The purpose of this work is to carry out investigations of a generalized two-phase model for porous media flow. The momentum balance equations account for fluid-rock resistance forces as well as fluid-fluid drag force effects, in addition, to internal viscosity through a Brinkmann type viscous term. We carry out detailed investigations of a one-dimensional version of the general model. Various a priori estimates are derived that give rise to an existence result. More precisely, we rely on the energy method and use compressibility in combination with the structure of the viscous term to obtain $ H^1 $-estimates as well upper and lower uniform bounds of mass variables. These a priori estimates imply existence of solutions in a suitable functional space for a global time $ T>0 $. We also derive discrete schemes both for the incompressible and compressible case to explore the role of the viscosity term (Brinkmann type) as well as the incompressible versus the compressible case. We demonstrate similarities and differences between a formulation that is based, respectively, on interstitial velocity and Darcy velocity in the viscous term. The investigations may suggest that interstitial velocity seems more natural to use in the formulation of momentum balance than Darcy velocity.
Citation: Yangyang Qiao, Huanyao Wen, Steinar Evje. Compressible and viscous two-phase flow in porous media based on mixture theory formulation[J]. Networks and Heterogeneous Media, 2019, 14(3): 489-536. doi: 10.3934/nhm.2019020
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The purpose of this work is to carry out investigations of a generalized two-phase model for porous media flow. The momentum balance equations account for fluid-rock resistance forces as well as fluid-fluid drag force effects, in addition, to internal viscosity through a Brinkmann type viscous term. We carry out detailed investigations of a one-dimensional version of the general model. Various a priori estimates are derived that give rise to an existence result. More precisely, we rely on the energy method and use compressibility in combination with the structure of the viscous term to obtain $ H^1 $-estimates as well upper and lower uniform bounds of mass variables. These a priori estimates imply existence of solutions in a suitable functional space for a global time $ T>0 $. We also derive discrete schemes both for the incompressible and compressible case to explore the role of the viscosity term (Brinkmann type) as well as the incompressible versus the compressible case. We demonstrate similarities and differences between a formulation that is based, respectively, on interstitial velocity and Darcy velocity in the viscous term. The investigations may suggest that interstitial velocity seems more natural to use in the formulation of momentum balance than Darcy velocity.
Chemotaxis is the property of cells to move in an oriented manner in response to an increasing concentration of chemo-attractant or decreasing concentration of chemo-repellent, where the former is referred to as attractive chemotaxis and the later to repulsive chemotaxis. To begin with, it is important to study the quasilinear Keller-Segel system as follows
$ {ut=∇⋅(D(u)∇u)−χ∇⋅(ϕ(u)∇v),x∈Ω,t>0,τvt=Δv−αv+βu,x∈Ω,t>0, $ | (1.1) |
subject to homogeneous Neumann boundary conditions, where the functions $ D(u) $ and $ \phi(u) $ denote the strength of diffusion and chemoattractant, respectively, and the function $ u = u(x, t) $ idealizes the density of cell, $ v = x(x, t) $ represents the concentration of the chemoattractant. Here the attractive (repulsive) chemotaxis corresponds to $ \chi > 0 $ ($ \chi < 0 $), and $ |\chi|\in\mathbb{R}\backslash\{0\} $ measures the strength of chemotactic response. The parameters $ \tau\in\{0, 1\} $, and $ \alpha, \beta > 0 $ denote the production and degradation rates of the chemical. The above system describes the chemotactic interaction between cells and one chemical signal (either attractive or repulsive), and it has been investigated quite extensively on the existence of global bounded solutions or the occurrence of blow-up in finite time in the past four decades. In particular, the system (1.1) is the prototypical Keller-Segel model [1] when $ D(u) = 1, \phi(u) = u $. In the case $ \tau = 1 $, there are many works to show that the solution is bounded [2,3,4,5], and blow-up in finite time [6,7,8,9,10,11]. If the cell's movement is much slower than the chemical signal diffusing, the second equation of (1.1) is reduced to $ 0 = \Delta v-M+u $, where $ M: = \frac{1}{|\Omega|}\int_{\Omega}u(x, t)dx $ and the simplified system has many significant results [12,13,14,15].
For further information concerning nonlinear signal production, when the chemical signal function is denoted by $ g(u) $, authors derived for more general nonlinear diffusive system as follows
$ {ut=∇⋅(D(u)∇u)−∇⋅(ϕ(u)∇v),x∈Ω,t>0,0=Δv−M+g(u),x∈Ω,t>0, $ | (1.2) |
where $ M: = \frac{1}{|\Omega|}\int_{\Omega}g(u(x, t))dx $. Recently, when $ D(u) = u^{-p}, \phi(u) = u $ and $ g(u) = u^{l} $, it has been shown that all solutions are global and uniformly bounded if $ p+l < \frac{2}{n} $, whereas $ p+l > \frac{2}{n} $ implies that the solution blows up in finite time [16]. What's more, there are many significant works [17,18,19] associated with this system.
Subsequently, the attraction-repulsion system has been introduced in ([20,21]) as follows
$ {ut=Δu−χ∇⋅(u∇v)+ξ∇⋅(u∇w),x∈Ω,t>0,τ1vt=Δv+αu−βv,x∈Ω,t>0,τ2wt=Δw+δu−γw,x∈Ω,t>0, $ | (1.3) |
subject to homogeneous Neumann boundary conditions, where $ \chi, \xi, \alpha, \beta, \delta, \gamma > 0 $ are constants, and the functions $ u(x, t), v(x, t) $ and $ w(x, t) $ denote the cell density, the concentration of the chemoattractant and chemorepellent, respectively. The above attraction-repulsion chemotaxis system has been studied actively in recent years, and there are many significant works to be shown as follows.
For example, if $ \tau_{1} = \tau_{2} = 0 $, Perthame [22] investigated a hyperbolic Keller-Segel system with attraction and repulsion when $ n = 1 $. Subsequently, Tao and Wang [23] proved that the solution of (1.3) is globally bounded provided $ \xi\gamma-\chi\alpha > 0 $ when $ n\geq2 $, and the solution would blow up in finite time provided $ \xi\gamma-\chi\alpha < 0, \alpha = \beta $ when $ n = 2 $. Then, there is a blow-up solution when $ \chi\alpha-\xi\gamma > 0, \delta\geq\beta $ or $ \chi\alpha\delta-\xi\gamma\beta > 0, \delta < \beta $ for $ n = 2 $ [24]. Moreover, Viglialoro [25] studied the explicit lower bound of blow-up time when $ n = 2 $. In another hand, if $ \tau_{1} = 1, \tau_{2} = 0 $, Jin and Wang [26] showed that the solution is bounded when $ n = 2 $ with $ \xi\gamma-\chi\alpha\geq0 $, and Zhong et al. [27] obtained the global existence of weak solution when $ \xi\gamma-\chi\alpha\geq0 $ for $ n = 3 $. Furthermore, if $ \tau_{1} = \tau_{2} = 1 $, Liu and Wang [28] obtained the global existence of solutions, and Jin et al. [29,30,31] also showed a uniform-in-time bound for solutions. In addition, there are plenty of available results of the attraction-repulsion system with logistic terms [32,33,34,35,36,37,38,39,40], and for further information concerning (1.3) based on the nonlinear signal production, it was used to model the aggregation patterns formed by some bacterial chemotaxis in [41,42,43].
We turn our eyes into a multi-dimensional attraction-repulsion system
$ {ut=Δu−χ∇⋅(ϕ(u)∇v)+ξ∇⋅(ψ(u)∇w),x∈Ω,t>0,τ1vt=Δv−μ1(t)+f(u),x∈Ω,t>0,τ2wt=Δw−μ2(t)+g(u),x∈Ω,t>0, $ | (1.4) |
where $ \Omega\in\mathbb{R}^{n}(n\geq2) $ is a bounded domain with smooth boundary, $ \mu_{1}(t) = \frac{1}{|\Omega|}\int_{\Omega}f(u)dx, \mu_{2}(t) = \frac{1}{|\Omega|}\int_{\Omega}g(u)dx $ and $ \tau_{1}, \tau_{2}\in\{0, 1\} $. Later on, the system (1.4) has attracted great attention of many mathematicians. In particular, when $ \phi(u) = \psi(u) = u, f(u) = u^{k} $ and $ g(u) = u^{l} $, Liu and Li [44] proved that all solutions are bounded if $ k < \frac{2}{n} $, while blow-up occurs for $ k > l $ and $ k > \frac{2}{n} $ in the case $ \tau_{1} = \tau_{2} = 0 $.
Inspired by the above literature, we are devoted to deal with the quasilinear attraction-repulsion chemotaxis system
$ {ut=∇⋅(D(u)∇u)−χ∇⋅(u∇v)+ξ∇⋅(u∇w),x∈Ω,t>0,0=Δv−μ1(t)+f1(u),x∈Ω,t>0,0=Δw−μ2(t)+f2(u),x∈Ω,t>0,∂u∂ν=∂v∂ν=∂w∂ν=0,x∈∂Ω,t>0,u(x,0)=u0(x),x∈Ω, $ | (1.5) |
in a bounded domain $ \Omega\subset\mathbb{R}^n, n\geq2 $ with smooth boundary, where $ \frac{\partial}{\partial\nu} $ denotes outward normal derivatives on $ \partial\Omega $. The function $ u(x, t) $ denotes the cell density, $ v(x, t) $ represents the concentration of an attractive signal (chemo-attractant), and $ w(x, t) $ is the concentration of a repulsive signal (chemo-repellent). The parameters satisfy $ \chi, \xi\geq0 $, which denote the strength of the attraction and repulsion, respectively. Here $ \mu_{1}(t) = \frac{1}{|\Omega|}\int_{\Omega}f_{1}(u(x, t))dx $, $ \mu_{2}(t) = \frac{1}{|\Omega|}\int_{\Omega}f_{2}(u(x, t))dx $, and $ f_{1}, f_{2} $ are nonnegative Hölder continuous functions.
In the end, we propose the following assumptions on $ D, f_{1}, f_{2} $ and $ u_{0} $ for the system $ (1.5) $.
$ (I_{1}) \; {\rm\ The \ nonlinear \ diffusivity} \ D \ {\rm is \ positive \ function\ satisfying} $
$ D∈C2([0,∞)). $ | (1.6) |
$(I_{2}) \; \ {\rm The\ function\ } f_{i} \ {\rm is\ nonnegative\ and\ nondecreasing\ and \ satisfies } $
$ fi∈⋃θ∈(0,1)Cθloc([0,∞))∩C1((0,∞)) $ | (1.7) |
with $ i\in\{1, 2\} $.
$ (I_{3}) \; {\rm \ The\ initial\ datum } $
$ u0∈⋃θ∈(0,1)Cθ(¯Ω) is nonnegative and radially decreasing,∂u0∂ν=0 on ∂Ω. $ | (1.8) |
The goal of the article is twofold. On the one hand, we need to find out the mutual effect of the nonlinear diffusivity $ D(u) $ and the nonlinear signal production $ f_{i}(u) (i = 1, 2) $. On the other hand, we need to make a substantial step towards the dynamic of blowing up in finite time. Hence, we draw our main results concerning (1.5) read as follows.
Theorem 1.1. Let $ n\geq2 $, $ R > 0 $ and $ \Omega = B_{R}(0)\subset\mathbb{R}^n $ be a ball, and suppose that the function $ D $ satisfies (1.6) and $ f_{1}, f_{2} $ are assumed to fulfill (1.7) as well as
$ D(u)≤d(1+u)m−1, f1(u)≥k1(1+u)γ1, f2(u)≤k2(1+u)γ2 for all u≥0, $ |
with $ m\in\mathbb{R} $, $ k_{1}, k_{2}, \gamma_{1}, \gamma_{2}, d > 0 $ and
$ γ1>γ2 and 1+γ1−m>2n. $ | (1.9) |
For any $ M > 0 $ there exist $ \varepsilon = \varepsilon(\gamma_{1}, M, R)\in(0, M) $ and $ r^{\ast} = r^{\ast}(\gamma_{1}, M, R)\in(0, R) $ such that if $ u_{0} $ satisfies (1.8) with
$ ∫Ωu0=M and ∫Br∗(0)u0≥M−ε, $ |
then the corresponding solution of the system (1.5) blows up in finite time.
Theorem 1.2. Let $ n\geq2 $, $ \Omega\subset\mathbb{R}^n $ be a smooth bounded domain, and suppose that the function $ D $ satisfies (1.6) and $ f_{1}, f_{2} $ are assumed to fulfill (1.7) as well as
$ D(u)≥d(1+u)m−1, f1(u)≤k1(1+u)γ1, f2(u)=k2(1+u)γ2 for all u≥0, $ |
with $ m\in\mathbb{R} $, $ k_{1}, k_{2}, \gamma_{1}, \gamma_{2}, d > 0 $ and
$ γ2<1+γ1<2n+m. $ | (1.10) |
Then for each $ u_{0}\in\bigcup_{\theta\in(0, 1)}C^{\theta}\big(\overline\Omega\big) $, $ u_{0}\geq0 $ with $ \frac{\partial u_{0}}{\partial\nu} = 0 $ on $ \partial\Omega $, and the system (1.5) admits a unique global classical solution $ (u, v, w) $ with
$ u,v,w∈C2,1(¯Ω×(0,∞))∩C0(¯Ω×[0,∞)). $ |
Furthermore, $ u, v $ and $ w $ are all non-negative and bounded.
The structure of this paper reads as follows: In section $ 2 $, we will show the local-in-time existence of a classical solution to the system $ (1.5) $ and some lemmas that we will use later. In section $ 3 $, we will prove Theorem $ 1.1 $ by establishing a superlinear differential inequality. In section $ 4 $, we will solve the boundedness of $ u $ in $ L^{\infty} $ and prove Theorem $ 1.2 $.
Firstly, we state one result concerning local-in-time existence of a classical solution to the system $ (1.5) $. Then, we denote some new variables to transfer the original equations in $ (1.5) $ to a new system according to the ideas in [19,20,21,22,23,24,25]. In addition, in order to prove the main result, we will state some lemmas which will be needed later.
Lemma 2.1. Let $ \Omega\subset\mathbb{R}^n $ with $ n\geq2 $ be a bounded domain with smooth boundary. Assume that $ D $ fulfills (1.6), $ f_1, f_2 $ satisfy (1.7) and $ u_{0}\in\bigcup_{\theta\in(0, 1)}C^{\theta}\big(\overline\Omega\big) $ with $ \frac{\partial u_{0}}{\partial\nu} = 0 $ on $ \partial\Omega $ as well as $ u_{0}\geq0 $, then there exist $ T_{max}\in(0, \infty] $ and a classical solution $ (u, v, w) $ to (1.5) uniquely determined by
$ {u∈C0(¯Ω×[0,Tmax))∩C2,1(¯Ω×(0,Tmax)),v∈∩q>nL∞((0,Tmax);W1,q(Ω))∩C2,0(¯Ω×(0,Tmax)),w∈∩q>nL∞((0,Tmax);W1,q(Ω))∩C2,0(¯Ω×(0,Tmax)). $ |
In addition, the function $ u\geq0 $ in $ \Omega\times (0, T_{max}) $ and if $ T_{max} < \infty $ then
$ limt↗Tmaxsup‖u(⋅,t)‖L∞(Ω)=∞. $ | (2.1) |
Moreover,
$ ∫Ωv(⋅,t)=0,∫Ωw(⋅,t)=0 for all t∈(0,Tmax). $ | (2.2) |
Finally, the solution $ (u, v, w) $ is radially symmetric with respect to $ |x| $ if $ u_{0} $ satisfies (1.8).
Proof. The proof of this lemma needs to be divided into four steps. Firstly, the method to solve the local time existence of the classical solution to the problem (1.5) is based on a standard fixed point theorem. Next, we will use the standard extension theorem to obtain (2.1). Then, we are going to use integration by parts to deduce (2.2). Finally, we would use the comparison principle to conclude that the solution is radially symmetric. For the details, we refer to [45,46,47,48].
For the convenience of analysis and in order to prove Theorem 1.1, we set $ h = \chi v-\xi w $, then the system (1.5) is rewritten as
$ {ut=∇⋅(D(u)∇u)−∇⋅(u∇h),x∈Ω,t>0,0=Δh−μ(t)+f(u),x∈Ω,t>0,∂u∂ν=∂h∂ν=0,x∈∂Ω,t>0,u(x,0)=u0(x),x∈Ω, $ | (2.3) |
where $ \mu(t) = \chi\mu_{1}(t)-\xi\mu_{2}(t) = \frac{1}{|\Omega|}\int_{\Omega}f(u(x, t))dx $ and $ f(u) = \chi f_{1}(u)-\xi f_{2}(u) $.
For the same reason, we will convert the system $ (2.3) $ into a scalar equation. Let us assume $ \Omega = B_{R}(0) $ with some $ R > 0 $ is a ball and the initial data $ u_{0} = u_{0}(r) $ with $ r = |x|\in[0, R] $ satisfies $ (1.8) $. In the radial framework, the system $ (2.3) $ can be transformed into the following form
$ {rn−1ut=(rn−1D(u)ur)r−(rn−1uhr)r,r∈(0,R),t>0,0=(rn−1hr)r−rn−1μ(t)+rn−1f(u),r∈(0,R),t>0,ur=hr=0, r=R,t>0,u(r,0)=u0(r),r∈(0,R). $ | (2.4) |
Lemma 2.2. Let us introduce the function
$ U(s,t)=n∫s1n0ρn−1u(ρ,t)dρ,s=rn∈[0,Rn], t∈(0,Tmax), $ |
then
$ Us(t)=u(s1n,t), Uss(t)=1ns1n−1ur(s1n,t), $ | (2.5) |
and
$ Ut(s,t)=n2s2−2nD(Us)Uss−sμ(t)Us+Us⋅∫s0f(Us(σ,t))dσ. $ | (2.6) |
Proof. Firstly, integrating the second equation of $ (2.4) $ over $ (0, r) $, we have
$ rn−1hr(r,t)=rnnμ(t)−∫r0ρn−1f(u(ρ,t))dρ, $ |
so
$ s1−1nhr(s1n,t)=snμ(t)−1n∫s0f(u(σ1n,t))dσ,∀s∈(0,Rn), t∈(0,Tmax). $ |
Then, a direct calculation yields
$ Us(s,t)=u(s1n,t),∀s∈(0,Rn), t∈(0,Tmax), $ |
and
$ Uss(s,t)=1ns1n−1ur(s1n,t),∀s∈(0,Rn), t∈(0,Tmax), $ |
as well as
$ Ut(s,t)=n∫s1n0ρn−1ut(ρ,t)dρ=n2s2−2nD(Us)Uss−ns1−1nUshr=n2s2−2nD(Us)Uss−sμ(t)Us+Us⋅∫s0f(Us(σ,t))dσ $ |
for all $ s\in(0, R^{n}) $ and $ t\in(0, T_{max}) $.
Furthermore, by a direct calculation and (1.7), we know that the functions $ U $ and $ f $ satisfy the following results
$ {Us(s,t)=u(s1n,t)>0,s∈(0,Rn),t∈(0,Tmax),U(0,t)=0,U(Rn,t)=nωn∫Ωu(⋅,t)=nMωn,t∈[0,Tmax),|f(s)|,f1(s),f2(s)≤C0,0≤s≤A,C0=C0(A)>0, $ | (2.7) |
where $ \omega_{n} = n|B_{1}(0)| $ and $ A $ is a positive constant.
Lemma 2.3. Suppose that (1.7), (1.8) and (2.7) hold, then we have
$ hr(r,t)=1nμ(t)r−r1−n∫r0ρn−1f(u(ρ,t))dρfor all r∈(0,R),t∈(0,Tmax). $ |
In particular,
$ hr(r,t)≤1n(μ(t)+C0)r. $ | (2.8) |
Proof. By integration the second equation in (2.4), we obtain that
$ rn−1hr=μ(t)⋅∫r0ρn−1dρ−∫r0ρn−1f(u(ρ,t))dρ for all r∈(0,R),t∈(0,Tmax). $ |
According to (1.9), we can easily get that $ f(u)\geq0 $ if $ u\geq C^{\ast} = \max\{0, (\frac{k_{2}\xi}{k_{1}\chi})^{\frac{1}{\gamma_{1}-\gamma_{2}}}-1\}, $ and split
$ ∫r0ρn−1f(u(ρ,t))dρ=∫r0χ{u(⋅,t)≥C∗}(ρ)⋅ρn−1f(u(ρ,t))dρ+∫r0χ{u(⋅,t)<C∗}(ρ)⋅ρn−1f(u(ρ,t))dρ. $ |
Combining these we have
$ hr=1nμ(t)r−r1−n∫r0χ{u(⋅,t)≥C∗}(ρ)⋅ρn−1f(u(ρ,t))dρ−r1−n∫r0χ{u(⋅,t)<C∗}(ρ)⋅ρn−1f(u(ρ,t))dρ≤1nμ(t)r−r1−n∫r0χ{u(⋅,t)<C∗}(ρ)⋅ρn−1f(u(ρ,t))dρ≤1nμ(t)r+C0r1−n∫r0χ{u(⋅,t)<C∗}(ρ)⋅ρn−1dρ≤1nμ(t)r+C0r1−n∫r0ρn−1dρ≤1n(μ(t)+C0)r, $ |
so we complete this proof.
To show the existence of a finite-time blow-up solution of (2.4), we need to prove that $ U_{ss} $ is nonpositive by the following lemma. The proof follows the strategy in [48].
Lemma 2.4. Suppose that $ D, f $ and $ u_{0} $ satisfy $ (I_{1}), (I_{2}) $ and $ (I_{3}) $ respectively. Then
$ ur(r,t)≤0 for all r∈(0,R),t∈(0,Tmax). $ | (2.9) |
Moreover,
$ Uss(s,t)≤0 for all r∈(0,R),t∈(0,Tmax). $ | (2.10) |
Proof. Without loss of generality we may assume that $ \nonumber u_{0}\in C^{2} $ $ ([0, \infty)) $ and $ f\in C^{2}([0, \infty)) $. Applying the regularity theory in ([49,50]), we all know that $ u $ and $ u_{r} $ belong to $ C^{0}([0, R]\times[0, T))\cap C^{2, 1}((0, R)\times(0, T)) $ and we fixed $ T\in(0, T_{max}) $. From (2.4), we have for $ r\in(0, R) $ and $ t\in(0, T) $
$ hrr+n−1rhr=μ(t)−f(u), $ | (2.11) |
and from (2.4) we obtain
$ urt=((D(u)ur)r+n−1rD(u)ur+uf(u)−uμ(t)−urhr)r=(D(u)ur)rr+a1(D(u)ur)r+a2urr+bur, $ |
for all $ r\in(0, R) $ and $ t\in(0, T) $, where
$ a1(r,t)=n−1r,a2(r,t)=−hr,b(r,t)=−n−1r2D(u)−μ(t)−hrr+f(u)+uf′(u), $ |
for all $ r\in(0, R) $ and $ t\in(0, T) $. Moreover, we have $ h_{r}\leq\frac{r}{n}(\mu(t)+C_{0}) $ by (2.8) and from (2.11) such that
$ −hrr=n−1rhr−μ(t)+f(u)≤n−1nμ(t)+n−1nC0−μ(t)+f(u)≤f(u)+C0for all r∈(0,R) and t∈(0,T), $ |
then setting $ c_{1}: = \sup_{(r, t)\in(0, R)\times(0, T)}(2f(u)+uf'(u)+C_{0}) $, we obtain
$ b(r,t)≤c1for all r∈(0,R) and t∈(0,T), $ |
and we introduce
$ c2:=sup(r,t)∈(0,R)×(0,T)((D(u))rr+a1(D(u))r)<∞, $ |
and set $ c_{3} = 2(c_{1}+c_{2}+1) $. Since $ u_{r}(r, t) = 0 $ for $ r\in\{0, R\}, t\in(0, T) $ (because $ u $ is radially symmetric) and $ u_{0_{r}}\leq 0 $, the function $ y:[0, R]\times[0, T]\rightarrow \mathbb{R} $, $ (r, t)\mapsto u_{r}(r, t)-\varepsilon e^{c_{3}t} $ belongs to $ C^{0}([0, R]\times[0, T]) $ and fulfills
$ {yt=(D(u)(y+εec3t))rr+a1(D(u)(y+εec3t))r+a2yr+b(y+εec3t)−c3εec3t=(D(u)y)rr+a1(D(u)y)r+a2yr+by+εec3t((D(u))rr+a1(D(u))r+b−c3)≤(D(u)y)rr+a1(D(u)y)r+a2yr+by+εec3t(c1+c2−c3),in (0,R)×(0,T),y<0,on {0,R}×(0,T),y(⋅,0)<0,in (0,R). $ | (2.12) |
By the estimate for $ y(\cdot, 0) $ in $ (2.12) $ and continuity of $ y $, the time $ t_{0}: = \sup\{t\in(0, T):y\leq 0\ { \rm in} \ [0, R]\times (0, T)\} \in (0, T] $ is defined. Suppose that $ t_{0} < T $, then there exists $ r_{0}\in [0, R] $ such that $ y(r_{0}, t_{0}) = 0 $ and $ y(r, t)\leq 0 $ for all $ r\in[0, R] $ and $ t\in[0, t_{0}] $; hence, $ y_{t}(r_{0}, t_{0})\geq 0 $. As $ D\geq 0 $ in $ [0, \infty) $, not only $ y(\cdot, t_{0}) $ but also $ z:(0, R)\rightarrow \mathbb{R}, r\longmapsto D(u(r, t_{0}))y(r, t_{0}) $ attains its maximum $ 0 $ at $ r_{0} $. Since the second equality in $ (2.12) $ asserts $ r_{0}\in(0, R) $, we conclude $ z_{rr}(r_{0})\leq 0, z_{r}(r_{0}) = 0 $ and $ y_{r}(r_{0}, t_{0}) = 0 $. Hence, we could obtain the contradiction
$ 0≤yt(r0,t0)≤zrr(r0)+a1(r0,t0)zr(r0)+a2(r0,t0)yr(r0,t0)+b(r0,t0)y(r0,t0)+εec3t0(c1+c2−c3)≤−c32εec3t0<0, $ |
since we have
$ c1+c2≤c32. $ |
So that $ t_{0} = T, $ implying $ y\leq 0 $ in $ [0, R]\times [0, T] $ and hence $ u_{r}\leq\varepsilon e^{c_{3}t} $ in $ [0, R]\times [0, T] $. Letting first $ \varepsilon\searrow 0 $ and then $ T\nearrow T_{max} $, this proves that $ u_{r}\leq 0 $ in $ [0, R]\times [0, T_{max}) $, and we have $ U_{ss}\leq0 $ because of (2.5).
In this section our aim is to establish a function and to select appropriate parameters such that the function satisfies ODI, which means finiteness of $ T_{max} $ by counter evidence. Firstly, we introduce a moment-like functional as follows
$ ϕ(t):=∫s00s−γ(s0−s)U(s,t)ds,t∈[0,Tmax), $ | (3.1) |
with $ \gamma\in(-\infty, 1) $ and $ s_{0}\in(0, R^{n}) $. As a preparation of the subsequent analysis of $ \phi $, we denote
$ Sϕ:={t∈(0,Tmax)|ϕ(t)≥nM−s0(1−γ)(2−γ)ωn⋅s2−γ0}. $ | (3.2) |
The following lemma provides a lower bound for $ U $.
Lemma 3.1. Let $ \gamma\in(-\infty, 1) $ and $ s_{0}\in(0, R^{n}) $, then
$ U(s02,t)≥1ωn⋅(nM−4s02γ(3−γ)). $ | (3.3) |
Proof. If (3.3) was false for some $ t\in S_{\phi} $ such that $ U(\frac{s_{0}}{2}, t) < \frac{1}{\omega_{n}}\cdot(nM-\frac{4s_{0}}{2^{\gamma}(3-\gamma)}) $, then necessarily $ \delta: = \frac{4s_{0}}{2^{\gamma}(3-\gamma)} < nM $. By the monotonicity of $ U(\cdot, t) $ we would obtain that $ U(s, t) < \frac{nM-\delta}{\omega_{n}} $ for all $ s\in(0, \frac{s_{0}}{2}) $. Since $ U(s, t) < \frac{nM}{\omega_{n}} $ for all $ s\in(0, R^{n}) $, we have
$ ϕ(t)<nM−δωn⋅∫s020s−γ(s0−s)ds+nMωn⋅∫s0s02s−γ(s0−s)ds=nMωn⋅∫s00s−γ(s0−s)ds−δωn⋅∫s020s−γ(s0−s)ds $ |
$ =nMωn⋅s2−γ0(1−γ)(2−γ)−δωn⋅2γ(3−γ)s2−γ04(1−γ)(2−γ). $ |
In view of the definition of $ S_{\phi} $, we find that $ nM-s_{0} < nM-\frac{2^{\gamma}(3-\gamma)\delta}{4} $, which contradicts our definition of $ \delta $.
An upper bound for $ \mu $ is established by the following lemma.
Lemma 3.2. Let $ \gamma\in(-\infty, 1) $ and $ s_{0} > 0 $ such that $ s_{0}\leq\frac{R^{n}}{6} $. Then the function $ \mu(t) $ has property that
$ μ(t)≤C1+12s∫s0f(Us(σ,t))dσ for all s∈(0,s0) and any t∈Sϕ, $ | (3.4) |
where $ C_{1} = \frac{\frac{\chi}{2}C_{0}+C_{0}+C_{2}}{3}+C_{3} = \frac{1}{3}\bigg(\frac{\chi}{2}C_{0}+C_{0}+\frac{\chi k_{1}(\gamma_{1}-\gamma_{2})}{2\gamma_{2}}(\frac{2\xi k_{2}\gamma_{2}}{\chi k_{1}\gamma_{1}})^{\frac{\gamma_{1}}{\gamma_{1}-\gamma_{2}}}\bigg)+\chi f_{1}\bigg(\frac{2\delta}{\omega_{n}s_{0}}\bigg) $.
Proof. First for any fixed $ t\in S_{\phi} $, we may invoke Lemma 3.1 to see that
$ U(s02,t)≥nM−δωn, $ |
and thus, as $ U\leq\frac{nM}{\omega_{n}} $,
$ U(s0,t)−U(s02,t)s02≤nMωn−nM−δωns02=2δωns0. $ |
However, by concavity of $ U(\cdot, t) $, as asserted by Lemma 2.4,
$ U(s0,t)−U(s02,t)s02≥Us(s0,t)≥Us(s,t) for all s∈(s0,Rn). $ |
Then let $ s_{0}\in(0, R^{n}) $, we know that
$ μ(t)=1Rn∫s00f(Us(σ,t))dσ+1Rn∫Rns0f(Us(σ,t))dσ=1Rn∫s0f(Us(σ,t))dσ+1Rn∫s0sf(Us(σ,t))dσ+1Rn∫Rns0f(Us(σ,t))dσ,∀t∈(0,Tmax). $ | (3.5) |
Since $ \gamma_{1} > \gamma_{2} $ and Young's inequality such that $ \xi f_{2}(u)\leq\xi k_{2}(1+u)^{\gamma_{2}}\leq\frac{\chi k_{1}}{2}(1+u)^{\gamma_{1}}+C_{2}\leq\frac{\chi}{2}f_{1}(u)+C_{2} $ with $ C_{2} = \frac{\chi k_{1}(\gamma_{1}-\gamma_{2})}{2\gamma_{2}}(\frac{2\xi k_{2}\gamma_{2}}{\chi k_{1}\gamma_{1}})^{\frac{\gamma_{1}}{\gamma_{1}-\gamma_{2}}} $ for $ u\geq0 $, then for all $ s\in(0, R^{n}) $ and $ t\in(0, T_{max}) $ we show that
$ χ2f1(Us(s,t))−C2≤f(Us(s,t))≤χf1(Us(s,t)). $ | (3.6) |
Accordingly, by the monotonicity of $ U_{s}(\cdot, t) $ along with (1.7) and (3.6), we have
$ ∫s0sf(Us(σ,t))dσ≤∫s0sχf1(Us(σ,t))dσ≤∫s0sχf1(Us(s,t))dσ≤s0χf1(Us(s,t)),∀s∈(0,s0), t∈(0,Tmax). $ |
Since the condition of (2.7) implies that
$ ∫s0f(Us(σ,t))dσ=∫s0χ{Us(⋅,t)≥1}(σ)⋅f(Us(σ,t)dσ+∫s0χ{Us(⋅,t)<1}(σ)⋅f(Us(σ,t)dσ≥∫s0χ{Us(⋅,t)≥1}(σ)⋅(χ2f1(Us(σ,t))−C2)dσ−C0s≥∫s0χ{Us(⋅,t)≥1}(σ)⋅χ2f1(Us(σ,t))dσ−(C0+C2)s=∫s0χ2f1(Us(σ,t))dσ−∫s0χ{Us(⋅,t)<1}(σ)⋅χ2f1(Us(σ,t))dσ−(C0+C2)s≥∫s0χ2f1(Us(s,t))dσ−χ2C0s−(C0+C2)s≥sχ2f1(Us(s,t))−(χ2C0+C0+C2)s. $ |
Therefore, we obtain
$ ∫s0sf(Us(σ,t))dσ≤2s0s∫s0f(Us(σ,t))dσ+2(χ2C0+C0+C2)s0. $ |
Since (3.5) we have for all $ s\in(0, s_{0}) $
$ 1Rn∫s0f(Us(σ,t))dσ+1Rn∫s0sf(Us(σ,t))dσ≤1Rn∫s0f(Us(σ,t))dσ+2s0Rns∫s0f(Us(σ,t))dσ+2(χ2C0+C0+C2)s0Rn, $ | (3.7) |
where $ s_{0}\leq\frac{R^{n}}{6} $ such that $ \frac{1}{R^{n}}\leq\frac{1}{6s_{0}}\leq\frac{1}{6s}, $ $ \frac{2s_{0}}{R^{n}s}\leq\frac{1}{3s} $ and $ \frac{s_{0}}{R^{n}}\leq\frac{1}{6} $ for all $ s\in(0, s_{0}) $. Finally, we estimate the last summand of (3.5)
$ 1Rn∫Rns0f(Us(σ,t))dσ≤1Rn∫Rns0χf1(Us(σ,t))dσ≤χf1(2δωns0)=C3. $ | (3.8) |
Together with (3.5), (3.7) and (3.8) imply (3.4).
Lemma 3.3. Assume that $ \gamma\in(-\infty, 1) $ satisfying
$ γ<2−2n, $ |
and $ s_{0}\in(0, \frac{R^{n}}{6}] $. Then the function $ \phi:[0, T_{max})\rightarrow \mathbb{R} $ defined by (3.1) belongs to $ C^{0}([0, T_{max}))\cap C^{1}((0, T_{max})) $ and satisfies
$ ϕ′(t)≥n2∫s00s2−2n−γ(s0−s)UssD(Us)ds+12∫s00s−γ(s0−s)Us⋅{∫s0f(Us(σ,t))dσ}ds−C1∫s00s1−γ(s0−s)Usds=:J1(t)+J2(t)+J3(t), $ | (3.9) |
for all $ t\in[0, T_{max}) $, where $ C_{1} $ is defined in Lemma 3.2.
Proof. Combining (2.6) and (3.4) we have
$ Ut(s,t)=n2s2−2nD(Us)Uss−sμ(t)Us+Us⋅∫s0f(Us(σ,t))dσ≥n2s2−2nUssD(Us)+12Us⋅∫s0f(Us(σ,t))dσ−C1sUs. $ |
Notice $ \phi(t) $ conforms $ \phi(t) = \int^{s_{0}}_{0}s^{-\gamma}(s_{0}-s)U(s, t)ds $. So $ (3.9) $ is a direct consequence.
Lemma 3.4. Let $ s_{0}\in (0, \frac{R^{n}}{6}] $, and $ \gamma\in(-\infty, 1) $ satisfying $ \gamma < 2-\frac{2}{n}. $ Then $ J_{1}(t) $ in $ (3.9) $ satisfies
$ J1(t)≥−I, $ | (3.10) |
where
$ I:={−n2dm(2−2n−γ)∫s00s1−2n−γ(s0−s),m<0,n2d(2−2n−γ)∫s00s1−2n−γ(s0−s)ln(Us+1),m=0,n2dm(2−2n−γ)∫s00s1−2n−γ(s0−s)(Us+1)m,m>0, $ | (3.11) |
for all $ t\in S_{\phi} $.
Proof. Since $ D\in C^{2}([0, \infty)) $, suppose that
$ G(τ)=∫τ0D(δ)dδ, $ |
then
$ 0<G(τ)≤d∫τ0(1+δ)m−1dδ≤{−dm,m<0,dln(τ+1),m=0,dm(τ+1)m,m>0. $ |
Here integrating by parts we obtain
$ J1(t)=n2∫s00s2−2n−γ(s0−s)dG(Us)=n2s2−2n−γ(s0−s)G(Us)|s00+n2∫s00s2−2n−γG(Us)ds−n2(2−2n−γ)∫s00s1−2n−γ(s0−s)G(Us)ds. $ |
Hence a direct calculation yields
$ J1(t)≥{n2dm(2−2n−γ)∫s00s1−2n−γ(s0−s),m<0,−n2d(2−2n−γ)∫s00s1−2n−γ(s0−s)ln(Us+1),m=0,−n2dm(2−2n−γ)∫s00s1−2n−γ(s0−s)(Us+1)m,m>0, $ |
for all $ t\in S_{\phi} $. We conclude (3.10).
Lemma 3.5. Assume that $ \gamma\in(-\infty, 1) $ satisfying $ \gamma < 2-\frac{2}{n} $ and $ s_{0}\in(0, \frac{R^{n}}{6}] $. Then we have
$ J2(t)+J3(t)≥k1χ4∫s00s1−γ(s0−s)U1+γ1sds−C4∫s00s1−γ(s0−s)Usds $ | (3.12) |
for all $ t\in S_{\phi} $, where $ C_{4} = C_{1}+\frac{(\frac{\chi}{2}C_{0}+C_{0}+C_{2})}{2} $.
Proof. Since Lemma 3.2 we have
$ ∫s0f(Us(σ,t))dσ≥s2χf1(Us(s,t))−(χ2C0+C0+C2)sfor all s∈(0,s0) and t∈(0,Tmax). $ |
Therefore,
$ J2(t)=12∫s00s−γ(s0−s)Us⋅{∫s0f(Us(σ,t))dσ}ds≥χ4∫s00s1−γ(s0−s)Usf1(Us(s,t))ds−(χ2C0+C0+C2)2∫s00s1−γ(s0−s)Usds≥k1χ4∫s00s1−γ(s0−s)U1+γ1sds−(χ2C0+C0+C2)2∫s00s1−γ(s0−s)Usds, $ |
where $ f_{1}(U_{s}(s, t))\geq k_{1}(1+U_{s})^{\gamma_{1}}\geq k_{1}(U_{s})^{\gamma_{1}} $. Combining these inequalities we can deduce (3.12).
Lemma 3.6. Let $ \gamma_{1} > \max\big\{0, m-1\big\} $. For any $ \gamma\in(-\infty, 1) $ satisfying
$ γ∈min{2−2n⋅1+γ1γ1, 2−2n⋅1+γ11+γ1−m}, $ | (3.13) |
and $ s_{0}\in(0, \frac{R^{n}}{6}], $ the function $ \phi:[0, T_{max})\rightarrow \mathbb{R} $ defined in (3.1) satisfies
$ ϕ′(t)≥{Cψ(t)−Cs3−γ−2n⋅1+γ1γ10,m≤1,Cψ(t)−Cs3−γ−2n⋅1+γ11+γ1−m0,m>1, $ | (3.14) |
with $ C > 0 $ for all $ t\in S_{\phi} $, where $ \psi(t): = \int^{s_{0}}_{0}s^{1-\gamma}(s_{0}-s)U_{s}^{1+\gamma_{1}}ds $.
Proof. From (3.10) and (3.12) we have
$ ϕ′(t)≥k1χ4ψ(t)−I−C4∫s00s1−γ(s0−s)Usds, $ | (3.15) |
for all $ t\in S_{\phi} $ and $ I $ is given by (3.11). In the case $ m < 0 $,
$ −n2dm(2−2n−γ)∫s00s1−2n−γ(s0−s)ds≤−n2dm(2−2n−γ)s0∫s00s1−2n−γds=−n2dms3−2n−γ0. $ |
If $ m = 0 $, we use the fact that $ \frac{\ln(1+x)}{x} < 1 $ for any $ x > 0 $ and Hölder's inequality to estimate
$ n2d(2−2n−γ)∫s00s1−2n−γ(s0−s)ln(Us+1)ds=n2d(2−2n−γ)∫s00[s1−γ(s0−s)U1+γ1s]11+γ1⋅s1−2n−γ−1−γ1+γ1(s0−s)1−11+γ1ln(1+Us)Usds≤n2d(2−2n−γ){∫s00s1−γ(s0−s)U1+γ1sds}11+γ1⋅{∫s00(s1−2n−γ−1−γ1+γ1(s0−s)γ11+γ1)1+γ1γ1ds}γ11+γ1≤n2d(2−2n−γ)sγ11+γ10{∫s00s(1−2n−γ)γ1−2nγ1ds}γ11+γ1ψ11+γ1(t)=C5ψ11+γ1(t)s(3−2n−γ)γ1−2n1+γ10, $ |
for all $ t\in S_{\phi} $ with $ C_{5}: = n^{2}d(2-\frac{2}{n}-\gamma)\cdot\Big(\frac{1}{2-\gamma-\frac{2}{n}\cdot\frac{1+\gamma_{1}}{\gamma_{1}}}\Big)^{\frac{\gamma_{1}}{1+\gamma_{1}}} > 0 $ by (3.13). In the case $ m > 0 $, by using the elementary inequality $ (a+b)^{\alpha}\leq2^{\alpha}(a^{\alpha}+b^{\alpha}) $ for all $ a, b > 0 $ and every $ \alpha > 0 $, we obtain
$ n2dm(2−2n−γ)∫s00s1−2n−γ(s0−s)(Us+1)mds≤2mn2dm(2−2n−γ)∫s00s1−2n−γ(s0−s)Umsds+2mn2dm(2−2n−γ)∫s00s1−2n−γ(s0−s)ds, $ | (3.16) |
for all $ t\in S_{\phi} $, and we first estimate the second term on the right of (3.16)
$ 2mn2dm(2−2n−γ)∫s00s1−2n−γ(s0−s)ds≤2mn2dms3−2n−γ0. $ |
Since $ \gamma_{1} > m-1 $ and by Hölder's inequality we deduce that
$ 2mn2dm(2−2n−γ)∫s00s1−2n−γ(s0−s)Umsds=2mn2dm(2−2n−γ)∫s00s(1−γ)⋅m1+γ1(s0−s)m1+γ1Ums⋅s1−2n−γ−(1−γ)⋅m1+γ1(s0−s)1−m1+γ1ds≤2mn2dm(2−2n−γ){∫s00[s(1−γ)⋅m1+γ1(s0−s)m1+γ1Ums]1+γ1mds}m1+γ1×{∫s00[s1−2n−γ−(1−γ)⋅m1+γ1(s0−s)1−m1+γ1]1+γ11+γ1−mds}1+γ1−m1+γ1≤2mn2dm(2−2n−γ)ψm1+γ1(t)s1+γ1−m1+γ10⋅{∫s00s(1+γ1−m)(1−γ)−2n(1+γ1)1+γ1−mds}1+γ1−m1+γ1≤C6ψm1+γ1(t)s(1+γ1−m)(3−γ)−2n(1+γ1)1+γ10, $ |
for all $ t\in S_{\phi} $ with $ C_{6} = 2^{m}\frac{n^{2}d}{m}(2-\frac{2}{n}-\gamma)\Big(\frac{1}{2-\gamma-\frac{2}{n}\cdot\frac{1+\gamma_{1}}{1+\gamma_{1}-m}}\Big)^{\frac{1+\gamma_{1}-m}{1+\gamma_{1}}} > 0 $ where $ \gamma < 2-\frac{2}{n}\cdot\frac{1+\gamma_{1}}{1+\gamma_{1}-m} $ from (3.13).
Next, we can estimate the third expression on the right of (3.15) as follows
$ C4∫s00s1−γ(s0−s)Usds=C4∫s00[s1−γ(s0−s)U1+γ1s]11+γ1⋅s1−γ−1−γ1+γ1(s0−s)1−11+γ1ds≤C4{∫s00s1−γ(s0−s)U1+γ1sds}11+γ1⋅{∫s00[s1−γ−1−γ1+γ1(s0−s)1−11+γ1]1+γ1γ1ds}γ11+γ1≤C4ψ11+γ1(t)sγ11+γ10{∫s00s1−γds}γ11+γ1=C7ψ11+γ1(t)s(3−γ)γ11+γ10, $ |
where $ C_{7} = C_{4}\big(\frac{1}{2-\gamma}\big)^{\frac{\gamma_{1}}{1+\gamma_{1}}} $ for all $ t\in S_{\phi} $. By (3.15) and collecting the estimates above we have
$ ϕ′(t)≥{k1χ4ψ(t)+n2dms3−2n−γ0−C7ψ11+γ1(t)s(3−γ)γ11+γ10,m<0, t∈Sϕ,k1χ4ψ(t)−C5ψ11+γ1(t)s(3−2n−γ)γ1−2n1+γ10−C7ψ11+γ1(t)s(3−γ)γ11+γ10,m=0, t∈Sϕ,k1χ4ψ(t)−2mn2dms3−2n−γ0−C6ψm1+γ1(t)s(1+γ1−m)(3−γ)−2n(1+γ1)1+γ10−C7ψ11+γ1(t)s(3−γ)γ11+γ10,m>0, t∈Sϕ. $ |
If $ m = 0 $, by Young's inequality we can find positive constants $ C_{8}, C_{9} $ such that
$ C5ψ11+γ1(t)s(3−2n−γ)γ1−2n1+γ10≤k1χ16ψ(t)+C8s3−γ−2n⋅1+γ1γ10,∀t∈Sϕ, $ |
while as $ m > 0 $ we have
$ C6ψm1+γ1(t)s(1+γ1−m)(3−γ)−2n(1+γ1)1+γ10≤k1χ16ψ(t)+C9s3−γ−2n⋅1+γ11+γ1−m0,∀t∈Sϕ. $ |
On the other hand, we use Young's inequality again
$ C7ψ11+γ1(t)s(3−γ)γ11+γ10≤k1χ16ψ(t)+C10s3−γ0,∀t∈Sϕ. $ |
In the case $ m < 0 $, because of $ s_{0}\in(0, \frac{R^{n}}{6}] $, we have
$ s3−2n−γ0=s3−γ−2n⋅1+γ1γ10⋅s2nγ10≤(Rn6)2nγ1s3−γ−2n⋅1+γ1γ10, $ |
when $ m > 0 $ we have
$ s3−2n−γ0=s3−γ−2n⋅1+γ11+γ1−m0⋅s2mn(1+γ1−m)0≤(Rn6)2mn(1+γ1−m)s3−γ−2n⋅1+γ11+γ1−m0. $ |
All in all, we have
$ ϕ′(t)≥{k1χ8ψ(t)−C11s3−γ−2n⋅1+γ1γ10−C10s3−γ0,m≤0,k1χ8ψ(t)−C12s3−γ−2n⋅1+γ11+γ1−m0−C10s3−γ0,m>0, $ | (3.17) |
for all $ t\in S_{\phi} $ with $ C_{11} = C_{8}-\frac{n^{2}d}{m}\big(\frac{R^{n}}{6}\big)^{\frac{2}{n\gamma_{1}}} $ and $ C_{12} = C_{9}+\frac{2^{m}n^{2}d}{m}\big(\frac{R^{n}}{6}\big)^{\frac{2m}{n(1+\gamma_{1}-m)}} $. When $ 0 < m\leq1 $, we have $ \frac{1+\gamma_{1}}{1+\gamma_{1}-m}\leq\frac{1+\gamma_{1}}{\gamma_{1}} $ such that $ s_{0}^{3-\gamma-\frac{2}{n}\cdot\frac{1+\gamma_{1}}{1+\gamma_{1}-m}} = s_{0}^{3-\gamma-\frac{2}{n}\cdot\frac{1+\gamma_{1}}{\gamma_{1}}}s_{0}^{\frac{2}{n}(\frac{1+\gamma_{1}}{\gamma_{1}}-\frac{1+\gamma_{1}}{1+\gamma_{1}-m})} \leq\big(\frac{R^{n}}{6}\big)^{\frac{2(1-m)(1+\gamma_{1})}{n\gamma_{1}(1+\gamma_{1}-m)}}s_{0}^{3-\gamma-\frac{2}{n}\cdot\frac{1+\gamma_{1}}{\gamma_{1}}} $. In the case $ m\leq1 $
$ s3−γ0=s2n⋅1+γ1γ10⋅s3−γ−2n⋅1+γ1γ10≤(Rn6)2n⋅1+γ1γ1⋅s3−γ−2n⋅1+γ1γ10, $ |
and if $ m > 1 $ we have
$ s3−γ0=s2n⋅1+γ11+γ1−m0⋅s3−γ−2n⋅1+γ11+γ1−m0≤(Rn6)2n⋅1+γ1(1+γ1−m)⋅s3−γ−2n⋅1+γ11+γ1−m0. $ |
Thus (3.17) turns into (3.14).
Next, we need to build a connection between $ \phi(t) $ and $ \psi(t) $. Let us define
$ Sψ:={t∈(0,Tmax)|ψ(t)≥s3−γ0}. $ | (3.18) |
Lemma 3.7. Let $ \gamma\in(-\infty, 1) $ satisfying $ \gamma > 1-\gamma_{1} $ and $ (3.13) $. Then for any choice of $ s_{0}\in(0, \frac{R^{n}}{6}] $, the following inequality
$ ϕ′(t)≥{Cs−γ1(3−γ)0ϕ1+γ1(t)−Cs3−γ−2n⋅1+γ1γ10,m≤1,Cs−γ1(3−γ)0ϕ1+γ1(t)−Cs3−γ−2n⋅1+γ11+γ1−m0,m>1, $ | (3.19) |
holds for all $ t\in S_{\phi}\cap S_{\psi} $ with $ C > 0 $.
Proof. We first split
$ U(s,t)=∫s0Us(σ,t)dσ=∫s0χ{Us(⋅,t)<1}(σ)⋅Us(σ,t)dσ+∫s0χ{Us(⋅,t)≥1}(σ)⋅Us(σ,t)dσ≤s+∫s0χ{Us(⋅,t)≥1}(σ)⋅{σ1−γ(s0−σ)U1+γ1s}11+γ1⋅σ−1−γ1+γ1(s0−σ)−11+γ1dσ≤s+(s0−s)−11+γ1ψ11+γ1(t)⋅{∫s0σ−1−γ1+γ1⋅1+γ1γ1dσ}γ11+γ1=s+(γ1γ+γ1−1)γ11+γ1(s0−s)−11+γ1sγ+γ1−11+γ1ψ11+γ1(t), $ | (3.20) |
for all $ s\in(0, s_{0}) $ and $ t\in(0, T_{max}) $ where $ \frac{\gamma_{1}}{\gamma+\gamma_{1}-1} > 0 $. According to the definition of $ S_{\psi} $, we can find
$ s(s0−s)−11+γ1sγ+γ1−11+γ1ψ11+γ1(t)=s2−γ1+γ1(s0−s)11+γ1ψ−11+γ1(t)≤s2−γ1+γ10⋅s11+γ10⋅(s3−γ0)−11+γ1=1, $ | (3.21) |
for all $ s\in(0, s_{0}) $ and $ t\in S_{\psi} $. Combining (3.20) and (3.21) we have
$ U(s,t)≤C1sγ+γ1−11+γ1(s0−s)−11+γ1ψ11+γ1(t), $ |
where $ C_{1} = 1+\Big(\frac{\gamma_{1}}{\gamma+\gamma_{1}-1}\Big)^{\frac{\gamma_{1}}{1+\gamma_{1}}} $ for all $ s\in(0, s_{0}) $ and $ t\in S_{\psi} $. Invoking Hölder's inequality, we get
$ ϕ(t)=∫s00s−γ(s0−s)U(s,t)ds≤C1∫s00s−γ+γ+γ1−11+γ1(s0−s)1−11+γ1ds⋅ψ11+γ1(t)≤C1sγ11+γ10∫s00s−γ+γ+γ1−11+γ1ds⋅ψ11+γ1(t)=C2sγ1(3−γ)1+γ10⋅ψ11+γ1(t), $ | (3.22) |
where $ C_{2} = C_{1}\frac{1+\gamma_{1}}{\gamma_{1}(2-\gamma)} $ for all $ s\in(0, s_{0}) $ and $ t\in S_{\psi} $. Employing these conclusion we deduce (3.19).
These preparations above will enable us to establish a superlinear ODI for $ \phi $ as mentioned earlier, and we prove our main result on blow-up based on a contradictory argument.
Proof of Theorem 1.1. Step 1. Assume on the contrary that $ T_{max} = +\infty $, and we define the function
$ S:={T∈(0,+∞)|ϕ(t)>nM−s0ωn(1−γ)(2−γ)⋅s2−γ0 for all t∈[0,T]}. $ | (3.23) |
Let us choose $ s_{0} > 0 $ such that
$ s0≤min{Rn6,nM2,nMγ12(1−γ)ωn[(C3+1)(1+γ1)−1]}, $ | (3.24) |
where $ M $ and $ \omega_{n} $ were defined in $ (2.7) $ and $ C_{3} = \big(\frac{\gamma_{1}}{\gamma+\gamma_{1}-1}\big)^{\frac{\gamma_{1}}{1+\gamma_{1}}} $ has been mentioned in (3.20). Then we pick $ 0 < \varepsilon(\gamma_{1}, M, R) = \varepsilon < \frac{s_{0}}{n} $ and $ s^{\star}(\gamma_{1}, M, R)\in(0, s_{0}) $ wtih $ r^{\star}(\gamma_{1}, M, R) = (s^{\star})^{\frac{1}{n}}\in(0, R) $ such that
$ U(s,0)≥U(s⋆,0)=nωn∫Br⋆(0)u0dx≥nωn(M−ε),∀s∈(s⋆,Rn). $ |
Therefore it is possible to estimate
$ ϕ(0)=∫s00s−γ(s0−s)U(s,0)ds≥∫s00s−γ(s0−s)U(s⋆,0)ds>nωn(M−s0n)∫s00s−γ(s0−s)ds=nM−s0ωn(1−γ)(2−γ)⋅s2−γ0. $ | (3.25) |
Then $ S $ is non-empty and denote $ T = \sup S\in(0, \infty] $. Next, we need to prove $ (0, T)\subset S_{\phi}\cap S_{\psi}\neq\emptyset $. Note that
$ ϕ(t)>nM−s0ωn(1−γ)(2−γ)⋅s2−γ0,∀t∈(0,T), $ | (3.26) |
we obtain $ (0, T)\subset S_{\phi} $. From (3.20) we have
$ ϕ(t)≤∫s00s−γ(s0−s)[s+C3sγ+γ1−11+γ1(s0−s)−11+γ1ψ11+γ1(t)]ds≤s0∫s00s1−γds+C3∫s00s−γ+γ+γ1−11+γ1(s0−s)γ11+γ1ψ11+γ1(t)ds=s3−γ02−γ+C3(1+γ1)γ1(2−γ)sγ1(3−γ)1+γ10⋅ψ11+γ1(t). $ |
It follows from $ (3.24) $ and $ (3.26) $ that
$ ϕ(t)≥nM2(1−γ)(2−γ)ωn⋅s2−γ0for all t∈(0,T). $ | (3.27) |
Then
$ C3(1+γ1)γ1(2−γ)sγ1(3−γ)1+γ10⋅ψ11+γ1(t)≥nM2(1−γ)(2−γ)ωn⋅s2−γ0−s3−γ02−γ. $ |
Note that (3.24) implies
$ nMγ12C3(1−γ)ωn(1+γ1)s0−γ1C3(1+γ1)≥1, $ |
then we have
$ ψ(t)≥[(nMs2−γ02(1−γ)(2−γ)ωn−s3−γ02−γ)⋅γ1(2−γ)C3(1+γ1)s−γ1(3−γ)1+γ10]1+γ1≥[nMγ12C3(1−γ)ωn(1+γ1)s0−γ1C3(1+γ1)]1+γ1⋅s3−γ0≥s3−γ0. $ |
Therefore, $ (0, T)\subset S_{\phi}\cap S_{\psi}\neq\emptyset. $
Step 2. Applying Lemma 3.7 we can find $ \gamma\in(-\infty, 1) $ and $ C_{1}, C_{2} > 0 $ such that for all $ s_{0}\in(0, \frac{R^{n}}{6}] $
$ ϕ′(t)≥{C1s−γ1(3−γ)0ϕ1+γ1(t)−C2s3−γ−2n⋅1+γ1γ10,m≤1,C1s−γ1(3−γ)0ϕ1+γ1(t)−C2s3−γ−2n⋅1+γ11+γ1−m0,m>1, $ |
for all $ t\in S_{\phi}\cap S_{\psi} $ and with (3.22) we have
$ ψ(t)≥C3s−γ1(3−γ)0ϕ1+γ1(t),∀t∈Sψ. $ |
To specify our choice of $ s_{0} $, for given $ M > 0 $ we choose $ s_{0}\in(0, \frac{R^{n}}{6}] $ small enough such that
$ s0≤nM2, $ | (3.28) |
and also
$ sγ10<TC1γ14(nM2ωn(1−γ)(2−γ))γ1, $ | (3.29) |
as well as
$ s1+γ10≤C3(nM2ωn(1−γ)(2−γ))1+γ1. $ | (3.30) |
From (3.23), (3.28) and (3.30) we have
$ ψ(t)≥C3s−γ1(3−γ)0ϕ1+γ1(t)>C3(nM−s0ωn(1−γ)(2−γ)⋅1s0)1+γ1⋅s3−γ0≥s3−γ0,∀t∈Sψ, $ |
which shows that $ S\subset S_{\phi}\cap S_{\psi} $. Since $ 1+\gamma_{1}-m > \frac{2}{n} $, we have $ (1+\gamma_{1})(1-\frac{2}{n(1+\gamma_{1}-m)}) > 0 $ if $ m > 1 $ so that we can choose $ s_{0} $ sufficiently small satisfying $ (3.28) $–$ (3.30) $ such that
$ s(1+γ1)(1−2n(1+γ1−m))0≤C12C2(nM2ωn(1−γ)(2−γ))1+γ1, $ |
while in the case $ m\leq1 $, the condition $ \gamma_{1} > m-1+\frac{2}{n}\geq\frac{2}{n} $ which infers that $ (1+\gamma_{1})(1-\frac{2}{n\gamma_{1}}) > 0 $ and we select $ s_{0} $ small enough fulfilling $ (3.28) $–$ (3.30) $ such that
$ s(1+γ1)(1−2nγ1)0≤C12C2(nM2ωn(1−γ)(2−γ))1+γ1. $ |
It is possible to obtain
$ C12s−γ1(3−γ)0ϕ1+γ1(0)C2s3−γ−2n⋅1+γ11+γ1−m0≥C12C2(nM2ωn(1−γ)(2−γ))1+γ1⋅s−(1+γ1)+2n⋅1+γ1(1+γ1−m)0≥1,∀m>1, $ |
and we have
$ C12s−γ1(3−γ)0ϕ1+γ1(0)C2s3−γ−2n⋅1+γ1γ10≥C12C2(nM2ωn(1−γ)(2−γ))1+γ1⋅s−(1+γ1)+2n⋅1+γ1γ10≥1,∀m≤1. $ |
All in all, for any $ m\in\mathbb{R} $, we apply an ODI comparison argument to obtain that
$ ϕ′(t)≥C12s−γ1(3−γ)0ϕ1+γ1(t),∀t∈(0,T). $ |
By a direct calculation we obtain
$ −1γ1(1ϕγ1(t)−1ϕγ1(0))≥C12s−γ1(3−γ)0t,∀t∈(0,T). $ |
Hence, according to (3.25) and (3.29) we conclude
$ t<2C1γ1sγ10(2ωn(1−γ)(2−γ)nM)γ1≤T2, $ |
for all $ t\in(0, T) $. As a consequence, we infer that $ T_{max} $ must be finite.
In this section, we are preparing to prove Theorem $ 1.2 $ by providing the $ L^{p} $ estimate of $ u $ and the Moser-type iteration.
Lemma 4.1. Let $ (u, v, w) $ be a classical solution of the system (1.5) under the condition of Theorem $ 1.2 $. Suppose that
$ γ2<1+γ1<2n+m. $ | (4.1) |
Then for any $ p > \max\big\{1, 2-m, \gamma_{2}\big\} $, there exists $ C = C(p) > 0 $ such that
$ ∫Ω(1+u)p(x,t)dx≤Con (0,Tmax). $ | (4.2) |
Proof. Notice $ f_{1}(u)\leq k_{1}(1+u)^{\gamma_{1}}, \ f_{2}(u) = k_{2}(1+u)^{\gamma_{2}} $ for all $ u\geq0 $. Multiplying the first equation of (1.5) by $ p(1+u)^{p-1} $ and integrating by parts with the boundary conditions for $ u, v $ and $ w $, we have
$ ddt∫Ω(1+u)pdx+p(p−1)∫Ω(1+u)p−2D(u)|∇u|2dx=χp(p−1)∫Ωu(1+u)p−2∇u⋅∇vdx−ξp(p−1)∫Ωu(1+u)p−2∇u⋅∇wdx=−χ(p−1)∫Ω(1+u)pΔvdx+χp∫Ω(1+u)p−1Δvdx+ξ(p−1)∫Ω(1+u)pΔwdx−ξp∫Ω(1+u)p−1Δwdx≤χ(p−1)∫Ω(1+u)pf1(u)dx+χp∫Ω(1+u)p−1μ1(t)dx+ξ(p−1)∫Ω(1+u)pμ2(t)dx−ξ(p−1)∫Ω(1+u)pf2(u)dx+ξp∫Ω(1+u)p−1f2(u)dx≤k1χ(p−1)∫Ω(1+u)p+γ1dx+χp∫Ω(1+u)p−1μ1(t)dx+ξ(p−1)∫Ω(1+u)pμ2(t)dx−k2ξ(p−1)∫Ω(1+u)p+γ2dx+k2ξp∫Ω(1+u)p+γ2−1dx,∀t∈(0,Tmax). $ | (4.3) |
Firstly,
$ p(p−1)∫Ω(1+u)p−2D(u)|∇u|2dx≥dp(p−1)∫Ω(1+u)p+m−3|∇u|2dx=4dp(p−1)(p+m−1)2∫Ω|∇(1+u)p+m−12|2dx. $ |
By Young's inequality and Hölder's inequality, we obtain
$ χp∫Ω(1+u)p−1μ1(t)dx≤C1∫Ω(1+u)p+γ1dx+C2μp+γ11+γ11(t)=C1∫Ω(1+u)p+γ1dx+C2(1|Ω|∫Ωf1(u)dx)p+γ11+γ1≤C1∫Ω(1+u)p+γ1dx+C3(∫Ω(1+u)1+γ1dx)p+γ11+γ1≤C1∫Ω(1+u)p+γ1dx+C3{(∫Ω(1+u)p+γ1dx)1+γ1p+γ1⋅|Ω|p−1p+γ1}p+γ11+γ1=C1∫Ω(1+u)p+γ1dx+C3|Ω|p−11+γ1∫Ω(1+u)p+γ1dx. $ |
for all $ t\in(0, T_{max}) $. Then by Hölder's inequality we obtain
$ ξ(p−1)∫Ω(1+u)pμ2(t)dx=k2ξ(p−1)|Ω|∫Ω(1+u)γ2dx∫Ω(1+u)pdx≤k2ξ(p−1)|Ω|{∫Ω(1+u)p+γ2dx}γ2p+γ2|Ω|pp+γ2×{∫Ω(1+u)p+γ2dx}pp+γ2|Ω|γ2p+γ2=k2ξ(p−1)∫Ω(1+u)p+γ2dx,∀t∈(0,Tmax). $ |
Furthermore, by using Young's inequality and (4.1) we have
$ k2ξp∫Ω(1+u)p+γ2−1dx≤C4∫Ω(1+u)p+γ1dx+C5, $ |
for all $ t\in(0, T_{max}) $. Therefore, combining these we conclude
$ ddt∫Ω(1+u)pdx+4dp(p−1)(p+m−1)2∫Ω|∇(1+u)p+m−12|2dx≤C6∫Ω(1+u)p+γ1dx+C5,∀t∈(0,Tmax), $ |
where $ C_{6} = C_{1}+C_{3}|\Omega|^{\frac{p-1}{1+\gamma_{1}}}+C_{4}+k_{1}\chi(p-1) $. By means of Gagliardo-Nirenberg inequality we can find $ C_{7} $ such that
$ \begin{align} C_{6}\int_{\Omega}(1+u)^{p+\gamma_{1}}dx& = C_{6}\|(1+u)^{\frac{p+m-1}{2}}\|^{\frac{2(p+\gamma_{1})}{p+m-1}}_{L^{\frac{2(p+\gamma_{1})}{p+m-1}}(\Omega)}\\ &\leq C_{7}\|\nabla(1+u)^{\frac{p+m-1}{2}}\|^{\frac{2(p+\gamma_{1})}{p+m-1}\cdot a}_{L^{2}(\Omega)}\cdot \|(1+u)^{\frac{p+m-1}{2}}\|^{\frac{2(p+\gamma_{1})}{p+m-1}\cdot(1-a)}_{L^{\frac{2}{p+m-1}}(\Omega)}\\ & \quad +C_{7}\|(1+u)^{\frac{p+m-1}{2}}\|^{\frac{2(p+\gamma_{1})}{p+m-1}}_{L^{\frac{2}{p+m-1}}(\Omega)} \end{align} $ |
for all $ t\in(0, T_{max}) $, where
$ \begin{equation} \nonumber a = \frac{\frac{p+m-1}{2}-\frac{p+m-1}{2(p+\gamma_{1})}}{\frac{p+m-1}{2}-(\frac{1}{2}-\frac{1}{n})}\in(0,1). \end{equation} $ |
Since $ 1-m+\gamma_{1} < \frac{2}{n} $, we have $ \frac{2(p+\gamma_{1})}{p+m-1}\cdot a < 2 $, and we use Young's inequality to see that for all $ t\in(0, T_{max}) $
$ \begin{equation} \nonumber C_{6}\int_{\Omega}(1+u)^{p+\gamma_{1}}dx\leq\frac{2dp(p-1)}{(p+m-1)^{2}}\|\nabla(1+u)^{\frac{p+m-1}{2}}\|^{2}_{L^{2}(\Omega)}+C_{8}. \end{equation} $ |
In quite a similar manner, we obtain $ C_{9} = C_{9}(p) > 0 $ fulfilling
$ \begin{equation} \nonumber\int_{\Omega}(1+u)^{p}dx\leq\frac{2dp(p-1)}{(p+m-1)^{2}}\|\nabla(1+u)^{\frac{p+m-1}{2}}\|^{2}_{L^{2}(\Omega)}+C_{9} \quad {\rm for\ all\ } t\in(0,T_{max}). \end{equation} $ |
Finally, combining these to (4.3) we obtain
$ \begin{equation} \nonumber\frac{d}{dt}\int_{\Omega}(1+u)^{p}dx+\int_{\Omega}(1+u)^{p}dx\leq C_{5}+C_{8}+C_{9} \quad {\rm for\ all\ } t\in(0,T_{max}). \end{equation} $ |
Thus,
$ \begin{equation} \nonumber\int_{\Omega}(1+u)^{p}dx\leq\max\Big\{\int_{\Omega}(1+u_{0})^{p}dx,C_{5}+C_{8}+C_{9}\Big\} \quad {\rm for\ all\ } t\in(0,T_{max}). \end{equation} $ |
We have done the proof.
Under the condition of Lemma 4.1 we can use the above information to prove Theorem 1.2.
Proof of Theorem 1.2. From Lemma 4.1, we let $ p > \max\big\{\gamma_{1}n, \gamma_{2}n, 1\big\} $. By the elliptic $ L^{p} $-estimate to the two elliptic equations in (1.5), we get that for all $ t\in(0, T_{max}) $ there exists some $ C_{10}(p) > 0 $ such that
$ \begin{equation} \|v(\cdot,t)\| _{w^{2,\frac{p}{\gamma_{1}}}(\Omega)}\leq C_{10}(p), \quad \|w(\cdot,t)\| _{w^{2,\frac{p}{\gamma_{2}}}(\Omega)}\leq C_{10}(p), \end{equation} $ | (4.4) |
and hence, by the Sobolev embedding theorem, we get
$ \begin{equation} \|v(\cdot,t)\|_{C^{1}(\overline{\Omega})}\leq C_{10}(p), \quad \|w(\cdot,t)\|_{C^{1}(\overline{\Omega})}\leq C_{10}(p). \end{equation} $ | (4.5) |
Now the Moser iteration technique ([3,51]) ensures that $ \|u(\cdot, t)\|_{L^{\infty}(\Omega)}\leq C $ for any $ t\in(0, T_{max}) $.
This concludes by Lemma 2.1 that $ T_{max} = \infty $.
The paper is supported by the Research and Innovation Team of China West Normal University (CXTD2020–5).
The authors declare that there is no conflict of interest.
1. | Bruno Telch, Genyle Nascimento, Global existence and boundedness in a two‐dimensional parabolic chemotaxis system with competing attraction and repulsion effects, 2024, 0170-4214, 10.1002/mma.10550 | |
2. | Zhan Jiao, Irena Jadlovská, Tongxing Li, Finite-time blow-up and boundedness in a quasilinear attraction–repulsion chemotaxis system with nonlinear signal productions, 2024, 77, 14681218, 104023, 10.1016/j.nonrwa.2023.104023 |
Water fractional flow function
Upper row: Results produced by the discrete scheme described in Appendix D (incompressible model). Three kinds of curves are plotted including the case without viscous effect, i.e.,
Simulation results with smaller viscous parameters after 10 days of water flooding. Three kinds of curves are compared: zero viscous effect, Darcy velocity
The results after 10 days with initial data are shown in Fig. 1 with interstitial velocity in viscous term. Four curves are compared: the one with large values of
Initial water saturation profile from Coclite et al. [9]
Left: The results from Coclite et al. [9] based on Darcy velocity in viscous term. Right: Numerical scheme (after 8 days) which uses interstitial velocity in viscous term with different viscous values
Comparison between the compressible model and the incompressible model for water-oil flow with
The water pressure evolution in the compressible model for the case with Darcy velocity in viscous term (left figure) and the case with interstitial velocity in viscous term (right figure). Water pressure increases with time in the water displacing part of the reservoir layer which leads to a compression effect where the magnitude of the viscous terms increase and thereby slows down the displacement of the water front
Left: Comparison of saturation profiles for water injection and gas injection, respectively, after the same time period (10 days) in the compressible model using interstitial velocity in viscous term (