# Blowing Up for thep-Laplacian Parabolic Equation with Logarithmic Nonlinearity

Authors: Asma Alharbi
Journal: Advances in Mathematical Physics
 (2021)

---

## Abstract

In this article, we are concerned with a problem for thep-Laplacian parabolic equation with logarithmic nonlinearity; the blow-up result of the solution is proven. This work is completed Boulaaras’ work in Math. Methods Appl. Sci., (2020), where the author did not study the blowup of the solution.


---

## Body

## 1. Introduction

In the current manuscript, we consider the following initial-boundary value problem for a nonlinearp-Laplacian equation:
(1)ut−div∇up−2∇u+up−2u=up−2ulnu,x∈Ω,t>0,ux,0=u0x,x∈Ω,ux,t=0,x∈∂Ω,t≥0,where Ω⊂Rn is a bounded domain with smooth boundary ∂Ω and u0 is the initial data p satisfying
(2)2<p<∞,ifn≤p,2<p<npn−p,ifn>p.The terminology of nonlinear polynomials is among the work that researchers have focused on recently. For example, it is found in edge detection and optical elasticity, materials science, engineering, physics, and photonics. In addition, many works and problems in applied sciences have been designed and proposed by means of partial differential equations, including the modeling of some dynamic systems in physics and engineering ([1–13]).The same is said for the evolutionary partial differential equations associated withpx-Laplacian (see [8, 14, 15]).We also note that logarithmic nonlinearity has been concerned by many scientists and researchers, and it has introduced many issues, including the wave equation (see [3, 16–18]).And for more information on some of the other works to which this term was introduced, we refer the reader to [13, 14, 16–24].Later on, in [25], the authors by the multiplier method gave the energy decay of the solution of the following problem:
(3)utt−div∇up−2∇u−Δut+utq−1ut=up−1u.In addition, the authors in [14] proved the decay rate of solutions (exponential and polynomial) by using the inequality of Nakao for the seminar problem (3).On the other hand, for the Laplacian parabolic equation with the logarithmic source term in [21], Chen et al. studied the following problem:
(4)ut−Δu−Δut=ulnu.Then, in [23], the authors proved the global existence, the decay, and the blowup of the solutions of the problem:
(5)ut−div∇up−2∇u−Δut=up−2ulnu,where p>2.Also, in [14], the authors established the global boundedness and the blowup of the solution of the problem (5) for 1<p<2.Motivated by the last recent mentioned works, here, we investigated problem (1) with the nonlinear diffusion Δp=div∇up−2∇u and logarithmic nonlinearity up−2ulnu which extends problem in [14]. Our goal is to blow up solutions for problem (1) in order to put some preliminaries. More precisely, we give the blow-up result.


## 2. Preliminaries

As a starting point, we gave some essential definitions and lemmas.(6)up=uLpΩ,u1,p=uW01,pΩ=up+∇up1/p,for 1<p<∞, and we symbolize the positive constants by C and Ci (i=1,2,⋯).Lemma 1 [7] (logarithmic Sobolev inequality).
Letu be all function u∈W01,pRn\0. Then, for p>1, μ>0,
(7)p∫RnuplnuuLpRndx≤μ∫Rn∇updx−nplnpμenLp∫Rnupdx,where
(8)Lp=pnp−1ep−1π−p/2Γn/2+1Γnp−1/p+1p/n.Remark 2.
Letu∈W01,pΩ\0, and by defining ux=0 for x∈Rn\Ω, we can write
(9)p∫ΩuplnuuLpΩdx≤μ∫Ω∇updx−nplnpμenLp∫Ωupdx.


## 3. Blowup

In this third section, we gave the proof of blowup of solution of our problem.Theorem 3.
For any initial datau0∈H, the problem (1) has a unique weak solution:
(10)u∈C0,T;H,for some T>0.First, we introduce the energy functional in the following lemma.Lemma 4.
Letut be a solution of (1), then Et is nonincreasing; that is,
(11)Et=1p∇upp−1p∫Ωlnuupdx+p+1p2uppsatisfies
(12)E′t=−ut22.Proof.
Multiplying (1) by ut and integrating on Ω, we have
(13)−∫Ωdiv∇up−2∇uutdx+∫Ωup−2uutdx+∫Ωututdx=∫Ωup−2ulnuutdx,ddt1p∇upp+1pupp−1p∫Ωlnuupdx+1p2upp=−ut2.
Thus,(14)E′t=−ut2.
☐To get to our goal of proving the main result, we define the functional(15)Ht=−Et=−1p∇upp+1p∫Ωlnuupdx−p+1p2upp.Theorem 5.
Assume thatE0<0, then the solution of problem (1) blows up in finite time.Proof.
From (12), we have
(16)Et≤E0≤0.
Hence,(17)H′t=−E′t=ut22≥0,0≤H0≤Ht≤1p∫Ωlnuupdx.
We set(18)Kt=H1−α+ε2∫Ωu2dx,where ε>0 and
(19)0<α<p−2p<1.
Multiplying (1) by u and the derivative of (18) gives
(20)K′−αH′t−ε∇upp−ε∇upp+ε∫Ωuplnudx.
Adding and subtractingεδHt into (20) (δ>0), we obtain
(21)K′−αH′t+εδ−pp∇upp+εδ−pp+1p2upp−εδ−pp∫Ωlnuupdx+εδHt.
Applying the logarithmic Sobolev inequality gives(22)K′−αH′t+εδHt+εδ−pp1−μp∇upp+εδ−pp.1+δpδ−p−lnup+np2lnpμenLpupp.
Settingμ=p/2 and taking δ>p give
(23)1+δpδ−p−lnup+np2lnp2e2nLp>0,since
(24)up>e1+δpδ−pp2e2nLpn/p2.
Consequently, for someβ>0, inequality (25) gives
(25)K′t≥βHt+upp+∇upp,(26)Kt≥K0>0,t>0.
Next, by (18), we have
(27)Kt=H1−α+ε2∫Ωu2dx≤H1−α+εCup2≤H1−α+εCupp2/p.
Therefore,(28)K1/1−αt≤H1−α+εCupp2/p1−α,where 0<2/p1−α<1,
(29)up22/p1−α≤Cuppp+Ht.
Hence,(30)K1/1−αt≤C1Ht+upp≤C1Ht+∇upp+upp.
According to (25) and (30), we get
(31)K′t≥λK1/1−αt,where λ=C1/β>0, depending only on β and C1.
Finally, by integrating (31), we obtain
(32)Kα/1−αt≥1K−α/1−α0−λα/1−αt.
Hence,Kt blows up in time:
(33)T≤T∗=1−αλαKα/1−α0.
As a result, the proof is completed. ☐



---

*Source: 4275898-2021-07-27.xml*