dde {ddesolve} | R Documentation |
A solver for systems of delay differential equations based off numerical routines from Simon Wood's solv95 program. This solver is also capable of solving systems of ordinary differential equations.
Please see the included demos for examples of how to use dde
.
To view available demos run demo(package="ddesolve")
. The supplied
demos require that the package PBSmodelling be installed.
dde(y, times, func, parms=NULL, switchfunc=NULL, mapfunc=NULL, tol=1e-08, dt=0.1, hbsize=10000)
y |
vector of initial values of the DDE system. The size of the supplied vector determines the number of variables in the system. |
times |
numeric vector of specific times to solve. |
func |
a user supplied function that computes the gradients in the DDE
system at time t . The function must be defined using the arguments:
(t,y) or (t,y,parms) , where t is the current time
in the integration, y is a vector of the current estimated variables
of the DDE system, and parms is any R object representing
additional parameters (optional).
The argument func must return one of the two following return types:
1) a vector containing the calculated gradients for each variable; or
2) a list with two elements - the first a vector of calculated gradients,
the second a vector (possibly named) of values for a variable specified by
the user at each point in the integration.
|
parms |
any constant parameters to pass to func , switchfunc ,
and mapfunc .
|
switchfunc |
an optional function that is used to manipulate state
values at given times. The switch function takes the arguments (t,y) or
(t,y,parms) and must return a numeric vector. The size of the vector
determines the number of switches used by the model. As values of
switchfunc pass through zero (from positive to negative), a corresponding
call to mapfunc is made, which can then modify any state value.
|
mapfunc |
if switchfunc is defined, then a map function must also
be supplied with arguments (t,y,switch_id) or t,y,switch_id,parms) ,
where t is the time, y are the current state values, switch_id
is the index of the triggered switch, and parms are additional constant parameters.
|
tol |
maximum error tolerated at each time step (as a proportion of the state variable concerned) |
dt |
maximum initial time step |
hbsize |
history buffer size required for solving DDEs) |
The user supplied function func
can access past values (lags) of
y
by calling the pastvalue
function. Past gradients
are accessible by the pastgradient
function. These functions
can only be called from func
and can only be passed values of
t
greater or equal to the start time, but less than the current time
of the integration point. For example, calling pastvalue(t)
is not
allowed, since these values are the current values which are passed in as
y
.
A data frame with one column for t
, a column for every variable in the system,
and a column for every additional value that may (or may not) have been returned
by func
in the second element of the list.
If the initial y
values parameter was named, then the solved values column
will use the same names. Otherwise y1
, y2
, ... will be used.
If func
returned a list, with a named vector as the second element, then
those names will be used as the column names. If the vector was not named, then
extra1
, extra2
, ... will be used.
################################################## # This is just a single example of using dde. # For more examples see demo(package="ddesolve") # the demos require the package PBSmodelling ################################################## #create a func to return dde gradient require(ddesolve) yprime <- function(t,y,parms) { if (t < parms$tau) lag <- parms$initial else lag <- pastvalue(t - parms$tau) y1 <- parms$a * y[1] - (y[1]^3/3) + parms$m * (lag[1] - y[1]) y2 <- y[1] - y[2] return(c(y1,y2)) } #define initial values and parameters yinit <- c(1,1) parms <- list(tau=3, a=2, m=-10, initial=yinit) # solve the dde system yout <- dde(y=yinit,times=seq(0,30,0.1),func=yprime,parms=parms) # and display the results plot(yout$t, yout$y1, type="l", col="red", xlab="t", ylab="y", ylim=c(min(yout$y1, yout$y2), max(yout$y1, yout$y2))) lines(yout$t, yout$y2, col="blue") legend("topleft", legend = c("y1", "y2"),lwd=2, lty = 1, xjust = 1, yjust = 1, col = c("red","blue"))