Proving $S_1-S_2$ is monotone decreasing

Question

I have system number (1) as:

$S_1’=-bS_1I_1+(1-c)aI_1$

$I_1’=bS_1I_1-aI_1$

And on the other hand I have system number (2) as:

$S_2’=-bS_2I_2+aI_2$

$I_2’=bS_2I_2-aI_2$

All parameters $a,b,c>0$.

Both systems have the same initial conditions as in $S_1(0)=S_2(0)=s_0>0$ and $I_1(0)=I_2(0)=i_0>0$.

I want to prove that $S_1-S_2$ is decreasing for $s_0>\frac{a}{b}$. What I proved so far:

1. I proved $S_1-S_2$ and $I_1-I_2$ are negative for all $t>0$.
2. Using the initial conditions then I have $S_1= \frac{(1-c)a}{b}+(s_0- \frac{(1-c)a}{b}$$ )e^{-\int_{0}^{t} bI_1(t) \ dt}$.
3. Similarly, I obtained $S_2= \frac{a}{b}+(s_0- \frac{a}{b}$$ )e^{-\int_{0}^{t} bI_2(t) \ dt}$.
4. From 2 and 3 we conclude that $S_1,S_2$ are monotone decreasing and that for $s_0>\frac{a}{b}$ we have $\frac{-ca}{b}<S_1-S_2<0$.
5. Initially we have $(S_1-S_2)’<0$ and because $S_1,S_2$ are $C^\infty$ then $(S_1-S_2)$ decreases initially and $$\lim_{t\rightarrow \infty}(S_1-S_2)’(t)=0.$$
6. I expressed $(S_1-S_2)’=F(t)(S_1-S_2)+G(t)$ where $G(t),F(t)$ are negative ($F(t)$ starts being negative from a certain point $t_0$ in case that $F(0)>0$) and descending toward finite limit.

Despite all that I can’t seem to prove that $(S_1-S_2)(t)$ is decreasing for $t>0$. I checked this via simulations and indeed it is the case but I’m trying by contradiction and can’t find something that can lead to a contradiction in anything. Help is appreciated!

Answer 1

It looks like it is false in general. A particular counterexample is $b=1, a=30, c=0.5, s_0=31, i_0=1$ where the monotonicity reverses for $t>0.95$. Below I'll try to explain how I found that and what is behind that effect.

We will change the variables like in my answer to Brenda's previous post and work with the "implicit solution" $$ S_\sigma(\xi)=-\sigma+(s+\sigma)e^{-\xi}, \\ I_\sigma(\xi)=i-\sigma\xi+(s+\sigma)(1-e^{-\xi}), \\ t_\sigma(\xi)=\int_0^\xi\frac{d\eta}{i-\sigma\eta+(s+\sigma)(1-e^{-\eta})}\,, $$ where the meaningful values of $\xi$ are those for which $I_\sigma>0$ on $[0,\xi]$.

The current conjecture is equivalent to the statement that for $s,\sigma>0$, we have $$ -\frac{dS_{\sigma}}{dt_{\sigma}}\ge -\frac{dS_0}{dt_0} $$ when $t_{\sigma}=t_0$.

The last condition (the agreement of time) is equivalent to the differential relation between the parameters $\xi_\sigma$ and $\xi_0$, which is $$ \frac{d\xi_\sigma}{i-\sigma\xi_\sigma+(s+\sigma)(1-e^{-\xi_\sigma})}= \frac{d\xi_0}{i+s(1-e^{-\xi_0})},. $$ Introducing new variables $y=1-e^{-\xi_\sigma}$ and $x=1-e^{-\xi_0}$ and playing with the chain rule and elementary algebra a bit more, we arrive at the following equivalent reformulation of the conjecture:

If $y(0)=0$ and $y'=F(x,y)=\frac{i+\sigma\log(1-y)+(s+\sigma)y}{i+sx}\frac{1-y}{1-x}$ on $[0,1)$, then $\frac{s+\sigma}s y'=\frac{s+\sigma}sF(x,y)\ge 1$ on $[0,1)$.

Since $\psi(x)=-\log(1-y)-y\ge 0$, it is easy to see that $y$ lags behind $x$ all the time and as $x\to 1$, we have $y\to y_1$ where $y_1$ solves the equation $i+sy_1-\sigma\psi(y_1)=0$. Clearly, $y_1<1$.

We now will consider the "endgame" for $x$ close to $1$. Then $F(x,y)\approx \frac{(\sigma\psi'(y_1)-s)(1-y_1)}{i+s} \frac{y_1-y}{1-x}$. If the first fraction equals $\rho>1$, then in the endgame we have, roughly speaking, $y_1-y\approx (1-x)^\rho$ and $F(x,y)\to 0$ as $x\to 1$. It is fairly clear that we can achieve it by increasing $\sigma>0$ to some large value keeping other parameters fixed because $\psi'(y)/\psi(y)\to+\infty$ as $y\to 0$ and that what the above example (in which $s=i=1, \sigma=15$, so $y_1\approx 0.3686$ and $\rho\approx 2.45$) is based upon.

I do not know whether $\rho>1$ is a necessary condition for breaking the monotonicity though.