fixed some small things

latex/main paper/main.tex

@@ -1128,7 +1128,7 @@ \subsection{Initial Value Problems}
  \begin{equation}
  y(t) \cong Y_{j}(t) = y(t_{j}) + (t-t_{j})Y_{j}((t-t_{j})U+t_{j}), \quad t>t_{j}.
  \end{equation}
- A problem with these RRVEs is that we do not know $y(t_{j})$.
+ The problem with these RRVEs is that we do not know $y(t_{j})$.
  Instead, we can replace it with an unbiased estimate $y_{j}$
  which we keep fixed in the inner recursion:
  \begin{align}
@@ -1159,7 +1159,7 @@ \subsection{Initial Value Problems}
  y_{t}(t)= A(t)y(t)+g(t), y(t_{0}),
  \end{equation}
  with $A$ a matrix and $g$ a vector function, both
- once differentiable with the corresponding RRVE:
+ once Lipschitz differentiable with the corresponding RRVE:
  \begin{equation}
  y(t) \cong Y(t) = y(t_{0}) + h B \left( \frac{t-t_{0}}{h}\right)
  A(S) Y(S) + g(S),
@@ -1380,32 +1380,34 @@ \subsection{Heat Equation}
  Again by applying (\ref{eq:poisson main}) with $\sigma = \frac{2}{\Delta x^{2}} +a_{0}$
  we obtain at an interior point:
  \begin{equation} \label{eq:int semi heat source}
- u =
- \int_{0}^{e^{-t \sigma }} u_{0} d\tau + \int_{e^{-t \sigma }}^{1}
+ \tilde{u} =
+ \int_{0}^{e^{-t \sigma }} \tilde{u}_{0} d\tau + \int_{e^{-t \sigma }}^{1}
  \sigma^{-1}
  \left(
- \frac{u_{+} + u_{-}}{\Delta x^{2}} +(a(x,s)+ a_{0}) u+ f
+ \frac{\tilde{u}_{+} + \tilde{u}_{-}}{\Delta x^{2}} +(a(x,s)+ a_{0}) \tilde{u}+ f
  \right)
  d\tau
  .
  \end{equation}
+
  This can be turned into a estimator $Y(j \Delta x,t,\Delta x ) \cong \tilde{u}(j \Delta x,t)$
  by sampling $\tau \sim U$.
  To avoid branching recursion we sample
  based on the magnitude of coefficients of the recursive terms, let $a_m$ be approximately the magnitude of
- $(a(x,s)+ a_{0})$. We chose to sample the $(a(x,s)+ a_{0})u$ term and $f$
+ $(a(x,s)+ a_{0})$. We chose to sample the $(a(x,s)+ a_{0})\tilde{u}$ term and $f$
  with probability $p_{\text{source}} = \frac{a_m}{a_m + \frac{2}{\Delta x^{2}}}$.
- As $\Delta x \rightarrow 0$ the $\frac{u_{+} + u_{-}}{\Delta x^{2}}$ term is
+ As $\Delta x \rightarrow 0$ the $\frac{\tilde{u}_{+} + \tilde{u}_{-}}{\Delta x^{2}}$ term is
  the main contribution to the second integral therefore
  being sampled almost always.
  In total for $1$ fully recursed sample,
- the $f(x,s)$ and the $(a(x,s)+ a_{0})u$ term only
+ the $f(x,s)$ and the $(a(x,s)+ a_{0})\tilde{u}$ term only
  is sampled a few times and this does not scale with $\Delta x$. We
- sampled $f$ together with $(a(x,s)+ a_{0})u$ for simplicity.
+ sampled $f$ together with $(a(x,s)+ a_{0})\tilde{u}$ for simplicity.
+
 \end{example}
 
 \begin{julia}[Implementation of estimator of (\ref{eq:int semi heat source})]\label{jl:point estimator heat source}
- Because of the high chance of sampling the $\frac{u_{+} + u_{-}}{\Delta x^{2}}$ it is efficient
+ Because of the high chance of sampling the $\frac{\tilde{u}_{+} + \tilde{u}_{-}}{\Delta x^{2}}$ it is efficient
  to sample the interarrival until not sampling it which is geometrically distributed.
 
  \juliacode{julia_code/pest_heat_varcoef.jl}
@@ -1447,7 +1449,7 @@ \subsection{Heat Equation}
 \begin{julia}[\ref{jl:point estimator heat source} using precomputed paths]\label{jl:path point estimator heat source}
  We implement \ref{jl:point estimator heat source} with a precomputed path from \ref{tech:presampling heat} using tail
  recursion. Note that $(x,t)$ is
- never explicitly used only for figuring out how we exited the domain. The whole calculation is linear in
+ never explicitly used, only for figuring out how we exited the domain. The whole calculation is linear in
  $a,f$ evaluated in the source points and $u_\text{bound}$ evaluated in the exit points so there are various
  ways to optimize it for example using unbiased estimates instead of exact evaluations.