© Wuhan University 2025

0 Introduction

The definite integration plays an important role in calculus. It is usually calculated by the well-known Newton-Leibniz formula. However, when the integrand cannot be represented by fundamental functions, the primitive function is complicated or the integrand has no analytic expression (e.g., some discrete point values are known), numerical integration becomes a practical necessity.

To date, the common numerical integrations are based on the composite Newton-Cotes formula. If the integrand is sufficiently smooth, the error of the composite trapezoidal and Simpson rule can attain $O(h^{\mathrm{2}})$ and $O(h^{\mathrm{4}})$, where $h$ is the cell size. Otherwise, the composite Newton-Cotes formula based on the uniform mesh will produce very large errors.

Consider the numerical approximation for the following definite integration

$I(u)={\int }_{\mathrm{0}}^{\mathrm{1}}u(x)\mathrm{d}x$(1)

Assume that the function $u(x)$ can be decomposed as

$u(x)=\overline{u}(x)+u_{\epsilon ,l}(x)+u_{\epsilon ,r}(x),x\in (\mathrm{0,1})$(2)

where $\overline{u}(x)$ is the regular component, $u_{\epsilon ,l}(x)$ and $u_{\epsilon ,r}(x)$ are the two boundary layer components, which satisfy

$\{\begin{array}{l}\left|{\overline{u}}^{\left(j\right)}\left(x\right)\right|\le C,\\ \left|u_{\epsilon ,l}^{\left(j\right)}\left(x\right)\right|\le C{\epsilon }^{-j/\mathrm{2}}{\mathrm{e}}^{-\beta x/\sqrt[]{\epsilon }},\\ \left|u_{\epsilon ,r}^{\left(j\right)}\left(x\right)\right|\le C{\epsilon }^{-j/\mathrm{2}}{\mathrm{e}}^{-\beta \left(\mathrm{1}-x\right)/\sqrt[]{\epsilon }},\end{array}$(3)

for any $x\in [\mathrm{0,1}],$where $\epsilon \ll \mathrm{1}$ is a small positive parameter, $\beta$ is a given positive constant and $j\ge \mathrm{0}$ is an integer. It is obvious that the function $u(x)$ exhibits a typical boundary layer, namely it has large gradients at the boundary ends $x=\mathrm{0}$ and $x=\mathrm{1}$.

Such function of (2) can be viewed as the solution of the singularly perturbed reaction-diffusion problem

$-\epsilon u^{″}(x)+b(x)u(x)=f(x),u(\mathrm{0})=u(\mathrm{1})=\mathrm{0}$(4)

where $b(x)>{\beta }^{\mathrm{2}}>\mathrm{0},$$b(x)$ and $f(x)$ are smooth functions. The problem (4) has wide applications in such fields as optimal control, chemical reactions, fluid dynamics and so on. It is of great scientific significance to study the corresponding numerical solutions^[1].

Regarding the solution to the singularly perturbed convection-diffusion problem, there are abundant results on the numerical integration. For example, it was shown in Refs. [2-3] that the composite trapezoidal and Simpson rule on the uniform mesh will lead to an integration error of order $O(h)$. Ref. [4] studied the Lagrange interpolation of piecewise polynomials of degree $k$ and the composite Newton-Cotes formula on a piecewise-uniform Shishkin mesh^[5]. The convergence rate is $O((N^{-\mathrm{1}}{\mathrm{l}\mathrm{n}N)}^{k+\mathrm{1}})$, where $N$ is the number of total elements. In Ref. [6], the results were extended to the Bakhvalov mesh^[7-8], the interpolation error was proved to be $O(N^{-(k+\mathrm{1})})$ and the integration error was $O(N^{-k})$. Furthermore, a new $k$ point interpolation formula was discussed in Ref. [9]. The interpolation error was proved to be $O(h^k)$ when $\epsilon =\mathrm{1}$, but increased to $O(h^{k-\mathrm{1}})$ if $\epsilon$ is small. Ref. [10] showed the error of the Gauss integral formula using $k$ nodes on the Shishkin mesh, the convergence rate is $O(N^{-\mathrm{2}k})$.

However, all the above results were obtained for the integrand with one-sided boundary layer. There are very few results on the numerical integration for the solution of the singularly perturbed reaction-diffusion problem (4) with twin boundary layers. In this paper, we address this problem and present three types of numerical integrations on the Shishkin mesh. We derive some error estimates that are uniform in the perturbation parameter $\epsilon$. The contributions of this paper are the following:

1) We prove that the Lagrange interpolation on the Shishkin mesh converges at a rate of order $O((N^{-\mathrm{1}}{\mathrm{l}\mathrm{n}N)}^{k+\mathrm{1}})$, when piecewise polynomials of degree $k$ and total element number $N$ are used. Consequently, we obtain an integral error $O((N^{-\mathrm{1}}{\mathrm{l}\mathrm{n}N)}^{k+\mathrm{1}})$ for the Newton-Cotes formula on the Shishkin mesh, see Theorems 1 and 2.

2) Based on the local $L^{\mathrm{2}}$ projection for the integrand, we obtain an integral error of order $O((N^{-\mathrm{1}}{\mathrm{l}\mathrm{n}N)}^{k+\mathrm{1}})$, see Theorems 3 and 4.

3) We prove that the Gauss integral formula using $k$ nodes on the Shishkin mesh can attain a convergence order $O(N^{-\mathrm{2}k})$, see Theorem 5.

1 Shishkin Mesh

Define the mesh transition parameter

$\sigma =\mathrm{m}\mathrm{i}\mathrm{n}\left\{\frac{\mathrm{1}}{\mathrm{4}},\frac{\tau \sqrt[]{\epsilon }}{\beta }\mathrm{l}\mathrm{n}N\right\}$(5)

where $\tau >\mathrm{0}$ is a user-chosen parameter whose value will be discussed later. The mesh points are given by

$x_i=\{\begin{array}{l}\mathrm{4}\sigma \frac{i}{N},i=\mathrm{0,1},\cdots ,\frac{N}{\mathrm{4}},\\ \sigma +\mathrm{2}\left(\mathrm{1}-\mathrm{2}\sigma \right)\left(\frac{i}{N}-\frac{\mathrm{1}}{\mathrm{4}}\right),i=\frac{N}{\mathrm{4}}+\mathrm{1},\frac{N}{\mathrm{4}}+\mathrm{2},\cdots ,\frac{\mathrm{3}N}{\mathrm{4}},\\ \mathrm{1}-\mathrm{4}\sigma \left(\mathrm{1}-\frac{i}{N}\right),i=\frac{\mathrm{3}N}{\mathrm{4}}+\mathrm{1},\frac{\mathrm{3}N}{\mathrm{4}}+\mathrm{2},\cdots ,N.\end{array}$

Set the Shishkin mesh as ${\mathrm{\Omega }}_N=\{I_i{\}}_{i=\mathrm{1}}^N$ with element $I_i=\left[x_{i-\mathrm{1}},x_i\right]$. Set the mesh size $h_i=x_i-x_{i-\mathrm{1}}$ and then we have

$h_i=\{\begin{array}{l}h:=\frac{\mathrm{4}\sigma }{N},i=\mathrm{1,2},\cdots ,\frac{N}{\mathrm{4}},\frac{\mathrm{3}N}{\mathrm{4}}+\mathrm{1},\cdots ,N,\\ H:=\frac{\mathrm{2}(\mathrm{1}-\mathrm{2}\sigma )}{N},i=\frac{N}{\mathrm{4}}+\mathrm{1},\frac{N}{\mathrm{4}}+\mathrm{2},\cdots ,\frac{\mathrm{3}N}{\mathrm{4}}.\end{array}$(6)

Denote ${\mathrm{\Omega }}_N={\mathrm{\Omega }}^c\bigcup {\mathrm{\Omega }}^f$, where ${\mathrm{\Omega }}^c:=[\sigma ,\mathrm{1}-\sigma ]$ is the coarse domain and ${\mathrm{\Omega }}^f:=\left[\mathrm{0},\sigma \right]\bigcup \left[\mathrm{1}-\sigma ,\mathrm{1}\right]$ is the refined domain. Figure 1 shows a Shishkin mesh for $N=\mathrm{16}$.

Fig. 1 The Shishkin mesh ($\mathit{N}=\mathrm{16}$)

2 The Newton-Cotes Formula and Its Error Estimate

The common numerical integration is based on the Newton-Cotes formula using the Lagrange interpolation for the integrand. However, for the function (2) with boundary layer, the traditional method on the uniform mesh will result in a large error. This section presents the Lagrange interpolation and the Newton-Cotes formula on the Shishkin mesh. We will establish optimal-order error estimates that are independent of the small parameter $\epsilon$.

2.1 The Lagrange Interpolation

Assume that $N$ is a multiple of $\mathrm{4}k$, and divide the domain $[\mathrm{0,1}]$ into $N/k$ subintervals

$\left[\mathrm{0,1}\right]={\displaystyle \underset{i=\mathrm{0},k}{\overset{N-k}{\cup }}}{K}_i:={\displaystyle \underset{i=\mathrm{0},k}{\overset{N-k}{\cup }}}\left[x_i,x_{i+k}\right]$

where ${\displaystyle \underset{i=\mathrm{0},k}{\overset{N-k}{\cup }}}{K}_i$ represents the set of cell $K_{\mathrm{0}},K_k,K_{\mathrm{2}k},\cdots ,$

$K_{N-k}$. On each interval $K_i=[x_i,x_{i+k}]$, we define the Lagrange interpolation of $u(x)$ by

$L_{i,k}\left(u\right)={\displaystyle \sum _{n=i}^{i+k}}{u}_n{P}_{n,i}\left(x\right):={\displaystyle \sum _{n=i}^{i+k}}{u}_n{\displaystyle \prod _{\begin{array}{l}j=i\\ j\ne n\end{array}}^{i+k}}\frac{x-x_j}{x_n-x_j}$(7)

where $P_{n,i}(x)$ is the base function with degree k.

Define the Lebesgue constant of $K_i$ as

${\lambda }_{i,k+\mathrm{1}}=\underset{x\in K_i}{\mathrm{m}\mathrm{a}\mathrm{x}}{\displaystyle \sum _{n=i}^{i+k}}\left|P_{n,i}\left(x\right)\right|$(8)

Since $N$ is a multiple of $\mathrm{4}k$, we have $K_i\subseteq {\mathrm{\Omega }}^c$ or $K_i\subseteq {\mathrm{\Omega }}^f$. Therefore, the subinterval $K_i=[x_i,x_{i+k}]$ is uniform. From Ref. [11], one has

${\mathrm{2}}^{k-\mathrm{2}}{\left(k+\mathrm{1}\right)}^{-\mathrm{3}/\mathrm{2}}\le {\lambda }_{i,k+\mathrm{1}}\le {\mathrm{2}}^k$(9)

Lemma 1   Let $h_i$ be the mesh size of the interval $K_i$. Denote ${\omega }_{k+\mathrm{1}}(x)=(x-x_i)(x-x_{i+\mathrm{1}})\cdots (x-x_{i+k})$, then one has

$\left|{\omega }_{k+\mathrm{1}}\left(x\right)\right|\le \frac{\mathrm{1}}{\mathrm{4}}{h}_i^{k+\mathrm{1}}k!,\forall x\in K_i$(10)

Proof   Without loss of generality, let $x\in K_i=[x_i,x_{i+k}]$. For any $j=i+\mathrm{2},i+\mathrm{3},\cdots ,i+k,$ one has $|x-x_j|\le (j-i)h_i$ and

$\begin{array}{l}\left|\left(x-x_i\right)\left(x-x_{i+\mathrm{1}}\right)\right|=\left(x-x_i\right)\left(x_{i+\mathrm{1}}-x\right)\\ \le {\left(\frac{x-x_i+x_{i+\mathrm{1}}-x}{\mathrm{2}}\right)}^{\mathrm{2}}=\frac{h_i^{\mathrm{2}}}{\mathrm{4}},\end{array}$

thus $\left|{\omega }_{k+\mathrm{1}}\left(x\right)\right|\le \frac{h_i^{\mathrm{2}}}{\mathrm{4}}\times \mathrm{2}{h}_i\times \mathrm{3}{h}_i\times \cdots \times k{h}_i=\frac{\mathrm{1}}{\mathrm{4}}{h_i}^{k+\mathrm{1}}k!$.

This finishes the proof.

2.2 Error Estimate of the Lagrange Interpolation

Lemma 2   One has the following interpolation error

$\left|u\left(x\right)-L_{i,k}\left(u\right)\right|\le \frac{h_i^{k+\mathrm{1}}}{\mathrm{4}\left(k+\mathrm{1}\right)}\underset{x\in [x_i,x_{i+k}]}{\mathrm{m}\mathrm{a}\mathrm{x}}\left|u^{\left(k+\mathrm{1}\right)}\left(x\right)\right|$(11)

Proof   From Ref. [12], one has

$\left|u\left(x\right)-L_{i,k}\left(u\right)\right|\le \frac{\left|{\omega }_{k+\mathrm{1}}\left(x\right)\right|}{\left(k+\mathrm{1}\right)!}\underset{x\in \left[x_i,x_{i+k}\right]}{\mathrm{m}\mathrm{a}\mathrm{x}}\left|u^{\left(k+\mathrm{1}\right)}\left(x\right)\right|$(12)

Then (11) follows from Lemma 1.

Theorem 1   Suppose that $u(x)$ satisfies (2) and (3) and set $\tau =k+\mathrm{1}$. Then one has the following interpolation error estimates.

If $\sigma <\mathrm{1}/\mathrm{4}$ and $K_i=\left[x_i,x_{i+k}\right]\subseteq {\mathrm{\Omega }}^f$, then

$\left|u\left(x\right)-L_{i,k}\left(u\right)\right|\le C{\left(N^{-\mathrm{1}}\mathrm{l}\mathrm{n}N\right)}^{k+\mathrm{1}}$(13)

If $\sigma <\mathrm{1}/\mathrm{4}$ and $K_i=\left[x_i,x_{i+k}\right]\subseteq {\mathrm{\Omega }}^c$, then

$\left|u\left(x\right)-L_{i,k}\left(u\right)\right|\le C{N}^{-\left(k+\mathrm{1}\right)}$(14)

If $\sigma =\mathrm{1}/\mathrm{4}$, then

$\begin{array}{l}\left|u\left(x\right)-L_{i,k}\left(u\right)\right|\\ \le C\mathrm{m}\mathrm{i}\mathrm{n}\left\{{\left(\sqrt[]{\epsilon }N\right)}^{-\left(k+\mathrm{1}\right)},{\left(N^{-\mathrm{1}}\mathrm{l}\mathrm{n}N\right)}^{k+\mathrm{1}}\right\}\end{array}$(15)

Here the bounding constant $C>\mathrm{0}$ is independent of $\epsilon$ and $N$.

Proof   From (3) and (11), one has

$\left|\overline{u}\left(x\right)-L_{i,k}\left(\overline{u}\right)\right|\le C{h}_i^{k+\mathrm{1}},x\in K_i$

then

$\left|\overline{u}\left(x\right)-L_{i,k}\left(\overline{u}\right)\right|\le C{N}^{-\left(k+\mathrm{1}\right)},x\in K_i$(16)

To bound the interpolation error for the left boundary layer component $u_{\epsilon ,l}(x),$we proceed from the following two situations.

Case 1 $\sigma <\mathrm{1}/\mathrm{4}$

For $K_i\subseteq {\mathrm{\Omega }}^f$, it follows from (3) and (6) that

$h_i=h=\frac{\mathrm{4}\sigma }{N}=\frac{\mathrm{4}\left(k+\mathrm{1}\right)\sqrt[]{\epsilon }}{\beta }\frac{\mathrm{l}\mathrm{n}N}{N}$(17)

Noticing

$\left|u_{\epsilon ,l}^{\left(k+\mathrm{1}\right)}\left(x\right)\right|\le C{\epsilon }^{-(k+\mathrm{1})/\mathrm{2}}{\mathrm{e}}^{-\beta x/\sqrt[]{\epsilon }},x\in \left[x_i,x_{i+k}\right]$(18)

and Lemma 2, one obtains

$\left|u_{\epsilon ,l}\left(x\right)-L_{i,k}\left(u_{\epsilon ,l}\right)\right|\le C{\left(h/\sqrt[]{\epsilon }\right)}^{k+\mathrm{1}}{\mathrm{e}}^{-\beta x/\sqrt[]{\epsilon }}$(19)

namely,

$\left|u_{\epsilon ,l}\left(x\right)-L_{i,k}\left(u_{\epsilon ,l}\right)\right|\le C{\left(N^{-\mathrm{1}}\mathrm{l}\mathrm{n}N\right)}^{k+\mathrm{1}}$(20)

For $K_i\subseteq {\mathrm{\Omega }}^c$, due to (3) and $\tau =k+\mathrm{1}$, one obtains

$\left|u_{\epsilon ,l}\left(x\right)\right|\le C{\mathrm{e}}^{-\beta \sigma /\sqrt[]{\epsilon }}=C{N}^{-\left(k+\mathrm{1}\right)}$(21)

From (7), (8) and (21), one gets

$\left|L_{i,k}\left(u_{\epsilon ,l}\right)\right|\le C{N}^{-\left(k+\mathrm{1}\right)}{\lambda }_{i,k+\mathrm{1}}$

Hence

$\begin{array}{l}\left|u_{\epsilon ,l}\left(x\right)-L_{i,k}\left(u_{\epsilon ,l}\right)\right|\le \left|u_{\epsilon ,l}\left(x\right)\right|+\left|L_{i,k}\left(u_{\epsilon ,l}\right)\right|\\ \le C{N}^{-\left(k+\mathrm{1}\right)}\left(\mathrm{1}+{\lambda }_{i,k+\mathrm{1}}\right).\end{array}$

Using (9) yields

$\left|u_{\epsilon ,l}\left(x\right)-L_{i,k}\left(u_{\epsilon ,l}\right)\right|\le C{N}^{-\left(k+\mathrm{1}\right)}$(22)

Case 2 $\sigma =\mathrm{1}/\mathrm{4}$

The partition ${\mathrm{\Omega }}_N$ is uniform and $h_i=N^{-\mathrm{1}}$. By (11), one has

$\left|u_{\epsilon ,l}\left(x\right)-L_{i,k}\left(u_{\epsilon ,l}\right)\right|\le C{N}^{-(k+\mathrm{1})}{\epsilon }^{-\left(k+\mathrm{1}\right)/\mathrm{2}}=C{\left(\sqrt[]{\epsilon }N\right)}^{-\left(k+\mathrm{1}\right)}$

If $\sqrt[]{\epsilon }\ge \frac{\beta }{\mathrm{4}\left(k+\mathrm{1}\right)\mathrm{l}\mathrm{n}N}$, then

$\left|u_{\epsilon ,l}\left(x\right)-L_{i,k}\left(u_{\epsilon ,l}\right)\right|\le C\mathrm{m}\mathrm{i}\mathrm{n}\left\{{\left(\sqrt[]{\epsilon }N\right)}^{-\left(k+\mathrm{1}\right)},{\left(N^{-\mathrm{1}}\mathrm{l}\mathrm{n}N\right)}^{k+\mathrm{1}}\right\}$

In a similar manner, one can bound the interpolation error for the right boundary layer component $u_{\epsilon ,r}\left(x\right)$. This finishes the proof.

2.3 Error Estimate of the Newton-Cotes Formula

Let

$I\left(u\right)={\int }_{\mathrm{0}}^{\mathrm{1}}u\left(x\right)\mathrm{d}x={\displaystyle \sum _{i=\mathrm{0},k}^{N-k}}{\int }_{x_i}^{x_{i+k}}u\left(x\right)\mathrm{d}x={\displaystyle \sum _{i=\mathrm{0},k}^{N-k}}{I}_{i,k}\left(u\right)$(23)

where $u(x)$ satisfies (2) and (3). On the Shishkin mesh ${\mathrm{\Omega }}_N$ with $\tau =k+\mathrm{1}$, we introduce the Newton-Cotes formula with $k+\mathrm{1}$ nodes

$S_k\left(u\right)={\displaystyle \sum _{i=\mathrm{0},k}^{N-k}}{S}_{i,k}\left(u\right)={\displaystyle \sum _{i=\mathrm{0},k}^{N-k}}{\int }_{x_i}^{x_{i+k}}{L}_{i,k}\left(u\right)\mathrm{d}x$(24)

Theorem 2   Suppose that $u(x)$ satisfies (2) and (3) and set $\tau =k+\mathrm{1}$. Then one has the following integral error estimates.

If $\sqrt[]{\epsilon }<\frac{\beta }{\mathrm{4}\left(k+\mathrm{1}\right)\mathrm{l}\mathrm{n}N}$, then

$\left|I\left(u\right)-S_k\left(u\right)\right|\le C\left[N^{-\left(k+\mathrm{1}\right)}+\sqrt[]{\epsilon }(N^{-\mathrm{1}}{\mathrm{l}\mathrm{n}N)}^{k+\mathrm{1}}\right]$(25)

If $\sqrt[]{\epsilon }\ge \frac{\beta }{\mathrm{4}\left(k+\mathrm{1}\right)\mathrm{l}\mathrm{n}N}$, then

$\left|I\left(u\right)-S_k\left(u\right)\right|\le C\mathrm{m}\mathrm{i}\mathrm{n}\left\{{\left(\sqrt[]{\epsilon }N\right)}^{-\left(k+\mathrm{1}\right)},{\left(N^{-\mathrm{1}}\mathrm{l}\mathrm{n}N\right)}^{k+\mathrm{1}}\right\}$(26)

Proof   We proceed from the following two situations.

Case 1 $\sqrt[]{\epsilon }<\frac{\beta }{\mathrm{4}\left(k+\mathrm{1}\right)\mathrm{l}\mathrm{n}N}$

Now one has $\sigma =\mathrm{m}\mathrm{i}\mathrm{n}\left\{\frac{\mathrm{1}}{\mathrm{4}},\frac{\tau \sqrt[]{\epsilon }}{\beta }\mathrm{l}\mathrm{n}N\right\}=\frac{\tau \sqrt[]{\epsilon }}{\beta }\mathrm{l}\mathrm{n}N<\frac{\mathrm{1}}{\mathrm{4}}$.

By (16), one has

${\displaystyle \sum _{i=\mathrm{0},k}^{N-k}}{\int }_{x_i}^{x_{i+k}}\left|\overline{u}\left(x\right)-L_{i,k}\left(\overline{u}\right)\right|\mathrm{d}x\le {\displaystyle \sum _{i=\mathrm{0},k}^{N-k}}C{h}_i{N}^{-\left(k+\mathrm{1}\right)}\le C{N}^{-\left(k+\mathrm{1}\right)}$

Using (19) and (22) yields

$\begin{array}{l}{\displaystyle \sum _{i=\mathrm{0},k}^{N-k}}{\int }_{x_i}^{x_{i+k}}\left|u_{\epsilon ,l}\left(x\right)-L_{i,k}\left(u_{\epsilon ,l}\right)\right|\mathrm{d}x\\ \le {\displaystyle \sum _{i=\mathrm{0},k}^{N/\mathrm{4}-k}}C{\left(N^{-\mathrm{1}}\mathrm{l}\mathrm{n}N\right)}^{k+\mathrm{1}}{\int }_{x_i}^{x_{i+k}}{\mathrm{e}}^{-\beta x/\sqrt[]{\epsilon }}\mathrm{d}x+{\displaystyle \sum _{i=N/\mathrm{4},k}^{\mathrm{3}N/\mathrm{4}-k}}{\int }_{x_i}^{x_{i+k}}C{N}^{-\left(k+\mathrm{1}\right)}\mathrm{d}x\\ +{\displaystyle \sum _{i=\mathrm{3}N/\mathrm{4},k}^{N-k}}C{\left(N^{-\mathrm{1}}\mathrm{l}\mathrm{n}N\right)}^{k+\mathrm{1}}{\int }_{x_i}^{x_{i+k}}{\mathrm{e}}^{-\beta x/\sqrt[]{\epsilon }}\mathrm{d}x\\ \le C\sqrt[]{\epsilon }{\left(N^{-\mathrm{1}}\mathrm{l}\mathrm{n}N\right)}^{k+\mathrm{1}}+C{N}^{-\left(k+\mathrm{1}\right)}.\end{array}$

Analogously,

$\begin{array}{l}{\displaystyle \sum _{i=\mathrm{0},k}^{N-k}}{\int }_{x_i}^{x_{i+k}}\left|u_{\epsilon ,r}\left(x\right)-L_{i,k}\left(u_{\epsilon ,r}\right)\right|\mathrm{d}x\\ \le C\sqrt[]{\epsilon }{\left(N^{-\mathrm{1}}\mathrm{l}\mathrm{n}N\right)}^{k+\mathrm{1}}+C{N}^{-\left(k+\mathrm{1}\right)},\end{array}$

Collecting the above estimates gives

$\left|I\left(u\right)-S_k\left(u\right)\right|\le C\left[N^{-\left(k+\mathrm{1}\right)}+\sqrt[]{\epsilon }{\left(N^{-\mathrm{1}}\mathrm{l}\mathrm{n}N\right)}^{k+\mathrm{1}}\right]$(27)

Case 2 $\sqrt[]{\epsilon }\ge \frac{\beta }{\mathrm{4}\left(k+\mathrm{1}\right)\mathrm{l}\mathrm{n}N}$

Now one has $\sigma =\mathrm{m}\mathrm{i}\mathrm{n}\left\{\frac{\mathrm{1}}{\mathrm{4}},\frac{\tau \sqrt[]{\epsilon }}{\beta }\mathrm{l}\mathrm{n}N\right\}=\frac{\mathrm{1}}{\mathrm{4}}.$

Using (15), one obtains

$\begin{array}{l}\left|I\left(u\right)-S_k\left(u\right)\right|\le {\displaystyle \sum _{i=\mathrm{0},k}^{N-k}}\left|I_{i,k}\left(u\right)-S_{i,k}\left(u\right)\right|\\ \le {\displaystyle \sum _{i=\mathrm{0},k}^{N-k}}k{h}_i\underset{x\in \left[x_i,x_{i+k}\right]}{\mathrm{m}\mathrm{a}\mathrm{x}}\left|u\left(x\right)-L_{i,k}\left(u\right)\right|\\ \le C{N}^{-\left(k+\mathrm{1}\right)}\mathrm{m}\mathrm{i}\mathrm{n}\left\{{\epsilon }^{-\left(k+\mathrm{1}\right)/\mathrm{2}},\mathrm{l}{\mathrm{n}}^{k+\mathrm{1}}N\right\}\end{array}$(28)

This finishes the proof.

3 Numerical Integration Based on Local ${\mathit{L}}^{\mathrm{2}}$ Projection

It is common to use the projection polynomials to approximate the integrand. Local $L^{\mathrm{2}}$ projection can realize element orthogonality and is widely used in the approximation theory and finite element error analyses. In this section, we discuss the numerical integration using the local $L^{\mathrm{2}}$ projection for the integrand.

3.1 Local ${\mathit{L}}^{\mathrm{2}}$ Projection

Let $P^k(I_i)$ be the space composed with piecewise degree of at most $k$ on the subinterval $I_i=\left[x_{i-\mathrm{1}},x_i\right]$. For any $z\in L^{\mathrm{2}}(I_i)$, define $\mathrm{\pi }z\in P^k(I_i)$ by

${\int }_{I_i}\mathrm{\pi }z\cdot v\mathrm{d}x={\int }_{I_i}z\cdot v\mathrm{d}x,\forall v\in P^k\left(I_i\right).$

Denote ${\Vert \cdot \Vert }_{L^{\mathrm{\infty }}\left(I_i\right)}$ by the usual maximum-norm on the cell $I_i$. We have the following projection error estimates on the Shishkin mesh.

Lemma 3^[13] There exists a constant $C>\mathrm{0}$ independent of $u$ and $h_i$ such that

${\Vert u-\mathrm{\pi }u\Vert }_{L^{\mathrm{\infty }}\left(I_i\right)}\le C{h}_i^{k+\mathrm{1}}{\Vert u^{\left(k+\mathrm{1}\right)}\Vert }_{L^{\mathrm{\infty }}\left(I_i\right)}$(29)

${\Vert u-\mathrm{\pi }u\Vert }_{L^{\mathrm{\infty }}\left(I_i\right)}\le C{\Vert u\Vert }_{L^{\mathrm{\infty }}\left(I_i\right)}$(30)

3.2 Error Estimate

Theorem 3   Suppose that $u(x)$ satisfies (2) and (3) and set $\tau =k+\mathrm{1}$. Then one has the following projection error estimates.

If $\sigma <\mathrm{1}/\mathrm{4}$ and $I_i=\left[x_{i-\mathrm{1}},x_i\right]\subseteq {\mathrm{\Omega }}^f$, then

${\Vert u-\mathrm{\pi }u\Vert }_{L^{\mathrm{\infty }}\left(I_i\right)}\le C{\left(N^{-\mathrm{1}}\mathrm{l}\mathrm{n}N\right)}^{k+\mathrm{1}}$(31)

If $\sigma <\mathrm{1}/\mathrm{4}$ and $I_i=\left[x_{i-\mathrm{1}},x_i\right]\subseteq {\mathrm{\Omega }}^c$, then

${\Vert u-\mathrm{\pi }u\Vert }_{L^{\mathrm{\infty }}\left(I_i\right)}\le C{N}^{-\left(k+\mathrm{1}\right)}$(32)

If $\sigma =\mathrm{1}/\mathrm{4}$, then

${\Vert u-\mathrm{\pi }u\Vert }_{L^{\mathrm{\infty }}\left(I_i\right)}\le C\mathrm{m}\mathrm{i}\mathrm{n}\left\{{\left(\sqrt[]{\epsilon }N\right)}^{-\left(k+\mathrm{1}\right)},{\left(N^{-\mathrm{1}}\mathrm{l}\mathrm{n}N\right)}^{k+\mathrm{1}}\right\}$(33)

Here the bounding constant $C>\mathrm{0}$ is independent of $\epsilon$ and $N$.

Proof   Recall $u(x)=\overline{u}(x)+u_{\epsilon ,l}(x)+u_{\epsilon ,r}(x)$. From (3) and (29), one has

${\Vert \overline{u}-\mathrm{\pi }\overline{u}\Vert }_{L^{\mathrm{\infty }}\left(I_i\right)}\le C{h}_i^{k+\mathrm{1}}{\Vert {\overline{u}}^{\left(k+\mathrm{1}\right)}\Vert }_{L^{\mathrm{\infty }}\left(I_i\right)}\le C{N}^{-\left(k+\mathrm{1}\right)}$(34)

Now we turn to bound the projection error for the left boundary layer component $u_{\epsilon ,l}\left(x\right)$.

Case 1 $\sigma <\mathrm{1}/\mathrm{4}$

For $I_i=\left[x_{i-\mathrm{1}},x_i\right]\subseteq {\mathrm{\Omega }}^f$, it follows from (17) and (29) that

$\begin{array}{l}{\Vert u_{\epsilon ,l}-\mathrm{\pi }{u}_{\epsilon ,l}\Vert }_{L^{\mathrm{\infty }}\left(I_i\right)}\le C{h}_i^{k+\mathrm{1}}{\Vert u_{\epsilon ,l}^{(k+\mathrm{1})}\Vert }_{L^{\mathrm{\infty }}\left(I_i\right)}\\ \le C{\left(N^{-\mathrm{1}}\mathrm{l}\mathrm{n}N\right)}^{k+\mathrm{1}}{\mathrm{e}}^{-\beta x/\sqrt[]{\epsilon }},\end{array}$(35)

namely,

${\Vert u_{\epsilon ,l}-\mathrm{\pi }{u}_{\epsilon ,l}\Vert }_{L^{\mathrm{\infty }}\left(I_i\right)}\le C{\left(N^{-\mathrm{1}}\mathrm{l}\mathrm{n}N\right)}^{k+\mathrm{1}}$(36)

For $I_i=\left[x_{i-\mathrm{1}},x_i\right]\subseteq {\mathrm{\Omega }}^c$, the inequalities (21), (30) and $\tau =k+\mathrm{1}$ give

${\Vert u_{\epsilon ,l}-\mathrm{\pi }{u}_{\epsilon ,l}\Vert }_{L^{\mathrm{\infty }}\left(I_i\right)}\le C{\Vert u_{\epsilon ,l}\Vert }_{L^{\mathrm{\infty }}\left(I_i\right)}\le C{\mathrm{e}}^{-\beta \sigma /\sqrt[]{\epsilon }}=C{N}^{-\left(k+\mathrm{1}\right)}$(37)

Case 2 $\sigma =\mathrm{1}/\mathrm{4}$

The partition ${\mathrm{\Omega }}_N$ is uniform and $h_i=N^{-\mathrm{1}}$. By (11), one has

${\Vert u_{\epsilon ,l}-\mathrm{\pi }{u}_{\epsilon ,l}\Vert }_{L^{\mathrm{\infty }}\left(I_i\right)}\le C{h}_i^{k+\mathrm{1}}{\Vert u_{\epsilon ,l}^{(k+\mathrm{1})}\Vert }_{L^{\mathrm{\infty }}\left(I_i\right)}\le C{\left(\sqrt[]{\epsilon }N\right)}^{-\left(k+\mathrm{1}\right)}$

Note that $\sigma =\mathrm{1}/\mathrm{4}$ implies $\sqrt[]{\epsilon }\ge \frac{\beta }{\mathrm{4}\left(k+\mathrm{1}\right)\mathrm{l}\mathrm{n}N}$, then one gets

${\Vert u_{\epsilon ,l}-\mathrm{\pi }{u}_{\epsilon ,l}\Vert }_{L^{\mathrm{\infty }}\left(I_i\right)}\le C{N}^{-\left(k+\mathrm{1}\right)}\mathrm{m}\mathrm{i}\mathrm{n}\left\{{\epsilon }^{-\left(k+\mathrm{1}\right)/\mathrm{2}},{\left(\mathrm{l}\mathrm{n}N\right)}^{k+\mathrm{1}}\right\}$(38)

In a similar manner, one can bound the projection error for the right boundary layer component $u_{\epsilon ,r}\left(x\right)$. This finishes the proof.

Theorem 4   Define $T\left(u\right)={\displaystyle \sum _{i=\mathrm{1}}^N}{\int }_{I_i}\mathrm{\pi }u(x)\mathrm{d}x.$ Suppose that $u(x)$ satisfies (2) and (3) and set $\tau =k+\mathrm{1}$. Then one has the following integral error estimates.

If $\sqrt[]{\epsilon }<\frac{\beta }{\mathrm{4}\left(k+\mathrm{1}\right)\mathrm{l}\mathrm{n}N}$, then

$\left|I\left(u\right)-T\left(u\right)\right|\le C\left[N^{-\left(k+\mathrm{1}\right)}+\sqrt[]{\epsilon }(N^{-\mathrm{1}}{\mathrm{l}\mathrm{n}N)}^{k+\mathrm{1}}\right]$(39)

If$\sqrt[]{\epsilon }\ge \frac{\beta }{\mathrm{4}\left(k+\mathrm{1}\right)\mathrm{l}\mathrm{n}N}$, then

$\left|I\left(u\right)-T\left(u\right)\right|\le C\mathrm{m}\mathrm{i}\mathrm{n}\left\{{\left(\sqrt[]{\epsilon }N\right)}^{-\left(k+\mathrm{1}\right)},{\left(N^{-\mathrm{1}}\mathrm{l}\mathrm{n}N\right)}^{k+\mathrm{1}}\right\}$(40)

Proof   We proceed from the following two situations.

Case 1 $\sqrt[]{\epsilon }<\frac{\beta }{\mathrm{4}\left(k+\mathrm{1}\right)\mathrm{l}\mathrm{n}N}$

Now one has

$\sigma =\mathrm{m}\mathrm{i}\mathrm{n}\left\{\frac{\mathrm{1}}{\mathrm{4}},\frac{\tau \sqrt[]{\epsilon }}{\beta }\mathrm{l}\mathrm{n}N\right\}=\frac{(k+\mathrm{1})\sqrt[]{\epsilon }}{\beta }\mathrm{l}\mathrm{n}N<\frac{\mathrm{1}}{\mathrm{4}}$

From (34), one obtains

${\displaystyle \sum _{i=\mathrm{1}}^N}{\int }_{x_{i-\mathrm{1}}}^{x_i}\left|\overline{u}\left(x\right)-L_{i,k}\left(\overline{u}\right)\right|\mathrm{d}x\le {\displaystyle \sum _{i=\mathrm{1}}^N}C{N}^{-\left(k+\mathrm{1}\right)}{h}_i\le C{N}^{-\left(k+\mathrm{1}\right)}$

From (35) and (37), one has

$\begin{array}{l}{\displaystyle \sum _{i=\mathrm{1}}^N}{\int }_{x_{i-\mathrm{1}}}^{x_i}\left|u_{\epsilon ,l}\left(x\right)-L_{i,k}\left(u_{\epsilon ,l}\right)\right|\mathrm{d}x\\ \le {\displaystyle \sum _{i=\mathrm{1}}^{N/\mathrm{4}}}C{\left(N^{-\mathrm{1}}\mathrm{l}\mathrm{n}N\right)}^{k+\mathrm{1}}{\int }_{x_{i-\mathrm{1}}}^{x_i}{\mathrm{e}}^{-\beta x/\sqrt[]{\epsilon }}\mathrm{d}x+{\displaystyle \sum _{i=N/\mathrm{4}+\mathrm{1}}^{\mathrm{3}N/\mathrm{4}}}{\int }_{x_{i-\mathrm{1}}}^{x_i}C{N}^{-\left(k+\mathrm{1}\right)}\mathrm{d}x\\ +{\displaystyle \sum _{i=\mathrm{3}N/\mathrm{4}+\mathrm{1}}^N}C{\left(N^{-\mathrm{1}}\mathrm{l}\mathrm{n}N\right)}^{k+\mathrm{1}}{\int }_{x_{i-\mathrm{1}}}^{x_i}{\mathrm{e}}^{-\beta x/\sqrt[]{\epsilon }}\mathrm{d}x\\ \le C\sqrt[]{\epsilon }{\left(N^{-\mathrm{1}}\mathrm{l}\mathrm{n}N\right)}^{k+\mathrm{1}}+C{N}^{-\left(k+\mathrm{1}\right)}.\end{array}$

Likewise, one gets

$\begin{array}{l}{\displaystyle \sum _{i=\mathrm{1}}^N}{\int }_{x_{i-\mathrm{1}}}^{x_i}\left|u_{\epsilon ,r}\left(x\right)-L_{i,k}\left(u_{\epsilon ,r}\right)\right|\mathrm{d}x\\ \le C\sqrt[]{\epsilon }{\left(N^{-\mathrm{1}}\mathrm{l}\mathrm{n}N\right)}^{k+\mathrm{1}}+C{N}^{-\left(k+\mathrm{1}\right)}.\end{array}$

Collecting the above estimates gives

$\left|I\left(u\right)-T\left(u\right)\right|\le C\left[N^{-\left(k+\mathrm{1}\right)}+\sqrt[]{\epsilon }{\left(N^{-\mathrm{1}}\mathrm{l}\mathrm{n}N\right)}^{k+\mathrm{1}}\right]$(41)

Case 2 $\sqrt[]{\epsilon }\ge \frac{\beta }{\mathrm{4}\left(k+\mathrm{1}\right)\mathrm{l}\mathrm{n}N}$

Now one has $\sigma =\mathrm{m}\mathrm{i}\mathrm{n}\left\{\frac{\mathrm{1}}{\mathrm{4}},\frac{\tau \sqrt[]{\epsilon }}{\beta }\mathrm{l}\mathrm{n}N\right\}=\frac{\mathrm{1}}{\mathrm{4}}.$ It follows from (33) that