Add example of eigenfunction of Koopman operator

docs/Project.toml

@@ -18,6 +18,7 @@ MultivariatePolynomials = "102ac46a-7ee4-5c85-9060-abc95bfdeaa3"
 MutableArithmetics = "d8a4904e-b15c-11e9-3269-09a3773c0cb0"
 PermutationGroups = "8bc5a954-2dfc-11e9-10e6-cd969bffa420"
 Plots = "91a5bcdd-55d7-5caf-9e0b-520d859cae80"
+QuadGK = "1fd47b50-473d-5c70-9696-f719f8f3bcdc"
 RecipesBase = "3cdcf5f2-1ef4-517c-9805-6587b60abb01"
 SpecialFunctions = "276daf66-3868-5448-9aa4-cd146d93841b"
 SumOfSquares = "4b9e565b-77fc-50a5-a571-1244f986bda1"

docs/src/tutorials/Systems and Control/koopman_eigenfunctions.jl

@@ -0,0 +1,107 @@
+# # Eigenfunctions of the Koopman operator
+
+#md # [![](https://mybinder.org/badge_logo.svg)](@__BINDER_ROOT_URL__/generated/Systems and Control/koopman_eigenfunctions.ipynb)
+#md # [![](https://img.shields.io/badge/show-nbviewer-579ACA.svg)](@__NBVIEWER_ROOT_URL__/generated/Systems and Control/koopman_eigenfunctions.ipynb)
+# **Adapted from**: Example 2.6 of [MAI20]
+#
+# [MAI20] Mauroy, Alexandre, Aivar Sootla, and Igor Mezić.
+# *Koopman framework for global stability analysis.*
+# The Koopman Operator in Systems and Control: Concepts, Methodologies, and Applications (2020): 35-58.
+
+using Test #src
+using DynamicPolynomials
+@polyvar x[1:2]
+
+a = 1
+I = 0.05
+ε = 0.08
+γ = 1
+F0 = [-x[2] - x[1] * (x[1] - 1) * (x[1] - a) + I, ε * (x[1] - γ * x[2])]
+
+# We move equilibrium `(0.0256, 0.0256)` to the origin
+
+x1 = x2 = 0.0256
+F = [f(x => [x[1] + x1, x[2] + x2]) for f in F0]
+
+# We compute the Jacobian at the equilibrium
+
+J = [j(x => zeros(2)) for j in differentiate(F, x)]
+
+# We see that its eigenvalues indeed have negative real part:
+using LinearAlgebra
+E = eigen(J)
+
+# We set `w` as its dominant eigenvector:
+
+λ = E.values[end]
+w = E.vectors[:, end]
+
+using SumOfSquares
+r = 0.3
+X = @set x[1]^2 + x[2]^2 ≤ r^2
+
+# We define the the program for the FitzHugh-Nagumo problem below:
+# `N` is the degree of `ϕN` as defined in [MAI20, p. 50] and `M` is the
+# degree of the multipliers for the constraints of `X`.
+# As `maxdegree` corresponds to the degree of the multiplier multiplied
+# by `r^2 - x[1]^2 - x[2]^2`, we set it to `M + 2`.
+
+function fitzhugh_nagumo(solver, N, M)
+    model = SOSModel(solver)
+    @variable(model, γ)
+    @objective(model, Min, γ)
+    @variable(model, ϕN, Poly(monomials(x, 2:N)))
+    ϕ = w ⋅ x + ϕN
+    ∇ϕ = differentiate(ϕ, x)
+    @constraint(model, -γ ≤ F ⋅ ∇ϕ - λ * ϕ, domain = X, maxdegree = M + 2)
+    @constraint(model, F ⋅ ∇ϕ - λ * ϕ ≤ γ, domain = X, maxdegree = M + 2)
+    optimize!(model)
+    return ϕ, model
+end
+
+# In [MAI20, p. 50], we read that the result is obtained with `N = 10` and
+# `M = 20`.
+
+import CSDP
+ϕ, model = fitzhugh_nagumo(CSDP.Optimizer, 10, 20)
+solution_summary(model)
+
+# The optimal value of `ϕ` is obtained as follows:
+
+ϕ_opt = value(ϕ)
+
+# Its plot is given below:
+
+using Plots
+x1s = x2s = range(-0.3, stop = 0.3, length = 40)
+ϕs = ϕ_opt.(x1s', x2s)
+contour(x1s, x2s, ϕs, levels=[-0.2, -0.1, 0, 0.1, 0.2, 0.5, 0.7])
+θ = range(0, stop = 2π, length = 100)
+plot!(r * cos.(θ), r * sin.(θ), label = "")
+scatter!([0], [0], label = "")
+
+# We can compute the Laplace average as follows:
+
+using DifferentialEquations
+function S(t, x1, x2, solver = DifferentialEquations.Tsit5())
+    tspan = (0.0, t)
+    prob = DifferentialEquations.ODEProblem((v, p, t) -> [f(x => v) for f in F], [x1, x2], tspan)
+    traj = DifferentialEquations.solve(prob, solver, reltol=1e-4, abstol=1e-4)
+    return traj[end]
+end
+
+using QuadGK
+function laplace_average(f, λ, x1, x2, T = 10, args...)
+    v, _ = quadgk(0, T, rtol=1e-3) do t
+        s = S(t, x1, x2, args...)
+        return f(S(t, x1, x2, args...)) * exp(-λ * t)
+    end
+    return v / T
+end
+
+lap(x1, x2) = laplace_average(v -> ϕ_opt(x => v), λ, x1, x2, 10)
+laplace = lap.(x1s', x2s)
+
+# The error is given by:
+
+norm(laplace - ϕs)