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Global boundedness and blow-up in a repulsive chemotaxis-consumption system in higher dimensions

Jaewook Ahn [email protected] Kyungkeun Kang [email protected] Dongkwang Kim [email protected] Department of Mathematics, Dongguk University, Seoul, Republic of Korea School of Mathematics & Computing(Mathematics), Yonsei University, Seoul, Republic of Korea Department of Mathematical Sciences, Ulsan National Institute of Science and Technology (UNIST), Ulsan, Republic of Korea
Abstract

This paper investigates the repulsive chemotaxis-consumption model

tu\displaystyle\partial_{t}u =(D(u)u)+(uv),\displaystyle=\nabla\cdot(D(u)\nabla u)+\nabla\cdot(u\nabla v),
0\displaystyle 0 =Δvuv,\displaystyle=\Delta v-uv,

in an nn-dimensional ball, n3n\geq 3, where the diffusion coefficient DD is an appropriate extension of the function 0ξ(1+ξ)m10\leq\xi\mapsto(1+\xi)^{m-1} for some m>0m>0. Under the boundary conditions

ν(D(u)u+uv)=0 and v=M>0,\nu\cdot(D(u)\nabla u+u\nabla v)=0\quad\text{ and }\quad v=M>0,

we first demonstrate that for m>1m>1, or m=1m=1 with 0<M<2/(n2)0<M<2/(n-2), the system admits globally defined classical solutions that are uniformly bounded in time for any choice of sufficiently smooth radial initial data. This result is further extended to the case 0<m<10<m<1 when MM is chosen to be sufficiently small, depending on the initial conditions. In contrast, it is shown that for 0<m<2n0<m<\frac{2}{n}, the system exhibits blow-up behavior for sufficiently large MM.

keywords:
repulsive chemotaxis-consumption system, blow-up, global boundedness
MSC:
[2010] 35B44 , 35K51 , 92C17

1 Introduction

Since the pioneering work of Keller and Segel in the early 1970s [10, 11], the mathematical investigation of chemotaxis – the directed movement of organisms in response to chemical gradients – has significantly enhanced the understanding of various biological mechanisms. Of particular interest is the chemotactic behavior of aerobic bacteria that consume oxygen as a nutrient, a subject that has attracted considerable attention due to its complex dynamics observed in both experimental and simulation settings [6, 20]. Accordingly, the following equations have been derived and subjected to rigorous analysis as a fundamental model of this nutrient-based chemotaxis interaction [15]:

tu=(D(u)u)(uS(u,v)v),tv=Δvuf(v),\begin{split}\partial_{t}u&=\nabla\cdot(D(u)\nabla u)-\nabla\cdot(uS(u,v)\nabla v),\\ \partial_{t}v&=\Delta v-uf(v),\end{split} (1)

where uu denotes the density of the organisms and vv stands for the concentration of nutrients, such as oxygen.

When DD, SS, and ff are positive and appropriately chosen, it is well-known that the Neumann problem for (1) possesses an energy dissipation structure, arising from the interplay between the consumption term uf(v)-uf(v) and the chemotaxis term (uS(u,v)v)-\nabla\cdot(uS(u,v)\nabla v), of the form in its simplest prototype case where D(u)=1D(u)=1, S(u,v)=1S(u,v)=1, and f(v)=vf(v)=v:

ddt{Ωulnu+2Ω|v|2}+Ω|u|2u+Ωv|2lnv|20.\frac{d}{dt}\left\{\int_{\Omega}u\ln u+2\int_{\Omega}|\nabla\sqrt{v}|^{2}\right\}+\int_{\Omega}\frac{|\nabla u|^{2}}{u}+\int_{\Omega}v|\nabla^{2}\ln v|^{2}\leq 0. (2)

The energy dissipation structure (2) has played a crucial role in demonstrating global-in-time boundedness, stability, and asymptotic behaviors of solutions as well as global solvability (c.f. [15]).

Specifically, beyond the small initial data case where the global existence and asymptotic behaviors converging to a constant steady state have been proved using weighted LpL^{p}-estimates [16, 31], global large-data classical and weak solutions have been achieved via exploiting (2) in the two and three dimensions, respectively [25] (see [9, 18] for eventual smoothness and asymptotic behaviors of such weak solutions in three dimensions).

Moreover, variants of the dissipation structure (2), capable of allowing singular behaviors of S(v)=1vαS(v)=\frac{1}{v^{\alpha}}, have been employed to see the global solvability and asymptotic behaviors of solutions. Indeed, the global existence of the classical solution has been proved for general ff and SS involving f(v)=vf(v)=v and S(v)=1vαS(v)=\frac{1}{v^{\alpha}} with 0<α<12+12(82+117)120.39270<\alpha<-\frac{1}{2}+\frac{1}{2}\left(\frac{8\sqrt{2}+11}{7}\right)^{\frac{1}{2}}\cong 0.3927 for n=2n=2 [2]. Later, this result has been improved to 0<α<12+12n+8n0<\alpha<-\frac{1}{2}+\frac{1}{2}\sqrt{\frac{n+8}{n}} for n=2,3n=2,3 [12]. For 0<α<10<\alpha<1 and n=2n=2, weak solutions have been constructed when the initial data is sufficiently small [21]. In the case of α=1\alpha=1, another dissipation structure that provides weak regularity than (2) has been employed to prove global solvability of generalized solutions for n=2n=2 [26] (c.f. [14]), and of renormalized solutions for n2n\geq 2 [27].

Furthermore, similar quantitative results have been considered when DD is of porous medium type, um1u^{m-1}, or its nondegenerate form, (u+1)m1(u+1)^{m-1} with m>1m>1, by using this enhancing diffusivity or variants of (2). Exploiting the diffusivity of DD, in presence of general S0S\geq 0 and f>0f>0, the existence of global bounded solutions has been shown for m>1m>1 and n=2n=2 in a weak sense [19], and for m>22nm>2-\frac{2}{n} and n2n\geq 2 in the classical sense [23]. In the case of S=1S=1, the global solvability has been shown for m>26n+4m>2-\frac{6}{n+4} and n3n\geq 3 using a variant of (2) [22]. Later, this result has been improved by [7], where the global well-posedness and asymptotic behaviors are established for m>321nm>\frac{3}{2}-\frac{1}{n} and n3n\geq 3 [7]. In the case of S(v)=1vS(v)=\frac{1}{v}, the global solvability has been proven for m>1+4nm>1+\frac{4}{n} and n2n\geq 2 [13], and for m>321nm>\frac{3}{2}-\frac{1}{n} and n2n\geq 2 in a generalized concept [29]. The latter has been extended to the tensor-valued S(u,v,x)S(u,v,x) satisfying |S(u,v,x)|1vα|S(u,v,x)|\leq\frac{1}{v^{\alpha}} with α[0,1)\alpha\in[0,1) for m>321nm>\frac{3}{2}-\frac{1}{n} [28].

For parabolic-elliptic simplification of (1) with D=1D=1, S=1S=1, and f(v)=vf(v)=v, the global well-posedness and asymptotic behavior have been shown under Robin boundary condition by Fuest-Lankeit-Mizukami [8] when n1n\geq 1 (see also [1]). We refer to [30] for its Dirichlet counterpart.

Remarkably, the parabolic-elliptic counterpart of (1) may exhibit blow-up behavior when SS is of repulsion type, S<0S<0. Indeed, under No-flux/Dirichlet boundary conditions, finite-time blow-up solutions have been found for D(u)(u+1)m1D(u)\leq(u+1)^{m-1}, 0<m<10<m<1, and S(v)=1vS(v)=-\frac{1}{v} by Wang-Winkler [24] when n2n\geq 2. This result has also been verified for S=1S=-1 and n=2n=2 [3]. Additionally, in the latter case, the global existence of bounded solutions has been established when DD has a positive lower bound. However, to the best of our knowledge, there are no results concerning the existence or blow-up of solutions for S=1S=-1 and n3n\geq 3.  

Our main objective in this paper is to either establish the global existence of bounded solutions or construct unbounded solutions for the repulsion-type chemotaxis-consumption system in dimensions three and higher. To be more precise, we consider the following initial-boundary value problem

{tu=(D(u)u)+(uv),xΩ,t>0,0=Δvuv,xΩ,t>0,u(,t)|t=0=u0,xΩ,ν(D(u)u+uv)=0andv=M,xΩ,t>0,\begin{cases}\partial_{t}u=\nabla\cdot(D(u)\nabla u)+\nabla\cdot(u\nabla v),&x\in\Omega,\,t>0,\\ \quad 0=\Delta v-uv,&x\in\Omega,\,t>0,\\ u(\cdot,t)\rvert_{t=0}=u_{0},&x\in\Omega,\\ \nu\cdot(D(u)\nabla u+u\nabla v)=0\quad\text{and}\quad v=M,&x\in\partial\Omega,\,t>0,\end{cases} (3)

In our analysis, the spatial domain Ω\Omega is specifically chosen as BRB_{R}, a ball of radius R>0R>0 centered at the origin in n\mathbb{R}^{n}. The diffusion coefficient DD is chosen to extend the prototype choice as follows:

D(ξ)=(ξ+1)m1forξ0 and m>0.D(\xi)=(\xi+1)^{m-1}\quad\text{for}\quad\xi\geq 0\,\text{ and }\,m>0.

Throughout this paper, the initial function u0u_{0} is assumed to satisfy

0u0W1,(Ω),u00,and is radially symmetric.0\not\equiv u_{0}\in W^{1,\infty}(\Omega),\quad u_{0}\geq 0,\quad\text{and is radially symmetric}. (4)

Before presenting the main results, we begin by outlining the local existence and blow-up criteria for (3). Although these topics have been widely explored in the literature for more generalized frameworks [5], this paper specifically addresses our particular case.

Proposition 1.

Let n3n\geq 3, Ω=BRn\Omega=B_{R}\subset\mathbb{R}^{n} with R>0R>0, and M>0M>0. Assume that

DC2([0,)) fulfills D(ξ)>0 for all ξ0,D\in C^{2}([0,\infty))\text{ fulfills }D(\xi)>0\text{ for all }\xi\geq 0, (5)

and that the initial function u0u_{0} satisfies (4). Then, there exists a maximal time Tmax(0,]T_{max}\in(0,\infty] for which the problem (3) possesses a unique classical solution (u,v)(u,v) in Ω×(0,Tmax)\Omega\times(0,T_{max}), which is radially symmetric and positive over Ω¯×(0,Tmax)\overline{\Omega}\times(0,T_{max}), with the following regularity properties:

{uq>nC([0,Tmax);W1,q(Ω))C2,1(Ω¯×(0,Tmax)),vC2,0(Ω¯×(0,Tmax)).\begin{cases}u\in\bigcup_{q>n}C([0,T_{max});W^{1,q}(\Omega))\cap C^{2,1}(\overline{\Omega}\times(0,T_{max})),\\ v\in C^{2,0}(\overline{\Omega}\times(0,T_{max})).\\ \end{cases}

Furthermore, if Tmax<T_{max}<\infty, then uu blows up in the sense that

u(,t)L(Ω)as tTmax.\|u(\cdot,t)\|_{L^{\infty}(\Omega)}\rightarrow\infty\quad\textrm{as }t\rightarrow T_{max}.

Additionally, the mass of uu is conserved over time:

Ωu(x,t)𝑑x=Ωu0(x)𝑑xfor all t(0,Tmax),\int_{\Omega}u(x,t)dx=\int_{\Omega}u_{0}(x)dx\quad\text{for all }t\in(0,T_{max}), (6)

and vv is uniformly bounded in time:

v(,t)L(Ω)Mfor all t(0,Tmax).\|v(\cdot,t)\|_{L^{\infty}(\Omega)}\leq M\quad\text{for all }t\in(0,T_{max}). (7)

The first objective of this study is to investigate conditions on the diffusion rate that ensure the existence of globally bounded regular solutions. The following theorem demonstrates that, within this radially symmetric setting, the existence of a global-in-time bounded solution is assured for certain values of mm, specifically m>1m>1 or m=1m=1 with a constraint on MM.

Theorem 1.

Let n3n\geq 3, Ω=BRn\Omega=B_{R}\subset\mathbb{R}^{n} with R>0R>0, M>0M>0, and kD>0k_{D}>0. Assume that DD satisfies (5) and

D(ξ)kD(1+ξ)m1 for all ξ>0,D(\xi)\geq k_{D}(1+\xi)^{m-1}\quad\text{ for all }\,\xi>0, (8)

and that either

m>1, or m=1 and M<2n2kDm>1,\quad\text{ or }\quad m=1\,\text{ and }\,M<\frac{2}{n-2}k_{D}

holds. Then, for any choice of the initial function satisfying (4), the problem (3) admits a classical radial solution (u,v)(u,v) in Ω×(0,)\Omega\times(0,\infty) which is uniformly bounded in time in a way that

supt0u(,t)L(Ω)C\sup_{t\geq 0}\|u(\cdot,t)\|_{L^{\infty}(\Omega)}\leq C

with some C>0C>0.

A subsequent natural question is under what conditions the boundedness or unboundedness of the solution is determined for the case 0<m<10<m<1. It might be expected to experts that small data implies the existence of global bounded solutions. However, we can not find it in the literature and therefore, for clarity, we first demonstrate that bounded solutions can be constructed globally in time when the Dirichlet data MM is sufficiently small.

Theorem 2.

Let n3n\geq 3 and Ω=BRn\Omega=B_{R}\subset\mathbb{R}^{n} with R>0R>0. Assume that

D(ξ)=(1+ξ)m1 for all ξ>0 with  0<m<1.D(\xi)=(1+\xi)^{m-1}\quad\text{ for all }\,\xi>0\quad\text{ with }\,0<m<1. (9)

Then, given initial function u0u_{0} satisfying (4), there exists M=M(R,u0)>0M_{*}=M_{*}(R,u_{0})>0 such that for any choice of M(0,M]M\in(0,M_{*}], the corresponding problem (3) possesses a classical radial solution (u,v)(u,v) in Ω×(0,)\Omega\times(0,\infty) which is uniformly bounded in time in a way that

supt0u(,t)L(Ω)C\sup_{t\geq 0}\|u(\cdot,t)\|_{L^{\infty}(\Omega)}\leq C

with some C=C(M)>0C=C(M)>0.

Finally, we establish that when MM is sufficiently large, blow-up of the solution may occur, in particular, in a more restricted range m(0,2n)m\in(0,\frac{2}{n}).

Theorem 3.

Let n3n\geq 3, Ω=BRn\Omega=B_{R}\subset\mathbb{R}^{n} with R>0R>0, and KD>0K_{D}>0. Assume that DD satisfies (5) and

D(ξ)KD(1+ξ)m1 for all ξ>0 with  0<m<2n.D(\xi)\leq K_{D}(1+\xi)^{m-1}\quad\text{ for all }\,\xi>0\quad\text{ with }\,0<m<\frac{2}{n}. (10)

Then, given initial function u0u_{0} satisfying (4), there exists M=M(R,u0)>0M^{*}=M^{*}(R,u_{0})>0 such that for any choice of MMM\geq M^{*}, the corresponding solution (u,v)(u,v) of (3) blows up in finite or infinite time; that is,

u(,t)L(Ω)as tTmax.\|u(\cdot,t)\|_{L^{\infty}(\Omega)}\rightarrow\infty\quad\textrm{as }t\rightarrow T_{max}.

In particular, when n=3n=3, the blow-up time is finite, i.e., Tmax<T_{max}<\infty.

Remark 1.

We remind that in two dimensions, blow-up solutions have been constructed for any m(0,1)m\in(0,1) [3]. We may also expect that the blow-up could occur in the range m[2n,1)m\in[\frac{2}{n},1) for n3n\geq 3 as well. We leave it as an open question.

Plan of the paper Theorem 1 extends the result in [3] to higher dimensions. Whereas the authors in [3] exploited the uniform L2L^{2} smallness of v\nabla v near the origin to ensure uniform L1L^{1} boundedness of ulnuu\ln u, thereby achieving global boundedness of uu for n=2n=2, our proof begins with a pointwise estimate of v\nabla v near the origin (Lemma 2). This estimate is derived from the uniform boundedness of vv and the assumption of radial symmetry. Employing this approach, we can control the mass accumulation function UU near the origin through a comparison method for the equation of UU (Lemma 4). This control leads to an improved regularity of v\nabla v (Lemma 5) near the origin, which, when combined with estimates away from the origin (Lemma 1), ensures global boundedness of uu through LpL^{p}-estimates in higher dimensions (Lemma 6).

For Theorem 2, we first exploit the smallness assumption on MM, depending on u0u_{0}, to obtain the behavior of UU via a comparison argument similar to that used in Lemma 4 (Lemma 7). This behavior of UU, in conjunction with a Hardy-type inequality (Lemma 8) and a one-dimensional variant of the Gagliardo-Nirenberg inequality, allows us to derive LpL^{p}-estimates for uu (Lemma 9).

In Theorem 3, the method follows the approach employed in [3] (see also [24]). This method involves tracking the time evolution of the quantity ϕ(t):=0RsαU(s,t)𝑑s\phi(t):=\int_{0}^{R}s^{-\alpha}U(s,t)ds with appropriately chosen α\alpha. Obtaining a suitable lower bound for v\nabla v is crucial in this process. The authors in [3] provided this lower bound by appropriately estimating lnv\nabla\ln v and vv, respectively. However, while the lower bound for lnv\nabla\ln v can be efficiently estimated even in higher dimensions using a variant of the ODE comparison method (Lemma 10, and see [24, Lemma 3.1] for the proof), demonstrating the polynomial decay of vv, which was straightforward in dimensions two [3], seems difficult to demonstrate for dimensions three and higher. As a result, it is unlikely to obtain ϕ(t)ϕ(t)\phi^{\prime}(t)\gtrsim\phi(t) as in [3]. To overcome the difficulty, instead, we make full use of the second equation of (3) and the estimate of vv at the spatial origin (Lemma 11) to obtain an appropriate ordinary differential inequality (ODI) for ϕ(t)\phi(t) (Lemma 12), which is a technical novelty compare to [3]. It turns out that such ODI satisfies ϕ(t)1\phi^{\prime}(t)\gtrsim 1, namely, ϕ(t)\phi(t) has a linear growth in time, which implies blow-up in finite or infinite time. On the other hand, in three dimensions, since it is shown that ϕ(t)\phi(t) is uniformly bounded, blow-up must occur in a finite time.

For clarity throughout this paper, we abbreviate Lp(Ω)\|\cdot\|_{L^{p}(\Omega)} as p\|\cdot\|_{p} when no confusion arises. Constants are denoted by CiC_{i} with i=1,2,i=1,2,..., where the subscript serves to distinguish these constants within each chapter. Additionally, C(x,y,)C(x,y,...) indicates a dependency of the constant CC on variables x,yx,y, and others. Finally, we denote by σn\sigma_{n} the measure of the n1n-1 dimensional unit sphere B1\partial B_{1}.

2 Global boundedness for m1m\geq 1

Hereafter, we fix n3n\geq 3, Ω=BRn\Omega=B_{R}\subset\mathbb{R}^{n} with R>0R>0, and assume that DD satisfies (5). we denote by (u,v)(u,v) the solution in Ω×(0,Tmax)\Omega\times(0,T_{max}) as in Proposition 1, and denote

L:=1σnΩu0(x)𝑑x=0Rsn1u0(s)𝑑s.L:=\frac{1}{\sigma_{n}}\int_{\Omega}u_{0}(x)dx=\int_{0}^{R}s^{n-1}u_{0}(s)ds.

Utilizing the radial symmetry, the equations in (3) can be rewritten as the following scalar equations with a slight abuse of notation (u,v)(x,t)=(u,v)(|x|,t)(u,v)(x,t)=(u,v)(|x|,t), where |x|=s(0,R)|x|=s\in(0,R) and t(0,Tmax)t\in(0,T_{max}):

tu=s1ns(sn1D(u)su)+s1ns(sn1usv),\partial_{t}u=s^{1-n}\partial_{s}(s^{n-1}D(u)\partial_{s}u)+s^{1-n}\partial_{s}(s^{n-1}u\partial_{s}v), (11)
0=s1ns(sn1sv)uv.0=s^{1-n}\partial_{s}(s^{n-1}\partial_{s}v)-uv. (12)

Hereafter, we fix n3n\geq 3, Ω=BRn\Omega=B_{R}\subset\mathbb{R}^{n} with R>0R>0, and assume that DD satisfies (5).

2.1 Estimates of v\nabla v

In the subsequent two lemmas, pointwise estimates for v\nabla v are presented. The estimates are two-fold: (Lemma 1) away from the spatial origin and (Lemma 2) near the origin. The first estimate indicates that the origin is the only possible blow-up point for uu, while the second leads to Lemma 4, the behavior of uu near the origin.

We first observe that sv\partial_{s}v is positive and bounded away from the origin.

Lemma 1.

Let (u,v)(u,v) be a solution given in Proposition 1. It holds that sv\partial_{s}v is positive in (0,R)×(0,Tmax)(0,R)\times(0,T_{max}). Moreover, for any choice of δ0(0,R)\delta_{0}\in(0,R), we have

sv(s,t)δ0n+1LMfor all (s,t)[δ0,R)×(0,Tmax).\partial_{s}v(s,t)\leq\delta_{0}^{-n+1}LM\quad\text{for all }(s,t)\in[\delta_{0},R)\times(0,T_{max}).
Proof.

Since uu and vv are positive, the positivity of sv\partial_{s}v immediately follows from (12). Let us fix δ0(0,R)\delta_{0}\in(0,R). Through integration by parts, it holds from (12) and (7) that

L0sρn1u(ρ,t)𝑑ρ\displaystyle L\geq\int_{0}^{s}\rho^{n-1}u(\rho,t)d\rho =0sρ(ρn1ρv(ρ,t))v(ρ,t)𝑑ρ\displaystyle=\int_{0}^{s}\frac{\partial_{\rho}(\rho^{n-1}\partial_{\rho}v(\rho,t))}{v(\rho,t)}d\rho
=0sρn1(ρv(ρ,t))2v2(ρ,t)𝑑ρ+sn1sv(s,t)v(s,t)\displaystyle=\int_{0}^{s}\frac{\rho^{n-1}(\partial_{\rho}v(\rho,t))^{2}}{v^{2}(\rho,t)}d\rho+\frac{s^{n-1}\partial_{s}v(s,t)}{v(s,t)}
δ0n1sv(s,t)M\displaystyle\geq\delta_{0}^{n-1}\frac{\partial_{s}v(s,t)}{M}

for all (s,t)[δ0,R)×(0,Tmax)(s,t)\in[\delta_{0},R)\times(0,T_{max}), as desired. ∎

The following lemma shows that sv1/s\partial_{s}v\lesssim 1/s near the origin uniformly in time. Our basic approach involves employing a Newtonian kernel-type test function φ(x)=|x|2nR2n\varphi(x)=|x|^{2-n}-R^{2-n}.

Lemma 2.

Let (u,v)(u,v) be a solution given in Proposition 1. For any C>1C_{*}>1, it follows that

sv(s,t)(n2)MCsfor all (s,t)(0,R(11/C)1n2]×(0,Tmax).\partial_{s}v(s,t)\leq\frac{(n-2)MC_{*}}{s}\quad\text{for all }(s,t)\in(0,R(1-1/C_{*})^{\frac{1}{n-2}}]\times(0,T_{max}).
Proof.

In light of the positivity of uu and vv, it holds through integration by parts and the positivity of vv that

0\displaystyle 0 1σnΩBsΔv(x,t)(|x|2nR2n)𝑑x\displaystyle\leq\frac{1}{\sigma_{n}}\int_{\Omega\setminus B_{s}}\Delta v(x,t)(|x|^{2-n}-R^{2-n})dx
=sRρ(ρn1ρv(ρ,t))(ρ2nR2n)dρ\displaystyle=\int_{s}^{R}\partial_{\rho}(\rho^{n-1}\partial_{\rho}v(\rho,t))(\rho^{2-n}-R^{2-n})d\rho
=(n2)sRρv(ρ,t)dρsn1(s2nR2n)sv(s,t)\displaystyle=(n-2)\int_{s}^{R}\partial_{\rho}v(\rho,t)d\rho-s^{n-1}(s^{2-n}-R^{2-n})\partial_{s}v(s,t)
(n2)Msn1(s2nR2n)sv(s,t)\displaystyle\leq(n-2)M-s^{n-1}(s^{2-n}-R^{2-n})\partial_{s}v(s,t)

for all (s,t)(0,R)×(0,Tmax)(s,t)\in(0,R)\times(0,T_{max}). Let C>1C_{*}>1 be arbitrarily chosen. Since sn2Rn2(11C)s^{n-2}\leq R^{n-2}(1-\frac{1}{C_{*}}) implies s2nC(s2nR2n)s^{2-n}\leq C_{*}(s^{2-n}-R^{2-n}) and sv\partial_{s}v is positive as shown in Lemma 1, we obtain the desired result. ∎

2.2 On the mass accumulation function of uu

We now examine the mass accumulation function of uu given by

U(s,t):=1σnBsu(x,t)𝑑x=0sρn1u(ρ,t)𝑑ρU(s,t):=\frac{1}{\sigma_{n}}\int_{B_{s}}u(x,t)dx=\int_{0}^{s}\rho^{n-1}u(\rho,t)d\rho (13)

for (s,t)[0,R]×[0,Tmax)(s,t)\in[0,R]\times[0,T_{max}). Clearly, UU is nonnegative and nondecreasing. Based on the observation that blow-up of the solution can only occur near the origin (Lemma 1), our analysis primarily focuses on the behavior of U(s,t)U(s,t) near the origin. As a preparation, the following simple calculation is needed.

Lemma 3.

Let (u,v)(u,v) be a solution given in Proposition 1 and UU be defined in (13). For any C>1C_{*}>1,

UC0([0,Tmax);C1([0,R]))C2,1((0,R]×(0,Tmax))U\in C^{0}([0,T_{max});C^{1}([0,R]))\cap C^{2,1}((0,R]\times(0,T_{max}))

satisfies

tU(s,t)\displaystyle\partial_{t}U(s,t) (U(s,t))\displaystyle\leq\mathcal{L}(U(s,t)) (14)
:=sn1D(sn+1sU(s,t))s(sn+1sU(s,t))\displaystyle:=s^{n-1}D(s^{-n+1}\partial_{s}U(s,t))\partial_{s}(s^{-n+1}\partial_{s}U(s,t))
+χ(n2)MCs1sU(s,t)\displaystyle\quad+\chi(n-2)MC_{*}s^{-1}\partial_{s}U(s,t)

for all (s,t)(0,R(11/C)1n2]×(0,Tmax)(s,t)\in(0,R(1-1/C_{*})^{\frac{1}{n-2}}]\times(0,T_{max}).

Proof.

The regularity of UU is directly derived from the properties of uu in Proposition 1. From the definition of UU, it is easily seen from (11) that

tU(s,t)=sn1D(sn+1sU(s,t))s(sn+1sU(s,t))+sU(s,t)sv(s,t)\partial_{t}U(s,t)=s^{n-1}D(s^{-n+1}\partial_{s}U(s,t))\partial_{s}(s^{-n+1}\partial_{s}U(s,t))+\partial_{s}U(s,t)\partial_{s}v(s,t) (15)

for all (s,t)(0,R)×(0,Tmax)(s,t)\in(0,R)\times(0,T_{max}). Hence, (14) can be straightforwardly deduced from (15) and Lemma 2. ∎

In the next lemma, we show that when m=1m=1, the inequality (14) provides the behavior of UU near the origin.

Lemma 4.

Let (u,v)(u,v) be a solution given in Proposition 1 and UU be defined in (13). Assume that DD satisfies (8) with some kD>0k_{D}>0, and that

m=1 and M<2n2kD.m=1\quad\text{ and }\quad M<\frac{2}{n-2}k_{D}.

Then, one can find β=β(n,kD,M)(0,2)\beta=\beta(n,k_{D},M)\in(0,2) for which it holds that

U(s,t)Z(s):=Lδ1nβsnβ for all (s,t)(0,δ1)×(0,Tmax)U(s,t)\leq Z(s):=\frac{L}{\delta_{1}^{n-\beta}}s^{n-\beta}\quad\text{ for all }(s,t)\in(0,\delta_{1})\times(0,T_{max}) (16)

with some δ1=δ1(β,R,M,u0)(0,R)\delta_{1}=\delta_{1}(\beta,R,M,u_{0})\in(0,R).

Proof.

Given the condition M<2kD/(n2)M<2k_{D}/(n-2), we first fix β(M(n2)kD,2)\beta\in\left(\frac{M(n-2)}{k_{D}},2\right), choose C(1,kDβM(n2))C_{*}\in\left(1,\frac{k_{D}\beta}{M(n-2)}\right), and define δ1\delta_{1} as

δ1=min{R(11C)1n2,(nLu0)1n,((LM)nnσn(2β))1n(n2)}.\delta_{1}=\min\left\{R\left(1-\frac{1}{C_{*}}\right)^{\frac{1}{n-2}},\left(\frac{nL}{\|u_{0}\|_{\infty}}\right)^{\frac{1}{n}},\left(\frac{(LM)^{n}}{n\sigma_{n}(2-\beta)}\right)^{\frac{1}{n(n-2)}}\right\}.

Let (s,t)(0,δ1)×(0,Tmax)(s,t)\in(0,\delta_{1})\times(0,T_{max}). By performing a fundamental calculation on Z(s)Z(s), we obtain

(Z(s))=sn1D(sn+1sZ(s))s(sn+1sZ(s))+M(n2)Cs1sZ(s)Lβ(nβ)kDδ1nβsnβ2+M(n2)CL(nβ)δ1nβsnβ2=L(nβ)kDδ1nβ[β+M(n2)CkD]snβ20.\begin{split}\mathcal{L}(Z(s))&=s^{n-1}D(s^{-n+1}\partial_{s}Z(s))\partial_{s}(s^{-n+1}\partial_{s}Z(s))+M(n-2)C_{*}s^{-1}\partial_{s}Z(s)\\ &\leq-\frac{L\beta(n-\beta)k_{D}}{\delta_{1}^{n-\beta}}s^{n-\beta-2}+M(n-2)C_{*}\frac{L(n-\beta)}{\delta_{1}^{n-\beta}}s^{n-\beta-2}\\ &=\frac{L(n-\beta)k_{D}}{\delta_{1}^{n-\beta}}\left[-\beta+\frac{M(n-2)C_{*}}{k_{D}}\right]s^{n-\beta-2}\\ &\leq 0.\end{split}

In addition, we note that U(s,t)Z(s)U(s,t)\leq Z(s) along the parabolic boundary of (0,δ1)×(0,Tmax)(0,\delta_{1})\times(0,T_{max}). Indeed, it is trivial that U(0,t)=0=Z(0)U(0,t)=0=Z(0) and U(δ1,t)L=Z(δ1)U(\delta_{1},t)\leq L=Z(\delta_{1}) for all t(0,Tmax)t\in(0,T_{max}). Furthermore, at t=0t=0, we find that

U(s,0)u0nsnu0nδ1βsnβ=u0δ1nnLZ(s)Z(s)U(s,0)\leq\frac{\|u_{0}\|_{\infty}}{n}s^{n}\leq\frac{\|u_{0}\|_{\infty}}{n}\delta_{1}^{\beta}s^{n-\beta}=\frac{\|u_{0}\|_{\infty}\delta_{1}^{n}}{nL}Z(s)\leq Z(s)

for all s(0,δ1)s\in(0,\delta_{1}). Here, the last inequality of the above holds due to the fact that u0δ1nnL\|u_{0}\|_{\infty}\delta_{1}^{n}\leq nL. We now define an auxiliary function Y(s,t)=et(Z(s)U(s,t))Y(s,t)=e^{-t}(Z(s)-U(s,t)) over the domain (0,δ1)×(0,Tmax)(0,\delta_{1})\times(0,T_{max}). Then, we have

tY(s,t)+Y(s,t)=ettU(s,t)et((Z(s))(U(s,t))).\partial_{t}Y(s,t)+Y(s,t)=-e^{-t}\partial_{t}U(s,t)\geq e^{-t}(\mathcal{L}(Z(s))-\mathcal{L}(U(s,t))).

To prove the inequality in (16), we employ a variant of the comparison principle: Let ϵ>0\epsilon>0 and define the set

𝒜ϵ:={t(0,Tmax):Y(s,t)ϵ for some s(0,δ1)}.\mathcal{A}_{\epsilon}:=\left\{t\in(0,T_{max}):Y(s,t)\leq-\epsilon\,\text{ for some }s\in(0,\delta_{1})\right\}.

Assume, for the sake of contradiction, that 𝒜ϵ\mathcal{A}_{\epsilon}\neq\emptyset. Let inf𝒜ϵ=tϵ.\inf\mathcal{A}_{\epsilon}=t_{\epsilon}. Then, tϵ(0,Tmax)t_{\epsilon}\in(0,T_{max}) due to Y0Y\geq 0 on the parabolic boundary of (0,δ1)×(0,Tmax)(0,\delta_{1})\times(0,T_{max}). Thus we can choose sϵ(0,δ1)s_{\epsilon}\in(0,\delta_{1}) fulfilling Y(sϵ,tϵ)=ϵY(s_{\epsilon},t_{\epsilon})=-\epsilon. Note that sY(sϵ,tϵ)=0\partial_{s}Y(s_{\epsilon},t_{\epsilon})=0, s2Y(sϵ,tϵ)0\partial_{s}^{2}Y(s_{\epsilon},t_{\epsilon})\geq 0, and tY(sϵ,tϵ)0\partial_{t}Y(s_{\epsilon},t_{\epsilon})\leq 0. Therefore, it holds that

ϵ\displaystyle-\epsilon tY(sϵ,tϵ)+Y(sϵ,tϵ)\displaystyle\geq\partial_{t}Y(s_{\epsilon},t_{\epsilon})+Y(s_{\epsilon},t_{\epsilon})
=D(sϵn+1sZ(sϵ,tϵ))s2Y(sϵ,tϵ)0,\displaystyle=D(s_{\epsilon}^{-n+1}\partial_{s}Z(s_{\epsilon},t_{\epsilon}))\partial_{s}^{2}Y(s_{\epsilon},t_{\epsilon})\geq 0,

which contradicts ϵ>0\epsilon>0. This implies that 𝒜ϵ=\mathcal{A}_{\epsilon}=\emptyset. Since ϵ>0\epsilon>0 can be taken arbitrarily small, it follows that Y(s,t)0Y(s,t)\geq 0 for every (s,t)(0,δ1)×(0,Tmax)(s,t)\in(0,\delta_{1})\times(0,T_{max}), thereby proving (16). ∎

2.3 Proof of Theorem 1

The upper estimate of UU established in Lemma 4 leads to a more refined behavior for v\nabla v near the origin compared to that in Lemma 2.

Lemma 5.

Let (u,v)(u,v) be a solution given in Proposition 1. Assume that DD satisfies (8) with some kD>0k_{D}>0, and that

m=1 and M<2n2kD.m=1\quad\text{ and }\quad M<\frac{2}{n-2}k_{D}.

Then, there exists β(0,2)\beta\in(0,2) such that for some C>0C>0

sv(s,t)Csβ+1for all (s,t)(0,R)×(0,Tmax).\partial_{s}v(s,t)\leq Cs^{-\beta+1}\quad\text{for all }(s,t)\in(0,R)\times(0,T_{max}). (17)

In particular, for any ϵ>0\epsilon>0, we can find δ=δ(ϵ)>0\delta=\delta(\epsilon)>0 satisfying

v(,t)Ln(Bδ)ϵfor all t(0,Tmax).\|\nabla v(\cdot,t)\|_{L^{n}(B_{\delta})}\leq\epsilon\quad\text{for all }t\in(0,T_{max}). (18)
Proof.

Note that it is enough to consider the case ϵ(0,1)\epsilon\in(0,1). Let ϵ(0,1)\epsilon\in(0,1). We fix β(0,2)\beta\in(0,2) and δ1\delta_{1} as in Lemma 4, and then define

C1=((LM)nnσn(2β))1n(n2)C_{1}=\left(\frac{(LM)^{n}}{n\sigma_{n}(2-\beta)}\right)^{\frac{1}{n(n-2)}}

and

δ=ϵ12βC1n22βδ1nβ2β.\delta=\epsilon^{\frac{1}{2-\beta}}C_{1}^{-\frac{n-2}{2-\beta}}\delta_{1}^{\frac{n-\beta}{2-\beta}}.

Since the definition of δ1\delta_{1} yields δ1C1C1ϵ1n2\delta_{1}\leq C_{1}\leq C_{1}\epsilon^{-\frac{1}{n-2}}, it holds that

δδ1=ϵ12βC1n22βδ1n22β1,\frac{\delta}{\delta_{1}}=\epsilon^{\frac{1}{2-\beta}}C_{1}^{-\frac{n-2}{2-\beta}}\delta_{1}^{\frac{n-2}{2-\beta}}\leq 1,

that is, δδ1\delta\leq\delta_{1}. For fixed (s,t)(0,δ)×(0,Tmax)(s,t)\in(0,\delta)\times(0,T_{max}), it follows from (12) and (7) that

sv(s,t)=sn+10sρn1uv𝑑sMsn+1U(s,t)\partial_{s}v(s,t)=s^{-n+1}\int_{0}^{s}\rho^{n-1}uvds\leq Ms^{-n+1}U(s,t) (19)

which, along with (16), proves (17). Furthermore, (17) leads to (18),

v(,t)Ln(Bδ)=(1σn0δsn1(sv)n𝑑s)1/nC1n2δ1nβδ2β=ϵ.\|\nabla v(\cdot,t)\|_{L^{n}(B_{\delta})}=\left(\frac{1}{\sigma_{n}}\int_{0}^{\delta}s^{n-1}(\partial_{s}v)^{n}ds\right)^{1/n}\leq\frac{C_{1}^{n-2}}{\delta_{1}^{n-\beta}}\delta^{2-\beta}=\epsilon.

This completes the proof. ∎

We are ready to prove the boundedness of uu in LpL^{p} for large pp.

Lemma 6.

Let (u,v)(u,v) be a solution given in Proposition 1. Assume that DD satisfies (8) with some kD>0k_{D}>0, and that either

m>1, or m=1andM<2n2kDm>1,\quad\text{ or }\quad m=1\quad\text{and}\quad M<\frac{2}{n-2}k_{D}

holds. For any p>max{1,m1}p>\max\left\{1,m-1\right\}, there exists C=C(p)>0C=C(p)>0 such that

supt(0,Tmax)u(,t)pC.\sup_{t\in(0,T_{max})}\|u(\cdot,t)\|_{p}\leq C. (20)
Proof.

Let us fix p>max{1,m1}p>\max\left\{1,m-1\right\}. We first consider the case that

m=1 and M<2n2kD.m=1\quad\text{ and }\quad M<\frac{2}{n-2}k_{D}.

Let ϵ=kD8pB\epsilon=\frac{k_{D}}{8pB}, where B=B(n,Ω)>0B=B(n,\Omega)>0 is the constant appeared in Sobolev embedding theorem

f2nn2B(f2+f2) for all fH1(Ω).\|f\|_{\frac{2n}{n-2}}\leq B(\|\nabla f\|_{2}+\|f\|_{2})\quad\text{ for all }f\in H^{1}(\Omega).

For such ϵ\epsilon, we take δ>0\delta>0 as in Lemma 5. By applying Lemma 1 with δ0=δ\delta_{0}=\delta, we obtain that sv\partial_{s}v is bounded in [δ,R)×(0,Tmax)[\delta,R)\times(0,T_{max}). Thus, it holds that

v(,t)Ln(Bδ)ϵ and v(,t)L(Ω\Bδ)C2for all t(0,Tmax)\|\nabla v(\cdot,t)\|_{L^{n}(B_{\delta})}\leq\epsilon\,\text{ and }\,\|\nabla v(\cdot,t)\|_{L^{\infty}(\Omega\backslash B_{\delta})}\leq C_{2}\quad\text{for all }\,t\in(0,T_{max}) (21)

with some C2=C2(δ)>0C_{2}=C_{2}(\delta)>0. For fixed t(0,Tmax)t\in(0,T_{max}), we now observe standard LpL^{p}-estimates for uu:

ddtu(,t)pp+4kD(p1)pup2222(p1)up2up2v1=2(p1)(up2up2vL1(Bδ)+up2up2vL1(Ω\Bδ))=:I1+I2.\begin{split}\frac{d}{dt}\|u(\cdot,t)\|_{p}^{p}&+\frac{4k_{D}(p-1)}{p}\|\nabla u^{\frac{p}{2}}\|_{2}^{2}\\ &\leq 2(p-1)\|u^{\frac{p}{2}}\nabla u^{\frac{p}{2}}\cdot\nabla v\|_{1}\\ &=2(p-1)\left(\|u^{\frac{p}{2}}\nabla u^{\frac{p}{2}}\cdot\nabla v\|_{L^{1}(B_{\delta})}+\|u^{\frac{p}{2}}\nabla u^{\frac{p}{2}}\cdot\nabla v\|_{L^{1}(\Omega\backslash B_{\delta})}\right)\\ &=:I_{1}+I_{2}.\end{split}

By Sobolev embedding, the Gagliardo-Nirenberg inequality, and Young’s inequality, we have

up22nn2\displaystyle\|u^{\frac{p}{2}}\|_{\frac{2n}{n-2}} B(up22+up22)\displaystyle\leq B(\|\nabla u^{\frac{p}{2}}\|_{2}+\|u^{\frac{p}{2}}\|_{2})
B(up22+C3(up22θu1p2(1θ)+u1p2))\displaystyle\leq B(\|\nabla u^{\frac{p}{2}}\|_{2}+C_{3}(\|\nabla u^{\frac{p}{2}}\|_{2}^{\theta}\|u\|_{1}^{\frac{p}{2}(1-\theta)}+\|u\|_{1}^{\frac{p}{2}}))
2Bup22+C4,\displaystyle\leq 2B\|\nabla u^{\frac{p}{2}}\|_{2}+C_{4},

where C3>0C_{3}>0, C4=C4(p)>0C_{4}=C_{4}(p)>0, and θ=p1p1+2/n(0,1)\theta=\frac{p-1}{p-1+2/n}\in(0,1). Hence, it follows from (21) and Young’s inequality that for some C5=C5(p)>0C_{5}=C_{5}(p)>0

I12(p1)up22up22nn2vLn(Bδ)8B(p1)ϵup222+C5kD(p1)pup222+C5.\begin{split}I_{1}&\leq 2(p-1)\|\nabla u^{\frac{p}{2}}\|_{2}\|u^{\frac{p}{2}}\|_{\frac{2n}{n-2}}\|\nabla v\|_{L^{n}(B_{\delta})}\\ &\leq 8B(p-1)\epsilon\|\nabla u^{\frac{p}{2}}\|_{2}^{2}+C_{5}\\ &\leq\frac{k_{D}(p-1)}{p}\|\nabla u^{\frac{p}{2}}\|_{2}^{2}+C_{5}.\end{split}

Similarly,

I22(p1)|up2|up21vL(Ω\Bδ)2(p1)C2up22up222(p1)C2C3up22(up22θu1p2(1θ)+u1p2)kD(p1)pup222+C6,\begin{split}I_{2}&\leq 2(p-1)\||\nabla u^{\frac{p}{2}}|u^{\frac{p}{2}}\|_{1}\|\nabla v\|_{L^{\infty}(\Omega\backslash B_{\delta})}\\ &\leq 2(p-1)C_{2}\|\nabla u^{\frac{p}{2}}\|_{2}\|u^{\frac{p}{2}}\|_{2}\\ &\leq 2(p-1)C_{2}C_{3}\|\nabla u^{\frac{p}{2}}\|_{2}(\|\nabla u^{\frac{p}{2}}\|_{2}^{\theta}\|u\|_{1}^{\frac{p}{2}(1-\theta)}+\|u\|_{1}^{\frac{p}{2}})\\ &\leq\frac{k_{D}(p-1)}{p}\|\nabla u^{\frac{p}{2}}\|_{2}^{2}+C_{6},\end{split}

where C6=C6(p)>0C_{6}=C_{6}(p)>0 and θ=p1p1+2/n(0,1)\theta=\frac{p-1}{p-1+2/n}\in(0,1). Therefore, it follows from the above estimates that

ddtu(,t)pp+2kD(p1)pup222C(p)for all t(0,Tmax).\frac{d}{dt}\|u(\cdot,t)\|_{p}^{p}+\frac{2k_{D}(p-1)}{p}\|\nabla u^{\frac{p}{2}}\|_{2}^{2}\leq C(p)\quad\text{for all }\,t\in(0,T_{max}). (22)

Next, in the case of

m>1,m>1,

we use Lemma 1 and Lemma 2 to control v\nabla v. In a similar way, let t(0,Tmax)t\in(0,T_{max}), and we see that

ddtu(,t)pp+kDp(p1)Ω(1+u)m1up2|u|2𝑑xp(p1)up1uv1.\frac{d}{dt}\|u(\cdot,t)\|_{p}^{p}+k_{D}p(p-1)\int_{\Omega}(1+u)^{m-1}u^{p-2}|\nabla u|^{2}dx\leq p(p-1)\|u^{p-1}\nabla u\cdot\nabla v\|_{1}. (23)

We first note that

4kDp(p1)(p+m1)2Ω|up+m12|2𝑑xkDp(p1)Ω(1+u)m1up2|u|2𝑑x.\frac{4k_{D}p(p-1)}{(p+m-1)^{2}}\int_{\Omega}|\nabla u^{\frac{p+m-1}{2}}|^{2}dx\leq k_{D}p(p-1)\int_{\Omega}(1+u)^{m-1}u^{p-2}|\nabla u|^{2}dx. (24)

On the one hand, a simple calculation and Hölder’s inequality yields

p(p1)up1uv1=2p(p1)p+m1upm+12up+m12v12p(p1)p+m1up+m122upm+12qvq,\begin{split}p(p-1)\|u^{p-1}\nabla u\cdot\nabla v\|_{1}&=\frac{2p(p-1)}{p+m-1}\|u^{\frac{p-m+1}{2}}\nabla u^{\frac{p+m-1}{2}}\cdot\nabla v\|_{1}\\ &\leq\frac{2p(p-1)}{p+m-1}\|\nabla u^{\frac{p+m-1}{2}}\|_{2}\|u^{\frac{p-m+1}{2}}\|_{q}\|\nabla v\|_{q^{\prime}},\end{split}

where q=(p+m1)pm+12nn2q=\frac{(p+m-1)}{p-m+1}\cdot\frac{2n}{n-2} and qq^{\prime} is such that 1q+1q=12.\frac{1}{q}+\frac{1}{q^{\prime}}=\frac{1}{2}. By Sobolev embedding, it follows that

upm+12q\displaystyle\|u^{\frac{p-m+1}{2}}\|_{q} =up+m122nn2pm+1p+m1\displaystyle=\|u^{\frac{p+m-1}{2}}\|_{\frac{2n}{n-2}}^{\frac{p-m+1}{p+m-1}}
B(up+m122pm+1p+m1+up+m122pm+1p+m1)\displaystyle\leq B(\|\nabla u^{\frac{p+m-1}{2}}\|_{2}^{\frac{p-m+1}{p+m-1}}+\|u^{\frac{p+m-1}{2}}\|_{2}^{\frac{p-m+1}{p+m-1}})
B(up+m122pm+1p+m1\displaystyle\leq B(\|\nabla u^{\frac{p+m-1}{2}}\|_{2}^{\frac{p-m+1}{p+m-1}}
+C7(up+m122pm+1p+m1θu1pm+12(1θ)+u1pm+12))\displaystyle\quad+C_{7}(\|\nabla u^{\frac{p+m-1}{2}}\|_{2}^{\frac{p-m+1}{p+m-1}\theta}\|u\|_{1}^{\frac{p-m+1}{2}(1-\theta)}+\|u\|_{1}^{\frac{p-m+1}{2}}))
2Bup+m1222pp+m11+C8,\displaystyle\leq 2B\|\nabla u^{\frac{p+m-1}{2}}\|_{2}^{\frac{2p}{p+m-1}-1}+C_{8},

where C7>0C_{7}>0, C8=C8(p)>0C_{8}=C_{8}(p)>0, and θ=p+m2p+m2+2/n(0,1)\theta=\frac{p+m-2}{p+m-2+2/n}\in(0,1). From Lemma 1 and Lemma 2, we can show that ssvs\partial_{s}v is bounded in (0,R)×(0,Tmax)(0,R)\times(0,T_{max}), which induces that v(,t)q\|\nabla v(\cdot,t)\|_{q^{\prime}} is bounded in (0,Tmax)(0,T_{max}) due to the fact that

1q=121q=12pm+1p+m1n22n>12n22n=1n.\frac{1}{q^{\prime}}=\frac{1}{2}-\frac{1}{q}=\frac{1}{2}-\frac{p-m+1}{p+m-1}\cdot\frac{n-2}{2n}>\frac{1}{2}-\frac{n-2}{2n}=\frac{1}{n}.

Thus, noticing that 2pp+m1<2\frac{2p}{p+m-1}<2, we obtain through Young’s inequality that

p(p1)up1uv12kDp(p1)(p+m1)2up+m1222+C9p(p-1)\|u^{p-1}\nabla u\cdot\nabla v\|_{1}\leq\frac{2k_{D}p(p-1)}{(p+m-1)^{2}}\|\nabla u^{\frac{p+m-1}{2}}\|_{2}^{2}+C_{9} (25)

with some C9=C9(p)>0C_{9}=C_{9}(p)>0. Hence, combining (24) and (25) into (23), we see that

ddtu(,t)pp+2kDp(p1)(p+m1)2up+m1222C9for all t(0,Tmax).\frac{d}{dt}\|u(\cdot,t)\|_{p}^{p}+\frac{2k_{D}p(p-1)}{(p+m-1)^{2}}\|\nabla u^{\frac{p+m-1}{2}}\|_{2}^{2}\leq C_{9}\quad\text{for all }\,t\in(0,T_{max}). (26)

In addition, the Gagliardo-Nirenberg inequality, Young’s inequality, and (6) yield that for any μ>0\mu>0 and m1m\geq 1, we can find C(p,μ)>0C(p,\mu)>0 such that

up222μup+m1222+C(p,μ).\|u^{\frac{p}{2}}\|_{2}^{2}\leq\mu\|\nabla u^{\frac{p+m-1}{2}}\|_{2}^{2}+C(p,\mu).

Therefore, writing y(t):=up2(,t)22y(t):=\|u^{\frac{p}{2}}(\cdot,t)\|_{2}^{2}, both (22) and (26) can be rewritten by

y(t)+y(t)C(p),y^{\prime}(t)+y(t)\leq C(p),

which induces (20). ∎

Proof of Theorem 1.

Note that the standard elliptic regularity theory and Lemma 6 with p>np>n show v(,t)\|\nabla v(\cdot,t)\|_{\infty} is bounded uniformly in time. Thus, as in (23) and (24), we have for some C>0C>0

ddtu(,t)pp\displaystyle\frac{d}{dt}\|u(\cdot,t)\|_{p}^{p} +4kDp(p1)(p+m1)2Ω|up+m12|2𝑑x\displaystyle+\frac{4k_{D}p(p-1)}{(p+m-1)^{2}}\int_{\Omega}|\nabla u^{\frac{p+m-1}{2}}|^{2}dx
=2p(p1)p+m1Ωupm+12up+m12vdx\displaystyle=-\frac{2p(p-1)}{p+m-1}\int_{\Omega}u^{\frac{p-m+1}{2}}\nabla u^{\frac{p+m-1}{2}}\cdot\nabla vdx
2kDp(p1)(p+m1)2Ω|up+m12|2𝑑x+Cp(p1)Ωupm+1𝑑x\displaystyle\leq\frac{2k_{D}p(p-1)}{(p+m-1)^{2}}\int_{\Omega}|\nabla u^{\frac{p+m-1}{2}}|^{2}dx+Cp(p-1)\int_{\Omega}u^{p-m+1}dx

for all p>max{1,m1}p>\max\left\{1,m-1\right\}. Now, employing a standard iteration method (see e.g., [4, 17]), we can show that u(,t)\|u(\cdot,t)\|_{\infty} is uniformly bounded in (0,Tmax)(0,T_{max}), which proves Theorem 1 with the help of blow-up criteria in Proposition 1. ∎

3 Global boundedness for 0<m<10<m<1

Lemma 7.

Let (u,v)(u,v) be a solution given in Proposition 1. Assume that DD satisfies (9). For any γ(0,22m)\gamma\in(0,\frac{2}{2-m}), there exists M=M(γ,u0)>0M_{*}=M_{*}(\gamma,u_{0})>0 such that if MMM\leq M_{*}, then UU defined in (13) satisfies

U(s,t)W(s):=ηsnγ for all (s,t)(0,R)×(0,Tmax),U(s,t)\leq W(s):=\eta s^{n-\gamma}\quad\text{ for all }(s,t)\in(0,R)\times(0,T_{max}), (27)

where η=η(u0,γ):=supr(0,R)rn+γ0rρn1u0(ρ)𝑑ρ\eta=\eta(u_{0},\gamma):=\sup_{r\in(0,R)}r^{-n+\gamma}\int_{0}^{r}\rho^{n-1}u_{0}(\rho)d\rho.

Proof.

We begin by noting that η>0\eta>0 is finite since u0u_{0} satisfies (4). For fixed (s,t)(0,R)×(0,Tmax)(s,t)\in(0,R)\times(0,T_{max}), since (15) and (19) are still valid, we have

tU(s,t)\displaystyle\partial_{t}U(s,t) (U(s,t))\displaystyle\leq\mathcal{H}(U(s,t))
:=sn1(1+sn+1sU(s,t))m1s(sn+1sU(s,t))\displaystyle:=s^{n-1}(1+s^{-n+1}\partial_{s}U(s,t))^{m-1}\partial_{s}(s^{-n+1}\partial_{s}U(s,t))
+Msn+1U(s,t)sU(s,t).\displaystyle\quad+Ms^{-n+1}U(s,t)\partial_{s}U(s,t).

We again use the comparison principle, as in the proof of Lemma 4; we first observe that

(W(s,t))\displaystyle\mathcal{H}(W(s,t)) =ηγ(nγ)(1+η(nγ)sγ)m1snγ2+Mη2(nγ)sn2γ\displaystyle=-\eta\gamma(n-\gamma)(1+\eta(n-\gamma)s^{-\gamma})^{m-1}s^{n-\gamma-2}+M\eta^{2}(n-\gamma)s^{n-2\gamma}
=η(nγ)sn2γ[Mηγ(sγ+η(nγ))m1sγ(2m)2].\displaystyle=\eta(n-\gamma)s^{n-2\gamma}\left[M\eta-\gamma(s^{\gamma}+\eta(n-\gamma))^{m-1}s^{\gamma(2-m)-2}\right].

Thus, for any

MM:=γη(Rγ+η(nγ))1mR2γ(2m),M\leq M_{*}:=\frac{\gamma}{\eta(R^{\gamma}+\eta(n-\gamma))^{1-m}R^{2-\gamma(2-m)}},

we have (W(s,t))0.\mathcal{H}(W(s,t))\leq 0. Furthermore, it follows from the definition of γ\gamma that for all t(0,Tmax)t\in(0,T_{max}) and r(0,R)r\in(0,R),

U(R,t)W(R) and U(r,0)W(r), as well as U(0,t)=0=W(0).U(R,t)\leq W(R)\,\text{ and }\,U(r,0)\leq W(r),\,\text{ as well as }\,U(0,t)=0=W(0).

Consequently, as in the proof of Lemma 4, we can conclude (27). ∎

We now proceed to the standard LqL^{q}- estimates for uu. As a preliminary step, we introduce a simplified version of the Hardy inequality on balls.

Lemma 8.

For any radially symmetric function f(H1Ll)(BR)f\in(H^{1}\cap L^{l})(B_{R}), l>0l>0, there exists C>0C>0 such that

BR(f(x))2|x|2𝑑xC(f22+fl2).\int_{B_{R}}\frac{(f(x))^{2}}{|x|^{2}}dx\leq C(\|\nabla f\|_{2}^{2}+\|f\|_{l}^{2}). (28)
Proof.

By applying integration by parts and Young’s inequality, we can derive

0Rsn3f2(s)𝑑s\displaystyle\int_{0}^{R}s^{n-3}f^{2}(s)ds =2n20Rsn2f(s)sf(s)ds+1n2Rn2f2(R)\displaystyle=-\frac{2}{n-2}\int_{0}^{R}s^{n-2}f(s)\partial_{s}f(s)ds+\frac{1}{n-2}R^{n-2}f^{2}(R) (29)
120Rsn3f2(s)𝑑s+C10Rsn1(sf(s))2𝑑s\displaystyle\leq\frac{1}{2}\int_{0}^{R}s^{n-3}f^{2}(s)ds+C_{1}\int_{0}^{R}s^{n-1}(\partial_{s}f(s))^{2}ds
+1(n2)RBRf2(x)𝑑S\displaystyle\quad+\frac{1}{(n-2)R}\int_{\partial B_{R}}f^{2}(x)dS

with some C1>0C_{1}>0. Note that it is enough to consider the case l<2l<2. By utilizing the standard trace embedding W1,2(Ω)L2(Ω)W^{1,2}(\Omega)\hookrightarrow L^{2}(\partial\Omega) and subsequently applying the Gagliardo-Nirenberg inequality along with Young’s inequality, we obtain

fL2(BR)2\displaystyle\|f\|_{L^{2}(\partial B_{R})}^{2} C2(f22+f22)\displaystyle\leq C_{2}(\|\nabla f\|_{2}^{2}+\|f\|_{2}^{2}) (30)
C3(f22+f22θfl2(1θ)+fl2)\displaystyle\leq C_{3}(\|\nabla f\|_{2}^{2}+\|\nabla f\|_{2}^{2\theta}\|f\|_{l}^{2(1-\theta)}+\|f\|_{l}^{2})
C4(f22+fl2),\displaystyle\leq C_{4}(\|\nabla f\|_{2}^{2}+\|f\|_{l}^{2}),

where Ci>0C_{i}>0, i=2,3,4i=2,3,4, and θ=1l121l12+1n(0,1)\theta=\frac{\frac{1}{l}-\frac{1}{2}}{\frac{1}{l}-\frac{1}{2}+\frac{1}{n}}\in(0,1). In consequence, by employing (29) and (30) together, we have (28). ∎

Lemma 9.

Let (u,v)(u,v) be a solution given in Proposition 1. Under the same assumptions as in Lemma 7, if MMM\leq M_{*}, where MM_{*} is defined in Lemma 7, then there exist qq with q>(n1)(2m)m+1m\displaystyle q>\frac{(n-1)(2-m)}{m}+1-m and C=C(q)>0C=C(q)>0 such that

supt(0,Tmax)u(,t)qC.\sup_{t\in(0,T_{max})}\|u(\cdot,t)\|_{q}\leq C. (31)
Proof.

For given γ(0,22m)\gamma\in(0,\frac{2}{2-m}), we pick q>1q>1 such that

(n1)(2m)m+1m<q<n1(γ1)++1m.\frac{(n-1)(2-m)}{m}+1-m<q<\frac{n-1}{(\gamma-1)_{+}}+1-m. (32)

Then, we choose k>1k>1 to satisfy

1m<1k<1(n1)(2m)q+m1.1-m<\frac{1}{k}<1-\frac{(n-1)(2-m)}{q+m-1}. (33)

For fixed t(0,Tmax)t\in(0,T_{max}), we multiply the first equation in (3) by q(u+1)q1q(u+1)^{q-1} and integrate by parts, then add Ω(u+1)q𝑑x\int_{\Omega}(u+1)^{q}dx to both sides to obtain

ddtΩ(u+1)q𝑑x+\displaystyle\frac{d}{dt}\int_{\Omega}(u+1)^{q}dx+ Ω(u+1)qdx+4q(q1)(m+q1)2Ω|(u+1)q+m12|2dx\displaystyle\int_{\Omega}(u+1)^{q}dx+\frac{4q(q-1)}{(m+q-1)^{2}}\int_{\Omega}|\nabla(u+1)^{\frac{q+m-1}{2}}|^{2}dx (34)
=q(q1)Ω(u+1)q2uuvdx+Ω(u+1)q𝑑x.\displaystyle=-q(q-1)\int_{\Omega}(u+1)^{q-2}u\nabla u\cdot\nabla vdx+\int_{\Omega}(u+1)^{q}dx.

Define a nonnegative function F(ξ)=0ξ(ζ+1)q2ζ𝑑ζ.F(\xi)=\int_{0}^{\xi}(\zeta+1)^{q-2}\zeta d\zeta. Given the nonnegativity of νv|Ω\partial_{\nu}v|_{\partial\Omega} and the increasing nature of the function [0,)ζ(ζ+1)q2ζ[0,\infty)\ni\zeta\mapsto(\zeta+1)^{q-2}\zeta, it follows from integration by parts again, uniform boundedness of vv as established in (7), and the Sobolev embedding that

Ω(u+1)q2uuvdx\displaystyle-\int_{\Omega}(u+1)^{q-2}u\nabla u\cdot\nabla vdx =ΩF(u)uv𝑑xΩF(u)νvdS\displaystyle=\int_{\Omega}F(u)uvdx-\int_{\partial\Omega}F(u)\partial_{\nu}vdS (35)
MΩ(u+1)q2u3𝑑x\displaystyle\leq M\int_{\Omega}(u+1)^{q-2}u^{3}dx
Mσn0Rsn1(u(s,t)+1)q+1𝑑s.\displaystyle\leq M\sigma_{n}\int_{0}^{R}s^{n-1}(u(s,t)+1)^{q+1}ds.

For convenience, we define w(s,t):=(u(s,t)+1)q+m12w(s,t):=(u(s,t)+1)^{\frac{q+m-1}{2}} and z(s,t):=sn12w(s,t)z(s,t):=s^{\frac{n-1}{2}}w(s,t). By combining (34) and (35), and then applying Young’s inequality and Hölder inequality, we obtain

ddt(u+1)qq\displaystyle\frac{d}{dt}\|(u+1)\|_{q}^{q} +(u+1)qq+w22\displaystyle+\|(u+1)\|_{q}^{q}+\|\nabla w\|_{2}^{2} (36)
C50Rs(n1)(2m)q+m1(z(s,t))2(q+1)q+m1𝑑s+C5\displaystyle\leq C_{5}\int_{0}^{R}s^{-\frac{(n-1)(2-m)}{q+m-1}}(z(s,t))^{\frac{2(q+1)}{q+m-1}}ds+C_{5}
C5zL2(q+1)kq+m1((0,R))2(q+1)q+m1(0Rs(n1)(2m)q+m1kk1𝑑s)k1k+C5\displaystyle\leq C_{5}\|z\|_{L^{\frac{2(q+1)k}{q+m-1}}((0,R))}^{\frac{2(q+1)}{q+m-1}}\left(\int_{0}^{R}s^{-\frac{(n-1)(2-m)}{q+m-1}\cdot\frac{k}{k-1}}ds\right)^{\frac{k-1}{k}}+C_{5}

with some C5>0C_{5}>0, where kk is given by (33). Since

(n1)(2m)q+m1kk1<1\frac{(n-1)(2-m)}{q+m-1}\cdot\frac{k}{k-1}<1

due to (33), it follows that

(0Rs(n1)(2m)q+m1kk1ds)k1kC6,\bigr{(}\int_{0}^{R}s^{-\frac{(n-1)(2-m)}{q+m-1}\cdot\frac{k}{k-1}}ds\bigr{)}^{\frac{k-1}{k}}\leq C_{6}, (37)

where C6>0C_{6}>0. Next, using a one-dimensional version of the Gagliardo-Nirenberg inequality, we observe that

zL2(q+1)kq+m1((0,R))C7szL2((0,R))θzL2q+m1((0,R))1θ+zL2q+m1((0,R))\|z\|_{L^{\frac{2(q+1)k}{q+m-1}}((0,R))}\leq C_{7}\|\partial_{s}z\|_{L^{2}((0,R))}^{\theta}\|z\|_{L^{\frac{2}{q+m-1}}((0,R))}^{1-\theta}+\|z\|_{L^{\frac{2}{q+m-1}}((0,R))} (38)

with some C7>0C_{7}>0, where

θ=q+m1q+m1(q+1)kq+m(0,1).\theta=\frac{q+m-1-\frac{q+m-1}{(q+1)k}}{q+m}\in(0,1).

We note that

2(q+1)q+m1θ=2(11k+m1q+m)<2\frac{2(q+1)}{q+m-1}\theta=2\left(1-\frac{\frac{1}{k}+m-1}{q+m}\right)<2

due to (33). In addition, we have

szL2((0,R))20Rsn1(sw(s,t))2𝑑s+(n1)240Rsn3w2(s,t)𝑑s,\|\partial_{s}z\|_{L^{2}((0,R))}^{2}\leq\int_{0}^{R}s^{n-1}(\partial_{s}w(s,t))^{2}ds+\frac{(n-1)^{2}}{4}\int_{0}^{R}s^{n-3}w^{2}(s,t)ds,

which, with an application of Lemma 8 with f(r)=w(r,t)f(r)=w(r,t) and l=2q+m1l=\frac{2}{q+m-1}, entails that

szL2((0,R))C8(w2+1)\|\partial_{s}z\|_{L^{2}((0,R))}\leq C_{8}(\|\nabla w\|_{2}+1) (39)

with some C8>0C_{8}>0. Moreover, we use integration by parts to estimate

zL2q+m1((0,R))2q+m1\displaystyle\|z\|_{L^{\frac{2}{q+m-1}}((0,R))}^{\frac{2}{q+m-1}} =0Rsn1q+m1(u(s,t)+1)𝑑s\displaystyle=\int_{0}^{R}s^{\frac{n-1}{q+m-1}}(u(s,t)+1)ds (40)
=0Rsn+1+n1q+m1sU(s,t)ds+R\displaystyle=\int_{0}^{R}s^{-n+1+\frac{n-1}{q+m-1}}\partial_{s}U(s,t)ds+R
=C9(0Rsn+n1q+m1U(s,t)𝑑s+1),\displaystyle=C_{9}(\int_{0}^{R}s^{-n+\frac{n-1}{q+m-1}}U(s,t)ds+1),

where C9>0C_{9}>0. Since n1q+m1γ>1\frac{n-1}{q+m-1}-\gamma>-1 by (32), applying (27) to (40) allows us to deduce for some C10>0C_{10}>0

zL2q+m1((0,R))2q+m1C9η0Rsn1q+m1γ𝑑s+C9C10,\|z\|_{L^{\frac{2}{q+m-1}}((0,R))}^{\frac{2}{q+m-1}}\leq C_{9}\eta\int_{0}^{R}s^{\frac{n-1}{q+m-1}-\gamma}ds+C_{9}\leq C_{10}, (41)

where η\eta is a number defined in Lemma 7. Therefore, by combining (39) and (41) with (38), we infer that

zL2(q+1)kq+m1((0,R))2(q+1)q+m1C11(w22(q+1)q+m1θ+1),\|z\|_{L^{\frac{2(q+1)k}{q+m-1}}((0,R))}^{\frac{2(q+1)}{q+m-1}}\leq C_{11}(\|\nabla w\|_{2}^{\frac{2(q+1)}{q+m-1}\theta}+1), (42)

where C11>0C_{11}>0. Consequently, by plugging (37) and (42) into (36), we can choose C12>0C_{12}>0 using Young’s inequality that

ddt(u+1)qq+(u+1)qq+w22\displaystyle\frac{d}{dt}\|(u+1)\|_{q}^{q}+\|(u+1)\|_{q}^{q}+\|\nabla w\|_{2}^{2} C5C6C11(w22(q+1)q+m1θ+1)+C5\displaystyle\leq C_{5}C_{6}C_{11}(\|\nabla w\|_{2}^{\frac{2(q+1)}{q+m-1}\theta}+1)+C_{5}
12w22+C12.\displaystyle\leq\frac{1}{2}\|\nabla w\|_{2}^{2}+C_{12}.

Therefore, the function y(t):=(u(,t)+1)qqy(t):=\|(u(\cdot,t)+1)\|_{q}^{q} satisfies

y(t)+y(t)C(q),y^{\prime}(t)+y(t)\leq C(q),

thereby leading to the result in (31). ∎

Proof of Theorem 2.

Let q0=qq_{0}=q satisfy (32). By Lemma 9, there exists C>0C>0 such that u(,t)q0C\|u(\cdot,t)\|_{q_{0}}\leq C for all t(0,Tmax)t\in(0,T_{max}). Let pp be such that pmax{n1,q01}p\geq\max\left\{n-1,q_{0}-1\right\}. For fixed t(0,Tmax)t\in(0,T_{max}), we employ again (34) and (35) to obtain, for some C>0C>0,

ddt(u+1)pp+(u+1)pp+(u+1)p+m1222C((u+1)p+1p+1+1).\frac{d}{dt}\|(u+1)\|_{p}^{p}+\|(u+1)\|_{p}^{p}+\|\nabla(u+1)^{\frac{p+m-1}{2}}\|_{2}^{2}\leq C(\|(u+1)\|_{p+1}^{p+1}+1). (43)

By the Gagliardo-Nirenberg inequality, we see that for some C>0C>0

(u+1)p+1p+1\displaystyle\|(u+1)\|_{p+1}^{p+1} =(u+1)p+m122(p+1)p+m12(p+1)p+m1\displaystyle=\|(u+1)^{\frac{p+m-1}{2}}\|_{\frac{2(p+1)}{p+m-1}}^{\frac{2(p+1)}{p+m-1}} (44)
C((u+1)p+m1222(p+1)p+m1θ(u+1)p+m122q0p+m12(p+1)p+m1(1θ)\displaystyle\leq C\biggr{(}\|\nabla(u+1)^{\frac{p+m-1}{2}}\|_{2}^{\frac{2(p+1)}{p+m-1}\theta}\|(u+1)^{\frac{p+m-1}{2}}\|_{\frac{2q_{0}}{p+m-1}}^{\frac{2(p+1)}{p+m-1}(1-\theta)}
+(u+1)p+m122q0p+m12(p+1)p+m1)\displaystyle\quad+\|(u+1)^{\frac{p+m-1}{2}}\|_{\frac{2q_{0}}{p+m-1}}^{\frac{2(p+1)}{p+m-1}}\biggr{)}
=C((u+1)p+m1222(p+1)p+m1θu+1q0(p+1)(1θ)+u+1q0p+1).\displaystyle=C(\|\nabla(u+1)^{\frac{p+m-1}{2}}\|_{2}^{\frac{2(p+1)}{p+m-1}\theta}\|u+1\|_{q_{0}}^{(p+1)(1-\theta)}+\|u+1\|_{q_{0}}^{p+1}).

Here, due to the condition on pp, we have θ=p+m12q0p+m12(p+1)p+m1n(2m)12+1n(0,1).\displaystyle\theta=\frac{\frac{p+m-1}{2q_{0}}-\frac{p+m-1}{2(p+1)}}{\frac{p+m-1}{n(2-m)}-\frac{1}{2}+\frac{1}{n}}\in(0,1). Furthermore, since q0q_{0} fulfills q0>(n1)(2m)m+1m>n2(2m)q_{0}>\frac{(n-1)(2-m)}{m}+1-m>\frac{n}{2}(2-m), it follows that (p+1)θp+m1<1.\frac{(p+1)\theta}{p+m-1}<1. Thus, substituting (44) into (43) and applying Young’s inequality, we establish the LpL^{p}-boundedness of uu for any pmax{n1,q01}p\geq\max\left\{n-1,q_{0}-1\right\}. Hence, similar to the proof of Theorem 1, we can apply the standard iteration method to achieve the uniform boundedness of u(,t)\|u(\cdot,t)\|_{\infty} in (0,Tmax)(0,T_{max}). ∎

4 Blow-up in a finite or infinite time for 0<m<2n0<m<\frac{2}{n}

This section aims to detect the blow-up phenomena that occur due to the competition of diffusive and cross-diffusive dynamics. This will be verified through the analysis of a temporal evolution involving mass accumulation function UU.

4.1 Estimates for vv at the origin

We begin by addressing the lower bound for lnv\nabla\ln v. For a detailed proof, we refer to [24, Lemma 3.1].

Lemma 10.

Let (u,v)(u,v) be a solution given in Proposition 1 and UU be defined in (13). It holds that

s(lnv)(s,t)sn+1U(s,t)1+0sρn+1U(ρ,t)𝑑ρfor all (s,t)(0,R)×(0,Tmax).\partial_{s}(\ln v)(s,t)\geq\frac{s^{-n+1}U(s,t)}{1+\int_{0}^{s}\rho^{-n+1}U(\rho,t)d\rho}\quad\text{for all }\,(s,t)\in(0,R)\times(0,T_{max}). (45)

As a result of Lemma 10, the value of vv at the origin can be controlled by MM with a quantity involving UU.

Lemma 11.

Let (u,v)(u,v) be a solution given in Proposition 1 and UU be defined in (13). It holds that

v(0,t)M1+ψ(t) for all t(0,Tmax),v(0,t)\leq\frac{M}{1+\psi(t)}\quad\text{ for all }t\in(0,T_{max}), (46)

where ψ(t):=0Rρn+1U(ρ,t)𝑑ρ\psi(t):=\int_{0}^{R}\rho^{-n+1}U(\rho,t)d\rho.

Proof.

From (45), we have

lnMv(0,t)=0Rs(lnv(s,t))ds0Rs(ln(1+0sρn+1U(ρ,t)𝑑ρ))ds=ln(1+0Rρn+1U(ρ,t)𝑑ρ),\begin{split}\ln\frac{M}{v(0,t)}&=\int_{0}^{R}\partial_{s}(\ln v(s,t))ds\\ &\geq\int_{0}^{R}\partial_{s}(\ln(1+\int_{0}^{s}\rho^{-n+1}U(\rho,t)d\rho))ds\\ &=\ln(1+\int_{0}^{R}\rho^{-n+1}U(\rho,t)d\rho),\end{split}

which immediately gives (46). ∎

4.2 ODI for a functional involving UU

We derive an ODI for ϕ(t)\phi(t), which is crucial for our blow-up argument.

Lemma 12.

Let (u,v)(u,v) be a solution given in Proposition 1 and UU be defined in (13). Assume that DD satisfies (10) with some KD>0K_{D}>0, and that

0<m<2n.0<m<\frac{2}{n}.

Then, for any fixed α(n3,n(1m)1)\alpha\in(n-3,n(1-m)-1), a quantity

ϕ(t):=0RsαU(s,t)𝑑s\phi(t):=\int_{0}^{R}s^{-\alpha}U(s,t)ds

belongs to C0([0,Tmax))C1((0,Tmax))C^{0}([0,T_{max}))\cap C^{1}((0,T_{max})), and satisfies

tϕ(t)C1Lm+C2(Mv(0,t))2M\partial_{t}\phi(t)\geq-C_{1}L^{m}+C_{2}\frac{(M-v(0,t))^{2}}{M} (47)

with some C1>0C_{1}>0 and C2>0C_{2}>0.

Proof.

Let t(0,Tmax)t\in(0,T_{max}). Obviously, ϕ\phi is well-defined for such α\alpha, and the regularity property of ϕ\phi is guaranteed via the dominated convergence theorem. In order to see (47), we first define

D~(ξ):=0ξD(σ)𝑑σ for ξ0.\tilde{D}(\xi):=\int_{0}^{\xi}D(\sigma)d\sigma\quad\text{ for }\xi\geq 0. (48)

We recall that UUsatisfies the following equation (15) in (0,R)×(0,Tmax)(0,R)\times(0,T_{max}):

tU(s,t)=sn1D(sn+1sU(s,t))s(sn+1sU(s,t))+sU(s,t)sv(s,t).\partial_{t}U(s,t)=s^{n-1}D(s^{-n+1}\partial_{s}U(s,t))\partial_{s}(s^{-n+1}\partial_{s}U(s,t))+\partial_{s}U(s,t)\partial_{s}v(s,t).

Then, ϕ(t)\phi(t) is formulated as the following:

tϕ(t)=0Rsn1αs(D~(sn+1sU(s,t)))ds+0RsαsU(s,t)sv(s,t)ds=:I1(t)+I2(t).\begin{split}\partial_{t}\phi(t)=&\int_{0}^{R}s^{n-1-\alpha}\partial_{s}(\tilde{D}(s^{-n+1}\partial_{s}U(s,t)))ds\\ &+\int_{0}^{R}s^{-\alpha}\partial_{s}U(s,t)\partial_{s}v(s,t)ds\\ =:&I_{1}(t)+I_{2}(t).\end{split}

From the assumption (10) and (48), we have

D~(ξ)KD0ξ(1+σ)m1𝑑σKDmξm,\tilde{D}(\xi)\leq K_{D}\int_{0}^{\xi}(1+\sigma)^{m-1}d\sigma\leq\frac{K_{D}}{m}\xi^{m},

which entails, via integration by parts, that

I1(t)=(n1α)0Rsn2αD~(sn+1sU(s,t))𝑑s+{sn1αD~(sn+1sU(s,t))}|s=0RKD(n1α)m0Rsn2αm(n1)(sU(s,t))m𝑑s,\begin{split}I_{1}(t)&=-(n-1-\alpha)\int_{0}^{R}s^{n-2-\alpha}\tilde{D}(s^{-n+1}\partial_{s}U(s,t))ds\\ &+\left\{s^{n-1-\alpha}\tilde{D}(s^{-n+1}\partial_{s}U(s,t))\right\}\bigg{|}_{s=0}^{R}\\ &\geq-\frac{K_{D}(n-1-\alpha)}{m}\int_{0}^{R}s^{n-2-\alpha-m(n-1)}(\partial_{s}U(s,t))^{m}ds,\end{split} (49)

where we used the condition α<n(1m)1\alpha<n(1-m)-1 to see that

lims0|sn1αD~(sn+1sU(s,t))|lims0sn1αu(,t)m=0.\lim_{s\rightarrow 0}|s^{n-1-\alpha}\tilde{D}(s^{-n+1}\partial_{s}U(s,t))|\leq\lim_{s\rightarrow 0}s^{n-1-\alpha}\|u(\cdot,t)\|_{\infty}^{m}=0.

Applying Hölder inequality and integration by parts, we further compute the rightmost term of (49) as

0Rsn2αm(n1)(sU(s,t))mds(0Rsn2αm(n1)1m𝑑s)1m(0RsU(s,t)ds)m.\begin{split}\int_{0}^{R}&s^{n-2-\alpha-m(n-1)}(\partial_{s}U(s,t))^{m}ds\\ &\leq\left(\int_{0}^{R}s^{\frac{n-2-\alpha-m(n-1)}{1-m}}ds\right)^{1-m}\left(\int_{0}^{R}\partial_{s}U(s,t)ds\right)^{m}.\end{split} (50)

Here, the first integral on the right-hand side is finite because α<n(1m)1\alpha<n(1-m)-1 implies

n2αm(n1)1m>1.\frac{n-2-\alpha-m(n-1)}{1-m}>-1.

Hence, in light of (49) and (50), one can find C1>0C_{1}>0 such that

I1(t)C1(U(R,t)U(0,t))m=C1Lm.I_{1}(t)\geq-C_{1}(U(R,t)-U(0,t))^{m}=-C_{1}L^{m}.

In order to estimate I2(t)I_{2}(t), we first observe from (12) that

sU(s,t)=s(sn1sv(s,t))vs(sn1slnv(s,t)).\partial_{s}U(s,t)=\frac{\partial_{s}(s^{n-1}\partial_{s}v(s,t))}{v}\geq\partial_{s}(s^{n-1}\partial_{s}\ln v(s,t)). (51)

for all s(0,R)s\in(0,R). Using (51), vs0v_{s}\geq 0, and integration by parts, we find

I2(t)0Rsαs(sn1slnv(s,t))sv(s,t)ds=α0Rsnα2slnv(s,t)sv(s,t)ds0Rsnα1slnv(s,t)ssv(s,t)ds+{snα1slnv(s,t)sv(s,t)}|s=0R.\begin{split}I_{2}(t)&\geq\int_{0}^{R}s^{-\alpha}\partial_{s}(s^{n-1}\partial_{s}\ln v(s,t))\partial_{s}v(s,t)ds\\ &=\alpha\int_{0}^{R}s^{n-\alpha-2}\partial_{s}\ln v(s,t)\partial_{s}v(s,t)ds\\ &\quad-\int_{0}^{R}s^{n-\alpha-1}\partial_{s}\ln v(s,t)\partial_{ss}v(s,t)ds\\ &\quad+\left\{s^{n-\alpha-1}\partial_{s}\ln v(s,t)\partial_{s}v(s,t)\right\}\bigg{|}_{s=0}^{R}.\end{split} (52)

We employ (12) again to see that

0Rsnα1slnv(s,t)ssv(s,t)ds=0Rsnα1slnv(s,t)(u(s,t)v(s,t)n1ssv(s,t))ds=I2(t)+(n1)0Rsnα2slnv(s,t)sv(s,t)ds,\begin{split}&-\int_{0}^{R}s^{n-\alpha-1}\partial_{s}\ln v(s,t)\partial_{ss}v(s,t)ds\\ &\quad=-\int_{0}^{R}s^{n-\alpha-1}\partial_{s}\ln v(s,t)\left(u(s,t)v(s,t)-\frac{n-1}{s}\partial_{s}v(s,t)\right)ds\\ &\quad=-I_{2}(t)+(n-1)\int_{0}^{R}s^{n-\alpha-2}\partial_{s}\ln v(s,t)\partial_{s}v(s,t)ds,\end{split}

which yields, due to the nonnegativity of the last term in (52) and (7), that

I2(t)n+α120Rsnα2slnv(s,t)sv(s,t)dsn+α12M0Rsnα2(sv(s,t))2𝑑s.\begin{split}I_{2}(t)&\geq\frac{n+\alpha-1}{2}\int_{0}^{R}s^{n-\alpha-2}\partial_{s}\ln v(s,t)\partial_{s}v(s,t)ds\\ &\geq\frac{n+\alpha-1}{2M}\int_{0}^{R}s^{n-\alpha-2}(\partial_{s}v(s,t))^{2}ds.\end{split} (53)

We observe from Hölder inequality that

(Mv(0,t))2=(0Rsv(s,t)ds)2(0Rsnα2(sv(s,t))2𝑑s)(0Rsn+α+2𝑑s),\begin{split}(M-v(0,t))^{2}&=\left(\int_{0}^{R}\partial_{s}v(s,t)ds\right)^{2}\\ &\leq\left(\int_{0}^{R}s^{n-\alpha-2}(\partial_{s}v(s,t))^{2}ds\right)\cdot\left(\int_{0}^{R}s^{-n+\alpha+2}ds\right),\end{split} (54)

where the last integral is finite due to the condition n3<αn-3<\alpha. Therefore, combining (53) and (54), it follows that

I2(t)C2(Mv(0,t))2MI_{2}(t)\geq C_{2}\frac{(M-v(0,t))^{2}}{M}

with some C2>0C_{2}>0. Hence, we can conclude (47). ∎

We are ready to present the proof of Theorem 3.

Proof of Theorem 3..

Let m(0,2n)m\in(0,\frac{2}{n}), and fix α\alpha as in Lemma 12. Plugging (46) into (47) shows that

tϕ(t)C1Lm+C2Mψ2(t)(1+ψ(t))2\partial_{t}\phi(t)\geq-C_{1}L^{m}+C_{2}M\frac{\psi^{2}(t)}{(1+\psi(t))^{2}} (55)

for all t(0,Tmax)t\in(0,T_{max}), where C1C_{1} and C2C_{2} are constants specified in Lemma 12. For M>0M>0 and ξ0\xi\geq 0, let us define

ω(ξ,M):=2C1Lm+C2Mξ2(1+ξ)2,\omega(\xi,M):=-2C_{1}L^{m}+C_{2}M\frac{\xi^{2}}{(1+\xi)^{2}},

and fix a constant depending u0u_{0}

CC(u0):=12Rn+α+1ϕ(0).C^{*}\equiv C^{*}(u_{0}):=\frac{1}{2}R^{-n+\alpha+1}\phi(0).

We choose M=M(u0)>0M^{*}=M^{*}(u_{0})>0 such that

M2C1LmC2(1+CC)2.M^{*}\geq\frac{2C_{1}L^{m}}{C_{2}}\cdot\left(\frac{1+C^{*}}{C^{*}}\right)^{2}.

Since 0xx2/(1+x)20\leq x\mapsto x^{2}/(1+x)^{2} is increasing, it holds that for ξC\xi\geq C^{*} and MMM\geq M^{*}

ω(ξ,M)2C1Lm+C2M(C1+C)20.\omega(\xi,M)\geq-2C_{1}L^{m}+C_{2}M^{*}\left(\frac{C^{*}}{1+C^{*}}\right)^{2}\geq 0. (56)

Upon fixing MMM\geq M^{*}, and defining ϕ\phi and ψ\psi as in Lemma 12, the continuity of ϕ\phi allows us to find T>0T>0 such that ϕ(t)>ϕ(0)2\phi(t)>\frac{\phi(0)}{2} for all t(0,T)t\in(0,T). This establishes the existence of

T=sup{T(0,Tmax):ϕ(t)>ϕ(0)2 for all t(0,T)}.T^{*}=\sup\left\{T\in(0,T_{max}):\phi(t)>\frac{\phi(0)}{2}\,\text{ for all }\,t\in(0,T)\right\}.

We show T=TmaxT^{*}=T_{max}. Indeed, since it holds that

ψ(t)Rn+α+1ϕ(t)12Rn+α+1ϕ(0)=C\psi(t)\geq R^{-n+\alpha+1}\phi(t)\geq\frac{1}{2}R^{-n+\alpha+1}\phi(0)=C^{*}

for all t(0,T)t\in(0,T^{*}), it follows, in conjunction with (55) and (56), that

tϕ(t)ω(ψ(t),M)+C1LmC1Lm\partial_{t}\phi(t)\geq\omega(\psi(t),M)+C_{1}L^{m}\geq C_{1}L^{m} (57)

for all t(0,T)t\in(0,T^{*}). Solving the ODI (57) yields

ϕ(t)ϕ(0)+C3t for all t(0,T)\phi(t)\geq\phi(0)+C_{3}t\quad\text{ for all }t\in(0,T^{*}) (58)

with some C3>0C_{3}>0, necessitating T=TmaxT^{*}=T_{max} because if T<TmaxT^{*}<T_{max}, the continuity of ϕ\phi would imply ϕ(T)=ϕ(0)2\phi(T^{*})=\frac{\phi(0)}{2}, contradicting the growth established in (58). Hence, in accordance with (58), we obtain

C3t+ϕ(0)ϕ(t)Rnα+1u(,t),C_{3}t+\phi(0)\leq\phi(t)\leq R^{n-\alpha+1}\|u(\cdot,t)\|_{\infty},

which gives the desired results. In particular, in the three-dimensional case, since α\alpha can be set to satisfy 0<α<min{3(1m)1,1}0<\alpha<\min\left\{3(1-m)-1,1\right\}, it holds that

C3t+ϕ(0)ϕ(t)=0RsαU(s,t)𝑑sL1αR1αC_{3}t+\phi(0)\leq\phi(t)=\int_{0}^{R}s^{-\alpha}U(s,t)ds\leq\frac{L}{1-\alpha}R^{1-\alpha}

for all t(0,Tmax)t\in(0,T_{max}), which shows that TmaxT_{max} must be finite, as desired. ∎

Acknowledgement

J. Ahn is supported by NRF grant No. RS-2024-00336346. K. Kang is supported by NRF grant No. RS-2024-00336346. D. Kim is supported by the National Research Foundation of Korea(NRF) grant funded by the Korea government(MSIT)(grant No. 2022R1A4A1032094).

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