I'm studying the SIR Model:
$$\left\{\begin{aligned} &\frac{dS}{dt} = -\beta S(t)I(t)\\ &\frac{dI}{dt} = \beta S(t)I(t)-\gamma I(t)\\ &\frac{dR}{dt} = \gamma I(t)\\ \end{aligned}\right.$$
In particular, I am trying to deduce the exact equation for $S(t)$, $I(t)$ and $R(t)$. I have:
$$S(t)=S_0\cdot e^{\frac{R(t)-R_0}{\gamma}}$$
From the inegration of $$\frac{dS}{dR}=-\frac{S(t)}{\gamma}$$
Now, $S(t)$ is written still in terms of the function $R(t)$, so we want to find $R(t)$. In particular, we know that:
$$\frac{dR}{dt}=\beta\cdot I(t)=\beta\cdot(N-R(t)-S(t))$$
And, from what I have said:
$$\frac{dR}{dt}=\beta(N-R(t)-c_0\cdot e^{\frac{R(t)-*R_0}{\gamma}}$$
Here I am stuck and I don't know how to proceed. I think we have to use Taylor expansion of $e^{-x}$ in $I_{\delta}(0)$. Is it correct? Or we just have a trascendental equation, so we can't solve it?
Also, if we don't know about $S(t)$ or $R(t)$, how can we compute the number of infected or susceptible people at the end of the epidemic. In other words, how can we compute:
$$\lim_{t\rightarrow +\infty} S(t)$$ and $$\lim_{t\rightarrow +\infty} R(t)$$
Knowing only $I(S_{(t)})$ and $I(R_{(t)})$, how can do this? Here a graph of what I mean:
Thank you.
The problem is that there's a complicated integral no matter how you spin it. Using The first and third ODE's we can find $R(S)$ \begin{align} \dot R&=-\frac{\gamma \dot S}{\beta S}\\ R&=-\frac{\gamma}{\beta}\ln\left(\frac{S}{S_0}\right)\ \ \ \ \ \ \text{(taking $R_0=0$).} \tag{*} \end{align}
Then, using the chain rule and the first and second ODE's \begin{align} \frac{\mathrm dS}{\mathrm dt}=\frac{\mathrm dS}{\mathrm dI}\frac{\mathrm dI}{\mathrm dt}=I(\beta S-\gamma)\frac{\mathrm dS}{\mathrm dI}, \end{align} so that \begin{align} \frac{\beta S-\gamma}{S}\frac{\mathrm dS}{\mathrm dI}+\beta=0\\ \beta (S-S_0)-\gamma\ln\left(\frac{S}{S_0}\right)+\beta(I-I_0)=0\\ I=N+\frac{\gamma}{\beta}\ln\left(\frac{S}{S_0}\right)-S.\tag{**} \end{align} So if you could get $S(t)$ you'd have all three unknown functions. Substituting $I(S)$ into your first ODE gives that \begin{align} \frac{\mathrm dS}{\mathrm dt}=S\left[\beta(S-N)-\gamma\ln\left(\frac{S}{S_0}\right)\right],\end{align} and \begin{align} \int\frac{\mathrm dS}{S\left[S-\frac{\gamma}{\beta}\ln\left(\frac{S}{S_0}\right)-N\right]}=\beta t+c, \end{align} or in terms of $R$ instead, using equation (*), \begin{align} \int\frac{\mathrm dR}{R+S_0\text{exp}\left(-\frac{\beta R}{\gamma}\right)-N}=-\gamma t+c_2. \tag{$\dagger$} \end{align} there is no elementary solution to this integral, and in my meager ODE experience, if you reduce your ODE to an integral, there are no more tricks to be had to get around them.
You can use the Taylor series expansion of the integrand in equation ($\dagger$) to get an answer. Then the challenge becomes inverting the resultant power series (can't imagine you can do this analytically) to get $R(t)$, which you can substitute into equations (*) and (**) (finding $I(t)$ would involve the Lambert-W function) to get your solutions (numerically).
Edit You need one functions limit to get the others. Thankfully you put that graph up, as I believe it's fair to say that $\lim_{t\rightarrow\infty}I(t)=0$.
Taking the limit as $t\rightarrow\infty$ of equation (**) gives \begin{align} \lim_{t\rightarrow\infty}I(t)=\lim_{t\rightarrow\infty}\left(N+\frac{\gamma}{\beta}\ln\left(\frac{S}{S_0}\right)-S.\right)\\ 0=N+\frac{\gamma}{\beta}\ln\left(\frac{S_{\infty}}{S_0}\right)-S_\infty. \end{align} where I've denoted $\lim_{t\rightarrow\infty}S(t)=S_{\infty}$.
Solving for $S_\infty$ gives that \begin{align} \lim_{t\rightarrow\infty}S(t)=-\frac{\gamma}{\beta}W\left(-\frac{\beta}{\gamma}e^{\frac{\beta}{\gamma}(\ln S_0-N)}\right) \end{align} where $W(x)$ is the Lambert-W function, and assuming the argument is greater than $-1/e$, or put another way, \begin{align} S_0<\left(\frac{\gamma}{\beta}\right)^{\gamma/\beta}e^{N-\gamma/\beta} \end{align} which seems reasonable.
$\lim_{t\rightarrow\infty}R(t)$ can be found using equation (*), \begin{align} \lim_{t\rightarrow\infty}R(t)=-\frac{\gamma}{\beta}\ln\left(-\frac{\gamma}{S_0\beta}W\left(-\frac{\beta}{\gamma}e^{\frac{\beta}{\gamma}(\ln S_0-N)}\right)\right) \end{align}
Hopefully assuming $\lim_{t\rightarrow\infty}I(t)=0$ wasn't false!