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Consider the following paper:

"A note on global stability for a tuberculosis model" by Gao and Huang: https://doi.org/10.1016/j.aml.2017.05.004

The methodology is understood in this paper apart from some technical things; How did the author arrive at constants $m_i, i=1,...,5$? They achieved this by killing the variable terms ($xu, xv, z, u, v, yu, yv$), however when I try solving the system, it outputs all $m_i$'s equalling zero. Can someone show a detailed answer how they arrived at these constants and how they got their final form of the Lyapunov function (page 168)?

Code:

sol = Solve[{-m1 \[Beta] is + m3 l \[Beta] ss is/ls + 
     m4 (1 - l) \[Beta] ss == 
    0, -m1 \[Rho]1 \[Beta] ts + m3 \[Rho]1 l \[Beta] ss ts/ls + 
     m4 (1 - l) \[Beta] \[Rho]1 ss ts/is == 
    0, -(\[Mu] + \[Delta]) m3 + m4 \[Delta] ls/is == 0, 
   m1 \[Beta] is + m2 \[Rho]2 \[Beta] is - 
     m4 (\[Mu] + \[Alpha] + \[Gamma]) + m5 (\[Mu] + \[Rho]) == 0, 
   m1 \[Rho]1 \[Beta] ts + m2 \[Rho]1 \[Rho]2 \[Beta] ts + 
     m3 \[Rho] ts/ls - m5 (\[Mu] + \[Rho]) == 
    0, -m2 \[Rho]2 \[Beta] is + m3 \[Rho]2 \[Beta] vs is/ls == 
    0, -m2 \[Rho]1 \[Rho]2 \[Beta] ts + 
     m3 \[Rho]1 \[Rho]2 \[Beta] vs ts/ls == 0 }, {m1, m2, m3, m4, m5}]

EDIT:

Trying with the authors co-efficients, I managed to get:

m1 = ss;
m2 = (\[Delta] vs )/(((1 - l) (\[Mu] + \[Delta]) + \[Delta] l));
m3 = (\[Delta] ls )/(((1 - l) (\[Mu] + \[Delta]) + \[Delta] l));
m4 = ((\[Mu] + \[Delta]) is )/(((1 - 
         l) (\[Mu] + \[Delta]) + \[Delta] l));
m5 = (1/(\[Mu] + \[Rho])) (\[Rho]1 \[Beta] ss ts + \[Rho]1 \[Rho]2 \
\[Beta] ((\[Delta] vs ts)/((1 - 
             l) (\[Mu] + \[Delta]) + \[Delta] l)) + ((\[Rho] \[Delta] \
ts)/((1 - l) (\[Mu] + \[Delta]) + \[Delta] l)));

$xu$ term:

FullSimplify[-m1 \[Beta] is + m3 l \[Beta] ss is /ls + 
  m4 (1 - l) \[Beta] ss]

(* 0 *)

$xv$ term:

FullSimplify[-m1 \[Rho]1 \[Beta] ts + m3 \[Rho]1 l \[Beta] ss ts/ls + 
  m4 (1 - l) \[Beta] \[Rho]1 ss ts/is]

(* 0 *)

$z$ term:

FullSimplify[-(\[Mu] + \[Delta]) m3 + m4  \[Delta] ls/is]

(* 0 *)

$u$ term:

    FullSimplify[
     m1 \[Beta] is + m2 \[Rho]2 \[Beta] is - 
      m4 (\[Mu] + \[Alpha] + \[Gamma]) + m5 (\[Mu] + \[Rho])]

(* (-is \[Alpha] (\[Delta] + \[Mu]) - 
 is (\[Gamma] + \[Mu]) (\[Delta] + \[Mu]) + 
 is ss \[Beta] (\[Delta] + \[Mu] - l \[Mu]) - (-1 + 
    l) ss ts \[Beta] \[Mu] \[Rho]1 + is vs \[Beta] \[Delta] \[Rho]2 + 
 ts \[Delta] (\[Rho] + \[Beta] \[Rho]1 (ss + 
       vs \[Rho]2)))/(\[Delta] + \[Mu] - l \[Mu]) *)

$v$ term:

FullSimplify[
 m1 \[Rho]1 \[Beta] ts + \[Rho]2 \[Beta] \[Rho]1 ts m2 + 
  m3 \[Rho] ts/ls - m5 (\[Mu] + \[Rho])]

(* 0 *)

$yu$ term:

FullSimplify[m3 \[Rho]2 \[Beta] vs is/ls - \[Rho]2 \[Beta] is m2]

 (* 0 *)

$yv$ term:

FullSimplify[
 m3 \[Rho]2 \[Rho]1 \[Beta] vs ts/ls - \[Rho]2 \[Rho]1 \[Beta] ts m2]

 (* 0 *)

The $u$ term is causing some problems here, why?

Improve this question
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6
  • $\begingroup$ they need $dH_1/dt\leq 0$, and that fixes these coefficients in the expression for $H_1$. $\endgroup$
    – Carlo Beenakker
    Feb 14, 2022 at 14:22
  • $\begingroup$ @CarloBeenakker I understand we need the lyapunov function to be negative semi definite but this doesn't explain why they picked these coefficients in particular. Granted these coefficients worked, but why these? or better, how were they deduced? $\endgroup$
    – Math
    Feb 14, 2022 at 14:54
  • $\begingroup$ looks like a contrived problem:the solution was the written down first and then the problem was created $\endgroup$
    – Piyush Grover
    Feb 15, 2022 at 15:49
  • $\begingroup$ @PiyushGrover I believe the the author makes the variable terms($xu, xv, z, u, v, yu, yv$) equal to zero at the top of page 167 in journal. when I try solving these, it outputs $m_i$'s equal to zero.. $\endgroup$
    – Math
    Feb 17, 2022 at 13:28
  • $\begingroup$ I looked at that paper and the equations make little sense to me; consider the second equality on page 167, there is the term $-(m_1\Lambda/S^\ast)x^{-1}$. It is the only term $\propto 1/x$. In the next equality they set $m_1=S^\ast$, so that term should just read $-\Lambda/x$. Instead, it has expanded to 10 (!) terms... $\endgroup$
    – Carlo Beenakker
    Feb 21, 2022 at 18:45

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