Solution of $R(t)$ in SIR model

Question

Consider the SIR model,

$$ \begin{align} \frac{dS}{dt}&= -rSI \tag 1\\ \frac{dI}{dt}&= rSI-\gamma I\tag 2\\ \frac{dR}{dt}&= \gamma I\tag 3 \end{align} $$

from $(1)$ and $(3)$ we get, $\frac{dS}{S}+\frac{dR}{\rho}=0$ and the solution $S(R)=S_0 e^{-\frac{R}{\rho}}$

So that there will always be susceptible in the population and some individuals will always escape infection. (Why?)

Now, $$\frac{dR}{dt}=\gamma \left(N-R-S_0 e^{-\frac{R}{\rho}}\right)$$

The equation can't be solved explicitly. However, if the epidemic is not very large (that is, $\rho$ is large), then $R / \rho$ is small (certainly $R / \rho<1$ ) and by Taylor series $$ e^{-R / \rho}=1-R / \rho+1 / 2(R / \rho)^2-\ldots \ldots $$ Then $$ \begin{aligned} \frac{d R}{d t} &=\gamma\left[N-R-S_0\left[1-R / \rho+1 / 2(R / \rho)^2-\ldots \ldots\right]\right.\\ &=\gamma\left[N-S_0+\left(\frac{S_0}{\rho}-1\right) R-\frac{S_0}{2}(R / \rho)^2+\ldots \ldots\right] . \end{aligned} $$ The solution of the equation

$$ R(t)=\frac{\rho^2}{S_0}\left[\left(\frac{S_0}{\rho}-1\right)+\alpha \tanh \left(\frac{1}{2} \alpha v t-\phi\right)\right] \tag 4 $$

where $\alpha=\left[\left(\frac{S_0}{\rho}-1\right)^2+\frac{2 S_0\left(N-S_0\right)}{\rho^2}\right]^{1 / 2}, \tanh z=\frac{e^z-e^{-z}}{e^z+e^{-z}}$ and $\phi=\tanh ^{-1}\left[\frac{1}{\alpha}\left(\frac{S_0}{\rho}-1\right)\right]$. As $t \rightarrow \infty,(4)$ gives $$ R_{\infty}=\frac{\rho^2}{S_0}\left(\frac{S_0}{\rho}-1+\alpha\right) \approx 2 \rho\left(1-\frac{\rho}{S_0}\right) . $$ This gives the ultimate size of the epidemic. Hence $$ \frac{d R}{d t}=\frac{\gamma \alpha^2 \rho^2}{2 S_0} \sec h^2\left(\frac{1}{2} \alpha \gamma t-\phi\right) \tag 5 $$

Equation $(5)$ defines a symmetric bell shaped curve in the $t − dR/dt$ plane. This curve is called the epidemic curve of the diseases.

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I couldn't understand how they get the solution $(4)$. Another thing is, How to know that the equation $(5)$ was a symmetric bell shaped curve in the $t − dR/dt$ plane?

Any help will be appreciated. Thanks in advance.

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Try to complete the square as @LutzLehmann suggest,

$$ \begin{align} &\gamma\left[N-S_0+\left(\frac{S_0}{\rho}-1\right) R-\frac{S_0}{2}(R / \rho)^2\right]\\ &\gamma(N-S_0)+\gamma \left(\frac{S_0}{\rho}-1\right) R-\frac{\gamma S_0}{2 \rho^2}R^2\\ &-\frac{\gamma S_0}{2 \rho^2}\left( R^2-\frac{2\rho^2}{S_0}\left(\frac{S_0}{\rho}-1\right)R-\frac{2\rho^2}{S_0}(N-S_0) \right)\\ &-\frac{\gamma S_0}{2 \rho^2} \left(R^2 - 2R\frac{2\rho^2}{S_0}\left(\frac{S_0}{\rho}-1\right)+\left(\frac{2\rho^2}{S_0}\left(\frac{S_0}{\rho}-1\right)\right)^2-\left(\frac{2\rho^2}{S_0}\left(\frac{S_0}{\rho}-1\right)\right)^2-\frac{2\rho^2}{S_0}(N-S_0)\right)\\ & -\frac{\gamma S_0}{2 \rho^2}\left(\left( R-\frac{2\rho^2}{S_0}\left(\frac{S_0}{\rho}-1\right)\right)^2-\left(\frac{2\rho^2}{S_0}\left(\frac{S_0}{\rho}-1\right)\right)^2-\frac{2\rho^2}{S_0}(N-S_0)\right) \end{align} $$

Answer 1

Starting from $$ R'=γ(-aR^2+bR+c) $$ I would first multiply with $a$ and use $aR$ as the dependent function $$ (aR)'=γ[-(aR)^2+b(aR)+ac]. $$ Now we can comfortably switch to $u=aR-b/2$ to get $$ u'=γ[ac+b^2/4-u^2]. $$ Set $4ac+b^2=α^2$ and apply one of the solution formulas, I like the Riccati substitution $u=\frac{v'}{γv}$ so that $$ \frac{v''}{γv}-\frac{v'^2}{γv^2}+γ\left(\frac{v'}{γv}\right)^2=\frac{γα^2}4\\~\\ v''=\frac{(γα)^2}4 v\implies v=Ae^{γαt/2}+Be^{-γαt/2} $$ and inserting backwards $$ u=\frac{α}{2}\frac{Ae^{γαt/2}-Be^{-γαt/2}}{Ae^{γαt/2}+Be^{-γαt/2}}\\~\\ R=\frac{b}{2a}+\frac{α}{2a}\frac{Ae^{γαt/2}-Be^{-γαt/2}}{Ae^{γαt/2}+Be^{-γαt/2}} $$ Now because $R_0=0$ ..., so that $A$ and $B$ have the same sign, ..., the symmetrized constant can be pushed into the exponent, so that $$ R=\frac1{2a}\left(b+α\tanh(\tfrac12γαt-ϕ)\right) $$

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The constants are $$ 2a=\frac{S_0}{ρ^2}, ~~ b=\frac{S_0}{ρ}-1, ~~ c=N-S_0. $$