Ode error message
See inline remarks.
On 16-07-2013, at 10:07, Rapha?lle Carraud <raphaelle.carraud at oc-metalchem.com> wrote:
Hello, I am creating a program with R to solve a differential equation system. However, I get the following message I do not understand :
out <- ode(y = state, times = z, func = liquide, parms = 0, atol = 0)
DLSODA- EWT(I1) is R1 .le. 0.0 In above message, I1 = 1 In above message, R1 = 0 Error in lsoda(y, times, func, parms, ...) : illegal input detected before taking any integration steps - see written message or this one when I tried modifying the atoll value :
out <- ode(y = state, times = z, func = liquide, parms = 0, atol = 10^-14)
DLSODA- Warning..Internal T (=R1) and H (=R2) are
such that in the machine, T + H = T on the next step
(H = step size). Solver will continue anyway.
In above message, R1 = 0, R2 = 0
DINTDY- T (=R1) illegal
In above message, R1 = 1
T not in interval TCUR - HU (= R1) to TCUR (=R2)
In above message, R1 = 0, R2 = 0
DINTDY- T (=R1) illegal
In above message, R1 = 2
T not in interval TCUR - HU (= R1) to TCUR (=R2)
In above message, R1 = 0, R2 = 0
DLSODA- Trouble in DINTDY. ITASK = I1, TOUT = R1
In above message, I1 = 1
In above message, R1 = 2
Here is my program. I also tried changing the initial values but it does not work well.
liquide <- function(z, state, parameters) {
with(as.list(c(state,parameters)),{
# rate of change
Tr <- 273+90
Why are you defining Tr? It is not used anywhere
C <- CA + CB + CC + CD + CE + CI + CG + CJ + CK + CH
Same thing. Not used.
K32 <- 6.54*10^4 K33 <- 1.37*10^4 K34 <- 330 K35 <- 5.81*10^4 kf2 <- 1.37*10^3 kf3 <- 1.37*10^3 kf4 <- 8.68*10^5 kf5 <- 157.2 K2 <- 10^1.37 K3 <- 10^(-3.35) r1 <- kf4*CD - kf4/K34*CE^2 r2 <- kf3*CA*CB - kf3/K33*CD r3 <- kf2*CA^2 - kf2/K32*CC r4 <- kf5*CC - kf5/K35*CE*CI^2 dCA <- -r2 # dNO/dt dCB <- -r3 - r2 # dNO2/dt dCC <- r3/2 - r4 # dN2O4/dt dCD <- r2 - r1 # dN2O3/dt dCE <- 2*r1 + r4 # dHNO2/dt dCI <- r4 # dHNO3/dt dCG <- -r4 - r1 # dH2O/dz dCH <- (dCE + dCI)/((K2 + K3)*(CE + CI)) # dH/dz dCJ <- (CH*dCI - CI*dCH)/(K3*CH^2) # dNO3-/dz
You are dividing by CH, which is 0 initially. So what value does dCH then get?
dCK <- (CH*dCE - CE*dCH)/(K2*CH^2) # dNO2-/dz
Same thing.
list(c(dCA, dCB, dCC, dCD, dCE, dCI, dCG, dCH, dCJ, dCK)) }) # end with(as.list ... } Ti <- 273+90 # K
You are not using Ti.
Ct <- 5100 # mol/m^3
And Ct is also not used.
state <-c(CA = 0, # mol/m^3 NO2
CB = 0, # mol/m^3 NO
CC = 0, # mol/m^3 N2O4
CD = 0, # mol/m^3 N2O3
CE = 50, # mol/m^3 HNO2
CI = 50, # mol/m^3 HNO3
CG = 5000, # mol/m^3 H2O
CH = 0, # mol/m^3 H+
0!!
CJ = 0, # mol/m^3 NO3-
CK = 0) # mol/m^3 NO2-
parameters <- c(Ct = 5100)
Why not parameters <- c(Ct = Ct)?
z <- seq(0, 15, by = 1) # en seconde library(deSolve) out <- ode(y = state, times = z, func = liquide, parms = 0, atol = 10^-14) head(out) plot(out)
You will still get messages. You should really learn to de elementary debugging. Such as inserting liquide(0,state,parameters) after defining state and parameters to check and test. Berend
Thank you
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