Rs中的Lsoda函数

Question

To calculate the behaviour of my system of 2 equations, I use the lsoda function in R. The parameters a varies over time : is 1 when time is even and 0 otherwise. Here is a minimal reproducible example :

mo=function(t,x,m){
if(t%%2==0){
   a=m
}else{
   a=0
}
dx=rep(0,2)
dx[1]=-a*x[1]
dx[2]=a*x[1]
res=dx
return(list(dx))
}
xs=c(10,0)
ti=1:100
m=1
data1=as.data.frame(lsoda(xs,ti,mo,m))

The problem is that when you investigate the results, you see that x[1] remains at 10 while x[2] remains at 0, meaning that lsoda always take the value 0 for a (and never 1). Is it an issue with the modulo?

Answer 1

lsoda chooses its own internal time step, so there is no guarantee that t%%2 == 0 will ever be true even if you ask for output when this condition is true (eg, t=2 ). My understanding is that lsoda will interpolate between solution times to get output at the times requested by the user and not necessarily solve the model at those times. As a rule of thumb, hard temporal or spatial interfaces are a Bad Thing™ when it comes to numerical ODE solvers.