Arbitrary-order monotonic finite-volume schemes for 1D elliptic problems

Abstract

When solving numerically an elliptic problem, it is important in most applications that the scheme used preserves the positivity of the solution. When using finite volume schemes on deformed meshes, the question has been solved rather recently. Such schemes are usually (at most) second-order convergent, and non-linear. On the other hand, many high-order schemes have been proposed that do not ensure positivity of the solution. In this paper, we propose a very high-order monotonic (that is, positivity preserving) numerical method for elliptic problems in 1D. We prove that this method converges to an arbitrary order (under reasonable assumptions on the mesh) and is indeed monotonic. We also show how to handle discontinuous sources or diffusion coefficients, while keeping the order of convergence. We assess the new scheme, on several test problems, with arbitrary (regular, distorted, and random) meshes.

Appendices

A Dirichlet boundary conditions

Consider first the right boundary of the domain. The adaptation to the left boundary is straightforward. The kth-order Taylor expansion in the neighborhood of $x_{n+\frac{1}{2}}$ gives

$$\begin{aligned} \forall x,\quad \bar{u}(x) = \bar{u}\left( x_{n+\frac{1}{2}}\right) + \sum _{\ell =1}^{k}\frac{\left( x-x_{n+\frac{1}{2}}\right) ^{\ell }}{\ell !}\frac{{\text {d}}^{\ell }\bar{u}}{{\text {d}}x^{\ell }}\left( x_{n+\frac{1}{2}}\right) + \mathcal {O}\left( \left( x-x_{n+\frac{1}{2}}\right) ^{k+1}\right) . \end{aligned}$$

Here again, we integrate this expression to use mean values. This gives

$$\begin{aligned} \frac{1}{h_{n}}\int _{x_{n-\frac{1}{2}}}^{x_{n+\frac{1}{2}}}\bar{u}(x){\text {d}}x = \bar{u}\left( x_{n+\frac{1}{2}}\right) + \frac{1}{h_{n}}\sum _{\ell =1}^{k}\int _{x_{n-\frac{1}{2}}}^{x_{n+\frac{1}{2}}}\frac{\left( x-x_{n+\frac{1}{2}}\right) ^{\ell }}{\ell !}\frac{{\text {d}}^{\ell }\bar{u}}{{\text {d}}x^{\ell }}\left( x_{n+\frac{1}{2}}\right) dx +\mathcal {O}\left( h_{n}^{k+1}\right) , \end{aligned}$$

that is to say

$$\begin{aligned} \bar{u}_{n} = \bar{u}\left( x_{n+\frac{1}{2}}\right) + \frac{1}{h_n}\sum _{\ell =1}^{k}\left[ \frac{\left( x-x_{n+\frac{1}{2}}\right) ^{\ell +1}}{(\ell +1)!}\right] _{x_{n-\frac{1}{2}}}^{x_{n+\frac{1}{2}}}\frac{{\text {d}}^{\ell }\bar{u}}{{\text {d}}x^{\ell }}\left( x_{n+\frac{1}{2}}\right) +\mathcal {O}\left( h_{n}^{k+1}\right) , \end{aligned}$$

from which we obtain

$$\begin{aligned} \frac{{\text {d}}\bar{u}}{{\text {d}}x}\left( x_{n+\frac{1}{2}}\right) = \frac{2}{h_{n}}\left( \bar{u}\left( x_{n+\frac{1}{2}}\right) -\bar{u}_{n}\right) + 2\sum _{\ell =2}^{k}\frac{(-1)^{\ell } h_{n}^{\ell -1}}{(\ell +1)!}\frac{{\text {d}}^{\ell }\bar{u}}{{\text {d}}x^{\ell }}\left( x_{n+\frac{1}{2}}\right) + \mathcal {O}\left( h_{n}^{k}\right) . \end{aligned}$$

The numerical flux is obtained by approximating the derivatives of $\bar{u}$ at $x_{n+\frac{1}{2}}$ using a polynomial reconstruction of the solution

$$\begin{aligned} \mathcal {F}_{n+\frac{1}{2}}(\textbf{u}) = \kappa _{n+\frac{1}{2}}\left( \frac{2}{h_{n}}\left( u_{n+\frac{1}{2}}-u_{n}\right) + r_{n+\frac{1}{2}}(\textbf{u})\right) . \end{aligned}$$

The trick of Sect. 1.3 can be applied to ensure monotonicity, that is in the non-symmetric version

$$\begin{aligned} \mathcal {F}_{n+\frac{1}{2}}(\textbf{u}) = \kappa _{n+\frac{1}{2}}\left[ \left( \frac{2}{h_{n}} + \frac{r_{n+\frac{1}{2}}^+(\textbf{u})}{u_{n+\frac{1}{2}}}\right) u_{n+\frac{1}{2}} - \left( \frac{2}{h_{n}} + \frac{r_{n+\frac{1}{2}}^-(\textbf{u})}{u_n}\right) u_{n}\right] , \end{aligned}$$

and in the symmetric version

$$\begin{aligned} \mathcal {F}_{n+\frac{1}{2}}(\textbf{u}) = \kappa _{n+\frac{1}{2}}\left[ \left( \frac{2}{h_{n}} + \frac{r_{n+\frac{1}{2}}^+(\textbf{u}) + s_{n+\frac{1}{2}}(\textbf{u})}{u_{n+\frac{1}{2}}}\right) u_{n+\frac{1}{2}} - \left( \frac{2}{h_{n}} + \frac{r_{n+\frac{1}{2}}^-(\textbf{u}) + s_{n+\frac{1}{2}}(\textbf{u})}{u_n}\right) u_{n}\right] , \end{aligned}$$

(68)

with

$$\begin{aligned} s_{n+\frac{1}{2}}(\textbf{u}) = \dfrac{u_n r_{n+\frac{1}{2}}^+(\textbf{u}) - u_{n+\frac{1}{2}}r_{n+\frac{1}{2}}^-(\textbf{u})}{u_{n+\frac{1}{2}} - u_n}. \end{aligned}$$

To preserve positivity, a condition similar to (19) must be satisfied for the symmetric version of the scheme

$$\begin{aligned} \frac{\frac{2}{h_{n}}\left( u_{n+\frac{1}{2}}-u_n\right) + r_{n+\frac{1}{2}}(\textbf{u})}{u_{n+\frac{1}{2}}-u_n} \ge 0, \end{aligned}$$

that is to say that $u_{n+\frac{1}{2}}-u_n$ and $\mathcal {F}_{n+\frac{1}{2}}(\textbf{u})$ must have the same sign. As above, this condition seems natural, because if $\displaystyle \frac{{\text {d}}\bar{u}}{{\text {d}}x}(x_{n+\frac{1}{2}}) \ge 0$ (resp. $\le 0$), then $\bar{u}$ is locally increasing (resp. decreasing), so $\bar{u}_{n+\frac{1}{2}} \ge \bar{u}_n$ (resp. $\bar{u}_{n+\frac{1}{2}} \le \bar{u}_n$).

Applying the boundary condition, (68) becomes

$$\begin{aligned} \mathcal {F}_{n+\frac{1}{2}}(\textbf{u}) = \kappa _{n+\frac{1}{2}}\left[ \left( \frac{2}{h_{n}} + \frac{r_{n+\frac{1}{2}}^+(\textbf{u}) + s_{n+\frac{1}{2}}(\textbf{u})}{g\left( x_{n+\frac{1}{2}}\right) }\right) g\left( x_{n+\frac{1}{2}}\right) - \left( \frac{2}{h_{n}} + \frac{r_{n+\frac{1}{2}}^-(\textbf{u}) + s_{n+\frac{1}{2}}(\textbf{u})}{u_n}\right) u_{n}\right] . \end{aligned}$$

(69)

For the left boundary, we obtain similarly

$$\begin{aligned} \mathcal {F}_{\frac{1}{2}}(\textbf{u}) = \kappa _{\frac{1}{2}}\left[ \left( \frac{2}{h_{1}} + \frac{r_{\frac{1}{2}}^+(\textbf{u}) + s_{\frac{1}{2}}(\textbf{u})}{u_1}\right) u_1 - \left( \frac{2}{h_{1}} + \frac{r_{\frac{1}{2}}^-(\textbf{u}) + s_{\frac{1}{2}}(\textbf{u})}{g\left( x_{\frac{1}{2}}\right) }\right) g\left( x_{\frac{1}{2}}\right) \right] . \end{aligned}$$

(70)

B Exactness for polynomials of degree k

To simplify the calculation, let us take a polynomial of degree k centered on $x_{i+\frac{1}{2}}$ as an exact solution to demonstrate that the approximation of $\displaystyle \frac{{\text {d}}\bar{u}}{{\text {d}}x}(x_{i+\frac{1}{2}})$ is exact for polynomials of degree k. For

$$\begin{aligned} \bar{u}(x) = \sum _{p=0}^{k}a_{i+\frac{1}{2},p}\left( x-x_{i+\frac{1}{2}}\right) ^p; \end{aligned}$$

we obtain

$$\begin{aligned} \frac{{\text {d}}^{\ell }\bar{u}}{{\text {d}}x^{\ell }}(x) = \sum _{p=\ell }^{k}\frac{p!}{(p-\ell )!} a_{i+\frac{1}{2},p}\left( x-x_{i+\frac{1}{2}}\right) ^{p-\ell }, \end{aligned}$$

that is

$$\begin{aligned} \frac{{\text {d}}^{\ell }\bar{u}}{{\text {d}}x^{\ell }}\left( x_{i+\frac{1}{2}}\right) = \ell !a_{i+\frac{1}{2},\ell }. \end{aligned}$$

Besides, mean values were used to estimate the values of u at the centers of the cells, so

$$\begin{aligned} \begin{aligned} \bar{u}_{i+1}=\frac{1}{h_{i+1}}\int _{x_{i+\frac{1}{2}}}^{x_{i+\frac{3}{2}}}\sum _{p=0}^{k}a_{i+\frac{1}{2},p}\left( x-x_{i+\frac{1}{2}}\right) ^p=\sum _{p=0}^{k}a_{i+\frac{1}{2},p}\frac{h_{i+1}^{p}}{p+1}, \end{aligned} \end{aligned}$$

and

$$\begin{aligned} \begin{aligned} \bar{u}_{i}&= \frac{1}{h_{i}}\int _{x_{i-\frac{1}{2}}}^{x_{i+\frac{1}{2}}}\sum _{p=0}^{k}a_{i+\frac{1}{2},n}\left( x-x_{i+\frac{1}{2}}\right) ^{p}=\sum _{p=0}^{k}a_{i+\frac{1}{2},p}\frac{(-1)^{p} h_{i}^p}{p+1}. \end{aligned} \end{aligned}$$

The flux is

$$\begin{aligned} \mathcal {F}_{i+\frac{1}{2}}(\bar{\textbf{u}})=\frac{\kappa _{i+\frac{1}{2}}}{h_{i+\frac{1}{2}}}\left[ \bar{u}_{i+1}-\bar{u}_{i}- \sum _{p=2}^{k}\frac{h_{i+1}^{p}+(-1)^{p+1} h_{i}^{p}}{(p+1)!}\frac{{\text {d}}^pP}{{\text {d}}x^p}\left( x_{i+\frac{1}{2}}\right) \right] , \end{aligned}$$

where P is an interpolation polynomial of $\bar{u}$. Besides, $P = \bar{u}$ in that case, since $\bar{u}$ is a polynomial of degree k and P leaves invariant polynomials of degree k. The flux becomes

$$\begin{aligned} \mathcal {F}_{i+\frac{1}{2}}(\bar{\textbf{u}}) = \frac{\kappa _{i+\frac{1}{2}}}{h_{i+\frac{1}{2}}}\left( \left[ \sum _{p=0}^{k}a_{i+\frac{1}{2},p}\frac{h_{i+1}^{p}}{p+1}-\sum _{p=0}^{k}a_{i+\frac{1}{2},p}\frac{(-1)^{p} h_{i}^{p}}{p+1}\right] - \sum _{p=2}^{k} \frac{h_{i+1}^{p}+(-1)^{p+1} h_{i}^{p}}{(p+1)!}p!a_{i+\frac{1}{2},p}\right) , \end{aligned}$$

that is to say

$$\begin{aligned} \mathcal {F}_{i+\frac{1}{2}}(\bar{\textbf{u}}) = \kappa _{i+\frac{1}{2}}\left( a_{i+\frac{1}{2},1} + \sum _{p=2}^{k}a_{i+\frac{1}{2},p}\frac{h_{i+1}^{p}+(-1)^{p+1} h_{i}^{p}}{h_{i+\frac{1}{2}}(p+1)}- \sum _{p=2}^{k} \frac{h_{i+1}^{p}+(-1)^{p+1} h_{i}^{p}}{h_{i+\frac{1}{2}}(p+1)}a_{i+\frac{1}{2},p}\right) =\kappa _{i+\frac{1}{2}}a_{i+\frac{1}{2},1}. \end{aligned}$$

The flux is exact for polynomials of degree k.