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

Abstract

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 $N_{r} \times N_{t}$ 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.

1 Introduction

Singularly perturbed parabolic partial differential equations have received an increasing research attention over the past few decades (see [1–9] 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 [13–16] 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.

In [19], the authors considered second order parabolic delay differential equations with integral boundary condition of the form

$$ \textstyle\begin{cases} -k u_{rr}(r,t) +u_{t}(r,t) =-d(r,t)u(r,t)+b(r,t)u(r,t-\tau ), \quad (r,t) \in (0,\pi )\times [0,T], \\ u(r,t)=\phi (r,t), \quad t\in [-\tau ,0],y\in (0,\pi ), \\ u_{r}(0,t)=0,\quad t\in [0,T], \\ \int _{0}^{\pi }u(r,t)\,dr=\psi (t),\quad t\in [0,T], \end{cases} $$

and proved the existence and uniqueness of the solution.

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 $\psi (t)$ in the right. In the past few years interest has substantially increased in solving singularly perturbed differential equations with integral boundary conditions (see [20–25]). 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.

2 Preliminaries

We consider singularly perturbed parabolic partial delay differential equations with integral boundary condition described by

$$ \textstyle\begin{cases} \mathcal{L}u_{\varepsilon }(r,t)= (-\varepsilon \frac{\partial ^{2}}{\partial r^{2}} +\frac{\partial }{\partial t} +a(r,t) )u_{\varepsilon }(r,t)+b(r,t)u_{\varepsilon }(r-1,t)=f(r,t), \quad (r,t) \in D, \\ u_{\varepsilon }(r,t)=\phi _{l} (r,t), \quad \phi _{l} (r,t)\in \Gamma _{l}= \{(r,t); -1 \leq r \leq 0 \text{ and } 0 \leq t \leq T\}, \\ \mathcal{K}u_{\varepsilon }(r,t)=u_{\varepsilon }(2,t)-\varepsilon \int _{0}^{2} g(r)u_{\varepsilon }(r,t)\,dr=\phi _{\bar{r}}(r,t), \\ \quad \phi _{\bar{r}} (r,t)\in \Gamma _{\bar{r}}=\{(2,t);0\leq t \leq T\}, \\ u_{\varepsilon }(r,t)=\phi _{b} (r,t),\quad \phi _{b} (r,t)\in \Gamma _{b}= \{(r,0);0\leq r \leq 2\}, \end{cases} $$

(1)

where $(r, t) \in D = (0, 2)\times (0, T ]$, $\bar{D} = [0, 2]\times [0, T ]$, $D_{1} = (0, 1)\times [0, T ]$, $D_{2} = (1, 2)\times [0, T ]$, $D^{*}=D_{1}\cup D_{2}$ and ε is a small positive parameter $(0<\varepsilon < <1)$. Assume that $a(r, t) \geq \alpha > 0$, $b(r, t) \leq \beta < 0$, $\alpha +\beta >0$, $f(r,t)$, $\phi _{l}$, $\phi _{\tilde{r}}$, $\phi _{b}$ are sufficiently smooth and $g(r)$ is a monotonically nonnegative function and satisfies $\int _{0}^{2}g(r)\,dr<1$.

Problem (1) is equivalent to

$$ \mathcal{L}u_{\varepsilon }(r,t)=F(r,t), $$

where

$$\begin{aligned}& \mathcal{L}u_{\varepsilon }(r,t) \\& \quad = \textstyle\begin{cases} \mathcal{L}_{1}u_{\varepsilon }(r,t)= (-\varepsilon \frac{\partial ^{2}}{\partial r^{2}} +\frac{\partial }{\partial t} +a(r,t) )u_{\varepsilon }(r,t),& (r,t)\in D_{1} , \\ \mathcal{L}_{2}u_{\varepsilon }(r,t)= (-\varepsilon \frac{\partial ^{2}}{\partial r^{2}} +\frac{\partial }{\partial t} +a(r,t) )u_{\varepsilon }(r,t)+b(r,t)u_{\varepsilon }(r-1,t), &(r,t)\in D_{2}, \end{cases}\displaystyle \end{aligned}$$

(2)

$$\begin{aligned}& F(r,t)= \textstyle\begin{cases} f(r,t)-b(r,t)\phi _{l}(r-1,t),&(r,t)\in D_{1}, \\ f(r,t),&(r,t)\in D_{2}, \end{cases}\displaystyle \end{aligned}$$

(3)

with boundary conditions

$$ \textstyle\begin{cases} u_{\varepsilon }(r,t)=\phi _{l} (r,t), \quad \phi _{l} (r,t)\in \Gamma _{l}= \{(r,t); -1 \leq r \leq 0 \text{ and } 0 \leq t \leq T\}, \\ u_{\varepsilon }(1^{-},t)=u_{\varepsilon }(1^{+},t), \qquad \frac{\partial u_{\varepsilon }}{\partial z}(1^{-},t)= \frac{\partial u_{\varepsilon }}{\partial z}(1^{+},t), \\ \mathcal{K}u_{\varepsilon }(r,t)=u_{\varepsilon }(2,t)-\varepsilon \int _{0}^{2} g(r)u_{\varepsilon }(r,t)\,dr=\phi _{\tilde{r}}(r,t), \\ \quad \phi _{\bar{r}} (r,t)\in \Gamma _{\bar{r}}=\{(2,t);0\leq t \leq T\}, \\ u_{\varepsilon }(r,t)=\phi _{b} (r,t), \quad \phi _{b} (r,t)\in \Gamma _{b}= \{(r,0);0\leq r \leq 2\}. \end{cases} $$

(4)

3 The analytical problem

Lemma 1

(Maximum principle)

If $\Psi (r,t)\in C^{(0,0)}(\bar{D})\cap C^{(1,0)}(D)\cap C^{(2,1)}(D_{1} \cup D_{2})$ such that $\Psi (0,t)\geq 0$, $\Psi (r,0)\geq 0$, $\mathcal{K}\Psi (2,t)\geq 0$, $\mathcal{L}_{1}\Psi (r,t)\geq 0$, $\forall (r,t)\in D_{1}$, $\mathcal{L}_{2}\Psi (r,t)\geq 0$, $\forall (r,t)\in D_{2}$ and $[\Psi _{r}](1,t)=\Psi _{r}(1^{+},t)- \Psi _{r}(1^{-},t)\leq 0$, then $\Psi (r,t)\geq 0$ for all $(r,t)\in \bar{D}$.

Proof

Define a test function

$$ s(r,t)= \textstyle\begin{cases} \frac{1}{8}+\frac{r}{2}, &(r,t)\in D_{1}, \\ \frac{3}{8}+\frac{r}{4}, &(r,t)\in D_{2}. \end{cases} $$

(5)

Note that $s(r,t)>0$, $\forall (r,t) \in \bar{D}$, $\mathcal{L} s(r,t)>0$, $\forall (r,t)\in D_{1}\cup D_{2}$, $s(0,t)>0$, $s(r,0)>0$, $\mathcal{K}s(2,t)>0$, and $[s_{r}](1,t)<0$. Let

$$ \mu _{1}=\max \biggl\{ \frac{-\Psi (r,t)}{s(r,t)}:(r,t)\in \bar{D}\biggr\} . $$

Then there exists $(r_{0},t_{0})\in \bar{D}$ such that $\Psi (r_{0},t_{0})+\mu _{1} s(r_{0},t_{0})=0$ and $\Psi (r,t)+\mu _{1} s(r,t)\geq 0$, $\forall (r,t)\in \bar{D}$. Then the function is minimum at $(r,t)=(r_{0},t_{0})$, and the proof is completed. Suppose $\mu _{1}>0$, we need a contradiction.

Case (i): $(r_{0},t_{0})=(0,t_{0})$

$$ 0< (\Psi +\mu _{1} s) (0,t)=\Psi (0,t_{0})+\mu _{1} s(0,t_{0})=0. $$

Case (ii): $(r_{0},t_{0})\in D_{1}$

$$\begin{aligned} 0 < &\mathcal{L}_{1}(\Psi +\mu _{1} s) (r_{0},t_{0}) \\ =&-\varepsilon (\Psi + \mu _{1} s)_{rr}(r_{0},t_{0})+( \Psi +\mu _{1} s)_{t}(r_{0},t_{0})+a(r_{0},t_{0}) (\Psi +\mu _{1} s) (r_{0},t_{0})\leq 0. \end{aligned}$$

Case (iii): $(r_{0},t_{0})=(1,t_{0})$

$$ 0\leq \bigl[(\Psi +\mu _{1} s)'\bigr](1,t_{0})= \bigl[\Psi '\bigr](1,t_{0})+\mu _{1} \bigl[s'\bigr](1,t_{0})< 0. $$

Case (iv): $(r_{0},t_{0})\in D_{2}$

$$\begin{aligned} 0 < &\mathcal{L}_{2}(\Psi +\mu _{1} s) (r_{0},t_{0}) \\ =&-\varepsilon ( \Psi +\mu _{1} s)_{rr}(r_{0},t_{0})+( \Psi +\mu _{1} s)_{t}(r_{0},t_{0}) \\ &{}+a(r_{0},t_{0}) (\Psi +\mu _{1} s) (r_{0},t_{0})+b(r_{0},t_{0}) (\Psi + \mu _{1} s) (r_{0}-1,t_{0})\leq 0. \end{aligned}$$

Case (v): $(r_{0},t_{0})=(2,t_{0})$

$$ 0< \mathcal{K}(\Psi +\mu _{1} s) (2,t_{0})=(\Psi +\mu _{1} s) (2,t_{0})- \varepsilon \int _{0}^{2} g(r) (\Psi +\mu _{1} s) (r,t)\,dr\leq 0. $$

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

$$ \Vert u \Vert _{\bar{D}} \leq C \max \bigl\{ \Vert u \Vert _{\Gamma _{l}}, \Vert u \Vert _{ \Gamma _{b}}, \Vert \mathcal{K}u \Vert _{\Gamma _{\bar{r}}}, \Vert \mathcal{L}u \Vert _{D^{*}} \bigr\} ,\quad (r,t)\in \bar{D}. $$

Proof

It can be easily proved by using the maximum principle (Lemma 1) and the barrier functions $\theta ^{\pm }(r,t)=CMs(r,t)\pm u(r,t)$, $(r,t)\in \bar{D}$, where $M= \max \{ \|u\|_{\Gamma _{l}},\|u\|_{\Gamma _{b}},\| \mathcal{K}u\|_{\Gamma _{\bar{r}}}, \|\mathcal{L}u\|_{D^{*}} \}$ and $s(r,t)$ is the test function as in (5). □

3.1 Compatibility conditions

For problem (2)–(4) to have existence and uniqueness of a solution, we assume that the coefficients of the problem are Hölder continuous and also impose proper compatibility conditions at $(0, 0)$, $(2, 0)$, $(-1, 0)$, and $(1, 0)$ [26]. Necessarily the following conditions are satisfied:

$$ \phi _{b}(0,0)=\phi _{l}(0,0),\qquad \phi _{b}(2,0)=\phi _{\bar{r}}(2,0) $$

(6)

and

$$\begin{aligned}& -\varepsilon \frac{\partial ^{2} \phi _{b}(0,0)}{\partial r^{2}}+a(0,0) \phi _{b}(0,0)+ \frac{\partial \phi _{l}(0,0)}{\partial t}+b(0,0) \phi _{l}(-1,0) = f(0,0), \\& -\varepsilon \frac{\partial ^{2} \phi _{b}(2,0)}{\partial r^{2}}+a(2,0) \phi _{b}(2,0)+ \frac{\partial \phi _{\tilde{r}}(2,0)}{\partial t}+b(2,0) \phi _{b}(1,0) = f(2,0). \end{aligned}$$

(7)

Note that $\phi _{l}(r,t)$, $\phi _{\bar{r}}(r,t)$, and $\phi _{b}(r,t)$ are assumed to be smooth for (7) to make sense, namely $\phi _{l}(r,t)\in C^{1}([0,T])$, $\phi _{\bar{r}}(r,t)\in C^{1}([0,T])$, and $\phi _{b}(r,t)\in C^{(2,1)}(\Gamma _{b})$.

The following classical theorem gives sufficient conditions for the existence of a unique solution of problem (1).

Theorem 1

Suppose $a(r,t), b(r,t), f(r,t)\in C^{(\alpha _{1},\alpha _{1}/2)}(\bar{D})$ and $\phi _{l}\in C^{1+\alpha _{1}/2}([0, T ])$, $\phi _{b}\in C^{(2+\alpha _{1} ,1+ \alpha _{1}/2)}(\Gamma _{b})$, $\phi _{\bar{r}}\in C^{1+ \alpha _{1}/2}([0, T ])$, $\alpha _{1}\in (0,1)$, and assume that the compatibility conditions (6) and (7) are satisfied. Then problem (2)–(4) has a unique solution $u_{\varepsilon }\in C^{(2+ \alpha _{1} ,1+ \alpha _{1} /2)}(\bar{D})$.

Proof

See [26]. □

3.2 The solution and its derivatives are bounded

Theorem 2

Suppose $a(r,t), b(r,t), f(r,t) \in C^{(2+\alpha _{1},1+{\alpha _{1}}/{2})}( \bar{D})$ and $\phi _{l}\in C^{(2+{\alpha _{1}}/{2})}([0,T])$, $\phi _{b}\in C^{(4+\alpha _{1},2+{\alpha _{1}}/{2})}(\Gamma _{b})$, $\phi _{\bar{r}}\in C^{(2+{\alpha _{1}}/{2})}([0,T])$, where $\alpha _{1}\in (0,1)$. Then problem (2)–(4) has a unique solution, which satisfies $u_{\varepsilon }\in C^{(4+\alpha _{1},2+{\alpha _{1}}/{2})}(\bar{D})$. Also, the derivatives of the solution $u_{\varepsilon }$ are bounded, $\forall {i,j}\in {\mathbb{Z}\geq 0}$ such that $0\leq i+2j \leq 4$,

$$ \biggl\Vert \frac{\partial ^{i+j}u_{\varepsilon }}{\partial r^{i} \partial t^{j}} \biggr\Vert \leq C\varepsilon ^{\frac{-i}{2}}. $$

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 transformation $\tilde{r}=\frac{r}{\sqrt{\varepsilon }}$, problem (1) can be rewritten as follows:

$$ \textstyle\begin{cases} \mathcal{L}\tilde{u}(\tilde{r},t)= (- \frac{\partial ^{2}}{\partial \tilde{r}^{2}}+ \frac{\partial }{\partial t} +\tilde{a}(\tilde{r},t) )\tilde{u}( \tilde{r},t)+\tilde{b}(\tilde{r},t)\tilde{u}(\tilde{r}-1, t)=f( \tilde{r} , t) , \quad (\tilde{r},t)\in \tilde{D}_{\varepsilon }, \\ \tilde{u} (\tilde{r},t)=\phi _{l} (\tilde{r},t), \quad (\tilde{r},t)\in \tilde{\Gamma }_{l}, \\ \mathcal{K}\tilde{u}(\tilde{r},t)=\tilde{u}(2,t)-\varepsilon \int _{0}^{2} g(r)\tilde{u}(\tilde{r},t)\,dr=\phi _{\bar{r}}( \tilde{r},t),\quad (\tilde{r},t)\in \tilde{\Gamma }_{\bar{r}}, \\ \tilde{u}(\tilde{r},t)=\phi _{b} (\tilde{r},t), \quad (\tilde{r},t)\in \tilde{\Gamma }_{b}. \end{cases} $$

(8)

The equivalent problem is

$$\begin{aligned}& \textstyle\begin{cases} \mathcal{L}_{1}\tilde{u}(\tilde{r},t)= (- \frac{\partial ^{2}}{\partial \tilde{r}^{2}}+ \frac{\partial }{\partial t}+\tilde{a}(\tilde{r},t) )\tilde{u}( \tilde{r},t) =f(\tilde{x},t)-\tilde{b}(\tilde{r},t)\tilde{\phi _{l}}( \tilde{r}-1,t),\quad (\tilde{r},t)\in \tilde{D}_{1,\varepsilon }, \\ \mathcal{L}_{2}\tilde{u}(\tilde{r},t)= (- \frac{\partial ^{2}}{\partial \tilde{r}^{2}}+ \frac{\partial }{\partial t}+\tilde{a}(\tilde{r},t) )\tilde{u}( \tilde{r},t) +\tilde{b}(\tilde{r},t)\tilde{u}(\tilde{r}-1,t)=f( \tilde{x},t),\quad (\tilde{r},t)\in \tilde{D}_{2,\varepsilon }, \end{cases}\displaystyle \end{aligned}$$

(9)

$$\begin{aligned}& \textstyle\begin{cases} \tilde{u} (\tilde{r},t)=\phi _{l} (\tilde{r},t), \quad (\tilde{r},t)\in \tilde{\Gamma }_{l}, \\ \tilde{u}(1^{-},t)=\tilde{u}(1^{+},t),\qquad \frac{\partial \tilde{u}}{\partial z}(1^{-},t)= \frac{\partial \tilde{u}}{\partial z}(1^{+},t), \\ \mathcal{K}\tilde{u}(\tilde{r},t)=\tilde{u}(2,t)-\varepsilon \int _{0}^{2} g(r)\tilde{u}(\tilde{r},t)\,dr=\phi _{\bar{r}}( \tilde{r},t), \quad (\tilde{r},t)\in \tilde{\Gamma }_{\bar{r}}, \\ \tilde{u}(\tilde{r},t)=\phi _{b} (\tilde{r},t), \quad (\tilde{r},t)\in \tilde{\Gamma }_{b}, \end{cases}\displaystyle \end{aligned}$$

(10)

where $\tilde{D}_{\varepsilon }=(0,\frac{2}{\sqrt{\varepsilon }})\times (0,T]$ and the boundary conditions Γ̃ to Γ.

Note that (8) is independent of ε. Then, using the idea of estimation (10.6) from [26, p. 352], we get

$$ \biggl\Vert \frac{\partial ^{i+j}\tilde{u}}{\partial \tilde{r}^{i} \partial t^{j}} \biggr\Vert _{\tilde{N}_{\delta }}\leq C \bigl(1+ \Vert \tilde{u} \Vert _{\tilde{N}_{2 \delta }}\bigr) $$

for all $\tilde{N}_{\delta }$ in $\tilde{D}_{\varepsilon }$. Here, $\tilde{N}_{\delta }$, $\delta >0$ is a neighborhood with diameter δ in $\tilde{D}_{\varepsilon }$. Returning to the original variable

$$ \biggl\Vert \frac{\partial ^{i+j}u_{\varepsilon }}{\partial r^{i} \partial t^{j}} \biggr\Vert _{\bar{D}}\leq C \varepsilon ^{\frac{-i}{2}}\bigl(1+ \Vert u_{\varepsilon } \Vert _{\bar{D}}\bigr). $$

The proof is completed by using the bound on $u_{\varepsilon }$ in Lemma 2. □

3.3 Decomposition of the solution

The solution $u_{\varepsilon }$ of (2)–(4) is decomposed into $v_{\varepsilon }$-smooth and $w_{\varepsilon }$-singular components.

Further, $v_{\varepsilon }$ is separated into

$$ v_{\varepsilon }=v_{0}+\varepsilon v_{1}, $$

where $v_{0}$ and $v_{1}$ are solutions of the following differential equations respectively:

$$ \textstyle\begin{cases} \frac{\partial v_{0}}{\partial t} (r,t)+a(r,t)v_{0}(r,t)+b(r,t)v_{0}(r-1,t)=f(r,t),\quad (r,t)\in D, \\ v_{0}(r,t)=0,\quad (r,t)\in \Gamma _{b}, \end{cases} $$

(11)

and

$$ \textstyle\begin{cases} \mathcal{L}v_{1}(r,t)=\frac{\partial ^{2} v_{0}}{\partial r^{2}}(r,t), \quad (r,t)\in D, \\ v_{1} (r,t)=0,\quad (r,t)\in \Gamma _{l}, \\ \mathcal{K} v_{1} (r,t)=0,\quad (r,t)\in \Gamma _{\bar{r}}, \\ v_{1}(r,t)=0, \quad (r,t)\in \Gamma _{b}. \end{cases} $$

(12)

Then $v_{\varepsilon }$ satisfies the condition

$$ \textstyle\begin{cases} \mathcal{L}v_{\varepsilon }(r,t)=f(r,t), \quad (r,t)\in D, \\ v_{\varepsilon }(r,t)=v_{0} (r,t),\quad (r,t)\in \Gamma _{l}, \\ \mathcal{K}v_{\varepsilon }(r,t)=\mathcal{K}v_{0}(2,t), \quad (r,t)\in \Gamma _{\bar{r}}, \\ v_{\varepsilon }(r,t)=\phi _{b} (r,t),\quad (r,t)\in \Gamma _{b}. \end{cases} $$

(13)

The singular component $w_{\varepsilon }$ is determined from

$$ \textstyle\begin{cases} \mathcal{L}_{1}w_{\varepsilon }(r,t)=0, \quad (r,t)\in D_{1}, \\ w_{\varepsilon }(r,t)=\phi _{l}(r,t)-v_{0}(r,t), \quad (r,t)\in \Gamma _{l}, \\ [w](r,t)=-[v](r,t),\quad (r,t)\in \Gamma _{\bar{r}}, \\ \text{i,e) }w_{r}(1^{+},t)-w_{r}(1^{-},t)=-(v_{r}(1^{+},t)-v_{r}(1^{-},t)), \\ w_{\varepsilon }(r,t)=0,\quad (r,t)\in \Gamma _{b}, \end{cases} $$

(14)

and

$$ \textstyle\begin{cases} \mathcal{L}_{2}w_{\varepsilon }(r,t)=0,\quad (r,t)\in D_{2}, \\ [w](r,t)=-[v](r,t), \quad (r,t)\in \Gamma _{l}, \\ \text{i,e) }w_{r}(1^{+},t)-w_{r}(1^{-},t)=-(v_{r}(1^{+},t)-v_{r}(1^{-},t)), \\ \mathcal{K} w_{\varepsilon }(r,t)=\mathcal{K} u_{\varepsilon }(r,t)- \mathcal{K}v_{0}(r,t), \quad (r,t)\in \Gamma _{\bar{r}}, \\ w_{\varepsilon }(r,t)=0, \quad (r,t)\in \Gamma _{b}. \end{cases} $$

(15)

The reaction diffusion problem has boundary layers on both boundaries $\Gamma _{l}$ and $\Gamma _{\bar{r}}$. Separate $w_{\varepsilon }$ as $w_{\varepsilon }=w_{l}+w_{\tilde{r}}$, where $w_{l}$ and $w_{\tilde{r}}$ satisfy the following problems:

$$\begin{aligned}& \textstyle\begin{cases} \mathcal{L}_{1}w_{l}(r,t)=0,\quad (r,t)\in D_{1}, \\ w_{l} (r,t)=\phi _{l} (r,t)-v_{0}(r,t), \quad (r,t)\in \Gamma _{l}, \\ w_{l} (r,t)=0, \quad (r,t)\in \Gamma _{\bar{r}}, \\ w_{l} (r,t)=0, \quad (r,t)\in \Gamma _{b}, \end{cases}\displaystyle \end{aligned}$$

(16)

$$\begin{aligned}& \textstyle\begin{cases} \mathcal{L}_{2}w_{l}(r,t)=0,\quad (r,t)\in D_{2}, \\ w_{l} (r,t)=A, \quad (r,t)\in \Gamma _{l}, \\ w_{l} (r,t)=0, \quad (r,t)\in \Gamma _{\bar{r}}, \\ w_{l} (r,t)=0,\quad (r,t)\in \Gamma _{b}, \end{cases}\displaystyle \end{aligned}$$

(17)

and

$$\begin{aligned}& \textstyle\begin{cases} \mathcal{L}_{1}w_{\tilde{r}}(r,t)=0, \quad (r,t)\in D_{1}, \\ w_{\tilde{r}} (r,t)=0, \quad (r,t)\in \Gamma _{l}, \\ w_{\tilde{r}} (r,t)=A, \quad (r,t)\in \Gamma _{\bar{r}}, \\ w_{\tilde{r}} (r,t)=0, \quad (r,t)\in \Gamma _{b}, \end{cases}\displaystyle \end{aligned}$$

(18)

$$\begin{aligned}& \textstyle\begin{cases} \mathcal{L}_{2}w_{\tilde{r}}(r,t)=0,\quad (r,t)\in D_{2}, \\ w_{\tilde{r}} (r,t)=0,\quad (r,t)\in \Gamma _{l}, \\ \mathcal{K}w_{\tilde{r}} (r,t)=\mathcal{K}w(r,t),\quad (r,t)\in \Gamma _{ \bar{r}}, \\ w_{\tilde{r}} (r,t)=0,\quad (r,t)\in \Gamma _{b}. \end{cases}\displaystyle \end{aligned}$$

(19)

Here, we choose A to be a suitable constant, then the solution is continuous at $r=1$.

Theorem 3

Suppose $a(r,t),b(r,t),f(r,t) \in C^{(4+\alpha _{1},2+{\alpha _{1}}/{2})}( \bar{D})$ and $\phi _{l}\in C^{(3+\alpha _{1}/2)}([0,T])$, $\phi _{b}\in C^{(6+\alpha _{1},3+{\alpha _{1}}/{2})}(\Gamma _{b})$, $\phi _{\tilde{r}}\in C^{(3+{\alpha _{1}}/{2})}([0,T])$, where $\alpha _{1}\in (0,1)$. Then we have

$$\begin{aligned}& \biggl\Vert \frac{\partial ^{i+j}v_{\varepsilon }}{\partial r^{i} \partial t^{j}} \biggr\Vert _{\bar{D}}\leq C \bigl(1+\varepsilon ^{1-i/{2}}\bigr), \end{aligned}$$

(20)

$$\begin{aligned}& \biggl\vert \frac{\partial ^{i+j}w_{l}(r,t)}{\partial r^{i} \partial t^{j}} \biggr\vert \leq \textstyle\begin{cases} C \varepsilon ^{\frac{-i}{2}} e^{\frac{-r}{\sqrt{\varepsilon }}}, &(r,t) \in D_{1}, \\ C \varepsilon ^{\frac{-i}{2}} e^{\frac{-(r-1)}{\sqrt{\varepsilon }}}, &(r,t) \in D_{2}, \end{cases}\displaystyle \end{aligned}$$

(21)

$$\begin{aligned}& \biggl\vert \frac{\partial ^{i+j}w_{\tilde{r}}(r,t)}{\partial r^{i} \partial t^{j}} \biggr\vert \leq \textstyle\begin{cases} C \varepsilon ^{\frac{-i}{2}} e^{\frac{-(1-r)}{\sqrt{\varepsilon }}}, &(r,t) \in D_{1}, \\ C \varepsilon ^{\frac{-i}{2}} e^{\frac{-(2-r)}{\sqrt{\varepsilon }}}, &(r,t) \in D_{2}, \end{cases}\displaystyle \end{aligned}$$

(22)

where C is a constant independent of the parameter ε, $(r,t)\in \bar{D}$, $i,j\geq 0$, $0\leq i+2j \leq 4$.

Proof

The existence and smooth component result follow from [26, Chap. 4, p. 320]. The derivatives of smooth component functions are bounded and derived as follows.

First, we estimate that the reduced problem solution $v_{0}$ is bounded and its derivative is bounded

$$ \biggl\Vert \frac{\partial ^{i+j}v_{0}}{\partial r^{i} \partial t^{j}} \biggr\Vert _{\bar{D}}\leq C. $$

(23)

Using Theorem 2, we estimate that the derivative of the solution $v_{1}$ becomes

$$ \biggl\Vert \frac{\partial ^{i+j}v_{1}}{\partial r^{i} \partial t^{j}} \biggr\Vert _{\bar{D}}\leq C \varepsilon ^{\frac{-i}{2}} $$

(24)

since

$$ \frac{\partial ^{i+j}v_{\varepsilon }}{\partial r^{i} \partial t^{j}}= \frac{\partial ^{i+j}v_{0}}{\partial r^{i} \partial t^{j}}+ \varepsilon \frac{\partial ^{i+j}v_{1}}{\partial r^{i} \partial t^{j}}. $$

By (23) and (24), we establish smooth component $v_{\varepsilon }$ estimates.

To prove inequality (21), following the procedure adapted in [12], it is easy to find that

$$ \bigl\vert w_{l}(r,t) \bigr\vert \leq C e^{\frac{-r}{\sqrt{\varepsilon }}} ,\quad (r,t)\in D_{1}. $$

Now, we derive the bound on $w_{l}$ on $D_{2}$. From the defining equations for $w_{l}$, we have

$$ \mathcal{L}_{2} w_{l}(r,t) = -\varepsilon w_{rr} (r,t)+w_{t} (r,t)+a(r,t)w_{l}(r,t)+b(r,t)w_{l}(r-1,t)=0 $$

or

$$\begin{aligned}& \mathcal{L}_{2} w_{l}(r,t) = -\varepsilon w_{rr} (r,t)+w_{t} (r,t)+a(r,t)w_{l}(r,t)=-b(r,t)w_{l}(r-1,t), \\& \mathcal{L}_{2} w_{l}(r,t) = \mathcal{L}_{1} w_{l}(r,t)=-b(r,t)w_{l}(r-1,t), \\& \bigl\vert \mathcal{L}_{2} w_{l}(r,t) \bigr\vert \leq C e^{ \frac{-(r-1)}{\sqrt{\varepsilon }}} \end{aligned}$$

for a choice of suitable $C>0$. Using the stability result (Lemma 2), we have $| w_{l}(r,t)|\leq C e^{\frac{-(r-1)}{\sqrt{\varepsilon }}}$.

Hence,

$$ \bigl|w_{l}(r,t)\bigr|\leq \textstyle\begin{cases} C e^{\frac{-(r)}{\sqrt{\varepsilon }}}, &(r,t)\in D_{1}, \\ C e^{\frac{-(r-1)}{\sqrt{\varepsilon }}}, &(r,t)\in D_{2}. \end{cases} $$

Let $\tilde{r}=\frac{r}{\sqrt{\varepsilon }}$. Clearly, under the transformation, stretched variable r̃ domain is $(0,\frac{2}{\sqrt{\varepsilon }})$, note that $(r,t) \Rightarrow (\tilde{r},t)$ and the parameter ε is independent of problem (16)–(17), then suitable estimates in [26, Sect. 4.10] are the solution of $\tilde{w}_{l}$. The position of r̃ argument is divided into two cases. In the first case, for a neighborhood of $\tilde{N}_{\delta }$ in $(0,\frac{2}{\sqrt{\varepsilon }})\times (0,T]$, apply [26, Sect. 4.10], then

$$ \biggl\Vert \frac{\partial ^{i+j}\tilde{w}_{l}}{\partial \tilde{r}^{i} \partial t^{j}} \biggr\Vert \leq C\bigl( \Vert \tilde{w}_{l} \Vert _{\tilde{N}_{2\delta }}\bigr) $$

and derive the essential bound of r, the variable r̃ changing into variable r, using the bound of $w_{l}$.

Similarly, for any neighborhood $\tilde{N}_{\delta }$ in $(0,2]\times (0,T]$, apply [26, Sect. 4.10], then

$$ \biggl\Vert \frac{\partial ^{i+j}\tilde{w}_{l}}{\partial \tilde{r}^{i} \partial t^{j}} \biggr\Vert \leq C\bigl(1+ \Vert \tilde{w}_{l} \Vert _{\tilde{N}_{2\delta }}\bigr), $$

and derive the essential bound of r, the variable r̃ changing into variable r, using the $w_{l}$ bound, and observe $e^{\frac{-r}{\sqrt{\varepsilon }}}\geq e^{-2}=C$ for $\tilde{r}\leq 2$.

Now to prove (22), consider

$$ \theta ^{\pm }(r,t)=C \textstyle\begin{cases} e^{\alpha ^{*}t} e^{\frac{-(1-r)}{\sqrt{\varepsilon }}}, &(r,t)\in D_{1} \\ e^{\beta ^{*}t} e^{\frac{-(2-r)}{\sqrt{\varepsilon }}}, &(r,t)\in D_{2} \end{cases}\displaystyle \pm w_{\tilde{r}}(r,t), $$

where $\alpha ^{*}=\max \{0, 1-\min_{(r,t)\in \bar{D}}a(r,t)\}$ and $\beta ^{*}=\max \{0, 1-\min_{(r,t)\in \bar{D}}(a(r,t)+b(r,t)) \}$.

Note that $\theta ^{\pm }(0,t)\geq 0$,

$$\begin{aligned}& \begin{aligned} \mathcal{L}_{1}\theta ^{\pm }(r,t)&=-\varepsilon \theta ^{\pm }_{rr} (r,t)+ \theta ^{\pm }_{t} (r,t)+a(r,t) \theta ^{\pm }(r,t) \\ &= C\bigl(\alpha ^{*}-1+a(r,t)\bigr)e^{\frac{-(1-r)}{\sqrt{\varepsilon }}}e^{ \alpha ^{*}t} \\ &\geq 0, \end{aligned} \\& \begin{aligned} \mathcal{L}_{2}\theta ^{\pm }(r,t)&=-\varepsilon \theta ^{\pm }_{rr} (r,t)+ \theta ^{\pm }_{t} (r,t)+a(r,t) \theta ^{\pm }(r,t)+b(r,t) \theta ^{\pm }(r-1,t) \\ &=C\bigl(\beta ^{*}-1+a(r,t)+b(r,t)e^{\frac{-1}{\sqrt{\varepsilon }}} \bigr)e^{ \frac{-(2-r)}{\sqrt{\varepsilon }}} e^{ \beta ^{*}t } \\ &\geq C\bigl(\alpha ^{*}-1+a(r,t)+b(r,t)\bigr)e^{ \frac{-(2-r)}{\sqrt{\varepsilon }}}e^{ \alpha ^{*}t } \\ &\geq 0, \end{aligned} \end{aligned}$$

and

$$\begin{aligned} \mathcal{K}\theta ^{\pm }(2,t) =&\theta ^{\pm }(2,t)- \varepsilon \int _{0}^{2}g(r)\theta ^{\pm }(r,t)\,dr \\ =&C-C\varepsilon \int _{0}^{1}g(r)\theta ^{\pm }(r,t)\,dr-C \varepsilon \int _{1}^{2} g(r) \theta ^{\pm }(r,t)\,dr \\ =&C-C\varepsilon \int _{0}^{1}g(r)e^{ \frac{-(1-r)}{\sqrt{\varepsilon }}}\,dr-C \varepsilon \int _{1}^{2} g(r) e^{\frac{-(2-r)}{\sqrt{\varepsilon }}}\,dr \\ \geq & C\biggl(1-\varepsilon \int _{0}^{2}g(r)\,dr\biggr) \\ \geq & 0 \end{aligned}$$

for a choice of suitable $C>0$. Using the maximum principle (Lemma 1), we have

$$ \bigl\vert w_{\tilde{r}}(r,t) \bigr\vert \leq \textstyle\begin{cases} C e^{\frac{-(1-r)}{\sqrt{\varepsilon }}}, &(r,t)\in D_{1}, \\ C e^{\frac{-(2-r)}{\sqrt{\varepsilon }}}, &(r,t)\in D_{2}. \end{cases} $$

Let $\tilde{r}=\frac{r}{\sqrt{\varepsilon }}$. Clearly, under the transformation, the stretched variable r̃ domain is $(0,\frac{2}{\sqrt{\varepsilon }})$, note that $(r,t) \Rightarrow (\tilde{r},t)$ and the parameter ε is independent of problem (18)–(19), then suitable estimates in [26, Sect. 4.10] are the solution of $\tilde{w}_{\tilde{r}}$. The position of r̃ argument is divided into two cases. In the first case, for the neighborhood of $\tilde{N}_{\delta }$ in $(0,\frac{2}{\sqrt{\varepsilon }})\times (0,T]$, applying [26, Sect. 4.10], we get

$$ \biggl\Vert \frac{\partial ^{i+j}\tilde{w}_{\tilde{r}}}{\partial \tilde{r}^{i} \partial t^{j}} \biggr\Vert \leq C\bigl( \Vert \tilde{w}_{\tilde{r}} \Vert _{\tilde{N}_{2\delta }}\bigr) $$

and derive the essential bound of r, variable r̃ changing into variable r, using the $w_{\tilde{r}}$ bound.

Similarly, for any neighborhood $\tilde{N}_{\delta }$ in $(0,2]\times (0,T]$ applying [26, Sect. 4.10], we get

$$ \biggl\Vert \frac{\partial ^{i+j}\tilde{w}_{\tilde{r}}}{\partial \tilde{r}^{i} \partial t^{j}} \biggr\Vert \leq C\bigl(1+ \Vert \tilde{w}_{\tilde{r}} \Vert _{\tilde{N}_{2\delta }}\bigr) $$

and derive the essential bound of r; moreover, the variable r̃ changing into variable r and using bound of $w_{\tilde{r}}$, observe $e^{\frac{-r}{\sqrt{\varepsilon }}}\geq e^{-2}=C$ for $\tilde{r}\leq 2$. Hence proved. □

4 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 ($\delta _{r}^{2}U$), and the first order time derivative ($u_{t}$) is replaced by the backward difference scheme ($D_{t}^{-}U$). In r-direction the interval $\Theta =[0,2]$ is divided into $\bar{\Theta }_{1}=[0,1]$ and $\bar{\Theta }_{2}=[1,2]$ with $\frac{N}{2}$ equal mesh points. Furthermore, the piecewise uniform mesh (Shishkin mesh) $\bar{\Theta }_{1}=[0,1]$ is divided into three subintervals

$$ \bar{\Theta }_{1}=\bar{\Theta }_{l}\cup \bar{\Theta }_{c}\cup \bar{\Theta }_{\tilde{r}}, $$

where $\Theta _{l}=(0,\mu )$, $\Theta _{c}=(\mu ,1-\mu )$, $\Theta _{\tilde{r}}=(1-\mu ,1)$.

Similarly, the interval $\bar{\Theta }_{2}=[1,2]$ is divided into three subintervals

$$ \bar{\Theta }_{2}=\bar{\Theta }_{l}\cup \bar{\Theta }_{c}\cup \bar{\Theta }_{\tilde{r}}, $$

where $\Theta _{l}=(1,1+\mu )$, $\Theta _{c}=(1+\mu ,2-\mu )$, $\Theta _{\tilde{r}}=(2-\mu ,2)$. Furthermore, μ is called a fitting factor, and it satisfies the following condition:

$$ \mu =\min \biggl\{ \frac{1}{4},2{\sqrt{\varepsilon }}\ln ({N_{r}})\biggr\} , $$

where $N_{r}$ denotes the number of mesh elements in the r-direction.

A piecewise uniform mesh $\Theta _{\mu }^{N_{r}}$ is on Θ with $N_{r}$ mesh elements. Both intervals $\Theta _{l}$ and $\Theta _{\tilde{r}}$ are uniform meshes with $\frac{N_{r}}{8}$ elements and $\Theta _{c}$ is also a uniform mesh with $\frac{N_{r}}{4}$ mesh element, and thus we divide r-direction. Uniform meshes on t-direction on step size Δt and the number mesh denotes $N_{t}$ in t-direction. $D_{\mu }^{N}=\Theta _{\mu }^{N_{r}}\times \Theta ^{N_{t}}$ and the boundary analogues $\Gamma _{\mu }^{N}$ of $D_{\mu }^{N}$ are $\Gamma _{\mu }^{N}=\bar{D}_{\mu }^{N}\cap \Gamma$. We put $\Gamma _{l,\mu }^{N} =\Gamma _{\mu }^{N}\cap \Gamma _{l}$, $\Gamma _{r,\mu }^{N}=\Gamma _{\mu }^{N}\cap \Gamma _{\bar{r}}$, and $\Gamma _{b,\mu }^{N}=\Gamma _{\mu }^{N}\cap \Gamma _{b}$.

Using the finite difference method for the continuous problem (2)–(4).

$$ \mathcal{L}^{N} U_{\varepsilon }(r_{i},t_{j})=F(r_{i},t_{j}), $$

where

$$\begin{aligned}& \mathcal{L}^{N} U_{\varepsilon }(r_{i},t_{j})= \textstyle\begin{cases} \mathcal{L}_{1}^{N}U_{\varepsilon }(r_{i},t_{j})= (-\varepsilon \delta _{r}^{2}U_{\varepsilon }+ D_{t}^{-}U_{\varepsilon }+aU_{\varepsilon } )(r_{i},t_{j}), & (r_{i},t_{j})\in D_{1}^{N} , \\ \mathcal{L}_{2}^{N}U_{\varepsilon }(r_{i},t_{j})= (-\varepsilon \delta _{r}^{2}U_{\varepsilon }+ D_{t}^{-}U_{\varepsilon }+aU_{\varepsilon } )(r_{i},t_{j}) \\ \hphantom{\mathcal{L}_{2}^{N}U_{\varepsilon }(r_{i},t_{j})={}}{}+bU_{\varepsilon }(r_{i-\frac{N}{2}} ,t_{j}), & (r_{i},t_{j})\in D_{2}^{N} , \end{cases}\displaystyle \end{aligned}$$

(25)

$$\begin{aligned}& F(r_{i},t_{j})= \textstyle\begin{cases} f(r_{i},t_{j})-b(r_{i},t_{j})\phi _{l}(r_{i-\frac{N}{2}},t_{j}),&(r_{i},t_{j}) \in D_{1}^{N}, \\ f(r_{i},t_{j}),&(r_{i},t_{j})\in D_{2}^{N}, \end{cases}\displaystyle \end{aligned}$$

(26)

with boundary conditions

$$ \textstyle\begin{cases} U_{\varepsilon }(r_{i},t_{j})=\phi _{l}(r_{i},t_{j}), \quad \phi _{l}(r_{i},t_{j}) \in \Gamma ^{N}_{l}, \\ \mathcal{K}^{N} U_{\varepsilon }(r_{N},t_{j})=U_{\varepsilon }(r_{N},t_{j})- \varepsilon \sum_{i=1}^{N} \frac{g(r_{i-1})U_{\varepsilon }(r_{i-1},t_{j})+g(r_{i})U_{\varepsilon }(r_{i},t_{j})}{2}h_{i} \\ \hphantom{\mathcal{K}^{N} U_{\varepsilon }(r_{N},t_{j})}= \phi _{\tilde{r}}, \quad \phi _{\tilde{r}}(r_{i},t_{j})\in \Gamma ^{N}_{\tilde{r}}, \\ D_{r}^{-}U_{\varepsilon }(r_{\frac{N}{2}},t_{j})=D_{r}^{+}U_{\varepsilon }(r_{\frac{N}{2}},t_{j}), \\ U_{\varepsilon }(r_{i},t_{j})=\phi _{b} (r_{i},t_{j}), \quad \phi _{b}(r_{i},t_{j}) \in \Gamma ^{N}_{b}, \text{where } i, j=1,2,\dots , N. \end{cases} $$

(27)

The differential operator notation is

$$ \textstyle\begin{cases} \delta _{r}^{2}U_{i,j}= \frac{(D_{r}^{+} -D_{r}^{-})U_{i,j}}{(r_{i+1}-r_{i-1})/2} \\ \text{with} \\ D_{r}^{+} U_{i,j}=\frac{U_{i+1,j}-U_{i,j}}{r_{i+1}-r_{i}},\qquad D_{r}^{-} U_{i,j}= \frac{U_{i,j}-U_{i-1,j}}{r_{i}-r_{i-1}} \end{cases} $$

and an analogous definition of $D_{t}^{-}$.

Lemma 3

(Discrete maximum principle)

Let Z be any mesh function satisfying $Z(r_{0},t_{j})\geq 0$, $Z(r_{i},t_{0})\geq 0$, $\mathcal{K}^{N}Z(r_{N},t_{j})\geq 0$, $\mathcal{L}_{1}^{N}Z(r_{i},t_{j})\geq 0$, $\forall (r_{i},t_{j})\in D_{1}^{N}$, $\mathcal{L}_{2}^{N}Z(r_{i},t_{j})\geq 0$, $\forall (r_{i},t_{j})\in D_{2}^{N}$, and $[D_{r}]Z(r_{\frac{N}{2}},t_{j})=D^{+}_{r}Z(r_{\frac{N}{2}},t_{j})-D^{-}_{r}Z(r_{ \frac{N}{2}},t_{j})\leq 0$. Then $Z(r_{i},t_{j})\geq 0$ for all $(r_{i},t_{j})\in \bar{D}^{N}$.

Proof

Define a test function $S(r_{i},t_{j})$ as

$$ S(r_{i},t_{j})= \textstyle\begin{cases} \frac{1}{8}+\frac{r_{i}}{2}, &(r_{i},t_{j})\in D_{1}^{N}, \\ \frac{3}{8}+\frac{r_{i}}{4}, &(r_{i},t_{j})\in D_{2}^{N}. \end{cases} $$

(28)

Note that $S(r_{i},t_{j})>0$, $\forall (r_{i},t_{j}) \in \bar{D}^{N}$, $\mathcal{L}^{N} S(r_{i},t_{j})>0$, $\forall (r_{i},t_{j})\in D_{1}^{N} \cup D_{2}^{N}$, $S(r_{0},t_{j})>0$, $S(r_{i},t_{0})>0$, $\mathcal{K}^{N}S(r_{N},t_{j})>0$, and $[D_{r}]S(r_{\frac{N}{2}},t_{j})<0$. Let

$$ \eta =\max \biggl\{ \frac{-r(r_{i},t_{j})}{S(r_{i},t_{j})}:(r_{i},t_{j}) \in \bar{D}^{N}\biggr\} . $$

Then there exists $(r^{*},t^{*})\in \bar{D}^{N}$ such that $Z(r^{*},t^{*})+\eta S(r^{*},t^{*})=0$ and $Z(r_{i},t_{j})+\eta S(r_{i},t_{j})\geq 0$, $\forall (r_{i},t_{j})\in \bar{D}^{N}$. Then the function is minimum at $(r,t)=(r^{*},t^{*})$ and this proof is completed. Suppose $\eta >0$, we need a contradiction.

Case (i): $(r^{*},t^{*})=(r_{0},t^{*})$

$$ 0< (r+\eta S) \bigl(r_{0},t^{*}\bigr)=Z \bigl(r_{0},t^{*}\bigr)+\eta S\bigl(r_{0},t^{*} \bigr)=0. $$

Case (ii): $(r^{*},t^{*})\in D_{1}^{N}$

$$ 0< \mathcal{L}_{1}^{N}(r+\eta S) \bigl(r^{*},t^{*} \bigr)= \bigl(-\varepsilon \delta _{r}^{2} + D_{t}+a \bigr) (r+\eta S) \bigl(r^{*},t^{*} \bigr)\leq 0. $$

Case (iii): $(r^{*},t^{*})=(r_{\frac{N}{2}},t^{*})$

$$ 0\leq \bigl[D_{r}(\Psi +\eta s)\bigr]\bigl(r_{\frac{N}{2}},t^{*} \bigr)< 0. $$

Case (iv): $(r^{*},t^{*})\in D^{N}_{2}$

$$\begin{aligned} 0 < &\mathcal{L}_{2}^{N}(r+\eta S) \bigl(r^{*},t^{*} \bigr) \\ =& \bigl(-\varepsilon \delta _{r}^{2}+D_{t}+a \bigr) (r+\eta S) \bigl(r^{*},t^{*}\bigr) +b(r+\eta S) \bigl(r^{*}-r_{\frac{N}{2}},t^{*} \bigr)\leq 0. \end{aligned}$$

Case (v): $(r^{*},t^{*})=(r_{N},t^{*})$

$$ 0< \mathcal{K}^{N}(r+\eta S) \bigl(r_{N},t^{*} \bigr)=(r+\eta S) \bigl(r_{N},t^{*}\bigr)- \varepsilon \sum_{i=1}^{N} \frac{g_{i-1}(r+\eta S)_{i-1,j}+g_{i}(r+\eta S)_{i,j}}{2}h_{i} \leq 0. $$

In all the cases we arrived at a contradiction. Hence proved. □

Lemma 4

Prove that

$$ \Vert Z \Vert _{\bar{D}^{N}} \leq C \max \Bigl\{ \Vert Z \Vert _{\Gamma _{l}^{N}}, \Vert Z \Vert _{\Gamma _{b}^{N}}, \bigl\Vert \mathcal{K}^{N}Z \bigr\Vert _{\Gamma _{\bar{r}}^{N}}, \max _{(r_{i},t_{j})\in D_{1}^{N}\cup D_{2}^{N}} \bigl\Vert \mathcal{L}^{N}Z \bigr\Vert \Bigr\} ,\quad (r_{i},t_{j})\in \bar{D}^{N}. $$

Proof

It can be easily proved using the discrete maximum principle (Lemma 3) and the barrier functions $\Xi ^{\pm }(r_{i},t_{j})=CMS(r_{i},t_{j})\pm Z(r_{i},t_{j})$, $(r_{i},t_{j}) \in \bar{D}^{N}$, where

$$ M= \max \Bigl\{ \Vert Z \Vert _{\Gamma _{l}^{N}}, \Vert Z \Vert _{\Gamma _{b}^{N}}, \bigl\Vert \mathcal{K}^{N}Z \bigr\Vert _{\Gamma _{\bar{r}}^{N}},\max_{(r_{i},t_{j})\in D_{1}^{N} \cup D_{2}^{N}} \bigl\Vert \mathcal{L}^{N}Z \bigr\Vert \Bigr\} , $$

(29)

and $S(r_{i},t_{j})$ is the test function as in Lemma 3. □

5 Error estimate

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

Theorem 4

Let $u_{\varepsilon }$ and $U_{\varepsilon }$ be solutions of problems (2)–(4) and (25)–(27). Assume that the coefficients $a(r,t),b(r,t),f(r,t) \in C^{(4+\alpha _{1},2+{\alpha _{1}}/{2})}( \bar{D})$ and the boundary conditions satisfy $\phi _{l}\in C^{(3+\alpha _{1}/2)}([0,T])$, $\phi _{b}\in C^{(6+\alpha _{1},3+{\alpha _{1}}/{2})}(\Gamma _{b})$, $\phi _{\tilde{r}}\in C^{(3+{\alpha _{1}}/{2})}([0,T])$, where $\alpha _{1}\in (0,1)$. Then we have

$$ \sup_{0< \varepsilon \leq 1} \Vert U_{\varepsilon }-u_{\varepsilon } \Vert _{ \bar{D}^{N}}\leq C\bigl(N_{r}^{-2}\ln ^{2} N_{r}+N_{t}^{-1}\bigr). $$

Proof

Since the continuous problem solution $u_{\varepsilon }$ of $\mathcal{L}_{\varepsilon }$ is decomposed into smooth and singular components, by the same process, the discrete problem solution $U_{\varepsilon }$ of $\mathcal{L}_{\varepsilon }^{N}$ is decomposed into smooth and singular components as

$$ U_{\varepsilon }=V_{\varepsilon }+W_{\varepsilon }, $$

where the $V_{\varepsilon }$-smooth component is the solution of the following problem:

$$ \textstyle\begin{cases} \mathcal{L}^{N} V_{\varepsilon }(r_{i},t_{j})=f(r_{i},t_{j}),\quad (r_{i},t_{j}) \in D^{N}, \\ V_{\varepsilon }(r_{i},t_{j})=\phi _{0} (r_{i},t_{j}),\quad (r_{i},t_{j}) \in \Gamma _{l}^{N}, \\ \mathcal{K}^{N} V_{\varepsilon }(r_{N},t_{j})=\mathcal{K}^{N}V_{0}(r_{N},t_{j}), \quad (r_{i},t_{j})\in \Gamma _{\bar{r}}^{N}, \\ V_{\varepsilon }(r_{i},t_{j})=\phi _{b} (r_{i},t_{j}), \quad (r_{i},t_{j}) \in \Gamma _{b}^{N}, \end{cases} $$

and therefore $W_{\varepsilon }$ must satisfy

$$ \textstyle\begin{cases} \mathcal{L}^{N} W_{\varepsilon }(r_{i},t_{j})=0, \quad (r_{i},t_{j})\in D^{N}, \\ W_{\varepsilon }(r_{i},t_{j})=U_{\varepsilon }(r_{i},t_{j})-V_{\varepsilon }(r_{i},t_{j}), \quad (r_{i},t_{j})\in \Gamma _{l}^{N}, \\ \mathcal{K}^{N} W_{\varepsilon }(r_{N},t_{j})=\mathcal{K}^{N} U_{\varepsilon }(r_{N},t_{j})- \mathcal{K}^{N} V_{0}(r_{N},t_{j}), \quad (r_{i},t_{j}) \in \Gamma _{\bar{r}}^{N}, \\ W_{\varepsilon }(r_{i},t_{j})=0,\quad (r_{i},t_{j})\in \Gamma _{b}^{N}. \end{cases} $$

The error can be written in the form

$$ U_{\varepsilon }-u_{\varepsilon }=(V_{\varepsilon }-v_{\varepsilon })+ (W_{\varepsilon }-w_{\varepsilon }), $$

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 $r_{i}=r_{N}$,

$$\begin{aligned}& \begin{aligned} \mathcal{K}^{N}(V_{\varepsilon }-v_{\varepsilon }) (r_{N},t_{j})&=\mathcal{K}^{N}V_{\varepsilon }(r_{N}, t_{j})-\mathcal{K}^{N}v_{\varepsilon }(r_{N},t_{j} ) \\ &=\phi _{\tilde{r}}-\mathcal{K}^{N}v_{\varepsilon }(r_{N},t_{j}) \\ &=\mathcal{K}v_{\varepsilon }(r_{N},t)-\mathcal{K}^{N}v_{\varepsilon }(r_{N},t_{j}) \\ &=v_{\varepsilon }(r_{N},t)- \int _{r_{0}}^{r_{N}}g(r)v_{\varepsilon }(r,t)\,dr-v_{\varepsilon }(r_{N}, t_{j})+\sum_{i=1}^{N} \frac{g_{i-1} v_{i-1,j}+g_{i}v_{i,j}}{2}h_{i} \\ &=\frac{g_{0}v_{0,j}+g_{1}v_{1,j}}{2}h_{1}+\cdots + \frac{g_{N-1}v_{N-1,j}+g_{N}v_{N,j}}{2}h_{N} \\ &\quad {}- \int _{ r_{0} } ^{r_{1}}g(r)v_{\varepsilon }(r,t)\,dr- \cdots - \int _{r_{N-1}}^{r_{N}} g(r)v_{\varepsilon }(r,t)\,dr \\ &=-\frac{h_{1}^{3}}{12}g''(\chi _{1})v_{\varepsilon }''(\chi _{1},t)- \cdots -\frac{h_{2N}^{3}}{12}g''( \chi _{N}) v_{\varepsilon }''(\chi _{N},t), \end{aligned} \\& \bigl\vert \mathcal{K}^{N}(V_{\varepsilon }-v_{\varepsilon }) (r_{N},t_{j}) \bigr\vert = \bigl\vert C \bigl(h_{1}^{3}v_{\varepsilon }''( \chi _{1},t)+\cdots +h_{N} ^{3}v_{\varepsilon }''( \chi _{N}, t)\bigr) \bigr\vert , \\& \begin{aligned} \bigl\vert \mathcal{K}^{N}(V_{\varepsilon }-v_{\varepsilon }) (r_{N},t_{j}) \bigr\vert &\leq C \varepsilon ( \bigl(h_{1}^{3}v_{\varepsilon }''( \chi _{1}, t) +\cdots +h_{N} ^{3}v_{\varepsilon }''( \chi _{N},t)\bigr) \\ &\leq C\varepsilon ^{-1}\bigl(e^{-\chi _{1}\sqrt{ \frac{\alpha }{\varepsilon }}} h_{1}^{3}+\cdots +e^{-\chi _{N}\sqrt{ \frac{\alpha }{\varepsilon }}}h_{N}^{3} \bigr) \\ &\leq C\varepsilon \bigl(h_{1}^{3}+\cdots +h_{N}^{3}\bigr) \\ &\leq CN_{r}^{-2}, \end{aligned} \end{aligned}$$

where $r_{i-1}\leq \chi _{i}\leq r_{i}$, $1\leq i\leq N$, for a proper choice of C.

From the difference and differential equations, it is not hard to see that

$$ \mathcal{L}_{1}^{N}(V_{\varepsilon }-v_{\varepsilon })=f- \mathcal{L}_{1}^{N}v_{\varepsilon }=\bigl( \mathcal{L}_{1}-\mathcal{L}_{1}^{N} \bigr)v_{\varepsilon }. $$

Then

$$ \mathcal{L}_{1}^{N}(V_{\varepsilon }-v_{\varepsilon })=- \varepsilon \biggl(\frac{\partial ^{2}}{\partial r^{2}} -\delta _{r}^{2} \biggr)v_{\varepsilon }+ \biggl(\frac{\partial }{\partial t}-D_{t}^{-} \biggr)v_{\varepsilon }. $$

It follows from classical estimates [12], at each point $(r_{i},t_{j})$ in $D_{1}^{N}$,

$$ \bigl\vert \mathcal{L}_{1}^{N}(V_{\varepsilon }-v_{\varepsilon }) (r_{i},t_{j}) \bigr\vert \leq C \textstyle\begin{cases} (\sqrt{\varepsilon }N_{r}^{-1}+N_{t}^{-1}),&\text{if } r_{i}=\mu \text{ or } r_{i}=1-\mu , \\ (N_{r}^{-2}+N_{t}^{-1}),& \text{otherwise}. \end{cases} $$

Similarly,

$$ \bigl\vert \mathcal{L}_{2}^{N}(V_{\varepsilon }-v_{\varepsilon }) (r_{i},t_{j}) \bigr\vert \leq C \textstyle\begin{cases} (\sqrt{\varepsilon }N_{r}^{-1}+N_{t}^{-1}),&\text{if } r_{i}=1+\mu \text{ or } r_{i}=2-\mu , \\ (N_{r}^{-2}+N_{t}^{-1}),& \text{otherwise}. \end{cases} $$

Define

$$ \phi (r_{i},t_{j})=C \textstyle\begin{cases} \frac{\mu }{\sqrt{\varepsilon }} \theta _{1}(r_{i})N_{r}^{-2}+(1+t_{j})N_{r}^{-2}+t_{j}N_{t}^{-1},&(r_{i},t_{j}) \in D_{1}^{N}, \\ \frac{(1+\mu )}{\sqrt{\varepsilon }} \theta _{2}(r_{i})N_{r}^{-2}+(1+t_{j})N_{r}^{-2}+t_{j}N_{t}^{-1},&(r_{i},t_{j}) \in D_{2}^{N}, \end{cases} $$

where $\theta _{1}$ and $\theta _{2}$ are piecewise linear polynomials

$$ \theta _{1}= \textstyle\begin{cases} \frac{r}{\mu } & \text{for } 0\leq z\leq \mu , \\ 1 & \text{for } \mu \leq z\leq 1-\mu , \\ \frac{1-r}{\mu } & \text{for } 1-\mu \leq z\leq 1, \end{cases} $$

and

$$ \theta _{2}= \textstyle\begin{cases} \frac{r}{1+\mu } & \text{for } 1\leq z\leq 1+\mu , \\ 1 & \text{for } 1+\mu \leq z\leq 2-\mu , \\ \frac{3-r}{1+\mu } & \text{for } 2-\mu \leq z\leq 2. \end{cases} $$

Then, for all $(r_{i},t_{j})\in \bar{D}^{N}$,

$$ 0\leq \phi (r_{i},t_{j})\leq C\bigl(N_{r}^{-2} \ln N_{r}+N_{t}^{-1}\bigr), $$

and also

$$ \mathcal{L}_{1}^{N}\phi (r_{i},t_{j}) \geq \textstyle\begin{cases} C(\sqrt{\varepsilon }N_{r}^{-1}+N_{r}^{-2}+N_{t}^{-1}),&\text{if } r_{i}= \mu \text{ or } 1-\mu , \\ C(N_{r}^{-2}+N_{t}^{-1}),&\text{otherwise}, \end{cases} $$

note that $\frac{\mu }{\sqrt{\varepsilon }}\leq 2\ln N$ and

$$ \mathcal{L}_{1}^{N}\theta (r_{i})= \textstyle\begin{cases} \frac{\varepsilon N_{r}}{\mu }+a(r_{i}),&\text{if } r_{i}=\mu \text{ or } r_{i}=1-\mu , \\ a(r_{i})\theta (r_{i}),&\text{otherwise}. \end{cases} $$

Similarly,

$$ \mathcal{L}_{2}^{N}\phi (r_{i},t_{j}) \geq \textstyle\begin{cases} C(\sqrt{\varepsilon }N_{r}^{-1}+N_{r}^{-2}+N_{t}^{-1}),&\text{if } r_{i}=1+ \mu \text{ or } 2-\mu , \\ C(N_{r}^{-2}+N_{t}^{-1}),&\text{otherwise}, \end{cases} $$

note that $\frac{1+\mu }{\sqrt{\varepsilon }}\leq 2\ln N$ and

$$ \mathcal{L}_{2}^{N}\theta (r_{i})= \textstyle\begin{cases} \frac{\varepsilon N_{r}}{1+\mu }+a(r_{i})+b(r_{i}),&\text{if } r_{i}=1+ \mu \text{ or } r_{i}=2-\mu , \\ a(r_{i})\theta (r_{i})+b(r_{i})\theta (r_{i-\frac{N}{2}}),&\text{otherwise}, \end{cases} $$

and $\forall (r_{i},t_{j})\in \Gamma ^{N}$, then $\phi (r_{i},t_{j})\geq 0$.

Observe that $\mathcal{K}^{N}\theta (r_{N},t_{j})\geq 0$.

Define the barrier functions

$$ \Xi ^{\pm }(r_{i},t_{j})=\phi (r_{i},t_{j})\pm (V_{\varepsilon }-v_{\varepsilon }) (r_{i},t_{j}), $$

it follows from $\forall (r_{i},t_{j})\in D_{1}^{N}$, $\forall (r_{i},t_{j})\in D_{2}^{N}$,

$$ \mathcal{L}_{1}^{N}\Xi ^{\pm }(r_{i},t_{j}) \geq 0\quad \text{and}\quad \mathcal{L}_{2}^{N}\Xi ^{\pm }(r_{i},t_{j})\geq 0. $$

Then, from the discrete maximum principle,

$$ \Xi ^{\pm }(r_{i},t_{j})\geq 0,\quad (r_{i},t_{j})\in \bar{D}^{N}. $$

Then we have

$$\begin{aligned}& \bigl\vert (V_{\varepsilon }-v_{\varepsilon }) (r_{i},t_{j}) \bigr\vert \leq \phi (r_{i},t_{j}) \leq C \bigl(N_{r}^{-2}\ln N_{r}+N_{t}^{-1} \bigr), \\& \bigl\vert (V_{\varepsilon }-v_{\varepsilon }) (r_{i},t_{j}) \bigr\vert \leq C\bigl(N_{r}^{-2}\ln N_{r}+N_{t}^{-1}\bigr). \end{aligned}$$

Next, we derive an error estimate for the singular component. We decompose $W_{\varepsilon }$ into $W_{l}$ and $W_{\tilde{r}}$, where

$$\begin{aligned}& \textstyle\begin{cases} \mathcal{L}_{1}^{N}W_{l}(r_{i},t_{j})=0, \quad (r_{i},t_{j})\in D_{1}^{N}, \\ W_{l} (r_{i},t_{j})=\phi _{l} (r_{i},t_{j})-v_{0}(r_{i},t_{j}), \quad (r_{i},t_{j}) \in \Gamma _{l}^{N}, \\ W_{l} (r_{i},t_{j})=0, \quad (r_{i},t_{j})\in \Gamma _{\bar{r}}^{N}, \\ W_{l} (r_{i},t_{j})=0, \quad (r_{i},t_{j})\in \Gamma _{b}^{N}, \end{cases}\displaystyle \\& \textstyle\begin{cases} \mathcal{L}_{2}^{N}W_{l}(r_{i},t_{j})=0, \quad (r_{i},t_{j})\in D_{2}^{N}, \\ W_{l} (r_{i},t_{j})=A,\quad (r_{i},t_{j})\in \Gamma _{l}^{N}, \\ \mathcal{K}^{N}W_{l} (r_{i},t_{j})=0, \quad (r_{i},t_{j})\in \Gamma _{ \bar{r}}^{N}, \\ W_{l} (r_{i},t_{j})=0, \quad (r_{i},t_{j})\in \Gamma _{b}^{N}, \end{cases}\displaystyle \end{aligned}$$

and

$$\begin{aligned}& \textstyle\begin{cases} \mathcal{L}_{1}^{N}W_{\tilde{r}}(r_{i},t_{j})=0, \quad (r_{i},t_{j})\in D_{1}^{N}, \\ W_{\tilde{r}} (r_{i},t_{j})=0,\quad (r_{i},t_{j})\in \Gamma _{l}^{N}, \\ W_{\tilde{r}} (r_{i},t_{j})=A, \quad (r_{i},t_{j})\in \Gamma _{\bar{r}}^{N}, \\ W_{\tilde{r}} (r_{i},t_{j})=0, \quad (r_{i},t_{j})\in \Gamma _{b}^{N}, \end{cases}\displaystyle \\& \textstyle\begin{cases} \mathcal{L}_{2}^{N}W_{\tilde{r}}(r,t)=0, \quad (r_{i},t_{j})\in D_{2}^{N}, \\ W_{\tilde{r}} (r_{i},t_{j})=0, \quad (r_{i},t_{j})\in \Gamma _{l}^{N}, \\ \mathcal{K}^{N}W_{\tilde{r}} (r_{i},t_{j})=\mathcal{K}^{N}W(r_{i},t_{j}),\quad (r_{i},t_{j})\in \Gamma _{\bar{r}}^{N}, \\ W_{\tilde{r}} (r_{i},t_{j})=0,\quad (r_{i},t_{j})\in \Gamma _{b}^{N}. \end{cases}\displaystyle \end{aligned}$$

The singular component error is equivalent to

$$\begin{aligned}& W_{\varepsilon }-w_{\varepsilon }=(W_{l}-w_{l})+(W_{\tilde{r}}-w_{ \tilde{r}}), \\& \mathcal{L}^{N}(W_{\varepsilon }-w_{\varepsilon })=-\varepsilon \biggl( \frac{\partial ^{2}}{\partial r^{2}} -\delta _{r}^{2} \biggr)w_{\varepsilon }+ \biggl(\frac{\partial }{\partial t}-D_{t}^{-} \biggr)w_{\varepsilon }. \end{aligned}$$

It follows from classical estimates [12], at each point $(r_{i},t_{j})$ in $D_{1}^{N}$,

$$ \bigl\vert \mathcal{L}^{N}(W_{\varepsilon }-w_{\varepsilon }) (r_{i},t_{j}) \bigr\vert \leq C \textstyle\begin{cases} ((N_{r}^{-1}\ln N_{r})^{2}+N_{t}^{-1}),&\text{if } (r_{i},t_{j}) \in D_{1}^{N}, \\ ((N_{r}^{-1}\ln N_{r})^{2}+N_{t}^{-1}),&\text{if } (r_{i},t_{j}) \in D_{2}^{N}. \end{cases} $$

First, the estimate for $W_{l}-w_{l}$ is given. The argument depends on whether $\mu =\frac{1}{4}$ or $\mu =2\sqrt{{\varepsilon }}\ln N$

Case(i): $\mu =\frac{1}{4}$.

In this case the mesh is uniform and $2\sqrt{{\varepsilon }}\ln N\geq \frac{1}{4}$. It is clear that $r_{i}-r_{i-1}=N^{-1}$ and $\varepsilon ^{-\frac{1}{2}}\leq C\ln N$. By [12] we have

$$\begin{aligned}& \begin{aligned} \mathcal{K}^{N}(W_{l}-w_{l}) (r_{N},t_{j})&=\mathcal{K}^{N}W_{l}(r_{N},t_{j})- \mathcal{K}^{N}w_{l}(r_{N} ,t_{j}) \\ &=\phi _{\tilde{r}}-\mathcal{K}^{N}w_{l}(r_{N},t_{j}) \\ &=\mathcal{K}w_{l}(r_{N},t)-\mathcal{K}^{N}w_{l}(r_{N},t_{j}), \end{aligned} \\& \begin{aligned} \bigl\vert \mathcal{K}^{N}(W_{l}-w_{l}) (r_{N}) \bigr\vert &\leq C\varepsilon (\bigl(h_{1}^{3}w_{l}''( \chi _{1},t_{j})+\cdots +h_{N}^{3}w_{l} ''(\chi _{N},t_{j})\bigr) \\ &\leq C\varepsilon ^{-1}\bigl(h_{1}^{3}+ \cdots +h_{N}^{3}\bigr) \\ &\leq CN_{r}^{-2}, \end{aligned} \end{aligned}$$

where $r_{i-1}\leq \chi _{i}\leq r_{i}$. Using Lemma 4 to the function $(W_{l}-w_{l})(r_{i})$ gives

$$\begin{aligned} \bigl\vert (W_{l}-w_{l}) (r_{i}) \bigr\vert \leq C\bigl(N_{r}^{-2}\ln ^{2} N_{r}\bigr). \end{aligned}$$

Case(ii): $\mu <\frac{1}{4}$.

Since the mesh is piecewise uniform with the subinterval $[\mu , 1-\mu ]$, mesh elements are $4(1-2\mu )/N$ and the rest of the intervals $[0, \mu ]$ and $[1-\mu , 1]$ with $8\mu /N$ mesh elements.

By [12], we have

$$ \bigl\vert \mathcal{K}^{N}(W_{l}-w_{l}) (r_{i}) \bigr\vert \leq C\bigl(N_{r}^{-2} \ln ^{2} N_{r}\bigr) $$

and

$$\begin{aligned} \bigl\vert \mathcal{K}^{N}(W_{l}-w_{l}) (r_{N}) \bigr\vert \leq & \varepsilon \bigl\vert C \bigl(h_{1}^{3}w''( \chi _{1})+\cdots +h_{N}^{3}w''( \chi _{N})\bigr) \bigr\vert \\ \leq &C\bigl(h_{1}^{3}+\cdots +h_{N}^{3} \bigr) \\ \leq &CN_{r}^{-2}, \end{aligned}$$

where $r_{i-1}\leq \chi _{i}\leq r_{i}$. Using Lemma 4 to the function $(W_{l}-w_{l})(r_{i})$ gives

$$\begin{aligned} \bigl\vert (W_{l}-w_{l}) (r_{i}) \bigr\vert \leq C\bigl(N_{r}^{-2}\ln ^{2} N_{r}+N_{t}^{-1}\bigr). \end{aligned}$$

Analogous arguments are used to establish the error estimate for $W_{R}$. Hence proved. □

6 Numerical examples

As the exact solutions of these problems are not known, to compute the error estimate, the double mesh principle is used, which is stated as follows:

$$ E_{\varepsilon }^{N,\Delta t}=\max_{(r_{i},t_{j})\in \bar{D}^{N}} \bigl\vert U^{N, \Delta t} (r_{i},t_{j}) -U^{2N,\Delta t/2}(r_{i},t_{j}) \bigr\vert . $$

We determine the uniform error $E^{N,\Delta t}=\max_{\varepsilon }E^{N,\Delta t}$ and the rate of convergence as $p^{N,\Delta t}=\log _{2} (\frac{E^{N,\Delta t}}{E^{2N,\Delta t/2}})$.

Example 6.1

$$ \textstyle\begin{cases} (-\varepsilon \frac{\partial ^{2} u_{\varepsilon }}{\partial r^{2}} + \frac{\partial u_{\varepsilon }}{\partial t} +5u_{\varepsilon } )(r,t)-u_{\varepsilon }(r-1,t)=e^{-r},\quad (r,t)\in (0,2)\times (0,2], \\ u_{\varepsilon }(r,t)=0, \quad \forall (r,t)\in \Gamma _{l}, \\ \mathcal{K}u_{\varepsilon }(2,t)=u_{\varepsilon }(2,t)-\varepsilon \int _{0}^{2} \frac{r}{3}u_{\varepsilon }(r,t)\,dr=0, \quad \forall (r,t) \in \Gamma _{\bar{r}}, \\ u_{\varepsilon }(r,t)=0,\quad \forall (r,t)\in \Gamma _{b}. \end{cases} $$

Example 6.2

$$ \textstyle\begin{cases} (-\varepsilon \frac{\partial ^{2} u_{\varepsilon }}{\partial r^{2}} + \frac{\partial u_{\varepsilon }}{\partial t} +5u_{\varepsilon } )(r,t)-ru_{\varepsilon }(r-1,t)=1, \quad (r,t)\in (0,2)\times (0,2], \\ u_{\varepsilon }(r,t)=0, \quad \forall (r,t)\in \Gamma _{l}, \\ \mathcal{K}u_{\varepsilon }(2,t)=u_{\varepsilon }(2,t)-\varepsilon \int _{0}^{2} \frac{1}{6}u_{\varepsilon }(r,t)\,dr=0,\quad \forall (r,t) \in \Gamma _{\bar{r}}, \\ u_{\varepsilon }(r,t)=\sin (\pi r), \quad \forall (r,t)\in \Gamma _{b}. \end{cases} $$

7 Discussion

In the literature, so far, no one has considered singularly perturbed parabolic partial differential equation with nonlocal boundary condition. We considered a class of singularly perturbed partial differential equations with delay in space and integral boundary condition. A finite difference method and a trapezoidal rule were proposed. We showed that the order of convergence and the error is of order $O(N_{r}^{-2} \ln N_{r}^{2} +N_{t}^{-1})$. The theory has been validated with two examples. Our numerical results reflect the theoretical estimates. Maximum point-wise errors and the order of convergence of Example 6.1 and Example 6.2 are given in Tables 1 and 2 respectively. The numerical solution of Examples 6.1 and 6.2 are plotted in Figs. 1 and 2.

Figure 1

Solution graph of Example 6.1

Figure 2

Solution graph of Example 6.2

Table 1 Example 6.1 corresponding maximum pointwise errors and the rate of convergence such as $E_{\varepsilon }^{N,\Delta t}$, $E^{N,\Delta t}$, $p^{N,\Delta t}$

Table 2 Example 6.2 corresponding maximum pointwise errors and the rate of convergence such as $E_{\varepsilon }^{N,\Delta t}$, $E^{N,\Delta t}$, $p^{N,\Delta t}$