An accelerated technique for solving a coupled system of differential equations for a catalytic converter in interphase heat transfer

JID:YJMAA

AID:20628 /FLA

Doctopic: Applied Mathematics

[m3L; v1.184; Prn:11/08/2016; 13:53] P.1 (1-19)

J. Math. Anal. Appl. ••• (••••) •••–•••

Contents lists available at ScienceDirect

Journal of Mathematical Analysis and Applications www.elsevier.com/locate/jmaa

An accelerated technique for solving a coupled system of differential equations for a catalytic converter in interphase heat transfer Linia Anie Sunny, Rupsha Roy, V. Antony Vijesh ∗ School of Basic Sciences, Indian Institute of Technology Indore, Indore – 453552, India

a r t i c l e

i n f o

a b s t r a c t

Article history: Received 5 January 2016 Available online xxxx Submitted by B. Kaltenbacher Keywords: Catalytic converter Coupled system Finite difference Monotone iterations Quasilinearization Successive approximation

This paper deals with an accelerated iterative procedure for a coupled system of partial differential equations arising from a catalytic converter model. Based on the main theorem, a finite difference based numerical method is developed. The monotone property as well as the convergence analysis and the error estimate of the proposed finite difference scheme are proved theoretically. The efficiency of the proposed scheme is illustrated by providing a comparative numerical study with the existing method. The proposed iterative procedure reduces the number of iterations at least by forty percent for certain benchmarks available in the literature. © 2016 Elsevier Inc. All rights reserved.

1. Introduction Catalytic converter is a reliable emissions control device that converts toxic pollutants in exhaust gas to less toxic pollutants and is located in the exhaust system of automobiles. The increasing concern about the atmospheric pollution caused due to the harmful emissions from the vehicles leads to the development of various mathematical models for the study of interphase heat-transfer problem in catalytic converter [3–6, 8,10,12,13]. One of such models is studied in [7] where the vehicle and converter temperatures are governed by a coupled system of a first order partial differential equation and an ordinary differential equation. After suitable simplifications [1,2,9], the problem reduces to the following system. ⎧ ⎪ ⎨ ⎪ ⎩

∂u ∂t ∂v ∂t

+ a ∂u ∂x + cu = cv,

t > 0, 0 < x ≤ l

+ bv = bu + λ exp(v),

t > 0, 0 < x ≤ l

u(0, t) = η, u(x, 0) = u0 (x), v(x, 0) = v0 (x), t > 0, 0 < x ≤ l.

* Corresponding author. E-mail address: [email protected] (V.A. Vijesh). http://dx.doi.org/10.1016/j.jmaa.2016.07.066 0022-247X/© 2016 Elsevier Inc. All rights reserved.

(1.1)

JID:YJMAA

AID:20628 /FLA

Doctopic: Applied Mathematics

[m3L; v1.184; Prn:11/08/2016; 13:53] P.2 (1-19)

L.A. Sunny et al. / J. Math. Anal. Appl. ••• (••••) •••–•••

2

The existence and uniqueness of the classical solution of the above coupled system has been proved using the contraction principle in [1]. Later by coupling successive iteration and monotone method, the existence and uniqueness as well as the blowup property of the solution have been discussed in [2]. Based on the main theorem in [2], a finite difference based iterative procedure has been developed in [9] to solve the coupled system numerically. The study in [9] has also proved that the finite difference scheme preserves the monotone property. It is important to note that for the numerical method in [9] based on the successive approximation discussed in [2], the performance of the numerical scheme is slow. In this paper to accelerate the iterative procedure, a modification to the iterative scheme in [2] is proposed. More specifically, by combining successive iteration and quasilinearization together with monotone method, an accelerated iterative procedure is proposed. The first part of the paper discusses about the convergence analysis, error estimate as well as the monotone property of the proposed accelerated iterative procedure for the continuous case. In the second part, based on this iterative procedure, a new iterative scheme based on finite difference method is proposed to solve the coupled system numerically. This part also proves the convergence and the monotone property of the discretized version of the iterative procedure. Moreover, a detailed error estimate is also derived. In the proposed iterative scheme, at each step one has to solve a system with variable coefficients distinct from [2] and [9] where constant coefficients are only dealt with. Consequently in the discretized case, a new comparison theorem is developed to obtain the monotone property of the sequences. This paper is organised as follows. Section 2 provides certain basic results that are used in the following sections. In Section 3, the existence and uniqueness of the coupled system (1.1) is proved via the new accelerated iterative scheme. This section also provides the error estimate for the iterative procedure. Section 4 gives the convergence analysis as well as the error estimate for the proposed numerical scheme. The convergence of the finite difference solution to the continuous solution as the mesh sizes tend to zero is obtained in Section 5. Some numerical results are given in Section 6 to illustrate the efficiency of the proposed scheme. A comparative study is also provided in this section. 2. Preliminaries In this section, some basic results are stated that will be used to obtain the results in the following sections. Definition 2.1. [11] An n × n real matrix A = (ai,j ) is said to be a Z-matrix if ai,j ≤ 0 for all i = j; 1 ≤ i, j ≤ n. An n × n matrix A that can be expressed in the form A = sI − B where B = (bi,j ) with bi,j ≥ 0 for all 1 ≤ i, j ≤ n and s ≥ ρ(B), the maximum of the moduli of the eigenvalues of B is called an M-matrix. Note: If A is an n × n real Z-matrix, then the following statements are equivalent to A being a nonsingular M-matrix. • All the principal minors of A are positive. • A is inverse positive; A−1 exists and is positive. • A is monotone; Ax ≥ 0 ⇒ x ≥ 0. For more details on M-matrix, one can refer to [11]. The existence and uniqueness theorem for (1.1) using contraction principle discussed in [1] can be stated as follows Theorem 2.1. (Theorem 6; [1]) Suppose u0 (x) = u(x, 0) ∈ C 1 [0, l] and v0 (x) = v(x, 0) ∈ C 1 [0, l] with u0 (0) = η. There is a constant tmax > 0 such that [0, tmax ) is the maximal time interval for the unique solution (u, v) of the differential equation (1.1) on the interval [0, l] × [0, tmax ).

JID:YJMAA

AID:20628 /FLA

Doctopic: Applied Mathematics

[m3L; v1.184; Prn:11/08/2016; 13:53] P.3 (1-19)

L.A. Sunny et al. / J. Math. Anal. Appl. ••• (••••) •••–•••

3

In the first part of the paper, the following lemmas are used to obtain the monotone property of the sequences. Lemma 2.1. (Lemma 1; [2]) If w ∈ C 1 (Q) satisfies the inequalities ⎧ ⎪ ⎨ ⎪ ⎩

∂w ∂t

+ a ∂w ∂x + bw ≥ 0;

w(0, t) ≥ 0

t ∈ [0, T ],

w(x, 0) ≥ 0

x ∈ [0, l],

(x, t) ∈ Q,

where a ≥ 0 and b > 0 are constants, then w ≥ 0 on Q. Lemma 2.2. Let v ∈ C(Q) be continuously differentiable with respect to t such that ∂v − f (x, t)v ≥ 0, ∂t where f (x, t) is a continuous function defined on Q with v(x, 0) ≥ 0 for 0 < x ≤ l. Then v(x, t) ≥ 0 on Q. 3. Convergence analysis for the continuous case This section provides a modification to the iterative procedure discussed in [2] which deals with variable coefficients unlike that in [2]. It also proves the convergence, error and the monotone property of the new iterative scheme. Let Q denote (0, l] × (0, T ] and let Q denote [0, l] × [0, T ] where l and T are arbitrary positive constants. Definition 3.1. A function (α, β) ∈ C 1 (Q) × C 1 (Q) is called an upper solution of (1.1) if it satisfies ⎧ ⎪ ⎪ ⎨

∂α ∂t ∂β ∂t

+ a ∂α ∂x + cα ≥ cβ,

t > 0, 0 < x ≤ l

+ bβ ≥ bα + λ exp(β), t > 0, 0 < x ≤ l ⎪ ⎪ ⎩ α(0, t) ≥ η, α(x, 0) ≥ u (x), β(x, 0) ≥ v (x), t > 0, 0 < x ≤ l. 0 0

(3.1)

Similarly (α, β) ∈ C 1 (Q) × C 1 (Q) is called a lower solution if it satisfies (3.1) with the inequalities reversed. For a given pair of ordered lower and upper solutions of (1.1), set S = {(u, v) ∈ C(Q) × C(Q) : (α, β) ≤ (u, v) ≤ (α, β)}. Using (α, β) and (α, β) respectively as the initial iterations (u0 , v 0 ), two sequences can be computed. Applying the successive approximation method to the first equation and the quasilinearization technique to the second in (1.1), an iterative scheme can be obtained as follows. ⎧ ⎪ ⎪ ⎨

∂un+1 ∂t ∂v n+1 ∂t n+1

⎪ ⎪ ⎩u

n+1

+ a ∂u∂x

+ cun+1 = cv n ,

+ (b − λ exp(v ))v n

n+1

(0, t) = η, u

n+1

= bu

t > 0, 0 < x ≤ l n+1

(x, 0) = u0 (x), v

+ λ exp(v n )(1 − v n ), n+1

t > 0, 0 < x ≤ l

(3.2)

(x, 0) = v0 (x), t > 0, 0 < x ≤ l,

where n = 0, 1, 2, · · · . Denote the sequence generated from the lower solution by {(αn+1 , β n+1 )} and the n+1 upper solution by {(αn+1 , β )} and refer to them as minimal and maximal sequences respectively. The iterative schemes respectively are given by

JID:YJMAA

AID:20628 /FLA

Doctopic: Applied Mathematics

[m3L; v1.184; Prn:11/08/2016; 13:53] P.4 (1-19)

L.A. Sunny et al. / J. Math. Anal. Appl. ••• (••••) •••–•••

4

⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩

∂αn+1 ∂t ∂β n+1 ∂t n+1

α

n+1

+ a ∂α∂x

+ (b − λ exp(β ))β n

(0, t) = η, α

∂αn+1 ∂t n+1

∂β ∂t

n+1

+ a ∂α∂x

t > 0, 0 < x ≤ l

+ cαn+1 = cβ n , n+1

n+1

= bα

(x, 0) = u0 (x), β n

+ (b − λ exp(β n ))β

n+1

t > 0, 0 < x ≤ l

(3.3)

(x, 0) = v0 (x), t > 0, 0 < x ≤ l,

t > 0, 0 < x ≤ l

+ cαn+1 = cβ , n+1

+ λ exp(β n ) − λ exp(β n )β n ,

n+1

n

n

= bαn+1 + λ exp(β ) − λ exp(β n )β ,

αn+1 (0, t) = η, αn+1 (x, 0) = u0 (x), β

n+1

t > 0, 0 < x ≤ l

(3.4)

(x, 0) = v0 (x), t > 0, 0 < x ≤ l.

Remark 3.1. When the lower and upper solutions (α, β), (α, β) ∈ C 1 (Q) × C 1 (Q), the iterative procedures (3.3) and (3.4) are well defined. For more details, one can refer to [1]. Define L1 u =

∂u ∂t

+ a ∂u ∂x + cu and L2 v =

∂v ∂t

+ bv. Then (3.2) can be rewritten as

⎧ n+1 = cv n , t > 0, 0 < x ≤ l ⎪ ⎨ L1 u L2 v n+1 − λ exp(v n )v n+1 = bun+1 + λ exp(v n )(1 − v n ), t > 0, 0 < x ≤ l ⎪ ⎩ n+1 u (0, t) = η, un+1 (x, 0) = u0 (x), v n+1 (x, 0) = v0 (x), t > 0, 0 < x ≤ l, where n = 0, 1, 2, · · · . The monotone property of both the sequences (3.3) and (3.4) and their convergence to the unique solution of (1.1) is given in the following theorem. Theorem 3.1. Let (α, β) and (α, β) be a pair of ordered lower and upper solutions of (1.1). Then n

(i) The minimal sequence {(αn , β n )} and the maximal sequence {(αn , β )} converge monotonically to the unique solution (u∗ , v ∗ ) of (1.1) in S. (ii) The relation (α, β) ≤ (αn , β n ) ≤ (αn+1 , β n+1 ) ≤ (α∗ , β ∗ ) ≤ ∗

(α∗ , β ) ≤ (αn+1 , β

n+1

n

) ≤ (αn , β ) ≤ (α, β)

(3.5)

holds for n = 1, 2, · · · . Proof. The monotone property of both the minimal and maximal sequences is obtained first. Let w0 = 0 1 1 α0 − α1 = α − α1 and z 0 = β − β = β − β . L1 w0 = L1 α − L1 α1 ≥ cβ − cβ = 0 w0 (0, t) = α(0, t) − η ≥ 0; w0 (x, 0) = α(x, 0) − u0 (x) ≥ 0. By Lemma 2.1, w0 ≥ 0 on Q. Hence α ≥ α1 . Also 1

1

L2 z 0 − λ exp(β)z 0 = (L2 β − λ exp(β)β) − (L2 β − λ exp(β)β )   1  ∂β ∂β 1 1 + bβ − λ exp(β)β − + bβ − λ exp(β)β = ∂t ∂t



 ≥ bα + λ exp(β) − λ exp(β)β − bα1 + λ exp(β) − λ exp(β)β = bα − bα1 ≥ 0 z 0 (x, 0) = β(x, 0) − v0 (x) ≥ 0.

JID:YJMAA

AID:20628 /FLA

Doctopic: Applied Mathematics

[m3L; v1.184; Prn:11/08/2016; 13:53] P.5 (1-19)

L.A. Sunny et al. / J. Math. Anal. Appl. ••• (••••) •••–•••

5

1

Hence z 0 ≥ 0 and thus (α, β) ≥ (α1 , β ) on Q. Similarly (α, β) ≤ (α1 , β 1 ). Now let w1 = α1 − α1 and 1 z1 = β − β1. L1 w1 = L1 α1 − L1 α1 = cβ − cβ ≥ 0 and w1 (0, t) = 0; w1 (x, 0) = 0. By Lemma 2.1, w1 ≥ 0 on Q. Hence α1 ≥ α1 . Also 1

1

L2 z 1 − λ exp(β)z 1 = (L2 β − λ exp(β)β ) − (L2 β 1 − λ exp(β)β 1 )   1  ∂β ∂β 1 1 1 + bβ − λ exp(β)β + bβ 1 − λ exp(β)β 1 − = ∂t ∂t  1 

1 = bα + λ exp(β) − λ exp(β)β − bα + λ exp(β) − λ exp(β)β ˜ ≥ λ exp(β)(β − β) − λ exp(β)(β − β);

β ≤ β˜ ≤ β

L2 z 1 − λ exp(β)z 1 ≥ 0 1

together with z 1 (x, 0) = 0 conclude that z 1 ≥ 0 and thus (α1 , β 1 ) ≤ (α1 , β ) on Q. The above conclusions show that 1

(α, β) ≤ (α1 , β 1 ) ≤ (α1 , β ) ≤ (α, β). Assume that n

(αn−1 , β n−1 ) ≤ (αn , β n ) ≤ (αn , β ) ≤ (αn−1 , β for some n > 1. Clearly (αn+1 , β n+1 ) and (αn+1 , β

n+1

n−1

) n

) exist. Define wn = αn − αn+1 and z n = β − β

L1 wn = L1 αn − L1 αn+1 = cβ

n−1

n+1

.

n

− cβ ≥ 0

with wn (0, t) = 0; wn (x, 0) = 0. By Lemma 2.1, wn ≥ 0 on Q. Hence αn ≥ αn+1 . Also n

n

n+1

n+1

L2 z n − λ exp(β n )z n = (L2 β − λ exp(β n )β ) − (L2 β − λ exp(β n )β )

n−1 n n−1 n = bαn + λ exp(β ) + λ exp(β n−1 )(β − β ) − λ exp(β n )β

n n − bαn+1 + λ exp(β ) − λ exp(β n )β ≥ λ exp(β

n−1

ˆ = λ exp(β)(β

n

) + λ exp(β n−1 )(β − β

n−1

n

n−1

− β ) − λ exp(β n−1 )(β

n

) − λ exp(β )

n−1

n

− β );

n

β ≤ βˆ ≤ β

n−1

L2 z n − λ exp(β n )z n ≥ 0 n

n+1

together with z n (x, 0) = 0 conclude that z n ≥ 0 and thus (αn , β ) ≥ (αn+1 , β ) on Q. A similar reasoning using the property of lower solution gives (αn , β n ) ≤ (αn+1 , β n+1 ). Now let wn+1 = αn+1 − αn+1 n+1 and z n+1 = β − β n+1 . n

L1 wn+1 = L1 αn+1 − L1 αn+1 = cβ − cβ n ≥ 0 with wn+1 (0, t) = 0; wn+1 (x, 0) = 0. By Lemma 2.1, wn+1 ≥ 0 on Q. Hence αn+1 ≥ αn+1 . Also

JID:YJMAA

AID:20628 /FLA

Doctopic: Applied Mathematics

[m3L; v1.184; Prn:11/08/2016; 13:53] P.6 (1-19)

L.A. Sunny et al. / J. Math. Anal. Appl. ••• (••••) •••–•••

6

L2 z n+1 − λ exp(β n )z n+1 = (L2 β

n+1

− λ exp(β n )β

n+1

) − (L2 β n+1 − λ exp(β)n β n+1 ) n

= b(αn+1 − αn+1 ) + λ exp(β ) − λ exp(β n ) +λ exp(β n )β n − λ exp(β n )β

n

L2 z n+1 − λ exp(β n )z n+1 ≥ 0 together with z n+1 (x, 0) = 0, conclude that z n+1 ≥ 0 and thus (αn+1 , β n+1 ) ≤ (αn+1 , β (α, β) ≤ (αn , β n ) ≤ (αn+1 , β n+1 ) ≤ (αn+1 , β

n+1

n+1

) on Q. Thus

n

) ≤ (αn , β ) ≤ (α, β)

for all n and this guarantees the existence of the limits lim (αn , β n ) = (α∗ , β ∗ );

n→∞

n

∗

lim (αn , β ) = (α∗ , β )

(3.6)

n→∞

∗

Moreover both the limits are solutions of (1.1). The uniqueness of the solution (α∗ , β ∗ ) = (α∗ , β ) = (u∗ , v ∗ ) follows from Theorem 6 of [1]. 2 Theorem 3.2. For all n ∈ N, the following error estimates hold. u∗ − αn+1  ≤ u∗ − αn 

 v ∗ − β n+1  ≤ C u∗ − αn+1  + v ∗ − β n 2

(3.7) (3.8)

for some positive constant C. Proof. From the monotone property, (3.7) trivially holds. Now define en+1 = u∗ (x, t) − αn+1 (x, t) and en+1 = v ∗ (x, t) − β n+1 (x, t). en+1 = v ∗ (x, t) − v n+1 (x, t) t

t exp(b(τ − t))e

n+1

=b

dτ + λ

0

exp(b(τ − t)) (exp(v ∗ (x, τ ) − exp(v n (x, τ ))) dτ

0

t −λ

exp(b(τ − t)) exp(v n (x, τ ))(v n+1 (x, τ ) − v n (x, τ ))dτ 0

t

t exp(b(τ − t))e

n+1

=b

exp(b(τ − t)) exp(ˆ v (x, τ ))en dτ

dτ + λ

0

0

t exp(b(τ − t)) exp(v n (x, τ ))en+1 dτ

+λ 0

t −λ

exp(b(τ − t)) exp(v n (x, τ ))en dτ ;

v n ≤ vˆ ≤ v ∗

0

t en+1 ≤ b

t exp(b(τ − t))en+1 dτ + λ

0

exp(b(τ − t)) exp(v n (x, τ ))en+1 dτ 0

JID:YJMAA

AID:20628 /FLA

Doctopic: Applied Mathematics

[m3L; v1.184; Prn:11/08/2016; 13:53] P.7 (1-19)

L.A. Sunny et al. / J. Math. Anal. Appl. ••• (••••) •••–•••

t +λ

7

exp(b(τ − t))(exp(v ∗ (x, τ )) − exp(v n (x, τ )))en dτ

0

t n+1

e

≤b

t exp(b(τ − t))e

n+1

dτ + λ

0

exp(b(τ − t)) exp(v ∗ (x, τ ))en+1 dτ

0

t +λ

exp(b(τ − t)) exp(v ∗ (x, τ ))(en )2 dτ

0

T ≤b

t exp(b(τ − t))en+1 dτ + λ

0

exp(b(τ − t)) exp(v ∗ (x, τ ))en+1 dτ

0

T +λ

exp(b(τ − t)) exp(v ∗ (x, τ ))(en )2 dτ

0

t ≤ bK1 e

n+1

 + λK2 e  + λ n 2

| exp(b(τ − t)) exp(v ∗ (x, τ ))|en+1 dτ

0

 en+1 ≤ K3 en+1  + en 2 + λ

t

| exp(b(τ − t)) exp(v ∗ (x, τ ))|en+1 dτ

0

where K3 = max{bK1 , λK2 } and K1 and K2 are positive constants. Applying Gronwall’s inequality,

e

n+1

≤ K3 en+1  + en 2

⎧ ⎨

t 1+λ

⎩

| exp(b(τ − t)) exp(v ∗ (x, τ ))|

0

⎛ t ⎞ ⎫  ⎬ exp ⎝ exp(b(s − t)) exp(v ∗ (x, s))ds⎠ dτ ⎭ τ

e

n+1



  ≤ K3 en+1  + en 2 (1 + K4 ) = C en+1  + en 2

where C = K3 (1 + K4 ) and K4 is a positive constant.

2

Remark 3.2. Similar error estimate can be obtained in the case of maximal sequence given by (3.4) also. 4. Convergence analysis for the discretized case In this section, a finite difference system is developed using Theorem 3.1 for solving the coupled equations (1.1) numerically. More specifically, the derivative terms in the iterative procedure are discretized using backward finite difference formula. This section discusses the monotone property as well as the convergence analysis of the proposed finite difference scheme. Unlike [9], variable coefficients are dealt with at each step and as a result, a new comparison lemma is obtained to prove the monotone property of the proposed scheme. The error estimate for the iterative procedure is also derived. T Let h = x = Ml , k = k = N be the space and time increment and let xi = ih, tj = jk be a mesh point in [0, l] × [0, T ] where M and N are the total numbers of intervals in [0, l] and [0, T ] respectively. The ¯ Define sets of mesh points (xi , tj ) in (0, l] × (0, T ] and [0, l] × [0, T ] are denoted respectively by Λ and Λ.

JID:YJMAA

AID:20628 /FLA

Doctopic: Applied Mathematics

[m3L; v1.184; Prn:11/08/2016; 13:53] P.8 (1-19)

L.A. Sunny et al. / J. Math. Anal. Appl. ••• (••••) •••–•••

8

ui,j = u(xi , tj ) and vi,j = v(xi , tj ). Using the backward implicit approximation for first order differential equations, (1.1) is approximated by the finite difference system ⎧ ⎪ ⎨ ⎪ ⎩

ui,j −ui,j−1 k vi,j −vi,j−1 k

+a

ui,j −ui−1,j h

+ cui,j = cvi,j ,

+ bvi,j = bui,j + λ exp(vi,j );

u0,j = ηj , ui,0 = ψi , vi,0 = φi ; i = 1, 2, . . . , M, j = 1, 2, . . . , N.

This can be rewritten as ⎧ 1 ak ⎪ ⎨ A ui,j = ckvi,j + ui,j−1 + h ui−1,j A2 vi,j = vi,j−1 + bkui,j + kλ exp(vi,j ), ⎪ ⎩ u0,j = ηj , ui,0 = ψi , vi,0 = φi ; i = 1, 2, . . . , M, j = 1, 2, . . . , N with A1 = 1 +

ak h

(4.1)

+ ck, A2 = 1 + bk.

Definition 4.1. A function (αi,j , βi,j ) defined on Λ is called a lower solution of (4.1) if it satisfies ⎧ 1 ak ⎪ ⎨ A αi,j ≤ ckβi,j + αi,j−1 + h αi−1,j A2 βi,j ≤ βi,j−1 + bkαi,j + kλ exp(βi,j ), ⎪ ⎩ α0,j ≤ ηj , αi,0 ≤ ψi , βi,0 ≤ φi ; i = 1, 2, . . . , M, j = 1, 2, . . . , N.

(4.2)

Similarly (αi,j , β i,j ) is called an upper solution if it satisfies (4.2) with inequalities reversed. For a given pair of ordered lower and upper solutions (αi,j , βi,j ) and (αi,j , β i,j ) of (4.1), set S = {(ui,j , vi,j ) ∈ R2 : (αi,j , βi,j ) ≤ (ui,j , vi,j ) ≤ (αi,j , β i,j )}. 0 As explained in Section 3, using (αi,j , βi,j ) and (αi,j , β i,j ) respectively as the initial iterations (u0i,j , vi,j ), two sequences can be constructed by applying successive approximation and quasilinearization technique to the first and second equations of (4.1) respectively which yields

⎧ ⎪ ⎪ ⎨

n+1 un+1 i,j −ui,j−1 k n+1 n+1 vi,j −vi,j−1 ⎪ k ⎪ ⎩ n+1 u0,j = ηj ,

+a

n+1 un+1 i,j −ui−1,j h

n + cun+1 i,j = cvi,j ,

n+1  n+1 n n n + bvi,j = bun+1 i,j + λ exp(vi,j ) + λ exp(vi,j ) vi,j − vi,j , n+1 un+1 i,0 = ψi , vi,0 = φi ; i = 1, 2, . . . , M, j = 1, 2, . . . , N ,

with n = 0, 1, . . . where ηj = η(tj ), ψi = ψ(xi ) and φi = φ(xi ). The above system can be written in the form ⎧ n+1 ak n+1 n ⎪ A1 un+1 i,j − ckvi,j = ui,j−1 + h ui−1,j ⎪ ⎨ n+1 n n+1 n n n (4.3) −bkun+1 i,j + Bi,j vi,j = vi,j−1 + kλ exp(vi,j ) − kλ exp(vi,j )vi,j , ⎪ ⎪ ⎩ n+1 n+1 u0,j = ηj , un+1 i,0 = ψi , vi,0 = φi ; i = 1, 2, . . . , M, j = 1, 2, . . . , N ,

 n n with n = 0, 1, . . . and Bi,j = A2 − kλ exp(vi,j ) . Equivalently,  n+1  ⎧ ak n+1 n un+1 ⎪ i,j−1 + h ui−1,j + ckvi,j ⎨ An ui,j = n+1 n+1 n n n vi,j vi,j−1 + kλ exp(vi,j ) − kλ exp(vi,j )vi,j ⎪ ⎩ n+1 n+1 n+1 u0,j = ηj , ui,0 = ψi , vi,0 = φi ; i = 1, 2, . . . , M, j = 1, 2, . . . , N ,

(4.4)

JID:YJMAA

AID:20628 /FLA

Doctopic: Applied Mathematics

[m3L; v1.184; Prn:11/08/2016; 13:53] P.9 (1-19)

L.A. Sunny et al. / J. Math. Anal. Appl. ••• (••••) •••–•••

 with n = 0, 1, · · · and An =

A1 0 n −bk Bi,j

9

. Throughout this section, assume that the time step k satisfies

the condition (3.7) in [9], i.e.; a 1 > max{b − c − , c − b + λ exp(ξ ∗ )} k h

(4.5)

¯ Denote the sequence generated from the lower solution by {(αn+1 , β n+1 )} where ξ ∗ = max{β i,j : (i, j) ∈ Λ}. i,j i,j n+1

and the upper solution by {(αn+1 i,j , β i,j )} and refer to them as minimal and maximal sequences respectively.

2 n n With Bi,j = A − kλ exp(βi,j ) , the iterative schemes respectively are constructed by ⎧ 1 n+1 n+1 ak n+1 n ⎪ ⎨ A αi,j = ckβi,j + αi,j−1 + h αi−1,j n+1 n+1 n n+1 n n n Bi,j βi,j = bkαi,j + βi,j−1 + kλ exp(βi,j ) − kλ exp(βi,j )βi,j , ⎪ ⎩ n+1 n+1 n+1 α0,j = ηj , αi,0 = ψi , βi,0 = φi ; i = 1, 2, . . . , M, j = 1, 2, . . . , N , ⎧ n n+1 ak n+1 A1 αn+1 ⎪ i,j = ckβ i,j + αi,j−1 + h αi−1,j ⎪ ⎨ n+1 n n n n+1 n Bi,j β i,j = bkαn+1 i,j + β i,j−1 + kλ exp(β i,j ) − kλ exp(βi,j )β i,j , ⎪ ⎪ n+1 ⎩ n+1 α0,j = ηj , αn+1 i,0 = ψi , β i,0 = φi ; i = 1, 2, . . . , M, j = 1, 2, . . . , N.

(4.6)

(4.7)

The following lemma is vital to prove the monotone property of the sequences. ¯ b1 , b2 , c1 and c2 ≥ 0. If wi,j and zi,j satisfy Lemma 4.1. Let a1 , ai,j > 0 for all (i, j) ∈ Λ, a1 wi,j − b1 wi−1,j − c1 wi,j−1 ≥ 0;

(i, j) ∈ Λ

(4.8)

ai,j zi,j − b2 wi,j − c2 zi,j−1 ≥ 0;

(i, j) ∈ Λ

(4.9)

with w0,j ≥ 0, wi,0 ≥ 0, zi,0 ≥ 0

(4.10)

¯ then wi,j ≥ 0 and zi,j ≥ 0 for all (i, j) ∈ Λ. Proof. The proof is by an induction process. (4.8) gives a1 wi,j ≥ b1 wi−1,j + c1 wi,j−1 Let i = 1 and using (4.10), a1 w1,j ≥ b1 w0,j + c1 w1,j−1 a1 w1,j ≥ c1 w1,j−1

(4.11)

Now for i = 1 and using (4.11) in (4.9), a1,j z1,j ≥

b2 c1 w1,j−1 + c2 z1,j−1 a1

Hence for j = 1 and using (4.10), one can conclude that,

(4.12)

JID:YJMAA

AID:20628 /FLA

Doctopic: Applied Mathematics

[m3L; v1.184; Prn:11/08/2016; 13:53] P.10 (1-19)

L.A. Sunny et al. / J. Math. Anal. Appl. ••• (••••) •••–•••

10

a1,1 z1,1 ≥

b2 c1 w1,0 + c2 z1,0 ≥ 0 ⇒ z1,1 ≥ 0 a1

Similarly for j = 1 in (4.11), w1,1 ≥ 0. Assume that w1,j ≥ 0 and z1,j ≥ 0 for j = n − 1. For j = n, from (4.12) a1,n z1,n ≥

b2 c1 w1,n−1 + c2 z1,n−1 ≥ 0 ⇒ z1,n ≥ 0 a1

Similarly for j = n in (4.11), w1,n ≥ 0. Thus w1,j ≥ 0 and z1,j ≥ 0 for all j. Now assume that wi,j ≥ 0 and zi,j ≥ 0 for all j and i = n − 1. For i = n in (4.8) a1 wn,j ≥ b1 wn−1,j + c1 wn,j−1 a1 wn,j ≥ c1 wn,j−1

(4.13)

For i = n in (4.9) and using (4.13), an,j zn,j ≥

b2 c1 wn,j−1 + c2 zn,j−1 a1

(4.14)

For j = 1 in (4.14) and by using (4.10), one can conclude that zn,1 ≥ 0. Similarly, wn,1 ≥ 0. Assume that wn,j ≥ 0 and zn,j ≥ 0 for j = k − 1. For j = k from (4.14), an,k zn,k ≥

b2 c1 wn,k−1 + c2 zn,k−1 ≥ 0 ⇒ zn,k ≥ 0 a1

Similarly, from (4.13), a1 wn,k ≥ c1 wn,k−1 ⇒ wn,k ≥ 0. ¯ 2 Thus wi,j ≥ 0 and zi,j ≥ 0 for all (i, j) ∈ Λ. Theorem 4.1. Let (αi,j , βi,j ) and (αi,j , β i,j ) be a pair of ordered lower and upper solutions of (4.1). Then the following statements hold: (i) The iterative schemes (4.6) and (4.7) are well defined. n n n (ii) The minimal sequence {(αi,j , βi,j )} and the maximal sequence {(αni,j , β i,j )} converge monotonically to ∗ the unique solution (u∗i,j , vi,j ) of (4.1) in S. (iii) The relation n+1 n+1 n n ∗ ∗ (αi,j , βi,j ) ≤ (αi,j , βi,j ) ≤ (αi,j , βi,j ) ≤ (αi,j , βi,j )≤ ∗

n+1

n

n (α∗i,j , β i,j ) ≤ (αn+1 i,j , β i,j ) ≤ (αi,j , β i,j ) ≤ (αi,j , β i,j )

(4.15)

holds for every (i, j) ∈ Λ¯ and n = 1, 2, · · · . Proof. The proof is given by induction on n. For n = 0, (αi,j , βi,j ) ∈ S and using (4.5) one can conclude 1 1 1 that A0 is invertible. Hence (αi,j , βi,j ) and (α1i,j , β i,j ) exist for all i and j. Note that since (αi,j , βi,j ) ∈ S, 0 0 1 (4.5) ensures that Bi,j = A2 − kλ exp(βi,j ) > A2 − kλ exp(ξ ∗ ) > 0 for all i and j. Let wi,j = αi,j − αi,j and 0 1 zi,j = βi,j − βi,j . Consider

JID:YJMAA

AID:20628 /FLA

Doctopic: Applied Mathematics

[m3L; v1.184; Prn:11/08/2016; 13:53] P.11 (1-19)

L.A. Sunny et al. / J. Math. Anal. Appl. ••• (••••) •••–•••

ak 0 ak 1 ak 1 w α αi−1,j = A1 αi,j − A1 αi,j − + h i−1,j h i−1,j h ak 1 ak 1 ak 1 α α αi−1,j = ckβi,j + αi,j−1 + − A1 αi,j − + h i−1,j h i−1,j h ak 0 ak 1 wi−1,j = ckβi,j + αi,j−1 αi−1,j − − A1 αi,j + h h

11

0 A1 wi,j −

0 A1 wi,j

(4.16)

and 0 0 1 1 0 Bi,j zi,j = bkαi,j + βi,j−1 + kλ exp(βi,j ) (1 − βi,j ) − Bi,j βi,j .

(4.17)

From (4.16) for j = 1, ak 0 ak 1 w αi−1,1 = ckβi,1 + αi,0 − A1 αi,1 + h i−1,1 h ak αi−1,1 ≥ ckβi,1 + αi,0 − A1 αi,1 + h ak 0 0 w A1 wi,1 − ≥ 0. h i−1,1 0 A1 wi,1 −

(4.18)

From (4.17) for j = 1, 0 0 1 1 0 Bi,1 zi,1 = bkαi,1 + βi,0 + kλ exp(βi,1 ) (1 − βi,1 ) − Bi,1 βi,1 1 ≥ bkαi,1 + bkαi,1 − bkαi,1 + βi,0 + kλ exp(βi,1 ) (1 − βi,1 ) − B 0 βi,1 0 0 0 Bi,1 zi,1 ≥ bkwi,1 .

(4.19)

0 1 0 1 From the boundary and initial conditions, w0,1 = α0,1 −α0,1 = η1 −α0,1 ≥ 0; zi,0 = βi,0 −βi,0 = φi −βi,0 ≥ 0 0 1 and wi,0 = αi,0 − αi,0 = ψi − αi,0 ≥ 0. For i = 1, (4.18) and (4.19) give

ak 0 0 w ≥ 0 ⇒ w1,1 ≥ 0, h 0,1 0 0 ≥ bkw1,1 ⇒ z1,1 ≥ 0.

0 A1 w1,1 ≥ 0 0 B1,1 z1,1

For i = 2, (4.18) and (4.19) give ak 0 0 w ≥ 0 ⇒ w2,1 ≥ 0, h 1,1 0 0 ≥ bkw2,1 ⇒ z2,1 ≥ 0.

0 A1 w2,1 ≥ 0 0 B2,1 z2,1

0 0 1 1 Proceeding like this, one can prove that wi,1 ≥ 0 and zi,1 ≥ 0 for all i, i.e., αi,1 ≤ αi,1 and βi,1 ≤ βi,1 for all i. From (4.16) for j = 2,

ak 0 ak 1 w αi−1,2 = ckβi,2 + αi,1 − A1 αi,2 + h i−1,2 h ak αi−1,2 ≥ ckβi,2 + αi,1 − A1 αi,2 + h ak 0 0 w A1 wi,2 − ≥ 0. h i−1,2 0 A1 wi,2 −

From (4.17) for j = 2,

(4.20)

JID:YJMAA

AID:20628 /FLA

Doctopic: Applied Mathematics

[m3L; v1.184; Prn:11/08/2016; 13:53] P.12 (1-19)

L.A. Sunny et al. / J. Math. Anal. Appl. ••• (••••) •••–•••

12

0 0 1 1 0 Bi,2 zi,2 = bkαi,2 + βi,1 + kλ exp(βi,2 ) (1 − βi,2 ) − Bi,2 βi,2 1 0 ≥ bkαi,2 + bkαi,2 − bkαi,2 + βi,1 + kλ exp(βi,2 ) (1 − βi,2 ) − Bi,2 βi,2 0 0 0 Bi,2 zi,2 ≥ bkwi,2 .

(4.21)

0 0 0 From the boundary and initial conditions, w0,2 ≥ 0; zi,0 ≥ 0 and wi,0 ≥ 0. For i = 1, (4.20) and (4.21) give

ak 0 0 w ≥ 0 ⇒ w1,2 ≥ 0, h 0,2 0 0 ≥ bkw1,2 ⇒ z1,2 ≥ 0.

0 A1 w1,2 ≥ 0 0 z1,2 B1,2

For i = 2, (4.20) and (4.21) give ak 0 0 w ≥ 0 ⇒ w2,2 ≥ 0, h 1,2 0 0 ≥ bkw2,2 ⇒ z2,2 ≥ 0.

0 A1 w2,2 ≥ 0 0 z2,2 B2,2

0 0 1 1 Proceeding like this, one can prove that wi,2 ≥ 0 and zi,2 ≥ 0 for all i, i.e., αi,2 ≤ αi,2 and βi,2 ≤ βi,2 for 1 1 ¯ all i. Repeating the similar argument for j = 3, . . . , N leads to αi,j ≤ αi,j and βi,j ≤ βi,j for all (i, j) ∈ Λ. 1

1

1 1 1 1 Similarly, αi,j ≥ α1i,j and β i,j ≥ β i,j for every i and j. Now let wi,j = α1i,j − αi,j and zi,j = β i,j − βi,j .

1 A1 wi,j = ckβ i,j + α1i,j−1 + 1 1 Thus A1 wi,j − wi,j−1 −

ak 1 h wi−1,j

ak 1 ak 1 1 α αi−1,j − ckβi,j − αi,j−1 − h h i−1,j

≥ 0. Now

 0 1 1 1 Bi,j zi,j = bkwi,j + zi,j−1 + kλ exp(β i,j ) − kλ exp(βi,j ) − kλ exp(βi,j ) β i,j − βi,j  



0 1 1 1 Bi,j zi,j = bkwi,j + zi,j−1 + kλ exp(˜ vi,j ) β i,j − βi,j − kλ exp(βi,j ) β i,j − βi,j ; βi,j ≤ v˜i,j ≤ β i,j 0 1 1 1 1 1 1 1 Thus Bi,j zi,j − bkwi,j − zi,j−1 ≥ 0. Note that w0,j = 0; zi,0 = 0; wi,0 = 0. Hence by Lemma 4.1, αi,j ≤ α1i,j 1 ¯ The above conclusions show that and β 1 ≤ β for all (i, j) ∈ Λ. i,j

i,j

1

1 1 (αi,j , βi,j ) ≤ (αi,j , βi,j ) ≤ (α1i,j , β i,j ) ≤ (αi,j , β i,j ).

(4.22)

n

n n Assume by induction that (αi,j , βi,j ) and (αni,j , β i,j ) exist and n

n−1

n−1 n−1 n n (αi,j , βi,j ) ≤ (αi,j , βi,j ) ≤ (αi,j , βi,j ) ≤ (αni,j , β i,j ) ≤ (αn−1 i,j , β i,j ) ≤ (αi,j , β i,j )

(4.23)

for some n > 1. From induction hypothesis and using (4.5), one can conclude that An is invertible. Consen+1 n+1 n+1 quently, (αi,j , βi,j ) and (αn+1 i,j , β i,j ) exist for all i and j. The induction hypothesis and (4.5) also confirm n+1 n+1 n n n n n that Bi,j > 0 for all i and j. Let wi,j = αi,j − αi,j and zi,j = βi,j − βi,j . Consider n A1 wi,j −

n A1 wi,j −

ak n ak n+1 ak n n+1 n w α α = A1 αi,j − A1 αi,j − + h i−1,j h i−1,j h i−1,j n+1 n−1 n n = ckβi,j + αi,j−1 − ckβi,j − αi,j−1 ak n n w ≥ wi,j−1 h i−1,j

JID:YJMAA

AID:20628 /FLA

Doctopic: Applied Mathematics

[m3L; v1.184; Prn:11/08/2016; 13:53] P.13 (1-19)

L.A. Sunny et al. / J. Math. Anal. Appl. ••• (••••) •••–•••

n Thus A1 wi,j −

ak n h wi−1,j

13

n ¯ Moreover − wi,j−1 ≥ 0 for all (i, j) ∈ Λ.

 n+1 n+1 n+1 n n n A2 zi,j = bkαi,j + βi,j−1 + kλ exp(βi,j ) 1 + βi,j − βi,j

 n−1 n−1 n n n − βi,j−1 − kλ exp(βi,j ) 1 + βi,j − βi,j − bkαi,j n−1 n n n n n = bkwi,j + zi,j−1 + kλ exp(βi,j )zi,j + kλ exp(βi,j ) − kλ exp(βi,j ) 

n−1 n−1 n − kλ exp(βi,j ) βi,j − βi,j

n  n−1 n n n n = bkwi,j + zi,j−1 + kλ exp(βi,j )zi,j + kλ exp(˜ vi,j ) βi,j − βi,j

n  n−1 n−1 n−1 n ) βi,j − βi,j ; βi,j ≤ v˜i,j ≤ βi,j − kλ exp(βi,j

n A2 zi,j

n n n n Bi,j zi,j ≥ bkwi,j + zi,j−1 n n n n ¯ Moreover, wn = 0; z n = 0; wn = 0. Hence by Thus Bi,j zi,j − bkwi,j − zi,j−1 ≥ 0 for all (i, j) ∈ Λ. 0,j i,0 i,0 n+1 n+1 n n n n ¯ A similar Lemma 4.1, wi,j ≥ 0 and zi,j ≥ 0 or equivalently αi,j ≤ αi,j and βi,j ≤ βi,j for all (i, j) ∈ Λ. n+1

n

n+1

n+1 n+1 n+1 n argument gives (αn+1 i,j , β i,j ) ≤ (αi,j , β i,j ) and (αi,j , βi,j ) ≤ (αi,j , β i,j ). Thus n+1

n

n+1 n+1 n n n (αi,j , βi,j ) ≤ (αi,j , βi,j ) ≤ (αi,j , βi,j ) ≤ (αn+1 i,j , β i,j ) ≤ (αi,j , β i,j ) ≤ (αi,j , β i,j )

this guarantees the existence of the limits n

n n ∗ ∗ lim (αi,j , βi,j ) = (αi,j , βi,j );

∗

lim (αni,j , β i,j ) = (α∗i,j , β i,j )

n→∞

(4.24)

n→∞

Hence (4.15) holds and in the limiting case, both the limits are solutions of (1.1). The proof for the uniqueness of the solution follows from Theorem 3.1 in [9]. 2 The following remark is similar to Remark 3.1(b) in [9]. Remark 4.1. Theorem 4.1 holds true for the more general system ⎧ ⎪ ⎨ ⎪ ⎩

∂u ∂t ∂v ∂t

+ a ∂u ∂x + cu = f (x, t, u, v), + bv = g(x, t, u, v),

t > 0, 0 < x ≤ l

t > 0, 0 < x ≤ l

u(0, t) = η(t), u(x, 0) = u0 (x), v(x, 0) = v0 (x), t > 0, 0 < x ≤ l,

where f (x, t, u, v) and g(x, t, u, v) are continuous and C 1 -functions of (u, v) such that fu (x, t, u, v), fv (x, t, u, v), gu (x, t, u, v) are positive and gv (x, t, u, v) is nondecreasing and Lipschitz with respect to u and v for α ≤ u ≤ α and β ≤ v ≤ β. The monotone iteration process is given by ⎧ ⎪ ⎪ ⎨

∂un+1 ∂t ∂v n+1 ∂t n+1

⎪ ⎪ ⎩u

n+1

+ a ∂u∂x

+ cun+1 = f (x, t, un , v n ),

t > 0, 0 < x ≤ l

+ bv n+1 = g(x, t, un+1 , v n ) + gv (x, t, un+1 , v n )(v n+1 − v n ), n+1

(0, t) = η, u

(x, 0) = u0 (x), v

n+1

t > 0, 0 < x ≤ l

(x, 0) = v0 (x), t > 0, 0 < x ≤ l.

∗ Theorem 4.2. If (u∗i,j , vi,j ) for all (i, j) ∈ Λ¯ is the solution of (4.1), then there exists a positive constant C such that         en+1 + ak en+1     en+1  ckeni,j       i,j i,j−1 h i−1,j

 ≤ C +       n+1 n 2 ∗       en+1 e kλ exp(ξ ) e i,j i,j−1 i,j ∞

∞

n+1 n+1 n+1 ∗ ∗ ¯ where en+1 and ξ ∗ = max{β i,j : (i, j) ∈ Λ}. i,j = ui,j − αi,j ; ei,j = vi,j − βi,j

∞

JID:YJMAA

AID:20628 /FLA

Doctopic: Applied Mathematics

[m3L; v1.184; Prn:11/08/2016; 13:53] P.14 (1-19)

L.A. Sunny et al. / J. Math. Anal. Appl. ••• (••••) •••–•••

14

Proof. From (4.6), n+1 n+1 n A1 αi,j = ckβi,j + αi,j−1 +

ak n+1 α h i−1,j

(4.25)



n+1 n+1 n+1 n+1 n n n = bkαi,j + βi,j−1 + kλ exp(βi,j ) + kλ exp(βi,j ) βi,j − βi,j . A2 βi,j

(4.26)

∗ If (u∗i,j , vi,j ) is the solution of (4.1) in S, then

∗ A1 u∗i,j = ckvi,j + u∗i,j−1 +

ak ∗ u h i−1,j

(4.27)

∗ ∗ ∗ A2 vi,j = bku∗i,j + vi,j−1 + kλ exp(vi,j ).

(4.28)

Hence, n+1 A1 en+1 i,j − ei,j−1 −

ak n+1 e = ckeni,j . h i−1,j

Similarly,



n+1  n+1 n+1 ∗ n n n A2 en+1 i,j = bkei,j + ei,j−1 + kλ exp(vi,j ) − exp(βi,j ) − kλ exp(βi,j ) βi,j − βi,j 

∗ n+1 n+1 n ˆ A2 en+1 i,j = bkei,j + ei,j−1 + kλ exp(βi,j ) vi,j − βi,j 

n+1 n ∗ ∗ n n ∗ ; βi,j ) βi,j − vi,j + vi,j − βi,j ≤ βˆi,j ≤ vi,j − kλ exp(βi,j n+1 n+1 n n n ˆ B n en+1 i,j = bkei,j + ei,j−1 + kλ exp(βi,j )ei,j − kλ exp(βi,j )ei,j

 n+1 n ∗ n ≤ bken+1 i,j + ei,j−1 + kλei,j exp(vi,j ) − exp(βi,j )

n 2 n+1 n+1 ∗ B n en+1 i,j ≤ bkei,j + ei,j−1 + kλ exp(ξ ) ei,j

Thus, 

en+1 i,j en+1 i,j

A

n



 −

ak n+1 en+1 i,j−1 + h ei−1,j en+1 i,j−1



 ≤

ckeni,j

2 kλ exp(ξ ∗ ) eni,j

(4.29)

Note that for each n ∈ N, An is a nonsingular M-matrix. Hence (4.29) can be written as 

en+1 i,j en+1 i,j



 n −1

≤ (A )



ak n+1 en+1 i,j−1 + h ei−1,j en+1 i,j−1

 +

ckeni,j

2 kλ exp(ξ ∗ ) eni,j



Consequently,    en+1    i,j     en+1 i,j

∞

   en+1 + ak en+1    i,j−1 i−1,j h ≤C   n+1   ei,j−1

∞

     ckeni,j  

 +  2  kλ exp(ξ ∗ ) eni,j  ∞

1 b where C = max{1, 1+bk−kλe 2 ξ∗ + c }.

Remark 4.2. Similar error estimate can be obtained in the case of maximal sequence given by (4.7) also.

JID:YJMAA

AID:20628 /FLA

Doctopic: Applied Mathematics

[m3L; v1.184; Prn:11/08/2016; 13:53] P.15 (1-19)

L.A. Sunny et al. / J. Math. Anal. Appl. ••• (••••) •••–•••

15

5. Convergence of finite difference solutions ∗ In this section, the convergence of (u∗i,j , vi,j ) to the continuous solution (u∗ (xi , tj ), v ∗ (xi , tj )) as the mesh size tends to zero is obtained. The following theorem is similar to the Theorem 5.1 in [9]. ∗ ) be the respective solutions of (1.1) and (4.1) respectively Theorem 5.1. Let (u∗ (x, t), v ∗ (x, t)) and (u∗i,j , vi,j ¯ = [0, l] × [0, T ]. Then and let Λ¯ be a given partition of Q ∗ (u∗i,j , vi,j ) → (u∗ (xi , tj ), v ∗ (xi , tj )) as h + k → 0

¯ at every mesh point (xi , tj ) in Λ. Proof. To prove this theorem, for given any > 0 it has to be shown that there exists δ > 0 such that ∗ |u∗ (xi , tj ) − u∗i,j | + |v ∗ (xi , tj ) − vi,j | < when h + k < δ.

(5.1)

0 0 Let (α0 , β 0 ) = (αi,j , βi,j ) = (0, 0) for both the minimal sequences given by (3.2) and (4.3) respectively. By Theorems 3.1 and 4.1, there exists an integer n = n∗ ( ) such that

3

n+1 − βi,j |< 3

|u∗ − αn+1 | + |v ∗ − β n+1 | < n+1 ∗ | + |vi,j |u∗i,j − αi,j

where (i, j) ∈ Λ¯ for all n ≥ n∗ . Note that ∗

∗

∗

∗

∗

∗

∗

n n |u∗ (xi , tj ) − u∗i,j | ≤ |u∗ (xi , tj ) − αn (xi , tj )| + |αn (xi , tj ) − αi,j | + |αi,j − u∗i,j | ∗

∗ n n ∗ | ≤ |v ∗ (xi , tj ) − β n (xi , tj )| + |β n (xi , tj ) − βi,j | + |βi,j − vi,j |. |v ∗ (xi , tj ) − vi,j

Hence the proof is complete if one can prove that ∗

∗

∗

∗

n n |αn (xi , tj ) − αi,j | + |β n (xi , tj ) − βi,j |<

¯ ; (i, j) ∈ Λ. 3

(5.2)

From (3.2) and (4.3), it can be seen that (αn+1 (xi , tj ), β n+1 (xi , tj )) satisfies the equations ⎧ 1 n+1 n+1 (xi , tj ) − ckβ n (xi , tj ) = αn+1 (xi , tj−1 ) + ak (xi−1 , tj ) + o(h, k) A α ⎪ h α ⎪ ⎪ ⎪ ⎨ B n β n+1 (xi , tj ) − bkαn+1 (xi , tj ) = β n+1 (xi , tj−1 ) + ⎪ kλ exp(β n (xi , tj )) (1 − β n (xi , tj )) + o(h, k) ⎪ ⎪ ⎪ ⎩ n+1 α (0, tj ) = ηj , αn+1 (xi , 0) = ψ(xi ), β n+1 (xi , 0) = φ(xi )

(5.3)

where o(h, k) → 0 as h + k → 0. Let n+1 n+1 n+1 n+1 en+1 (xi , tj ) − αi,j ; en+1 (xi , tj ) − βi,j . i,j = α i,j = β

(5.4)

Subtracting (4.3) from (5.3) and using mean value theorem, ⎧ 1 n+1 n+1 ak n+1 n ⎪ ⎨ A ei,j − ckei,j = ei,j−1 + h ei−1,j + o(h, k) n+1 n+1 ˆn ˆn n B n en+1 i,j − bkei,j = ei,j−1 − kλβi,j exp(βi,j )ei,j + o(h, k) ⎪ ⎩ n+1 n+1 e0,j = 0, en+1 i,0 = 0, ei,0 = 0

(5.5)

JID:YJMAA

AID:20628 /FLA

Doctopic: Applied Mathematics

[m3L; v1.184; Prn:11/08/2016; 13:53] P.16 (1-19)

L.A. Sunny et al. / J. Math. Anal. Appl. ••• (••••) •••–•••

16

n n where βˆi,j is an intermediate value between β n (xi , tj ) and βi,j . Define column vectors Ejn+1 and E n+1 j respectively by

⎧

T ⎪ ⎨ E n+1 = en+1 , en+1 , . . . , en+1 j 1,j 2,j M,j

T ⎪ n+1 n+1 ⎩ E n+1 = en+1 j 1,j , e2,j , . . . , eM,j

(5.6)

where (·)T denotes the transpose of a row vector. Let A be an M × M bidiagonal matrix and Djn a diagonal matrix given by ⎞

⎛

A1 ⎜ − ak h ⎜ ⎜ ⎜ A = (ajk ) = ⎜ ⎜ ⎜ ⎜ ⎝

A1 − ak h

A .

1

. .

. . − ak h

⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ . ⎠ A1



n n n n n n Djn = diag −βˆ1,j exp(βˆ1,j ), −βˆ2,j exp(βˆ2,j ), . . . , −βˆM,j exp(βˆM,j ) Using the same argument in [9], one can conclude that A−1 exists and is a nonnegative matrix. Also A has the positive smallest eigenvalue μ0 = A1 . (5.5) can be written as ⎧ n+1 n+1 = Ej−1 + ckE nj + O(h, k) ⎪ ⎨ AEj n+1 B n E n+1 = E n+1 + kλDjn E nj + O(h, k) j j−1 + bkEj ⎪ ⎩ n+1 E0 = 0, E n+1 =0 0 where O(h, k) → 0 for any suitable norm  ·  in RM as h + k → 0. Define σ1 = ξ ∗ exp(ξ ∗ ), σ0 = kλσ1 ˜ = 1 + bk − kλ exp(ξ ∗ ) where ξ ∗ = max{β i,j : (i, j) ∈ Λ}. ¯ Proceeding as in Theorem 5.1 in [9], one and B will end up with  n+1  ⎧ n+1 n 2 ⎪ ⎨ Ej  ≤ μ0  Ej−1  + ckE j  + O(h, k)  n+1 E n+1  ≤ B1˜ E n+1  + σ0 E nj  + O(h, k) j j−1  + bkEj ⎪ ⎩ n+1 E0  = E n+1 =0 0

(5.7)

2ck σ0 Define Sjn+1 = Ejn+1  + E n+1 , γ1 = max{ μ20 , B1˜ }, γ2 = bk ˜ and γ3 = max{ μ0 , B ˜ }. Since h + k → 0, one j B n+1 ∗ can choose k such that kλ exp(ξ ) < 1. Consequently, γ2 < 1. Note that S0 = Sj0 = 0 for all n and j. Proceeding similar to Theorem 5.1 in [9], for given any 1 > 0, there exists δ1 > 0 such that



n+1 (1 − γ2 ) Sjn+1 ≤ γ1 Sj−1 + γ3 Sjn + 1 when h + k < δ1

S0n+1 = Sj0 = 0; j = 1, 2, . . . , N, n = 0, 1, . . . , n∗ Define β ∗ = 1 +

γ3 1−γ2

γ3 γ3 + ( 1−γ )2 + . . . + ( 1−γ )n 2 2

Sjn+1

β∗ ≤ (1 − γ2 )



γ1 β ∗ 1 − γ2

∗

−1

. An induction argument in j leads to

j−1

 +

γ1 β ∗ 1 − γ2

j−2

 γ1 β ∗ + ··· + + 1 1 1 − γ2

(5.8)

JID:YJMAA

AID:20628 /FLA

Doctopic: Applied Mathematics

[m3L; v1.184; Prn:11/08/2016; 13:53] P.17 (1-19)

L.A. Sunny et al. / J. Math. Anal. Appl. ••• (••••) •••–•••

17

 ∗ N −1 ∗ N −2 ∗ β∗ γ1 β γ1 β for all n ≤ n∗ , j ≤ N . Thus Sjn ≤ K 1 for all j = 1, 2, . . . , N where K = (1−γ + 1−γ + 1−γ2 2) 2  ∗ ∗ ∗ γ1 β ∗ n , there exists δ > 0 such that Sjn = Ejn  + E j  < 3 when · · · + 1−γ2 + 1 . For the choice of 1 < 3K h + k < δ. This leads to (5.2). Thus (5.1) holds and hence the theorem. 2 6. Numerical examples In this section, the accelerated iterative technique is illustrated by applying to different examples. The existence and uniqueness of the solution and the convergence of the proposed examples are followed by Theorem 4.1 and Theorem 5.1 respectively. The iterative schemes taken for the numerical solution of the examples are same as the iterative scheme discussed in Remark 4.1. Throughout this section, n denotes the number of iterations required for the stopping criteria   n+1 n n max |un+1 i,j − ui,j | + |vi,j − vi,j | ≤ . (i,j)

Example 6.1. Consider the following differential system discussed in [9]. ⎧ ⎪ ⎪ ⎪ ⎪ ⎨

∂u ∂t ∂v ∂t

+

∂u ∂x

+ u = v + q1 (x, t),

0 < x ≤ 1, 0 < t ≤ T 0 < x ≤ 1, 0 < t ≤ T

+ v = u + λ exp(v) + q2 (x, t),

⎪ u(0, t) = 2 − exp(−t), 0 < t ≤ T ⎪ ⎪ ⎪ ⎩ u(x, 0) = 1 − x2 , v(x, 0) = 1 − x2 ,

(6.1)

0 < x ≤ 1,

where λ > 0 is considered as a parameter with q1 (x, t) = (1 − x)2 exp(−t) and q2 (x, t) = (1 + x2 ) exp(−t) − λ exp(2 − (1 + x2 ) exp(−t)). The solution of (6.1) is given by u(x, t) = v(x, t) = 2 − (1 + x2 ) exp(−t). Numerical results for the minimal solution and the exact solution are given in Table 1 and Table 2. From Table 1 and Table 2, one can conclude that the proposed scheme performs faster than the scheme in [9]. Here T = 1, λ = 0.05, h = k = 10−3 and = 2 × 10−5 . Example 6.2. Consider the following differential system discussed in [9]. ⎧ ⎪ ⎪ ⎪ ⎪ ⎨

∂u ∂t ∂v ∂t

+

∂u ∂x

+ u = λ1 v p1 + q1 (x, t),

+ v = u + λ2 v

p2

+ q2 (x, t),

0 < x ≤ 1, 0 < t ≤ T 0 < x ≤ 1, 0 < t ≤ T

⎪ u(0, t) = 2 − exp(−t), 0 < t ≤ T ⎪ ⎪ ⎪ ⎩ u(x, 0) = 1 − x, v(x, 0) = 1 − x, 0 < x ≤ 1,

(6.2)

where λi , pi > 0 for i = 1, 2 with q1 (x, t) = (2 − exp(−t)) − λ1 [2 − (1 + x) exp(−t)]p1 and q2 (x, t) = (1 + x) exp(−t) −λ2 [2 −(1 +x) exp(−t)]p2 . The solution of (6.2) is given by u(x, t) = v(x, t) = 2 −(1 +x) exp(−t). Numerical results for the minimal solution and the exact solution are given in Table 3 and Table 4 for the 1 choice of T = 1, h = k = 10−3 , p1 = 2, p2 = 3, λ1 = 14 , λ2 = 8e and = 2 × 10−5 . From Table 3 and Table 4, one can conclude that the proposed scheme outperforms the scheme in [9]. 7. Conclusion The existence and uniqueness of the solution for a coupled system of partial differential equations is established through an accelerated iterative scheme. The monotonicity, convergence and the error estimate of the sequences obtained from both continuous and discrete cases are also obtained. The efficiency of the proposed scheme is proved by comparing with the existing scheme available in the literature.

JID:YJMAA 18

AID:20628 /FLA

Doctopic: Applied Mathematics

[m3L; v1.184; Prn:11/08/2016; 13:53] P.18 (1-19)

L.A. Sunny et al. / J. Math. Anal. Appl. ••• (••••) •••–•••

Table 1 Numerical solution of u(x, t) for Example 6.1. Grid point

Exact

(xi , tj ) (0.4, 0.2) (0.8, 0.2) (0.4, 0.4) (0.8, 0.4) (0.4, 0.6) (0.8, 0.6) (0.4, 0.8) (0.8, 0.8) (0.4, 1) (0.8, 1)

1.050272326 0.657281565 1.222428747 0.900675125 1.363378502 1.099948917 1.478778402 1.263100499 1.573259848 1.396677716

Successive [9]

Proposed scheme

n = 10

n=6

1.050008303 0.656981110 1.221989786 0.900170790 1.362970024 1.099303519 1.478408732 1.262364658 1.572925710 1.395978881

1.050008298 0.656981108 1.221989780 0.900170784 1.362970010 1.099303500 1.478408701 1.262364626 1.572925661 1.395978804

Table 2 Numerical solution of v(x, t) for Example 6.1. Grid point

Exact

(xi , tj ) (0.4, 0.2) (0.8, 0.2) (0.4, 0.4) (0.8, 0.4) (0.4, 0.6) (0.8, 0.6) (0.4, 0.8) (0.8, 0.8) (0.4, 1) (0.8, 1)

1.050272326 0.657281565 1.222428747 0.900675125 1.363378502 1.099948917 1.478778402 1.263100499 1.573259848 1.396677716

Successive [9]

Proposed scheme

n = 10

n=6

1.050149607 0.657115969 1.222179075 0.900349866 1.363022778 1.099477231 1.478349819 1.262497886 1.572783575 1.395970568

1.050149601 0.657115962 1.222179049 0.900349838 1.363022712 1.099477158 1.478349717 1.262497772 1.572783541 1.395970586

Table 3 Numerical solution of u(x, t) for Example 6.2. Grid point

Exact

(xi , tj ) (0.4, 0.2) (0.8, 0.2) (0.4, 0.4) (0.8, 0.4) (0.4, 0.6) (0.8, 0.6) (0.4, 0.8) (0.8, 0.8) (0.4, 1) (0.8, 1)

0.853776946 0.526284644 1.061551936 0.793423917 1.231663710 1.012139055 1.370939450 1.191207865 1.484968782 1.337817006

Successive [9]

Proposed scheme

n = 10

n=5

0.853665705 0.526141887 1.061374624 0.793190963 1.231492264 1.011846944 1.370774925 1.190878991 1.484809943 1.337493421

0.853665703 0.526141886 1.061374625 0.793190960 1.231492262 1.011846935 1.370774903 1.190878980 1.484809874 1.337493371

Table 4 Numerical solution of v(x, t) for Example 6.2. Grid point

Exact

(xi , tj ) (0.4, 0.2) (0.8, 0.2) (0.4, 0.4) (0.8, 0.4) (0.4, 0.6) (0.8, 0.6) (0.4, 0.8) (0.8, 0.8) (0.4, 1) (0.8, 1)

0.853776946 0.526284644 1.061551936 0.793423917 1.231663710 1.012139055 1.370939450 1.191207865 1.484968782 1.337817006

Successive [9]

Proposed scheme

n = 10

n=5

0.853650111 0.526122512 1.061322831 0.793132909 1.231358465 1.011746640 1.370581962 1.190735411 1.484575963 1.337284272

0.853650112 0.526122511 1.061322829 0.793132909 1.231358458 1.011746639 1.370581955 1.190735413 1.484575964 1.337284280

JID:YJMAA

AID:20628 /FLA

Doctopic: Applied Mathematics

[m3L; v1.184; Prn:11/08/2016; 13:53] P.19 (1-19)

L.A. Sunny et al. / J. Math. Anal. Appl. ••• (••••) •••–•••

19

Acknowledgments The authors would like to thank the referees for their valuable suggestions and comments to improve the paper. References [1] Y.H. Chang, G.C. Jau, The behaviour of interphase heat transfer for fast-igniting catalytic converters, Taiwanese J. Math. 10 (3) (2006) 807–827. [2] Y.H. Chang, G.C. Jau, C.V. Pao, Blowup and global existence of solutions for a catalytic converter in interphase heat transfer, Nonlinear Anal. Real World Appl. 9 (2008) 822–829. [3] S. Chauhan, V.K. Srivastava, Modeling exhaust gas pollution abatement: part I, single hydrocarbon propylene, Comput. Math. Appl. 55 (2008) 319–330. [4] S.N. Ha, S.W. Roh, J. Park, Numerical study of an optimal control problem for a catalytic converter function, Comput. Math. Appl. 41 (2001) 893–902. [5] J.D. Hoernel, A non stationary problem coupling PDE’s and ODE’s modeling automobile catalytic converter, Appl. Anal. 84 (2005) 617–630. [6] G.C. Jau, Y.H. Chang, The upper-lower solution method for the coupled system of first order nonlinear PDEs, J. Math. Anal. Appl. 401 (2013) 367–378. [7] D.T. Leighton, H.C. Chang, A theory of fast-igniting catalytic converters, AIChE J. 41 (1995) 1898–1914. [8] S.H. Oh, E.T. Bissett, P.A. Battiston, Mathematical modeling of electrically heated monolith converters: model formulation, numerical methods and experimental verification, Ind. Eng. Chem. Res. 32 (8) (1993) 1560–1567. [9] C.V. Pao, Y.H. Chang, G.C. Jau, Numerical methods for a coupled system of differential equations arising from a thermal ignition problem, Numer. Methods Partial Differential Equations 29 (2013) 251–279. [10] C.P. Please, P.S. Hagan, D.W. Schwendeman, Light-off behavior of catalytic converters, SIAM J. Appl. Math. 54 (1994) 72–92. [11] R.J. Plemmons, M-matrix characterizations. I – Nonsingular M-matrices, Linear Algebra Appl. 18 (1977) 175–188. [12] K. Ramanathan, V. Balakotaiah, Ignition criterion for general kinetics in a catalytic monolith, AIChE J. 52 (2006) 1623–1629. [13] S. Siemund, J.P. Leclerc, D. Schweich, M. Prigent, F. Castagna, Three way monolith converter: simulations versus experiments, Chem. Eng. Sci. 51 (1996) 3709–3720.