# Stability and Error Analysis of the Semidiscretized Fractional Nonlocal Thermistor Problem

Authors: M. R. Sidi Ammi; A. El Hachimi
Journal: Conference Papers in Mathematics
 (2013)

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## Abstract

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.


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## Body

## 1. Introduction

LetΩ be a bounded domain in ℝN 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)∂αu∂tα-Δu=λf(u)(∫Ω‍f(u)dx)2,inQT=Ω×(0,T),∂u∂ν=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 [1–4].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 [5–8]. In [4], 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 [9] and given by
(3)∂αu(x,t)∂tα=1Γ(1-α)∫0t‍∂u(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 α→1 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 [10].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 [11]). In the sequel, we will assume the following assumptions: (H1)
f:ℝ→ℝ is a positive Lipshitzian continuous function;(H2)
there exist positive constantsc and α such that for all ξ∈ℝ we have
(4)c≤f(ξ)≤c|ξ|α+1+c,(H3)
u0∈L∞(Ω).It can be shown (e.g., see [12, 13]) that the quantity
(5)∥v∥*=(∥v∥22+α0∥dudx∥22)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. Letδ=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 0≤k≤N-1,
(6)∂αu(x,t)∂tα=1Γ(1-α)∑j=0k‍∫tjtj+1‍∂u(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 [14] that the truncation error verifies
(7)rδk+1≲cuδ2-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)δ∫tjtj+1‍ds(tk+1-s)α=1Γ(1-α)∑j=0k‍u(x,tj+1)-u(x,tj)δ∫tk-jtk+1-j‍dttα=1Γ(1-α)∑j=0k‍u(x,tk+1-j)-u(x,tk-j)δ∫tjtj+1‍dttα=1Γ(2-α)∑j=0k‍u(x,tk+1-j)-u(x,tk-j)δα×{(j+1)1-α-(j)1-α}.Let us denotebj=(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+1∂n=0,
and the initial condition u0=u0, noting that
(14)bj>0,j=0,1,…k,1=b0>b1>⋯>bk,bk⟶0ask⟶∞,∑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 k≥1. 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 thatuk+1 is a weak solution of (11) if
(21)〈uk+1,v〉+α0∫Ω‍∇uk+1∇vdx=(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 solutionuk of (12), such that
(22)uk∈H1(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 constantc independent of k, such that
(23)∥uk+1∥H1(Ω)≤c.Proof.
We prove this result by recurrence. First, whenk=0, we have, for v∈H01(Ω),
(24)∫Ω‍u1vdx+α0∫Ω‍∇u1∇vdx=∫Ω‍u0vdx+λα0(∫Ω‍f(u1)dx)2(f(u1),v).
Notice that u0∈L∞(Ω)⊂L2(Ω). Taking v=u1, we have
(25)∥u1∥22+α0∥∇u1∥22=∫Ω‍u0u1dx+λα0∫Ω‍f(u1)u1dx(∫Ω‍f(u1)dx)2≤∥u1∥2∥u0∥2+cα0∫Ω‍f(u1)u1dx≤c∥u1∥2+cα0∫Ω‍f(u1)u1dx,
or
(26)∫Ω‍f(u1)u1dx≤c∫Ω‍(|u|α+1+c)|u1|dx≤c∥u1∥1≤c∥u1∥2≤c∥u1∥H1(Ω)≤12∥u1∥H1(Ω)2+c.
Then
(27)∥u1∥22+α0∥∇u1∥22≤c.
Hence, since the standard H1-norm and the norm ∥·∥* defined by (5) are equivalent, we have
(28)∥u1∥H1(Ω)≤c.
Suppose now that we have
(29)∥uj∥H1(Ω)≤c,j=1,2,…,k,
and prove that ∥uk+1∥H1(Ω)≤c. Multiplying (16) by v=uk+1 and using the fact that fk∈H1(Ω), we obtain
(30)∥uk+1∥22+α0∥∇uk+1∥22=∫Ω‍fkuk+1dx+λα0∫Ω‍f(uk+1)uk+1dx(∫Ω‍f(uk+1)dx)2≤∥fk∥2∥uk+1∥2+λα0∫Ω‍f(uk+1)uk+1dx(∫Ω‍f(uk+1)dx)2≤c∥uk+1∥2+λα0∫Ω‍f(uk+1)uk+1dx(∫Ω‍f(uk+1)dx)2.
Following the same as for the case j=1 with respect to the nonlocal term λα0∫Ω‍f(uk+1)uk+1dx/(∫Ω‍f(uk+1)dx)2, we then have
(31)∥uk+1∥22+α0∥∇uk+1∥22≤c.
Hence,
(32)∥uk+1∥H1(Ω)≤c.


## 3.1. Existence of the Semidiscrete Scheme

Definition 1.
We say thatuk+1 is a weak solution of (11) if
(21)〈uk+1,v〉+α0∫Ω‍∇uk+1∇vdx=(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 solutionuk of (12), such that
(22)uk∈H1(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 constantc independent of k, such that
(23)∥uk+1∥H1(Ω)≤c.Proof.
We prove this result by recurrence. First, whenk=0, we have, for v∈H01(Ω),
(24)∫Ω‍u1vdx+α0∫Ω‍∇u1∇vdx=∫Ω‍u0vdx+λα0(∫Ω‍f(u1)dx)2(f(u1),v).
Notice that u0∈L∞(Ω)⊂L2(Ω). Taking v=u1, we have
(25)∥u1∥22+α0∥∇u1∥22=∫Ω‍u0u1dx+λα0∫Ω‍f(u1)u1dx(∫Ω‍f(u1)dx)2≤∥u1∥2∥u0∥2+cα0∫Ω‍f(u1)u1dx≤c∥u1∥2+cα0∫Ω‍f(u1)u1dx,
or
(26)∫Ω‍f(u1)u1dx≤c∫Ω‍(|u|α+1+c)|u1|dx≤c∥u1∥1≤c∥u1∥2≤c∥u1∥H1(Ω)≤12∥u1∥H1(Ω)2+c.
Then
(27)∥u1∥22+α0∥∇u1∥22≤c.
Hence, since the standard H1-norm and the norm ∥·∥* defined by (5) are equivalent, we have
(28)∥u1∥H1(Ω)≤c.
Suppose now that we have
(29)∥uj∥H1(Ω)≤c,j=1,2,…,k,
and prove that ∥uk+1∥H1(Ω)≤c. Multiplying (16) by v=uk+1 and using the fact that fk∈H1(Ω), we obtain
(30)∥uk+1∥22+α0∥∇uk+1∥22=∫Ω‍fkuk+1dx+λα0∫Ω‍f(uk+1)uk+1dx(∫Ω‍f(uk+1)dx)2≤∥fk∥2∥uk+1∥2+λα0∫Ω‍f(uk+1)uk+1dx(∫Ω‍f(uk+1)dx)2≤c∥uk+1∥2+λα0∫Ω‍f(uk+1)uk+1dx(∫Ω‍f(uk+1)dx)2.
Following the same as for the case j=1 with respect to the nonlocal term λα0∫Ω‍f(uk+1)uk+1dx/(∫Ω‍f(uk+1)dx)2, we then have
(31)∥uk+1∥22+α0∥∇uk+1∥22≤c.
Hence,
(32)∥uk+1∥H1(Ω)≤c.


## 4. Stability and Error Analysis

### 4.1. Stability Result

The weak formulation of (16) is for all k≥1 and v∈H1(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 allδ>0 it holds
(34)∥uk+1∥H1(Ω)≤∥u0∥2+c.Proof.
We prove this result by recurrence. First, whenk=0, we have, for v∈H1(Ω),
(35)(u1,v)+α0(-Δu1,v)=(u0,v)+α0λ(f(u1),v)(∫Ω‍f(u1)dx)2.
On other terms
(36)∫Ω‍u1vdx+α0∫Ω‍∇u1∇vdx=∫Ω‍u0vdx+α0λ∫Ω‍f(uk+1)vdx(∫Ω‍f(uk+1)dx)2.
Taking v=u1 in (36), we have
(37)∫Ω‍|u1|2dx+α0∫Ω‍|∇u1|2dx=∫Ω‍u0u1dx+α0λ∫Ω‍f(uk+1)u1dx(∫Ω‍f(uk+1)dx)2.
In a similar way, we have
(38)α0λ∫Ω‍f(uk+1)u1dx(∫Ω‍f(uk+1)dx)2≤c∥uk+1∥H1(Ω).
We also have
(39)∫Ω‍u0u1dx≤∥u0∥2∥u1∥2≤∥u0∥2∥u1∥2≤∥u0∥2∥u1∥H1(Ω).
We then obtain by (5) and (36) that
(40)∥u1∥H1(Ω)2≤(∥u0∥2+c)∥u1∥H1(Ω).
Dividing both sides of the previous inequality (40) by ∥u1∥H1(Ω), we get
(41)∥u1∥H1(Ω)≤∥u0∥2+c.
Suppose now that we have
(42)∥uj∥H1(Ω)≤∥u0∥2+c,j=1,2,…,k,
and prove that ∥uk+1∥H1(Ω)≤∥u0∥2+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+1∥H1(Ω)2≤(1-b1)∥uk∥2∥uk+1∥2+∑j=1k-1‍(bj-bj+1)∥uk-j∥2∥uk+1∥2+bk∥u0∥2∥uk+1∥2+α0(λf(uk+1)(∫Ω‍f(uk+1)dx)2,uk+1)≤{(1-b1)+∑j=1k-1‍(bj-bj+1)+bk}×(∥u0∥2+c)∥uk+1∥2+α0(λf(uk+1)(∫Ω‍f(uk+1)dx)2,uk+1)≤{(1-b1)+∑j=1k-1‍(bj-bj+1)+bk}×(∥u0∥2+c)∥uk+1∥H1(Ω)+α0(λf(uk+1)(∫Ω‍f(uk+1)dx)2,uk+1)≤(∥u0∥2+c)∥uk+1∥H1(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)≤c∥uk+1∥H1(Ω).
Then
(46)∥uk+1∥H1(Ω)≤∥u0∥2+c.
We have the following error analysis for the solution of the semidiscretized problem.Theorem 5.
Letu 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)-uj∥H1(Ω)≤cu,αTαδ2-α,j=1,…,N,where0≤α<1 and cu,α=cu/(1-α); cu is a constant depending on u.(b) when α→1,
(48)∥u(tj)-uj∥H1(Ω)≤cuTδ,j=1,…,N.Proof.
Letek=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 when0≤α<1. For j=1, by gathering equations corresponding to exact and discrete solutions, the error equation reads
(49)(e1,v)+α0∫Ω‍∇e1∇vdx=(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)∥e1∥22+α0∥∇e1∥22≤∥r1∥2∥e1∥2+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.
Letui,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)≤c∥w∥22,
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)≤c∥w∥22+∫Ω‍(f(u2)-f(u1))dx∫Ω‍(f(u2)-f(u1))dx(∫Ω‍f(u1)dx)2(∫Ω‍f(u2)dx)2×(f(u2),w)≤c∥w∥22+c∥w∥2∥w∥1≤c∥w∥22.
The proof of Lemma 6 is now completed.
Now, we continue the proof of Theorem5. Using (50), it follows that
(54)∥e1∥22+α0∥∇e1∥22≤∥r1∥2∥e1∥2+c∥e1∥22≤(c+ε)∥e1∥22+cε∥r1∥22.
Then, by (5), we have
(55)∥e1∥H1(Ω)2≤(c+ε)∥e1∥22+cε∥r1∥22.
It follows that
(56)(1-(c+ε))∥e1∥H1(Ω)2≤cε∥r1∥22.
For a good choice of ε and using (20) and b0=1, we obtain
(57)∥u(t1)-u1∥1≤cub0-1δ2.
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+1∥22+α0∥∇ek+1∥22≤(1-b1)∥ek∥2∥ek+1∥2+∑j=1k-1‍(bj-bj+1)∥ek-j∥2∥ek+1∥2+bk∥e0∥2∥ek+1∥2+∥rk+1∥2∥ek+1∥2+c∥ek+1∥H1(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+1∥22+α0∥∇ek+1∥22≤{(1-b1)bk-1-1+∑j=1k-1‍(bj-bj+1)bk-j-1-1}×cuδ2∥ek+1∥2+c∥ek+1∥H1(Ω)2≤{(1-b1)+∑j=1k-1‍(bj-bj+1)+bk}×cubk-1-1δ2∥ek+1∥H1(Ω)+c∥ek+1∥H1(Ω)2.
We then have
(61)∥ek+1∥22+α0∥∇ek+1∥22≤cubk-1-1δ2∥ek+1∥H1(Ω)+c∥ek+1∥H1(Ω)2,
since (1-b1)+∑j=1k-1‍(bj-bj+1)+bk=1. Then
(62)∥ek+1∥H1(Ω)2≤cubk-1-1δ2∥ek+1∥H1(Ω)+c∥ek+1∥H1(Ω)2.
By using Young’s inequality, we get
(63)∥ek+1∥H1(Ω)2≤(c+ε)∥ek+1∥H1(Ω)2+(cεcubk-1-1δ2)2.
Hence,
(64)(1-(c+ε))∥ek+1∥H1(Ω)2≤(cεcubk-1-1δ2)2.
For a suitable choice of ε and dividing both sides by ∥ek+1∥H1(Ω), we get
(65)∥ek+1∥H1(Ω)≤cubk-1-1δ2.
One can show easily that
(66)k-αbk-1-1≤11-α,k=1,…,N.
Hence, we have, for all k, such that kδ≤T,
(67)∥u(tk)-uk∥H1(Ω)≤cubk-1-1δ2=cuk-αbk-1-1kαδ2≤cu11-α(kδ)αδ2-α≤cu1-αTαδ2-α.
(b) We are now interested in the case α→1. We will derive again the following estimation by induction:
(68)∥u(tj)-uj∥1≤cujδ2,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+1∥22+α0∥∇ek+1∥22≤(1-b1)∥ek∥2∥ek+1∥2+∑j=1k-1‍(bj-bj+1)∥ek-j∥2∥ek+1∥2+bk∥e0∥2∥ek+1∥2+∥rk+1∥2∥ek+1∥2+c∥ek+1∥H1(Ω)2≤{∑j=1k-1‍(1-b1)(cukδ2)+∑j=1k-1‍(bj-bj+1)(cu(k-j)δ2)+cuδ2}∥ek+1∥2+c∥ek+1∥H1(Ω)2≤{(1-b1)kk+1+∑j=1k-1‍(bj-bj+1)k-jk+1+1k+1}×cu(k+1)δ2∥ek+1∥2+c∥ek+1∥H1(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)δ2∥ek+1∥2+c∥ek+1∥H1(Ω)2.
Notice that
(70)(1-b1)1k+1+∑j=1k-1‍(bj-bj+1)j+1k+1+bk≥1k+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+1∥H1(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+1∥H1(Ω)≤cu(k+1)δ2.
Then the estimate (b) is proved. This completes the proof of the theorem.


## 4.1. Stability Result

The weak formulation of (16) is for all k≥1 and v∈H1(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 allδ>0 it holds
(34)∥uk+1∥H1(Ω)≤∥u0∥2+c.Proof.
We prove this result by recurrence. First, whenk=0, we have, for v∈H1(Ω),
(35)(u1,v)+α0(-Δu1,v)=(u0,v)+α0λ(f(u1),v)(∫Ω‍f(u1)dx)2.
On other terms
(36)∫Ω‍u1vdx+α0∫Ω‍∇u1∇vdx=∫Ω‍u0vdx+α0λ∫Ω‍f(uk+1)vdx(∫Ω‍f(uk+1)dx)2.
Taking v=u1 in (36), we have
(37)∫Ω‍|u1|2dx+α0∫Ω‍|∇u1|2dx=∫Ω‍u0u1dx+α0λ∫Ω‍f(uk+1)u1dx(∫Ω‍f(uk+1)dx)2.
In a similar way, we have
(38)α0λ∫Ω‍f(uk+1)u1dx(∫Ω‍f(uk+1)dx)2≤c∥uk+1∥H1(Ω).
We also have
(39)∫Ω‍u0u1dx≤∥u0∥2∥u1∥2≤∥u0∥2∥u1∥2≤∥u0∥2∥u1∥H1(Ω).
We then obtain by (5) and (36) that
(40)∥u1∥H1(Ω)2≤(∥u0∥2+c)∥u1∥H1(Ω).
Dividing both sides of the previous inequality (40) by ∥u1∥H1(Ω), we get
(41)∥u1∥H1(Ω)≤∥u0∥2+c.
Suppose now that we have
(42)∥uj∥H1(Ω)≤∥u0∥2+c,j=1,2,…,k,
and prove that ∥uk+1∥H1(Ω)≤∥u0∥2+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+1∥H1(Ω)2≤(1-b1)∥uk∥2∥uk+1∥2+∑j=1k-1‍(bj-bj+1)∥uk-j∥2∥uk+1∥2+bk∥u0∥2∥uk+1∥2+α0(λf(uk+1)(∫Ω‍f(uk+1)dx)2,uk+1)≤{(1-b1)+∑j=1k-1‍(bj-bj+1)+bk}×(∥u0∥2+c)∥uk+1∥2+α0(λf(uk+1)(∫Ω‍f(uk+1)dx)2,uk+1)≤{(1-b1)+∑j=1k-1‍(bj-bj+1)+bk}×(∥u0∥2+c)∥uk+1∥H1(Ω)+α0(λf(uk+1)(∫Ω‍f(uk+1)dx)2,uk+1)≤(∥u0∥2+c)∥uk+1∥H1(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)≤c∥uk+1∥H1(Ω).
Then
(46)∥uk+1∥H1(Ω)≤∥u0∥2+c.
We have the following error analysis for the solution of the semidiscretized problem.Theorem 5.
Letu 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)-uj∥H1(Ω)≤cu,αTαδ2-α,j=1,…,N,where0≤α<1 and cu,α=cu/(1-α); cu is a constant depending on u.(b) when α→1,
(48)∥u(tj)-uj∥H1(Ω)≤cuTδ,j=1,…,N.Proof.
Letek=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 when0≤α<1. For j=1, by gathering equations corresponding to exact and discrete solutions, the error equation reads
(49)(e1,v)+α0∫Ω‍∇e1∇vdx=(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)∥e1∥22+α0∥∇e1∥22≤∥r1∥2∥e1∥2+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.
Letui,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)≤c∥w∥22,
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)≤c∥w∥22+∫Ω‍(f(u2)-f(u1))dx∫Ω‍(f(u2)-f(u1))dx(∫Ω‍f(u1)dx)2(∫Ω‍f(u2)dx)2×(f(u2),w)≤c∥w∥22+c∥w∥2∥w∥1≤c∥w∥22.
The proof of Lemma 6 is now completed.
Now, we continue the proof of Theorem5. Using (50), it follows that
(54)∥e1∥22+α0∥∇e1∥22≤∥r1∥2∥e1∥2+c∥e1∥22≤(c+ε)∥e1∥22+cε∥r1∥22.
Then, by (5), we have
(55)∥e1∥H1(Ω)2≤(c+ε)∥e1∥22+cε∥r1∥22.
It follows that
(56)(1-(c+ε))∥e1∥H1(Ω)2≤cε∥r1∥22.
For a good choice of ε and using (20) and b0=1, we obtain
(57)∥u(t1)-u1∥1≤cub0-1δ2.
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+1∥22+α0∥∇ek+1∥22≤(1-b1)∥ek∥2∥ek+1∥2+∑j=1k-1‍(bj-bj+1)∥ek-j∥2∥ek+1∥2+bk∥e0∥2∥ek+1∥2+∥rk+1∥2∥ek+1∥2+c∥ek+1∥H1(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+1∥22+α0∥∇ek+1∥22≤{(1-b1)bk-1-1+∑j=1k-1‍(bj-bj+1)bk-j-1-1}×cuδ2∥ek+1∥2+c∥ek+1∥H1(Ω)2≤{(1-b1)+∑j=1k-1‍(bj-bj+1)+bk}×cubk-1-1δ2∥ek+1∥H1(Ω)+c∥ek+1∥H1(Ω)2.
We then have
(61)∥ek+1∥22+α0∥∇ek+1∥22≤cubk-1-1δ2∥ek+1∥H1(Ω)+c∥ek+1∥H1(Ω)2,
since (1-b1)+∑j=1k-1‍(bj-bj+1)+bk=1. Then
(62)∥ek+1∥H1(Ω)2≤cubk-1-1δ2∥ek+1∥H1(Ω)+c∥ek+1∥H1(Ω)2.
By using Young’s inequality, we get
(63)∥ek+1∥H1(Ω)2≤(c+ε)∥ek+1∥H1(Ω)2+(cεcubk-1-1δ2)2.
Hence,
(64)(1-(c+ε))∥ek+1∥H1(Ω)2≤(cεcubk-1-1δ2)2.
For a suitable choice of ε and dividing both sides by ∥ek+1∥H1(Ω), we get
(65)∥ek+1∥H1(Ω)≤cubk-1-1δ2.
One can show easily that
(66)k-αbk-1-1≤11-α,k=1,…,N.
Hence, we have, for all k, such that kδ≤T,
(67)∥u(tk)-uk∥H1(Ω)≤cubk-1-1δ2=cuk-αbk-1-1kαδ2≤cu11-α(kδ)αδ2-α≤cu1-αTαδ2-α.
(b) We are now interested in the case α→1. We will derive again the following estimation by induction:
(68)∥u(tj)-uj∥1≤cujδ2,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+1∥22+α0∥∇ek+1∥22≤(1-b1)∥ek∥2∥ek+1∥2+∑j=1k-1‍(bj-bj+1)∥ek-j∥2∥ek+1∥2+bk∥e0∥2∥ek+1∥2+∥rk+1∥2∥ek+1∥2+c∥ek+1∥H1(Ω)2≤{∑j=1k-1‍(1-b1)(cukδ2)+∑j=1k-1‍(bj-bj+1)(cu(k-j)δ2)+cuδ2}∥ek+1∥2+c∥ek+1∥H1(Ω)2≤{(1-b1)kk+1+∑j=1k-1‍(bj-bj+1)k-jk+1+1k+1}×cu(k+1)δ2∥ek+1∥2+c∥ek+1∥H1(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)δ2∥ek+1∥2+c∥ek+1∥H1(Ω)2.
Notice that
(70)(1-b1)1k+1+∑j=1k-1‍(bj-bj+1)j+1k+1+bk≥1k+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+1∥H1(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+1∥H1(Ω)≤cu(k+1)δ2.
Then the estimate (b) is proved. This completes the proof of the theorem.



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*Source: 454329-2013-09-08.xml*