How extreme values of a functional can be found using R? - r

Question

I have a functional like this :
(LaTex formula: $v[y]=\int_0^2 (y'^2+23yy'+12y^2+3ye^{2t})dt$)
with given start and end conditions y(0)=-1, y(2)=18.
How can I find extreme values of this functional in R? I realize how it can be done for example in Excel but didn't find appropriate solution in R.

Answer 1

Before trying to solve such a task in a numerical setting, it might be better to lean back and think about it for a moment.
This is a problem typically treated in the mathematical discipline of "variational calculus". A necessary condition for a function y(t) to be an extremum of the functional (ie. the integral) is the so-called Euler-Lagrange equation, see
Calculus of Variations at Wolfram Mathworld.
Applying it to f(t, y, y') as the integrand in your request, I get (please check, I can easily have made a mistake)
y'' - 12*y + 3/2*exp(2*t) = 0
You can go now and find a symbolic solution for this differential equation (with the help of a textbook, or some CAS), or solve it numerically with the help of an R package such as 'deSolve'.
PS: Solving this as an optimization problem based on discretization is possible, but may lead you on a long and stony road. I remember solving the "brachistochrone problem" to a satisfactory accuracy only by applying several hundred variables (not in R).

Answer 2

Here is a numerical solution in R. First the functional:
f<-function(y,t=head(seq(0,2,len=length(y)),-1)){
len<-length(y)-1
dy<-diff(y)*len/2
y0<-(head(y,-1)+y[-1])/2
2*sum(dy^2+23*y0*dy+12*y0^2+3*y0*exp(2*t))/len
}
Now the function that does the actual optimization. The best results I got were using the BFGS optimization method, and parametrizing using dy rather than y:
findMinY<-function(points=100, ## number of points of evaluation
boundary=c(-1,18), ## boundary values
y0=NULL, ## optional initial value
method="Nelder-Mead", ## optimization method
dff=T) ## if TRUE, optimizes based on dy rather than y
{
t<-head(seq(0,2,len=points),-1)
if(is.null(y0) || length(y0)!=points)
y0<-seq(boundary[1],boundary[2],len=points)
if(dff)
y0<-diff(y0)
else
y0<-y0[-1]
y0<-head(y0,-1)
ff<-function(z){
if(dff)
y<-c(cumsum(c(boundary[1],z)),boundary[2])
else
y<-c(boundary[1],z,boundary[2])
f(y,t)
}
res<-optim(y0,ff,control=list(maxit=1e9),method=method)
cat("Iterations:",res$counts,"\n")
ymin<-res$par
if(dff)
c(cumsum(c(boundary[1],ymin)),boundary[2])
else
c(boundary[1],ymin,boundary[2])
}
With 500 points of evaluation, it only takes a few seconds with BFGS:
> system.time(yy<-findMinY(500,method="BFGS"))
Iterations: 90 18
user system elapsed
2.696 0.000 2.703
The resulting function looks like this:
plot(seq(0,2,len=length(yy)),yy,type='l')

Answer 3

And now a solution that numerically integrates the Euler equation.
As #HansWerner pointed out, this problem boils down to applying the Euler-Lagrange equation to the integrand in OP's question, and then solving that differential equation, either analytically or numerically. In this case the relevant ODE is
y'' - 12*y = 3/2*exp(2*t)
subject to:
y(0) = -1
y(2) = 18
So this is a boundary value problem, best approached using bvpcol(...) in package bvpSolve.
library(bvpSolve)
F <- function(t, y.in, pars){
dy <- y.in[2]
d2y <- 12*y.in[1] + 1.5*exp(2*t)
return(list(c(dy,d2y)))
}
init <- c(-1,NA)
end <- c(18,NA)
t <- seq(0, 2, by = 0.01)
sol <- bvpcol(yini = init, yend = end, x = t, func = F)
y = function(t){ # analytic solution...
b <- sqrt(12)
a <- 1.5/(4-b*b)
u <- exp(2*b)
C1 <- ((18*u + 1) - a*(exp(4)*u-1))/(u*u - 1)
C2 <- -1 - a - C1
return(a*exp(2*t) + C1*exp(b*t) + C2*exp(-b*t))
}
par(mfrow=c(1,2))
plot(t,y(t), type="l", xlim=c(0,2),ylim=c(-1,18), col="red", main="Analytical Solution")
plot(sol[,1],sol[,2], type="l", xlim=c(0,2),ylim=c(-1,18), xlab="t", ylab="y(t)", main="Numerical Solution")
It turns out that in this very simple example, there is an analytical solution:
y(t) = a * exp(2*t) + C1 * exp(sqrt(12)*t) + C2 * exp(-sqrt(12)*t)
where a = -3/16 and C1 and C2 are determined to satisfy the boundary conditions. As the plots show, the numerical and analytic solution agree completely, and also agree with the solution provided by #mrip