swh:1:snp:bbc0b94b1c0463480b615c86f1ae2df3fcc700c0
Tip revision: 6a2a464
PDE.Rnw
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\usepackage{amsmath}

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\newcommand{\rt}{\textbf{\textsf{ReacTran }}}
\newcommand{\rs}{\textbf{\textsf{rootSolve }}}
\newcommand{\R}{\proglang{R }}
\newcommand{\ds}{\textbf{\textsf{deSolve }}}
\newcommand{\dd}{\textbf{\textsf{ddesolve }}}
\newcommand{\rb}{\raisebox{1.5ex}{#1}}

\title{Solving partial differential equations, using  \R package \rt}
\Plaintitle{Solving partial differential equations using R package ReacTran}

\Keywords{Partial Differential Equations, hyperbolic, parabolic,
elliptic, \R}

\Plainkeywords{Partial Differential Equations, hyperbolic, parabolic,
elliptic, R}

\author{Karline Soetaert and Filip Meysman\\
Royal Netherlands Institute of Sea Research (NIOZ)\\
Yerseke\\
The Netherlands
}

\Plainauthor{Karline Soetaert and Filip Meysman}

\Abstract{
\R-package \rt \citep{ReacTran_paper} contains functions to solve reactive-transport equations, as used e.g. in the environmental sciences.

Essentially, it
\begin{enumerate}
\item Provides functions that subdivide the spatial extent into a number of discrete grid cells.
\item Approximates the advective-diffusive transport term by finite differences or finite volumes.
\end{enumerate}
The main package vignette \citep{ReacTran-vignette} explains how \rt can be used to model reaction-transport phenomena.

However, the functions from \rt can be use to solve more general types of partial differential equations ($\leq$ order 2).

In this vignette, show how the package can be used to solve partial differential equations of the parabolic, hyperbolic and elliptic type, providing one example each.
}

%% The address of (at least) one author should be given
%% in the following format:
Karline Soetaert\\
Royal Netherlands Institute of Sea Research (NIOZ)\\
4401 NT Yerseke, Netherlands\\
E-mail: \email{karline.soetaert@nioz.nl}\\
URL: \url{http://www.nioz.nl}\\
\\
Filip Meysman\\
Royal Netherlands Institute of Sea Research (NIOZ)\\
4401 NT Yerseke, Netherlands\\
E-mail: \email{filip.meysman@nioz.nl}\\
URL: \url{http://www.nioz.nl}\\
}

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%\VignetteIndexEntry{Solving parabolic, hyperbolic and elliptic partial differential equations in R}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% Begin of the document
\begin{document}
\SweaveOpts{engine=R,eps=FALSE}
\SweaveOpts{keep.source=TRUE}
<<preliminaries,echo=FALSE,results=hide>>=
library(ReacTran)
options(prompt = " ")
@

\maketitle

\section{Partial differential equations}
In \textbf{partial differential equations }(PDE), the function has several
independent variables (e.g. time and depth) and contains their partial
derivatives.

A first step to solve partial differential equations (PDE), is to discretise one or more of the independent variables.

Usually, the independent variable time'' is not discretised, while other variables (e.g. spatial axes) are discetised, so that a set of ODE is obtained, which can be solved with appropriate initial values solvers from package \ds \citep{deSolve}.
This technique, the method-of-lines, applies to hyperbolic and parabolic PDEs.

For time-invariant problems, usually all independent variables are discretised, and the resulting algebraic equations solved with root-solving functions from package \rs
\citep{rootSolve}.

Functions \code{tran.1D}, \code{tran.2D}, and \code{tran.3D} from
\R package \pkg{ReacTran} implement finite difference approximations of the
general diffusive-advective transport equation, which for 1-D is:

\begin{eqnarray*}
-\frac{1}{A_x \xi_x}\cdot (\frac{\partial}{\partial x}
A_x \cdot (-  D  \cdot \frac{\partial \xi_x C}{\partial x})
- \frac{\partial}{\partial x}
(A_x \cdot v  \cdot \xi_x C))
\end{eqnarray*}
Here $D$ is the "diffusion coefficient", $v$ is the "advection rate", and $A_x$ and $\xi$ are the surface area and volume fraction respectively.

Assuming that $A$, $\xi$, $D$ and $v$ are constant along $x$, we can rewrite this in a more general form:
\begin{eqnarray*}
D \frac{\partial^2C}{\partial x^2} - u \frac{\partial C}{\partial x}
\end{eqnarray*}

In which the first term is a second-order, the second term a first-order derivative.

\R-function \code{tran.1D} is defined as (simplified):
\begin{verbatim}
tran.1D <- function (C, C.up = C, C.down = C[length(C)], flux.up = NULL,
flux.down = NULL, a.bl.up = NULL, a.bl.down = NULL, D = 0,
v = 0, AFDW = 1, VF = 1, A = 1, dx, ...)
\end{verbatim}

where \code{C.up} and \code{C.down} are the upstream and downstream boundary values, \code{flux.up} and \code{flux.down} are the upstream and downstream fluxes, \code{v} and \code{D} are the advection and diffusion coefficient respectively, \code{A} is the surface area, \code{x} contains the grid, and \code{VF} is the volume fraction ($\xi$).
For the other arguments, see the help file of \code{tran.1D}.

A suitable \emph{grid} can be generated with functions \code{setup.grid.1D} and
\code{setup.grid.2D} (there is no corresponding 3D function), while a \emph{property} can be added to this grid using functions \code{setup.prop.1D}, and \code{setup.prop.2D}.
These latter two functions are useful to have the variable surface areas, volume fractions, advection and diffusion rates being defined at the correct places of the grid.

These functions are defined as (simplified):
\begin{verbatim}
setup.grid.1D <- function(x.up = 0, x.down = NULL, L = NULL, N = NULL,
dx.1 = NULL, ...)

setup.grid.2D <- function(x.grid = NULL, y.grid = NULL)

setup.prop.1D <- function (func = NULL, value = NULL, xy = NULL,
interpolate = "spline", grid, ...)

setup.prop.2D <- function (func = NULL, value = NULL, grid, y.func = func,
y.value = value, ...)
\end{verbatim}

\clearpage
\section{A parabolic PDE}
As an example of the parabolic type, consider the 1-D diffusion-reaction model, in spherical,
cylindrical and cartesian coordinates, defined for $r$ in $[0, 10]$:

\begin{eqnarray*}
\frac{\partial C}{\partial t} &=& \frac{1}{r^2}\cdot \frac{\partial}{\partial r}
\left(r^2 \cdot D  \cdot \frac{\partial C}{\partial r}\right)  - Q \\
\frac{\partial C}{\partial t} &=& \frac{1}{r}\cdot \frac{\partial}{\partial r}
\left(r \cdot D  \cdot \frac{\partial C}{\partial r}\right)  - Q \\
\frac{\partial C}{\partial t} &=& \frac{\partial}{\partial r}
\left(D  \cdot \frac{\partial C}{\partial r}\right)  - Q
\end{eqnarray*}

with $t$ the time, $r$ the (radial) distance from the origin,
$Q$, the consumption rate, and with boundary conditions
(values at the model edges):

\begin{eqnarray*}
\frac{\partial C}{\partial r}_{r=0} &=& 0 \\
C_{r=10} &=& Cext
\end{eqnarray*}

To solve this model in \R{}, first the 1-D model \code{Grid} is
defined; it divides 10 cm (\code{L}) into 1000 equally-sized boxes (\code{N}).

<<>>=
Grid <- setup.grid.1D(N = 1000, L = 10)
@

Next the properties $r$ and $r^2$ are defined on this grid:
\SweaveOpts{keep.source=TRUE}

<<>>=
r  <- setup.prop.1D(grid = Grid, func = function(r) r)
r2 <- setup.prop.1D(grid = Grid, func = function(r) r^2)
@
\SweaveOpts{keep.source=FALSE}

The model equation includes a transport term, approximated by
\pkg{ReacTran} function \code{tran.1D} and a consumption term
(\code{Q}); only the downstream boundary condition, prescribed as a
concentration (\code{C.down}) needs to be specified, as the
zero-gradient at the upstream boundary is the default:

<<>>=
library(ReacTran)
pde1D <-function(t, C, parms, A = 1)  {
tran   <- tran.1D (C = C, A = A, D = D, C.down = Cext, dx = Grid)$dC list(tran - Q) # the return value: rate of change } @ The model parameters are defined: <<>>= D <- 1 # diffusion constant Q <- 1 # uptake rate Cext <- 20 @ \subsection{Steady-state solution} In a first application, the model is solved to \emph{steady-state}, which retrieves the condition where the concentrations are invariant, e.g. for the cylindrical coordinate case: $0 = \frac{1}{r^2}\cdot \frac{\partial}{\partial r} \left(r^2 \cdot D \cdot \frac{\partial C}{\partial r}\right) - Q$ In \R, steady-state conditions can be estimated using functions from package \rs which implement amongst others a Newton-Raphson algorithm \citep{Press92}. For 1-dimensional models, \code{steady.1D} should be used. The initial guess'' of the steady-state solution (\code{y}) is unimportant; here we take simply \code{N} random numbers. Argument \code{nspec = 1} informs the solver that only one component is described. Although a system of 1000 equations needs to be solved, this takes only a fraction of a second: \SweaveOpts{keep.source=TRUE} <<>>= library(rootSolve) Cartesian <- steady.1D(y = runif(Grid$N),
func = pde1D, parms = NULL, nspec = 1, A = 1)
Cylindrical <- steady.1D(y = runif(Grid$N), func = pde1D, parms = NULL, nspec = 1, A = r) @ <<>>= print(system.time( Spherical <- steady.1D(y = runif(Grid$N),
func = pde1D, parms = NULL, nspec = 1, A = r2)
))
@

The values of the state-variables (\code{y}) are plotted against the
radial distance, in the middle of the grid cells (\code{Grid\$x.mid}). We use \code{rootSolve}'s plot method to do so. This function accepts several steady-state outputs at once: <<label=pde, include=FALSE>>= plot(Cartesian, Cylindrical, Spherical, grid = Grid$x.mid,
main = "steady-state PDE", xlab = "x", ylab = "C",
col = c("darkgreen", "blue", "red"), lwd = 3, lty = 1:3)

legend("bottomright", c("cartesian", "cylindrical", "spherical"),
col = c("darkgreen", "blue", "red"), lwd = 3, lty = 1:3)
@

\setkeys{Gin}{width=0.6\textwidth}
\begin{figure}
\begin{center}
<<label=figpde, fig=TRUE, echo=FALSE>>=
<<pde>>
@
\end{center}
\caption{Steady-state solution of the 1-D diffusion-reaction model}
\label{fig:pde}
\end{figure}

The analytical solutions compare well with the numerical approximation
for all three cases:
\SweaveOpts{keep.source=TRUE}

<<>>=
max(abs(Q/6/D*(r2$mid - 10^2) + Cext - Spherical$y))
max(abs(Q/4/D*(r2$mid - 10^2) + Cext - Cylindrical$y))
max(abs(Q/2/D*(r2$mid - 10^2) + Cext - Cartesian$y))
@
Note that there is no automatic error checking/control here, so to reduce this error, the number of boxes can be increased.
\subsection{The method of lines}
Next the model (for spherical coordinates) is run dynamically for 100
time units using \ds function \code{ode.1D}, and starting with an initially uniform distribution (\code{y = rep(1, Grid$N)}): <<>>= require(deSolve) times <- seq(0, 100, by = 1) system.time( out <- ode.1D(y = rep(1, Grid$N), times = times, func = pde1D,
parms = NULL, nspec = 1, A = r2)
)
@

Here, \code{out} is a matrix, whose $1^{st}$ column contains the
output times, and the next columns the values of the state variables
in the different boxes:

<<>>=
tail(out[, 1:4], n = 3)
@

We plot the result using a blue-yellow-red color scheme, and using deSolve's
S3 method \code{image}:

<<label=yy, include=FALSE>>=
image(out, grid = Grid$x.mid, xlab = "time, days", ylab = "Distance, cm", main = "PDE", add.contour = TRUE) @ \setkeys{Gin}{width=0.6\textwidth} \begin{figure} %%<<label = yyfig, fig=TRUE, echo=FALSE>>= %%<<yy>> %%@ \begin{center} \includegraphics{diffusion.jpeg} % to make it smaller \end{center} \caption{Dynamic solution of the 1-D diffusion-reaction model} \label{fig:yy} \end{figure} \clearpage \section{A hyperbolic PDE} The equation for a wave travelling in one direction ($x) is given by: \begin{align} \frac{\partial^2 u}{\partial t^2} &= c^2 \frac{\partial^2 u}{\partial x^2} \end{align} wherec$is the propagation speed of the wave, and$u$is the variable that changes as the wave passes. This equation is second-order in both$t$and$x. The wave equation is the prototype of a hyperbolic'' partial differential equation. For it to be solved in \R, the equation is rewritten as two coupled equations, first-order in time: \begin{align} \frac{\partial u_1}{\partial t} &= u_2\\ \frac{\partial u_2}{\partial t} &= c^2 \frac{\partial^2 u_1}{\partial x^2} \end{align} We solve the equation with the following initial and boundary conditions: \begin{align*} u_1(0,x) &= \exp ^{-0.05 x^2}\\ u_2(0,x) &= 0\\ u_{t,-\infty} &= 0\\ u_{t,\infty} &= 0 \label{wavecond} \end{align*} where the first condition represents a Gaussian pulse. The implementation in \R starts with defining the box size \code{dx} and the grid, \code{xgrid}. To comply with the boundary conditions (which are defined at\infty$), the grid needs to be taken large enough such that$u$remains effectively$0$at the boundaries, for all run times. Here, the grid extends from -100 to 100: <<>>= dx <- 0.2 xgrid <- setup.grid.1D(-100, 100, dx.1 = dx) x <- xgrid$x.mid
N     <- xgrid$N @ The initial condition, \code{yini} and output \code{times} are defined next: <<>>= uini <- exp(-0.05 * x^2) vini <- rep(0, N) yini <- c(uini, vini) times <- seq (from = 0, to = 50, by = 1) @ The wave equation derivative function first extracts, from state variable vector \code{y} the two properties \code{u1, u2}, both of length \code{N}, after which \pkg{ReacTran} function \code{tran.1D} performs transport of \code{u1}. The squared velocity ($c^2$) is taken as 1 (\code{D=1}): <<>>= wave <- function (t, y, parms) { u1 <- y[1:N] u2 <- y[-(1:N)] du1 <- u2 du2 <- tran.1D(C = u1, C.up = 0, C.down = 0, D = 1, dx = xgrid)$dC
return(list(c(du1, du2)))
}
@

The wave equation can be solved efficiently with a non-stiff solver
such as the Runge-Kutta method \code{ode45}.
<<>>=
out <- ode.1D(func = wave, y = yini, times = times, parms = NULL,
nspec = 2, method = "ode45", dimens = N, names = c("u", "v"))
@

We now plot the results (Fig. \ref{fig:wave}) using \code{deSolve}s function
\code{matplot.1D}; intial condition in
black, the values for selected time points in \code{darkgrey}; a
\code{legend} with times is written.

<<label=wave,include=FALSE>>=
matplot.1D(out, which = "u", subset = time %in% seq(0, 50, by = 10),
type = "l", col = c("black", rep("darkgrey", 5)), lwd = 2,
grid = x, xlim = c(-50,50))
legend("topright", lty = 1:6, lwd = 2, col = c("black", rep("darkgrey", 5)),
legend = paste("t = ",seq(0, 50, by = 10)))
@
\setkeys{Gin}{width=0.6\textwidth}
\begin{figure}[h]
\begin{center}
<<label=wave,fig=TRUE,echo=FALSE,height=5>>=
<<wave>>
@
\end{center}
\caption{The 1-D wave equation; black = initial condition; grey: several time
lines}
\label{fig:wave}
\end{figure}

S3-method \code{image} can also be used to generate \code{persp}-like plots:

<<label=waveimg,include=FALSE>>=
par(mar=c(0,0,0,0))
image(out, which = "u", method = "persp", main = "",
border = NA, col = "lightblue", box = FALSE,
shade = 0.5, theta = 0, phi = 60)
@
% note: load stored - persp
\setkeys{Gin}{width=0.6\textwidth}
\begin{figure}[h]
\begin{center}
\includegraphics{perspwave.jpeg}
\end{center}
\caption{The 1-D wave equation as a persp plot}
\label{fig:waveimg}
\end{figure}

You may also want to try the following "movie":
\begin{verbatim}
plot.1D(out, grid = x, which = "u", type = "l",
lwd = 2, ylim = c(0, 1), ask = TRUE)
\end{verbatim}

\clearpage
\section{An elliptic PDE}

The final example describes again a diffusion-reaction system with
production $p$, consumption $r C$, and diffusive transport (diffusion coefficients $Dx, Dy$) of a substance $C$ in 2 dimensions $(x, y)$; the
boundaries are prescribed as zero-gradient (the default).

\begin{equation}
\frac{\partial C}{\partial t} = \frac{\partial}{\partial x}
[D_x  \cdot \frac{\partial C}{\partial x} ] +
\frac{\partial} {\partial y}
[D_y  \cdot \frac{\partial C}{\partial y}]
- r C + p_{xy}
\label{eq:tran}
\end{equation}
The parameter $p_{xy}$ is the production rate, which is zero everywhere
except for 50 randomly positioned spots where it is $1$.

Transport is performed by \code{ReacTran} function \code{tran.2D}; the
state variable vector (\code{y}) is recast in matrix form
(\code{CONC}) before it is transported. The first-order consumption
rate \code{-r * CONC} is added to the rate of change due to transport
(\code{Tran\$dC}). Production, \code{p} is added to 50 cells indexed by \code{ii}, and which are randomly selected from the grid. The function returns a list, containing the derivatives, as a vector: <<>>= require(ReacTran) pde2D <- function (t, y, parms) { CONC <- matrix(nr = n, nc = n, y) Tran <- tran.2D(CONC, D.x = Dx, D.y = Dy, dx = dx, dy = dy) dCONC <- Tran$dC - r * CONC
dCONC[ii]<- dCONC[ii] + p
return(list(as.vector(dCONC)))
}
@
Before running the model, the grid sizes (\code{dx, dx}), diffusion
coefficients (\code{Dx, Dy}), $1^{st}$ order consumption rate (\code{r})
are defined. There
are 100 boxes in \code{x}- and \code{y} direction (\code{n}).
Furthermore, we assume that the substance is produced in 50 randomly
chosen cells (\code{ii}) at a constant rate (\code{p}):
\SweaveOpts{keep.source=FALSE}
<<>>=
n  <- 100
dy <- dx <- 1
Dy <- Dx <- 2
r  <- 0.001
p  <- runif(50)
ii <- trunc(cbind(runif(50)*n, runif(50)*n) + 1)
@
takes as arguments amongst other the dimensionality of the problem
(\code{dimens}) and the length of the work array needed by the solver
(\code{lrw = 600000}).  It takes less than 0.5 seconds to solve this
10000 state variable model.

<<>>=
require(rootSolve)
Conc0 <- matrix(nr = n, nc = n, 10.)
print(system.time(
ST <- steady.2D(y = Conc0, func = pde2D, parms = NULL, dimens = c(n, n),
lrw = 600000)
))
@
<<label=D2, include=FALSE, keep.source=TRUE>>=
image(ST, main = "steady-state 2-D PDE")
@

\setkeys{Gin}{width=1.0\textwidth}
\begin{figure}
<<label=D2fig, fig=TRUE, echo=FALSE>>=
<<D2>>
@
\caption{Steady-state solution of the 2-D diffusion-reaction model}
\label{fig:D2}
\end{figure}

\bibliography{bibs}

\end{document}