Engee documentation

Code Optimization for Differential Equations

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:

using DifferentialEquations, BenchmarkTools
u0 = [1.0; 0.0; 0.0]
tspan = (0.0, 100.0)
prob = ODEProblem(lorenz, u0, tspan)
@btime solve(prob, Tsit5());
  3.795 ms (101102 allocations: 7.82 MiB)

The BenchmarkTools.jl package’s @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.

@btime solve(prob, Tsit5(), save_everystep = false);
  3.209 ms (89529 allocations: 6.83 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.

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 = ODEProblem(lorenz!, u0, tspan)
@btime solve(prob, Tsit5());
  762.725 μs (11415 allocations: 996.33 KiB)
@btime solve(prob, Tsit5(), save_everystep = false);
  338.451 μs (49 allocations: 4.00 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 = ODEProblem(lorenz!, u0, tspan)
@btime solve(prob, Tsit5(), save_everystep = false);
  2.093 ms (49 allocations: 4.00 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:

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

Notice that the 3 after SVector gives the size of the SVector. It cannot be changed. Additionally, SVectors are immutable, so we have to create a new 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]
    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 = SA[1.0, 0.0, 0.0]
tspan = (0.0, 100.0)
prob = ODEProblem(lorenz_static, u0, tspan)
@btime solve(prob, Tsit5());
  355.294 μs (1293 allocations: 387.39 KiB)
@btime solve(prob, Tsit5(), save_everystep = false);
  233.695 μs (22 allocations: 2.25 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):

₁

Given that these equations are stiff, non-stiff ODE solvers like Tsit5 or 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:

using DifferentialEquations
using 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 = ODEProblem(rober!, [1.0, 0.0, 0.0], (0.0, 1e5), [0.04, 3e7, 1e4])
sol = solve(prob)
plot(sol, tspan = (1e-2, 1e5), xscale = :log10)

using BenchmarkTools
@btime solve(prob);
  121.170 μs (675 allocations: 58.75 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 (Rosenbrock23 or Rodas5) for smaller (<50 ODEs) problems, ESDIRK methods for slightly larger (TRBDF2 or KenCarp4 for <2000 ODEs), and 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 Rosenbrock23() for stiff ODEs at default tolerances:

@btime solve(prob, Rosenbrock23());
  100.705 μs (500 allocations: 40.64 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 ODEFunction. First we have to derive the Jacobian 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! = ODEFunction(rober!, jac = rober_jac!)
prob_jac = 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
timespan: (0.0, 100000.0)
u0: 3-element Vector{Float64}:
 1.0
 0.0
 0.0
@btime solve(prob_jac, Rosenbrock23());
  86.306 μs (422 allocations: 34.64 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.

using ModelingToolkit
de = modelingtoolkitize(prob)
Model ##MTKizedODE#15999 with 3 equations
States (3):
  x₁(t) [defaults to 1.0]
  x₂(t) [defaults to 0.0]
  x₃(t) [defaults to 0.0]
Parameters (3):
  α₁ [defaults to 0.04]
  α₂ [defaults to 3.0e7]
  α₃ [defaults to 10000.0]

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

ModelingToolkit.generate_jacobian(de)[2] # Second is in-place
:(function (ˍ₋out, ˍ₋arg1, ˍ₋arg2, t)
      #= /root/.julia/packages/SymbolicUtils/YVse6/src/code.jl:373 =#
      #= /root/.julia/packages/SymbolicUtils/YVse6/src/code.jl:374 =#
      #= /root/.julia/packages/SymbolicUtils/YVse6/src/code.jl:375 =#
      begin
          begin
              begin
                  #= /root/.julia/packages/Symbolics/rvztO/src/build_function.jl:537 =#
                  #= /root/.julia/packages/SymbolicUtils/YVse6/src/code.jl:422 =# @inbounds begin
                          #= /root/.julia/packages/SymbolicUtils/YVse6/src/code.jl:418 =#
                          ˍ₋out[1] = (*)(-1, ˍ₋arg2[1])
                          ˍ₋out[2] = ˍ₋arg2[1]
                          ˍ₋out[3] = 0
                          ˍ₋out[4] = (*)(ˍ₋arg2[3], ˍ₋arg1[3])
                          ˍ₋out[5] = (+)((*)((*)(-2, ˍ₋arg2[2]), ˍ₋arg1[2]), (*)((*)(-1, ˍ₋arg2[3]), ˍ₋arg1[3]))
                          ˍ₋out[6] = (*)((*)(2, ˍ₋arg2[2]), ˍ₋arg1[2])
                          ˍ₋out[7] = (*)(ˍ₋arg2[3], ˍ₋arg1[2])
                          ˍ₋out[8] = (*)((*)(-1, ˍ₋arg2[3]), ˍ₋arg1[2])
                          ˍ₋out[9] = 0
                          #= /root/.julia/packages/SymbolicUtils/YVse6/src/code.jl:420 =#
                          nothing
                      end
              end
          end
      end
  end)

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

prob_jac2 = ODEProblem(de, [], (0.0, 1e5), jac = true)
ODEProblem with uType Vector{Float64} and tType Float64. In-place: true
timespan: (0.0, 100000.0)
u0: 3-element Vector{Float64}:
 1.0
 0.0
 0.0
@btime solve(prob_jac2);
  107.581 μs (608 allocations: 59.53 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:

using 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
    SA[du1, du2, du3]
end
prob = ODEProblem(rober_static, SA[1.0, 0.0, 0.0], (0.0, 1e5), SA[0.04, 3e7, 1e4])
sol = solve(prob, Rosenbrock23())
retcode: Success
Interpolation: specialized 2nd order "free" stiffness-aware interpolation
t: 61-element Vector{Float64}:
      0.0
      3.196206628740808e-5
      0.00014400709336278452
      0.00025605212043816096
      0.00048593871402339607
      0.0007179482102678373
      0.0010819240251828343
      0.0014801655107859655
      0.0020679567717440095
      0.002843584518457066
      ⋮
  25371.93159838571
  30784.11718374498
  37217.42390396605
  44850.61094811346
  53893.688830057334
  64593.73530179436
  77241.71691097679
  92180.81843146283
 100000.0
u: 61-element Vector{SVector{3, Float64}}:
 [1.0, 0.0, 0.0]
 [0.9999987215181657, 1.2780900152625978e-6, 3.9181897521319503e-10]
 [0.9999942397329006, 5.7185104612947566e-6, 4.175663804739006e-8]
 [0.9999897579688383, 9.992106612572491e-6, 2.49924549040571e-7]
 [0.9999805626683271, 1.7833623941038088e-5, 1.6037077316934769e-6]
 [0.9999712826607852, 2.403488562731424e-5, 4.682453587410618e-6]
 [0.9999567250114038, 3.0390689334989113e-5, 1.2884299261094982e-5]
 [0.9999407986095145, 3.388427339038224e-5, 2.531711709494679e-5]
 [0.9999172960310598, 3.583508669306405e-5, 4.686888224684217e-5]
 [0.9998862913763157, 3.6412401619257426e-5, 7.729622206475401e-5]
 ⋮
 [0.05563508171413305, 2.3546322394505495e-7, 0.9443646828226426]
 [0.047925352159210115, 2.012149653947115e-7, 0.9520744466258239]
 [0.04123342367542567, 1.7192789116847091e-7, 0.9587664043966821]
 [0.03543700020207701, 1.4688362762022196e-7, 0.9645628529142937]
 [0.03042537309965345, 1.2546809864160592e-7, 0.9695745014322467]
 [0.026099133126498978, 1.071560882169501e-7, 0.9739007597174127]
 [0.022369692367946337, 9.149845157822593e-8, 0.977630216133602]
 [0.019158563494465593, 7.811096455346351e-8, 0.9808413583945704]
 [0.017827893845894716, 7.258919980139027e-8, 0.982172033564906]

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

@btime sol = solve(prob, Rosenbrock23());
  82.839 μs (807 allocations: 46.66 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.

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:

where , , and are matrices. Here, we will use the simplified version where is the tridiagonal stencil ], i.e. it’s the 2D discretization of the Laplacian. The native code would be something along the lines of:

using DifferentialEquations, LinearAlgebra, BenchmarkTools
# 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(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 = ODEProblem(basic_version!, r0, (0.0, 0.1), p)
ODEProblem with uType Array{Float64, 3} and tType Float64. In-place: true
timespan: (0.0, 0.1)
u0: 100×100×2 Array{Float64, 3}:
[:, :, 1] =
 11.0297  11.0356  11.0428  11.0467  …  11.0225  11.0233  11.038   11.0872
 11.0549  11.0961  11.0168  11.096      11.0164  11.0972  11.0107  11.0099
 11.015   11.0057  11.0514  11.0409     11.0082  11.023   11.0189  11.0118
 11.0015  11.0915  11.0032  11.003      11.0114  11.0573  11.0665  11.0138
 11.0602  11.0724  11.0952  11.0322     11.0306  11.0269  11.0808  11.0369
 11.0913  11.011   11.0755  11.0714  …  11.0866  11.0974  11.0936  11.0343
 11.0029  11.0355  11.093   11.0127     11.0059  11.0543  11.0804  11.096
 11.0688  11.0521  11.0789  11.0599     11.0039  11.0845  11.0969  11.0574
 11.0029  11.0872  11.0116  11.0584     11.0567  11.0728  11.0656  11.0137
 11.0224  11.0414  11.0153  11.0092     11.0163  11.0839  11.0493  11.018
  ⋮                                  ⋱
 11.0642  11.0425  11.0113  11.0711     11.0482  11.0105  11.0953  11.0753
 11.0039  11.0969  11.0645  11.0514     11.023   11.0244  11.0764  11.0472
 11.0903  11.0955  11.0155  11.0446     11.048   11.0918  11.0061  11.0577
 11.0058  11.0153  11.0618  11.0181     11.0111  11.0618  11.0591  11.0679
 11.0696  11.0355  11.0647  11.0514  …  11.0199  11.0414  11.0463  11.0582
 11.0742  11.0107  11.0072  11.0221     11.0272  11.0425  11.0576  11.0433
 11.091   11.0258  11.065   11.0152     11.0767  11.0518  11.0037  11.0915
 11.1     11.0123  11.0879  11.0601     11.0283  11.0435  11.0621  11.0687
 11.0969  11.071   11.0128  11.05       11.0859  11.0866  11.0665  11.057

[:, :, 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.

@btime solve(prob, Tsit5());
  64.948 ms (7349 allocations: 186.83 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}:
 0.233875535898035
 2.0
 0.08848357049758093
 0.41287083506350053

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 = ODEProblem(gm2!, r0, (0.0, 0.1), p)
@btime solve(prob, Tsit5());
  59.937 ms (5879 allocations: 119.71 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 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]
    mul!(Ayu, Ay, u)
    mul!(uAx, u, Ax)
    mul!(Ayv, Ay, v)
    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 = ODEProblem(gm3!, r0, (0.0, 0.1), p)
@btime solve(prob, Tsit5());
  44.045 ms (4703 allocations: 29.97 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]
    mul!(Ayu, Ay, u)
    mul!(uAx, u, Ax)
    mul!(Ayv, Ay, v)
    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 = ODEProblem(gm4!, r0, (0.0, 0.1), p)
@btime solve(prob, Tsit5());
  37.362 ms (1028 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 = ODEProblem(fast_gm!, r0, (0.0, 0.1), p)
@btime solve(prob, Tsit5());
  7.813 ms (440 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:

using ModelingToolkit
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 = ODEProblem(basic_version!, r0, (0.0, 0.1), p)
de = modelingtoolkitize(prob)

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

fastprob = ODEProblem(de, [], (0.0, 0.1), jac = true, sparse = true)
ODEProblem with uType Vector{Float64} and tType Float64. In-place: true
timespan: (0.0, 0.1)
u0: 20000-element Vector{Float64}:
 11.00511096022425
 11.02154866401249
 11.093170305404836
 11.054832027492262
 11.07510929869822
 11.080173590728513
 11.02244672066665
 11.074092575697362
 11.073488249322759
 11.034784443024972
  ⋮
 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 Tsit5() as our test algorithm, but in reality this problem is a stiff PDE discretization and thus one recommendation is to use 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 , where is read by the linear index (i.e. down columns). But since the variables depend on the , the band size here is large, and thus this will not do well with a Banded Jacobian solver. Instead, we utilize sparse Jacobian algorithms. CVODE_BDF allows us to use a sparse Newton-Krylov solver by setting linear_solver = :GMRES.

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 = ODEProblem(fast_gm!, r0, (0.0, 10.0), p)
@btime solve(prob, Tsit5());
  279.728 s (39314 allocations: 2.76 GiB)
using Sundials
@btime solve(prob, CVODE_BDF(linear_solver = :GMRES));
  1.230 s (13791 allocations: 121.90 MiB)
prob = ODEProblem(fast_gm!, r0, (0.0, 100.0), p)
# Will go out of memory if we don't turn off `save_everystep`!
@btime solve(prob, Tsit5(), save_everystep = false);
  7.607 s (68 allocations: 2.90 MiB)
@btime solve(prob, CVODE_BDF(linear_solver = :GMRES), save_everystep = false);
  3.360 s (33712 allocations: 2.39 MiB)
prob = ODEProblem(fast_gm!, r0, (0.0, 500.0), p)
@btime solve(prob, CVODE_BDF(linear_solver = :GMRES), save_everystep = false);
  6.055 s (55128 allocations: 3.33 MiB)

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

Why is 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 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.