# 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:

`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, $t_n = t_0 + n*dt$ (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:

## 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".