Uniform-in-time convergence of numerical methods for non-linear degenerate parabolic equations

Abstract

Gradient schemes is a framework that enables the unified convergence analysis of many numerical methods for elliptic and parabolic partial differential equations: conforming and non-conforming finite element, mixed finite element and finite volume methods. We show here that this framework can be applied to a family of degenerate non-linear parabolic equations (which contain in particular the Richards’, Stefan’s and Leray–Lions’ models), and we prove a uniform-in-time strong-in-space convergence result for the gradient scheme approximations of these equations. In order to establish this convergence, we develop several discrete compactness tools for numerical approximations of parabolic models, including a discontinuous Ascoli–Arzelà theorem and a uniform-in-time weak-in-space discrete Aubin–Simon theorem. The model’s degeneracies, which occur both in the time and space derivatives, also requires us to develop a discrete compensated compactness result.

Appendix: Uniform-in-time compactness results for time-dependent problems

We establish in this appendix some generic results, unrelated to the framework of gradient schemes, that form the starting point for our uniform-in-time convergence results.

Solutions of numerical schemes for parabolic equations are usually piecewise constant, and therefore not continous, in time. As their jumps nevertheless tend to become small as the time step goes to 0, it is possible to establish uniform-in-time convergence properties using a generalisation to non-continuous functions of the classical Ascoli–Arzelà theorem.

Definition 6.1

If $(K,d_K)$ and $(E,d_E)$ are metric spaces, we denote by $\mathcal F(K,E)$ the space of functions $K\rightarrow E$ endowed with the uniform metric $d_\mathcal F(v,w)=\sup _{s\in K}d_E(v(s),w(s))$ (note that this metric may take infinite values).

Theorem 6.2

(Discontinuous Ascoli–Arzelà’s theorem) Let $(K,d_K)$ be a compact metric space, $(E,d_E)$ be a complete metric space and $(\mathcal F(K,E),d_{\mathcal F})$ be as in Definition 6.1.

Let $(v_m)_{m\in \mathbb N}$ be a sequence in $\mathcal F(K,E)$ such that there exists a function $\omega :K\times K\rightarrow [0,\infty ]$ and a sequence $(\delta _m)_{m\in \mathbb N} \subset [0,\infty )$ satisfying

$$\begin{aligned} \begin{array}{l} \displaystyle \lim _{d_K(s,s')\rightarrow 0}\omega (s,s')=0,\quad \lim _{m\rightarrow \infty }\delta _m=0,\\ \displaystyle \forall (s,s')\in K^2,\;\forall m\in \mathbb N,\; d_E(v_m(s),v_m(s'))\le \omega (s,s')+\delta _m. \end{array} \end{aligned}$$

(79)

We also assume that, for all $s\in K$, $\{v_m(s)\,:\,m\in \mathbb N\}$ is relatively compact in $(E,d_E)$.

Then $(v_m)_{m\in \mathbb N}$ is relatively compact in $(\mathcal F(K,E),d_{\mathcal F})$ and any adherence value of $(v_m)_{m\in \mathbb N}$ in this space is continuous $K\rightarrow E$.

Proof

Let us first notice that the last conclusion of the theorem, i.e. that any adherence value v of $(v_m)_{m\in \mathbb N}$ in $\mathcal F(K,E)$ is continuous, is trivially obtained by passing to the limit in (79), which shows that the modulus of continuity of v is bounded above by $\omega$.

The proof of the compactness result is an easy generalisation of the proof of the classical Ascoli–Arzelà theorem. We start by taking a countable dense subset $\{s_l\,:\,l\in \mathbb N\}$ in K (the existence of this set is ensured since K is compact metric). Since each set $\{v_m(s_l)\,:\,m\in \mathbb N\}$ is relatively compact in E, by diagonal extraction we can select a subsequence of $(v_m)_{m\in \mathbb N}$, denoted the same way, such that, for any $l\in \mathbb N$, $(v_m(s_l))_{m\in \mathbb N}$ converges in E. We then proceed to show that $(v_m)_{m\in \mathbb N}$ is a Cauchy sequence in $(\mathcal F(K,E),d_{\mathcal F})$. Since this space is complete, this will prove that this sequence converges in this space, which will complete the proof.

Let $\varepsilon >0$ and, using (79), take $\rho >0$ and $M\in \mathbb N$ such that $\omega (s,s')\le \varepsilon$ whenever $d_K(s,s')\le \rho$ and $\delta _m\le \varepsilon$ whenever $m\ge M$. Select a finite set $\{s_{l_1},\ldots ,s_{l_N}\}$ such that any $s\in K$ is within distance $\rho$ of a $s_{l_i}$. Then for any $m,m'\ge M$

$$\begin{aligned} d_E(v_m(s),v_{m'}(s))\le & {} d_E(v_m(s),v_m(s_{l_i}))+d_E(v_m(s_{l_i}),v_{m'}(s_{l_i})) +d_E(v_{m'}(s_{l_i}),v_{m'}(s))\\\le & {} \omega (s,s_{l_i})+\delta _m + d_E(v_m(s_{l_i}),v_{m'}(s_{l_i})) + \omega (s,s_{l_i})+\delta _{m'}\\\le & {} 4\varepsilon + d_E(v_m(s_{l_i}),v_{m'}(s_{l_i})). \end{aligned}$$

Since $\{(v_m(s_{l_i}))_{m\in \mathbb N}\,:\,i=1,\ldots ,N\}$ forms a finite number of converging sequences in E, we can find $M'\ge M$ such that, for all $m,m'\ge M'$ and all $i=1,\ldots ,N$, $d_E(v_m(s_{l_i}),v_{m'}(s_{l_i}))\le \varepsilon$. This shows that, for all $m,m'\ge M'$ and all $s\in K$, $d_E(v_m(s),v_{m'}(s))\le 5\varepsilon$ and concludes the proof that $(v_m)_{m\in \mathbb N}$ is a Cauchy sequence in $(\mathcal F(K,E),d_{\mathcal F})$. $\square$

Remark 6.3

Conditions (79) are usually the most practical when $(v_m)_{m\in \mathbb N}$ are piecewise constant-in-time solutions to numerical schemes (see e.g. the proof of Theorem 3.1). Here, $\omega$ is expected to measure the size of the cumulated jumps of $v_m$ between s and $s'$, and $\delta _m$ accounts for boundary effects which may occur in the small time intervals containing s and $s'$.

It is easy to see that (79) can be replaced with

$$\begin{aligned} d_E(v_m(s),v_m(s')) \rightarrow 0, \text{ as } m\rightarrow \infty \text{ and } d_K(s,s')\rightarrow 0 \end{aligned}$$

(80)

(under this condition, the proof can be carried out by selecting $M\in \mathbb N$ and $\rho >0$ such that $d_E(v_m(s),v_m(s'))\le \varepsilon$ whenever $m\ge M$ and $d_K(s,s')\le \rho$). It turns out that (80) is actually a necessary and sufficient condition for the theorem’s conclusions to hold true.

The following lemma states an equivalent condition for the uniform convergence of functions, which proves extremely useful to establish uniform-in-time convergence of numerical schemes for parabolic equations when no smoothness is assumed on the data.

Lemma 6.4

Let $(K,d_K)$ be a compact metric space, $(E,d_E)$ be a metric space and $(\mathcal F(K,E),d_{\mathcal F})$ as in Definition 6.1. Let $(v_m)_{m\in \mathbb N}$ be a sequence in $\mathcal F(K,E)$ and $v:K\mapsto E$ be continuous.

Then $v_m\rightarrow v$ for $d_{\mathcal F}$ if and only if, for any $s\in K$ and any sequence $(s_m)_{m\in \mathbb N}\subset K$ converging to s for $d_K$, we have $v_m(s_m)\rightarrow v(s)$ for $d_E$.

Proof

If $v_m\rightarrow v$ for $d_{\mathcal F}$ then for any sequence $(s_m)_{m\in \mathbb N}$ converging to s

$$\begin{aligned} d_E(v_m(s_m),v(s))\le & {} d_E(v_m(s_m),v(s_m))+d_E(v(s_m),v(s))\\\le & {} d_{\mathcal F}(v_m,v)+d_E(v(s_m),v(s)). \end{aligned}$$

The right-hand side tends to 0 by definition of $v_m\rightarrow v$ for $d_{\mathcal F}$ and by continuity of v, which shows that $v_m(s_m)\rightarrow v(s)$ for $d_E$.

Let us now prove the converse by contradiction. If $(v_m)_{m\in \mathbb N}$ does not converge to v for $d_{\mathcal F}$ then there exists $\varepsilon >0$ and a subsequence $(v_{m_k})_{k\in \mathbb N}$, such that, for any $k\in \mathbb N$, $\sup _{s\in K} d_E(v_{m_k}(s),v(s))\ge \varepsilon$. We can then find a sequence $(r_k)_{k\in \mathbb N}\subset K$ such that, for any $k\in \mathbb N$,

$$\begin{aligned} d_E(v_{m_k}(r_k),v(r_k))\ge \varepsilon /2. \end{aligned}$$

(81)

K being compact, up to another subsequence denoted the same way, we can assume that $r_k$ converges as $k\rightarrow \infty$ to some s in K. It is then trivial to construct a sequence $(s_m)_{m\in \mathbb N}$ converging to s and such that $s_{m_k}=r_k$ (just take $s_m=s$ when m is not an $m_k$). We then have $v_m(s_m)\rightarrow v(s)$ in E and, by continuity of v, $v(s_m)\rightarrow v(s)$ in E. This shows that $d_E(v_m(s_m),v(s_m))\rightarrow 0$, which contradicts (81) and concludes the proof. $\square$

The next result is classical. Its short proof is recalled for the reader’s convenience.

Proposition 6.5

Let E be a closed bounded ball in $L^2(\Omega )$ and let $(\varphi _l)_{l\in \mathbb N}$ be a dense sequence in $L^2(\Omega )$. Then, on E, the weak topology of $L^2(\Omega )$ is the topology given by the metric

$$\begin{aligned} d_E(v,w)=\sum _{l\in \mathbb N} \frac{\min (1,|\langle v-w,\varphi _l\rangle _{L^2(\Omega )}|)}{2^l}. \end{aligned}$$

(82)

Moreover, a sequence of functions $u_m:[0,T]\rightarrow E$ converges uniformly-in-time to $u:[0,T]\rightarrow E$ for the weak topology of $L^2(\Omega )$ (see Definition 2.11) if and only if, as $m\rightarrow \infty$, $d_E(u_m,u):[0,T]\rightarrow [0,\infty )$ converges uniformly to 0.

Proof

The sets $E_{\varphi ,\varepsilon }= \{v\in E\,:\,|\langle v,\varphi \rangle _{L^2(\Omega )}|<\varepsilon \}$, for $\varphi \in L^2(\Omega )$ and $\varepsilon >0$, define a basis of neighborhoods of 0 for the weak $L^2(\Omega )$ topology on E, and a basis of neighborhoods of any other point is obtained by translation. If R is the radius of the ball E then for any $\varphi \in L^2(\Omega )$, $l\in \mathbb N$ and $v\in E$ we have

$$\begin{aligned} |\langle v,\varphi \rangle _{L^2(\Omega )}|\le R||\varphi -\varphi _l||_{L^2(\Omega )} +|\langle v,\varphi _l\rangle _{L^2(\Omega )}|. \end{aligned}$$

By density of $(\varphi _l)_{l\in \mathbb N}$ we can select $l\in \mathbb N$ such that $||\varphi -\varphi _l||_{L^2(\Omega )}<\varepsilon /(2R)$ and we then see that $E_{\varphi _l,\varepsilon /2}\subset E_{\varphi ,\varepsilon }$. Hence, a basis of neighborhoods of 0 in E for the weak $L^2(\Omega )$ is also given by $(E_{\varphi _l,\varepsilon })_{l\in \mathbb N,\,\varepsilon >0}$.

From the definition of $d_E$ we see that, for any $l\in \mathbb N$, $\min (1,|\langle v,\varphi _l\rangle _{L^2(\Omega )}|)\le 2^l d_E(0,v)$. If $d_E(0,v)<2^{-l}$ this shows that $|\langle v,\varphi _l\rangle _{L^2(\Omega )}|\le 2^l d_E(0,v)$ and therefore that

$$\begin{aligned} B_{d_E}(0,\min (2^{-l},\varepsilon 2^{-l}))\subset E_{\varphi _l,\varepsilon }. \end{aligned}$$

Hence, any neighborhood of 0 in E for the $L^2(\Omega )$ weak topology is a neighborhood of 0 for $d_E$. Conversely, for any $\varepsilon >0$, selecting $N\in \mathbb N$ such that $\sum _{l\ge N+1}2^{-l}<\varepsilon /2$ gives, from the definition (82) of $d_E$,

$$\begin{aligned} \bigcap _{l=1}^N E_{\varphi _l,\varepsilon /4}\subset B_{d_E}(0,\varepsilon ). \end{aligned}$$

Hence, any ball for $d_E$ centered at 0 is a neighborhood of 0 for the $L^2(\Omega )$ weak topology. Since $d_E$ and the $L^2(\Omega )$ weak neighborhoods are invariant by translation, this concludes the proof that this weak topology is identical to the topology generated by $d_E$.

The conclusion on weak uniform convergence of sequences of functions follows from the preceding result, and more precisely by noticing that all previous inclusions are, when applied to $u_m(t)-u(t)$, uniform with respect to $t\in [0,T]$. $\square$

The following lemma has been initially established in [35, Proposition 9.3].

Lemma 6.6

Let $(t^{(n)})_{n\in \mathbb Z}$ be a stricly increasing sequence of real values such that ${\delta t}^{(n+{\frac{1}{2}})} := t^{(n+1)} - t^{(n)}$ is uniformly bounded by ${\delta t}>0$, $\displaystyle \lim \nolimits _{n\rightarrow -\infty } t^{(n)} = -\infty$ and $\displaystyle \lim \nolimits _{n\rightarrow \infty } t^{(n)} = \infty$. For all $t\in \mathbb R$, we denote by n(t) the element $n\in \mathbb Z$ such that $t\in (t^{(n)},t^{(n+1)}]$. Let $(a^{(n)})_{n\in \mathbb Z}$ be a family of non negative real numbers with a finite number of non zero values. Then

$$\begin{aligned} \int _{\mathbb R} \sum _{n=n(t)+1}^{n(t+\tau )} ({\delta t}^{(n+{\frac{1}{2}})} a^{(n+1)}) \mathrm{d}t = \tau \sum _{n\in \mathbb Z} ({\delta t}^{(n+{\frac{1}{2}})} a^{(n+1)}), \quad \forall \tau >0, \end{aligned}$$

(83)

and

$$\begin{aligned} \int _{\mathbb R} \left( \sum _{n=n(t)+1}^{n(t+\tau )}{\delta t}^{(n+{\frac{1}{2}})}\right) a^{n(t+s)+1} \mathrm{d}t \le (\tau + {\delta t}) \sum _{n\in \mathbb Z} {\delta t}^{(n+{\frac{1}{2}})} a^{(n+1)}, \quad \forall \tau >0, \ \forall s\in \mathbb R. \end{aligned}$$

(84)

Proof

Let us define $\chi$ by $\chi (t,n,\tau ) = 1$ if $t^{(n)}\in [t,t+\tau )$, otherwise $\chi (t,n,\tau ) = 0$. We have

$$\begin{aligned} \int _{\mathbb R} \sum _{n=n(t)+1}^{n(t+\tau )}({\delta t}^{(n+{\frac{1}{2}})} a^{(n+1)}) \mathrm{d}t&= \int _{\mathbb R} \sum _{n\in \mathbb Z} ({\delta t}^{(n+{\frac{1}{2}})} a^{(n+1)} \chi (t,n,\tau ))\mathrm{d}t\\&= \sum _{n\in \mathbb Z} \left( {\delta t}^{(n+{\frac{1}{2}})} a^{(n+1)} \int _{\mathbb R} \chi (t,n,\tau )\mathrm{d}t\right) . \end{aligned}$$

Since $\int _{\mathbb R} \chi (t,n,\tau )\mathrm{d}t = \int _{t^{(n)}-\tau }^{t^{(n)}} \mathrm{d}t = \tau$, Relation (83) is proved.

We now turn to the proof of (84). We define $\widetilde{\chi }$ by $\widetilde{\chi }(n,t) = 1$ if $n(t) = n$, otherwise $\widetilde{\chi }(n,t) = 0$. We have

$$\begin{aligned}&\int _{\mathbb R} \left( \sum _{n=n(t)+1}^{n(t+\tau )}{\delta t}^{(n+{\frac{1}{2}})}\right) a^{(n(t+s)+1)} \mathrm{d}t \\&\quad = \int _{\mathbb R} \left( \sum _{n=n(t)+1}^{n(t+\tau )}{\delta t}^{(n+{\frac{1}{2}})}\right) \sum _{m\in \mathbb Z}a^{(m+1)}\widetilde{\chi }(m,t+s) \mathrm{d}t, \end{aligned}$$

which yields

$$\begin{aligned} \int _{\mathbb R} \left( \sum _{n=n(t)+1}^{n(t+\tau )}{\delta t}^{(n+{\frac{1}{2}})}\right) a^{(n(t+s)+1)} \mathrm{d}t \!=\! \sum _{m\in \mathbb Z} a^{(m+1)} \int _{t^{(m)}-s}^{t^{(m+1)}-s} \!\left( \sum _{n=n(t)+1}^{n(t+\tau )}{\delta t}^{(n+{\frac{1}{2}})}\right) \mathrm{d}t. \end{aligned}$$

(85)

Since

$$\begin{aligned} \sum _{n=n(t)+1}^{n(t+\tau )}{\delta t}^{(n+{\frac{1}{2}})} = \sum _{n\in \mathbb Z,\ t\le t^{(n)} < t+\tau } (t^{(n+1)} - t^{(n)}) \le \tau + {\delta t}, \end{aligned}$$

we deduce from (85) that

$$\begin{aligned} \displaystyle \int _{\mathbb R} \left( \sum _{n=n(t)+1}^{n(t+\tau )}{\delta t}^{(n+{\frac{1}{2}})}\right) a^{(n(t+s)+1)} \mathrm{d}t&\le (\tau + {\delta t}) \sum _{m\in \mathbb Z} a^{(m+1)} \int _{t^{(m)}-s}^{t^{(m+1)}-s} \mathrm{d}t \\&= (\tau + {\delta t})\sum _{m\in \mathbb Z} a^{(m+1)} {\delta t}^{(m+{\frac{1}{2}})}, \end{aligned}$$

which is exactly (84). $\square$