# Problem

Everything needed to solve a PDE in `FourierFlows.jl`

is gathered in a composite type named `Problem`

. `Problem`

contains various other composite types (see `Problem`

for details).

Here, we demonstrate how we can construct a `Problem`

to solve the simple 1D equation:

\[\partial_t u(x, t) = - \alpha \, u(x, t) ,\]

on domain $x \in [-1, 1]$.

First, we construct our grid

```
using FourierFlows
nx, Lx = 32, 2.0
grid = OneDGrid(; nx, Lx)
```

```
OneDimensionalGrid
├─────────── Device: CPU
├──────── FloatType: Float64
├────────── size Lx: 2.0
├──── resolution nx: 32
├── grid spacing dx: 0.0625
├─────────── domain: x ∈ [-1.0, 0.9375]
└─ aliased fraction: 0.3333333333333333
```

Our problem has a parameter $\alpha$. Thus, we create a `Params`

as:

```
struct Params <: AbstractParams
α :: Float64
end
```

and then we use the `Params`

's constructor to populate our `params`

with the parameter value, e.g., $\alpha = 0.1$:

```
α = 0.1
params = Params(α)
```

```
Parameters
└───── parameter: α -> Float64
```

The particular equation is so simple that it makes no difference performance-wise whether we time-step it in physical or in wavenumber space. For PDEs with nonlinear terms, time-stepping in wavenumber space is much more efficient. Thus, for demonstration purposes, we will time-step the equation in wavenumber space, i.e.,

\[\partial_t \hat{u}(k, t) = - \alpha \, \hat{u}(k, t) .\]

The variables involved are $u$ and its Fourier transform $\hat{u}$. Thus, we construct the `vars`

as:

```
struct Vars <: AbstractVars
u :: Array{Float64, 1}
uh :: Array{Complex{Float64}, 1}
end
```

and, like before, we use the `Vars`

's constructor to populate the `vars`

with zero arrays,

`vars = Vars(zeros(Float64, (grid.nx,)), zeros(Complex{Float64}, (grid.nkr,)))`

```
Variables
├───── variable: u -> 32-element Vector{Float64}
└───── variable: uh -> 17-element Vector{ComplexF64}
```

Note that the Fourier transform of a real-valued array `u`

is complex-valued. Also because we use the real Fourier transform, the array `uh`

is smaller.

In this example our state variable is simply `uh`

, i.e., `sol = uh`

.

Next we need to construct the equation. The equation contains the linear coefficients for the linear part of the PDE, stored in an array `L`

, and the function `calcN!()`

that calculates the nonlinear terms from the state variable `sol`

. In our case, our equation is linear and, therefore,

`L = - params.α * ones(grid.nkr)`

and

```
function calcN!(N, sol, t, clock, vars, params, grid)
@. N = 0
return nothing
end
```

Note that `calcN!()`

needs to have the above argument structure. With `L`

and `calcN!`

in hand we can construct our problem's equation:

`equation = FourierFlows.Equation(L, calcN!, grid)`

```
Equation
├──────── linear coefficients: L
│ ├───type: Float64
│ └───size: (17,)
├───────────── nonlinear term: calcN!()
└─── type of state vector sol: ComplexF64
```

Last, we have to pick a time-stepper and a time-step `dt`

and gather everything a FourierFlows's `Problem`

.

Time-steppers are prescribed via a string. Here we choose `"ForwardEuler"`

time-stepping scheme with a time step of 0.02.

```
stepper, dt = "ForwardEuler", 0.02
prob = FourierFlows.Problem(equation, stepper, dt, grid, vars, params)
```

```
Problem
├─────────── grid: grid (on CPU)
├───── parameters: params
├────── variables: vars
├─── state vector: sol
├─────── equation: eqn
├────────── clock: clock
│ └──── dt: 0.02
└──── timestepper: ForwardEulerTimeStepper
```

For more information and a list of implemented time-stepping schemes see the the following Time-stepping section.

By default, the `Problem`

constructor takes `sol`

a complex valued array filled with zeros with same size as `L`

.

The `prob.clock`

contains the time-step `dt`

and the current `step`

and time `t`

of the simulation:

`prob.clock`

```
Clock
├─── timestep dt: 0.02
├────────── step: 0
└──────── time t: 0.0
```

Let's initiate our problem with, e.g., $u(x, 0) = \cos(\pi x)$, integrate up to $t = 4$ and compare our numerical solution with the analytical solution $u(x, t) = e^{-\alpha t} \cos(\pi x)$.

```
u0 = @. cos(π * grid.x)
using LinearAlgebra: mul!
mul!(prob.sol, grid.rfftplan, u0)
```

Since our time-step is chosen `dt = 0.02`

, we need to step forward `prob`

for $200$ time-steps to reach $t = 4$.

`stepforward!(prob, 200)`

Now let's transform our state vector `sol`

back in physical space

```
using LinearAlgebra: ldiv!
ldiv!(prob.vars.u, grid.rfftplan, prob.sol)
```

and finally, let's plot our solution and compare with the analytical solution:

```
using CairoMakie, Printf
fig = Figure()
ax = Axis(fig[1, 1], xlabel = "x", title = @sprintf("u(x, t=%1.2f)", prob.clock.t))
x = range(-Lx/2, Lx/2, length=200)
lines!(ax, x, @. cos(π * x) * exp(-prob.params.α * 4); label = "analytical")
lines!(ax, x, @. cos(π * x); linestyle = :dash, color = :gray, label = "initial condition")
scatter!(ax, grid.x, prob.vars.u; markersize = 14, color = :salmon, label = "numerical")
axislegend()
```

A good practice is to encompass all functions and type definitions related with a PDE under a single module, e.g.,

```
module MyPDE
...
end # end module
```

For a more elaborate example we urge you to have a look at the `Diffusion`

module located at `src/diffusion.jl`

and also the modules included in the child package GeophysicalFlows.jl.