Postprocessing technique of the discontinuous Galerkin method for solving delay differential equations

Abstract

We introduce an innovative postprocessing technique aimed at refining the accuracy of the discontinuous Galerkin method for solving linear delay differential equations (DDEs) with vanishing delays. The fundamental idea behind this postprocessing technique is to enhance the discontinuous Galerkin solution of degree k by incorporating a generalized Jacobi polynomial of degree $k+1$. We demonstrate that this postprocessing step enhances convergence by one order under the $L^\infty$-norm. Moreover, we apply this technique to both nonlinear DDEs with vanishing delays and linear DDEs with non-vanishing delays. We further validated the theoretical results through a series of numerical examples.

Some proofs

1.1 Proof of Lemma 2.1

Proof

Since the $L^\infty$-error estimate (2.4) has been established in [16, Corollary 3.4], our task is now reduced to verifying the $L^2$- and $H^1$-error estimates.

Step 1. We will show the $L^2$-error estimate in (2.3). Utilizing (1.1) and (2.2), it follows that

$$\begin{aligned} \int _{I_{n}}e'\varphi dt+[e]_{n-1}\varphi ^+_{n-1}=\int _{I_{n}}(ae+be(\theta ))\varphi dt,\quad \forall \varphi \in P_{k}(I_{n}). \end{aligned}$$

(A.1)

Since $e = \xi + \eta$, using (A.1), (2.12), integration by parts, and the fact $\xi ^-_0=0$, we derive

$$\begin{aligned} \int _{I_n}\xi '\varphi dt+[\xi ]_{n-1}\varphi _{n-1}^+= \int _{I_n}(ae+be(\theta ))\varphi dt, \quad \forall \displaystyle \varphi \in P_k(I_n). \end{aligned}$$

(A.2)

Further application of integration by parts yields

$$\begin{aligned} \xi _n^-\varphi _n^--\int _{I_n}\xi \varphi ^{\prime }dt=\xi _{n-1}^-\varphi _{n-1}^++\int _{I_n}(ae+be(\theta ))\varphi dt, \quad \forall \displaystyle \varphi \in P_k(I_n). \end{aligned}$$

(A.3)

By selecting $\varphi = \xi$ in (A.2), we obtain

$$\begin{aligned} \begin{aligned} \frac{1}{2}\left( |\xi _n^-|^2 +|\xi _{n-1}^+|^2\right)&=\xi _{n-1}^-\xi _{n-1}^++\int _{I_n}(ae+be(\theta ))\xi dt\\&\le \frac{1}{2}|\xi _{n-1}^-|^2+\frac{1}{2}|\xi _{n-1}^+|^2+C\left( \Vert e\Vert _{L^2(I_n)}^2+\Vert \xi \Vert _{L^2(I_n)}^2\right) \\&\quad +C\left( \Vert e(\theta )\Vert _{L^2(I_n)}^2+\Vert \xi \Vert _{L^2(I_n)}^2\right) , \end{aligned} \end{aligned}$$

which implies

$$\begin{aligned} \begin{aligned} |\xi _n^-|^2&\le |\xi _{n-1}^-|^2+C\left( \Vert e\Vert _{L^2(I_n)}^2+\Vert \xi \Vert _{L^2(I_n)}^2\right) +C\left( \Vert e(\theta )\Vert _{L^2(I_n)}^2+\Vert \xi \Vert _{L^2(I_n)}^2\right) \\&\le |\xi _{n-1}^-|^2+C\Vert \eta \Vert _{L^2(I_n)}^2+C\Vert \xi \Vert _{L^2(I_n)}^2+C\Vert e(\theta )\Vert _{L^2(I_n)}^2. \end{aligned} \end{aligned}$$

(A.4)

Summing (A.4) over $I_i$ for $1\le i \le n$, then utilizing

$$\begin{aligned} \Vert e(\theta )\Vert _{L^2(0,t_n)}^2=\int _0^{t_n}|e(\theta )|^2dt\le \frac{1}{q_0}\int _0^{\theta (t_n)}|e(s)|^2ds\le \frac{1}{q_0}\Vert e\Vert _{L^2(0,t_n)}^2,\quad 1\le n \le N, \end{aligned}$$

(A.5)

we obtain

$$\begin{aligned} \begin{aligned} |\xi _n^-|^2\le C\big (\Vert \eta \Vert _{L^2(0,t_n)}^2+\Vert \xi \Vert _{L^2(0,t_n)}^2\big ). \end{aligned} \end{aligned}$$

(A.6)

For $k=0$, we have $\xi '=0$. Utilizing (A.6), we obtain

$$\begin{aligned} \Big (\int _{I_n}\xi dt\Big )^2=\Big (\int _{I_n}\xi _n^-dt\Big )^2=h_n^2|\xi _n^-|^2\le Ch_n^2\big (\Vert \xi \Vert _{L^2(0,t_n)}^2+\Vert \eta \Vert _{L^2(0,t_n)}^2\big ). \end{aligned}$$

(A.7)

For $k>0$, selecting $\varphi = t_{n-1}-t$ in (A.3), we obtain

$$\begin{aligned} \begin{aligned} \int _{I_n}\xi dt=h_n\xi _n^-+\int _{I_n}(ae+be(\theta )(t_{n-1}-t)dt. \end{aligned} \end{aligned}$$

(A.8)

Then, using (A.6) gives

$$\begin{aligned} \begin{aligned} \left( \int _{I_n}\xi dt\right) ^2&\le Ch_n^2|\xi _n^-|^2+C\Big (\int _{I_n}|e|\cdot |t_{n-1}-t|dt\Big )^2+C\Big (\int _{I_n}|e(\theta )|\cdot |t_{n-1}-t|dt\Big )^2\\&\le Ch_n^2\big (\Vert \xi \Vert _{L^2(0,t_n)}^2+\Vert \eta \Vert _{L^2(0,t_n)}^2\big )+Ch_n^3\Vert e\Vert _{L^2(I_n)}^2+Ch_n^3\Vert e\Vert _{L^2(0,t_n)}^2\\&\le Ch_n^2\big (\Vert \xi \Vert _{L^2(0,t_n)}^2+\Vert \eta \Vert _{L^2(0,t_n)}^2\big ). \end{aligned} \end{aligned}$$

(A.9)

Here, we have utilized the fact

$$\begin{aligned} \left\| e(\theta )\right\| _{L^2(I_n)}^2\le \int _0^{t_n}|e(\theta )|^2dt\le \frac{1}{q_0}\int _0^{\theta (t_n)}|e|^2ds\le \frac{1}{q_0}\Vert e\Vert _{L^2(0,t_n)}^2. \end{aligned}$$

(A.10)

Combining (A.7) and (A.9), we conclude

$$\begin{aligned} \Big (\int _{I_n}\xi (t)dt\Big )^2\le Ch_n^2\big (\Vert \xi \Vert _{L^2(0,t_n)}^2+\Vert \eta \Vert _{L^2(0,t_n)}^2\big ),\quad k\ge 0. \end{aligned}$$

(A.11)

For $k>0$, selecting $\varphi =\xi ^{\prime }(t)(t-t_{n-1})$ in (A.2) and then utilizing (A.10) and the Cauchy-Schwarz inequality, we have

$$\begin{aligned} \begin{aligned}&\int _{I_n}(t-t_{n-1})|\xi '|^2dt \le Ch_n^{\frac{1}{2}}\Vert e\Vert _{L^2(I_n)}\Big \{\int _{I_n}(t-t_{n-1})|\xi ^{\prime }|^2dt\Big \}^{\frac{1}{2}}\\&\quad +Ch_n^{\frac{1}{2}}\Vert e\Vert _{L^2(0,t_n)}\Big \{\int _{I_n}(t-t_{n-1})|\xi ^{\prime }|^2dt\Big \}^{\frac{1}{2}}, \end{aligned} \end{aligned}$$

and thus,

$$\begin{aligned} \begin{aligned} \int _{I_n}(t-t_{n-1})|\xi '|^2d\le Ch_n(\Vert \xi \Vert ^2_{L^2(0,t_n)}+\Vert \eta \Vert ^2_{L^2(0,t_n)}). \end{aligned} \end{aligned}$$

(A.12)

For $k=0$, where $\xi '=0$, (A.12) still holds.

Recalling the inequality (see [26, Lemma 2.4])

$$\begin{aligned} \int _{I_n}|\varphi |^2dt\le \frac{1}{h_n}\Big (\int _{I_n}\varphi (t)dt\Big )^2+\frac{1}{2}\int _{I_n}(t_n-t)(t-t_{n-1})|\varphi '|^2dt, \quad \forall \varphi \in P_k(I_n), \ k \ge 0. \end{aligned}$$

(A.13)

Then, using (A.11) and (A.12), we obtain

$$\begin{aligned} \begin{aligned} \Vert \xi \Vert _{L^{2}(I_{n})}^{2}&=\int _{I_{n}}|\xi |^{2}dt\le \frac{1}{h_{n}}\Big (\int _{I_{n}}\xi dt\Big )^{2}+\frac{1}{2}\int _{I_{n}}(t_n-t)(t-t_{n-1})|\xi '|^{2}dt\\&\le Ch_n\Vert \xi \Vert _{L^2(0,t_n)}^2+Ch_n\Vert \eta \Vert _{L^2(0,t_n)}^2\\&\le Ch_{n}\Vert \xi \Vert _{L^{2}(0,t_{n-1})}^{2}+Ch_{n}\Vert \xi \Vert _{L^{2}(I_{n})}^{2}+Ch_{n}\Vert \eta \Vert _{L^{2}(0,t_{n})}^{2}, \end{aligned} \end{aligned}$$

which implies that for $h_n$ sufficiently small

$$\begin{aligned} \frac{\Vert \xi \Vert _{L^2(I_n)}^2}{h_n}\le C\Vert \eta \Vert _{L^2(0,t_n)}^2+C\sum _{i=1}^{n-1}h_i\frac{\Vert \xi \Vert _{L^2(I_i)}^2}{h_i}. \end{aligned}$$

Therefore, applying the discrete Gronwall inequality yields

$$\begin{aligned} \Vert \xi \Vert _{L^2(I_n)}^2\le Ch_n\Vert \eta \Vert _{L^2(0,t_n)}^2. \end{aligned}$$

(A.14)

Combining (2.18) and (A.14), we obtain

$$\begin{aligned} \begin{aligned} \Vert e\Vert _{L^{2}(I)}^{2}&\le 2\Vert \eta \Vert _{L^2(I)}^2+2\sum _{n=1}^N\Vert \xi \Vert _{L^2(I_n)}^2 \le C\Vert \eta \Vert _{L^2(I)}^2 \le Ch^{2k+2}\Vert u\Vert _{H^{k+1}(I)}^2. \end{aligned} \end{aligned}$$

(A.15)

This concludes the proof of the $L^2$-estimate in (2.3).

Step 2. In this step, we establish the $H^1$-error estimate in (2.3). Given that $\xi |_{I_n} \in P_{k}(I_{n})$, we utilize the inverse inequality (see, e.g., [11])

$$\begin{aligned} \Vert \xi ^{\prime }\Vert _{L^{\infty }(I_{n})}\le Ch_{n}^{-\frac{1}{2}}\Vert \xi ^{\prime }\Vert _{L^{2}(I_{n})}, \end{aligned}$$

(A.16)

and the fact $\Vert L_{n,k}\Vert _{L^{2}(I_{n})}=\left( \displaystyle \frac{h_n}{2k+1}\right) ^\frac{1}{2} \le C h_n^{\frac{1}{2}}$ to obtain

$$\begin{aligned} \left\| \xi '-(-1)^{k}\xi '(t_{n-1}^{+})L_{n,k}\right\| _{L^{2}(I_{n})}\le \Vert \xi '\Vert _{L^{2}(I_{n})}+ Ch_{n}^{-\frac{1}{2}}\Vert \xi ^{\prime }\Vert _{L^{2}(I_{n})}h_n^{\frac{1}{2}} \le C \Vert \xi '\Vert _{L^{2}(I_{n})}. \end{aligned}$$

(A.17)

Choosing $\varphi =\xi ^{\prime }-(-1)^{k}\xi ^{\prime }(t_{n-1}^{+})L_{n,k}$ in (A.2) and utilizing (A.17) as well as the properties $\varphi _{n-1}^+=0$, we arrive at

$$\begin{aligned} \Vert \xi ^{\prime }\Vert _{L^{2}(I_{n})}^{2}\le & {} C\left( \Vert e\Vert _{L^{2}(I_{n})}+\Vert e(\theta )\Vert _{L^{2}(I_{n})}\right) \left\| \xi '-(-1)^{k}\xi '(t_{n-1}^{+})L_{n,k}\right\| _{L^{2}(I_{n})}\\\le & {} C\left( \Vert e\Vert _{L^2(I_n)} + \Vert e(\theta )\Vert _{L^2(I_n)}\right) \Vert \xi '\Vert _{L^{2}(I_{n})}, \end{aligned}$$

which leads to

$$\begin{aligned} \Vert \xi '\Vert _{L^2(I_n)}\le C\Vert e\Vert _{L^2(I_n)}+C\Vert e(\theta )\Vert _{L^2(I_n)}. \end{aligned}$$

(A.18)

Squaring and summing (A.18) over $I_n$ for $1\le n \le N$, then utilizing (A.5), we have

$$\begin{aligned} \Vert \xi '\Vert _{L^2(I)}^2\le C\Vert e\Vert _{L^2(I)}^2+C\Vert e(\theta )\Vert _{L^2(I)}^2 \le C\Vert e\Vert _{L^2(I)}^2. \end{aligned}$$

(A.19)

Combining (A.19), (2.18), and (2.3), we obtain

$$\begin{aligned} \Vert e'\Vert ^2_{L^2(I)} \le C\left( \Vert \eta '\Vert ^2_{L^2(I)}+\Vert \xi '\Vert ^2_{L^2(I)}\right) \le C \left( \Vert \eta '\Vert ^2_{L^2(I)}+\Vert e\Vert ^2_{L^2(I)}\right) \le Ch^{2k}\Vert u\Vert ^2_{H^{k+1}(I)}, \end{aligned}$$

which along with (A.15) implies the desired $H^1$-estimate. $\square$

1.2 Proof of Lemma 2.2

Proof

We begin by constructing the following problem: seek v so that

$$\begin{aligned} \left\{ \begin{array}{ll} v'+av+\widetilde{b}v(\theta ^{-1}(t))=0,\quad t\in [t_0,t_n),\\ v(t_{n})=e(t_n^-) \end{array}\right. \end{aligned}$$

(A.20)

for $1\le n\le N$, and $\widetilde{b}$ is given as follows:

$$\begin{aligned} \widetilde{b}:=\left\{ \begin{aligned}&b(\theta ^{-1}(t))\left( \theta ^{-1}(t)\right) ',\quad t_0 \le t \le \theta (t_n),\\&\ 0,\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \theta (t_n)<t\le t_n. \end{aligned} \right. \end{aligned}$$

Here, the functions a, b, and $\theta$ are given by (1.1).

Because of the potential discontinuity of $\widetilde{b}$ at the points $\theta (t_n)$, the derivative $v'$ may also exhibit discontinuity at these points. We make the assumption that the functions a and b belong to $C^k(I)$ such that v satisfies the inequality

$$\begin{aligned} |v(\theta (t_n))|+|v'(\theta (t_n))|+|v^{(i)}(t)|\le C|e_n^-| \end{aligned}$$

(A.21)

for $i=0,1,...,k+1$, where $t \in [0,\theta (t_n))\cup (\theta (t_n),t_n]$.

For convenience, let’s define

$$\begin{aligned} \rho _{n}:=\int _{I_{n}}\left( e'-ae-be(\theta )\right) vdt+[e]_{n-1}v_{n-1}^{+}. \end{aligned}$$

(A.22)

Utilizing the technique of integration by parts, yields

$$\begin{aligned} \begin{aligned} \rho _{n}&=\int _{I_{n}}e'vdt-\int _{I_{n}}aevdt-\int _{I_{n}}be(\theta )vdt+[e]_{n-1}v_{n-1}^{+} \\&=\left( e_{n}^{-} v_{n}^{-}-e_{n-1}^{+}v_{n-1}^{+}-\int _{I_{n}}v'edt\right) -\int _{I_{n}}aevdt -\int _{I_{n}}be(\theta )vdt+ [e]_{n-1}v_{n-1}^+ \\&=-\int _{I_{n}}\left( v'+av\right) edt-\int _{I_{n}}be(\theta )vdt +e_{n}^{-} v_{n}^{-}-e_{n-1}^{-}v_{n-1}^{+}. \end{aligned}\end{aligned}$$

(A.23)

By summing up (A.23) over $I_i$, $1 \le i \le n$, and utilizing (A.20), the initial condition $e_0^-=u_0-U_0^-=0$, and the fact $v\in H^1(0, t_n)$, we get

$$\begin{aligned} \sum _{i=1}^n\rho _i= & {} \displaystyle -\int _{0}^{t_{n}}\left( v'+av\right) edt-\int _{\theta (0)}^{\theta (t_n)}b(\theta ^{-1})ev(\theta ^{-1}) \left( \theta ^{-1}\right) 'dt +\displaystyle \sum _{i=1}^{n}(e_{i}^{-}v_{i}^{-}-e_{i-1}^{-}v_{i-1}^{+}) \nonumber \\= & {} \displaystyle -\int _{0}^{t_{n}}\left( v'+av\right) edt-\int _{0}^{\theta (t_n)}\widetilde{b}ev(\theta ^{-1})dt-\displaystyle \int _{\theta (t_n)}^{t_n}\widetilde{b}ev(\theta ^{-1})dt \nonumber \\{} & {} +\sum _{i=1}^{n}\left( e_{i}^{-}v_{i}^{-}-e_{i-1}^{-}v_{i-1}^{+})\right) \nonumber \\= & {} \displaystyle -\int _{0}^{t_n}(v'+av+\widetilde{b}v(\theta ^{-1}))edt+\sum _{i=1}^{n}\left( e_{i}^{-}v_{i}^{-}-e_{i-1}^{-}v_{i-1}^{+})\right) \nonumber \\= & {} \displaystyle \sum _{i=1}^{n}\left( e_{i}^{-}v(t_i)-e_{i-1}^{-}v(t_{i-1})\right) =e_n^-v(t_n) \nonumber \\= & {} \displaystyle |e_n^-|^2. \end{aligned}$$

(A.24)

Since $e = \xi + \eta$ and using (2.18), (A.18), (A.10), and (2.3), we have

$$\begin{aligned} \begin{aligned} \Vert e'\Vert ^2_{L^2(I_{n})}&\le C\left( \Vert \eta '\Vert ^2_{L^2(I_{n})}+\Vert \xi '\Vert ^2_{L^2(I_{n})}\right) \le Ch_n^{2k}\Vert u\Vert ^2_{H^{k+1}(I_n)}\\&\quad +C\Vert e\Vert ^2_{L^2(I_n)}+C\Vert e(\theta (t))\Vert ^2_{L^2(I_n)}\\&\le Ch_n^{2k+1}\Vert u\Vert ^2_{W^{k+1,\infty }(I_n)}+C\Vert e\Vert ^2_{L^2(I)}\\&\le Ch^{2k+1}\Vert u\Vert ^2_{W^{k+1,\infty }(I)}. \end{aligned} \end{aligned}$$

(A.25)

Now, considering (A.22), (A.1), and employing (2.13), we find that

$$\begin{aligned} \begin{aligned} \sum _{i=1}^{n}\rho _{i}&=\sum _{i=1}^{n}\int _{I_{i}}(e'-ae-be(\theta ))(v-\hat{\pi }{^{k}_{I_{i}}}v)dt+\sum _{i=1}^{n}\int _{I_{i}}(e'-ae-be(\theta ))\hat{\pi }{^{k}_{I_{i}}}vdt\\&\quad +\sum _{i=1}^{n}[e]_{i-1}v_{i-1}^{+} \\&= \sum _{i=1}^{n}\int _{I_{i}}(e'-ae-be(\theta ))(v-\hat{\pi }{^{k}_{I_{i}}}v)dt+\sum _{i=1}^{n}[e]_{i-1}\left( v_{i-1}^{+}-(\hat{\pi }_{I_i}^{k}v)_{i-1}^{+}\right) \\&= \sum _{i=1}^{n}\int _{I_{i}}(e'-ae-be(\theta ))(v-\hat{\pi }{^{k}_{I_{i}}}v)dt, \end{aligned} \end{aligned}$$

(A.26)

where $\hat{\pi }_{I_n}^kv \in P_k(I_n)$ denote the projection of v as defined by (2.13).

Let $n^*$ be an integer such that $\theta (t_n)$ falls within $[t_{n^*-1},t_{n^*+1}]$, where $1\le n^*\le n-1$. By (A.26) and(A.24), we have

$$\begin{aligned} |e^-_n|^2:=A_1+A_2+A_3, \end{aligned}$$

(A.27)

with

$$\begin{aligned} A_1= \sum _{i=1}^{n^*-1}\int _{I_i}(e'-ae-be(\theta ))(v-\hat{\pi }{^{k}_{I_{i}}}v)dt, \\ A_2= \sum _{i=n^*}^{n^*+1}\int _{I_i}(e'-ae-be(\theta ))(v-\hat{\pi }{^{k}_{I_{i}}}v)dt, \\ A_3= \sum _{i=n^*+2}^{n}\int _{I_i}(e'-ae-be(\theta ))(v-\hat{\pi }{^{k}_{I_{i}}}v)dt. \end{aligned}$$

Applying the Cauchy-Schwarz inequality, (A.25), (2.3), and (2.20), we get

$$\begin{aligned} A_1\le & {} \sum _{i=1}^{n^*-1}\left\| e'-ae-be(\theta ))\right\| _{L^2(I_i)}\left\| v-\hat{\pi }_{I_i}^{k}v\right\| _{L^2(I_i)} \nonumber \\\le & {} C\sum _{i=1}^{n^*-1}\left( \Vert e'\Vert _{L^2(I_i)}+\Vert e\Vert _{L^2(I)}\right) h_i^{k+1}\Vert v\Vert _{H^{k+1}(I_i)} \nonumber \\\le & {} C\sum _{i=1}^{n^*-1}\left( h^{k+\frac{1}{2}}\Vert u\Vert _{W^{k+1,\infty }(I)}+h^{k+1}\Vert u\Vert _{W^{k+1,\infty }(I)}\right) h_i^{k+\frac{3}{2}}\Vert v\Vert _{W^{k+1,\infty }(I_i)}\nonumber \\\le & {} C\sum _{i=1}^{n^*-1}\left( h^{k+\frac{1}{2}}h_i^{k+\frac{3}{2}}+h^{k+1}h_{i}^{k+\frac{3}{2}}\right) \Vert u\Vert _{W^{k+1,\infty }(I)}\Vert v\Vert _{W^{k+1,\infty }(I_{i})} \nonumber \\\le & {} Ch^{2k+1}\Vert u\Vert _{W^{k+1,\infty }(I)}|e_n^-|. \end{aligned}$$

(A.28)

It’s worth noting that v may possess lower regularity on the intervals $I_{j}$, $j=n^{*},n^{*}+1$. Similar to the derivation of $A_1$, we obtain

$$\begin{aligned} \begin{aligned} A_2&\le \sum _{i=n^*}^{n^*+1}\left\| e'-ae-be(\theta ))\right\| _{L^2(I_i)}\left\| v-\hat{\pi }_{I_i}^{k}v\right\| _{L^2(I_i)} \\&\le C\sum _{i=n^*}^{n^*+1}\left( \Vert e\Vert _{H^1(I_i)}+\Vert e\Vert _{L^2(I)}\right) h_i\Vert v\Vert _{H^{1}(I_i)} \\&\le C\sum _{i=n^*}^{n^*+1}\left( h^{k+\frac{1}{2}}\Vert u\Vert _{W^{k+1,\infty }(I)}+h^{k+1}\Vert u\Vert _{W^{k+1,\infty }(I)}\right) h_i^{\frac{3}{2}}\Vert v\Vert _{W^{1,\infty }(I_i)}\\&\le C\sum _{i=n^*}^{n^*+1}\left( h^{k+\frac{1}{2}}h_i^{\frac{3}{2}}+h^{k+1}h_{i}^{\frac{3}{2}}\right) \Vert u\Vert _{W^{k+1,\infty }(I)}\Vert v\Vert _{W^{1,\infty }(I_{i})}\\&\le Ch^{k+2}\Vert u\Vert _{W^{k+1,\infty }(I)}|e_n^-|. \end{aligned} \end{aligned}$$

(A.29)

Moreover, we can estimate $A_3$ in a similar manner to $A_1$, obtaining

$$\begin{aligned} A_3\le Ch^{2k+1}\Vert u\Vert _{W^{k+1,\infty }(I)}|e_n^-|. \end{aligned}$$

(A.30)

Finally, by combining (A.27)-(A.30), we derive the desired estimate

$$\begin{aligned} |e_n^-|\le Ch^{\min \{2k+1, k+2\}}\Vert u\Vert _{W^{k+1,\infty }(I)}, \end{aligned}$$

which implies (2.5). $\square$