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Moreover, this region is forward invariant. To prove that, we can use Nagumo's theorem, which states that it is sufficient to prove that the vector field is pointing inward into your set on its border. We can look at the $1$D-dynamics of $S$.

The equations for these infectious compartments can be written in the general form:

• $\mathbf{f}(S, \mathbf{x})$: the nonlinear coupling for a given number of susceptible $S$, and is the function defined by $$ \mathbf{f}(S, \mathbf{x}) = \begin{bmatrix} p \beta_1 S I_2 + q \beta_2 S J + r \beta_3 S A \\ (1-p) \beta_1 S I_2 + (1 - q) \beta_2 S J + (1 - r) \beta_3 S A \\ 0 \\ 0 \end{bmatrix} = S \begin{bmatrix} p \beta_1 I_2 + q \beta_2 J + r \beta_3 A \\ (1-p) \beta_1 I_2 + (1 - q) \beta_2 J + (1 - r) \beta_3 A \\ 0 \\ 0 \end{bmatrix} = S \begin{bmatrix} 0 & p\beta_1 & q\beta_2 & r\beta_3 \\ 0 & (1-p)\beta_1 & (1-q)\beta_2 & (1-r)\beta_3 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix}\begin{bmatrix} I_1 \\ I_2 \\ J \\ A \end{bmatrix} := S \mathbf{F} \mathbf{x}.$$ Interestingly, it is linear in $S$.

• $\mathbf{V}\mathbf{x}$: all other linear dynamics, $$ \begin{bmatrix} b_1 I_1 - \xi_1 J \\ b_2 I_2 - \epsilon I_1 - \xi_2 J \\ b_3 J - p_1 I_2 \\ b_4 A - p_2 J \end{bmatrix} = \begin{bmatrix} b_1 & 0 & -\xi_1 & 0 \\ -\epsilon & b_2 & -\xi_2 & 0 \\ 0 & -p_1 & b_3 & 0 \\ 0 & 0 & -p_2 & b_4 \end{bmatrix} \begin{bmatrix} I_1 \\ I_2 \\ J \\ A \end{bmatrix} := \mathbf{V} \mathbf{x} $$

Remark: I didn't prove it theoretically, but I used symbolic calculus to confirm that. You will need to prove that theoretically, but it's fastidious in terms of notation.

$$ \mathbf{f}(S^*, \mathbf{x}^*) - \mathbf{V} \mathbf{x}^* = 0 \Leftrightarrow S^* \mathbf{F} \mathbf{x}^* = \mathbf{V} \mathbf{x}^* \Leftrightarrow \mathbf{x}^* = S^* \mathbf{V}^{-1} \mathbf{F} \mathbf{x}^* $$

Then, the DFE is locally stable if the maximal eigenvalue of $\mathbf{V}^{-1}\mathbf{F} <1 \Leftrightarrow \mathcal{R}_0 <1$.

Global Asymptotic Stability of DFE

1. Differentiate the Lyapunov function: $$ \dot{L}(\mathbf{x}) = \mathbf{w}^\top \mathbf{V}^{-1} \dot{\mathbf{x}} $$

   Now, substitute the system dynamics by writing the new infection term as $\mathbf{f}(S, \mathbf{x}) = \frac{S(t)}{S^*} \mathbf{F}\mathbf{x}$. The full system for the compartments is $\dot{\mathbf{x}} = \frac{S(t)}{S^*} \mathbf{F}\mathbf{x} - \mathbf{V}\mathbf{x}$. Then, \begin{align*} \dot{L}(\mathbf{x}) &= \mathbf{w}^\top \mathbf{V}^{-1} \left( \frac{S(t)}{S^*} \mathbf{F}\mathbf{x} - \mathbf{V}\mathbf{x} \right) = \frac{S(t)}{S^*} \mathbf{w}^\top \mathbf{V}^{-1} \mathbf{F}\mathbf{x} - \mathbf{w}^\top (\mathbf{V}^{-1} \mathbf{V})\mathbf{x} \\ &= \frac{S(t)}{S^*} \mathbf{w}^\top (\mathbf{V}^{-1} \mathbf{F})\mathbf{x} - \mathbf{w}^\top \mathbf{x} \end{align*}

   Now we substitute the property of our weight vector, $\mathbf{w}^\top (\mathbf{V}^{-1}\mathbf{F}) = \mathcal{R}_0 \mathbf{w}^\top$, into the equation. $$ \dot{L}(\mathbf{x}) = \frac{S(t)}{S^*} (\mathcal{R}_0 \mathbf{w}^\top) \mathbf{x} - \mathbf{w}^\top \mathbf{x} = \left( \frac{S(t)}{S^*} \mathcal{R}_0 - 1 \right) \mathbf{w}^\top \mathbf{x} $$

2. An attracting region where the derivative is always negative: From the first equation of your system, $$\dot{S} = \mu - (\text{infection terms}) - \nu S.$$

   Since the infection terms are non-negative, we have the inequality

   $$\dot{S} \le \mu - \nu S.$$ Using Gronwall implies $S(t) \leq S^* = \mu/\nu$ for large $t$. i.e., the region $S \le S^*$ is attracting.

   Let $$\mathcal{S} = \left\{ (S, \mathbf{x}) \in \mathbb{R}_+^5 \mid S(t) \le S^* \right\}.$$

   Then, for any $(S,\mathbf{x}) \in \mathcal{S}$, we have $\frac{S(t)}{S^*} \le 1$.

   If we assume $\mathcal{R}_0 \le 1$, then the product $\frac{S(t)}{S^*} \mathcal{R}_0 \le 1 \cdot \mathcal{R}_0 \le 1$. Therefore, the term $\left( \frac{S(t)}{S^*} \mathcal{R}_0 - 1 \right)$ must be less than or equal to zero. Since we already established that $\mathbf{w}^\top \mathbf{x} \ge 0$, we have proven that, for any $(S, \mathbf{x}) \in \mathcal{S}$, $$ \dot{L}(\mathbf{x}) \leq 0.$$ This proves that the DFE is locally stable.

3. Apply LaSalle's Invariance Principle for global stability: To prove asymptotic stability, we must identify the largest invariant set in $\mathcal{S}$ where $\dot{L}(\mathbf{x}) = 0$.

   $\dot{L}(\mathbf{x}) = 0$ requires that $\left( \frac{S(t)}{S^*} \mathcal{R}_0 - 1 \right) \mathbf{w}^\top \mathbf{x} = 0$. This means that on the invariant set, for all time, either $\mathbf{x}(t)=\mathbf{0}$ or $\frac{S(t)}{S^*} \mathcal{R}_0 = 1$.

   • Case 1: $\mathcal{R}_0 < 1$. Since $S(t) \le S^*$, we have $\frac{S(t)}{S^*} \mathcal{R}_0 < 1 $. The condition $\frac{S(t)}{S^*} \mathcal{R}_0 = 1$ is impossible. Therefore, the only possibility for $\dot{L}(\mathbf{x})=0$ is that $\mathbf{w}^\top \mathbf{x} = 0$. Since $\mathbf{w}$ and $\mathbf{x}$ are non-negative vectors and $\mathbf{w}$ is non-zero, this forces $\mathbf{x}=\mathbf{0}$ (i.e., $I_1=I_2=J=A=0$). If the system is in a state where all infectious compartments are zero, the first ODE becomes $\dot{S} = \mu - \nu S$, whose only equilibrium is $S(t) = \mu/\nu = S^*$. Thus, the only invariant set is the DFE, $e_1$.

   • Case 2: $\mathcal{R}_0 = 1$. Now, $\dot{L} = 0$ implies either $\mathbf{x}(t)=\mathbf{0}$ (which leads to the DFE as above) or $S(t)=S^*$. Let's analyze the second possibility. If a trajectory stays in the set where $S(t)=S^*$ for all $t$, then we must have $\dot{S}=0$. Plugging $S=S^*$ into the first ODE gives $$0 = \mu - (\beta_1 S^* I_2 + \beta_2 S^* J + \beta_3 S^* A) - \nu S^*.$$

   Since $\mu = \nu S^*$, this simplifies to $$\beta_1 S^* I_2 + \beta_2 S^* J + \beta_3 S^* A = 0.$$

   As all parameters and variables are non-negative, this forces $I_2 = J = A = 0$. With these compartments being zero, the system for $I_1$ becomes $\dot{I_1} = -b_1 I_1$, which implies $I_1(t) \to 0$. Therefore, any trajectory must approach a state where all infectious compartments are zero. Again, the only invariant set is the DFE, $e_1$.

By LaSalle's Invariance Principle, since the only invariant set contained in $\mathcal{S}$ where $\{\dot{L}=0\}$ is the DFE, all solutions in the feasible region converge to the DFE, $e_1$, when $\mathcal{R}_0 \le 1$. Thus, the DFE is globally asymptotically stable.

Moreover, this region is forward invariant. To prove that, we can use Nagumo's theorem, which states that it is sufficient to prove that the vector field is pointing inward into your set on its border. We can look at the $1$D-dynamics of $S$ over the interval $[0, S^*]$. At the edges, one has

The equations for these infectious compartments can be written in the general form, for any $(S,\mathbf{x}) \in \mathcal{S}$,

• $\mathbf{f}(S, \mathbf{x})$: the nonlinear coupling between $S$ and the other quantities. For a given number of susceptible $S$, one can rewrite the function as, $$ \mathbf{f}(S, \mathbf{x}) = %\begin{bmatrix} %p \beta_1 S I_2 + q \beta_2 S J + r \beta_3 S A \\ %(1-p) \beta_1 S I_2 + (1 - q) \beta_2 S J + (1 - r) \beta_3 S A \\ %0 \\ %0 %\end{bmatrix} = S \begin{bmatrix} p \beta_1 I_2 + q \beta_2 J + r \beta_3 A \\ (1-p) \beta_1 I_2 + (1 - q) \beta_2 J + (1 - r) \beta_3 A \\ 0 \\ 0 \end{bmatrix} = S \begin{bmatrix} 0 & p\beta_1 & q\beta_2 & r\beta_3 \\ 0 & (1-p)\beta_1 & (1-q)\beta_2 & (1-r)\beta_3 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix}\begin{bmatrix} I_1 \\ I_2 \\ J \\ A \end{bmatrix} := S\, \mathbf{F} \mathbf{x}.$$

• $\mathbf{V}\mathbf{x}$: all other linear dynamics, $$ \begin{bmatrix} b_1 I_1 - \xi_1 J \\ b_2 I_2 - \epsilon I_1 - \xi_2 J \\ b_3 J - p_1 I_2 \\ b_4 A - p_2 J \end{bmatrix} = \begin{bmatrix} b_1 & 0 & -\xi_1 & 0 \\ -\epsilon & b_2 & -\xi_2 & 0 \\ 0 & -p_1 & b_3 & 0 \\ 0 & 0 & -p_2 & b_4 \end{bmatrix} \begin{bmatrix} I_1 \\ I_2 \\ J \\ A \end{bmatrix} := \mathbf{V} \mathbf{x} $$

Remark: I didn't prove it theoretically, but I used symbolic calculus to confirm that. You will need to prove that theoretically, but it's fastidious (due to the notation), so I skipped that. I used Python's library Sympy. P.S.: I can think about an efficient proof.

$$ \mathbf{f}(S^*, \mathbf{x}^*) - \mathbf{V} \mathbf{x}^* = 0 \Leftrightarrow S^* \mathbf{F} \mathbf{x}^* = \mathbf{V} \mathbf{x}^* \Leftrightarrow \mathbf{x}^* = S^* \mathbf{V}^{-1} \mathbf{F} \mathbf{x}^* \Leftrightarrow 0 = (S^* \mathbf{V}^{-1} \mathbf{F} - \mathbf{I}) \mathbf{x}^* $$

Then, the DFE is locally stable if the maximal eigenvalue of $\mathbf{V}^{-1}\mathbf{F} <1 \Leftrightarrow \mathcal{R}_0 <1$ by the Linearization principle.

Global Asymptotic Stability of DFE via Lyapunov argument

1. Differentiate the Lyapunov function: $$ \dot{L}(\mathbf{x}) = \mathbf{w}^\top \mathbf{V}^{-1} \dot{\mathbf{x}} $$

   Now, substitute the system dynamics by writing the new infection term as $\mathbf{f}(S, \mathbf{x}) = \frac{S(t)}{S^*} \mathbf{F}\mathbf{x}$. The full system for the compartments is $\dot{\mathbf{x}} = \frac{S(t)}{S^*} \mathbf{F}\mathbf{x} - \mathbf{V}\mathbf{x}$. Then, \begin{align*} \dot{L}(\mathbf{x}) &= \mathbf{w}^\top \mathbf{V}^{-1} \left( \frac{S(t)}{S^*} \mathbf{F}\mathbf{x} - \mathbf{V}\mathbf{x} \right) = \frac{S(t)}{S^*} \mathbf{w}^\top \mathbf{V}^{-1} \mathbf{F}\mathbf{x} - \mathbf{w}^\top (\mathbf{V}^{-1} \mathbf{V})\mathbf{x} \\ &= \frac{S(t)}{S^*} \mathbf{w}^\top (\mathbf{V}^{-1} \mathbf{F})\mathbf{x} - \mathbf{w}^\top \mathbf{x} \end{align*}

   Now we substitute the property of our weight vector, $\mathbf{w}^\top (\mathbf{V}^{-1}\mathbf{F}) = \mathcal{R}_0 \mathbf{w}^\top$, into the equation. $$ \dot{L}(\mathbf{x}) = \frac{S(t)}{S^*} (\mathcal{R}_0 \mathbf{w}^\top) \mathbf{x} - \mathbf{w}^\top \mathbf{x} = \left( \frac{S(t)}{S^*} \mathcal{R}_0 - 1 \right) \mathbf{w}^\top \mathbf{x} $$

2. An attracting region where the derivative is always negative:

   We know from the previous discussion that $$\mathcal{S} = \left\{ (S, \mathbf{x}) \in \mathbb{R}_+^5 \mid S(t) \le S^* \right\},$$

   is 'globally' attracting and forward invariant. Then, there exists some time $t\geq 0$ such that $S(t) \in \mathcal{S}$. (We have to be cautious here, need to think more.) We then only need to look at the behaviors of our Lyapunov function on $\mathcal{S}$. For any $S \in \mathcal{S}$, we have $\frac{S}{S^*} \le 1$.

   If we assume $\mathcal{R}_0 \le 1$, then the product $\frac{S}{S^*} \mathcal{R}_0 \leq \mathcal{R}_0 \leq 1$. Therefore, the term $\left( \frac{S(t)}{S^*} \mathcal{R}_0 - 1 \right)$ must be less than or equal to zero. Since we already established that $\mathbf{w}^\top \mathbf{x} \ge 0$, we have proven that, for any $(S, \mathbf{x}) \in \mathcal{S}$, $$ \dot{L}(\mathbf{x}) \leq 0.$$

3. Apply LaSalle's Invariance Principle for global stability: To prove asymptotic stability, we must identify the largest invariant set in $\mathcal{S}$ where $\dot{L}(\mathbf{x}) = 0$.

   $\dot{L}(\mathbf{x}) = 0$ requires that $\left( \frac{S(t)}{S^*} \mathcal{R}_0 - 1 \right) \mathbf{w}^\top \mathbf{x} = 0$. This means that on the invariant set, for all time, either $\mathbf{x}(t)=\mathbf{0}$ or $\frac{S(t)}{S^*} \mathcal{R}_0 = 1$.

   • Case 1: $\mathcal{R}_0 < 1$. Since $S(t) \le S^*$, we have $\frac{S(t)}{S^*} \mathcal{R}_0 < 1 $. The condition $\frac{S(t)}{S^*} \mathcal{R}_0 = 1$ is impossible. Therefore, the only possibility for $\dot{L}(\mathbf{x})=0$ is that $\mathbf{w}^\top \mathbf{x} = 0$. Since $\mathbf{w}$ and $\mathbf{x}$ are non-negative vectors and $\mathbf{w}$ is non-zero, this forces $\mathbf{x}=\mathbf{0}$ (i.e., $I_1=I_2=J=A=0$). If the system is in a state where all infectious compartments are zero, the first ODE becomes $\dot{S} = \mu - \nu S$, whose only equilibrium is $S(t) = \mu/\nu = S^*$. Thus, the only invariant set is the DFE, $e_1$.

   • Case 2: $\mathcal{R}_0 = 1$. Now, $\dot{L} = 0$ implies either $\mathbf{x}(t)=\mathbf{0}$ (which leads to the DFE as above) or $S(t)=S^*$. Let's analyze the second possibility. If a trajectory stays in the set where $S(t)=S^*$ for all $t$, then we must have $\dot{S}=0$. Plugging $S=S^*$ into the first ODE gives $$0 = \mu - (\beta_1 S^* I_2 + \beta_2 S^* J + \beta_3 S^* A) - \nu S^*.$$

   Since $\mu = \nu S^*$, this simplifies to $$\beta_1 S^* I_2 + \beta_2 S^* J + \beta_3 S^* A = 0.$$

   As all parameters and variables are non-negative, this forces $I_2 = J = A = 0$. With these compartments being zero, the system for $I_1$ becomes $\dot{I_1} = -b_1 I_1$, which implies $I_1(t) \to 0$. Therefore, any trajectory must approach a state where all infectious compartments are zero. Again, the only invariant set is the DFE, $e_1$.

   By LaSalle's Invariance Principle, since the only invariant set contained in $\mathcal{S}$ where $\{\dot{L}=0\}$ is the DFE, all solutions in the feasible region converge to the DFE, $e_1$, when $\mathcal{R}_0 \le 1$. Thus, the DFE is globally asymptotically stable.

import sympy   mu, nu = sympy.symbols('mu nu', positive=True) epsilon, p1, p2 = sympy.symbols('varepsilon p_1 p_2', positive=True) alpha, xi1, xi2 = sympy.symbols('alpha xi_1 xi_2', positive=True) b1 = epsilon + mu b2 = p1 + nu b3 = xi1 + xi2 + p2 + nu b4 = alpha + nu V = sympy.Matrix([ [b1, 0, -xi1, 0], [-epsilon, b2, -xi2, 0], [0, -p1, b3, 0], [0, 0, -p2, b4] ]) adj_V = V.adjugate() all_elements_non_negative = True for i in range(adj_V.rows): for j in range(adj_V.cols): element = sympy.expand(adj_V[i, j]) is_non_neg = element.is_nonnegative print(f"Element ({i},{j}) is non-negative? {is_non_neg}") if not is_non_neg: all_elements_non_negative = False print(f" --> Element expression: {element}") print(f"\nIs the entire adjugate matrix non-negative? {all_elements_non_negative}") 

General remarks

Usually, when I tackle this kind of problem, I'm looking at what part of the space is forward invariant first. It allows me to better understand what I have to consider in my analysis. Here, the SIS-like models are usually confined in $\mathbb{R}^N_+$ since the quantities are the number of individuals in each category.

An attracting region of interest:

Let $$\mathcal{S} = \left\{ (S, \mathbf{x}) \in \mathbb{R}_+^5 \mid S(t) \le S^* \right\} = \left[ 0 , S^* \right] \times \mathbb{R}_+^4.$$

From the first equation of your system, $$\dot{S} = \mu - (\text{infection terms}) - \nu S.$$

Since the infection terms are non-negative, we have the inequality $$\dot{S} \le \mu - \nu S.$$

Using Gronwall implies $S(t) \leq S^* = \mu/\nu$ for large $t$. i.e., the region $\mathcal{S}$ is attracting.

Moreover, this region is forward invariant. To prove that, we can use Nagumo's theorem, which states that it is sufficient to prove that the vector field is pointing inward into your set on its border. We can look at the $1$D-dynamics of $S$.

\begin{align*} \dot{S}_{|S = S^*} &= \mu - \beta_1 S^* I_2 - \beta_2 S^* J - \beta_3 S^* A - \nu S^*\\ &\leq \mu - \nu S^* = \mu - \nu \frac{\mu}{\nu} = 0. \end{align*} and \begin{align*} \dot{S}_{|S = 0} &= \mu - \beta_1 S^* I_2 - \beta_2 S^* J - \beta_3 S^* A - \nu S^*\\ &= \mu \geq 0. \end{align*}

Then, $\mathcal{S}$ is attracting and is forward invariant.

This fact allows us to simplify our life and only study the dynamics on $\mathcal{S}$. We will, in the following, even remove $S$ explicitly from our study since we will restrict ourselves to $\mathcal{S}$.

import sympy mu, nu = sympy.symbols('mu nu', positive=True) epsilon, p1, p2 = sympy.symbols('varepsilon p_1 p_2', positive=True) alpha, xi1, xi2 = sympy.symbols('alpha xi_1 xi_2', positive=True) b1 = epsilon + mu b2 = p1 + nu b3 = xi1 + xi2 + p2 + nu b4 = alpha + nu   V = sympy.Matrix([ [b1, 0, -xi1, 0], [-epsilon, b2, -xi2, 0], [0, -p1, b3, 0], [0, 0, -p2, b4] ])   adj_V = V.adjugate()   all_elements_non_negative = True for i in range(adj_V.rows): for j in range(adj_V.cols): element = sympy.expand(adj_V[i, j]) is_non_neg = element.is_nonnegative print(f"Element ({i},{j}) is non-negative? {is_non_neg}") if not is_non_neg: all_elements_non_negative = False print(f" --> Element expression: {element}")   print(f"Is the entire adjugate matrix non-negative? {all_elements_non_negative}") 

1. Differentiate the Lyapunov function: $$ \dot{L}(\mathbf{x}) = \mathbf{w}^\top \mathbf{V}^{-1} \dot{\mathbf{x}} $$

   Now, substitute the system dynamics by writing the new infection term as $\mathbf{f}(S, \mathbf{x}) = \frac{S(t)}{S^*} \mathbf{F}\mathbf{x}$. The full system for the compartments is $\dot{\mathbf{x}} = \frac{S(t)}{S^*} \mathbf{F}\mathbf{x} - \mathbf{V}\mathbf{x}$. Then, \begin{align*} \dot{L}(\mathbf{x}) &= \mathbf{w}^\top \mathbf{V}^{-1} \left( \frac{S(t)}{S^*} \mathbf{F}\mathbf{x} - \mathbf{V}\mathbf{x} \right) = \frac{S(t)}{S^*} \mathbf{w}^\top \mathbf{V}^{-1} \mathbf{F}\mathbf{x} - \mathbf{w}^\top (\mathbf{V}^{-1} \mathbf{V})\mathbf{x} \\ &= \frac{S(t)}{S^*} \mathbf{w}^\top (\mathbf{V}^{-1} \mathbf{F})\mathbf{x} - \mathbf{w}^\top \mathbf{x} \end{align*}

   Now we substitute the property of our weight vector, $\mathbf{w}^\top (\mathbf{V}^{-1}\mathbf{F}) = \mathcal{R}_0 \mathbf{w}^\top$, into the equation. $$ \dot{L}(\mathbf{x}) = \frac{S(t)}{S^*} (\mathcal{R}_0 \mathbf{w}^\top) \mathbf{x} - \mathbf{w}^\top \mathbf{x} = \left( \frac{S(t)}{S^*} \mathcal{R}_0 - 1 \right) \mathbf{w}^\top \mathbf{x} $$

2. An attracting region where the derivative is always negative: From the first equation of your system, $$\dot{S} = \mu - (\text{infection terms}) - \nu S.$$

   Since the infection terms are non-negative, we have the inequality

   $$\dot{S} \le \mu - \nu S.$$ Using Gronwall implies $S(t) \leq S^* = \mu/\nu$ for large $t$. i.e., the region $S \le S^*$ is attracting.

   Let $$\mathcal{S} = \left\{ (S, \mathbf{x}) \in \mathbb{R}_+^5 \mid S(t) \le S^* \right\}.$$

   Then, for any $(S,\mathbf{x}) \in \mathcal{S}$, we have $\frac{S(t)}{S^*} \le 1$.

   If we assume $\mathcal{R}_0 \le 1$, then the product $\frac{S(t)}{S^*} \mathcal{R}_0 \le 1 \cdot \mathcal{R}_0 \le 1$. Therefore, the term $\left( \frac{S(t)}{S^*} \mathcal{R}_0 - 1 \right)$ must be less than or equal to zero. Since we already established that $\mathbf{w}^\top \mathbf{x} \ge 0$, we have proven that, for any $(S, \mathbf{x}) \in \mathcal{S}$, $$ \dot{L}(\mathbf{x}) \leq 0.$$ This proves that the DFE is locally stable.

3. Apply LaSalle's Invariance Principle for global stability: To prove asymptotic stability, we must identify the largest invariant set in $\mathcal{S}$ where $\dot{L}(\mathbf{x}) = 0$.

   $\dot{L}(\mathbf{x}) = 0$ requires that $\left( \frac{S(t)}{S^*} \mathcal{R}_0 - 1 \right) \mathbf{w}^\top \mathbf{x} = 0$. This means that on the invariant set, for all time, either $\mathbf{x}(t)=\mathbf{0}$ or $\frac{S(t)}{S^*} \mathcal{R}_0 = 1$.

   • Case 1: $\mathcal{R}_0 < 1$. Since $S(t) \le S^*$, we have $\frac{S(t)}{S^*} \mathcal{R}_0 < 1 $. The condition $\frac{S(t)}{S^*} \mathcal{R}_0 = 1$ is impossible. Therefore, the only possibility for $\dot{L}(\mathbf{x})=0$ is that $\mathbf{w}^\top \mathbf{x} = 0$. Since $\mathbf{w}$ and $\mathbf{x}$ are non-negative vectors and $\mathbf{w}$ is non-zero, this forces $\mathbf{x}=\mathbf{0}$ (i.e., $I_1=I_2=J=A=0$). If the system is in a state where all infectious compartments are zero, the first ODE becomes $\frac{dS}{dt} = \mu - \nu S$, whose only equilibrium is $S(t) = \mu/\nu = S^*$. Thus, the only invariant set is the DFE, $e_1$.

   • Case 2: $\mathcal{R}_0 = 1$. Now, $\dot{L} = 0$ implies either $\mathbf{x}(t)=\mathbf{0}$ (which leads to the DFE as above) or $S(t)=S^*$. Let's analyze the second possibility. If a trajectory stays in the set where $S(t)=S^*$ for all $t$, then we must have $\dot{S}=0$. Plugging $S=S^*$ into the first ODE gives $$0 = \mu - (\beta_1 S^* I_2 + \beta_2 S^* J + \beta_3 S^* A) - \nu S^*.$$

   Since $\mu = \nu S^*$, this simplifies to $$\beta_1 S^* I_2 + \beta_2 S^* J + \beta_3 S^* A = 0.$$

   As all parameters and variables are non-negative, this forces $I_2 = J = A = 0$. With these compartments being zero, the system for $I_1$ becomes $\dot{I_1} = -b_1 I_1$, which implies $I_1(t) \to 0$. Therefore, any trajectory must approach a state where all infectious compartments are zero. Again, the only invariant set is the DFE, $e_1$.

1. Differentiate the Lyapunov function: $$ \dot{L}(\mathbf{x}) = \mathbf{w}^\top \mathbf{V}^{-1} \dot{\mathbf{x}} $$

   Now, substitute the system dynamics by writing the new infection term as $\mathbf{f}(S, \mathbf{x}) = \frac{S(t)}{S^*} \mathbf{F}\mathbf{x}$. The full system for the compartments is $\dot{\mathbf{x}} = \frac{S(t)}{S^*} \mathbf{F}\mathbf{x} - \mathbf{V}\mathbf{x}$. Then, \begin{align*} \dot{L}(\mathbf{x}) &= \mathbf{w}^\top \mathbf{V}^{-1} \left( \frac{S(t)}{S^*} \mathbf{F}\mathbf{x} - \mathbf{V}\mathbf{x} \right) = \frac{S(t)}{S^*} \mathbf{w}^\top \mathbf{V}^{-1} \mathbf{F}\mathbf{x} - \mathbf{w}^\top (\mathbf{V}^{-1} \mathbf{V})\mathbf{x} \\ &= \frac{S(t)}{S^*} \mathbf{w}^\top (\mathbf{V}^{-1} \mathbf{F})\mathbf{x} - \mathbf{w}^\top \mathbf{x} \end{align*}

   Now we substitute the property of our weight vector, $\mathbf{w}^\top (\mathbf{V}^{-1}\mathbf{F}) = \mathcal{R}_0 \mathbf{w}^\top$, into the equation. $$ \dot{L}(\mathbf{x}) = \frac{S(t)}{S^*} (\mathcal{R}_0 \mathbf{w}^\top) \mathbf{x} - \mathbf{w}^\top \mathbf{x} = \left( \frac{S(t)}{S^*} \mathcal{R}_0 - 1 \right) \mathbf{w}^\top \mathbf{x} $$

2. An attracting region where the derivative is always negative: From the first equation of your system, $$\dot{S} = \mu - (\text{infection terms}) - \nu S.$$

   Since the infection terms are non-negative, we have the inequality

   $$\dot{S} \le \mu - \nu S.$$ Using Gronwall implies $S(t) \leq S^* = \mu/\nu$ for large $t$. i.e., the region $S \le S^*$ is attracting.

   Let $$\mathcal{S} = \left\{ (S, \mathbf{x}) \in \mathbb{R}_+^5 \mid S(t) \le S^* \right\}.$$

   Then, for any $(S,\mathbf{x}) \in \mathcal{S}$, we have $\frac{S(t)}{S^*} \le 1$.

   If we assume $\mathcal{R}_0 \le 1$, then the product $\frac{S(t)}{S^*} \mathcal{R}_0 \le 1 \cdot \mathcal{R}_0 \le 1$. Therefore, the term $\left( \frac{S(t)}{S^*} \mathcal{R}_0 - 1 \right)$ must be less than or equal to zero. Since we already established that $\mathbf{w}^\top \mathbf{x} \ge 0$, we have proven that, for any $(S, \mathbf{x}) \in \mathcal{S}$, $$ \dot{L}(\mathbf{x}) \leq 0.$$ This proves that the DFE is locally stable.

3. Apply LaSalle's Invariance Principle for global stability: To prove asymptotic stability, we must identify the largest invariant set in $\mathcal{S}$ where $\dot{L}(\mathbf{x}) = 0$.

   $\dot{L}(\mathbf{x}) = 0$ requires that $\left( \frac{S(t)}{S^*} \mathcal{R}_0 - 1 \right) \mathbf{w}^\top \mathbf{x} = 0$. This means that on the invariant set, for all time, either $\mathbf{x}(t)=\mathbf{0}$ or $\frac{S(t)}{S^*} \mathcal{R}_0 = 1$.

   • Case 1: $\mathcal{R}_0 < 1$. Since $S(t) \le S^*$, we have $\frac{S(t)}{S^*} \mathcal{R}_0 < 1 $. The condition $\frac{S(t)}{S^*} \mathcal{R}_0 = 1$ is impossible. Therefore, the only possibility for $\dot{L}(\mathbf{x})=0$ is that $\mathbf{w}^\top \mathbf{x} = 0$. Since $\mathbf{w}$ and $\mathbf{x}$ are non-negative vectors and $\mathbf{w}$ is non-zero, this forces $\mathbf{x}=\mathbf{0}$ (i.e., $I_1=I_2=J=A=0$). If the system is in a state where all infectious compartments are zero, the first ODE becomes $\dot{S} = \mu - \nu S$, whose only equilibrium is $S(t) = \mu/\nu = S^*$. Thus, the only invariant set is the DFE, $e_1$.

   • Case 2: $\mathcal{R}_0 = 1$. Now, $\dot{L} = 0$ implies either $\mathbf{x}(t)=\mathbf{0}$ (which leads to the DFE as above) or $S(t)=S^*$. Let's analyze the second possibility. If a trajectory stays in the set where $S(t)=S^*$ for all $t$, then we must have $\dot{S}=0$. Plugging $S=S^*$ into the first ODE gives $$0 = \mu - (\beta_1 S^* I_2 + \beta_2 S^* J + \beta_3 S^* A) - \nu S^*.$$

   Since $\mu = \nu S^*$, this simplifies to $$\beta_1 S^* I_2 + \beta_2 S^* J + \beta_3 S^* A = 0.$$

   As all parameters and variables are non-negative, this forces $I_2 = J = A = 0$. With these compartments being zero, the system for $I_1$ becomes $\dot{I_1} = -b_1 I_1$, which implies $I_1(t) \to 0$. Therefore, any trajectory must approach a state where all infectious compartments are zero. Again, the only invariant set is the DFE, $e_1$.