# A Nonlinear ODE Model for a Consumeristic Society

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## Abstract

**:**

## 1. Introduction

## 2. General Results on the Solutions

**Proposition 1.**

**Proof.**

**Proposition 2.**

**Proof.**

**Proposition 3.**

**Proof.**

**Proposition 4.**

**Proof.**

- Suppose $w\left(t\right)\ge \frac{\delta y\left(t\right)+\delta z\left(t\right)}{\sigma}$ for all $t\ge T$. In this case, we have$${w}^{\prime}=m(x,w)(\delta y+\delta z-\sigma w)\le 0,$$
- Suppose now that there exists ${t}_{1}\ge T$ such that $w\left({t}_{1}\right)<\frac{\delta y\left({t}_{1}\right)+\delta z\left({t}_{1}\right)}{\sigma}$. Let $t>{t}_{1}$. If $w\left(t\right)\le \frac{\delta y\left(t\right)+\delta z\left(t\right)}{\sigma}$, then $w\left(t\right)<1$. Otherwise, if $w\left(t\right)>\frac{\delta y\left(t\right)+\delta z\left(t\right)}{\sigma}$, define$${t}_{2}=inf\left\{\tau \in ({t}_{1},t)\phantom{\rule{0.277778em}{0ex}}|\phantom{\rule{0.277778em}{0ex}}w\left(s\right)>\frac{\delta y\left(s\right)+\delta z\left(s\right)}{\sigma},\phantom{\rule{0.277778em}{0ex}}\forall s\in (\tau ,t)\right\}.$$We have that ${t}_{2}>{t}_{1}$. Because of continuity we also have $w\left(s\right)>\frac{\delta y\left(s\right)+\delta z\left(s\right)}{\sigma}\phantom{\rule{0.277778em}{0ex}}\forall s\in ({t}_{2},t)$ and $w\left({t}_{2}\right)=\frac{\delta y\left({t}_{2}\right)+\delta z\left({t}_{2}\right)}{\sigma}<1$. Hence we have ${w}^{\prime}\left(s\right)<0$ in $({t}_{2},t)$ implying that $w\left(s\right)$ is decreasing in $({t}_{2},t)$ and thus $w\left(t\right)\le w\left({t}_{2}\right)<1$. In conclusion, $w\left(t\right)<1$ for every $t\ge {t}_{1}$, and therefore w is bounded on $[0,+\infty )$.

**Proposition 5.**

**Proof.**

**Proposition 6.**

**Proof.**

## 3. Steady States If $\mathit{x}\le \mathit{w}$

**(i)**- $w=0$. In this case we obtain $x=0$, and from (4) we have two possibilities: either $y=0$ or $y=\lambda $. There is no constraint on z, so we have two families of critical points:$$\left\{(0,0,Z,0),\phantom{\rule{0.277778em}{0ex}}Z\ge 0\right\}.$$$$\left\{(0,\lambda ,Z,0),\phantom{\rule{0.277778em}{0ex}}Z\ge 0\right\}.$$We remark that the points (7) correspond to what is called “desert state” in [6], while the points (8) correspond to what is called “nature state” in the same paper. We also notice that these points are not in ${\mathcal{C}}_{0}$, and the vector field F is not defined on them. However, it is interesting to study what we can improperly call their stability, that is to know whether the trajectories starting close to them are attracted or repelled by them, because this says to us if a society with declining population and wealth can recover or not.
**(ii)**- $w\ne 0$. In this case we have$$\begin{array}{cc}\hfill \left(3\right)\Rightarrow x=0,\phantom{\rule{1.em}{0ex}}& \mathrm{or}\phantom{\rule{1.em}{0ex}}c=d\frac{\mu xw}{{w}^{2}+{\mu}^{2}{x}^{2}},\hfill \\ \hfill \left(4\right)\Rightarrow y=0,\phantom{\rule{1.em}{0ex}}& \mathrm{or}\phantom{\rule{1.em}{0ex}}\gamma \lambda -\gamma y-\delta w=0,\hfill \\ \hfill \left(5\right)\Rightarrow z=0,\phantom{\rule{1.em}{0ex}}& \mathrm{or}\phantom{\rule{1.em}{0ex}}\delta {w}^{2}-kw+\delta =0,\hfill \end{array}$$We note that, if $x\ne 0$ from (3) we have$$d\frac{\mu xw}{{w}^{2}+{\mu}^{2}{x}^{2}}=c\phantom{\rule{0.277778em}{0ex}}\Rightarrow \phantom{\rule{0.277778em}{0ex}}d\frac{\frac{w}{\mu x}}{{\left(\frac{w}{\mu x}\right)}^{2}+1}=c\phantom{\rule{0.277778em}{0ex}}\Rightarrow \phantom{\rule{0.277778em}{0ex}}d\frac{t}{{t}^{2}+1}=c\phantom{\rule{0.277778em}{0ex}}\Rightarrow \phantom{\rule{0.277778em}{0ex}}c{t}^{2}-dt+c=0,$$$$\begin{array}{cc}\hfill {\displaystyle {t}_{1}}& =\frac{d-\sqrt{{d}^{2}-4{c}^{2}}}{2c}\in (0,1),\hfill \\ \hfill {\displaystyle {t}_{2}}& =\frac{d+\sqrt{{d}^{2}-4{c}^{2}}}{2c}>1.\hfill \end{array}$$Notice that the hypothesis $d>2c$ rules out the case ${t}_{1}={t}_{2}=1$.We then obtain two possible values for x given by$${x}_{1}=\frac{w}{\mu {t}_{1}},\phantom{\rule{1.em}{0ex}}{x}_{2}=\frac{w}{\mu {t}_{2}},$$If $y\ne 0$ then, from (4), $\gamma \lambda -\gamma y-\delta w=0$, i.e.,$$y=\lambda -\frac{\delta}{\gamma}w.$$Finally, if $z\ne 0$ then$${w}_{1}=\frac{k-\sqrt{{k}^{2}-4{\delta}^{2}}}{2\delta},\phantom{\rule{1.em}{0ex}}{w}_{2}=\frac{k+\sqrt{{k}^{2}-4{\delta}^{2}}}{2\delta},\phantom{\rule{0.277778em}{0ex}}\mathrm{with}\phantom{\rule{0.277778em}{0ex}}k\ge 2\delta .$$We can now write down a complete list of equilibrium points, in the half-space $x\le w$, when $w>0$.
**(ii.1)**- First, we observe that if $x=y=z=0$, then from (6) we obtain $w=0$, hence here we exclude this case, because we have already dealt with it (see above the case $w=0$).
**(ii.2)**- If $x=y=0$ and $z\ne 0$ from (5) we have $w={w}_{1,2}$ and from (6) $z}_{1,2}=\frac{\sigma}{\delta}{w}_{1,2$. Therefore, we have the two equilibrium points$${P}_{i}=\left(0,0,\frac{\sigma}{\delta}{w}_{i},{w}_{i}\right),\phantom{\rule{0.277778em}{0ex}}i=1,2,\phantom{\rule{0.277778em}{0ex}}k\ge 2\delta ,$$
**(ii.3)**- If $x=z=0,\phantom{\rule{0.277778em}{0ex}}y=\lambda -\frac{\delta}{\gamma}w$, then from (6) we have $w=\frac{\delta}{\sigma}y$ so that$$w=\frac{\gamma \lambda \delta}{\gamma \sigma +{\delta}^{2}},\phantom{\rule{1.em}{0ex}}y=\frac{\gamma \lambda \sigma}{\gamma \sigma +{\delta}^{2}},$$$$P=\left(0,\frac{\gamma \lambda \sigma}{\gamma \sigma +{\delta}^{2}},0,\frac{\gamma \lambda \delta}{\gamma \sigma +{\delta}^{2}}\right).$$
**(ii.4)**- If $x=0,\phantom{\rule{0.277778em}{0ex}}y=\lambda -\frac{\delta}{\gamma}w$, $z\ne 0$, we obtain $w={w}_{1,2}$, hence from (6) we deduce$$z=-\lambda +\left(\frac{\delta}{\gamma}+\frac{\sigma}{\delta}\right)w.$$We then obtain two equilibrium points$${P}_{i}=\left(0,\lambda -\frac{\delta}{\gamma}{w}_{i},\phantom{\rule{0.166667em}{0ex}}-\lambda +\left(\frac{\delta}{\gamma}+\frac{\sigma}{\delta}\right){w}_{i},{w}_{i}\right),\phantom{\rule{0.277778em}{0ex}}i=1,2,\phantom{\rule{0.277778em}{0ex}}k\ge 2\delta .$$
**(ii.5)**- If $x>0$ and $y=z=0$, from (6) we obtain $w=0$ and we exclude this case because we are assuming $x\le w$.
**(ii.6)**- If $x>0$, $y=0$ and $z>0$, from (5) we obtain $w={w}_{1,2}$, with $k\ge 2\delta $. For x we obtain (10) with $d\ge 2c$, and from (6) $z=\frac{\sigma w}{\delta}$. We have then 4 equilibrium points$${P}_{i,j}=\left({\displaystyle \frac{{w}_{j}}{\mu {t}_{i}},0,\frac{\sigma {w}_{j}}{\delta},{w}_{j}}\right),\phantom{\rule{0.277778em}{0ex}}i,j=1,2,\phantom{\rule{0.277778em}{0ex}}k\ge 2\delta ,\phantom{\rule{0.277778em}{0ex}}d\ge 2c.$$
**(ii.7)**- If $x>0$, $y>0$, $z=0$, from (6) we deduce for y and w the same values as in (14). Also, x is as in (10), and we obtain the two equilibrium points$${P}_{i}=\left(\frac{\delta \gamma \lambda}{\mu {t}_{i}(\gamma \sigma +{\delta}^{2})},\frac{\gamma \lambda \sigma}{\gamma \sigma +{\delta}^{2}},0,\frac{\delta \gamma \lambda}{\gamma \sigma +{\delta}^{2}}\right),\phantom{\rule{0.277778em}{0ex}}i=1,2,\phantom{\rule{0.277778em}{0ex}}d\ge 2c.$$
**(ii.8)**- If $x>0,\phantom{\rule{0.277778em}{0ex}}y>0,\phantom{\rule{0.277778em}{0ex}}z>0$, then $\lambda -\frac{\delta}{\gamma}w$ and arguing as in [ii.7] we obtain 4 equilibrium points$${P}_{i,j}=\left(\frac{{w}_{j}}{\mu {t}_{i}},\lambda -\frac{\delta}{\gamma}{w}_{j},-\lambda +\left(\frac{\delta}{\gamma}+\frac{\sigma}{\delta}\right){w}_{j},{w}_{j}\right),\phantom{\rule{0.277778em}{0ex}}i,j=1,2,\phantom{\rule{0.277778em}{0ex}}k\ge 2\delta ,\phantom{\rule{0.277778em}{0ex}}d\ge 2c.$$

## 4. Instability in the Case $\mathit{w}=\mathbf{0}$

**Proposition 7.**

**Proof.**

**Proposition 8.**

**Proof.**

## 5. Stability of Equilibria in the Case $\mathit{w}\ne \mathbf{0}$

**Proposition 9.**

**Proof.**

- (
**ii.2)** - As we just said, an eigenvalue of the Jacobian matrix $J\left({P}_{i}\right)$ is given by ${\rho}_{1}=-c<0$. So, we have to study the eigenvalues of the submatrix$$\left(\begin{array}{ccc}\gamma \lambda -\delta w& 0& 0\\ \\ 0& {\displaystyle \frac{k{w}^{2}}{{w}^{2}+1}-\delta w}& {\displaystyle \frac{2kwz}{{({w}^{2}+1)}^{2}}-\delta z}\\ \\ \delta w& \delta w& \delta z-2\sigma w\end{array}\right).$$We note that we have the eigenvalue ${\rho}_{2}=\gamma \lambda -\delta w$ and, for ${w}_{1},{w}_{2}$ as in (12), with $k\ge 2\delta $, $\delta \to {0}^{+}$, we obtain the following expansions$$\begin{array}{cc}\hfill {w}_{1}& =\frac{\delta}{k}+\mathcal{O}\left({\delta}^{3}\right),\hfill \end{array}$$$$\begin{array}{cc}\hfill {w}_{2}& =\frac{k}{\delta}-\frac{\delta}{k}+\mathcal{O}\left({\delta}^{3}\right).\hfill \end{array}$$We need to distinguish the two cases ${P}_{1}=\left(0,0,\frac{\sigma}{\delta}{w}_{1},{w}_{1}\right)$ and ${P}_{2}=\left(0,0,\frac{\sigma}{\delta}{w}_{2},{w}_{2}\right)$. Because of the continuous dependence of the eigenvalues on the coefficients of the matrix, we can conclude that the point ${P}_{1}$ is unstable as $\delta \to {0}^{+}$, because $J\left({P}_{1}\right)$ has the eigenvalue ${\rho}_{2}=\gamma \lambda -\delta {w}_{1}>0$, as $\delta {w}_{1}\to {0}^{+}$. We have then proved the following proposition.
**Proposition 10.**The point ${P}_{1}=\left(0,0,\frac{\sigma}{\delta}{w}_{1},{w}_{1}\right)$, with ${w}_{1}$ as in (12), is an unstable equilibrium point as $\delta \to {0}^{+}$.As to ${P}_{2}$ we notice that $\delta {w}_{2}\to k$ for $\delta \to {0}^{+}$, and therefore we must take in account the sign of $\gamma \lambda -k$. If $\gamma \lambda -k>0$ we have that the matrix $J\left({P}_{2}\right)$ has an eigenvalue ${\rho}_{2}={\rho}_{2}\left(\delta \right)$, which, by continuous dependence of the eigenvalues of a matrix on the entries, tends to the value $\gamma \lambda -k>0$ as $\delta \to {0}^{+}$. We then obtain, in this case, the instability for the point ${P}_{2}$ as $\delta \to {0}^{+}$. Let us now assume $\gamma \lambda -k<0$. We consider the submatrix $\widehat{J}\left({P}_{2}\right)$, defined as follows$$\widehat{J}\left({P}_{2}\right)=\left(\begin{array}{cc}{\displaystyle \frac{k{w}_{2}^{2}}{{w}_{2}^{2}+1}-\delta {w}_{2}}& {\displaystyle \frac{2k{w}_{2}z}{{({w}_{2}^{2}+1)}^{2}}-\delta z}\\ \\ \delta {w}_{2}& \delta z-2\sigma {w}_{2}\end{array}\right)=\left(\begin{array}{cc}\mathcal{O}\left({\delta}^{2}\right)& {\displaystyle -\frac{k\sigma}{\delta}+\mathcal{O}\left(\delta \right)}\\ \\ k+\mathcal{O}\left({\delta}^{2}\right)& {\displaystyle -\frac{k\sigma}{\delta}+\mathcal{O}\left(\delta \right)}\end{array}\right).$$We easily obtain that, as $\delta \to {0}^{+}$, the trace of the matrix $\widehat{J}\left({P}_{2}\right)$ becomes negative and its determinant positive. This implies that the eigenvalues of $\widehat{J}\left({P}_{2}\right)$ have strictly negative real parts. Now, the eigenvalues of the matrix $J\left({P}_{2}\right)$ are given by ${\rho}_{1}=-c<0,\phantom{\rule{0.166667em}{0ex}}{\rho}_{2}=\gamma \lambda -k$ and by the two eigenvalues of $\widehat{J}\left({P}_{2}\right)$. Hence, we have proved the following proposition.**Proposition 11.**As $\delta \to {0}^{+}$, the point ${P}_{2}=\left(0,0,\frac{\sigma}{\delta}{w}_{2},{w}_{2}\right)$, with ${w}_{2}$ as in (12), is an unstable equilibrium point if $\gamma \lambda -k>0$, and it is an asymptotically stable equilibrium point if $\gamma \lambda -k<0$. - (
**ii.3)** - We refer to P as (15). We consider $J\left(P\right)$ for $\delta \to {0}^{+}$ and we obtain$$J\left(P\right)=\left(\begin{array}{cccc}{\displaystyle -c}& 0& 0& 0\\ \\ 0& -\gamma \lambda +\mathcal{O}\left({\delta}^{2}\right)& 0& -\lambda \delta +\mathcal{O}\left({\delta}^{3}\right)\\ \\ 0& 0& {\displaystyle \frac{k\lambda -\sigma}{{\sigma}^{2}}\lambda {\delta}^{2}+\mathcal{O}\left({\delta}^{4}\right)}& 0\\ \\ 0& {\displaystyle \frac{\lambda}{\sigma}{\delta}^{2}+\mathcal{O}\left({\delta}^{4}\right)}& {\displaystyle \frac{\lambda}{\sigma}{\delta}^{2}+\mathcal{O}\left({\delta}^{4}\right)}& {\displaystyle -\lambda \delta +\mathcal{O}\left({\delta}^{3}\right)}\end{array}\right).$$With some lengthy but standard computations we obtain the following eigenvalues$$\begin{array}{ccc}& {\rho}_{1}=-c<0,\phantom{\rule{0.277778em}{0ex}}\hfill & \hfill {\rho}_{2}=-\lambda \delta +\mathcal{O}\left({\delta}^{2}\right)<0,\\ & {\rho}_{3}=-\gamma \lambda +\mathcal{O}\left(\delta \right)<0,\phantom{\rule{0.277778em}{0ex}}\hfill & \hfill {\displaystyle {\rho}_{4}=\frac{\lambda (k\lambda -\sigma )}{{\sigma}^{2}}{\delta}^{2}+\mathcal{O}\left({\delta}^{4}\right).}\end{array}$$As a consequence, we have the following result.
**Proposition 12.**The point $P=\left(0,\frac{\gamma \lambda \sigma}{\gamma \sigma +{\delta}^{2}},0,\frac{\gamma \lambda \delta}{\gamma \sigma +{\delta}^{2}}\right)$ is an unstable equilibrium point if $k\lambda -\sigma >0$ and is an asymptotically stable equilibrium point if $k\lambda -\sigma <0$ as $\delta \to {0}^{+}$. - (
**ii.4)** - The Jacobian matrix calculated at the points ${P}_{i},\phantom{\rule{0.277778em}{0ex}}i=1,2$ is as follows$$J\left({P}_{i}\right)=\left(\begin{array}{cccc}{\displaystyle -c}& 0& 0& 0\\ \\ 0& -\gamma \lambda +\delta w& 0& -\delta y\\ \\ 0& 0& 0& {\displaystyle \delta z\left(\frac{2\delta}{kw}-1\right)}\\ \\ 0& \delta w& \delta w& -\sigma w\end{array}\right),\phantom{\rule{1.em}{0ex}}i=1,2.$$We use the Routh–Hurwitz criterion (see [11]). As ${\rho}_{1}=-c$ is an eigenvalue, we can write the characteristic polynomial of $J\left({P}_{i}\right),\phantom{\rule{0.166667em}{0ex}}i=1,2$, in the following form$${p}_{4}\left(\rho \right)=(\rho +c){p}_{3}\left(\rho \right).$$$$\begin{array}{cc}\hfill \phantom{\rule{1.em}{0ex}}& {a}_{1}>0\hfill \\ \hfill \phantom{\rule{1.em}{0ex}}& {a}_{1}{a}_{2}>{a}_{3}>0.\hfill \end{array}$$Let us study the stability of the point ${P}_{1}$ obtained by setting $w={w}_{1}$ as in (12). After some lengthy but standard computations, we obtain$$\begin{array}{cc}& {a}_{1}=\gamma \lambda +\mathcal{O}\left(\delta \right)>0\hfill \\ & {a}_{2}=\frac{\gamma \lambda \sigma}{k}\delta +\mathcal{O}\left({\delta}^{3}\right)>0\hfill \\ & {\displaystyle {a}_{3}=\frac{\gamma \lambda (k\lambda -\sigma )}{{k}^{2}}{\delta}^{3}+\mathcal{O}\left({\delta}^{5}\right)}\hfill \\ & {a}_{1}\phantom{\rule{0.166667em}{0ex}}{a}_{2}-{a}_{3}=\frac{{\gamma}^{2}{\lambda}^{2}\sigma}{k}\delta +\mathcal{O}\left({\delta}^{2}\right)>0.\hfill \end{array}$$Hence, we obtain instability when $k\lambda -\sigma <0$, stability when $k\lambda -\sigma >0$. However, the asymptotic development of the z component of ${P}_{1}$ gives$$z=-\lambda +\left(\frac{\delta}{\gamma}+\frac{\sigma}{\delta}\right){w}_{1}=-\frac{1}{k}\phantom{\rule{0.166667em}{0ex}}\left(k\lambda -\sigma \right)+\mathcal{O}\left({\delta}^{3}\right),$$
**Proposition 13.**The point ${P}_{1}=\left(0,\lambda -\frac{\delta}{\gamma}{w}_{1},\phantom{\rule{0.166667em}{0ex}}-\lambda +\left(\frac{\delta}{\gamma}+\frac{\sigma}{\delta}\right){w}_{1},{w}_{1}\right),\phantom{\rule{0.277778em}{0ex}}k\ge 2\delta$, with ${w}_{1}$ as in (12), is an unstable equilibrium as $\delta \to {0}^{+}\phantom{\rule{4pt}{0ex}}$ if $k\lambda -\sigma <0$. If $k\lambda -\sigma >0$ the point ${P}_{1}$ is an asymptotically stable equilibrium of the ODE system (1), but it is not in $\overline{\mathcal{C}}$, as $\delta \to {0}^{+}$.We now study the point ${P}_{2}$ with $w={w}_{2}$ as in (12). After some computations, we obtain the following results for the coefficients of ${p}_{3}$:$$\begin{array}{cc}& {a}_{1}=\frac{k\sigma}{\delta}+\mathcal{O}\left(1\right)>0\hfill \\ & {a}_{2}=\frac{k\gamma \lambda \sigma}{\delta}+\mathcal{O}\left(\delta \right)>0\hfill \\ & {\displaystyle {a}_{3}=\frac{{k}^{2}\sigma (\gamma \lambda -k)}{\delta}+\mathcal{O}\left(\delta \right)}\hfill \\ & {a}_{1}\phantom{\rule{0.166667em}{0ex}}{a}_{2}-{a}_{3}=\frac{\gamma \lambda {\sigma}^{2}{k}^{2}}{{\delta}^{2}}+\mathcal{O}\left({\delta}^{-1}\right)>0.\hfill \end{array}$$Hence, by applying the Routh–Hurwitz criterion again, we obtain that, as $\delta \to {0}^{+}$, the critical point ${P}_{2}$ is stable when $\gamma \lambda -k>0$, unstable when $\gamma \lambda -k<0$. But in ${P}_{2}$ we have$$y=\lambda -\frac{\delta}{\gamma}{w}_{2}=\lambda -\frac{k}{\gamma}+\mathcal{O}\left({\delta}^{2}\right).$$Hence, in the instability case $\gamma \lambda -k<0$, we have $y<0$ as $\delta \to {0}^{+}$, and ${P}_{2}$ is not in $\overline{\mathcal{C}}$. We have then proved the following proposition.**Proposition 14.**The point ${P}_{2}=\left(0,\lambda -\frac{\delta}{\gamma}{w}_{2},\phantom{\rule{0.166667em}{0ex}}-\lambda +\left(\frac{\delta}{\gamma}+\frac{\sigma}{\delta}\right){w}_{2},{w}_{2}\right),\phantom{\rule{0.277778em}{0ex}}k\ge 2\delta$, with ${w}_{2}$ as in (12), is an asymptotically stable equilibrium point if $\gamma \lambda -k>0$ as $\delta \to {0}^{+}$. When $\gamma \lambda -k<0$${P}_{2}$ is an unstable equilibrium point of the ODE system (1), but it is not in $\overline{\mathcal{C}}$, as $\delta \to {0}^{+}$. - (
**ii.6)** - In this case we have 4 critical points ${P}_{i,j}=P({t}_{i},{w}_{j}),\phantom{\rule{0.166667em}{0ex}}i,j=1,2$ as in (17) with ${t}_{i},{w}_{j}$ given in (9) and (12), respectively. For sake of simplicity, we report the expression of the Jacobian matrix calculated at these points.$$J\left({P}_{i,j}\right)=\left(\begin{array}{cccc}{\displaystyle -c+2d\mu \frac{x{w}^{3}}{{({w}^{2}+{\mu}^{2}{x}^{2})}^{2}}}& 0& 0& {\displaystyle \frac{d\mu {x}^{2}({\mu}^{2}{x}^{2}-{w}^{2})}{{({w}^{2}+{\mu}^{2}{x}^{2})}^{2}}}\\ \\ 0& \gamma \lambda -\delta w& 0& 0\\ \\ 0& 0& 0& {\displaystyle \frac{2kwz}{{({w}^{2}+1)}^{2}}-\delta z}\\ \\ 0& \delta w& \delta w& -\sigma w\end{array}\right).$$Based on what is established in (20), the following proposition holds.Let us now begin the study in the case $t={t}_{1}=\frac{d-\sqrt{{d}^{2}-4{c}^{2}}}{2c}$ where ${\rho}_{1}\left({t}_{1}\right)<0$ as seen in (21).Let us then analyze the eigenvalue ${\rho}_{2}=\gamma \lambda -\delta w$. We have$$\begin{array}{cc}\hfill {\rho}_{2}& =\gamma \lambda -k+\mathcal{O}\left({\delta}^{2}\right)\phantom{\rule{1.em}{0ex}}\mathrm{if}\phantom{\rule{0.277778em}{0ex}}w={w}_{2}\hfill \\ \hfill {\rho}_{2}& =\gamma \lambda +\mathcal{O}\left({\delta}^{2}\right)\phantom{\rule{1.em}{0ex}}\mathrm{if}\phantom{\rule{0.277778em}{0ex}}w={w}_{1}.\hfill \end{array}$$Therefore, the Jacobian calculated at point ${P}_{1,1}$ has a positive eigenvalue for $\delta \to {0}^{+}$ and thus the equilibrium ${P}_{1,1}$ is unstable. As for point ${P}_{1,2}$, we must distinguish between two cases. If $\gamma \lambda -k>0$ then this equilibrium will also be unstable, so let us see what happens if $\gamma \lambda -k<0$. In this case the eigenvalues of $J\left({P}_{1,2}\right)$ are as follows. A first one is given by ${\rho}_{1}\left({t}_{1}\right)$ as in (21), a second one is ${\rho}_{2}=\gamma \lambda -k+\mathcal{O}\left({\delta}^{2}\right)<0$. The last two are the eigenvalues of the following matrix $\widehat{J}\left({P}_{1,2}\right)$:$$\widehat{J}\left({P}_{1,2}\right)=\left(\begin{array}{cc}0& {\displaystyle \frac{2k{w}_{2}z}{{({w}_{2}^{2}+1)}^{2}}-\delta z}\\ \\ \delta {w}_{2}& -\sigma {w}_{2}\end{array}\right)=\left(\begin{array}{cc}0& {\displaystyle -\frac{k\sigma}{\delta}+\mathcal{O}\left(\delta \right)}\\ \\ k+\mathcal{O}\left({\delta}^{2}\right)& {\displaystyle -\frac{k\sigma}{\delta}+\mathcal{O}\left(\delta \right)}\end{array}\right).$$As $\delta \to {0}^{+}$, this matrix has strictly negative trace and strictly positive determinant, so its eigenvalues have strictly negative real parts. Thus we have proved that, in this case, the point ${P}_{1,2}$ is a stable equilibrium. Let us summarize everything in the following proposition.
**Proposition 16.**The point ${P}_{1,1}=\left(\frac{{w}_{1}}{\mu {t}_{1}},0,\frac{\sigma {w}_{1}}{\delta},{w}_{1}\right),k\ge 2\delta ,d\ge 2c$, with ${t}_{1}$ as in (9) and ${w}_{1}$ as in (12), is an unstable equilibrium. The point ${P}_{1,2}=\left(\frac{{w}_{2}}{\mu {t}_{1}},0,\frac{\sigma {w}_{2}}{\delta},{w}_{2}\right),k\ge 2\delta ,d>2c$, with ${t}_{1}$ as in (9) and ${w}_{2}$ as in (12), is an unstable equilibrium if $\gamma \lambda -k>0$ and is an asymptotically stable equilibrium if $\gamma \lambda -k<0$. - (
**ii.7)** - Based on what is established in (20), the following proposition holds.
**Proposition 17.**The point ${P}_{2}=\left(\frac{\delta \gamma \lambda}{\mu {t}_{2}(\gamma \sigma +{\delta}^{2})},\frac{\gamma \lambda \sigma}{\gamma \sigma +{\delta}^{2}},0,\frac{\gamma \lambda \delta}{\gamma \sigma +{\delta}^{2}}\right)$ is an unstable equilibrium point.Let us then analyze the behavior of the Jacobian matrix calculated at point ${P}_{1}=\left(\frac{\delta \gamma \lambda}{\mu {t}_{1}(\gamma \sigma +{\delta}^{2})},\frac{\gamma \lambda \sigma}{\gamma \sigma +{\delta}^{2}},0,\frac{\delta \gamma \lambda}{\gamma \sigma +{\delta}^{2}}\right).$ In this case, the Jacobian matrix is almost the same one computed at point P in case (ii.3). Indeed, the two points differ only for their first entry, and the two Jacobian matrices only for the entries ${j}_{1,1}$ (first row, first column). Hence, the two matrices have different ${\rho}_{1}$ eigenvalues (${\rho}_{1}=-c$ in case (ii.3), ${\rho}_{1}=-\frac{c}{d}\sqrt{{d}^{2}-4{c}^{2}}$ in the present case), but both are negative, while the other three eigenvalues are the same as in the case (ii.3). Using the results we have obtained in that case, we obtain the following proposition.**Proposition 18.**The point ${P}_{1}=\left(\frac{\delta \gamma \lambda}{\mu {t}_{1}(\gamma \sigma +{\delta}^{2})},\frac{\gamma \lambda \sigma}{\gamma \sigma +{\delta}^{2}},0,\frac{\gamma \lambda \delta}{\gamma \sigma +{\delta}^{2}}\right)$ is an unstable equilibrium if $k\lambda -\sigma >0$ and is an asymptotically stable equilibrium if $k\lambda -\sigma <0$ as $\delta \to {0}^{+}$. - (
**ii.8)** - In this case we have 4 critical points ${P}_{i,j}=P({t}_{i},{w}_{j}),\phantom{\rule{0.166667em}{0ex}}i,j=1,2$ as in (19) with ${t}_{i},{w}_{j}$ given in (9) and (12), respectively. Based on what is established in (20), the following proposition holds.
**Proposition 19.**The points ${P}_{2,j}=\left(\frac{{w}_{j}}{\mu {t}_{2}},\lambda -\frac{\delta}{\gamma}{w}_{j},-\lambda +\left(\frac{\delta}{\gamma}+\frac{\sigma}{\delta}\right){w}_{j},{w}_{j}\right),$$j=1,2,\phantom{\rule{0.277778em}{0ex}}k\ge 2\delta ,\phantom{\rule{0.277778em}{0ex}}d>2c$ with ${t}_{2}$ as in (9) and ${w}_{j}$ as in (12), are unstable equilibria.Regarding the points ${P}_{1,j}$ ($j=1,2$) we argue as we have just done for the previous case (ii.7). Indeed, in this case, the Jacobian matrix is almost the same one computed at point ${P}_{2}$ in case (ii.4): the two points differ only for their first entry, and the two Jacobian matrices only for the entries ${j}_{1,1}$, hence the two matrices have different ${\rho}_{1}$ eigenvalues, which are both negative, while the other three eigenvalues are the same in the two cases. Using the arguments and the results of case (ii.4), we obtain the following proposition.**Proposition 20.**The point ${P}_{1,1}=\left(\frac{{w}_{1}}{\mu {t}_{1}},\lambda -\frac{\delta}{\gamma}{w}_{1},-\lambda +\left(\frac{\delta}{\gamma}+\frac{\sigma}{\delta}\right){w}_{1},{w}_{1}\right)$, with ${w}_{1}$ as in (12), is an unstable equilibrium as $\delta \to {0}^{+}\phantom{\rule{4pt}{0ex}}$ if $k\lambda -\sigma <0$. If $k\lambda -\sigma >0$ the point ${P}_{1,1}$ is an asymptotically stable equilibrium of the ODE system (1), but it is not in $\overline{\mathcal{C}}$, as $\delta \to {0}^{+}$. The point ${P}_{1,2}=\left(\frac{{w}_{2}}{\mu {t}_{1}},\lambda -\frac{\delta}{\gamma}{w}_{2},\phantom{\rule{0.166667em}{0ex}}-\lambda +\left(\frac{\delta}{\gamma}+\frac{\sigma}{\delta}\right){w}_{2},{w}_{2}\right),\phantom{\rule{0.277778em}{0ex}}k\ge 2\delta $, with ${w}_{2}$ as in (12) is an asymptotically stable equilibrium if $\gamma \lambda -k>0$ as $\delta \to {0}^{+}$. When $\gamma \lambda -k<0$, ${P}_{1,2}$ is an unstable equilibrium of ODE system (1), but it is not in $\overline{\mathcal{C}}$, as $\delta \to {0}^{+}$.

## 6. Fixed Points and Stability If $\mathit{x}>\mathit{w}$

**(iii.1)**- $y=0$ and $z\ne 0$. From (30) we obtain $z=\frac{\sigma}{\delta}w$. From (29), we obtain$$\begin{array}{cc}\hfill {w}_{1}\left(t\right)& =\frac{k\mu t-\sqrt{{k}^{2}{\mu}^{2}{t}^{2}-4{\delta}^{2}t\mu}}{2\delta},\hfill \\ \hfill {w}_{2}\left(t\right)& =\frac{k\mu t+\sqrt{{k}^{2}{\mu}^{2}{t}^{2}-4{\delta}^{2}t\mu}}{2\delta},\phantom{\rule{2.em}{0ex}}{k}^{2}\mu t-4{\delta}^{2}\ge 0.\hfill \end{array}$$Let us set ${w}_{i,j}={w}_{j}\left({t}_{i}\right),$ with $i,j=1,2,$ and $\phantom{\rule{0.277778em}{0ex}}{t}_{i}$ as in (9). We have the expression for x from (31). Hence, we obtain the following equilibria$${P}_{i,j}=\left(\frac{{w}_{i,j}}{\mu {t}_{i}},0,\frac{\sigma}{\delta}{w}_{i,j},{w}_{i,j}\right)\phantom{\rule{2.em}{0ex}}i,j=1,2.$$
**(iii.2)**- $z=0$ and $y\ne 0$. From (30) and (31), we have $y=\frac{\sigma}{\delta}w=\sigma {t}_{i}\mu x/\delta $ and, substituting in (28), this gives us the expression for x$$x=\frac{\delta \gamma \lambda}{{\delta}^{2}+\gamma \mu \sigma {t}_{i}},\phantom{\rule{2.em}{0ex}}i=1,2.$$Then we have two equilibrium points$${P}_{i}=\left(\frac{\delta \gamma \lambda}{{\delta}^{2}+\gamma \mu \sigma {t}_{i}},\frac{\gamma \lambda \mu \sigma {t}_{i}}{{\delta}^{2}+\gamma \mu \sigma {t}_{i}},0,\frac{\delta \gamma \lambda \mu {t}_{i}}{{\delta}^{2}+\gamma \mu \sigma {t}_{i}}\right),\phantom{\rule{2.em}{0ex}}i=1,2.$$
**(iii.3)**- $y\ne 0,\phantom{\rule{0.277778em}{0ex}}z\ne 0$. From (29) we obtain the expression (32) for w and from (31) the expression for x. Then, (28) gives us the expression for y and finally (30) the value of z. So we have four equilibrium points$${P}_{i,j}=\left(\frac{{w}_{i,j}}{\mu {t}_{i}},\phantom{\rule{0.277778em}{0ex}}\lambda -\frac{\delta}{\gamma \mu {t}_{i}}{w}_{i,j},\phantom{\rule{0.277778em}{0ex}}-\lambda +\left(\frac{\delta}{\gamma \mu {t}_{i}}+\frac{\sigma}{\delta}\right){w}_{i,j},\phantom{\rule{0.277778em}{0ex}}{w}_{i,j}\right),\phantom{\rule{2.em}{0ex}}i,j=1,2.$$Next, we will study the stability of these equilibria. For this, we compute the Jacobian matrix of the differential model (1) with $x>w$. Recalling that $c=d\frac{t}{{t}^{2}+1}$ and considering (31) we obtain$$J\left(X\right)=\left(\begin{array}{cccc}{\displaystyle \frac{dt({t}^{2}-1)}{{({t}^{2}+1)}^{2}}}& 0& 0& {\displaystyle -\frac{d({t}^{2}-1)}{\mu {({t}^{2}+1)}^{2}}}\\ \\ -\delta y& \gamma \lambda -2\gamma y-\delta x& 0& 0\\ \\ {\displaystyle \frac{kwz}{{(xw+1)}^{2}}-\delta z}& 0& {\displaystyle \frac{kxw}{xw+1}-\delta x}& {\displaystyle \frac{kxz}{{(xw+1)}^{2}}}\\ \\ \delta y+\delta z-\sigma w& \delta x& \delta x& -\sigma x\end{array}\right),$$

**(iii.1)**- As $y=0$, in this case the entry ${j}_{2,2}=\gamma \lambda -\delta x$ is an eigenvalue, with $x={x}_{i,j}={w}_{i,j}/\mu {t}_{i}$. Let us call ${\rho}_{1}$ this eigenvalue. Considering the asymptotic for $\delta \to {0}^{+}$ we obtain$$\begin{array}{cc}\hfill {w}_{1}\left(t\right)& =\frac{1}{k}\delta +\mathcal{O}\left({\delta}^{3}\right)\hfill \\ \hfill {w}_{2}\left(t\right)& =\frac{k\mu t}{\delta}+\mathcal{O}\left(\delta \right),\hfill \end{array}$$$$\begin{array}{ccc}\hfill {\rho}_{1}& =\gamma \lambda +\mathcal{O}\left({\delta}^{2}\right)\phantom{\rule{2.em}{0ex}}\hfill & \hfill \mathrm{if}\phantom{\rule{0.277778em}{0ex}}w={w}_{i,1},\phantom{\rule{0.166667em}{0ex}}i=1,2,\\ \hfill {\rho}_{1}& =\gamma \lambda -k+\mathcal{O}\left({\delta}^{2}\right)\phantom{\rule{2.em}{0ex}}\hfill & \hfill \mathrm{if}\phantom{\rule{0.277778em}{0ex}}w={w}_{i,2},\phantom{\rule{0.166667em}{0ex}}i=1,2.\end{array}$$Therefore, the Jacobian matrix calculated at the points ${P}_{i,1}$ for $\delta \to {0}^{+}$ has at least one positive eigenvalue, and therefore the equilibria ${P}_{i,1}$ are unstable. We thus state the following proposition.The behavior of the points ${P}_{i,2}$ depends instead on the sign of $\gamma \lambda -k$. If $\gamma \lambda -k>0$ these equilibria are unstable. We then study the behavior of such points under the assumption $\gamma \lambda -k<0.$ We examine the asymptotic behavior for $\delta \to {0}^{+}$. The asymptotic of the Jacobian matrix $J\left({P}_{i,2}\right)$ becomes$$J\left({P}_{i,2}\right)=\left(\begin{array}{cccc}{\displaystyle \frac{dt({t}^{2}-1)}{{({t}^{2}+1)}^{2}}}& 0& 0& {\displaystyle -\frac{d({t}^{2}-1)}{\mu {({t}^{2}+1)}^{2}}}\\ \\ 0& \gamma \lambda -k+\mathcal{O}\left({\delta}^{2}\right)& 0& 0\\ \\ {\displaystyle -\frac{k\mu \sigma t}{\delta}+\mathcal{O}\left(\delta \right)}& 0& 0& {\displaystyle \frac{\sigma}{k\mu t}\delta +\mathcal{O}\left({\delta}^{3}\right)}\\ \\ 0& k+\mathcal{O}\left({\delta}^{2}\right)& k+\mathcal{O}\left({\delta}^{2}\right)& {\displaystyle -\frac{k\sigma}{\delta}+\mathcal{O}\left(\delta \right)}\end{array}\right).$$We use the Routh–Hurwitz criterion. We introduce the characteristic polynomial of $J\left({P}_{i,2}\right)$$${p}_{4}\left(\rho \right)=(\rho -\gamma \lambda +k+\mathcal{O}\left({\delta}^{2}\right))({\rho}^{3}+{a}_{1}{\rho}^{2}+{a}_{2}\rho +{a}_{3}).$$Through some lengthy but standard computations we obtain$$\begin{array}{cc}\hfill {a}_{1}& {\displaystyle =\frac{k\sigma}{\delta}+\mathcal{O}\left(1\right)}\hfill \\ \hfill {a}_{2}& =\frac{k\sigma dt(1-{t}^{2})}{{({t}^{2}+1)}^{2}\delta}+\mathcal{O}\left(\delta \right)\hfill \\ \hfill {a}_{3}& {\displaystyle =\frac{(1-{t}^{2})d{k}^{2}\sigma t}{{({t}^{2}+1)}^{2}\delta}+\mathcal{O}\left(\delta \right).}\hfill \end{array}$$Let us recall that the necessary and sufficient condition for the roots of the polynomial ${p}_{3}\left(\rho \right)={\rho}^{3}+{a}_{1}{\rho}^{2}+{a}_{2}\rho +{a}_{3}$ to have strictly negative real parts is that ${a}_{1},{a}_{3}>0$ e ${a}_{1}{a}_{2}-{a}_{3}>0$. We recall that ${t}_{1}\in (0,1)$ and ${t}_{2}>1$ as given in (9). Therefore, ${a}_{2}<0$ and ${a}_{3}<0$ if $t={t}_{2}$ making ${P}_{2,2}$ unstable. Conversely, ${a}_{2}>0$ and ${a}_{3}>0$ if $t={t}_{1}$ with$${a}_{1}{a}_{2}-{a}_{3}=\frac{{k}^{2}{\sigma}^{2}dt(1-{t}^{2})}{{({t}^{2}+1)}^{2}{\delta}^{2}}+\mathcal{O}\left({\delta}^{-1}\right)>0.$$We can therefore summarize in the following proposition.
**Proposition 22.**The equilibrium point ${P}_{2,2}=\left(\frac{{w}_{2,2}}{\mu {t}_{2}},0,\frac{\sigma}{\delta}{w}_{2,2},{w}_{2,2}\right),$ with ${w}_{2,2}$ given in (32) and ${t}_{2}$ given in (9), is unstable for $\delta \to {0}^{+}$. The equilibrium point ${P}_{1,2}=\left(\frac{{w}_{1,2}}{\mu {t}_{1}},0,\frac{\sigma}{\delta}{w}_{1,2},{w}_{1,2}\right),$ with ${w}_{1,2}$ given in (32) and ${t}_{1}$ given in (9), is unstable for $\delta \to {0}^{+}$ if $\gamma \lambda -k>0$. The equilibrium ${P}_{1,2}$ is asymptotically stable for $\delta \to {0}^{+}$ if $\gamma \lambda -k<0$. **(iii.2)**- In this case, we have two critical points ${P}_{i},\phantom{\rule{0.166667em}{0ex}}i=1,2$ as in (35). Computing the asymptotic expansions for $\delta \to {0}^{+}$ we obtain$$\begin{array}{cc}\hfill {x}_{i}& =\frac{\delta \gamma \lambda}{{\delta}^{2}+\gamma \mu \sigma {t}_{i}}=\frac{\lambda}{\mu \sigma {t}_{i}}\delta +\mathcal{O}\left({\delta}^{3}\right)\hfill \\ \hfill {y}_{i}& =\frac{\gamma \lambda \mu \sigma {t}_{i}}{{\delta}^{2}+\gamma \mu \sigma {t}_{i}}=\lambda +\mathcal{O}\left({\delta}^{2}\right)\hfill \\ \hfill {z}_{i}& =0\hfill \\ \hfill {w}_{i}& {\displaystyle =\frac{\delta \gamma \lambda \mu {t}_{i}}{{\delta}^{2}+\gamma \mu \sigma {t}_{i}}=\frac{\lambda}{\sigma}\delta +\mathcal{O}\left({\delta}^{3}\right),}\hfill \end{array}$$$$J\left({P}_{i}\right)=\left(\begin{array}{cccc}{\displaystyle \frac{dt({t}^{2}-1)}{{({t}^{2}+1)}^{2}}}& 0& 0& {\displaystyle -\frac{d({t}^{2}-1)}{\mu {({t}^{2}+1)}^{2}}}\\ \\ -\lambda \delta +\mathcal{O}\left({\delta}^{3}\right)& -\gamma \lambda +\mathcal{O}\left({\delta}^{2}\right)& 0& 0\\ \\ 0& 0& {\displaystyle \frac{\lambda (k\lambda -\sigma )}{\mu {\sigma}^{2}t}{\delta}^{2}+\mathcal{O}\left({\delta}^{4}\right)}& 0\\ \\ 0& {\displaystyle \frac{\lambda}{\mu \sigma t}{\delta}^{2}+\mathcal{O}\left({\delta}^{4}\right)}& {\displaystyle \frac{\lambda}{\mu \sigma t}{\delta}^{2}+\mathcal{O}\left({\delta}^{4}\right)}& {\displaystyle -\frac{\lambda}{\mu t}\delta +\mathcal{O}\left({\delta}^{3}\right)}\end{array}\right),$$$${\rho}_{1}=\frac{dt({t}^{2}-1)}{{({t}^{2}+1)}^{2}}.$$We have repeatedly noted that ${\rho}_{1}>0$ if $t={t}_{2}$ and ${\rho}_{1}<0$ if $t={t}_{1}$. Due to the continuity of eigenvalues with respect to the elements of the matrix, in the case $t={t}_{2}$ for $\delta \to {0}^{+}$, there is an eigenvalue of the Jacobian matrix that tends to a positive value, and therefore the critical point ${P}_{2}$ is unstable. If $t={t}_{1}$, we obtain instead ${\rho}_{1}<0$, hence we must apply the Routh–Hurwitz criterion to the characteristic polynomial ${p}_{4}\left(\rho \right)$. The criterion states, in this case, that the necessary and sufficient conditions for all roots of characteristic polynomial ${p}_{4}\left(\rho \right)={\rho}^{4}+{a}_{1}{\rho}^{3}+{a}_{2}{\rho}^{2}+{a}_{3}\rho +{a}_{4}$ to have a strictly negative real part are the following:$$\left\{\begin{array}{c}{a}_{i}>0\phantom{\rule{2.em}{0ex}}i=0,\cdots ,4\hfill \\ {a}_{1}{a}_{2}-{a}_{3}>0\hfill \\ {a}_{1}{a}_{2}{a}_{3}-{a}_{1}^{2}{a}_{4}-{a}_{3}^{2}>0.\hfill \end{array}\right.$$After lengthy computations we obtain the following asymptotic developments of the coefficients of the polynomial ${p}_{4}\left(\rho \right)$ as $\delta \to {0}^{+}$$$\begin{array}{cc}\hfill \phantom{\rule{1.em}{0ex}}& {a}_{4}=-\frac{(k\lambda -\sigma )\gamma {\lambda}^{3}(1-{t}^{2})d}{{\mu}^{2}{\sigma}^{2}t{({t}^{2}+1)}^{2}}{\delta}^{3}+\mathcal{O}\left({\delta}^{5}\right)\hfill \\ \hfill \phantom{\rule{1.em}{0ex}}& {a}_{3}=\frac{\gamma {\lambda}^{2}d(1-{t}^{2})}{\mu {({t}^{2}+1)}^{2}}\delta +\mathcal{O}\left({\delta}^{2}\right)\hfill \\ \hfill \phantom{\rule{1.em}{0ex}}& {a}_{2}=\frac{\gamma \lambda (1-{t}^{2})dt}{{({t}^{2}+1)}^{2}}+\mathcal{O}\left(\delta \right)\hfill \\ \hfill \phantom{\rule{1.em}{0ex}}& {a}_{1}=\gamma \lambda +\frac{dt(1-{t}^{2})}{{({t}^{2}+1)}^{2}}+\mathcal{O}\left(\delta \right)\hfill \\ \hfill \phantom{\rule{1.em}{0ex}}& {a}_{1}{a}_{2}-{a}_{3}={a}_{1}{a}_{2}+\mathcal{O}\left(\delta \right)\hfill \\ \hfill \phantom{\rule{1.em}{0ex}}& {a}_{1}{a}_{2}{a}_{3}-{a}_{1}^{2}{a}_{4}-{a}_{3}^{2}={a}_{1}{a}_{2}{a}_{3}+\mathcal{O}\left({\delta}^{2}\right).\hfill \end{array}$$Recalling that $1-{t}_{1}^{2}>0$, we have that ${a}_{1},{a}_{2},{a}_{3}>0$ for $t={t}_{1}$. Also, we have ${a}_{4}>0$ if $k\lambda -\sigma <0$. Analyzing the orders of magnitude of the coefficients, the last two conditions are definitely met when the leading terms of the coefficients of the characteristic polynomial are positive. On the other hand, ${p}_{4}\left(\rho \right)$ obviously has a positive root if ${a}_{4}<0$. We can therefore conclude with the following proposition
**Proposition 23.**The point ${P}_{2}=\left(\frac{\delta \gamma \lambda}{{\delta}^{2}+\gamma \mu \sigma {t}_{2}},\frac{\gamma \lambda \mu \sigma {t}_{2}}{{\delta}^{2}+\gamma \mu \sigma {t}_{2}},0,\frac{\delta \gamma \lambda \mu {t}_{2}}{{\delta}^{2}+\gamma \mu \sigma {t}_{2}}\right)$, for ${t}_{2}$ defined in (9), is an unstable equilibrium point. The point ${P}_{1}=\left(\frac{\delta \gamma \lambda}{{\delta}^{2}+\gamma \mu \sigma {t}_{1}},\frac{\gamma \lambda \mu \sigma {t}_{1}}{{\delta}^{2}+\gamma \mu \sigma {t}_{1}},0,\frac{\delta \gamma \lambda \mu {t}_{1}}{{\delta}^{2}+\gamma \mu \sigma {t}_{1}}\right)$, for ${t}_{1}$ defined in (9), is an asymptotically stable equilibrium point if $k\lambda -\sigma <0$ and is unstable if $k\lambda -\sigma >0$. **(iii.3)**- In this case, we have four equilibrium points varying with the values of ${w}_{i,j}$ see (32) and ${t}_{i}$, see (9). We start with ${w}_{i,1}$ for $i=1,2$. Considering the asymptotic expansions of the variables for $\delta \to {0}^{+}$, we obtain$$\begin{array}{cc}\hfill {x}_{i,1}& {\displaystyle =\frac{{w}_{i,1}}{\mu {t}_{i}}=\frac{\delta}{k\mu {t}_{i}}+\mathcal{O}\left({\delta}^{3}\right)}\hfill \\ \hfill {y}_{i,1}& =\lambda -\frac{\delta}{\gamma}{x}_{i,1}=\lambda +\mathcal{O}\left({\delta}^{2}\right)\hfill \\ \hfill {z}_{i,1}& {\displaystyle =-{y}_{i,1}+\frac{\sigma}{\delta}{w}_{i,1}=\frac{\sigma}{k}-\lambda +\mathcal{O}\left({\delta}^{2}\right)}\hfill \\ \hfill {w}_{i,1}& {\displaystyle =\frac{k\mu {t}_{i}-\sqrt{{k}^{2}{\mu}^{2}{t}_{i}^{2}-4{\delta}^{2}{t}_{i}\mu}}{2\delta}=\frac{\delta}{k}+\mathcal{O}\left({\delta}^{3}\right).}\hfill \end{array}$$Notice that, when $\lambda k-\sigma >0$, we obtain ${z}_{i,1}<0$ as $\delta \to {0}^{+}$, hence in this case the points ${P}_{i,1}$ are not in $\overline{\mathcal{C}}$.The developments above, when substituted into the Jacobian matrix, lead to$$J\left({P}_{i,1}\right)=\left(\begin{array}{cccc}{\displaystyle \frac{dt({t}^{2}-1)}{{({t}^{2}+1)}^{2}}}& 0& 0& {\displaystyle -\frac{d({t}^{2}-1)}{\mu {({t}^{2}+1)}^{2}}}\\ \\ -\lambda \delta +\mathcal{O}\left({\delta}^{3}\right)& -\gamma \lambda +\mathcal{O}\left({\delta}^{2}\right)& 0& 0\\ \\ \mathcal{O}\left({\delta}^{3}\right)& 0& 0& {\displaystyle \frac{\sigma -k\lambda}{k\mu t}\delta +\mathcal{O}\left({\delta}^{3}\right)}\\ \\ 0& {\displaystyle \frac{{\delta}^{2}}{kt\mu}+\mathcal{O}\left({\delta}^{4}\right)}& {\displaystyle \frac{{\delta}^{2}}{kt\mu}+\mathcal{O}\left({\delta}^{4}\right)}& {\displaystyle -\frac{\sigma}{kt\mu}\delta +\mathcal{O}\left({\delta}^{3}\right)}\end{array}\right).$$We now argue as in the previous case, noticing that, when $\delta =0$, this matrix reduces to a diagonal matrix with an eigenvalue$${\rho}_{1}=\frac{dt({t}^{2}-1)}{{({t}^{2}+1)}^{2}}.$$As we know, this value satisfies ${\rho}_{1}>0$ if $t={t}_{2}>1$ and ${\rho}_{1}<0$ if $t={t}_{1}\in (0,1).$ Therefore, when $t={t}_{2}$, due to the continuity of eigenvalues with respect to the entries of the matrix, there exists at least one positive eigenvalue as $\delta \to {0}^{+}$, and thus the equilibrium point ${P}_{2,1}$ is unstable.We now proceed by studying the stability of the point ${P}_{1,1}$. We apply again the Routh–Hurwitz criterion, and we study the characteristic polynomial ${p}_{4}\left(\rho \right)$ of the matrix and its coefficients ${a}_{i}$, $i=1,..,4$. We want to study what happens for $\delta \to {0}^{+}$, so we look at the asymptotic behavior of the ${a}_{i}$s. We start noticing that$$\begin{array}{cc}\hfill \phantom{\rule{1.em}{0ex}}& {a}_{4}=\frac{(k\lambda -\sigma )\gamma \lambda (1-{t}^{2})d}{{k}^{2}{\mu}^{2}t{({t}^{2}+1)}^{2}}{\delta}^{3}+\mathcal{O}\left({\delta}^{5}\right)\hfill \\ \hfill \phantom{\rule{1.em}{0ex}}& {a}_{3}=\frac{\gamma \lambda \sigma (1-{t}^{2})dt}{k\mu t{({t}^{2}+1)}^{2}}\delta +\mathcal{O}\left({\delta}^{3}\right)\hfill \\ \hfill \phantom{\rule{1.em}{0ex}}& {a}_{2}=\frac{\gamma \lambda (1-{t}^{2})dt}{{({t}^{2}+1)}^{2}}+\mathcal{O}\left(\delta \right)\hfill \\ \hfill \phantom{\rule{1.em}{0ex}}& {a}_{1}=\frac{\gamma \lambda {(1+{t}^{2})}^{2}+dt(1-{t}^{2})}{{({t}^{2}+1)}^{2}}+\mathcal{O}\left(\delta \right)\hfill \\ \hfill \phantom{\rule{1.em}{0ex}}& {a}_{1}{a}_{2}-{a}_{3}={a}_{1}{a}_{2}+\mathcal{O}\left(\delta \right)\hfill \\ \hfill \phantom{\rule{1.em}{0ex}}& {a}_{1}{a}_{2}{a}_{3}-{a}_{1}^{2}{a}_{4}-{a}_{3}^{2}={a}_{1}{a}_{2}{a}_{3}+\mathcal{O}\left({\delta}^{2}\right).\hfill \end{array}$$We have that ${a}_{1},{a}_{2},{a}_{3}>0$ for $t={t}_{1}$ and ${a}_{4}>0$ for $t={t}_{1}$ and $k\lambda -\sigma >0.$ Under these conditions, the ${a}_{1}{a}_{2}-{a}_{3}$ and ${a}_{1}{a}_{2}{a}_{3}-{a}_{1}^{2}{a}_{4}-{a}_{3}^{2}$ are also positive, as $\delta \to {0}^{+}$.We recall that the equilibria ${P}_{i,1}$ are not in $\overline{\mathcal{C}}$ when $k\lambda -\sigma >0$. We collect our results in the following proposition:
**Proposition 24.**${P}_{2,1}$ is an unstable equilibrium. If $k\lambda -\sigma >0$ the equilibrium ${P}_{1,1}$ is asymptotically stable but is not in $\overline{\mathcal{C}}$, together with ${P}_{2,1}$, as $\delta \to {0}^{+}$. If $k\lambda -\sigma <0$ the two equilibria are in $\mathcal{C}$ and are unstable, as $\delta \to {0}^{+}$.We now proceed with the study of the stability of the points ${P}_{i,2}.$ The asymptotic developments in this case are$$\begin{array}{cc}\hfill {x}_{i,2}& {\displaystyle =\frac{{w}_{i,2}}{\mu {t}_{i}}=\frac{k}{\delta}+\mathcal{O}\left(\delta \right)}\hfill \\ \hfill {y}_{i,2}& =\lambda -\frac{\delta}{\gamma}{x}_{i,2}=\lambda -\frac{k}{\gamma}+\mathcal{O}\left({\delta}^{2}\right)\hfill \\ \hfill {z}_{i,2}& {\displaystyle =-{y}_{i,2}+\frac{\sigma}{\delta}{w}_{i,2}=\frac{\sigma k\mu {t}_{i}}{{\delta}^{2}}+\mathcal{O}\left(1\right)}\hfill \\ \hfill {w}_{i,2}& {\displaystyle =\frac{k\mu {t}_{i}+\sqrt{{k}^{2}{\mu}^{2}{t}_{i}^{2}-4{\delta}^{2}{t}_{i}\mu}}{2\delta}=\frac{k\mu {t}_{i}}{\delta}+\mathcal{O}\left(\delta \right),}\hfill \end{array}$$$$J\left({P}_{i,2}\right)=\left(\begin{array}{cccc}{\displaystyle \frac{dt({t}^{2}-1)}{{({t}^{2}+1)}^{2}}}& 0& 0& {\displaystyle -\frac{d({t}^{2}-1)}{\mu {({t}^{2}+1)}^{2}}}\\ \\ {\displaystyle \left(-\lambda +\frac{k}{\gamma}\right)\delta +\mathcal{O}\left({\delta}^{3}\right)}& -\gamma \lambda +k+\mathcal{O}\left({\delta}^{2}\right)& 0& 0\\ \\ {\displaystyle -\frac{\sigma k\mu t}{\delta}+\mathcal{O}\left(\delta \right)}& 0& \mathcal{O}\left({\delta}^{6}\right)& {\displaystyle \frac{\sigma \delta}{k\mu t}+\mathcal{O}\left({\delta}^{3}\right)}\\ \\ 0& k+\mathcal{O}\left({\delta}^{2}\right)& k+\mathcal{O}\left({\delta}^{2}\right)& {\displaystyle -\frac{\sigma k}{\delta}+\mathcal{O}\left(\delta \right)}\end{array}\right).$$We now apply the Routh–Hurwitz criterion to the characteristic polynomial ${p}_{4}\left(\rho \right)={\rho}^{4}+{a}_{1}{\rho}^{3}+{a}_{2}{\rho}^{2}+{a}_{3}\rho +{a}_{1}$ with$$\begin{array}{cc}\hfill \phantom{\rule{1.em}{0ex}}& {a}_{4}=\frac{(\gamma \lambda -k)(1-{t}^{2})d{k}^{2}\sigma t}{{({t}^{2}+1)}^{2}\delta}+\mathcal{O}\left(\delta \right)\hfill \\ \hfill \phantom{\rule{1.em}{0ex}}& {a}_{3}=\frac{\gamma \lambda \sigma k(1-{t}^{2})dt}{{({t}^{2}+1)}^{2}\delta}+\mathcal{O}\left(\delta \right)\hfill \\ \hfill \phantom{\rule{1.em}{0ex}}& {a}_{2}=k\sigma \left[(\gamma \lambda -k)+\frac{dt(1-{t}^{2})}{{({t}^{2}+1)}^{2}}\right]\frac{1}{\delta}+\mathcal{O}\left(1\right)\hfill \\ \hfill \phantom{\rule{1.em}{0ex}}& {a}_{1}=\frac{\sigma k}{\delta}+\mathcal{O}\left(1\right).\hfill \end{array}$$If we assume $\gamma \lambda -k<0$ and $t={t}_{1}\in (0,1)$, we have ${a}_{4}<0$, while if $t={t}_{2}>1$ it holds ${a}_{2},{a}_{3}<0$, and in both cases we obtain that there is an eigenvalue with strictly positive real part. Hence, in this case, the two equilibrium points we are dealing with are unstable. Notice also that from $\gamma \lambda -k<0$, we derive ${y}_{i,2}<0$ as $\delta \to {0}^{+}$, so in this case the two points are not in the positive cone.Let us now assume $\gamma \lambda -k>0$. If $t={t}_{2}>1$ we obtain ${a}_{4}<0$ and the equilibrium is unstable. If $t={t}_{1}\in (0,1)$, all the coefficients ${a}_{i}$ are positive. For the stability we have then to verify that, at least for $\delta \to {0}^{+}$, it holds$${a}_{1}{a}_{2}>{a}_{3},\phantom{\rule{1.em}{0ex}}\phantom{\rule{1.em}{0ex}}{a}_{1}{a}_{2}{a}_{3}>{{a}_{1}}^{2}{a}_{4}+{{a}_{3}}^{2}.$$We have that ${a}_{1}{a}_{2}$ is led by a term of the type $c/{\delta}^{2}$, with $c>0$, while ${a}_{3}=\mathcal{O}\left({\delta}^{-1}\right)$, so the first inequality above is verified for small $\delta $s. Regarding the last condition, that is ${a}_{1}{a}_{2}{a}_{3}-{a}_{1}^{2}{a}_{4}-{a}_{3}^{2}>0$, we notice that ${a}_{3}^{2}=\mathcal{O}\left({\delta}^{-2}\right)$ while ${a}_{1}{a}_{2}{a}_{3}=\mathcal{O}\left({\delta}^{-3}\right)$ and ${a}_{1}^{2}{a}_{4}=\mathcal{O}\left({\delta}^{-3}\right).$ Therefore, the last condition is simplified by requiring ${a}_{2}{a}_{3}-{a}_{1}{a}_{4}>0.$ With simple calculations, we obtain$${a}_{2}{a}_{3}-{a}_{1}{a}_{4}=\frac{{\sigma}^{2}{k}^{2}dt(1-{t}^{2})}{{({t}^{2}+1)}^{2}}\left[{(\gamma \lambda -k)}^{2}+\gamma \lambda \frac{dt(1-{t}^{2})}{{({t}^{2}+1)}^{2}}\right]\frac{1}{{\delta}^{2}}+\mathcal{O}\left({\delta}^{-1}\right),$$**Proposition 25.**If $\gamma \lambda -k<0$ the two points ${P}_{i,2}$ are both unstable equilibria of the system (1), but are not in $\overline{\mathcal{C}}$, as $\delta \to {0}^{+}$. If $\gamma \lambda -k>0$ and $\delta \to {0}^{+}$, the point ${P}_{2,2}$ is an unstable equilibrium, while the point ${P}_{1,2}$ is an asymptotically stable equilibrium (and both are in $\mathcal{C}$).

## 7. Simulation Results

`ode45`solver, an explicit adaptive Runge–Kutta method known for its broad applicability in solving ordinary differential equations (ODEs). While our study utilizes

`ode45`for its efficiency and general applicability, it is worth acknowledging the existence of specialized methods for particular types of nonlinear ODEs, see for example [12], with an iterative finite difference method for nonlinear ODEs.

## 8. Conclusions: Main Results and Future Steps

## Author Contributions

## Funding

## Data Availability Statement

## Conflicts of Interest

## References

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**Figure 1.**Case (i). Scenario around the unstable critical points. (

**a**) $Q=(0,0,1,0)$. (

**b**) $R=(0,10,1,0)$.

**Figure 2.**Case (ii.2), analysis of the points ${P}_{1}$ and ${P}_{2}$, with $\delta \to {0}^{+}$. (

**a**) ${P}_{1}=(0,0,9.4054,0.9387)$. (

**b**) ${P}_{2}=(0,0,10.6748,1.0653)$.

**Figure 3.**Case (ii.3), analysis of the point P with $\delta =10$. (

**a**) $k\lambda -\sigma =95$. (

**b**) $k\lambda -\sigma =-2$.

**Figure 4.**Case (iii.1) analysis of the point ${P}_{12}$ with $\gamma \lambda -k>0$. (

**a**) $\delta =0.1$. (

**b**) $\delta =0.4360$.

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**MDPI and ACS Style**

Badiale, M.; Cravero, I.
A Nonlinear ODE Model for a Consumeristic Society. *Mathematics* **2024**, *12*, 1253.
https://doi.org/10.3390/math12081253

**AMA Style**

Badiale M, Cravero I.
A Nonlinear ODE Model for a Consumeristic Society. *Mathematics*. 2024; 12(8):1253.
https://doi.org/10.3390/math12081253

**Chicago/Turabian Style**

Badiale, Marino, and Isabella Cravero.
2024. "A Nonlinear ODE Model for a Consumeristic Society" *Mathematics* 12, no. 8: 1253.
https://doi.org/10.3390/math12081253