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consider: \begin{align} \dot S &= A-\beta S(I+\rho_1 T)-(\mu+p)S\\[1ex] \dot V &=pS-\rho_2 \beta V(I+\rho_1 T)-\mu V\\[1ex] \dot L &=l \beta S(I+\rho_1 T)+\rho_2 \beta V(I+\rho_1 T)-(\mu+\delta)L +\rho T\\[1ex] \dot I &=(1-l)\beta S(I+\rho_1 T)+\delta L-(\mu+\alpha+\gamma)I\\[1ex] \dot T &=\gamma I-(\mu+\rho)T \end{align} with $N=S+V+L+I+T$.

to find the equilibriums, we set the above system to $0$: \begin{align} A-\beta S(I+\rho_1 T)-(\mu+p)S&=0\\[1ex] pS-\rho_2 \beta V(I+\rho_1 T)-\mu V&=0\\[1ex] l \beta S(I+\rho_1 T)+\rho_2 \beta V(I+\rho_1 T)-(\mu+\delta)L +\rho T&=0\\[1ex] (1-l)\beta S(I+\rho_1 T)+\delta L-(\mu+\alpha+\gamma)I&=0\\[1ex] \gamma I-(\mu+\rho)T&=0 \end{align}

How we solve this for $S,V,L,T$ in terms of $I$?

For example we see: $$T^*=\frac{\gamma}{\mu+\rho}I^*$$

I tried using solve but this isn't working. This is the code I used:

Solve[A - \[Beta] S (i + \[Rho]1 T) - (\[Mu] + p) S == 0,   p S - \[Rho]2 \[Beta] V (i + \[Rho]1 T) - \[Mu] V == 0,   l \[Beta] S (i + \[Rho]1 T) + \[Rho]2 \[Beta] V (i + \[Rho]1 T) - (\[Mu] + \[Delta]) L + \[Rho] T ==    0, (1 - l) \[Beta] S (i + \[Rho]1 T) + \[Delta] L - (\[Mu] + \[Alpha] + \[Gamma]) i == 0, \[Gamma] i - (\[Mu] + \[Rho]) T == 0, {S,    V, L, T}]

Reference(page 4 in pdf, page 839 in journal): https://reader.elsevier.com/reader/sd/pii/S0895717711001932?token=4C8B07AF574B6CAFE11E8CD5FEA143D0E498B5427696EE44C35371A30602EB3CB0825235CB97498301A86775EDC25D17&originRegion=eu-west-1&originCreation=20220124124053

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  • 2
    $\begingroup$ "I tried using solve but this isn't working." Please post the code you tried. $\endgroup$ Jan 24 at 14:52
  • $\begingroup$ @RohitNamjoshi Solve[A - \[Beta] S (i + \[Rho]1 T) - (\[Mu] + p) S == 0, p S - \[Rho]2 \[Beta] V (i + \[Rho]1 T) - \[Mu] V == 0, l \[Beta] S (i + \[Rho]1 T) + \[Rho]2 \[Beta] V (i + \[Rho]1 T) - (\ \[Mu] + \[Delta]) L + \[Rho] T == 0, (1 - l) \[Beta] S (i + \[Rho]1 T) + \[Delta] L - (\[Mu] + \ \[Alpha] + \[Gamma]) i == 0, \[Gamma] i - (\[Mu] + \[Rho]) T == 0, {S, V, L, T}] $\endgroup$
    – Math
    Jan 24 at 15:02
  • $\begingroup$ This is not a linear system. There are terms involving products like S * I, both of which are solution variables. $\endgroup$ Jan 24 at 15:21
  • $\begingroup$ @SjoerdSmit My bad, I rushed the title. Fixed now. $\endgroup$
    – Math
    Jan 24 at 15:23
  • $\begingroup$ added the code from the comment to the OP $\endgroup$
    – kcr
    Jan 24 at 15:52

2 Answers 2

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Clear["Global`*"]

You have five equations with only four unknowns. Use the option MaxExtraConditions. The use of FullSimplify slows this down considerably.

sol = Solve[{
    A - β S (i + ρ1 T) - (μ + p) S == 0,
    p S - ρ2 β V (i + ρ1 T) - μ V == 0, 
    l β S (i + ρ1 T) + ρ2 β V (i + ρ1 T) - (μ + δ) L + ρ T == 0,
    (1 - l) β S (i + ρ1 T) + δ L - (μ + α + γ) i == 0,
    γ i - (μ + ρ) T == 0},
   {S, V, L, T}, MaxExtraConditions -> 1] // FullSimplify

enter image description here

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  • $\begingroup$ How long did it take to run this on your computer? $\endgroup$
    – Math
    Jan 24 at 17:02
  • $\begingroup$ Using version 13.0.0 for Mac OS X x86 (64-bit) (December 3, 2021) the AbsoluteTiming is 2.41174 $\endgroup$
    – Bob Hanlon
    Jan 24 at 17:05
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Use lower case names, otherwise there is danger of a conflict with built in names. You use "I" what in MMA is the imaginary unit. I therefore replaced it by W. In addition using the hint from Bob Halon:

sol = Solve[{A - β S (W + ρ1 T) - (μ + p) S == 0, 
    p S - ρ2 β V (W + ρ1 T) - μ V == 0, 
    l β S (W + ρ1 T) + ρ2 β V (W + ρ1 T) - \
(μ + δ) L + ρ T == 
     0, (1 - l) β S (W + ρ1 T) + δ L - (μ + \
α + γ) W == 0, γ W - (μ + ρ) T == 
     0}, {S, V, L, T}, MaxExtraConditions -> 1] // FullSimplify

enter image description here

Update

To bring this into more readable form:

Grid[Transpose[{{"S= ", "V= ", "L= ", "T= "}, {S, V, L, T} /. 
    sol[[1]]}] , Alignment -> Left]

enter image description here

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  • $\begingroup$ both solutions are nice, however, can we have it in the form like in the paper? $\endgroup$
    – Math
    Jan 25 at 13:12
  • $\begingroup$ See the update. $\endgroup$ Jan 25 at 13:32
  • $\begingroup$ what I meant was, if you can access the paper, you'll notice his solutions are in a "nice form" using one another, do you know how to do this? $\endgroup$
    – Math
    Jan 25 at 14:56

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