(Experimental) Modeling Discrete Systems

In this example, we will use the new DiscreteSystem API to create an SIR model.

using ModelingToolkit
using ModelingToolkit: t_nounits as t
using OrdinaryDiffEq: solve, FunctionMap

@inline function rate_to_proportion(r, t)
    1 - exp(-r * t)
end
@parameters c δt β γ
@constants h = 1
@variables S(t) I(t) R(t)
k = ShiftIndex(t)
infection = rate_to_proportion(
    β * c * I(k - 1) / (S(k - 1) * h + I(k - 1) + R(k - 1)), δt * h) * S(k - 1)
recovery = rate_to_proportion(γ * h, δt) * I(k - 1)

# Equations
eqs = [S(k) ~ S(k - 1) - infection * h,
    I(k) ~ I(k - 1) + infection - recovery,
    R(k) ~ R(k - 1) + recovery]
@mtkbuild sys = DiscreteSystem(eqs, t)

u0 = [S(k - 1) => 990.0, I(k - 1) => 10.0, R(k - 1) => 0.0]
p = [β => 0.05, c => 10.0, γ => 0.25, δt => 0.1]
tspan = (0.0, 100.0)
prob = DiscreteProblem(sys, u0, tspan, p)
sol = solve(prob, FunctionMap())

retcode: Success
Interpolation: left-endpoint piecewise constant
t: 101-element Vector{Float64}:
   0.0
   1.0
   2.0
   3.0
   4.0
   5.0
   6.0
   7.0
   8.0
   9.0
   ⋮
  92.0
  93.0
  94.0
  95.0
  96.0
  97.0
  98.0
  99.0
 100.0
u: 101-element Vector{Vector{Float64}}:
 [989.5051237293776, 10.24797539090573, 0.24690087971667385]
 [988.9982323978576, 10.501843308492768, 0.49992429364961877]
 [988.479053494723, 10.761730776476044, 0.7592157288009717]
 [987.9473092971556, 11.027766894444877, 1.0249238083995655]
 [987.4027168203793, 11.300082836466316, 1.2972003431544508]
 [986.8449877701768, 11.578811847349312, 1.5762003824739328]
 [986.2738284979789, 11.864089236428692, 1.8620822655923719]
 [985.6889399587333, 12.15605236872322, 2.1550076725435083]
 [985.090017671768, 12.454840653316563, 2.45514167491548]
 [984.4767516848771, 12.76059552880404, 2.7626527863189656]
 ⋮
 [843.9317615809414, 77.24103252919139, 78.82720588986679]
 [840.6787392619304, 78.58696696003413, 80.73429377803495]
 [837.3819011012713, 79.94348599302349, 82.67461290570472]
 [834.0414203767291, 81.31015501563621, 84.64842460763423]
 [830.6575018520915, 82.6865186599478, 86.6559794879602]
 [827.2303822969169, 84.07210079533738, 88.69751690774523]
 [823.7603309652128, 85.46640456244165, 90.77326447234499]
 [820.2476500305196, 86.86891244986607, 92.88343751961379]
 [816.692674974925, 88.27908641507038, 95.02823861000405]

All shifts must be non-positive, i.e., discrete-time variables may only be indexed at index k, k-1, k-2, .... If default values are provided, they are treated as the value of the variable at the previous timestep. For example, consider the following system to generate the Fibonacci series:

@variables x(t) = 1.0
@mtkbuild sys = DiscreteSystem([x ~ x(k - 1) + x(k - 2)], t)

\[ \begin{align} \mathrm{Shift(t, 1)}\left( \mathrm{Shift(t, -1)}\left( x\left( t \right) \right) \right) =& x\left( t \right) \\ \mathrm{Shift(t, 1)}\left( x\left( t \right) \right) =& \mathrm{Shift(t, -1)}\left( x\left( t \right) \right) + x\left( t \right) \end{align} \]

The "default value" here should be interpreted as the value of x at all past timesteps. For example, here x(k-1) and x(k-2) will be 1.0, and the inital value of x(k) will thus be 2.0. During problem construction, the past value of a variable should be provided. For example, providing [x => 1.0] while constructing this problem will error. Provide [x(k-1) => 1.0] instead. Note that values provided during problem construction do not apply to the entire history. Hence, if [x(k-1) => 2.0] is provided, the value of x(k-2) will still be 1.0.