Expansion of SI model solution

09 May 2020 - viridi - Sparisoma Viridi

McLaurien series expansion of SI epidemic model.

Solution of SI model

It has been shonwn in previous post that

$$\begin{equation} \label{egn:model-si-normalized-solution} i = \frac{i_0 e^{\beta \lt k \gt t}}{1 + i_0 \left(e^{\beta \lt k \gt t} - 1 \right)} \end{equation}$$

is the solution of SI (susceptible-infectious) model, whose

$$\begin{equation} \label{egn:model-si-differential-equation} \frac{di}{dt} = \beta \lt k \gt (1- i) i, \end{equation}$$

is the differential equation.

Simplify solution terms

We can define $a \equiv i_0$, $b \equiv \beta \lt k \gt$ and $c = (1- a)$, which reduce the symbols in Eqn $\eqref{egn:model-si-normalized-solution}$ to be

$$\begin{equation} \label{egn:model-si-normalized-solution-abc} i = \frac{a e^{b t}}{1 + a \left(e^{b t} - 1 \right)} = \frac{a e^{b t}}{(1 - a) + a e^{b t}} = \frac{a e^{b t}}{a e^{b t} + c}. \end{equation}$$

Then we can also define

$$\begin{equation} \label{egn:model-si-normalized-definition-f} f \equiv f(t) = ae^{bt} + c, \end{equation}$$

which turn Eqn $\eqref{egn:model-si-normalized-solution-abc}$ into

$$\begin{equation} \label{egn:model-si-normalized-solution-f} i = (f - c) f^{-1}, \end{equation}$$

that is more simple, hopefully.

Time derivative of $f^{-1}$

The $n$-th time derivatif of $f^{-1}$ is denoted as $f^{-1}_{n}$. Then

$$\begin{equation} \label{egn:model-si-normalized-f_1-derivative-1} f^{-1}_1 = \frac{d}{dt} \left( \frac{1}{ae^{bt} + c} \right) = -f^{-2} ab e^{bt} \end{equation}$$

is the 1-st time derivative. The 2-nd will be

$$\begin{equation} \label{egn:model-si-normalized-f_1-derivative-2} f^{-1}_2 = \frac{d}{dt} \left[ -\frac{ab e^{bt}}{\left( ae^{bt} + c \right)^2} \right] = 2f^{-3} a^2 b^ 2 e^{2bt} - f^{-2} a b^2 e^{bt}, \end{equation}$$

the 3-rd

$$\begin{equation} \label{egn:model-si-normalized-f_1-derivative-3} f^{-1}_3 = \frac{d}{dt} \left[ \frac{2 a^2 b^2 e^{2bt}}{\left( ae^{bt} + c \right)^3} - \frac{a b^2 e^{bt}}{\left( ae^{bt} + c \right)^2} \right] = -6f^{-4} a^3 b^3 e^{3bt} + 6f^{-3} a^2 b^3 e^{2bt} - f^{-2} ab^3 e^{bt}. \end{equation}$$

And for the $n$-th, it could be

$$\begin{equation} \label{egn:model-si-normalized-f_1-derivative-n} f^{-1}_n = (-1)^n n! f^{-1-n} a^n b^n e^{nbt} + (-1)^{n-1} f^{-1-n+1} a^{n-1} b^n e^{(n-1)bt} \dots - f^{-2} a b^n e^{bt}. \end{equation}$$

Coefient of the first and last term are already clear, which are $(-1)^n n!$ and $-1$, but the others must be still discovered. A home work for me, .

McLaurien series

Eqn $\eqref{egn:model-si-normalized-solution-f}$ will have expansion

$$\begin{equation} \label{egn:model-si-normalized-solution-maclaurin-1} i(t) = [f(0) - c] f^{-1}(0) + \left\{f_1(0) f^{-1}(0) + [f(0) - c] f^{-1}_1 \right\} t + \dots, \end{equation}$$

that leads to

$$\begin{equation} \label{egn:model-si-normalized-solution-maclaurin-2} i(t) = a + ba(1- a) t + \dots. \end{equation}$$

Further discussion will be postoned for now due to current limited knowledge and time.