Unified Analysis of Any Order Spectral Volume Methods for Diffusion Equations

Abstract

In this paper, a new class of spectral volume (SV) methods are proposed, analyzed, and implemented for diffusion equations, with the viscous flux taken as an interior penalty or direct discontinuous Galerkin formulation. The control volumes are constructed by using four kinds of special points (including Legendre–Gauss, Legendre–Gauss–Lobatto, right Legendre–Gauss–Radau and left Legendre–Gauss–Radau points) in subintervals of the underlying meshes, which leads to four different SV schemes. A framework for the stability analysis and error estimates of the four SV schemes is established. In particular, the influence of the choice of the parameters in the numerical fluxes on the convergence rate and the optimal choices of coefficients for each SV scheme are discussed and provided. Numerical experiments are presented to demonstrate the stability and accuracy of the four SV schemes for both linear and nonlinear diffusion equations.

Appendix

1.1 Proof of Lemma 3.1

We begin with some preliminaries.

First, for any function $w\in U_h$, let $w^*={{\mathcal {F}}}w$ be defined in (3.4) with the coefficients given by (3.5). Then a direct calculation from (3.5) yields

$$\begin{aligned} \begin{aligned}&[w^{\tau ,*}_{k,j}]^y=A_j^{\tau ,y} \big (w_y-A_{k+1}^{\tau ,x}w_{xy}\big )(g_{k+1,j}^{\tau }),&\\&[w^{\tau ,*}_{i,k}]^x =A_i^{\tau ,x}\big (w_x- A_{k+1}^{\tau ,y}w_{xy}\big )(g_{i,k+1}^{\tau }),&\\&w^{\tau ,*}_{0,k}=\big (w+A_0^{\tau ,x}w_x-A_{k+1}^{\tau ,y}w_y-A_{0,k+1}^\tau w_{xy}\big )(g_{0,k+1}^{\tau }),&\\&w^{\tau ,*}_{k,0}=\big (w+A_0^{\tau ,y}w_y-A_{k+1}^{\tau ,x}w_x-A_{k+1,0}^\tau w_{xy}\big )(g_{k+1,0}^{\tau }),&\\&w^{\tau ,*}_{k,k}=\big (w-A_{k+1}^{\tau ,x}w_x-A_{k+1}^{\tau ,y}w_y+A_{k+1,k+1}^{\tau }w_{xy}\big )(g_{k+1,k+1}^{\tau }).&\end{aligned} \end{aligned}$$

(7.1)

Second, denoting

$$\begin{aligned}{} & {} E_1(v,w)=A_{0}^{\tau ,x}\big (\int _{g_0^{\tau ,y}}^{g_{k+1}^{\tau ,y}} (w_xv)(g_0^{\tau ,x},y)dy+ R_\tau ^y(w_{xy}\partial _y^{-1}v)(g^{\tau ,x}_0)\big ),\\{} & {} E_2(v,w)=A_{k+1}^{\tau ,x}\big (\int _{g_0^{\tau ,y}}^{g_{k+1}^{\tau ,y}} (w_xv)(g_{k+1}^{\tau ,x},y)dy+ R_\tau ^y(w_{xy}\partial _y^{-1}v)(g^{\tau ,x}_{k+1})\big ). \end{aligned}$$

Recalling the definition of $\langle ,\rangle _{\partial \tau _y}$ in (3.8), we have

$$\begin{aligned} \langle v,w^*\rangle _{\partial \tau _y}=\sum _{j=0}^{k} w_{k,j}^{\tau ,*} \int _{g_j^{\tau ,y}}^{g_{j+1}^{\tau ,y}} v(g_{k+1}^{\tau ,x},y)dy-\sum _{j=0}^{k} w_{0,j}^{\tau ,*} \int _{g_j^{\tau ,y}}^{g_{j+1}^{\tau ,y}} v(g_{0}^{\tau ,x},y)dy:=I_2-I_1. \end{aligned}$$

Then a direct calculation from (3.5) and (7.1) leads to

$$\begin{aligned} I_1= & {} -\sum _{j=1}^{k} [w^{\tau ,*}_{0,j}]^y \partial _y^{-1}v(g_0^{\tau ,x},g_j^{\tau ,y})+w_{0,k}^{\tau ,*}\partial _y^{-1}v(g_0^{\tau ,x},g_{k+1}^{\tau ,y})-w_{0,0}^{\tau ,*}\partial _y^{-1}v(g_0^{\tau ,x},g_{0}^{\tau ,y})\\= & {} -\sum _{j=0}^{k+1}\left( (A_j^{\tau ,y}w_y+A_{0,j}^\tau w_{xy})\partial _y^{-1}v\right) (g_{0,j}^{\tau })+(w\partial _y^{-1}v)|_{g_{0,0}^{\tau }}^{g_{0,k+1}^{\tau }}+A_0^{\tau ,x}(w_x\partial _y^{-1}v)|^{g_{0,k+1}^\tau }_{g_{0,0}^\tau }. \end{aligned}$$

By using (3.2)–(3.3) and the integration by parts,

$$\begin{aligned} I_1=\int _{g_0^{\tau ,y}}^{g_{k+1}^{\tau ,y}} (wv)(g_0^{\tau ,x},y)dy +R_\tau ^y(w_y\partial _y^{-1}v)(g^{\tau ,x}_0)+E_1(v,w). \end{aligned}$$

Following the same argument, we can prove that

$$\begin{aligned} I_2=\int _{g_0^{\tau ,y}}^{g_{k+1}^{\tau ,y}} (wv)(g_{k+1}^{\tau ,x},y)dy +R_\tau ^y(w_y\partial _y^{-1}v)(g^{\tau ,x}_{k+1})-E_2(v,w). \end{aligned}$$

Consequently,

$$\begin{aligned} \langle v,w^*\rangle _{\partial \tau _y} = \langle v,w\rangle _{\partial \tau _y} +R_\tau ^y(w_y\partial _y^{-1}v)|_{g^{\tau ,x}_0}^{g^{\tau ,x}_{k+1}}-E_{2}(v,w)-E_1(v,w). \end{aligned}$$

(7.2)

Now we are ready to prove Lemma 3.1.

Proof

Denoting $I=(({\widehat{\alpha v_x}})_x, w^*)_\tau$. In light of (7.2), we have for all $\tau =\tau _m^x\times \tau _n^y$ that

$$\begin{aligned} I= & {} ( (\alpha v_{x})_x, w^*)_\tau + \langle {\widehat{\alpha v_x}}-\alpha v_x,w^*\rangle _{\partial \tau _y}\\= & {} ( (\alpha v_{x})_x, w^*)_\tau +\langle {\widehat{\alpha v_x}}-\alpha v_x,w\rangle _{\partial \tau _y}+R_\tau ^y(w_y\partial _y^{-1} (\widehat{\alpha v_x}-\alpha v_x))|_{g^{\tau ,x}_0}^{g^{\tau ,x}_{k+1}}\\{} & {} -(E_2+E_1)(\widehat{\alpha v_x}-\alpha v_x, w). \end{aligned}$$

Using (3.6) and the integration by parts yields

$$\begin{aligned} I= & {} -( \alpha v_{x}, w_x)_\tau +\langle \widehat{\alpha v_x},w\rangle _{\partial \tau _y}+ {{\mathcal {E}}}_{\tau }^1( (\alpha v_{x})_x,w)+R_\tau ^y(w_y\partial _y^{-1} (\widehat{\alpha v_x}-\alpha v_x))|_{g^{\tau ,x}_0}^{g^{\tau ,x}_{k+1}}\\{} & {} -(E_2+E_1)(\widehat{\alpha v_x}-\alpha v_x, w). \end{aligned}$$

Then (3.9) follows by summing up all $\tau$ and using the following equation

$$\begin{aligned} (\alpha v_{x}, w_x)_\tau =\int _{\tau _n^y}\left( \sum _{i=0}^{k+1}A_{i}^{\tau ,x}\alpha v_xw_x(g_{i}^{\tau ,x},y)+R_{\tau }^x(\alpha v_xw_x)(y)\right) dy. \end{aligned}$$

Using the fact the $k+2$ Gauss–Lobatto and $k+1$ left and right Radau numerical quadrature is exact for polynomials of degree not more than 2k, we easily get

$$\begin{aligned} {\mathcal {E}}^1_{\tau }(v_{xx},v)+{\mathcal {E}}^2_{\tau }(v,v)=0,\ \ \forall v\in U_h, \end{aligned}$$

and thus (3.11) holds true for GLSV, RRSV and LRSV. As for GSV, by using the weight and error of Gauss numerical quadrature in Table 1, we have for all $\tau =\tau ^x_m\times \tau ^y_n\in {{\mathcal {T}}}_h$

$$\begin{aligned}{} & {} {\mathcal {E}}^1_{\tau }(v_{xx},w)=\int _{\tau _m^x}R^y_\tau (w_{y}\partial _y^{-1}v_{xx})dx=-\int _{\tau _m^x}R^y_\tau (w_{yx}\partial _y^{-1}v_{x})dx+R^y_\tau (w_{y}\partial _y^{-1}v_{x})|_{x_{m-\frac{1}{2}}}^{x_{m+\frac{1}{2}}}, \\{} & {} {\mathcal {E}}^2_{\tau }(v,w)=R^y_\tau (w_{y}\partial _y^{-1}(\widehat{v_x}-v_{x}))|_{x_{m-\frac{1}{2}}}^{x_{m+\frac{1}{2}}}. \end{aligned}$$

Consequently, we use the error of Gauss numerical quadrature in Table 1 to derive

$$\begin{aligned} {\mathcal {E}}^1_{\tau }(v_{xx},w)+{\mathcal {E}}^2_{\tau }(v,w)= & {} -\int _{\tau _m^x}R^y_\tau (w_{yx}\partial _y^{-1}v_{x})dx+R^y_\tau (w_{y}\partial _y^{-1}\widehat{v_x})|_{x_{m-\frac{1}{2}}}^{x_{m+\frac{1}{2}}}\\= & {} -c_kh^{2k+1}\left( \int _{\tau _m^x} (\partial _y^kv_{x}\partial _y^{k}v_{x})(x,\eta _n)dx- \partial _y^{k}v\partial _y^{k}\widehat{v_x}|_{(x_{m-\frac{1}{2}},\eta _n)}^{(x_{m+\frac{1}{2}},\eta _n)}\right) \\= & {} -c_kh^{2k} \left( (\partial _y^kv_x,\partial _y^k v_x)_{\tau }-\langle \partial _y^{k}{{\widehat{v}}}_{x}, \partial _y^{k} v\rangle _{\partial \tau _y}\right) , \end{aligned}$$

where $c_k$ is a constant only dependent on k, and $\eta _n\in (y_{n-\frac{1}{2}}, y_{n+\frac{1}{2}})$ is an arbitrary point, and in the last step, we have used the fact that $\partial _y^k v$ for any $v\in U_h$ is a constant function about y. By denoting $\zeta =c_k^{\frac{1}{2}}h^k\partial _y^kv$, we easily obtain

$$\begin{aligned} {\mathcal {E}}^1_{\tau }(v_{xx},v)+{\mathcal {E}}^2_{\tau }(v,v)=-(\zeta _x,\zeta _x)_{\tau }+\langle {{\widehat{\zeta }}}_x,\zeta \rangle _{\partial \tau _y}. \end{aligned}$$

This finishes the proof of (3.11). $\square$

1.2 Proof of Lemma 3.2

Proof

First, for all $v\in {\mathbb {P}}_{k-1}([-1,1])$, suppose

$$\begin{aligned} v(s)=\sum _{i=0}^{k-1} \sqrt{\frac{2i+1}{2}}a_iL_i(s),\ \ s\in [-1,1]. \end{aligned}$$

Then a direct calculation yields

$$\begin{aligned} v(1)=\sum _{i=0}^{k-1}a_i\sqrt{\frac{2i+1}{2}},\ \ \partial _sv(1)=\sum _{i=0}^{k-1}a_i\sqrt{\frac{2i+1}{2}} \frac{i(i+1)}{2},\ \ \Vert v\Vert ^2_{0,[-1,1]}=\sum _{i=0}^{k-1}|a_i|^2. \end{aligned}$$

Using the Cauchy-Schwarz inequality, we get

$$\begin{aligned} (\gamma v(1)-4\beta _1\partial _sv(1))^2\le & {} \sum _{i=0}^{k-1} a_i^2\sum _{i=0}^{k-1}\frac{2i+1}{2}(\gamma -2\beta _1 i (i+1))^2 \\= & {} \frac{k^2}{2}(\gamma ^2-\beta _1\gamma (k^2-1)+\frac{\beta _1^2}{3}(k^2-1)^2)\Vert v\Vert ^2_{0,[-1,1]}. \end{aligned}$$

Then the first equation of (3.14) follows.

We next estimate $\Gamma _2$. For any $\theta \in {\mathbb {P}}_{k-1}$, let

$$\begin{aligned} \theta (s)=\sum _{i=1}^k c_iL_{i-1}(s),\ \ \textbf{c}:=\textbf{c}_k=(c_1,\ldots ,c_k)^T. \end{aligned}$$

Then a direct calculation yields

$$\begin{aligned} |\theta (1)|^2=\textbf{c}^T A\textbf{c},\ \ |\theta (-1)|^2=\textbf{c}^T B\textbf{c} \end{aligned}$$

with

$$\begin{aligned} A=a^Ta, \ B=b^Tb, \ a=(1,1,\ldots , 1)^T,\ b=(1,-1,\ldots , (-1)^{k-1})^T. \end{aligned}$$

Consequently,

$$\begin{aligned} I_k:=|\theta (1)|^2+|\theta (-1)|^2=\textbf{c}^T (A+B)\textbf{c}=\textbf{c}^T M_k\textbf{c}, \end{aligned}$$

where

$$\begin{aligned} M_k=\left( \begin{array}{ccccc} 2&{} 0&{} 2&{} \cdots &{} 1+(-1)^{k-1} \\ 0 &{} 2&{} 0&{} \cdots &{} 1+(-1)^{k}\\ \cdots &{}\cdots &{} \cdots &{} \cdots \\ 1+(-1)^{k-1}&{}1+(-1)^k&{} 1+(-1)^{k+1}&{}\cdots &{} 2 \end{array} \right) . \end{aligned}$$

By using the method of mathematical induction, we can obtain that

$$\begin{aligned} I_k\le \frac{(k+1)^2}{2}\sum _{i=1}^{k}\frac{2}{2i-1}c_i^2= \frac{(k+1)^2}{2}\Vert \theta \Vert _0^2, \end{aligned}$$

which yields the second equation of (3.14). The proof is complete. $\square$