【學術】重構具有時間延遲相互作用的動力學網絡

Reconstruction of dynamic networks with time-delayed interactions in presence of fast-varying noises

Zhaoyang Zhang	Yang Chen	Yuanyuan Mi	Gang Hu
Ningbo University	中科院腦網中心和國家模式識別實驗室	Chongqing University	Beijing Normal University

Dated: April 1, 2019

理論推導

考慮$N$個節點的動力學系統
$\dot{x}_i=F_i[x_i(t)]+\sum_{j=1,j\neq i}^N\Phi_{ij}[x_i(t),x_j(t-\tau_{ij})]+\eta_i(t)+\Gamma_i(t),i=1,2,...,N$

其中$\eta_i(t)$爲色噪聲，$\Gamma_i(t)$爲白噪聲，滿足
$<\eta_i(t)>=0,<\eta_i(t)\eta_i(t+t')>=P_{ij}e^{-\frac{|t'|}{\tau_C}}$

$<\Gamma_i(t)>=0,<\Gamma_i(t)\Gamma_i(t+t')>=Q_i\delta_{ij}\delta(t')$

$<\Gamma_i(t)\eta_j(t)>=0$

$F_i$和$\Phi_{ij}$可以爲線性或者非線性。需要設計算法，從數據中重構出網絡連接。假設在整個網絡中我們只能夠測量兩個節點$A$和$B$，並且有充足的數據。
$x_A(t)=[x_A(t_1),x_A(t_2)....,x_A(t_k),...,x_A(t_L)]$

$x_B(t)=[x_Bt_1),x_B(t_2)....,x_B(t_k),...,x_B(t_L)]$

$0<\Delta t=t_{k+1}-t_k\ll 1,L\gg 1$

對$x_i(t)$求時間的二階段導可得
$\ddot{x}_i(t)=\frac{\partial F_i[x_i(t)]}{\partial x_i(t)}\dot{x}_i(t)+\sum_{j=1,j\neq i}^N\frac{\partial\Phi_{ij}[x_i(t),x_j(t-\tau_{ij})]}{\partial x_i(t)}\dot{x}_i(t)+\sum_{j=1,j\neq i}^N\frac{\partial\Phi_{ij}[x_i(t),x_j(t-\tau_{ij})]}{\partial x_j(t-\tau_{ij})}\dot{x}_j(t-\tau_{ij})+\dot{\eta}_i(t)+\dot{\Gamma}_i(t)$

其中高階導數可以使用前向差分進行計算
$\dot{x}_i(t_k)=\frac{x_i(t_{k+1})-x_i(t_{k})}{\Delta t},\ddot{x}_i(t_k)=\frac{\dot{x}_i(t_{k+1})-\dot{x}_i(t_{k})}{\Delta t}$

噪聲的導數被定義爲
$\dot{\eta}_i(t_k)=\frac{\eta_i(t_{k+1})-\eta_i(t_{k})}{\Delta t},\dot{\Gamma}_i(t_k)=\frac{\Gamma_i(t_{k+1})-\Gamma_i(t_{k})}{\Delta t}$

根據前面的公式可知
$\ddot{x}_A(t_k)=\frac{1}{2}\frac{\partial F_A[x_A(t_k)]}{\partial x_i(t_k)}[\dot{x}_A(t_k) +\dot{x}_A(t_{k+1})]+\frac{1}{2}\sum_{j=1,j\neq A}^N\frac{\partial\Phi_{Aj}[x_A(t),x_j(t_k-\tau_{Aj})]}{\partial x_A(t_k)}[\dot{x}_A(t_k)+\dot{x}_A(t_{k+1})] +\frac{1}{2}\sum_{j=1,j\neq A}^N\frac{\partial\Phi_{Aj}[x_i(t),x_j(t-\tau_{Aj})]}{\partial x_j(t-\tau_{Aj})}[\dot{x}_j(t_k-\tau_{Aj})+\dot{x}_j(t_{k+1})] +\frac{\eta_A(t_{k+1-\tau_{Aj}})-\eta_A(t_{k})}{\Delta t} +\frac{\Gamma_A(t_{k+1})-\Gamma_A(t_{k})}{\Delta t}$

然後對方程左右兩邊同乘$x_B(t_k+\Delta t)$，計算每項的關聯可得
$R_{AB}=<\ddot{x}_A(t_k)x_B(t_k+\Delta t)>$

$=<\frac{1}{2}\frac{\partial F_A[x_A(t_k)]}{\partial x_i(t_k)}[\dot{x}_A(t_k)x_B(t_k+\Delta t)+\dot{x}_A(t_{k+1}x_B(t_k+\Delta t))]>$

$+\frac{1}{2}\sum_{j=1,j\neq A}^N\frac{\partial\Phi_{Aj}[x_A(t),x_j(t_k-\tau_{Aj})]}{\partial x_A(t_k)}[\dot{x}_A(t_k)x_B(t_k+\Delta t)+\dot{x}_A(t_{k+1})x_B(t_k+\Delta t)]$

$+\frac{1}{2}\sum_{j=1,j\neq A}^N\frac{\partial\Phi_{Aj}[x_i(t),x_j(t-\tau_{Aj})]}{\partial x_j(t-\tau_{Aj})}[\dot{x}_j(t_k-\tau_{Aj})x_B(t_k+\Delta t))+\dot{x}_j(t_{k+1})x_B(t_k+\Delta t))]$

$+\frac{\eta_A(t_{k+1-\tau_{Aj}})x_B(t_k+\Delta t)-\eta_A(t_{k})x_B(t_k+\Delta t)}{\Delta t}+\frac{\Gamma_A(t_{k+1})x_B(t_k+\Delta t)-\Gamma_A(t_{k})x_B(t_k+\Delta t)}{\Delta t}$

記該公式爲方程1。因爲有白噪聲的存在，對$<\dot{x}_i(t)\dot{x}_j(t+t')>$積分在$t'=0$處有個階躍，所以
$<\dot{x}_i(t)x_j(t+\Delta t)>-<\dot{x}_i(t)x_j(t)>=Q_j\delta_{ij}$

定義$n_{AB}=\frac{\tau_{AB}}{\Delta t}$，$V_i(\Delta k)=<\dot{x}_i(t_k)x_i(t_{k+\Delta k})>$，在$\Delta k=0$到$\Delta k=1$處有
$V_i(1)-V_i(0)=Q_i$

由於$<\dot{x}_B(t_k-\tau_{AB})x_B(t_{k+\Delta k})>$和$<\dot{x}_B(t_{k+1}-\tau_{AB})x_B(t_{k+\Delta k})>$在$k=-n_{AB}$處不聯繫，可以得到$n_{AB}$即$\tau_{AB}$
根據$<\dot{x}_i(t)x_j(t+\Delta t)>-<\dot{x}_i(t)x_j(t)>=Q_j\delta_{ij}$的性質可知方程1右邊除了$<\dot{x}_B(t_k-\tau_{AB})x_B(t_{k+\Delta k})>$和$<\dot{x}_B(t_{k+1}-\tau_{AB})x_B(t_{k+\Delta k})>$沒有不連續的項，所以從方程1可以得到
$D_{AB}=R_{AB}(-n_{AB}+2)-R_{AB}(-n_{AB})=<\ddot{x}_A(t_k)x_B(t_{t_k-n_{AB}+2})>-<\ddot{x}_A(t_k)x_B(t_{t_k-n_{AB}})>$

$=<\frac{\partial\Phi_{Aj}[x_i(t),x_j(t-\tau_{Aj})]}{\partial x_B(t-\tau_{AB})}>Q_B$

定義$M_{AB}=<\frac{\partial\Phi_{Aj}[x_i(t),x_j(t-\tau_{Aj})]}{\partial x_B(t-\tau_{AB})}>$，可得
$M_{AB}=\frac{D_{AB}}{Q_B}=J_{AB}$

數值仿真

定義均方根誤差$E$作爲衡量$M_{AB}$和$J_{AB}$之間的誤差，$T$表示測量時長
對於線性系統：

對於擴散耦合的FHN網絡（上），和Rossler網絡（下）

對於基因調控網絡

總結

文章的核心亮點是重構有時滯作用的系統，trick是利用白噪聲的性質。