Solving a system of differential equations in R - r

Question

I have a simple flux model in R. It boils down to two differential equations that model two state variables within the model, we'll call them A and B. They are calculated as simple difference equations of four component fluxes flux1-flux4, 5 parameters p1-p5, and a 6th parameter, of_interest, that can take on values between 0-1.
parameters<- c(p1=0.028, p2=0.3, p3=0.5, p4=0.0002, p5=0.001, of_interest=0.1)
state <- c(A=28, B=1.4)
model<-function(t,state,parameters){
with(as.list(c(state,parameters)),{
#fluxes
flux1 = (1-of_interest) * p1*(B / (p2 + B))*p3
flux2 = p4* A #microbial death
flux3 = of_interest * p1*(B / (p2 + B))*p3
flux4 = p5* B
#differential equations of component fluxes
dAdt<- flux1 - flux2
dBdt<- flux3 - flux4
list(c(dAdt,dBdt))
})
I would like to write a function to take the derivative of dAdt with respect to of_interest, set the derived equation to 0, then rearrange and solve for the value of of_interest. This will be the value of the parameter of_interest that maximizes the function dAdt.
So far I have been able to solve the model at steady state, across the possible values of of_interest to demonstrate there should be a maximum.
require(rootSolve)
range<- seq(0,1,by=0.01)
for(i in range){
of_interest=i
parameters<- c(p1=0.028, p2=0.3, p3=0.5, p4=0.0002, p5=0.001, of_interest=of_interest)
state <- c(A=28, B=1.4)
ST<- stode(y=y,func=model,parms=parameters,pos=T)
out<- c(out,ST$y[1])
Then plotting:
plot(out~range, pch=16,col='purple')
lines(smooth.spline(out~range,spar=0.35), lwd=3,lty=1)
How can I analytically solve for the value of of_interest that maximizes dAdt in R? If an analytical solution is not possible, how can I know, and how can I go about solving this numerically?
Update: I think this problem can be solved with the deSolve package in R, linked here, however I am having trouble implementing it using my particular example.

Answer 1

Your equation in B(t) is just-about separable since you can divide out B(t), from which you can get that
B(t) = C * exp{-p5 * t} * (p2 + B(t)) ^ {of_interest * p1 * p3}
This is an implicit solution for B(t) which we'll solve point-wise.
You can solve for C given your initial value of B. I suppose t = 0 initially? In which case
C = B_0 / (p2 + B_0) ^ {of_interest * p1 * p3}
This also gives a somewhat nicer-looking expression for A(t):
dA(t) / dt = B_0 / (p2 + B_0) * p1 * p3 * (1 - of_interest) *
exp{-p5 * t} * ((p2 + B(t) / (p2 + B_0)) ^
{of_interest * p1 * p3 - 1} - p4 * A(t)
This can be solved by integrating factor (= exp{p4 * t}), via numerical integration of the term involving B(t). We specify the lower limit of the integral as 0 so that we never have to evaluate B outside the range [0, t], which means the integrating constant is simply A_0 and thus:
A(t) = (A_0 + integral_0^t { f(tau; parameters) d tau}) * exp{-p4 * t}
The basic gist is B(t) is driving everything in this system -- the approach will be: solve for the behavior of B(t), then use this to figure out what's going on with A(t), then maximize.
First, the "outer" parameters; we also need nleqslv to get B:
library(nleqslv)
t_min <- 0
t_max <- 10000
t_N <- 10
#we'll only solve the behavior of A & B over t_rng
t_rng <- seq(t_min, t_max, length.out = t_N)
#I'm calling of_interest ttheta
ttheta_min <- 0
ttheta_max <- 1
ttheta_N <- 5
tthetas <- seq(ttheta_min, ttheta_max, length.out = ttheta_N)
B_0 <- 1.4
A_0 <- 28
#No sense storing this as a vector when we'll only ever use it as a list
parameters <- list(p1 = 0.028, p2 = 0.3, p3 = 0.5,
p4 = 0.0002, p5 = 0.001)
From here, the basic outline is:
Given the parameter values (in particular ttheta), solve for BB over t_rng via non-linear equation solving
Given BB and the parameter values, solve for AA over t_rng by numerical integration
Given AA and your expression for dAdt, plug & maximize.
derivs <-
sapply(tthetas, function(th){
#append current ttheta
params <- c(parameters, ttheta = th)
#declare a function we'll use to solve for B (see above)
b_slv <- function(b, t)
with(params, b - B_0 * ((p2 + b)/(p2 + B_0)) ^
(ttheta * p1 * p3) * exp(-p5 * t))
#solving point-wise (this is pretty fast)
# **See below for a note**
BB <- sapply(t_rng, function(t) nleqslv(B_0, function(b) b_slv(b, t))$x)
#this is f(tau; params) that I mentioned above;
# we have to do linear interpolation since the
# numerical integrator isn't constrained to the grid.
# **See below for note**
a_int <- function(t){
#approximate t to the grid (t_rng)
# (assumes B is monotonic, which seems to be true)
# (also, if t ends up negative, just assign t_rng[1])
t_n <- max(1L, which.max(t_rng - t >= 0) - 1L)
idx <- t_n:(t_n+1)
ts <- t_rng[idx]
#distance-weighted average of the local B values
B_app <- sum((-1) ^ (0:1) * (t - ts) / diff(ts) * BB[idx])
#finally, f(tau; params)
with(params, (1 - ttheta) * p1 * p3 * B_0 / (p2 + B_0) *
((p2 + B_app)/(p2 + B_0)) ^ (ttheta * p1 * p3 - 1) *
exp((p4 - p5) * t))
}
#a_int only works on scalars; the numeric integrator
# requires a version that works on vectors
a_int_v <- function(t) sapply(t, a_int)
AA <- exp(-params$p4 * t_rng) *
sapply(t_rng, function(tt)
#I found the subdivisions constraint binding in some cases
# at the default value; no trouble at 1000.
A_0 + integrate(a_int_v, 0, tt, subdivisions = 1000L)$value)
#using the explicit version of dAdt given as flux1 - flux2
max(with(params, (1 - ttheta) * p1 * p3 * BB / (p2 + BB) - p4 * AA))})
Finally, simply run `tthetas[which.max(derivs)]` to get the maximizer.
Note:
This code is not optimized for efficiency. There are a few places where there are some potential speed-ups:
probably faster to run the equation solver recursively, as it'll converge faster with better initial guesses -- using the previous value instead of the initial value is surely better
Will be faster to simply use Riemann sums to integrate; the tradeoff is in accuracy, but should be fine if you have a dense enough grid. One beauty of Riemann is you won't have to interpolate at all, and numerically they're simple linear algebra. I ran this with t_N == ttheta_N == 1000L and it ran within a few minutes.
Probably possible to vectorize a_int directly instead of just sapplying on it, which concomitant speed-up by more direct appeal to BLAS.
Loads of other small stuff. Pre-compute ttheta * p1 * p3 since it's re-used so much, etc.
I didn't bother including any of that stuff, though, because you're honestly probably better off porting this to a faster language -- Julia is my own pet favorite, but of course R speaks well with C++, C, Fortran, etc.