数学建模:微分方程模型— SIR 传染病模型

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模型假设

  1. 被研究人群是封闭的, 总人为



    N

    N






    N





    ,病人、健康人和移出者 (有免疫能力, 移出感染系统的人, 病人治愈后为移出者) 的比例分别为



    i

    (

    t

    )

    ,

    s

    (

    t

    )

    ,

    r

    (

    t

    )

    i (t ), s (t ), r (t )






    i


    (


    t


    )


    ,




    s


    (


    t


    )


    ,




    r


    (


    t


    )




  2. 病人的日接触率



    λ

    \lambda






    λ





    , 日治愈率



    μ

    \mu






    μ





    , 接触数



    σ

    =

    λ

    μ

    \sigma = \displaystyle{\frac{\lambda}{\mu}}






    σ




    =




















    μ














    λ

























模型建立





s

(

t

)

+

i

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t

)

+

r

(

t

)

=

1

,

(

1

)

s (t ) + i(t ) + r (t ) = 1, \quad (1)






s


(


t


)




+








i


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t


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+








r


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)




=








1


,






(


1


)









N

[

i

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Δ

t

)

i

(

t

)

]

=

λ

N

s

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t

)

i

(

t

)

Δ

t

μ

N

i

(

t

)

Δ

t

,

(

2

)

N[i(t+\Delta t)-i(t)]=\lambda N s(t) i(t) \Delta t-\mu N i(t) \Delta t, \quad (2)






N


[


i


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t




+








Δ


t


)













i


(


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]




=








λ


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s


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i


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Δ


t













μ


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i


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Δ


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,






(


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)









N

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s

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)

s

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]

=

λ

N

s

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t

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i

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)

Δ

t

,

(

3

)

N[s(t+\Delta t)-s(t)]=-\lambda N s(t) i(t) \Delta t, \quad (3)






N


[


s


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t




+








Δ


t


)













s


(


t


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]




=











λ


N


s


(


t


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i


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Δ


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,






(


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)








\Longrightarrow















{

d

i

d

t

=

λ

s

i

μ

i

d

s

d

t

=

λ

s

i

i

(

0

)

=

i

0

,

 

s

(

0

)

=

s

0

i

0

+

s

0

1

(

 通常 

r

(

0

)

=

r

0

 很小 

)

\begin{cases} {\frac{d i}{d t}}=\lambda s i-\mu i \\ {\frac{d s}{d t}}=-\lambda s i \\ i(0)=i_{0}, \ s(0)=s_{0} \\ i_{0}+s_{0} \approx 1\left(\text { 通常 } r(0)=r_{0} \text { 很小 }\right) \end{cases}


































































































































d


t
















d


i
























=




λ


s


i









μ


i





















d


t
















d


s
























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λ


s


i








i


(


0


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=





i











0



















,






s


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0


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=





s











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+





s











0


























1





(





通常





r


(


0


)




=





r











0






















很小





)


























无法求出



i

(

t

)

,

s

(

t

)

i(t ), s(t )






i


(


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,




s


(


t


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的解析解, 在相平面



s

t

s-t






s













t





上研究解的性质.



相轨线



i

(

s

)

i ( s )






i


(


s


)





及其分析

消去



d

t

d t






d


t





,



σ

=

λ

μ

\sigma=\displaystyle{\frac{\lambda}{\mu}}






σ




=




















μ














λ
























,



\Longrightarrow

















{

d

i

d

s

=

1

σ

s

1

i

s

=

s

0

=

i

0

\begin{cases} \displaystyle{\frac{d i}{d s}=\frac{1}{\sigma \cdot s}-1} \\ \left.i\right|_{s=s_{0}}=i_{0} \end{cases}











































































d


s














d


i






















=















σ









s














1



























1













i















s


=



s











0






































=





i











0













































\Longrightarrow












相轨线:





i

(

s

)

=

(

s

0

+

i

0

)

s

+

1

σ

ln

s

s

0

i(s)=\left(s_{0}+i_{0}\right)-s+\frac{1}{\sigma} \ln \frac{s}{s_{0}}






i


(


s


)




=









(



s











0





















+





i











0



















)














s




+



















σ














1






















ln
















s











0































s























定义域



D

=

{

(

s

,

i

)

s

0

,

i

0

,

s

+

i

1

}

D=\{(s, i) \mid s \geq 0, i \geq 0, s+i \leq 1\}






D




=








{



(


s


,




i


)













s













0


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i













0


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+








i













1


}




2022-03-22-18-41-45




s

(

t

)

s(t)






s


(


t


)





单调减.



t

t \rightarrow \infty






t



















时,



i

0

i \rightarrow 0






i













0





.



\Longrightarrow















t

t \rightarrow \infty






t

























s

0

+

i

0

s

+

1

σ

ln

s

s

0

=

0

s_{0}+i_{0}-s_{\infty}+\frac{1}{\sigma} \ln \frac{s_{\infty}}{s_{0}}=0







s











0





















+









i











0































s

































+



















σ














1






















ln
















s











0
































s



















































=








0








P

1

:

s

0

>

1

σ

P_{1}: s_{0}>\displaystyle{\frac{1}{\sigma}}







P











1





















:









s











0





















>




















σ














1



























i

(

t

)

\Longrightarrow i(t)















i


(


t


)





先升后降至



0

0






0








\Longrightarrow












传染病蔓延;




P

2

:

s

0

<

1

σ

P_{2}: s_{0}<\displaystyle{\frac{1}{\sigma}}







P











2





















:









s











0





















<




















σ














1



























i

(

t

)

\Longrightarrow i(t)















i


(


t


)





单调降至



0

0






0








\Longrightarrow












传染病不蔓延.




1

σ

\displaystyle{\frac{1}{\sigma}}


















σ














1
























—— 阈值



预防传染病蔓延的手段

  • 提高阈值



    1

    σ

    \displaystyle{\frac{1}{\sigma }}\Longrightarrow


















    σ














    1





























    降低



    σ

    (

    =

    λ

    μ

    )

    λ

    ,

     

    μ

    \sigma \left(=\displaystyle{\frac{\lambda}{\mu }} \right) \Longrightarrow \lambda \downarrow, \ \mu \uparrow






    σ






    (



    =
















    μ














    λ






















    )















    λ











    ,






    μ












    • λ

      \lambda






      λ





      (日接触率)



       

      \downarrow \ \Longrightarrow





















      卫生水平



      \uparrow














    • μ

      \mu






      μ





      (日治愈率)



      \uparrow















      \Longrightarrow












      医疗水平



      \uparrow











  • 降低



    s

    0

    s_{0}







    s











    0

























    s

    0

    +

    i

    0

    +

    r

    0

    =

    1

    \displaystyle{\xRightarrow{s_{0}+i_{0}+r_{0}=1}}

















    s











    0



















    +



    i











    0



















    +



    r











    0



















    =


    1





























    提高



    r

    0

    r_{0}







    r











    0

























    \Longrightarrow












    群体免疫






σ

\sigma






σ





的估计




s

0

+

i

0

s

+

1

σ

ln

s

S

0

=

0

\displaystyle{s_{0}+i_{0}-s_{\infty}+\frac{1}{\sigma} \ln \frac{s_{\infty}}{S_{0}}=0}








s











0





















+





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0



























s

































+















σ














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ln
















S











0
































s



















































=




0






, 忽略



i

0

i_0







i










0
























\Longrightarrow

















σ

=

ln

s

0

ln

s

s

0

s

\sigma=\frac{\ln s_{0}-\ln s_{\infty}}{s_{0}-s_{\infty}}






σ




=




















s











0



























s











































ln





s











0


























ln





s






















































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