Description Usage Arguments Details Value Note Author(s) See Also Examples
Solves a system of ordinary differential equations resulting from 1Dimensional partial differential equations that have been converted to ODEs by numerical differencing.
1 2 3 4 5 
y 
the initial (state) values for the ODE system, a vector. If

times 
time sequence for which output is wanted; the first
value of 
func 
either an Rfunction that computes the values of the
derivatives in the ODE system (the model definition) at time
If The return value of If 
parms 
parameters passed to 
nspec 
the number of species (components) in the model. If

dimens 
the number of boxes in the model. If 
method 
the integrator. Use Method 
names 
the names of the components; used for plotting. 
bandwidth 
the number of adjacent boxes over which transport occurs.
Normally equal to 1 (box i only interacts with box i1, and i+1).
Values larger than 1 will not work with 
restructure 
whether or not the Jacobian should be restructured.
Only used if the 
... 
additional arguments passed to the integrator. 
This is the method of choice for multispecies 1dimensional models, that are only subjected to transport between adjacent layers.
More specifically, this method is to be used if the state variables are arranged per species:
A[1], A[2], A[3],.... B[1], B[2], B[3],.... (for species A, B))
Two methods are implemented.
The default method rearranges the state variables as A[1], B[1], ... A[2], B[2], ... A[3], B[3], .... This reformulation leads to a banded Jacobian with (upper and lower) half bandwidth = number of species.
Then the selected integrator solves the banded problem.
The second method uses lsodes
. Based on the dimension
of the problem, the method first calculates the sparsity pattern
of the Jacobian, under the assumption that transport is only
occurring between adjacent layers. Then lsodes
is called to
solve the problem.
As lsodes
is used to integrate, it may be necessary to
specify the length of the real work array, lrw
.
Although a reasonable guess of lrw
is made, it is possible
that this will be too low. In this case, ode.1D
will
return with an error message telling the size of the work array
actually needed. In the second try then, set lrw
equal to
this number.
For instance, if you get the error:
DLSODES RWORK length is insufficient to proceed. Length needed is .ge. LENRW (=I1), exceeds LRW (=I2) In above message, I1 = 27627 I2 = 25932
set lrw
equal to 27627 or a higher value
If the model is specified in compiled code (in a DLL), then option 2,
based on lsodes
is the only solution method.
For singlespecies 1D models, you may also use ode.band
.
See the selected integrator for the additional options.
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 second element of the return from func
, plus an
additional column (the first) for the time value. There will be one
row for each element in times
unless the integrator returns
with an unrecoverable error. If y
has a names attribute, it
will be used to label the columns of the output value.
The output will have the attributes istate
, and rstate
,
two vectors with several useful elements. The first element of istate
returns the conditions under which the last call to the integrator
returned. Normal is istate = 2
. If verbose = TRUE
, the
settings of istate and rstate will be written to the screen. See the
help for the selected integrator for details.
It is advisable though not mandatory to specify both
nspec
and dimens
. In this case, the solver can check
whether the input makes sense (i.e. if nspec * dimens ==
length(y)
).
Karline Soetaert <karline.soetaert@nioz.nl>
ode
for a general interface to most of the ODE solvers,
ode.band
for integrating models with a banded Jacobian
ode.2D
for integrating 2D models
ode.3D
for integrating 3D models
lsodes
,lsode
, lsoda
,
lsodar
,vode
for the integration options.
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  ## =======================================================================
## example 1
## a predator and its prey diffusing on a flat surface
## in concentric circles
## 1D model with using cylindrical coordinates
## LotkaVolterra type biology
## =======================================================================
## ================
## Model equations
## ================
lvmod < function (time, state, parms, N, rr, ri, dr, dri) {
with (as.list(parms), {
PREY < state[1:N]
PRED < state[(N+1):(2*N)]
## Fluxes due to diffusion
## at internal and external boundaries: zero gradient
FluxPrey < Da * diff(c(PREY[1], PREY, PREY[N]))/dri
FluxPred < Da * diff(c(PRED[1], PRED, PRED[N]))/dri
## Biology: LotkaVolterra model
Ingestion < rIng * PREY * PRED
GrowthPrey < rGrow * PREY * (1PREY/cap)
MortPredator < rMort * PRED
## Rate of change = Flux gradient + Biology
dPREY < diff(ri * FluxPrey)/rr/dr +
GrowthPrey  Ingestion
dPRED < diff(ri * FluxPred)/rr/dr +
Ingestion * assEff  MortPredator
return (list(c(dPREY, dPRED)))
})
}
## ==================
## Model application
## ==================
## model parameters:
R < 20 # total radius of surface, m
N < 100 # 100 concentric circles
dr < R/N # thickness of each layer
r < seq(dr/2,by = dr,len = N) # distance of center to midlayer
ri < seq(0,by = dr,len = N+1) # distance to layer interface
dri < dr # dispersion distances
parms < c(Da = 0.05, # m2/d, dispersion coefficient
rIng = 0.2, # /day, rate of ingestion
rGrow = 1.0, # /day, growth rate of prey
rMort = 0.2 , # /day, mortality rate of pred
assEff = 0.5, # , assimilation efficiency
cap = 10) # density, carrying capacity
## Initial conditions: both present in central circle (box 1) only
state < rep(0, 2 * N)
state[1] < state[N + 1] < 10
## RUNNING the model:
times < seq(0, 200, by = 1) # output wanted at these time intervals
## the model is solved by the two implemented methods:
## 1. Default: banded reformulation
print(system.time(
out < ode.1D(y = state, times = times, func = lvmod, parms = parms,
nspec = 2, names = c("PREY", "PRED"),
N = N, rr = r, ri = ri, dr = dr, dri = dri)
))
## 2. Using sparse method
print(system.time(
out2 < ode.1D(y = state, times = times, func = lvmod, parms = parms,
nspec = 2, names = c("PREY","PRED"),
N = N, rr = r, ri = ri, dr = dr, dri = dri,
method = "lsodes")
))
## ================
## Plotting output
## ================
# the data in 'out' consist of: 1st col times, 2N+1: the prey
# N+2:2*N+1: predators
PREY < out[, 2:(N + 1)]
filled.contour(x = times, y = r, PREY, color = topo.colors,
xlab = "time, days", ylab = "Distance, m",
main = "Prey density")
# similar:
image(out, which = "PREY", grid = r, xlab = "time, days",
legend = TRUE, ylab = "Distance, m", main = "Prey density")
image(out2, grid = r)
# summaries of 1D variables
summary(out)
# 1D plots:
matplot.1D(out, type = "l", subset = time == 10)
matplot.1D(out, type = "l", subset = time > 10 & time < 20)
## =======================================================================
## Example 2.
## Biochemical Oxygen Demand (BOD) and oxygen (O2) dynamics
## in a river
## =======================================================================
## ================
## Model equations
## ================
O2BOD < function(t, state, pars) {
BOD < state[1:N]
O2 < state[(N+1):(2*N)]
## BOD dynamics
FluxBOD < v * c(BOD_0, BOD) # fluxes due to water transport
FluxO2 < v * c(O2_0, O2)
BODrate < r * BOD # 1st order consumption
## rate of change = flux gradient  consumption + reaeration (O2)
dBOD < diff(FluxBOD)/dx  BODrate
dO2 < diff(FluxO2)/dx  BODrate + p * (O2satO2)
return(list(c(dBOD = dBOD, dO2 = dO2)))
}
## ==================
## Model application
## ==================
## parameters
dx < 25 # grid size of 25 meters
v < 1e3 # velocity, m/day
x < seq(dx/2, 5000, by = dx) # m, distance from river
N < length(x)
r < 0.05 # /day, firstorder decay of BOD
p < 0.5 # /day, airsea exchange rate
O2sat < 300 # mmol/m3 saturated oxygen conc
O2_0 < 200 # mmol/m3 riverine oxygen conc
BOD_0 < 1000 # mmol/m3 riverine BOD concentration
## initial conditions:
state < c(rep(200, N), rep(200, N))
times < seq(0, 20, by = 0.1)
## running the model
## step 1 : model spinup
out < ode.1D(y = state, times, O2BOD, parms = NULL,
nspec = 2, names = c("BOD", "O2"))
## ================
## Plotting output
## ================
## select oxygen (first column of out:time, then BOD, then O2
O2 < out[, (N + 2):(2 * N + 1)]
color = topo.colors
filled.contour(x = times, y = x, O2, color = color, nlevels = 50,
xlab = "time, days", ylab = "Distance from river, m",
main = "Oxygen")
## or quicker plotting:
image(out, grid = x, xlab = "time, days", ylab = "Distance from river, m")

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