求解R中的微分方程组

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:

1. Given the parameter values (in particular ttheta ), solve for BB over t_rng via non-linear equation solving
2. Given BB and the parameter values, solve for AA over t_rng by numerical integration
3. 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 sapply ing 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.