Integration of interaction tensor - julia

Question

I need to calculate the following integral several thousand times per time step:
where and:
, and
So far I have implemented in Julia as:
using StaticArrays
function interactiontensor(C, a1, a2, a3, ϕ, θ)
n1,n2 = 100,50
T = fill(0.0,3,3,3,3)
Av = zeros(4,4)
invAv = similar(Av)
xi = Vector{Float64}(undef, 3)
#inbounds for p ∈ 1:n1
sinθp = sind(θ[p])
cosθp = cosd(θ[p])
for q ∈ 1:n2
sinϕq = sind(ϕ[q])
cosϕq = cosd(ϕ[q])
# -- Director cosines
xi[1] = sinθp*cosϕq/a1
xi[2] = sinθp*sinϕq/a2
xi[3] = cosθp/a3
Christoffel!(Av,C,xi)
fillAv!(Av, xi)
invAv = inv(SMatrix{4,4}(Av))
tensorT!(T,invAv,xi,sinθp)
surface += sinθp
end
end
return T ./= surface
end
#inline function Christoffel!(Av,C,xi)
#inbounds for t ∈ 1:3, r ∈ 1:3
aux = zero(eltype(C))
for u ∈ 1:3, s ∈ 1:3
aux += C[r, s, t, u] * xi[s] * xi[u]
end
Av[r, t] = aux
end
end
#inline function tensorT!(T,invAv,xi,sinθp)
#inbounds for k ∈ 1:3, i ∈ 1:3
aux = invAv[i, k]
for l ∈ 1:3, j ∈ 1:3
T[i, j, k, l] += aux * xi[j] * xi[l] * sinθp
end
end
end
#inline function fillAv!(Av, xi)
#inbounds for i ∈ 1:3
xi0 = xi[i]
Av[i, 4] = xi0
Av[4, i] = xi0
end
end
with
n1,n2 = 100,100
step = π/n1
dθ,dϕ = π/n1, 2π/n2
θ = rad2deg.(range(dθ, stop = pi, length = n1))
ϕ = rad2deg.(range(dϕ, stop = 2pi, length = n2))
C = #SArray rand(3,3,3,3)
#btime interactiontensor($C, $10.0, $5.0, $1.0, $ϕ, $θ);
# 544.795 μs (4 allocations: 1.08 KiB)
Given the number of times I ideally need to compute this integral, is there any optimization to my implementation, or an alternative approach, to considerably reduce the computational cost?

Answer 1

Here are some suggestions:
sinθp, cosθp = sincosd(θ[p]), i.e. computing sine and cosine in one step.
Initializing xi = #SVector zeros(3) as a static vector and then use Setfield.jl to assign the values in each iteration, i.e. #set x[1] = sinθp*cosϕq/a1.
Load the package LoopVectorization.jl and use the #avx macro (very roughly speaking similar to #simd) to speed up the loops in Christoffel!, tensorT! and fillAv!.
On my machine I find that these changes reduce the computation time by more than a factor of 5 (relative to the original function in the OP). The biggest chunk is due to #avx, the second point above amounts to about ~30%.
julia> #btime interactiontensor_original($C, $10.0, $5.0, $1.0, $ϕ, $θ);
661.655 μs (5 allocations: 1.28 KiB)
julia> #btime interactiontensor_optimized($C, $10.0, $5.0, $1.0, $ϕ, $θ);
125.352 μs (4 allocations: 1.17 KiB)
Here the full modified code (note that I commented out the lines involving surface which isn't specified in the OP):
using StaticArrays, Setfield, LoopVectorization
function interactiontensor_optimized(C, a1, a2, a3, ϕ, θ)
n1,n2 = 100,50
T = fill(0.0,3,3,3,3)
Av = zeros(4,4)
invAv = similar(Av)
xi = #SVector zeros(3)
#inbounds for p ∈ 1:n1
sinθp, cosθp = sincosd(θ[p])
for q ∈ 1:n2
sinϕq, cosϕq = sincosd(ϕ[q])
# -- Director cosines
#set xi[1] = sinθp*cosϕq/a1
#set xi[2] = sinθp*sinϕq/a2
#set xi[3] = cosθp/a3
Christoffel!(Av,C,xi)
fillAv!(Av, xi)
invAv = inv(SMatrix{4,4}(Av))
tensorT!(T,invAv,xi,sinθp)
# surface += sinθp
end
end
return T #./= surface
end
#inline function Christoffel!(Av,C,xi)
#avx for t ∈ 1:3, r ∈ 1:3
aux = zero(eltype(C))
for u ∈ 1:3, s ∈ 1:3
aux += C[r, s, t, u] * xi[s] * xi[u]
end
Av[r, t] = aux
end
end
#inline function tensorT!(T,invAv,xi,sinθp)
#avx for k ∈ 1:3, i ∈ 1:3
aux = invAv[i, k]
for l ∈ 1:3, j ∈ 1:3
T[i, j, k, l] += aux * xi[j] * xi[l] * sinθp
end
end
end
#inline function fillAv!(Av, xi)
#avx for i ∈ 1:3
xi0 = xi[i]
Av[i, 4] = xi0
Av[4, i] = xi0
end
end