Discrete Problems

Mathematical Specification of a Discrete Problem

To define an Discrete Problem, you simply need to give the function $f$ and the initial condition $u₀$ which define a function map:

\[u_{n+1} = f(u,p,t_{n+1})\]

f should be specified as f(u,p,t) (or in-place as f(du,u,p,t)), and u₀ should be an AbstractArray (or number) whose geometry matches the desired geometry of u. Note that we are not limited to numbers or vectors for u₀; one is allowed to provide u₀ as arbitrary matrices / higher dimension tensors as well. t_{n+1} is the current time at which the map is applied. For a FunctionMap with defaults, t_n = t0 + n*dt (with dt=1 being the default). For continuous-time Markov chains this is the time at which the change is occuring.

Note that if the discrete solver is set to have scale_by_time=true, then the problem is interpreted as the map:

\[u_{n+1} = u_n + dt f(u,p,t_n)\]

Problem Type

Constructors

DiscreteProblem{isinplace}(f::ODEFunction,u0,tspan,p=NullParameters();kwargs...) : Defines the discrete problem with the specified functions.
DiscreteProblem{isinplace}(f,u0,tspan,p=NullParameters();kwargs...) : Defines the discrete problem with the specified functions.
DiscreteProblem{isinplace}(u0,tspan,p=NullParameters();kwargs...) : Defines the discrete problem with the identity map.

Parameters are optional, and if not given then a NullParameters() singleton will be used which will throw nice errors if you try to index non-existent parameters. Any extra keyword arguments are passed on to the solvers. For example, if you set a callback in the problem, then that callback will be added in every solve call.

For specifying Jacobians and mass matrices, see the DiffEqFunctions page.

Fields

f: The function in the map.
u0: The initial condition.
tspan: The timespan for the problem.
p: The parameters for the problem. Defaults to NullParameters
kwargs: The keyword arguments passed onto the solves.

Note About Timing

Note that if no dt and not tstops is given, it's assumed that dt=1 and thus tspan=(0,n) will solve for n iterations. If in the solver dt is given, then the number of iterations will change. And if tstops is not empty, the solver will revert to the standard behavior of fixed timestep methods, which is "step to each tstop".