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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}
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\author{ {\fontspec{Trebuchet MS}Marcin Chrz\k{a}szcz} (Universit\"{a}t Z\"{u}rich)}
\institute{UZH}
\title[Partial Differential Equation Solving]{Partial Differential Equation Solving}
\date{\fixme}
\begin{document}
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\begin{center}
\begin{center}
\begin{columns}
\begin{column}{0.9\textwidth}
\flushright\fontspec{Trebuchet MS}\bfseries \Huge {Partial Differential Equation Solving, vol 2.}
\end{column}
\begin{column}{0.2\textwidth}
%\includegraphics[width=\textwidth]{SHiP-2}
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\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}}
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\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, \\ 12 May, 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{Dirichlet conditions:expected number of steps}
\begin{minipage}{\textwidth}
\begin{footnotesize}
\ARROW find the function $u(x_1,x_2,...,x_k)$ such that if fulfils the Laplace equation:
\begin{align*}
\dfrac{\partial^2 u }{\partial x_1^2} + \dfrac{\partial^2 u }{\partial x_2^2}+...+\dfrac{\partial^2 u }{\partial x_k^2}=0,~~~(x_1,x_2,...,x_k) \in D \subset \mathbb{R}^k
\end{align*}
In the domain $D$, on the the $\Gamma(D)$ the $u$ function is given by:
\begin{align*}
U(x_1,x_2,...,x_k)=f(x_1,x_2,...,x_k),~~~~(x_1,x_2,...,x_k) \in \Gamma( D )
\end{align*}
\ARROW Now lets assume that the domain $D$ is a hyperball:
\begin{align*}
0 \leq \sum_{i=1}^k x_i^2 \leq r^2,~~~r={\rm const}
\end{align*}
\ARROW Now $\pi_{\nu}(x_1,x_2,...,x_k)$ is a probability that a particle starting from $(x_1,x_2,...,x_k)$ will end up on the edge after $\nu$ steps. The $\kappa(x_1,x_2,...,x_k)$ is the estimated number of steps for this trajectory.
\begin{tiny}
\begin{align}
\pi_0=\begin{cases}
1,~~& (x_1,x_2,...,x_k) \in \Gamma(D)\\
0,~~& (x_1,x_2,...,x_k) \in D
\end{cases}
\label{eq1}
\end{align}
\begin{align}
\pi_{\nu}=\frac{1}{2k} \sum^{ \prime} \pi_{\nu}(x_1\prime,x_2\prime,...,x_k\prime)
\label{eq2}
\end{align}
\end{tiny}
\end{footnotesize}
\end{minipage}
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}\frametitle{Dirichlet conditions:expected number of steps}
\begin{minipage}{\textwidth}
\begin{footnotesize}
\ARROW From Eq.~\ref{eq1} and \ref{eq2} one gets:
\begin{align*}
\kappa(x_1,x_2,...,x_k)=\sum_{\nu=1}^{\infty} \nu \pi_{\nu}(x_1,x_2,...,x_k)
\end{align*}
one gets:
\begin{align*}
\kappa(x_1,x_2,...,x_k)=\frac{1}{2k}\sum_{\nu=1}^{\infty} \left[ \nu \sum^{\prime}\pi_{\nu -1 }(x_1,x_2,...,x_k) \right]\\ = \frac{1}{2k} \sum_{\nu =1 }^{\infty} \left[ (\nu-1)\sum^{\prime} \pi_{\nu-1}(x_1\prime,x_2\prime,...,x_k\prime) \right] + \frac{1}{2k} \sum_{\nu=1}^{\infty} \sum^{\prime} \pi_{\nu-1}x_1\prime,x_2\prime,...,x_k\prime)
\end{align*}
\ARROW From which we get:
\begin{align*}
\kappa(x_1,x_2,...,x_k)=\frac{1}{2k}\sum^{\prime}\kappa(x_1\prime,x_2\prime,...,x_k\prime) +1
\end{align*}
\ARROW Now this is equivalent of the Poisson differential equation:
\begin{align*}
\frac{\partial^2 \kappa}{\partial x_1^2}+\frac{\partial^2 \kappa}{\partial x_2^2}+...+\frac{\partial^2 \kappa}{\partial x_k^2} = -2k,~{\rm b.~con.}~~ \kappa(x_1,x_2,...,x_k)=0,~~(x_1,x_2,...,x_k) \in \Gamma(D)
\end{align*}
\end{footnotesize}
\end{minipage}
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}\frametitle{Dirichlet conditions:expected number of steps}
\begin{minipage}{\textwidth}
\begin{footnotesize}
\ARROW From previous equation: $\kappa(x_1,x_2,...,x_k)=\psi(x_1,x_2,...,x_k)-\sum_{i=1}^k x_i^2$
we get the for the $\psi$ function the Laplace equation:
\begin{align*}
\dfrac{\partial^2 \psi }{\partial x_1^2} + \dfrac{\partial^2 \psi }{\partial x_2^2}+...+\dfrac{\partial^2 \psi }{\partial x_k^2}=0
\end{align*}
because on the border ($\Gamma(D)$):
\begin{align*}
\psi(x_1,x_2,...,x_k)=r^2= {\rm const}
\end{align*}
so also inside the $D$:
$\psi(x_1,x_2,...,x_k)=r^2= {\rm const}$
\ARROW From which we can estimate the number steps in the random walk:
\begin{align*}
\kappa(x_1,x_2,...,x_k)=r^2-\sum_{i=1}^k\leq r^2
\end{align*}
\begin{alertblock}{Important conclusion:}
The expected number of steps in the random walk (the time of walk) from the point $(x_1,x_2,...,x_k)$ till the edge od the domain can be estimated by $r$ number (the LINEAR! size). It is completly independent of the $k$!
\end{alertblock}
\end{footnotesize}
\end{minipage}
\end{frame}
\begin{frame}\frametitle{Dirichlet conditions as linear system}
\begin{minipage}{\textwidth}
\begin{footnotesize}
\ARROW In the discrete form we can write the Dirichlet conditions as (2-dim case):
\begin{align*}
u(x,y) & =\frac{1}{4} \left[ u(x-1,y)+u(x+1,y)+u(x,y-1)+u(x,y+1) \right],~~(x,y) \in D \\
u(x,y) & =f(x,y),~~~(x,y) \in \Gamma(D)
\end{align*}
\ARROW Now we can order the grid ($(x,y) \in D \cup D$), we can represente the above equations as a linear system:
\begin{align*}
u_i=a_i+\sum_{j=1}^n h_{i j}u_j, ~~~~i=1,2,....,n
\end{align*}
\begin{exampleblock}{The trick:}
So to solve a differential equation with Dirichlet boundary condition we can use all the methods of solving linear equation systems such as Neumann-Ulam or Wassow.
\end{exampleblock}
\end{footnotesize}
\end{minipage}
\end{frame}
\begin{frame}\frametitle{Dirichlet conditions as linear system - example}
\begin{minipage}{\textwidth}
\begin{footnotesize}
\begin{columns}
\column{3in}
\begin{itemize}
\item To do this we act as following: we number separately the points inside the $D$ domain and on the border $\Gamma(D)$.
\item We write for each point inside the domain the Laplace equation as system of linear equations:
\end{itemize}
\column{2in}
\includegraphics[width=0.99\textwidth]{images/grid1.png}
\end{columns}
\begin{tiny}
\begin{align*}
&u_1 & -u_2/4 & &-u_4/4 & & & & = (f_1+f_{10})/4\\
&-u_1/4 & u_2 & -u_3/4 & & -u_5/4 & & & = (f_2)/4\\
& & -u_2/4 & ~~~~u_3 & & & -u_6/4 & & = (f_3+f_4)/4\\
&-u_1/4 & & & u_4 & -u_5/4 & & & = (f_8+f_9)/4\\
&-u_1 /4 & & & -u_4/4 &~~~~ u_5 & -u_6/4 & -u_7/4 & = 0\\
& & & -u_3/4 & & -u_5/4 & u_6 & & = (f_5+f_6)/4\\
& & & & & -u_5/4 & &~~~~~ u_7 & = (f_5+f_6)/4
\end{align*}
\end{tiny}
%-u_1/4 & u_2 & -u_3/4 & & -u_5/4 & & & = (f_1+f_10)/4\\
\end{footnotesize}
\end{minipage}
\end{frame}
\begin{frame}\frametitle{Dirichlet conditions as linear system - example}
\begin{minipage}{\textwidth}
\begin{footnotesize}
\ARROW The above equation we can transform the above equation into the iterative representation:
\begin{align*}
\overrightarrow{u}=\overrightarrow{a}+\textbf{H}\overrightarrow{u}
\end{align*}
where $\overrightarrow{u}=(u_1,u_2,...,u_7)$ is the vector which represent the values of the function inside the $D$ domain, $\overrightarrow{a}$ is the linear combinations of the boundary values. In our example:
\begin{columns}
\column{2in}
\begin{align*}
\textbf{H}=\begin{pmatrix}
0 & \dfrac{1}{4} & 0 & \dfrac{1}{4} & 0 & 0 & 0\\
\dfrac{1}{4} & 0 & \dfrac{1}{4} & 0 & \dfrac{1}{4} & 0& 0\\
0 & \dfrac{1}{4} & 0 & 0 & 0 & \dfrac{1}{4} & 0\\
\dfrac{1}{4} & 0 & 0 & 0 & \dfrac{1}{4} & 0 & 0\\
0 & \dfrac{1}{4} & 0 & \dfrac{1}{4} & 0 & \dfrac{1}{4} & \dfrac{1}{4}\\
0 & 0 & \dfrac{1}{4} & 0 & \dfrac{1}{4} & 0 & 0\\
0 & 0 & 0 & 0 & \dfrac{1}{4} & 0 & 0\\
\end{pmatrix}
\end{align*}
\column{3in}
\ARROW To find the solution to aka $\overline{u}$ one can use the methods we already know: Neumann-Ulam and Wasow, etc.\\
\ARROW There are tricks and tips one can use to make this problem faster as each of the entry is $\frac{1}{4}$.
\end{columns}
\end{footnotesize}
\end{minipage}
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}\frametitle{Neumann-Ulam method}
\begin{minipage}{\textwidth}
\begin{footnotesize}
\ARROW We put the particle in $(x,y)$. \\
\ARROW We observe the trajectory of the particle until it reaches the boundary. Point $P_k$ is the last point before hitting the boundary.\\
\ARROW For each trajectory we assign a value that the arithmetical mean of the boundary points that are neighbours of the point $P_k$.\\
\ARROW We repeat the above $n$ times and calculate the mean.\\
\ARROW The example solution for $20$ trajectories:
\begin{align*}
u(2,2)=1.0500\pm 0.2756
\end{align*}
\ARROW E 10.1 Solve the above linear system using the Neumann-Ulam method for an assumed boundary conditions.
\end{footnotesize}
\end{minipage}
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}\frametitle{Dual Wasow method}
\begin{minipage}{\textwidth}
\begin{footnotesize}
\ARROW We choose the boundary conditions with arbitrary chosen probability \pdf~$p(Q)$ the starting point.
\ARROW We choose with equal probability the point inside $D$ where the particle goes.\\
\ARROW With equal probability we choose the next positions and so on until the particle hits the boundary in the point $Q^{\prime}$.\\
\ARROW We count all trajectories $N((x_1,x_2,x_3,...,x_k)$ that that have passed the point $(x_1,x_2,x_3,...,x_k)$.
\ARROW For the point $(x_1,x_2,...,x_k)$ we calculate:
\begin{align*}
w(x_1,x_2,...,x_k)=\frac{1}{2k}N(x_1,x_2,...,x_k)\frac{f(Q)}{p(Q)}
\end{align*}
\ARROW The above steps we repeat $N$ times.\\
\ARROW After that we take the arithmetic mean of $w$.
\end{footnotesize}
\end{minipage}
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}\frametitle{Random walk with different step size}
\begin{minipage}{\textwidth}
\begin{footnotesize}
\ARROW If $u(x,y)$ is a harmonic function that obeys the Laplace equation and $S_r(x,y)$ is a circle in with the middle point $(x,y)$ and radius $r$. Then a theorem states:
\begin{align*}
S_r(x,y)=\frac{1}{2 \pi } \int_0^{2\pi} u(x+r \cos \phi ,y+ r \sin \phi) d \phi
\end{align*}
\ARROW The above is true for in all the dimensions.\\
\ARROW The E.Muller method:
\begin{itemize}
\item At the begging we set the point in the initial point: $(x_1,x_2,...,x_k)$.
\item We construct a $k$ dimensional sphere with center $(x_1,x_2,...,x_k)$ and radius $r$. The $r$ has to be choosen in a way that the whole is inside the $D$: $S_r(\overrightarrow{x}) \in D$. We choose a random point from $\mathcal{U}(0,2\pi)$ on the sphere which is our new point.
\item We stop the walk when the point is on $\Gamma(D)$.
\end{itemize}
\ARROW We repeat this $N$ times.\\
\ARROW The final result if the arithmetical mean of all trajectories and is equal of the $u(x_1,x_2,...,x_k)$.
\end{footnotesize}
\end{minipage}
\end{frame}
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mchrzasz