Consider the following paper:
From my understanding figure 2(on page 5) is a graph of $s(t), i(t)$ and $r(t)$ by the function in the appendix(page 6) for case 1.
However, when I try plotting this function, I get the wrong out output, why?
Plot[{ 1 -
0.36*t + (0.72*10^-1)*t^2 - (0.96*10^-2)*t^3 + (0.96*10^-3)*
t^4 - (0.768*10^-4)*t^5 + (0.5688888892*10^-6)*t^6,
0.36*t - (0.72*10^-1)*t^2 + (0.96*10^-2)*t^3 - (0.96*10^-3)*
t^4 + (0.768*10^-4)*t^5 - (0.512*10^-5)*t^6}, {t, 0, 10}]
The details from the paper since it was pointed out it cant be accessed:
Consider the system:
$$\frac{ds}{dt}=(1-p)\mu -\beta s i -\mu s \quad (1)$$ $$\frac{di}{dt}=\beta s i -(\gamma+\mu) i \;\;\;\;\;\;\ \;\;\quad (2)$$ $$\frac{dr}{dt}=p\mu +\gamma i-\mu r \;\;\;\;\;\;\;\;\;\;\ \quad(3)$$
using Adomian decomposition technique we have:
$$s(t) = s(0) +(1-p)\mu t -\beta \int_0^t si\; dt -\mu\int_0^t s \;dt \;\;\;\;\quad (4)$$ $$i(t) = i(0) +\beta \int_0^t si\; dt -(\gamma+\mu)\int_0^t i \;dt \quad \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;(5)$$ $$r(t) = r(0) +p\mu t +\gamma \int_0^t i\; dt -\mu\int_0^t r \;dt \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\quad(6)$$
From Adomian decomposition euations (4)-(6) are considered the sum of the following:
$$s=\sum_{n=0}^\infty s_n,\quad i=\sum_{n=0}^\infty i_n, \quad r=\sum_{n=0}^\infty r_n \qquad (7)$$
Then we approximate non-linear terms by:
$$si=\sum_{n=0}^\infty F_n \quad (8)$$
where
$$F_n=\frac{1}{n!} \left[ \frac{d^n\left(\sum_{k=0}^\infty s_k \lambda^k\right)\left(\sum_{k=0}^\infty i_k \lambda^k\right)}{d\lambda^n} \right]_{\lambda=0} \qquad (9)$$
The nonlinear functions $F_n$ are the Adomian polynomials. Substituting (7)-(9) into (4)-(6) we have
$$\sum_{n=0}^\infty s_n = s(0) +(1-p)\mu t -\beta \int_0^t \sum_{n=0}^\infty F_n\; dt -\mu\int_0^t \sum_{n=0}^\infty s_n \;dt \;\;\;\;\quad (10)$$ $$\sum_{n=0}^\infty i_n = i(0) +\beta \int_0^t \sum_{n=0}^\infty F_n\; dt -(\gamma+\mu)\int_0^t \sum_{n=0}^\infty i_n \;dt \quad \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;(11)$$ $$\sum_{n=0}^\infty r_n = r(0) +p\mu t +\gamma \int_0^t \sum_{n=0}^\infty i_n\; dt -\mu\int_0^t \sum_{n=0}^\infty r_n \;dt \;\;\;\;\;\;\;\;\;\;\;\;\;\quad(12)$$
from (10)-(12) we define the following scheme:
$$ s_0 = s(0)+(1-p)\mu t, \quad i_0=i(0), \quad r_0=r(0)+p\mu t \qquad\; (13)$$ $$s_{n+1} = -\beta \int_0^t F_n\; dt -\mu\int_0^t s_n \;dt \quad \;\;\;\;\;\;\;\;\;\;\;\;\quad \quad \qquad \quad \;\;\;(14)$$ $$i_{n+1} = \beta \int_0^t F_n\; dt -(\gamma+\mu)\int_0^t i_n \;dt \quad \;\;\;\;\;\;\;\;\;\;\;\;\quad \quad \quad \;\;\;\;\;(15)$$ $$r_{n+1} = \gamma \int_0^t i_n\; dt -\mu\int_0^t r_n \;dt \quad \;\;\;\;\;\;\;\;\;\;\;\;\quad \quad \quad \;\;\;\;\;\quad \quad \;\;\;(16)$$
using (9) we compute some Adomian polynomials as follows:
$$F_0 = s_0 i_0, \quad F_1 = s_0 i_0 +s_1 i_0, F_2 = s_0 i_2 +s_1 i_1 +s_2 i_0,$$ $$F_3 = s_0 i_3+s_1 i_2 +s_2 i_1 +s_3 i_0,\dots \quad \quad \qquad \qquad \quad (17)$$
Substituting (13)-(17) into (10)-(12) we should get this:
$$s_N = \sum_{n=0}^N s_n, \quad i_N = \sum_{n=0}^N i_n, \quad r_N = \sum_{n=0}^N r_n \quad \quad \quad \;\;(18)$$
where $$ s(t) =\lim_{N\rightarrow \infty} s_N, \quad i(t) =\lim_{N\rightarrow \infty} i_N, \quad r(t) =\lim_{N\rightarrow \infty} r_N \quad \quad (19) $$