Basic Usage

Example 1: three-species single-element problem

The simplest non-trivial Gibbs problem: three species sharing one conserved quantity.

using OptimaSolver
using LinearAlgebra

μ⁰ = [0.0, 1.0, 2.0]

G(n, p)    = sum(n[i] * (p.μ⁰[i] + log(n[i])) for i in eachindex(n))
∇G!(g,n,p) = for i in eachindex(n); g[i] = p.μ⁰[i] + log(n[i]) + 1; end

A = ones(1, 3)
b = [1.0]
prob = OptimaProblem(A, b, G, ∇G!; lb=fill(1e-16, 3), p=(μ⁰=μ⁰,))
result = solve(prob)

println(result.n)   # ≈ [0.6652, 0.2447, 0.0900]
@assert result.converged

[0.6652409557238408, 0.24472847106840734, 0.09003057320775174]

Analytical check:  .

n_exact = exp.(-μ⁰) ./ sum(exp.(-μ⁰))
@assert maximum(abs, result.n .- n_exact) < 1e-7

Example 2: two elements, four species

A system with two conserved elements and four species.

# Conservation matrix: each column = elemental composition of one species
# A[i,k] = number of atoms of element i in species k
A2 = Float64[2 1 1 2;   # element X
              1 0 1 0]  # element Y
b2 = [4.0, 1.0]        # 4 units of X, 1 unit of Y

μ⁰4 = [0.0, 0.5, 1.0, 1.5]

G4(n, p)     = sum(n[i] * (p.μ⁰[i] + log(n[i])) for i in eachindex(n))
∇G4!(g, n, p) = for i in eachindex(n); g[i] = p.μ⁰[i] + log(n[i]) + 1; end

prob4 = OptimaProblem(A2, b2, G4, ∇G4!;
                      lb = fill(1e-16, 4),
                      p  = (μ⁰ = μ⁰4,))
result4 = solve(prob4)

@assert result4.converged
@assert norm(A2 * result4.n .- b2) < 1e-8

Reusing a Canonicalizer for temperature scans

When is fixed across a series of solves (e.g. varying temperature so only changes), build the Canonicalizer once and pass it to solve. This skips the QR decomposition on every call and reduces overhead for large :

# Build canonicalizer once (expensive QR + LU)
can = Canonicalizer(A2)

# Stand-in for whatever supplies standard chemical potentials at a temperature.
# It is a placeholder, not thermodynamics: what this example is about is reusing
# the factorisation, not where μ⁰(T) comes from.
μ⁰_at_temperature(T_K) = μ⁰4 .* (298.15 / T_K)

# Solve at many temperatures, reusing the LU factorisation
for T_K in range(298.15, 400.0; step=5.0)
    μ⁰_T = μ⁰_at_temperature(T_K)
    prob_T = OptimaProblem(A2, b2, G4, ∇G4!;
                           lb = fill(1e-16, 4),
                           p  = (μ⁰ = μ⁰_T,))
    result_T = solve(prob_T, can)     # QR not repeated
    # ... process result_T
end

Verbose iteration log

Set verbose=true in OptimaOptions to see a per-iteration summary:

result = solve(prob, OptimaOptions(tol=1e-12, verbose=true))

OptimaResult{Float64}([0.6652409557747402, 0.24472847105481932, 0.09003057317044014], [-0.592394035555374], 46, true, 1.6076305905894033e-16, 3.3306690738754696e-16, 3.3306690738754696e-16)

Each row shows the iteration count, the current barrier weight , the optimality and feasibility residuals, the accepted line-search step length and the step norms. The log above is the real one: this block is executed when the documentation is built, so what you read is what the solver prints, not a transcription of it.

Lower and upper bounds

Both lower and upper bounds can be set per species:

# Aqueous species: lb = 1e-16 (trace amount), no upper bound
# Solid species:   lb = 0, ub = total moles available
prob_mixed = OptimaProblem(A, b, G, ∇G!;
                           lb = [1e-16, 1e-16, 0.0],
                           ub = [Inf,   Inf,   2.5],
                           p  = (μ⁰ = μ⁰,))

OptimaProblem{Float64, typeof(Main.G), typeof(Main.∇G!)}([1.0 1.0 1.0], [1.0], Main.G, Main.∇G!, 3, 1, [1.0e-16, 1.0e-16, 0.0], [Inf, Inf, 2.5], (μ⁰ = [0.0, 1.0, 2.0],))

For solids and gases with zero curvature ( ), enable finite-difference Hessian computation:

opts = OptimaOptions(tol=1e-10, use_fd_hessian=true)
result_mixed = solve(prob_mixed, opts)
println(result_mixed.n)

[0.6652409557238409, 0.24472847106840737, 0.09003057320775174]