mebdfi | R Documentation |
Solves either:
a system of ordinary differential equations (ODE) of the form
y' = f(t,y,...)
or
a system of differential algebraic equations (DAE) of the form
F(t,y,y') = 0
or
a system of linearly implicit DAES in the form
M y' = f(t,y)
using the Modified Extended Backward Differentiation formulas for stiff fully implicit inital value problems
These formulas increase the absolute stability regions of the classical BDFs.
The orders of the implemented formulae range from 1 to 8.
The R function mebdfi
provides an interface to the Fortran DAE
solver of the same name, written by T.J. Abdulla and J.R. Cash.
The system of DE's is written as an R function or can be defined in compiled code that has been dynamically loaded.
mebdfi(y, times, func = NULL, parms, dy = NULL, res = NULL,
nind=c(length(y),0,0), rtol = 1e-6, atol = 1e-6, jacfunc = NULL,
jacres = NULL, jactype = "fullint", mass = NULL, verbose = FALSE,
tcrit = NULL, hini = 0, ynames = TRUE, maxord = 7, bandup = NULL,
banddown = NULL, maxsteps = 5000, dllname = NULL,
initfunc = dllname, initpar = parms, rpar = NULL,
ipar = NULL, nout = 0, outnames = NULL,
forcings=NULL, initforc = NULL, fcontrol=NULL, ...)
y |
the initial (state) values for the DE system. If |
times |
time sequence for which output is wanted; the first
value of |
func |
cannot be used if the model is a DAE system. If an ODE
system,
The return value of Note that it is not possible to define |
parms |
vector or list of parameters used in |
dy |
the initial derivatives of the state variables of the DE system. Ignored if an ODE. |
res |
if a DAE system: either an R-function that computes the
residual function F(t,y,y') of the DAE system (the model
defininition) at time If Here The return value of If |
nind |
if a DAE system: a three-valued vector with the number of variables of index 1, 2, 3 respectively. The equations must be defined such that the index 1 variables precede the index 2 variables which in turn precede the index 3 variables. The sum of the variables of different index should equal N, the total number of variables. |
rtol |
relative error tolerance, either a scalar or a vector, one value for each y, |
atol |
absolute error tolerance, either a scalar or a vector, one value for each y. |
jacfunc |
if not If the Jacobian is a full matrix, If the Jacobian is banded, |
jacres |
If If the Jacobian is a full matrix, If the Jacobian is banded, |
jactype |
the structure of the Jacobian, one of
|
mass |
the mass matrix.
If not If |
verbose |
if TRUE: full output to the screen, e.g. will
print the |
tcrit |
the Fortran routine |
hini |
initial step size to be attempted; if 0, the initial step
size is set to 1e-6, but it may be better to set it equal to |
ynames |
logical, if |
maxord |
the maximum order to be allowed, an integer between 2 and 7.
The default is |
bandup |
number of non-zero bands above the diagonal, in case
the Jacobian is banded (and |
banddown |
number of non-zero bands below the diagonal, in case
the Jacobian is banded (and |
maxsteps |
maximal number of steps per output interval taken by the solver. |
dllname |
a string giving the name of the shared library
(without extension) that contains all the compiled function or
subroutine definitions referred to in |
initfunc |
if not |
initpar |
only when ‘dllname’ is specified and an
initialisation function |
rpar |
only when ‘dllname’ is specified: a vector with
double precision values passed to the dll-functions whose names are
specified by |
ipar |
only when ‘dllname’ is specified: a vector with
integer values passed to the dll-functions whose names are specified
by |
nout |
only used if ‘dllname’ is specified and the model is
defined in compiled code: the number of output variables calculated
in the compiled function |
outnames |
only used if ‘dllname’ is specified and
|
forcings |
only used if ‘dllname’ is specified: a list with
the forcing function data sets, each present as a two-columned matrix,
with (time,value); interpolation outside the interval
[min( See package vignette |
initforc |
if not |
fcontrol |
A list of control parameters for the forcing functions.
vignette |
... |
additional arguments passed to |
The mebdfi solver uses modified extended backward differentiation
formulas of orders one through eight (specified with maxord
)
to solve either:
an ODE system of the form
y' = f(t,y,...)
for y = Y, or
a DAE system of the form
F(t,y,y') = 0
for y = Y and y' = YPRIME.
The recommended value of maxord
is eight, unless it is believed
that there are severe stability problems in which case maxord
=
4 or 5 should be tried instead.
ODEs are specified in func
, DAEs are specified in res
.
If a DAE system, Values for Y and YPRIME at the initial time must be given as input. Ideally,these values should be consistent, that is, if T, Y, YPRIME are the given initial values, they should satisfy F(T,Y,YPRIME) = 0.
The form of the Jacobian can be specified by
jactype
. This is one of:
a full Jacobian, calculated internally
by mebdfi
, the default,
a full Jacobian, specified by user
function jacfunc
or jacres
,
a banded Jacobian, specified by user
function jacfunc
or jacres
; the size of the bands
specified by bandup
and banddown
,
a banded Jacobian, calculated by
mebdfi
; the size of the bands specified by bandup
and
banddown
.
If jactype
= "fullusr" or "bandusr" then the user must supply a
subroutine jacfunc
.
If jactype = "fullusr" or "bandusr" then the user must supply a
subroutine jacfunc
or jacres
.
The input parameters rtol
, and atol
determine the
error control performed by the solver. If the request for
precision exceeds the capabilities of the machine, mebdfi will return
an error code.
res and jacres may be defined in compiled C or Fortran code, as
well as in an R-function. See deSolve's vignette "compiledCode"
for details. Examples
in Fortran are in the ‘dynload’ subdirectory of the
deSolve
package directory.
The diagnostics of the integration can be printed to screen
by calling diagnostics
. If verbose
= TRUE
,
the diagnostics will written to the screen at the end of the integration.
See vignette("deSolve") for an explanation of each element in the vectors containing the diagnostic properties and how to directly access them.
A matrix of class deSolve
with up to as many rows as elements in
times
and as many
columns as elements in y
plus the number of "global" values
returned in the next elements of the return from func
or
res
, plus an additional column (the first) for the time value.
There will be one row for each element in times
unless the
Fortran routine ‘mebdfi’ returns with an unrecoverable error. If
y
has a names attribute, it will be used to label the columns
of the output value.
Karline Soetaert <karline.soetaert@nioz.nl>
Jeff Cash
J. R. Cash, The integration of stiff initial value problems in O.D.E.S using modified extended backward differentiation formulae, Comp. and Maths. with applics., 9, 645-657, (1983).
J.R. Cash and S. Considine, an MEBDF code for stiff initial value problems, ACM Trans Math Software, 142-158, (1992).
J.R. Cash, Stable recursions with applications to the numerical solution of stiff systems, Academic Press,(1979).
gamd
and bimd
two other DAE solvers,
daspk
another DAE solver from package deSolve
,
diagnostics
to print diagnostic messages.
## =======================================================================
## Coupled chemical reactions including an equilibrium
## modeled as (1) an ODE and (2) as a DAE
##
## The model describes three chemical species A,B,D:
## subjected to equilibrium reaction D <- > A + B
## D is produced at a constant rate, prod
## B is consumed at 1s-t order rate, r
## Chemical problem formulation 1: ODE
## =======================================================================
## Dissociation constant
K <- 1
## parameters
pars <- c(
ka = 1e6, # forward rate
r = 1,
prod = 0.1)
Fun_ODE <- function (t, y, pars)
{
with (as.list(c(y, pars)), {
ra <- ka*D # forward rate
rb <- ka/K *A*B # backward rate
## rates of changes
dD <- -ra + rb + prod
dA <- ra - rb
dB <- ra - rb - r*B
return(list(dy = c(dA, dB, dD),
CONC = A+B+D))
})
}
## =======================================================================
## Chemical problem formulation 2: DAE
## 1. get rid of the fast reactions ra and rb by taking
## linear combinations : dD+dA = prod (res1) and
## dB-dA = -r*B (res2)
## 2. In addition, the equilibrium condition (eq) reads:
## as ra = rb : ka*D = ka/K*A*B = > K*D = A*B
## =======================================================================
Res_DAE <- function (t, y, yprime, pars)
{
with (as.list(c(y, yprime, pars)), {
## residuals of lumped rates of changes
res1 <- -dD - dA + prod
res2 <- -dB + dA - r*B
## and the equilibrium equation
eq <- K*D - A*B
return(list(c(res1, res2, eq),
CONC = A+B+D))
})
}
times <- seq(0, 100, by = 1)
## Initial conc; D is in equilibrium with A,B
y <- c(A = 2, B = 3, D = 2*3/K)
## ODE model solved with mebdfi
ODE <- as.data.frame(mebdfi(y = y, times = times, func = Fun_ODE,
parms = pars, atol = 1e-8, rtol = 1e-8))
## Initial rate of change
dy <- c(dA = 0, dB = 0, dD = 0)
## DAE model solved with mebdfi
DAE <- as.data.frame(mebdfi(y = y, dy = dy, times = times,
res = Res_DAE, parms = pars, atol = 1e-8, rtol = 1e-8))
## =======================================================================
## Chemical problem formulation 3: Mass * Func
## Based on the DAE formulation
## =======================================================================
Mass_FUN <- function (t, y, pars)
{
with (as.list(c(y, pars)), {
## as above, but without the
f1 <- prod
f2 <- - r*B
## and the equilibrium equation
f3 <- K*D - A*B
return(list(c(f1, f2, f3),
CONC = A+B+D))
})
}
Mass <- matrix(nr=3, nc=3, byrow = TRUE,
data=c(1, 0, 1, # dA + 0 + dB
-1, 1, 0, # -dA + dB +0
0, 0, 0)) # algebraic
times <- seq(0, 100, by = 2)
## Initial conc; D is in equilibrium with A,B
y <- c(A = 2, B = 3, D = 2*3/K)
## ODE model solved with daspk
ODE <- as.data.frame(daspk(y = y, times = times, func = Fun_ODE,
parms = pars, atol = 1e-10, rtol = 1e-10))
## Initial rate of change
dy <- c(dA = 0, dB = 0, dD = 0)
## DAE model solved with daspk
DAE <- as.data.frame(daspk(y = y, dy = dy, times = times,
res = Res_DAE, parms = pars, atol = 1e-10, rtol = 1e-10))
MASS<- mebdfi(y = y, times = times, func = Mass_FUN,
parms = pars, mass = Mass)
## ================
## plotting output
## ================
opa <- par(mfrow = c(2, 2))
for (i in 2:5)
{
plot(ODE$time, ODE[, i], xlab = "time",
ylab = "conc", main = names(ODE)[i], type = "l")
points(DAE$time, DAE[,i], col = "red")
}
legend("bottomright",lty = c(1,NA),pch = c(NA,1),
col = c("black","red"),legend = c("ODE","DAE"))
# difference between both implementations:
max(abs(ODE-DAE))
par(mfrow = opa)
## =============================================================================
##
## Example 3: higher index DAE
##
## Car axis problem, index 3 DAE, 8 differential, 2 algebraic equations
## from
## F. Mazzia and C. Magherini. Test Set for Initial Value Problem Solvers,
## release 2.4. Department
## of Mathematics, University of Bari and INdAM, Research Unit of Bari,
## February 2008.
## Available at http://www.dm.uniba.it/~testset.
## =============================================================================
# car returns the residuals of the implicit DAE
car <- function(t, y, dy, pars){
with(as.list(c(pars, y)), {
f <- rep(0, 10)
yb <- r*sin(w*t)
xb <- sqrt(L*L - yb*yb)
Ll <- sqrt(xl^2 + yl^2)
Lr <- sqrt((xr-xb)^2 + (yr-yb)^2)
f[1:4] <- y[5:8]
k <- M*eps*eps/2
f[5] <- (L0-Ll)*xl/Ll + lam1*xb+2*lam2*(xl-xr)
f[6] <- (L0-Ll)*yl/Ll + lam1*yb+2*lam2*(yl-yr)-k*g
f[7] <- (L0-Lr)*(xr-xb)/Lr - 2*lam2*(xl-xr)
f[8] <- (L0-Lr)*(yr-yb)/Lr - 2*lam2*(yl-yr)-k*g
f[9] <- xb*xl+yb*yl
f[10] <- (xl-xr)^2+(yl-yr)^2-L*L
delt <- dy-f
delt[5:8] <- k*dy[5:8]-f[5:8]
delt[9:10] <- -f[9:10]
list(delt=delt,f=f)
})
}
# parameters
pars <- c(eps = 1e-2, M = 10, L = 1, L0 = 0.5,
r = 0.1, w = 10, g = 1)
# initial conditions: state variables
yini <- with (as.list(pars),
c(xl = 0, yl = L0, xr = L, yr = L0, xla = -L0/L,
yla = 0, xra = -L0/L, yra = 0, lam1 = 0, lam2 = 0)
)
# initial conditions: derivates
dyini <- rep(0, 10)
FF <- car(0, yini, dyini, pars)
dyini[1:4] <- yini[5:8]
dyini[5:8] <- 2/pars["M"]/(pars["eps"])^2*FF$f[5:8]
# check consistency of initial condition: delt should be = 0.
car(0, yini, dyini, pars)
# running the model
times <- seq(0, 3, by = 0.01)
nind <- c(4, 4, 2) # index 1, 2 and 3 variables
out <- mebdfi(y = yini, dy = dyini, times, res = car, parms = pars,
nind = nind, rtol = 1e-5, atol = 1e-5)
plot(out, which = 1:4, type = "l", lwd=2)
mtext(outer = TRUE, side = 3, line = -0.5, cex = 1.5, "car axis")
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