A Generalized Framework for Direct Discontinuous Galerkin Methods for Nonlinear Diffusion Equations

Abstract

In this study, we propose a unified, general framework for the direct discontinuous Galerkin methods. In the new framework, the antiderivative of the nonlinear diffusion matrix is not needed. This allows a simple definition of the numerical flux, which can be used for general diffusion equations with no further modification. We also present the nonlinear stability analyses of the new direct discontinuous Galerkin methods and perform several numerical experiments to evaluate their performance. The numerical tests show that the symmetric and the interface correction versions of the method achieve optimal convergence and are superior to the nonsymmetric version, which demonstrates optimal convergence only for problems with diagonal diffusion matrices but loses order for even degree polynomials with a non-diagonal diffusion matrix. Singular or blow up solutions are also well captured with the new direct discontinuous Galerkin methods.

A. Important Inequalities

In this section, we discuss important inequalities used in the proofs of the stability analysis for symmetric DDG and DDGIC methods.

Lemma 6

(Young’s inequality) Suppose that $a,b \ge 0$, $1<p,q,\infty$, and that $\frac{1}{p} + \frac{1}{q}=1$. Then, we have that

$$\begin{aligned} ab\le \frac{a^p}{p} + \frac{b^q}{q}. \end{aligned}$$

A corollary to Lemma 6 can be obtained by considering $ab=(a\epsilon ^{1/p})(b/\epsilon ^{1/p})$ for $\epsilon >0$.

Corollary 7

Suppose that $a,b \ge 0$, $1<p,q,\infty$, and that $\frac{1}{p} + \frac{1}{q}=1$. Furthermore, if $\epsilon >0$, then

$$\begin{aligned} ab\le \frac{\epsilon a^p}{p} + \frac{b^q}{q\epsilon ^{q/p}}. \end{aligned}$$

We will also recall the following lemma due to Ern and Guermord [50]:

Lemma 8

Let K be an element in the mesh partition ${\mathcal {T}}_h$ of the computational domain $\Omega \subset {\mathbb {R}}^d$, $v\in {\mathbb {V}}_h^k$ and $l\in {\mathbb {N}}$. There exists a constant $C>0$ for any non-negative integer $m \le l$ such that

$$\begin{aligned} \left| v \right| _{W^{l,p}(K)} \le Ch_K^{m-l+d\left( \frac{1}{p}-\frac{1}{r}\right) }\left| v \right| _{W^{m,r}(K)}, \quad \forall p,r\in [1,\infty ]. \end{aligned}$$

Proof

See the proof of Lemma 12.1 in [50]. $\square$

Next, we will derive a series of essential inequalities used in the proof of the stability results for symmetric DDG and DDGIC methods.

Lemma 9

Assume that $A(u)\in {\mathbb {R}}^{2\times 2}$ is positive definite with positive eigenvalues and there exist $\gamma ,\gamma ^*\in {\mathbb {R}}^{+}$ such that the eigenvalues $({\gamma _1(u),\gamma _2(u)})$ of A(u) lie between $[\gamma ,\gamma ^*]$ for $\forall u\in {\mathbb {R}}$. If $\varvec{\xi }(u)$ is given as in Eq. (7), then $\forall {\textbf{x}}\in {\mathbb {R}}^2$ there holds

$$\begin{aligned} \left| \varvec{\xi }(u)\cdot {\textbf{x}}\right| \le \gamma ^*\left\| {\textbf{x}}\right\| , \end{aligned}$$

where we denote by $\left\| \cdot \right\|$ the Euclidean norm in ${\mathbb {R}}^2$.

Proof

Using the Scharwz inequality, we have

$$\begin{aligned} \left| \varvec{\xi }(u)\cdot {\textbf{x}}\right| \le \left\| \varvec{\xi }(u) \right\| \left\| {\textbf{x}} \right\| . \end{aligned}$$

Let ${\textbf{e}}_1,{\textbf{e}}_2$ be the orthonormal eigenvectors corresponding to the eigenvalues $\gamma \text { and } \gamma ^*$, respectively. Since A(u) is positive-definite, its eigenvectors form a basis of ${\mathbb {R}}^2$. Then, the unit normal vector can be written as ${\textbf{n}}=n_1 {\textbf{e}}_1 + n_2 {\textbf{e}}_2$, and then, it follows that

and the conclusion holds. $\square$

Lemma 10

Suppose that $A(u)\in {\mathbb {R}}^{2\times 2}$ is positive definite with positive eigenvalues and there exist $\gamma ,\gamma ^*\in {\mathbb {R}}^{+}$ such that the eigenvalues $({\gamma _1(u),\gamma _2(u)})$ of A(u) lie between $[\gamma ,\gamma ^*]$ for $\forall u\in {\mathbb {R}}$. If $\beta _0 \ge 0$, then we have that

Proof

Recall the definition of the new direction vector . Then,

Since , the conclusion follows. $\square$

Lemma 11

Suppose that $A(u)\in {\mathbb {R}}^{2\times 2}$ is positive definite with positive eigenvalues and there exist $\gamma ,\gamma ^*\in {\mathbb {R}}^{+}$ such that the eigenvalues $({\gamma _1(u),\gamma _2(u)})$ of A(u) lie between $[\gamma ,\gamma ^*]$ for $\forall u\in {\mathbb {V}}_h^k$. If $\beta _0 \ge 0$, then there exists a constant $C>0$ such that

Furthermore, under the same assumptions, there also holds

Proof

Note that . Then, we have

where we have invoked Lemma 9 in the last step. Moreover, by Corollary 7 with $\epsilon =\frac{\gamma \beta _0}{\gamma ^*h}$, we obtain

$$\begin{aligned} \gamma ^*\int _{e} \left| \llbracket u \rrbracket \right| \, \left\| (\nabla u)^\pm \right\| ~ds \le \frac{\gamma \beta _0}{2h}\left\| \llbracket u \rrbracket \right\| ^2_{L^2(e)} + \frac{(\gamma ^*)^2h}{2\gamma \beta _0}\left\| (\nabla u)^\pm \right\| ^2_{L^2(e)}. \end{aligned}$$

(29)

So that we have

(30)

Note that the last two terms above are simply summations over individual edges in the triangulation ${\mathcal {T}}_h$, and each summation is responsible for accumulating $\left\| (\nabla u)^\pm \right\| ^2_{L^2(e)}$ only from one side of the edge. In the global sense, these summations accumulate $\left\| (\nabla u)^{interior} \right\| ^2_{L^2(\partial K)}$ for each cell K in the domain. Thus, they can be converted into a single summation over cells. That is,

$$\begin{aligned} \sum _{e\in {\mathcal {E}}_h} \frac{(\gamma ^*)^2h}{4\gamma \beta _0}\left\| (\nabla u)^+ \right\| ^2_{L^2(e)}+\sum _{e\in {\mathcal {E}}_h} \frac{(\gamma ^*)^2h}{4\gamma \beta _0}\left\| (\nabla u)^- \right\| ^2_{L^2(e)} = \sum _{K\in {\mathcal {T}}_h} \frac{(\gamma ^*)^2h}{4\gamma \beta _0}\left\| \nabla u \right\| ^2_{L^2(\partial K)}.\nonumber \\ \end{aligned}$$

(31)

This expression is useful since we can invoke the trace inequality

$$\begin{aligned} \sum _{K\in {\mathcal {T}}_h} \frac{(\gamma ^*)^2h}{4\gamma \beta _0}\left\| \nabla u \right\| ^2_{L^2(\partial K)} \le \sum _{K\in {\mathcal {T}}_h} C\frac{(\gamma ^*k)^2}{4\gamma \beta _0}\left\| \nabla u \right\| ^2_{L^2(K)}, \end{aligned}$$

(32)

for some constant $C>0$. Thus, substituting Eqs. (31) and (32) into Eq. (30) completes the first part of the proof. The second part follows after following the same steps but using $\epsilon =\frac{\gamma \beta _0}{2\gamma ^*h}$ in Eq. (29) instead. $\square$

Lemma 12

Suppose that $A(u)\in {\mathbb {R}}^{2\times 2}$ is positive definite and there exist $\gamma ,\gamma ^*\in {\mathbb {R}}^{+}$ such that the eigenvalues $({\gamma _1(u),\gamma _2(u)})$ of A(u) lie between $[\gamma ,\gamma ^*]$ for $\forall u\in {\mathbb {V}}_h^k$. If $\beta _0 \ge 0$, then there exists a constant $C>0$ such that

Proof

By convention, the outward unit normal vector ${\textbf{n}}$ is understood as ${\textbf{n}}={\textbf{n}}^+$. Also, it can be understood in terms of the inward unit normal vector as ${\textbf{n}}=-{\textbf{n}}^-$. Therefore, the jump term for the second derivatives can be rewritten as

$$\begin{aligned} \llbracket \nabla (\nabla u\cdot {\textbf{n}})\rrbracket = \left( \nabla (\nabla u\cdot {\textbf{n}})\right) ^+ + \left( \nabla (\nabla u\cdot {\textbf{n}})\right) ^-. \end{aligned}$$

With this understanding, we have that

Note that we have invoked Lemma 9 and triangle inequality in the last step. Furthermore, by Corollary 7 with $\epsilon =\frac{\gamma \beta _0}{2\gamma ^*\beta _1h^2}$, we obtain

$$\begin{aligned} \gamma ^*\int _{e} \beta _1 h \left| \llbracket u \rrbracket \right| \left\| (\nabla (\nabla u\cdot {\textbf{n}}) )^\pm \right\| ds \le \frac{\gamma \beta _0}{4h}\left\| \llbracket u \rrbracket \right\| ^2_{L^2(e)} + \frac{(\gamma ^*\beta _1)^2h^3}{\gamma \beta _0} \left\| (\nabla (\nabla u\cdot {\textbf{n}}) )^\pm \right\| ^2_{L^2(e)}. \end{aligned}$$

Thus,

(33)

As in the proof of Lemma 11, we convert the summations over edges to a summation over cells:

$$\begin{aligned} \begin{aligned}&\sum _{e\in {\mathcal {E}}_h}\frac{(\gamma ^*\beta _1)^2h^3}{\gamma \beta _0} \left\| (\nabla (\nabla u\cdot {\textbf{n}}) )^+ \right\| ^2_{L^2(e)} + \sum _{e\in {\mathcal {E}}_h}\frac{(\gamma ^*\beta _1)^2h^3}{\gamma \beta _0} \left\| (\nabla (\nabla u\cdot {\textbf{n}}) )^- \right\| ^2_{L^2(e)} \\&\quad = \sum _{K\in {\mathcal {T}}_h} \frac{(\gamma ^*\beta _1)^2h^3}{\gamma \beta _0} \left\| \nabla (\nabla u\cdot {\textbf{n}})\right\| ^2_{L^2(\partial K)}. \end{aligned} \end{aligned}$$

(34)

At this point, it might be tempting to invoke the trace theorem for the norm on the right-hand side of the above equation. However, a more useful inequality can be obtained by considering the Euclidean norm $\left\| \nabla (\nabla u\cdot {\textbf{n}}) \right\| ^2$. We first note that

$$\begin{aligned} \begin{aligned} \left\| \nabla (\nabla u\cdot {\textbf{n}}) \right\| ^2&=(u_{xx}n_1 + u_{yx}n_2)^2 + (u_{xy}n_1 + u_{yy}n_2)^2 \\&= (u_{xx}^2+u_{xy})^2n_1^2 + (u_{yx}^2+u_{yy})^2n_2^2 + 2n_1n_2(u_{xx}u_{yx} + u_{xy}u_{yy}). \end{aligned} \end{aligned}$$

For the cross-product term above, we invoke Lemma 6 with $p=q=2$

$$\begin{aligned} \begin{aligned} 2n_1n_2(u_{xx}u_{yx} + u_{xy}u_{yy})&= 2\left( (n_2u_{xx})(n_1u_{yx})+(n_2u_{xy})(n_1u_{yy})\right) \\&\le (u_{xx}^2+u_{xy})^2n_2^2 + (u_{yx}^2+u_{yy})^2n_1^2. \end{aligned} \end{aligned}$$

Since $n_1^2+n_2^2=1$, we obtain

$$\begin{aligned} \left\| \nabla (\nabla u\cdot {\textbf{n}}) \right\| ^2 \le u_{xx}^2 + u_{xy}^2 + u_{yx}^2 + u_{yy}^2. \end{aligned}$$

Using this and the trace inequality gives

$$\begin{aligned} \begin{aligned} \left\| \nabla (\nabla u\cdot {\textbf{n}})\right\| ^2_{L^2(\partial K)}&= \int _{\partial K} \left\| \nabla (\nabla u\cdot {\textbf{n}}) \right\| ^2 ~ds \\&\le \int _{\partial K} \left( u_{xx}^2 + u_{xy}^2 + u_{yx}^2 + u_{yy}^2\right) ~ds \\&\le C \frac{k^2}{h} \int _{K} \left( u_{xx}^2 + u_{xy}^2 + u_{yx}^2 + u_{yy}^2\right) ~dxdy = C \frac{k^2}{h} \left| u \right| ^2_{H^2(K)}. \end{aligned} \end{aligned}$$

By Lemma 8 with $d=l=p=r=2$ and $m=1$, we have

$$\begin{aligned} \left| u \right| _{H^2(K)} \le \frac{C}{h} \left| u \right| _{H^1(K)}=\frac{C}{h} \left\| \nabla u \right\| _{L^2(K)}, \end{aligned}$$

which leads to

$$\begin{aligned} \left\| \nabla (\nabla u\cdot {\textbf{n}})\right\| ^2_{L^2(\partial K)} \le C \frac{k^2}{h^3} \left\| \nabla u \right\| ^2_{L^2(K)}. \end{aligned}$$

(35)

Finally, substituting Eqs. (34) and (35) in Eq. (33) leads to the desired result. $\square$