henry R - 7 days ago 5x

R Question

Hey I am trying to use the deSolve package for a set of ODE's with equations as auxillary variables. I keep getting the error where the number of derivatives isn't the same length as the initial conditions vector. What should I change?

`rm(list=ls())`

library(deSolve)

exponential=function(t,state,parameters){ with(as.list( c(state,parameters)), {

#Aux. Var.

fX2 = pmax(0,1-(1-(d2/r12)*(X2/K2)))

fX1 = X1/(X1+k1);

# equations (ODE)

dX1 = C-((d1)*(X1))-(r12)*(X2)*fX2*fX1 # differential equaion

dX2 = r12*(X2)*fX2*fX1-((d2)*(X2))

return(list(c(dX1, dX2)))

})

}

# -- RUN INFORMATION

# Set Initial Values and Simulation Time

state = c(X1=2,X2=0.01,K2= 10)

times=0:100

# Assign Parameter Values

parameters = c(d1=0.001, d2=0.008, r12=0.3,C=0.5,k1= 0.001)

for (i in 1:length(times)){

out= ode(y=state,times=times,func=exponential,parms=parameters)

}

**> for (i in 1:length(times)){

+ out= ode(y=state,times=times,func=exponential,parms=parameters)

+ }

Error in checkFunc(Func2, times, y, rho) :

The number of derivatives returned by func() (2) must equal the length of the initial conditions vector (3)**

Answer

The error comes from the `return`

in your defined function:
Your input parameter `y`

has length 3, but you only return 2 values back, that's the error. You can solve your problem with

```
return(list(c(X1, X2, K2)))
```

Another possibility is to take `K2`

to the parameters, then your old `return`

was right. You have to decide if `K2`

is a variable or a parameter.

And BTW: Why a for loop with the time? In my opinion that is not necessary, because the ODEs are solved in the timeinterval you submitted to the `ode`

function.

```
out= ode(y=state,times=times,func=exponential,parms=parameters)
```

Source (Stackoverflow)

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