數學建模 - SEIR模型
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###### tags: `數學建模`

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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. $S = S_0,\ E= E_0,\ I=I_0,\ R=R_0 \enspace at \enspace t = 0$
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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. $\dfrac{d^2I}{dt^2} + (\gamma+\alpha)\dfrac{dI}{dt} + \alpha\dfrac{dS}{dt} + \alpha\gamma I = 0$
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Rewrite (1)
8. $I = -\dfrac{1}{\beta}\dfrac{d\ln S}{dt}$
Substituted into (7)
9. $\begin{multline*}\dfrac{d^3\ln S}{dt^3} + (\gamma+\alpha)\dfrac{d^2\ln S}{dt^2} \\ - \alpha\beta\dfrac{dS}{dt} + \alpha\gamma \dfrac{d\ln S}{dt} = 0\end{multline*}$
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(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. $C = \alpha\gamma\ln (S_0) - \alpha\beta(E_0 + I_0 + S_0)$
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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. $f = \ln S_0,\ \dfrac{df}{dt} = -\beta I_0 \enspace at \enspace t = 0$
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We know
14. $S= e^f$
The solution for $I$ follows directly from (8) and (14) as
15. $I = -\dfrac{1}{\beta}\dfrac{df}{dt}$
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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. $E = E_0 + I_0 + S_0 + R_0 − I − S − R$
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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. $\tilde a_{n+1>0} = \dfrac{1}{n+1}\sum^n_{j=0} (n − j + 1) a_{n-j+1} \tilde a_j,\ \tilde a_0 = S_0$
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* 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
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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. $f \in (-\infty ,\ \ln \gamma/\beta )$
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11. $C = \alpha\gamma\ln (S_0) - \alpha\beta(E_0 + I_0 + S_0)$
12. $\dfrac{d^2f}{dt^2} + (\gamma+\alpha)\dfrac{df}{dt} - \alpha\beta e^f + \alpha\gamma f = C$
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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. $\dfrac{d^2g}{dt^2} + (\gamma+\alpha)\dfrac{dg}{dt}+(\alpha\gamma-\alpha\beta e^{f_{\infty}})g = 0$
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11. $C = \alpha\gamma\ln (S_0) - \alpha\beta(E_0 + I_0 + S_0)$
12. $\dfrac{d^2f}{dt^2} + (\gamma+\alpha)\dfrac{df}{dt} - \alpha\beta e^f + \alpha\gamma f = C$
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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)
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23. $f \in (-\infty ,\ \ln \gamma/\beta )$
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Thus the long-time asymptotic behavior of $f$ is given by
28. $f \sim f_{\infty} + \epsilon_1e^{\lambda_1t},\ t→ ∞$
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* 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.
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* 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>
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* $g = \epsilon_1e^{\lambda_1t}+\epsilon_2e^{\lambda_2t}$
* $f \sim f_{\infty} + \epsilon_1e^{\lambda_1t},\ t→ ∞$
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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. $\tilde a_{n+1>0} = \dfrac{1}{n+1}\sum^n_{j=0} (n − j + 1) a_{n-j+1} \tilde a_j,\ \tilde a_0 = S_0$
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* 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
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29. $f_{A,N} = f_{\infty} + e^{\lambda_1t}\dfrac{\sum^{N/2}_{n=0}A_nt^n}{1+\sum^{N/2}_{n=1}B_nt^n},\ N \enspace even$
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* <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.
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* 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. $\sum^N_{n=0}\left[ \sum^n_{j=0} \dfrac{(-\lambda_1)^j}{j!}\tilde a_{n-j} \right]t^n = \dfrac{\sum^{N/2}_{n=0}A_nt^n}{1+\sum^{N/2}_{n=1}B_nt^n}$
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Rewrite (28)
* $\epsilon_1 = (f - f_{\infty})e^{-\lambda_1t},\ t→ ∞$
* $f = \sum^{\infty}_{n=0}a_nt^n$
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$S_0 = 0.88$

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$S_{A,18}$

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$\alpha = 0.466089$
$\beta = 0.2$
$\gamma = 0.1$
$S_0 = 0.88$
$E_0 = 0.07$
$I_0 = 0.05$
$R_0 = 0$
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中國雲南 January 28, 2020. $S_0 = 144$

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中國雲南 January 28, 2020. $S_{A,22}$

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$\alpha = 0.395031$
$\beta = 0.00333$
$\gamma = 0.0553093$
$S_0 = 144$
$E_0 = 0$
$I_0 = 44$
$R_0 = 0$
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瑞典 March 21, 2020. $S_0 = 50306$

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瑞典 March 21, 2020. $S_{A,52}$

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$\alpha = 0.041281$
$\beta = 1.513332*10^{-6}$
$\gamma = 0.004407$
$S_0 = 50306$
$E_0 = 10015$
$I_0 = 1743$
$R_0 = 20$
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日本 April 1, 2020. $S_0 = 15442$

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日本 April 1, 2020. $S_{A,42}$

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$\alpha = 0.2332207$
$\beta = 2.040015*10^{-5}$
$\gamma = 0.034334$
$S_0 = 15442$
$E_0 = 0$
$I_0 = 1694$
$R_0 = 529$
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臺灣 March 1, 2020. $S_0 = 438$

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臺灣 March 1, 2020. $S_{A,60}$

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$\alpha = 0.395031$
$\beta = 0.000513$
$\gamma = 0.02756093$
$S_0 = 438$
$E_0 = 0$
$I_0 = 30$
$R_0 = 10$
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* 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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