How does this expression follow algebraically from the last one? (continued)

Question

Continuing from here: How does this expression follow algebraically from the last one?

The "new" system: \begin{align*} \dot S &= \Lambda - (\beta_1 S I_2 +\beta_2 S J+ \beta_3 S A )-\mu S \\ \dot I_1 &= p\beta_1 S I_2 +q\beta_2 S J +r \beta_3 S A +\xi_1 J -b_1 I_1\\ \dot I_2 &= (1-P)\beta_1 S I_2 +(1-q)\beta_2 S J+(1-r) \beta_3 S A +\epsilon I_1 +\xi_2 J -b_2 I_2\\ \dot J &= p_1 I_2 -b_3 J\\ \dot A &= p_2 J - b_4 A \end{align*}

Reproduction number:

\begin{align} \mathcal{R}_0 &= \frac{\Lambda\left[ \beta_1 b_3 b_4 \left( \epsilon p +b_1(\left( 1-p\right)\right)+\beta_2 p_1 b_4 \left( \epsilon q +b_1(\left( 1-q\right)\right)+\beta_3 p_1 p_2 \left( \epsilon r +b_1(\left( 1-r\right)\right) \right]}{\mu b_4 \left[ b_1 b_2 b_3 - p_1\left( \epsilon \xi_1 +b_1 \xi_2 \right) \right] } \end{align}

Equilibrium point:

\begin{align} S^* &= \frac{\Lambda}{\mu \mathcal{R}_0}\\[2ex] I_1^* &= \frac{1}{b_1}\left[ p \beta_1 \frac{\Lambda b_3 b_4}{\left(\beta_1 b_3 b_4+\beta_2 p_1 b_4+\beta_3 p_1 p_2 \right)J^* +\mu b_4 p_1}\right.\\[1ex] &\quad\;\;+\left. q \beta_2 \frac{\Lambda b_4 p_1}{\left(\beta_1 b_3 b_4+\beta_2 p_1 b_4+\beta_3 p_1 p_2 \right)J^* +\mu b_4 p_1}\right.\\[1ex] &\quad\;\;+\left. r \beta_3 \frac{\Lambda p_1 p_2}{\left(\beta_1 b_3 b_4+\beta_2 p_1 b_4+\beta_3 p_1 p_2 \right)J^* +\mu b_4 p_1}+\xi_1\right]J^*\\[2ex] I_2^* &= \frac{b_3}{p_1}J^*\\[2ex] J^*&= \frac{\mu b_4 p_1 }{\beta_1 b_3 b_4 +\beta_2 p_1 b_4 +\beta_3 p_1 p_2}\left(\mathcal{R}_0-1\right)\\[2ex] A^*&=\frac{p_2}{b_4}J^* \end{align}

Theorem: If $p=q=r$ and $\mathcal{R}_0 >1$ then the above equilibrium point is globally stable.

To prove this:

Define the following Lyapunov function

\begin{align} V &= S-S^* \ln S+ B\left( I_1-{I_1}^* \ln I_1\right) +C\left(I_2-{I_2}^* \ln I_2\right) + D\left( J -J^* \ln J\right)\\ &+E\left(A-A^* \ln A\right). \end{align}

The derivative of $V$

\begin{align*} \dot V &=\left(1-\frac{S^*}{S}\right) \dot S + B\left(1-\frac{{I_1}^*}{I_1} \right)\dot I_1 + C\left(1-\frac{{I_2}^*}{I_2}\right)\dot I_2+D\left(1-\frac{J^*}{J} \right)\dot J\\[1ex] &+ E\left(1-\frac{A^*}{A}\right) \dot A \end{align*}

where: \begin{align*} B &= \frac{\epsilon}{\epsilon p +b_1(1-p)}\\[1ex] C&= \frac{b_1}{\epsilon p +b_1(1-p)}\\[1ex] D &= \frac{b_1 b_2}{p_1[\epsilon p +b_1(1-p)]} -\frac{\beta_1 S^*}{p_1}\\[1ex] E &= \frac{\beta_3 S^*}{b4} \end{align*}

We get to the following:

\begin{align*} \dot V &= -\mu S^* \frac{\left(1-x\right)^2}{x}+ \left[ \beta_1 S^* {I_2}^*+\beta_2 S^* J^*+ \beta_3 S^* A^* +B p\beta_1 S^* {I_2}^* + B q \beta_2 S^* J^*\right.\\[1ex] &+\left. B r \beta_3 S^* A^*+ B \xi_1 J^*+ C(1-p)\beta_1 S^* {I_2}^*+C(1-q)\beta_2 S^* J^*+C(1-r)\beta_3 S^* A^*\right.\\[1ex] &+\left. C \epsilon {I_1}^*+C \xi_2 J^*+D p_1 {I_2}^*+ E p_2 J^* \right]-x\left[ C(1-p)\beta_1 S^* {I_2}^* \right] -\frac{xz}{y}B p\beta_1 S^* {I_2}^*\\[1ex] & -\frac{xu}{y}B q \beta_2 S^* J^* -\frac{xv}{y}B r \beta_3 S^* A^*-\frac{u}{y}B \xi_1 J^* - \frac{xu}{z}C(1-q)\beta_2 S^* J^*\\[1ex] &- \frac{xv}{z}C(1-r)\beta_3 S^* A^* - \frac{y}{z}C \epsilon {I_1}^* - \frac{u}{z}C \xi_2 J^*-\frac{z}{u}D p_1 {I_2}^*-\frac{u}{v}E p_2 J^*\\[1ex] &- \frac{1}{x}\left[\beta_1 S^* {I_2}^*+ \beta_2 S^* J^*+\beta_3 S^* A^*\right]\\[2ex] \end{align*}

How do we get this expression to the 'final form' like in the question I referred at the beginning of this question?

Answer 1

The "final form" expression in the previous question leverages the arithmetic-geometric mean inequality to guarantee that signs of various $xyzu$ expressions are non-positive (which seems to important to the argument in the cited article). AM-GM tells us that $$\sqrt[n]{x_1x_2\cdots x_n}\;\leq\;\frac{x_1+x_2+\cdots+x_n}{n}$$ When, in particular, we have terms whose product is $1$, we can write $$n\leq x_1+x_2+\cdots+x_n\quad\to\quad n-x_1-x_2-\cdots-x_n \leq 0 \tag{1}$$ So, the article's authors augment certain expressions with terms like $1-1/x$, $1-y/z$, $1-z/u$ as needed, dutifully also subtracting-off the terms for balance, to get the form $(1)$. For instance, since $\frac{xz}{y}\cdot\frac{1}{x}\cdot\frac{y}{z}=1$, we may be inspired to write $$\begin{align} 1-\frac{xz}{y} \quad&=\quad 1-\frac{xz}{y}+\color{red}{1-\frac1x}+\color{blue}{1-\frac{y}{z}} \quad - \left(\color{red}{1-\frac1x}\right)-\left(\color{blue}{1-\frac{y}{z}}\right) \\[0.5em] &=\quad\underbrace{3-\frac{xz}{y}-\frac1x-\frac{y}{z}}_{\text{non-positive, by $(1)$}} \quad\underbrace{- \left(\color{red}{1-\frac1x}\right)-\left(\color{blue}{1-\frac{y}{z}}\right)}_{\text{hope these cancel elsewhere}} \tag2 \end{align}$$ It's a clever strategy, and it works surprisingly well: all those subtracted-off bits actually do cancel. The strategy works in this case, too.

---

To start, let's make a first pass at grouping terms:

$$\begin{align} \dot V = &-\mu S^* \frac{(1-x)^2}{x} \\[0.5em] &+ B \left(\begin{array}{l} \phantom{+}p\beta_1 S^* I_2^* \left(1-\frac{xz}{y}\right) \\ + q \beta_2 S^* J^*\left(1-\frac{xu}{y}\right) \\[1ex] + r \beta_3 S^* A^*\left(1-\frac{xv}{y}\right)\\ + \xi_1 J^* \left(1-\frac{u}{y}\right)\end{array}\right) + C\left(\begin{array}{l} \phantom{+}(1-p)\beta_1 S^* I_2^* (1-x)\\ + (1-q)\beta_2 S^* J^* \left(1-\frac{xu}{z}\right)\\ + (1-r)\beta_3 S^* A^* \left(1-\frac{xv}{z}\right)\\[1ex] + \epsilon I_1^*\left(1- \frac{y}{z}\right) \\ + \xi_2 J^*\left(1- \frac{u}{z}\right) \end{array}\right)\\[0.5em] &+D p_1 I_2^* \left(1-\frac{z}{u}\right) + E p_2 J^* \left(1-\frac{u}{v}\right) \\[0.5em] &+\left(\beta_1 S^* I_2^*+ \beta_2 S^* J^*+\beta_3 S^* A^*\right)\left(1-\frac1x\right) \end{align}\tag{3}$$

As our second pass, we'll augment the $xyzuv$ factors into form $(1)$ with $1-\frac1x$, $1 - \frac{u}{v}$, $1-\frac{z}{u}$, and $1-\frac{y}{z}$. We'll gather the subtracted-off multiples together with already-existing terms having those factors.

\begin{align*} \dot V = &-\mu S^* \frac{(1-x)^2}{x} \tag{4.0}\\[0.5em] &+ B \left(\begin{array}{l} \phantom{+}p\beta_1 S^* I_2^* \left(3-\frac{xz}{y}-\frac1x-\frac{y}{z}\right) \\ + q \beta_2 S^* J^*\left(4-\frac{xu}{y}-\frac{1}{x}-\frac{y}{z}-\frac{z}{u}\right) \\[1ex] + r \beta_3 S^* A^*\left(5-\frac{xv}{y}-\frac{1}{x}-\frac{y}{z}-\frac{z}{u}-\frac{u}{v}\right)\\ + \xi_1 J^* \left(3-\frac{u}{y}-\frac{z}{u}-\frac{y}{z}\right)\end{array}\right) \tag{4.1}\\[0.5em] &+ C\left(\begin{array}{l} \phantom{+}(1-p)\beta_1 S^* I_2^* \left(2-x-\frac1x\right)\\ + (1-q)\beta_2 S^* J^* \left(3-\frac{xu}{z}-\frac1x-\frac{z}{u}\right)\\ + (1-r)\beta_3 S^* A^* \left(4-\frac{xv}{z}-\frac1x-\frac{z}{u}-\frac{u}{v}\right) \\[1ex] + \xi_2 J^*\left(2- \frac{u}{z}-\frac{z}{u}\right) \end{array}\right) \tag{4.2} \\[0.5em] &+\left(1-\frac1x\right)S^*\left(\begin{array}{l} \phantom{-}\beta_1 I_2^*+ \beta_2 J^*+\beta_3 A^* \\ -Bp\beta_1 I_2^* - Bq \beta_2 J^* - Br \beta_3 A^* \\ -C(1-p)\beta_1 I_2^*- C(1-q)\beta_2 J^* \\-C(1-r)\beta_3 A^* \end{array}\right) \tag{4.3} \\[0.5em] &+\left(1 - \frac{u}{v}\right)\left(\begin{array}{l} \phantom{-}E p_2 J^* \\ - Br \beta_3 S^* A^* - C (1-r)\beta_3 S^* A^* \end{array}\right) \tag{4.4}\\[0.5em] &+\left(1-\frac{z}{u}\right)\left(\begin{array}{l} \phantom{-}D p_1 I_2^* \\ - Bq \beta_2 S^* J^* - Br \beta_3 S^* A^* - B\xi_1 J^* \\ - C (1-q)\beta_2 S^* J^* - C(1-r)\beta_3 S^* A^* - C\xi_2 J^* \end{array}\right) \tag{4.5}\\[0.5em] &+\left(1-\frac{y}{z}\right)\left(\begin{array}{l} \phantom{-}C\epsilon I_1^* \\ -Bp\beta_1 S^* I_2^*-Bq \beta_2 S^* J^*-Br \beta_3 S^* A^*-B\xi_1 J^*\end{array}\right) \tag{4.6} \end{align*}

From here, the goal is to show each of the sub-expressions $(4.3)$-$(4.6)$ vanish under the assumption $p=q=r$. Note that this assumption, along with the definitions of $B$ and $C$, implies $$B p + C (1-p) = B q + C (1-q) = B r + C (1-r ) = 1 \tag{$\star$}$$ Also, $$\begin{align} R_0 &= \frac{\Lambda ( \epsilon p + b_1 (1 - p))( \beta_1 b_3 b_4 + \beta_2 b_4 p_1 + \beta_3 p_1 p_2)}{ \mu b_4 (b_1 b_2 b_3 - p_1 (\epsilon \xi_1 + b_1 \xi_2))} \\[1em] \implies \quad S^* &= \frac{ b_4 (C b_2 b_3 - p_1 (B\xi_1 + C \xi_2))}{\beta_1 b_3 b_4 + \beta_2 b_4 p_1 + \beta_3 p_1 p_2} \tag{$\star\star$} \end{align}$$ Moreover, $$\begin{align} I_1^* &= J^* \frac{1}{b_1} \left(\frac{p \beta_1 b_3 b_4 + q \beta_2 b_4 p_1 + r \beta_3 p_1 p_2}{b_4 p_1} S^* + \xi_1\right) \tag{$\star\star\star$} \end{align}$$ Relation $(\star)$ makes light work of $(4.3)$: $$\begin{align} &\phantom{-}(\beta_1 I_2^* + \beta_2 J^* + \beta_3 A^*) \\ &-(B p + C (1 - p)) \beta_1 I_2^* - (B q + C (1 - q)) \beta_2 J^* - (B r + C (1 - r)) \beta_3 A^* \\[0.5em] =&\phantom{-}(\beta_1 I_2^* + \beta_2 J^* + \beta_3 A^*) \\ &-1\cdot \beta_1 I_2^* - 1\cdot \beta_2 J^* - 1\cdot \beta_3 A^* \\[0.5em] =&\phantom{-}0 \end{align} \tag{5}$$ And just a little more effort handles $(4.4)$: $$\begin{align} &\phantom{-}E p_2 J^* - (B r + C (1 - r)) \beta_3 S^* A^* \\[0.5em] =&\phantom{-}\frac{\beta_3}{b_4} S^* \cdot p_2 J^* - 1\cdot\beta_3 S^* \cdot \frac{p_2}{b_4} J^*\\[0.5em] =&\phantom{-}0 \end{align} \tag{6}$$ Then, $(4.5)$ is a bit more involved, but vanishes by $(\star\star)$: $$\begin{align} &\phantom{-}D p_1 I_2^* \\ &- (B q + C (1 - q)) \beta_2 S^* J^* -(B r + C (1 - r)) \beta_3 S^* A^* -B \xi_1 J^* -C \xi_2 J^*\\[0.5em] =&\phantom{-}J^* \left(Db_3 - 1\cdot \beta_2 S^* -1\cdot \beta_3 S^* \frac{p_2}{b_4} -B \xi_1 -C \xi_2 \right) \\ =&\phantom{-}J^* \left( \frac{1}{p_1}\left(C b_2 - \beta_1 S^*\right) b_3 - \beta_2 S^* - \beta_3 S^* \frac{p_2}{b_4} - B \xi_1 - C \xi_2 \right) \\ =&\phantom{-}J^* \frac{1}{b_4p_1}\left( b_4 ( C b_2 b_3 - p_1(B \xi_1 +C \xi_2)) - S^* ( \beta_1 b_3 b_4 + \beta_2 b_4 p_1 +\beta_3 p_1 p_2 ) \right) \\ =&\phantom{-}0 \end{align} \tag{7}$$ Finally, $(\star\star\star)$, and the observation that $C\epsilon/b_1 = B$, brings $(4.6)$ home: $$\begin{align} &\phantom{-}C \epsilon I_1^* - B\left( p \beta_1 S^* I_2^* + q \beta_2 S^* J^* + r \beta_3 S^* A^* + \xi_1 J^* \right) \\[0.5em] =&\phantom{-} C \frac{\epsilon}{b_1} (\cdots) J^* - B \left( p \beta_1 S^* \frac{b_3}{p_1}J^* + q \beta_2 S^* J^* + r \beta_3 S^* \frac{p_2}{b_4}J^* + \xi_1 J^* \right)\\[0.5em] =&\phantom{-} B J^*\left( (\cdots) - \left(\frac{1}{b_4p_1} S^* ( p \beta_1 b_3 b_4 + q \beta_2 b_4 p_1 + r \beta_3 p_1 p_2 ) + \xi_1 \right)\right) \\[0.5em] =&\phantom{-}0 \end{align} \tag{8}$$

Consequently, sub-expressions $(4.0)$-$(4.2)$ alone would appear to constitute an appropriate counterpart of the "final form" expression from the previous question.

(The reader is invited to apply the augmentation strategy and show that the subtracted-off bits cancel explicitly, instead of resorting to brute-force symbol-crunching for Mathematica (as I did), to verify the equality in the previous question.)