Code Optimization for Differential Equations

Note

See this FAQ for information on common pitfalls and how to improve performance.

Code Optimization in Julia

Before starting this tutorial, we recommend the reader to check out one of the many tutorials for optimization Julia code. The following is an incomplete list:

User-side optimizations are important because, for sufficiently difficult problems, most time will be spent inside your f function, the function you are trying to solve. “Efficient” integrators are those that reduce the required number of f calls to hit the error tolerance. The main ideas for optimizing your DiffEq code, or any Julia function, are the following:

  • Make it non-allocating
  • Use StaticArrays for small arrays
  • Use broadcast fusion
  • Make it type-stable
  • Reduce redundant calculations
  • Make use of BLAS calls
  • Optimize algorithm choice

We'll discuss these strategies in the context of differential equations. Let's start with small systems.

Example Accelerating a Non-Stiff Equation: The Lorenz Equation

Let's take the classic Lorenz system. Let's start by naively writing the system in its out-of-place form:

function lorenz(u, p, t)
    dx = 10.0 * (u[2] - u[1])
    dy = u[1] * (28.0 - u[3]) - u[2]
    dz = u[1] * u[2] - (8 / 3) * u[3]
    [dx, dy, dz]
end
lorenz (generic function with 1 method)

Here, lorenz returns an object, [dx,dy,dz], which is created within the body of lorenz.

This is a common code pattern from high-level languages like MATLAB, SciPy, or R's deSolve. However, the issue with this form is that it allocates a vector, [dx,dy,dz], at each step. Let's benchmark the solution process with this choice of function:

import DifferentialEquations as DE, BenchmarkTools as BT
u0 = [1.0; 0.0; 0.0]
tspan = (0.0, 100.0)
prob = DE.ODEProblem(lorenz, u0, tspan)
BT.@btime DE.solve(prob, DE.Tsit5());
  3.282 ms (192104 allocations: 7.44 MiB)

The BenchmarkTools.jl package's BT.@benchmark runs the code multiple times to get an accurate measurement. The minimum time is the time it takes when your OS and other background processes aren't getting in the way. Notice that in this case it takes about 5ms to solve and allocates around 11.11 MiB. However, if we were to use this inside of a real user code, we'd see a lot of time spent doing garbage collection (GC) to clean up all the arrays we made. Even if we turn off saving, we have these allocations.

BT.@btime DE.solve(prob, DE.Tsit5(); save_everystep = false);
  2.767 ms (169897 allocations: 6.49 MiB)

The problem, of course, is that arrays are created every time our derivative function is called. This function is called multiple times per step and is thus the main source of memory usage. To fix this, we can use the in-place form to ***make our code non-allocating***:

function lorenz!(du, u, p, t)
    du[1] = 10.0 * (u[2] - u[1])
    du[2] = u[1] * (28.0 - u[3]) - u[2]
    du[3] = u[1] * u[2] - (8 / 3) * u[3]
    nothing
end
lorenz! (generic function with 1 method)

Here, instead of creating an array each time, we utilized the cache array du. When the in-place form is used, DifferentialEquations.jl takes a different internal route that minimizes the internal allocations as well.

Note

Notice that nothing is returned. When in in-place form, the ODE solver ignores the return. Instead, make sure that the original du array is mutated instead of constructing a new array

When we benchmark this function, we will see quite a difference.

u0 = [1.0; 0.0; 0.0]
tspan = (0.0, 100.0)
prob = DE.ODEProblem(lorenz!, u0, tspan)
BT.@btime DE.solve(prob, DE.Tsit5());
  755.276 μs (23214 allocations: 1013.34 KiB)
BT.@btime DE.solve(prob, DE.Tsit5(); save_everystep = false);
  362.438 μs (71 allocations: 3.89 KiB)

There is a 16x time difference just from that change! Notice there are still some allocations and this is due to the construction of the integration cache. But this doesn't scale with the problem size:

tspan = (0.0, 500.0) # 5x longer than before
prob = DE.ODEProblem(lorenz!, u0, tspan)
BT.@btime DE.solve(prob, DE.Tsit5(); save_everystep = false);
  1.872 ms (71 allocations: 3.89 KiB)

Since that's all setup allocations, the user-side optimization is complete.

Further Optimizations of Small Non-Stiff ODEs with StaticArrays

Allocations are only expensive if they are “heap allocations”. For a more in-depth definition of heap allocations, there are many sources online. But a good working definition is that heap allocations are variable-sized slabs of memory which have to be pointed to, and this pointer indirection costs time. Additionally, the heap has to be managed, and the garbage controllers has to actively keep track of what's on the heap.

However, there's an alternative to heap allocations, known as stack allocations. The stack is statically-sized (known at compile time) and thus its accesses are quick. Additionally, the exact block of memory is known in advance by the compiler, and thus re-using the memory is cheap. This means that allocating on the stack has essentially no cost!

Arrays have to be heap allocated because their size (and thus the amount of memory they take up) is determined at runtime. But there are structures in Julia which are stack-allocated. structs for example are stack-allocated “value-type”s. Tuples are a stack-allocated collection. The most useful data structure for DiffEq though is the StaticArray from the package StaticArrays.jl. These arrays have their length determined at compile-time. They are created using macros attached to normal array expressions, for example:

import StaticArrays
A = StaticArrays.SA[2.0, 3.0, 5.0]
typeof(A) # StaticArrays.SVector{3, Float64} (alias for StaticArrays.SArray{Tuple{3}, Float64, 1, 3})
SVector{3, Float64} (alias for StaticArraysCore.SArray{Tuple{3}, Float64, 1, 3})

Notice that the 3 after StaticArrays.SVector gives the size of the StaticArrays.SVector. It cannot be changed. Additionally, StaticArrays.SVectors are immutable, so we have to create a new StaticArrays.SVector to change values. But remember, we don't have to worry about allocations because this data structure is stack-allocated. SArrays have numerous extra optimizations as well: they have fast matrix multiplication, fast QR factorizations, etc. which directly make use of the information about the size of the array. Thus, when possible, they should be used.

Unfortunately, static arrays can only be used for sufficiently small arrays. After a certain size, they are forced to heap allocate after some instructions and their compile time balloons. Thus, static arrays shouldn't be used if your system has more than ~20 variables. Additionally, only the native Julia algorithms can fully utilize static arrays.

Let's ***optimize lorenz using static arrays***. Note that in this case, we want to use the out-of-place allocating form, but this time we want to output a static array:

function lorenz_static(u, p, t)
    dx = 10.0 * (u[2] - u[1])
    dy = u[1] * (28.0 - u[3]) - u[2]
    dz = u[1] * u[2] - (8 / 3) * u[3]
    StaticArrays.SA[dx, dy, dz]
end
lorenz_static (generic function with 1 method)

To make the solver internally use static arrays, we simply give it a static array as the initial condition:

u0 = StaticArrays.SA[1.0, 0.0, 0.0]
tspan = (0.0, 100.0)
prob = DE.ODEProblem(lorenz_static, u0, tspan)
BT.@btime DE.solve(prob, DE.Tsit5());
  268.098 μs (2615 allocations: 392.71 KiB)
BT.@btime DE.solve(prob, DE.Tsit5(); save_everystep = false);
  193.299 μs (25 allocations: 2.03 KiB)

And that's pretty much all there is to it. With static arrays, you don't have to worry about allocating, so use operations like * and don't worry about fusing operations (discussed in the next section). Do “the vectorized code” of R/MATLAB/Python and your code in this case will be fast, or directly use the numbers/values.

Example Accelerating a Stiff Equation: the Robertson Equation

For these next examples, let's solve the Robertson equations (also known as ROBER):

\[\begin{aligned} \frac{dy_1}{dt} &= -0.04y₁ + 10^4 y_2 y_3 \\ \frac{dy_2}{dt} &= 0.04 y_1 - 10^4 y_2 y_3 - 3*10^7 y_{2}^2 \\ \frac{dy_3}{dt} &= 3*10^7 y_{2}^2 \\ \end{aligned}\]

Given that these equations are stiff, non-stiff ODE solvers like DE.Tsit5 or DE.Vern9 will fail to solve these equations. The automatic algorithm will detect this and automatically switch to something more robust to handle these issues. For example:

import DifferentialEquations as DE
import Plots
function rober!(du, u, p, t)
    y₁, y₂, y₃ = u
    k₁, k₂, k₃ = p
    du[1] = -k₁ * y₁ + k₃ * y₂ * y₃
    du[2] = k₁ * y₁ - k₂ * y₂^2 - k₃ * y₂ * y₃
    du[3] = k₂ * y₂^2
    nothing
end
prob = DE.ODEProblem(rober!, [1.0, 0.0, 0.0], (0.0, 1e5), [0.04, 3e7, 1e4])
sol = DE.solve(prob)
Plots.plot(sol, tspan = (1e-2, 1e5), xscale = :log10)
Example block output
import BenchmarkTools as BT
BT.@btime DE.solve(prob);
  103.470 μs (1739 allocations: 124.88 KiB)

Choosing a Good Solver

Choosing a good solver is required for getting top-notch speed. General recommendations can be found on the solver page (for example, the ODE Solver Recommendations). The current recommendations can be simplified to a Rosenbrock method (DE.Rosenbrock23 or DE.Rodas5) for smaller (<50 ODEs) problems, ESDIRK methods for slightly larger (DE.TRBDF2 or DE.KenCarp4 for <2000 ODEs), and DE.QNDF for even larger problems. lsoda from LSODA.jl is sometimes worth a try for the medium-sized category.

More details on the solver to choose can be found by benchmarking. See the SciMLBenchmarks to compare many solvers on many problems.

From this, we try the recommendation of DE.Rosenbrock23() for stiff ODEs at default tolerances:

BT.@btime DE.solve(prob, DE.Rosenbrock23());
  75.070 μs (635 allocations: 31.66 KiB)

Declaring Jacobian Functions

In order to reduce the Jacobian construction cost, one can describe a Jacobian function by using the jac argument for the DE.ODEFunction. First we have to derive the Jacobian $\frac{df_i}{du_j}$ which is J[i,j]. From this, we get:

function rober_jac!(J, u, p, t)
    y₁, y₂, y₃ = u
    k₁, k₂, k₃ = p
    J[1, 1] = k₁ * -1
    J[2, 1] = k₁
    J[3, 1] = 0
    J[1, 2] = y₃ * k₃
    J[2, 2] = y₂ * k₂ * -2 + y₃ * k₃ * -1
    J[3, 2] = y₂ * 2 * k₂
    J[1, 3] = k₃ * y₂
    J[2, 3] = k₃ * y₂ * -1
    J[3, 3] = 0
    nothing
end
f! = DE.ODEFunction(rober!, jac = rober_jac!)
prob_jac = DE.ODEProblem(f!, [1.0, 0.0, 0.0], (0.0, 1e5), (0.04, 3e7, 1e4))
ODEProblem with uType Vector{Float64} and tType Float64. In-place: true
Non-trivial mass matrix: false
timespan: (0.0, 100000.0)
u0: 3-element Vector{Float64}:
 1.0
 0.0
 0.0
BT.@btime DE.solve(prob_jac, DE.Rosenbrock23());
  54.980 μs (561 allocations: 29.08 KiB)

Automatic Derivation of Jacobian Functions

But that was hard! If you want to take the symbolic Jacobian of numerical code, we can make use of ModelingToolkit.jl to symbolic-ify the numerical code and do the symbolic calculation and return the Julia code for this.

import ModelingToolkit as MTK
de = MTK.complete(MTK.modelingtoolkitize(prob))

\[ \begin{align} \frac{\mathrm{d} \mathtt{x_1}\left( t \right)}{\mathrm{d}t} &= - \mathtt{x_1}\left( t \right) \mathtt{\alpha_1} + \mathtt{x_2}\left( t \right) \mathtt{x_3}\left( t \right) \mathtt{\alpha_3} \\ \frac{\mathrm{d} \mathtt{x_2}\left( t \right)}{\mathrm{d}t} &= \mathtt{x_1}\left( t \right) \mathtt{\alpha_1} - \left( \mathtt{x_2}\left( t \right) \right)^{2} \mathtt{\alpha_2} - \mathtt{x_2}\left( t \right) \mathtt{x_3}\left( t \right) \mathtt{\alpha_3} \\ \frac{\mathrm{d} \mathtt{x_3}\left( t \right)}{\mathrm{d}t} &= \left( \mathtt{x_2}\left( t \right) \right)^{2} \mathtt{\alpha_2} \end{align} \]

We can tell it to compute the Jacobian if we want to see the code:

MTK.generate_jacobian(de)[2] # Second is in-place
:(function (ˍ₋out, __mtk_arg_1, ___mtkparameters___, t)
      #= /root/.cache/julia-buildkite-plugin/depots/0185fce3-4489-413a-a934-123dd653ef61/packages/Symbolics/egyPz/src/build_function.jl:368 =# @inbounds begin
              #= /root/.cache/julia-buildkite-plugin/depots/0185fce3-4489-413a-a934-123dd653ef61/packages/Symbolics/egyPz/src/build_function.jl:368 =#
              begin
                  #= /root/.cache/julia-buildkite-plugin/depots/0185fce3-4489-413a-a934-123dd653ef61/packages/SymbolicUtils/aooYZ/src/code.jl:409 =#
                  #= /root/.cache/julia-buildkite-plugin/depots/0185fce3-4489-413a-a934-123dd653ef61/packages/SymbolicUtils/aooYZ/src/code.jl:410 =#
                  #= /root/.cache/julia-buildkite-plugin/depots/0185fce3-4489-413a-a934-123dd653ef61/packages/SymbolicUtils/aooYZ/src/code.jl:411 =#
                  begin
                      __mtk_arg_2 = ___mtkparameters___[1]
                      __mtk_arg_3 = ___mtkparameters___[2]
                      begin
                          #= /root/.cache/julia-buildkite-plugin/depots/0185fce3-4489-413a-a934-123dd653ef61/packages/SymbolicUtils/aooYZ/src/code.jl:464 =# @inbounds begin
                                  #= /root/.cache/julia-buildkite-plugin/depots/0185fce3-4489-413a-a934-123dd653ef61/packages/SymbolicUtils/aooYZ/src/code.jl:460 =#
                                  ˍ₋out[1] = (*)(-1, __mtk_arg_2[2])
                                  ˍ₋out[2] = __mtk_arg_2[2]
                                  ˍ₋out[3] = 0
                                  ˍ₋out[4] = (*)(__mtk_arg_1[3], __mtk_arg_2[1])
                                  ˍ₋out[5] = (+)((*)((*)(-2, __mtk_arg_1[2]), __mtk_arg_2[3]), (*)((*)(-1, __mtk_arg_1[3]), __mtk_arg_2[1]))
                                  ˍ₋out[6] = (*)((*)(2, __mtk_arg_1[2]), __mtk_arg_2[3])
                                  ˍ₋out[7] = (*)(__mtk_arg_1[2], __mtk_arg_2[1])
                                  ˍ₋out[8] = (*)((*)(-1, __mtk_arg_1[2]), __mtk_arg_2[1])
                                  ˍ₋out[9] = 0
                                  #= /root/.cache/julia-buildkite-plugin/depots/0185fce3-4489-413a-a934-123dd653ef61/packages/SymbolicUtils/aooYZ/src/code.jl:462 =#
                                  ˍ₋out
                              end
                      end
                  end
              end
          end
  end)

Now let's use that to give the analytical solution Jacobian:

prob_jac2 = DE.ODEProblem(de, [], (0.0, 1e5); jac = true)
ODEProblem with uType Vector{Float64} and tType Float64. In-place: true
Initialization status: FULLY_DETERMINED
Non-trivial mass matrix: false
timespan: (0.0, 100000.0)
u0: 3-element Vector{Float64}:
 1.0
 0.0
 0.0
BT.@btime DE.solve(prob_jac2);
  93.719 μs (1491 allocations: 110.03 KiB)

See the ModelingToolkit.jl documentation for more details.

Accelerating Small ODE Solves with Static Arrays

If the ODE is sufficiently small (<20 ODEs or so), using StaticArrays.jl for the state variables can greatly enhance the performance. This is done by making u0 a StaticArray and writing an out-of-place non-mutating dispatch for static arrays, for the ROBER problem, this looks like:

import StaticArrays
function rober_static(u, p, t)
    y₁, y₂, y₃ = u
    k₁, k₂, k₃ = p
    du1 = -k₁ * y₁ + k₃ * y₂ * y₃
    du2 = k₁ * y₁ - k₂ * y₂^2 - k₃ * y₂ * y₃
    du3 = k₂ * y₂^2
    StaticArrays.SA[du1, du2, du3]
end
prob = DE.ODEProblem(rober_static, StaticArrays.SA[1.0, 0.0, 0.0], (0.0, 1e5), StaticArrays.SA[0.04, 3e7, 1e4])
sol = DE.solve(prob, DE.Rosenbrock23())
retcode: Success
Interpolation: specialized 2nd order "free" stiffness-aware interpolation
t: 61-element Vector{Float64}:
      0.0
      3.196206628740808e-5
      0.00014400709669791316
      0.00025605212710841824
      0.00048593872520561496
      0.0007179482298405134
      0.0010819240431281
      0.0014801655696439716
      0.0020679567767069723
      0.0028435846274559506
      ⋮
  25371.934242574636
  30784.11997687391
  37217.42637183451
  44850.61378119757
  53893.69155452613
  64593.73799781129
  77241.71960460565
  92180.81944906656
 100000.0
u: 61-element Vector{StaticArraysCore.SVector{3, Float64}}:
 [1.0, 0.0, 0.0]
 [0.9999987215181657, 1.2780900152625978e-6, 3.9181897521319503e-10]
 [0.9999942397327672, 5.718510591908378e-6, 4.175664083814195e-8]
 [0.9999897579685716, 9.992106860321925e-6, 2.4992456809854223e-7]
 [0.99998056266788, 1.7833624284594367e-5, 1.6037078354141815e-6]
 [0.9999712826600025, 2.403488607694387e-5, 4.6824539206438825e-6]
 [0.9999567250106862, 3.0390689558961382e-5, 1.2884299754832173e-5]
 [0.999940798607161, 3.388427373871301e-5, 2.531711910029394e-5]
 [0.9999172960308617, 3.583508668595173e-5, 4.686888245237194e-5]
 [0.9998862913719596, 3.641240165367128e-5, 7.729622638673522e-5]
 ⋮
 [0.055635077310009065, 2.3546320422346706e-7, 0.944364687226786]
 [0.0479253487640473, 2.012149504322489e-7, 0.9520744500210007]
 [0.041233421490035026, 1.719278816689166e-7, 0.9587664065820823]
 [0.03543699837030739, 1.4688361975260092e-7, 0.9645628547460712]
 [0.030425371816163987, 1.2546809318472718e-7, 0.9695745027157405]
 [0.026099132201150083, 1.0715608431717981e-7, 0.9739007606427643]
 [0.02236969169455009, 9.149844876161108e-8, 0.9776302168069992]
 [0.01915856331353317, 7.811096380138723e-8, 0.9808413585755008]
 [0.01782789385075171, 7.258919982166322e-8, 0.9821720335600463]

If we benchmark this, we see a really fast solution with really low allocation counts:

BT.@btime sol = DE.solve(prob, DE.Rosenbrock23());
  18.279 μs (151 allocations: 18.90 KiB)

This version is thus very amenable to multithreading and other forms of parallelism.

Example Accelerating Linear Algebra PDE Semi-Discretization

In this tutorial, we will optimize the right-hand side definition of a PDE semi-discretization.

Note

We highly recommend looking at the Solving Large Stiff Equations tutorial for details on customizing DifferentialEquations.jl for more efficient large-scale stiff ODE solving. This section will only focus on the user-side code.

Let's optimize the solution of a Reaction-Diffusion PDE's discretization. In its discretized form, this is the ODE:

\[\begin{align} du &= D_1 (A_y u + u A_x) + \frac{au^2}{v} + \bar{u} - \alpha u\\ dv &= D_2 (A_y v + v A_x) + a u^2 + \beta v \end{align}\]

where $u$, $v$, and $A$ are matrices. Here, we will use the simplified version where $A$ is the tridiagonal stencil $[1,-2,1]$, i.e. it's the 2D discretization of the Laplacian. The native code would be something along the lines of:

import DifferentialEquations as DE, LinearAlgebra as LA, BenchmarkTools as BT
# Generate the constants
p = (1.0, 1.0, 1.0, 10.0, 0.001, 100.0) # a,α,ubar,β,D1,D2
N = 100
Ax = Array(LA.Tridiagonal([1.0 for i in 1:(N - 1)], [-2.0 for i in 1:N],
    [1.0 for i in 1:(N - 1)]))
Ay = copy(Ax)
Ax[2, 1] = 2.0
Ax[end - 1, end] = 2.0
Ay[1, 2] = 2.0
Ay[end, end - 1] = 2.0

function basic_version!(dr, r, p, t)
    a, α, ubar, β, D1, D2 = p
    u = r[:, :, 1]
    v = r[:, :, 2]
    Du = D1 * (Ay * u + u * Ax)
    Dv = D2 * (Ay * v + v * Ax)
    dr[:, :, 1] = Du .+ a .* u .* u ./ v .+ ubar .- α * u
    dr[:, :, 2] = Dv .+ a .* u .* u .- β * v
end

a, α, ubar, β, D1, D2 = p
uss = (ubar + β) / α
vss = (a / β) * uss^2
r0 = zeros(100, 100, 2)
r0[:, :, 1] .= uss .+ 0.1 .* rand.()
r0[:, :, 2] .= vss

prob = DE.ODEProblem(basic_version!, r0, (0.0, 0.1), p)
ODEProblem with uType Array{Float64, 3} and tType Float64. In-place: true
Non-trivial mass matrix: false
timespan: (0.0, 0.1)
u0: 100×100×2 Array{Float64, 3}:
[:, :, 1] =
 11.0886  11.0786  11.0876  11.0217  …  11.0523  11.0373  11.0243  11.0999
 11.0746  11.0673  11.0905  11.0824     11.0455  11.0571  11.0862  11.0484
 11.0923  11.045   11.0701  11.0496     11.0909  11.0165  11.0746  11.075
 11.0155  11.0871  11.0278  11.0711     11.0109  11.0387  11.0671  11.0264
 11.0923  11.0996  11.0562  11.0558     11.0856  11.0343  11.0768  11.0819
 11.0865  11.0403  11.0934  11.0425  …  11.0298  11.0932  11.0801  11.0745
 11.0552  11.0952  11.0063  11.0121     11.0107  11.0978  11.0955  11.0877
 11.0946  11.0648  11.0794  11.0866     11.0947  11.0357  11.057   11.0341
 11.0161  11.0767  11.0142  11.0633     11.0305  11.0452  11.0171  11.0426
 11.0474  11.0006  11.0878  11.0218     11.0388  11.0377  11.0815  11.0919
  ⋮                                  ⋱                             
 11.0828  11.0039  11.0124  11.0457     11.011   11.0874  11.0819  11.0694
 11.0949  11.0947  11.0563  11.0404     11.0112  11.0149  11.076   11.036
 11.0984  11.092   11.0281  11.0916     11.0121  11.0652  11.015   11.0808
 11.0343  11.0863  11.0754  11.0081     11.0171  11.0664  11.0381  11.0615
 11.0185  11.0344  11.0767  11.0488  …  11.0808  11.0607  11.0933  11.0326
 11.0801  11.0879  11.0086  11.071      11.0358  11.0452  11.0339  11.0946
 11.0634  11.0753  11.0467  11.036      11.0359  11.0823  11.0124  11.0952
 11.0509  11.0376  11.0957  11.0209     11.0955  11.0669  11.0132  11.0445
 11.0442  11.0738  11.0814  11.0758     11.0465  11.0457  11.0883  11.0938

[:, :, 2] =
 12.1  12.1  12.1  12.1  12.1  12.1  …  12.1  12.1  12.1  12.1  12.1  12.1
 12.1  12.1  12.1  12.1  12.1  12.1     12.1  12.1  12.1  12.1  12.1  12.1
 12.1  12.1  12.1  12.1  12.1  12.1     12.1  12.1  12.1  12.1  12.1  12.1
 12.1  12.1  12.1  12.1  12.1  12.1     12.1  12.1  12.1  12.1  12.1  12.1
 12.1  12.1  12.1  12.1  12.1  12.1     12.1  12.1  12.1  12.1  12.1  12.1
 12.1  12.1  12.1  12.1  12.1  12.1  …  12.1  12.1  12.1  12.1  12.1  12.1
 12.1  12.1  12.1  12.1  12.1  12.1     12.1  12.1  12.1  12.1  12.1  12.1
 12.1  12.1  12.1  12.1  12.1  12.1     12.1  12.1  12.1  12.1  12.1  12.1
 12.1  12.1  12.1  12.1  12.1  12.1     12.1  12.1  12.1  12.1  12.1  12.1
 12.1  12.1  12.1  12.1  12.1  12.1     12.1  12.1  12.1  12.1  12.1  12.1
  ⋮                             ⋮    ⋱         ⋮                      
 12.1  12.1  12.1  12.1  12.1  12.1     12.1  12.1  12.1  12.1  12.1  12.1
 12.1  12.1  12.1  12.1  12.1  12.1     12.1  12.1  12.1  12.1  12.1  12.1
 12.1  12.1  12.1  12.1  12.1  12.1     12.1  12.1  12.1  12.1  12.1  12.1
 12.1  12.1  12.1  12.1  12.1  12.1     12.1  12.1  12.1  12.1  12.1  12.1
 12.1  12.1  12.1  12.1  12.1  12.1  …  12.1  12.1  12.1  12.1  12.1  12.1
 12.1  12.1  12.1  12.1  12.1  12.1     12.1  12.1  12.1  12.1  12.1  12.1
 12.1  12.1  12.1  12.1  12.1  12.1     12.1  12.1  12.1  12.1  12.1  12.1
 12.1  12.1  12.1  12.1  12.1  12.1     12.1  12.1  12.1  12.1  12.1  12.1
 12.1  12.1  12.1  12.1  12.1  12.1     12.1  12.1  12.1  12.1  12.1  12.1

In this version, we have encoded our initial condition to be a 3-dimensional array, with u[:,:,1] being the A part and u[:,:,2] being the B part.

BT.@btime DE.solve(prob, DE.Tsit5());
  53.268 ms (8903 allocations: 186.88 MiB)

While this version isn't very efficient,

We recommend writing the “high-level” code first, and iteratively optimizing it!

The first thing that we can do is get rid of the slicing allocations. The operation r[:,:,1] creates a temporary array instead of a “view”, i.e. a pointer to the already existing memory. To make it a view, add @view. Note that we have to be careful with views because they point to the same memory, and thus changing a view changes the original values:

A = rand(4)
@show A
B = @view A[1:3]
B[2] = 2
@show A
4-element Vector{Float64}:
 0.1670872839659684
 2.0
 0.5386024761612592
 0.8129095006896223

Notice that changing B changed A. This is something to be careful of, but at the same time we want to use this since we want to modify the output dr. Additionally, the last statement is a purely element-wise operation, and thus we can make use of broadcast fusion there. Let's rewrite basic_version! to ***avoid slicing allocations*** and to ***use broadcast fusion***:

function gm2!(dr, r, p, t)
    a, α, ubar, β, D1, D2 = p
    u = @view r[:, :, 1]
    v = @view r[:, :, 2]
    du = @view dr[:, :, 1]
    dv = @view dr[:, :, 2]
    Du = D1 * (Ay * u + u * Ax)
    Dv = D2 * (Ay * v + v * Ax)
    @. du = Du + a .* u .* u ./ v + ubar - α * u
    @. dv = Dv + a .* u .* u - β * v
end
prob = DE.ODEProblem(gm2!, r0, (0.0, 0.1), p)
BT.@btime DE.solve(prob, DE.Tsit5());
  46.830 ms (6698 allocations: 119.66 MiB)

Now, most of the allocations are taking place in Du = D1*(Ay*u + u*Ax) since those operations are vectorized and not mutating. We should instead replace the matrix multiplications with LA.mul!. When doing so, we will need to have cache variables to write into. This looks like:

Ayu = zeros(N, N)
uAx = zeros(N, N)
Du = zeros(N, N)
Ayv = zeros(N, N)
vAx = zeros(N, N)
Dv = zeros(N, N)
function gm3!(dr, r, p, t)
    a, α, ubar, β, D1, D2 = p
    u = @view r[:, :, 1]
    v = @view r[:, :, 2]
    du = @view dr[:, :, 1]
    dv = @view dr[:, :, 2]
    LA.mul!(Ayu, Ay, u)
    LA.mul!(uAx, u, Ax)
    LA.mul!(Ayv, Ay, v)
    LA.mul!(vAx, v, Ax)
    @. Du = D1 * (Ayu + uAx)
    @. Dv = D2 * (Ayv + vAx)
    @. du = Du + a * u * u ./ v + ubar - α * u
    @. dv = Dv + a * u * u - β * v
end
prob = DE.ODEProblem(gm3!, r0, (0.0, 0.1), p)
BT.@btime DE.solve(prob, DE.Tsit5());
  42.356 ms (3758 allocations: 29.86 MiB)

But our temporary variables are global variables. We need to either declare the caches as const or localize them. We can localize them by adding them to the parameters, p. It's easier for the compiler to reason about local variables than global variables. ***Localizing variables helps to ensure type stability***.

p = (1.0, 1.0, 1.0, 10.0, 0.001, 100.0, Ayu, uAx, Du, Ayv, vAx, Dv) # a,α,ubar,β,D1,D2
function gm4!(dr, r, p, t)
    a, α, ubar, β, D1, D2, Ayu, uAx, Du, Ayv, vAx, Dv = p
    u = @view r[:, :, 1]
    v = @view r[:, :, 2]
    du = @view dr[:, :, 1]
    dv = @view dr[:, :, 2]
    LA.mul!(Ayu, Ay, u)
    LA.mul!(uAx, u, Ax)
    LA.mul!(Ayv, Ay, v)
    LA.mul!(vAx, v, Ax)
    @. Du = D1 * (Ayu + uAx)
    @. Dv = D2 * (Ayv + vAx)
    @. du = Du + a * u * u ./ v + ubar - α * u
    @. dv = Dv + a * u * u - β * v
end
prob = DE.ODEProblem(gm4!, r0, (0.0, 0.1), p)
BT.@btime DE.solve(prob, DE.Tsit5());
  41.603 ms (1247 allocations: 29.66 MiB)

We could then use the BLAS gemmv to optimize the matrix multiplications some more, but instead let's devectorize the stencil.

p = (1.0, 1.0, 1.0, 10.0, 0.001, 100.0, N)
function fast_gm!(du, u, p, t)
    a, α, ubar, β, D1, D2, N = p

    @inbounds for j in 2:(N - 1), i in 2:(N - 1)

        du[i, j, 1] = D1 *
                      (u[i - 1, j, 1] + u[i + 1, j, 1] + u[i, j + 1, 1] + u[i, j - 1, 1] -
                       4u[i, j, 1]) +
                      a * u[i, j, 1]^2 / u[i, j, 2] + ubar - α * u[i, j, 1]
    end

    @inbounds for j in 2:(N - 1), i in 2:(N - 1)

        du[i, j, 2] = D2 *
                      (u[i - 1, j, 2] + u[i + 1, j, 2] + u[i, j + 1, 2] + u[i, j - 1, 2] -
                       4u[i, j, 2]) +
                      a * u[i, j, 1]^2 - β * u[i, j, 2]
    end

    @inbounds for j in 2:(N - 1)
        i = 1
        du[1, j, 1] = D1 *
                      (2u[i + 1, j, 1] + u[i, j + 1, 1] + u[i, j - 1, 1] - 4u[i, j, 1]) +
                      a * u[i, j, 1]^2 / u[i, j, 2] + ubar - α * u[i, j, 1]
    end
    @inbounds for j in 2:(N - 1)
        i = 1
        du[1, j, 2] = D2 *
                      (2u[i + 1, j, 2] + u[i, j + 1, 2] + u[i, j - 1, 2] - 4u[i, j, 2]) +
                      a * u[i, j, 1]^2 - β * u[i, j, 2]
    end
    @inbounds for j in 2:(N - 1)
        i = N
        du[end, j, 1] = D1 *
                        (2u[i - 1, j, 1] + u[i, j + 1, 1] + u[i, j - 1, 1] - 4u[i, j, 1]) +
                        a * u[i, j, 1]^2 / u[i, j, 2] + ubar - α * u[i, j, 1]
    end
    @inbounds for j in 2:(N - 1)
        i = N
        du[end, j, 2] = D2 *
                        (2u[i - 1, j, 2] + u[i, j + 1, 2] + u[i, j - 1, 2] - 4u[i, j, 2]) +
                        a * u[i, j, 1]^2 - β * u[i, j, 2]
    end

    @inbounds for i in 2:(N - 1)
        j = 1
        du[i, 1, 1] = D1 *
                      (u[i - 1, j, 1] + u[i + 1, j, 1] + 2u[i, j + 1, 1] - 4u[i, j, 1]) +
                      a * u[i, j, 1]^2 / u[i, j, 2] + ubar - α * u[i, j, 1]
    end
    @inbounds for i in 2:(N - 1)
        j = 1
        du[i, 1, 2] = D2 *
                      (u[i - 1, j, 2] + u[i + 1, j, 2] + 2u[i, j + 1, 2] - 4u[i, j, 2]) +
                      a * u[i, j, 1]^2 - β * u[i, j, 2]
    end
    @inbounds for i in 2:(N - 1)
        j = N
        du[i, end, 1] = D1 *
                        (u[i - 1, j, 1] + u[i + 1, j, 1] + 2u[i, j - 1, 1] - 4u[i, j, 1]) +
                        a * u[i, j, 1]^2 / u[i, j, 2] + ubar - α * u[i, j, 1]
    end
    @inbounds for i in 2:(N - 1)
        j = N
        du[i, end, 2] = D2 *
                        (u[i - 1, j, 2] + u[i + 1, j, 2] + 2u[i, j - 1, 2] - 4u[i, j, 2]) +
                        a * u[i, j, 1]^2 - β * u[i, j, 2]
    end

    @inbounds begin
        i = 1
        j = 1
        du[1, 1, 1] = D1 * (2u[i + 1, j, 1] + 2u[i, j + 1, 1] - 4u[i, j, 1]) +
                      a * u[i, j, 1]^2 / u[i, j, 2] + ubar - α * u[i, j, 1]
        du[1, 1, 2] = D2 * (2u[i + 1, j, 2] + 2u[i, j + 1, 2] - 4u[i, j, 2]) +
                      a * u[i, j, 1]^2 - β * u[i, j, 2]

        i = 1
        j = N
        du[1, N, 1] = D1 * (2u[i + 1, j, 1] + 2u[i, j - 1, 1] - 4u[i, j, 1]) +
                      a * u[i, j, 1]^2 / u[i, j, 2] + ubar - α * u[i, j, 1]
        du[1, N, 2] = D2 * (2u[i + 1, j, 2] + 2u[i, j - 1, 2] - 4u[i, j, 2]) +
                      a * u[i, j, 1]^2 - β * u[i, j, 2]

        i = N
        j = 1
        du[N, 1, 1] = D1 * (2u[i - 1, j, 1] + 2u[i, j + 1, 1] - 4u[i, j, 1]) +
                      a * u[i, j, 1]^2 / u[i, j, 2] + ubar - α * u[i, j, 1]
        du[N, 1, 2] = D2 * (2u[i - 1, j, 2] + 2u[i, j + 1, 2] - 4u[i, j, 2]) +
                      a * u[i, j, 1]^2 - β * u[i, j, 2]

        i = N
        j = N
        du[end, end, 1] = D1 * (2u[i - 1, j, 1] + 2u[i, j - 1, 1] - 4u[i, j, 1]) +
                          a * u[i, j, 1]^2 / u[i, j, 2] + ubar - α * u[i, j, 1]
        du[end, end, 2] = D2 * (2u[i - 1, j, 2] + 2u[i, j - 1, 2] - 4u[i, j, 2]) +
                          a * u[i, j, 1]^2 - β * u[i, j, 2]
    end
end
prob = DE.ODEProblem(fast_gm!, r0, (0.0, 0.1), p)
BT.@btime DE.solve(prob, DE.Tsit5());
  6.620 ms (659 allocations: 29.62 MiB)

Notice that in this case fusing the loops and avoiding the linear operators is a major improvement of about 10x! That's an order of magnitude faster than our original MATLAB/SciPy/R vectorized style code!

Since this is tedious to do by hand, we note that ModelingToolkit.jl's symbolic code generation can do this automatically from the basic version:

import ModelingToolkit as MTK
function basic_version!(dr, r, p, t)
    a, α, ubar, β, D1, D2 = p
    u = r[:, :, 1]
    v = r[:, :, 2]
    Du = D1 * (Ay * u + u * Ax)
    Dv = D2 * (Ay * v + v * Ax)
    dr[:, :, 1] = Du .+ a .* u .* u ./ v .+ ubar .- α * u
    dr[:, :, 2] = Dv .+ a .* u .* u .- β * v
end

a, α, ubar, β, D1, D2 = p
uss = (ubar + β) / α
vss = (a / β) * uss^2
r0 = zeros(100, 100, 2)
r0[:, :, 1] .= uss .+ 0.1 .* rand.()
r0[:, :, 2] .= vss

prob = DE.ODEProblem(basic_version!, r0, (0.0, 0.1), p)
de = MTK.complete(MTK.modelingtoolkitize(prob))

# Note jac=true,sparse=true makes it automatically build sparse Jacobian code
# as well!

fastprob = DE.ODEProblem(de, [], (0.0, 0.1); jac = true, sparse = true)
ODEProblem with uType Vector{Float64} and tType Float64. In-place: true
Initialization status: FULLY_DETERMINED
Non-trivial mass matrix: false
timespan: (0.0, 0.1)
u0: 20000-element Vector{Float64}:
 11.072743449748454
 11.04257855137012
 11.096653907931728
 11.073235079302975
 11.047123031742554
 11.015740617568705
 11.041975897417185
 11.009592847121269
 11.088245855435948
 11.027388828293535
  ⋮
 12.100000000000001
 12.100000000000001
 12.100000000000001
 12.100000000000001
 12.100000000000001
 12.100000000000001
 12.100000000000001
 12.100000000000001
 12.100000000000001

Lastly, we can do other things like multithread the main loops. LoopVectorization.jl provides the @turbo macro for doing a lot of SIMD enhancements, and @tturbo is the multithreaded version.

Optimizing Algorithm Choices

The last thing to do is then ***optimize our algorithm choice***. We have been using DE.Tsit5() as our test algorithm, but in reality this problem is a stiff PDE discretization and thus one recommendation is to use Sundials.CVODE_BDF(). However, instead of using the default dense Jacobian, we should make use of the sparse Jacobian afforded by the problem. The Jacobian is the matrix $\frac{df_i}{dr_j}$, where $r$ is read by the linear index (i.e. down columns). But since the $u$ variables depend on the $v$, the band size here is large, and thus this will not do well with a Banded Jacobian solver. Instead, we utilize sparse Jacobian algorithms. Sundials.CVODE_BDF allows us to use a sparse Newton-Krylov solver by setting linear_solver = :GMRES.

Note

The Solving Large Stiff Equations tutorial goes through these details. This is simply to give a taste of how much optimization opportunity is left on the table!

Let's see how our fast right-hand side scales as we increase the integration time.

prob = DE.ODEProblem(fast_gm!, r0, (0.0, 10.0), p)
BT.@btime DE.solve(prob, DE.Tsit5());
  718.500 ms (60116 allocations: 2.76 GiB)
import Sundials
BT.@btime DE.solve(prob, Sundials.CVODE_BDF(; linear_solver = :GMRES));
  264.165 ms (7181 allocations: 51.98 MiB)
prob = DE.ODEProblem(fast_gm!, r0, (0.0, 100.0), p)
# Will go out of memory if we don't turn off `save_everystep`!
BT.@btime DE.solve(prob, DE.Tsit5(); save_everystep = false);
  3.999 s (90 allocations: 2.90 MiB)
BT.@btime DE.solve(prob, Sundials.CVODE_BDF(; linear_solver = :GMRES); save_everystep = false);
  1.731 s (39961 allocations: 2.34 MiB)
prob = DE.ODEProblem(fast_gm!, r0, (0.0, 500.0), p)
BT.@btime DE.solve(prob, Sundials.CVODE_BDF(; linear_solver = :GMRES); save_everystep = false);
  2.339 s (53665 allocations: 2.83 MiB)

Notice that we've eliminated almost all allocations, allowing the code to grow without hitting garbage collection and slowing down.

Why is Sundials.CVODE_BDF doing well? What's happening is that, because the problem is stiff, the number of steps required by the explicit Runge-Kutta method grows rapidly, whereas Sundials.CVODE_BDF is taking large steps. Additionally, the GMRES linear solver form is quite an efficient way to solve the implicit system in this case. This is problem-dependent, and in many cases using a Krylov method effectively requires a preconditioner, so you need to play around with testing other algorithms and linear solvers to find out what works best with your problem.

Now continue to the Solving Large Stiff Equations tutorial for more details on optimizing the algorithm choice for such codes.