# Blow-Up Phenomena for Porous Medium Equation with Nonlinear Flux on the Boundary

Authors: Yan Hu; Jing Li; Liangwei Wang
Journal: Journal of Applied Mathematics
 (2013)

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## Abstract

We investigate the blow-up phenomena for nonnegative solutions of porous medium equation with Neumann boundary conditions. We find that the absorption and the nonlinear flux on the boundary have some competitions in the blow-up phenomena.


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## Body

## 1. Introduction

In this paper, we are concerned with the blow-up of solutions of porous medium equations with nonlinear flux on the boundary. Consider(1)ut=Δum-f(u),(x,t)∈Ω×(0,t*),(2)∂um∂ν=g(u),(x,t)∈∂Ω×(0,t*),(3)u(x,0)=u0(x),x∈Ω,
where m>1, the nonnegative initial value u0(x)∈C(Ω)⋂‍L∞(Ω), Ω is a bounded region in ℝN (N≥2) with the sufficiently smooth boundary ∂Ω, ν is the unit normal vector on ∂Ω, t* is the blow-up time if blow-up occurs, or else t*=∞.The blow-up phenomena for the nonnegative solutions of the heat equation with nonlinear sources (m=1 and f(u)=-up in (1)) in the whole space ℝN was first found by Fujita in 1966, see [1]. He proved the following results:  (a)
if1<p<1+(2/N), then (1) has no global positive solutions;(b)
ifp>1+(2/N), then there exist global positive solutions.The critical casep=1+(2/N) was proved to belong to the blow-up case in 1970’s by several authors [2–4]. In 1980, Galaktionov and others [5] considered the nonnegative solutions of (1) (with m>1 and f(u)=-up) in whole space ℝN. They found some results similar to those for the heat equation (m=1) as follows(a)
if1<p<m+(2/N), then (1) has no global solutions;(b)
ifp>m+(2/N), then there exist global positive solutions that decay like t-1/(p-1).In [6, 7], Galaktionov, Mochizuki and Suzuki, had also revealed that the critical case p=m+(2/N) belongs to the blow-up case, see also [8, 9].In 2010, Payne et al. [10] considered a semilinear heat equation with nonlinear boundary condition (m=1 in (1)) and established conditions on nonlinearities sufficient to guarantee that u(x,t) exists for all time t>0 as well as conditions on data forcing the solution u(x,t) to blow up at some finite time t*. When N=1, the blow-up phenomena for the solutions of the porous medium equation with nonlinear flux on the boundary had also been studied by several authors [11, 12]. For other interesting results on the large time behavior on the solutions of the porous medium equation, we refer the reader to papers [13–16].Inspired by the above papers, we will study the blow-up phenomena for the solutions of the porous medium equation with nonlinear flux on the boundary in higher dimensional space (N≥2). In fact, we find that if the absorption is more powerful than the boundary flux, then the solutions of the problem (1)–(3) exist for all time on a bounded star-shaped region. On the other hand, if the boundary flux is more powerful, then the solutions of the problem (1)–(3) blow-up at a finite time. Moreover, we will give the upper-bound estimates for the blow-up time.The paper is organized as follows. In Section2, we concentrate our attention on the conditions of the global existence for the solutions of the problem (1)–(3). Section 3 is devoted to the investigation of the blow-up phenomena for the solutions of the problem (1)–(3).


## 2. Criterion for Global Existence

In this section, we investigate the global solutions of problem (1)–(3). The main result of this section is the following theorem.Theorem 1.
LetΩ be a bounded star-shaped region and assume that q>m satisfy
(4)2q<m+p.
If f and g satisfy the following conditions:
(5)f(ξ)≥k1ξp,ξ≥0,(6)0≤g(ξ)≤k2ξq,ξ≥0,
where k1, k2 are nonnegative constants, then the nonnegative solutions u(x,t) of the problem (1)–(3) do not blow up.Proof.
Let(7)Φ(t)=∫Ωu2dx.
Differentiating (7) and making use of (1), we obtain that
(8)Φ′(t)=2∫Ωuutdx=2∫Ωu[Δum-f(u)]dx.
From the hypothesis (5), we get
(9)Φ′(t)≤2∫Ωu(Δum-k1up)dx.
By (2), (6) and the divergence theorem, we have
(10)∫ΩuΔumdx=∫∂Ωu∇um·νds-∫Ω∇u·∇umdx=∫∂Ωu∂um∂νds-∫Ω∇u·∇umdx≤k2∫∂Ωuq+1ds-∫Ω∇u·∇umdx.
Here we used the identity div(u∇um)=uΔum+∇u·∇um. By the divergence theorem again, we get
(11)∫∂Ω(uq+1x)·νds=∫Ωdiv(uq+1x)dx.
Let
(12)ρ0=min∂Ω(x·ν),d=max∂Ω|x|.
Point out that ρ0 is positive because Ω is star-shaped by hypothesis. Notice also that
(13)div(uq+1x)=div[(u(m+1)/2)2(q+1)/(m+1)x]=Nuq+1+2(q+1)m+1u(2q-m+1)/2(x·∇u(m+1)/2).
We thus have
(14)∫∂Ωuq+1ds≤Nρ0∫Ωuq+1dx+2d(q+1)ρ0(m+1)×∫Ω‍u(2q-m+1)/2|∇u(m+1)/2|dx.
On the another hand
(15)∇u·∇um=∇u·(mum-1∇u)=m(u(m-1)/2∇u)(u(m-1)/2∇u)=4m(m+1)2|∇u(m+1)/2|2.
Therefore, from (10)–(15), we have
(16)Φ′(t)≤2k2∫∂Ωuq+1ds-2∫Ω∇u·∇umdx-2k1∫Ωup+1dx≤2k2Nρ0∫Ωuq+1dx+4k2d(q+1)ρ0(m+1)∫Ωu(2q-m+1)/2|∇u(m+1)/2|dx-8m(m+1)2∫Ω|∇u(m+1)/2|2dx-2k1∫Ωup+1dx.
We obtain from the Young inequality that
(17)∫Ωu(2q-m+1)/2|∇u(m+1)/2|dx≤σ2∫Ωu2q-m+1dx+12σ∫Ω|∇u(m+1)/2|2dx,
where
(18)σ=k2d(q+1)(m+1)4mρ0.
This σ leads to
(19)4k2d(q+1)ρ0(m+1)∫Ωu(2q-m+1)/2|∇u(m+1)/2|dx≤8m(m+1)2σ2∫Ωu2q-m+1dx+8m(m+1)2∫Ω|∇u(m+1)/2|2dx.
Combining this with (16), we get
(20)Φ′(t)≤2k2Nρ0∫Ωuq+1dx+8m(m+1)2σ2×∫Ωu2q-m+1dx-2k1∫Ωup+1dx.
Let
(21)α=q-mp-q.
Therefore, the hypotheses that q>m and 2q<m+p imply that
(22)0<α<1.
So, by Hölder's inequality, we have
(23)∫Ωu2q-m+1dx≤(∫Ωuq+1dx)α(∫Ωup+1dx)1-α.
For ϵ>0, we obtain from (23) that
(24)∫Ωu2q-m+1dx≤(ϵ∫Ωup+1dx)1-α(ϵ(α-1)/α∫Ωuq+1dx)α≤(1-α)ϵ∫Ωup+1dx+αϵ(α-1)/α∫Ωuq+1dx.
Thus, inserting (24) in (20), we obtain
(25)Φ′(t)≤2k2Nρ0∫Ωuq+1dx+8m(m+1)2σ2×{(1-α)ϵ∫Ωup+1dx+αϵ(α-1)/α∫Ωuq+1dx}-2k1∫Ωup+1dx=(2k2Nρ0+8mσ2α(m+1)2ϵ(α-1)/α)×∫Ωuq+1dx+(8mσ2ϵ(1-α)(m+1)2-2k1)×∫Ωup+1dx=M1∫Ωuq+1dx-M2∫Ωup+1dx,
where
(26)M1=2k2Nρ0+8mσ2α(m+1)2ϵ(α-1)/α>0,M2=2k1-8mσ2ϵ(1-α)(m+1)2,
and let ϵ be sufficiently small to ensure M2>0. By Hölder's inequality again, we have
(27)∫Ωuq+1dx≤(∫Ωup+1dx)(q+1)/(p+1)|Ω|(p-q)/(p+1),
where we assume throughout the paper that |Ω|=∫Ωdx is the measure of Ω. Using (25) and (27), we obtain
(28)Φ′(t)≤M1(∫Ωup+1dx)(q+1)/(p+1)×{|Ω|(p-q)/(p+1)-M2M1(∫Ωup+1dx)(p-q)/(p+1)}.
Moreover, using Hölder's inequality once more, we have
(29)Φ(t)=∫Ωu2dx≤(∫Ωup+1dx)2/(p+1)|Ω|(p-1)/(p+1),
that is,
(30)(∫Ωup+1dx)(p-q)/(p+1)≥Φ(t)(p-q)/2|Ω|(1-p)(p-q)/2(p+1).
Finally, from (28) and (30), we obtain
(31)Φ′(t)≤M1(∫Ωup+1dx)(q+1)/(p+1)×{M2M1|Ω|(p-q)/(p+1)hhhhh-M2M1Φ(t)(p-q)/2|Ω|(1-p)(p-q)/2(p+1)}.
We deduced from (31) that Φ(t)≤max{Φ(0),(M2/M1)2/(q-p)|Ω|}. On the other hand, Φ(t) is nonnegative function by assumption. So that Φ(t) keeps bounded continuously under the conditions given in Theorem 1, the solutions exsit for all time t>0. That is, we find that the global solution exists when the absorption is more powerful than the nonlinear boundary flux and this accomplishes the proof of Theorem 1.


## 3. Criterion for Blow-Up

In this section, we concentrate on the finite timet* on which blow-up occurs. We construct two auxiliary functions to redefine f and g, then the nonlinear boundary-flux is more powerful than the absorption, and we obtain the following result.Theorem 2.
Suppose(32)0≤α≤β.
Let
(33)F(ξ)=∫0ξf(η)dη-m(m-1)2∫0ξ|∇η|2ηm-2dη,G(ξ)=∫0ξg(η)dη,Ψ(t)=2∫∂ΩG(u)ds-∫Ω∇u·∇umdx-2∫ΩF(u)dx.
If
(34)Ψ(0)>0,ξf(ξ)≤2(1+α)F(ξ),ξ≥0,ξg(ξ)≥2(1+β)G(ξ),ξ≥0,
then the solutions u(x,t) of the problem (1)–(3) blow up at time t*<T with
(35)T=Φ(0)2β(1+β)Ψ(0).
Here Φ(t) is defined in (7). Moreover, if β=0, then T=∞.Proof.
Differentiating (7) and using the hypothesis (33), we have
(36)Φ′(t)=2∫Ωuutdx=2∫Ωu[Δum-f(u)]dx=2∫∂Ωug(u)ds-2∫Ω∇u·∇umdx-2∫Ωuf(u)dx≥2(1+β)Ψ(t).
Differentiating (33), we thus obtain from (15) that
(37)Ψ′(t)=2∫∂Ωg(u)utds-4m(m+1)2∫Ω(|∇u(m+1)/2|2)tdx-2∫Ω‍f(u)utdx+m(m-1)∫Ω‍|∇u|2um-2utdx.
Note the identity that
(38)(|∇u(m+1)/2|2)t=(m+1)2(m-1)4×|∇u|2um-2ut+(m+1)22um-1∇u·∇ut.
So, from (37), we get
(39)Ψ′(t)=2∫∂Ωg(u)utds-2∫Ω∇ut·∇umdx-2∫Ωf(u)utdx.
Therefore,
(40)Ψ′(t)=2∫Ωut2dx>0.
Here, we have used the identities
(41)div(ut∇um)=utΔum+∇ut·∇um,∫Ω∇ut·∇umdx=∫∂Ωut∇um·νds-∫ΩutΔumdx.
So, the hypothesis Ψ(0)>0 implies that for all t∈(0,t*), the following inequality holds  (t)>0:
(42)Ψ(t)>0.
By the Schwarz inequality, we have
(43)(Φ′(t))2=4(∫Ωuutdx)2≤2Φ(t)Ψ′(t).
Together with (36), we have
(44)Φ(t)Ψ′(t)≥12[Φ′(t)]2≥(1+β)Φ′(t)Ψ(t).
That is,
(45)(ΨΦ-(1+β))′≥0.
Integrating this from 0 to t, we obtain
(46)Ψ(t)(Φ(t))-(1+β)≥Ψ(0)(Φ(0))-(1+β)=M.
Substituting (46) in (36) we obtain the differential inequality
(47)Φ′(t)≥2(1+β)Ψ≥2(1+β)MΦ1+β.
If β>0, then
(48)(Φ(t))-β≤(Φ(0))-β-2β(1+β)Mt.
This leads to
(49)t*≤T=12β(1+β)M(Φ(0))-β=Φ(0)2β(1+β)Ψ(0).
If α=β=0, then
(50)Φ(t)≥Φ(0)e2Mt
holds for t>0. This implies that t*=∞ and completes the proof of Theorem 2.



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*Source: 952126-2013-11-13.xml*