Description Usage Arguments Details Value Author(s) References See Also Examples
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.
1 2 3 4 5 6 7 8  mebdfi(y, times, func = NULL, parms, dy = NULL, res = NULL,
nind=c(length(y),0,0), rtol = 1e6, atol = 1e6, 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 Rfunction 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 threevalued 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 1e6, 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 nonzero bands above the diagonal, in case
the Jacobian is banded (and 
banddown 
number of nonzero 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 dllfunctions whose names are
specified by 
ipar 
only when ‘dllname’ is specified: a vector with
integer values passed to the dllfunctions 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 twocolumned 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 Rfunction. 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 <[email protected]>
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, 645657, (1983).
J.R. Cash and S. Considine, an MEBDF code for stiff initial value problems, ACM Trans Math Software, 142158, (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.
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160 161 162 163 164 165 166 167 168 169 170 171 172 173 174 175 176 177 178 179 180 181 182 183 184 185 186 187 188 189 190 191 192 193 194 195 196 197 198 199 200 201 202 203 204 205 206 207 208 209 210  ## =======================================================================
## 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 1st 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
## dBdA = 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 = 1e8, rtol = 1e8))
## 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 = 1e8, rtol = 1e8))
## =======================================================================
## 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 = 1e10, rtol = 1e10))
## 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 = 1e10, rtol = 1e10))
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(ODEDAE))
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((xrxb)^2 + (yryb)^2)
f[1:4] < y[5:8]
k < M*eps*eps/2
f[5] < (L0Ll)*xl/Ll + lam1*xb+2*lam2*(xlxr)
f[6] < (L0Ll)*yl/Ll + lam1*yb+2*lam2*(ylyr)k*g
f[7] < (L0Lr)*(xrxb)/Lr  2*lam2*(xlxr)
f[8] < (L0Lr)*(yryb)/Lr  2*lam2*(ylyr)k*g
f[9] < xb*xl+yb*yl
f[10] < (xlxr)^2+(ylyr)^2L*L
delt < dyf
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 = 1e2, 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 = 1e5, atol = 1e5)
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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