Numerical approximation of modified non-linear SIR model of computer viruses

TL;DR

This paper compares semi-analytical methods for numerically approximating a modified non-linear SIR model of computer viruses, demonstrating the effectiveness of the Laplace-Adomian decomposition method over others.

Contribution

It introduces a modified non-linear SIR model for computer viruses and compares the performance of DTM, LADM, and HATM methods for its numerical approximation.

Findings

LADM outperforms HATM in accuracy.

Residual error graphs validate the effectiveness of LADM.

Phase portraits illustrate solution behaviors.

Abstract

In this paper, the non-linear modified epidemiological model of computer viruses is illustrated. For this aim, two semi-analytical methods, the differential transform method (DTM) and the Laplace-Adomian decomposition method (LADM) are applied. The numerical results are estimated for different values of iterations and compared to the results of the LADM and the homotopy analysis transform method (HATM). Also, graphs of residual errors and phase portraits of approximate solutions for $n=5,10,15$ are demonstrated. The numerical approximations show the performance of the LADM in comparison to the LADM and the HATM.

Tables

Table 1. Table 1: List of parameters and functions.

Parameters	Meaning	Values
Functions
$S\left(t\right)$	Susceptible computers at time $t$	$S\left(0\right)=20$
$I\left(t\right)$	Infected computers at time $t$	$I\left(0\right)=15$
$R\left(t\right)$	Recovered computers at time $t$	$R\left(0\right)=10$
$f_1,f_2,f_3$	Rate of external computers connected to the network	$f_1=f_2=f_3=0$
$\lambda$	Rate of infecting for susceptible computer	$\lambda =0.001$
$\epsilon$	Rate of recovery for infected computers	$\epsilon =0.1$
$d$	Rate of removing from the network	$d=0.1$

Table 2. Table 2: Main operations of DTM.

Original functions	Transformed functions
$f\left(t\right)=u\left(t\right)\pm v\left(t\right)$	$F\left(k\right)=U\left(k\right)\pm V\left(k\right)$
$f\left(t\right)=\beta u\left(t\right)$	$F\left(k\right)=\beta U\left(k\right)$
$f\left(t\right)=u\left(t\right)v\left(t\right)$	$F\left(k\right)={\sum }_{i=0}^{k}U\left(k\right)V_{k-s}\left(k\right)$
$f\left(t\right)=\frac{du\left(t\right)}{dt}$	$F\left(k\right)=\left(k+1\right)U\left(k+1\right)$
$f\left(t\right)=\frac{d^{m}u\left(t\right)}{d{t}^m}$	$F\left(k\right)=\left(k+1\right)\left(k+2\right)\cdots \left(k+m\right)U\left(k+m\right)$
$f\left(t\right)={\int }_{t_0}^{t}u\left(\xi \right)𝑑\xi$	$F\left(k\right)=\frac{U\left(k-1\right)}{k},k\ge 1$
$f\left(t\right)=t^m$	$F\left(k\right)=\delta \left(k-m\right)$
$f\left(t\right)=\exp \left(\lambda t\right)$	$F\left(k\right)=\frac{{\lambda }^k}{k!}$
$f\left(t\right)=\sin \left(\omega t+\alpha \right)$	$F\left(k\right)=\frac{{\omega }^k}{k!}\sin \left(\frac{\pi k}{2}+\alpha \right)$
$f\left(t\right)=\cos \left(\omega t+\alpha \right)$	$F\left(k\right)=\frac{{\omega }^k}{k!}\cos \left(\frac{\pi k}{2}+\alpha \right)$

Table 3. Table 3: Numerical comparison of residual error E n , S ​ ( t ) subscript 𝐸 𝑛 𝑆 𝑡 E_{n,S}(t) between LADM, DTM and HATM for n = 5 , 10 𝑛 5 10 n=5,10 .

$t$	$E_{5,S}\left(t\right)$-LADM	$E_{5,S}\left(t\right)$-DTM	$E_{5,S}\left(t\right)$-HATM	$E_{10,S}\left(t\right)$-LADM	$E_{10,S}\left(t\right)$-DTM	$E_{10,S}\left(t\right)$-HATM
$0.0$	$0$	$0$	$0$	$0$	$0$	$0$
$0.2$	$1.98295\times {10}^{-11}$	$6.22316\times {10}^{-8}$	$0.0000116327$	$4.44089\times {10}^{-16}$	$4.44089\times {10}^{-16}$	$4.22713\times {10}^{-11}$
$0.4$	$4.38413\times {10}^{-10}$	$9.99398\times {10}^{-7}$	$0.0000188901$	$0$	$1.77636\times {10}^{-15}$	$8.71449\times {10}^{-10}$
$0.6$	$1.87637\times {10}^{-9}$	$5.07724\times {10}^{-6}$	$0.0000214386$	$1.77636\times {10}^{-15}$	$5.9508\times {10}^{-14}$	$4.95175\times {10}^{-9}$
$0.8$	$1.93464\times {10}^{-9}$	$0.0000161$	$0.00015492$	$2.57572\times {10}^{-14}$	$7.94032\times {10}^{-13}$	$1.57635\times {10}^{-8}$
$1.0$	$1.18760\times {10}^{-8}$	$0.0000394305$	$0.000429509$	$2.30038\times {10}^{-13}$	$5.97122\times {10}^{-12}$	$3.43742\times {10}^{-8}$

Table 4. Table 4: Numerical comparison of residual error E n , I ​ ( t ) subscript 𝐸 𝑛 𝐼 𝑡 E_{n,I}(t) between LADM, DTM and HATM for n = 5 , 10 𝑛 5 10 n=5,10 .

$t$	$E_{5,I}\left(t\right)$-LADM	$E_{5,I}\left(t\right)$-DTM	$E_{5,I}\left(t\right)$-HATM	$E_{10,I}\left(t\right)$-LADM	$E_{10,I}\left(t\right)$-DTM	$E_{10,I}\left(t\right)$-HATM
$0.0$	$0$	$0$	$7.10543\times {10}^{-15}$	$0$	$0$	$0$
$0.2$	$2.08857\times {10}^{-10}$	$7.88132\times {10}^{-8}$	$0.000013859$	$4.44089\times {10}^{-16}$	$4.44089\times {10}^{-16}$	$4.24451\times {10}^{-11}$
$0.4$	$6.48732\times {10}^{-9}$	$1.26470\times {10}^{-6}$	$0.000036861$	$4.44089\times {10}^{-16}$	$1.77636\times {10}^{-15}$	$8.8009\times {10}^{-10}$
$0.6$	$4.78103\times {10}^{-8}$	$6.42035\times {10}^{-6}$	$0.0000397647$	$1.77636\times {10}^{-15}$	$4.44089\times {10}^{-14}$	$5.06861\times {10}^{-9}$
$0.8$	$1.95500\times {10}^{-7}$	$0.0000203449$	$8.51494\times {10}^{-6}$	$3.15303\times {10}^{-14}$	$5.96412\times {10}^{-13}$	$1.65033\times {10}^{-8}$
$1.0$	$5.78838\times {10}^{-7}$	$0.000049794$	$0.000140914$	$2.89546\times {10}^{-13}$	$4.50395\times {10}^{-12}$	$3.7476\times {10}^{-8}$

Table 5. Table 5: Numerical comparison of residual error E n , R ​ ( t ) subscript 𝐸 𝑛 𝑅 𝑡 E_{n,R}(t) between LADM, DTM and HATM for n = 5 , 10 𝑛 5 10 n=5,10 .

$t$	$E_{5,R}\left(t\right)$-LADM	$E_{5,R}\left(t\right)$-DTM	$E_{5,R}\left(t\right)$-HATM	$E_{10,R}\left(t\right)$-LADM	$E_{10,R}\left(t\right)$-DTM	$E_{10,R}\left(t\right)$-HATM
$0.0$	$0$	$0$	$0$	$0$	$0$	$0$
$0.2$	$5.69638\times {10}^{-10}$	$6.38387\times {10}^{-8}$	$1.95392\times {10}^{-6}$	$1.11022\times {10}^{-16}$	$3.33067\times {10}^{-16}$	$1.04154\times {10}^{-12}$
$0.4$	$1.82284\times {10}^{-8}$	$1.02142\times {10}^{-6}$	$0.0000285358$	$3.33067\times {10}^{-16}$	$1.22125\times {10}^{-15}$	$8.17403\times {10}^{-12}$
$0.6$	$1.38422\times {10}^{-7}$	$5.17094\times {10}^{-6}$	$0.000112891$	$7.21645\times {10}^{-16}$	$4.56857\times {10}^{-14}$	$1.57476\times {10}^{-10}$
$0.8$	$5.83309\times {10}^{-7}$	$0.0000163427$	$0.000290976$	$9.43690\times {10}^{-15}$	$6.10734\times {10}^{-13}$	$1.69486\times {10}^{-9}$
$1.0$	$1.78012\times {10}^{-6}$	$0.0000398992$	$0.000601621$	$8.77631\times {10}^{-14}$	$4.55164\times {10}^{-12}$	$8.65934\times {10}^{-9}$

Equations

$$\begin{array}[]{l}\displaystyle\frac{dS(t)}{dt}=f_{1}-\lambda S(t)I(t)-dS(t),\\ \\ \displaystyle\frac{dI(t)}{dt}=f_{2}+\lambda S(t)I(t)-\varepsilon I(t)-dR(t),\\ \\ \displaystyle\frac{dR(t)}{dt}=f_{3}+\varepsilon I(t)-dR(t),\\ \end{array}$$

$$S(0)=S_0(t),I(0)=I_0(t),R(0)=R_0(t),$$

$$F(k)=\frac{1}{k!}{\left[\frac{d^{k}f(t)}{d{t}^k}\right]}_{t=t_0},$$

$$f(t)=\sum _{k=0}^{\infty }F(k)(t-t_0{)}^k$$

$$f(t)=\sum _{k=0}^{N}F(k)(t-t_0{)}^k$$

$$\begin{array}[]{l}\displaystyle S_{k+1}=\frac{1}{k+1}\left[f_{1}-\lambda\sum_{i=0}^{k}S_{i}I_{k-i}-dS_{k}\right],\\ \\ \displaystyle I_{k+1}=\frac{1}{k+1}\left[f_{2}-\lambda\sum_{i=0}^{k}S_{i}I_{k-i}-\varepsilon I_{k}-dR_{k}\right],\\ \\ \displaystyle R_{k+1}=\frac{1}{k+1}\bigg{[}f_{3}-\varepsilon I_{k}-dR_{k}\bigg{]}.\\ \end{array}$$

$$\begin{array}[]{l}\displaystyle S(t)=\sum_{j=0}^{n}S_{j}t^{j},\\ \\ \displaystyle I(t)=\sum_{j=0}^{n}I_{j}t^{j},\\ \\ \displaystyle R(t)=\sum_{j=0}^{n}R_{j}t^{j},\\ \end{array}$$

$$\begin{array}[]{l}\displaystyle\mathcal{L}[S(t)]=\frac{S(0)}{z}+\frac{\mathcal{L}[f_{1}]}{z}-\frac{\lambda}{z}\mathcal{L}[S(t)I(t)]-\frac{d}{z}\mathcal{L}[S(t)],\\ \\ \displaystyle\mathcal{L}[I(t)]=\frac{I(0)}{z}+\frac{\mathcal{L}[f_{2}]}{z}+\frac{\lambda}{z}\mathcal{L}[S(t)I(t)]-\frac{\varepsilon}{z}\mathcal{L}[I(t)]-\frac{d}{z}\mathcal{L}[R(t)],\\ \\ \displaystyle\mathcal{L}[R(t)]=\frac{R(0)}{z}+\frac{\mathcal{L}[f_{3}]}{z}+\frac{\varepsilon}{z}\mathcal{L}[I(t)]-\frac{d}{z}\mathcal{L}[R(t)].\\ \end{array}$$

$$\begin{array}[]{l}\displaystyle\mathcal{L}[S(t)]=\frac{S(0)}{z}+\frac{f_{1}}{z^{2}}-\frac{\lambda}{z}\mathcal{L}[A]-\frac{d}{z}\mathcal{L}[S(t)],\\ \\ \displaystyle\mathcal{L}[I(t)]=\frac{I(0)}{z}+\frac{f_{2}}{z^{2}}+\frac{\lambda}{z}\mathcal{L}[A]-\frac{\varepsilon}{z}\mathcal{L}[I(t)]-\frac{d}{z}\mathcal{L}[R(t)],\\ \\ \displaystyle\mathcal{L}[R(t)]=\frac{R(0)}{z}+\frac{f_{3}}{z^{2}}+\frac{\varepsilon}{z}\mathcal{L}[I(t)]-\frac{d}{z}\mathcal{L}[R(t)],\\ \end{array}$$

$$S=\sum _{j=0}^{\infty }{S}_j,I=\sum _{j=0}^{\infty }{I}_j,R=\sum _{j=0}^{\infty }{R}_j.$$

$$A=\sum _{j=0}^{\infty }{A}_j,$$

$$\begin{array}[]{l}A_{0}=S_{0}I_{0},\\ \\ A_{1}=S_{0}I_{1}+S_{1}I_{0},\\ \\ A_{2}=S_{0}I_{2}+S_{1}I_{1}+S_{2}I_{0},\\ \\ A_{3}=S_{0}I_{3}+S_{1}I_{2}+S_{2}I_{1}+S_{3}I_{0},\\ \\ A_{4}=S_{0}I_{4}+S_{1}I_{3}+S_{2}I_{2}+S_{3}I_{1}+S_{4}I_{0},\\ ~{}~{}~{}\vdots\end{array}$$

$$\begin{array}[]{l}\displaystyle\mathcal{L}\left[\sum_{j=0}^{\infty}S_{j}\right]=\frac{S(0)}{z}+\frac{f_{1}}{z^{2}}-\frac{\lambda}{z}\mathcal{L}\left[\sum_{j=0}^{\infty}A_{j}\right]-\frac{d}{z}\mathcal{L}\left[\sum_{j=0}^{\infty}S_{j}\right],\\ \\ \displaystyle\mathcal{L}\left[\sum_{j=0}^{\infty}I_{j}\right]=\frac{I(0)}{z}+\frac{f_{2}}{z^{2}}+\frac{\lambda}{z}\mathcal{L}\left[\sum_{j=0}^{\infty}A_{j}\right]-\frac{\varepsilon}{z}\mathcal{L}\left[\sum_{j=0}^{\infty}I_{j}\right]-\frac{d}{z}\mathcal{L}\left[\sum_{j=0}^{\infty}R_{j}\right],\\ \\ \displaystyle\mathcal{L}\left[\sum_{j=0}^{\infty}R_{j}\right]=\frac{R(0)}{z}+\frac{f_{3}}{z^{2}}+\frac{\varepsilon}{z}\mathcal{L}\left[\sum_{j=0}^{\infty}I_{j}\right]-\frac{d}{z}\mathcal{L}\left[\sum_{j=0}^{\infty}R_{j}\right].\\ \end{array}$$

$$\begin{array}[]{l}\displaystyle\mathcal{L}[S_{0}]=\frac{S(0)}{z}+\frac{f_{1}}{z^{2}},\\ \\ \displaystyle\mathcal{L}[I_{0}]=\frac{I(0)}{z}+\frac{f_{2}}{z^{2}},\\ \\ \displaystyle\mathcal{L}[R_{0}]=\frac{R(0)}{z}+\frac{f_{3}}{z^{2}},\\ \\ \displaystyle\mathcal{L}[S_{1}]=\frac{\lambda}{z}\mathcal{L}\left[A_{0}\right]-\frac{d}{z}\mathcal{L}\left[S_{0}\right],\\ \\ \displaystyle\mathcal{L}[I_{1}]=\frac{\lambda}{z}\mathcal{L}\left[A_{0}\right]-\frac{\varepsilon}{z}\mathcal{L}\left[I_{0}\right]-\frac{d}{z}\mathcal{L}\left[R_{0}\right],\\ \\ \displaystyle\mathcal{L}[R_{1}]=\frac{\varepsilon}{z}\mathcal{L}\left[I_{0}\right]-\frac{d}{z}\mathcal{L}\left[R_{0}\right],\end{array}$$

$$\begin{array}[]{l}\displaystyle\mathcal{L}[S_{j}]=\frac{\lambda}{z}\mathcal{L}\left[A_{j-1}\right]-\frac{d}{z}\mathcal{L}\left[S_{j-1}\right],\\ \\ \displaystyle\mathcal{L}[I_{j}]=\frac{\lambda}{z}\mathcal{L}\left[A_{j-1}\right]-\frac{\varepsilon}{z}\mathcal{L}\left[I_{j-1}\right]-\frac{d}{z}\mathcal{L}\left[R_{j-1}\right],\\ \\ \displaystyle\mathcal{L}[R_{j}]=\frac{\varepsilon}{z}\mathcal{L}\left[I_{j-1}\right]-\frac{d}{z}\mathcal{L}\left[R_{j-1}\right].\end{array}$$

$$\begin{array}[]{l}S_{0}=S(0)+f_{1}t,\\ \\ I_{0}=I(0)+f_{2}t,\\ \\ R_{0}=R(0)+f_{3}t,\end{array}$$

$$\begin{array}[]{ll}\displaystyle\mathcal{L}[S_{1}]&\displaystyle=\frac{\lambda}{z}\left(\frac{S(0)I(0)}{z}+\frac{S(0)f_{2}}{z^{2}}+\frac{I(0)f_{1}}{z^{2}}+\frac{2f_{1}f_{2}}{z^{3}}\right)-\frac{d}{z}\left(\frac{S(0)}{z}+\frac{f_{1}}{z^{2}}\right),\\ \\ \displaystyle\mathcal{L}[I_{1}]&\displaystyle=\frac{\lambda}{z}\left(\frac{S(0)I(0)}{z}+\frac{S(0)f_{2}}{z^{2}}+\frac{I(0)f_{1}}{z^{2}}+\frac{2f_{1}f_{2}}{z^{3}}\right)-\frac{\varepsilon}{z}\left(\frac{I(0)}{z}+\frac{f_{2}}{z^{2}}\right)\\ \\ &\displaystyle-\frac{d}{z}\left(\frac{R(0)}{z}+\frac{f_{3}}{z^{2}}\right),\\ \\ \displaystyle\mathcal{L}[R_{1}]&\displaystyle=\frac{\varepsilon}{z}\left(\frac{I(0)}{z}+\frac{f_{2}}{z^{2}}\right)-\frac{d}{z}\left(\frac{R(0)}{z}+\frac{f_{3}}{z^{2}}\right),\end{array}$$

$$S_n=\sum _{j=0}^n{S}_j,I_n=\sum _{j=0}^n{I}_j,R_n=\sum _{j=0}^n{R}_j,$$

$$\begin{array}[]{l}S_{5}(t)=20-2.3t+0.15425t^{2}-0.00790458t^{3}+0.000309711t^{4},\\ \\ I_{5}(t)=15-2.2t+0.04575t^{2}+0.00573792t^{3}-0.000406169t^{4},\\ \\ R_{5}(t)=10+0.5t-0.135t^{2}+0.006025t^{3}-7.17708\times 10^{-6}t^{4},\end{array}$$

$$\begin{array}[]{l}S_{10}(t)=20-2.3t+0.15425t^{2}-0.00790458t^{3}+0.000309711t^{4}\\ \\ ~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}-7.74864\times 10^{-6}t^{5}-1.35996\times 10^{-8}t^{6}+1.41005\times 10^{-8}t^{7}\\ \\ ~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}-8.93373\times 10^{-10}t^{8}+3.32927\times 10^{-11}t^{9},\\ \\ I_{10}(t)=15-2.2t+0.04575t^{2}+0.00573792t^{3}-0.000406169t^{4}\\ \\ ~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}+9.82135\times 10^{-6}t^{5}+1.12052\times 10^{-7}t^{6}-1.97453\times 10^{-8}t^{7}\\ \\ ~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}+9.96904\times 10^{-10}t^{8}-3.2067\times 10^{-11}t^{9},\\ \\ R_{10}(t)=10+0.5t-0.135t^{2}+0.006025t^{3}-7.17708\times 10^{-6}t^{4}\\ \\ ~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}-7.97984\times 10^{-6}t^{5}+2.96686\times 10^{-7}t^{6}-2.63764\times 10^{-9}t^{7}\\ \\ ~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}-2.13846\times 10^{-10}t^{8}+1.34528\times 10^{-11}t^{9},\end{array}$$

$$\begin{array}[]{l}S_{15}(t)=20-2.3t+0.15425t^{2}-0.00790458t^{3}+0.000309711t^{4}\\ \\ ~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}-7.74864\times 10^{-6}t^{5}-1.35996\times 10^{-8}t^{6}+1.41005\times 10^{-8}t^{7}\\ \\ ~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}-8.93373\times 10^{-10}t^{8}+3.32927\times 10^{-11}t^{9}-5.68453\times 10^{-13}t^{10}\\ \\ ~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}-2.34166\times 10^{-14}t^{11}+2.59234\times 10^{-15}t^{12}-1.28786\times 10^{-16}t^{13}+3.78868\times 10^{-18}t^{14},\\ \\ I_{15}(t)=15-2.2t+0.04575t^{2}+0.00573792t^{3}-0.000406169t^{4}\\ \\ ~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}+9.82135\times 10^{-6}t^{5}+1.12052\times 10^{-7}t^{6}-1.97453\times 10^{-8}t^{7}\\ \\ ~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}+9.96904\times 10^{-10}t^{8}-3.2067\times 10^{-11}t^{9}+4.21668\times 10^{-13}t^{10}\\ \\ ~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}+2.88892\times 10^{-14}t^{11}-2.70438\times 10^{-15}t^{12}+1.28307\times 10^{-16}t^{13}-3.62709\times 10^{-18}t^{14},\\ \\ R_{15}(t)=10+0.5t-0.135t^{2}+0.006025t^{3}-7.17708\times 10^{-6}t^{4}\\ \\ ~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}-7.97984\times 10^{-6}t^{5}+2.96686\times 10^{-7}t^{6}-2.63764\times 10^{-9}t^{7}\\ \\ ~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}-2.13846\times 10^{-10}t^{8}+1.34528\times 10^{-11}t^{9}-4.55198\times 10^{-13}t^{10}\\ \\ ~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}+7.97151\times 10^{-15}t^{11}+1.74314\times 10^{-16}t^{12}-2.21438\times 10^{-17}t^{13}+1.07465\times 10^{-18}t^{14}.\end{array}$$

$$\begin{array}[]{lll}S_{0}(t)=20,&I_{0}(t)=15,&R_{0}(t)=10,\\ \\ S_{1}(t)=-2.3t,&I_{1}(t)=-2.2t,&R_{1}(t)=0.5t,\\ \\ S_{2}(t)=0.15425t^{2},&I_{2}(t)=0.04575t^{2},&R_{2}(t)=-0.135t^{2},\\ \\ ~{}~{}~{}~{}~{}~{}~{}~{}\vdots&~{}~{}~{}~{}~{}~{}~{}~{}\vdots&~{}~{}~{}~{}~{}~{}~{}~{}\vdots\\ \\ S_{10}(t)=-5.68453\times 10^{-13}t^{10},&I_{10}(t)=4.21668\times 10^{-13}t^{10},&R_{10}(t)=-4.55198\times 10^{-13}t^{10},\\ \\ ~{}~{}~{}~{}~{}~{}~{}~{}\vdots&~{}~{}~{}~{}~{}~{}~{}~{}\vdots&~{}~{}~{}~{}~{}~{}~{}~{}\vdots\end{array}$$

$$\begin{array}[]{l}\displaystyle S_{10}(t)=\sum_{j=0}^{10}S_{j}(t)=20-2.3t+0.15425t^{2}-0.00790458t^{3}+0.000309711t^{4}\\ \\ ~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}-7.74864\times 10^{-6}t^{5}-1.35996\times 10^{-8}t^{6}+1.41005\times 10^{-8}t^{7}\\ \\ ~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}-8.93373\times 10^{-10}t^{8}+3.32927\times 10^{-11}t^{9}-5.68453\times 10^{-13}t^{10},\\ \\ \displaystyle I_{10}(t)=\sum_{j=0}^{10}I_{j}(t)=15-2.2t+0.04575t^{2}+0.00573792t^{3}-0.000406169t^{4}\\ \\ ~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}+9.82135\times 10^{-6}t^{5}+1.12052\times 10^{-7}t^{6}-1.97453\times 10^{-8}t^{7}\\ \\ ~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}+9.96904\times 10^{-10}t^{8}-3.2067\times 10^{-11}t^{9}+4.21668\times 10^{-13}t^{10},\\ \\ \displaystyle R_{10}(t)=\sum_{j=0}^{10}R_{j}(t)=10+0.5t-0.135t^{2}+0.006025t^{3}-7.17708\times 10^{-6}t^{4}\\ \\ ~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}-7.97984\times 10^{-6}t^{5}+2.96686\times 10^{-7}t^{6}-2.63764\times 10^{-9}t^{7}\\ \\ ~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}-2.13846\times 10^{-10}t^{8}+1.34528\times 10^{-11}t^{9}-4.55198\times 10^{-13}t^{10}.\end{array}$$

$$\begin{array}[]{l}\displaystyle E_{n,S}(t)=S^{\prime}_{n}(t)-f_{1}+\lambda S_{n}(t)I_{n}(t)+dS_{n}(t),\\ \\ \displaystyle E_{n,I}(t)=I^{\prime}_{n}(t)-f_{2}-\lambda S_{n}(t)I_{n}(t)+\varepsilon I_{n}(t)+dR_{n}(t),\\ \\ \displaystyle E_{n,R}(t)=R^{\prime}_{n}(t)-f_{3}-\varepsilon I_{n}(t)+dR_{n}(t).\\ \end{array}$$

$$\begin{array}[]{c}\includegraphics[width=180.67499pt]{1.eps}\\ \\ \includegraphics[width=180.67499pt]{2.eps}\\ \\ \includegraphics[width=180.67499pt]{3.eps}\end{array}$$

$$\begin{array}[]{c}\includegraphics[width=180.67499pt]{4.eps}\\ \\ \includegraphics[width=180.67499pt]{5.eps}\\ \\ \includegraphics[width=180.67499pt]{6.eps}\end{array}$$

$$\begin{array}[]{c}\includegraphics[width=180.67499pt]{7.eps}\\ \\ \includegraphics[width=180.67499pt]{8.eps}\\ \\ \includegraphics[width=180.67499pt]{9.eps}\end{array}$$

$$\begin{array}[]{ccc}\includegraphics[width=144.54pt]{10.eps}&{}{}{}{}{}{}{}{}{}{}{}\hfil&\includegraphics[width=144.54pt]{11.eps}\\ \\ \includegraphics[width=144.54pt]{12.eps}&{}{}{}{}{}{}{}{}{}{}{}\hfil&\includegraphics[width=144.54pt]{13.eps}\\ \end{array}$$

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Full text

keywords: Non-linear Susceptible-Infected-Recovered model, Differential transform method, Laplace transformations, Adomian decomposition method.

Department of Mathematics, Central Tehran Branch, Islamic Azad University, Tehran, Iran.

1 Introduction

The computer viruses are malware programs that have been able to infect thousands of computers and have hurt billions dollar in computers around the world. The virus should never be considered to be harmless and remain in the system. There are several types of viruses that can be categorized according to their source, technique, file type that infects, where they are hiding, the type of damage they enter, the type of operating system, or the design on which they are attacking. We can introduce some of famous and malicious viruses such as ILOVEYOU, Melissa, My Doom, Code Red, Sasser, Stuxnet and so on. Therefore, it is important that we study the methods to analyze, track, model, and protect against viruses. In recent years, many scientists have been illustrated the epidemiological models of computer viruses [2, 14, 17, 21, 32, 33, 36, 37, 45, 46, 47]. These models have been estimated by many mathematical methods such as collocation method [30], homotopy analysis method [5, 27], variational iteration method [23] and others [33, 41].

One of applicable and important models is the classical Susceptible-Infected-Recovered (SIR) computer virus propagation model [21, 22, 33] which is presented in the following form:

[TABLE]

where

[TABLE]

are the initial conditions of non-linear system of Eqs. (1). Functions and initial values of system (1) are given in Table 1.

Recently, several numerical and semi-analytical methods are introduced to solve the mathematical and engineering problems [9, 10, 11, 24, 25, 26, 48] that we can apply them to solve the non-linear model (1). The DTM and the LADM are two important and efficient tools to solve the linear and non-linear problems arising in the mathematics, physics and engineering [6, 7, 29, 34, 35, 40, 43]. Specially, the LADM [12, 15, 31] obtained by combining the Adomian decomposition method [1, 3, 4, 16, 39, 44] and the Laplace transformations [18] similar to the HATM [8, 24, 27, 28], Laplace homotopy perturbation method [13, 38, 42] and so on.

The aim of this paper is to apply the DTM and the LADM to find the approximate solution of non-linear epidemiological system of Eqs. (1). The numerical results are compared with the HATM [8, 24, 27, 28] by plotting the residual errors function for different iterations. Also, the phase portraits of approximate solutions for $n=10$ and different functions of $S(t),I(t)$ and $R(t)$ are presented. The numerical results show the abilities and capabilities of the LADM in comparison to the DTM and the HATM.

2 Differential transform method

Transformation of the $k$-th derivative of a function in one variable is as follows

[TABLE]

and the inverse transformation is defined by

[TABLE]

where $F(k)$ is the differential transform of $f(t)$. In actual applications, the function $f(t)$ is expressed by a finite series and Eq. (4) can be rewritten as follows:

[TABLE]

where $N$ is decided by the convergence of natural frequency. The fundamental operations of DTM have been given in Table 3.

By applying the presented method to system of Eqs. (1), we get

[TABLE]

The differential transform method series solution for system (1) can be obtained as

[TABLE]

3 Laplace-Adomian decomposition method

We apply the Laplace transformation $\mathcal{L}$ as

[TABLE]

By putting the initial conditions we have

[TABLE]

where $A=SI$ and

[TABLE]

Also, the non-linear operator $A$ is called the Adomian polynomials and it is presented as

[TABLE]

where

[TABLE]

By substituting series (10) and (11) into (9) we get

[TABLE]

Now, the following relations can be obtained:

[TABLE]

and for term $j$ we have

[TABLE]

Applying the inverse Laplace transformation $\mathcal{L}^{-1}$ for first equations of (14) as follows

[TABLE]

By putting $S_{0},I_{0},R_{0}$ in second equations of (14) and using the Laplace transformations we have

[TABLE]

and by applying the inverse Laplace transform $\mathcal{L}^{-1}$ we can find $S_{1},I_{1}$ and $R_{1}$. By repeating above process, the other terms $S_{2},\cdots,S_{j},I_{2},\cdots,I_{j},R_{2},\cdots,R_{j}$ can be obtained. By using the relations

[TABLE]

the $n$-th order approximate solutions can be estimated.

4 Numerical Illustration

In this section, the numerical results of the DTM and the LADM for solving the system of Eqs. (1) are presented. The approximate solutions for $n=5$ by using the DTM are obtained in the following form

[TABLE]

and for $n=10$ we have

[TABLE]

and finally for $n=15$ the approximate solutions are obtained as

[TABLE]

Now, by applying the LADM we get

[TABLE]

and finally the approximate solution of epidemiological model of computer viruses (1) for $n=10$ is in the following form

[TABLE]

In order to show the accuracy of the presented methods, following residual errors are presented. Also, the numerical results are compared to the obtained results of the HATM for $n=5,10$. The results are presented in Tables 3, 4 and 5.

[TABLE]

The comparative graphs between the residual errors of the LADM, the DTM and the HATM for $n=5,10,15$ are demonstrated in Figs. 1, 2 and 3. Also, phase portraits of $S-I,S-R,I-R$ and $S-I-R$ which are obtained by $10$-th order approximation of the DTM and the LADM are presented in Figs. 4 and 5. According to the generated results, the LADM has suitable scheme than the DTM and the HATM.

5 Conclusion

In this study, two robust and applicable methods, the DTM and the LADM were applied to solve the non-linear epidemiological model of computer viruses. In order to show the efficiency and accuracy of presented method, the residual errors for different iterations were presented based on the LADM, DTM and HATM. Also, the graphs of residual error were demonstrated to show the abilities of the LADM than the other methods.