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Diffstat (limited to 'main.tex')
| -rwxr-xr-x | main.tex | 212 |
1 files changed, 126 insertions, 86 deletions
@@ -149,8 +149,8 @@ \begin{itemize} \item Equilibrium vs nonequilibrium dynamics; \item Definition and computation of the mobility; - \item Computation of other transport coefficients. - \item Error analysis + \item Computation of other transport coefficients; + \item Error analysis. \end{itemize} \end{frame} @@ -435,14 +435,14 @@ The \textbf{minorization condition} is satisfied. Indeed for $t > 0$ \begin{align*} - p(x, A) - &= \expect \left[ q_t \in A \, \middle| \, q_0 = x \right] - = \expect \left[ \mathds 1_{A} \left(x + W_t \right) M_t \right] - && M_t = \text{Girsanov weight} \\ - &= \proba \left[ x + W_t \in A \right] \expect \left[ M_t \, | \, \{x + W_t \in A\} \right] \\ - &\geq C \proba \left[ x + W_t \in A \right] \geq C \lambda(A) && \lambda := \text{Lebesgue measure}. + \mathcal P^{\dagger}\mu (A) + &= \expect \left[ q_t \in A \, \middle| \, q_0 \sim \mu \right] + = \int_{\mathcal E} \int_{A} p_t(x, y) \, \mu(\d x) \, + && p_t = \text{transition pdf} \\ + &\geq \left( \inf_{(x,y) \in \mathcal E^2} p_t(x, y) \right) \lambda(A) && \lambda := \text{Lebesgue measure}. \end{align*} - and additionally ${\rm Law} (q_t)$ is smooth by parabolic regularity. + The infimum is achieved by parabolic regularity, + and achieved by {\blue Harnack's inequality}. \item \textbf{Decay of the semigroup}: For $t \in [0, \infty)$ and $\varphi \in L^{\infty}_*$, it holds that @@ -604,51 +604,90 @@ \begin{frame} {Elements of proof} - Let us introduce + Let $\Pi_0$ denote the following projection operator \[ - H^1_{p}(\psi_0) = - \Bigl\{ \varphi \in L^2(\psi_0) : \grad_p \varphi \in L^2(\psi) \Bigr\}, - \qquad \| \varphi \|_{H^1_{p}(\psi_0)}^2 = \| \varphi \|_{L^2(\psi_0)}^2 + \| \nabla_p \varphi \|_{L^2(\psi_0)}^2. + \Pi_0 f := f - \int_{\mathcal E} f \, \psi_0 \] \vspace{-.3cm} \begin{itemize} \itemsep.2cm - \item - The operator {\blue $\widetilde {\mathcal L}^*\colon H^1_p(\psi_0) \to L^2_0(\psi)$} is well-defined and bounded. - Indeed + \item + The operator $\mathcal L_0^{-1}$ is a well defined bounded operator on $L_0^2(\psi_0)$ \\ + + ({\red Hypocoercivity} + {\red hypoelliptic regularization}) + + \item Since $\dps \gamma \| \nabla_p \varphi \|^2_{L^2(\psi_0)} = -\langle \mathcal L_0 \varphi,\varphi \rangle_{L^2(\psi_0)}$, it follows that + \vspace{-0.2cm} \[ - \lVert \widetilde {\mathcal L}^* \varphi \rVert_{L^2_0(\psi_0)}^2 - = \ip{\nabla_p^* F \varphi}{\nabla_p^* F \varphi}_{L^2_0(\psi_0)} - \leq \lVert \varphi \rVert_{H^1_p(\psi_0)}^2 + \| \widetilde {\mathcal L} \varphi \|^2_{L^2(\psi_0)} \leq \| \nabla_p \varphi \|^2_{L^2(\psi_0)} \leq \frac{1}{\gamma} \| \mathcal L_0 \varphi \|_{L^2(\psi_0)} \| \varphi \|_{L^2(\psi_0)} \] - and + Thus {\blue $\Pi_0 \widetilde {\mathcal L} \mathcal L_0^{-1}$ is bounded on $L^2_0(\psi_0)$}. \[ - \int_{\mathcal E} \widetilde {\mathcal L}^* \phi \, \psi_0 - = \int_{\mathcal E} \nabla_p^* (F \phi) \, \psi_0 = 0. + \| \widetilde {\mathcal L} \mathcal L_0^{-1} \varphi \|^2_{L^2(\psi_0)}\leq \frac{\beta}{\gamma} \| \varphi \|_{L^2(\psi_0)} \| \mathcal L_0^{-1} \varphi \|_{L^2(\psi_0)}. \] - \item - The operator {\blue $(\mathcal L_0^*)^{-1} \colon L^2_0(\psi_0) \to H^1_p(\psi_0)$} is well-defined and bounded, - by {\red hypocoercivity} and {\red hypoelliptic regularization}. - % In particular, for $\phi = (\mathcal L_0^*)^{-1} \varphi$ - % \begin{align*} - % \| \phi \|_{L^2(\psi_0)}^2 - % + \| \nabla_p \phi \|_{L^2(\psi_0)}^2 - % &= \|(\mathcal L_0^*)^{-1} \varphi \|_{L^2(\psi_0)}^2 - % + \frac{1}{\gamma} \ip{-\mathcal L_0^* \phi}{\phi}_{L^2(\psi_0)} \\ - % &\leq \frac{1}{\gamma} \norm{(\mathcal L_0^*)^{-1}}_{\mathcal B\bigl(L^2(\psi_0)\bigr)}^2 - % \norm{\varphi}_{L^2(\psi_0)} - % \end{align*} + + + \item It follows that $(\widetilde {\mathcal L} \mathcal L_0^{-1})^* \Pi_0 = (\widetilde {\mathcal L} \mathcal L_0^{-1})^*$ is also bounded on $L^2_0(\psi_0)$ + + \medskip \item Invariance of $f_\eta$ by $\mathcal L_\eta^* = \mathcal L_0^* + \eta \wcL^*$ \vspace{-0.2cm} \[ - \mathcal L_\eta^* f_\eta = \mathcal L_0^* \left(1 + \eta (\mathcal L_0^*)^{-1} \widetilde {\mathcal L}^* \right) f_\eta = \mathcal L_0^* \mathbf{1} = 0. + \mathcal L_\eta^* f_\eta = \mathcal L_0^* \left(1 + \eta (\wcL \mathcal L_0^{-1})^* \right) f_\eta = \mathcal L_0^* \mathbf{1} = 0 \] \item {\red Prove that $f_\eta \geq 0$}. \end{itemize} \end{frame} +% \begin{frame} +% {Elements of proof} +% Let us introduce +% \[ +% H^1_{p}(\psi_0) = +% \Bigl\{ \varphi \in L^2(\psi_0) : \grad_p \varphi \in L^2(\psi) \Bigr\}, +% \qquad \| \varphi \|_{H^1_{p}(\psi_0)}^2 = \| \varphi \|_{L^2(\psi_0)}^2 + \| \nabla_p \varphi \|_{L^2(\psi_0)}^2. +% \] +% \vspace{-.3cm} +% \begin{itemize} +% \itemsep.2cm +% \item +% The operator {\blue $\widetilde {\mathcal L}^*\colon H^1_p(\psi_0) \to L^2_0(\psi)$} is well-defined and bounded. +% Indeed +% \[ +% \lVert \widetilde {\mathcal L}^* \varphi \rVert_{L^2_0(\psi_0)}^2 +% = \ip{\nabla_p^* F \varphi}{\nabla_p^* F \varphi}_{L^2_0(\psi_0)} +% \leq \lVert \varphi \rVert_{H^1_p(\psi_0)}^2 +% \] +% and +% \[ +% \int_{\mathcal E} \widetilde {\mathcal L}^* \phi \, \psi_0 +% = \int_{\mathcal E} \nabla_p^* (F \phi) \, \psi_0 = 0. +% \] +% \item +% The operator {\blue $(\mathcal L_0^*)^{-1} \colon L^2_0(\psi_0) \to H^1_p(\psi_0)$} is well-defined and bounded, +% by {\red hypocoercivity} and {\red hypoelliptic regularization}. +% % In particular, for $\phi = (\mathcal L_0^*)^{-1} \varphi$ +% % \begin{align*} +% % \| \phi \|_{L^2(\psi_0)}^2 +% % + \| \nabla_p \phi \|_{L^2(\psi_0)}^2 +% % &= \|(\mathcal L_0^*)^{-1} \varphi \|_{L^2(\psi_0)}^2 +% % + \frac{1}{\gamma} \ip{-\mathcal L_0^* \phi}{\phi}_{L^2(\psi_0)} \\ +% % &\leq \frac{1}{\gamma} \norm{(\mathcal L_0^*)^{-1}}_{\mathcal B\bigl(L^2(\psi_0)\bigr)}^2 +% % \norm{\varphi}_{L^2(\psi_0)} +% % \end{align*} +% +% \item Invariance of $f_\eta$ by $\mathcal L_\eta^* = \mathcal L_0^* + \eta \wcL^*$ +% \vspace{-0.2cm} +% \[ +% \mathcal L_\eta^* f_\eta = \mathcal L_0^* \left(1 + \eta (\mathcal L_0^*)^{-1} \widetilde {\mathcal L}^* \right) f_\eta = \mathcal L_0^* \mathbf{1} = 0. +% \] +% +% \item {\red Prove that $f_\eta \geq 0$}. +% \end{itemize} +% \end{frame} + \begin{frame} {Perturbation expansion for {\yellow $\eta$ sufficiently small} (3/3)} @@ -1064,7 +1103,7 @@ The Hamiltonian of the system is the sum of the potential and kinetic energies: \begin{frame} - {Shear viscosity in fluids (2)} + {Shear viscosity in fluids (3)} Assume pairwise interactions \[ V(q) = \sum_{1 \leq i < j \leq N} \mathcal V(\abs{q_i - q_j}). @@ -1092,7 +1131,7 @@ The Hamiltonian of the system is the sum of the potential and kinetic energies: \begin{frame} - {Shear viscosity in fluids (3)} + {Shear viscosity in fluids (4)} \bu {\blue Linear response}: \[ @@ -1362,58 +1401,10 @@ We verify the error estimate for $\varphi \in \mathrm{Ran}(P_\dt-\I)$. \begin{block}{} Suggests $f_{p+1} = -\left( \mathcal A_1^* \right)^{-1} g_{p+1}$ \end{block} - \end{frame} \begin{frame} - {Numerical estimators and associated challenges} - \begin{itemize} - \item - Estimator of linear response (observable~$R$ with equilibrium average~0) - \[ - \widehat{A}_{\eta,t} = \frac{1}{\eta t}\int_0^t R(q_s^\eta,p_s^\eta) \, ds \xrightarrow[t\to+\infty]{\mathrm{a.s.}} - \alpha_\eta := \frac1\eta \int_{\mathcal E} R \, f_\eta \, d\mu = \alpha + \mathcal O(\eta) - \] - {\bf Issues with linear response methods:} - \begin{itemize} - \item Statistical error with {\red asymptotic variance $\mathcal O(\eta^{-2})$} - \item Bias $\mathcal O(\eta)$ due to $\eta \neq 0$ - \item Bias from finite integration time - \end{itemize} - - \end{itemize} -\end{frame} - -\begin{frame}\frametitle{Analysis of variance / finite integration time bias} - - \bu {\bf Statistical error} dictated by {\blue Central Limit Theorem}: - \[ - \sqrt{t} \left(\widehat{A}_{\eta,t} - \alpha_\eta \right) \xrightarrow[t \to +\infty]{\mathrm{law}} \mathcal{N}\left(0,\frac{\sigma_{R,\eta}^2}{\eta^2}\right), - \qquad - \sigma_{R,\eta}^2 = \sigma_{R,0}^2 + \mathcal O(\eta) - \] - so $\dps \widehat{A}_{\eta,t} = \alpha_\eta + \mathcal O_{\rm P}\left(\frac{1}{\eta \sqrt{t}}\right)$ $\to$ requires {\red long simulation times} $t \sim \eta^{-2}$ - - \bigskip - - \bu {\bf Finite time integration bias}: $\dps \left| \mathbb{E}\left(\widehat{A}_{\eta,t}\right) - \alpha_\eta \right| \leq \frac{K}{\eta t}$ \\ - Bias due to $t < +\infty$ is $\dps \mathcal O\left(\frac{1}{\eta t}\right)$ $\to$ typically {\red smaller than statistical error} - -%\bigskip - %\bu Bias~$\mathcal O(\eta)$ and statistical error equilibrated for~$t \sim \eta^{-3}$ - -\bigskip - -\bu Key equality for the proofs: introduce $\dps -\left(\mathcal{L}+\eta\widetilde{\mathcal{L}}\right) \mathscr{R}_\eta = R - \int_\mathcal{E} R f_\eta \, d\mu$ -\[ -\widehat{A}_{\eta,t} - \frac1\eta \!\int_{\mathcal{E}} \!R f_\eta \, d\mu = \frac{\mathscr{R}_\eta(q_0^\eta,p_0^\eta) - \mathscr{R}_\eta(q_t^\eta,p_t^\eta)}{\eta t} + \frac{\sqrt{2\gamma}}{\eta t\sqrt{\beta}} \int_0^t \!\!\nabla_p \mathscr{R}_\eta(q_s^\eta,p_s^\eta)^T dW_s -\] - -\end{frame} - - - -\begin{frame}\frametitle{Examples of splitting schemes for Langevin dynamics (1)} + {Examples of splitting schemes for Langevin dynamics (1)} \bu Example: Langevin dynamics, discretized using a {\blue splitting} strategy \[ @@ -1475,6 +1466,55 @@ p^{n+1} & = \alpha_{\dt/2} \widetilde{p}^{n+1} + \sqrt{\frac{1-\alpha_{\dt}}{\be \end{frame} +\begin{frame} + {Numerical estimators and associated challenges} + \begin{itemize} + \item + Estimator of linear response (observable~$R$ with equilibrium average~0) + \[ + \widehat{A}_{\eta,t} = \frac{1}{\eta t}\int_0^t R(q_s^\eta,p_s^\eta) \, ds \xrightarrow[t\to+\infty]{\mathrm{a.s.}} + \alpha_\eta := \frac1\eta \int_{\mathcal E} R \, f_\eta \, d\mu = \alpha + \mathcal O(\eta) + \] + {\bf Issues with linear response methods:} + \begin{itemize} + \item Statistical error with {\red asymptotic variance $\mathcal O(\eta^{-2})$} + \item Bias $\mathcal O(\eta)$ due to $\eta \neq 0$ + \item Bias from finite integration time + \end{itemize} + + \end{itemize} +\end{frame} + +\begin{frame}\frametitle{Analysis of variance / finite integration time bias} + + \bu {\bf Statistical error} dictated by {\blue Central Limit Theorem}: + \[ + \sqrt{t} \left(\widehat{A}_{\eta,t} - \alpha_\eta \right) \xrightarrow[t \to +\infty]{\mathrm{law}} \mathcal{N}\left(0,\frac{\sigma_{R,\eta}^2}{\eta^2}\right), + \qquad + \sigma_{R,\eta}^2 = \sigma_{R,0}^2 + \mathcal O(\eta) + \] + so $\dps \widehat{A}_{\eta,t} = \alpha_\eta + \mathcal O_{\rm P}\left(\frac{1}{\eta \sqrt{t}}\right)$ $\to$ requires {\red long simulation times} $t \sim \eta^{-2}$ + + \bigskip + + \bu {\bf Finite time integration bias}: $\dps \left| \mathbb{E}\left(\widehat{A}_{\eta,t}\right) - \alpha_\eta \right| \leq \frac{K}{\eta t}$ \\ + Bias due to $t < +\infty$ is $\dps \mathcal O\left(\frac{1}{\eta t}\right)$ $\to$ typically {\red smaller than statistical error} + +%\bigskip + %\bu Bias~$\mathcal O(\eta)$ and statistical error equilibrated for~$t \sim \eta^{-3}$ + +\bigskip + +\bu Key equality for the proofs: introduce $\dps -\left(\mathcal{L}+\eta\widetilde{\mathcal{L}}\right) \mathscr{R}_\eta = R - \int_\mathcal{E} R f_\eta \, d\mu$ +\[ +\widehat{A}_{\eta,t} - \frac1\eta \!\int_{\mathcal{E}} \!R f_\eta \, d\mu = \frac{\mathscr{R}_\eta(q_0^\eta,p_0^\eta) - \mathscr{R}_\eta(q_t^\eta,p_t^\eta)}{\eta t} + \frac{\sqrt{2\gamma}}{\eta t\sqrt{\beta}} \int_0^t \!\!\nabla_p \mathscr{R}_\eta(q_s^\eta,p_s^\eta)^T dW_s +\] + +\end{frame} + + + + \begin{frame}\frametitle{Error estimates on linear response} \begin{block}{Error estimates for nonequilibrium dynamics} |