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Nov 29, 2012 - Ilia Vekua Institute of Applied Mathematics, Ivane Javakhishvili Tbilisi ... Faculty of Exact and Natural Sciences, Ivane Javakhishvili Tbilisi State ...
Finite element approximations of a nonlinear diffusion model with memory

Temur Jangveladze, Zurab Kiguradze, Beny Neta & Simeon Reich

Numerical Algorithms ISSN 1017-1398 Volume 64 Number 1 Numer Algor (2013) 64:127-155 DOI 10.1007/s11075-012-9658-7

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Author's personal copy Numer Algor (2013) 64:127–155 DOI 10.1007/s11075-012-9658-7 ORIGINAL PAPER

Finite element approximations of a nonlinear diffusion model with memory Temur Jangveladze · Zurab Kiguradze · Beny Neta · Simeon Reich

Received: 4 November 2011 / Accepted: 18 October 2012 / Published online: 29 November 2012 © Springer Science+Business Media New York 2012

Abstract The convergence of a finite element scheme approximating a nonlinear system of integro-differential equations is proven. This system arises in mathematical modeling of the process of a magnetic field penetrating into a substance. Properties of existence, uniqueness and asymptotic behavior of the solutions are briefly described. The decay of the numerical solution is compared with both the theoretical and finite difference results. Keywords System of nonlinear integro-differential equations · Finite element scheme Mathematics Subject Classifications (2010) 45K05 · 65N30 · 35K55

T. Jangveladze · Z. Kiguradze Ilia Vekua Institute of Applied Mathematics, Ivane Javakhishvili Tbilisi State University, 2 University Street, 0186 Tbilisi, Georgia T. Jangveladze · Z. Kiguradze Faculty of Exact and Natural Sciences, Ivane Javakhishvili Tbilisi State University, 3 Chavchavdze Ave., 0179 Tbilisi, Georgia T. Jangveladze Caucasus University, 77 Kostava Ave., 0175 Tbilisi, Georgia B. Neta () Department of Applied Mathematics, Naval Postgraduate School, Monterey, CA 93943, USA e-mail: [email protected] S. Reich Department of Mathematics, The Technion–Israel Institute of Technology, 32000 Haifa, Israel

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1 Introduction Integro-differential models arise in many scientific and engineering disciplines. Such a model arises, for instance, in mathematical modeling of the process of a magnetic field penetrating into a substance. If the coefficient of thermal heat capacity and electroconductivity of the substance is highly dependent on the temperature, then the corresponding Maxwell system [1] can be rewritten in the following form [2]:     t ∂W |∇ × W |2 dτ ∇ × W , = −∇ × a ∂t 0 where W = (W1 , W2 , W3 ) is the vector of the magnetic field and the function a = a(σ ) is defined for σ ∈ [0, ∞). If the magnetic field has the form W = (0, u1 , u2 ) and ui = ui (x, t), i = 1, 2, then we have      ∂ ∂u1 ∂ ∂u2 ∇ × (a(σ )∇ × W ) = 0, − a(σ ) , − a(σ ) . ∂x ∂x ∂x ∂x Therefore, we obtain the following system of nonlinear integro-differential equations:      

t ∂u2 2 ∂ ∂ui ∂u1 2 ∂ui + = a dτ , i = 1, 2. (1.1) ∂t ∂x ∂x ∂x ∂x 0 Note that the (1.1)-type model is complex, but special cases of it were investigated; see [2–8]. The existence of global solutions to initial-boundary value problems for such models has been proven in [2–5, 8] by using some modifications of the Galerkin method and compactness arguments [9, 10]. For solvability and uniqueness properties of initial-boundary value problems for (1.1)-type models, see also [6, 7] as well as many other scientific works. Assume the temperature of the considered body is constant throughout the material, i.e., dependent on time, but independent of the space coordinates. If the magnetic field again has the form W = (0, u1 , u2 ) and ui = ui (x, t), i = 1, 2, then the same process of the magnetic field penetrating into the material is modeled by the following system of integro-differential equations:   

   t 1 ∂u2 2 ∂ui ∂u1 2 ∂ 2 ui + , i = 1, 2. (1.2) =a dxdτ ∂t ∂x ∂x ∂x 2 0 0 The existence and uniqueness of the solutions to (1.2)-type scalar models were studied in [8]. The asymptotic behavior of the solutions to the initial-boundary value problem for the (1.1) and (1.2)-type models have also been the subject of intensive research; see [8, 11–16]. For system (1.2) this issue is studied in [15]. Note that in [12, 16–19] and in a number of other works difference schemes for (1.1) and (1.2)-type models were investigated. Difference schemes and finite element approximations for a nonlinear parabolic integro-differential scalar model

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similar to (1.1) were studied in [20] and [21]. Finite difference schemes and finite element approximations for the scalar equation of (1.2)-type with a(σ ) = 1 + σ were studied in [16] and [22], respectively. The convergence of the finite difference approximations of system (1.2) for the case a(σ ) = 1 + σ was studied in [19]. Our main goals in the present paper are to study the finite element approximations of system (1.2), as well as to discuss the existence, uniqueness and asymptotic behavior of its solutions, to observe the asymptotic behavior obtained by our numerical experiments, and to carry out a comparative analysis of the finite element and finite difference methods. The rest of the paper is organized as follows. In the next section we briefly discuss the existence, uniqueness and asymptotic behavior of solutions to the initial boundary value problem. In Section 3 a variational formulation of the problem is presented. In Section 4 a finite element scheme for (1.2) is investigated. We close with a section on numerical implementation, where we present numerical results and compare the decay rate to the theoretical results and to the outcome of the finite difference scheme.

2 Statement of problem. Existence, uniqueness and asymptotic behavior of solutions Consider the following initial-boundary value problem: ∂ui ∂ 2 ui = (1 + σ ) 2 , ∂t ∂x

(x, t) ∈ (0, 1) × (0, ∞),

ui (0, t) = ui (1, t) = 0, ui (x, 0) = ui0 (x),

(2.1)

t ≥ 0,

(2.2)

x ∈ [0, 1],

(2.3)

i = 1, 2, where σ (t) =

 t 0

1 0



∂u1 ∂x

2

 +

∂u2 ∂x

2 dxdτ

and ui0 = ui0 (x), i = 1, 2 are given functions. We use the usual spaces C k , Lp , H k and H0k . Let us assume that ui = ui (x, t), i = 1, 2 is a solution of problem (2.1)–(2.3) 2 i (·,t) ∂ui (·,t) ∂ ui (·,t) , ∂t , ∂t ∂x , i = 1, 2 are all in C 0 ([0, ∞); L2 (0, 1)), such that ui (·, t), ∂u∂x 2

while ∂ u∂ti (·,t) , i = 1, 2 are in L2 ((0, ∞); L2 (0, 1)). 2 It is easy to obtain the continuous dependence of solutions on initial data. Indeed, by multiplying equation (2.1) by ui , i = 1, 2, after simple transformations, we get the following estimate u1  + u2  ≤ u10  + u20 .

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Below we will show that stronger estimates guaranteeing the continuous dependence of solution on initial data are valid (see Theorems 2.1 and 2.2 below). The following theorem holds [15]. Theorem 2.1 If ui0 ∈ H01 (0, 1), i = 1, 2, then for the solution to problem (2.1)–(2.3) the following estimate is true   ∂u1 ∂u2 t u1  + + u2  + ≤ C exp − . ∂x ∂x 2 Remark 1 Here and below in this section, C stands for positive constants which depend on ui0 , i = 1, 2 but are independent of t. Note that Theorem 2.1 gives exponential stabilization of the solution to problem (2.1)–(2.3) in the norm of the space H 1 (0, 1). The stabilization is also achieved in the norm of the space C 1 (0, 1). In particular, we now show that the following theorem holds [15]. Theorem 2.2 If ui0 ∈ H 4 (0, 1)∩H01 (0, 1), i = 1, 2, then for the solution to problem (2.1)–(2.3) the following relations hold:     ∂ui (x, t) ≤ C exp − t , ∂ui (x, t) ≤ C exp − t , i = 1, 2. ∂t ∂x 2 2 Here we will give a schematic proof of Theorem 2.2, but first we state and prove an auxiliary lemma [15]. Lemma 2.1 For the solution of problem (2.1)–(2.3) the following estimate holds:   ∂u1 (x, t) ∂u2 (x, t) + ≤ C exp − t . ∂t ∂t 2 Proof Differentiating equation (2.1) with respect to t for i = 1 and multiplying by ∂u1 /∂t, we deduce after some transformations that   1  2 2  1 ∂u1 2 ∂ u1 d dx + (1 + σ ) dx dt 0 ∂t ∂x∂t 0

     2  1   1 ∂u1 2 ∂u2 2 ∂u1 2 −1 ≤ (1 + σ ) dx + dx. (2.4) ∂x ∂x ∂x 0 0 Using Poincar´e’s inequality, Theorem 2.1, the nonnegativity of σ (t) and relation (2.4), we arrive at   ∂u1 ≤ C exp − t . ∂t 2 Analogously,   ∂u2 ≤ C exp − t . ∂t 2 Now we turn to the proof of Theorem 2.2.

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Proof First we estimate ∂ 2 u1 /∂x 2 in the norm of the space L1 (0, 1). From (2.1) for i = 1 we have ∂ 2 u1 ∂u1 = (1 + σ )−1 . (2.5) ∂t ∂x 2 Integrating (2.5) on (0, 1), using the Cauchy–Schwarz inequality, applying Lemma 2.1 and taking into account the nonnegativity of σ (t), we derive

 0



1 ∂ 2u 1 ∂x 2 dx

  t ≤ C exp − . 2

From this, taking into account the relation ∂u1 (x, t) = ∂x



1 0

∂u1 (y, t) dy + ∂y



1 x

0

y

∂ 2 u1 (ξ, t) dξ dy ∂ξ 2

and the boundary conditions (2.2), it follows that  1 2   ∂u1 (x, t) ∂ u1 (y, t) ≤ dy ≤ C exp − t . ∂x ∂y 2 2 0 Analogously,

  ∂u2 (x, t) ≤ C exp − t . ∂x 2

Next, we estimate ∂u1 /∂t in the norm of the space C 1 (0, 1). First we multiply (2.1) for i = 1 by ∂ 3 u1 /∂x 2 ∂t. Using Theorem 2.1, relation (2.5) and Lemma 2.1, after some transformations we arrive at ∂u1 2 ∂ 2 u1 2 ∂ 2 u1 2 + + u1 2 + ∂x ∂x 2 ∂x∂t ≤ C exp(−t). From this, taking into account Lemma 2.1 once again, it follows that   1 x 2 1 ∂u (y, t) ∂u1 (x, t) ∂ u (ξ, t) 1 1 = dy + dξ dy 0 ∂t ∂t ∂ξ ∂t 0 y   1/2    1 2 1 ∂u (x, t) 2 ∂ u1 (y, t) 1 dy ≤ C exp − t . dx + ≤ ∂y∂t ∂t 2 0 0 Analogously,

  ∂u2 (x, t) ≤ C exp − t . ∂t 2

This completes the proof of Theorem 2.2.

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Remark 2 The existence of globally defined solutions of problems (2.1)–(2.3) can be obtained by a routine procedure. One first establishes the existence of local solutions on a maximal time interval and then uses the derived a priori estimates to show that the solutions cannot escape in finite time. This approach is used very often; see, for example, [9] and [10]. The uniqueness of solutions of problem (2.1)–(2.3) can be proven as well. Indeed, let ui , u¯ i , i = 1, 2, be two solutions of problem (2.1)–(2.3) and zi (x, t) = ui (x, t) − u¯ i (x, t). We have ∂zi ∂ 2 ui ∂ 2 u¯ i = [1 + σ (t)] − + σ ¯ (t)] , [1 ∂t ∂x 2 ∂x 2 where σ¯ (t) =

 t 0

1 0



∂ u¯ 1 ∂x



2 +

∂ u¯ 2 ∂x

(2.6)

2 dxdτ.

Multiplying (2.6) by zi and integrating, we get 

1

   1 ∂ u¯ i ∂ui ∂ u¯ i ∂ui ∂zi dx + − − dx ∂t ∂x ∂x ∂x ∂x 0    1 ∂ui ∂ u¯ i ∂ui ∂ u¯ i + σ (t) − σ¯ (t) − dx ∂x ∂x ∂x ∂x 0     1   1 1 1 ∂ui ∂ u¯ i 2 1 d ∂zi 2 2 − z dx + dx + dx = [σ (t) + σ¯ (t)] 2 dt 0 i ∂x 2 0 ∂x ∂x 0      1 1 ∂ui 2 ∂ u¯ i 2 dx. + − [σ (t) − σ¯ (t)] 2 0 ∂x ∂x zi

0

Integrating with respect to t, we get the inequality 2  2  t  1   1 ∂u ∂ u ¯ i i zi2 dx + − dxdτ ≤ 0, [σ (τ ) − σ¯ (τ )] ∂x ∂x 0 0 0

i = 1, 2.

Summing these inequalities, we obtain  1  t  d z12 + z22 dx + [σ (τ ) − σ¯ (τ )] [σ (τ ) − σ¯ (τ )] dτ ≤ 0, dτ 0 0 or



1 0

 1 z12 + z22 dx + [σ (t) − σ¯ (t)]2 ≤ 0. 2

From this we immediately get zi (x, t) ≡ 0, i = 1, 2, which proves the uniqueness of the solution.

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3 Variational formulation Consider the following initial-boundary value problem: ∂ui ∂ 2 ui = (1 + σ (t)) 2 + fi (x, t), ∂t ∂x

(x, t) ∈ (0, 1) × (0, T ),

ui (0, t) = ui (1, t) = 0,

0 ≤ t ≤ T,

ui (x, 0) = ui0 (x),

x ∈ [0, 1],

i = 1, 2, where σ (t) =

 t 0

1 0



∂u1 ∂x

(3.1) 

2 +

∂u2 ∂x

2 dxdτ ,

T = Const. > 0 and ui0 = ui0 (x), i = 1, 2 are given functions. One of the ingredients of the finite element method is a variational formulation of the problem. To provide this variational formulation, let us denote by H the linear space of functions ui satisfying the boundary conditions in (3.1) and ||ui (·, t)||1 < ∞, where ||ui (·, t)||r =

⎧ ⎨ ⎩

⎤ ⎫1/2 2 r j ⎬  ∂ u (x, t) i ⎦ dx ⎣|ui (x, t)|2 + , ∂x j ⎭ ⎡

1 0

i = 1, 2.

j =1

The variational formulation of the problem can now be stated as follows: Find a pair of functions ui (x, t) ∈ H for which 

   ∂ui ∂ui ∂vi vi , + (1 + σ (t)) , = < fi , vi >, ∂t ∂x ∂x

∀vi ∈ H,

(3.2)

and < vi , ui (x, 0) > = < vi , ui0 (x) >, where



∀vi ∈ H,

i = 1, 2,

(3.3)

1

< p(x), q(x) > =

p(x)q(x)dx. 0

To approximate the solution of (3.2) and (3.3) we require that ui and vi lie in a finite-dimensional subspace Sh of H for each t and i = 1, 2. The following property concerning approximability in Sh can be readily verified for finite element spaces; see [23].

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Approximation property There is an integer r ≥ 2 and positive numbers C0 , C1 independent of h such that for any v ∈ H , there exists a point v h ∈ Sh satisfying ||v − v h || ≤ C hj − ||v||j

for 0 ≤  ≤ 1,

 < j ≤ r.

(3.4)

The approximation uhi ∈ Sh to ui is defined by the following variational analog of (3.2), (3.3): Find a pair uhi ∈ Sh such that     h ∂uhi ∂vih h ∂ui vi , + (1 + σh (t)) , = < fi , vih >, ∀vih ∈ Sh , (3.5) ∂t ∂x ∂x and ∀vih ∈ Sh ,

< vih , uhi (x, 0) >=< vih , ui0 (x) >, where σh (t) =

 t 0

0

1

⎡

∂uh1 ⎣ ∂x

2

 +

∂uh2 ∂x

i = 1, 2,

(3.6)

2 ⎤ ⎦ dxdτ.

Once a basis has been selected for Sh , (3.5) and (3.6) are equivalent to a set of N integro-differential equations, where N is the dimension of Sh . The solution of such a system will be discussed in Section 5.

4 Error estimates In this section we shall estimate the error in the finite element approximation using the norm ⎡ ⎛ ⎞ ⎤1/2  T  1  r j ∂ E(x, t) 2 ⎠ ⎝ ⎦ . |||E|||r = ⎣ ∂x j dxdt 0

0

j =0

Whenever r = 0 we will omit the subscript for this norm as well. Theorem 4.1 The error in the f inite element approximation uhi generated by (3.5), (3.6) satisf ies the inequality

2 h j −1 2 2 2 ∂ui |||ui − ui |||1 ≤ h c1 h ||ui0 || + c2 h ∂t + c3 |||ui |||2 + c4 h2(j −1)  + c5

2  m=1

2 

|||um |||2

m=1 ⎫1/2  2⎬  , c6 hj −1 |||um |||2 + c7 []um [] ⎭

j > 1,

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where 

T

[]u[] =



0

1

|u| dxdτ

0

and ci , i = 1, 2, ..., 7, denote various positive constants. Proof Subtracting (3.5) from (3.2) with vih instead of vi , we obtain  vih ,

       ∂uhi ∂uhi ∂vih ∂ui ∂vih h ∂ui + (1 + σh (t)) , = vi , + (1 + σ (t)) , , ∂t ∂x ∂x ∂t ∂x ∂x ∀vih ∈ Sh ,

i = 1, 2.

Let u˜ hi be any function in Sh . Then 

 $ %   ∂ uhi − u˜ hi ∂uhi ∂ u˜ hi ∂vih + (1 + σh (t)) − (1 + σ˜ h (t)) , ∂t ∂x ∂x ∂x     $ % h h h ∂ u − u ˜ ∂ u ˜ ∂v ∂u i i i = vih , + (1 + σ (t)) − (1 + σ˜ h (t)) i , i , ∂t ∂x ∂x ∂x

vih ,

∀vih ∈ Sh ,

i = 1, 2,

where σ˜ h (t) =

 t 0

0

1

⎡

∂ u˜ h1 ⎣ ∂x

(4.1)



2 +

∂ u˜ h2 ∂x

2 ⎤ ⎦ dxdτ.

Define the errors as follows: ei (x, t) = uhi (x, t) − u˜ hi (x, t), Ei (x, t) = ui (x, t) − u˜ hi (x, t),

i = 1, 2.

(4.2)

Since ei ∈ Sh , we can let vih = ei and (4.1) becomes       ∂uhi ∂ u˜ hi ∂ei ∂ei ∂ei ∂ei + , + σh (t) − σ˜ h (t) , ei , ∂t ∂x ∂x ∂x ∂x ∂x       ∂ u˜ hi ∂Ei ∂Ei ∂ei ∂ui ∂ei = ei , + , + σ (t) − σ˜ h (t) , , ∂t ∂x ∂x ∂x ∂x ∂x

i = 1, 2. (4.3)

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Considering the last term on the left-hand side of (4.3) and denoting, ωi =

∂uhi , ∂x

ηi =

∂ u˜ hi , ∂x

we have 

 ∂uhi ∂ u˜ hi ∂ei σh (t) − σ˜ h (t) , ∂x ∂x ∂x     t  1& ' ' t 1& 1 2 2 2 2 2 ω1 + ω2 dξ dτ + η1 + η2 dξ dτ, (ωi − ηi ) = 2 0 0 0 0     t  1& ' ' t 1& 1 2 2 2 2 2 2 ω1 + ω2 dξ dτ − η1 + η2 dξ dτ, ωi − ηi + 2 0 0 0 0    t  1     1 1 2 2 2 2 ω1 + ω2 − η1 − η2 dξ dτ ωi2 − ηi2 dx. ≥ 2 0 0 0

Now ω12 − η12 + ω22 − η22 ≥ 2 min (ωj2 − ηj2 ) ≡ 2(ωk2 − ηk2 ). Also ωi2 − ηi2 ≥ j =1,2

min

j =1, 2

(ωj2

− ηj2 ). 

So,

 ∂uhi ∂ u˜ hi ∂ei σh (t) − σ˜ h (t) , ∂x ∂x ∂x ≥

 t 0

  ωk2 − ηk2 dξ dτ

1 0

1 0

 1 dφk2 ωk2 − ηk2 dx = , 2 dt

where φk (t) ≡

 t 0

0

1$

% ωk2 − ηk2 dξ dτ.

Therefore the left-hand side of (4.3) can be rewritten as follows: 

     ∂uhi ∂ u˜ hi ∂ei ∂ei ∂ei ∂ei ei , + , + σh (t) − σ˜ h (t) , ∂t ∂x ∂x ∂x ∂x ∂x 2 2 ∂ei 1 d + 1 dφk , ||ei ||2 + i = 1, 2. ≥ 2 dt ∂x 2 dt

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Now consider the last term on the right-hand side of (4.3). Substituting for (4.2), we have

∂ui ∂x

from

∂ u˜ h ∂ui − σ˜ h (t) i ∂x ∂x ⎡

2 

2 ⎤  t 1 h h ∂ u˜ ∂ u˜ ∂Ei ⎣ ∂E1 + 1 + ∂E2 + 2 ⎦ dxdτ = ∂x 0 0 ∂x ∂x ∂x ∂x

σ (t)

∂ u˜ h + i ∂x

 t 1 0

0

∂E1 ∂x

2   ∂E2 2 ∂E1 ∂ u˜ h1 ∂E2 ∂ u˜ h2 + +2 +2 dxdτ, i = 1, 2. ∂x ∂x ∂x ∂x ∂x

Taking this into account in the right-hand side of (4.3), we get ⎡ ⎤       t  1  h 2  h 2  ∂ u ˜ ∂ u ˜ ∂Ei ∂ei ∂Ei ∂ei ∂Ei 1 2 ⎣ ⎦ dxdτ + , + , ei , + ∂t ∂x ∂x ∂x ∂x 0 0 ∂x ∂x   t  1     ∂Ei ∂ei ∂E2 2 ∂E1 2 , + + dxdτ ∂x ∂x 0 0 ∂x ∂x         t 1 ∂ u˜ hi ∂ei ∂E2 2 ∂E1 2 , + dxdτ + ∂x ∂x ∂x ∂x 0 0    

  t 1 ∂ u˜ hi ∂ei ∂E1 ∂ u˜ h1 ∂E2 ∂ u˜ h2 ∂Ei ∂Ei + , 2 + dxdτ ≤ ei , + ∂x ∂x ∂x ∂x ∂x ∂x ∂x ∂t 0 0 ⎧ ⎡ ⎤    h 2   h 2  t 1  ∂Ei ∂ei ⎨ ∂ u˜ 1 ∂ u˜ 2 1 1 ⎣ ⎦ dxdτ + 1+ + , 1+ 1+ ⎩ ∂x ∂x

1 ∂x

2 ∂x 0 0 

+

 t 0

0

1





∂E1 (1 + 1 ) ∂x

2



∂E2 + (1 + 2 ) ∂x 

   ∂ u˜ h ∂e  t  1  ∂E 2  ∂E 2 1 2 i i + + dxdτ , ∂x ∂x ∂x ∂x 0 0

   ∂ u˜ h1 t 1 ∂E1 dxdτ + 2 sup x,t ∂x 0 0 ∂x 

   ∂ u˜ h2 t 1 ∂E2 + 2 sup dxdτ , x,t ∂x 0 0 ∂x where 1 and 2 come from the Cauchy–Schwarz inequality.

2

 dxdτ

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Now incorporate these estimates into (4.3) to obtain ∂ei (·, t) 2 1 dφk2 1 d + ||ei (·, t)||2 + 2 dt ∂x 2 dt     ∂Ei (·, t) ∂ei (·, t) ∂Ei (·, t) + , ≤ ei (·, t), ∂t ∂x ∂x ⎧ ⎡ ⎤   h 2    h 2  t 1  ⎨ ∂ u˜ 1 ∂ u˜ 2 1 1 ⎣ 1+ ⎦ dxdτ + 1+ × 1+ ⎩

1 ∂x

2 ∂x 0 0       t  1 ∂E1 2 ∂E2 2 + (1 + 1 ) dxdτ + (1 + 2 ) ∂x ∂x 0 0    ∂ u˜ h ∂e  t  1  ∂E 2  ∂E 2 1 2 i i + + dxdτ , ∂x ∂x ∂x ∂x 0 0 

  ∂ u˜ h1 t 1 ∂E1 dxdτ + 2 sup x,t ∂x 0 0 ∂x 

   ∂ u˜ h2 t 1 ∂E2 + 2 sup dxdτ . (4.4) x,t ∂x 0 0 ∂x Integrating (4.4) with respect to t, we have ∂ei (·, t) 2 1 2 ∂x dt + 2 φk (T ) 0  T 1  T  1 ∂Ei ∂ei ∂Ei 1 dxdt + ei ≤ ||ei (·, 0)||2 + ∂x ∂x dx 2 ∂t 0 0 0 0 ⎧ ⎡ ⎤   h 2    h 2  t 1  ⎨ ∂ u˜ 1 ∂ u˜ 2 1 1 ⎣ 1+ ⎦ dxdτ + 1+ × 1+ ⎩

1 ∂x

2 ∂x 0 0       t  1 ∂E1 2 ∂E2 2 + (1 + 2 ) + (1 + 1 ) dxdτ dt ∂x ∂x 0 0        T  1 h t 1 ∂E1 2 ∂E2 2 ∂ u˜ i ∂ei dxdτ + + dx ∂x ∂x 0 0 ∂x ∂x 0 0

   ∂ u˜ h1 t 1 ∂E1 + 2 sup dxdτ x,t ∂x 0 0 ∂x 

   ∂ u˜ h2 t 1 ∂E2 + 2 sup dxdτ dt. (4.5) x,t ∂x 0 0 ∂x

1 ||ei (·, T )||2 + 2



T

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We can estimate the following factor in the last integral: 

     ∂ u˜ h1 t 1 ∂E1 ∂E2 2 ∂E1 2 + dxdτ + 2 sup dxdτ x,t ∂x 0 0 ∂x ∂x ∂x 0 

  ∂E1 2 ∂E2 2 ∂ u˜ h2 t 1 ∂E2 dxdτ ≤ + + 2 sup x,t ∂x 0 0 ∂x ∂x ∂x  



   ∂ u˜ h1 ∂E1 ∂ u˜ h2 ∂E2 + 2 sup + 2 sup . x,t ∂x x,t ∂x ∂x ∂x

 t 0

1



Denoting the right-hand side of this inequality by I, the last term of (3.5) becomes 

T

I 0

 ∂ u˜ hi 1 ∂ei dxdt ≤ sup x ∂x 0 ∂x

( ) ) I*

T 0

( ) ) ≤ I*

( )  2 h 2 1 ∂e ) T ∂ u ˜ i * i dt sup ∂x dx dt x ∂x 0 0

0

2 ∂ei ∂ u˜ hi dt sup x ∂x ∂x



T

T

3 2 ≤ I 2

0

2 ∂ u˜ hi 1 sup dt + x ∂x 2 3

∂ei 2 ∂x .

Thus, (4.5) yields  2

∂ei 2 ≤ 1 ||ei (·, 0)||2 + 1 4 |||ei |||2 + 1 ∂Ei ∂x 2 2

4 ∂t  

2  ∂Em 2 ∂Ei 2 ∂ei 2 1 1 + L+ (1 + m )

5 + 2 ∂x ∂x

5 ∂x m=1



3 + 2 

m=1

T

×

  h    2 2  ∂Em 2 ∂ u ˜ ∂E m m ∂x + 2 sup ∂x ∂x x,t

0

2 ∂ u˜ hi 1 sup dt + x ∂x 2 3

where L=1+

 t 0

0

⎡ 1



⎣ 1+ 1

1



∂ei 2 ∂x ,

∂ u˜ h1 ∂x

2



1 + 1+

2



∂ u˜ h2 ∂x

2 ⎤ ⎦ dxdτ.

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Since Ei is the interpolation error, we have from (3.4), |||Ei ||| ∂Ei ∂x ∂Ei ∂t   ∂Ei ∂x

≤ C0 hj |||ui ||| , ≤ C1 hj −1 |||ui ||| , ∂ui , ≤ C2 h ∂t j

≤ C3 hj −1 []ui [],

i = 1, 2.

Therefore, 2 ∂ei 2 ≤ 1 ||ei (·, 0)||2 + 4 |||ei |||2 + 1 C 2 h2j ∂ui 2 ∂x 2 2 2 4 ∂t

 2  1 2 2(j −1) 2 + L+ (1 + m )C1 h |||um ||| 2 m=1 

1 ∂ei 2 2 2(j −1) 2 × 5 C1 h |||ui ||| +

5 ∂x 

3 + 2

2  

C12 h2j −2 |||um |||2

m=1



T

× 0

2 ∂ u˜ hi 1 sup dt + x ∂x 2 3

 h   2 ∂ u ˜ m j −1 C3 h + 2 sup [] um [] ∂x x,t

∂ei 2 ∂x .

Now we collect terms with norms of the error ei to obtain 

2  ∂ei 2 4 1 1 2 (1 + m )C12 h2(j −1) |||um |||2 ∂x − 2 |||ei ||| − 2 L +

5 m=1

∂ei 2 ∂x

1 ∂ei 2 1 1 2 2j ∂ui 2 2 − ≤ ||ei (·, 0)|| + C h 2 3 ∂x 2 2 4 2 ∂t 

2  1 + L+ (1 + m )C12 h2(j −1) |||um |||2 5 C12 h2(j −1) |||ui |||2 2 m=1

3 + 2 



C12 h2j −2 |||um |||2

m=1

T

× 0

2  

2 ∂ u˜ hi sup dt. x ∂x

 2  h  ∂ u˜ m j −1 + 2 sup C3 h [] um [] ∂x x,t

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Next we use the Poincar´e inequality ∂ei ∂x ≥ Cp ||ei || ∂ei ∂x ≥ Cp |||ei ||| .

to show that

So, the last relation becomes 

2 ∂ei 2 L 1  1

4 2 2(j −1) 2 − − (1 + m )C1 h |||um ||| − 1− 2 5 2 5 2 3 ∂x 2Cp2 m=1 1 1 2 2j ∂ui 2 ≤ ||ei (·, 0)||2 + C2 h 2 2 4 ∂t 

2  1 2 2(j −1) 2 2 2(j −1) 2 + 5 C1 h |||ui ||| L + (1 + m )C1 h |||um ||| 2 m=1  2   2  h  ∂ u ˜

3  m 2 2j −2 2 j −1 C3 h |||um ||| + 2 sup C1 h + [] um [] 2 ∂x x,t m=1 2  T ∂ u˜ hi × dt. sup 0 x ∂x 2 i Now choose 3 , 4 and 5 so that the coefficient of ∂e ∂x is positive, say C4 . The right-hand side depends on the grid size h and the known error at time t = 0, i.e., ||ei (·, 0)|| ≤ C5 hj ||ui0 || . Thus,

∂ei 2 C52 2 C22 2 ∂ui 2 2j −2 2 ≤ h ||u || h h i0 + ∂x 2C4 2C4 4 ∂t 

2 

5 C12 2 2 2(j −1) 2 + |||ui ||| L+ (1 + m )C1 h |||um ||| 2C4 m=1  2   h   2

3  ∂ u˜ m 2 j −1 2 |||um ||| +2 sup C3 []um [] + C1 h 2C4 ∂x x,t m=1 2 ⎫  T ⎬ h ∂ u˜ × sup i dt . ⎭ 0 x ∂x Recall that |||ei |||21

 ∂ei 2 1 ∂ei 2 = |||ei ||| + ≤ 1 + 2 , ∂x ∂x Cp 2

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so taking into account the last estimate we obtain

C52 2 C22 2 ∂ui 2 2 2j −2 2 |||ei |||1 ≤ h C6 h ||ui0 || + h 2C4 2C4 4 ∂t 

2 

5 C12 2 2 2(j −1) 2 + |||ui ||| L+ (1+ m )C1 h |||um ||| 2C4 m=1  2   h   2

3  ∂ u˜ m 2 j −1 2 |||um ||| +2 sup C3 []um [] + C1 h 2C4 ∂x x,t m=1 2 ⎫  T ⎬ ∂ u˜ hi × sup dt , ⎭ 0 x ∂x $ where C6 = 1 +

1 Cp2

%

.

From this, using the triangle inequality |||ui − uhi |||21 = |||ui − u¯ hi + u¯ hi − uhi |||21 ≤ 2|||Ei |||21 + 2|||ei |||21 and estimating |||Ei |||21

∂Ei 2 ≤ C 2 h2j |||ui |||2 + C 2 h2j −2 |||ui |||2 = |||Ei ||| + 0 1 ∂x   = C02 h2 + C12 h2j −2 |||ui |||2 , 2

we finally get

|||ui − uhi |||1

≤h

j −1

2 2 ∂ui

c1 h ||ui0 || + c2 h ∂t 2

2

+ c3 |||ui |||2 + c4 h2(j −1)

2 

|||um |||2

m=1

 + c5

⎫1/2 2   2⎬  c6 hj −1 |||um |||2 + c7 []um [] , ⎭

m=1

where c1 =

C52 C6 , C4

c2 =

C22 C6 , C4 4

c3 =

5 C14 C6 , C4  h  ∂ u˜ m c7 = 2 sup C3 . ∂x x,t

c4 = (1 + max{ 1 , 2 }) c6 = C12 ,

5 C12 C6 L + 2C02 h2 + 2C12 , C4 2  ∂ u˜ hi

3 C6 T c5 = dt, sup C4 0 x ∂x

This completes the proof of Theorem 4.1.

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5 Numerical solution For the numerical solution of (3.5) and (3.6), we let φ1 (x), . . . , φN (x) be a basis for Sh (where N is the dimension of Sh ). Thus any uhi ∈ Sh can be represented as follows: uhi (x, t) =

N 

(5.1)

uij (t)φj (x).

j =1

Since (3.5) and (3.6) are valid for all vih ∈ Sh , one can let vih = φk . Using (5.1), we are led to the following system for the vectors of weights ui (t) = (ui1 (t), ui2 (t), . . . , uiN (t)): Mu˙ i + K(u1 , u2 )ui = Fi , Mui (0) = Wi ,

i = 1, 2,

(5.2)

i = 1, 2,

(5.3)

where Mj k =< φk , φj >,

(5.4)

K(u1 , u2 )j k =< (1 + σh (t))φk , φj >,

(5.5)

Fik =< φk , fi >,

Wik =< φk , ui0 > .

(5.6)

Now we can evaluate σh (t) as follows: ⎡

2  N

2 ⎤  t  1  N  ⎣ σh (t) = u1 φ + u2 φ ⎦ dxdτ 0

=

 t 0

=

0

1

=1



0

=1

N  N 

N  N  =1 m=1

φm

+

=1 m=1

N  N  

t

N  N  =1 m=1



1

(u1 u1m + u2 u2m )

=1 m=1 0

=

u1 u1m φ

K˜ m

+ 

t

0

u2 u2m φ φm

dxdτ



φ φm dx dτ

,-

=K˜ m

.

(u1 u1m + u2 u2m ) dτ.

(5.7)

0

The time integral can be approximated by the trapezoidal rule using the equally spaced point tp , where t0 = 0, tn = t and t = ti − ti−1 , i = 1, . . . , n:  t (u1 u1m + u2 u2m ) dτ 0

=

n  p=0

  $ % tζp u1 (tp )u1m (tp ) + u2 (tp )u2m (tp ) + O ( t)2 ,

(5.8)

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where ζp = 1/2 for p = 0, n and ζp = 1 for p = 1, . . . , n − 1. Combining (5.8) and (5.7) with (5.5), we get K(u1 , u2 )j k ⎛ = ⎝1 +

N  N 

K˜ m

=1 m=1

⎛ = ⎝1 +

N N  

= ⎝1 +

n 

tζp

p=0

K˜ m

=1 m=1



n 

n 

tζp

$

$

⎞  % u1 (tp )u1m (tp ) + u2 (tp )u2m (tp ) ⎠ φk , φj

⎞ % u1 (tp )u1m (tp ) + u2 (tp )u2m (tp ) ⎠ K˜ j k

p=0



tζp v(tp )⎠ K˜ j k ,

(5.9)

p=0

˜ 1 (t) + uT (t)Ku ˜ 2 (t). To solve the system (5.2) and (5.3), we where v(t) = uT1 (t)Ku 2 use Taylor’s series. Let   1 ui (t + t) = ui (t) + t u˙ i (t) + ( t)2 u¨ i (t) + O ( t)3 . 2

(5.10)

Differentiating (5.2) with respect to t, we obtain ˙ i = F˙ i , M u¨ i + K(u1 , u2 )u˙ i + Ku

i = 1, 2,

(5.11)

where the matrices M and F are defined by (5.4), (5.6) and / 0 K˙ kj = σ˙ h φj , φk ⎡  N

2  N

2 ⎤  1   ⎣ ⎦ u1 (t)φ + u2 (t)φ = φj , φk 0

=

=1

N N  

=1 m=1 N  N 

= 

=1



1

(u1 (t)u1m (t) + u2 (t)u2m (t)) 0

φ φm

K˜ kj

(u1 (t)u1m (t) + u2 (t)u2m (t)) K˜ m K˜ kj

=1 m=1

 ˜ 1 (t) + uT (t)Ku ˜ 2 (t) K˜ kj = v(t)K˜ kj . = uT1 (t)Ku 2

(5.12)

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Now multiplying (5.10) by M and using (5.2), (5.11) and (5.12), and after dropping terms of order higher than the second, we obtain   ui (t + t)−ui (t) 1 M = Mu˙ i (t)+ tMu¨ i (t) t 2 2 1 1 ˙ i = [Fi −K(u1 , u2 )ui ] + t F˙ i −K(u1 , u2 )u˙ i − Ku 2 = [Fi −K(u1 , u2 )ui ] ' 1 & ˙ i + t F˙ i −K(u1 , u2 )M−1 (Fi −K(u1 , u2 )ui )− Ku 2   1 ˙ 1 −1 = Fi + t Fi − tK(u1 , u2 )M Fi 2 2   1 1 ˙ i. − K(u1 , u2 ) ui − tM−1 K(u1 , u2 )ui − t Ku 2 2 (5.13) ˙ from (5.9) and (5.12), we get from (5.13), Substituting for K and K   ui (t + t)−ui (t) M t ⎛ ⎡ ⎤ ⎞ n  1 1 ˜ −1 Fi ⎦ = ⎣Fi + t F˙ i − t ⎝1+ tζp v(tp )⎠ KM 2 2 p=0 ⎛ ⎞ n  − t ⎝1 + tζp v(tp )⎠ ⎡

p=0



˜ ⎣ui − 1 t M−1 ⎝1+ ×K 2

n  p=0





˜ i . (5.14) ˜ i ⎦ − 1 t v(t)Ku tζp v(tp )⎠ Ku 2

If we take t = tn as in (5.8) and denote uni = ui (tn ), then (5.14) can be written as follows: ⎛ ⎞ ⎤

⎡  n n  un+1 −u 1 1 i i ˜ −1 Fn ⎦ = ⎣Fni + t F˙ ni − t ⎝1+ M tζp v p ⎠ KM i t 2 2 p=0 ⎛ ⎞ n  ˜ − ⎝1+ tζp v p ⎠ K p=0

⎛ ⎞ ⎤ n  1 ˜ n ⎦ − 1 t v n Ku ˜ n. tζp v p ⎠ Ku × ⎣uni − tM−1 ⎝1+ i i 2 2 ⎡

p=0

(5.15)

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Now let us denote φn = 1 +

n 

tζp v p .

(5.16)

p=0

Using notation (5.16) from (5.15), we have 

− uni un+1 i M t



  1 1 n −1 n ˜ n ˜ ˜ n = −φ K ui − t M φ Kui − t v n Ku i 2 2   1 1 ˜ −1 Fn , + Fni + ( t)F˙ ni − t φ n KM i = 1, 2. i 2 2 (5.17) n

Note that in (5.17), φ n and v n depend on both u1 and u2 . We can update both φ n and v n after we solve the two systems or we can update them after solving each system. In our first numerical experiment we have chosen the right-hand side so that the exact solution is given by u1 (x, t) = x(1 − x) sin(x + t) and u2 (x, t) = x(1 − x) cos(x + t) . In this case the right-hand side is   11 f1 (x, t) = x(1 − x) cos(x + t)− 1+ t (−2 sin(x + t) + 2(1 − x) cos(x + t) 30 − 2x cos(x + t) − x(1 − x) sin(x + t)) and   11 f2 (x, t) = −x(1 − x) sin(x + t)− 1+ t (−2 cos(x + t) − 2(1 − x) sin(x + t) 30 + 2x sin(x + t) − x(1 − x) cos(x + t)) . The parameters used are N = 100 which dictates h = 0.01. In the next two figures we plotted the numerical solution (marked with ∗) and the exact solution at t = 0.5 (Figs. 1 and 2) and t = 1.0 (Figs. 3 and 4). It is clear that the two solutions are almost identical (Table 1). Note that the energy norm of the error decreases linearly with h as expected by Theorem 4.1.

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147 Time t =0.5

0.25

0.2

0.15

0.1

0.05

0

0

20

40

60

80

100

120

Fig. 1 The solution at t = 0.5. The exact solution u1 is a solid line and the numerical solution is marked by ∗ Time t =0.5 0.16

0.14

0.12

0.1

0.08

0.06

0.04

0.02

0

0

20

40

60

80

100

120

Fig. 2 The solution at t = 0.5. The exact solution u2 is a solid line and the numerical solution is marked by ∗

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Numer Algor (2013) 64:127–155 Time t =1 0.25

0.2

0.15

0.1

0.05

0

0

20

40

60

80

100

120

Fig. 3 The solution at t = 1.0. The exact solution u1 is a solid line and the numerical solution is marked by ∗ Time t =1 0.06

0.05

0.04

0.03

0.02

0.01

0

−0.01

−0.02

−0.03

−0.04

0

20

40

60

80

100

120

Fig. 4 The solution at t = 1.0. The exact solution u2 is a solid line and the numerical solution is marked by ∗

Author's personal copy Numer Algor (2013) 64:127–155 Table 1 The empirical rate of convergence in the energy norm when integrating up to t = .5 and t = 1

149

t = .5 h

t =1 Energy

Rate

h

norm

Energy

Rate

norm

.2

.023124

.934145

.2

.030767

.941436

.04

.005142

.983124

.04

.006762

.988143

.02

.0026012

.992146

.02

.003409

.991434

.01

.0013077

.01

.001715

In our second experiment we have taken a zero right-hand side and initial data given by u10 (x) = u1 (x, 0) = x(1 − x) sin(8π x),

u20 (x) = u2 (x, 0) = x(1 − x) cos(4π x) .

In this case, we know (Theorem 2.2) that the solution will decay in time. The parameters N, h are as before. In Fig. 5 we plotted the initial data and in Fig. 6 we show the numerical solution at four different times. In both figures the top subplot is for u1 and the bottom subplot is for u2 . It is clear that the numerical solution is approaching zero for all x. We have also plotted the maximum norm of the par∂u2 −t/2 . Figure 7 shows that the 1 tial derivatives ∂u ∂x and ∂x versus the exponential e ∂u2 1 maximum norm of ∂u ∂x (top) and ∂x (bottom) decays exponentially. Therefore the numerical approximation of the x-derivative of the solutions obtained in our numerical experiment fully agrees with the theoretical results given in Theorem 2.2. Note that the line for the derivative is very close to the origin which means that the constant in Theorem 2.2 is very small. To test the numerical data for a longer time run, we have run the first example up to t = 5. The exact solution is plotted with the numerical solution at t = 5 in Figs. 8 and 9. Note the agreement between the finite element and exact solution at the end of the run. We have experimented with several other initial solutions, and in all cases we observed an agreement with the exact solution. Remark At each time step one can advance u1 and then advance u2 using either the previous u1 or the most current one. As a result, the matrices are smaller and banded.

5.1 Comparison with the finite difference method The authors of [19] have developed a finite difference scheme to solve the system (3.1). In the finite difference method, the system is nonlinear and Newton’s method was used at each time step. This required computing and storing the Jacobian. Therefore the system becomes dense. On the other hand, in the finite element method we did not have to solve a nonlinear system. Therefore the system is banded and the bandwidth is dictated by the degree of the elements used. This fact had already been discussed by Neta and Igwe [21].

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0.2

0.15

0.1

0.05

0

−0.05

−0.1

−0.15

−0.2

−0.25

0

20

40

60

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100

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Time t = 0 0.25

0.2

0.15

0.1

0.05

0

−0.05

−0.1

−0.15

−0.2

0

20

40

60

Fig. 5 The initial data u10 (x) = x(1 − x) sin(8πx) (top) and u20 (x) = x(1 − x) cos(4πx) (bottom) for Example 2

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151

−5

x 10

8

t=0.1 t=0.2 t=0.3 t=0.4

6

4

2

0

−2

−4

−6

−8

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

−3

0

x 10

t=0.1 t=0.2 t=0.3 t=0.4

−0.2

−0.4

−0.6

−0.8

−1

−1.2

−1.4

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

Fig. 6 The numerical solution at t = 0.1, 0.2, 0.3, 0.4 for u1 (top) and u2 (bottom)

1

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0.9

0.8

0.7

0.6

0.5

0.4

0.3

0.2

0.1

0

0

0.5

1.0

1.5

2.0

2.5

3.0

3.5

4.0

4.5

0

0.5

1.0

1.5

2.0

2.5

3.0

3.5

4.0

4.5

1

0.9

0.8

0.7

0.6

0.5

0.4

0.3

0.2

0.1

0

∂u2 1 Fig. 7 The maximum norm of the numerical solution for ∂u ∂x (top) and ∂x (bottom) (Example 2) and ∂u2 1 e−t/2 . A solid line is used for ∂u and and a line marked with ∗ for the exponential ∂x ∂x

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153 Time t =5

0

−0.02

−0.04

−0.06

−0.08

−0.1

−0.12

−0.14

−0.16

−0.18

−0.2

0

20

40

60

80

100

120

Fig. 8 The solution at t = 5.0. The exact solution u1 is a solid line and the numerical solution is marked by ∗ Time t =5 0.2

0.18

0.16

0.14

0.12

0.1

0.08

0.06

0.04

0.02

0

0

20

40

60

80

100

120

Fig. 9 The solution at t = 5.0. The exact solution u2 is a solid line and the numerical solution is marked by ∗

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We have recorded the CPU time it took our PC to solve the second example for N = 25 and t = 1. The finite difference solution took more than twice the time it took to solve the same problem using finite (linear) elements. The accuracy is the same.

6 Conclusions In this paper we have used the finite element method to solve a system of two nonlinear integro-differential equations. We have shown that it gives the same accuracy as the finite difference scheme without the need to compute the Jacobian at each time step. We have also established the rate of convergence of the finite element method in an appropriate energy norm.

References 1. Landau, L., Lifschitz, E.: Electrodynamics of Continuous Media. Course of Theoretical Physics, Vol. 8. Pergamon Press, Oxford-London-New York-Paris; Addison-Wesley Publishing Co., Inc., Reading, Mass, 1960, Russian original: Gosudarstv. Izdat. Tehn-Teor. Lit., Moscow (1957, translated from the Russian) 2. Gordeziani, D., Dzhangveladze, T., Korshia, T.: Existence and uniqueness of the solution of a class of nonlinear parabolic problems. Differ. Uravn. 19, 1197–1207 (1983, Russian). Diff. Equ. 19, 887–895 (1983, English translation) 3. Dzhangveladze, T.A.: First boundary value problem for a nonlinear equation of parabolic type. Dokl. Akad. Nauk SSSR 269, 839–842 (1983, Russian). Sov. Phys. Dokl. 28, 323–324 (1983, English translation) 4. Dzhangveladze, T.: An Investigation of the First Boundary Value Problem for Some Nonlinear Integro-differential Equations of Parabolic Type. Tbilisi State University, (Russian) (1983) 5. Dzhangveladze, T.A.: A nonlinear integro-differential equation of parabolic type. Differ. Uravn. 21, 41–46 (1985, Russian). Diff. Equ. 21, 32–36 (1985, English translation) 6. Laptev, G.: Quasilinear parabolic equations which contain in coefficients Volterra’s operator. Math. Sbornik 136, 530–545 (1988, Russian). Sbornik Math. 64, 527–542 (1989, English translation) 7. Lin, Y., Yin, H.M.: Nonlinear parabolic equations with nonlinear functionals. J. Math. Anal. Appl. 168, 28–41 (1992) 8. Jangveladze, T.: On one class of nonlinear integro-differential equations. Semin. I. Vekua Inst. Appl. Math. 23, 51–87 (1997) 9. Vishik, M.: Solvability of boundary-value problems for quasi-linear parabolic equations of higher orders (Russian). Math. Sb. (N.S.) 59(101), suppl. 289–325 (1962) 10. Lions, J.-L.: Quelques M´ethodes de R´esolution des Probl`emes aux Limites Non-lin´eaires. Dunod, Gauthier-Villars, Paris (1969) 11. Kiguradze, Z.: The asymptotic behavior of the solutions of one nonlinear integro-differential model. Semin. I. Vekua Inst. Appl. Math. Reports 30, 21–32 (2004) 12. Kiguradze, Z.V.: Asymptotic behavior and numerical solution of the system of nonlinear integrodifferential equations. Rep. Enl. Sess. Sem. I. Vekua Inst. Appl. Math. 19, 58–61 (2004) 13. Jangveladze, T.A., Kiguradze, Z.V.: Asymptotics of a solution of a nonlinear system of diffusion of a magnetic field into a substance. Sib. Mat. Z. 47, 1058–1070 (2006, Russian). Sib. Math. J. 47, 867–878 (2006, English translation) 14. Jangveladze, T., Kiguradze, Z., Neta, B.: Large time behavior of solutions to a nonlinear integrodifferential system. J. Math. Anal. Appl. 351, 382–391 (2009) 15. Jangveladze, T., Kiguradze, Z., Neta, B.: Large time asymptotic and numerical solution of a nonlinear diffusion model with memory. Comput. Math. Appl. 59, 254–273 (2010)

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