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세미나

세미나

원드라이브

$$ \begin{align*} {\frac{ \partial u }{ \partial x }} + {\frac{ \partial v }{ \partial y }} =& 0 \\ {\frac{ \partial u }{ \partial t }} + u {\frac{ \partial u }{ \partial x }} + v {\frac{ \partial u }{ \partial y }} =& {\frac{ \partial^{2} u }{ \partial x^{2} }} + {\frac{ \partial^{2} u }{ \partial y^{2} }} - u - u \sqrt{u^{2} + v^{2}} \\ {\frac{ \partial v }{ \partial t }} + u {\frac{ \partial v }{ \partial x }} + v {\frac{ \partial v }{ \partial y }} =& {\frac{ \partial^{2} v }{ \partial x^{2} }} + {\frac{ \partial^{2} v }{ \partial y^{2} }} - v - v \sqrt{u^{2} + v^{2}} \\ {\frac{ \partial \theta }{ \partial t }} + u {\frac{ \partial \theta }{ \partial x }} + v {\frac{ \partial \theta }{ \partial y }} =& {\frac{ \partial^{2} \theta }{ \partial x^{2} }} + {\frac{ \partial^{2} \theta }{ \partial y^{2} }} \end{align*} $$

Paper

Claim: Same subsystem $\iff MSE \approx 0$

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

Let $X_{1}, X_{2}$ be two disjoint datasets of a non-smooth dynamical system and $\Xi_{1}$ and $\Xi_{2}$ represent the minimizers of $\left\| \dot{X}_{1} - \Theta \left( X_{1} \right) \Xi \right\|$ and $\left\| \dot{X}_{2} - \Theta \left( X_{2} \right) \Xi \right\|$, respectively.

Denote the concatenation of two matrices $M_{1}$ and $M_{2}$ as $$ \begin{align*} M_{12} =& \begin{bmatrix} M_{1} \\ M_{2} \end{bmatrix} \\ \Theta \left( M_{12} \right) =& \begin{bmatrix} \Theta \left( M_{1} \right) \\ \Theta \left( M_{2} \right) \end{bmatrix} \end{align*} $$

Let $A, B$ be matrices the difference of $\Xi_{1}$ and $\Xi_{2}$ from $\Xi_{12}$, respectively. $$ \Xi_{12} = \Xi_{1} - A = \Xi_{2} - B $$

Then MSE of the concatenated dataset $X_{12}$ is given by $$ \begin{align*} \varepsilon =& \left\| \dot{X}_{12} - \Theta \left( X_{12} \right) \Xi_{12} \right\| \\ =& \left\| \begin{bmatrix} \dot{X}_{1} \\ \dot{X}_{2} \end{bmatrix} - \begin{bmatrix} \Theta \left( X_{1} \right) \\ \Theta \left( X_{2} \right) \end{bmatrix} \Xi_{12} \right\| \\ =& \left\| \begin{bmatrix} \dot{X}_{1} \\ \dot{X}_{2} \end{bmatrix} - \begin{bmatrix} \Theta \left( X_{1} \right) \Xi_{12} \\ \Theta \left( X_{2} \right) \Xi_{12} \end{bmatrix} \right\| \\ =& \left\| \begin{bmatrix} \dot{X}_{1} - \Theta \left( X_{1} \right) \Xi_{12} \\ \dot{X}_{2} - \Theta \left( X_{2} \right) \Xi_{12} \end{bmatrix} \right\| \\ =& \left\| \begin{bmatrix} \dot{X}_{1} - \Theta \left( X_{1} \right) \Xi_{1} + \Theta \left( X_{1} \right) A \\ \dot{X}_{2} - \Theta \left( X_{2} \right) \Xi_{2} + \Theta \left( X_{2} \right) B \end{bmatrix} \right\| \\ =& \left\| \begin{bmatrix} \Theta \left( X_{1} \right) A \\ \Theta \left( X_{2} \right) B \end{bmatrix} \right\| \end{align*} $$ If $X_{1}$ and $X_{2}$ are from the same subsystem, then $\Xi_{12} = \Xi_{1} = \Xi_{2} \implies A = B = O$ and $\left\| \dot{X}_{12} - \Theta \left( X_{12} \right) \Xi_{12} \right\| = 0$, that is, $\Xi_{12}$ can be minimizer both datasets.

Method

  1. Let $s = 1$.
  2. Build the whole dataset $X$ by vertical concatenating all datasets and calculate MSE $\varepsilon = \left\| \dot{X} - \Theta \left( X \right) \Xi \right\|$. This MSE will be used as a threshold for stopping the algorithm.
    1. Take an arbitrary dataset $X_{i}$ and calculate MSE $\varepsilon_{ik} = \left\| \dot{X}_{ik} - \Theta \left( X_{ik} \right) \Xi \right\|$ for all indices $k$. Note that $\varepsilon_{ik}$ indicates how much the $k$-th dataset is different from the $i$-th dataset.
    2. Find $j = \argmax_{k} \varepsilon_{ik}$ and calculate MSE $\varepsilon_{jk} = \left\| \dot{X}_{jk} - \Theta \left( X_{jk} \right) \Xi \right\|$ for all indices $k$. If $\varepsilon_{jk} < \varepsilon$, then we conclude that the most different dataset $X_{j}$ is similar to the $X_{i}$ so there is no more need to check the $j$-th dataset.
    3. Compare $\varepsilon_{ik}$ and $\varepsilon_{jk}$ for all indices $k$. If $\varepsilon_{ik} < \varepsilon_{jk}$, then $i$-th dataset is a candidate dataset from the $s$-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 $l = \argmax_{k} \varepsilon_{ik}$ and calculate $\varepsilon_{lj}$ for all indices $j$. If $\varepsilon_{il} > \min_{j} \left\{ \varepsilon_{lj} \right\}$, then non-candidates group has at least one dataset that is more similar than the $i$-th dataset and $l$-th dataset is not a candidate dataset anymore. We repeat this step until $\varepsilon_{il} < \min_{j} \left\{ \varepsilon_{lj} \right\}$.
      2. If $\varepsilon_{il} < \min_{j} \left\{ \varepsilon_{lj} \right\}$, then we conclude that the datasets in candidates group are from $s$-th subsystem and remove the datasets.
  3. Update $s \gets s + 1$ and repeat step 2 until all datasets are labeled.