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-
- \author{ {\fontspec{Trebuchet MS}Marcin Chrz\k{a}szcz} (Universit\"{a}t Z\"{u}rich)}
- \institute{UZH}
- \title[Specific \pdf~generation]{Specific \pdf~generation}
- \date{\fixme}
- \newcommand*{\QEDA}{\hfill\ensuremath{\blacksquare}}%
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-
- \author{ {\fontspec{Trebuchet MS}Marcin Chrz\k{a}szcz} (Universit\"{a}t Z\"{u}rich)}
- \institute{UZH}
- \title[Matrix inversion and Partial Differential Equation Solving]{Matrix inversion and Partial Differential Equation Solving}
- \date{\fixme}
-
-
- \begin{document}
- \tikzstyle{every picture}+=[remember picture]
-
- {
- \setbeamertemplate{sidebar right}{\llap{\includegraphics[width=\paperwidth,height=\paperheight]{bubble2}}}
- \begin{frame}[c]%{\phantom{title page}}
- \begin{center}
- \begin{center}
- \begin{columns}
- \begin{column}{0.9\textwidth}
- \flushright\fontspec{Trebuchet MS}\bfseries \Huge {Matrix inversion and Partial Differential Equation Solving}
- \end{column}
- \begin{column}{0.2\textwidth}
- %\includegraphics[width=\textwidth]{SHiP-2}
- \end{column}
- \end{columns}
- \end{center}
- \quad
- \vspace{3em}
- \begin{columns}
- \begin{column}{0.44\textwidth}
- \flushright \vspace{-1.8em} {\fontspec{Trebuchet MS} \Large Marcin Chrząszcz\\\vspace{-0.1em}\small \href{mailto:mchrzasz@cern.ch}{mchrzasz@cern.ch}}
-
- \end{column}
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- \includegraphics[height=1.3cm]{uzh-transp}
- \end{column}
- \end{columns}
-
- \vspace{1em}
- % \footnotesize\textcolor{gray}{With N. Serra, B. Storaci\\Thanks to the theory support from M. Shaposhnikov, D. Gorbunov}\normalsize\\
- \vspace{0.5em}
- \textcolor{normal text.fg!50!Comment}{Monte Carlo methods, \\ 28 April, 2016}
- \end{center}
- \end{frame}
- }
-
- \begin{frame}\frametitle{Announcement}
-
- \begin{Large}
- There will be no lectures and class on 19$^{th}$ of May
- \end{Large}
-
- \end{frame}
-
-
- \begin{frame}\frametitle{Matrix inversion}
- \begin{minipage}{\textwidth}
- \begin{footnotesize}
-
- \ARROW The last time we discussed the method of linear equations solving. The same methods can be used for matrix inversions! The columns of inverse matrix can be found solving:
- \begin{align*}
- \textbf{A}\overrightarrow{x}= \hat{e}_i,~~~i=1,2,...,n
- \end{align*}
- %where $\hat{e}_i$ is the $i^{th}$ versor. \\
- \ARROW In order to determine the inverse of a matrix $\textbf{A}$ we need to choose a temprorary matrix $\textbf{M}$ such that:
- \begin{align*}
- \textbf{H}=\textbf{I}-\textbf{M}\textbf{A}
- \end{align*}
- with the normalization condition:
- \begin{align*}
- \Vert \textbf{H} \Vert = \max_{1 \leq i \leq n} \sum_{j=1}^n \vert h_{ij} \vert < 1
- \end{align*}
- where $\textbf{I}$ is a unit matrix.\\
- \ARROW Next we Neumann expand the $(\textbf{MA})^{-1}$ matrix:
- \begin{align*}
- (\textbf{MA})^{-1}=(\textbf{I}-\textbf{H})^{-1}=\textbf{I}+\textbf{H}
- +\textbf{H}^2+....
- \end{align*}
- \ARROW The inverse matrix we get from the equation:
- \begin{align*}
- \textbf{A}^{-1}=\textbf{A}^{-1} \textbf{M}^{-1} \textbf{M}=(\textbf{MA})^{-1}\textbf{M}
- \end{align*}
-
-
- \end{footnotesize}
-
- \end{minipage}
-
- \end{frame}
- %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
-
- \begin{frame}\frametitle{Matrix inversion, basic method}
- \begin{minipage}{\textwidth}
- \begin{footnotesize}
- \ARROW For the $(i,j)$ element of the matrix $(MA)^{-1}$ we have:
- \begin{align*}
- (MA)^{-1}_{ij} = \delta_{ij} + h_{ij} + \sum_{i_1=1}^n h_{i i_1} h_{i_1 j} + \sum_{i_1=1}^n \sum_{i_2=1}^n h_{i i_1} h_{i_1 i_2}h_{i_2 j} + ...
- \end{align*}
- \ARROW The algorithm:
- We choose freely a probability matrix $P=(p_{ij})$ with the conditions:
- \begin{align*}
- p_{i,j}\geq 0,~~~~ p_{ij}=0 \Leftrightarrow h_{ij}=0,~~~~p_{i,0}=1-\sum_{j=1}^np_{ij} >0
- \end{align*}
- \ARROW We construct a random walk for the state set $\lbrace 0,1,2,3...,n \rbrace$:
- \begin{enumerate}
- \item In the initial moment $(t=0)$ we start in the state $i_0=i$.
- \item If in the moment $t$ the point is in the $i_t$ state, then in the time $t+1$ he will be in state $i_{t+1}$ with the probability $p_{i_t,t_{t+1}}$.
- \item We stop the walk if we end up in the state $0$.
- \end{enumerate}
- \end{footnotesize}
-
- \end{minipage}
-
- \end{frame}
-
-
- \begin{frame}\frametitle{Matrix inversion, basic method}
- \begin{minipage}{\textwidth}
- \begin{footnotesize}
- \ARROW For the observed trajectory $\gamma_k=(i,i_1,..,j_k,0)$ we assign the value of:
- \begin{align*}
- X(\gamma_k)=\frac{ h_{ii_1} h_{i_1 i_2}... h_{i_{k-1} i_k} ~~ \delta_{i_k j }}{ p_{ii_1} p_{i_1 i_2}... p_{i_{k-1} i_k} ~~p_{i_k 0} }
- \end{align*}
- \ARROW The mean is the of all observed $X(\gamma_k)$ is an unbiased estimator of the $(MA)^{-1}_{ij}$.\\
- \begin{exampleblock}{Prove:}
- \begin{itemize}
- \item The probability of observing the $\gamma_k$ trajectory:
- \begin{align*}
- P(\gamma_k) = p_{i i_1} p_{i_1 i_2}... p_{i_{k-1} i_k} p_{i_k 0}
- \end{align*}
- \item Form this point we follow the prove of the previous lecture (Neumann-Ulan) and prove that:
- \begin{align*}
- E \lbrace X(\gamma_k) \rbrace = (MA)^{-1}
- \end{align*}
-
- \end{itemize}
- \end{exampleblock}
- \ARROW A different estimator for the $(MA)^{-1}_{ij}$ element is the Wasow estimator:
- \begin{align*}
- X^{\ast} (\gamma_k) = \sum_{m=0}^k \frac{ h_{ii_1} h_{i_1 i_2}... h_{i_{m-1} i_m} } { p_{ii_1} p_{i_1 i_2}... p_{i_{m-1} i_m} } \delta_{i_m j}
- \end{align*}
-
-
- \end{footnotesize}
-
- \end{minipage}
-
- \end{frame}
-
-
-
- \begin{frame}\frametitle{Matrix inversion, dual method}
- \begin{minipage}{\textwidth}
- \begin{footnotesize}
- \ARROW On the set of states $\lbrace 0, 1, 2,...,n \rbrace$ we set a binned \pdf~
- \begin{align*}
- q_1,q_2,...,q_n~{ \rm such~that~}q_i>0,~i=1,2,3...n{\rm ~and~} \sum_{i=1}^n q_i =1.
- \end{align*}
- \ARROW Then choose arbitrary the probability matrix $P$ (usual restrictions apply):
- \begin{itemize}
- \item The initial point we choose with the probability ${q_i}$.
- \item If in the moment $t$ the point is in the $i_t$ state, then in the time $t+1$ he will be in state $i_{t+1}$ with the probability $p_{i_t,t_{t+1}}$.
- \item The walk ends when we reach $0$ state.
- \item For the trajectory we assign a matrix:
- \end{itemize}
- \begin{align*}
- Y(\gamma_k)=\frac{ h_{i_1 i} h_{i_2 i_1}... h_{i_k i_{k-1}} }{ p_{i_1 i} p_{i_2 i_1}... p_{i_k i_{k-1}} } \frac{1}{q_{i_0}p_{i_k 0} } e_{i_k i_0} \in \mathbb{R}^n \times\mathbb{R}^n
- \end{align*}
- \ARROW The mean of $Y(\gamma)$ is an unbiased estimator of the $(MA)^{-1}$ matrix.\\
- \ARROW The Wasow estimator reads:
- \begin{align*}
- Y^{\ast}=\sum_{m=0}^k \frac{ h_{i_1 i} h_{i_2 i_1}... h_{i_m i_{m-1}} }{ p_{i_1 i} p_{i_2 i_1}... p_{i_m i_{m-1}} } e_{i_m i_0} \in \mathbb{R}^n \times\mathbb{R}^n
- \end{align*}
- \end{footnotesize}
-
- \end{minipage}
-
- \end{frame}
-
-
-
-
- \begin{frame}\frametitle{Partial differential equations, intro}
- \begin{minipage}{\textwidth}
- \begin{footnotesize}
- \ARROW Let's say we are want to describe a point that walks on the $\mathbb{R}$ axis:
- \begin{itemize}
- \item At the beginning $(t=0)$ the particle is at $x=0$
- \item If in the $t$ the particle is in the $x$ then in the time $t+1$ it walks to $x+1$ with the known probability $p$ and to the point $x-1$ with the probability $q=1-p$.
- \item The moves are independent.
- \end{itemize}
- \ARROW So let's try to described the motion of the particle. \\
- \ARROW The solution is clearly a probabilistic problem. Let $\nu(x,t)$ be a probability that at time $t$ particle is in position $x$. We get the following equation:
- \begin{align*}
- \nu(x,t+1)=p \nu(x-1,t)+q \nu(x+1,t)
- \end{align*}
- with the initial conditions:
- \begin{align*}
- \nu(0,0)=1,~~~~~\nu(x,0)=0~~{\rm if~}x \neq 0.
- \end{align*}
- \ARROW The above functions describes the whole system (every $(t,x)$ point).
- \end{footnotesize}
-
- \end{minipage}
-
- \end{frame}
-
-
-
- \begin{frame}\frametitle{Partial differential equations, intro}
- \begin{minipage}{\textwidth}
- \begin{tiny}
- \ARROW Now in differential equation language we would say that the particle walks in steps of $\Delta x$ in times: $k\Delta t$, $k=1,2,3....$:
- \begin{align*}
- \nu(x,t+\Delta t)=p\nu(x-\Delta x,t)+q\nu(x+\Delta x,t).
- \end{align*}
- \ARROW To solve this equation we need to expand the $\nu(x,t)$ funciton in the Taylor series:
- \begin{align*}
- \nu(x,t) + \frac{\partial \nu(x,t)}{\partial t} \Delta t = p \nu(x,t) - p \frac{\partial\nu(x,t) }{\partial x} \Delta x + \frac{1}{2} p \frac{\partial^2 \nu(x,t)}{\partial x^2} (\Delta x)^2\\ + q \nu(x,t) + q \frac{\partial\nu(x,t) }{\partial x} \Delta x + \frac{1}{2} q \frac{\partial^2 \nu(x,t)}{\partial x^2} (\Delta x)^2
- \end{align*}
- \ARROW From which we get:
- \begin{align*}
- \frac{\partial \nu(x,t)}{\partial t} \Delta t = -(p-q) \frac{\partial \nu(x,t) }{\partial x}\Delta x + \frac{1}{2} \frac{\partial^2 \nu(x,t) }{\partial x^2}(\Delta x)^2
- \end{align*}
- \ARROW Now We divide the equation by $\Delta t$ and take the $\Delta t \to 0$:
- \begin{align*}
- (p-q) \frac{\Delta x }{\Delta t} \to 2 c,~~~~~~\frac{ (\Delta x)^2}{\Delta t } \to 2D,
- \end{align*}
- \ARROW We get the Fokker-Planck equation for the diffusion with current:
- \begin{align*}
- \frac{\partial \nu(x,t)}{\partial t } = -2c \frac{\partial \nu(x,t) }{\partial x} + D \frac{\partial^2 \nu(x,t)}{\partial x^2}
- \end{align*}
- \ARROW The $D$ is the diffusion coefficient, $c$ is the speed of current. For $c=0$ it is a symmetric distribution.
-
- \end{tiny}
-
- \end{minipage}
-
- \end{frame}
-
-
-
-
-
-
- \begin{frame}\frametitle{Laplace equation, Dirichlet boundary conditions}
- \begin{minipage}{\textwidth}
- \begin{footnotesize}
- \ARROW The aforementioned example shows the way to solve the partial differential equation using Markov Chain MC. \\
- \ARROW We will see how different classes of partial differential equations can be approximated with a Markov Chain MC, whose expectation value is the solution of the equation.
- \ARROW The Laplace equation:
- \begin{align*}
- \frac{\partial^2 u }{\partial x_1^2 } +\frac{\partial^2 u }{\partial x_2^2 }+...+\frac{\partial^2 u }{\partial x_k^2 }=0
- \end{align*}
- The $u(x_1,x_2,...,x_k)$ function that is a solution of above equation we call harmonic function. If one knows the values of the harmonic function on the edges $\Gamma(D)$ of the $D$ domain one can solve the equation.\\
- \begin{exampleblock}{The Dirichlet boundary conditions:}
- Find the values of $u(x_1,x_2,...,x_k)$ inside the $D$ domain knowing the values of the edge are given with a function:
- \begin{align*}
- u(x_1,x_2,...,x_k)=f(x_1,x_2,...,x_k) \in \Gamma(D)
- \end{align*}
- \end{exampleblock}
- \ARROW Now I am lazy so I put $k=2$ but it's the same for all k!
-
- \end{footnotesize}
-
- \end{minipage}
-
- \end{frame}
-
-
-
-
- \begin{frame}\frametitle{Laplace equation, Dirichlet boundary conditions}
- \begin{minipage}{\textwidth}
- \begin{footnotesize}
- \begin{columns}
- \column{0.1in}
- {~}\\
- \column{3in}
- \ARROW We will put the Dirichlet boundary condition as a discrete condition:\\
- \begin{itemize}
- \item The domain $D$ we put a lattice with distance $h$.
- \item Some points we treat as inside {\color{green}(denoted with circles)}. Their form a set denoted $D^{\ast}$.
- \item The other points we consider as the boundary points and they form a set $\Gamma(D)$.
- \end{itemize}
-
- \column{2in}
- \begin{center}
- \includegraphics[width=0.95\textwidth]{images/dir.png}
- \end{center}
-
- \end{columns}
- \ARROW We express the second derivatives with the discrete form:
- \begin{align*}
- \frac{ \frac{u(x+h)-u(x)}{h} -\frac{u(x)-u(x-h) }{h} }{h} = \frac{u(x+h)-2u(x)+u(x-h)}{h^2}
- \end{align*}
- \ARROW Now we choose the units so $h=1$.
-
- \end{footnotesize}
- \end{minipage}
- \end{frame}
-
-
- \begin{frame}\frametitle{Laplace equation, Dirichlet boundary conditions}
- \begin{minipage}{\textwidth}
- \begin{footnotesize}
- \begin{exampleblock}{The Dirichlet condition in the discrete form:}
- Find the $u^{\ast}$ function which obeys the differential equation:
- \begin{align*}
- U^{\ast}(x,y)=\frac{1}{4}\left[ u^{\ast}(x-1,y)+u^{\ast}(x+1,y)+u^{\ast}(x,y-1)+u^{\ast}(x,y+1) \right]
- \end{align*}
- in all points $(x,y) \in D^{\ast}$ with the condition:
- \begin{align*}
- u^{\ast}(x,y)=f^{\ast}(x,y),~~~(x,y) \in \Gamma(D^{\ast})
- \end{align*}
- where $f^{\ast}(x,y)$ is the discrete equivalent of $f(x,y)$ function.
- \end{exampleblock}
- \ARROW We consider a random walk over the lattice $D^{\ast} \cup \Gamma(D^{\ast})$.
- \begin{itemize}
- \item In the $t=0$ we are in some point $(\xi,\eta) \in D^{\ast})$
- \item If at the $t$ the particle is in $(x,y)$ then at $t+1$ it can go with equal probability to any of the four neighbour lattices: $(x-1,y)$, $(x+1,y)$, $(x,y-1)$, $(x,y+1)$.
- \item If the particle at some moment gets to the edge $\Gamma(D^{\ast}$ then the walk is terminated.
- \item For the particle trajectory we assign the value of: $\nu(\xi,\eta)=f^{\ast}(x,y)$, where $(x,y)\in \Gamma(D^{\ast})$.
- \end{itemize}
- \end{footnotesize}
- \end{minipage}
- \end{frame}
-
-
-
-
- \begin{frame}\frametitle{Laplace equation, Dirichlet boundary conditions}
- \begin{minipage}{\textwidth}
- \begin{footnotesize}
- \ARROW Let $p_{\xi,\eta}(x,y)$ be the probability of particle walk that starting in $(\xi,\eta)$ to end the walk in $(x,y)$.\\
- \ARROW The possibilities:
- \begin{enumerate}
- \item The point $(\xi,\eta) \in \Gamma(D^{\ast})$. Then:
- \begin{align}
- p_{\xi,\eta}(x,y)=\begin{cases}
- 1,~~(x,y)=\xi,\eta)\\
- 0,~~(x,y)\neq \xi,\eta)
- \end{cases}\label{eq:trivial}
- \end{align}
- \item The point $(\xi,\eta) \in D^{\ast}$:
- \begin{align}
- p_{\xi,\eta}(x,y) = \frac{1}{4}\left[ p_{\xi-1,\eta}(x,y) + p_{\xi+1,\eta}(x,y)+ p_{\xi,\eta-1}(x,y)+ p_{\xi,\eta+1}(x,y) \right]
- \label{eq:1}
- \end{align}
-
-
- \end{enumerate}
- this is because to get to $(x,y)$ the particle has to walk through one of the neighbours: $(x-1,y)$, $(x+1,y)$, $(x,y-1)$, $(x,y+1)$.\\
- \ARROW The expected value of the $\nu(\xi,\eta)$ is given by equation:
- \begin{align}
- E(\xi,\eta)=\sum_{(x,y)\in \Gamma^{\ast}} p_{\xi,\eta}(x,y) f^{\ast}(x,y)\label{eq:2}
- \end{align}
- where the summing is over all boundary points
- \end{footnotesize}
- \end{minipage}
- \end{frame}
-
-
-
-
-
- \begin{frame}\frametitle{Laplace equation, Dirichlet boundary conditions}
- \begin{minipage}{\textwidth}
- \begin{footnotesize}
- \ARROW Now multiplying the \ref{eq:1} by $f^{\ast}(x,y)$ and summing over all edge points $(x,y)$:
- \begin{align*}
- E(\xi,\eta)=\frac{1}{4}\left[ E(\xi-1,\eta) + E(\xi+1,\eta) + E(\xi,\eta-1) + E(\xi,\eta+1) \right]
- \end{align*}
- \ARROW Putting now \ref{eq:trivial} to \ref{eq:2} one gets:
- \begin{align*}
- E(x,y)=f^{\ast}(x,y),~~(\xi,\eta) \in \Gamma(D^{\ast})
- \end{align*}
- \ARROW Now the expected value solves identical equation as our $u^{\ast}(x,y)$ function. From this we conclude:
- \begin{align*}
- E(x,y)=u^{\ast}(x,y)
- \end{align*}
- \ARROW The algorithm:
- \begin{itemize}
- \item We put a particle in $(x,y)$.
- \item We observe it's walk up to the moment when it's on the edge $\Gamma(D^{\ast})$.
- \item We calculate the value of $f^{\ast}$ function in the point where the particle stops.
- \item Repeat the walk $N$ times taking the average afterwards.
- \end{itemize}
- \begin{alertblock}{Important:}
- One can show the the error does not depend on the dimensions!
- \end{alertblock}
-
- \end{footnotesize}
- \end{minipage}
- \end{frame}
-
-
-
- \begin{frame}\frametitle{Example}
-
- \begin{minipage}{\textwidth}
- \begin{footnotesize}
- Let function $u(x,y)$ be a solution of Laplace equation in the square: $0 \leq x,y \leq 4$ with the boundary conditions:
- \begin{align*}
- u(x,0)=0,~~~u(4,y)=y,~~~u(x,4)=x,~~~x(0,y)=0
- \end{align*}
- \ARROWR Find the $u(2,2)$!\\
- \ARROW The exact solution: $u(x,y)=xy/4$ so $u(2,2)=1$.
- \begin{columns}
- \column{0.1in}
- {~}\\
- \column{3in}
- \begin{itemize}
- \item We transform the continues problem to a discrete one with $h=1$.
- \item Perform a random walk starting from $(2,2)$ which ends on the edge assigning as a result the appropriative values of the edge conditions as an outcome.
- \end{itemize}
-
- \column{2in}
- \begin{center}
- \includegraphics[width=0.95\textwidth]{images/problem1.png}
- \end{center}
-
- \end{columns}
- \ARROW E9.1 Implement the above example and find $u(2,2)$.
-
- \end{footnotesize}
- \end{minipage}
- \end{frame}
-
-
-
- \begin{frame}\frametitle{Parabolic equation}
-
- \begin{minipage}{\textwidth}
- \begin{footnotesize}
- \ARROW We are looking for a function $u(x_1,x_2,...,x_k,t)$, which inside the $D \subset \mathbb{R}^k$ obeys the parabolic equation:
- \begin{align*}
- \frac{\partial^2 u }{\partial x_1^2 } +\frac{\partial^2 u }{\partial x_2^2 }+...+\frac{\partial^2 u }{\partial x_k^2 }=c \frac{\partial u}{\partial t}
- \end{align*}
- with the boundary conditions:
- \begin{align*}
- u(x_1,x_2,...,x_k,t)=g(x_1,x_2,...,x_k,t),~~~(x_1,x_2,x_3,...,x_k)\in \Gamma(D)
- \end{align*}
- and with the initial conditions:
- \begin{align*}
- u(x_1,x_2,...,x_k,0)=h(x_1,x_2,...,x_k,t),~~~(x_1,x_2,x_3,...,x_k)\in D
- \end{align*}
- \ARROW In the general case the boundary conditions might have also the derivatives. \\
- \ARROW We will find the solution to the above problem using random walk starting from 1-dim case and then generalize it for n-dim.
- \end{footnotesize}
- \end{minipage}
- \end{frame}
-
-
- \begin{frame}\frametitle{Parabolic equation, 1-dim}
-
- \begin{minipage}{\textwidth}
- \begin{footnotesize}
- \ARROW We are looking for a function $u(x,t)$, which satisfies the equation:
- \begin{align*}
- \frac{\partial^2 u }{\partial x^2 } = c \frac{\partial u}{\partial t}
- \end{align*}
- with the boundary conditions:
- \begin{align*}
- u(0,t)=f_1(t),~~u(a,t)=f_2(t)
- \end{align*}
- and with the initial conditions:
- \begin{align*}
- u(x,0)=g(x).
- \end{align*}
- \ARROW The above equation can be seen as describing the temperature of a line with time. We know the initial temperature in different points and we know that the temperature on the end points is know.\\
- \ARROW The above problem can be discreteized:
- \begin{align*}
- x=kh,~~h=\frac{a}{n},~k=1,2,...n~~~~~~~~t=jl,~j=0,1,2,3...,~l={\rm const}
- \end{align*}
- \ARROW The differential equation:
- \begin{align*}
- \frac{u(x+h,t-l) -2u(x,t-l)+u(x-h,t-l}{h^2})=c \frac{u(x,t)-u(x,t-l)}{l}
- \end{align*}
-
-
- \end{footnotesize}
- \end{minipage}
- \end{frame}
-
-
-
-
- \begin{frame}\frametitle{Parabolic equation, 1-dim}
-
- \begin{minipage}{\textwidth}
- \begin{footnotesize}
- \ARROW The steps we choose such that: $c h^2 = 2l$.\\
- \ARROW Then we obtain the equation:\\
- \begin{align*}
- u(x,t)=\frac{1}{2}u(x+h,t-l)+\frac{1}{2}u(x-h,t-l)
- \end{align*}
- \ARROW The value of function $u$ in the point $x$ and $t$ can be evaluated with the arithmetic mean form points: $x+h$ and $x-h$ in the previous time step.
- \ARROW The algorithm estimating the function in the time $\tau$ and point $\xi$:
- \begin{itemize}
- \item The particle we put in the point $\xi$ and a ''weight'' equal $\tau$.
- \item If in a given time step $t$ particle is at $x$ then with $50:50$ chances it can go to $x-h$ or $x+h$ and time $t-l$.
- \item The particle ends the walk in two situations:
- \begin{itemize}
- \item If it reaches the $x=0$ or $x=a$. In this case we assign to a given trajectory a value of $f_1(t)$ or $f_2(t)$, where $t$ is the actuall ''weight''.
- \item If the ''weight'' of the particle is equal zero. in this case we assign as a value of the trajectory the $g(x)$, where $x$ is the actual position of the particle.
- \end{itemize}
- \end{itemize}
-
- \end{footnotesize}
- \end{minipage}
- \end{frame}
-
-
-
- \begin{frame}\frametitle{Parabolic equation, 1-dim}
-
- \begin{minipage}{\textwidth}
- \begin{footnotesize}
- \ARROW Repeat the above procedure $N$ times. The expected value of a function $u$ in $(\xi,\tau)$ point is the mean of observed values.
-
- \begin{exampleblock}{Digression:}
- The 1-dim calse can be treated as a 2-dim $(x,t)$, where the area is unbounded in the $t$ dimension. The walk is terminated after maximum $\tau/l$ steps.
-
- \end{exampleblock}
- \begin{center}
- \includegraphics[width=0.6\textwidth]{images/par.png}
- \end{center}
-
- \end{footnotesize}
- \end{minipage}
- \end{frame}
-
-
-
-
-
- \begin{frame}\frametitle{Parabolic equation, 1-dim}
-
- \begin{minipage}{\textwidth}
- \begin{footnotesize}
- \ARROW Repeat the above procedure $N$ times. The expected value of a function $u$ in $(\xi,\tau)$ point is the mean of observed values.
-
- \begin{exampleblock}{Digresion:}
- The 1-dim calse can be treated as a 2-dim $(x,t)$, where the area is unbounded in the $t$ dimension. The walk is terminated after maximum $\tau/l$ steps.
-
- \end{exampleblock}
- \begin{center}
- \includegraphics[width=0.6\textwidth]{images/par.png}
- \end{center}
-
- \end{footnotesize}
- \end{minipage}
- \end{frame}
-
-
- \begin{frame}\frametitle{Parabolic equation, n-dim generalization}
-
- \begin{minipage}{\textwidth}
- \begin{footnotesize}
- \ARROW We still choose the $k$ and $l$ values accordingly to:
- \begin{align*}
- \frac{ch^2}{l}=2k
- \end{align*}
- where $k$ is the number of space dimensions.\\
- \ARROW We get:
- \begin{align*}
- u(x_1,x_2,...,x_k)=\frac{1}{2k} \lbrace u(x_1+h,x_2,..,x_k,t-l) - u(x_1-h,x_2,..,x_k,t-l) \\ +...+u(x_1,x_2,..,x_k+h,t-l)+u(x_1,x_2,..,x_k-h,t-l) \rbrace
- \end{align*}
- \ARROW The k dimension problem we can solve in he same way as 1dim.\\
- \ARROW In each point we have $2k$ possibility to move(left-right) in each of the dimensions. The probability has to be $\frac{1}{2k}$.
-
-
- \end{footnotesize}
- \end{minipage}
- \end{frame}
-
-
-
-
- \backupbegin
-
- \begin{frame}\frametitle{Backup}
-
-
- \end{frame}
-
- \backupend
-
- \end{document}