Superconvergence of UWLDG Method for One-Dimensional Linear Sixth-Order Equations

Abstract

This paper concerns the superconvergence property of the ultraweak-local discontinuous Galerkin (UWLDG) method for one-dimensional linear sixth-order equations. The crucial technique is the construction of a special projection. We will discuss in three different situations according to the remainder of k, the highest degree of polynomials in the function space, divided by 3. We can prove the $(2k-1)$ th-order superconvergence for the cell averages when $k\equiv 0$ or 2 (mod 3). But if $k\equiv$1 (mod 3), we can only prove a $(2k-2)$ th-order superconvergence. The same superconvergence orders can also be gained for the errors of numerical fluxes. We will also prove the superconvergence of order $k+2$ at some special quadrature points. Some numerical examples are given at the end of this paper.

Appendices

Proof of Lemma 1

We will prove this lemma by induction. Since $\omega _u^{(0)},\,\omega _v^{(0)}\perp \mathcal {P}^{k-3}(I_j)$, we can easily deduce from the definitions of $\omega _u^{(i)}$ and $\omega _v^{(i)}$ that

$$\begin{aligned} \omega _u^{(1)},\,\omega _v^{(1)}\perp \mathcal {P}^{k-6}(I_j). \end{aligned}$$

Thus, we can write

$$\begin{aligned} {\omega _u^{(1)}\mid _{I_j}=\sum _{m=k-5}^{k} c_{j,m}^1 L_{j,m},\ \omega _v^{(1)}\,\Bigg |\,_{I_j}=\sum _{m=k-5}^{k} d_{j,m}^1 L_{j,m}.} \end{aligned}$$

Then, we are going to estimate $c^1_{j,m}$ and $d^1_{j,m}$. If $f\perp \mathcal {P}^0(I_j)$, we can define $D^{-1}f(x)=\overline{h_j}\int _{x_{j-\frac{1}{2}}}^x f(t)\ \textrm{d}t,\,x\in I_j$, where $\overline{h_j}=\frac{h_j}{2}$. And $D^{-n}=(D^{-1})^n$ for any $n\in \mathbb {N}$. Taking $\varphi ={D^{-3}L_{j,m}},\ m=k-5,\,k-4,\,k-3$, respectively, in (11), we have

$$\begin{aligned} \begin{aligned} \frac{4}{h_j^2}\frac{2}{2m+1}c^1_{j,m}&=-\int _{I_j}\omega _v^{(0)}D^{-3}L_{j,m}\ \textrm{d}x \\&\leqslant \Vert \omega _v^{(0)}\Vert _{I_j}\Vert D^{-3}L_{j,m}\Vert _{I_j} \\&\lesssim h^{k+1+\frac{1}{2}}\Vert u\Vert _{k+4,I_j}. \end{aligned} \end{aligned}$$

Thus,

$$\begin{aligned} \vert c^1_{j,m}\vert \lesssim h^{k+\frac{7}{2}}\Vert u\Vert _{k+4,I_j},\quad m=k-5,\,k-4,\,k-3. \end{aligned}$$

Similarly, we have

$$\begin{aligned} \vert d^1_{j,m}\vert \lesssim h^{k+\frac{7}{2}}\Vert u\Vert _{k+4,I_j},\quad m=k-5,\,k-4,\,k-3. \end{aligned}$$

From (11), we can also obtain

$$\begin{aligned} \sum _{m=k-2}^{k}L_m(-1)c_{j,m}^1=\sum _{m=k-5}^{k-3}L_m(-1)c_{j,m}^1, \\ \sum _{m=k-2}^{k}L_m^\prime (-1)c_{j,m}^1=\sum _{m=k-5}^{k-3}L_m^\prime (-1)c_{j,m}^1, \\ \sum _{m=k-2}^{k}L_m^{\prime \prime }(-1)c_{j,m}^1=\sum _{m=k-5}^{k-3}L_m^{\prime \prime }(-1)c_{j,m}^1. \end{aligned}$$

Thus,

$$\begin{aligned} \sum _{j=1}^N\sum _{m=k-2}^k \left( c_{j,m}^1\right) ^2\lesssim \sum _{j=1}^N\sum _{m=k-5}^{k-3} \left( c_{j,m}^1\right) ^2\lesssim h^{2k+7}\Vert u\Vert _{k+4}^2. \end{aligned}$$

Thus,

$$\begin{aligned} \begin{aligned} \Vert \omega _u^{(1)}\Vert&=\left( \sum _{j=1}^N \left\| \sum _{m=k-5}^{k} c_{j,m}^1 L_{j,m}\right\| ^2_{I_j} \right) ^{\frac{1}{2}} \\&\lesssim \left( \sum _{j=1}^N\sum _{m=k-5}^k h\left( c_{j,m}^1\right) ^2\right) ^{\frac{1}{2}} \\&\lesssim h^{k+4}. \end{aligned} \end{aligned}$$

Similarly, we have the estimate for $\omega _v^{(1)}$. And we take the time derivate on both sides of (11) and (12), we can obtain the estimates for $\partial _t^n\omega _u^{(1)}$ and $\partial _t^n\omega _v^{(1)}$. We can prove (14) for all $1\leqslant i\leqslant l$ by the recursion formula.

Proof of Lemma 3

Integrating (20) by part, we have

$$\begin{aligned} (f,\varphi )_j-\left( \xi _{xxx},\varphi \right) _j-\llbracket \xi _{xx}\rrbracket \varphi ^+\vert _{j-\frac{1}{2}}+\llbracket \xi _{x}\rrbracket \varphi _x^+\vert _{j-\frac{1}{2}}-\llbracket \xi \rrbracket \varphi _{xx}^+\vert _{j-\frac{1}{2}}=0. \end{aligned}$$

(B1)

First, we take

$$\begin{aligned} \varphi (x)\vert _{I_j}=\xi _{xxx}+aL_k(\theta )+bL_{k-1}(\theta )+cL_{k-2}(\theta ), \end{aligned}$$

where $L_k$ is the standard Legendre polynomial with degree k, $\theta =\frac{2}{h_j}\left( x-x_{j}\right)$ with $x_j=\frac{x_{j-\frac{1}{2}}+x_{j+\frac{1}{2}}}{2}$, and $a,\,b,$ and c are the solutions of the equations:

$$\begin{aligned} \left\{ \begin{aligned}&aL_k(-1)+bL_{k-1}(-1)+cL_{k-2}(-1)+\xi _{xxx}\left( x_{j-\frac{1}{2}}^+\right) =0, \\&aL^\prime _k(-1)+bL^\prime _{k-1}(-1)+cL^\prime _{k-2}(-1)+\frac{h_j}{2}\xi _{xxxx}\left( x_{j-\frac{1}{2}}^+\right) =0, \\&aL^{\prime \prime }_k(-1)+bL^{\prime \prime }_{k-1}(-1)+cL^{\prime \prime }_{k-2}(-1)+\frac{h_j^2}{4}\xi _{xxx}\left( x_{j-\frac{1}{2}}^+\right) =0. \end{aligned} \right. \end{aligned}$$

(B2)

That is,

$$\begin{aligned} \left\{ \begin{aligned} a&=\frac{(-1)^k(k-2)(k-1)^2}{4(2k-1)}\xi _{xxx}\left( x_{j-\frac{1}{2}}^+\right) +\frac{(-1)^k(k-2)}{2k-1}\frac{h_j}{2}\xi _{xxxx}\left( x_{j-\frac{1}{2}}^+\right) \\&\qquad -\frac{2(-1)^k}{k(k-2)}\frac{h_j^2}{4}\xi _{xxxxx}\left( x_{j-\frac{1}{2}}^+\right) , \\ b&=\frac{(-1)^k(k-2)(k+1)}{4}\xi _{xxx}\left( x_{j-\frac{1}{2}}^+\right) +(-1)^k\frac{h_j}{2}\xi _{xxxx}\left( x_{j-\frac{1}{2}}^+\right) \\&\qquad -\frac{2(-1)^k}{k(k-1)}\frac{h_j^2}{4}\xi _{xxxxx}\left( x_{j-\frac{1}{2}}^+\right) , \\ c&=\frac{(-1)^k k^2(k+1)}{4(2k-1)}\xi _{xxx}\left( x_{j-\frac{1}{2}}^+\right) +\frac{(-1)^k(k+1)}{2k-1}\frac{h_j}{2}\xi _{xxxx}\left( x_{j-\frac{1}{2}}^+\right) \\&\qquad -\frac{2(-1)^k}{(2k-1)(k-1)}\frac{h_j^2}{4}\xi _{xxxxx}\left( x_{j-\frac{1}{2}}^+\right) . \end{aligned} \right. \end{aligned}$$

By the inverse inequality, we have

$$\begin{aligned} \vert a\vert ,\ \vert b\vert ,\ \vert c\vert \lesssim h^{-\frac{1}{2}}\Vert \xi _{xxx}\Vert _{I_j}. \end{aligned}$$

Clearly, there holds $\varphi \left( x_{j-\frac{1}{2}}^+\right) =0$, $\varphi _x\left( x_{j-\frac{1}{2}}^+\right) =0$, and $\varphi _{xx}\left( x_{j-\frac{1}{2}}^+\right) =0$. So, by (B1) we have

$$\begin{aligned} \begin{aligned} \Vert \xi _{xxx}\Vert _{I_j}^2&=(f,\varphi )_j \\&\leqslant \Vert f\Vert _{I_j}\left( \Vert \xi _{xxx}\Vert _{I_j}+\vert a\vert \Vert L_k(\theta )\Vert _{ I_j}+\vert b\vert \Vert L_{k-1}(\theta )\Vert _{I_j}+\vert c\vert \Vert L_{k-2}(\theta )\Vert _{I_j}\right) \\&\leqslant C\Vert f\Vert _{I_j}\Vert \xi _{xxx}\Vert _{I_j}, \end{aligned} \end{aligned}$$

thus

$$\begin{aligned} \Vert \xi _{xxx}\Vert _{I_j}\lesssim \Vert f\Vert _{I_j}. \end{aligned}$$

Next, we take $\varphi =1$ in (B1) to obtain

$$\begin{aligned} (f,1)_j-(\xi _{xxx},1)_j-\llbracket \xi _{xx}\rrbracket \vert _{j-\frac{1}{2}}=0, \end{aligned}$$

thus

$$\begin{aligned} \begin{aligned} \vert \llbracket \xi _{xx}\rrbracket _{j-\frac{1}{2}}\vert&\leqslant h^{\frac{1}{2}}(\Vert f\Vert _{I_j}+\Vert \xi _{xxx}\Vert _{I_j}) \\&\lesssim h^{\frac{1}{2}}\Vert f\Vert _{I_j}. \end{aligned} \end{aligned}$$

Next, we take $\varphi =\frac{2(x-x_j)}{h_j}$ in (B1) to obtain

$$\begin{aligned} (f,\varphi )_j-(\xi _{xxx},\varphi )_j-\llbracket \xi _{xx}\rrbracket \vert _{j-\frac{1}{2}}+\frac{2}{h_j}\llbracket \xi _{x}\rrbracket \vert _{j-\frac{1}{2}}=0, \end{aligned}$$

thus

$$\begin{aligned} \begin{aligned} \vert \llbracket \xi _{x}\rrbracket _{j-\frac{1}{2}}\vert&\leqslant \frac{h_j}{2}\Vert \varphi \Vert _{I_j}(\Vert f\Vert _{ I_j}+\Vert \xi _{xxx}\Vert _{I_j})+\frac{h_j}{2}\vert \llbracket \xi _{xx}\rrbracket _{j-\frac{1}{2}}\vert \\&\lesssim h^{\frac{3}{2}}\Vert f\Vert . \end{aligned} \end{aligned}$$

Finally, we take $\varphi =\left( \frac{2(x-x_j)}{h_j}\right) ^2$ in (B1) to obtain

$$\begin{aligned} (f,\varphi )_j-(\xi _{xxx},\varphi )_j-\llbracket \xi _{xx}\rrbracket \vert _{j-\frac{1}{2}}+\frac{4}{h_j}\llbracket \xi _{x}\rrbracket \vert _{j-\frac{1}{2}}-\frac{8}{h^2_j}\llbracket \xi \rrbracket \vert _{j-\frac{1}{2}}=0, \end{aligned}$$

thus

$$\begin{aligned} \begin{aligned} \vert \llbracket \xi \rrbracket _{j-\frac{1}{2}}\vert&\leqslant \frac{h^2_j}{8}\Vert \varphi \Vert _{I_j}(\Vert f\Vert _{I_j}+\Vert \xi _{xxx}\Vert _{I_j})+\frac{h^2_j}{8}\vert \llbracket \xi _{xx}\rrbracket _{j-\frac{1}{2}}\vert +\frac{h_j}{2}\vert \llbracket \xi _{x}\rrbracket _{j-\frac{1}{2}}\vert \\&\lesssim h^{\frac{5}{2}}\Vert f\Vert . \end{aligned} \end{aligned}$$

This finishes our proof.

Proof of Lemma 4

Since $f_1\perp \mathcal {P}^0$, so we can define $D^{-1}f_1(x)=\overline{h_j}\int _{x_{j-\frac{1}{2}}}^x f_1(t)\ \textrm{d}t,\,x\in I_j$, where $\overline{h_j}=\frac{h_j}{2}$. It’s easy to check that $D^{-1}f_1\left( x_{j\pm \frac{1}{2}}^{\mp }\right) =0$ and $\Vert D^{-1} f_1\Vert _{I_j}\lesssim \Vert f_1\Vert _{I_j}$. Thus, we have $(f_1,\varphi )_j=-(\overline{h_j}D^{-1}f_1,\varphi _x)_j$.

Integrating the left-hand side of (22) by parts twice, we have

$$\begin{aligned} A_j(\xi ,\varphi )=(\xi _{xx},\varphi _x)_j-\llbracket \xi \rrbracket \varphi _{xx}^+\vert _{j-\frac{1}{2}}+\llbracket \xi _x\rrbracket \varphi _{x}^+\vert _{j-\frac{1}{2}}-\xi _{xx}^-\varphi ^-\vert _{j+\frac{1}{2}}+\xi _{xx}^-\varphi ^+\vert _{j-\frac{1}{2}}. \end{aligned}$$

Taking $\varphi =\xi _x$ and summing over j, we have

$$\begin{aligned}{} & {} \Vert \xi _{xx}\Vert ^2-\sum _{j=1}^N \llbracket \xi \rrbracket \xi _{xxx}^+\vert _{j-\frac{1}{2}}+\sum _{j=1}^N \llbracket \xi _x\rrbracket \xi _{xx}^+\vert _{j-\frac{1}{2}}+\sum _{j=1}^N \llbracket \xi _x\rrbracket \xi _{xx}^-\vert _{j-\frac{1}{2}}\\ ={} & {} \sum _{j=1}^N-(\overline{h_j}D^{-1}f_1,\xi _{xx})_j+(f_2,\xi _x)_j, \end{aligned}$$

thus

$$\begin{aligned} \Vert \xi _{xx}\Vert ^2{} & {} \leqslant h^{-\frac{1}{2}}\Vert \xi _{xxx}\Vert \left( \sum _{j=1}^N\llbracket \xi \rrbracket _{j+\frac{1}{2}}^2\right) ^{\frac{1}{2}}+h^{-\frac{1}{2}}\Vert \xi _{xx}\Vert \left( \sum _{j=1}^N\llbracket \xi _x\rrbracket _{j+\frac{1}{2}}^2\right) ^{\frac{1}{2}}\\{} & {} \quad +\,\,h\Vert f_1\Vert \Vert \xi _{xx}\Vert +\Vert f_2\Vert \Vert \xi _x\Vert . \end{aligned}$$

By Lemma 3, we have $\Vert \xi _{xxx}\Vert \lesssim \Vert f_1\Vert +\Vert f_2\Vert$, $h^{-\frac{1}{2}}\left( \sum _{j=1}^N\llbracket \xi \rrbracket _{j+\frac{1}{2}}^2\right) ^{\frac{1}{2}}\lesssim h^2(\Vert f_1\Vert +\Vert f_2\Vert )$, $h^{-\frac{1}{2}}\left( \sum _{j=1}^N\llbracket \xi _x\rrbracket _{j+\frac{1}{2}}^2\right) ^{\frac{1}{2}}\lesssim h(\Vert f_1\Vert +\Vert f_2\Vert )$, and by the discrete Poincaré inequalities, we also have

$$\begin{aligned} \Vert \xi _x\Vert \lesssim \Vert \xi _{xx}\Vert +h^{-\frac{1}{2}}\left( \sum _{j=1}^N\llbracket \xi _x\rrbracket _{j+\frac{1}{2}}^2\right) ^{\frac{1}{2}}+h^{-\frac{1}{2}}\left( \sum _{j=1}^N\llbracket \xi \rrbracket _{j+\frac{1}{2}}^2\right) ^{\frac{1}{2}}. \end{aligned}$$

(C1)

Thus,

$$\begin{aligned} \Vert \xi _{xx}\Vert ^2\lesssim h^2(\Vert f_1\Vert +\Vert f_2\Vert )+h(\Vert f_1\Vert +\Vert f_2\Vert )\Vert \xi _{xx}\Vert +\Vert f_2\Vert \Vert \xi _{xx}\Vert +h\Vert f_2\Vert (\Vert f_1\Vert +\Vert f_2\Vert ), \end{aligned}$$

thus

$$\begin{aligned} \Vert \xi _{xx}\Vert \lesssim h\Vert f_1\Vert +\Vert f_2\Vert . \end{aligned}$$

And, by (C1), we have

$$\begin{aligned} \Vert \xi _{x}\Vert \lesssim h\Vert f_1\Vert +\Vert f_2\Vert . \end{aligned}$$

Proof of Lemma 5

For $u\in L^2(I_j)$, we can have the Legendre expansion

$$\begin{aligned} u(x)=\sum _{m=0}^{\infty } u_{j,m}L_{j,m}(x). \end{aligned}$$

(D1)

In [10], the following expression of $u_{j,m}$ was given:

$$\begin{aligned} u_{j,m}=\frac{(-1)^l(2m+1)}{2^{m+1}m!}\int _{-1}^1\frac{\textrm{d}^l}{\textrm{d}\theta ^l}\hat{u}_j(\theta )\frac{\textrm{d}^{m-l}}{\textrm{d}\theta ^{m-l}}(\theta ^2-1)^m \textrm{d}\theta , \end{aligned}$$

where $\theta =\frac{2}{h_j}(x-x_j)$ as in the proof of Lemma 3, and $\hat{u}(\theta )=u(x(\theta ))$. If $u\in W^{l,\infty }(I_j)$, we have

$$\begin{aligned} \vert u_{j,m}\vert \leqslant Ch^l\vert u\vert _{W^{l,\infty }(I_j)}, \ \forall l, 0\leqslant l\leqslant m. \end{aligned}$$

(D2)

As to $P_h^+ u$, we can also express it in the Legendre basis:

$$\begin{aligned} P_h^+ u=\sum _{m=0}^k \tilde{u}_{j,m}L_{j,m}(x), \end{aligned}$$

(D3)

we can easily deduce from the definition of $P_h^+$ (9) that

$$\begin{aligned} \tilde{u}_{j,m}=u_{j,m},\ \forall m \leqslant k-3. \end{aligned}$$

Substituting (D1) and (D3) into (10), we have

$$\begin{aligned} \mathcal {M}_j \left( \begin{aligned} u_{j,k-2}&-\tilde{u}_{j,k-2}\\ u_{j,k-1}&-\tilde{u}_{j,k-1}\\ u_{j,k}&-\tilde{u}_{j,k} \end{aligned} \right) =\sum _{m=k+1}^{\infty } u_{j,m} \left( \begin{aligned}&L_{j,m}\left( x_{j-\frac{1}{2}}\right) \\ \frac{2}{h_j}&\frac{\textrm{d}}{\textrm{d}x}L_{j,m}\left( x_{j-\frac{1}{2}}\right) \\ \left( \frac{2}{h_j}\right) ^2&\frac{\textrm{d}^2}{\textrm{d}x^2}L_{j,m}\left( x_{j-\frac{1}{2}}\right) \end{aligned} \right) , \end{aligned}$$

(D4)

where

$$\begin{aligned} \mathcal {M}_j= \begin{pmatrix} L_{j,k-2}\left( x_{j-\frac{1}{2}}\right) &{} L_{j,k-1}\left( x_{j-\frac{1}{2}}\right) &{} L_{j,k}\left( x_{j-\frac{1}{2}}\right) \\ \frac{2}{h_j}\frac{\textrm{d}}{\textrm{d}x}L_{j,k-2}\left( x_{j-\frac{1}{2}}\right) &{} \frac{2}{h_j}\frac{\textrm{d}}{\textrm{d}x}L_{j,k-1}\left( x_{j-\frac{1}{2}}\right) &{} \frac{2}{h_j}\frac{\textrm{d}}{\textrm{d}x}L_{j,k}\left( x_{j-\frac{1}{2}}\right) \\ \left( \frac{2}{h_j}\right) ^2\frac{\textrm{d}^2}{\textrm{d}x^2}L_{j,k-2}\left( x_{j-\frac{1}{2}}\right) &{} \left( \frac{2}{h_j}\right) ^2\frac{\textrm{d}^2}{\textrm{d}x^2}L_{j,k-1}\left( x_{j-\frac{1}{2}}\right) &{} \left( \frac{2}{h_j}\right) ^2\frac{\textrm{d}^2}{\textrm{d}x^2}L_{j,k}\left( x_{j-\frac{1}{2}}\right) \end{pmatrix}. \end{aligned}$$

With further calculations, we know that $\mathcal {M}_j$ is independent of h and is identical for all j, so we drop the subscript j and denote it by $\mathcal {M}$. The reversibility of $\mathcal {M}$ enables us to rewrite (D4) into

$$\begin{aligned} \left( \begin{aligned} u_{j,k-2}&-\tilde{u}_{j,k-2}\\ u_{j,k-1}&-\tilde{u}_{j,k-1}\\ u_{j,k}&-\tilde{u}_{j,k} \end{aligned} \right) =\sum _{m=k+1}^{\infty } u_{j,m} \mathcal {M}^{-1}\left( \begin{aligned}&L_{j,m}\left( x_{j-\frac{1}{2}}\right) \\ \frac{2}{h_j}&\frac{\textrm{d}}{\textrm{d}x}L_{j,m}\left( x_{j-\frac{1}{2}}\right) \\ \left( \frac{2}{h_j}\right) ^2&\frac{\textrm{d}^2}{\textrm{d}x^2}L_{j,m}\left( x_{j-\frac{1}{2}}\right) \end{aligned} \right) . \end{aligned}$$

Then on $I_j$, we have

$$\begin{aligned} \begin{aligned} u-P_h^+u&=\sum _{m=k-2}^k (u_{j,m}-\tilde{u}_{j,m})L_{j,m}+\sum _{m=k+1}^{\infty }u_{j,m}L_{j,m}\\&=\sum _{m=k+1}^{\infty } u_{j,m} R_{j,m}, \end{aligned} \end{aligned}$$

(D5)

where

$$\begin{aligned} R_{j,m}=L_{j,m}+(L_{j,k-2},L_{j,k-1},L_{j,k})\mathcal {M}^{-1}\left( \begin{aligned}&L_{j,m}\left( x_{j-\frac{1}{2}}\right) \\ \frac{2}{h_j}&\frac{\textrm{d}}{\textrm{d}x}L_{j,m}\left( x_{j-\frac{1}{2}}\right) \\ \left( \frac{2}{h_j}\right) ^2&\frac{\textrm{d}^2}{\textrm{d}x^2}L_{j,m}\left( x_{j-\frac{1}{2}}\right) \end{aligned} \right) . \end{aligned}$$

Taking $u=L_{j,k+1}$ in (D5), we can see that $R_{j,k+1}$ we define here is identical with the one in (31). Therefore, from (D2), we have for $x\in D^s_j$,

$$\begin{aligned} \frac{\partial ^s}{\partial x^s}(u-P^+_h)(x)=\sum _{m=k+2}^{\infty } u_{j,m}\frac{\textrm{d}^s}{\textrm{d}x^s}R_{j,m}\leqslant Ch^{k+2-s}\vert u\vert _{W^{k+2,\infty }(I_j)}. \end{aligned}$$

Discrete Poincaré Inequality

Let

$$\begin{aligned} H^1(\Omega ,\mathcal {T})=\{u\in L^2(\Omega )\!: u\vert _{I_j}\in H^1(I_j),\; \forall j \in Z_N \} \end{aligned}$$

be the piecewise $H^1$ functions on $\mathcal {T}$. We have the following discrete Poincaré inequality.

Lemma E1

Suppose $u\in H^1(\Omega ,\mathcal {T})$, and u is a periodic function with a period of $2\uppi$. Then, we have

$$\begin{aligned} \Vert u \Vert \lesssim \Vert u_x \Vert +h^{-\frac{1}{2}} \left( \sum _{j=1}^N \llbracket u\rrbracket _{j+\frac{1}{2}}^2\right) ^{\frac{1}{2}}+\left| \int _{\Omega } u \ \textrm{d}x \right| . \end{aligned}$$

Proof

Let $V=\{ u\in L^2(\Omega )\!:\ u\vert _{I_j} \in \mathcal {P}^1(I_j)\ \textrm{and} \ u \ \mathrm {is \ continue}\}$, we introduce an interpolation operator $\mathcal {I}$: $H^1(\Omega ,\mathcal {T}) \rightarrow V$

$$\begin{aligned} \mathcal {I}u\left( x_{j+\frac{1}{2}}\right) =\frac{1}{2}\left( u^+\left( x_{j+\frac{1}{2}}\right) +u^-\left( x_{j+\frac{1}{2}}\right) \right) ,\; \forall j\in Z_N. \end{aligned}$$

(E1)

Let $\uppi _{I_j}\!:\!H^1(I_j)\rightarrow \mathcal {P}^1(I_j)$ be the local interpolation operator defined by

$$\begin{aligned} \uppi _{I_j}u\left( x_{j\pm \frac{1}{2}}\right) =u\left( x_{j\pm \frac{1}{2}}\right) . \end{aligned}$$

(E2)

We have the following well-known estimates:

$$\begin{aligned} \Vert u-\uppi _{I_j}u \Vert _{I_j}^2+h_j^2 \Vert (\uppi _{I_j}u)_x \Vert _{I_j}^2 \lesssim h_j^2 \Vert u_x \Vert _{I_j}^2. \end{aligned}$$

(E3)

From (E1) and (E2), we have on $I_j$

$$\begin{aligned} (\mathcal {I}u-\uppi _{I_j}u)\left( x_{j\pm \frac{1}{2}}\right) =\pm \llbracket u\rrbracket _{j\pm \frac{1}{2}}. \end{aligned}$$

It is easy to check that

$$\begin{aligned} \begin{aligned}&\Vert \mathcal {I}u-\uppi _{I_j}u \Vert _{I_j}^2 \lesssim h_j\left( \llbracket u\rrbracket _{j+\frac{1}{2}}^2+\llbracket u\rrbracket _{j-\frac{1}{2}}^2\right) , \\&\Vert (\mathcal {I}u-\uppi _{I_j}u)_x \Vert _{I_j}^2 \lesssim h_j^{-1}\left( \llbracket u\rrbracket _{j+\frac{1}{2}}^2+\llbracket u\rrbracket _{j-\frac{1}{2}}^2\right) . \end{aligned} \end{aligned}$$

Along with (E3), we have

$$\begin{aligned} \left\{ \begin{array}{l} \Vert u-\mathcal {I}u \Vert ^2 \lesssim h^2\Vert u_x \Vert ^2+h \sum _{j=1}^N \llbracket u\rrbracket _{j+\frac{1}{2}}^2, \\ \Vert (\mathcal {I}u)_x \Vert ^2 \lesssim \Vert u_x \Vert ^2+h^{-1} \sum _{j=1}^N \llbracket u\rrbracket _{j+\frac{1}{2}}^2. \end{array}\right. \end{aligned}$$

(E4)

Since $\mathcal {I}u\in H^1(\Omega )$, we can use the usual Poincaré inequality. Thus,

$$\begin{aligned} \begin{aligned} \Vert \mathcal {I}u \Vert ^2&\lesssim \Vert (\mathcal {I}u)_x \Vert ^2+\left| \int _{\Omega } \mathcal {I}u \ \textrm{d}x \right| ^2 \\&\lesssim \Vert (\mathcal {I}u)_x \Vert ^2+\left| \int _{\Omega } u \ \textrm{d}x \right| ^2+\left| \int _{\Omega } u-\mathcal {I}u \ \textrm{d}x \right| ^2 \\&\lesssim \Vert (\mathcal {I}u)_x \Vert ^2+\left| \int _{\Omega } u \ \textrm{d}x \right| ^2+\Vert u-\mathcal {I}u \Vert ^2. \end{aligned} \end{aligned}$$

(E5)

From (E4) and (E5), we have

$$\begin{aligned} \begin{aligned} \Vert u \Vert ^2&\lesssim \Vert u-\mathcal {I}u \Vert ^2+\Vert \mathcal {I}u \Vert ^2\\&\lesssim \Vert u_x \Vert ^2+h^{-1} \sum _{j=1}^N \llbracket u\rrbracket _{j+\frac{1}{2}}^2+\left| \int _{\Omega } u \ \textrm{d}x \right| ^2. \end{aligned} \end{aligned}$$

(E6)