# 僵尸的悲惨故事

## 物理模型

$$Q(t) = \sum_m P_m(t) .$$

$$P_\xi(t) = \sum_\mu R_{\xi\mu} \tau P_\mu(t-\tau)$$

$$\sum_{\xi} R_{\xi\mu}\tau = 1$$

$$P_{\xi}(t) = \sum_{\mu} R_{\mu\xi}\tau P_{\xi} (t) .$$

$$\frac{d}{dt}P_m = F(P_{n+1} + P_{n-1}) - 2F P_m .$$

$$\frac{d}{dt}P_m = F(P_{n+1} + P_{n-1}) - 2F P_m - C \sum_r \delta_{m,r}P_m .$$

## 小喵解方程

$$\frac{d}{dt}P_m = F(P_{n+1} + P_{n-1} - 2 P_m) - C \delta_{m,s}P_m .$$

$$\frac{d}{dt}P_m = F(P_{n+1} + P_{n-1} - 2 P_m).$$

Fourier transform，小喵心想。于是他快速的在大脑中两边乘以 $e^{ikm}$ 并对 $m$ 求和。

$$\frac{d}{dt} P^k = F(e^{ikm} P_{m+1} + e^{ikm} P_{m-1} -2 P^k).$$

$$P^k(t) = P^k(0) e^{-4F \sin^2\frac{k}{2} t}.$$

\begin{align} P_m(t) & = \frac{1}{N} \sum_ {k} P^k(t) e^{-i km}
& = \frac{1}{N} \sum_ {k} P^k(0) e^{-4F \sin^2\frac{k}{2} t} e^{-i km} . \end{align}

$$P^k(0) = \sum_n P_n(0)e^{ikn}.$$

\begin{align} P_m(t) & = \frac{1}{N} \sum_ {k} \sum_n P_n(0)e^{ikn} e^{-4F \sin^2\frac{k}{2} t} e^{-i km}
&= \sum_n \Pi_{m-n}(t) P_n(0) , \end{align}

$$\Pi_{m-n}(t) = \frac{1}{N} \sum_ {k} e^{-4F \sin^2\frac{k}{2} t} e^{-i k(m-n)} .$$

$$\tilde \eta_m = \sum_{n} \tilde \Pi_{m-n} \tilde P_n,$$

$$\tilde P_m = \tilde \eta - C \tilde \Pi_{m-r} \tilde P_r .$$

$$\tilde P_r = \frac{\tilde \eta}{1+C\tilde \Pi_0}.$$

$$\tilde P_m = \tilde \eta_m - \frac{\tilde \Pi_{m-r} \tilde \eta_r}{1/C + \tilde \Pi_0}.$$

$$Q(t) = \sum_m P_m(t).$$

$$\tilde Q = \frac{1}{\epsilon} \left( 1 - \frac{\tilde \eta_r}{1/C + \tilde \Pi_0} \right).$$

$$\frac{d}{dt}Q(t) = - \int_0^t \mathscr M(t-t’)\eta(t’) dt’,$$

$$\mathscr M(t-t’) = \frac{1}{1/C + \tilde \Pi_0} .$$

By OctoMiao

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