Automatic Bifurcation diagram computation

Thanks to the functionality presented in this part, we can compute the bifurcation diagram of a system recursively and fully automatically. More precisely, the function bifurcationdiagram allows to:

  • compute a branch $\gamma$ of equilibria
  • detect all bifurcations on the branch
  • recursively compute the branches emanating from branch points on $\gamma$.

Pitfalls

For now, there is no way to decide if two branches $\gamma_1,\gamma_2$ are the same. As a consequence:

  • there is no loop detection. Hence, if the branch $\gamma$ has a component akin to a circle, you may experience a large number of branches
  • if the bifurcation diagram itself has loops (see example below), you may experience a large number of branches
Memory

The whole diagram is stored in RAM and you might be careful computing it on GPU. We'll add a file system for this in the future.

Basic example with simple branch points

using Revise, Plots
using BifurcationKit

Fbp(u, p) = @. -u * (p + u * (2-5u)) * (p -.15 - u * (2+20u))

# bifurcation problem
prob = BifurcationProblem(Fbp, [0.0], -0.2,
	# specify the continuation parameter
	(@lens _);
	record_from_solution = (x, p) -> x[1])

# options for newton
# we reduce a bit the tolerances to ease automatic branching
opt_newton = NewtonPar(tol = 1e-9)

# options for continuation
opts_br = ContinuationPar(dsmin = 0.001, dsmax = 0.005, ds = 0.001,
	newton_options = opt_newton,
	nev = 1,
	# parameter interval
	p_min = -1.0, p_max = .3,
	# detect bifurcations with bisection method
	# we increase here the precision for the detection of
	# bifurcation points
	n_inversion = 8)

diagram = bifurcationdiagram(prob, PALC(),
	# very important parameter. This specifies the maximum amount of recursion
	# when computing the bifurcation diagram. It means we allow computing branches of branches
	# at most in the present case.
	2,
	opts_br,
)

# You can plot the diagram like
plot(diagram; putspecialptlegend=false, markersize=2, plotfold=false, title = "#branches = $(size(diagram))")
Example block output

This gives

diagram
[Bifurcation diagram]
 ┌─ From 0-th bifurcation point.
 ├─ Children number: 4
 └─ Root (recursion level 1)
      ┌─ Curve type: EquilibriumCont
      ├─ Number of points: 76
      ├─ Type of vectors: Vector{Float64}
      ├─ Parameter p starts at -0.2, ends at 0.3
      ├─ Algo: PALC
      └─ Special points:

- #  1,       bp at p ≈ +0.00000281 ∈ (-0.00000065, +0.00000281), |δp|=3e-06, [converged], δ = ( 1,  0), step =  31
- #  2,       bp at p ≈ +0.15000016 ∈ (+0.14999995, +0.15000016), |δp|=2e-07, [   guessL], δ = (-1,  0), step =  53
- #  3, endpoint at p ≈ +0.30000000,                                                                     step =  75

Example with nonsimple branch points

To show the ability of the branch switching method to cope with non simple branch points, we look at the normal form of the Pitchfork with D6 symmetry which occurs frequently in problems with hexagonal symmetry. You may want to look at Bratu–Gelfand problem for a non trivial example of use.

using Revise, Plots
using BifurcationKit
const BK = BifurcationKit

function FbpD6(x, p)
	return [ p.μ * x[1] + (p.a * x[2] * x[3] - p.b * x[1]^3 - p.c*(x[2]^2 + x[3]^2) * x[1]),
		p.μ * x[2] + (p.a * x[1] * x[3] - p.b * x[2]^3 - p.c*(x[3]^2 + x[1]^2) * x[2]),
		p.μ * x[3] + (p.a * x[1] * x[2] - p.b * x[3]^3 - p.c*(x[2]^2 + x[1]^2) * x[3])]
end

# model parameters
pard6 = (μ = -0.2, a = 0.3, b = 1.5, c = 2.9)

# problem
prob = BifurcationProblem(FbpD6, zeros(3), pard6, (@lens _.μ);
		record_from_solution = (x, p) -> (n = norminf(x)))

# newton options
opt_newton = NewtonPar(tol = 1e-9, max_iterations = 20)

# continuation options
opts_br = ContinuationPar(
	# we limit the step size to have smooth branches
	dsmax = 0.005, ds = 0.001,
	# parameter interval
	p_max = 0.4, p_min = -0.25,
	# number of eigenvalues to be computed
	nev = 3,
	newton_options = opt_newton,
	max_steps = 1000,
	# increased precision for bifurcation points
	n_inversion = 4, max_bisection_steps = 20)

diagram = bifurcationdiagram(prob, PALC(), 3,
	opts_br;
	normC = norminf)
[Bifurcation diagram]
 ┌─ From 0-th bifurcation point.
 ├─ Children number: 3
 └─ Root (recursion level 1)
      ┌─ Curve type: EquilibriumCont
      ├─ Number of points: 89
      ├─ Type of vectors: Vector{Float64}
      ├─ Parameter μ starts at -0.2, ends at 0.4
      ├─ Algo: PALC
      └─ Special points:

- #  1,       nd at μ ≈ +0.00019961 ∈ (-0.00024233, +0.00019961), |δp|=4e-04, [converged], δ = ( 3,  0), step =  31
- #  2, endpoint at μ ≈ +0.40000000,                                                                     step =  88

We can now plot the result:

plot(diagram; putspecialptlegend =false, markersize=2, plotfold=false, title="#branch = $(size(diagram))")
Example block output

We can access the different branches with BK.getBranch(diagram, (1,)). Alternatively, you can plot a specific branch:

Finally, you can resume the computation of the bifurcation diagram if not complete by using the syntax

BK.bifurcationdiagram!(prob,
	# this resume the computation of the diagram from the 2nd node
	# diagram is written inplace
	get_branch(diagram, (2,)), 6,
	(args...) -> opts_br)
[Bifurcation diagram]
 ┌─ From 1-th bifurcation point.
 ├─ Children number: 12
 └─ Root (recursion level 2)
      ┌─ Curve type: EquilibriumCont from NonSimpleBranchPoint bifurcation point.
      ├─ Number of points: 77
      ├─ Type of vectors: Vector{Float64}
      ├─ Parameter μ starts at 0.00019960742344131129, ends at 0.4
      ├─ Algo: PALC
      └─ Special points:

- #  1,       bp at μ ≈ -0.00000923 ∈ (-0.00000923, -0.00000870), |δp|=5e-07, [converged], δ = ( 1,  0), step =   1
- #  2,       bp at μ ≈ -0.00308222 ∈ (-0.00308222, -0.00308221), |δp|=1e-08, [converged], δ = (-1,  0), step =   6
- #  3,       nd at μ ≈ +0.27097788 ∈ (+0.27087371, +0.27097788), |δp|=1e-04, [converged], δ = ( 2,  0), step =  56
- #  4, endpoint at μ ≈ +0.40000000,                                                                     step =  76

Printing the structure of the diagram

It is sometimes useful to have a global representation of the bifurcation diagram. Here, we provide a text representation

using AbstractTrees

AbstractTrees.children(node::BK.BifDiagNode) = node.child

## Things that make printing prettier
AbstractTrees.printnode(io::IO, node::BifDiagNode) = print(io, "$(node.code) [ $(node.level)]")

print_tree(diagram)
0 [ 1]
├─ 1 [ 2]
│  ├─ 2 [ 3]
│  ├─ 2 [ 3]
│  ├─ 4 [ 3]
│  ├─ 4 [ 3]
│  ├─ 4 [ 3]
│  ├─ 4 [ 3]
│  ├─ 4 [ 3]
│  └─ 4 [ 3]
├─ 1 [ 2]
│  ├─ 3 [ 3]
│  │  ├─ 3 [ 4]
│  │  └─ 3 [ 4]
│  ├─ 3 [ 3]
│  ├─ 3 [ 3]
│  ├─ 3 [ 3]
│  │  ├─ 3 [ 4]
│  │  └─ 3 [ 4]
│  ├─ 3 [ 3]
│  │  ├─ 3 [ 4]
│  │  └─ 3 [ 4]
│  ├─ 3 [ 3]
│  ├─ 3 [ 3]
│  │  ├─ 3 [ 4]
│  │  └─ 3 [ 4]
│  ├─ 3 [ 3]
│  ├─ 3 [ 3]
│  ├─ 3 [ 3]
│  │  ├─ 3 [ 4]
│  │  └─ 3 [ 4]
│  ├─ 3 [ 3]
│  │  ├─ 3 [ 4]
│  │  └─ 3 [ 4]
│  └─ 3 [ 3]
└─ 1 [ 2]
   ├─ 2 [ 3]
   └─ 2 [ 3]

Plotting the structure of the diagram

We can also use GraphPlot to plot the tree underlying the bifurcation diagram:

using LightGraphs, MetaGraphs, GraphPlot

function graphFromDiagram!(_graph, diagram, indp)
	# ind is the index of the parent node
	# add vertex and associated information
	add_vertex!(_graph)
	set_props!(_graph, nv(_graph), Dict(:code => diagram.code, :level => diagram.level))
	if nv(_graph) > 1
		add_edge!(_graph, indp, nv(_graph))
	end
	if length(diagram.child) > 0
		# we now run through the children
		new_indp = nv(_graph)
		for diag in diagram.child
			graphFromDiagram!(_graph, diag, new_indp)
		end
	end
end

function graphFromDiagram(diagram) 
	_g = MetaGraph()
	graphFromDiagram!(_g, diagram, 1)
	return _g
end

_g = graphFromDiagram(bdiag)

gplot(_g, nodelabel = [props(_g, ve)[:code] for ve in vertices(_g)])

which gives the following picture. The node label represent the index of the bifurcation point from which the branch branches.

Using GraphRecipes

Another solution is to use GraphRecipes and

using GraphRecipes

graphplot(_g, 
	node_weights = ones(nv(_g)).*10, 
	names=[props(_g, ve)[:code] for ve in vertices(_g)], 
	curvature_scalar=0.)