﻿ A Numerical Approach to a Nonlinear and Degenerate Parabolic Problem by Regularization Scheme Journal of Applied Mathematics and Physics, 2014, 2, 88-93 Published Online April 2014 in SciRes. http://www.scirp.org/journal/jamp http://dx.doi.org/10.4236/jamp.2014.25012 How to cite this paper: Cao, H.T. (2014) A Numerical Approach to a Nonlinear and Degenerate Parabolic Problem by Regu-larization Scheme. Journal of Applied Mathematics and Physics, 2, 88-93. http://dx.doi.org/10. 4236/jam p. 2014.25012 A Numerical Approach to a Nonlinear and Degenerate Parabolic Problem by Regularization Scheme Haitao Cao Department of Mathematics and Physics, Changzhou Campus, Hohai University, Changzhou, China Email: 2010400 7004@s uda.edu .cn Received October 2013 Abstract In this work we propose a numerical scheme for a nonlinear and degenerate parabolic problem having application in petroleum reservoir and groundwater aquifer simulation. The degeneracy of the equation includes both locally fast and slow diffusion (i.e. the diffusion coefficients may ex- plode or vanish in some point). The main difficulty is that the true solution is typically lacking in regularity. Our numerical approach includes a regularization step and a standard discretization procedure by means of C0-piecewise linear finite elements in space and backward-differences in time. Within this frame work, we analyze the accuracy of the scheme by using an integral test function and obtain several error estimates in suitable norms. Keywords Nonlinear Degenerate Parabolic Equation, Finite Element Method, Regularization, Implicit Scheme 1. Introduction Denoting nRΩ⊂ (n ≥ 1) as the domain occupied by the porous medium with Lipschitz boundary and (0, T] (0 < T < +∞) as the time interval. In this work we propose a numerical method to the following nonlinear and de- generate parabolic equation. (]tb(u)div(A(x)u + g(x, b(u))) = 0, (x, t) Ω 0,T∂ −∇∈× (1.1) with the boundary and initial conditions u = 0 on Ω (0,T],u(x,0)=u00 in Ω,∂× ≥ (1.2) where the function b(s) is uniformly bounded, continuous and strictly increasing in s. But the degeneracy condi- tions, b′(s) = 0 and b′(s) = +∞ for some point, are included. It means that the equation includes both locally fast and slow diffusion. Throughout this paper we assume, without loss of generality, that b′(s) = 0, s → +∞ and b′(0+) = +∞. Thus the problem (1.1)-(1.2) often used to describe the flows in porous media, infiltration and multi-phase change see - H. T. Cao 89  etc. The existence of variational solution of (1.1)-(1.2) has been studied in   etc. Here we point out that the Equation (1.1) is usually obtained after using the Kirchhoff transformation. Such problems can be inves-tigated through a parabolic regularization of function b(u) or by perturbing the boundary and initial data such that the corresponding solutions do not take the degenerate points. In the past several decades, there are numer-ous papers about discussing this problem. For example, in  , the authors consider a rather general class of singular parabolic problems; two-phase Stefan problem and porous medium equation are included. By using a regularization scheme and a parabolic duality technique, a fully discrete numerical approach is proposed and analyzed. In  , Richards’ equation is analyzed by a numerical approach also consisting in a regularization procedure and discretization by means of C0-piecewise linear finite elements (or mixed finite elements) in space and backward-differences in time. In , basing on the maximum principle, the authors considered the porous medium equation by perturbing the boundary and initial data to overcome the degeneracy. On the other hand, there are many other papers dealing with such problem by some (linear) relaxation schemes or imposing some suitable conditions to deal with the degeneracy and nonlinearity  1213]. However, to our knowledge, most of papers have considered one of the degenerate cases, i. e. b′(s) = 0 for some point but b is Lipchtize continuous (such as the Richards’ equation) or b′(s) > 0 but discontinuous at one point 0 (such as Stefan problem or porous-medium equation). In our paper, we will deal with the two degenerate cases simultaneously and present the error estimates of several unknowns by using an integral test function. Due to the singularity and degeneracy of b, solutions of (1.1) may not be classical; therefore they must be understood in the sense of distribution . The proposed method in the paper, we first replace b by a regular function bϵ (whose derivative is bounded by two values depending on regularized parameter ϵ > 0). The second step is a standard discretization procedure by means of C0-piecewise linear finite elements in space and backward-dif- ferences in time. Within this frame work, we analyze the accuracy of the scheme by using an integral test func- tion and obtain several error estimates in suitable norms. The layout of the paper is as follows. In Section 2, we give out the numerical formulation and main result. In Section 3, we will prove a prior estimate and the main result of our paper. Notations: 2 1-1T0Ω = Ω (0, T), L(0, T;H () ,H()× ΩΩ is de dual space of 10H. Other Soblev space can be referred to . From now on, C will denote a generic positive constant which is independent of ϵ. 2. Problem Setting and Main Result For the problem (1.1)-(1.2), we make the following hypotheses upon the data. Assumption: (H1) The function b mapping [0, ] [0,1]+∞  is continuous and strictly increasing. And b′(s) = 0, s → +∞, b′(0) = +∞. (H2) (())(1,) :nnijAAxi jnRM=≤≤  is continuous and satisfies 221 2120, ||||,.nARλλ λξξξλξξ∃<≤≤⋅ ≤∈ (H3) nnig = (g(x, s))(1 in) : R R R≤≤× → is continuous in s and fulfill the following identities: 2 |g(y, b(u))g(y, b(v))| C|b(u) b(v)||u v|. − ≤−− (H4) 2u0 L()().H∞∈ Ω∩ Ω Remark 2.1. Due to the maximum principle, the solution of the problem (1.1)-(1.2) is great or equal to zero  . In physical model, the function b(s) usually presents the enthalpy of Stefan model or denotes the re-duced saturation of fluid in porous media. According to Alt and Luckhaus , we have at least that 2-12 1t0b(u)L(0, T;H()), uL(0, T;H()).∂ ∈Ω∈Ω Because b is uniformly bounded, therefore we conclude that 0 -1b(u) C(0, T;H()).∈Ω This gives us b(u(·,t)) point wise for every t ∈ (0, T]. Let HT be a decomposition of Ω into a regular conforming finite element mesh with maximal element di- ameter H. Denote HKTK∈Ω=∪. Then we define finite element space HX as: 0{( );|0}.HXCislinearfor allKandχχ χ∂Ω=∈Ω = Let N be an integer, τ = T/N, τiti=. Our numerical method reads as: find , ()HHi Hiiv Xbvεη∈= (i = 1,···, N), for all HXχ∈ H. T. Cao 90 100(,)(()( ,),)0,,HHii ii HAxvgxvI uηη χ ηχτ−−+ ∇+∇== (2.3) where HI is the C0-piecewise linear interpolant operator to HX and satisfies: for 2()uH∈Ω 2() (),1 2,ssHLHI uuCHusΩΩ− ≤≤≤ (2.4) and ()bsε is defined as 111,[0,( )]()() .( ),((),]ss bbbsbsss bεεεεεε−−−∈=+ ∈+∞ (2.5) In the above terms, ϵ is small parameter. It is easy to get 1| ()()|max{,||},'(),()bsbssbsbεεεεε εε−−≤≤ ≤ (2. 6) The Equation (2.3) is a general nonlinear elliptic problem. It can be numerically solved by relaxation iteration scheme or to apply a linearization scheme first. We point out that the regularization (2.5) of a degenerate prob- lem is not necessary in implementing numerical analysis with suitable assumptions . Our main result of the numerical approach is present as the following: Theorem 2.2. Let , ()u buθ= and Hkv be the solution of (1.1) and (2.3) respectively. Suppose (H1) -(H4) hold, then there exist 11222222011()()((),)()() ()n kkkknnt ttHH HHkk kkttkkLLbv uvuvuvCHθ τε−−Ω== ΩΩ−−+∇−+ ∇−≤++∑∑∫ ∫∫ where the constant C is independent of τ, ϵ,H. Remark 2.3. Due to the lacking regularity of solution, the result of Theorem 2.2 is not optimal with respect to time discretization. On the other hand, if b′ is positive and bounded, we can get the classical results. 3. Proof of Main Result Because of the lacking regularity, the solution of (1.1)-(1.2) must be understood in terms of distributions, as proposed in . Firstly, we formulate the weak solution of (1.1)-(1.2) as follows: Definition3.1. We say that u is a weak solution of problem (1.1)-(1.2), if it satisfies the following two identi- ties: (1) 2Tb(u) L(Ω )∈ and2 -1tb(u) L(0,T;H() ) ∂∈Ω with 0ttb(u) dxdt(b(u(t))-b(u)) dxdt0, TTϕϕΩΩ∂ +∂=∫∫ (3. 7) for every2 112 (0,;())(0,;())L THHTLϕ∈ Ω∩ Ω with φ(T) = 0. (2) For 21 L(0,T;H() ) ϕ∈Ω tb(u) dxdt(()(,()))dxdt0,TTAxu gxbuϕϕΩΩ∂+∇ +⋅∇=∫∫ (3.8) According to , it follows that Lemma3.2. Assuming (H1)-(H4) hold, if u is a solution of (3.7)-(3.8), we have 1221()( )(0, ;())[0, ]max (())()TtLLL THtTButbuu C−ΩΩΩ∈+∂+∇ ≤ where ()0( ())()( ())()utBut utbutbsds= −∫. For the discrete scheme (2.3), we also have Lemma3.3. Assuming (H1)-(H4) hold, if for all 1iN=, Hiv solves problem (2.3), for any 0kN<<, we have 221()( ()).kHHkiiLB bvdxvCετΩ=Ω+∇ ≤∑∫ H. T. Cao 91 Proof: Let Hivχ= in (2.3), we have 1(,)(()() ,)0H HHiiiii ivAx vgxvηη τη−Ω Ω−+∇+∇ = Summing 1ik=, we get 001 111(,) (,)(,)(()(,),)0kkH HHHHHkki iiiiiiivvvvAxvgxvηη ητηΩΩ −−ΩΩ= =−−−+∇+∇ =∑∑ Denote the above terms by 1230.III++= 1221 0001222311() ()(,)(,)()(( ))(()),, ||.2HiHikvHHH Hkk kvikkHHiiiiLLIvvbsdsdxBbv dxBbv dxIvIC vεεεηηττ−ΩΩΩ ΩΩ== =ΩΩ≥−−= +≥ ∇≤+∇∑∫∫ ∫∫∑∑ Combining all the above terms, we get the conclusion. Proof of Theorem 2.2: Let(( ))bu tθ=, we integrate (1.1) over time from 1it− to it, for every HXχ∈, to get 111(,)((),)((),) 0,iiiittii ttA xudtgdtθθ χχθχ−−−ΩΩ Ω−+∇∇+∇ =∫∫ (3.9) Subtract (2.3) from (3.9), we get 1111((),)(()(),) (()(),)0iiiittHii iiiittA xuvdtggdtθη θηχχθηχ−−−− ΩΩΩ−−−+∇−∇+−∇=∫∫ Summing the equality from 1i= to k, we obtain 1100 11((),)(()(),)(() (),)0.iiiikkttHkki ittiiA xuvdtggdtθη θηχχθηχ−−Ω ΩΩ= =− −−+∇−∇+−∇=∑∑∫∫ Summing k from 1 to n again, we get 11001 1111((),)(()(),)(()(), )0.iiiin nknkttHkki ittk kikiA xuvdtggdtθη θηχχθηχ−−Ω ΩΩ== == =− −−+∇−∇+−∇=∑ ∑∑∑∑∫∫ Setting 1kktHHk HtI uv dtXχ−= −∈∫ and 0 000,( ),()uH Heu IububIuεθη=−= =, to gives 1 111111001 1111001(( ),())(()(),())(()() ,())((),)(()(),k ikk ikikikkikin nkt ttH HHkkki kt ttk kink ttHikttkinttHkk uittkuv dtAxuvdtuvdtggdtuvdte dtAxuvdtθη θηθηθη θη− −−−−−−ΩΩ== =Ω= =Ω=− −−−+∇−∇−+− ∇−=− −−+∇−∑ ∑∑∫ ∫∫∑∑ ∫∫∑∫∫1111111)(()() ,)kkikiknk tutkink ttiuttkie dtggdte dtθη−−−Ω= =Ω= =+−∑∑ ∫∑∑ ∫∫ Denoting the above formulation by 123 456,IIIIII++=++ we will deal with each terms in the following. 1111 110011 121111 11 1111 112 113(,) (,),((),) ((),)(,),kkkkk knnkkHHkk kkttkknn nk kktHHH HHkkkk kskt tttkk kIu vdtu vdtIIIbvuv dtbvuv dtdsuv dtIIIθη θηθ ηθ−−−− −ΩΩ= =Ω ΩΩ= ===−− −−−=+=− −+−−+∂−=++∑∑∫∫∑∑ ∑∫ ∫∫∫ It is easy to see that 1110.I≥ Use (2.6) and Lemma 3.3, it follows that H. T. Cao 92 2112222222112 ()11 1() ()||()( )()( )kkkknn nttHH Hkk kttLkk kLLI CvuvCuvδετδδεδ−−Ω= ==ΩΩ≤+∇−≤ +∇−∑∑ ∑∫∫ Considering that 2-121t0b(u)L(0, T;H()), uL(0, T;H())∂ ∈Ω∈Ω, we get 1110113 ()()|| .kktHtktHHIu vdtCτθ τ−−ΩΩ≤∂ −≤∫ So we get 0122211 1()((),)( )()().nkknttHH Hkk kttkLIbv uvdtCuvθδετδ−Ω=Ω≥ −−−+−∇−∑∫∫ Because 00 0000|||() ()||() ()|Hbu bubu bIuε εεθη−≤ −+−, we have 22121222224120 0() ()11()2241()| |()()()( )()().kkkknntHkLHtkkLntHktkLI CuHuuvC Huvδ τεδδε δ−−ΩΩ== Ω=Ω≤++ ∇−≤+ +∇−∑∑∫∑∫ Using the following equality 2211 112( )(),nk nnki k kki kkaa aa= ==== +∑∑ ∑∑ we can get 21122112222()11()22()11()11() ()() ()22(()() ).kkkkkkkknnttHHkkttLkkLnnttHHkkttLkkLIA xuvdtA xuvdtCu vdtu vdt−−−−Ω= =ΩΩ= =Ω=∇−+∇−≥∇−+ ∇−∑∑∫∫∑∑∫∫ For the term 3I, noticing ()Hiibvεη=, we use (H3), Cauchy inequality and the fact (2.6) to get 111 11122311 1122311 1() ()(()(()),())(()(()),()),||()()( ())()ikikikikikiknk nkttt tHH HHikii kttt tki kink nttH HHi kkttkik kLLIggbvdtuvdtg vg bdtuvdtICggb vdtuvdtvθηδ τθδετ−−−−−−= == =ΩΩ= ==ΩΩ=−∇− +−∇−≤−+∇−+∑∑ ∑∑∫∫∫ ∫∑∑ ∑∫∫201221()2211()( )((),)().kkknLnnttHH Hii kttkkLCb vuvdtuvdtδτθ δε−=Ω= =ΩΩ≤−− +∇−+∑∑∑∫∫ Similarly, the right hand terms 456,,III can be estimated as the following. Use the fact (2.4), we have 2 220122()220112221224()(,; ())00()22225(,; ())11() ()6||( ),||( )()( )(),||(( ),)knkLkknkknntkkukkuL LttLtkkLnnttHHku kLttLttkkLLHHiitIe dteCHIu vdtCeu vdtCHICbvuvθη τθηδδδδτθ−Ω−−ΩΩ= =ΩΩ= =ΩΩ≤−≤ −≤≤ ∇−+∇≤ ∇−+≤ −−∑∑∫∑∑∫∫2200022 22(,; ())11((),)().kknnnttHHu iiLttL tkkdteCb vuvdtCHε τθεΩ= =ΩΩ+∇+ ≤−−++∑∑∫∫ Combining all the terms and choosing δ properly, we have 0 112202211()()221((),) ()()(( ),)().n kkkkknnt ttHH HHkk kkt ttkkLLntHHiitkbvuvdtu vdtuvdtCb vuvdtCHθτ θτε−−Ω== ΩΩΩ=−−+∇−+ ∇−≤−−+ ++∑∑∫ ∫∫∑∫ H. T. Cao 93 Finally, we use Gronwall inequality to get the conclusion of the Theorem 2.2. Acknowledgements The author is supported by the Fundamental Research Funds for the Central Universities (NO.2013B10114). References  Alt, H.W. (1985) Nonsteady Fluid Flow through Porous Media. Applications and Theory, 3, 222-228.  Damlamina , A. (1977) Some Result in the Multiphase Stefan Problem. 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