(Experimental) Modeling Discrete Systems

In this example, we will use the System 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]
@mtkcompile sys = System(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, vcat(u0, p), tspan)
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}}:
 [0.0, 10.0, 990.0]
 [0.24690087971667385, 10.24797539090573, 989.5051237293776]
 [0.49992429364961877, 10.501843308492768, 988.9982323978576]
 [0.7592157288009717, 10.761730776476044, 988.479053494723]
 [1.0249238083995655, 11.027766894444877, 987.9473092971556]
 [1.2972003431544508, 11.300082836466316, 987.4027168203793]
 [1.5762003824739328, 11.578811847349312, 986.8449877701768]
 [1.8620822655923719, 11.864089236428692, 986.2738284979789]
 [2.1550076725435083, 12.15605236872322, 985.6889399587333]
 [2.45514167491548, 12.454840653316563, 985.090017671768]
 ⋮
 [76.95307768711389, 75.90609660457736, 847.1408257083083]
 [78.82720588986679, 77.24103252919139, 843.9317615809414]
 [80.73429377803495, 78.58696696003413, 840.6787392619304]
 [82.67461290570472, 79.94348599302349, 837.3819011012713]
 [84.64842460763423, 81.31015501563621, 834.0414203767291]
 [86.6559794879602, 82.6865186599478, 830.6575018520915]
 [88.69751690774523, 84.07210079533738, 827.2303822969169]
 [90.77326447234499, 85.46640456244165, 823.7603309652128]
 [92.88343751961379, 86.86891244986607, 820.2476500305196]

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
@mtkcompile sys = System([x ~ x(k - 1) + x(k - 2)], t)

\[ \begin{align} Shift(t, 1)\left( \mathtt{x_{t - 2}}\left( t \right) \right) &= \mathtt{x_{t - 1}}\left( t \right) \\ Shift(t, 1)\left( \mathtt{x_{t - 1}}\left( t \right) \right) &= x\left( t \right) \end{align} \]

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 initial 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.