Finite difference scheme for singularly perturbed reaction diffusion problem of partial delay differential equation with nonlocal boundary condition

This paper investigates singularly perturbed parabolic partial differential equations with delay in space, and the right end plane is an integral boundary condition on a rectangular domain. A small parameter is multiplied in the higher order derivative, which gives boundary layers, and due to the delay term, one more layer occurs on the rectangle domain. A numerical method comprising the standard finite difference scheme on a rectangular piecewise uniform mesh (Shishkin mesh) of Nr×Nt\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$N_{r} \times N_{t}$\end{document} elements condensing in the boundary layers is suggested, and it is proved to be parameter-uniform. Also, the order of convergence is proved to be almost two in space variable and almost one in time variable. Numerical examples are proposed to validate the theory.


Introduction
Singularly perturbed parabolic partial differential equations have received an increasing research attention over the past few decades (see [1][2][3][4][5][6][7][8][9] and the references therein). Ansari et al. [10] discussed singularly perturbed time delay partial differential equations with Dirichlet boundary conditions. Avudai Selvi and Ramanujam [11] discussed singularly perturbed parabolic partial time delay differential equations with Robin type boundary condition. The authors in [10] and [11] studied a finite difference scheme for solving singularly perturbed parabolic time delay differential equation with Dirichlet and Robin boundary conditions on Shishkin mesh. They made use of the results of Miller et al. [12]. In [13][14][15][16] researchers discussed a fitted operator method to solve singularly perturbed time delay partial differential equations. By the way, there are many methods available in the literature for time delay problems [17] and [18], but the study of problems with delay in space variable are still in the initial stage.
The above problem arises in the study of population density with time delay. There is no flux condition considered in the left, and the average population size is being controlled by the function ψ(t) in the right. In the past few years interest has substantially increased in solving singularly perturbed differential equations with integral boundary conditions (see [20][21][22][23][24][25]). Motivated by the above works, we have designed a finite difference scheme for a singularly perturbed reaction diffusion problem of partial delay differential equation with nonlocal boundary condition.
The remaining article is structured as follows: In Sect. 2, the model problem is stated and some preliminaries are presented. In Sect. 3, uniqueness and stability of the solution are established. Also we prove that the derivatives of the solution are bounded. A finite difference method for the continuous problem, the discrete maximum principle, and the stability result are discussed in Sect. 4. The convergence analysis of the finite difference method on Shishkin mesh is given in Sect. 5. The theoretical results are verified by numerical examples in Sect. 6. Finally, discussion is given in Sect. 7.
In all the cases we arrived at a contradiction, and thus the proof is completed.

Lemma 2 (Stability result) If u(r, t) satisfies problem (2)-(4), then the bound of u(r, t) is
Proof It can be easily proved by using the maximum principle (Lemma 1) and the barrier functions θ ± (r, t) = CMs(r, t) ± u(r, t), (r, t) ∈D, where M = max{ u l , u b , Ku r , Lu D * } and s(r, t) is the test function as in (5).

The solution and its derivatives are bounded
Proof The first part of the proof is given in Ladyzhenskaya [26, Chap. IV, p. 320]. The solution and its derivatives are bounded as follows. Under the stretched transformatioñ r = r √ ε , problem (1) can be rewritten as follows: The equivalent problem is whereD ε = (0, 2 √ ε ) × (0, T] and the boundary conditions˜ to . Note that (8) is independent of ε. Then, using the idea of estimation (10.6) from [26, p. 352], we get The proof is completed by using the bound on u ε in Lemma 2.

Decomposition of the solution
The solution u ε of (2)-(4) is decomposed into v ε -smooth and w ε -singular components.
where v 0 and v 1 are solutions of the following differential equations respectively: and Then v ε satisfies the condition The singular component w ε is determined from and The reaction diffusion problem has boundary layers on both boundaries l and r . Separate w ε as w ε = w l + wr, where w l and wr satisfy the following problems: and Here, we choose A to be a suitable constant, then the solution is continuous at r = 1.
where C is a constant independent of the parameter ε, (r, t) ∈D, i, j ≥ 0, 0 ≤ i + 2j ≤ 4. First, we estimate that the reduced problem solution v 0 is bounded and its derivative is bounded

Proof
Using Theorem 2, we estimate that the derivative of the solution v 1 becomes By (23) and (24), we establish smooth component v ε estimates.
To prove inequality (21), following the procedure adapted in [12], it is easy to find that Now, we derive the bound on w l on D 2 . From the defining equations for w l , we have or L 2 w l (r, t) = -εw rr (r, t) + w t (r, t) + a(r, t)w l (r, t) = -b(r, t)w l (r -1, t), for a choice of suitable C > 0. Using the stability result (Lemma 2), we have |w l (r, t)| ≤
Note that θ ± (0, t) ≥ 0, for a choice of suitable C > 0. Using the maximum principle (Lemma 1), we have Clearly, under the transformation, the stretched variabler domain is (0, 2 √ ε ), note that (r, t) ⇒ (r, t) and the parameter ε is independent of problem (18)- (19), then suitable estimates in [26,Sect. 4.10] are the solution ofwr. The position ofr argument is divided into two cases. In the first case, for the neighborhood ofÑ δ in (0, 2 and derive the essential bound of r; moreover, the variabler changing into variable r and using bound of wr, observe e -r √ ε ≥ e -2 = C forr ≤ 2. Hence proved.

The discretised problem
In this part, we apply the finite difference method to the continuous problem (2)-(4) on a piecewise uniform mesh. The second order space derivative (u rr ) is replaced by the central difference scheme (δ 2 r U), and the first order time derivative (u t ) is replaced by the backward difference scheme (Dt U). In r-direction the interval = [0, 2] is divided intō 1 = [0, 1] and¯ 2 = [1, 2] with N 2 equal mesh points. Furthermore, the piecewise uniform mesh (Shishkin mesh)¯ 1 = [0, 1] is divided into three subintervals Similarly, the interval¯ 2 = [1, 2] is divided into three subintervals , 2). Furthermore, μ is called a fitting factor, and it satisfies the following condition: where N r denotes the number of mesh elements in the r-direction. A piecewise uniform mesh N r μ is on with N r mesh elements. Both intervals l and r are uniform meshes with N r 8 elements and c is also a uniform mesh with N r 4 mesh element, and thus we divide r-direction. Uniform meshes on t-direction on step size t and the number mesh denotes N t in t-direction. D N μ = N r μ × N t and the boundary analogues Using the finite difference method for the continuous problem (2)- (4).
with boundary conditions (27) The differential operator notation is and an analogous definition of Dt .

Lemma 3 (Discrete maximum principle) Let Z be any mesh function satisfying Z(r
Proof Define a test function S(r i , t j ) as (28) Then there exists (r * , t * ) ∈D N such that Z(r * , t * ) + ηS(r * , t * ) = 0 and Z(r i , t j ) + ηS(r i , t j ) ≥ 0, ∀(r i , t j ) ∈D N . Then the function is minimum at (r, t) = (r * , t * ) and this proof is completed.
In all the cases we arrived at a contradiction. Hence proved.

Lemma 4 Prove that
Proof It can be easily proved using the discrete maximum principle (Lemma 3) and the barrier functions ± (r i , and S(r i , t j ) is the test function as in Lemma 3.

Error estimate
In this section, we prove parameter uniform convergence of the numerical solution.

Theorem 4 Let u ε and U ε be solutions of problems (2)-(4) and (25)-(27).
Assume that the coefficients a(r, t), b(r, t), f (r, t) ∈ C (4+α 1 ,2+α 1 /2) (D) and the boundary conditions satisfy Proof Since the continuous problem solution u ε of L ε is decomposed into smooth and singular components, by the same process, the discrete problem solution U ε of L N ε is decomposed into smooth and singular components as where the V ε -smooth component is the solution of the following problem: and therefore W ε must satisfy The error can be written in the form and we prove separately the error estimates of smooth and singular components. First, we derive the error estimate for the smooth component using the following classical argument.
At the point r i = r N , From the difference and differential equations, it is not hard to see that It follows from classical estimates [12], at each point (r i , t j ) in D N 1 , Similarly, where θ 1 and θ 2 are piecewise linear polynomials and Then, for all (r i , t j ) ∈D N , and also Similarly, note that 1+μ √ ε ≤ 2 ln N and Define the barrier functions Then, from the discrete maximum principle, Then we have Next, we derive an error estimate for the singular component. We decompose W ε into W l and Wr, where