bvpSolve (version 1.4.4.1)

plot.bvpSolve: Plot and Print Methods for Output of bvp solvers

Description

Plot the output of boundary value solver routines.

Usage

# S3 method for bvpSolve
plot(x, ..., which = NULL, ask = NULL, 
                        obs = NULL, obspar= list())
# S3 method for bvpSolve
print(x, ...)

Arguments

x

the output of bvpSolve, as returned by the boundary value solvers, and to be plotted.

It is allowed to pass several objects of class bvpSolve after x (unnamed) - see second example.

which

the name(s) or the index to the variables that should be plotted. Default = all variables, except the first column.

ask

logical; if TRUE, the user is asked before each plot, if NULL the user is only asked if more than one page of plots is necessary and the current graphics device is set interactive, see par(ask=.) and dev.interactive.

obs

a data.frame or matrix with "observed data" that will be added as points to the plots. obs can also be a list with multiple data.frames and/or matrices containing observed data.

By default the first column of an observed data set should contain the time-variable. The other columns contain the observed values and they should have names that are known in x.

If the first column of obs consists of factors or characters (strings), then it is assumed that the data are presented in long (database) format, where the first three columns contain (name, time, value).

If obs is not NULL and which is NULL, then the variables, common to both obs and x will be plotted.

obspar

additional graphics arguments passed to points, for plotting the observed data

...

additional arguments.

The graphical arguments are passed to plot.default.

The dots may also contain other objects of class bvpSolve, as returned by the boundary value solvers, and to be plotted on the same graphs as x - see second example. In this case, x and and these other objects should be compatible, i.e. the names should be the same and they should have same number of rows.

The arguments after ... must be matched exactly.

Details

print.bvpSolve prints the matrix without the attributes.

plot.bvpSolve plots multiple figures on a page.

The number of panels per page is automatically determined up to 3 x 3 (par(mfrow = c(3, 3))). This default can be overwritten by specifying user-defined settings for mfrow or mfcol. Set mfrow equal to NULL to avoid the plotting function to change user-defined mfrow or mfcol settings.

Other graphical parameters can be passed as well. Parameters are vectorized, either according to the number of plots (xlab, ylab, main, sub, xlim, ylim, log, asp, ann, axes, frame.plot,panel.first,panel.last, cex.lab,cex.axis,cex.main) or according to the number of lines within one plot (other parameters e.g. col, lty, lwd etc.) so it is possible to assign specific axis labels to individual plots, resp. different plotting style. Plotting parameter ylim, or xlim can also be a list to assign different axis limits to individual plots.

Similarly, the graphical parameters for observed data, as passed by obspar can be vectorized, according to the number of observed data sets.

See Also

diagnostics.bvpSolve, for a description of diagnostic messages.

Examples

## =============================================================================
## The example MUSN from Ascher et al., 1995.
## Numerical Solution of Boundary Value Problems for Ordinary Differential
## Equations, SIAM, Philadelphia, PA, 1995.
##
## The problem is
##      u' =  0.5*u*(w - u)/v
##      v' = -0.5*(w - u)
##      w' = (0.9 - 1000*(w - y) - 0.5*w*(w - u))/z
##      z' =  0.5*(w - u)
##      y' = -100*(y - w)
##   on the interval [0 1] and with boundary conditions:
##      u(0) = v(0) = w(0) = 1,  z(0) = -10,  w(1) = y(1)
## =============================================================================

musn <- function(t, Y, pars)  {
  with (as.list(Y),
  {
   du <- 0.5 * u * (w-u)/v
   dv <- -0.5 * (w-u)
   dw <- (0.9 - 1000 * (w-y) - 0.5 * w * (w-u))/z
   dz <- 0.5 * (w-u)
   dy <- -100 * (y-w)
   return(list(c(du, dv, dw, dz, dy)))
  })
}

#--------------------------------
# Boundaries
#--------------------------------
bound <- function(i,y,pars) {
  with (as.list(y), {
    if (i ==1) return (u-1)
    if (i ==2) return (v-1)
    if (i ==3) return (w-1)
    if (i ==4) return (z+10)
    if (i ==5) return (w-y)
 })
}

xguess <- seq(0, 1, len = 5)
yguess <- matrix(ncol = 5, (rep(c(1, 1, 1, -10, 0.91), times = 5)) )
rownames(yguess) <- c("u", "v", "w", "z", "y")

sol  <- bvpcol (bound = bound, x = seq(0, 1, by = 0.05), 
          leftbc = 4, func = musn, xguess = xguess, yguess = yguess)

mf <- par("mfrow")
plot(sol)
par(mfrow = mf)

## =============================================================================
## Example 2. Example Problem 31 from Jeff Cash's website
## =============================================================================

Prob31 <- function(t, Y, pars)  {
  with (as.list(Y), {
    dy    <- sin(Tet)
    dTet  <- M
    dM    <- -Q/xi
    T     <- 1/cos (Tet) +xi*Q*tan(Tet)
    dQ    <- 1/xi*((y-1)*cos(Tet)-M*T)
    list(c( dy, dTet, dM, dQ))
  })
}

ini <- c(y = 0, Tet = NA, M = 0, Q = NA)
end <- c(y = 0, Tet = NA, M = 0, Q = NA)

# run 1
xi <-0.1
twp  <- bvptwp(yini = ini, yend = end, x = seq(0, 1, by = 0.01),
               func = Prob31, atol = 1e-10)

# run 2
xi <- 0.05
twp2 <- bvptwp(yini = ini, yend = end, x = seq(0, 1, by = 0.01),
               func = Prob31, atol = 1e-10)

# run 3
xi <- 0.01
twp3 <- bvptwp(yini = ini, yend = end, x = seq(0, 1, by = 0.01),
               func = Prob31, atol = 1e-10)

# print all outputs at once
plot(twp, twp2, twp3, xlab = "x", ylab = names(ini))


# change type, colors, ...
plot(twp, twp2, twp3, type = c("l", "b", "p"), 
     main = paste ("State Variable", names(ini)), 
     col = c("red", "blue", "orange"), cex = 2)

## =============================================================================
## Assume we have two 'data sets':
## =============================================================================
# data set in 'wide' format
obs1 <- cbind(time = c(0, 0.5, 1), Tet = c(0.4, 0.0, -0.4))

# data set in 'long' format
obs2 <- data.frame(name = "Tet", time = c(0, 0.5, 1), value = c(0.35, 0.0, -0.35))

plot(twp, twp2, obs = obs1, obspar = list(pch = 16, cex = 1.5))

plot(twp, twp2, obs = list(obs1, obs2), 
    obspar = list(pch = 16, cex = 1.5))

plot(twp, twp2, obs = list(obs1, obs2), which = c("Tet", "Q"),
    obspar = list(pch = 16:17, cex = 1.5, col = c("red", "black"))
    )