标签：other review array number 需要 row begin 它的 end
$\mathbf{1.\,\text{公式}}$
假设$\,F(t)=\int_{a}^{b}f(t-\theta)d\theta,$ 那么可求得
\begin{equation}
F‘(t)=f(t-a)-f(t-b).\label{eq:1}
\end{equation}
事实上, 做变量代换$\,\eta=t-\theta$, 则$\,d\eta=-d\theta$, 从而
\[
F‘(t)=-\int_{t-a}^{t-b}f‘(\eta)d\eta=f(t-a)-f(t-b).
\]
根据上面的公式$\,($\ref{eq:1}$)$ 可知论文中
\[
\frac{d}{dt}\int_{0}^{\tau_{1}}H(x,t-\theta)N(x,t-\theta)d\theta=H(x,t)N(x,t)-H(x,t-\tau_{1})N(x,t-\tau_{1}),
\]
其中固定每个点$\,x$ 去算, 因为这个积分和求导都与$\,x$ 无关.
$\mathbf{2.\,(15)\Longrightarrow(16)}$
首先去掉拉普拉斯项, 其次去掉$\,\ln$ 项, 然后看带$\,H(x,t-\tau_{2})Y(x,t-\tau_{2})$
的项通过计算知道是平衡的, 再看带$\,Z$ 的项计算后发现也是平衡的, 接着$\,d_{1}N$ 项也是平衡的所以划掉. 然后从论文上面的三个恒等式稍加$\mathbf{\text{变形}}$知道
\begin{eqnarray*}
\kappa & = & dH_{1}+\beta_{2}H_{1}Y_{1}\\
d_{3} & = & \frac{\mu}{V_{1}}+\alpha_{3}\beta_{3}Y_{1}\\
\beta_{2}H_{1} & = & \frac{e^{a_{2}\tau_{2}}}{\alpha_{2}}\left(\beta_{3}V_{1}+d_{2}\right)
\end{eqnarray*}
现在来看$\,(15)$ 的第一项如下:
\begin{eqnarray*}
& & \left(1-\frac{H_{1}}{H}\right)[\kappa-dH-\beta_{1}HN-\beta_{2}HY]\\
& = & \left(1-\frac{H_{1}}{H}\right)[dH_{1}-dH+\beta_{2}H_{1}Y_{1}-\beta_{1}HN-\beta_{2}HY]\\
& = & \left(1-\frac{H_{1}}{H}\right)(dH_{1}-dH)+\beta_{2}H_{1}Y_{1}-\beta_{1}HN-\beta_{2}HY\\
& & -\beta_{2}\frac{H_{1}^{2}}{H}Y_{1}+\beta_{1}H_{1}N+\beta_{2}H_{1}Y\\
& = & \mathrm{I+II+III+IV}\\
& & +\mathrm{V+VI+VII}
\end{eqnarray*}
其中$\,\mathrm{I}$ 项与$\,(16)$ 的第一项抵消, $\,\mathrm{II}$ 项与$\,(16)$
的第五项抵消掉一半$\,(2$变成$1)$, $\mathrm{III}$ 项与$\,(15)$ 的第六项抵消, $\mathrm{IV}$
项与$\,(15)$ 的第七项抵消, $\mathrm{V}$ 项与$\,(16)$ 的第五项抵消, $\mathrm{VI}$
项与$\,(16)$ 的第二项抵消. 所以这表明$\,(15)$ 的第一项剩下$\,\mathrm{VII}$ 项, 即$\,\beta_{2}H_{1}Y.$
现在要证$\,(15)=(16)$ 只需要证
\begin{eqnarray*}
& & \beta_{2}H_{1}Y+\frac{e^{a_{2}\tau_{2}}}{\alpha_{2}}\left(1-\frac{Y_{1}}{Y}\right)[-\beta_{3}YV-d_{2}Y]\\
& & +\frac{e^{a_{2}\tau_{2}}}{\alpha_{2}\alpha_{3}}\left(1-\frac{V_{1}}{V}\right)[\mu+\alpha_{3}\beta_{3}YV-d_{3}V]\\
& = & \frac{\mu e^{a_{2}\tau_{2}}}{\alpha_{2}\alpha_{3}}\left(2-\frac{V_{1}}{V}-\frac{V}{V_{1}}\right)+\beta_{2}H_{1}Y_{1}
\end{eqnarray*}
即证
\begin{eqnarray}
& & \beta_{2}H_{1}(Y-Y_{1})+\frac{e^{a_{2}\tau_{2}}}{\alpha_{2}}\left(1-\frac{Y_{1}}{Y}\right)[-\beta_{3}YV-d_{2}Y]\nonumber \\
& & +\frac{e^{a_{2}\tau_{2}}}{\alpha_{2}\alpha_{3}}\left(1-\frac{V_{1}}{V}\right)[\alpha_{3}\beta_{3}YV-d_{3}V]\nonumber \\
& = & \frac{\mu e^{a_{2}\tau_{2}}}{\alpha_{2}\alpha_{3}}\left(1-\frac{V}{V_{1}}\right)\label{eq:2}
\end{eqnarray}
\[
\]
另一方面, 运用第三个恒等式有
\begin{eqnarray}
& & \beta_{2}H_{1}(Y-Y_{1})\nonumber \\
& = & \frac{e^{a_{2}\tau_{2}}}{\alpha_{2}}\left(\beta_{3}V_{1}+d_{2}\right)(Y-Y_{1})\nonumber \\
& = & \frac{e^{a_{2}\tau_{2}}}{\alpha_{2}\alpha_{3}}\left(\alpha_{3}\beta_{3}V_{1}Y+d_{2}\alpha_{3}Y-\alpha_{3}\beta_{3}V_{1}Y_{1}-d_{2}\alpha_{3}Y_{1}\right)\label{eq:3}
\end{eqnarray}
将$\,($\ref{eq:3}$)$ 代入$\,($\ref{eq:2}$)$ 且两边消去系数$\,\frac{e^{a_{2}\tau_{2}}}{\alpha_{2}\alpha_{3}}$,
$($\ref{eq:2}$)$ 变成
\begin{eqnarray*}
& & (\alpha_{3}\beta_{3}V_{1}Y+d_{2}\alpha_{3}Y-\alpha_{3}\beta_{3}V_{1}Y_{1}-d_{2}\alpha_{3}Y_{1})\\
& & +(-\alpha_{3}\beta_{3}YV-d_{2}\alpha_{3}Y+\alpha_{3}\beta_{3}Y_{1}V+d_{2}\alpha_{3}Y_{1})\\
& & +(\alpha_{3}\beta_{3}YV-d_{3}V-\alpha_{3}\beta_{3}YV_{1}+d_{3}V_{1})\\
& = & \mu(1-V/V_{1})
\end{eqnarray*}
将上式左端约去同类项后, 即
\begin{eqnarray*}
& & -\alpha_{3}\beta_{3}V_{1}Y_{1}+\alpha_{3}\beta_{3}Y_{1}V+d_{3}(V_{1}-V)\\
& = & \mu(1-V/V_{1})
\end{eqnarray*}
注意到第二个恒等式, 可知上式是恒成立的, 证毕！
$\mathbf{3.\,\text{矩阵}}$
矩阵$\,\lambda I+D\xi_{i}-J_{1}-J_{2}e^{-\lambda\tau_{1}}-J_{3}e^{-\lambda\tau_{2}}$
如下
\[
\left[\begin{array}{ccccc}
\lambda+D_{H}\xi_{i}+d+\beta_{1}N^{*}+\beta_{2}Y^{*} & \beta_{1}H^{*} & \beta_{2}H^{*} & 0 & 0\\
-\alpha_{1}\beta_{1}e^{-(\lambda+a_{1})\tau_{1}}N^{*} & \lambda+D_{N}\xi_{i}+d_{1}-\alpha_{1}\beta_{1}e^{-(\lambda+a_{1})\tau_{1}}H^{*} & 0 & 0 & 0\\
-\alpha_{2}\beta_{2}e^{-(\lambda+a_{2})\tau_{2}}Y^{*} & 0 & \lambda+D_{Y}\xi_{i}+\beta_{3}V^{*}+\beta_{4}Z^{*}+d_{2}-\alpha_{2}\beta_{2}e^{-(\lambda+a_{2})\tau_{2}}H^{*} & \beta_{3}Y^{*} & \beta_{4}Y^{*}\\
0 & 0 & -\alpha_{3}\beta_{3}V^{*} & \lambda+D_{V}\xi_{i}-\alpha_{3}\beta_{3}Y^{*}+d_{3} & 0\\
0 & 0 & -\alpha_{4}\beta_{4}Z^{*} & 0 & \lambda+D_{Z}\xi_{i}-\alpha_{4}\beta_{4}Y^{*}+d_{4}
\end{array}\right]
\]
对于$\,E_{0}=(H_{0},0,0,V_{0},0),$ 上面矩阵变成
\[
\left[\begin{array}{ccccc}
\lambda+D_{H}\xi_{i}+d & \beta_{1}H_{0} & \beta_{2}H_{0} & 0 & 0\\
0 & \lambda+D_{N}\xi_{i}+d_{1}-\alpha_{1}\beta_{1}e^{-(\lambda+a_{1})\tau_{1}}H_{0} & 0 & 0 & 0\\
0 & 0 & \lambda+D_{Y}\xi_{i}+\beta_{3}V_{0}+d_{2}-\alpha_{2}\beta_{2}e^{-(\lambda+a_{2})\tau_{2}}H_{0} & 0 & 0\\
0 & 0 & -\alpha_{3}\beta_{3}V_{0} & \lambda+D_{V}\xi_{i}+d_{3} & 0\\
0 & 0 & 0 & 0 & \lambda+D_{Z}\xi_{i}+d_{4}
\end{array}\right]
\]
那么按第一行展开后它的行列式就是
\begin{eqnarray*}
& = & (\lambda+D_{H}\xi_{i}+d)[\lambda+D_{N}\xi_{i}+d_{1}-\alpha_{1}\beta_{1}e^{-(\lambda+a_{1})\tau_{1}}H_{0}]\cdot[\lambda+D_{Y}\xi_{i}+\beta_{3}V_{0}+d_{2}-\alpha_{2}\beta_{2}e^{-(\lambda+a_{2})\tau_{2}}H_{0}](\lambda+D_{V}\xi_{i}+d_{3})(\lambda+D_{Z}\xi_{i}+d_{4})\\
& & -\beta_{1}H_{0}\cdot0\\
& & +\beta_{2}H_{0}\cdot0
\end{eqnarray*}
这正好是论文中的$\,(13)$ 式.
标签：other review array number 需要 row begin 它的 end
原文地址：https://www.cnblogs.com/zdzyh/p/12204197.html
