Numerical Scheme for Singularly Perturbed Mixed Delay Differential Equation on Shishkin Type Meshes

Abstract

Two non-uniform meshes used as part of the finite difference method to resolve singularly perturbed mixed-delay differential equations are studied in this article. The second-order derivative is multiplied by a small parameter which gives rise to boundary layers at $x=0$ and $x=3$ and strong interior layers at $x=1$ and $x=2$ due to the delay terms. We prove that this method is almost first-order convergent on Shishkin mesh and is first-order convergent on Bakhvalov–Shishkin mesh. Error estimates are derived in the discrete maximum norm. Some examples are provided to validate the theoretical result.

1. Introduction

Mixed-type functional differential equations are typically used to describe functional differential equations that include both delay and advanced argument. The initial motivation for studying these equations came from optimal control issues (see [1]). Numerous applications, including economic dynamics [2], travelling waves in a spatial lattice [3], and the theory of nerve conduction [4], produce functional differential equations of this sort.

Typically, singularly perturbed differential equations with delay and advanced argument are identified by the existence of a tiny positive parameter that multiplies some or all of the differential equation’s greatest derivatives. These equations appear in biological science and engineering mathematical models.

In the literature on diseases and population in [5], the justifications for a small delay problem can be discussed. Lange and Miura [6,7] addressed the issue of estimating the time at which random synaptic inputs in the dendrites will cause nerve cells to produce action potentials. When modelling the activation of a neuron, the generic boundary-value problem for the linear second-order differential-difference equation is given by

$({\beta }^2/2)z^{″}\left(t\right)+(\mu -t)z^{\prime }\left(t\right)+{\sigma }_{E}z(t+a_E)+{\sigma }_{I}z(t-a_I)-({\sigma }_E+{\sigma }_I)z\left(t\right)=-1$

where the estimated first exit time is z and the variance and drift parameters are $\beta$ and $\mu$. Between synaptic inputs, the first-order derivative term $-t{z}^{\prime }\left(t\right)$ represents the exponential decay. The undifferentiated terms are excitatory and inhibitory synaptic inputs that are described as Poisson processes with mean rates of ${\sigma }_E$ and ${\sigma }_I$. These inputs cause tiny jumps in the membrane potential of $a_E$ and $a_I$ that may be voltage dependent. The boundary condition is

$z\left(t\right)=0,\forall t\notin (t_1,t_2)$

where $t=t_1$ and $t=t_2$ represent the inhibitory reversal potential and the threshold value of the membrane potential, respectively, for the production of action potentials. The study of boundary value problem for differential equations with mixed delay is motivated by this biological problem. Intriguing phenomena include boundary and interior layers, quick oscillations, resonance, turning-point behaviour, non-uniqueness and/or non-existence for non-linear delay differential equations, etc., in the solutions of such problems. To promote the use of delay differential equation models in the physical and biological sciences, we are conducting these investigations of boundary value problems for singularly perturbed delay differential equations.

While examining boundary value problems for singularly perturbed ordinary differential equations with small delay and solutions displaying layer behaviour, new results have been discovered. The key findings are as follows: (i) Even if the changes are minor, they may have an impact on the leading-order approximations. (ii) The layer behaviour may alter and could lose its identity when the adjustments grow, yet stay tiny. (iii) The characteristic exponential polynomial may include zeros in the right half s-plane that extend to infinity for some situations, but the Laplace transform approach used to evaluate the boundary-layer solutions may still be appropriate.

The finite difference method (FDM) is an approximate method for solving ordinary differential equations. It has been used to solve a wide range of problems. These include linear and non-linear, time independent and dependent problems. This method can be applied to problems with different boundary shapes, different kinds of boundary conditions, and for a region containing numerous different materials. The application of FDM is not difficult, as it involves only simple arithmetic in the derivation of the discretisation equations and in writing the corresponding programs.

The distribution of inputs is assumed to be a Poisson process with exponential decay in Stein’s model [8,9]. In [8,10,11,12,13], various problems on singularly perturbed delay differential equations with integral boundary conditions are considered. Lange and Miura have addressed several cases of extremely minor changes, including [6,7,14,15,16]. Kadalbajoo and Sharma have discussed the finite difference method for singularly perturbed mixed small delay problems in [17,18,19]. Patitor and Sharma in [20] have developed the fitted operator approach for mixed small-delay problems. The above authors only discussed singularly perturbed mixed small-delay differential equations. Meanwhile, the authors of [21] developed a singularly perturbed convection-diffusion type with mixed large delay using hybrid difference technique and a finite difference method on Shishkin mesh. We studied singularly perturbed reaction–diffusion type with mixed large delay differential equations in this paper, which was inspired by the aforementioned publications. We also developed the finite difference technique on Shishkin mesh and Bakhvalov–Shishkin mesh.

In this article, we examine a fitted finite difference scheme on a piecewise uniform mesh for the numerical solution of singularly perturbed reaction–diffusion type with mixed large delay problem. The structure of the paper is as follows. Section 2 identifies the continuous problem. Section 3 presents the maximum principle, the stability finding, and suitable bounds for the derivatives of the problem’s solution. Section 4 explains the numerical approach. Section 5 provides an error analysis for an approximation of the solution. Numerical results are provided in Section 6. The conclusion is included in Section 7.

The following notations are used throughout our analysis: $\overline{\mathsf{\Omega }}=[0,3]$, $\mathsf{\Omega }=(0,3)$, ${\mathsf{\Omega }}_1=(0,1)$, ${\mathsf{\Omega }}_2=(1,2)$, ${\mathsf{\Omega }}_3=(2,3)$, ${\mathsf{\Omega }}^{*}={\mathsf{\Omega }}_1\cup {\mathsf{\Omega }}_2\cup {\mathsf{\Omega }}_3$. ${\overline{\mathsf{\Omega }}}^{3N}=\{0,1,2,\dots ,3N\}$ mesh points, ${\mathsf{\Omega }}_1^{3N}=\{1,2,\dots ,N-1\}$, ${\mathsf{\Omega }}_2^{3N}=\{N+1,\dots ,2N-1\}$, ${\mathsf{\Omega }}_3^{3N}=\{2N+1,\dots ,3N-1\}$. Assume the parameter $\epsilon$ and the number of mesh points $3N$ have no effect on the positive constants C, $C_1$, and $C_2$.

2. Problem Description

A class of singularly perturbed reaction–diffusion type with mixed large delay differential equations are

(1)$\left\{\begin{array}{c}Lz\left(t\right)=-\epsilon z^{″}\left(t\right)+p\left(t\right)z\left(t\right)+q\left(t\right)z(t-1)+r\left(t\right)z(t+1)=f\left(t\right),t\in {\mathsf{\Omega }}^{*},\hfill \\ z\left(t\right)=\varphi \left(t\right),t\in [-1,0],z\left(t\right)=\phi \left(t\right),t\in [3,4],\hfill \end{array}\right.$

where $p\left(t\right)$, $q\left(t\right)$, $r\left(t\right)$ and $f\left(t\right)$ are sufficiently smooth functions on $[0,3]$, satisfying $p\left(t\right)\ge {\alpha }_1>2\alpha >0$, $\beta \le q\left(t\right)<0$, $\gamma \le r\left(t\right)<0$; $\alpha +\beta >0$, $\alpha +\gamma >0$ and $\alpha +\beta +\gamma >0$ are smooth functions on $t\in \overline{\mathsf{\Omega }}$. The historical functions $\varphi \left(t\right)$ and $\phi \left(t\right)$ are smooth on $[-1,0]$ and $[3,4]$.

The assumptions above guarantee that $z\in X=C^0(\overline{\mathsf{\Omega }})\cap C^1\left(\mathsf{\Omega }\right)\cap C^2({\mathsf{\Omega }}_1\cup {\mathsf{\Omega }}_2\cup {\mathsf{\Omega }}_3).$

The above Equation (1) is the same as $Lz\left(t\right)=g\left(t\right)$, where

(2)$Lz\left(t\right)=\left\{\begin{array}{cc}{L}_{1}z\left(t\right)=-\epsilon z^{″}\left(t\right)+p\left(t\right)z\left(t\right)+r\left(t\right)z(t+1),\hfill & t\in {\mathsf{\Omega }}_1\hfill \\ L_{2}z\left(t\right)=-\epsilon z^{″}\left(t\right)+p\left(t\right)z\left(t\right)+q\left(t\right)z(t-1)+r\left(t\right)z(t+1),\hfill & t\in {\mathsf{\Omega }}_2\hfill \\ L_{3}z\left(t\right)=-\epsilon z^{″}\left(t\right)+p\left(t\right)z\left(t\right)+q\left(t\right)z(t-1),\hfill & t\in {\mathsf{\Omega }}_3\hfill \end{array}\right.$

(3)$g\left(t\right)=\left\{\begin{array}{cc}f\left(t\right)-q\left(t\right)\varphi (x-1),\hfill & t\in {\mathsf{\Omega }}_1,\hfill \\ f\left(t\right),\hfill & t\in {\mathsf{\Omega }}_2,\hfill \\ f\left(t\right)-r\left(t\right)\phi (x+1),\hfill & t\in {\mathsf{\Omega }}_3\hfill \end{array}\right.$

with

(4)$\left\{\begin{array}{c}z\left(t\right)=\varphi \left(t\right),t\in [-1,0],\hfill \\ z\left(1^{-}\right)=z\left(1^{+}\right),z^{\prime }\left(1^{-}\right)=z^{\prime }\left(1^{+}\right),\hfill \\ z\left(2^{-}\right)=z\left(2^{+}\right),z^{\prime }\left(2^{-}\right)=z^{\prime }\left(2^{+}\right),\hfill \\ z\left(t\right)=\phi \left(t\right),t\in [3,4].\hfill \end{array}\right.$

3. Stability Result

Lemma 1 (Maximum Principle).

If $\Xi \left(t\right)\in X$ satisfies $\Xi \left(0\right)\ge 0$, $\Xi \left(3\right)\ge 0$, $L_1\Xi \left(t\right)\ge 0$, $\forall t\in {\mathsf{\Omega }}_1$, $L_2\Xi \left(t\right)\ge 0$, $\forall t\in {\mathsf{\Omega }}_2$, $L_3\Xi \left(t\right)\ge 0$, $\forall t\in {\mathsf{\Omega }}_3$, $\left[{\Xi }^{\prime }\right]\left(1\right)\le 0$ and $\left[{\Xi }^{\prime }\right]\left(2\right)\le 0$ then $\Xi \left(t\right)\ge 0$, $\forall t\in \overline{\mathsf{\Omega }}$.

Proof. 

Consider a test function

(5)$s\left(t\right)=\left\{\begin{array}{cc}\frac{1}{12}+\frac{t}{4},\hfill & t\in [0,1],\hfill \\ \frac{2}{12}+\frac{t}{6},\hfill & t\in [1,2],\hfill \\ \frac{4}{12}+\frac{t}{12},\hfill & t\in [2,3].\hfill \end{array}\right.$

Note that $s\left(t\right)>0$, $\forall t\in \overline{\mathsf{\Omega }}$, $Ls\left(t\right)>0,\forall t\in {\mathsf{\Omega }}^{*}$, $s\left(0\right)>0$, $s\left(3\right)>0$, $\left[s^{\prime }\right]\left(1\right)<0$ and $\left[s^{\prime }\right]\left(2\right)<0$.

Let $\mu =max\{\frac{-\Xi \left(t\right)}{s\left(t\right)}:t\in \overline{\mathsf{\Omega }}\}$, clearly $\Xi \left(t_0\right)+\mu s\left(t_0\right)=0$ for some $t_0\in \overline{\mathsf{\Omega }}$, then $\Xi \left(t\right)+\mu s\left(t\right)\ge 0,\forall t\in \overline{\mathsf{\Omega }}$ attains its minimum at $t=t_0$. If $\mu <0$, then $\Xi \left(t\right)\ge 0$. Suppose $\mu >0$,

Case (i): $t_0=0,$

$0<(\Xi +\mu s)\left(0\right)=\Xi \left(0\right)+\mu s\left(0\right)=0,$

Case (ii): $t_0\in {\mathsf{\Omega }}_1,$

$\begin{array}{ccc}\hfill 0<L_1(\Xi +\mu s)\left(t_0\right)& =& -\epsilon {(\Xi +\mu s)}^{″}\left(t_0\right)+a\left(t_0\right)(\Xi +\mu s)\left(t_0\right)\hfill \\ & & +c\left(t_0\right)(\Xi +\mu s)(t_0+1)\le 0.\hfill \end{array}$

Case (iii): $t_0=1,$

$0\le \left[{(\Xi +\mu s)}^{\prime }\right]\left(1\right)=\left[{\Xi }^{\prime }\right]\left(1\right)+\mu \left[s^{\prime }\right]\left(1\right)<0.$

Case (iv): $t_0\in {\mathsf{\Omega }}_2,$

$\begin{array}{ccc}\hfill 0<L_2(\Xi +\mu s)\left(t_0\right)& =& -\epsilon {(\Xi +\mu s)}^{″}\left(t_0\right)+a\left(t_0\right)(\Xi +\mu s)\left(t_0\right)\hfill \\ & & +b\left(t_0\right)(\Xi +\mu s)(t_0-1)+c\left(t_0\right)(\Xi +\mu s)(t_0+1)\le 0.\hfill \end{array}$

Case (v): $t_0=2,$

$0\le \left[{(\Xi +\mu s)}^{\prime }\right]\left(2\right)=\left[{\Xi }^{\prime }\right]\left(2\right)+\mu \left[s^{\prime }\right]\left(2\right)<0.$

Case (vi): $t_0\in {\mathsf{\Omega }}_3,$

$\begin{array}{ccc}\hfill 0<L_3(\Xi +\mu s)\left(t_0\right)& =& -\epsilon {(\Xi +\mu s)}^{″}\left(t_0\right)+a\left(t_0\right)(\Xi +\mu s)\left(t_0\right)\hfill \\ & & +b\left(t_0\right)(\Xi +\mu s)(t_0-1)\le 0.\hfill \end{array}$

Case (vii): $t_0=3,$

$0\le (\Xi +\mu s)\left(3\right)=(\Xi )\left(3\right)+\mu \left(s\right)\left(3\right)=0.$

This is contradicted by every case. Hence, it is proved. □

Lemma 2 (Stability Result).

The problem (2)–(4) has the solution $z\left(t\right)$, which fulfils the bound.

$\left|z\left(t\right)\right|\le Cmax\left\{\left|z\left(0\right)\right|,\left|z\left(3\right)\right|,\underset{t\in {\mathsf{\Omega }}^{*}}{sup}\left|Lz\left(t\right)\right|\right\},t\in \overline{\mathsf{\Omega }}.$

Proof. 

The barrier functions and Lemma 1 can be used to demonstrate this theorem. ${\theta }^{\pm }\left(t\right)=CMs\left(t\right)\pm z\left(t\right),t\in \overline{\mathsf{\Omega }}$, where $M=max\left\{\left|z\left(0\right)\right|,\left|z\left(3\right)\right|,{sup}_{t\in {\mathsf{\Omega }}^{*}}\left|Lz\left(t\right)\right|\right\}$ and the Lemma 1 has the same definition as $s\left(t\right)$. □

Lemma 3.

The problem (2)–(4) has derivatives of solution $z\left(t\right)$, which fulfils the bound:

$\parallel z^{\left(k\right)}{\parallel }_{{\mathsf{\Omega }}^{*}}\le C{\epsilon }^{-k},fork=1,2,3.$

Proof. 

We examine, in order to constrain $z^{\prime }\left(t\right)$ on the interval ${\mathsf{\Omega }}_1$:

$L_{1}z\left(t\right)=-\epsilon z^{″}\left(t\right)+p\left(t\right)z\left(t\right)+r\left(t\right)z(t+1)=f\left(t\right)-q\left(t\right)\varphi (x-1).$

By integrating both sides of the aforementioned equation, we obtain

$\begin{array}{c}\hfill -\epsilon (z^{\prime }\left(t\right)-z^{\prime }\left(0\right))=-\underset{0}{\overset{x}{\int }}p\left(t\right)z\left(t\right)dt-\underset{0}{\overset{x}{\int }}r\left(t\right)z(t+1)dt+\underset{0}{\overset{x}{\int }}[f\left(t\right)-q\left(t\right)\varphi (t-1)]dt.\end{array}$

Therefore,

$\begin{array}{c}\hfill \epsilon z^{\prime }\left(0\right)=\epsilon z^{\prime }\left(t\right)-\underset{0}{\overset{x}{\int }}p\left(t\right)z\left(t\right)dt-\underset{0}{\overset{x}{\int }}r\left(t\right)z(t+1)dt+\underset{0}{\overset{x}{\int }}[f\left(t\right)-r\left(t\right)\varphi (t-1)]dt.\end{array}$

Then, according to the Mean Value Theorem, $z\in (0,\epsilon )$ exists in a way that $|\epsilon z^{\prime }\left(z\right)|\le C(\parallel z\left(t\right)\parallel ,\parallel f\left(t\right)\parallel ,\parallel \varphi \parallel )$ and $|\epsilon z^{\prime }\left(0\right)|\le C(\parallel z\left(t\right)\parallel +\parallel f\left(t\right)\parallel +\parallel \varphi \left(t\right)\parallel )$.

Hence,

$|\epsilon z^{\prime }\left(t\right)|\le C\max (\parallel z\left(t\right)\parallel ,\parallel f\left(t\right)\parallel ,\parallel \varphi \parallel ).$

Using a similar justification, to prove $z^{\prime }\left(t\right)$ on ${\mathsf{\Omega }}_2$ and ${\mathsf{\Omega }}_3$, as $|\epsilon z^{\prime }\left(t\right)|\le C$. Meanwhile, (2) and (3) imply that $\parallel z^{\left(k\right)}{\parallel }_{{\mathsf{\Omega }}^{*}}\le C{\epsilon }^{-k},k=2,3$. □

The Solution’s Decomposition

The solution $z\left(t\right)$ of (2)–(4) is decomposed of Shishkin as follows: $z\left(t\right)=v\left(t\right)\left(\mathrm{regular}\right)+w\left(t\right)\left(\mathrm{singular}\right)$ components. Additionally, $v\left(t\right)=v_0\left(t\right)+\epsilon v_1\left(t\right)$ where the reduced problem answer is $v_0\left(t\right)$, and the solutions to the subsequent issues are $v_1\left(t\right)$:

(6)$\begin{array}{ccc}\hfill L_1{v}_1\left(t\right)& =& v_0^{″}\left(t\right),t\in {\mathsf{\Omega }}_1,v_1\left(0\right)=0,v_1\left(1\right)=0.\hfill \\ \hfill L_2{v}_1\left(t\right)& =& v_0^{″}\left(t\right),t\in {\mathsf{\Omega }}_2,v_1\left(1\right)=0,v_1\left(2\right)=0.\hfill \\ \hfill L_3{v}_1\left(t\right)& =& v_0^{″}\left(t\right),t\in {\mathsf{\Omega }}_3,v_1\left(2\right)=0,v_1\left(3\right)=0.\hfill \end{array}$

Further, $w\left(t\right)$ satisfy the following problems:

(7)$\begin{array}{ccc}\hfill L_{1}w\left(t\right)& =& 0,t\in {\mathsf{\Omega }}_1,w\left(0\right)=z\left(0\right)-v\left(0\right),\left[w\right]\left(1\right)=-\left[v\right]\left(1\right).\hfill \\ \hfill L_{2}w\left(t\right)& =& 0,t\in {\mathsf{\Omega }}_2,\left[w^{\prime }\right]\left(1\right)=-\left[v^{\prime }\right]\left(1\right),\left[w\right]\left(2\right)=-\left[v\right]\left(2\right).\hfill \\ \hfill L_{3}w\left(t\right)& =& 0,t\in {\mathsf{\Omega }}_3,\left[w^{\prime }\right]\left(2\right)=-\left[v^{\prime }\right]\left(2\right),w\left(3\right)=z\left(3\right)-v\left(3\right).\hfill \end{array}$

We further decompose $w\left(t\right)$ as $w\left(t\right)=w_L\left(t\right)+w_R\left(t\right)$, where the function $w_L\left(t\right)$ are left-layers components given as $w_L\left(t\right)=w_{L_1}\left(t\right)+w_{L_2}\left(t\right)+w_{L_3}\left(t\right)$ and $w_R\left(t\right)$ are right-layer components given as $w_R\left(t\right)=w_{R_1}\left(t\right)+w_{R_2}\left(t\right)+w_{R_3}\left(t\right)$.

Furthermore, the solutions to the following problem are $w_L\left(t\right)$:

Find $w_L\left(t\right)\in X$, such that

(8)$\begin{array}{ccc}\hfill L_1{w}_{L_1}\left(t\right)& =& 0,t\in {\mathsf{\Omega }}_1,w_{L_1}\left(0\right)=w\left(0\right),w_{L_1}\left(1\right)=0.\hfill \\ \hfill L_2{w}_{L_2}\left(t\right)& =& 0,t\in {\mathsf{\Omega }}_2,w_{L_2}\left(1\right)=A,w_{L_2}\left(2\right)=0.\hfill \\ \hfill L_3{w}_{L_3}\left(t\right)& =& 0,t\in {\mathsf{\Omega }}_3,w_{L_3}\left(2\right)=B,w_{L_3}\left(3\right)=0.\hfill \end{array}$

Furthermore, the solutions of the following problem are $w_R\left(t\right)$:

Find $w_R\left(t\right)\in X$, such that

(9)$\begin{array}{ccc}\hfill L_1{w}_{R_1}\left(t\right)& =& 0,t\in {\mathsf{\Omega }}_1,w_{R_1}\left(0\right)=0,w_{R_1}\left(1\right)=A_1.\hfill \\ \hfill L_2{w}_{R_2}\left(t\right)& =& 0,t\in {\mathsf{\Omega }}_2,w_{R_2}\left(1\right)=0,w_{R_2}\left(2\right)=B_1.\hfill \\ \hfill L_3{w}_{R_3}\left(t\right)& =& 0,t\in {\mathsf{\Omega }}_3,w_{R_3}\left(3\right)=0,w_{R_3}\left(3\right)=w\left(3\right).\hfill \end{array}$

where the variables A, B, $A_1$, and $B_1$ must be selected to satisfy the jump requirements at $x=1$ and $x=2$.

Lemma 4.

For solutions $z\left(t\right)$, the regular component $v\left(t\right)$ and singular component $w\left(t\right)$ fulfil the following bounds.

(10) $\begin{array}{ccc}\hfill \parallel v^k{\left(t\right)\parallel }_{{\mathsf{\Omega }}^{*}}& \le & C(1+{\epsilon }^{1-\frac{k}{2}}),fork=0,1,2,3.\hfill \end{array}$

(11) $\begin{array}{c}\hfill |w_L^k\left(t\right)|\le C{\epsilon }^{-k/2}\left\{\begin{array}{cc}exp(\frac{-t\sqrt{\alpha }}{\sqrt{\epsilon }}),\hfill & t\in {\mathsf{\Omega }}_1,0\le k\le 3.\hfill \\ exp(\frac{-(t-1)\sqrt{\alpha }}{\sqrt{\epsilon }}),\hfill & t\in {\mathsf{\Omega }}_2,0\le k\le 3.\hfill \\ exp(\frac{-(t-2)\sqrt{\alpha }}{\sqrt{\epsilon }}),\hfill & t\in {\mathsf{\Omega }}_3,0\le k\le 3.\hfill \end{array}\right.\end{array}$

(12) $\begin{array}{c}\hfill |w_R^k\left(t\right)|\le C{\epsilon }^{-k/2}\left\{\begin{array}{cc}exp(\frac{-(1-t)\sqrt{\alpha }}{\sqrt{\epsilon }}),\hfill & t\in {\mathsf{\Omega }}_1,0\le k\le 3.\hfill \\ exp(\frac{-(2-t)\sqrt{\alpha }}{\sqrt{\epsilon }}),\hfill & t\in {\mathsf{\Omega }}_2,0\le k\le 3.\hfill \\ exp(\frac{-(3-t)\sqrt{\alpha }}{\sqrt{\epsilon }}),\hfill & t\in {\mathsf{\Omega }}_3,0\le k\le 3.\hfill \end{array}\right.\end{array}$

Proof. 

It is simple to demonstrate the inequality (4) by integrating the simplified problem of (2) and (7) and applying the stability result.

Now, to prove inequality (11), consider the barrier functions,

$\begin{array}{c}\hfill {\mathsf{\Phi }}^{\pm }\left(t\right)=C\left\{\begin{array}{cc}exp(\frac{-t\sqrt{\alpha }}{\sqrt{\epsilon }}),\hfill & t\in {\mathsf{\Omega }}_1\hfill \\ exp(\frac{-(t-1)\sqrt{\alpha }}{\sqrt{\epsilon }}),\hfill & t\in {\mathsf{\Omega }}_2\pm w_L\left(t\right).\hfill \\ exp(\frac{-(t-2)\sqrt{\alpha }}{\sqrt{\epsilon }}),\hfill & t\in {\mathsf{\Omega }}_3\hfill \end{array}\right.\end{array}$

Note that ${\mathsf{\Phi }}^{\pm }\left(0\right)\ge 0$, ${\mathsf{\Phi }}^{\pm }\left(3\right)\ge 0$ for an appropriate selection of $C>0$. Additionally,

$\begin{array}{ccc}\hfill L_1{\mathsf{\Phi }}^{\pm }\left(t\right)& =& C\left[-\alpha +p\left(t\right)+r\left(t\right)exp(-\frac{\sqrt{\alpha }}{\sqrt{\epsilon }})\right]exp(\frac{-x\sqrt{\alpha }}{\sqrt{\epsilon }})\pm L_1{w}_L\left(t\right),\hfill \\ & \ge & 0.\hfill \end{array}$

$\begin{array}{ccc}\hfill L_2{\mathsf{\Phi }}^{\pm }\left(t\right)& =& C\left[-\alpha +p\left(t\right)+q\left(t\right)exp(\frac{\sqrt{\alpha }}{\sqrt{\epsilon }})+r\left(t\right)exp(-\frac{\sqrt{\alpha }}{\sqrt{\epsilon }})\right]exp(\frac{-(x-1)\sqrt{\alpha }}{\sqrt{\epsilon }})\pm L_2{w}_L\left(t\right),\hfill \\ & \ge & 0.\hfill \end{array}$

$\begin{array}{ccc}\hfill L_3{\mathsf{\Phi }}^{\pm }\left(t\right)& =& C\left[-\alpha +p\left(t\right)+q\left(t\right)exp(\frac{\sqrt{\alpha }}{\sqrt{\epsilon }})\right]exp(\frac{-\sqrt{\alpha }(x-2)}{\sqrt{\epsilon }})\pm L_3{w}_L\left(t\right),\hfill \\ & =& C\left[(p\left(t\right)-\alpha )exp(-\frac{\sqrt{\alpha }}{\sqrt{\epsilon }})+q\left(t\right)\right]exp(\frac{\sqrt{\alpha }}{\sqrt{\epsilon }})exp(\frac{-\sqrt{\alpha }(x-2)}{\sqrt{\epsilon }})\pm L_3{w}_L\left(t\right),\hfill \\ & \ge & 0.\hfill \end{array}$

By the Lemma 1, we get left-layer bounds.

The estimations of $|w_L^{\prime }\left(t\right)|$ result from the integration of (9). The remaining derivative estimations (11) may be derived from the differential Equation (9).

The bounds for right-layer components (12) can be derived in a similar manner.

Hence, the proof is completed. □

In the following lemma, sharper estimates of the smooth component are presented.

Lemma 5.

The following constraints are satisfied by the solution’s regular component $v\left(t\right)$ of $z\left(t\right)$.

$\begin{array}{ccc}\hfill \parallel v^k{\left(t\right)\parallel }_{{\mathsf{\Omega }}^{*}}& \le & C{\epsilon }^{-k/2}\left\{\begin{array}{cc}(1+exp(\frac{-t\sqrt{\alpha }}{\sqrt{\epsilon }})+exp(\frac{-(1-t)\sqrt{\alpha }}{\sqrt{\epsilon }})),\hfill & t\in {\mathsf{\Omega }}_1\hfill \\ (1+exp(\frac{-(t-1)\sqrt{\alpha }}{\sqrt{\epsilon }})+exp(\frac{-(2-t)\sqrt{\alpha }}{\sqrt{\epsilon }})),\hfill & t\in {\mathsf{\Omega }}_2\hfill \\ (1+exp(\frac{-(t-2)\sqrt{\alpha }}{\sqrt{\epsilon }})+exp(\frac{-(3-t)\sqrt{\alpha }}{\sqrt{\epsilon }})),\hfill & t\in {\mathsf{\Omega }}_3\hfill \end{array}\right.\hfill \end{array}$

Proof. 

The barrier functions and Lemma 1 can be used to demonstrate this theorem.

$\begin{array}{c}\hfill {\mathsf{\Phi }}^{\pm }\left(t\right)=C{\epsilon }^{-k/2}\left\{\begin{array}{cc}(1+exp(\frac{-t\sqrt{\alpha }}{\sqrt{\epsilon }})+exp(\frac{-(1-t)\sqrt{\alpha }}{\sqrt{\epsilon }})),\hfill & t\in {\mathsf{\Omega }}_1\hfill \\ (1+exp(\frac{-(t-1)\sqrt{\alpha }}{\sqrt{\epsilon }})+exp(\frac{-(2-t)\sqrt{\alpha }}{\sqrt{\epsilon }})),\hfill & t\in {\mathsf{\Omega }}_2\pm v\left(t\right)\hfill \\ (1+exp(\frac{-(t-2)\sqrt{\alpha }}{\sqrt{\epsilon }})+exp(\frac{-(3-t)\sqrt{\alpha }}{\sqrt{\epsilon }})),\hfill & t\in {\mathsf{\Omega }}_3\hfill \end{array}\right.\end{array}$

□

4. The Discrete Problem

4.1. Shishkin Mesh

The continuous problem denoted by (2)–(4) displays strong inner layers (left and right) at $x=1$ and $x=2$, as well as strong boundary layers at $x=0$ and $x=3$. In order to create three piecewise uniform Shishkin meshes, the interval $[0,1]$ is divided as follows $[0,\rho ],(\rho ,1-\rho ],(1-\rho ,1]$, where the transitional parameter $\rho =min\{0.25,0.5\sqrt{(\frac{\epsilon }{\alpha }})logN\}.$

Likewise, $(1,2]$ is partitioned into $(1,1+\rho ],(1+\rho ,2-\rho ],(2-\rho ,2].$ and $(2,3]$ is partitioned into $(2,2+\rho ],(2+\rho ,3-\rho ],(3-\rho ,3].$

In the interval $(\rho ,1-\rho ],(1+\rho ,2-\rho ],(2+\rho ,3-\rho ]$, a uniform mesh with $\frac{N}{2}$ mesh points is placed, and a uniform mesh with $\frac{N}{4}$ mesh points is also placed in each of the subintervals $[0,\rho ],(1-\rho ,1],(1,1+\rho ],(2-\rho ,2],(2,2+\rho ],(3-\rho ,3]$.

The mesh ${\overline{\mathsf{\Omega }}}^{3N}=\{t_0,t_1,\cdots \cdots ,t_{3N}\}$ is defined by

$\begin{array}{cc}\hfill t_0=& 0\\ \hfill t_i=& t_0+ih,& i=1to\frac{N}{4}\hfill \\ \hfill t_{i+\frac{N}{4}}=& t_{\frac{N}{4}}+iH,& i=1to\prime \frac{N}{2}\hfill \\ \hfill t_{i+\frac{3N}{4}}=& t_{\frac{3N}{4}}+ih,& i=1to\frac{N}{4}\hfill \\ \hfill t_{i+N}=& t_N+ih,& i=1to\frac{N}{4}\hfill \\ \hfill t_{i+\frac{5N}{4}}=& t_{\frac{5N}{4}}+iH,& i=1to\frac{N}{2}\hfill \\ \hfill t_{i+\frac{7N}{4}}=& t_{\frac{7N}{4}}+ih,& i=1to\frac{N}{4}\hfill \\ \hfill t_{i+2N}=& t_{2N}+ih,& i=1to\frac{N}{4}\hfill \\ \hfill t_{i+\frac{9N}{4}}=& t_{\frac{9N}{4}}+iH,& i=1to\frac{N}{2}\hfill \\ \hfill t_{i+\frac{11N}{4}}=& t_{\frac{11N}{4}}+ih,& i=1to\frac{N}{4}\hfill \end{array}$

where $h=\frac{4\rho }{N}, H=\frac{2(1-2\rho )}{N}$.

4.2. Bakhvalov–Shishkin Mesh

The mesh ${\overline{\mathsf{\Omega }}}^{3N}=\{t_0,t_1,\cdots \cdots ,t_{3N}\}$ is defined by

$\begin{array}{c}\hfill \left\{\begin{array}{cc}2\sqrt{\frac{\epsilon }{\alpha }}{\varphi }_1\left(t_i\right),\hfill & i=0,1,\dots ,\frac{N}{4}\hfill \\ \rho +\frac{2}{N}(1-2\rho )(i-\frac{N}{4}),\hfill & i=\frac{N}{4}+1,\dots ,\frac{3N}{4}\hfill \\ 1-2\sqrt{\frac{\epsilon }{\alpha }}{\varphi }_2\left(t_i\right),\hfill & i=\frac{3N}{4}+1,\dots ,N\hfill \\ 1+2\sqrt{\frac{\epsilon }{\alpha }}{\varphi }_3\left(t_i\right),\hfill & i=N+1,\dots ,\frac{5N}{4}\hfill \\ 1+\rho +\frac{2}{N}(1-2\rho )(i-\frac{5N}{4}),\hfill & i=\frac{5N}{4}+1,\dots ,\frac{7N}{4}\hfill \\ 2-2\sqrt{\frac{\epsilon }{\alpha }}{\varphi }_4\left(t_i\right),\hfill & i=\frac{7N}{4}+1,\dots ,2N\hfill \\ 2+2\sqrt{\frac{\epsilon }{\alpha }}{\varphi }_5\left(t_i\right),\hfill & i=2N+1,\dots ,\frac{9N}{4}\hfill \\ 2+\rho +\frac{2}{N}(1-2\rho )(i-\frac{9N}{4}),\hfill & i=\frac{9N}{4}+1,\dots ,\frac{11N}{4}\hfill \\ 3-2\sqrt{\frac{\epsilon }{\alpha }}{\varphi }_6\left(t_i\right),\hfill & i=\frac{11N}{4}+1,\dots ,3N\hfill \end{array}\right.\end{array}$

where $t_i=\frac{i}{N}, {\varphi }_j=-\ln {\psi }_j, j=1,2,3,4,5,6$

$\begin{array}{ccc}\hfill {\psi }_1& =& 1-4(1-N^{-1})t\hfill \\ \hfill {\psi }_2& =& 1-2(1-N^{-1})(2-2t)\hfill \\ \hfill {\psi }_3& =& 1-2(1-N^{-1})(2t-1)\hfill \\ \hfill {\psi }_4& =& 1-2(1-N^{-1})(4-2t)\hfill \\ \hfill {\psi }_5& =& 1-2(1-N^{-1})(2t-4)\hfill \\ \hfill {\psi }_6& =& 1-4(1-N^{-1})(3-t)\hfill \end{array}$

On the layer portion of the Shishkin mesh, it is widely known that

$h_i\le C\epsilon N^{-1}lnN$

Then, we have the Bakhvalov–Shishkin mesh.

$h_i\le \left\{\begin{array}{cc}\left(\frac{\epsilon }{\alpha }\right)\left(N^{-1}max|{\psi }_1^{\prime }\right|)\left(\exp \right(\frac{\alpha }{\epsilon }\left(t_{i-1}\right)\left)\right),\hfill & i=0,1,\dots ,\frac{N}{4}\hfill \\ \left(\frac{\epsilon }{\alpha }\right)\left(N^{-1}max|{\psi }_1^{\prime }\right|)\left(\exp \right(\frac{\alpha }{\epsilon }(1-t_{i-1})\left)\right),\hfill & i=\frac{3N}{4}+1,\dots ,N\hfill \\ \left(\frac{\epsilon }{\alpha }\right)\left(N^{-1}max|{\psi }_1^{\prime }\right|)\left(\exp \right(\frac{\alpha }{\epsilon }(t_{i-1}-1)\left)\right),\hfill & i=N+1,\dots ,\frac{5N}{4}\hfill \\ \left(\frac{\epsilon }{\alpha }\right)\left(N^{-1}max|{\psi }_1^{\prime }\right|)\left(\exp \right(\frac{\alpha }{\epsilon }(2-t_{i-1})\left)\right),\hfill & i=\frac{7N}{4}+1,\dots ,2N\hfill \\ \left(\frac{\epsilon }{\alpha }\right)\left(N^{-1}max|{\psi }_1^{\prime }\right|)\left(\exp \right(\frac{\alpha }{\epsilon }(t_{i-1}-2)\left)\right),\hfill & i=2N+1,\dots ,\frac{9N}{4}\hfill \\ \left(\frac{\epsilon }{\alpha }\right)\left(N^{-1}max|{\psi }_1^{\prime }\right|)\left(\exp \right(\frac{\alpha }{\epsilon }(3-t_{i-1})\left)\right),\hfill & i=\frac{11N}{4}+1,\dots ,3N\hfill \end{array}\right.$

and

$\frac{h_i}{\epsilon }\le C{N}^{-1}max\left|{\varphi }^{\prime }\right|\le C.$

4.3. Discretisation of the Problem

According to the original problem (2)–(4), the discrete strategy is as follows:

(13)$\left\{\begin{array}{cc}{L}_1^N{U}_i\hfill & =f_i-b_i{\varphi }_{i-N},1\le i\le N-1,\hfill \\ L_2^N{U}_i\hfill & =f_i,N+1\le i\le 2N-1,\hfill \\ L_3^N{U}_i\hfill & =f_i-c_i{\phi }_{i+N},2N+1\le i\le 3N-1,\hfill \end{array}\right.$

subject to the restrictions of the boundaries:

(14)$\left\{\begin{array}{c}{U}_i={\varphi }_i,i=-N,-N+1,\dots ,0.\hfill \\ U_i={\phi }_i,i=3N,\dots ,4N.\hfill \\ D^{-}{U}_N=D^{+}{U}_N,D^{-}{U}_{2N}=D^{+}{U}_{2N}\hfill \end{array}\right.$

where

$L_1^N{U}_i=-\epsilon {\delta }^{2}z\left(t_i\right)+p\left(t_i\right)z\left(t_i\right)+r\left(t_i\right)z\left(t_{i+N}\right)$

$L_2^N{U}_i=-\epsilon {\delta }^{2}z\left(t_i\right)+p\left(t_i\right)z\left(t_i\right)+q\left(t_i\right)z\left(t_{i-N}\right)+r\left(t_i\right)z\left(t_{i+N}\right)$

$L_3^N{U}_i=-\epsilon {\delta }^{2}z\left(t_i\right)+p\left(t_i\right)z\left(t_i\right)+q\left(t_i\right)z\left(t_{i-N}\right)$

Lemma 6 (Discrete Maximum Principle).

Let $\Xi \left(t_i\right)$ satisfies $\Xi \left(t_0\right)\ge 0$, and $\Xi \left(t_{3N}\right)\ge 0$. Then, $L_1^N\Xi \left(t_i\right)\ge 0$, $L_2^N\Xi \left(t_i\right)\ge 0$, $L_3^N\Xi \left(t_i\right)\ge 0$, $D^{+}(\Xi \left(t_N\right))-D^{-}(\Xi \left(t_N\right))\le 0$ and $D^{+}(\Xi \left(t_{2N}\right))-D^{-}(\Xi \left(t_{2N}\right))\le 0$ imply that $\Xi \left(t_i\right)\ge 0$, $\forall t_i\in {\overline{\mathsf{\Omega }}}^{3N}$.

Proof. 

$\begin{array}{c}\hfill \mathrm{Define}s\left(t_i\right)=\left\{\begin{array}{cc}\frac{1}{12}+\frac{t_i}{4},\hfill & t_i\in [0,1]\cap {\overline{\mathsf{\Omega }}}^{3N},\hfill \\ \frac{2}{12}+\frac{t_i}{6},\hfill & t_i\in [1,2]\cap {\overline{\mathsf{\Omega }}}^{3N},\hfill \\ \frac{4}{12}+\frac{t_i}{12},\hfill & t_i\in [2,3]\cap {\overline{\mathsf{\Omega }}}^{3N}.\hfill \end{array}\right.\end{array}$

Note that $s\left(t_i\right)>0,\forall t_i\in {\overline{\mathsf{\Omega }}}^{3N}$, $Ls\left(t_i\right)>0,\forall t_i\in {\mathsf{\Omega }}_1^{3N}\cup {\mathsf{\Omega }}_2^{3N}\cup {\mathsf{\Omega }}_3^{3N}$, $s\left(0\right)>0$, $s\left(t_{3N}\right)>0$, $\left[s^{\prime }\right]\left(t_N\right)<0$ and $\left[s^{\prime }\right]\left(t_{2N}\right)<0$. Let

$\mu =min\{\frac{-\Xi \left(t_i\right)}{s\left(t_i\right)}:t_i\in {\overline{\mathsf{\Omega }}}^{3N}\}.$

Then, there exists $t_k\in {\overline{\mathsf{\Omega }}}^{3N}$, such that $\Xi \left(t_k\right)+\mu s\left(t_k\right)=0$ and $\Xi \left(t_i\right)+\mu s\left(t_i\right)\ge 0,\forall t_i\in {\overline{\mathsf{\Omega }}}^{3N}$. As a result, at $x=t_k$, the function $(\Xi +\mu s)$ reaches its maximum value. If the theorem is false, then $\mu >0$ will apply.

Case (i): $t_k=t_0$

$0<(\Xi +\mu s)\left(t_0\right)=0$

Case (ii): $t_k\in {\mathsf{\Omega }}_1^{3N}$

$\begin{array}{c}\hfill 0<L_1(\Xi +\mu s)\left(t_k\right)=-\epsilon {\delta }^2(\Xi +\mu s)\left(t_i\right)+p\left(t_i\right)(\Xi +\mu s)\left(t_i\right)+r\left(t_i\right)(\Xi +\mu s)\left(t_{i+N}\right)\le 0\end{array}$

Case (iii): $t_k=t_N$

$0\le \left[D\right](\Xi +\mu s)\left(t_N\right)<0$

Case (iv): $t_k\in {\mathsf{\Omega }}_2^{3N}$

$\begin{array}{ccc}\hfill 0<L_2(\Xi +\mu s)\left(t_k\right)& =& -\epsilon {\delta }^2(\Xi +\mu s)\left(t_i\right)+p\left(t_i\right)(\Xi +\mu s)\left(t_i\right)\hfill \\ & & +q\left(t_i\right)(\Xi +\mu s)\left(t_{i-N}\right)+r\left(t_i\right)(\Xi +\mu s)\left(t_{i+N}\right)\le 0\hfill \end{array}$

Case (v): $t_k=t_{2N}$

$0\le \left[D\right](\Xi +\mu s)\left(t_{2N}\right)<0$

Case (v): $t_k\in {\mathsf{\Omega }}_3^{3N}$

$\begin{array}{ccc}\hfill 0<L_3(\Xi +\mu s)\left(t_k\right)& =& -\epsilon {\delta }^2(\Xi +\mu s)\left(t_i\right)+p\left(t_i\right)(\Xi +\mu s)\left(t_i\right)+q\left(t_i\right)(\Xi +\mu s)\left(t_{i-N}\right)\le 0\hfill \end{array}$

Case (vi): $t_k=t_{3N}$

$0<(\Xi +\mu s)t_{3N}=0$

This is in conflict with every case. Hence, it offers proof. □

Lemma 7.

Consider any mesh function as $\Xi \left(t_i\right)$, for $0\le i\le 3N$. Then,

$|\Xi \left(t_i\right)|\le Cmax\left\{|\Xi \left(t_0\right)|,|\Xi \left(t_{3N}\right)|,\underset{i\in {\mathsf{\Omega }}_1^{3N}\cup {\mathsf{\Omega }}_2^{3N}\cup {\mathsf{\Omega }}_3^{3N}}{max}|L^N\Xi \left(t_i\right)|\right\}.$

Proof. 

The barrier functions and Lemma 6 can be used to demonstrate this theorem.

(15)${\theta }^{\pm }\left(t_i\right)=CM\pm \Xi \left(t_i\right),0\le i\le 3N,$

where

$M=max\{|\Xi \left(t_0\right)|,|\Xi \left(t_{3N}\right)|,\underset{i\in {\mathsf{\Omega }}_1^{3N}\cup {\mathsf{\Omega }}_2^{3N}\cup {\mathsf{\Omega }}_3^{3N}}{max}|L^N\Xi \left(t_i\right)|\}.$

□

5. Error Estimate

We deconstruct the discrete solution $U\left(t_i\right)$ as $U\left(t_i\right)=V\left(t_i\right)+W\left(t_i\right),$ where $V\left(t_i\right)$ and $W\left(t_i\right)$, such that

$\begin{array}{ccc}\hfill L_1^{N}V\left(t_i\right)& =& f_i-b_i{\varphi }_{i-N},i=1\mathrm{to}N-1,\hfill \\ \hfill V\left(t_0\right)& =& v\left(0\right),\hfill \\ \hfill V\left(t_{N-1}\right)& =& v\left(1^{-}\right).\hfill \end{array}$

$\begin{array}{ccc}\hfill L_2^{N}V\left(t_i\right)& =& f_i,i=N+1\mathrm{to}2N-1,\hfill \\ \hfill V\left(t_{N+1}\right)& =& v\left(1^{+}\right),\hfill \\ \hfill V\left(t_{2N-1}\right)& =& v\left(2^{-}\right).\hfill \end{array}$

$\begin{array}{ccc}\hfill L_3^{N}V\left(t_i\right)& =& f_i-{\phi }_{i+N},i=2N+1\mathrm{to}3N-1,\hfill \\ \hfill V\left(t_{2N+1}\right)& =& v\left(2^{+}\right),\hfill \\ \hfill V\left(t_{3N}\right)& =& v\left(3\right).\hfill \end{array}$

and

$\begin{array}{ccc}\hfill L_1^{N}W\left(t_i\right)& =& 0,i=1,\mathrm{to}N-1,\hfill \\ \hfill W\left(t_0\right)& =& w\left(0\right),\hfill \\ \hfill L_2^{N}W\left(t_i\right)& =& 0,i=N+1\mathrm{to}2N-1,\hfill \\ \hfill W\left(t_{2N}\right)& =& w\left(2\right),\hfill \\ \hfill L_3^{N}W\left(t_i\right)& =& 0,i=2N+1\mathrm{to}3N-1,\hfill \\ \hfill W\left(t_{3N}\right)& =& w\left(3\right),\hfill \end{array}$

At $i=N,2N$

$\begin{array}{c}V\left(t_{i+1}\right)+W\left(t_{i+1}\right)=V\left(t_{i-1}\right)+W\left(t_{i-1}\right)\\ D^{-}W\left(t_i\right)+D^{-}V\left(t_i\right)=D^{+}W\left(t_i\right)+D^{+}V\left(t_i\right).\end{array}$

Defines the nodal error $e\left(t_i\right)=U\left(t_i\right)-u\left(t_i\right)=(V\left(t_i\right)-v\left(t_i\right))+(W\left(t_i\right)-w\left(t_i\right)).$

5.1. Error Estimate for Shishkin Mesh

Theorem 1.

The smooth component $v\left(t\right)$ and $V\left(t_i\right)$ of (2)–(4) and (13)–(14), respectively, satisfy the following estimate

$\begin{array}{ccc}\hfill |(V-v)\left(t_i\right)|& \le & C{N}^{-1},0\le i\le 3N.\hfill \end{array}$

The singular component $w\left(t\right)$ and $W\left(t_i\right)$ of (2)–(4) and (13)–(14), respectively, satisfy the following estimate

$\begin{array}{ccc}\hfill |(W-w)\left(t_i\right)|& \le & C{N}^{-1}lnN,0\le i\le 3N.\hfill \end{array}$

Proof. 

Now,

$\begin{array}{ccc}\hfill |L_1^N(V-v)\left(t_i\right)|& \le & -\epsilon ({\delta }^2-\frac{d^2}{d{x}^2})v\left(t_i\right)+r\left(t_i\right)(v\left(t_{i+N}\right)-v(t_i+1))\hfill \\ \hfill |L_2^N(V-v)\left(t_i\right)|& \le & -\epsilon ({\delta }^2-\frac{d^2}{d{x}^2})v\left(t_i\right)+q\left(t_i\right)(v\left(t_{i-N}\right)-v(t_i-1))\hfill \\ \hfill & +& r\left(t_i\right)(v\left(t_{i+N}\right)-v(t_i+1))\hfill \\ \hfill |L_3^N(V-v)\left(t_i\right)|& \le & -\epsilon ({\delta }^2-\frac{d^2}{d{x}^2})v\left(t_i\right)+q\left(t_i\right)(v\left(t_{i-N}\right)-v(t_i-1))\hfill \end{array}$

Since $|v\left(t_{i-N}\right)-v(t_i-1)|\le C{N}^{-2}$ in [22], we find that $|L_k^N(V-v)\left(t_i\right)|\le C{N}^{-1},i\in {\overline{\mathsf{\Omega }}}^{3N}\backslash \{0,N,2N,3N\},k=1,2,3$

Then, according to Lemma 7, we have $|(V-v)\left(t_i\right)|\le C{N}^{-1},0\le i\le 3N,$

The necessary constraints are met by the equation generated for the local truncation error in W and estimates for the derivatives of the singular components, both of which take exactly the form provided in Chapter 6 of [23].

$|L_k^N(W-w)\left(t_i\right)|\le C{N}^{-1}lnN,i\in {\overline{\mathsf{\Omega }}}^{3N}\backslash \{0,N,2N,3N\},k=1,2,3$. Lemma 7 then gives us $|(W-w)\left(t_i\right)|\le C{N}^{-1}lnN,0\le i\le 3N.$ □

Theorem 2.

Let $u\left(t\right)$ represent the solution to problems (2)–(4) and $u\left(t_i\right)$ represent the solution to problems (13)–(14) for $0\le i\le 3N,$

$|(U-u)\left(t_i\right)|\le C{N}^{-1}lnN.$

5.2. Bakhvalov–Shishkin Mesh Error Approximate

Theorem 3.

The smooth component $v\left(t\right)$ and $V\left(t_i\right)$ of (2)–(4) and (13)–(14), respectively, satisfy the following estimate

$\begin{array}{ccc}\hfill |(V-v)\left(t_i\right)|& \le & C{N}^{-1},0\le i\le 3N.\hfill \end{array}$

The singular component $w\left(t\right)$ and $W\left(t_i\right)$ of (2)–(4) and (13)–(14), respectively, satisfy the following estimate

$\begin{array}{ccc}\hfill |(W-w)\left(t_i\right)|& \le & C{N}^{-1},0\le i\le 3N,\hfill \end{array}$

Proof. 

By adopting the methods of proof of Theorem 1, we can prove

$|(V-v)\left(t_i\right)|\le C{N}^{-1}.$

To calculate the solution’s singular component error:

If $t_i\in {\mathsf{\Omega }}_1^{3N}$, then $|L_1^N(W-w)\left(t_i\right)|\le C{h}_i^{2}max|w^{″}\left(t_i\right)|$

If $t_i\in [\rho ,1-\rho ]$ we have

$|L_1^N(W-w)\left(t_i\right)|\le C{N}^{-1}$

If $t_i\in [0,\rho ]\cup [1-\rho ,1]$ we have

$\begin{array}{ccc}\hfill |L_1^N(W-w)\left(t_i\right)|& \le & C{(\frac{\epsilon }{\alpha })}^2(N^{-1}max|{\psi }_1^{\prime }{|)}^2{(\exp (\frac{\alpha }{\epsilon }\left(t_{i-1}\right)))}^2\hfill \\ & \le & C{\epsilon }^2{N}^{-2}(max|\frac{{\psi }_1^{\prime }}{{\psi }_1}{|)}^2\hfill \\ & \le & C{\epsilon }^2[N^{-1}max|{\varphi }_1{|]}^2\hfill \\ & \le & C{N}^{-1}.\hfill \end{array}$

$\begin{array}{ccc}\hfill |L_1^N(W-w)\left(t_i\right)|& \le & C{(\frac{\epsilon }{\alpha })}^2(N^{-1}max|{\psi }_1^{\prime }{|)}^2{(\exp (\frac{\alpha }{\epsilon }(1-t_{i-1})))}^2\hfill \\ & \le & C{N}^{-1}.\hfill \end{array}$

Similarly $|L_k^N(W-w)\left(t_i\right)|\le C{N}^{-1}$, where k = 2,3. following Lemma 7

$|(W-w)\left(t_i\right)|\le C{N}^{-1},0\le i\le 3N$

□

Theorem 4.

Let $u\left(t_i\right)$ be the solution of the problem (2)–(4) and $U\left(t_i\right)$ be the solution of the problem (13)–(14). Then, for $0\le i\le 3N$,

$|(U-u)\left(t_i\right)|\le C{N}^{-1}.$

6. Numerical Experiments

In this section, two examples are provided to clarify the numerical approach mentioned earlier. The test issues’ precise solutions are unknown. As a result, we compute the experiment rate of convergence to the estimated solution and estimate the error using the double mesh concept. To this end, we put

$D_{\epsilon }^N=\underset{0\le i\le 3N}{max}|U_i^N-U_{2i}^{2N}|$

where ith components of the numerical solutions on N and $2N$ meshes, respectively, are $U_i^N$ and $U_{2i}^{2N}$. Calculations used to determine the uniform error and convergence rate include

$D^N=\underset{\epsilon }{max}{D}_{\epsilon }^N\mathrm{and}{P}^N=lo{g}_2(\frac{D^N}{D^{2N}}).$

For the values of the perturbation parameter $\epsilon \in \{2^{-2},2^{-3},\dots ,2^{-20}\}$ and $\epsilon \in \{2^{-11},2^{-12},\dots ,2^{-20}\}$ for Shishkin mesh and Bakhvalov–Shishkin mesh, respectively, the numerical results are reported.

Example 1.

$\begin{array}{c}-\epsilon z^{″}\left(t\right)+5z\left(t\right)-z(t-1)-z(t+1)=1,fort\in {\mathsf{\Omega }}^{*},\\ z\left(t\right)=1fort\in [-1,0],z\left(t\right)=1fort\in [3,4].\end{array}$

Example 2.

$\begin{array}{c}-\epsilon z^{″}\left(t\right)+(5+x^2)z\left(t\right)-z(t-1)-z(t+1)=x,fort\in {\mathsf{\Omega }}^{*},\\ z\left(t\right)=1fort\in [-1,0],z\left(t\right)=1fort\in [3,4].\end{array}$

7. Conclusions

We used the finite difference approach on the Shishkin mesh and Bakhvalov–Shishkin mesh to solve a class of singularly perturbed reaction–diffusion-type differential equations with delay and advanced argument (2)–(4). On Shishkin mesh and Bakhvalov–Shishkin mesh, the approach was demonstrated to be of order $O\left(N^{-1}lnN\right)$ and $O\left(N^{-1}\right)$ with regard to $\epsilon$. To demonstrate the numerical technique, two examples are provided. The theoretical estimations are reflected in our numerical findings. Table 1 and Table 2 provide the maximum pointwise errors and convergence order for Examples 1 and 2, respectively. Figure 1 shows the numerical solution to Example 1. Figure 2 displays the numerical solution to Example 2.

In future work, we will develop the finite element method for singularly perturbed non-linear partial mixed-delay differential equations with Dirichlet and Robin boundary conditions.

Author Contributions

Conceptualization, S.E. and B.U.; methodology, S.E.; software, B.U.; validation, B.U and S.E.; formal analysis, B.U.; investigation, S.E.; data curation, S.E.; writing—original draft preparation, B.U.; writing—review and editing, supervision, B.U. All authors have read and agreed to the published version of the manuscript.

Data Availability Statement

Data sharing is not applicable to this article as no datasets were generated or analyzed during the current study.

Acknowledgments

We thank the reviewers and editors for their valuable comments.

Conflicts of Interest

The authors declare that there are no conflict of interest regarding the publication of this paper.

Footnotes

Disclaimer/Publisher’s Note: The statements, opinions and data contained in all publications are solely those of the individual author(s) and contributor(s) and not of MDPI and/or the editor(s). MDPI and/or the editor(s) disclaim responsibility for any injury to people or property resulting from any ideas, methods, instructions or products referred to in the content.

Figures and Tables

Figure 1. The solution for Example 1.

Figure 2. The solution for Example 2.

Table 1

Derived $\epsilon$-uniform errors $D^N$ and orders of convergence $P^N$ for Example 1.

3N Is the Mesh Points
Shishkin Mesh
	32	64	128	256	512	1024	2048
$D^N$	$5.2513\times {10}^{-3}$	$3.4209\times {10}^{-3}$	$2.0832\times {10}^{-3}$	$1.1436\times {10}^{-3}$	$7.3014\times {10}^{-4}$	$3.7640\times {10}^{-4}$	$2.0644\times {10}^{-4}$
$P^N$	$6.1827\times {10}^{-1}$	$7.1558\times {10}^{-1}$	$8.6515\times {10}^{-1}$	$6.4740\times {10}^{-1}$	$9.5589\times {10}^{-1}$	$8.6652\times {10}^{-1}$
Bakhvalov–Shishkin Mesh
$D^N$	$2.9952\times {10}^{-3}$	$1.5057\times {10}^{-3}$	$7.5395\times {10}^{-4}$	$3.7712\times {10}^{-4}$	$1.8858\times {10}^{-4}$	$9.4293\times {10}^{-5}$	$4.7146\times {10}^{-5}$
$P^N$	$9.9221\times {10}^{-1}$	$9.9790\times {10}^{-1}$	$9.9944\times {10}^{-1}$	$9.9985\times {10}^{-1}$	$9.9995\times {10}^{-1}$	$9.9998\times {10}^{-1}$

Table 2

Derived $\epsilon$-uniform errors $D^N$ and orders of convergence $P^N$ for Example 2.

3N Is the Mesh Point Count
Shishkin Mesh
	32	64	128	256	512	1024	2048
$D^N$	$5.9350\times {10}^{-3}$	$3.9301\times {10}^{-3}$	$2.4026\times {10}^{-3}$	$1.3469\times {10}^{-3}$	$8.4650\times {10}^{-4}$	$4.3631\times {10}^{-4}$	$2.4257\times {10}^{-4}$
$P^N$	$5.9466\times {10}^{-1}$	$7.0994\times {10}^{-1}$	$8.3491\times {10}^{-1}$	$6.7014\times {10}^{-1}$	$9.5613\times {10}^{-1}$	$8.4693\times {10}^{-1}$
Bakhvalov–Shishkin Mesh
$D^N$	$3.3073\times {10}^{-3}$	$1.6716\times {10}^{-3}$	$8.3715\times {10}^{-4}$	$4.1850\times {10}^{-4}$	$2.0918\times {10}^{-4}$	$1.0457\times {10}^{-4}$	$5.2278\times {10}^{-5}$
$P^N$	$9.8440\times {10}^{-1}$	$9.9769\times {10}^{-1}$	$1.0002\times e^{+00}$	$1.0004\times e^{+00}$	$1.0003\times e^{+00}$	$1.0001\times e^{+00}$

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