1. Introduction
Singularly Perturbed Problems (SPPs) are widespread in nature. SPPs often occur in different applied fields such as control systems or fluid dynamics, among others [
1,
2,
3]. In particular, systems of Singularly Perturbed Partial Differential Equations (SPPDEs) frequently arise in the modeling of heat and mass transfer processes whenever the diffusion coefficients and the thermal conductivity are very small [
4]. In [
2,
3], some mathematical models are developed for generators and their controls in control systems using singular perturbation techniques. The models are based on a system of 
n SPPs, whose first 
m equations (with 
) have the higher order derivatives affected by a small parameter 
. It is well-known that classical numerical methods are not satisfactory for SPPs because of the multiscale behavior of the solutions [
1]. A similar type of non-SPPs has a wide range of real-time applications, for instance, the calcium distribution in nerve cells is a dynamical system that is a generalized two-dimensional space-time reaction-diffusion model [
5,
6].
Many unconventional numerical methods can be found in the literature to solve various types of SPPs [
7,
8,
9,
10,
11]. A numerical method composed of the backward Euler method together with the HODIE (High Order via Differential Identity Expansion) scheme, classical central difference scheme, and a Shishkin mesh has appeared recently in [
12] for a system of SPPDEs of Convection–diffusion (CD) type in which all the higher order spatial derivatives are multiplied by a parameter 
. In [
13], a numerical method which is a combination of the backward Euler scheme, the upwind finite difference scheme, and a nonuniform mesh based on a monitor function is devised for the aforementioned system with two equations in which the higher order spatial derivatives are multiplied by different perturbation parameters. For this same system, a numerical method consisting of the backward Euler method, a hybrid scheme, and a Shishkin mesh is developed in [
14] with higher order convergence in space.
A scheme based on standard finite difference operators and the condensing mesh technique is designed in [
15] to solve a system of two SPPDEs of Reaction–diffusion (RD) type in which all the higher order spatial derivatives are multiplied by a parameter 
. The aforementioned system in which the higher order spatial derivatives are multiplied by different parameters is solved by a numerical method in [
16], which consists of two additive schemes using a Shishkin mesh and a standard central difference operator. This problem is extended to a general system in [
17] and solved by a numerical method composed of classical finite difference operators and appropriate layer-adapted meshes. For this same type of system with two equations, the convergence in both time and space is improved in [
18] through a numerical method which consists of the Crank–Nicolson method, the classical central difference scheme, and a Shishkin mesh. The aforementioned problem is extended to a general system in [
19] in which only the higher order convergence in space is maintained through a numerical method composed of classical finite difference operators and a Shishkin mesh. For the same system, the higher order convergence in time is achieved with the Crank–Nicolson method in [
20].
In [
21], a first-order scheme for a system of two singularly perturbed RD equations was presented, where only the second derivative of the first equation was multiplied by a small perturbation parameter. For this system, a higher order convergent scheme was developed in [
22]. The article [
23] deals with the construction of a numerical method for a general system of 
n singularly perturbed RD equations in which only the first 
 equations are multiplied by different perturbation parameters. Inspired by [
21] and [
23], a numerical method is devised in [
24] for a similar system of SPPDEs with more assumptions. Motivated by [
2,
3,
15,
24], in this work, a system of 
n SPPDEs of RD type is considered in which only the first 
m equations (with 
) are multiplied by 
.
The structure of the paper is as follows. In 
Section 2, we state the problem to be solved. The existence, uniqueness, and stability properties of the continuous solution are presented in 
Section 3. In addition, the decomposition of the solution (into regular and singular components) and bounds on its derivatives are established. 
Section 4 contains the design of a finite difference scheme using a piecewise uniform Shishkin mesh to solve the discrete problem, which provides a robust numerical approximation to the solution. In 
Section 5, the error analysis of the numerical solution based on barrier function techniques is derived, which sets up the theoretical results. A numerical example is given in 
Section 6 to conform the applicability and expected rate of accuracy of the present method with the parameter-uniform maximum pointwise error, parameter-uniform error constant, and parameter-uniform rate of convergence in the tables and graphs, which validate the major findings of the paper.
  2. Continuous Problem
Let consider the singularly perturbed time-dependent RD system
      
      where 
  , with 
  and 
 For 
, we use the notations 
   . We note that 
 and 
E are square matrices of dimension 
n with 
, where 
 appears 
m times (
), being 
 It is assumed that for all 
 the elements 
  of 
 satisfy
      
      and
      
      for some constant 
. In operator form, the system (
1) and (2) is written as follows:
      with 
 given by
      
      where 
I is used to denote the identity matrix.
Taking 
 in (
1) and (2), the reduced problem is obtained, which is given for 
 by
      
      for 
,  
, with 
 on 
.
In general, 
 on 
 for 
. Hence, the components 
 for 
 of the solution 
 of (
1) and (2) exhibit boundary layers near 
 and 
, whereas the components 
 for 
 exhibit no layer throughout the domain.
In this paper, we adopt the following notations. For any continuous vector function  on   and , and for any mesh function   and  Throughout this article,  and  denote positive constants, which are independent of  and the discretization parameters N and 
  3. Analytical Results
Some of the analytical properties, maximum principles, stability results, and derivative bounds of the exact solution (
1) and (2) and its regular and singular components are estimated. The derivative bounds are necessary for the numerical analysis that is derived thoroughly in this section. The existence of the solution 
 of (
1) and (2) is stated in Theorem 1 without proof. For more details, the reader is referred to  [
25].
Theorem 1. Assume the components of B and  are sufficiently smooth functions,   , and the following compatibility conditions are satisfied at the corners  and  of Γ:and Then, there exists a solution  of (1) and (2) such that  where  is the space of Hölder continuous functions with .
 Lemma 1 (Maximum Principle)
. Let us assume that condition (4) holds and  is any function in the domain of  such that  on Γ and  on Ω. Then, it is  on   Proof.  Let 
 be such that 
. The result follows immediately if 
 Suppose 
 Then, 
  and 
 Let 
 We have that
        
Using (
4), we obtain 
, which is a contradiction. Thus, 
, and the lemma is proven.    □
 Lemma 2 (Stability Result)
. Let us assume that condition (4) holds and  is any function in the domain of  We find that for  and , it holds that  Proof.  Let 
 Define 
  , where 
. Then, 
 on 
 and using (
4), 
 on 
 Hence, from Lemma 1, 
 on 
 which proves the result.    □
 Lemma 3. Let us assume that conditions (3), (4), (6)–(10) hold. In this case, for , , , and , we get  Proof.  The bound on 
 follows from Lemma 2. Differentiating the 
 equation in (
1) partially with respect to 
t once and rearranging the terms, we obtain 
 Then, the bound on 
 follows by using Lemma 2. The bound on 
 can be derived in the same way. For each 
 consider a neighborhood 
 such that 
 Then, by using the mean value theorem with 
, we find that for 
, it is
        
        and thus,
        
Now, consider the relation
        
Using the bounds of 
   and 
 the bound of 
 follows from (
12). The bound on 
 follows by using a similar procedure considering the interval 
 where 
Further, the bounds of the mixed derivatives can also be derived by using this same technique. From (
1), we obtain
        
The bounds on 
 follow from (
14). Differentiating (
14) with respect to 
x once and using the bounds of 
    B and 
 the bound on 
 follow. In addition, differentiating (
14) with respect to 
x twice and using the bounds of 
    , 
, 
  and 
 the bound on 
 follows.    □
 Now, we decompose 
 into two parts 
 and 
 such that 
 with 
, where 
 is the solution of the reduced problem (
5), 
 is the solution of
      
      and the component 
 is the solution of
      
We can decompose the component  further as  and  such that , where  on   on   on  and  on   on   on .
Lemma 4. Let us assume that conditions (3), (4), (6)–(10) hold. In this case, for , ,   and , we have  Proof.  From (
5), it is not hard to see that for 
 and 
, it holds that
        
From (
15) and the above bounds, it is clear that 
 is the solution of a problem similar to (
1) and (2). Hence, the estimates in Lemma 3 hold for the components of 
 and their derivatives after replacing 
 by 
 and appropriate derivatives. Thus, the bounds of 
 and its derivatives follow from those of 
  and their derivatives.    □
 The layer functions 
 associated with the solution 
 of (
1) and (2) are defined for all 
 by
      
Lemma 5. Let us assume that conditions (3), (4), (6)–(10) hold. In this case, for  ,   , we have that Analogous results hold for  and their derivatives with x replaced by 
 Proof.  Define the functions 
, 
 and 
,
. Then, 
 on 
 for proper choices of 
 and 
 Now, for 
, we get
        
Using (
4), we find that 
 on 
 By using Lemma 1 with 
, the bounds of 
 and 
 follow. Now, we derive the bounds on 
 and 
 Define the functions 
, 
 and 
, 
. Differentiate the equation 
 partially with respect to 
t once and rearrange the terms to get
        
Using Lemmas 3 and 4, we obtain that 
. Thus, the bounds of 
 and 
 follow from using Lemma 1 applied to the functions 
. From Lemmas 3 and 4, we get
        
It is not hard to verify that 
 and 
≤
  and 
 Using the above estimates in (
17), it can be obtained that 
 Now, define the functions 
, 
 and 
, 
. Using Lemma 1 with the functions 
, the bounds of 
 and 
 follow. Using the mean value theorem with 
 and similar arguments to those used to get a bound of 
, we deduce that
        
Since 
 and 
 the bounds on 
 follow from (
20). Using a similar procedure, the bounds on 
 can be derived. The bounds on 
  follow from the equation 
 Differentiating the equation 
 once and twice partially with respect to 
x and using the above bounds, the bounds on 
 and 
  can be readily obtained.    □
   4. Discretization of the Domain and Discretized Problem
In this section, a finite difference scheme using a piecewise uniform Shishkin mesh with a well-suited transition parameter for the SPPDE (
1) and (2) is considered (see [
1] and references therein). A rectangular mesh 
 with 
 mesh regions is now constructed on the domain 
 The rectangular mesh 
 consists of a piecewise uniform Shishkin mesh on the spatial domain 
 and a uniform mesh on the temporal domain 
. Let 
 , 
 , 
 , 
 , 
 , 
  and 
 . On the spatial domain 
, a piecewise uniform Shishkin mesh 
 with 
N mesh intervals is constructed using a transition parameter 
 as follows. Thus, we split the interval [0,1] as
      
      where the parameter 
 is a function of 
, 
 and 
N, given by
      
Each of the sub-intervals 
 and 
 is divided into 
 subintervals using appropriate uniform mesh points, whereas 
 is divided into 
 sub-intervals using also appropriate uniform mesh points. On the temporal domain 
 a uniform mesh 
 with 
M mesh intervals is placed. The discrete problem corresponding to (
1) and (2) is
      
Here,
          and  such that  and 
In operator form, the problem (
21) is written as follows:
Thus, the operator  is defined by 
Lemma 6. [Discrete Maximum Principle] Let us assume that condition (4) holds and  is any mesh function such that  on  and,  on . Then, it holds that  on   Proof.  Take 
 for which 
. The result follows immediately if 
 Suppose 
 Then 
∉
, 
 and 
 Let 
 Now, we have
        
Using (
4), we get 
 which is a contradiction. Thus, 
≥ 0, and the lemma is proven.    □
 Lemma 7 (Discrete Stability Result)
. Let us assume that condition (4) holds and  is any mesh function. Then, for  and , we have that  Proof.  Let 
 Define 
 ±
, where 
 Then 
 on 
 and using (
4) we find that 
≥
 on 
 Hence, from Lemma 6, we have 
 on 
, which proves the result.    □
   5. Error Analysis
This section gives an error bound of the solution provided by the present method. We began by considering the decomposition of the solution of the discrete problem (
21). This is similar to the continuous case. The solution 
 of the problem (
21) can be decomposed into 
 and 
 such that 
, where 
 is the solution of
      
      and 
 is the solution of
      
It is noted that for any smooth function 
, we have
      
      and
      
      where, 
 and 
Theorem 2. Let  and  be the components of , and let  and  be the components of . Then, for all  and , it holds that  Proof.  From the defining equation for 
 and (
23), we have
        
        and from (
16) and (
24), we have
        
Note that for all 
  Since 
, for any choice of the parameter 
 from (
25), (26), and (
30), for 
 and 
, we have
        
        and
        
Then, using Lemma 4, (
28) follows from (
32) and (
33) for 
. Now, we estimate the error in 
 From (
31), we obtain
        
Here, we consider that the argument depends on whether 
 or 
 If 
, then 
, and it follows 
 in this case. Using (
25), (26), and (
34), for 
 and 
, we get
        
        and
        
Later, using Lemma 5, for 
, we get
        
Now, assume 
. Let us prove the results on the domains 
  and 
 separately. The spatial mesh spacing on both intervals 
 and 
 is 
 Using (
25), (26) and (
34) for both 
 and 
 and Lemma 5, for 
, we get
        
Finally, consider the interval 
 Here, the spatial mesh spacing is 
≤
 From (
25), (26), (
34) and using Lemma 5, for 
 and 
, we have
        
        and
        
From (
25), (
27), (
34) and using Lemma 5, for 
 and 
, we obtain
        
        and
        
It is not difficult to get
        
Now, we proceed with the proof by considering two instances: 
 and 
. In the first instance, using Lemma 5 and (
41) in (
37) and (
38), for 
, we get
        
Consider the second choice 
 In this case, using Lemma 5 and (
41) in (
39) and (
40), for 
, we have
        
Using similar arguments, we can estimate the error in  By combining the error estimates for  and  (29) holds in all cases for .    □
 Theorem 3. Let us denote by  the solution of (1) and (2) and by  the solution of (21). Then, for all  and , we have  Proof.  Note that for all 
 and 
,
        
Using Theorem 2 with (
43), one can obtain
        
Therefore, by using Lemma 7 with (
44), for all 
 and 
, we obtain the desired error bound, which is given by
        
This completes the proof.    □
 The obtained error bound provides the parameter uniform estimate as the constant C is independent of the perturbation parameter . Furthermore, the numerical solution converges to the continuous solution almost linearly.
  6. Numerical Illustrations
In this section, we consider a trial problem to investigate the effectiveness of the present method. The notations 
 , and 
 correspond to the parameter-uniform maximum pointwise errors, the parameter-uniform error constants, and the parameter-uniform rates of convergence, respectively. Let 
 be an approximation of the true solution 
 on the mesh 
, where 
M and 
N are the number of mesh points on each direction. We take 
 for the time-direction, and 
 for the space-direction. The exact solution for the considered test problem is not available, so for a finite set of values of the perturbation parameter 
, the parameter-uniform maximum errors are obtained as
      
From these values, one can obtain the parameter-uniform rate of convergence using
      
      and the parameter-uniform error constant given by
      
Example 1. Consider the following system of parabolic equation for with  on   on  and  on   It is to be noted that the above problem satisfies conditions (
3) and (
4) and all the compatibility conditions (
6)–(
10). For different values of 
 the values of 
 , and 
 in the variable 
x with a uniform mesh consist of 10 mesh intervals for the variable 
t given in 
Table 1, and for different values of 
 the values of 
 , and 
 in the variable 
t with a piecewise uniform Shishkin mesh consist of 128 mesh intervals for the variable 
x given in 
Table 2. From 
Table 1 and  
Table 2, it can be observed that the errors are independent of the singular perturbation parameter 
 and decrease as the numbers of mesh points 
M and 
N increase. Surface plots of the maximum error of the solution of the above test problem are presented in 
Figure 1 and 
Figure 2. For 
 , and 
 Figure 1 shows the numerical solution of the component 
 of the solution 
. It is observed that the component 
 changes rapidly near 
 and 
 (boundary layers) due to the presence of the perturbation parameter. 
Figure 2 shows the numerical solution of the component 
, but it varies smoothly throughout the domain. Furthermore, 
Figure 3 shows the numerical solution of 
  for the errors given in 
Figure 4 and 
Figure 5, which clearly indicate that the error bound is 
, as expected.