In this paper, we study porous media equation ut = ∆u m − u p with nonlinear boundary condition ∂u ∂ν = kuq . We determine some sufficient conditions for the occurrence of finite time blow-up or global existence. Moreover, lower and upper bounds for blow-up time are also derived by using various inequality techniques

Ran Xu

Ya Tian

Yao Qin

English

University of Szeged

Electronic Journal of Qualitative Theory of Differential Equations2020, No. 8, 1–11; https://doi.org/10.14232/ejqtde.2020.1.8 www.math.u-szeged.hu/ejqtde/Blow-up analysis for a porous media equation withnonlinear sink and nonlinear boundary conditionTian YaB , Xu Ran and Qin YaoSchool of Science, Chongqing University of Posts and Telecommunications,Chongqing, 400065, P.R. ChinaReceived 20 November 2020, appeared 7 February 2021Communicated by László SimonAbstract. In this paper, we study porous media equation ut = ∆um − up with nonlinearboundary condition ∂u∂ν = kuq. We determine some sufficient conditions for the occur-rence of finite time blow-up or global existence. Moreover, lower and upper bounds forblow-up time are also derived by using various inequality techniques.Keywords: porous media equation, nonlinear boundary condition, bounds for blow-uptime.2020 Mathematics Subject Classification: 35B44, 35K40.1 IntroductionIn this paper, we are interested in investigating the blow-up phenomena of the followingporous media equation with nonlinear sink and nonlinear boundary condition:ut = ∆um − up, (x, t) ∈ Ω× (0, t∗),∂u∂ν = kuq, (x, t) ∈ ∂Ω× (0, t∗),u(x, 0) = u0(x), x ∈ Ω,(1.1)where m > 1 and p > 1, q ≥ 1, k is a positive constant, Ω ⊂ R3 is a star-shaped domain withsmooth boundary. ν is the unit outward normal vector on ∂Ω, u0(x) > 0 is the initial value. t∗is the blow-up time if the solutions blow up. It is well known that the data m, p, q may greatlyaffect the behavior of u(x, t) as time evolves.The mathematical investigation of the phenomenon of blow-up of solutions to parabolicequations and systems has received much attention in the recent literature. We refer to thereaders the books of Straughan [14] and Quittner and Souplet [13], as well as papers ofWeissler [15, 16], and so on. The determination of sufficient conditions for blow-up and theexistence or nonexistence of global solution to problem, as well as bounds for the blow-uptime have been the focus of some of these studies [1, 5–7, 18].BCorresponding author. Email: tianya@cqupt.edu.cn2 Y. Tian, R. Xu and Y. QinFor the initial-boundary value problem of porous media equationut = ∆um + f (u) (1.2)where f (u) ≥ 0, Wu and Gao [17] established the blow-up criterion of equation (1.2) by usingthe method of energy. Besides, there are many references for the blow-up behavior of itssolutions [4, 8]. The methods used in the study of blow-up often lead to upper bound for theblow-up time when blow-up occur. However, in applied problems, because of the explosivenature of the solution, a lower bound on blow-up time is more important. Then, there aremany papers giving the estimate of the lower bound of blow-up time [2, 3, 9, 10, 12]. In [9],the authors gave the estimations of the lower bound for blow-up time for problem (1.2) underRobin boundary conditions, by using various inequalities. When m = 1, Payne and Philippinetc. [10] studied blow-up phenomena of the classical solution of the following initial-boundaryproblemut = ∆u− f (u) (1.3)under the help of energy method and Sobolev type inequality, they gave the lower bound ofblow-up time when condition for blow-up holds.However, to our best knowledge, there is no paper where the blow-up phenomenon isstudied with m > 1 and nonlinear sink as a reaction term. So, it is natural to consider problem(1.1). Methods used in this paper are motivated by the aforementioned papers. Because ofthe difference between the diffusion term and the reaction term, we will study the blow-upphenomena of (1.1) by modifying their techniques.In Sections 2 and 3, by using energy method and various inequality techniques, we deter-mine a criterion which implies blow-up, and drive upper and lower bounds for t∗; in Section 4,a criterion for boundedness of the solution in all time t > 0 is determined; In the last section,a relevant example will be listed to illustrate applications of our results.2 Blow-up and upper bound estimation of t∗In this section we establish a blow-up criterion for problem (1.1) and derive an upper boundfor blow-up time, by using the auxiliary function method.Theorem 2.1. Let u(x, t) be a nonnegative classical solution of problem (1.1) and assume m+ q− 1 ≥p. Then u(x, t) will blow-up in finite time t∗, andt∗ ≤2mϕ−a0(m + 1)2a(1 + a)M,where a, M and ϕ−a0 are some constants which will be given in the later proof.Proof. Denoteϕ(t) =∫Ωum+1dx. (2.1)Blow-up analysis for a porous media equation with nonlinear sink 3Taking the derivative of (2.1), we haveϕ′(t) = (m + 1)∫Ωumutdx= (m + 1)∫Ωum(∆um − up)dx= (m + 1)∫Ωum∆umdx− (m + 1)∫Ωum+pdx= (m + 1)mk∫∂Ωu2m+q−1ds− (m + 1)∫Ω|∇um|2dx− (m + 1)∫Ωum+pdx.(2.2)Moreover, by using the notationψ(t) =2m2k2m + q− 1∫∂Ωu2m+q−1ds−∫Ω|∇um|2dx− 2mm + p∫Ωum+pdx, (2.3)since m + q− 1 ≥ p, we haveϕ′(t) ≥ (m + 1)(2m + q− 1)2mψ(t). (2.4)From (2.3) one obtainsψ′(t) = 2m2k∫∂Ωu2m+q−2utds−∫Ω|∇um|2t dx− 2m∫Ωum+p−1utdx, (2.5)because∇(umt ∇um) = umt ∆um +12|∇um|2t . (2.6)Integrate both sides of (2.6), then we obtain∫Ω|∇um|2t dx = 2∫∂Ωumt∂um∂νds− 2∫Ωumt ∆umdx= 2m2k∫∂Ωu2m+q−2utds− 2m∫Ωum−1∆umutdx.(2.7)Substituting (2.7) into (2.5), we haveψ′(t) = 2m∫Ωum−1∆umutdx− 2m∫Ωum+p−1utdx= 2m∫Ωum−1u2t dx > 0.(2.8)Using Hölder’s inequality, we obtain(ϕ′(t))2 =[(m + 1)∫Ωumutdx]2≤ (m + 1)2∫Ωum+1dx∫Ωum−1u2t dx=(m + 1)22mϕ(t)ψ′(t).(2.9)Thus (2.4) implies(ϕ′(t))2 ≥ (m + 1)(2m + q− 1)2mϕ′(t)ψ(t). (2.10)4 Y. Tian, R. Xu and Y. QinWe get from (2.9) and (2.10)ϕ(t)ψ′(t) ≥ (2m + q− 1)m + 1ϕ′(t)ψ(t),by using the notation 2m+q−1m+1 = 1 + a, we findϕ(t)ψ′(t) ≥ (1 + a)ϕ′(t)ψ(t).From the above inequality, we obtain(ϕ−(1+a)ψ)′ ≥ 0,henceϕ−(1+a)ψ ≥ ϕ−(1+a)0 ψ0 = M,where ϕ0 = ϕ(0) and ψ0 = ψ(0). Combining the above formula with (2.10), we findϕ′(t) ≥ (m + 1)(2m + q− 1)2mψ(t) ≥ (m + 1)22m(1 + a)Mϕ1+a, (2.11)then we haveϕ−a(t) ≤ ϕ−a0 −(m + 1)22ma(1 + a)Mt. (2.12)Thereforet∗ ≤2mϕ−a0(m + 1)2a(1 + a)M. (2.13)3 Lower bound for the blow-up timeIn this section, we estimate the lower bound of the blow-up time by constructing some auxil-iary functions and using different inequality techniques, such as Sobolev type inequality andHölder inequality etc. Our theorem is given as follows.Theorem 3.1. Assume that u(x, t) is a nonnegative classical solution of problem (1.1), further it blowsup at finite time t∗. Thent∗ ≥∫ ∞φ(0)dηk1η32 + k2η3(n−m+1)n + k3η − k4ηn+p−1n,where n = 2(m + 2q− 3) and φ(0), k1, k2, k3, k4 are constants, defined in the proof later.Proof. We defineφ(t) =∫Ωu2(m+2q−3)dx =∫Ωundx. (3.1)The derivative of (3.1) w.r.t. t can be written as followsφ′(t) = n∫Ωun−1utdx= n∫Ωun−1(∆um − up)dx= n∫Ωun−1∆umdx− n∫Ωun+p−1dx= nmk∫∂Ωun+m+q−2ds− n(n− 1)m∫Ωun+m−3|∇u|2dx− n∫Ωun+p−1dx.(3.2)Blow-up analysis for a porous media equation with nonlinear sink 5To estimate∫∂Ω un+m+q−2ds, we can refer to the Lemma A.1 in [11], and obtain∫∂Ωun+m+q−2ds ≤ Nρ0∫Ωun+m+q−2dx +(n + m + q− 2)dρ0∫Ωun+m+q−3|∇u|dx, (3.3)whereρ0 = min∂Ω(x · ν), d = max∂Ω|x|.Note that ρ0 is positive since Ω is star-shaped by assumption.Applying the Hölder inequality, we have∫Ωun+m+q−2dx ≤(∫Ωundx) q−1m+2q−3(∫Ωun+m+2q−3dx) m+q−2m+2q−3=(∫Ωundx) 2(q−1)n(∫Ωu3n2 dx) 2(m+q−2)n≤ 2(q− 1)n∫Ωundx +2(m + q− 2)n∫Ωu3n2 dx.(3.4)Using Cauchy’s inequality with ε and inverse Hölder inequality, we get∫Ωun+m+q−3|∇u|dx ≤ 14ε∫Ωun+m+2q−3dx + ε∫Ωun+m−3|∇u|2dx, (3.5)and∫Ωun+p−1dx ≥ |Ω|1−pn(∫Ωundx) n+p−1n. (3.6)First taking (3.4) and (3.5) into (3.3), then taking (3.3) and (3.6) into (3.2), (3.2) becomesφ′(t) ≤[kmdε(n + m + q− 2)ρ0−m(n− 1)]n∫Ωun+m−3|∇u|2dx+[kmnd(n + m + q− 2)4ερ0+2kmN(m + q− 2)ρ0] ∫Ωu3n2 dx+2mkN(q− 1)ρ0∫Ωundx− n|Ω|1−pn(∫Ωundx) n+p−1n.(3.7)Now we estimate∫Ω u3n2 dx, using Sobolev type inequality (see (A.5) in [11]) which holds ifN = 3 and obtain for arbitrary µ > 0∫Ωu3n2 dx ≤ 1334[32ρ0∫Ωundx +n2(1 +dρ0) ∫Ωun−1|∇u|dx] 32≤ 212334{(32ρ0) 32(∫Ωundx) 32+[n2(1 +dρ0)] 32 14µ3|Ω|3(m−1)n(∫Ωundx) 3(n−m+1)n+[n2(1 +dρ0)] 32 3µ4∫Ωun+m−3|∇u|2dx}.(3.8)6 Y. Tian, R. Xu and Y. QinBy substituting (3.8) into (3.7), we obtainφ′(t) ≤{[kmndε(n + m + q− 2)ρ0−mn(n− 1)]+[kmnd(n + m + q− 2)4ερ0+2kmN(m + q− 2)ρ0]×[n2(1 +dρ0)] 32 3µ4212334} ∫Ωun+m−3|∇u|2dx+ k1φ32 + k2φ3(n−m+1)n + k3φ− k4φn+p−1n .(3.9)For ε > 0 small enough, choosing an appropriate µ > 0 such that k0 ≤ 0, this leads toφ′(t) ≤ k1φ32 + k2φ3(n−m+1)n + k3φ− k4φn+p−1n , (3.10)wherek0 ={[kmndε(n + m + q− 2)ρ0−mn(n− 1)]+[kmnd(n + m + q− 2)4ερ0+2kmN(m + q− 2)ρ0]×[n2(1 +dρ0)] 32 3µ4212334},k1 =[kmnd(n + m + q− 2)4ερ0+2kmN(m + q− 2)ρ0]212334(32ρ0) 32,k2 =[kmnd(n + m + q− 2)4ερ0+2kmN(m + q− 2)ρ0]212334[n2(1 +dρ0)] 32 14µ3|Ω|3(m−1)n ,k3 =2kmN(q− 1)ρ0,k4 = n|Ω|1−pn .Integrating (3.10) from 0 to t∗, we obtaint∗ ≥∫ ∞φ(0)dηk1η32 + k2η3(n−m+1)n + k3η − k4ηn+p−1n. (3.11)4 Non-existence of blow-upIn this section we show that if the classical solution exists then it may not blow-up when theexponents satisfy p > m + 2(q− 1). We defineϕ(t) =∫Ωum+1dx. (4.1)We establish the following theorem.Theorem 4.1. Let p > m + 2(q− 1), if u(x, t) is a classical solution of (1.1) for t < t∗ ≤ ∞ thenϕ(t) is bounded for all t < t∗.Blow-up analysis for a porous media equation with nonlinear sink 7Proof. Assume that u(x, t) is a classical solution of (1.1) for t < t∗ ≤ ∞. Taking the derivativeof (4.1), by (2.2) we haveϕ′(t) = (m + 1)mk∫∂Ωu2m+q−1ds− (m + 1)∫Ω|∇um|2dx− (m + 1)∫Ωum+pdx. (4.2)To estimate∫∂Ω u2m+q−1ds, we obtain∫∂Ωu2m+q−1ds ≤ Nρ0∫Ωu2m+q−1dx +(2m + q− 1)dρ0∫Ωu2m+q−2|∇u|dx=Nρ0∫Ωu2m+q−1dx +(2m + q− 1)dρ0m∫Ωum+q−1|∇um|dx.(4.3)Applying Cauchy’s inequality with β, we have∫Ωum+q−1|∇um|dx ≤ β∫Ωu2(m+q−1)dx +14β∫Ω|∇um|2dx. (4.4)Choosing β = (2m+q−1)kd4ρ0 , and inserting (4.4) into (4.3), then inserting (4.3) into (4.2), we getϕ′(t) ≤ (m + 1)[kmNρ0∫Ωu2m+q−1dx +kdβ(2m + q− 1)ρ0∫Ωu2(m+q−1)dx−∫Ωum+pdx]. (4.5)Using Hölder’s inequality and Young’s inequality with ε, we obtain∫Ωu2(m+q−1)dx ≤(∫Ωum+pdx)α (∫Ωu2m+q−1dx)1−α≤ αε 1α∫Ωum+pdx + (1− α)ε 1α−1∫Ωu2m+q−1dx,(4.6)where α = q−1p−(m+q−1) and 0 < α < 1 by the assumption of the theorem.Combining (4.6) with (4.5), we findϕ′(t) ≤ (m + 1)[H∫Ωu2m+q−1dx−W∫Ωum+pdx], (4.7)whereH =[(1− α)ε 1α−1 kdβ(2m + q− 1)ρ0+kmNρ0],W =[1− αε 1α kdβ(2m + q− 1)ρ0],we may choose ε so small that W > 0 holds.Using Hölder’s inequality again∫Ωu2m+q−1dx ≤ |Ω|p−(m+q−1)m+p(∫Ωum+pdx) 2m+q−1m+p, (4.8)8 Y. Tian, R. Xu and Y. Qinthus∫Ωum+pdx ≥ |Ω|−p+(m+q−1)2m+q−1(∫Ωu2m+q−1dx) m+p2m+q−1, (4.9)where |Ω| denotes the measure of Ω. Inserting (4.9) into (4.7), we haveϕ′(t) ≤ (m + 1)∫Ωu2m+q−1dxH −W|Ω|−p+(m+q−1)2m+q−1 (∫Ωu2m+q−1dx) p−(m+q−1)2m+q−1 . (4.10)Application of Hölder’s inequality leads toϕ(t) =∫Ωum+1dx ≤(∫Ωu2m+q−1dx) m+12m+q−1|Ω|m+q−22m+q−1 . (4.11)From the above equation, we obtain(|Ω|−(m+q−2)2m+q−1∫Ωum+1dx) 2m+q−1m+1≤∫Ωu2m+q−1dx.Thus from (4.10) we deriveϕ′(t) ≤ (m + 1)∫Ωu2m+q−1dx[H −W|Ω|m+q−1−pm+1 ϕ(t)p−(m+q−1)m+1]. (4.12)Since p > m + 2(q− 1) ≥ m + q− 1, from (4.12) one can conclude that ϕ(t) is bounded fort < t∗ ≤ +∞. In fact, if for some t0 < t∗, ϕ(t0) is so large that[H−W|Ω|m+q−1−pm+1 ϕ(t0)p−(m+q−1)m+1]is negative, then ϕ′(t) < 0 for all t0 < t < t∗ with the property ϕ(t) > ϕ(t0) since theexponent of ϕ(t) is positive. Consequently, the continuously differentiable function ϕ(t) is(strictly) monotone decreasing in [t0, t∗), thus ϕ(t) ≤ ϕ(t0) if t0 < t < t∗.Remark 4.2. For q = 1, we can see, p = m is the blow-up exponent. But for q > 1 andm + q− 1 < p < m + 2(q− 1), we do not assert whether the solutions blow-up in finite timewith nonlinear boundary condition. Due to technical reasons up to now, we can not give apositive or negative answer.5 Example and applicationsIn this part, we give an example to illustrate applications of Theorem 2.1 and Theorem 3.1.Example 5.1. Let u(x, t) is a solution of the following problemut = ∆u3 − u3, (x, t) ∈ Ω× (0, t∗),∂u∂ν = u2, (x, t) ∈ ∂Ω× (0, t∗),u(x, 0) = u0(x) = 0.5− |x|2 > 0, x ∈ Ω,where Ω = {x ∈ R3 | |x|2 = ∑3i=1 x2i < 0.0001} is a ball in R3. Now m = 3, q = 2, p = 3, k = 1,u0 = 0.5− |x|2, N = 3, ρ0 = 0.01, d = 0.01, |Ω| = 4.1888× 10−6.Blow-up analysis for a porous media equation with nonlinear sink 9First, we get the upper bound of blow-up time through the following calculationsϕ(0) =∫Ωum+10 dx=∫ 2π0dθ∫ π0sin ϕdϕ∫ 0.010(0.5− |r|2)4r2dr= 4π∫ 0.010(0.5− |r|2)4r2dr = 2.6167× 10−7,ψ(0) =2m2k2m + q− 1∫∂Ωu2m+q−10 ds−∫Ω|∇um0 |2dx−2mm + p∫Ωum+p0 dx=187∫ 2π0dθ∫ π0sin ϕdϕ∫ 0.010(0.5− |r|2)7r2dr− 9∫ 2π0dθ∫ π0sin ϕdϕ∫ 0.010(0.5− |r|2)4|∇(0.5− |r|2)|2r2dr−∫ 2π0dθ∫ π0sin ϕdϕ∫ 0.010(0.5− |r|2)6r2dr=727π∫ 0.010(0.5− |r|2)7r2dr− 144π∫ 0.010(0.5− |r|2)4r4dr− 4π∫ 0.010(0.5− |r|2)6r2dr = 1.8111× 10−8.Taking M and a into (2.13), thent∗ ≤ 2mϕ0(m + 1)2a(1 + a)ψ(0)= 4.1280. (5.1)Next, we obtain the lower bound of blow-up time by the following calculationsφ(0) =∫Ωu2(m+2q−3)0 dx=∫ 2π0dθ∫ π0sinϕdϕ∫ 0.010(0.5− |r|2)8r2dr= 4π∫ 0.010(0.5− |r|2)8r2dr = 1.6347× 10−8.We choose ε = 0.1, µ = 0.0022, and calculate thatk1 = 6.9069× 106, k2 = 1.8015× 108, k3 = 1800, k4 = 176.8348.Thent∗ ≥∫ ∞φ(0)dηk1η32 + k2η3(n−m+1)n + k3η − k4ηn+p−1n= 0.0012. (5.2)Therefore, combining (5.1) with (5.2), we get0.0012 ≤ t∗ ≤ 4.1280.10 Y. Tian, R. Xu and Y. QinReferences[1] J. T. Ding, H. J. Hu, Blow-up and global solutions for a class of nonlinear reaction diffu-sion equations under Dirichlet boundary conditions, J. Math. Anal. Appl. 433(2016), No. 2,1718–1735. https://doi.org/10.1016/j.jmaa.2015.08.046; MR3398787; Zbl 1327.35177[2] Z. B. Fang, Y. X. Wang, Blow-up analysis for a semilinear parabolic equation withtime dependent coefficients under nonlinear boundary flux, Z. Angew. Math. 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Blow-up analysis for a porous media equation with nonlinear sink and nonlinear boundary condition

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Blow-up analysis for a porous media equation with nonlinear sink and nonlinear boundary condition

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