text:
\section{Optimal quantization level for reducing overall communication cost}\label{sec:extension}
In this section, we aim to study the trade-off between number of iterations until achieving a target accuracy and quantization levels. Indeed, by increasing quantization levels the variance of quantization reduces and the total number of iterations to reach a specific accuracy decreases, but the communication overhead of each round is higher as we have to transmit more bits. Conversely, if we use a quantization with a small number of levels the communication cost per iteration will be low; however, the total number of iterations could be very large. The fundamental question here is how to choose the quantization levels to optimize the overall communication cost which is the product of number of iterations and communication cost of each iteration.

In this section, we only focus on unbiased quantizers for which the variance is proportionally bounded with the  squared norm of the quantizer's input vector, i.e., for any $\bx \in \mathbb{R}^p$ it holds that $\mathbb{E}\left[Q(\bx) | \bx \right] = \bx$ and $\mathbb{E} \left[\| Q(\bx) - \bx\|^2  | \bx \right] \leq \eta^2 \norm{\bx}^2$ for some fixed constant $\eta$. Theorem \ref{thm2} characterizes the (order-wise) convergence of the proposed algorithm considering this assumption. More precisely,  using the result in Theorem \ref{thm2} and \eqref{thm_2_claim} we can write for each node $i$:

\begin{align} \label{eq:minT}
    & \quad \bE \left[  \norm{\bx_{i,T} - \widetilde{\bx}^*}^2\right]  \nonumber\\
  & \leq \left[ \frac{4n c^2_2 D^2 \left(3+ 2L/\mu\right)^2 }{(1-\beta)^2}   + \frac{4 c_1 {  n}\widetilde{B}^2 \eta^2 \norm{W\!-\!W_D}^2}{\mu c_2}  \right] \frac{1}{T^{\delta}},
\end{align}

where the approximation is due to considering dominant terms in $B_1(T)$ and $B_2(T)$ (See Appendix \ref{app_proof_of_lemma_2} and \ref{app_proof_of_theorem_2} for notations and details of derivations). Therefore, given a target relative deviation error $\rho$ and using \eqref{eq:minT}
, the algorithm needs to iterate at least $T(\rho)$ where

\begin{align}\label{eq:T-eps}
     T(\rho) &:= \Bigg[ \frac{4n c^2_2 D^2 \left(3+ 2L/\mu\right)^2 }{(1-\beta)^2}  \nonumber\\
     &\quad\quad + \frac{4 c_1 {  n}\widetilde{B}^2 \eta^2 \norm{W-W_D}^2}{\mu c_2} \Bigg]^{1/\delta} \left( \frac{1}{\rho \norm{\widetilde{\bx}^*}^2} \right) ^{1/\delta}.
\end{align}

\iffalse
\subsection{Gradient sparsifier}
The gradient sparsifier denoted by $ Q^{\text{GS}}: \mathbb{R}^d \rightarrow \mathbb{R}^d$ is defined as
\begin{equation}
    Q^{\text{GS}}_i (\bx) =
\left\{
	\begin{array}{ll}
		x_i / p_i  & \mbox{w.p. } p_i \\
		0 & \mbox{oth},
	\end{array}
\right.
\end{equation}
where $p_i$ is probability that coordinate $i \in [d]$ is selected. It is easy to verify that this quantizer is unbiased, as for each $i$, $\mathbb{E}[Q^{\text{GS}}_i (\bx)] = x_i$. Moreover, one can show that the variance of this quantizer is bounded as follows,
\begin{equation}
    \mathbb{E}[\| Q^{\text{GS}}(\bx) - \bx\|^2] = \sum_{i=1}^{d} \left( \frac{1}{p_i} - 1\right) x_i^2 \leq \left( \frac{1}{p_\text{min}} - 1\right) \norm{\bx}^2,
\end{equation}
where $p_\text{min}$ denotes the minimum of probabilities $\{p_1,\cdots,p_d\}$.

Let each floating point scalar cost $b$ bits to communicate. Then, the expected communication cost of the quantized vector is
\begin{equation}
    \mathbb{E}[| \text{Code}(Q^{\text{GS}}(\bx)) |] = \sum_{i=1}^{d} p_i (b + \log d).
\end{equation}
\fi

It is shown in \cite{alistarh2017qsgd} that for the low-precision quantizer defined in \eqref{eq:low_per_def} and \eqref{eq:low_per_def_2} there exists an encoding scheme $\text{Code}_s$ such that for any $\bx \in \mathbb{R}^p$ and $s^2 + \sqrt{p} \leq p/2$, the communication cost of the quantized vector satisfies

\begin{align}\label{eq:code-length}
    &\mathbb{E} \left[| \text{Code}_s(Q^{\text{LP}}(\bx)) | \right]  \nonumber\\
    & \quad \quad   \leq b + \left( 3 + \frac{3}{2} \log^* \left( \frac{2(s^2+p)}{s^2+\sqrt{p}} \right) \right) (s^2+\sqrt{p}),
\end{align}

where $\log^*(x)=\log(x) + \log \log (x) + \cdots=(1+o(1))\log (x)$ and $b$ denotes the number bits for representing one floating point number ($b \in \{32,64\}$ are typical values). For large $s$, \cite{alistarh2017qsgd} also proposes a simple encoding scheme $\text{Code}'_s$ which is proved to impose no more than the following communication cost on the quantized vector

\begin{align}\label{eq:code-prime-length}
    &\mathbb{E} \left[| \text{Code}'_s(Q^{\text{LP}}(\bx)) | \right]  \nonumber\\
    & \quad \leq b +\left(\frac{5}{2}+ \frac{1}{2}  \log^* \left(1+ \frac{s^2+\min(d,s \sqrt{p})}{p} \right)\right) p.
\end{align}

Now we can easily derive the expected total communication cost (in bits) of a quantized decentralized consensus optimization in order for each agent to achieve a predefined target error. For instance, assume that the low-precision quantizer described above is employed for the quanization operations. Using this quantizer, the expected communication cost (in bits) for transmitting a single $p$-dimensional real vector is represented in (\ref{eq:code-length}) and (\ref{eq:code-prime-length}) for two sparsity regimes of the tuning parameter $s$.

On the other hand, in order for each agent to obtain a relative error $\rho$, the proposed algorithm iterates $T(\rho)$ times as denoted in (\ref{eq:T-eps}). Therefore, the total (expected) communication cost across all of the $n$ agents is upper-bounded by $n T(\rho)\cdot \mathbb{E} \left[| \text{Code}_s(Q^{\text{LP}}(\bx)) | \right] $ and  $n T(\rho) \cdot \mathbb{E} \left[| \text{Code}'_s(Q^{\text{LP}}(\bx)) | \right] $ for small and large $s$, respectively.

\begin{remark}
We can derive the total communication cost for the vanilla DGD method (\cite{yuan2016convergence}), as well. DGD method updates the iterations as follows:

\begin{equation}\label{eq:DGDwQ2}
    \bx_{i,t+1} =   w_{ii} \bx_{i,t} + \sum_{ j \in  \mathcal{N}_i } w_{ij} \bx_{j,t} - \al \gr f_i(\bx_{i,t}),
\end{equation}

where $\al = c/\sqrt{T}$ is the stepsize. DGD guarantees the following convergence rate for strongly convex objectives:

\begin{align}
    \norm{\bx_{i,T} - \widetilde{\bx}^*}^2
    &\leq
    \frac{(3 + 2L/\mu)^2 D^2}{(1-\beta)^2} \al^2 \nonumber\\
    &=
    \frac{c^2(3 + 2L/\mu)^2 D^2}{(1-\beta)^2} \frac{1}{T}.
\end{align}

Hence, to reach the $\rho$ approximation of the global optimal, DGD requires the total number of iterations

\begin{align}
    T_{\text{DGD}}(\rho) = \frac{c^2(3 + 2L/\mu)^2 D^2}{(1-\beta)^2} \frac{1}{\rho \norm{\widetilde{\bx}^*}^2}.
\end{align}

Given that each decision vector requires $b p$ number of bits in an implementation of DGD (without quantization), the  DGD method induces the communication cost of $n T_{\text{DGD}}(\rho) b p$.
\end{remark}

In the following, we numerically evaluate the communication cost of the proposed QDGD method for the following  least squares problem

\begin{equation} \label{eq:leasetsqr}
    \min_{\bx \in \mathbb{R}^p} f(\bx) = \sum_{i=1}^{n} \frac{1}{2} \norm{\bA_i \bx - \bb_i}^2.
\end{equation}

We assume that the network contains $n=50$ agents that collaboratively aim to solve problem (\ref{eq:leasetsqr}) over the real field of size $p=200$. The elements of the random matrices $\mathbf{A}_i\in \mathbb{R}^{p\times p}$ and the solution $\widetilde{\bx}^*$ are picked from the normal distribution $\mathcal{N}(0,1)$. Moreover, we let $\bb_i = \mathbf{A}_i \widetilde{\bx}^* +\mathcal{N}(0,0.1 I_p)$. All nodes update their local variables with respect to the proposed algorithm and send the quantized updates to the neighbors using a low-precision quantizer with $s$ quantization levels and $b=64$ bits for representing one floating point number, until they satisfy the predefined relative error $\rho = 10^{-2}$. The underlying graph is an  Erd\H{o}s-R\'enyi with edge probability $p_c=0.35$.  The edge weight matrix is picked as $W = I - \frac{2}{3 \lambda_{\text{max}}(\mathbf{L})} \mathbf{L}$ where $\mathbf{L}$ is the Laplacian with $\lambda_{\text{max}}(\mathbf{L})$ as its largest eigenvalue. We also set $\delta=0.1$.

\begin{center}
\begin{table}[t]
  \begin{tabular}{cccc}
    \toprule
            \begin{tabular}{@{}c@{}}\# quantization  \\ levels \end{tabular}
           &
           \begin{tabular}{@{}c@{}}\# iterations  \\ ($\times 10^2$)\end{tabular}
           &
           \begin{tabular}{@{}c@{}} code length   \\ per vector (bits)\end{tabular}
           &
           \begin{tabular}{@{}c@{}}communication cost   \\ (bits) ($\times 10^{7}$)\end{tabular} \\ \midrule
    $s=1$    & $10800$   & $216.9$   & $1171$  \\
    $s=50$   & $11.6$   & $949.8$   & $5.5$  \\
    $s^*=77$ & $9.91$   & $1062$   & $5.27$   \\
    $s=10^3$   & $8.79$  & $1793$  & $7.88$ \\
    $s=10^5$   & $8.78$  & $3122$  & $13.71$ \\
    $s=10^{10}$   & $8.78$  & $6443$  & $28.3$ \\
    $s=10^{15}$   & $8.78$  & $9765$  & $42.9$ \\
    $s=10^{19}$   & $8.78$  & $12420$  & $54.56$ \\ \bottomrule
  \end{tabular}
  \vspace{.2 cm}
  \caption{Quantization-communication trade-off for least squares problem}\label{tab:trade-off}
    \vspace{-.6 cm}
\end{table}
\end{center}

Table \ref{tab:trade-off} represents the total expected communication cost (in bits, as computed using (\ref{eq:T-eps}), (\ref{eq:code-length}) and (\ref{eq:code-prime-length}))   induced by the proposed algorithm to solve (\ref{eq:leasetsqr}) using the low-precision quantizer --as described above-- for four representative cases. As observed from this table and expected from the theoretical derivations, larger number of quantization levels translates to less noisy quantization and hence fewer iterations. Also, larger number of quantization levels induces more communication cost for each transmitted quantized data variable which results in larger code length per vector. However, the average total communication cost does not necessarily follow a monotonic trend. As Table \ref{tab:trade-off} shows, the optimal $s^*=77$ induces the smallest total communication cost among all levels $s \geq 1$. Moreover, Table \ref{tab:trade-off} demonstrates the significant gain of picking the optimal levels $s^*$ compared to the larger ones.

\iffalse
\begin{center}
\begin{table}[t]
  \begin{tabular}{cccc}
    \toprule
            \begin{tabular}{@{}c@{}}\# quantization  \\ levels \end{tabular}
           &
           \begin{tabular}{@{}c@{}}\# iterations  \\ ($\times 10^3$)\end{tabular}
           &
           \begin{tabular}{@{}c@{}} code length   \\ per vector (bits)\end{tabular}
           &
           \begin{tabular}{@{}c@{}}communication cost   \\ (bits) ($\times 10^8$)\end{tabular} \\ \midrule
    $s=1$    & $342.7$   & $216.9$   & $74.33$  \\
    $s=10$   & $16.4$   & $678.2$   & $11.12$  \\
    $s^*=34$ & $7.770$   & $860$   & $6.682$   \\
    $s=55$   & $7.188$  & $973.8$  & $7$ \\ \bottomrule
  \end{tabular}
  \vspace{.2 cm}
  \caption{Quantization-communication trade-off for least square problem}
    \vspace{-.6 cm}
\end{table}\label{tab:trade-off}
\end{center}
\fi


text:
\begin{document}

  \maketitle
  \subsection*{ABSTRACT}
  \input{abstract}

  \section{Introduction}\label{introduction}
  \input{introduction}

  \section{System Model}\label{model}
  \input{model}

 \section{Secret Generation}\label{generation}
 \input{generation}

  \section{Experimental Validation}\label{validation}
  \input{validation}

\setlength{\bibsep}{0pt plus 0.2ex}
{\small \bibliographystyle{acm}
  \bibliography{citations}}

\end{document}

config: arxiv


text:
We provide several experiments to demonstrate the superior capabilities of the proposed
physics-aware registration based auto-encoder.
In this section, we solve for the low-rank grid and test the
reducibility/compression of the input as well as its application in a neural
network architecture approximate the PDEs.
We attempt to demonstrate both compression and interpretability of our
results in these experiments.
For all these experiments, we resort to readily available optimization packages
capable of solving optimization with nonlinear constraints; such as
\verb|minimize| in \verb|scipy| for our Python implementation,
or interior-point method in \verb|fmincon| in our Matlab implementation.
Both of the implementations as well as the data and the results are made
available at \url{\mydepository}.

\subsection{Manifold learning in rotated character ``A''}
\label{sec:rotated_A}

Consider a computer vision task of learning the nonlinear transformation,
rotation, given a data-set comprised of a rotated character ``A''.
The image of character ``A'' is stored in a $50 \times 50$ matrix and is
rotated a total of 90 degrees with 3 degrees increments resulting in a snapshot
matrix of dimension $2500 \times 31$.
A representative sample of the snapshots is shown in
Fig.~\ref{fig:roated_A_HFM}, and a single POD mode reconstruction is
illustrated in Fig.~\ref{fig:roated_A_EPOD}.
In this problem, $\basisgridU$ is down-sampled to size of $7$, i.e. the total
of $49$ control points.
Moreover, $v_\text{min}=0$, $\bm{\Gamma}_1=100 \bm{D}_{x x}$ and
$\bm{\Gamma}_2=\left(100/\pi\right) \bm{D}_{\theta \theta}$, where
$\bm{D}_{x x}$ and
$\bm{D}_{\theta \theta}$ are the second derivative matrices in the spatial and
parameter space, respectively.
The boundary point constraints are removed for this particular problem.
The optimization problem of~\eqref{Eqn:ResMin_ALE_smooth} approximates the
rigid body rotation as in Fig.~\ref{fig:roated_A_MG}.
In Fig.~\ref{fig:roated_A_MGPOD}, the snapshots are approximated using a single
basis ($\sizeROM_r=1$) on the learned manifold of $\sizeGridROM=2$.
The reconstruction delivered by the learned grid, using the proposed approach,
is remarkably more accurate compared to the traditional POD approach.

\begin{figure}[!h]
	\centering

	\def\rotatingA{data/RotatingA/n_mag_1_n_pos_2_edit}

	\begin{subfigure}[h]{0.99\textwidth}
		\centering
		\includegraphics[trim=200pt 610pt 140pt
		75pt,clip,scale=0.35]{\rotatingA}
		\caption{The snapshots.}
		\label{fig:roated_A_HFM}
	\end{subfigure}

	\begin{subfigure}[h]{0.99\textwidth}
		\centering
		\includegraphics[trim=200pt 445pt 140pt 250pt,clip,scale=0.35]
		{\rotatingA}
		\caption{The rank-$1$ reconstruction of the snapshots ($\sizeROM=1$) on
		the POD subspace. }
		\label{fig:roated_A_EPOD}
	\end{subfigure}

	\begin{subfigure}[h]{0.99\textwidth}
		\centering

		\includegraphics[trim=200pt 110pt 140pt 570pt,clip,scale=0.35]
		{\rotatingA}
		\caption{The parameter-varying rank-$2$ grid corresponding to the
			learned manifold.}
		\label{fig:roated_A_MG}
	\end{subfigure}

	\begin{subfigure}[h]{0.99\textwidth}
		\centering
		\includegraphics[trim=200pt 280pt 140pt
		410pt,clip,scale=0.35]{\rotatingA}
		\caption{The rank-$1$ reconstruction of the snapshots ($\sizeROM_r=1$)
		on the learned manifold. }
		\label{fig:roated_A_MGPOD}
	\end{subfigure}

	\caption{90 degrees rotation of character ``A''.}
	\label{fig:rotated_A}
\end{figure}

\subsection{Manifold learning in two-dimensional fluid flows}
\label{sec:2DRiemann}

In this section, the proposed manifold learning approach is applied to the
two-dimensional Riemann problem of fluid dynamics,

$
\frac{\partial}{\partial t} \bm{q} +
\frac{\partial}{\partial x} \bm{f}_x +
\frac{\partial}{\partial y} \bm{f}_y = 0,
$

where $\bm{q} = [\rho, \rho u, \rho v, \rho e]$,
$\bm{f}_x = [\rho u , \rho u ^ 2 + p, \rho u v , \rho u H]$,
$\bm{f}_y = [\rho v , \rho u v + p, \rho v^2 + p , \rho v H]$,
and $H = e + p/\rho$ ,
$p = \rho (\gamma -1 ) (e - 0.5 (u^2 + v^2) )$
in the domain $(x,y,t) \in [0,1] \times [0,1] \times [0,t_\text{max}]$, with
initial conditions of configuration 3 and 12 as in~\cite{Lax1998}.
The snapshots of primitive variables are generated using a high-order
artificial viscosity scheme coupled with a $4^{\textit{th}}$-order Runge-Kutta
time discretization with $\Delta t = 5 \times 10^{-4}$ on a $150 \times 150$
grid.  Snapshots are collected at $\delta t=0.016$ and $\delta t=0.006$
intervals for simulation range of $t \in [0,0.80]$ and $t \in [0,0.25]$
for configuration 3 and 12, respectively.
The size of the snapshot matrices in both cases are $10000 \times 50$.
A rank-$2$ time-varying grid ($\sizeGridROM=2$) is learned
via~\eqref{Eqn:ResMin_ALE_smooth} setting $\sizeROM_r=4$, and
$\gamma_1 = \gamma_2 = 0.05$,
where $\bm{\Gamma}_1 = \gamma_1 \bm{D}_{x x} = \gamma_1 \bm{D}_{y y}$
is the second derivative matrix in the $x$ and $y$ directions and
$\bm{\Gamma}_2=\gamma_2  \bm{D}_{t t}$, where $\bm{D}_{t t}$ is the second
derivative matrix in time.
Also, $v_\text{min}=\Delta x_\text{min}\Delta y_\text{min}$,
where $\Delta x_\text{min}=\Delta y_\text{min}=6.7 \times 10^{-4}$.
The low-dimensional representation of density is compared on the constant grid
and the learned manifold in Fig.~\ref{fig:2DRie}.
On the learned manifold, traveling shocks are conserved and
free of non-physical oscillatory solutions, resulting in a
significant error reduction.

\begin{figure}[!h]
	\centering
	\def\FileAddTDRie{data/2DRiemann/config03/t0d80/}

	\begin{subfigure}[!h]{0.32\textwidth}
		\includegraphics[scale=0.11]{\FileAddTDRie/main-figure129.png}
		\caption{POD subspace($\sizeROM=8$)}
		\label{fig:2DRieROM_config03}
	\end{subfigure}

	\begin{subfigure}[!h]{0.32\textwidth}
		\includegraphics[scale=0.11]{\FileAddTDRie/main-figure127.png}
		\caption{Learned manifold ($\sizeROM_r=8$)}
		\label{fig:2DRieMGROM_config03}
	\end{subfigure}

	\begin{subfigure}[!h]{0.32\textwidth}
		\includegraphics[scale=0.68]{\FileAddTDRie/main-figure151.pdf}
		\caption{Error convergence}
		\label{fig:2DRie_config03_error}
	\end{subfigure}

	\centering
	\def\FileAddTDRie{data/2DRiemann/config12/t0d25/}

	\begin{subfigure}[h]{0.32\textwidth}
		\includegraphics[scale=0.11]{\FileAddTDRie/main-figure129.png}
		\caption{POD subspace ($\sizeROM=8$)}
		\label{fig:2DRieROM_config12}
	\end{subfigure}

	\begin{subfigure}[!h]{0.32\textwidth}
		\includegraphics[scale=0.11]{\FileAddTDRie/main-figure127.png}
		\caption{Learned manifold ($\sizeROM_r=8$)}
		\label{fig:2DRieMGROM_config12}
	\end{subfigure}

	\begin{subfigure}[!h]{0.32\textwidth}
		\includegraphics[scale=0.68]{\FileAddTDRie/main-figure152.pdf}
		\caption{Error convergence}
		\label{fig:2DRie_config12_error}
	\end{subfigure}

	\caption{The density snapshots of the two-dimensional Riemann problems at
	the last time step of the simulations and the computational grid
	representative of the learned manifolds.
	Configuration 3: Fig.~\ref{fig:2DRieROM_config03} to
	Fig.~\ref{fig:2DRie_config03_error};
	Configuration 12: Fig.~\ref{fig:2DRieROM_config12} to
	Fig.~\ref{fig:2DRie_config12_error};}
	\label{fig:2DRie}
\end{figure}

\subsection{Physics based auto-encoder in an LSTM architecture}
\label{sec:RNN1D}

In this section, the proposed method is implemented as an auto-encoder layer
wrapped around a traditional machine learning architecture to decrease the loss
in compression phase and subsequently to improve the predictive capabilities of
a recurrent neural network modeling the governing PDEs.
We employ a long short-term memory (LSTM) to approximate the dynamics of the
PDE on the learned manifold, a proven architecture to approximate
PDEs~\citep{Parish_CMAME_2020}.
To learn a low-dimensional manifold, a dense neural network with
linear activation is used. Finally, the proposed registration based
auto-encoder is added to the architecture.

Consider the scalar, one-dimensional nonlinear convection-diffusion equation,
known as viscous Burgers' equation,

$
\partial_t w(x,t)
+ w \partial_x w(x,t)
= \left(1/\Rey\right) \partial_{xx} w(x,t)
$

in the domain $(x,t) \in [x_a,x_b] \times [0,T]$, equipped with initial
conditions $w(x,0) = w_0(x)$, and Dirichlet boundary conditions at $x_a$, and
$x_b$, where $w(x,0)=0.8 + 0.5~e^{-(x-0.5)^2/0.1^2}$, $w(x_a,t)=w(x_b,t) = 0$,
for $(x,t) \in [0,2.5] \times [0,1]$.
An implicit second order time discretization is used with $\Delta
t=8\times10^{-3}$ and space is uniformly discretized where $\Delta
x=1\times10^{-2}$.
In the proposed architecture, the rank-$1$ time-varying grid
($\sizeGridROM=1$), representing the physics-aware auto-encoder, is learned as
in~\eqref{Eqn:ResMin_ALE_smooth} with $\sizeROM_r=4$.
In this problem, $v_\text{min}=10^{-3}$, $\bm{\Gamma}_1=\gamma_1
\bm{D}_{xx}$ and $\bm{\Gamma}_2=\gamma_2 \bm{D}_{tt}$, where
$\gamma_1=\gamma_2=1$, and $\bm{D}_{xx}$ and
$\bm{D}_{t t}$ are second derivative matrices in space and time.
Subsequently, the LSTM is trained to approximate $\mapEultoALE{\snapshot}$.
In the traditional architecture, the densely connected auto-encoder and the
LSTM are trained simultaneously (Fig.~\ref{fig:arch_conventional}).
In the proposed architecture (Fig.~\ref{fig:arch_physicsbased}), the manifold
and the neural network are trained separately.
The neural network architectures are deployed in Keras~\citep{chollet2015keras}.
The models are trained using the snapshot matrix of
$\snapshot\in\realset{N_x}{N_t}$.
In the proposed physics-aware AE, $\basisgridV$ is linearly extended.
The contour plots of the solutions are compared in Fig.~\ref{fig:Burgers_AERNN}
and Fig.~\ref{fig:Burgers_PAAERNN}.
As expected, the neural network AE cannot predict the convection underlying the
physics of the problem outside the training range
(Fig.~\ref{fig:Burgers_AERNN}); however, by levering the
physics of the problem, the LSTM trained on the low-dimensional manifold
realized by the registration-based AE leads to a solution much closer to the
solution of the PDE (Fig.~\ref{fig:Burgers_PAAERNN}).
The mean squared difference of the solutions, $\error$, for different sizes of
LSTM, $N_h$, are plotted in Fig.~\ref{fig:Burgers_RNN_error}.

\begin{figure}[!h]
	\centering
	\def\FileRNN{data/RNN/burgers}

	\begin{subfigure}[h]{0.32\textwidth}
		\includegraphics[trim={0 0 0 0},clip, scale=0.9]
		{\FileRNN/NNROM}
		\caption{Neural Network AE ($N_h=10$)}
		\label{fig:Burgers_AERNN}
	\end{subfigure}

	\begin{subfigure}[h]{0.32\textwidth}
		\includegraphics[trim={0 0 0 0},clip, scale=0.9]
		{\FileRNN/PBROM}
		\caption{Physics-aware AE ($N_h=10$)}
		\label{fig:Burgers_PAAERNN}
	\end{subfigure}

	\begin{subfigure}[h]{0.32\textwidth}
		\includegraphics[trim={0 0 0 0},clip, scale=0.78]
		{\FileRNN/error_pred}
		\caption{LSTM error}
		\label{fig:Burgers_RNN_error}
	\end{subfigure}

	\caption{LSTM approximating the Burgers' equation.}
	\label{fig:Burgers_RNN}
\end{figure}

Also, consider the one-dimensional wave equation,

$
\partial_{tt} w(x,t) - \partial_{xx} w(x,t) = 0,
$

in the domain $(x,t) \in [x_a,x_b] \times [0,T]$, equipped with initial
conditions $w(x,0) = w_0(x)$, and Dirichlet boundary conditions at $x_a$, and
$x_b$, where $w(x,0)= e^{-(x-0.5)^2/0.1^2}$, $w(x_a,t)=w(x_b,t) = 0$,
for $(x,t) \in [0,1] \times [0,1]$.
An implicit second-order time-discretization is used with $\Delta
t=2.5\times10^{-3}$ and space is uniformly discretized where $\Delta
x=10^{-2}$.
The architecture is set up similar to the Burgers' equation, with the following
parameters:
the time-varying grid is of rank-$2$ ($\sizeGridROM=2$),
the reconstruction on the learned manifold is of rank-$2$ ($\sizeROM_r=2$),
$v_\text{min}=10^{-3}$, $\gamma_1=\gamma_2=10$, and the size of the grid bases,
in both space and time, are down-sampled to $15$ control points.
The solution and error are plotted in Fig.~\ref{fig:wave_RNN}, showing the
performance of the proposed physics-aware AE compared to the traditional
architecture.

\begin{figure}[!h]
	\centering
	\def\FileRNN{data/RNN/wave}

	\begin{subfigure}[h]{0.32\textwidth}
		\includegraphics[trim={0 0 0 0},clip, scale=0.9]
		{\FileRNN/NNROM}
		\caption{Neural Network AE ($N_h=5$)}
		\label{fig:wave_AERNN}
	\end{subfigure}

	\begin{subfigure}[h]{0.32\textwidth}
		\includegraphics[trim={0 0 0 0},clip, scale=0.9]
		{\FileRNN/PBROM}
		\caption{Physics-aware AE ($N_h=5$)}
		\label{fig:wave_PAAERNN}
	\end{subfigure}

	\begin{subfigure}[h]{0.32\textwidth}
		\includegraphics[trim={0 0 0 0},clip, scale=0.78]
		{\FileRNN/error_train}
		\caption{LSTM error}
		\label{fig:wave_RNN_error}
	\end{subfigure}

	\caption{LSTM approximating the wave equation.}
	\label{fig:wave_RNN}
\end{figure}


TITLE: Some doubts on the relationship between Lie algebras and Lie groups
QUESTION [1 upvotes]: Let $(\mathbb G,*)$ be a Carnot group. Thus, by definition, $\mathbb G$ is a connected and simply connected Lie group whose Lie algebra $\mathfrak g$ admits a stratification, that is
$$\mathfrak g=V_1\oplus\dots\oplus V_r$$
such that
$$[V_1,V_{i}]=V_{i+1}$$
for $1\leq i\leq r-1$ and every other commutator of lenght > $r$ is zero (see here).
We set $m_i:=dim(V_i)$ and $h_0=0$, $h_{i+1}=h_i+m_{i+1}$ for $0\leq i\leq r-1$, thus $h_1+\dots+h_r=n$, where $n$ is the (topological) dimension of $\mathbb G$.
For these groups, we know that the $Exp\colon (\mathfrak g, \diamond)\to (\mathbb G,*)$ is an isomorphism of Lie groups, where with $\diamond$ I have denoted the Hausdorff-Campbell operation on $\mathfrak g$. Moreover, given a basis $\{Z_1,\dots,Z_n\}$ of $\mathfrak g$ adapted to the stratification, that is $\{Z_{h_i+1},\dots, Z_{h_{i+1}}\}$ is a basis for $V_{i+1}$, we can identify $\mathfrak g$ with $\mathbb R^n$ setting
$$\mathfrak g\ni\sum_{i=1}^nx_iZ_i\mapsto (x_1,\dots,x_n)\in \mathbb R^n.
$$
If we call the above map $\pi$, we get that $Exp\circ \pi^{-1}\colon \mathbb R^n\to \mathbb G$ is an isomorphism of Lie group if we endow $\mathbb R^n$ with the composition law $\bullet$ defined as
$$x\bullet y:=\pi(\sum x_i Z_i\diamond \sum_{i=1}^n y_i Z_i) \quad (*).$$
Now my doubts are:

In papers, I often read sentences like "Let $\mathfrak g$ be the Lie algebra generated, as a vector space by three vectors $X_1,X_2,X_3$ with the only non trivial commutation relation being $X_3=[X_1,X_2]$. Thus $\mathfrak g$ is a nilpotent algebra stratified by the strata $V_1=span\{X_1,X_2\}$ and $V_2=span\{X_3\}$. Let $(\mathbb G,*)$ be the only connected and simply connected Lie group whose Lie algebra is $\mathfrak g$."
Which theorem ensures us that there exists a connected and simply connected Lie group with $\mathfrak g$ as a Lie algebra and it is unique?
With the notation as in the previous question, suppose that there exists a change of basis $\phi$ that sends the basis $\{X_1,X_2,X_3\}$ in $\{Y_1,Y_2,Y_3\}$. A change of basis is not necessarily a Lie algebras isomorphism, thus $Y_1,Y_2,Y_3$ may have different commutation relations from just $[Y_1,Y_2]=Y_3$ and all the possible others zero. Nevertheless $\mathfrak g$ is spanned also by $\{Y_1,Y_2,Y_3\}$. Thus, the only connected and simply connected Lie group that has Lie algebra spanned by $\{Y_1,Y_2,Y_3\}$ is again $(\mathbb G,*)$, right?
Suppose that a Carnot group $(\mathbb G,*)$ is given. Suppose we have fixed a basis of its Lie algebra $\mathfrak g$ adapted to the stratification. Thus, if we use this basis, suppose $\bullet$ is the group law on $\mathbb R^n$ as in $(*)$. If we now use another basis adapted to the stratification, that may have different commutation relation from the former, we get a composition law $\odot$ on $\mathbb R^n$ that might be different from $\bullet$. Thus the groups $(\mathbb R^n,\bullet)$ and $(\mathbb R^n, \odot)$ are different Lie groups that are the representation in exponential coordinate of the same group $(\mathbb G,*)$. Is this also true?

Thanks a lot.

REPLY [3 votes]: Cartan-Lie theorem. Also known as Lie's third theorem - but more correctly: Cartan extended Lie's theorem to prove the result that you're looking for. Some links:

Lie's third theorem, wikipedia
Lie's three theorems at nLab

Commutation relations are definitely preserved under change of basis! Try it.
Answered by the answer to 2.


text:
\begin{document}

\title{Existence and Uniqueness of a Fractional Fokker-Planck Equation}

\author{Li Lin \footnote{E-mail address: linli@hust.edu.cn} \\

\small{ Center for Mathematical Sciences, Huazhong University of Science and Technology} \\
\small {Wuhan, 430074, China}\\}
\maketitle
\begin{abstract}
Stochastic differential equations with L\'{e}vy motion arise the mathematical models for various phenomenon in geophysical and biochemical sciences.
The Fokker-Planck equation for such a stochastic differential equations is a nonlocal partial differential equations. We prove the existence and uniqueness of the weak solution for this equation.

\end{abstract}

keywords: Fractional Laplacian operator, Fokker-Planck equation, Lax- Milgram theorem, Existence and  uniqueness

\section{introduction}
For a system described by a stochastic differential equation with a $\alpha-$stable L\'{e}vy motion $L_t^{\a}$ for $\a \in(0,2),$
$$dX_t=b(X_t)dt+adB_t+dL_t^{\a},$$
the corresponding Fokker-Planck equation \cite{D} contains a nonlocal Laplacian operator,
$$u_t=a\Delta u-(-\Delta)^{\a/2}u-div(b(x)\cdot u),$$
where $b$ is a two-dimensional vector function and  $-(-\Delta)^{\a/2}$ is a nonlocal Laplacian operator defined by
$$(-\Delta)^{\a/2}f(x)=\int_{\mathbb{R}^d/\{0\}}\frac{f(x)-f(y)}{|x-y|^{d+\a}}dy.$$

Fractional partial differential equations  arise the mathematical models for various phenomenon in physics and biology.
Such as anomalous diffusion of particles \cite{MM} and  the cell density evolution in certain biological processes \cite{JD}. Volume constraints are natural extensions to the fractional case of boundary conditions for differential equations \cite{QD}.

The Fokker-Planck equations have been used in the modeling of many physical phenomena, in particular, for the description of the evolution of plasmas \cite{HR}. Moreover, there are some settings in which particles may have long jumps\cite{DS}. Questions such as existence of solutions, hydrodynamic limits, and long-time behavior for the Fokker-Planck system has been extensively studied by many authors\cite{FB,NE,TG} .

There are few works dealing with the existence and uniqueness of the Fokker-Planck equations. Wei and Tian \cite{Wei}
obtain the existence and uniqueness of weak $L^p$ solution on the whole space. AcevesSanchez and Cesbron \cite{PA} studied the nonlocal Fokker-Planck problem on $\mathbb{R}^d$ in fractional Sobolev space with the drift term is a direct proportion function.  In \cite{He}, the weak solution is just considered in the Sobolev space $H^1.$ Moreover, there is no proof of the well-posedness for equation  with Laplace operator.

This paper is devoted to the study of the nonlocal Fokker-Planck equation on a bounded interval in $\mathbb{R}^2.$ There are two main results: For $a\geq0,$ we obtain the existence and uniqueness in the Sobolev space $H^1$ with respect to spatial variables; for $a=0,$ we reach the conclusion in a fractional sobolev space $H^{\a/2}$ with respect to spatial variables. We can see that, This is more accurate.
\section{Well-posedness of  Fokker-Planck equation}
In this section, we give some function spaces and obtain  the existence and uniqueness of
the weak solution of a fractional Fokker-Planck equation on a bounded domain in $\mathbb{R}^2.$
\subsection{Function spaces}
Denote $Q_T=D\times(0,T).$ Let $\partial_lQ_T$ and $\partial_pQ_T$ be the lateral boundary $D^c\times(0,T)$ and the parabolic boundary $\partial_lQ_T\cup \{(x,t); x\in \bar{D},t=0\}$ of $Q_T.$
Denote by $\mathring{W}, \dot{W}$ the set of all functions in $W$ vanish on $\partial_pQ_T$ and $\partial_lQ_T$ respectively.

For the integer $\beta, r \in \{0,1\},$ the set
$$\{u; \nabla^\beta u, \partial^r_t u \in L^2(Q_T)\}$$
endowed with the norm
$$||u||^2_{W_2^{1,1}(Q_T)}=\int\int_{Q_T}\sum\limits_{\beta\in\{0,1\}}|\nabla^\beta u|^2+\sum\limits_{r\in\{0,1\}}|\partial^r_t u|^2dxdt$$
is denoted by $W_2^{1,1}(Q_T).$
Denote
$$V(Q_T):=\{u\in \dot{W}_2^{1,1}(Q_T), \nabla u_t \in L^2(Q_T;\mathbb{R})\},$$
and define the inner product as
$$(u,v)_{V(Q_T)}=(u,v)_{W_2^{1,1}(Q_T)}+(\nabla u_t,\nabla v_t)_{L^2(Q_T)}.$$

\subsection{Main result}
We consider the following nonlocal Fokker-Planck equation
\begin{equation}{\label{1}}
  \begin{cases}
    u_t=a\Delta u-(-\Delta)^{\a/2}u-div(b(x)\cdot u)\qquad x\in D\subset \mathbb{R}^2,\;t\in(0,T),\\
    u|_{D^c}=0,\\
    u(x,0)=u_0(x),
  \end{cases}
\end{equation}
where $D=(0,1)$ is an bounded domain in $\mathbb{R}^2,$ and $D^c$ is the complement of D. We will prove the existence and uniqueness of the solution to the equation (\ref{1}).

\begin{remark}
  We define nonlocal divergence operator $\mathcal{D}$ on $\beta$ as
$$\mathcal{D}(\beta)(x):=\int_{\mathbb{R}^d}(\beta(x,z)+\beta(z,x))\cdot\gamma(x,z)dz  \qquad \text{for}\; x\in D.$$
For a function  $\phi(x)$, the adjoint operator $\mathcal{D^{*}}$ corresponding to $\mathcal{D}$ is the operator whose action on $\phi$ is given by
$$\mathcal{D^{*}}(\phi)(x,z)=-(\phi(z)-\phi(x))\gamma(x,z) \qquad \text{for} \; x,z\in D.$$
Here we take $\gamma(x,z)=\frac{1}{\sqrt{2}}(z-x)\frac{1}{|z-x|^{\frac{d+2+\a}{2}}}.$
we have
$$\mathcal{D}\mathcal{D^{*}}=-(-\Delta)^{\a/2}.$$
\end{remark}

\begin{definition}
  Consider $u_0$ in $L^2(D).$ We say that $u$ is a weak solution of equation (\ref{1}), if for any $\varphi\in C_c^\infty([0,T)\times D)$,
\begin{equation}
  \int\int_{Q_T}u(\partial_t \varphi +a\Delta \varphi-(-\Delta)^{\a/2}\varphi+b(x)\cdot\nabla\varphi)dxdt+ \int\int_{Q_T}u_0(x)\varphi(0,x)dx=0
\end{equation}
 \end{definition}

\begin{theorem}\label{t1}
Consider $u_0\in L^2(D)$,
\begin{enumerate}
  \item If $a=0$ and $div b(x)\geq0$, there exists a unique weak solution and this solution satisfies
$$f\in \mathcal{X}:=\{f: f\in L^2(Q_T), \frac{|f(t,x)-f(t,y)|}{|x-y|^{\frac{2+\a}{2}}}\in L^2(Q_T\times\mathbb{R}^2)\}.$$
  \item For $a\geq0$, there exist a unique weak solution in $\mathring{W}_2^{1,1}(Q_T).$
\end{enumerate}
\begin{remark}
  Note that this definition of $\mathcal{X}$ is equivalent to saying that it is the set of functions which are in $L^2([0,T))$ with respect to time  and in $H^{\alpha/2}(D)$ with respect to space.
\end{remark}
\begin{proof}
  $1.$ We consider the Hilbert space $\mathcal{X}$ provided with the norm
  $$||u||_{\mathcal{X}}^2=||u||^2_{L^2(Q_T)}+||D^*u||^2_{L^2(Q_T\times\mathbb{R})}.$$
  Let us denote $\mathcal{T}$ the operator, given by
  $$\mathcal{T}u=\partial_tu+div(b(x)\cdot u).$$
  Moreover, we define the Hilbert space $\mathcal{Y}$ as
  $$\mathcal{Y}=\{f\in \mathcal{X}: \mathcal{T}u\in \mathcal{X}^{'}\},$$
  where $\mathcal{X}^{'}$ is the dual of $\mathcal{X}.$
  From the fact that the space $C_c^\infty(Q_T)$ is a subspace of $\mathcal{X}$ with a  continuous injection, we define the pre-Hilbertian norm:
  $$|\varphi|_{C_c^\infty(Q_T)}=||\varphi||_{\mathcal{X}}^2+\frac{1}{2}||\varphi(0,x)||^2_{L^2(D)}.$$
  Now, we introduce the bilinear form $a: \mathcal{X}\times C_c^\infty(Q_T)\rightarrow \mathbb{R}$ as
  \begin{equation}\label{2}
  a(u,\varphi)=\int\int_{Q_T}-u\varphi_t+(\mathcal{D}^*u,\mathcal{D}^*\varphi)_{L^2(\mathbb{R})}-b(x)u\cdot\nabla \varphi dxdt.
  \end{equation}
  and $$L(\varphi)=-\int_{D}u_0(x)\varphi(0,x)dx.$$
  Then find a solution $u$ in $\mathcal{X}$ of $(\ref{1})$ is equivalent to finding a solution $u$ in $\mathcal{X}$ of $a(u,\varphi)=L(\varphi)$ for any $\varphi\in C_c^\infty(Q_T)$. First $a(u,\varphi)$ is continuous. Next we will obtain the coercivity of $a$.
  \begin{equation}\label{5}
  \begin{split}
  a(\varphi,\varphi)&=\int\int_{Q_T}-\varphi_t\varphi+(\mathcal{D}^*\varphi,\mathcal{D}^*\varphi)_{L^2{(\mathbb{R})}}-b(x)\varphi\cdot\nabla\varphi dxdt\\
  &=\frac{1}{2}\int_{D}\varphi^2(0,x)dx+\int\int_{Q_T}(\mathcal{D}^*\varphi,\mathcal{D}^*\varphi)_{L^2{(\mathbb{R})}}dxdt+\frac{1}{2}\int_{D}div b(x)\varphi^2dxdt.
  \end{split}
  \end{equation}

  Then, there exists a positive constant $\delta$ such that  $a(\varphi,\varphi)\geq \delta|\varphi|_{C_c^\infty(Q_T)}.$
  Thus the Lax-Milgram theorem implies the existence of $u$ in $\mathcal{X}$ satisfying equation (\ref{2}). That yields existence of a solution of equation (\ref{1}) with $a=0.$ For $u\in \mathcal{X},$ the linear bounded operator $\mathcal{T}$  maps $u\in \mathcal{T}$ to $-(-\Delta)^{\a/2}u \in \mathcal{X}^{'}$, hence
  the weak solution $u$ is in $\mathcal{Y}.$

  Since the equation (\ref{1}) is linear, to show the uniqueness, it is  enough to show the unique solution with zero initial is the function $u\equiv 0.$ Let $u$ be a solution of this problem on $\mathcal{Y}.$
  Through integration by parts we have
  \begin{equation*}
  \begin{split}
  2(\mathcal{T}u,u)_{\mathcal{X},\mathcal{X}^{'}}&=(\partial_tu+div(b(x)\cdot u),u)_{\mathcal{X},\mathcal{X}^{'}}\\
  &=\int_{D}u^2(T,x)dx+\int_{Q_T}divb(x)u^2dxdt\geq0.
  \end{split}
  \end{equation*}
  On the other hand, since $u$ satisfies equation (\ref{1}), $\mathcal{T}u=-(-\Delta)^{\a/2}$ in the weak sense, then
  $$(\mathcal{T}u,u)_{\mathcal{X},\mathcal{X}^{'}}=-(\mathcal{D}^*u,\mathcal{D}^*u)\leq0.$$
  That means $u\equiv0 $ a.e. on $Q_T$. The solution is unique.

$2.$ For the first part we set $a>0.$
To show the unique of the equation (\ref{1}), it is  enough to show the unique solution with zero initial is the function $w\equiv 0.$
\begin{equation}
  \begin{cases}
    w_t=a\Delta w-(-\Delta)^{\a/2}w-div(b(x)\cdot w) \qquad x\in D\subset \mathbb{R}^2,\\
    w|_{D^c}=0,\\
    w(x,0)=0.
  \end{cases}
\end{equation}

Set $B[u,v,t]=\int_Da\nabla u\cdot\nabla v+(D^*u,D^*v)_{L^2(\mathbb{R})}-b(x)\cdot u\cdot\nabla udx,$
then $$\frac{d}{dt}(\frac{1}{2}||w||^2_{L^2(D)})+B[w,w,t]=(w,w')+B[w,w,t]=0.$$
This means
\begin{equation}
\begin{split}
  a\int_{D}|\nabla w|^2dx&+(D^*w,D^*w)_{L^2(D\times\mathbb{R})}\\
  &=B[w,w,t]+\int_D b(x)\cdot w\cdot\nabla wdx\\
  &\leq B[w,w,t]+\varepsilon\int_{D}|\nabla w|^2dx+\frac{C}{4\varepsilon}\int_{D}|w|^2dx.
\end{split}
\end{equation}
From Poincar\'{e} inequality $$||w||_{L^2(D)}\leq C_1||D^*w||_{L^2(D\times\mathbb{R})}.$$
Then choose $\varepsilon$ small enough, then
$$\sigma||w||_{H^1}+\beta||w||_{H^{\a/2}}\leq B[w,w,t]+\gamma ||w||^2_{L^2(D)},$$
for positive constant $\sigma, \beta,$ and $\gamma\geq 0.$
We can see that $$B[w,w,t]\geq -\gamma ||w||^2_{L^2(D)}.$$
By Gronwall inequality, we obtain
$$w=0.$$ The solution is unique.

Next, we will prove the existence of the problem. Without loss of generality, we just consider the case $u_0=0.$
$$a(u,v)=\int\int_{Q_T}(u_tv_t+a\nabla u\nabla v_t+( D^*u, D^*v_t)_{L^2(\mathbb{R})}+ div(b(x)\cdot u)v_t)e^{-\theta t}dxdt, \theta>0.$$
Then $$|a(u,v)|\leq ||u||_{\dot{W}^{1,1}(Q_T)}||v||_{V(Q_T)}.$$
set $$a(v,v):=A+B+C+D$$
For $v\in V(Q_T)$,
\begin{equation}
  \begin{split}
   B&=a\int\int_{Q_T}\nabla v\cdot\nabla v_te^{-\theta t}dxdt
 =a\int\int_{Q_T}\frac{1}{2}\frac{\partial}{\partial_t}|\nabla v|^2e^{-\theta t}dxdt\\
 &=a\int\int_{Q_T}\frac{1}{2}\frac{\partial}{\partial_t}(|\nabla v|^2e^{-\theta t})dxdt+\frac{\theta}{2}\int\int_{Q_T}e^{-\theta t}|\nabla v|^2dxdt\\
 &=\frac{a}{2}e^{-\theta T}\int_{D}|\nabla v|^2|_{t=T}dx-\frac{a}{2}\int_{D}\gamma|\nabla v|^2|_{t=0}dx+\frac{a\theta}{2}\int\int_{Q_T}e^{-\theta t}|\nabla v|^2dxdt\\
  \end{split}
\end{equation}
Since $v\in V(Q_T)$, then $\gamma|\nabla v|^2_{t=0}=0,$ combine with Poincar\'{e} inequality
\begin{equation}
  \begin{split}
  B&\geq \frac{\theta}{2}\int\int_{Q_T}e^{-\theta t}|\nabla v|^2dxdt\\
  &\geq\frac{a\theta}{4}\int\int_{Q_T}e^{-\theta t}|\nabla v|^2dxdt+\frac{a\theta}{4\mu}\int\int_{Q_T}e^{-\theta t}v^2dxdt.
 \end{split}
\end{equation}

Similarly,
$$C\geq\frac{\theta}{2}\int\int_{Q_T}||\mathcal{D}^*v||_{L^2(\mathbb{R})}^2dxdt.$$
On the other hand, we have
\begin{equation}
  \begin{split}
  D&=\int\int_{Q_T} div(b(x)\cdot v)v_te^{-\theta t}dxdt\\
  &=\int\int_{Q_T}divb(x)\cdot v\cdot v_te^{-\theta t}dxdt+\int\int_{Q_T}b(x)\cdot\nabla v\cdot v_te^{-\theta t}dxdt\\
  &:=E+F,
   \end{split}
\end{equation}
where
$$|E|\leq \varepsilon\int\int_{Q_T}v_t^2e^{-\theta t}dxdt+\frac{C_2}{\varepsilon}\int_{Q_T}v^2e^{-\theta t}dxdt.$$
$$|F|\leq \varepsilon\int\int_{Q_T}v_t^2e^{-\theta t}dxdt+\frac{C_3}{\varepsilon}\int\int_{Q_T}|\nabla v|^2e^{-\theta t}dxdt.$$
Then we have
\begin{equation}
  \begin{split}
  a(v,v)&\geq(1-2\varepsilon)\int_{Q_T}v_t^2e^{-\theta t}dxdt+\frac{\theta}{2}\int_{Q_T}||D^*v||^2_{L^2(\mathbb{R})}e^{-\theta t}dxdt\\
  &(\frac{a\theta}{4}-\frac{C_4}{\varepsilon})\int_{Q_T}|\nabla v|^2e^{-\theta t}dxdt+(\frac{a\theta}{4\mu}-\frac{C_5}{\varepsilon})\int_{Q_T}v^2e^{-\theta t}dxdt
   \end{split}
\end{equation}
Choose $\varepsilon$ small enough and $\theta>0$ large enough, we have
$$ a(v,v)\geq\delta||v||_{W^{1,1}(Q_T)},$$
where $\delta$ is a positive constant.
Then the Lax-Milgram theorem implies the existence of equation (\ref{1}).

For $a=0,$ it is similar to the work that He and Duan did in \cite{He}.
\end{proof}
\end{theorem}

\section{Example}
Langevin equation provide models of a diffusing particle. We consider the following system of stochastic differential equations
\begin{equation}\label{3}
  \begin{cases}
    dx_t=v_tdt,\\
    mdv_t=-\gamma v_tdt+\sigma dL_t^{\a},
  \end{cases}
\end{equation}
where $m$ is the mass of the particle, $\gamma$ and $\sigma$ are the dissipation and
diffusion coefficient, respectively, and $L_t^{\a}$ is a L\'{e}vy process with generator
$-(-\Delta)^{\a/2}.$
Then the corresponding L\'{e}vy Fokker-Planck equation is
\begin{equation}\label{6}
  \begin{cases}
 u_t+v\cdot \nabla_xu=-\frac{\sigma}{m}(-\Delta_v)^{\a/2}u+\frac{\gamma}{m}div_v(v\cdot u)\qquad x,v\in D\subset \mathbb{R}^2,\;t\in(0,T),\\
    u|_{D^c}=0,\\
    u(x,0)=u_0(x)
  \end{cases}
\end{equation}
We can define the bilinear form as (\ref{2})
$$a_1(u,\varphi)=\int\int\int_{Q_T\times D}-u\varphi_t-vu\cdot\nabla_x\varphi+\frac{\sigma}{m}(\mathcal{D}_v^*u,\mathcal{D}_v^*\varphi)_{L^2(\mathbb{R})}+\frac{\gamma}{m}vu\cdot\nabla_v \varphi dxdtdv.$$
and from the fact $div_v(v)>0,$ and
$$\int\int\int_{Q_T\times D} v \varphi \cdot\nabla_x \varphi dxdtdv=0,$$
  Then, there exists a positive constant $\delta_1$ such that  $a(\varphi,\varphi)\geq \delta_1|\varphi|_{C_c^\infty(Q_T\times D)}.$
we verified that the equation (\ref{6}) has a unique weak solution in $\mathcal{X}_1$ by Theorem \ref{t1},
where, $$\mathcal{X}_1:=\{f: f\in L^2(Q_T\times D), \frac{|f(t,x,v)-f(t,x,w)|}{|v-w|^{\frac{2+\a}{2}}}\in L^2(Q_T\times D\times \mathbb{R}^2)\}.$$
Clearly, the result of the example agree with the theoretical finding in this study.
\section*{References}


TITLE: Apollonius ellipse equation $y^2=x \left( p-\frac{p}{2a}x \right) $ to standard form.
QUESTION [2 upvotes]: I am looking for a way to understand the last steps found at this site: http://mathdl.maa.org/mathDL/46/?pa=content&sa=viewDocument&nodeId=196&bodyId=203
The page finishes with showing in the modern style of Apollonius ellipse equation as
$$
y^2=x \left( p-\frac{p}{2a}x \right)
$$
together with the comment: "Can this be written in the standard form of an ellipse? Complete the square and see."
I try to get this to the standard form as in:
$$
\frac{y^2}{a^2}+\frac{x^2}{b^2}=1
$$
But the closest my algebra gets me is:
$$
\frac{y^2}{xp}+\frac{x}{2a}=1
$$
Can anyone show me how they are connected?

REPLY [2 votes]: $$ y^2=x \left( p-\frac{p}{2a}x \right)$$
$$2a\cdot y^2=p(2ax-x^2)=-p\{(x-a)^2-a^2\}$$
$$2a\cdot y^2+p(x-a)^2=p\cdot a^2$$
$$\frac{(x-a)^2}{a^2}+\frac{y^2}{\frac{pa}2}=1$$
$$\left(\frac{x-a}a\right)^2+\left(\frac y{\sqrt\frac{pa}2}\right)^2=1$$


TITLE: Is the differential form $\omega(x,y)=\frac{2x}{x^2+y^2-4}dx+\frac{2y}{x^2+y^2-4}dy$ exact on his natural domain?
QUESTION [1 upvotes]: The natural domain $D$ of the differential form $\omega$ is $D=D_1 \cup D_2$ where $D_1=B\bigl((0,0),2\bigr)$ is the open ball of center in the origin and radius $r=2$ and $D_2=\mathbb{R}^2-\bar{D_1}$. Since $\omega$ is closed on the simply connected $D_1$, we have that $\omega$ is exact on $D_1$. How can i prove if $\omega$ is exact on $D_2$ too? Thanks

REPLY [2 votes]: $\omega(x,y) = \mathrm{d} \log((x^2+y^2-4)^2)/2$, and is closed, being an exact form for all $(x,y)$ such that $x^2+y^2 \not=4$, i.e. on $D = D_1 \cup D_2$.


TITLE: Why isn't $\{(1, 1), (1, 2), (2, 1)\}$ transitive
QUESTION [1 upvotes]: I've been given a set of relations on $\{1,2,3,4\}$
\begin{align*}
R1 &= {(1, 1), (1, 2), (2, 1), (2, 2), (3, 4), (4, 1), (4, 4)}\\
R2 &= {(1, 1), (1, 2), (2, 1)}\\
R3 &= {(1, 1), (1, 2), (1, 4), (2, 1), (2, 2), (3, 3), (4, 1), (4, 4)}\\
R4 &= {(2, 1), (3, 1), (3, 2), (4, 1), (4, 2), (4, 3)}\\
R5 &= {(1, 1), (1, 2), (1, 3), (1, 4), (2, 2), (2, 3), (2, 4), (3, 3), (3, 4), (4, 4)}\\
R6 &= {(3, 4)}.
\end{align*}
And I've been told that $R4, R5, R6$ are transitive but the rest are not. I understand the nature of transitivity, if $(x,y)$ and $(y,z)$ are present then $x,z$ must be present, but I don't know how to apply this to examples given. $R6$ makes sense, there's no counterexample so it's already proven vacuously. But why isn't $R2$ transitive?
Anyone that can run down in detail each of the examples would help my understanding greatly.

REPLY [4 votes]: $R_2$ isn't transitive because $(2,1) \in R_2$ and $(1,2) \in R_2$, but $(2,2) \notin R_2$.
[Here, I've taken $(x,y) = (2,1)$, $(y,z) = (1,2)$, $(x,z) = (2,2)$.]
Another example: $R_1$ isn't transitive because $(3,4) \in R_1$ and $(4,1) \in R_1$, but $(3,1) \notin R_1$.

REPLY [2 votes]: R1: We have $(3,4)$ and $(4,1)$ but not $(3,1).$
R2: We have $(2,1)$ and $(1,2)$ but not $(2,2).$
R3: We have $(2,1)$ and $(1,4)$ but not $(2,4).$
R4: You can check by exhaustion or notice that this is the relation "$>$".
R5: You can check by exhaustion or notice that this is the relation "$\leq$".
R6: Vacuous, as you said.


TITLE: Characterization of fields in commutative rings.
QUESTION [2 upvotes]: For the following, by "ring" I will refer to a ring with identity, and by "rng" to a ring wihout identity. By a rng-homomorphism I mean a ring homomorphism that does not necessarily preserves the identity (if it exists).

I know that if $R$ is a commutative ring, then the following are true:

R is a field iff all non-zero rng-homomorphisms from it are injective.
R is a field iff the only ideals of $R$ are $0$ and $R$.

My question is, is it necessary for $R$ to be both commutative and a ring? I.e., what if $R$
is just:

A commutative rng.
A (not necessarily commutative) ring.
A (not necessarily commutative) rng.

Do $(1)$ and $(2$) still hold?

REPLY [1 votes]: Kind of reading between the lines here because the body of your question does not mention "fields" like the title does.  Anyhow, the two conditions are still equivalent even for (possibly noncommutative) rngs. Ideals correspond to kernels of departing homomorphisms, and the two points you listed are just two equivalent ways of saying that the only possibilities are the zero homomorphism or an injective homomorphism.
But of course they are not necessarily fields.
There are many examples of simple rings with identity that aren't fields, for example a square matrix ring over a field.
For an example of a noncommutative ring without identity that is simple, you could take the endomorphism ring $E$ of a countable dimensional vector space over a field, and then extract the subring $R$ of transformations with finite dimensional images.
Reasoning in that last example: the ring $E$ is known to have precisely three ideals, and you are selecting out the one nontrivial ideal.  Furthermore the ring is von Neumann regular, and it is known in such rings that "is an ideal" is a transitive relationship in the sense that if $I\lhd E$ and $J\lhd I$, then $J\lhd E$. So any ideal of $R$ would necessarily be an ideal of $E$ (and we know there are none between $R$ and $\{0\}$.)  It is not hard to produce two transformations that don't commute, and to see that the only candidate for the identity is the identity of $E$, which is absent owing to its infinite dimensional image.
There are also trivial examples of commutative simple rings without identity that (obviously) are not fields: say, the ideal $2\mathbb Z/4\mathbb Z$ of $\mathbb Z/4\mathbb Z$.
In short, the proposition you are looking at is interesting in the commutative case because it says "commutative simple rings are fields."  Simple rings are very diverse and interesting in general noncommutative algebra, but they are much more nicely behaved when commutativity is added.


TITLE: Operatornorm inequality
QUESTION [3 upvotes]: Let $$E:=(C^0([0,1]),\|\cdot\|_1) , \quad F:=(C^0([0,1]),\|\cdot\|_\infty),$$ with $$\|f\|_1:=\int_0^1\vert f(t)\vert \, dt,\quad \|f\|_\infty:=\sup_{t\in [0,1]}\vert f(t)\vert .$$ For $k \in C^0([0,1]\times[0,1])$ and $f\in E$ define $$K(f):=s\mapsto \int_0^1k(s,t)f(t)\,dt, s \in[0,1]$$
  Show that $K$ is a continuous linear transformation $E\to F$ and find the operatornorm $C\ge0$, so $x\in E$, so that $\|K(f)\|_\infty\le C\|f\|_1$

I already showed the linearity. So I can use a theorem:
Let $X$ and $Y$ be normed spaces and $f:X\to Y$ is linear, then the following are equivalent:
i) $f$ is continuous
ii) $\|f\|_{L(X,Y)}:=\sup_{\|x\|_X{\le1}}\|T_X\|_Y<\infty$
In this context is $\|f\|_{L(X,Y)}$ the Operatornorm $C$.
I tried to do the following:
I suppose $\|f\|_1=1$ should be one condition for $C$ in order to be the supremum in $$\|K(f)\|_{L(X,Y)}:=s\rightarrow \left|\int_0^1k(s,t)f(t)\,dt\right|, \quad s\in [0,1]$$
$$\le s\rightarrow\int_0^1|k(s,t)f(t)|\,dt,\quad s\in[0,1]$$ $$\le s\rightarrow\int_0^1|k(s,t)||f(t)|\,dt,\quad s\in [0,1]$$
My question is: Is there a way to get rid of the $f(t)$ in the equation? I guess the final C should be $$C=\sup_{s\in[0,1]}\int_0^1|k(s,t)|\,dt$$ But I don't know how to do the final step(s). Can someone help me?

REPLY [2 votes]: Let $C = \sup \limits_{s,t \in [0,1]} \lvert k(s,t)\rvert$. Note that for any $f \in E$ we have
$$
\|K(f) \|_\infty = \sup_{s \in [0,1]} \big\lvert \int_0^1 k(s,t) f(t) dt \big\rvert.
$$
For any $s \in [0,1]$ we have
$$
\big\lvert \int_0^1 k(s,t) f(t) dt \big\rvert \leq  \int_0^1 \lvert k(s,t) f(t) \rvert dt \leq \int_0^1 C \lvert f(t)\rvert dt= C \|f\|_1.
$$
So we have $\|K\|_{L(E,F)} \leq C$.


TITLE: Definition of Left Invariant Vector Fields
QUESTION [1 upvotes]: The definition (according to Marsden and Ratiu's $\textit{Introduction to Mechanics and Symmetry}$) of a left invariant vector field $X$ on a Lie group $G$ states that for each $g \in G$, $L_g^*X=X$. That is, $$(T_hL_g)X(h)=X(gh)$$ for every $h\in G$. Other sources put $(dL_g)_h$ in place of $T_h$ (Jack Lee's $\textit{Intro to Smooth Manifolds}$ follows this latter convention). My question is regarding the notation $T_h$ and the result $X(gh)$. Why do we write $T_h$ as the partial derivative w.r.t $h$ and what does it actually mean to take the derivative of the left action as we are doing in this definition? It seems we start with a vector at $h$ and map to a vector at $gh$ but these books just state it and move on so some clarification would be helpful.

REPLY [2 votes]: Why do we write $T_h$ as the partial derivative w.r.t $h$ and what does it actually mean to take the derivative of the left action as we are doing in this definition?

$T_h$ is not at all a partial derivative with respect to $h$; note that in the setting of general manifolds, you cannot use the naive difference-quotient definition of derivatives, because in general, you cannot subtract points on a manifold.
I suggest you take a look at Chapter $4$ on manifolds, in particular where they define tangent vectors. The authors define the tangent mapping between manifolds as follows: Given a smooth map $f: M \to N$ between two smooth manifolds, and a point $m \in M$, the tangent mapping at $m$ is denoted by $T_mf: T_mM \to T_{f(m)}N$, and is a linear map between tangent spaces defined by the rule
\begin{align}
(T_mf)([c]) := [f \circ c]
\end{align}
In words, it takes an equivalence class of curves $[c]$ in $T_mM$ and sends it to the equivalence class of curves $[f \circ c]$ in $T_{f(m)}N$ (recall that they define a tangent vector to be an equivalence class of smooth curves).
So, in your context, the notation $T_h$ does not mean partial derivatives at all. The symbol $T_h(L_g)$ means the tangent mapping at $h$ of the left-translation $L_g$, in the sense I described above. So, the way to "read" the formula
\begin{align}
(T_hL_g)X(h) = X(gh)
\end{align}
is as follows: first of all, $X$ is a vector field on the Lie group $G$, which means at each point $p\in G$, $X(p)$ is a tangent vector in the tangent space $T_pG$. Next, left-translation by $g$ is a smooth map $L_g:G \to G$, and we are taking its tangent at the point $h$. So, $T_h(L_g)$ is a linear mapping of $T_hG \to T_{L_g(h)}G = T_{gh}G$, hence $T_h(L_g)$ can be evaluated on the tangent vector $X(h)$. So, if we use a lot of brackets, we would write this as:
\begin{align}
\big(T_h(L_g) \big)(X(h))
\end{align}
but omitting some brackets, we just write $(T_hL_g)X(h)$, and in words it says "the tangent mapping at $h$ of the smooth map $L_g$, evaluated on the tangent vector $X(h)$". The left-invariance condition of a vector field requires that all of this be equal to the tangent vector $X(gh)$.

Like you observed, there is a lot of different notation (and terminology) around; some authors use $T_mf$ to mean the tangent mapping at $m$, while other authors (e.g Lee) use $df_m$ to mean the linear map from $T_mM \to T_{f(m)}N$. Another thing you may notice is that the actual construction of tangent space varies from author to author, this can be confusing but for starters, the key thing you should focus on is what the formulas are really trying to say geometrically.


TITLE: Hodge classes and Leray filtration
QUESTION [4 upvotes]: Let $f :X \to Y$ be a submersion between smooth projective varieties over $\mathbb{C}$ and let $\alpha \in Z^k(X)$ be an algebraic cycle of $X$. Is is true that for all odd numbers $p$ and $q$ such that $p+q = 2k$, $\alpha$ lives in the kernel of $H^{2k}(X,\mathbb{Q}) \to  H^p(Y,R^qf_*\mathbb{Q})$?
The reason why I ask this question is the following:
Since $f$ is a submersion, the class $[\alpha]$ in $H^p(Y,R^qf_*\mathbb{Q})$ can be represented by an element in
$$f^*\Omega^p_{Y,\mathbb{R}} \otimes \Omega^{q-p}_{X/Y,\mathbb{R}}.$$
However, it seems to me that any class which can be represented by one of above elements cannot be of type $(k,k)$. Can someone explain what happens here? Thank you.

REPLY [4 votes]: No, $\alpha$ doesn't have to lie in that kernel. To construct an explicit counterexample,
take a surface $X$ mapping to a curve $Y$, $k=1$, and let $\alpha$ be nontrivial divisor
on $X$ not supported on the fibres such that $\alpha\cdot (fibre)=0$.
Then $[\alpha]$ will be supported in $H^1(Y, R^1f_\ast\mathbb{Q})$. To make this more explicit, you can even take
$X$ to be a product of $Y$ with another curve.
To resolve the seeming paradox, note that $R^qf_\ast\mathbb{Q}$ carries a variation of Hodge structure, and its cohomology $H^p(Y, R^qf_\ast\mathbb{Q})$ carries a a Hodge structure of weight $p+q$ compatible with the one on $H^*(X)$. The Hodge filtration is not completely
obvious, and in particular,  if $p,q$ are odd and $k=(p+q)/2$, then the $(k,k)$-part need not vanish. The details of the construction can be found in the beginning sections of Zucker, Hodge theory with degenerating coefficients..., Annals (1979).


TITLE: Finding coefficient of $x^{100}$
QUESTION [0 upvotes]: What will be solution of this function for coefficient of $x^{100}$?
$$\displaystyle\frac{1}{\left ( 1-x^{10} \right )(1-x^{20})(1-x^{50})}$$
My solution:
$[x^{100}] \ \ \ \ \ \ \ \ \  \ $$\displaystyle \frac{1}{\left ( 1-x^{10} \right )(1-x^{20})(1-x^{50})}$
$[x^{100}] \ \ \ \ \ \ \ \ \  \ $ $(1-x^{10})^{-1}$$(1-x^{20})^{-1}$$(1-x^{50})^{-1}$
$\begin{aligned}\displaystyle &\binom{-n}{k} = (-1)^r\binom{n + k - 1}{k}\\
&=\displaystyle \color{red} {(-1)^{10}\binom{-1}{10}\binom{-1}{0}\binom{-1}{0}   + (-1)^5 \binom{-1}{0}\binom{-1}{5}\binom{-1}{0} + (-1)^2\binom{-1}{0}\binom{-1}{0}\binom{-1}{2}} \\ & + \color{blue}{ (-1)^8 (-1)^1 \binom{-1}{8}\binom{-1}{1}\binom{-1}{0} + (-1)^6 (-1)^2\binom{-1}{6}\binom{-1}{2}\binom{-1}{0} + (-1)^4 (-1)^3 \binom{-1}{4}\binom{-1}{3}\binom{-1}{0} + (-1)^2 (-1)^4\binom{-1}{2}\binom{-1}{4}\binom{-1}{0} }\\ & + \color{Orange}{ (-1)^5 (-1)^1 \binom{-1}{5}\binom{-1}{0}\binom{-1}{1} } \\ & + \color{maroon}{(-1)^3(-1)^1(-1)^1\binom{-1}{3}\binom{-1}{1}\binom{-1}{1} + (-1)^1 (-1)^2(-1)^3\binom{-1}{1}\binom{-1}{2}\binom{-1}{1} }\end{aligned} $
$ 1 – 1 + 1 + 1 – 1 + 1 – 1 – 1 + 1 – 1 = 0$
So coefficient of $x^{100}$ is $0$
There are many terms so to differentiate among them distinct colors are used
$\color{red}{\text{Red}}$ - Selecting $x$ from any one of $(1-x^{10})^{-1}$$(1-x^{20})^{-1}$$(1-x^{50})^{-1}$
$\color{Blue}{\text{Blue}}$ - Selecting from both $(1-x^{10})^{-1}$$(1-x^{20})^{-1}$
$\color{Orange}{\text{Orange}}$ - Selecting from both $(1-x^{10})^{-1}$$(1-x^{50})^{-1}$
$\color{maroon}{\text{Maroon}}$ - Selecting from all $(1-x^{10})^{-1}$$(1-x^{20})^{-1}$$(1-x^{50})^{-1}$

But answer is $10$,
is anything wrong with my solution?

REPLY [0 votes]: Your mistake is in calculating $\binom{-1}k$. You had
$$
(-1)^{10}\binom{-1}{10}+(-1)^5\binom{-1}{5}+\dots=(-1)^{10}\cdot 1+(-1)^5\cdot \color{red}1+\dots
$$
The one in red is a mistake. $\binom{-1}{5}=\frac{(-1)(-2)\cdots(-5)}{5!}=(-1)^5=-1$. In fact, your summation should just be a summation of ten ones, because each summand is a product of numbers of the form
$$
(-1)^k\binom{-1}{k}=(-1)^k(-1)^k=(-1)^{2k}=1
$$
The answer is then
$1+1+1+1+1+1+1+1+1+1=10$.