Energy-norm and balanced-norm supercloseness error analysis of a finite volume method on Shishkin meshes for singularly perturbed reaction–diffusion problems

Abstract

A singularly perturbed reaction–diffusion problem posed on the unit square in $\mathbb {R}^2$ is considered. To solve this problem numerically, a finite volume method (FVM) whose primal mesh is Shishkin is constructed; the FVM solution is piecewise bilinear on this mesh. Working in the standard energy norm, a superclose result (for the difference between the FVM solution and the Lagrange interpolant of the exact solution) is derived. This result yields an improved bound for the $L^2$ error of the FVM solution, and implies that a simple postprocessing of the FVM solution produces (in the energy norm) a higher-order approximation of the true solution. Next, we analyse errors in a balanced norm that is stronger than the energy norm; using a more complicated approximant of the exact solution from our piecewise bilinear space, we prove an optimal-order error bound and an associated supercloseness result showing that the difference between the FVM solution and our approximant is of higher order than the error itself. Finally, numerical experiments demonstrate the sharpness of our error bounds.

Appendix

Proof of Lemma 12

Assume that $b_1(x)\ge \beta _1>0$ and $b_2(y)\ge \beta _2>0$. We first prove an analogous result in 1D, then use this to deduce the result in 2D.

The 1D case: Recall that ${\bar{\varOmega }}_{0,x}$ is divided by a uniform primal mesh. Let $U_{h,\varOmega _{0,x}}$ be the space of piecewise linear functions defined on this mesh. Let $\varPi _{h,x}^*$ be the 1D analogue of the projector $\varPi _{h}^*$ defined in (6); this maps piecewise linears in $U_{h,\varOmega _{0,x}}$ to piecewise constants on the dual mesh restricted to ${\bar{\varOmega }}_{0,x}$. That is, if $w_h\in U_{h,\varOmega _{0,x}}$, then $\varPi _{h,x}^*w_h = w_h(x_i)$ on $[x_{i-1/2},x_{i+1/2}]$ (modified to $[x_{N/4},x_{N/4+1/2}]$ and $[x_{3N/4-1/2},x_{3N/4}]$ at the boundaries of $\varOmega _{0,x}$).

For each $v\in C({\bar{\varOmega }}_{0,x})$, define its weighted projection $\pi _{h,x} v \in U_{h,\varOmega _{0,x}}$ by

$$\begin{aligned} \langle \pi _{h,x} v, b_1 \varPi _{h,x}^* \phi _j\rangle _{{\bar{\varOmega }}_{0,x}} = \langle v, b_1 \varPi _{h,x}^* \phi _j\rangle _{{\bar{\varOmega }}_{0,x}} \ \text { for each } x_j\in {{\bar{\varOmega }}}_{0,x}, \end{aligned}$$

where $\phi _j$ is the piecewise linear “hat" basis function associated with the mesh point $x_j\in {\bar{\varOmega }}_{0,x}$. Writing $\pi _{h,x} v = \sum _{x_i\in {\bar{\varOmega }}_{0,x}} (\pi _{h,x} v)(x_i) \phi _i$, then the above definition becomes

$$\begin{aligned} \sum _{x_i\in {\bar{\varOmega }}_{0,x}}\langle \phi _i, b_1 \varPi _{h,x}^* \phi _j \rangle _{{\bar{\varOmega }}_{0,x}} \pi _{h,x} v(x_i) = \langle v, b_1 \varPi _{h,x}^* \phi _j\rangle _{{\bar{\varOmega }}_{0,x}}\ \end{aligned}$$

(50)

for each $x_j\in {{\bar{\varOmega }}}_{0,x}$.

We now compute the Gram matrix associated with (50). If $i = j$, then

$$\begin{aligned}{} & {} \langle \phi _j,b_1\varPi _{h,x}^* \phi _j\rangle _{{\bar{\varOmega }}_{0,x}} = \langle \phi _j,b_1\varPi _{h,x}^* \phi _j\rangle _{[x_{j-1/2},x_{j}]} +\langle \phi _j,b_1\varPi _{h,x}^* \phi _j\rangle _{[x_{j},x_{j+1/2}]}\\{} & {} \quad :=d_{j,l}+d_{j,r}, \end{aligned}$$

where one should delete $d_{j,l}$ if $j=N/4$ and delete $d_{j,r}$ if $j=3N/4$. If $i = j-1$, then

$$\begin{aligned} \langle \phi _{j-1},b_1\varPi _{h,x}^* \phi _j\rangle _{{\bar{\varOmega }}_{0,x}} =\langle \phi _{j-1},b_1\varPi _{h,x}^* \phi _j\rangle _{[x_{j-1/2},x_{j}]}:= e_{j,l}. \end{aligned}$$

If $i = j+1$, then

$$\begin{aligned} \langle \phi _{j+1},b_1\varPi _{h,x}^* \phi _j\rangle _{{\bar{\varOmega }}_{0,x}} =\langle \phi _{j+1},b_1\varPi _{h,x}^* \phi _j\rangle _{[x_{j},x_{j+1/2}]}:= e_{j,r}. \end{aligned}$$

For $i \notin \{j-1,j,j+1\}$ one has $\langle \phi _i,b_1\varPi _{h,x}^* \phi _j\rangle _{{\bar{\varOmega }}_{0,x}} = 0$. Rescaling, we divide (50) by

$$\begin{aligned} \langle \phi _j,\varPi _{h,x}^* \phi _j\rangle _{{\bar{\varOmega }}_{0,x}} = {\left\{ \begin{array}{ll} H &{}\text {if } \sigma<x_j<1-\sigma , \\ H/2 &{}\text {if } x_j=\sigma \text { or } x_j=1-\sigma . \end{array}\right. } \end{aligned}$$

This yields the linear system $Ax = g$ where $x:= \left( \pi _{h,x} v(x_j):x_j\in {{\bar{\varOmega }}}_{0,x}\right) ^T$ and

$$\begin{aligned} g:=\left( \frac{\langle v,b_1\varPi _{h,x}^* \phi _j\rangle _{{\bar{\varOmega }}_{0,x}}}{\langle \phi _j,\varPi _{h,x}^* \phi _j\rangle _{{\bar{\varOmega }}_{0,x}}}:x_j\in {{\bar{\varOmega }}}_{0,x}\right) ^T. \end{aligned}$$

Clearly $\Vert g\Vert _{\infty }\le C\Vert v\Vert _{L^\infty (\varOmega _{0,x})}$.

For $x_j\in \varOmega _{0,x}$, since $b_1\ge \beta _1>0$, $\phi _j>\phi _{j-1}$ on $(x_{j-1/2},x_j)$ and $\phi _j>\phi _{j+1}$ on $(x_{j},x_{j+1/2})$, a calculation yields $d_{j,l}-e_{j,l}\ge \beta _1 H/4$ and $d_{j,r}-e_{j,r}\ge \beta _1 H/4$ (if $j=N/4$ use only $d_{j,r}-e_{j,r}\ge \beta _1 H/4$; if $j = 3N/4$ use only $d_{j,l}-e_{j,l}\ge \beta _1 H/4$). Thus the matrix A is strictly diagonally dominant with $\min _j (|a_{j,j}|-\sum _{i\ne j}|a_{j,i}|) \ge \beta _1/2$, which is independent of H. Hence, by a well-known result [15, Theorem 1], one gets $\Vert A^{-1}\Vert _{\infty }\le 2/\beta _1$. It follows that there exists a constant $C_4$ such that

$$\begin{aligned} \Vert \pi _{h,x} v(x)\Vert _{L^\infty (\varOmega _{0,x})}\le C_4 \Vert v(x)\Vert _{L^\infty (\varOmega _{0,x})}\ \text { for all }v\in C({\bar{\varOmega }}_{0,x}). \end{aligned}$$

(51)

The 2D case: Now we return to the original 2D problem. Write

$$\begin{aligned} \pi _h v (x,y) = \sum _{(x_i,y_j)\in {{\bar{\varOmega }}}_{0}} \alpha _{i,j} \phi _i(x) \phi _j(y) \end{aligned}$$

for some $\alpha _{i,j}\in \mathbb {R}$. From the definition of $\pi _h$, for each $(x_i,y_j)\in {\varOmega }_{0}$ one has

$$\begin{aligned}&\hspace{-5mm}\int _{x = x_{i-1/2}}^{x_{i+1/2}} \left[ \int _{y = y_{j-1/2}}^{y_{j+1/2}} v (x,y) b_2(y)\varPi _{h,y}^*\phi _j(y)\,dy\right] b_1(x)\varPi _{h,x}^*\phi _i(x)\, dx\nonumber \\&=\int _{x = x_{i-1/2}}^{x_{i+1/2}}\left[ \int _{y = y_{j-1/2}}^{y_{j+1/2}} (\pi _h v) (x,y) b_2(y)\varPi _{h,y}^*\phi _j(y)\,dy\right] b_1(x)\varPi _{h,x}^*\phi _i(x)\, dx \end{aligned}$$

(52)

where the integrals in this formula are truncated when $(x_i,y_j)\in \partial \varOmega _0$ to avoid integrating outside $\varOmega _0$. For each j, define averages of v and $\pi _h v$ by

$$\begin{aligned} v_j(x)&:=\frac{1}{y_{j+1/2}-y_{j-1/2}} \int _{y_{j-1/2}}^{y_{j+1/2}}v(x,y)b_2(y)\varPi _{h,y}^*\phi _j(y)\,dy, \\ (\pi _h v)_j(x)&:=\frac{1}{y_{j+1/2}-y_{j-1/2}} \int _{y_{j-1/2}}^{y_{j+1/2}}(\pi _h v)(x,y)b_2(y)\varPi _{h,y}^*\phi _j(y)\,dy, \end{aligned}$$

where we again truncate integrals where necessary: if $j = N/4$, change $j-1/2$ to j, and if $j = 3N/4$, change $j+1/2$ to j. Then one can interpret (52) as a 1D weighted projection in the x-variable, where $v_j$ is projected onto $(\pi _h v)_j$. Thus, (51) implies that there exists a constant $C_5$ such that

$$\begin{aligned} \Vert (\pi _h v)_j\Vert _{L^\infty (\varOmega _{0,x})}\le C_4 \Vert v_j\Vert _{L^\infty (\varOmega _{0,x})}\le C_5 \Vert v\Vert _{L^\infty ({\varOmega }_0)} \ \text { for } j = N/4,...,3N/4,\nonumber \\ \end{aligned}$$

(53)

where the second inequality is valid because $0\le \varPi _{h,y}^*\phi _j\le 1$ and $b_2(y)$ is bounded. But for each j the function

$$\begin{aligned}&(\pi _h v)_j(x) \nonumber \\&=\sum _{x_i\in {\bar{\varOmega }}_{0,x}}\phi _i(x) \left[ \sum _{k = j-1}^{j+1}\alpha _{i,k} \frac{1}{y_{j+1/2}-y_{j-1/2}} \int _{y_{j-1/2}}^{y_{j+1/2}}\phi _k(y)b_2(y)\varPi _{h,y}^*\phi _j(y)\,dy\right] \end{aligned}$$

(54)

is piecewise linear, with modifications like above for certain j (when $j = N/4$ change the subscript $j-1/2$ to j and the sum lower limit $j-1$ to j; when $j = 3N/4$ change the subscript $j+1/2$ to j and the sum upper limit $j+1$ to j). Thus for $j = N/4+1,...,3N/4-1$ (modify this argument slightly if $j=N/4$ or 3N/4) one has

$$\begin{aligned}&\Vert (\pi _h v)_j\Vert _{L^\infty (\varOmega _{0,x})} \nonumber \\&\quad =\max _{i}\left| \sum _{k = j-1}^{j+1} \alpha _{i,k} \frac{1}{y_{j+1/2}-y_{j-1/2}} \int _{y_{j-1/2}}^{y_{j+1/2}}\phi _k(y)b_2(y)\varPi _{h,y}^*\phi _j(y)\,dy \right| \nonumber \\&\quad = \max _{i}\left| \gamma _{j-1}\alpha _{i,j-1}+\gamma _{j}\alpha _{i,j} +\gamma _{j+1}\alpha _{i,j+1}\right| \end{aligned}$$

(55)

where

$$\begin{aligned} \gamma _{j-1}:= & {} \frac{1}{y_{j+1/2}-y_{j-1/2}} \int _{y_{j-1/2}}^{y_{j+1/2}}\phi _{j-1}(y)b_2(y)\varPi _{h,y}^*\phi _j(y)\,dy, \\ \gamma _{j}:= & {} \frac{1}{y_{j+1/2}-y_{j-1/2}} \int _{y_{j-1/2}}^{y_{j+1/2}}\phi _{j}(y)b_2(y)\varPi _{h,y}^*\phi _j(y)\,dy, \\ \gamma _{j+1}:= & {} \frac{1}{y_{j+1/2}-y_{j-1/2}} \int _{y_{j-1/2}}^{y_{j+1/2}}\phi _{j+1}(y)b_2(y)\varPi _{h,y}^*\phi _j(y)\,dy \end{aligned}$$

satisfy

$$\begin{aligned} \gamma _{j-1}>0, \,\gamma _{j}>0, \,\gamma _{j+1}>0 \text { and } \gamma _{j}-(\gamma _{j-1}+\gamma _{j+1})\ge \beta _2/2 >0. \end{aligned}$$

(56)

Choose $(i^*,j^*)$ such that $|\alpha _{i^*,j^*}| = \max _{(x_i,y_j)\in {{\bar{\varOmega }}}_0}|\alpha _{i,j}|$. Then (53) and (55) give

$$\begin{aligned} C_5 \Vert v\Vert _{L^\infty ({\varOmega }_0)}&\ge \left| \gamma _{j^*-1}\alpha _{i^*,j^*-1}+\gamma _{j^*}\alpha _{i^*,j^*} +\gamma _{j^*+1}\alpha _{i^*,j^*+1}\right| \\&\ge \gamma _{j^*}\left| \alpha _{i^*,j^*}\right| -\gamma _{j^*-1}\left| \alpha _{i^*,j^*-1}\right| -\gamma _{j^*+1}\left| \alpha _{i^*,j^*+1}\right| \\&\ge (\gamma _{j^*}-\gamma _{j^*-1}-\gamma _{j^*+1})\left| \alpha _{i^*,j^*}\right| \\&\ge (\beta _2/2)\left| \alpha _{i^*,j^*}\right| , \end{aligned}$$

where we used (56) and the choice of $(i^*,j^*)$. Then we have

$$\begin{aligned} \Vert \pi _h v\Vert _{L^\infty ({\varOmega }_0)}&= \left\| \sum _{(x_i,y_j)\in {{\bar{\varOmega }}}_{0}} \alpha _{i,j} \phi _i(x) \phi _j(y)\right\| _{L^\infty ({\varOmega }_0)} \\&\le \max _{(x_i,y_j)\in {{\bar{\varOmega }}}_{0}}|\alpha _{i,j}| =\left| \alpha _{i^*,j^*}\right| \\&\le (2C_5/\beta _2) \Vert v\Vert _{L^\infty ({\varOmega }_0)}, \end{aligned}$$

as desired. $\square$