Nothing

```
#
# vim:set ff=unix expandtab ts=2 sw=2:
MeanAge=function# mean age for a general one pool model
### The function computes the mean age for one pool of a possibly nonlinear model
(IdotT, ##<< The inputrate as a function of time
OdotLin,##<< The outputrate of this pool as a linear operator (a function of y and t)
sol, ##<< The solution of the nonlinear equation for this pool as a function of time
times ##<< A vector containing the points in time where the solution is sought.
){
rho=function(a,tk){
startTime=tk-a
#startVal=IdotT(startTime)
startVal=sol(startTime)
if (startTime<0){
print("if")
print(paste("a=",a,"tk=",tk))
return(startVal)
}
times=c(startTime,tk,2*tk)
s=solver(times,OdotLin,startVal)
res=s[2]/sol(tk)
return(res)
}
E=function(tk){
E_integrand=function(Y,a){a*rho(a,tk)}
res=solver(c(0,tk*0.99),E_integrand,0)[2]
return(res)
}
res=sapply(times,E)
return(res)
### A vector containing the mean age for the specified times
}
########################################################
MeanAge2=function# mean age for a general one pool model
### The function computes the mean age for one pool of a possibly nonlinear model
(IdotT, ##<< The inputrate as a function of time
OdotLin,##<< The outputrate of this pool as a linear operator (a function of y and t)
sol, ##<< The solution of the nonlinear equation for this pool as a function of time
times ##<< A vector containing the points in time where the solution is sought.
){
maxage=max(times)-min(times)
fineTimes=(seq(sqrt(min(times)),sqrt(max(times)),sqrt(maxage)/10000))^2
#fineTimes=seq(min(times),max(times),maxage/100000)
startval=1
OofT=splinefun(fineTimes,solver(fineTimes,OdotLin,startval))
rho=function(a,tk){
s=OofT(tk)*IdotT(tk-a)/OofT(tk-a)
res=s/sol(tk)
return(res)
}
E=function(tk){
rho_tk=function(a){rho(a,tk)}
E_integrand=function(Y,a){a*rho(a,tk)}
res=solver(c(tk/1000,tk),E_integrand,0)[2]
return(res)
}
res=mcmapply(E,times,mc.cores=16)
return(res)
### A vector containing the mean age for the specified times
}
########################################################
MeanAge3=function# mean age for a general one pool model
### The function computes the mean age for one pool of a possibly nonlinear model
(IdotT, ##<< The inputrate as a function of time
OdotLin,##<< The outputrate of this pool as a linear operator (a function of y and t)
sol, ##<< The solution of the nonlinear equation for this pool as a function of time
times ##<< A vector containing the points in time where the solution is sought.
){
so=solver(times,OdotLin,1)
#we assume that the startvalue is the result of an input in the first timestep
# first we compute the value of the solution at the first time
# this is also very useful if the initial value is zero
# which would make our Linear Operator useless since it has a fixpoint at 0 (being linear)
deltaT=times[[2]]-times[[1]]
tstart=times[[2]]
startvalue=sol(tstart)
if(startvalue==0){stop("cannot compute the solution since we are in a fixed point")}
# now we ad an impulsive inputrate necessarry to achieve the startvalue
# nearly immidiately but possibly smooth for the integration
Id=function(t){IdotT(t)}
l=100
pdf(file="runit.MeanAge.Densities.pdf",paper="a4r")
#plot(times,so)
OofT=splinefun(times,so)
#plot(times,OofT(times))
s=function(a,tk){
OofT(tk)*Id(tk-a)/OofT(tk-a)
}
rho=function(a,tk){
res=s(a,tk)/sol(tk)
return(res)
}
E=function(tk){
l=100
#ages=log(seq(1,exp(tk*(1-1/l)),length=l))
ages=seq(0,tk-times[[2]],length=l)
s_tk=function(a){s(a,tk)}
ss_tk=splinefun(ages,mapply(s_tk,ages))
# we compute the probability for a partikle to
# have an age in the range of [0,tk-times[[2]]]
one=integrate(ss_tk,lower=0,upper=tk-times[[2]])$value/sol(tk)
# so error=1-one must be the probability to have an age in the range
#[tk-times[[2]],tk] so we estimate rho for this inteval as
rho_0=(1-one)/(deltaT)
Integrand=function(a){a*rho(a,tk)}
spIntegrand=splinefun(ages,sapply(ages,Integrand))
res=integrate(spIntegrand,lower=0,upper=max(ages))$value+rho_0*(tk-(deltaT)/2)
return(res)
}
times[[1]]<-times[[2]]
#times[[2]]<-times[[3]]
#print(times)
dev.off()
#lapply(times,E)
res=mclapply(times,E,mc.cores=16)
return(res)
### A vector containing the mean age for the specified times
}
```

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