令人头疼的科研工作

我们引用Tass的文章：

• Tass, P. A., et al. “Long-lasting desynchronization in rat hippocampal slice induced by coordinated reset stimulation.” Physical Review E 80.1 (2009): 011902.
• Hauptmann, Christian, and Peter A. Tass. “Restoration of segregated, physiological neuronal connectivity by desynchronizing stimulation.” Journal of neural engineering7.5 (2010): 056008.

Tam, H. C., Emily SC Ching, and Pik-Yin Lai. “Reconstructing networks from dynamics with correlated noise.” Physica A: Statistical Mechanics and its Applications 502 (2018): 106-122.

The time evolution of the node variables is given by a set of coupled differential equations:
$\frac{dx_i}{dt}=f_i(x_i)+\sum_{j\neq i}A_{ij}h(x_i,x_j)+\eta_i$

External influences on the system are modelled by a Gaussian correlated noise $\eta_i$ that is generated by the Ornstein-Uhlenbeck process:
$\tau_n\frac{d\eta_i}{dt}=-\eta_i+\xi_i$

where $\xi_i(t)$ is a zero-mean Gaussian white noise:
$<\xi_i(t)>=0,<\xi_i(t)\xi_j(t’)>=2D_{ij}\delta(t-t’)$

Thus we have
$<\eta_i(t)>=0,<\eta_i(t)\eta_j(t’)>=\frac{D_{ij}}{\tau_n}e^{-|t-t’|/\tau_n}$

线性化可得：
$\frac{d}{dt}\delta x=Q\delta x+\eta_i$

对上式积分可得：
$\delta x(t)=e^{Qt}\delta x(0)+\int_0^te^{(t-t’)Q}\eta_i(t’)dt’$

Hence we obtain
$K_{\tau}=<\delta x(t+\tau)\delta x(t)^T>$

$QK_0+K_0Q^T+D(I-\tau_nQ^T)^{-1}+(I-\tau_nQ)^{-1}D^T=0$

$K_{\tau}=e^{\tau Q}K_0+J(\tau)$

$J(\tau)=(e^{\tau Q}-e^{-\tau/\tau_n}I)U+\tau ^{\tau Q}V$

$U=\tau_nRD(I-\tau_nQ^T)^{-1}$

$V=[I-(I+\tau_nQ)R]RD(I-\tau_nQ^T)^{-1}$

$R=(I+\tau_nQ)^{-1}$

Replacing $\tau$ by $2\tau$ and then by $3\tau$ and after some algebra, we obtain
$K_{2\tau}=SK_{\tau}-TK_0$

$K_{3\tau}=SK_{2\tau}-TK_{\tau}$

$S=e^{Qt}+e^{-\tau/\tau_n}I$

$T=e^{-\tau/\tau_n}e^{Qt}$

Solving Eqs, we then obtain
$T=(K_{3\tau}-K_{2\tau}K_{\tau}^{-1}K_{2\tau})(K_{0}K_{\tau}^{-1}K_{2\tau}-K_{\tau})^{-1}$

$S=(K_{2\tau}+TK_))K_{\tau}^{-1}$

$S=e^{\tau/\tau_n}T+e^{-\tau/\tau_n}I$

to estimate the value of $e^{\tau/\tau_n}$ and thus $\tau_n$ by a least-square fit of the diagonal elements of $S$, denoted by $\mu$,

$Q=\frac{1}{\tau}log(\mu T)=\frac{1}{\tau}[log(\mu)+logT]$