A Numerical Comparison Between Degenerate Parabolic and Quasilinear Hyperbolic Models of Cell Movements Under Chemotaxis

Abstract

We consider two models which were both designed to describe the movement of eukaryotic cells responding to chemical signals. Besides a common standard parabolic equation for the diffusion of a chemoattractant, like chemokines or growth factors, the two models differ for the equations describing the movement of cells. The first model is based on a quasilinear hyperbolic system with damping, the other one on a degenerate parabolic equation. The two models have the same stationary solutions, which may contain some regions with vacuum. We first explain in details how to discretize the quasilinear hyperbolic system through an upwinding technique, which uses an adapted reconstruction, which is able to deal with the transitions to vacuum. Then we concentrate on the analysis of asymptotic preserving properties of the scheme towards a discretization of the parabolic equation, obtained in the large time and large damping limit, in order to present a numerical comparison between the asymptotic behavior of these two models. Finally we perform an accurate numerical comparison of the two models in the time asymptotic regime, which shows that the respective solutions have a quite different behavior for large times.

Appendix: Numerical Fluxes and Definition 1

Let us first recall the definition 1. A strongly consistent numerical flux $\displaystyle \mathcal {F}$ satisfies the two following conditions :

$$\begin{aligned} \left\{ \begin{array}{l} \hbox {if } \mathcal {F}^{\rho u}(r,0,R,0)=P(r), \hbox { then } r=R;\\ \\ \hbox {if } \mathcal {F}^{\rho u}(r,0,R,0)=P(R), \hbox { then } r=R.\\ \end{array}\right. \end{aligned}$$

(24)

We consider the following equations :

$$\begin{aligned} \mathcal {F}^{\rho u}(r_{i+1/2}^{n,-},0,r_{i+1/2}^{n,+},0)-\mathcal {F}^{\rho u}(r_{i-1/2}^{n,-},0,r_{i-1/2}^{n,+},0)= P\left( r_{i+1/2}^{n,-}\right) -P\left( r_{i-1/2}^{n,+}\right) , \end{aligned}$$

(25)

computed at the beginning of the proof of Theorem 1. In the following, we will prove that conditions (24) are necessary and sufficient conditions to ensure that the equalities $r_{i-1/2}^{n,-}=r_{i-1/2}^{n,+}$ for all $i$ are the unique solutions of equations (25). We will also show that the following classical fluxes : HLL, HLL-Roe and Suliciu relaxation flux adapted to vacuum are indeed strongly consistent fluxes.

Remark that conditions (24) have already been derived as necessary conditions on the flux in [9], where a sufficient condition on the flux is also given to ensure the uniqueness property for the solutions of equations (25). Since we are dealing here with the bounded domain case with boundary conditions (4), our computations are slightly different from the ones of [9] and we are able to prove that the conditions (24) are also sufficient conditions.

Indeed, considering equations (25) for all $i$, using that $r_{1/2}^{n,-}=r_{1/2}^{n,+}$ thanks to boundary conditions and using that the flux $\mathcal {F}$ is consistent, a straightforward induction implies that

$$\begin{aligned} \mathcal {F}^{\rho u}(r_{i+1/2}^{n,-},0,r_{i+1/2}^{n,+},0)= P\left( r_{i+1/2}^{n,-}\right) , \hbox { for all } i. \end{aligned}$$

(26)

Using condition (24), we obtain that $r_{i-1/2}^{n,-}=r_{i-1/2}^{n,+}$ for all $i$. Therefore, condition (24) is a necessary and sufficient condition to guarantee that the equalities $r_{i-1/2}^{n,-}=r_{i-1/2}^{n,+}$ for all $i$ are unique solutions of equations (25).

Now, let us show that HLL, HLL-Roe and Suliciu with vacuum fluxes satisfy conditions (24). We assume in the following that the functions $P$ and $P'$ are increasing, as satisfied by the pressure (2) we consider here.

1.1 HLL Flux

The definition of HLL flux is given at eq. (2.111) in Bouchut’s book [8] and we can compute

$$\begin{aligned} \mathcal {F}^{\rho u}(r,0,R,0)=\frac{c_{2}P(r)-c_{1}P(R)}{c_{2}-c_{1}}, \end{aligned}$$

with $\displaystyle c_{1}=\min (-\sqrt{P'(r)},-\sqrt{P'(R)} )$ and $\displaystyle c_{2}=\max (\sqrt{P'(r)},\sqrt{P'(R)} )$, that is to say

$$\begin{aligned} \mathcal {F}^{\rho u}(r,0,R,0)=\frac{P(r)+P(R)}{2}, \end{aligned}$$

which satisfies clearly conditions (24).

1.2 HLL-Roe Flux

In [18], we can find a version of the HLL flux adapted to vacuum. In that case,

$$\begin{aligned} \mathcal {F}^{\rho u}(r,0,R,0)=\frac{c_{2}P(r)-c_{1}P(R)}{c_{2}-c_{1}}, \end{aligned}$$

with $\displaystyle c_{1}\!=\!\min (-\sqrt{P'(r)},-\bar{c} )$ and $\displaystyle c_{2}\!=\!\max (\bar{c},\sqrt{P'(R)} )$, where $\displaystyle \bar{c}\!=\!\sqrt{\frac{\sqrt{R} P'(R)\!+\!\sqrt{r}P'(r)}{\sqrt{R}\!+\!\sqrt{r}}}$, that is to say

$$\begin{aligned} \mathcal {F}^{\rho u}(r,0,R,0)=\left\{ \begin{array}{ll} \displaystyle \frac{\sqrt{P'(R)} P(r)+\bar{c}P(R)}{\sqrt{P'(R)} +\bar{c}}, &{} \hbox { if } R>r, \\ \displaystyle \frac{\bar{c}P(r)+\sqrt{P'(r)}P(R)}{\bar{c}+\sqrt{P'(r)}}, &{} \hbox { if } r>R. \end{array}\right. \end{aligned}$$

From this expression, we conclude easily that HLL-Roe flux satisfies conditions (24).

1.3 Suliciu Flux Adapted to Vacuum

Now, we consider the Suliciu relaxation flux adapted to vacuum, which expression can be found in [8] at equations (2.133)–(2.136).

If $0<r<R$, a standard computation leads to

$$\begin{aligned} \mathcal {F}^{\rho u}(r,0,R,0)=\frac{c_{2}P(r)+c_{1}P(R)}{c_{1}+c_{2}}+\frac{Rc_{2}}{c_{1}+c_{2}}\times \frac{(P(r)-P(R))^2}{c_{2}(c_{1}+c_{2})+R(P(R)-P(r))}, \end{aligned}$$

with $\displaystyle c_{1}=r\sqrt{P'(r)}+\alpha r \left( \frac{P(R)-P(r)}{R\sqrt{P'(R)}}\right) >0$ and $\displaystyle c_{2}=R\sqrt{P'(R)}>0$.

If $r>R>0$, we obtain a similar formula, namely

$$\begin{aligned} \mathcal {F}^{\rho u}(r,0,R,0)=\frac{c_{2}P(r)+c_{1}P(R)}{c_{1}+c_{2}}+\frac{rc_{1}}{c_{1}+c_{2}}\times \frac{(P(r)-P(R))^2}{c_{1}(c_{1}+c_{2})+r(P(r)-P(R))}, \end{aligned}$$

with $\displaystyle c_{1}=r\sqrt{P'(r)}>0$ and $\displaystyle c_{2}=R\sqrt{P'(R)}+\alpha R \left( \frac{P(r)-P(R)}{r\sqrt{P'(r)}}\right) >0$.

Now, we consider the equation $\displaystyle \mathcal {F}^{\rho u}(r,0,R,0)=P(r)$. On the one hand, in the case $r<R$,

$$\begin{aligned} 0&= \mathcal {F}^{\rho u}(r,0,R,0)-P(r)\\&= \frac{c_{1}(P(R)-P(r))}{c_{1}+c_{2}}+\frac{Rc_{2}}{c_{1}+c_{2}}\times \frac{(P(r)-P(R))^2}{c_{2}(c_{1}+c_{2})+R(P(R)-P(r))}. \end{aligned}$$

Since the right-hand side of the last equation is the sum of two positive terms, it is straightforward that they are both null and that $P(R)=P(r)$, which leads to $R=r$. On the other hand, in the case $r>R$,

$$\begin{aligned} 0&= \mathcal {F}^{\rho u}(r,0,R,0)-P(r)\\&= \frac{c_{1}(P(R)-P(r))}{c_{1}+c_{2}}+\frac{rc_{1}}{c_{1}+c_{2}}\times \frac{(P(r)-P(R))^2}{c_{1}(c_{1}+c_{2})+r(P(r)-P(R))}. \end{aligned}$$

This equation can be simplified as :

$$\begin{aligned} c_{1}(c_{1}+c_{2})=0 \hbox { or }P(r)-P(R)=0. \end{aligned}$$

Since the first equality is impossible, we conclude that $r=R$.

Notice that the equation $\displaystyle \mathcal {F}^{\rho u}(r,0,R,0)=P(R)$ can be treated in a similar way. Therefore, we have proved that the Suliciu relaxation flux satisfies also the conditions (24).