CPIS.MATHEMATICS Conference Papers in Mathematics 2314-4777 Hindawi Publishing Corporation 454329 10.1155/2013/454329 454329 Conference Paper Stability and Error Analysis of the Semidiscretized Fractional Nonlocal Thermistor Problem Sidi Ammi M. R. Conference Presenter 1 El Hachimi A. 2 Frederico Gastão S. F. Martins Natália Torres Delfim F. M. Zaslavski Alexander J. 1 Department of Mathematics, AMNEA Group, Faculty of Sciences and Techniques, Moulay Ismail University, B.P. 509, Errachidia Morocco umi.ac.ma 2 Department of Mathematics, Faculty of Sciences, University Mohammed V, B.P. 1014, Rabat Morocco um5a.ac.ma 2013 8 9 2013 2013 09 06 2013 20 07 2013 2013 Copyright © 2013 M. R. Sidi Ammi and A. El Hachimi. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

A finite difference scheme is proposed for temporal discretization of the nonlocal time-fractional thermistor problem. Stability and error analysis of the proposed scheme are provided.

22 April 2013 - 25 April 2013 The Cape Verde International Days on Mathematics 2013 Praia, Cape Verde
1. Introduction

Let Ω be a bounded domain in N with a sufficiently smooth boundary Ω and let QT=Ω×(0,T). In this work, we propose a finite difference scheme for the following nonlocal time-fractional thermistor problem: (1)αutα-Δu=λf(u)(Ωf(u)dx)2,inQT=Ω×(0,T),uν=0,onST=Ω×(0,T),u(0)=u0,inΩ, where αu(x,t)/tα denotes the Caputo fractional derivative of order α(0<α<1), Δ is the Laplacian with respect to the spacial variables, f is supposed to be a smooth function prescribed next, and T is a fixed positive real. Here ν denotes the outward unit normal and /ν=ν· is the normal derivative on Ω. Such problems arise in many applications, for instance, in studying the heat transfer in a resistor device whose electrical conductivity f is strongly dependent on the temperature u. When α=1, (1) describes the diffusion of the temperature with the presence of a nonlocal term. Constant λ is a dimensionless parameter, which can be identified with the square of the applied potential difference at the ends of the conductor. Function β is the positive thermal transfer coefficient. The given value u0 is the temperature outside Ω. For the sake of simplicity, boundary conditions are chosen of homogeneous Neumann type. Mixed or more general boundary conditions which model the coupling of the thermistor to its surroundings appear naturally. u(x) is the temperature inside the conductor, and f(u) is the temperature dependent electrical conductivity. Recall that (1) is obtained from the so-called nonlocal thermistor problem by replacing the first-order time derivative with a fractional derivative of order α(0<α<1). For more description about the history of thermistors and more detailed accounts of their advantages and applications in industry, refer to .

In recent years, it has been turned out that fractional differential equations can be used successfully to model many phenomena in various fields as fluids mechanics, viscoelasticity, chemistry, and engineering . In , existence and uniqueness of a positive solution to a generalized nonlocal thermistor problem with fractional-order derivatives were proved. In this work, a finite difference method is proposed for solving the time-fractional nonlocal thermistor system. Stability and error analysis for this scheme are presented showing that the temporal accuracy is of 2-α order.

2. Formulation and Statement of the Problem

We consider the time-fractional thermistor problem (1), which is obtained from (2)u(x,t)t-Δu=λf(u)(Ωf(u)dx)2,inQT=Ω×(0,T), by replacing the first-order time derivative with a fractional derivative on Caputo sense as defined in  and given by (3)αu(x,t)tα=1Γ(1-α)0tu(x,s)sds(t-s)α,0<α<1, subject to the initial and homogenous boundary conditions and where α(0<α<1) is the order of the time-fractional derivative. (1) covers (2) and extends it to general cases. The classical nonlocal thermistor problem (2) with the time derivative of integer order can be obtained by taking the limit α1 in (1). While the case α=0 corresponds to the steady state thermistor problem, in the case 0<α<1, the Caputo fractional derivative depends on and uses the information of the solutions at all previous time levels (non-Markovian process). In this case, the physical interpretation of fractional derivative is that it represents a degree of memory in the diffusing material .

In the analysis of the numerical method, we will assume that problem (1) has a unique and sufficiently smooth solution which can be established by assuming more hypotheses and regularity on the data (see ). In the sequel, we will assume the following assumptions:

f: is a positive Lipshitzian continuous function;

there exist positive constants c and α such that for all ξ we have (4)cf(ξ)c|ξ|α+1+c,

u0L(Ω).

It can be shown (e.g., see [12, 13]) that the quantity (5)v*=(v22+α0dudx22)1/2, where α0 is given next, defines a norm on H1(Ω) which is equivalent to the ·H1(Ω) norm.

3. Time Discretization: A Finite Difference Scheme

We introduce a finite difference approximation to discretize the time-fractional derivative. Let δ=T/N be the length of each time step, for some large N. tk=kδ,k=0,1,,N. We use the following formulation: for all 0kN-1, (6)αu(x,t)tα=1Γ(1-α)j=0ktjtj+1u(x,s)sds(tk+1-s)α=1Γ(1-α)j=0ku(x,tj+1)-u(x,tj)δ×tjtj+1ds(tk+1-s)α+rδk+1, where rδk+1 is the truncation error. It can be seen from  that the truncation error verifies (7)rδk+1cuδ2-α, where cu is a constant depending only on u. On the other hand, by change of variables, we have (8)1Γ(1-α)j=0ku(x,tj+1)-u(x,tj)δtjtj+1ds(tk+1-s)α=1Γ(1-α)j=0ku(x,tj+1)-u(x,tj)δtk-jtk+1-jdttα=1Γ(1-α)j=0ku(x,tk+1-j)-u(x,tk-j)δtjtj+1dttα=1Γ(2-α)j=0ku(x,tk+1-j)-u(x,tk-j)δα×{(j+1)1-α-(j)1-α}.

Let us denote bj=(j+1)1-α-j1-α,j=0,1,k, and define the discrete fractional differential operator Ltα by (9)Ltαu(x,tk+1)=1Γ(2-α)j=0kbju(x,tk+1-j)-u(x,tk-j)δα. Then (6) becomes (10)αu(x,tk+1)tα=Ltαu(x,tk+1)+rδk+1. Using this approximation, it yields the following finite difference scheme to (1): for k=1,,N-1, (11)Ltαuk+1(x)-Δuk+1=λf(uk+1)(Ωf(uk+1)dx)2inΩ, where uk+1(x) are approximations to u(x,tk+1). Scheme (11) can be reformulated in the form (12)b0uk+1-Γ(2-α)δαΔuk+1=b0uk-j=1kbj{uk+1-j-uk-j}+Γ(2-α)δαλf(uk+1)(Ωf(uk+1)dx)2=b0uk-j=0k-1bj+1uk-j+j=1kbjuk-j+Γ(2-α)δαλf(uk+1)(Ωf(uk+1)dx)2=b0uk+j=0k-1(bj-bj+1)uk-j+Γ(2-α)δαλf(uk+1)(Ωf(uk+1)dx)2. To complete the semidiscrete problem, we consider the boundary conditions (13)uk+1n=0, and the initial condition u0=u0, noting that (14)bj>0,j=0,1,k,1=b0>b1>>bk,bk0ask,j=1k(bj-bj+1)+bk+1=(1-b1)+j=1k-1(bj-bj+1)+bk=1. If we set (15)α0=Γ(2-α)δα, then (12) can be rewritten into (16)uk+1-α0Δuk+1=(1-b1)uk+j=1k-1(bj-bj+1)uk-j+bku0+α0λf(uk+1)(Ωf(uk+1)dx)2, for all k1. When k=0, scheme (12) reads (17)u1-α0Δu1=u0+α0λf(u1)(Ωf(u1)dx)2. When k=1, scheme (12) becomes (18)u2-α0Δu2=(1-b1)u1+b1u0+α0λf(u2)(Ωf(u2)dx)2. We define the error term rk+1 by (19)rk+1=α0{αu(x,tk+1)tα-Ltαu(x,tk+1)}. Then we get from (7) that (20)|rk+1|=Γ(2-α)δα|rδk+1|cuδ2.

3.1. Existence of the Semidiscrete Scheme Definition 1.

We say that uk+1 is a weak solution of (11) if (21)uk+1,v+α0Ωuk+1vdx=(fk,v)+λf(uk+1)(Ωf(uk+1)dx)2, where fk=(1-b1)uk+j=1k-1(bj-bj+1)uk-j+bku0.

At each time step, we solve a discretized fractional thermistor problem.

Theorem 2.

Let hypotheses (H1)–(H3) be satisfied; then there exists at least a weak solution uk of (12), such that (22)ukH1(Ω).

Existence and uniqueness results follow from general results of elliptic problems [3, 4, 13]. From now on, we denote by c a generic constant which may not be the same at different occurrences.

3.2. A Priori Estimates

We search a priori estimates for solutions.

Lemma 3.

There exists a positive constant c independent of k, such that (23)uk+1H1(Ω)c.

Proof.

We prove this result by recurrence. First, when k=0, we have, for vH01(Ω), (24)Ωu1vdx+α0Ωu1vdx=Ωu0vdx+λα0(Ωf(u1)dx)2(f(u1),v). Notice that u0L(Ω)L2(Ω). Taking v=u1, we have (25)u122+α0u122=Ωu0u1dx+λα0Ωf(u1)u1dx(Ωf(u1)dx)2u12u02+cα0Ωf(u1)u1dxcu12+cα0Ωf(u1)u1dx, or (26)Ωf(u1)u1dxcΩ(|u|α+1+c)|u1|dxcu11cu12cu1H1(Ω)12u1H1(Ω)2+c. Then (27)u122+α0u122c. Hence, since the standard H1-norm and the norm ·* defined by (5) are equivalent, we have (28)u1H1(Ω)c. Suppose now that we have (29)ujH1(Ω)c,j=1,2,,k, and prove that uk+1H1(Ω)c. Multiplying (16) by v=uk+1 and using the fact that fkH1(Ω), we obtain (30)uk+122+α0uk+122=Ωfkuk+1dx+λα0Ωf(uk+1)uk+1dx(Ωf(uk+1)dx)2fk2uk+12+λα0Ωf(uk+1)uk+1dx(Ωf(uk+1)dx)2cuk+12+λα0Ωf(uk+1)uk+1dx(Ωf(uk+1)dx)2. Following the same as for the case j=1 with respect to the nonlocal term λα0Ωf(uk+1)uk+1dx/(Ωf(uk+1)dx)2, we then have (31)uk+122+α0uk+122c. Hence, (32)uk+1H1(Ω)c.

4. Stability and Error Analysis 4.1. Stability Result

The weak formulation of (16) is for all k1 and vH1(Ω):(33)(uk+1,v)+α0(-Δuk+1,v)=(1-b1)(uk,v)+j=1k-1(bj-bj+1)(uk-j,v)+bk(u0,v)+λα0(f(uk+1)(Ωf(uk+1)dx)2,v).We have the following unconditional stability result.

Theorem 4.

The semidiscretized problem is stable in the sense that for all δ>0 it holds (34)uk+1H1(Ω)u02+c.

Proof.

We prove this result by recurrence. First, when k=0, we have, for vH1(Ω), (35)(u1,v)+α0(-Δu1,v)=(u0,v)+α0λ(f(u1),v)(Ωf(u1)dx)2. On other terms (36)Ωu1vdx+α0Ωu1vdx=Ωu0vdx+α0λΩf(uk+1)vdx(Ωf(uk+1)dx)2. Taking v=u1 in (36), we have (37)Ω|u1|2dx+α0Ω|u1|2dx=Ωu0u1dx+α0λΩf(uk+1)u1dx(Ωf(uk+1)dx)2. In a similar way, we have (38)α0λΩf(uk+1)  u1dx(Ωf(uk+1)dx)2cuk+1H1(Ω). We also have (39)Ωu0u1dxu02u12u02u12u02u1H1(Ω). We then obtain by (5) and (36) that (40)u1H1(Ω)2(u02+c)u1H1(Ω). Dividing both sides of the previous inequality (40) by u1H1(Ω), we get (41)u1H1(Ω)u02+c. Suppose now that we have (42)ujH1(Ω)u02+c,  j=1,2,,k, and prove that uk+1H1(Ω)u02+c. Choosing v=uk+1 in (33), we obtain (43)(uk+1,uk+1)+α0(-Δuk+1,uk+1)=(1-b1)(uk,uk+1)+j=1k-1(bj-bj+1)(uk-j,uk+1)+bk(u0,uk+1)+α0(λf(uk+1)(Ωf(uk+1)dx)2,uk+1). Then using the recurrence hypothesis (42), we obtain (44)uk+1H1(Ω)2(1-b1)uk2uk+12+j=1k-1(bj-bj+1)uk-j2uk+12+bku02uk+12+α0(λf(uk+1)(Ωf(uk+1)dx)2,uk+1){(1-b1)+j=1k-1(bj-bj+1)+bk}×(u02+c)uk+12+α0(λf(uk+1)(Ωf(uk+1)dx)2,uk+1){(1-b1)+j=1k-1(bj-bj+1)+bk}×(u02+c)uk+1H1(Ω)+α0(λf(uk+1)(Ωf(uk+1)dx)2,uk+1)(u02+c)uk+1H1(Ω)+α0(λf(uk+1)(Ωf(uk+1)dx)2,uk+1), since (1-b1)+j=1k-1(bj-bj+1)+bk=1. Similarly to the case k=0, we have (45)α0(λf(uk+1)(Ωf(uk+1)dx)2,uk+1)cuk+1H1(Ω). Then (46)uk+1H1(Ω)u02+c. We have the following error analysis for the solution of the semidiscretized problem.

Theorem 5.

Let u be the exact solution of (1) and let (uj)j  be the time-discrete solution with the initial condition u0(x)=u(x,0). Then one has the following error estimates:

(a)(47)u(tj)-ujH1(Ω)cu,αTαδ2-α,j=1,,N,

where 0α<1 and cu,α=cu/(1-α); cu is a constant depending on u.

(b) when α1, (48)u(tj)-ujH1(Ω)cuTδ,j=1,,N.

Proof.

Let ek=u(x,tk)-uk(x) the difference between the exact solution of (1) and uk the solution of the time-discrete problem. Obviously e0=0.

(a) We will prove the result by induction. We begin with the first case when 0α<1. For j=1, by gathering equations corresponding to exact and discrete solutions, the error equation reads (49)(e1,v)+α0Ωe1vdx=(e0,v)+(r1,v)+α0(λf(u(x,t2))(Ωf(u(x,t2))dx)2,v)-α0(λf(u2)(Ωf(u2)dx)2,v)=(r1,v)+α0(λf(u(x,t2))(Ωf(u(x,t2))dx)2,v)-α0(λf(u2)(Ωf(u2)dx)2,v).   Choosing v=e1 in the previous equation, it yields that (50)e122+α0e122r12e12+α0(λf(u(x,t2))(Ωf(u(x,t2))dx)2,e1)-α0(λf(u2)(Ωf(u2)dx)2,e1).   To continue the proof, we will need the following lemma which is used in the sequel.

Lemma 6.

Let ui,i=1,2, be two weak solutions of (1). Assume that (H1)–(H3) hold. Then one has (51)(λf(u1)(Ωf(u1)dx)2,w)-(λf(u2)(Ωf(u2)dx)2,w)cw22, where w=u1-u2 and ε,c,  and cε are positive constants.

Proof.

We have (52)f(u1)(Ωf(u1)dx)2-f(u2)(Ωf(u2)dx)2=1(Ωf(u1)dx)2(f(u1)-f(u2))+(1(Ωf(u1)dx)2-1(Ωf(u2)dx)2)f(u2). If we multiply by w and integrate over Ω, we get (53)(f(u1)(Ωf(u1)dx)2-f(u2)(Ωf(u2)dx)2,w)cw22+Ω(f(u2)-f(u1))dx  Ω(f(u2)-f(u1))dx(Ωf(u1)dx)2(Ωf(u2)dx)2×(f(u2),w)cw22+cw2w1cw22.   The proof of Lemma 6 is now completed.

Now, we continue the proof of Theorem 5. Using (50), it follows that (54)e122+α0e122r12e12+ce122(c+ε)e122+cεr122. Then, by (5), we have (55)e1H1(Ω)2(c+ε)e122+cεr122. It follows that (56)(1-(c+ε))e1H1(Ω)2cεr122. For a good choice of ε and using (20) and b0=1, we obtain (57)u(t1)-u11cub0-1δ2. Then point (a) is verified for j=1. Suppose now that we have proven (a) for all k=1,,j, and prove it also for k=j+1. We have (58)(ek+1,v)+α0(-Δek+1,v)=(1-b1)(ek,v)+j=1k-1(bj-bj+1)(ek-j,v)+bk(e0,v)+(rk+1,v)+α0(λf(u(x,tk+1))(Ωf(u(x,tk+1))dx)2,v)-α0(λf(uk+1)(Ωf(uk+1)dx)2,v). Taking v=ek+1 in (58) and using Lemma 6, we then have (59)ek+122+α0ek+122(1-b1)ek2ek+12+j=1k-1(bj-bj+1)ek-j2ek+12+bke02ek+12+rk+12ek+12+cek+1H1(Ω)2. Using the induction assumption and the fact that bk-1/bk+1-1<1 for a positive integer k, we have (60)ek+122+α0ek+122{(1-b1)bk-1-1+j=1k-1(bj-bj+1)bk-j-1-1}×cuδ2ek+12+cek+1H1(Ω)2{(1-b1)+j=1k-1(bj-bj+1)+bk}×cubk-1-1δ2ek+1H1(Ω)+cek+1H1(Ω)2. We then have (61)ek+122+α0ek+122cubk-1-1δ2ek+1H1(Ω)+cek+1H1(Ω)2, since (1-b1)+j=1k-1(bj-bj+1)+bk=1. Then (62)ek+1H1(Ω)2cubk-1-1δ2ek+1H1(Ω)+cek+1H1(Ω)2. By using Young’s inequality, we get (63)ek+1H1(Ω)2(c+ε)ek+1H1(Ω)2+(cεcubk-1-1δ2)2. Hence, (64)(1-(c+ε))ek+1H1(Ω)2(cεcubk-1-1δ2)2. For a suitable choice of ε and dividing both sides by ek+1H1(Ω), we get (65)ek+1H1(Ω)cubk-1-1δ2. One can show easily that (66)k-αbk-1-111-α,k=1,,N. Hence, we have, for all k, such that kδT, (67)u(tk)-ukH1(Ω)cubk-1-1δ2=cuk-αbk-1-1kαδ2cu11-α(kδ)αδ2-αcu1-αTαδ2-α. (b) We are now interested in the case α1. We will derive again the following estimation by induction: (68)u(tj)-uj1cujδ2,j=1,2,,N. The previous inequality is obvious for j=1. Suppose now that (68) holds for all j=1,2,,k, and we need to prove that it holds also for j=k+1. Similarly to the previous case, by combining the corresponding equations of the exact and discrete solutions and taking v=ek+1 as a test function, it yields that (69)ek+122+α0ek+122(1-b1)ek2ek+12+j=1k-1(bj-bj+1)ek-j2ek+12+bke02ek+12+rk+12ek+12+cek+1H1(Ω)2{j=1k-1(1-b1)(cukδ2)+j=1k-1(bj-bj+1)(cu(k-j)δ2)+cuδ2}ek+12+cek+1H1(Ω)2{(1-b1)kk+1+j=1k-1(bj-bj+1)k-jk+1+1k+1}×cu(k+1)δ2ek+12+cek+1H1(Ω)2{(1-b1)+j=1k-1(bj-bj+1)-(1-b1)1k+1-j=1k-1(bj-bj+1)j+1k+1+1k+1}×cu(k+1)δ2ek+12+cek+1H1(Ω)2. Notice that (70)(1-b1)1k+1+j=1k-1(bj-bj+1)j+1k+1+bk1k+1{(1-b1)+j=1k-1(bj-bj+1)+bk}=1k+1. Then, similar to the earlier development, we have (71)(1-(c+ε))ek+1H1(Ω)2({(1-b1)+j=1k-1(bj-bj+1)+bk}cεcu(k+1)δ2)2=(cεcu(k+1)δ2)2. It follows, for an ε well chosen such that 1-(c+ε)>0, that (72)ek+1H1(Ω)cu(k+1)δ2. Then the estimate (b) is proved. This completes the proof of the theorem.

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