Let’s solve the linear ODE u'=1.01u
. First setup the
package:
<- diffeqr::diffeq_setup() de
Define the derivative function f(u,p,t)
.
<- function(u,p,t) {
f return(1.01*u)
}
Then we give it an initial condition and a time span to solve over:
<- 1/2
u0 <- c(0., 1.) tspan
With those pieces we define the ODEProblem
and
solve
the ODE:
= de$ODEProblem(f, u0, tspan)
prob = de$solve(prob) sol
This gives back a solution object for which sol$t
are
the time points and sol$u
are the values. We can treat the
solution as a continuous object in time via
and a high order interpolation will compute the value at
t=0.2
. We can check the solution by plotting it:
plot(sol$t,sol$u,"l")
linear_ode
Now let’s solve the Lorenz equations. In this case, our initial condition is a vector and our derivative functions takes in the vector to return a vector (note: arbitrary dimensional arrays are allowed). We would define this as:
<- function(u,p,t) {
f = p[1]*(u[2]-u[1])
du1 = u[1]*(p[2]-u[3]) - u[2]
du2 = u[1]*u[2] - p[3]*u[3]
du3 return(c(du1,du2,du3))
}
Here we utilized the parameter array p
. Thus we use
diffeqr::ode.solve
like before, but also pass in parameters
this time:
<- c(1.0,0.0,0.0)
u0 <- list(0.0,100.0)
tspan <- c(10.0,28.0,8/3)
p <- de$ODEProblem(f, u0, tspan, p)
prob <- de$solve(prob) sol
The returned solution is like before except now sol$u
is
an array of arrays, where sol$u[i]
is the full system at
time sol$t[i]
. It can be convenient to turn this into an R
matrix through sapply
:
<- sapply(sol$u,identity) mat
This has each row as a time series. t(mat)
makes each
column a time series. It is sometimes convenient to turn the output into
a data.frame
which is done via:
<- as.data.frame(t(mat)) udf
Now we can use matplot
to plot the timeseries
together:
matplot(sol$t,udf,"l",col=1:3)
timeseries
Now we can use the Plotly package to draw a phase plot:
::plot_ly(udf, x = ~V1, y = ~V2, z = ~V3, type = 'scatter3d', mode = 'lines') plotly