汇报

 

$$\begin{aligned} &\frac{d m}{d t}=-p_{g} m+p_{3} m_{e 2}\\ &\frac{d m_{e 2}}{d t}=p_{g} m-p_{2} m_{e 2}-p_{3} m_{e 2}=p_{g} m-\left(p_{2}+p_{3}\right) m_{e 2} \end{aligned}$$ [mathjax]

$$\frac{d}{d t}\left[\begin{array}{ccc} m \\ m_{e 2} \end{array}\right]=\left[\begin{array}{ccc} -p_{g} & p_{3} \\ p_{g} & -\left(p_{2}+p_{3}\right) \end{array}\right]\left[\begin{array}{ccc} m \\ m_{e 2} \end{array}\right]$$

方程的解为\(  \left[\begin{array}{cc} m \\ m_{e 2} \end{array}\right]=c_{1} e^{\lambda_{1} t} \mathbf{v_{1}}+c_{2} e^{\lambda_{2} t} \mathbf{v_{2}}\)

 

$$\begin{aligned}
&\lambda_{1}+\lambda_{2}=-\left(p_{2}+p_{3}+p_{g}\right)\\
&\lambda_{1} \lambda_{2}=p_{2} p_{g}
\end{aligned}$$

$$p_{2}=s\mathrm{exp}(\frac { -\Delta E }{ kT } )$$

$$F(R)=\frac{(1-R_{\infty})^{2}}{2 R_{\infty}} \propto \frac{\varepsilon c}{s}$$

$$F(R)=\frac{K}{S}=\frac{\left(1-R_{\infty}\right)^{2}}{2 R_{\infty}} \propto m$$

$$R_{\infty}=\frac {I_{r}  }{I_{in}  } $$

$$m(t)=cf(I_{r}(t),\, I_{in} )$$

$$F(R)=\frac{\left(1-R_{\infty}\right)^{2}}{2 R_{\infty}} \propto \frac{\sigma c}{s}$$

$$\lambda_{1} \lambda_{2}=p_{2} p_{g}=p_{g}\operatorname{sexp}\left(\frac{-\Delta E}{k T}\right)$$

$$g(T)=p_{g} \operatorname{sexp}\left(\frac{-\Delta E}{k T}\right)$$

$$\ln g(T)= \frac{-\Delta E}{k T}+ \ln p_{g} s$$

$$\frac{1}{k T}$$

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