Dynamic Mode Decomposition for Continuous Time Systems with the Liouville Operator

Abstract

Dynamic mode decomposition (DMD) has become synonymous with the Koopman operator, where continuous time dynamics are discretized and examined using Koopman (i.e. composition) operators. Using the newly introduced “occupation kernels,” the present manuscript develops an approach to DMD that treats continuous time dynamics directly through the Liouville operator. This manuscript outlines the technical and theoretical differences between Koopman-based DMD for discrete time systems and Liouville-based DMD for continuous time systems, which includes an examination of Koopman and Liouville operators over several reproducing kernel Hilbert spaces. While Liouville operators are modally unbounded, this manuscript introduces the concept of a scaled Liouville operator, which, for many dynamical systems, is a compact operator over the native space of the exponential dot product kernel. Compactness of scaled Liouville operators allows for norm convergence of Liouville-based DMD, which is a decided advantage over Koopman-based DMD.

Proofs of Theorem 1 and Proposition  2

Theorem 1 restated: Let $F^2({\mathbb {R}}^n)$ be the Bargmann-Fock space of real valued functions, which is the native space for the exponential dot product kernel, $K(x,y) = \exp (x^Ty)$, $a \in {\mathbb {R}}$ with $|a| < 1$, and let $A_{f,a}$ be the scaled Liouville operator with symbol $f:{\mathbb {R}}^n \rightarrow {\mathbb {R}}^n$. There exists a collection of coefficients, $\{ C_\alpha \}_{\alpha }$, indexed by the multi-index $\alpha$, such that if f is representable by a multi-variate power series, $f(x) = \sum _{\alpha } f_\alpha x^\alpha$, satisfying

$$\begin{aligned} \sum _{\alpha } |f_{\alpha }| C_\alpha < \infty , \end{aligned}$$

then $A_{f,a}$ is bounded and compact over $F^2({\mathbb {R}}^n)$.

Proof

The proof for the case $n = 1$ is presented to simplify the exposition. The case for $n > 1$ follows with some additional bookkeeping of the multi-index.

If $A_{x,a}$ is compact for all $|a| < 1$, then $A_{x^m,a} = A^{m}_{x,\root m \of {a}}$ is compact since products of compact operators are compact. If $f(x) = \sum _{m=0}^\infty f_m x^m$ is such that $\sum _{m=0}^\infty |f_m| \Vert A_{x^m,a}\Vert < \infty$, then $A_{f,a} = \lim _{m\rightarrow \infty } \sum _{m=0}^M f_m A_{x^m,a},$ with respect the operator norm via the triangle inequality, and $A_{f,a}$ is compact since it is the limit of compact operators. Thus, it is sufficient to demonstrate that $A_{x,a}$ is compact to prove the theorem.

Let $g \in F^{2}({\mathbb {R}})$, then $g(x) = \sum _{m=0}^\infty g_m \frac{x^m}{\sqrt{m!}}$ with norm $\Vert g \Vert _{F^{2}({\mathbb {R}})}^2 = \sum _{m=0}^\infty |g_m|^2 < \infty .$ Applying the scaled Liouville operator, $A_{x,a}$, yields

$$\begin{aligned}A_{x,a} g(x) = axg'(ax) = \sum _{m=0}^\infty g_{m} a^m m \frac{x^m}{\sqrt{m!}}. \end{aligned}$$

Hence, $\Vert A_{x,a} g \Vert _{F^({\mathbb {R}})}^2 = |a|^{2m} m^2 |g_{m}|^2 < \infty$ as for large enough m, $|a|^{2m} m^2 < 1$. Hence, $A_{x,a}$ is everywhere defined and by the closed graph theorem $A_{x,a}$ is bounded.

As $|a|^{m} m^2 \rightarrow 0$, there is an M such that for all $m > M$, $|a|^{m} m^2 < 1$. Let $P_M$ be the projection onto ${{\,\mathrm{span}\,}}\{ 1, x, x^2, \ldots , x^M\}$. Now consider

$$\begin{aligned} \Vert (A_{x,a} - A_{x,a} P_M) g\Vert ^2&= \sum _{m=M+1}^\infty |g_{m}|^2 |a|^{2m} m^2\\&\le \sum _{m=M+1}^\infty |g_{m}|^2 |a|^{m} \\&\le |a|^M \sum _{m=M+1}^\infty |g_{m}|^2 |a|^{m-M} \\&\le |a|^M \sum _{m=M+1}^\infty |g_{m}|^2 \le |a|^M \Vert g \Vert _{F^2({\mathbb {R}})}^2. \end{aligned}$$

Hence, the operator norm of $(A_{x,a} - A_{x,a} P_M)$ is bounded by $|a|^{M/2}$, and as $|a| < 1$, $A_{x,a} P_m \rightarrow A_{x,a}$ in the operator norm. $P_m$ is finite rank and therefore compact. It follows that $A_{x,a} P_m$ is compact, since compact operators form an ideal in the ring of bounded operators. Thus, $A_{x,a}$ is compact as it is the limit of compact operators. $\square$

Proposition 2restated: Let H be a RKHS of twice continuously differentiable functions over ${\mathbb {R}}^n$, f be Lipschitz continuous, and suppose that $\varphi _{i,a}$ is an eigenfunction of $A_{f,a}$ with eigenvalue $\lambda _{i,a}$. Let D be a compact subset of ${\mathbb {R}}^n$ that contains x(t) for all $0< t < T$. In this setting, if $\lambda _{i,a} \rightarrow \lambda _{i,1}$ and $\varphi _{i,a}(x(0)) \rightarrow \varphi _{i,1}(x(0))$ as $a \rightarrow 1^-$, then

$$\begin{aligned} \sup _{0 \le t \le T} \Vert \varphi _{i,a}(x(t)) - e^{\lambda _{i,a}t}\varphi _{i,a}(x(0))\Vert _2 \rightarrow 0. \end{aligned}$$

Proof

Suppose that x(t) remains in a compact set $D \subset {\mathbb {R}}^n$. Since $\phi _{m,a} \in H$ and H consists of twice continuously differentiable functions, there exists $M_1,M_2,F > 0$ such that

$$\begin{aligned} \sup _{x \in D} \Vert f(x) \Vert< F \quad \sup _{x \in D}, \Vert \nabla \phi _{m,a}(x) \Vert< M_{1,a}, \text { and} \quad \sup _{x \in D} \Vert \nabla ^2\phi _{m,a}(x) \Vert < M_{2,a}. \end{aligned}$$

First, it is necessary to demonstrate that $M_{1,a}$ and $M_{2,a}$ may be bounded independent of a. For each $i,j=1,\ldots ,n$ and $y \in {\mathbb {R}}^n$, the functionals $g \mapsto \frac{\partial }{\partial x_i} g(y)$ and $g \mapsto \frac{\partial ^2}{\partial x_i \partial x_j} g(y)$ are bounded (cf. Steinwart and Christmann 2008). Setting, $k_{y} = K(\cdot ,y)$, it can be seen that the functions $\frac{\partial }{\partial x_i} k_y$ and $\frac{\partial ^2}{\partial x_i \partial x_j} k_y$ are the unique functions that represent these functionals through the inner product of the RKHS (cf. Steinwart and Christmann 2008). As $\phi _{m,a}$ is a normal vector, $\Vert \phi _{m,a} \Vert _H = 1$, and by Cauchy-Schwarz

$$\begin{aligned} \Vert \nabla \phi _{m,a}(y) \Vert _2&= \sqrt{ \sum _{i=1}^n \left( \frac{\partial }{\partial x_i} \phi _{m,a}(y) \right) ^2}\nonumber \\&= \sqrt{\sum _{i=1}^n \left( \left\langle \phi _{m,a}, \frac{\partial }{\partial x_i} k_y \right\rangle _H \right) ^2}\nonumber \\&\le \sqrt{\sum _{i=1}^n \left\| \phi _{m,a}\right\| _H^2 \left\| \frac{\partial }{\partial x_i} k_y \right\| _H^2}\nonumber \\&= \sqrt{ \sum _{i=1}^n \left\| \frac{\partial }{\partial x_i} k_y \right\| _H^2 }. \end{aligned}$$

(13)

(13) is bounded over D as $x \mapsto \frac{\partial }{\partial x_i} k_y(x)$ is continuous. Thus, $M_{1,a}$ is bounded independent of a. A similar argument may be carried out for $M_{2,a}$. Let $M_1$ and $M_2$ be the respective bounding constants.

Note that

$$\begin{aligned} \frac{\partial }{\partial t} \phi _{m,a}(ax(t)) = a \nabla \phi _{m,a}(a x(t)) f(x(t)) = A_{f,a} \phi _{m,a}(x(t)) = \mu _{m,a} \phi _{m,a}(x(t)). \end{aligned}$$

Then by the mean value inequality, Cauchy-Schwarz, and the bounds given above,

$$\begin{aligned}&\left| \frac{\partial }{\partial t} \phi _{m,a}(ax(t)) - \frac{\partial }{\partial t} \phi _{m,a}(x(t))\right| \\&\quad = \left| a \nabla \phi _{m,a}(a x(t)) f(x(t)) - \nabla \phi _{m,a}(x(t)) f(x(t))\right| \\&\quad \le F\left\| a \nabla \phi _{m,a}(a x(t)) - a\nabla \phi _{m,a}(x(t)) + a\nabla \phi _{m,a}(x(t)) - \nabla \phi _{m,a}(x(t)) \right\| _2\\&\quad \le |a|F\left\| \nabla \phi _{m,a}(a x(t)) - \nabla \phi _{m,a}(x(t)) \right\| _2 + F|a-1| M_1 \Vert x(t)\Vert _2\\&\quad \le |a||a-1| M_2 F \Vert x(t)\Vert _2 + |a-1| M_1 F \Vert x(t)\Vert _2 = O(|a-1|). \end{aligned}$$

Setting $\epsilon _a(t) := \frac{\partial }{\partial t} \phi _{m,a}(ax(t)) - \frac{\partial }{\partial t} \phi _{m,a}( x(t))$, it follows that $\sup _{0 \le t \le T} \Vert \epsilon _a(t) \Vert _2 = O(|a-1|)$. Hence,

$$\begin{aligned} \mu _{m,a} \phi _{m,a}(x(t))&= \frac{\partial }{\partial t} \phi _{m,a}(a x(t))\\&= \frac{\partial }{\partial t} \phi _{m,a}(x(t)) + \epsilon (t), \end{aligned}$$

and

$$\begin{aligned} \phi _{m,a}(x(t)) = e^{\mu _{m,a}t} \phi _{m,a}(x(0)) - e^{\mu _{m,a}t} \int _0^t e^{-\mu _{m,a} \tau } \epsilon (\tau ) d\tau . \end{aligned}$$

As the time interval is fixed to [0, T], $e^{\mu _{m,a}t} \int _0^t e^{-\mu _{m,a} \tau } \epsilon (\tau ) d\tau = O(|a-1|),$ since $\mu _{m,a}$ is bounded with respect to a. $\square$

Theorem 2restated: Let $|a| < 1$. Suppose that $\{ \gamma _{i}:[0,T_i] \rightarrow {\mathbb {R}}^n \}_{i=1}^\infty$ is a sequence of trajectories satisfying ${\dot{\gamma }} = f(\gamma )$ for a dynamical system f corresponding to a compact scaled Liouville operator, $A_{f,a}$. If the collection of functions, $\{ \varGamma _{\gamma _i} \}_{i=1}^\infty$ is dense in the Bargmann-Fock space, then the sequence of operators $\{ P_{\alpha _M} A_{f,a} P_{\alpha _M} \}_{M=1}^\infty$ converges to $A_{f,a}$ in the norm topology, where $\alpha _M = \{ \varGamma _{\gamma _1}, \ldots , \varGamma _{\gamma _M} \}$.

Proof

The following proof is more general than what is indicated in the theorem statement of Theorem 2. In fact, for any compact operator, T, and any set $\{ g_i \}_{i=1}^\infty$ such that $\overline{{{\,\mathrm{span}\,}}(\{g_i\}_{i=1}^\infty )} = H$, the sequence of operators $P_{\alpha _M} T P_{\alpha _M} \rightarrow T$ in norm, where $P_{\alpha _M}$ is the projection onto ${{\,\mathrm{span}\,}}(\{g_i\}_{i=1}^M)$. Henceforth, it will be assumed that $\{ g_i \}_{i=1}^\infty$ is an orthonormal basis for H, since given any complete basis in H, an orthonormal basis may be obtained via the Gram-Schmidt process.

First note that every compact operator has a representation as $T = \sum _{i=1}^\infty \lambda _i \langle \cdot , v_i \rangle _H u_i$, where $\{ v_i \}$ and $\{ u_i \}$ are orthonormal collections of vectors (functions) in H, and $\{ \lambda _i \}_{i=1}^\infty \subset {\mathbb {C}}$ are the singular values of T. If $T_M := \sum _{i=1}^M \lambda _i \langle \cdot , v_i \rangle _H u_i$ then $T_M \rightarrow T$ as $M \rightarrow \infty$ in the operator norm.

Suppose that $\epsilon > 0$. Select M such that $\Vert T - T_M\Vert < \epsilon$, and select N such that for all $n > N$,

$$\begin{aligned} \Vert u_i - P_n u_i\Vert _H< \frac{\epsilon }{\sum _{i=1}^M |\lambda _i|^2} \text { and } \Vert v_i - P_n v_i\Vert _H < \frac{\epsilon }{\sqrt{M} \left( \sum _{i=1}^M |\lambda _i|^2\right) ^{1/2}} \end{aligned}$$

for all $i=1,\ldots ,M$. Let $g \in H$ be arbitrary.

Now consider,

$$\begin{aligned}&\Vert Tg - P_n T P_n g\Vert _H = \Vert Tg - T_M g + T_M g - P_n T P_n f\Vert _H\\&\quad \le \Vert T - T_M\Vert \Vert g\Vert _H + \Vert T_M g - P_n T P_n g\Vert _H \le \epsilon \Vert g\Vert _H + \Vert T_M g - P_n T P_n g\Vert _H. \end{aligned}$$

The second term after the inequality may be expanded as

$$\begin{aligned}&\Vert T_M g - P_n T P_n g\Vert _H \le \Vert T_M g - P_n T_M g\Vert _H + \Vert P_n T_M g - P_n T P_n g\Vert _H\\&\Vert T_M g - P_n T_M g\Vert _H + \Vert T_M g - T P_n g\Vert _H\\&\quad \le \Vert T_M g - P_n T_M g\Vert _H + \Vert T_M g - T_M P_n g\Vert _H + \Vert T_M P_n g - T P_n g\Vert _H\\&\quad \le \Vert T_M g - P_n T_M g\Vert _H + \Vert T_M g - T_M P_n g\Vert _H + \epsilon \Vert g\Vert _H. \end{aligned}$$

Now the objective is to demonstrate that both $\Vert T_M g - P_n T_M g\Vert _H$ and $\Vert T_M g - T_M P_n g\Vert _H$ are proportional to $\epsilon \Vert g\Vert _H$. Note that

$$\begin{aligned}&\Vert T_M g - P_n T_M g\Vert _H = \left\| \sum _{i=1}^M \lambda _i \langle g, v_i \rangle _H (u_i - P_n u_i)\right\| _H\\&\quad \le \sum _{i=1}^M |\lambda _i| |\langle g, v_i \rangle _H| \Vert u_i - P_n u_i \Vert _H\\&\quad \le \sqrt{\sum _{i=1}^M |\langle g, v_i\rangle _H|^2} \left( \sum _{i=1}^M |\lambda _i|^2 \Vert u_i - P_n u_i \Vert _H^2 \right) ^{1/2} \\&\quad \le \Vert g\Vert _H \left( \sum _{i=1}^M |\lambda _i|^2 \Vert u_i - P_n u_i \Vert _H^2 \right) ^{1/2} \le \epsilon \Vert g\Vert _H, \end{aligned}$$

and

$$\begin{aligned}&\Vert T_M(g - P_n g) \Vert _H = \left\| \sum _{i=1}^M \lambda _i \langle g - P_n g, v_i\rangle _H u_i \right\| \\&\quad \le \left\| \sum _{i=1}^M \lambda _i \left\langle \sum _{j=n+1}^\infty \langle g, g_j \rangle _H g_j, v_i \right\rangle _H u_i\right\| \\&\quad \le \sum _{j=n+1}^\infty |\langle g, g_j\rangle _H| \left( \sum _{i=1}^M |\lambda _i| |\langle g_j, v_i\rangle _H|\right) \\&\quad \le \left( \sum _{j=n+1}^\infty |\langle g, g_i\rangle _H|^2\right) ^{1/2} \left( \sum _{j=n+1}^\infty \left( \sum _{i=1}^M |\lambda _i| |\langle g_i, v_i \rangle _H|\right) ^{2}\right) ^{1/2}\\&\quad \le \Vert g\Vert _H \left( \sum _{j=n+1}^\infty \left( \sum _{i=1}^M |\lambda _i|^2 \right) \left( \sum _{i=1}^M |\langle g_j, v_i\rangle _H|^2\right) \right) ^{1/2}\\&\quad \le \Vert g\Vert _H \left( \sum _{i=1}^M |\lambda _i|^2\right) ^{1/2} \left( \sum _{i=1}^M \sum _{j=n+1}^\infty |\langle g, v_i\rangle _H|^2 \right) ^{1/2}\\&\quad = \Vert g\Vert _H \left( \sum _{i=1}^M |\lambda _i|^2\right) ^{1/2} \left( \sum _{i=1}^M \Vert v_i - P_n v_i\Vert ^2 \right) ^{1/2} \le \epsilon \Vert g\Vert _H. \end{aligned}$$

Thus, for every $\epsilon > 0$, there is an N such that for all $n > N$, $\Vert T g - P_n T P_n g\Vert _H \le 4\epsilon \Vert g\Vert _H.$ Hence, it follows that $\Vert T - P_n T P_n \Vert \le 4\epsilon$. Thus, as $n \rightarrow \infty$, $P_n T P_n \rightarrow T$ in the operator norm. $\square$