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세미나

세미나

원드라이브

∂u∂x+∂v∂y=0∂u∂t+u∂u∂x+v∂u∂y=∂2u∂x2+∂2u∂y2−u−uu2+v2∂v∂t+u∂v∂x+v∂v∂y=∂2v∂x2+∂2v∂y2−v−vu2+v2∂θ∂t+u∂θ∂x+v∂θ∂y=∂2θ∂x2+∂2θ∂y2 \begin{align*} {\frac{ \partial u }{ \partial x }} + {\frac{ \partial v }{ \partial y }} =& 0 \\ {\frac{ \partial u }{ \partial t }} + u {\frac{ \partial u }{ \partial x }} + v {\frac{ \partial u }{ \partial y }} =& {\frac{ \partial^{2} u }{ \partial x^{2} }} + {\frac{ \partial^{2} u }{ \partial y^{2} }} - u - u \sqrt{u^{2} + v^{2}} \\ {\frac{ \partial v }{ \partial t }} + u {\frac{ \partial v }{ \partial x }} + v {\frac{ \partial v }{ \partial y }} =& {\frac{ \partial^{2} v }{ \partial x^{2} }} + {\frac{ \partial^{2} v }{ \partial y^{2} }} - v - v \sqrt{u^{2} + v^{2}} \\ {\frac{ \partial \theta }{ \partial t }} + u {\frac{ \partial \theta }{ \partial x }} + v {\frac{ \partial \theta }{ \partial y }} =& {\frac{ \partial^{2} \theta }{ \partial x^{2} }} + {\frac{ \partial^{2} \theta }{ \partial y^{2} }} \end{align*}

Paper

Claim: Same subsystem   ⟺  MSE≈0\iff MSE \approx 0

Assume that the x˙=f(x)\dot{x} = f(x) is a smoooth dynamical system and (X˙,X)\left( \dot{X} , X \right) consists of finite subset of {(x˙(t),x(t))}t\left\{ \left( \dot{x}(t) , x(t) \right) \right\}_{t}. If Θ(X)\Theta \left( X \right) contains all terms(bases) of ff, then there exists Ξ\Xi so that ε=∥X˙−Θ(X)Ξ∥=0\varepsilon = \left\| \dot{X} - \Theta \left( X \right) \Xi \right\| = 0. Note that ε≫0\varepsilon \gg 0 means the violation of the assumption: ff is not a smooth function, or Θ\Theta is not a complete set of bases, or the noise of (X˙,X)\left( \dot{X} , X \right) is not negligible.

Let X1,X2X_{1}, X_{2} be two disjoint datasets of a non-smooth dynamical system and Ξ1\Xi_{1} and Ξ2\Xi_{2} represent the minimizers of ∥X˙1−Θ(X1)Ξ∥\left\| \dot{X}_{1} - \Theta \left( X_{1} \right) \Xi \right\| and ∥X˙2−Θ(X2)Ξ∥\left\| \dot{X}_{2} - \Theta \left( X_{2} \right) \Xi \right\|, respectively.

Denote the concatenation of two matrices M1M_{1} and M2M_{2} as M12=[M1M2]Θ(M12)=[Θ(M1)Θ(M2)] \begin{align*} M_{12} =& \begin{bmatrix} M_{1} \\ M_{2} \end{bmatrix} \\ \Theta \left( M_{12} \right) =& \begin{bmatrix} \Theta \left( M_{1} \right) \\ \Theta \left( M_{2} \right) \end{bmatrix} \end{align*}

Let A,BA, B be matrices the difference of Ξ1\Xi_{1} and Ξ2\Xi_{2} from Ξ12\Xi_{12}, respectively. Ξ12=Ξ1−A=Ξ2−B \Xi_{12} = \Xi_{1} - A = \Xi_{2} - B

Then MSE of the concatenated dataset X12X_{12} is given by ε=∥X˙12−Θ(X12)Ξ12∥=∥[X˙1X˙2]−[Θ(X1)Θ(X2)]Ξ12∥=∥[X˙1X˙2]−[Θ(X1)Ξ12Θ(X2)Ξ12]∥=∥[X˙1−Θ(X1)Ξ12X˙2−Θ(X2)Ξ12]∥=∥[X˙1−Θ(X1)Ξ1+Θ(X1)AX˙2−Θ(X2)Ξ2+Θ(X2)B]∥=∥[Θ(X1)AΘ(X2)B]∥ \begin{align*} \varepsilon =& \left\| \dot{X}_{12} - \Theta \left( X_{12} \right) \Xi_{12} \right\| \\ =& \left\| \begin{bmatrix} \dot{X}_{1} \\ \dot{X}_{2} \end{bmatrix} - \begin{bmatrix} \Theta \left( X_{1} \right) \\ \Theta \left( X_{2} \right) \end{bmatrix} \Xi_{12} \right\| \\ =& \left\| \begin{bmatrix} \dot{X}_{1} \\ \dot{X}_{2} \end{bmatrix} - \begin{bmatrix} \Theta \left( X_{1} \right) \Xi_{12} \\ \Theta \left( X_{2} \right) \Xi_{12} \end{bmatrix} \right\| \\ =& \left\| \begin{bmatrix} \dot{X}_{1} - \Theta \left( X_{1} \right) \Xi_{12} \\ \dot{X}_{2} - \Theta \left( X_{2} \right) \Xi_{12} \end{bmatrix} \right\| \\ =& \left\| \begin{bmatrix} \dot{X}_{1} - \Theta \left( X_{1} \right) \Xi_{1} + \Theta \left( X_{1} \right) A \\ \dot{X}_{2} - \Theta \left( X_{2} \right) \Xi_{2} + \Theta \left( X_{2} \right) B \end{bmatrix} \right\| \\ =& \left\| \begin{bmatrix} \Theta \left( X_{1} \right) A \\ \Theta \left( X_{2} \right) B \end{bmatrix} \right\| \end{align*} If X1X_{1} and X2X_{2} are from the same subsystem, then Ξ12=Ξ1=Ξ2  ⟹  A=B=O\Xi_{12} = \Xi_{1} = \Xi_{2} \implies A = B = O and ∥X˙12−Θ(X12)Ξ12∥=0\left\| \dot{X}_{12} - \Theta \left( X_{12} \right) \Xi_{12} \right\| = 0, that is, Ξ12\Xi_{12} can be minimizer both datasets.

Method

  1. Let s=1s = 1.
  2. Build the whole dataset XX by vertical concatenating all datasets and calculate MSE ε=∥X˙−Θ(X)Ξ∥\varepsilon = \left\| \dot{X} - \Theta \left( X \right) \Xi \right\|. This MSE will be used as a threshold for stopping the algorithm.
    1. Take an arbitrary dataset XiX_{i} and calculate MSE εik=∥X˙ik−Θ(Xik)Ξ∥\varepsilon_{ik} = \left\| \dot{X}_{ik} - \Theta \left( X_{ik} \right) \Xi \right\| for all indices kk. Note that εik\varepsilon_{ik} indicates how much the kk-th dataset is different from the ii-th dataset.
    2. Find j=arg max⁡kεikj = \argmax_{k} \varepsilon_{ik} and calculate MSE εjk=∥X˙jk−Θ(Xjk)Ξ∥\varepsilon_{jk} = \left\| \dot{X}_{jk} - \Theta \left( X_{jk} \right) \Xi \right\| for all indices kk. If εjk<ε\varepsilon_{jk} < \varepsilon, then we conclude that the most different dataset XjX_{j} is similar to the XiX_{i} so there is no more need to check the jj-th dataset.
    3. Compare εik\varepsilon_{ik} and εjk\varepsilon_{jk} for all indices kk. If εik<εjk\varepsilon_{ik} < \varepsilon_{jk}, then ii-th dataset is a candidate dataset from the ss-th subsystem. This step separates the datasets into two groups: candidates and non-candidates.
      1. In the candidate group, pick the dataset with the largest MSE l=arg max⁡kεikl = \argmax_{k} \varepsilon_{ik} and calculate εlj\varepsilon_{lj} for all indices jj. If εil>min⁡j{εlj}\varepsilon_{il} > \min_{j} \left\{ \varepsilon_{lj} \right\}, then non-candidates group has at least one dataset that is more similar than the ii-th dataset and ll-th dataset is not a candidate dataset anymore. We repeat this step until εil<min⁡j{εlj}\varepsilon_{il} < \min_{j} \left\{ \varepsilon_{lj} \right\}.
      2. If εil<min⁡j{εlj}\varepsilon_{il} < \min_{j} \left\{ \varepsilon_{lj} \right\}, then we conclude that the datasets in candidates group are from ss-th subsystem and remove the datasets.
  3. Update s←s+1s \gets s + 1 and repeat step 2 until all datasets are labeled.