# Blow-Up Solution of a Porous Medium Equation with Nonlocal Boundary Conditions

Authors: Huimin Tian; Lingling Zhang
Journal: Complexity
 (2020)

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## Abstract

In this paper, we devote to studying the blow-up phenomena for a porous medium equation under nonlocal boundary conditions. Based on auxiliary functions and differential inequality technique, we derive the sufficient conditions to guarantee the existence of blow-up solutions under different measures and obtain an upper bound for blow-up time. Moreover, we demonstrate the lower bounds for blow-up time under some appropriate measures inR3 and in the higher-dimensional space Rnn≥3, respectively. At last, two examples are given to illustrate the applications of main results.


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## Body

## 1. Introduction

As is known to all, porous medium equations usually describe processes involving fluid flow, heat transfer, or diffusion. There are a lot of applications in physical phenomena, mathematical biology, lubrication, boundary layer theory, and other fields [1, 2].The paper is focused on the following porous medium equation with time-dependent coefficient:(1)ut=Δum+axk1tfu,x,t∈Ω×0,t∗,subject to nonlocal boundary and initial conditions:(2)∂u∂ν=k2t∫Ωgudx,x,t∈∂Ω×0,t∗,(3)ux,0=u0x≥0,x∈Ω¯,where m>1 and Ω⊂Rnn≥3 is a bounded convex domain with smooth boundary ∂Ω. ∂u/∂ν is the outward normal derivative on ∂Ω,andt∗ is the maximal existence time of u. Set ℝ+=0,+∞. Throughout the paper, we assume that f,g∈C1ℝ+¯ are nonnegative, the weight function a∈CΩ¯ is positive, and k1,k2∈C1ℝ+¯ are positive functions. u0∈CΩ¯ is the initial value which satisfies compatibility conditions. By the degenerate parabolic theory in [3], one can deduce that the local classical solution of (1) exists uniquely and is nonnegative.During the past decades, there have been many works to study the blow-up phenomena of parabolic equations and systems. The related results include the existence of blow-up and global solutions, the bounds of the blow-up time, and blow-up rate. We refer to [4–12] and references therein. Porous medium equations as representative examples of parabolic equations have been widely investigated by many scholars [13–17]. In order to research problem (1), the recent papers have aroused our interest. Xiao and Fang [14] considered the following porous medium equation under nonlinear boundary conditions:(4)ut=Δum−ktfu,x,t∈Ω×0,t∗,∂um∂ν=gu,x,t∈∂Ω×0,t∗,ux,0=u0x≥0,x∈Ω¯,where m≥1andΩ⊂ℝnn≥2 is a bounded star-shaped region with smooth boundary ∂Ω. Using the auxiliary function method and modified differential inequality technique, they established some conditions on time-dependent coefficient and nonlinear functions to guarantee that the solution ux,t exists globally or blows up at some finite time. Moreover, the upper and lower bounds for the blow-up time were derived in the higher-dimensional space.In [17], the authors dealt with the blow-up problem of the following porous medium equation subject to nonlinear boundary and initial conditions:(5)ut=Δum+ktfu,x,t∈Ω×0,t∗,∂u∂ν=gu,x,t∈∂Ω×0,t∗,ux,0=u0x≥0,x∈Ω¯,where m>1andΩ⊂Rnn≥2 is a bounded convex region with smooth boundary ∂Ω. Under appropriate assumptions on the data, a criterion was given to guarantee that solution u blows up at finite time, and an upper bound and a lower bound for blow-up time were derived.Recently, there are some papers on the issue of studying reaction-diffusion equations under nonlocal boundary conditions. Marras andVernier Piro [18] studied the following reaction-diffusion equation under nonlocal boundary conditions:(6)ut=Δu+k1tfu,x,t∈Ω×0,t∗,∂u∂ν=k2t∫Ωgudx,x,t∈∂Ω×0,t∗,ux,0=u0x≥0,x∈Ω¯,where Ω is a bounded convex region in Rnn≥2 with smooth boundary ∂Ω. Under some conditions on data, the authors showed the solution must blow up at finite time t∗. At the same time, they obtained upper bounds of t∗. When Ω⊂R3, lower bounds of t∗ were derived.In [19], the authors were concerned about the following equations:(7)ut=∇⋅ρu∇u+k1tfu,x,t∈Ω×0,t∗,∂u∂ν=k2t∫Ωgudx,x,t∈∂Ω×0,t∗,ux,0=u0x≥0,x∈Ω¯,where Ω is a bounded convex region in Rnn≥2 and the boundary ∂Ω is smooth. By constructing some auxiliary functions and using differential inequality technique, they derived that the solution blows up at some finite time. Moreover, upper and lower bounds of the blow-up time were obtained.In this paper, our main interest lies in blow-up phenomena of problem (1). Obviously, boundary conditions (2) and (3) are different from the one in [14, 17], and the main part of equation (1) is different from the problem in [18, 19]. Therefore, we need to establish the new auxiliary functions to prove our main results. By means of new auxiliary functions and differential inequality technique, we prove the existence of blow-up solutions under different measures and obtain an upper bound for blow-up time. Moreover, we demonstrate the lower bounds for blow-up time under some appropriate measures in R3 and in the higher-dimensional space Rnn≥3, respectively.The rest of this paper is arranged as follows. In Section , a criterion for the blow-up solutions of problems (1)–(3) is established, and we obtain an upper bound for blow-up time. Section 3 shows that a lower bound is presented under two different measures when blow-up does occur. In Section 4, two examples are given to illustrate our main results.


## 2. Blow-Up Solution and Upper Bound fort∗

The section shows the blow-up solution for problem (1) in two different measures. And we get the sufficient conditions to ensure an upper bound for the blow-up time under two different measures. Suppose that the functions k1,a,f satisfy(8)fs≥s1/α+1,s>0,(9)k1t≥γ,t≥0,ax≥β,x∈Rn,where α,β,andγ are positive constants, and they satisfy(10)αm<α+1.To obtain our first result, we define the following auxiliary function:(11)Φt≔k1αt∫Ωω12udx.Letλ1 be the first positive eigenvalue and ω1 be the corresponding eigenfunction of the following problem:(12)Δω+λω=0,x∈Ω,ω=0,x∈∂Ω,with(13)∫Ωω2dx=1.Now, we show our theorem for the blow-up solution in the measureΦt.Theorem 1.
Letu be a nonnegative classical solution of problem (1). Assume that (8)–(10) and the following assumptions hold:(14)k1′tk1t≥η,t≥0,(15)βk1αm−α0Φ0α+1−αm/α−2λ1>0,(16)βΦ0α+1−αm/α−2λ1γαm−α>0,where η is a positive constant. Then, u blows up at some finite time t∗ in the measure Φt, and(17)t∗≤∫Φ0+∞dsαη0s+βsα+1/α−2λ1γα−αmsm.Proof.
Applying the divergence theorem and conditions (8)–(14), we have(18)Φ′t=αk1α−1tk1′t∫Ωω12udx+k1αt∫Ωω12utdx=αk1′tk1tk1αt∫Ωω12udx+k1αt∫Ωω12utdx≥αηΦt+k1αt∫Ωω12Δum+axk1tfudx≥αηΦt+k1αt∫ΩΔω12umdx+βk1α+1t∫Ωω12u1/α+1dx≥αηΦt+2k1αt∫Ωum∇ω12dx+2k1αt∫Ωω1umΔω1dx+βk1α+1t∫Ωω12u1/α+1dx.
From Hölder inequality, we get(19)∫Ωω12umdx≤∫Ωω12u1/α+1dxαm/α+1∫Ωω12dx1−αm/α+1=∫Ωω12u1/α+1dxαm/α+1.
Then, we rewrite (18) to gain(20)Φ′t≥αηΦt−2λ1k1αt∫Ωω12u1/α+1dxαm/α+1+βk1α+1t∫Ωω12u1/α+1dx≥αηΦt+k1αt∫Ωω12u1/α+1dxαm/α+1βk1t∫Ωω12u1/α+1dx1−αm/α+1−2λ1.
Using Hölder inequality again, we have(21)∫Ωω12udx≤∫Ωω12u1/α+1dxα/α+1∫Ωω12dx1/α+1=∫Ωω12u1/α+1dxα/α+1.
We substitute (21) into (20) to get(22)Φ′t≥αηΦt+k1αt∫Ωω12u1/α+1dxαm/α+1βk1t∫Ωω12udxα+1−αm/α−2λ1=αηΦt+k1αt∫Ωω12u1/α+1dxαm/α+1βk1αm−αtΦtα+1−αm/α−2λ1.
Now, we prove that(23)βk1αm−αtΦtα+1−αm/α−2λ1>0,t>0.
Suppose that (23) does not hold. Set t1=mint>0βk1αm−αtΦtα+1−αm/α−2λ1≤0. Then, for 0≤t<t1, we have(24)βk1αm−αtΦtα+1−αm/α−2λ1>0.
According to (22), for 0≤t<t1, we have Φ′t>0. Hence, (15) yields(25)Φt1>Φ0>2λ1βk1αm−α0α/α+1−αm,that is,(26)βk1αm−α0Φt1α+1−αm/α−2λ1>0.
Sincek1t is increasing, we get(27)βk1αm−αt1Φt1α+1−αm/α−2λ1>0.
There is a contradiction. Therefore, (23) holds. Similarly, from (16), we deduce(28)βΦtα+1/α−m−2λ1γα−αm>0.
Next, applying (21) to (22), we obtain(29)Φ′t≥αηΦt+k1α−αmtΦmtβk1αm−αtΦtα+1−αm/α−2λ1=αηΦt+βΦtα+1/α−2λ1k1α−αmtΦtm≥αηΦt+βΦtα+1/α−2λ1γα−αmΦtm=αηΦt+ΦtmβΦtα+1/α−m−2λ1γα−αm>0,t>0,where (23) and (28) are used. Integrating (29) from 0 to t yields(30)t≤∫Φ0Φtdsαηs+βsα+1/α−2λ1γα−αmsm.
From (30), we can conclude that the solution u blows up at some finite time t∗ in the measure Φt. In fact, assume that u remains global in the measure Φt; then, Φt<+∞,t>0, and(31)t≤∫Φ0Φtdsαηs+βsα+1/α−2λ1γα−αmsm<∫Φ0+∞dsαηs+βsα+1/α−2λ1γα−αmsm.
Whent⟶+∞, we have(32)∫Φ0+∞dsαηs+βsα+1/α−2λ1γα−αmsm=+∞.
However, sinceΦ0>0,α+1/α>m>1, we have(33)∫Φ0+∞dsαηs+βsα+1/α−2λ1γα−αmsm<+∞,that is, it is a contradiction. Then, the solution u blows up at some finite time t∗ in the measure Φt. Taking the limit as t⟶t∗ in (30), we obtain(34)t∗≤∫Φ0+∞dsαηs+βsα+1/α−2λ1γα−αmsm.
In the following corollary, we will present the blow-up solution for (1) in the measure Ψt, where(35)Ψt∫Ωω12udx.
From the proof similar to Theorem1, we get the following conclusion:Corollary 1.
Letu be a nonnegative classical solution of problem (1). Assume that (8)–(10) and the following assumptions hold:(36)βγΨ0α+1−αm/α−2λ1>0.
Then,u blows up at some finite time T in the measure Ψt, and(37)T≤∫Ψ0+∞dsβγsα+1/α−2λ1sm.


## 3. Lower Bound fort∗

The section presents lower bounds for blow-up timet∗ under two different measures when the solution of (1)–(3) blows up at some finite time. Firstly, we state two lemmas from [20] and some inequalities, which play a basic role in the process of our proof.Lemma 1.
LetΩ be a bounded star-shaped domain in RN,N≥2. Then, we have(38)∫∂Ωundσ≤Nρ0∫Ωundx+ndρ0∫Ωun−1∇udx,where u is a nonnegative C1-function, ρ0=min∂Ωx⋅ν, d=maxΩ¯x, and ν is the unit outward normal vector on ∂Ω.Lemma 2.
LetΩ be a bounded domain in R3 assumed to be star-shaped and convex in two orthogonal directions. Then, we have(39)∫Ωu3/2ndx≤32ρ0∫Ωundx+n21+dρ0∫Ωun−1∇udx3/2,where u is the nonnegative C1-function, n≥1. ρ0andd have the same meaning as in Lemma 1.In the proof process of the main results, we need to use the following inequality:(40)a+bl≤al+bl,a,b>0,0<l<1,and the following Sobolev inequality (n≥3) given in [21]:(41)uκ/2L2n/n−2Ω≤Cuκ/2W1,2Ω,where C is a constant depending on Ω and n, that is,(42)∫Ωuκ/22n/n−2dxn−2/2n≤C∫Ωuκdx+∫Ω∇uκ/22dx1/2,and Young’s inequality yields that(43)aτ1bτ2≤1−τ2ετ2/τ2−1aτ1/1−τ2+τ2εb,a,b>0,0<τ2<1,ε>0.For convenience, we suppose that the following assumptions on functionsa,f,g are satisfied:(44)ax≤b0,x∈Ω.(45)0<gs≤sr1,0<fs≤sr2,s>0.In the sequel, we restrict thatΩ⊂R3, and we define the new auxiliary function:(46)At=k1μt∫Ωuθdx,t≥0,where μ,θ are positive constants, and(47)θ>max32m−3,2m−2,m−1,m>3.Now, we show our main results.Theorem 2.
Letu be a nonnegative classical solution of problem (1). Suppose that (44)–(45) hold, and the following conditions are satisfied:(48)k1′tk1t≤η0,t>0.(49)r1<min2θ−8θ−3m+3m−22θ,θ−8θ−3m+3m−34θ,2θ−m+2,2θ+3−m2.(50)1<r2<min1+4θ28θ−3m+3,1+23θ.
Assume thatu becomes unbounded in the measure At at some finite time t∗. Then, t∗ is bounded below by(51)t∗≥∫0tF2se2∫0sF1τdτds−112A20,where(52)F1t=μη0+3m2θ−m−r1+22ρ03322m−2/θΩ1−r1/θ1ρ03m−2/θk2tk1−μm+r1−2/θt+4θ2−2θr1−8θ−3m+3m−24θ2−6θm−2⋅3m2θ−3m+62ρ0⋅θ+dθρ03m−2/θΩ3m−1m−2+2θθ−r1/2θ2⋅122m−2/θε1−3m−2/2θ−3m−2k2tk1−2μθr1+μ2θ−3m+3m−2/2θ2−3θm−2t+dmθ+m−22θ−m−2r1+34ρ0⋅332m−3/θ1ρ03m−3/2θΩ1−1/θε2−1k2tk1−μm+2r1−3/θt+4θ2−4θr1−8θ−3m+3m−34θ2−6θm−3⋅dmθ+m−22θ−3m+94ρ012m−3/θθ+dθρ03m−3/2θΩ3m−1m−3+4θ2−4θr1/4θ2ε3−2θ+3m−9/2θ−3m+9⋅k2tk1−4μθr1+μ2θ−3m+3m−3/2θ2−3θm−3t+2θ−r2+1b023322r2−1/θ1ρ03r2−1/θk11−μr2−1/θt+2θ−3r2+3b02⋅4θ2−8θ−3m+3r2−14θ2−6θr2−1122r2−1/θ⋅θ+dθρ03r2−1/θ⋅Ω3m−1r2−1/2θ2ε4−3r2−1/2θ−3r2−1k11−μ2θ−3m+3r2−1/2θ2−3θr2−1t,(53)F2t=3mm+r1−22ρ03322m−2/θΩ1−r1/θ1ρ03m−2/θk2tk1−μm+r1−2/θt+2θr1+2θ−3m+3m−24θ2−6θm−2⋅3m2θ−3m+62ρ0θ+dθρ03m−2/θΩ3m−1m−2+2θθ−r1/2θ2⋅ε1−3m−2/2θ−3m−2122m−2/θk2tk1−2μθr1+μ2θ−3m+3m−2/2θ2−3θm−2t+dmθ+m−2m+2r1−34ρ0⋅332m−3/θ1ρ03m−3/2θΩ1−1/θε2−1k2tk1−μm+2r1−3/θt+dmθ+m−22θ−3m+94ρ0⋅12m−3/θθ+dθρ03m−3/2θΩ3m−1m−3+4θ2−4θr1/4θ2ε3−2θ+3m−9/2θ−3m+94θr1+2θ−3m+3m−34θ2−6θm−3⋅k2tk1−4μθr1+μ2θ−3m+3m−3/2θ2−3θm−3t+r2−1b023322r2−1/θ1ρ03r2−1/θk11−μr2−1/θt+2θ−3r2+3b02⋅2θ−3m+3r2−14θ2−6θr2−1122r2−1/θθ1+dρ03r2−1/θ⋅Ω3m−1r2−1/2θ2ε4−3r2−1/2θ−3r2−1k11−μ2θ−3m+3r2−1/2θ2−3θr2−1t,where εii=1,2,3,4 are defined by the following equalities:(54)ε1=169m−22ρ0122m−2/θθ+dθρ03m−2/θΩ3m−1m−2+2θθ−r1/2θ2k2t−1θθ−1,ε2=16dθ+m−22ρ0332m−3/θ1ρ03m−2/θΩ1−1/θk2t−1θ−1,ε3=16dθ+m−22θ+3m−94ρ012m−3/θθ+dθρ03m−3/2θΩ3m−1m−3+4θ2−4θr1/4θ2k2t−1⋅θθ−1,ε4=123b0r2−12122r2−1/θθ1+dρ03r2−1/θΩ3m−1r2−1/2θ2k1t−1θθ−1m,Ω is the volume of Ω, ρ0=min∂Ωx⋅ν, and d=maxΩ¯x.Proof.
With the help of (44), (45), (48) and the divergence theorem, we have(55)A′t=μk1′tk1tk1μt∫Ωuθdx+θk1μt∫Ωuθ−1Δum+axk1tfudx≤μη0At+θmk2tk1μt∫∂Ωuθ+m−2dS∫Ωgudx−θθ−1mk1μt∫Ωuθ+m−3⋅∇u2dx+θb0k1μ+1t∫Ωuθ−1fudx≤μη0At+θmk2tk1μt∫∂Ωuθ+m−2dS∫Ωur1dx−θθ−1mk1μt∫Ωuθ+m−3⋅∇u2dx+θb0k1μ+1t∫Ωuθ+r2−1dx.
In the following, we deal with the second term of the right-hand side of (58). Owing to Lemma 1, we have(56)∫∂Ωuθ+m−2dS≤3ρ0∫Ωuθ+m−2dx+dθ+m−2ρ0∫Ωuθ+m−3∇udx.
From Hölder inequality, we have(57)∫Ωuθ+m−2dx≤∫Ωu3/2θdx2m−4/θ∫Ωuθdx1−2m−4/θ,where 0<2m−4/θ<1. According to Lemma 2, we have(58)∫Ωu3/2θdx≤32ρ0∫Ωuθdx+θ21+dρ0∫Ωuθ−1∇udx3/2.
The inequalitya+b3/2≤2a3/2+2b3/2 implies(59)∫Ωu3/2θdx≤232ρ03/2∫Ωuθdx3/2+2θ21+dρ03/2∫Ωuθ−1∇udx3/2≤232ρ03/2∫Ωuθdx3/2+2θ21+dρ03/2∫Ωuθ−m+1dx3/2∫Ωuθ+m−3∇u2dx3/4≤232ρ03/2∫Ωuθdx3/2+2θ21+dρ03/2Ω3m−1/4θ∫Ωuθdx3θ−m+1/4θ⋅∫Ωuθ+m−3|∇u|2dx3/2,where Hölder inequality is used. Then, after inserting (59) into (57), inequality (41) yields(60)∫Ωuθ+m−2dx≤3322m−2/θ1ρ03m−2/θ∫Ωuθdx1+m−2/θ+122m−2/θθ+dθρ03m−2/θ⋅Ω3m−1m−2+2θθ−r1/2θ2∫Ωuθdx1−θ+3m−3m−2/2θ2∫Ωuθ+m−3∇u2dx3m−2/θ.
Thanks to Hölder inequality, the last term on the right-hand side of (56) satisfies(61)∫Ωuθ+m−3∇udx≤∫Ωuθ+m−3dx1/2∫Ωuθ+m−3∇u2dx1/2.
Similarly, from (57)–(60), we can get(62)∫Ωuθ+m−3dx≤3322m−3/θ1ρ03m−3/θ∫Ωuθdx1+m−3/θ+122m−3/θθ+dθρ03m−3/θ⋅Ω3m−1m−3/2θ2∫Ωuθdx1−θ+3m−3m−3/2θ2∫Ωuθ+m−3∇u2dx3m−3/2θ.
Then, combining (61) and (62), we have(63)∫Ωuθ+m−3∇udx≤3322m−3/θ1ρ03m−3/θ∫Ωuθdx1+m−3/θ+122m−3/θθ+dθρ03m−3/θ⋅Ω3m−1m−3/2θ2∫Ωuθdx1−θ+3m−3m−3/2θ2⋅∫Ωuθ+m−3∇u2dx3m−3/2θ1/2⋅∫Ωuθ+m−3∇u2dx1/2≤332m−3/θ1ρ03m−3/2θ∫Ωuθdx1/2+m−3/2θ∫Ωuθ+m−3|∇u|2dx1/2+12m−3/θθ+dθρ03m−3/2θ|Ω|3m−1m−3/2θ2∫Ωuθdx1/2−θ+3m−3m−3/4θ2⋅∫Ωuθ+m−3∇u2dx3m−3/4θ+1/2,where inequality (41) is applied again. According to (60), (63), and Hölder inequality, the second term of the right-hand side of (55) is changed into(64)∫∂Ωuθ+m−2dS∫Ωur1dx≤3ρ0∫Ωuθ+m−2dx+dθ+m−2ρ0∫Ωuθ+m−3∇udx∫Ωuθdx1/θΩ1−1/θ≤3ρ03322m−2/θΩ1−1/θ1ρ03m−2/θ∫Ωuθdx1+m+r1−2/θ+3ρ0122m−2/θ⋅θ+dθρ03m−2/θΩ3m−1m−2+2θθ−r1/2θ2∫Ωuθdx1+2θr1−θ+3m−3m−2/2θ2⋅∫Ωuθ+m−3∇u2dx3m−2/θ+dθ+m−2ρ0332m−3/θ1ρ03m−3/2θ⋅Ω1−1/θ∫Ωuθdx1/2+m−3+2r1/2θ∫Ωuθ+m−3∇u2dx1/2+dθ+m−2ρ0⋅12m−3/θθ+dθρ03m−3/2θΩ3m−1m−3+4θ2−4θr1/4θ2∫Ωuθdx1/2−θ+3m−3m−3−4θr1/4θ2⋅∫Ωuθ+m−3∇u2dx3m−3/4θ+1/2.
Now, we estimate the last term of the right-hand side of (55). Similarly, from (57)–(60), we get(65)∫Ωuθ+r2−1dx≤3322r2−1/θ1ρ03r2−1/θ∫Ωuθdx1+r2−1/θ+122r2−1/θθ1+dρ03r2−1/θ⋅Ω3m−1r2−1/2θ2∫Ωuθdx1−θ+3m−3r2−1/2θ2∫Ωuθ+m−3∇u2dx3r2−1/2θ.
Then, we deal with the terms in (64) and (65) including ∫Ωuθ+m−3∇u2dx. It follows from (43) that(66)∫Ωuθdx1+2θr1−θ+3m−3m−2/2θ2∫Ωuθ+m−3∇u2dx3m−2/2θ≤2θ−3m−22θε1−3m−2/2θ−3m−2∫Ωuθdx1+2θr1+2θ−3m+3m−2/2θ2−3θm−2+3m−2ε12θ∫Ωuθ+m−3∇u2dx,(67)∫Ωuθdx1/2+m−3+2r1/2θ∫Ωuθ+m−3∇u2dx1/2≤12ε2−1∫Ωuθdx1+m+2r1−3/θ+12ε2∫Ωuθ+m−3∇u2dx,(68)∫Ωuθdx1/2+θ+3m−34θr1/4θ2∫Ωuθ+m−3∇u2dx3m−3/4θ+1/2≤2θ−3m−24θε1−2θ+3m−9/2θ−3m+9∫Ωuθdx1+4θr1+2θ−3m+3m−3/2θ2−3θm−3+2θ+3m−9ε32θ∫Ωuθ+m−3∇u2dx,(69)∫Ωuθdx1−θ+3m−3r2−1/2θ2⋅∫Ωuθ+m−3∇u2dx3r2−1/2θ≤2θ−3r2−12θε4−3r2−1/2θ−3r2−1∫Ωuθdx1+2θ−3m+3r2−1/2θ2−3θr2−1+3r2−1ε42θ∫Ωuθ+m−3∇u2dx,where εii=1,2,…,4 are defined by (56). Therefore, we substitute (64)–(69) into (55) and get(70)A′t≤μη0At+3θmρ03322m−2/θΩ1−1/θ1ρ03m−2/θk2tk1−μm+r1−2θtA1+m+r1−2/θt+3m2θ−3m+62ρ0122m−2/θθ+dθρ03m−2/θΩ3m−1m−2+2θθ−r1/2θ2ε1−3m−2/2θ−3m−2⋅k2tk1−2μθr1+μ2θ−3m+3m−2/2θ2−3θm−2tA1+2θr1+2θ−3m+3m−2/2θ2−3θm−2t+dθmθ+m−22ρ0332m−3/θ⋅1ρ03m−3/2θΩ1−1/θε2−1k2tk1−μm+2r1−3/θtAt1+m+2r1−3/θ+dmθ+m−22θ−3m+94ρ0⋅12m−3/θθ+dθρ03m−3/2θΩ3m−1m−3+4θ2−4θr1/4θ2ε3−2θ+3m−9/2θ−3m+9k2tk1−4μθr1+μ2θ−3m+3m−3/2θ2−3θm−3t⋅At1+4μθr1+μ2θ−3m+3m−3/2θ2−3θm−3+θb0k11−μr2−1/θt3322r2−1/θ1ρ03r2−1/θAt1+r2−1/θ+2θ−3r2+3b02122r2−1/θθ1+dρ03r2−1/θΩ3m−1r2−1/2θ2ε4−3r2−1/2θ−3r2−1⋅k11−μ2θ−3m+3r2−1/2θ2−3θr2−1tAt1+2θ−3m+3r2−1/2θ2−3θr2−1.
From (47)–(50), note that(71)0<m+r1−2θ,2θr1+2θ−3m+3m−22θ2−3θm−2,m+2r1−3θ<2,0<4θr1+2θ−3m+3m−32θ2−3θm−3,r2−1θ,2θ−3m+3r2−12θ2−3θr2−1<2.
Therefore, the inequalityA1+λt≤1−λ/2At+λ/2A3t,0<λ<2 implies that(72)A′t≤F1tAt+F2tA3t,where F1t,F2t are given by (52) and (53). Obviously, we can get(73)A−2te2∫0tF1sds′≥−2F2te2∫0tF1sds.
Integrating (73) from 0 to t, we obtain(74)A−2te2∫0tF1sds−A−20≥−2∫0tF2se2∫0sF1τdτds.
That is,(75)e2∫0tF1sdsA2t−1A20≥−2∫0tF2se2∫0sF1τdτds.
Taking the limit ast⟶t∗ in (75), we get(76)∫0t∗F2se2∫0sF1τdτds≥12A20.
Note that∫0tF2se2∫0sF1τdτds is a strictly increasing function, and we can conclude that(77)t∗≥∫0tF2se2∫0sF1τdτds−112A20.
Then, we finish the proof.
And we assume thatΩ⊂Rnn≥3 is a smooth bound domain. The functions k1 and k2 satisfy(78)k1t≤η1,k2t≤η2,t>0,where η1andη2 are some positive constants. Now, we define the new auxiliary function:(79)Bt=∫Ωuδdx,t≥0,where δ is a positive constant, and it satisfies(80)δ>max3−m,r1,nr2−m2,r2>1.Theorem 3.
Letu be a nonnegative classical solution of problem (1). Suppose that (44), (45), and (78) hold. Assume that u becomes unbounded in the measure Bt at some finite time t∗. Then, t∗ is bounded below by(81)t∗≥∫B0∞dτφτ,where(82)φτ=Q1τm+δ+r1−2/δ+Q2τδ+r2−1/δ+Q3τ2δδ+m−2+r1mn+2δ−nδ+m−1/2δ2δ+m−1+δnm−1+Q4τm+δ+2r1−3/δ+Q5τm+δ−1δδ+m−3+r1mn+2δ−n/δnm−1+δ2m+δ−1+Q6τmn+2δ−n2+r2−14δ+nm−14−n/mn+2δ−n2−n2m−1r2−1+Q7τmn+2δ−n−r2−1n−2/mn+2δ−nr2,where Qii=1,2,…,7 are defined by the following equalities:(83)Q1=δmnη2ρ0Cnm−1δ+m−2/δδ+m−1Ω1/δ+m−1+δ−r1/δ,Q2=δb0η1Cnr2−1/δ,Q3=2δδ+m−1+nm−1mn+2δ−nδ+m−1ε−nm−1δ+m−2/2δδ+m−1+nm−1C1,Q4=12ε−1C2,Q5=mn−n+δδ+m−1mn+2δ−nδ+m−1ε−nm−1δ+m−2+δδ+m−1/mn−n+δδ+m−1C3,Q6=1−n2m−1r2−1mn+2δ−n2ε−n2m−1r2−1/mn+2δ−n2−n2m−1r2−1C4,Q7=mn+2δ−nr2mn+2δ−nε−nr2−1/mn+2δ−nr2δb0η1C2nr2−1/mn+2δ−n.
Here,(84)C1=δmη2nρ02δ+nm−12δδ+m−2/δ+m−1C2nm−1δ+m−2/mn+2δ−nδ+m−1Ω1/δ+m−1+δ−r1/δ,C2=2dδmη2δ+m−2ρ0δ+m−1C2δ+nm−1δ+m−3/2δδ+m−1Ω1/δ+m−1+δ−r1/δ,C3=2dδmη2δ+m−2ρ0δ+m−12δ+nm−12δδ+m−3/2δ+m−1Cnm−1δ+m−3/mn+2δ−nδ+m−1Ω1/δ+m−1+δ−r1/δ,C4=δb0η12δ+nm−12δnr2−1/mn+2δ−nC4nr2−1mn+δ−n/mn+2δ−n2,ε=nm−1δ+m−2mn+2δ−nδ+m−1C1+12C2+nm−1δ+m−32mn+2δ−nδ+m−1+12C3+n2m−1r2−1mn+2δ−n2C4+nr2−1mn+2δ−n−1⋅4δδ−1mδ+m−12,Ω is the volume of Ω, ρ0=min∂Ωx⋅ν, and d=maxΩ¯x.Proof.
By assumptions (44), (45), and (78) and the divergence theorem, we obtain(85)B′t=δ∫Ωuδ−1utdx=δ∫Ωuδ−1Δum+axk1tfudx=δ∫Ωuδ−1Δumdx+δk1t∫Ωaxuδ−1fudx=δ∫Ω∇⋅uδ−1∇umdx−δ∫Ω∇uδ−1⋅∇umdx+δk1t∫Ωaxuδ−1fudx=δm∫∂Ωuδ+m−2∂u∂νdS−δδ−1m∫Ωuδ+m−3⋅∇u2dx+δk1t∫Ωaxuδ−1fudx=δmk2t∫∂Ωuδ+m−2dS∫Ωgudx−4δδ−1mδ+m−12∫Ω∇uδ+m−1/22dx+δk1t∫Ωaxuδ−1fudx≤δmη2∫∂Ωuδ+m−2dS∫Ωur1dx−4δδ−1mδ+m−12∫Ω∇uδ+m−1/22dx+δb0η1∫Ωuδ+r2−1dx.
Now, we estimate the first term of the right-hand side of (101). Note that 0<r1/δ<1, and Hölder inequality implies that(86)∫Ωur1dx≤∫Ωuδdxr1/δΩ1−r1/δ.
Owing to Lemma1, we have(87)∫∂Ωuδ+m−2dS≤nρ0∫Ωuδ+m−2dx+dδ+m−2ρ0∫Ωuδ+m−3∇udx.
From Hölder inequality, we have(88)∫Ωuδ+m−2dx≤∫Ωuδ+m−1dxδ+m−2/δ+m−1Ω1/δ+m−1,(89)∫Ωuδ+m−3∇udx≤∫Ωuδ+m−3dx1/2∫Ωuδ+m−3∇u2dx1/2≤∫Ωuδ+m−1dxδ+m−3/2δ+m−1Ω1/δ+m−1∫Ωuδ+m−3∇u2dx1/2,(90)∫Ωuδ+m−1dx≤∫Ωuδdx1−m−1n−2/mn+2δ−n∫Ωuδ+m−1/2⋅2n/n−2dxm−1n−2/mn+2δ−n,where 0<m−1n−2/mn+2δ−n<1. And it follows from Sobolev inequality (42) and inequality (41) that(91)∫Ωuδ+m−1dx≤∫Ωuδdx2m+δ−1/mn+2δ−n⋅C2nm−1/mn+2δ−n∫Ωuδ+m−1dx+∫Ω∇uδ+m−1/22dxnm−1/mn+2δ−n≤∫Ωuδdx2m+δ−1/mn+2δ−n⋅C2nm−1/mn+2δ−n∫Ωuδ+m−1dxnm−1/mn+2δ−n+∫Ωuδdx2m+δ−1/mn+2δ−n⋅C2nm−1/mn+2δ−n∫Ω∇uδ+m−1/22dxnm−1/mn+2δ−n=Cnm−1/δ⋅∫Ωuδdxm+δ−1/δ2δ/mn+2δ−n∫Ωuδ+m−1dxnm−1/mn+2δ−n+∫Ωuδdx2m+δ−1/mn+2δ−n⋅C2m+δ−1/mn+2δ−n∫Ω∇uδ+m−1/22dxnm−1/mn+2δ−n≤2δmn+2δ−nCnm−1/δ∫Ωuδdxm+δ−1/δ+nm−1mn+2δ−n∫Ωuδ+m−1dx+∫Ωuδdx2m+δ−1/mn+2δ−nC2nm−1/mn+2δ−n∫Ω∇uδ+m−1/22dxnm−1/mn+2δ−n,where Young’s inequality is also used. Then, it follows from (91) that(92)∫Ωuδ+m−1dx≤Cnm−1/δ∫Ωuδdxm+δ−1/δ+2δ+nm−12δC2nm−1/mn+2δ−n⋅∫Ωuδdx2m+δ−1/mn+2δ−n∫Ω∇uδ+m−1/22dxnm−1/mn+2δ−n.
Inserting (92) into (88) and (89), respectively, we have(93)∫Ωuδ+m−2dx≤Cnm−1/δ∫Ωuδdxnm−1/δ+2δ+nm−12δC2nm−1/mn+2δ−n⋅∫Ωuδdx2m+δ−1/mn+2δ−n∫Ω∇uδ+m−1/22dxnm−1/mn+2δ−nδ+m−2/δ+m−1Ω1/δ+m−1≤Cnm−1δ+m−2/δδ+m−1Ω1/δ+m−1∫Ωuδdxm+δ−2/δ+2δ+nm−12δδ+m−2/δ+m−1C2nm−1δ+m−2/mn+2δ−nδ+m−1⋅Ω1/δ+m−1∫Ωuδdx2m+δ−2/mn+2δ−n∫Ω∇uδ+m−1/22dxnm−1δ+m−2/mn+2δ−nδ+m−1,(94)∫Ωuδ+m−3∇udx≤Cnm−1/δ∫Ωuδdxm+δ−1/δ+2δ+nm−12δC2nm−1/mn+2δ−n∫Ωuδdx2m+δ−1/mn+2δ−n⋅∫Ω∇uδ+m−1/22dxnm−1/mn+2δ−nδ+m−3/2δ+m−1Ω1/δ+m−1∫Ωuδ+m−3∇u2dx1/2≤2δ+m−1Cnm−1δ+m−3/2δδ+m−1Ω1/δ+m−1∫Ωuδdxm+δ−3/2δ∫Ω∇uδ+m−1/22dx1/2+2δ+m−12δ+nm−12δδ+m−3/2δ+m−1Cnm−1δ+m−3/mn+2δ−nδ+m−1Ω1/δ+m−1∫Ωuδdxm+δ−3/mn+2δ−n⋅∫Ω∇uδ+m−1/22dxnm−1δ+m−3/mn+2δ−nδ+m−1+1/2,where inequality (41) is used again. Combining (86) and (87) and (93) and (94), we deduce(95)∫∂Ωuδ+m−2dS∫Ωur1dx≤∫Ωuδdx1/δΩ1−1/δnρ0∫Ωuδ+m−2dx+dδ+m−2ρ0∫Ωuδ+m−3∇udx≤nρ0Cnm−1δ+m−2/δδ+m−1Ω1/δ+m−1+δ−r1/δ∫Ωuδdxm+δ+r1−2/δ+nρ02δ+nm−12δδ+m−2/δ+m−1C2nm−1δ+m−2/mn+2δ−nδ+m−1Ω1/δ+m−1+δ−r1/δ⋅∫Ωuδdx2m+δ−2/mn+2δ−n+1/δ∫Ω∇uδ+m−1/22dxnm−1δ+m−2/mn+2δ−nδ+m−1+2dδ+m−2ρ0δ+m−1Cnm−1δ+m−3/2δδ+m−1Ω1/δ+m−1+δ−r1/δ∫Ωuδdxm+δ+2r1−3/2δ⋅∫Ω∇uδ+m−1/22dx1/2+2dδ+m−2ρ0δ+m−12δ+nm−12δδ+m−3/2δ+m−1⋅Cnm−1δ+m−3/mn+2δ−nδ+m−1Ω1/δ+m−1+δ−r1/δ∫Ωuδdxm+δ−3/mn+2δ−n+1/δ⋅∫Ω∇uδ+m−1/22dxnm−1δ+m−3/2mn+2δ−nδ+m−1+1/2.
Next, we deal with the last term of the right-hand side of (101). From Höldero¨ inequality, Sobolev inequality (42), and inequality (41), we obtain(96)∫Ωuδ+r2−1v≤∫Ωuδdx1−r2−1n−2/mn+2δ−n⋅∫Ωuδ+m−1/22n/n−2dxnr2−1/mn+2δ−n≤∫Ωuδdx1−r2−1n−2/mn+2δ−n⋅C2nr2−1/mn+2δ−n∫Ωuδ+m−1dx+∫Ω∇uδ+m−1/22dxnr2−1/mn+2δ−n≤∫Ωuδdx1−r2−1n−2/mn+2δ−n⋅C2nr2−1/mn+2δ−n∫Ωuδ+m−1dxnr2−1/mn+2δ−n+C2nr2−1/mn+2δ−n⋅∫Ωuδdx1−r2−1n−2/mn+2δ−n∫Ω∇uδ+m−1/22dxnr2−1/mn+2δ−n≤∫Ωuδdx1−r2−1n−2/mn+2δ−n⋅C2nr2−1/mn+2δ−nCm+δ−1/δ∫Ωuδdxm+δ−1/δ+2δ+nm−12δC2nr2−1/mn+2δ−n⋅∫Ωuδdx2m+δ−1/mn+2δ−n⋅∫Ω∇uδ+m−1/22dxnr2−1/mn+2δ−nnr2−1/mn+2δ−n+C2nr2−1/mn+2δ−n⋅∫Ωuδdx1−r2−1n−2/mn+2δ−n∫Ω∇uδ+m−1/22dxnr2−1/mn+2δ−n≤Cnr2−1/δ∫Ωuδdx1+r2−1/δ+2δ+nm−12δnr2−1/mn+2δ−nC4nr2−1mn+δ−n/mn+2δ−n2⋅∫Ωuδdx1+r2−14δ+nm−14−n/mn+2δ−n2⋅∫Ω∇uδ+m−1/22dxn2m−1r2−1/mn+2δ−n2+C2nr2−1/mn+2δ−n∫Ωuδdx1−r2−1n−2/mn+2δ−n⋅∫Ω∇uδ+m−1/22dxnr2−1/mn+2δ−n,where (92) is substituted. We insert (95) and (96) into (85), and we get(97)B′t≤Q1∫Ωuδdxm+δ+r1−2/δ+Q2∫Ωuδdx1+r2−1/δ+C1∫Ωuδdx2m+δ−2/mn+2δ−n+r1/δ⋅∫Ω∇uδ+m−1/22dxnm−1δ+m−2/mn+2δ−nδ+m−1+C2∫Ωuδdxm+δ+2r1−3/2δ⋅∫Ω∇uδ+m−1/22dx1/2+C3∫Ωuδdxm+δ−3/mn+2δ−n+r1/δ⋅∫Ω∇uδ+m−1/22dxnm−1δ+m−3/mn+2δ−nδ+m−1+1/2+C4∫Ωuδdx1+r2−14δ+nm−14−n/mn+2δ−n2⋅∫Ω∇uδ+m−1/22dxn2m−1r2−1/mn+2δ−n2+δb0η1C2nr2−1/mn+2δ−n∫Ωuδdx1−r2−1n−2/mn+2δ−n⋅∫Ω∇uδ+m−1/22dxnr2−1/mn+2δ−n−4δδ−1mδ+m−12∫Ω∇uδ+m−1/22dx,where Q1,Q2,Cii=1,2,3,4 are defined by (83) and (84). It follows from (43) that(98)∫Ωuδdx2m+δ−2/mn+2δ−n+r1/δ⋅∫Ω∇uδ+m−1/22dxnm−1δ+m−2/mn+2δ−nδ+m−1≤2δδ+m−1+nm−1mn+2δ−nδ+m−1ε−nm−1δ+m−2/2δδ+m−1+nm−1∫Ωuδdx2δδ+m−2+r1mn+2δ−nδ+m−1/2δ2δ+m−1+δnm−1+nm−1δ+m−2mn+2δ−nδ+m−1ε∫Ω∇uδ+m−1/22dx,(99)∫Ωuδdxm+δ+2r1−3/2δ⋅∫Ω∇uδ+m−1/22dx1/2≤12ε−1∫Ωuδdxm+δ+2r1−3/δ+12ε∫Ω∇uδ+m−1/22dx,(100)∫Ωuδdxm+δ−3/mn+2δ−n+r1/δ⋅∫Ω∇uδ+m−1/22dxnm−1δ+m−3/2mn+2δ−nδ+m−1+1/2≤mn−n+δδ+m−1mn+2δ−nδ+m−1ε−nm−1δ+m−2+δδ+m−1/mn−n+δδ+m−1∫Ωuδdxm+δ−1δδ+m−3+r1mn+2δ−n/δnm−1+δ2m+δ−1+nm−1δ+m−32mn+2δ−nδ+m−1+12ε∫Ω∇uδ+m−1/22dx,(101)∫Ωuδdx1+r2−14δ+nm−14−n/mn+2δ−n2∫Ω∇uδ+m−1/22dxn2m−1r2−1/mn+2δ−n2≤1−n2m−1r2−1mn+2δ−n2ε−n2m−1r2−1/mn+2δ−n2−n2m−1r2−1∫Ωuδdxmn+2δ−n2+r2−14δ+nm−14−n/mn+2δ−n2−n2m−1r2−1+n2m−1r2−1mn+2δ−n2ε∫Ω∇uδ+m−1/22dx,(102)∫Ωuδdx1−r2−1n−2/mn+2δ−n⋅∫Ω∇uδ+m−1/22dxnr2−1/mn+2δ−n≤mn+2δ−nr2mn+2δ−nε−nr2−1/mn+2δ−nr2∫Ωuδdxmn+2δ−n−r2−1n−2/mn+2δ−nr2+nr2−1mn+2δ−nε∫Ω∇uδ+m−1/22dx.
Therefore, by using (98)–(101), we rewrite (97) to get(103)B′t≤Q1∫Ωuδdxm+δ+r1−2/δ+Q2∫Ωuδdxδ+r2−1/δ+2δδ+m−1+nm−1mn+2δ−nδ+m−1ε−nm−1δ+m−2/2δδ+m−1+nm−1⋅C1∫Ωuδdx2δδ+m−2+r1mn+2δ−nδ+m−1/2δ2δ+m−1+δnm−1+12ε−1C2∫Ωuδdxm+δ+2r1−3/δ+mn−n+δδ+m−1mn+2δ−nδ+m−1ε−nm−1δ+m−2+δδ+m−1/mn−n+δδ+m−1C3∫Ωuδdxm+δ−1δδ+m−3+r1mn+2δ−n/δnm−1+δ2m+δ−1+1−n2m−1r2−1mn+2δ−n2ε−n2m−1r2−1/mn+2δ−n2−n2m−1r2−1C4∫Ωuδdxmn+2δ−n2+r2−14δ+nm−14−n/mn+2δ−n2−n2m−1r2−1+mn+2δ−nr2mn+2δ−nε−nr2−1/mn+2δ−nr2δb0η1C2nr2−1/mn+2δ−n∫Ωuδdxmn+2δ−n−r2−1n−2/mn+2δ−nr2+nm−1δ+m−2mn+2δ−nδ+m−1εC1+12εC2+nm−1δ+m−32mn+2δ−nδ+m−1+12εC3+n2m−1r2−1mn+2δ−n2εC4+nr2−1mn+2δ−nε−4δδ−1mδ+m−12⋅∫Ω∇uδ+m−1/22dx.
Because of the definition ofε, we get(104)B′t≤Q1Btm+δ+r1−2/δ+Q2Btδ+r2−1/δ+Q3Bt2δδ+m−2+r1mn+2δ−nδ+m−1/2δ2δ+m−1+δnm−1+Q4Btm+δ+r1−3/δ+Q5Btm+δ−1δδ+m−3+r1mn+2δ−n/δnm−1+δ2m+δ−1+Q6Btmn+2δ−n2+r2−14δ+nm−14−n/mn+2δ−n2−n2m−1r2−1+Q7Btmn+2δ−n−r2−1n−2/mn+2δ−nr2,where Qii=1,2,…,6 are given by (83). Integrating (104) from 0 to t, we deduce(105)∫B0Btdτφτ≤t.
Here,(106)φτ=Q1τm+δ+r1−2/δ+Q2τδ+r2−1/δ+Q3τ2δδ+m−2+r1mn+2δ−nδ+m−1/2δ2δ+m−1+δnm−1+Q4τm+δ+2r1−3/δ+Q5τm+δ−1δδ+m−3+r1mn+2δ−n/δnm−1+δ2m+δ−1.+Q6τmn+2δ−n2+r2−14δ+nm−14−n/mn+2δ−n2−δ2m−1r2−1+Q7τmn+2δ−n−r2−1n−2/mn+2δ−n−r2.
Hence, ifu blows up in the measure Bt at some finite time t∗, we pass to the limit as t⟶t∗ to obtain(107)t∗≥∫B0∞dτφτ.
The proof is complete.


## 4. Applications

In this section, some examples are given to demonstrate the applications of the main results.Example 1.
Letu be a nonnegative classical solution of the following equation:(108)ut=Δu4+27−x2et/2u5,x,t∈Ω×0,t∗,∂u∂ν=12601303πet/2u3,x,t∈∂Ω×0,t∗,ux,0=u0x=12+x2,x∈Ω¯,where Ω=x=x1,x2,x3x12+x22+x32<1.
Then,(109)m=4,ax=27−x2fu=u5,gu=u3,k1t=et/2,k2t=12601303πet/2,u0x=12+x2.
From (12) and (13), we get λ1=π2 and ω1=sinπx/2πx. Set α=1/4,β=26,η=1/2,andγ=1. Therefore,(110)Φt=e1/8t∫Ωsinπx2πx2udx.
We can compute(111)Φ0=∫Ωsinπx2πx2u0dx=∫0t∫02πsinπr2πr212+r2r2dφdr=0.782673,βk1αm−α0Φ0α+1−αm/α−2λ1=26×0.7826732−2π2=0.610282>0,βΦ0α+1−αmα−2λ1γαm−α=26×0.7826732−2π2=0.610282>0.
Then, conditions (8)–(14) are satisfied. By Theorem 1, we know that u blows up at a finite time t∗<T in the measure Φt. And(112)T=∫Φ0+∞dsαηs+βsα+1/α−2λ1γα−αmsm=∫0.7826732+∞ds1/8s+26s5−2π2s4=0.161936,which is an upper bound for the blow-up time.
Next, we will show a lower bound for the blow-up time. Setb0=27,r1=3,r2=5,η1=1/2,ρ0=d=1,andΩ=4/3π. We define(113)At=e1/2t∫Ωu10dx,t≥0,A0=∫Ωu010dx=∫Ω12+x210dx=46.1011.
Obviously, all conditions of Theorem2 are valid. And since u blows up at a finite time t∗ in the measure Φt, u can blow up in the measure At. Hence, according to Theorem 2, we can compute ε1=0.39897e−t/2,ε2=0.270855e−t/2,ε3=0.166525e−t/2,ε4=0.0410553e−t/2,(114)F1t=12+110.622e1/4t+463.634e3/10t+384.366e59/140t+531.608e13/20t+1702.72e269/340t+166040.1221e39/40t,F2t=36.874e1/4t+115.908e3/10t+92.2609e59/140t+286.25e13/20t+1062.25e269/340t+62980.7e39/40t.
Then,(115)t∗≥∫0tF2se2∫0sF1τdτds−112A20=3.641×10−9.
It follows from (112) and (115) that(116)3.641×10−9≤t∗≤0.161936.Example 2.
Letu be a nonnegative classical solution of the following equation:(117)ut=Δu3+734+120x22+1100e−tu4,x,t∈Ω×0,t∗,∂u∂ν=105736π5−e−t∫Ωu2dx,x,t∈∂Ω×0,t∗,ux,0=u0x=1+x2,x∈Ω¯,where Ω=x=x1,x2,x3x12+x22+x32<1.
Then,(118)m=3,ax=734+120x2,fu=u4,gu=u2,k1t=2+1100e−t,k2t=105736π5−e−t,u0x=1+x2.
From (12) and (13), we get λ1=π2 and ω1=sinπx/2πx. Set α=1/3,β=7.75,andγ=2. Therefore,(119)Ψt=∫Ωsinπx2πx2udx.
We can compute(120)Ψ0=∫Ωsinπx2πx2u0dx=∫01∫02πsinπr2πr21+r2r2dφdr=1.28267.βγΨ0α+1−αm/α−2λ1=7.75×2×1.28267−2π2=0.1422>0.
Then, conditions (9)–(14) are satisfied. By Corollary 1, we know that u blows up at a finite time t∗<T in the measure Φt. And(121)T=∫Ψ0+∞dsβγsα+1/α−2λ1sm=∫1.28267+∞ds14.5s4−2π2s3=0.107917,which is an upper bound for the blow-up time.
Next, we will show a lower bound for the blow-up time. Setb0=7.8,r1=2,r2=4,η1=2.01,η2=725/726π,ρ0=d=1,andΩ=4/3π. We define(122)Bt=∫Ωu5dx,t≥0,B0=∫Ωu05dx=∫Ω1+x25dx=54.3484.
Obviously, all conditions of Theorem3 are valid. And since u blows up at a finite time t∗ in the measure Φt, u can blow up in the measure Bt. It follows from [22] that the Sobolev embedding constant C=5.6958. According to (98) and (100), we can compute C1=135.5476,C2=45.7226,C3=31.8926,C4=1505.5976,andε=0.012. And we have(123)φs=545.771s8/5+1795.248/5+1623.15s161/95+1905.11s9/5+45305.8s399/205+3905.83s167/101+476413.607s13/6.
Then, by Theorem3, we get a lower bound of t∗ as follows:(124)t∗≥∫18.5105+∞dsφs=1.60861×10−8.
It follows from (121) and (124) that(125)1.60861×10−8≤t∗≤0.107917.



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