數學建模 - SEIR模型 === ###### tags: `數學建模` ![](https://i.imgur.com/1HybH6T.png) --- 1. $\dfrac{dS}{dt} = - \beta SI$ 2. $\dfrac{dE}{dt} = \beta SI - \alpha E$ 3. $\dfrac{dI}{dt} = \alpha E - \gamma I$ 4. $\dfrac{dR}{dt} = \gamma I$ 5. $S = S_0,\ E= E_0,\ I=I_0,\ R=R_0 \enspace at \enspace t = 0$ --- Equations (1) and (2) are added to obtain 6. $\dfrac{dS}{dt} + \dfrac{dE}{dt}= - \alpha E$ Solve (3) for E and substituting into (6) 7. $\dfrac{d^2I}{dt^2} + (\gamma+\alpha)\dfrac{dI}{dt} + \alpha\dfrac{dS}{dt} + \alpha\gamma I = 0$ --- Rewrite (1) 8. $I = -\dfrac{1}{\beta}\dfrac{d\ln S}{dt}$ Substituted into (7) 9. $\begin{multline*}\dfrac{d^3\ln S}{dt^3} + (\gamma+\alpha)\dfrac{d^2\ln S}{dt^2} \\ - \alpha\beta\dfrac{dS}{dt} + \alpha\gamma \dfrac{d\ln S}{dt} = 0\end{multline*}$ --- (9) may be integrated to yield 10. $\dfrac{d^2\ln S}{dt^2} + (\gamma+\alpha)\dfrac{d\ln S}{dt} - \alpha\beta S + \alpha\gamma\ln S = C$ where $C$ is 11. $C = \alpha\gamma\ln (S_0) - \alpha\beta(E_0 + I_0 + S_0)$ --- Substituted $f = \ln S$ in euqation (10) 12. $\dfrac{d^2f}{dt^2} + (\gamma+\alpha)\dfrac{df}{dt} - \alpha\beta e^f + \alpha\gamma f = C$ where from (5) and (8) 13. $f = \ln S_0,\ \dfrac{df}{dt} = -\beta I_0 \enspace at \enspace t = 0$ --- We know 14. $S= e^f$ The solution for $I$ follows directly from (8) and (14) as 15. $I = -\dfrac{1}{\beta}\dfrac{df}{dt}$ --- After substituting (15) into (4), integrating, and applying the constraint (13), R is expressed as: 16. $R = R_0 - \dfrac{\gamma}{\beta}(f - \ln S_0)$ By adding equations (1) through (4), integrating in t, and applying (5) 17. $E = E_0 + I_0 + S_0 + R_0 − I − S − R$ --- The series solution of (11)-(13) is given by 18. $f = \sum^{\infty}_{n=0}a_nt^n,\ a_0=\ln S_0,\ a_1=-\beta I_0$ 19. $a_2 = \left[ C - ( \alpha + \gamma )a_1 + \alpha\beta S_0 - \alpha\gamma a_0 \right] /2$ 20. $\begin{multline*}a_{n+2} = \dfrac{\alpha\beta \tilde a_n-(\gamma+\alpha )(n+1)a_{n+1} -\alpha\gamma a_n}{(n+2)(n+1)},\ \\ n>0\end{multline*}$ 21. $\tilde a_{n+1>0} = \dfrac{1}{n+1}\sum^n_{j=0} (n − j + 1) a_{n-j+1} \tilde a_j,\ \tilde a_0 = S_0$ --- * Although the <font color="#f00">series solution</font> given by (18)-(21) is an analytic solution to (11)-(13) , <font color="#3498DB">it is only valid within its radius of convergence</font> and is incapable of capturing the long-time behavior of the system. * The long-time asymptotic behavior of the system (18)-(21) is required to develop our asymptotic approximant --- It has been proven in prior literature that <font color="#f00">S approaches a limiting value, $S_∞$, as $t → ∞$</font>, and this corresponds to $I → 0$ in the same limit. Thus, $f$ approaches a limiting value, $f_∞ ≡ lnS_∞$, as $t → ∞$. The value of $f_∞$ satisfies the following equation 22. $e^{f_{\infty}}-\dfrac{\gamma}{\beta}(f_{\infty}-\ln S_0)-E_0-I_0-S_0 = 0$ in the interval 23. $f \in (-\infty ,\ \ln \gamma/\beta )$ ---- 11. $C = \alpha\gamma\ln (S_0) - \alpha\beta(E_0 + I_0 + S_0)$ 12. $\dfrac{d^2f}{dt^2} + (\gamma+\alpha)\dfrac{df}{dt} - \alpha\beta e^f + \alpha\gamma f = C$ --- We expand $f$ as $t → ∞$ as follows 24. $f \sim f_{\infty} + g(t) \enspace where \enspace g→0 \enspace as \enspace t→\infty$ Eq. (24) is substituted into (12) (with (11)), $e^g$ is replace with its power series expansion, and terms of $O(g^2)$ are neglected to achieve the following linearized equation 25. $\dfrac{d^2g}{dt^2} + (\gamma+\alpha)\dfrac{dg}{dt}+(\alpha\gamma-\alpha\beta e^{f_{\infty}})g = 0$ ---- 11. $C = \alpha\gamma\ln (S_0) - \alpha\beta(E_0 + I_0 + S_0)$ 12. $\dfrac{d^2f}{dt^2} + (\gamma+\alpha)\dfrac{df}{dt} - \alpha\beta e^f + \alpha\gamma f = C$ --- The general solution to (25) is 26. $g = \epsilon_1e^{\lambda_1t}+\epsilon_2e^{\lambda_2t}$ 27. $\lambda_{1,2} = \dfrac{1}{2}\left[-\alpha-\gamma\pm \sqrt{(\gamma-\alpha)^2+4\alpha\beta e^{f_{\infty}}} \right]$ where $\epsilon_1$ and $\epsilon_2$ are unknown constants and $\lambda_2 < \lambda_1 < 0$ since $e^{f_∞} < γ /β$ from (23) ---- 23. $f \in (-\infty ,\ \ln \gamma/\beta )$ --- Thus the long-time asymptotic behavior of $f$ is given by 28. $f \sim f_{\infty} + \epsilon_1e^{\lambda_1t},\ t→ ∞$ --- * Higher order corrections to the expansion (28) may be obtained by the method of <font color="#f00">dominant balance</font> as a series of more rapidly damped exponentials. * However, the pattern by which the corrections are asymptotically ordered is <font color="#3498DB">not as straightforward as that of the SIR model</font>, provided in Barlow and Weinstein. In that work, an asymptotic approximant is constructed as a series of exponentials that exactly mimics the long-time expansion. --- * In the SEIR model, complications in the higher-order asymptotic behavior arise from the competition between the two exponentials in (26). Here, we <font color="#3498DB">enforce the leading-order $t → ∞$ behavior given by (28) and make a more traditional choice for matching with the t = 0 expansion (18)-(21).</font> ---- * $g = \epsilon_1e^{\lambda_1t}+\epsilon_2e^{\lambda_2t}$ * $f \sim f_{\infty} + \epsilon_1e^{\lambda_1t},\ t→ ∞$ ---- 18. $f = \sum^{\infty}_{n=0}a_nt^n,\ a_0=\ln S_0,\ a_1=-\beta I_0$ 19. $a_2 = \left[ C - ( \alpha + \gamma )a_1 + \alpha\beta S_0 - \alpha\gamma a_0 \right] /2$ 20. $\begin{multline*}a_{n+2} = \dfrac{\alpha\beta \tilde a_n-(\gamma+\alpha )(n+1)a_{n+1} -\alpha\gamma a_n}{(n+2)(n+1)},\ \\ n>0\end{multline*}$ 21. $\tilde a_{n+1>0} = \dfrac{1}{n+1}\sum^n_{j=0} (n − j + 1) a_{n-j+1} \tilde a_j,\ \tilde a_0 = S_0$ --- * We create an approximant with an embedded rational function with equal-order numerator and denominator (i.e., <font color="#f00">a symmetric Padé approximant</font>), such that it <font color="#3498DB"> approaches the unknown constant $ε_1$ in (28) as $t → ∞$</font>, while converging to the intermediate behavior at shorter times. The assumed SEIR approximant is given by --- 29. $f_{A,N} = f_{\infty} + e^{\lambda_1t}\dfrac{\sum^{N/2}_{n=0}A_nt^n}{1+\sum^{N/2}_{n=1}B_nt^n},\ N \enspace even$ ---- * <font color="#3498DB">$A_n$ and $B_n$ coefficients are obtained such that the Taylor expansion of (29) about t = 0 is exactly (18)-(21).</font> * Note that, although a rational function is being used in (29), it is not a <font color="#f00">Padé approximant </font>itself. <font color="#f00">Padé approximants</font> are only capable of capturing $t^n$ behavior in the long-time limit, where $n$ is an integer. ---- * The pre-factor $e^{λ_1t}$ is required to make (29) an asymptotic approximant for the SEIR model. * However, we may still make use of <font color="#f00">fast Padé coefficient solvers</font> by recasting (29) as a <font color="#f00">Padé approximant</font> for the series that results from the <font color="#3498DB">Cauchy product between the expansions of $e^{−λ_1t}$ and $f − f_∞$, expressed as</font> 30. $\sum^N_{n=0}\left[ \sum^n_{j=0} \dfrac{(-\lambda_1)^j}{j!}\tilde a_{n-j} \right]t^n = \dfrac{\sum^{N/2}_{n=0}A_nt^n}{1+\sum^{N/2}_{n=1}B_nt^n}$ ---- Rewrite (28) * $\epsilon_1 = (f - f_{\infty})e^{-\lambda_1t},\ t→ ∞$ * $f = \sum^{\infty}_{n=0}a_nt^n$ --- $S_0 = 0.88$ ![](https://i.imgur.com/Uumj3Ek.png) --- $S_{A,18}$ ![](https://i.imgur.com/DKQnOrp.png) ---- $\alpha = 0.466089$ $\beta = 0.2$ $\gamma = 0.1$ $S_0 = 0.88$ $E_0 = 0.07$ $I_0 = 0.05$ $R_0 = 0$ --- 中國雲南 January 28, 2020. $S_0 = 144$ ![](https://i.imgur.com/sqbRsSh.png) --- 中國雲南 January 28, 2020. $S_{A,22}$ ![](https://i.imgur.com/VyiL0vG.png) ---- $\alpha = 0.395031$ $\beta = 0.00333$ $\gamma = 0.0553093$ $S_0 = 144$ $E_0 = 0$ $I_0 = 44$ $R_0 = 0$ --- 瑞典 March 21, 2020. $S_0 = 50306$ ![](https://i.imgur.com/W78CyrS.png) --- 瑞典 March 21, 2020. $S_{A,52}$ ![](https://i.imgur.com/9IIMgH0.png) ---- $\alpha = 0.041281$ $\beta = 1.513332*10^{-6}$ $\gamma = 0.004407$ $S_0 = 50306$ $E_0 = 10015$ $I_0 = 1743$ $R_0 = 20$ --- 日本 April 1, 2020. $S_0 = 15442$ ![](https://i.imgur.com/n7Cg2Bs.png) --- 日本 April 1, 2020. $S_{A,42}$ ![](https://i.imgur.com/LqB9VeH.png) ---- $\alpha = 0.2332207$ $\beta = 2.040015*10^{-5}$ $\gamma = 0.034334$ $S_0 = 15442$ $E_0 = 0$ $I_0 = 1694$ $R_0 = 529$ --- 臺灣 March 1, 2020. $S_0 = 438$ ![](https://i.imgur.com/Kq16jOC.png) --- 臺灣 March 1, 2020. $S_{A,60}$ ![](https://i.imgur.com/NvurNwd.png) ---- $\alpha = 0.395031$ $\beta = 0.000513$ $\gamma = 0.02756093$ $S_0 = 438$ $E_0 = 0$ $I_0 = 30$ $R_0 = 10$ --- * Our results demonstrate that <font color="#f00">an asymptotic approximant</font> can be used to provide <font color="#3498DB">accurate analytic solutions</font> to the SEIR model. * Future work should examine the ability of the asymptotic approximant technique to yield closed-form solutions for even more sophisticated epidemic models, as well as their endemic counterparts
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