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());
  3.301 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);
  2.794 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());
  735.216 μs (23214 allocations: 1013.34 KiB)
BT.@btime DE.solve(prob, DE.Tsit5(); save_everystep = false);
  353.008 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);
  1.828 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());
  271.418 μs (2615 allocations: 392.71 KiB)
BT.@btime DE.solve(prob, DE.Tsit5(); save_everystep = false);
  194.619 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);
  97.850 μs (1586 allocations: 121.69 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());
  69.930 μs (575 allocations: 29.78 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());
  54.680 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/lcQam/src/build_function.jl:368 =# @inbounds begin
              #= /root/.cache/julia-buildkite-plugin/depots/0185fce3-4489-413a-a934-123dd653ef61/packages/Symbolics/lcQam/src/build_function.jl:368 =#
              begin
                  #= /root/.cache/julia-buildkite-plugin/depots/0185fce3-4489-413a-a934-123dd653ef61/packages/SymbolicUtils/KmZ71/src/code.jl:409 =#
                  #= /root/.cache/julia-buildkite-plugin/depots/0185fce3-4489-413a-a934-123dd653ef61/packages/SymbolicUtils/KmZ71/src/code.jl:410 =#
                  #= /root/.cache/julia-buildkite-plugin/depots/0185fce3-4489-413a-a934-123dd653ef61/packages/SymbolicUtils/KmZ71/src/code.jl:411 =#
                  begin
                      __mtk_arg_2 = ___mtkparameters___[1]
                      __mtk_arg_3 = ___mtkparameters___[2]
                      begin
                          var"##cse#1" = (*)(-1, __mtk_arg_2[1])
                          var"##cse#2" = __mtk_arg_1[3]
                          var"##cse#3" = (*)(__mtk_arg_2[3], var"##cse#2")
                          var"##cse#4" = (*)((*)(-1, __mtk_arg_2[3]), var"##cse#2")
                          var"##cse#5" = __mtk_arg_1[2]
                          var"##cse#6" = (*)((*)(-2, var"##cse#5"), __mtk_arg_2[2])
                          var"##cse#7" = (+)(var"##cse#4", var"##cse#6")
                          var"##cse#8" = (*)((*)(2, var"##cse#5"), __mtk_arg_2[2])
                          var"##cse#9" = (*)(__mtk_arg_2[3], var"##cse#5")
                          var"##cse#10" = (*)((*)(-1, __mtk_arg_2[3]), var"##cse#5")
                          #= /root/.cache/julia-buildkite-plugin/depots/0185fce3-4489-413a-a934-123dd653ef61/packages/SymbolicUtils/KmZ71/src/code.jl:464 =# @inbounds begin
                                  #= /root/.cache/julia-buildkite-plugin/depots/0185fce3-4489-413a-a934-123dd653ef61/packages/SymbolicUtils/KmZ71/src/code.jl:460 =#
                                  ˍ₋out[1] = var"##cse#1"
                                  ˍ₋out[2] = __mtk_arg_2[1]
                                  ˍ₋out[3] = 0
                                  ˍ₋out[4] = var"##cse#3"
                                  ˍ₋out[5] = var"##cse#7"
                                  ˍ₋out[6] = var"##cse#8"
                                  ˍ₋out[7] = var"##cse#9"
                                  ˍ₋out[8] = var"##cse#10"
                                  ˍ₋out[9] = 0
                                  #= /root/.cache/julia-buildkite-plugin/depots/0185fce3-4489-413a-a934-123dd653ef61/packages/SymbolicUtils/KmZ71/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);
  103.549 μs (1561 allocations: 113.73 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());
  18.260 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] =
 11.0976  11.0574  11.0542  11.0787  …  11.053   11.0602  11.0578  11.0732
 11.0083  11.0693  11.0863  11.0896     11.045   11.0096  11.0492  11.0548
 11.0098  11.078   11.0357  11.0199     11.0312  11.0704  11.0617  11.0312
 11.0432  11.0078  11.0816  11.0445     11.0994  11.0315  11.066   11.0411
 11.0584  11.0708  11.0918  11.0452     11.0347  11.0822  11.0277  11.0886
 11.0925  11.0014  11.0186  11.0444  …  11.0557  11.0474  11.0298  11.0064
 11.0627  11.0505  11.004   11.099      11.0149  11.0495  11.0044  11.0571
 11.0839  11.0862  11.0259  11.0264     11.0099  11.0994  11.0498  11.0923
 11.0016  11.0157  11.0249  11.0667     11.0073  11.0254  11.0273  11.0448
 11.0466  11.0192  11.0878  11.0055     11.0828  11.0223  11.0361  11.0178
  ⋮                                  ⋱                             
 11.0709  11.0137  11.0017  11.0167     11.0791  11.078   11.0757  11.0199
 11.0116  11.0256  11.0803  11.0551     11.0416  11.0664  11.005   11.0352
 11.0815  11.099   11.0877  11.0219     11.0132  11.0475  11.0441  11.0191
 11.0165  11.0091  11.0915  11.0903     11.0311  11.0765  11.0312  11.0416
 11.0291  11.042   11.0297  11.0104  …  11.0388  11.0336  11.0416  11.0998
 11.0294  11.0688  11.0324  11.0347     11.008   11.0502  11.025   11.0879
 11.0289  11.0803  11.015   11.0305     11.0534  11.0655  11.0142  11.0549
 11.0997  11.055   11.0486  11.0271     11.0861  11.0874  11.0549  11.0525
 11.0705  11.0506  11.06    11.0443     11.0046  11.0168  11.025   11.0298

[:, :, 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());
  53.489 ms (8891 allocations: 186.87 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}:
 0.25297325433832274
 2.0
 0.9133253584747016
 0.3379348825174384

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());
  47.358 ms (6686 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());
  42.250 ms (3746 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());
  41.755 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());
  6.792 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}:
 11.01901345605872
 11.034249825487409
 11.064155612183436
 11.090328381606344
 11.099285903193104
 11.011293518162155
 11.02814662092186
 11.03777077540226
 11.016130703049532
 11.04194964655379
  ⋮
 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());
  684.301 ms (60116 allocations: 2.76 GiB)
import Sundials
BT.@btime DE.solve(prob, Sundials.CVODE_BDF(; linear_solver = :GMRES));
  276.585 ms (7564 allocations: 52.45 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);
  3.977 s (90 allocations: 2.90 MiB)
BT.@btime DE.solve(prob, Sundials.CVODE_BDF(; linear_solver = :GMRES); save_everystep = false);
  1.649 s (38215 allocations: 2.28 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);
  2.479 s (57577 allocations: 2.97 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.