## =============================================================================
## Example 1: simple standard problem
## solve the BVP ODE:
## d2y/dt^2=-3py/(p+t^2)^2
## y(t= -0.1)=-0.1/sqrt(p+0.01)
## y(t= 0.1)= 0.1/sqrt(p+0.01)
## where p = 1e-5
##
## analytical solution y(t) = t/sqrt(p + t^2).
##
## The problem is rewritten as a system of 2 ODEs:
## dy=y2
## dy2=-3p*y/(p+t^2)^2
## =============================================================================
#--------------------------------
# Derivative function
#--------------------------------
fun <- function(t, y, pars) {
dy1 <- y[2]
dy2 <- - 3 * p * y[1] / (p+t*t)^2
return(list(c(dy1,
dy2))) }
# parameter value
p <- 1e-5
# initial and final condition; second conditions unknown
init <- c(-0.1 / sqrt(p+0.01), NA)
end <- c( 0.1 / sqrt(p+0.01), NA)
# Solve bvp
sol <- bvpcol(yini = init, yend = end,
x = seq(-0.1, 0.1, by = 0.001), func = fun)
plot(sol, which = 1)
# add analytical solution
curve(x/sqrt(p+x*x), add = TRUE, type = "p")
diagnostics(sol)
zoom <- approx(sol, xout = seq(-0.005, 0.005, by = 0.0001))
plot(zoom, which = 1, main = "zoom in on [-0.0005,0.0005]")
## =============================================================================
## Example 1b:
## Same problem, now solved as a second-order equation
## and with different value of "p".
## =============================================================================
fun2 <- function(t, y, pars)
{ dy <- - 3 * p * y[1] / (p+t*t)^2
list(dy)
}
p <- 1e-4
sol2 <- bvpcol(yini = init, yend = end, order = 2,
x = seq(-0.1, 0.1, by = 0.001), func = fun2)
# plot both runs at once:
plot(sol, sol2, which = 1)
## =============================================================================
## Example 1c: simple
## solve d2y/dx2 + 1/x*dy/dx + (1-1/(4x^2)y = sqrt(x)*cos(x),
## on the interval [1,6] and with boundary conditions:
## y(1)=1, y(6)=-0.5
##
## Write as set of 2 odes
## dy/dx = y2
## dy2/dx = - 1/x*dy/dx - (1-1/(4x^2)y + sqrt(x)*cos(x)
## =============================================================================
f2 <- function(x, y, parms)
{
dy <- y[2]
dy2 <- -1/x * y[2]- (1-1/(4*x^2))*y[1] + sqrt(x)*cos(x)
list(c(dy, dy2))
}
x <- seq(1, 6, 0.1)
sol <- bvpcol(yini = c(1, NA), yend = c(-0.5, NA), bspline = TRUE,
x = x, func = f2)
plot(sol, which = 1)
# add the analytic solution
curve(0.0588713*cos(x)/sqrt(x) + 1/4*sqrt(x)*cos(x)+0.740071*sin(x)/sqrt(x)+
1/4*x^(3/2)*sin(x), add = TRUE, type = "l")
## =============================================================================
## Example 2. Uses continuation
## Test problem 24
## =============================================================================
Prob24<- function(t, y, ks) { #eps is called ks here
A <- 1+t*t
AA <- 2*t
ga <- 1.4
list(c(y[2],(((1+ga)/2 -ks*AA)*y[1]*y[2]-y[2]/y[1]-
(AA/A)*(1-(ga-1)*y[1]^2/2))/(ks*A*y[1])))
}
ini <- c(0.9129, NA)
end <- c(0.375, NA)
xguess <- c(0, 1)
yguess <- matrix(nrow = 2, ncol = 2, 0.9 )
# bvpcol works with eps NOT too small, and good initial condition ...
sol <- bvpcol(yini = ini, yend = end, x = seq(0, 1, by = 0.01),
xguess = xguess, yguess = yguess,
parms = 0.1, func = Prob24, verbose = FALSE)
# when continuation is used: does not need a good initial condition
sol2 <- bvpcol(yini = ini, yend = end, x = seq(0, 1, by = 0.01),
parms = 0.05, func = Prob24,
eps = 0.05)
#zoom <- approx(sol2, xout = seq(0.01, 0.02, by = 0.0001))
#plot(zoom, which = 1, main = "zoom in on [0.01, 0.02]")
sol3 <- bvpcol(yini = ini, yend = end, x = seq(0, 1, by = 0.01),
parms = 0.01, func = Prob24 , eps = 0.01)
sol4 <- bvpcol(yini = ini, yend = end, x = seq(0, 1, by = 0.01),
parms = 0.001, func = Prob24, eps = 0.001)
print(system.time(
sol5 <- bvpcol(yini = ini, yend = end, x = seq(0, 1, by = 0.01),
parms = 1e-4, func = Prob24, eps = 1e-4)
))
plot(sol, sol2, sol3, sol4, sol5, which = 1, main = "test problem 24",
lwd = 2)
legend("topright", col = 1:5, lty = 1:5, lwd = 2,
legend = c("0.05", "0.01", "0.001", "0.0001"), title = "eps")
## =============================================================================
## Example 3 - solved with specification of boundary, and jacobians
## d4y/dx4 =R(dy/dx*d2y/dx2 -y*dy3/dx3)
## y(0)=y'(0)=0
## y(1)=1, y'(1)=0
##
## dy/dx = y2
## dy2/dx = y3 (=d2y/dx2)
## dy3/dx = y4 (=d3y/dx3)
## dy4/dx = R*(y2*y3 -y*y4)
## =============================================================================
# derivative function: 4 first-order derivatives
f1st<- function(x, y, S) {
list(c(y[2],
y[3],
y[4],
1/S*(y[2]*y[3] - y[1]*y[4]) ))
}
# jacobian of derivative function
df1st <- function(x, y, S) {
matrix(nrow = 4, ncol = 4, byrow = TRUE, data = c(
0, 1, 0, 0,
0, 0, 1, 0,
0, 0, 0, 1,
-1*y[4]/S, y[3]/S, y[2]/S, -y[1]/S))
}
# boundary
g2 <- function(i, y, S) {
if (i == 1) return (y[1])
if (i == 2) return (y[2])
if (i == 3) return (y[1] - 1)
if (i == 4) return (y[2])
}
# jacobian of boundary
dg2 <- function(i, y, S) {
if (i == 1) return(c(1, 0, 0, 0))
if (i == 2) return(c(0, 1, 0, 0))
if (i == 3) return(c(1, 0, 0, 0))
if (i == 4) return(c(0, 1, 0, 0))
}
# we use posbound to specify the position of boundary conditions
# we can also use leftbc = 2 rather than posbound = c(0,0,1,1)
S <- 1/100
sol <- bvpcol(x = seq(0, 1, by = 0.01),
ynames = c("y", "dy", "d2y", "d3y"),
posbound = c(0, 0, 1, 1), func = f1st, parms = S, eps = S,
bound = g2, jacfunc = df1st, jacbound = dg2)
plot(sol)
## =============================================================================
## Example 3b - solved with specification of boundary, and jacobians
## and as a higher-order derivative
## d4y/dx4 =R(dy/dx*d2y/dx2 -y*dy3/dx3)
## y(0)=y'(0)=0
## y(1)=1, y'(1)=0
## =============================================================================
# derivative function: one fourth-order derivative
f4th <- function(x, y, S) {
list(1/S * (y[2]*y[3] - y[1]*y[4]))
}
# jacobian of derivative function
df4th <- function(x, y, S) {
matrix(nrow = 1, ncol = 4, byrow = TRUE, data = c(
-1*y[4]/S, y[3]/S, y[2]/S, -y[1]/S))
}
# boundary function - same as previous example
# jacobian of boundary - same as previous
# order = 4 specifies the equation to be 4th order
# solve with bspline false
S <- 1/100
sol <- bvpcol (x = seq(0, 1, by = 0.01),
ynames = c("y", "dy", "d2y", "d3y"),
posbound = c(0, 0, 1, 1), func = f4th, order = 4,
parms = S, eps = S, bound = g2, jacfunc = df4th,
jacbound = dg2 )
plot(sol)
# Use (manual) continuation to find solution of a more difficult example
# Previous solution collocation from sol passed ("guess = sol")
sol2 <- bvpcol(x = seq(0, 1, by = 0.01),
ynames = c("y", "dy", "d2y", "d3y"),
posbound = c(0, 0, 1, 1), func = f4th,
parms = 1e-6, order = 4, eps = 1e-6,
bound = g2, jacfunc = df4th, jacbound = dg2 )
# plot both at same time
plot(sol, sol2, lwd = 2)
legend("bottomright", leg = c(100, 10000), title = "R = ",
col = 1:2, lty = 1:2, lwd = 2)
## =============================================================================
## Example 4 - a multipoint bvp
## dy1 = (y2 - 1)/2
## dy2 = (y1*y2 - x)/mu
## over interval [0,1]
## y1(1) = 0; y2(0.5) = 1
## =============================================================================
multip <- function (x, y, p) {
list(c((y[2] - 1)/2,
(y[1]*y[2] - x)/mu))
}
bound <- function (i, y, p) {
if (i == 1) y[2] -1 # at x=0.5: y2=1
else y[1] # at x= 1: y1=0
}
mu <- 0.1
sol <- bvpcol(func = multip, bound = bound,
x = seq(0, 1, 0.01), posbound = c(0.5, 1))
plot(sol)
# check boundary value
sol[sol[,1] == 0.5,]Run the code above in your browser using DataLab