---
title: "Solving Ordinary Differential Equations (ODE) in R with diffeqr"
author: "Chris Rackauckas"
date: "`r Sys.Date()`"
output: rmarkdown::html_vignette
vignette: >
  %\VignetteIndexEntry{Solving Ordinary Differential Equations (ODE) in R with diffeqr}
  %\VignetteEngine{knitr::rmarkdown}
  %\VignetteEncoding{UTF-8}
---

```{r setup, include = FALSE}
knitr::opts_chunk$set(
  collapse = TRUE,
  comment = "#>"
)
```

### 1D Linear ODEs

Let's solve the linear ODE `u'=1.01u`. First setup the package:

```R
de <- diffeqr::diffeq_setup()
```

Define the derivative function `f(u,p,t)`.

```R
f <- function(u,p,t) {
  return(1.01*u)
}
```

Then we give it an initial condition and a time span to solve over:

```R
u0 <- 1/2
tspan <- c(0., 1.)
```

With those pieces we define the `ODEProblem` and `solve` the ODE:

```R
prob = de$ODEProblem(f, u0, tspan)
sol = de$solve(prob)
```

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:

```R
plot(sol$t,sol$u,"l")
```

![linear_ode](https://user-images.githubusercontent.com/1814174/39011970-e04f1fe8-43c7-11e8-8da3-848362691783.png)

### Systems of ODEs

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:

```R
f <- function(u,p,t) {
  du1 = p[1]*(u[2]-u[1])
  du2 = u[1]*(p[2]-u[3]) - u[2]
  du3 = u[1]*u[2] - p[3]*u[3]
  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:

```R
u0 <- c(1.0,0.0,0.0)
tspan <- list(0.0,100.0)
p <- c(10.0,28.0,8/3)
prob <- de$ODEProblem(f, u0, tspan, p)
sol <- de$solve(prob)
```

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`:

```R
mat <- sapply(sol$u,identity)
```

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:

```R
udf <- as.data.frame(t(mat))
```

Now we can use `matplot` to plot the timeseries together:

```R
matplot(sol$t,udf,"l",col=1:3)
```

![timeseries](https://user-images.githubusercontent.com/1814174/39012314-ef7a8fe2-43c8-11e8-9dde-1a8b87d3cfa4.png)

Now we can use the Plotly package to draw a phase plot:

```R
plotly::plot_ly(udf, x = ~V1, y = ~V2, z = ~V3, type = 'scatter3d', mode = 'lines')
```

![plotly_plot](https://user-images.githubusercontent.com/1814174/39012384-27ee7262-43c9-11e8-84d2-1edf937288ae.png)

Plotly is much prettier!

### Option Handling

If we want to have a more accurate solution, we can send `abstol` and `reltol`. Defaults are `1e-6` and `1e-3` respectively.
Generally you can think of the digits of accuracy as related to 1 plus the exponent of the relative tolerance, so the default is
two digits of accuracy. Absolute tolernace is the accuracy near 0.

In addition, we may want to choose to save at more time points. We do this by giving an array of values to save at as `saveat`.
Together, this looks like:

```R
abstol <- 1e-8
reltol <- 1e-8
saveat <- 0:10000/100
sol <- de$solve(prob,abstol=abstol,reltol=reltol,saveat=saveat)
udf <- as.data.frame(t(sapply(sol$u,identity)))
plotly::plot_ly(udf, x = ~V1, y = ~V2, z = ~V3, type = 'scatter3d', mode = 'lines')
```

![precise_solution](https://user-images.githubusercontent.com/1814174/39012651-e03124e6-43c9-11e8-8496-bbee87987a37.png)

We can also choose to use a different algorithm. The choice is done using a string that matches the Julia syntax. See
[the ODE tutorial for details](https://diffeq.sciml.ai/dev/tutorials/ode_example/#Choosing-a-Solver-Algorithm-1).
The list of choices for ODEs can be found at the [ODE Solvers page](https://diffeq.sciml.ai/dev/solvers/ode_solve/).
For example, let's use a 9th order method due to Verner:

```R
sol <- de$solve(prob,de$Vern9(),abstol=abstol,reltol=reltol,saveat=saveat)
```

Note that each algorithm choice will cause a JIT compilation

## Performance Enhancements

One way to enhance the performance of your code is to define the function in Julia
so that way it is JIT compiled. diffeqr is built using
[the JuliaCall package](https://github.com/Non-Contradiction/JuliaCall), and so
you can utilize the Julia JIT compiler. We expose this automatically over ODE
functions via `jitoptimize_ode`, like in the following example:

```R
f <- function(u,p,t) {
  du1 = p[1]*(u[2]-u[1])
  du2 = u[1]*(p[2]-u[3]) - u[2]
  du3 = u[1]*u[2] - p[3]*u[3]
  return(c(du1,du2,du3))
}
u0 <- c(1.0,0.0,0.0)
tspan <- c(0.0,100.0)
p <- c(10.0,28.0,8/3)
prob <- de$ODEProblem(f, u0, tspan, p)
fastprob <- diffeqr::jitoptimize_ode(de,prob)
sol <- de$solve(fastprob,de$Tsit5())
```

Note that the first evaluation of the function will have an ~2 second lag since
the compiler will run, and all subsequent runs will be orders of magnitude faster
than the pure R function. This means it's great for expensive functions (ex. large
PDEs) or functions called repeatedly, like during optimization of parameters.

We can also use the JuliaCall functions to directly define the function in Julia
to eliminate the R interpreter overhead and get full JIT compilation:

```R
julf <- JuliaCall::julia_eval("
function julf(du,u,p,t)
  du[1] = 10.0*(u[2]-u[1])
  du[2] = u[1]*(28.0-u[3]) - u[2]
  du[3] = u[1]*u[2] - (8/3)*u[3]
end")
JuliaCall::julia_assign("u0", u0)
JuliaCall::julia_assign("p", p)
JuliaCall::julia_assign("tspan", tspan)
prob3 = JuliaCall::julia_eval("ODEProblem(julf, u0, tspan, p)")
sol = de$solve(prob3,de$Tsit5())
```

To demonstrate the performance advantage, let's time them all:

```R
> system.time({ for (i in 1:100){ de$solve(prob    ,de$Tsit5()) }})
   user  system elapsed
   6.69    0.06    6.78
> system.time({ for (i in 1:100){ de$solve(fastprob,de$Tsit5()) }})
   user  system elapsed
   0.11    0.03    0.14
> system.time({ for (i in 1:100){ de$solve(prob3   ,de$Tsit5()) }})
   user  system elapsed
   0.14    0.02    0.15
```

This is about a 50x improvement!
