)B^1_n = 0, \\ B^1_n(r_0,t) = 0, \quad\frac{\partial^2 B^0_n}{\partial r^2}(R^0_\ast,t)R^1_\ast+\frac{\partial B^1_n}{\partial r}(R^0_\ast,t)+\sigma^0_\ast(R^0_\ast)b^1_n(t)+\sigma^1_\ast(R^0_\ast)b^0_n(t) = 0, \end{array}\right. \end{align} (3.51)
    \begin{align} &\left\{\begin{array}{l} -\frac{\partial^2 E^1_n}{\partial r^2}-\frac{1}{r}\frac{\partial E^1_n}{\partial r}+\frac{n^2}{r^2}E^1_n = \mu\frac{\partial \sigma^0_\ast}{\partial r}\frac{\partial E^0_n}{\partial r}+\mu\frac{\partial A_n^0}{\partial r}\frac{\partial p^0_\ast}{\partial r}-\mu\frac{\partial A_n^0}{\partial t}+\mu A_n^1, \\ \frac{\partial E^1_n}{\partial r}(r_0,t) = 0,\; E^1_n(R_\ast^0,t) = \gamma\frac{n^2-1}{(R_\ast^0)^2}a^1_n(t)-\frac{\partial E^0_n}{\partial r}(R_\ast^0,t)R_\ast^1-2\gamma\frac{n^2-1}{(R_\ast^0)^3}R_\ast^1a^0_n(t), \end{array}\right. \end{align} (3.52)
    \begin{align} &\left\{\begin{array}{l} -\frac{\partial^2 F^1_n}{\partial r^2}-\frac{1}{r}\frac{\partial F^1_n}{\partial r}+\frac{n^2}{r^2}F^1_n = \mu\frac{\partial \sigma^0_\ast}{\partial r}\frac{\partial F^0_n}{\partial r}+\mu\frac{\partial B_n^0}{\partial r}\frac{\partial p^0_\ast}{\partial r}-\mu\frac{\partial B_n^0}{\partial t}+\mu B_n^1, \\ \frac{\partial F^1_n}{\partial r}(r_0,t) = 0,\; F^1_n(R_\ast^0,t) = \gamma\frac{n^2-1}{(R_\ast^0)^2}b^1_n(t)-\frac{\partial F^0_n}{\partial r}(R_\ast^0,t)R_\ast^1-2\gamma\frac{n^2-1}{(R_\ast^0)^3}R_\ast^1b^0_n(t), \end{array}\right. \end{align} (3.53)
    \begin{align} &\frac{\mathrm{d}a_n^1(t)}{\mathrm{d}t} = -\frac{\partial^2 p_\ast^0}{\partial r^2}(R_\ast^0) a_n^1(t)-\frac{\partial^3 p_\ast^0}{\partial r^3}(R_\ast^0)R_\ast^1a_n^0(t) \\ &\qquad\qquad-\frac{\partial^2 p_\ast^1}{\partial r^2}(R_\ast^0) a_n^0(t)-\frac{\partial^2 E_n^0}{\partial r^2}(R_\ast^0,t)R_\ast^1-\frac{\partial E_n^1}{\partial r}(R_\ast^0,t), \end{align} (3.54)
    \begin{align} &\frac{\mathrm{d}b_n^1(t)}{\mathrm{d}t} = -\frac{\partial^2 p_\ast^0}{\partial r^2}(R_\ast^0) b_n^1(t)-\frac{\partial^3 p_\ast^0}{\partial r^3}(R_\ast^0)R_\ast^1b_n^0(t) \\ &\qquad\qquad-\frac{\partial^2 p_\ast^1}{\partial r^2}(R_\ast^0) b_n^0(t)-\frac{\partial^2 F_n^0}{\partial r^2}(R_\ast^0,t)R_\ast^1-\frac{\partial F_n^1}{\partial r}(R_\ast^0,t). \end{align} (3.55)

    To obtain the asymptotic behavior of a_n^1(t) as \infty , by (3.54) and the boundedness of the modified Bessel functions I_n(r) and K_n(r) on [r_0, R^0_\ast] , it suffices to analyze \frac{\partial E^1_n}{\partial r}(R^0_\ast, t) . For this purpose, in view of (3.52), we first compute A^1_n(r, t) . Solving (3.50) yields

    \begin{align} A_n^1(r,t) = \frac{\bar{\sigma}h_n(r_0,R^0_\ast)}{R^0_\ast} \frac{I_n(r_0)K_n(r)-K_n(r_0)I_n(r)}{I_0(r_0)K_1(R^0_\ast)+I_1(R^0_\ast)K_0(r_0)} V_n(r_0,R^0_\ast,R^1_\ast,a^0_n(t),a^1_n(t)) \end{align} (3.56)

    with

    \begin{align*} V_n(r_0,R^0_\ast,R^1_\ast,a^0_n(t),a^1_n(t)) = &a_n^0(t)R^1_\ast h_n(r_0,R^0_\ast)g_n(r_0,R^0_\ast)-a_n^1(t) \\ &+a_n^0(t)R^1_\ast\frac{I_0(R^0_\ast)K_0(r_0)-I_0(r_0)K_0(R^0_\ast)} {I_0(r_0)K_1(R^0_\ast)+I_1(R^0_\ast)K_0(r_0)}, \end{align*}

    where we have employed (2.8)–(2.10), (3.19), (3.36) and (3.37). Furthermore,

    \begin{align} \frac{\partial A_n^1}{\partial r}(r,t) = \frac{\bar{\sigma}h_n(r_0,R^0_\ast)V_n(r_0,R^0_\ast,R^1_\ast,a^0_n(t),a^1_n(t))} {R^0_\ast[I_0(r_0)K_1(R^0_\ast)+I_1(R^0_\ast)K_0(r_0)]} \frac1{h_n(r_0,r)}. \end{align} (3.57)

    Next, being similar to the computation of E^0_n , we set \eta^1_n = E^1_n+\mu A^1_n , then we derive from (3.50), (3.52) and (3.56) that

    \begin{align} &\left\{\begin{array}{l} -\frac{\partial^2 \eta^1_n}{\partial r^2}-\frac{1}{r}\frac{\partial \eta^1_n}{\partial r}+\frac{n^2}{r^2}\eta^1_n = \mu\frac{\partial \sigma^0_\ast}{\partial r}\frac{\partial E^0_n}{\partial r}+\mu\frac{\partial A^0_n}{\partial r}\frac{\partial p^0_\ast}{\partial r}-\mu\frac{\partial A^0_n}{\partial t}, \\ \frac{\partial \eta^1_n}{\partial r}(r_0,t) = \mu\frac{\partial A^1_n}{\partial r}(r_0,t),\quad \eta^1_n(R^0_\ast,t) = E^1_n(R^0_\ast,t)+\mu A^1_n(R^0_\ast,t). \end{array}\right. \end{align} (3.58)

    For brevity, we introduce the differential operator L_n = -\partial_{rr}-\frac1r\partial_r+\frac{n^2}{r^2} and write \eta^1_n = u^{(1)}_n+u^{(2)}_n+u^{(3)}_n+u^{(4)}_n , where u^{(1)}_n , u^{(2)}_n , u^{(3)}_n and u^{(4)}_n solve the following problems, respectively:

    \begin{align} &\left\{\begin{array}{l} L_nu^{(1)}_n = \mu\frac{\partial\sigma^0_\ast}{\partial r}\frac{\partial E^0_n}{\partial r}, \\ \frac{\partial u^{(1)}_n}{\partial r}(r_0,t) = 0, \quad u^{(1)}_n(R^0_\ast,t) = 0; \end{array}\right. \end{align} (3.59)
    \begin{align} &\left\{\begin{array}{l} L_nu^{(2)}_n = \mu\frac{\partial A^0_n}{\partial r}\frac{\partial p^0_\ast}{\partial r}, \\ \frac{\partial u^{(2)}_n}{\partial r}(r_0,t) = 0, \quad u^{(2)}_n(R^0_\ast,t) = 0; \end{array}\right. \end{align} (3.60)
    \begin{align} &\left\{\begin{array}{l} L_nu^{(3)}_n = -\mu\frac{\partial A^0_n}{\partial t}, \\ \frac{\partial u^{(3)}_n}{\partial r}(r_0,t) = 0, \quad u^{(3)}_n(R^0_\ast,t) = 0; \end{array}\right. \end{align} (3.61)
    \begin{align} &\left\{\begin{array}{l} L_nu^{(4)}_n = 0, \\ \frac{\partial u^{(4)}_n}{\partial r}(r_0,t) = \mu\frac{\partial A_n^1}{\partial r}(r_0,t), \quad u^{(4)}_n(R^0_\ast,t) = E^1_n(R^0_\ast,t)+\mu A^1_n(R^0_\ast,t). \end{array}\right. \end{align} (3.62)

    Let us first estimate u^{(1)}_n . By (3.18), (3.41) and (3.59), we have

    \begin{align} L_nu^{(1)}_n = &\mu\bar{\sigma}a_n^0(t)\frac{K_1(R^0_\ast)I_1(r)-I_1(R^0_\ast)K_1(r)}{I_0(r_0)K_1(R^0_\ast)+I_1(R^0_\ast)K_0(r_0)}\bigg\{n(n^2-1)\gamma(R^0_\ast)^{n-2}\frac{r^{n-1}-r_0^{2n}r^{-n-1}}{(R^0_\ast)^{2n}+r_0^{2n}} \\ &\; \; \; \; \; \; \; \; \; \; \; +\frac{\mu\bar{\sigma}h_n(r_0,R^0_\ast)}{R^0_\ast[I_0(r_0)K_1(R^0_\ast)+I_1(R^0_\ast)K_0(r_0)]} \bigg[\frac1{h_n(r_0,r)}+r_0^n\frac{r^{n-1}+(R^0_\ast)^{2n}r^{-n-1}}{(R^0_\ast)^{2n}+r_0^{2n}} \\ &\; \; \; \; \; \; \; \; \; \; \; +n(R^0_\ast)^n\frac{r^{n-1}-r_0^{2n}r^{-n-1}}{(R^0_\ast)^{2n}+r_0^{2n}}(I_n(R^0_\ast)K_n(r_0)-I_n(r_0)K_n(R^0_\ast))\bigg]\bigg\}. \end{align} (3.63)

    Based on the properties of the modified Bessel functions I_n(r) and K_n(r) , the righthand side of (3.63) is less than Q(n)a^0_n(t) when r_0 \leq r < R^0_\ast . Here, Q(n) denotes a polynomial function of n . Similar estimates can be established for u^{(2)}_n and u^{(3)}_n by (3.60) and (3.61).

    Lemma 3.2. Consider the elliptic problem

    \begin{align} &-\Delta\omega(x,t)+\frac{n^2}{|x|^2}\omega(x,t) = b(x,t), \qquad x\in\Omega_R, \end{align} (3.64)
    \begin{align} &\partial_{\vec n}\omega\big|_{|x| = r_0} = 0, \quad \omega\big|_{|x| = R} = 0, \end{align} (3.65)

    where \Omega_R = \big\{x\in \mathbb{R}^2: r_0 < |x| < R\big\} . If b(x, t) = b(|x|, t) and b(\cdot, t)\in L^2(\Omega_R) , then the problem (3.64) and (3.65) admits a unique solution \omega in H^2(\Omega_R) with estimates

    \begin{align} \|\omega(\cdot,t)\|_{H^2(\Omega_R)}\leq C\bigg(\int^R_{r_0}|b(r,t)|^2rdr\bigg)^{1/2}; \end{align} (3.66)
    \begin{align} \big\|\partial_{\vec\nu}\omega(\cdot,t)\big\|_{L^\infty(\partial B_R)}\leq C\bigg(\int^R_{r_0}|b(r,t)|^2rdr\bigg)^{1/2}, \end{align} (3.67)

    where the constant C in (3.66) and (3.67) is independent of n .

    The lemma can be proven by combining the proofs of [16, Lemma 4.6] and [17, Lemma 3.2]. The details are omitted here.

    Lemma 3.2 ensures the existence and uniqueness of u^{(k)}_n in H^2(\Omega{_\ast}) for k = 1, 2, 3 . Furthermore, there holds

    \begin{equation} \bigg|\frac{\partial u^{(1)}_n}{\partial r}(R^0_\ast,t)\bigg|+\bigg|\frac{\partial u^{(2)}_n}{\partial r}(R^0_\ast,t)\bigg|+\bigg|\frac{\partial u^{(3)}_n}{\partial r}(R^0_\ast,t)\bigg|\leq Ce^{-\delta t}. \end{equation} (3.68)

    Obviously, the solution u^{(4)}_n to the problem (3.62) has the form:

    \begin{equation} u^{(4)}_n(r,t) = C_5(t)r^n+C_6(t)r^{-n}, \end{equation} (3.69)

    where C_5(t) and C_6(t) are determined by the boundary conditions in (3.62). Using (2.10), (3.41), (3.56), (3.57) and the boundary conditions in (3.52), we get

    \begin{align} C_5(t) = &\frac{a_n^1(t)}{(R^0_\ast)^{2n}+r_0^{2n}}\bigg\{\frac{\mu\bar{\sigma}h_n(r_0,R^0_\ast)}{R^0_\ast[I_0(r_0)K_1(R^0_\ast)+I_1(R^0_\ast)K_0(r_0)]}\bigg[\frac{r_0^n}{n}+(R^0_\ast)^n(K_n(r_0)I_n(R^0_\ast) \\ &\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; -I_n(r_0)K_n(R^0_\ast))\bigg]+\gamma (n^2-1)(R^0_\ast)^{n-2}\bigg\}+H_1(r_0,R^0_\ast,R^1_\ast)a_n^0(t), \end{align} (3.70)
    \begin{align} C_6(t) = &\frac{a_n^1(t)r_0^n(R^0_\ast)^n}{(R^0_\ast)^{2n}+r_0^{2n}}\bigg\{\frac{\mu\bar{\sigma}h_n(r_0,R^0_\ast)}{R^0_\ast[I_0(r_0)K_1(R^0_\ast)+I_1(R^0_\ast)K_0(r_0)]}\bigg[-\frac{(R^0_\ast)^n}{n}+r_0^n(K_n(r_0)I_n(R^0_\ast) \\ &\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; -I_n(r_0)K_n(R^0_\ast))\bigg]+\frac{\gamma (n^2-1)r_0^n}{(R^0_\ast)^2}\bigg\}+H_2(r_0,R^0_\ast,R^1_\ast)a_n^0(t), \end{align} (3.71)

    where H_1 , H_2 are functions of r_0 , R^0_\ast and R^1_\ast .

    Now, since

    \begin{align*} E^1_n(r,t) = \eta^1_n-\mu A^1_n = u^{(1)}_n+u^{(2)}_n+u^{(3)}_n+u^{(4)}_n-\mu A^1_n, \end{align*}

    we derive

    \begin{align} &\frac{\partial E^1_n}{\partial r}(R^0_\ast,t) = \frac{\partial u^{(1)}_n}{\partial r}(R^0_\ast,t)+\frac{\partial u^{(2)}_n}{\partial r}(R^0_\ast,t)+\frac{\partial u^{(3)}_n}{\partial r}(R^0_\ast,t)+\frac{\partial u^{(4)}_n}{\partial r}(R^0_\ast,t)-\mu\frac{\partial A^1_n}{\partial r}(R^0_\ast,t). \end{align} (3.72)

    By (3.33), (3.57) and (3.69)–(3.72), we obtain from (3.54) that

    \begin{align*} \frac{\mathrm{d}a^1_n(t)}{\mathrm{d}t} = &-\frac{\partial^2 p^0_\ast}{\partial r^2}(R^0_\ast)a^1_n(t)-\frac{\partial^3 p^0_\ast}{\partial r^3}(R^0_\ast)R^1_\ast a^0_n(t)-\frac{\partial^2 p^1_\ast}{\partial r^2}(R^0_\ast)a^0_n(t)-\frac{\partial^2 E^0_n}{\partial r^2}(R^0_\ast,t)R^1_\ast-\frac{\partial E^1_n}{\partial r}(R^0_\ast,t) \\ = &a_n^1(t)\bigg\{\frac{\mu\bar{\sigma}h_n(r_0,R^0_\ast)} {I_0(r_0)K_1(R^0_\ast)+I_1(R^0_\ast)K_0(r_0)} \bigg[\frac{n}{(R^0_\ast)^2}\frac{(R^0_\ast)^{2n}-r_0^{2n}}{(R^0_\ast)^{2n} +r_0^{2n}}(I_n(r_0)K_n(R^0_\ast)-I_n(R^0_\ast)K_n(r_0)) \\ &\; \; \; \; \; \; \; \; \; \; \; +\frac{2 r_0(I_1(r_0)K_1(R^0_\ast)-I_1(R^0_\ast)K_1(r_0))} {(R^0_\ast)^2-r_0^2}\frac1{h_n(r_0,R^0_\ast)}-\frac{2r_0^n(R^0_\ast)^{n-2}}{(R^0_\ast)^{2n}+r_0^{2n}}\bigg] \\ &\; \; \; \; \; \; \; \; \; \; \; -\frac{\gamma n(n^2-1)}{(R^0_\ast)^3}\frac{(R^0_\ast)^{2n}-r_0^{2n}}{(R^0_\ast)^{2n}+r_0^{2n}}\bigg\}+\tilde{H}(n,r_0,R^0_\ast,R^1_\ast)a_n^0(t) \\ &-\frac{\partial u^{(1)}_n}{\partial r}(R^0_\ast,t)-\frac{\partial u^{(2)}_n}{\partial r}(R^0_\ast,t)-\frac{\partial u^{(3)}_n}{\partial r}(R^0_\ast,t) \\ = &a_n^1(t)U_n(r_0,R^0_\ast)-\frac{\partial u^{(1)}_n}{\partial r}(R^0_\ast,t)-\frac{\partial u^{(2)}_n}{\partial r}(R^0_\ast,t)-\frac{\partial u^{(3)}_n}{\partial r}(R^0_\ast,t) +\tilde{H}(n,r_0,R^0_\ast,R^1_\ast)a_n^0(t), \end{align*}

    where \tilde{H} is a known function of n , r_0 , R^0_\ast , R^1_\ast and satisfies

    \begin{equation} |\tilde{H}(n,r_0,R^0_\ast,R^1_\ast)|\leq C. \end{equation} (3.73)

    Thus, using Lemma 3.1, (3.68) and (3.73) gives

    \begin{align} \bigg|&\frac{\mathrm{d}a_n^1(t)}{\mathrm{d}t}-a_n^1(t)U_n(r_0,R^0_\ast)\bigg| \\ \leq& |\tilde{H}(n,r_0,R^0_\ast,R^1_\ast)a_n^0(t)|+\bigg|\frac{\partial u^{(1)}_n}{\partial r}(R^0_\ast,t)\bigg|+\bigg|\frac{\partial u^{(2)}_n}{\partial r}(R^0_\ast,t)\bigg|+\bigg|\frac{\partial u^{(3)}_n}{\partial r}(R^0_\ast,t)\bigg| \\ \leq& Ce^{-\delta t}. \end{align} (3.74)

    In addition, for n\geq0 , Lemma 3.1 implies

    \begin{equation*} -U_n(r_0,R^0_\ast) > \delta > 0. \end{equation*}

    Therefore, applying [16, Lemma 4.7] to (3.74) yields

    \begin{align} |a^1_n(t)|\leq Ce^{-\delta t}, \qquad \quad t > 0, \end{align} (3.75)

    i.e., a^1_n(t) decays exponentially as t\to\infty . Noticing that b_n(t) and a_n(t) have the same asymptotic behavior, we also have

    \begin{align} |b^0_n(t)|+|b^1_n(t)|\leq Ce^{-\delta t},\qquad t > 0. \end{align} (3.76)

    Proof of Theorem 1.2 The desired result (1.14) follows from Lemma 3.1, (3.75) and (3.76). The proof is complete.

    Remark 3.2. The results on tumor cord without time delays in [14] show that the radially symmetric stationary solution is asymptotically stable under nonradially symmetric perturbations. Here, our Theorem 1.2 says that such asymptotic stability does not be affected by small time delay.

    In this paper, we have investigated the effects of a time delay in cell proliferation on the growth of tumor cords, where the domain is a bounded subset in \mathbb{R}^2 and its boundary consists of two disjoint closed curves, one fixed and the other moving and a priori unknown. The existence, uniqueness and linear stability of the radially symmetric stationary solution were studied.

    Here are some interesting findings. 1) Adding the time delay would not change the stability of the radially symmetric stationary solution when compared with the same system without delay [14], but adding the time delay would result in a larger stationary tumor. The bigger the tumor growth intensity \mu is, the greater impact that time delay has on the size of the stationary tumor. 2) By the result of [16], we know that for tumor spheroids with the same time delay, there exists a threshold \mu_\ast > 0 for the tumor aggressiveness constant \mu such that only for \mu < \mu_\ast , the radially symmetric stationary solution is linearly stable under non-radial perturbations. For tumor cords, however, from Theorem 1.2 we saw that the radially symmetric stationary solution is always linearly stable, regardless of the value of \mu . It showed that there is an essential difference between tumor cords and tumor spheroids with the same time delay.

    We think that the linear stability analysis for the full system without quasi-steady state simplification, i.e., c > 0 , may be very challenging, which we expect to solve in future work.

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

    The authors would like to thank the reviewers for their valuable suggestions. This work was partly supported by the National Natural Science Foundation of China (No. 12161045 and No. 12261047), Jiangxi Provincial Natural Science Foundation (No. 20224BCD41001 and No. 20232BAB201010), the Science and Technology Planning Project from Educational Commission of Jiangxi Province, China (No. GJJ2200319), the Scientific Research Project of Hunan Provincial Education Department, China (No. 22B0725, No. 23B0670 and No. 23C0234), and the Research Initiation Project of Hengyang Normal University, China (No. 2022QD01).

    The authors declare there is no conflict of interest.



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