解下面的 ODE 方程组
\[
\left\{
\begin{array}{rcl}
\dfrac{di}{dt}&=&\lambda si-\mu i,\\
\dfrac{ds}{dt}&=&-\lambda si,\\
s(t)+i(t)+r(t)&=&1,\\
i(0)&=&i_0,\\
s(0)&=&s_0,\\
r(0)&=&0.
\end{array}\right.
\]
可以先解出
\[
s(t)=s_0 e^{-\sigma r(t)},
\]
这里 $\sigma:=\frac{\lambda}{\mu}$. 从而有
\[
\frac{dr}{dt}=\mu(1-r-s_0 e^{-\sigma r}).
\]
References:
姜启源、谢金星、叶俊 编 《数学模型》（第四版）P.140
1
由 $s(t)+i(t)+r(t)=1$ 推出 $s'(t)+i'(t)+r'(t)=0$. 故
\[
\frac{dr}{dt}=-\frac{ds}{dt}-\frac{di}{dt}=-(-\lambda si)-(\lambda si-\mu i)=\mu i=\mu(1-r(t)-s(t)).
\]
于是
\[
\frac{ds}{dr}=\frac{\frac{ds}{dt}}{\frac{dr}{dt}}=\frac{-\lambda si}{\mu i}=-\frac{\lambda}{\mu}s=-\sigma s.
\]
通过分离变量, 解得 $s=C\cdot e^{-\sigma r}$. 将初始值代入, 得 $s_0=C\cdot e^{-\sigma r_0}$, 注意 $r_0=0$, 因此 $C=s_0$. 故
\[
s(t)=s_0 e^{-\sigma r(t)}.
\]
为了方便做计算, 只取 $e^{-\sigma r(t)}$ Taylor 展式的前三项, 即 $1+(-\sigma r)+\frac{1}{2}(\sigma r)^2$. 因此, $\frac{dr}{dt}$ 满足
\[
\begin{split}
\frac{dr}{dt}&=\mu\Bigl(1-r-s_0(1-\sigma r+\frac{1}{2}\sigma^2 r^2)\Bigr)\\
&=\mu\Bigl[1-r-s_0+s_0\sigma r-\frac{1}{2}s_0\sigma^2 r^2\Bigr]\\
&=\mu\Bigl[(1-s_0)+(s_0\sigma-1)r-\frac{1}{2}s_0\sigma^2 r^2\Bigr]\\
&=\mu(1-s_0)+\mu(s_0\sigma-1)r-\frac{1}{2}\mu s_0\sigma^2 r^2.
\end{split}
\]
仿照 问题1959 的做法, 通过变量代换 $\gamma=r-m$, 使得成为 Bernoulli 方程.
令
\[
\mu(1-s_0)+\mu(s_0\sigma-1)m-\frac{1}{2}\mu s_0\sigma^2 m^2=0,
\]
化为
\[
s_0\sigma^2 m^2-2(s_0\sigma-1)m+2(s_0-1)=0.
\]
于是解为
\[
\begin{split}
m&=\dfrac{2(s_0\sigma-1)\pm\sqrt{4(s_0\sigma-1)^2-4s_0\sigma^2 2(s_0-1)}}{2s_0\sigma^2}\\
&=\dfrac{(s_0\sigma-1)\pm\sqrt{(s_0\sigma-1)^2-s_0\sigma^2 2(s_0-1)}}{s_0\sigma^2}\\
&=\dfrac{(s_0\sigma-1)\pm\sqrt{(s_0\sigma-1)^2+2s_0 i_0\sigma^2}}{s_0\sigma^2}
\end{split}
\]
最后一个等号用到了 $r_0=0$. 记 $\alpha^2=(s_0\sigma-1)^2+2s_0 i_0\sigma^2$, $\alpha$ 符号可正可负, 故不妨写
\[
m=\frac{(s_0\sigma-1)+\alpha}{s_0\sigma^2}.
\]
于是方程化为
\[
\begin{split}
\frac{d\gamma}{dt}&=\Bigl(\mu(s_0\sigma-1)+2m\cdot(-\frac{1}{2})\mu s_0\sigma^2\Bigr)\gamma-\frac{1}{2}\mu s_0\sigma^2 \gamma^2\\
&=-\alpha\mu\gamma-\frac{1}{2}\mu s_0\sigma^2 \gamma^2
\end{split}
\]
此时, 化为关于 $\gamma$ 的 Bernoulli 方程. 两边除以 $-\gamma^2$, 得
\[
-\frac{1}{\gamma^2}\gamma'=\alpha\mu\frac{1}{\gamma}+\frac{1}{2}\mu s_0\sigma^2,
\]
即
\[
(\frac{1}{\gamma})'-\alpha\mu\cdot\frac{1}{\gamma}=\frac{1}{2}\mu s_0\sigma^2,
\]
令 $u(t)=\frac{1}{\gamma(t)}$, 得
\[
u'(t)-\alpha\mu u(t)=\frac{1}{2}\mu s_0\sigma^2.
\]
解得
\[
\begin{split}
u(t)&=e^{-\int(-\alpha\mu)dt}\cdot\biggl[\int\frac{1}{2}\mu s_0 \sigma^2\cdot e^{\int(-\alpha\mu)dt}dt+C\biggr]\\
&=e^{\alpha\mu t}\cdot\biggl[\int\frac{1}{2}\mu s_0 \sigma^2 e^{-\alpha\mu t}+C\biggr]\\
&=e^{\alpha\mu t}\cdot\biggl[-\frac{s_0\sigma^2}{2\alpha}e^{-\alpha\mu t}+C\biggr]\\
&=-\frac{s_0\sigma^2}{2\alpha}+Ce^{\alpha\mu t},
\end{split}
\]
令 $t=0$, 得 $u(0)=-\frac{s_0\sigma^2}{2\alpha}+C$, 推出 $C=u(0)+\frac{s_0\sigma^2}{2\alpha}=\frac{1}{\gamma(0)}+\frac{s_0\sigma^2}{2\alpha}$. 这里
\[
\gamma(0)=r(0)-m=0-\frac{(s_0\sigma-1)+\alpha}{s_0\sigma^2},\quad \alpha^2=(s_0\sigma-1)^2+2s_0 i_0\sigma^2
\]
故
\[C=-\frac{s_0\sigma^2}{(s_0\sigma-1)+\alpha}+\frac{s_0\sigma^2}{2\alpha}=s_0\sigma^2\frac{s_0\sigma-1-\alpha}{2\alpha(s_0\sigma-1+\alpha)},\]
因此
\[
\gamma(t)=\frac{1}{u(t)}=\frac{1}{-\frac{s_0\sigma^2}{2\alpha}+s_0\sigma^2\frac{s_0\sigma-1-\alpha}{2\alpha(s_0\sigma-1+\alpha)}e^{\alpha\mu t}}
\]