daes
A differential-algebraic equation is defined by an implicit function
f(du,u,p,t)=0. All of the controls are the same as the
other examples, except here you define a function which returns the
residuals for each part of the equation to define the DAE. The initial
value u0 and the initial derivative du0 are
required, though they do not necessarily have to satisfy f
(known as inconsistent initial conditions). The methods will
automatically find consistent initial conditions. In order for this to
occur, differential_vars must be set. This vector states
which of the variables are differential (have a derivative term), with
false meaning that the variable is purely algebraic.
This example shows how to solve the Robertson equation:
f <- function (du,u,p,t) {
resid1 = - 0.04*u[1] + 1e4*u[2]*u[3] - du[1]
resid2 = + 0.04*u[1] - 3e7*u[2]^2 - 1e4*u[2]*u[3] - du[2]
resid3 = u[1] + u[2] + u[3] - 1.0
c(resid1,resid2,resid3)
}
u0 <- c(1.0, 0, 0)
du0 <- c(-0.04, 0.04, 0.0)
tspan <- c(0.0,100000.0)
differential_vars <- c(TRUE,TRUE,FALSE)
prob <- de$DAEProblem(f,du0,u0,tspan,differential_vars=differential_vars)
sol <- de$solve(prob)
udf <- as.data.frame(t(sapply(sol$u,identity)))
plotly::plot_ly(udf, x = sol$t, y = ~V1, type = 'scatter', mode = 'lines') %>%
plotly::add_trace(y = ~V2) %>%
plotly::add_trace(y = ~V3)Additionally, an in-place JIT compiled form for f can be
used to enhance the speed:
f = JuliaCall::julia_eval("function f(out,du,u,p,t)
out[1] = - 0.04u[1] + 1e4*u[2]*u[3] - du[1]
out[2] = + 0.04u[1] - 3e7*u[2]^2 - 1e4*u[2]*u[3] - du[2]
out[3] = u[1] + u[2] + u[3] - 1.0
end")
u0 <- c(1.0, 0, 0)
du0 <- c(-0.04, 0.04, 0.0)
tspan <- c(0.0,100000.0)
differential_vars <- c(TRUE,TRUE,FALSE)
JuliaCall::julia_assign("du0", du0)
JuliaCall::julia_assign("u0", u0)
JuliaCall::julia_assign("p", p)
JuliaCall::julia_assign("tspan", tspan)
JuliaCall::julia_assign("differential_vars", differential_vars)
prob = JuliaCall::julia_eval("DAEProblem(f, du0, u0, tspan, p, differential_vars=differential_vars)")
sol = de$solve(prob)
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