# Lower Bound for the Blow-Up Time for the Nonlinear Reaction-Diffusion System in High Dimensions

Authors: Baiping Ouyang; Wei Fan; Yiwu Lin
Journal: Discrete Dynamics in Nature and Society
 (2020)

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## Abstract

In this paper, we study the blow-up phenomenon for a nonlinear reaction-diffusion system with time-dependent coefficients under nonlinear boundary conditions. Using the technique of a first-order differential inequality and the Sobolev inequalities, we can get the energy expression which satisfies the differential inequality. The lower bound for the blow-up time could be obtained if blow-up does really occur in high dimensions.


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## Body

## 1. Introduction

During the past decades, the blow-up phenomena for the solutions to the parabolic problems have been widely concerned. It is important in practice that how to determine the bound of the blow-up timet∗ of the solutions about the parabolic equations and systems. Their applications are included in physics, chemistry, astronomy, biology, and population dynamics [1, 2]. Actually, when the blow-up occurs at t∗, it is difficult to get the exact value of t∗. We mainly focus on estimating its bounds. At present, the studies on the blow-up phenomena of parabolic problems mainly focus on homogeneous Dirichlet boundary condition and homogeneous Neumann and Robin boundary conditions [3–12]. There are also some works under nonlinear boundary conditions [13–15]. Most of these articles are focused on R3. There are only a few papers dealing with a lower bound for the blow-up time in high dimensions (see [16–18]). Recently, some scholars have started to investigate the blow-up problems with time-dependent coefficients [19–21]. In paper [21], the authors considered the following nonlinear reaction-diffusion system with time-dependent coefficients:(1)ut=△u+k1tupvq,x,t∈Ω×0,t∗,vt=△v+k2tvrus,x,t∈Ω×0,t∗,ux,t=vx,t=0,x,t∈∂Ω×0,t∗,ux,0=u0x,vx,0=v0x,x∈Ω.The authors obtained the lower and upper bounds for the blow-up time when the blow-up occurred. In this paper, we further consider the blow-up phenomena for the following system with time-dependent coefficients under nonlinear boundary conditions in high dimensions:(2)ut=Δum+k1tupvq,vt=Δvl+k2tvrus,x,t∈Ω×0,t∗,∂u∂n=g1u,∂v∂n=g2v,x,t∈∂Ω×0,t∗,ux,0=u0x≥0,vx,0=v0x≥0,x∈Ω.We assume thatgiζ are continuous, and α,β,p,q,r,s,a1,a2,m,l,σ, and ζ satisfy(3)g1ζ≤a1ζα,g2ζ≤a2ζβ,ζ>0,α>1,β>1,p>1,q>0,r>1,s>0,a1>0,a2>0,maxσp+q−1σ−q−2σn+1,1<m<σp+q−qσ−q,maxσr+s−1σ−s−2σn+1,1<l<σr+s−sσ−s,where σ is a positive constant to be defined later.Our goal in this paper is to obtain a lower bound for the blow-up time of the solutions to systems (2) and (3) in Rn for any n>3. The nonlinear terms Δum and Δvl and the boundary conditions are difficult to tackle. We cannot get the result by following the method proposed in [21], so we must use a new method to overcome these difficulties. To the best of our knowledge, no results exist in that direction, and we think our result is new and interesting.In the further discussions, we will use the following Hölder inequality:(4)∫Ωwx1+x2dx≤∫Ωwx1/n1dx1/n1∫Ωwx2/n2dx1/n2,where w is a nonnegative function and x1, x2, n1, and n2 are positive constants satisfying 1/n1+1/n2=1.We also need the following Sobolev inequality [22]:(5)∫Ωuσ+m−1n/n−2dx≤C2n/n−22n/n−2−1∫Ωuσ+m−1dxn/n−2+∫Ω∇uσ+m−1/22dxn/n−2,(6)∫Ωvσ+l−1n/n−2dx≤C2n/n−22n/n−2−1∫Ωvσ+l−1dxn/n−2+∫Ω∇vσ+l−1/22dxn/n−2,with C=Cn,Ω which is a Sobolev embedding constant depending on n and Ω.And the classical (or elementary) inequality is(7)a+bw≤aw+bw,where a, b, and w are positive constants, and w satisfies 0<w≤1.


## 2. Lower Bound for the Blow-Up Time

In this part, we define an auxiliary function of the form(8)ϕt=k1δt∫Ωuσdx+k2χt∫Ωvσdx,where δ=σ−qmn−n+2σ/2σp+q−1, χ=σ−sln−n+2σ/2σr+s−1, and σ>maxα−1n,β−1n,q,s.We establish the following theorem:Theorem 1.
Letux,t be the weak solution of problems (1)–(3) in a bounded convex domain ΩΩ∈Rnn>3. Then, the quantity ϕt defined in (8) satisfies the integral inequality(9)Θt∗≥∫ϕ∞1η+ηξ1+ηξ2+ηξ3+ηξ4=S,which follows that the blow-up time t∗ is bounded below. We have(10)t∗≥Θ−1S,where Θ, ξ1, ξ2, ξ3, and ξ4 will be defined later.Now, we prove Theorem1. For simplicity, assume that the solution is classical of problems (1)–(3). The general case can be done by approximation. Differentiating ϕt, we have(11)ϕ′t=δk1δ−1tk1′t∫Ωuσdx+σk1δt∫Ωuσ−1utdx+χk2χ−1tk2′t∫Ωvσdx+σk2χt∫Ωvσ−1vtdx≤Lϕt+σk1δt∫Ωuσ−1△umdx+σk1δ+1t∫Ωuσ+p−1vqdx+σk2χt∫Ωvσ−1△vldx+σk2χ+1t∫Ωvσ+r−1usdx,where L=maxδk1′t/k1t,χk2′t/k2t.For the second term on the right side of (11), we apply the divergence theorem, the L1 trace embedding, and (3) to get(12)∫Ωuσ−1△umdx=∫∂Ωuσ−1∂um∂ndA−∫Ω∇uσ−1⋅∇umdx≤m∫∂Ωuσ+m−2∂u∂ndA−mσ−1∫Ωuσ+m−3∇u2dx≤ma1∫∂Ωuσ+m+α−2dA−mσ−1∫Ωuσ+m−3∇u2dx≤ma1nρ0∫Ωuσ+m+α−2dx+ma1σ+m+α−2dρ0∫Ωuσ+m+α−3∇udx−4mσ−1σ+m−12∫Ω∇uσ+m−1/22dx,where ρ0:=min∂Ωx⋅n⟶, n⟶ is the outward normal vector of ∂Ω and d:=max∂Ωx.For the second term on the right side of (12), using (11), we obtain(13)∫Ωuσ+m+α−3∇udx≤12ε1∫Ωuσ+m+2α−3dx+2ε1σ+m−12∫Ω∇uσ+m−1/22dx,where ε1 is a positive constant which will be defined later. Using (4), we have(14)∫Ωuσ+m+α−2dx≤∫Ωuσ+m+2α−3dxx10∫Ωuσdxx20≤x10∫Ωuσ+m+2α−3dx+x20∫Ωuσdx,where x10=m+α−2/m+2α−3andx20=α−1/m+2α−3.Choosingx11=m+2α−2n−2/m−1n+2σ and x21=m−1n+2σ−m+2α−2n−2/m−1n+2σ and using (4), (5), and (7), we have(15)∫Ωuσ+m+2α−3dx≤∫Ωuσ+m−1n/n−2dxx11∫Ωuσdxx21≤r1nx11n−2∫Ωuσ+m−1dx+r1n−2−nx11n−21+ε2−nx11/n−2−nx11∫Ωuσdxx21n−2/n−2−nx11+r1nx11n−2ε2∫Ω∇uσ+m−1/22dx,where r1=C2n/n−22n/n−2−1x11 and ε2 is a positive constant which will be defined later.For the first term on the right side of (15), using (4) and Young’s inequality, we have(16)∫Ωuσ+m−1dx≤∫Ωuσ+m+2α−3dxx12∫Ωuσdxx22≤x12ε3∫Ωuσ+m+2α−3dx+x22ε3−m−1/2α−2∫Ωuσdx,where x12=m−1/m+2α−3, x22=2α−2/m+2α−3, and ε3 is a positive constant which will be defined later.Combining (15) and (16), if we choose suitable ε3 such that r1nx11x12ε3/n−2=1/2, we have(17)∫Ωuσ+m+2α−3dx≤r2∫Ωuσdx+r3∫Ωuσdxx21n−2/n−2−nx11+r4∫Ω∇uσ+m−1/22dx,where r2=2r1nx11/n−2x22ε3−m−1/2α−2, r3=2r1n−2−nx11/n−21+ε2−nx11/n−2−nx11, and r4=2r1nx11/n−2ε2.Combining (12), (13), (14), and (17), we have(18)∫Ωuσ−1△umdx≤r5r2+ma1nx20ρ0∫Ωuσdx+r5r3∫Ωuσdxx21n−2/n−2−nx11+r6∫Ω∇uσ+m−1/22dx,where r5=ma1nx10/ρ0+ma1σ+m+α−2d/2ε1ρ0 and r6=ma1σ+m+α−2d/ρ0⋅2ε1/σ+m−12−4mσ−1/σ+m−12.Similarly, for the fourth term on the right side of (11), using the divergence theorem and (3), we have(19)∫Ωvσ−1△vldx=∫∂Ωvσ−1∂vl∂ndA−∫Ω∇vσ−1⋅∇vldx≤la2nρ0∫Ωvσ+l+β−2dx+la2σ+l+β−2dρ0∫Ωvσ+l+β−3∇vdx−4lσ−1σ+l−12∫Ω∇vσ+v−1/22dx.For the second term on the right side of (19), using (4), we obtain(20)∫Ωvσ+l+β−3∇vdx≤12ε4∫Ωvσ+l+2β−3dx+2ε4σ+l−12∫Ω∇vσ+l−1/22dx,where ε4 is a positive constant which will be defined later. Similarly, we have(21)∫Ωvσ+l+β−2dx≤∫Ωvσ+l+2β−3dxy10∫Ωvσdxy20≤y10∫Ωvσ+l+2β−3dx+y20∫Ωvσdx,where y10=l+β−2/l+2β−3 and y20=β−1/l+2β−3.Choosingy11=l+2β−2n−2/l−1n+2σ and y21=l−1n+2σ−l+2β−3n−2/l−1n+2σ and using (4), (6), and (7), we have(22)∫Ωvσ+l+2β−3dx≤∫Ωvσ+l−1n/n−2dxy11∫Ωvσdxy21≤r7ny11n−2∫Ωvσ+l−1dx+r7n−2−ny11n−21+ε5−ny11/n−2−ny11∫Ωvσdxy21n−2/n−2−ny11+r7ny11n−2ε5∫Ω∇vσ+l−1/22dx,where r7=C2n/n−22n/n−2−1y11 and ε5 is a positive constant which will be defined later.For the first term on the right side of (22), using (4), we have(23)∫Ωvσ+l−1dx≤∫Ωvσ+l+2β−3dxy12∫Ωvσdxy22≤y12ε6∫Ωvσ+l+2β−3dx+y22ε6−l−1/2β−2∫Ωvσdx,where y12=l−1/l+2β−3, y22=2β−2/l+2β−3, and ε6 is a positive constant which will be defined later.Combining (22) and (23), if we choose suitable ε6 such that r7ny11y12ε6/n−2=1/2, we have(24)∫Ωvσ+l+2β−3dx≤r8∫Ωvσdx+r9∫Ωvσdxy21n−2/n−2−ny11+r10∫Ω∇vσ+l−1/22dx,where r8=2r7ny11y22/n−2ε6−l−1/2β−2,r9=2r7n−2−ny11/n−21+ε5−ny11/n−2−ny11,andr10=2r7ny11/n−2ε5.Combining (19)–(21) and (24), we have(25)∫Ωvσ−1△vldx≤r11r8+la2ny20ρ0∫Ωvσdx+r11r9∫Ωvσdxy21n−2/n−2−ny11+r12∫Ω∇vσ+l−1/22dx,where r11=la2ny10/ρ0+la2σ+l+β−2d/2ε4ρ0 and r12=la2σ+l+β−2d/ρ0⋅2ε4/σ+l−12−4lσ−1/σ+l−12.For the third term on the right side of (11), using Hölder inequality and Young’s inequality, we have(26)σk1δ+1t∫Ωuσ+p−1vqdx≤σk1δ+1t∫Ωuσ+p−1σ/σ−qdxσ−q/σ∫Ωvσdxq/σ≤σ−qk1δ+1t∫Ωuσ+p−1σ/σ−qdx+qk1δ+1t∫Ωvσdx.For the first term on the right side of (26), using (4), (5), and (7) and taking care of the given condition δ=σ−qmn−n+2σ/2σp+q−1, we have(27)k1δ+1t∫Ωuσ+p−1σ/σ−qdx≤k1δ+1t∫Ωuσ+m−1n/n−2dxx13∫Ωuσdxx23≤k1δ+1tC2n/n−22n/n−2−1x13∫Ωuσ+m−1dxx13n/n−2+∫Ω∇uσ+m−1/22dxx13n/n−2∫Ωuσdxx23=λ1k1δ+1t∫Ωuσ+m−1dxx13n/n−2∫Ωuσdxx23+λ1k1δ+1t∫Ω∇uσ+m−1/22dxx13n/n−2∫Ωuσdxx23≤λ1x13nn−2k1δ+1t∫Ωuσ+m−1dx+λ1n−2−nx13n−2k11−2x13δ/n−2−nx13+ε7−nx13δ/n−2−nx13⋅k1δt∫Ωuσdxx23n−2/n−2−nx13+λ1x13nε7n−2k1δt∫Ω∇uσ+m−1/22dx,where λ1=C2n/n−22n/n−2−1x13, x13=σp+q−1n−2/σ−qmn−n+2σ, and x23=σ−qmn−n+2σ−σp+q−1n−2/σ−qmn−n+2σ.For the first term on the right side of (27), using (4) and Young’s inequality, we get(28)∫Ωuσ+m−1dx≤x14ε7∫Ωuσ+p−1σ/σ−q+x24ε7−x14/x24∫Ωuσdx,where x14=m−1σ−q/σσ+q−1, x24=σσ+q−1−m−1σ−q/σσ+q−1, and ε7 is a positive constant which will be defined later.Combining (27) and (28), if we choose suitable ε7 such that λ1nx13x14ε7/n−2=1/2, we have(29)σ−qk1δ+1t∫Ωuσ+p−1σ/σ−qdx≤K1tk1δt∫Ωuσdx+K2tk1δt∫Ωuσdxx23n−2/n−2−nx13+λ2k1δt∫Ω∇uσ+m−1/22dx,where K1t=2σ−qλ1nx13x24/n−2ε7−x14/x24k1t, K2t=2σ−qλ1n−2−nx13/n−2k11−2x13δ/n−2−nx13t+ε7−nx13/n−2−nx13, and λ2=2σ−qλ1nx13ε7/n−2.Combining (26) and (29), we obtain(30)σk1δ+1t∫Ωuσ+p−1vqdx≤K1tk1δt∫Ωuσdx+K2tk1δt∫Ωuσdxx23n−2/n−2−nx13+λ2k1δt∫Ω∇uσ+m−1/22dx+qk1δ+1t∫Ωvσdx.By the same way, for the fifth term on the right side of (11), using (4), we have(31)σk2χ+1t∫Ωvσ+r−1usdx≤σ−sk2χ+1t∫Ωvσ+r−1σ/σ−sdx+sk2χ+1t∫Ωuσdx.For the first term on the right side of (31), using (4), (6), and (7) and taking care of the given condition χ=σ−sln−n+2σ/2σr+s−1, we have(32)k2χ+1t∫Ωvσ+r−1σ/σ−sdx≤k2χ+1t∫Ωvσ+l−1n/n−2dxy13∫Ωvσdxy23≤λ3y13nn−2k2χ+1t∫Ωvσ+l−1dx+λ3n−2−ny13n−2k21−2y13χ/n−2−ny13t+ε8−ny13χ/n−2−ny13⋅k2χt∫Ωvσdxy23n−2/n−2−ny13+λ3y13nε8n−2k2χt∫Ω∇vσ+l−1/22dx,where λ3=C2n/n−22n/n−2−1y13, y13=σr+s−1n−2/σ−sln−n+2σ, y23=σ−sln−n+2σ−σr+s−1n−2/σ−sln−n+2σ, and ε8 is a positive constant which will be defined later.For the first term on the right side of (32), using (4), we get(33)∫Ωvσ+l−1dx≤y14ε9∫Ωvσ+r−1σ/σ−s+y24ε9−y14/y24∫Ωvσdx,where y14=l−1σ−s/σσ+s−1, y24=σσ+s−1−l−1σ−s/σσ+s−1, and ε9 is a positive constant which will be defined later.Combining (32) and (33), if we choose suitable ε9 such that λ2ny13y14ε9/n−2=1/2, we have(34)σ−sk2χ+1t∫Ωvσ+r−1σ/σ−sdx≤K3tk2χt∫Ωvσdx+K4tk2χt∫Ωvσdxy23n−2/n−2−ny13+λ4k2χt∫Ω∇vσ+l−1/22dx,where K3t=2σ−sλ3ny13y24/n−2ε9−y14/y24k2t, K4t=2σ−sλ3n−2−ny13/n−2k21−2y13δ/n−2−ny13t+ε9−ny13/n−2−ny13, and λ4=2σ−sλ3ny13ε9/n−2.Combining (31) and (34), we obtain(35)σk2χ+1t∫Ωvσ+r−1usdx≤K3tk2χt∫Ωvσdx+K4tk2χt∫Ωvσdxy23n−2/n−2−ny13+λ4k2χt∫Ω∇vσ+l−1/22dx+sk2χ+1t∫Ωuσdx.Combining (11), (18), (25), (30), and (35), we have(36)ϕ′t≤L+K˜1+K˜2t+K˜3tϕt+K˜4tk1δt∫Ωuσdxx21n−2/n−2−nx11+K˜5tk2χt∫Ωvσdxy21n−2/n−2−ny11+r6σ+λ2k1δt∫Ω∇uσ+m−1/22dx+r12σ+λ4k2χt∫Ω∇vσ+l−1/22dx+K2tk1δt∫Ωuσdxx23n−2/n−2−nx13+K4tk2χt∫Ωvσdxy23n−2/n−2−ny13,where K˜1=maxσr5r2+ma1nx20ρ0−1,σr11r8+la2ny20ρ0−1,K˜2t=K1t+K3t, K˜3t=maxqk1δ+1tk2−χt,sk2χ+1tk1−δt, K˜4t=r5r3σk1−2x11δ/n−2−nx11t, and K˜5t=r11r9σk2−2y11δ/n−2−ny11t.If we choose suitableε1, ε4, ε7, and ε9 such that r6σ+λ2≤0,r12σ+λ4≤0, we can rewrite (36) as(37)ϕ′t≤K˜6tϕt+K˜4tϕtξ1+K˜5tϕtξ2+K2tϕtξ3+K4tϕtξ4,where ξ1=1+2x11/n−2−nx11,ξ2=1+2y11/n−2−ny11, ξ3=1+2x13/n−2−nx13, ξ4=1+2y13/n−2−ny13, and K˜6=L+K˜1+K˜2t+K˜3t.Let(38)Θt∗=∫0t∗Kτdτ,where Kt=K˜6t+K˜4t+K˜5t+K2t+K4t.Integrating (37) from 0 to t∗, we have(39)Θt∗≥∫ϕ0∞1η+ηξ1+ηξ2+ηξ3+ηξ4dη=S.Consideringξi>1i=1,2,3,4, the integration of the right side of (39) exists. It is clear that Θt∗ is an increasing function. So, we can get(40)t∗≥Θ−1S,where Θ−1 is the inverse function of Θ.The proof of Theorem1 is complete.



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