gradient {rootSolve}

Estimates the gradient matrix for a simple function

Description

Given a vector of variables (x), and a function (f) that estimates one function value or a set of function values (f(x)), estimates the gradient matrix, containing, on rows i and columns j

d(f(x)_i)/d(x_j)

The gradient matrix is not necessarily square.

Usage

gradient(f, x, centered = FALSE, pert = 1e-8, ...)

Arguments

f

function returning one function value, or a vector of function values.

x

either one value or a vector containing the x-value(s) at which the gradient matrix should be estimated.

centered

if TRUE, uses a centered difference approximation, else a forward difference approximation.

pert

numerical perturbation factor; increase depending on precision of model solution.

...

other arguments passed to function f.

Details

the function f that estimates the function values will be called as f(x, ...). If x is a vector, then the first argument passed to f should also be a vector.

The gradient is estimated numerically, by perturbing the x-values.

Value

The gradient matrix where the number of rows equals the length of f and the number of columns equals the length of x.

the elements on i-th row and j-th column contain: d((f(x))_i)/d(x_j)

Note

gradient can be used to calculate so-called sensitivity functions, that estimate the effect of parameters on output variables.

Author(s)

Karline Soetaert <[e-mail]>

References

Soetaert, K. and P.M.J. Herman (2008). A practical guide to ecological modelling - using R as a simulation platform. Springer.

See Also

jacobian.full, for generating a full and square gradient (jacobian) matrix and where the function call is more complex

hessian, for generating the Hessian matrix

Examples

## =======================================================================
## 1. Sensitivity analysis of the logistic differential equation
## dN/dt = r*(1-N/K)*N  , N(t0)=N0.
## =======================================================================

# analytical solution of the logistic equation:
logistic <- function (x, times) {

 with (as.list(x),
 {
  N <- K / (1+(K-N0)/N0*exp(-r*times))
  return(c(N = N))
  })
}

# parameters for the US population from 1900
x <- c(N0 = 76.1, r = 0.02, K = 500)

# Sensitivity function: SF: dfi/dxj at
# output intervals from 1900 to 1950
SF <- gradient(f = logistic, x, times = 0:50)

# sensitivity, scaled for the value of the parameter:
# [dfi/(dxj/xj)]= SF*x (columnise multiplication)
sSF <- (t(t(SF)*x))
matplot(sSF, xlab = "time", ylab = "relative sensitivity ",
        main = "logistic equation", pch = 1:3)
legend("topleft", names(x), pch = 1:3, col = 1:3)

# mean scaled sensitivity
colMeans(sSF)

## =======================================================================
## 2. Stability of the budworm model, as a function of its
## rate of increase.
##
## Example from the book of Soetaert and Herman(2009)
## A practical guide to ecological modelling,
## using R as a simulation platform. Springer
## code and theory are explained in this book
## =======================================================================

r   <- 0.05
K   <- 10
bet <- 0.1
alf <- 1

# density-dependent growth and sigmoid-type mortality rate
rate <- function(x, r = 0.05) r*x*(1-x/K) - bet*x^2/(x^2+alf^2)

# Stability of a root ~ sign of eigenvalue of Jacobian 
stability <- function (r)  {
  Eq <- uniroot.all(rate, c(0, 10), r = r)
  eig  <- vector()
  for (i in 1:length(Eq)) 
      eig[i] <- sign (gradient(rate, Eq[i], r = r))
  return(list(Eq = Eq, Eigen = eig))
}

# bifurcation diagram
rseq <- seq(0.01, 0.07, by = 0.0001)

plot(0, xlim = range(rseq), ylim = c(0, 10), type = "n",
     xlab = "r", ylab = "B*", main = "Budworm model, bifurcation",
     sub = "Example from Soetaert and Herman, 2009")

for (r in rseq) {
  st <- stability(r) 
  points(rep(r, length(st$Eq)), st$Eq, pch = 22,
         col = c("darkblue", "black", "lightblue")[st$Eigen+2],
         bg = c("darkblue", "black", "lightblue")[st$Eigen+2]) 
}

legend("topleft", pch = 22, pt.cex = 2, c("stable", "unstable"),
        col = c("darkblue","lightblue"), 
        pt.bg = c("darkblue", "lightblue"))