Dynamics of the competition between two languages

Abstract

This paper reports a new kind of mathematical model for language competition dynamics using compartmental epidemiological modeling approach. The model describes the competition between two languages, say the predominant one called language 1, and the less spoken language 2, both in the same community. We distinguish three groups of population building the concerned community: one majority speaking language 1, one minority speaking language 2 and a last minority speaking neither language 1 nor language 2. The study of the proposed model includes an analysis of the evolution of the number of speakers over time. This model predicts that language 2 can inevitably disappear if the threshold parameter ${\mathcal {R}}_0$ is less than one. We show that our model is well-posed mathematically and linguistically. We also show that the model has basically two linguistic equilibrium points. A monolingual equilibrium point that corresponds to the case where only one language is spoken: some individuals speak only the language 1 and some others do speak neither language 1, nor language 2. A bilingual equilibrium point which corresponds to the case where all the languages are spoken: some individuals speak both languages 1 and 2, some others speak only language 1, while a third group does speak only language 2, and the last group speaks none of the two languages. In this bilingual equilibrium point, both languages coexist. Depending on the threshold parameter, we demonstrate the stability of these equilibria. The model presents on one hand, a direct bifurcation phenomenon, in which we have a stable equilibrium point without bilingual speakers when the associated basic reproduction number is less than one. On the other hand, it presents a stable bilingual equilibrium point when the number of associated basic reproductions is greater than one. The analysis of the overall sensitivity of the model is performed and the impact of the system parameters on the basic reproduction number was performed using sensitivity analysis to determine the impact of each parameter of the system on bilingualism. The numerical simulations carried out are in agreement with the presented theory

Appendix: Proof of proposition 4:

Let us proceed to the following substitutions:

$$\begin{aligned} B= & {} x_1; \qquad X=x_2; \qquad Y=x_3; \qquad Z=x_4; \qquad N=x_1+x_2+x_3+x_4\\ \lambda= & {} \displaystyle \frac{\beta (x_2 + \varepsilon x_3)}{N} \qquad \rho =\displaystyle \frac{\beta (x_4 + (1-\varepsilon ) x_3)}{N} \end{aligned}$$

Using the vector $x=(x_1, x_2, x_3, x_4)$, then we can write our model as $\dot{x}=f(x)$ where $f=(f_1,f_2,f_3,f_4)$ is defined below:

$$\begin{aligned} \left\{ \begin{array}{ccl} \dot{x}_1 &{}=&{}f_1 = \Lambda -(\rho +\lambda +\alpha +\mu )x_1,\\ \\ \dot{x}_2 &{}=&{}f_2 = \gamma x_3 - \displaystyle \frac{\beta x_2 x_4}{N} +(\lambda +\alpha )x_1 - \mu x_2,\\ \\ \dot{x}_3&{}=&{}f_3 = \displaystyle \frac{\beta x_2 x_4}{N} + (\lambda +\sigma )x_4 - (\gamma + \mu + \eta ) x_3,\\ \\ \dot{x}_4&{}=&{}f_4 =\rho x_1 + \eta x_3 - (\lambda +\sigma +\mu )x_4 ,\\ \end{array} \right. \end{aligned}$$

(26)

The bilingual-free equilibrium is $Q^0 = (B, X^0, Y^0, Z^0),$

where

$$\begin{aligned} B=\displaystyle \frac{2\Lambda }{\alpha +\mu +\beta +\sqrt{\Delta }} ,\qquad X^0=\displaystyle \frac{\Lambda (\alpha -\mu +\beta +\sqrt{\Delta })}{\mu (\alpha +\mu +\beta +\sqrt{\Delta })}, \qquad Y^0=0 \qquad \text{ and } \qquad Z^0=0. \end{aligned}$$

The Jacobian matrix of model system (26), applied to the bilingual-free equilibrium is given by:

$$\begin{aligned} J(Q^0)=\left( \begin{array}{cc} J_1 &{} J_2 \\ \\ 0 &{} J_3 \\ \end{array}\right) , \end{aligned}$$

where

$$\begin{aligned} \begin{array}{ll}J_{1}=\left( \begin{array}{cc} -\beta (x_{20})^2-(\alpha +\mu ) &{} -\beta (x_{10})^2 \\ \\ \beta (x_{20})^2+\alpha &{} -\mu +\beta (x_{10})^2 \\ \end{array}\right) ,\\ \\ J_{2}=\left( \begin{array}{cc} -\beta (x_{10})^2&{} -\beta (x_{10})^2\\ \\ -\beta x_{10}x_{20}+\beta \varepsilon x_{10} +\gamma &{} -\beta ( 1+ x_{10})x_{20}\\ \end{array}\right) , \qquad \text{ and }\\ \\ J_{3}=\left( \begin{array}{cc} -(\gamma +\mu +\eta )&{}\sigma +2\beta x_{20} \\ \\ \eta +\beta (1-\varepsilon )x_{10} &{} -(\mu +\sigma )- \beta ( x_{20}- x_{10}) \\ \end{array}\right) . \end{array} \end{aligned}$$

with

$$\begin{aligned} x_{10}=\displaystyle \frac{2\mu }{\alpha +\beta +\mu +\sqrt{\Delta }} \qquad and \qquad x_{20}=\displaystyle \frac{\alpha +\beta -\mu +\sqrt{\Delta }}{\alpha +\beta +\mu +\sqrt{\Delta }}. \end{aligned}$$

Here

$$\begin{aligned} x_{10}=\displaystyle \frac{B}{N}=b_0 ,\qquad x_{20}=\displaystyle \frac{X^0}{N}=x_0, \qquad x_{30}=\displaystyle \frac{Y^0}{N} \qquad \text{ and } \qquad x_{40}=\displaystyle \frac{Z^0}{N}. \end{aligned}$$

Since $\beta > \mu ,$ it implies $x_{20} > x_{10}$ and the basic reproduction number is the same as the original system. Consequently, the bifurcation parameter is chosen as $\eta .$ Resolving ${\mathcal {R}}_0=1,$ we find

$$\begin{aligned} \eta ^*=\displaystyle \frac{(\gamma +\mu )[\sigma +\mu +\beta (x_{20}-x_{10})]-\beta (1-\varepsilon )x_{10}[\sigma +2\beta x_{20}]}{[\sigma +2\beta x_{20}]-[\sigma +\mu +\beta (x_{20}-x_{10})]}. \end{aligned}$$

The Jacobian $J(Q^0)$ is evaluated at $\eta =\eta ^*.$ It admits 0 as eigenvalue. The others eigenvalues have real negative parts. Consequently, to prove proposition 4:, the center manifold theory can be used to analyze the dynamic of the system and determine if there is a possibility to find $Z\ne 0,$ when ${\mathcal {R}}_0\le 1$.

The theory of center manifold [2]

We consider a general Ordinary Differential Equations with a parameter $\Phi$ as:

$$\begin{aligned} \displaystyle \frac{dx}{dt}=f(x,\Phi ), \qquad \qquad f:{\mathcal {R}}^n \times {\mathcal {R}} \rightarrow {\mathcal {R}}^n \qquad and \qquad f\in {\mathcal {C}}^2({\mathcal {R}}^n, {\mathcal {R}}) \end{aligned}$$

It is assumed that 0 is an equilibrium point of the System, without loss of generality (i.e., f(0,$\Phi )=0$ for all values of the parameter $\Phi$), by using the following hypothesis:

• 1. $A=D_xf(0,0)=(\displaystyle \frac{\partial f_i}{\partial x_j}(0,0))$ is the linearized matrix of the system near point 0, the equilibrium point with $\Phi$ evaluated at 0. The simple eigenvalue of A is Zero and A has negative real parts for all other eigenvalues;

• 2. Matrix A has a nonnegative right eigenvector u and a left eigenvector v corresponding to the zero eigenvalue.

  Let $f_m$ be the $m^{th}$ component of f and

  $$\begin{aligned} a=\sum _{m,i,j=1}^{n}v_m u_i u_j\displaystyle \frac{\partial ^2 f_m}{\partial x_i \partial x_j}(0,0), \end{aligned}$$
  $$\begin{aligned} b=\sum _{i,k=1}^{n}v_m u_i\displaystyle \frac{\partial ^2 f_m}{\partial x_i \partial \phi }(0,0). \end{aligned}$$

  The local dynamics of the system near 0 are totally determined by a and b.

  • (a) $a>0$ and $b>0$ when $\Phi <0$ and $|\Phi |\ll 1$, 0 is locally asymptotically stable, and it reveals the existence of a positive unstable equilibrium; when $0<\Phi \ll 1$, 0 is unstable and we can notice the existence of a negative and locally asymptotically stable equilibrium;

  • (b) $a<0$ and $b<0$ when $\Phi <0$ and $|\Phi |\ll 1$, 0 is unstable; when $0<\Phi \ll 1$, 0 is locally asymptotically stable, and it reveals the existence of a positive unstable equilibrium;

  • (c) $a>0$ and $b<0$ when $\Phi <0$ and $|\Phi |\ll 1$, 0 is unstable, and there exists a locally asymptotically stable negative equilibrium; when $0<\Phi \ll 1$, 0 is stable, and it appears a positive unstable equilibrium;

  • (d) $a<0$ and $b>0$ When $\Phi$ moves from negative to positive, 0 moves its stability from stable to unstable. Accordingly a negative unstable equilibrium becomes positive and locally asymptotically stable.

To apply this theorem, the following calculations are necessary(here, we use $\eta ^*$ as parameter of bifurcation to the place of $\Phi$).

Eigenvectors of $J_0$. In the case where ${\mathcal {R}}_0=1$, we can show that the jacobian of the system has a nonnegative right eigenvector u and a left eigenvector v associated to the eigenvalue 0 and given by:

$$\begin{aligned} u_1=\Gamma _1u_3; \qquad u_2=\Gamma _2u_3; \qquad u_3>0; \qquad u_4=\Gamma _3u_3, \end{aligned}$$

and

$$\begin{aligned} v_1=0; \qquad v_2=0; \qquad v_3>0; \qquad v_4=\Gamma _4v_3 \end{aligned}$$

with

$$\begin{aligned} \Gamma _1= & {} \displaystyle \frac{x_{10}^2[\beta x_{10}^2+\beta (x_{20}-\varepsilon )-\mu +\gamma ]}{\mu (x_{20}^2-x_{10}^2)}+\displaystyle \frac{x_{10}^2[ (\beta x_{10}^2-\mu )x_{10}^2+ \beta (1+x_{10})x_{20}]\Gamma _3}{\mu (x_{20}^2-x_{10}^2)};\\ \qquad \Gamma _2= & {} +\displaystyle \frac{\beta (\beta x_{20}^2+\alpha )x_{10}^2+(\beta x_{20}^2+\alpha +\mu )[\beta (x_{20}-\varepsilon )x_{10}+\gamma ]}{\beta \mu (x_{20}^2-x_{10}^2)}\\&+\displaystyle \frac{[x_{10}^2(\beta x_{20}^2+\alpha )+(\beta x_{20}^2+\alpha +\mu )(1+x_{10})x_{20}]\Gamma _3}{\mu (x_{20}^2-x_{10}^2)}; \\ \qquad \Gamma _3= & {} \displaystyle \frac{\eta +\beta (1-\varepsilon )x_{10}}{\sigma +\mu + \beta (x_{20}-x_{10})}; \qquad {and} \qquad \Gamma _4=\displaystyle \frac{\sigma +2\beta x_{20}}{\sigma +\mu + \beta (x_{20}-x_{10})}. \end{aligned}$$

Algebraic calculations yield

$$\begin{aligned}&\displaystyle \frac{\partial ^2 f_3}{\partial x_1 \partial x_4}=-\displaystyle \frac{2\beta x_{20}}{(x_1^*+ x_2^*)}; \qquad \displaystyle \frac{\partial ^2 f_3}{\partial x_2 \partial x_4}=\displaystyle \frac{2\beta }{(x_1^*+ x_2^*)}-\displaystyle \frac{2\beta x_{20}}{(x_1^*+ x_2^*)};\\&\displaystyle \frac{\partial ^2 f_3}{\partial x_3 \partial x_4}=-\displaystyle \frac{2\beta x_{20}}{(x_1^*+ x_2^*)}+\displaystyle \frac{\beta \varepsilon }{(x_1^*+ x_2^*)}; \qquad \displaystyle \frac{\partial ^2 f_3}{\partial x_4^2}=-\displaystyle \frac{4\beta x_{20}}{(x_1^*+ x_2^*)};\\&\displaystyle \frac{\partial ^2 f_4}{\partial x_1 \partial x_3}=\displaystyle \frac{\beta (1-\varepsilon )}{(x_1^*+ x_2^*)}-\displaystyle \frac{\beta (1-\varepsilon )x_{10}}{(x_1^*+ x_2^*)};\\&\qquad \displaystyle \frac{\partial ^2 f_4}{\partial x_2 \partial x_3}=-\displaystyle \frac{\beta (1-\varepsilon )x_{10}}{(x_1^*+ x_2^*)}; \qquad \displaystyle \frac{\partial ^2 f_4}{\partial x_3^2}=-\displaystyle \frac{2\beta (1-\varepsilon )x_{10}}{(x_1^*+ x_2^*)};\\&\displaystyle \frac{\partial ^2 f_4}{\partial x_1 \partial x_4}=\displaystyle \frac{\beta }{(x_1^*+ x_2^*)}-\displaystyle \frac{\beta x_{10}}{(x_1^*+ x_2^*)}+\displaystyle \frac{\beta x_{20}}{(x_1^*+ x_2^*)};\\&\qquad \displaystyle \frac{\partial ^2 f_4}{\partial x_2 \partial x_4}=-\displaystyle \frac{\beta }{(x_1^*+ x_2^*)}-\displaystyle \frac{\beta x_{10}}{(x_1^*+ x_2^*)}+\displaystyle \frac{\beta x_{20}}{(x_1^*+ x_2^*)};\\&\qquad \displaystyle \frac{\partial ^2 f_4}{\partial x_3 \partial x_4}=-\displaystyle \frac{\beta \varepsilon }{(x_1^*+ x_2^*)}-\displaystyle \frac{\beta (2-\varepsilon ) x_{10}}{(x_1^*+ x_2^*)}+\displaystyle \frac{\beta x_{20}}{(x_1^*+ x_2^*)};\\&\qquad \displaystyle \frac{\partial ^2 f_4}{\partial x_4^2}=-\displaystyle \frac{2\beta x_{10}}{(x_1^*+ x_2^*)}+\displaystyle \frac{2\beta x_{20}}{(x_1^*+ x_2^*)}\\&\displaystyle \frac{\partial ^2 f_3}{\partial x_3 \partial \eta ^*}=-1, \qquad {and} \qquad \displaystyle \frac{\partial ^2 f_4}{\partial x_4 \partial \eta ^*}=1, \end{aligned}$$

where $x_1^*=B \qquad and \qquad x_2^*=X^0.$

The rest of the second derivatives for a and b are all zero. Hence,

$$\begin{aligned} a=-\displaystyle \frac{2v_3u_3^2}{x_1^*+x_2^*}[m_1\Gamma _4\beta (1-\varepsilon )x_{10}+ m_2\Gamma _3\Gamma _4\beta x_{10}+m_3\beta +m_4\Gamma _4\beta \varepsilon +m_5\beta x_{20}] \end{aligned}$$

where

$$\begin{aligned} m_1= & {} \Gamma _1+\Gamma _2+\Gamma _3+1; \qquad m_2=2\Gamma _2+\Gamma _3+1; \qquad m_3=\Gamma _1\Gamma _4+\Gamma _2\Gamma _3+0.5\varepsilon \Gamma _3;\\ m_4= & {} \Gamma _1+\Gamma _3; \qquad {and} \qquad m_5=2(\Gamma _1+\Gamma _2+\Gamma _3+1)\Gamma _3+(\Gamma _1+\Gamma _3+\Gamma _3^2)\Gamma _4. \end{aligned}$$

With this, we can notice that $a<0$ and

$$\begin{aligned} b=(\Gamma _4-1)v_3u_3>0 \qquad \text{ since } \qquad x_{20}+x_{10}>\displaystyle \frac{\mu }{\beta }. \end{aligned}$$