Uniform-in-time bounds for approximate solutions of the drift–diffusion system

Abstract

In this paper, we consider a numerical approximation of the Van Roosbroeck’s drift–diffusion system given by a backward Euler in time and finite volume in space discretization, with Scharfetter–Gummel fluxes. We first propose a proof of existence of a solution to the scheme which does not require any assumption on the time step. The result relies on the application of a topological degree argument which is based on the positivity and on uniform-in-time upper bounds of the approximate densities. Secondly, we establish uniform-in-time lower bounds satisfied by the approximate densities. These uniform-in-time upper and lower bounds ensure the exponential decay of the scheme towards the thermal equilibrium as shown in Bessemoulin-Chatard (Numer Math 25(3):147–168, 2016).

Appendix: Some technical results

In this Appendix, we detail some technical results which we use in the paper. There are first functional inequalities and then some properties of the numerical fluxes.

We define $x^+=\max (x,0)$ and $x^-=\min (x,0)$ for all $x\in {\mathbb {R}}$. Let us first recall some elementary inequalities:

$$\begin{aligned}&\forall x,y\in {\mathbb {R}},\forall q\ge 1,\quad x\left( (x^+)^q-(y^+)^q\right) \ge x^+\left( (x^+)^q-(y^+)^q\right) , \end{aligned}$$

(66)

$$\begin{aligned}&\forall x,y\ge 0,\forall \alpha ,\beta >0\quad (y^{\alpha }-x^{\alpha })(y^{\beta }-x^{\beta })\ge \frac{4\alpha \beta }{(\alpha +\beta )^2}\left( y^{\frac{\alpha +\beta }{2}}-x^{\frac{\alpha +\beta }{2}}\right) ^2, \end{aligned}$$

(67)

$$\begin{aligned}&\forall x,y\ge 0,\forall q\ge 2,\quad y^q-x^q\ge q x^{q-1}(y-x) \end{aligned}$$

(68)

For ${{\overline{m}}}\in {\mathbb {R}}$, we can define the function $w_{{{\overline{m}}}}$ by $w_{{{\overline{m}}}} (x)= (\log x+{{\overline{m}}})^-$ for all $x\in {\mathbb {R}}_+^*$. This function is widely used for the proof of the uniform-in-time positive lower bound in Sect. 3. We give in Lemma 3 some properties of the function $w_{{\overline{m}}}$.

Lemma 3

Let ${{\overline{m}}}\in {\mathbb {R}}$, the function $w_{{{\overline{m}}}}$ defined by $w_{{{\overline{m}}}} (x)= -(\log x+{{\overline{m}}})^-$ for all $x\in {\mathbb {R}}_+^*$ verifies the following inequalities:

• For all $q\ge 2$, for all $x,\, y>0$,

  $$\begin{aligned} -\frac{x-y}{x}\bigl (w_{{\overline{m}}}(x)\bigl )^{q-1}\ge \frac{1}{q}\Bigl (\bigl (w_{{\overline{m}}}(x)\bigl )^q-\bigl (w_{{\overline{m}}}(y)\bigl )^q\Bigl ). \end{aligned}$$

  (69)

• For all $q\ge 2$, for all $x,\, y>0$,

  $$\begin{aligned} x\left( \frac{\left( w_{{\overline{m}}}(y)\right) ^{q-1}}{y}-\frac{\left( w_{{\overline{m}}}(x)\right) ^{q-1}}{x}\right)&\ge \left( w_{{\overline{m}}}(y)\right) ^{q-1}-\left( w_{{\overline{m}}}(x)\right) ^{q-1} \nonumber \\&\quad +\frac{1}{q}\left( \left( w_{{\overline{m}}}(y)\right) ^{q}-\left( w_{{\overline{m}}}(x)\right) ^{q}\right) . \end{aligned}$$

  (70)

Proof

We start with the proof of (69). It is trivial when $x=y$. We consider the case where $x\ne y$. We remark that:

$$\begin{aligned}&-(x-y)\bigl (w_{{\overline{m}}}(x)\bigl )^{q-1} \\&\quad =-\displaystyle \frac{x-y}{\log x -\log y}\bigl (\log x-\log y\bigl )\bigl (w_{{\overline{m}}}(x)\bigl )^{q-1}\\&\quad =\displaystyle \frac{x-y}{\log x -\log y}\Bigl (-(\log x+{{\overline{m}}})^+ +w_{{\overline{m}}}(x)+(\log y+{{\overline{m}}})^+-w_{{\overline{m}}}(y)\Bigl )\bigl (w_{{\overline{m}}}(x)\bigl )^{q-1}. \end{aligned}$$

Therefore,

$$\begin{aligned} -(x-y)\bigl (w_{{\overline{m}}}(x)\bigl )^{q-1}\ge \displaystyle \frac{x-y}{\log x -\log y}\bigl (w_{{\overline{m}}}(x)-w_{{\overline{m}}}(y)\bigl )\bigl (w_{{\overline{m}}}(x)\bigl )^{q-1}. \end{aligned}$$

But, on one hand the function $w_{{\overline{m}}}$ is a nonincreasing function and on the other hand, we have:

$$\begin{aligned} \displaystyle \frac{1}{x}\frac{x-y}{\log x -\log y} > 1 \Longleftrightarrow 0<x< y. \end{aligned}$$

This yields

$$\begin{aligned} -\frac{x-y}{x}\bigl (w_{{\overline{m}}}(x)\bigl )^{q-1}\ge \bigl (w_{{\overline{m}}}(x)-w_{{\overline{m}}}(y)\bigl )\bigl (w_{{\overline{m}}}(x)\bigl )^{q-1}. \end{aligned}$$

The inequality (69) is then deduced from (68).

Let us now prove (70). We first remark that

$$\begin{aligned} x\left( \frac{\left( w_{{\overline{m}}}(y)\right) ^{q-1}}{y}-\frac{\left( w_{{\overline{m}}}(x) \right) ^{q-1}}{x}\right) = \left( w_{{\overline{m}}}(y)\right) ^{q-1}-\left( w_{{\overline{m}}}(x)\right) ^{q-1}+\left( w_{{\overline{m}}}(y) \right) ^{q-1}\left( \frac{x}{y}-1\right) . \end{aligned}$$

Thus, we just need to prove:

$$\begin{aligned} \left( w_{{\overline{m}}}(y)\right) ^{q-1}\left( \frac{x}{y}-1\right) \ge \frac{1}{q}\Bigl (\left( w_{{\overline{m}}}(y)\right) ^{q}-\left( w_{{\overline{m}}}(x)\right) ^{q}\Bigl ),\ \forall x,y>0. \end{aligned}$$

(71)

If $y\ge e^{-{{\overline{m}}}}$, $w_{{\overline{m}}}(y)=0$ and the result holds for all $x>0$. We consider now the case where $y<e^{-{{\overline{m}}}}$. As a direct consequence of (68), we get:

$$\begin{aligned} \Bigl (\bigl (w_{{\overline{m}}}(y)\bigl )^q-\bigl (w_{{\overline{m}}}(x))^q\bigl )\Bigl )\le q \bigl (w_{{\overline{m}}}(y)\bigl )^{q-1}\bigl (w_{{\overline{m}}}(y)-w_{{\overline{m}}}(x)\bigl ). \end{aligned}$$

But, for all $x>0$, we have:

$$\begin{aligned} w_{{\overline{m}}}(y)-w_{{\overline{m}}}(x)\le -\log y+\log x\le \displaystyle \frac{x-y}{y}, \end{aligned}$$

which yields (71) and therefore (70). $\square$

We now establish some properties satisfied by the numerical fluxes. Lemma 4 is crucial for the proof of Proposition 4, while Lemma 5 is used in the proof of Proposition 5.

Lemma 4

Let $q\ge 1$. The numerical fluxes defined by (15), (16) and (17) verify that for all $K\in {\mathcal {T}}$, for all $\sigma \in {\mathcal {E}}_K$, for all $n\ge 0$,

$$\begin{aligned} {{\mathcal {F}}}_{K,\sigma }^{n+1} D_{K,\sigma }(N_M^{n+1})^q&\le -\frac{4q}{(q+1)^2} \tau _{\sigma }B(D_{\sigma }\Psi ^{n+1}) \left( D_{K,\sigma }(N_M^{n+1})^{\frac{q+1}{2}}\right) ^2\nonumber \\&\quad +\frac{q}{q+1} \tau _{\sigma }D_{K,\sigma }\Psi ^{n+1}D_{K,\sigma }(N_M^{n+1})^{q+1} \nonumber \\&\quad +M\tau _{\sigma }D_{K,\sigma }\Psi ^{n+1}D_{K,\sigma }(N_M^{n+1})^q, \end{aligned}$$

(72a)

$$\begin{aligned} {{\mathcal {G}}}_{K,\sigma }^{n+1}D_{K,\sigma }(P_M^{n+1})^q&\le -\frac{4q}{(q+1)^2} \tau _{\sigma }B(D_{\sigma }\Psi ^{n+1}) \left( D_{K,\sigma }(P_M^{n+1})^{\frac{q+1}{2}}\right) ^2\nonumber \\&\quad -\frac{q}{q+1} \tau _{\sigma }D_{K,\sigma }\Psi ^{n+1}D_{K,\sigma }(P_M^{n+1})^{q+1} \nonumber \\&\quad -M\tau _{\sigma }D_{K,\sigma }\Psi ^{n+1}D_{K,\sigma }(P_M^{n+1})^q. \end{aligned}$$

(72b)

Proof

We prove only inequality (72a) since (72b) can be deduced by replacing $D_{K,\sigma }\Psi ^{n+1}$ by $- D_{K,\sigma }\Psi ^{n+1}$. Using the property $B(x)-B(-x)=-x$ satisfied by the function B, we can rewrite the fluxes ${\mathcal F}_{K,\sigma }^{n+1}$ under two different forms:

$$\begin{aligned} {\mathcal F}_{K,\sigma }^{n+1}&=\tau _{\sigma }\left( D_{K,\sigma }\Psi ^{n+1}N_{K}^{n+1} -B(D_{K,\sigma }\Psi ^{n+1})D_{K,\sigma }N^{n+1}\right) \end{aligned}$$

(73a)

$$\begin{aligned}&=\tau _{\sigma }\left( D_{K,\sigma }\Psi ^{n+1}N_{K,\sigma }^{n+1} -B(-D_{K,\sigma }\Psi ^{n+1})D_{K,\sigma }N^{n+1}\right) . \end{aligned}$$

(73b)

With the formulation (73a), we write:

$$\begin{aligned} {{\mathcal {F}}}_{K,\sigma }^{n+1} D_{K,\sigma }(N_M^{n+1})^q&=\tau _{\sigma }\left( D_{K,\sigma }\Psi ^{n+1}(N_{K}^{n+1}-M)D_{K,\sigma }(N_M^{n+1})^q\right. \\&\quad \qquad + MD_{K,\sigma }\Psi ^{n+1}D_{K,\sigma }(N_M^{n+1})^q \\&\quad \qquad \left. -B(D_{K,\sigma }\Psi ^{n+1})D_{K,\sigma }N^{n+1}D_{K,\sigma }(N_M^{n+1})^q\right) . \end{aligned}$$

But, using (66) and (67), we get:

$$\begin{aligned} (N_{K}^{n+1}-M)D_{K,\sigma }(N_M^{n+1})^q&\le \frac{q}{q+1} D_{K,\sigma }(N_M^{n+1})^{q+1},\\ DN_{K,\sigma }^{n+1}D_{K,\sigma }(N_M^{n+1})^q&\ge \frac{4q}{(q+1)^2} \left( D_{K,\sigma }(N_M^{n+1})^{\frac{q+1}{2}}\right) ^2. \end{aligned}$$

Moreover, B is a nonnegative function. Then, we deduce (72a) if $D_{K,\sigma }\Psi ^{n+1}\ge 0$. The same result is obtained when $D_{K,\sigma }\Psi ^{n+1}\le 0$ but starting with (73b) instead of (73a). $\square$

Lemma 5

Let $q\ge 2$. Let ${{\overline{m}}}\in {\mathbb {R}}$, we set $w_K^{n+1}=w_{{{\overline{m}}}}(N_K^{n+1})$ for all $K\in {\mathcal {T}}$, for all $n\ge 0$. The numerical fluxes defined by (15) and (17) verify that for all $K\in {\mathcal {T}}$, for all $\sigma \in {\mathcal {E}}_K$, for all $n\ge 0$,

$$\begin{aligned}&{{\mathcal {F}}}_{K,\sigma }^{n+1}D_{K,\sigma }\left( (w^{n+1})^{q-1}\frac{1}{N^{n+1}}\right) \nonumber \\&\quad \ge \tau _{\sigma }D_{K,\sigma }\Psi ^{n+1}\left( D_{K,\sigma }(w^{n+1})^{q-1} +\frac{1}{q}D_{K,\sigma }(w^{n+1})^{q}\right) \nonumber \\&\qquad +\tau _{\sigma }B\left( D_{\sigma }\Psi ^{n+1}\right) \left( \frac{4(q-1)}{q^{2}} \left( D_{\sigma }(w^{n+1})^{q/2}\right) ^{2} +\frac{1}{(q+1)^{2}}\left( D_{\sigma }(w^{n+1})^{\frac{q+1}{2}}\right) ^{2}\right) . \end{aligned}$$

(74)

Proof

We first assume that $D_{K,\sigma }\Psi ^{n+1}\ge 0$. Using formulation (73a), we write

$$\begin{aligned} {{\mathcal {F}}}_{K,\sigma }^{n+1}D_{K,\sigma }\left( (w^{n+1})^{q-1}\frac{1}{N^{n+1}}\right) =R_{1}+R_{2}, \end{aligned}$$

with

$$\begin{aligned} R_{1}&=-\tau _{\sigma }B\left( D_{\sigma }\Psi ^{n+1}\right) D_{K,\sigma }N^{n+1} D_{K,\sigma }\left( (w^{n+1})^{q-1}\frac{1}{N^{n+1}}\right) ,\\ R_{2}&=\tau _{\sigma }D_{K,\sigma }\Psi ^{n+1}N_K^{n+1}D_{K,\sigma }\left( (w^{n+1})^{q-1} \frac{1}{N^{n+1}}\right) . \end{aligned}$$

We treat $R_{1}$ following the same computations as those used in [17, proof of Theorem 4] for the diffusion term. More precisely, we have

$$\begin{aligned} R_{1}=\tau _{\sigma }B\left( D_{\sigma }\Psi ^{n+1}\right) (R_{11}+R_{12}), \end{aligned}$$

with

$$\begin{aligned} R_{11}&=-D_{K,\sigma }N^{n+1}\frac{1}{2}D_{K,\sigma }\left( (w^{n+1})^{q-1} \right) \left( \frac{1}{N_K^{n+1}}+\frac{1}{N_{K,\sigma }^{n+1}}\right) ,\\ R_{12}&=-D_{K,\sigma }N^{n+1}\frac{1}{2}D_{K,\sigma }\left( \frac{1}{N^{n+1}} \right) \left( (w_{K}^{n+1})^{q-1}+(w_{K,\sigma }^{n+1})^{q-1}\right) . \end{aligned}$$

According to [17, Lemma 7], we can rewrite $R_{11}$ and $R_{12}$ respectively as:

$$\begin{aligned} \begin{aligned} R_{11}&=-f\left( D_{K,\sigma }\left( \log N^{n+1}\right) \right) D_{K,\sigma } \left( \log N^{n+1}+{\overline{m}}\right) D_{K,\sigma }(w^{n+1})^{q-1},\\ R_{12}&=g\left( D_{K,\sigma }\left( \log N^{n+1}\right) \right) \left( D_{K,\sigma } \left( \log N^{n+1}+{\overline{m}}\right) \right) ^{2}\frac{1}{2} \left( (w_{K}^{n+1})^{q-1}+(w_{K,\sigma }^{n+1})^{q-1}\right) , \end{aligned} \end{aligned}$$

with

$$\begin{aligned} \begin{aligned} f(z)&=\frac{(e^{z}-1)(e^{-z}+1)}{2z}\ge 1,\quad \forall z\in {\mathbb {R}},\\ g(z)&=-\frac{(e^{z}-1)(e^{-z}-1)}{z^{2}}\ge 1,\quad \forall z\in {\mathbb {R}}. \end{aligned} \end{aligned}$$

Moreover, using the definition of $w_{{\mathcal {T}}}^{n+1}$ and (67) with $\alpha =1$ and $\beta =q-1$, we have:

$$\begin{aligned} -D_{K,\sigma }\left( \log N^{n+1}+{\overline{m}}\right) D_{K,\sigma }(w^{n+1})^{q-1}&=-D_{K,\sigma }\left( (\log N^{n+1}+{\overline{m}})^{+}\right) D_{K,\sigma }(w^{n+1})^{q-1}\\&\quad +D_{K,\sigma }(w^{n+1})D_{K,\sigma }(w^{n+1})^{q-1}\\&\ge \frac{4(q-1)}{q^{2}}\left( D_{K,\sigma }(w^{n+1})^{q/2}\right) ^{2}. \end{aligned}$$

Then we get

$$\begin{aligned} R_{11}\ge \frac{4(q-1)}{q^{2}}\left( D_{K,\sigma }(w^{n+1})^{q/2}\right) ^{2}. \end{aligned}$$

(75)

We also have

$$\begin{aligned} R_{12}\ge \frac{1}{2}\left( D_{K,\sigma }w^{n+1}\right) ^{2} \left( (w_{K}^{n+1})^{q-1}+(w_{K,\sigma }^{n+1})^{q-1}\right) , \end{aligned}$$

and since for all $x,\,y\ge 0$ we have, as shown in [17, Lemma 6],

$$\begin{aligned} (x-y)^{2}(x^{q-1}+y^{q-1})\ge \frac{2}{(q+1)^{2}}\left( x^{\frac{q+1}{2}}-y^{\frac{q+1}{2}}\right) ^{2}, \end{aligned}$$

which yields

$$\begin{aligned} R_{12}\ge \frac{1}{(q+1)^{2}}\left( D_{K,\sigma }(w^{n+1})^{\frac{q+1}{2}}\right) ^{2}. \end{aligned}$$

(76)

Gathering (75) and (76), we finally deduce that

$$\begin{aligned} R_{1}\ge & {} \tau _{\sigma }B\left( D_{\sigma }\Psi ^{n+1}\right) \nonumber \\&\left( \frac{4(q-1)}{q^{2}} \left( D_{K,\sigma }(w^{n+1})^{q/2}\right) ^{2}+ \frac{1}{(q+1)^{2}}\left( D_{K,\sigma }(w^{n+1})^{\frac{q+1}{2}}\right) ^{2}\right) . \end{aligned}$$

(77)

The term $R_{2}$ is now treated using (70) from Lemma 3 with $x=N_K^{n+1}$ and $y=N_{K,\sigma }^{n+1}$ (we still assume that $D_{K,\sigma }\Psi ^{n+1}\ge 0$). We get:

$$\begin{aligned} R_{2}\ge \tau _{\sigma }D_{K,\sigma }\Psi ^{n+1}\left( D_{K,\sigma }(w^{n+1})^{q-1} +\frac{1}{q}D_{K,\sigma }(w^{n+1})^{q}\right) . \end{aligned}$$

(78)

Gathering (77) and (78) yields the result if $D_{K,\sigma }\Psi ^{n+1}\ge 0$. The case $D_{K,\sigma }\Psi ^{n+1}\le 0$ can be treated exactly in the same way, starting from the expression (73b) of the flux instead of (73a). $\square$