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An Introduction to Numerical Continuation Methods
with Applications
Eusebius Doedel
IIMAS-UNAM
July 28 - August 1, 2014
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Persistence of Solutions
• Newton’s method for solving a nonlinear equation [B83]1
G(u) = 0 , G(·) , u ∈ Rn ,
may not converge if the “initial guess” is not close to a solution.
• However, one can put a homotopy parameter in the equation.
• Actually, most equations already have parameters.
• We will discuss persistence of solutions to such equations.
1 See Page 83+ of the Background Notes on Elementary Numerical Methods.
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The Implicit Function Theorem
Let G : Rn × R → Rn satisfy
(i) G(u0, λ0) = 0 , u0 ∈ Rn , λ0 ∈ R .
(ii) Gu(u0, λ0) is nonsingular (i.e., u0 is an isolated solution) ,
(iii) G and Gu are smooth near u0 .
Then there exists a unique, smooth solution family u(λ) such that
• G(u(λ), λ) = 0 , for all λ near λ0 ,
• u(λ0) = u0 .
NOTE : The IFT also holds in more general spaces · · ·
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EXAMPLE : A Simple Homotopy .
( Course demo : Simple-Homotopy2)
Let
g(u, λ) = (u2 − 1) (u2 − 4) + λ u2 e110u .
When λ = 0 the equationg(u, 0) = 0 ,
has four solutions , namely,
u = ± 1 , and u = ± 2 .
We have
gu(u, λ)∣∣∣λ=0
≡ d
du(u, λ)
∣∣∣λ=0
= 4u3 − 10u .
2 http://users.encs.concordia.ca/ doedel/
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Sincegu(u, 0) = 4u3 − 10u ,
we have
gu(−1, 0) = 6 , gu( 1, 0) = −6 ,
gu(−2, 0) = −12 , gu( 2, 0) = 12 ,
which are all nonzero .
Thus each of the four solutions when λ = 0 is isolated .
Hence each of these solutions persists as λ becomes nonzero,
( at least for “small” values of | λ | · · · ).
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Solution families of g(u, λ) = 0 . Note the fold .
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NOTE :
• Each of the four solutions at λ = 0 is isolated .
• Thus each of these solutions persists as λ becomes nonzero.
• Only two of the four homotopies reach λ = 1 .
• The other two homotopies meet at a fold .
• IFT condition (ii) is not satisfied at the fold. (Why not? )
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In the equation
G(u, λ) = 0 , u , G(·, ·) ∈ Rn , λ ∈ R ,
let
x ≡(
uλ
).
Then the equation can be written
G(x) = 0 , G : Rn+1 → Rn .
DEFINITION :
A solution x0 of G(x) = 0 is regular if the matrix
G0x ≡ Gx(x0) , (with n rows and n+ 1 columns)
has maximal rank , i.e., if
Rank(G0x) = n .
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In the parameter formulation ,
G(u, λ) = 0 ,
we have
Rank(G0x) = Rank(G0
u | G0λ) = n ⇐⇒
(i) G0u is nonsingular,
or
(ii)
dim N (G0
u) = 1 ,andG0λ 6∈ R(G0
u) .
Here N (G0u) denotes the null space of G0
u ,
and
R(G0u) denotes the range of G0
u ,
i.e., R(G0u) is the linear space spanned by the n columns of G0
u .
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COROLLARY (to the IFT) : Let
x0 ≡ ( u0 , λ0 )
be a regular solution ofG(x) = 0 .
Then, near x0 , there exists a unique one-dimensional solution family
x(s) with x(0) = x0 .
PROOF : Since Rank( G0x ) = Rank( G0
u | G0λ ) = n , we have that
(i) either G0u is nonsingular and by the IFT we have
u = u(λ) near x0 ,
(ii) or else we can interchange colums in the Jacobian G0x to see that the
solution can locally be parametrized by one of the components of u .
Thus a (locally) unique solution family passes through x0 . QED !
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NOTE :
• Such a solution family is sometimes called a solution branch .
• Case (i) is where the IFT applies directly .
• Case (ii) is that of a simple fold .
• Thus even near a simple fold there is a unique solution family .
• However, near such a fold, the family cannot be parametrized by λ.
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More Examples of IFT Application
• We give examples where the IFT shows that a given solution persists
(at least locally ) when a problem parameter is changed.
• We also consider cases where the conditions of the IFT are not satisfied .
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EXAMPLE : The A→ B → C Reaction .
( Course demo : Chemical-Reactions/ABC-Reaction/Stationary )
u′1 = −u1 + D(1− u1)eu3 ,
u′2 = −u2 + D(1− u1)eu3 − Dσu2eu3 ,
u′3 = −u3 − βu3 + DB(1− u1)eu3 + DBασu2eu3 ,
where
1− u1 is the concentration of A , u2 is the concentration of B ,
u3 is the temperature, α = 1 , σ = 0.04 , B = 8 ,
D is the Damkohler number , β > 0 is the heat transfer coefficient .
NOTE : The zero stationary solution at D = 0 persists (locally), because
the Jacobian is nonsingular there, having eigenvalues −1, −1, and −(1 + β) .
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Families of stationary solutions of the A→ B → C reaction.
(From left to right : β = 1.1 , 1.3 , 1.5 , 1.6 , 1.7 , 1.8 . )
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NOTE :
In the preceding bifurcation diagram:
• ‖ u ‖ =√u2
1 + u22 + u2
3 .
• Solid/dashed curves denote stable/unstable solutions.
• The red squares are Hopf bifurcations .
From the basic theory of ODEs:
• u0 is a stationary solution of u′(t) = f(u(t)) if f(u0) = 0 .
• u0 is stable if all eigenvalues of fu(u0) are in the negative half-plane.
• u0 is unstable if one or more eigenvalues are in the positive half-plane.
• At a fold there is zero eigenvalue.
• At a Hopf bifurcation there is a pair of purely imaginary eigenvalues.
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EXAMPLE (of IFT application) : The Gelfand-Bratu Problem .
( Course demo : Gelfand-Bratu/Original )
The boundary value problem u′′(x) + λ eu(x) = 0 , ∀x ∈ [0, 1] ,
u(0) = u(1) = 0 ,
defines the stationary states of a solid fuel ignition model.
If λ = 0 then u(x) ≡ 0 is a solution.
This problem can be thought of as an operator equation G(u;λ) = 0 .
We can use (a generalized) IFT to prove that there is a solution family
u = u(λ) , for |λ| small .
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The linearization of G(u;λ) acting on v , i.e., Gu(u;λ)v , leads to
the homogeneous equation
v′′(x) + λ eu(x)v = 0 ,
v(0) = v(1) = 0 ,
which for the solution u(x) ≡ 0 at λ = 0 becomes
v′′(x) = 0 ,
v(0) = v(1) = 0 .
Since this equation only has the zero solution v(x) ≡ 0 , the IFT applies.
Thus (locally) a unique solution family passes through u(x) ≡ 0 , λ = 0 .
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In Course demo : Gelfand-Bratu/Original the BVP is implemented
as a first order system :
u′1(t) = u2(t) ,
u′2(t) = − λ eu1(t) ,
with boundary conditions
u1(0) = 0 ,
u1(1) = 0 .
A convenient solution measure in the bifurcation diagram is the value of∫ 1
0
u1(x) dx .
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Bifurcation diagram of the Gelfand-Bratu equation.
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Some solutions of the Gelfand-Bratu equation.(The solution at the fold is colored red ).
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EXAMPLE : A Boundary Value Problem with Bifurcations .
( Course demo : Basic-BVP/Nonlinear-Eigenvalue )
u′′ + λ u(1 − u) = 0 ,
u(0) = u(1) = 0 ,
has u(x) ≡ 0 as a solution for all λ .
QUESTION : Are there more solutions ?
Again, this problem corresponds to an operator equation G(u; λ) = 0 .
Its linearization acting on v leads to the equation Gu(u; λ)v = 0 , i.e.,
v′′ + λ (1− 2u)v = 0 ,
v(0) = v(1) = 0 .
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In particular, the linearization about the zero solution family u ≡ 0 is
v′′ + λ v = 0 ,
v(0) = v(1) = 0 ,
which for most values of λ only has the zero solution v(x) ≡ 0 .
However, whenλ = λk ≡ k2π2 ,
then there are nonzero solutions , namely,
v(x) = sin(kπx) ,
Thus the IFT does not apply at λk = k2π2 .
(We will see that these solutions are bifurcation points .)
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In the implementation we write the BVP as a first order system .
We also use a scaled version of λ.
The equations are then
u′1 = u2 ,
u′2 = λ2π2 u1 (1− u1) ,
with λ = λ2π2 .
A convenient solution measure in the bifurcation diagram is
γ ≡ u2(0) = u′1(0) .
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Solution families to the nonlinear eigenvalue problem.
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Some solutions to the nonlinear eigenvalue problem.
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Hopf Bifurcation
THEOREM : Suppose that along a stationary solution family (u(λ), λ) , of
u′ = f(u, λ) ,
a complex conjugate pair of eigenvalues
α(λ) ± i β(λ) ,
of fu(u(λ), λ) crosses the imaginary axis transversally , i.e., for some λ0 ,
α(λ0) = 0 , β(λ0) 6= 0 , and α(λ0) 6= 0 .
Also assume that there are no other eigenvalues on the imaginary axis .
Then there is a Hopf bifurcation, that is, a family of periodic solutions bifurcates
from the stationary solution at (u0, λ0) .
NOTE : The assumptions imply that f0u is nonsingular, so that the stationary
solution family is indeed (locally) a function of λ .
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EXAMPLE : The A→ B → C reaction .
( Course demo : Chemical-Reactions/ABC-Reaction/Homoclinic )
A stationary (blue) and a periodic (red) family of the A→ B → C reaction forβ = 1.2 . The periodic orbits are stable and terminate in a homoclinic orbit .
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The periodic family orbit family approaching a homoclinic orbit (black). Thered dot is the Hopf point; the blue dot is the saddle point on the homoclinic.
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Course demo : Chemical-Reactions/ABC-Reaction/Periodic
Bifurcation diagram for β = 1.1, 1.3, 1.5, 1.6, 1.7, 1.8 .
(For periodic solutions ‖ u ‖= 1T
∫ T0
√u2
1 + u22 + u2
3 dt , where T is the period.)
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EXAMPLE : A Predator-Prey Model .
( Course demo : Predator-Prey/ODE/2D )
u′1 = 3u1(1− u1)− u1u2 − λ(1− e−5u1 ) ,
u′2 = −u2 + 3u1u2 .
Here u1 may be thought of as “fish ” and u2 as “sharks ”, while the term
λ (1− e−5u1 ) ,
represents “fishing”, with “fishing-quota ” λ .
When λ = 0 the stationary solutions are
3u1(1− u1)− u1u2 = 0
−u2 + 3u1u2 = 0
⇒ (u1, u2) = (0, 0) , (1, 0) , (1
3, 2) .
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The Jacobian matrix is
Gu(u1, u2 ; λ) =
(3− 6u1 − u2 − 5λe
−5u1 −u1
3u2 −1 + 3u1
)so that
Gu(0, 0 ; 0) =
(3 00 −1
); real eigenvalues 3 , -1 (unstable)
Gu(1, 0 ; 0) =
( −3 −10 2
); real eigenvalues -3 , 2 (unstable)
Gu(1
3, 2 ; 0) =
( −1 −13
6 0
); complex eigenvalues − 1
2± 1
2
√7 i (stable)
All three Jacobians at λ = 0 are nonsingular .
Thus, by the IFT, all three stationary points persist for (small) λ 6= 0 .
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In this problem we can explicitly find all solutions:
Family 1 : (u1, u2) = (0 , 0)
Family 2 :
u2 = 0 , λ =3u1(1− u1)
1− e−5u1
( Note that limu1 → 0
λ = limu1 → 0
3(1− 2u1)
5e−5u1=
3
5)
Family 3 :
u1 =1
3,
2
3− 1
3u2 − λ(1− e−5/3) = 0 ⇒ u2 = 2−3λ(1− e−5/3)
These solution families intersect at two bifurcation points , one of which is
(u1, u2, λ) = (0 , 0 , 3/5) .
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quota0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9
sharks0.0
0.5
1.0
1.5
fish
0.00
0.25
0.50
0.75
Stationary solution families of the predator-prey model.Solid/dashed curves denote stable/unstable solutions.
Note the bifurcations and Hopf bifurcation (red square).
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0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9
quota
0.0
0.2
0.4
0.6
0.8
fish
Stationary solution families, showing fish versus quota.Solid/dashed curves denote stable/unstable solutions.
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Stability of Family 1 :
Gu(0, 0 ; λ) =
(3− 5λ 0
0 −1
); eigenvalues 3− 5λ, − 1 .
Hence the zero solution is :
unstable if λ < 3/5 ,
and
stable if λ > 3/5 .
Stability of Family 2 :
This family has no stable positive solutions.
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• Stability of Family 3 :
At λH ≈ 0.67 the complex eigenvalues cross the imaginary axis:
− This crossing is a Hopf bifurcation ,
− Beyond λH there are stable periodic solutions .
− Their period T increases as λ increases.
− The period becomes infinite at λ = λ∞ ≈ 0.70 .
− This final orbit is a heteroclinic cycle .
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0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0quota
−0.25
0.00
0.25
0.50
0.75
1.00
min
fish,
max
fish
Stationary (blue) and periodic (red) solution families of the predator-prey model.( For the periodic solution family both the maximum and minimum are shown. )
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Periodic solutions of the predator-prey model.The largest orbits are close to a heteroclinic cycle.
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The bifurcation diagram shows the solution behavior for (slowly) increasing λ :
• Family 3 is followed until λH ≈ 0.67 .
• Periodic solutions of increasing period until λ = λ∞ ≈ 0.70 .
• Collapse to trivial solution (Family 1).
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Continuation of Solutions
Parameter Continuation
Suppose we have a solution (u0, λ0) of
G(u, λ) = 0 ,
as well as the derivative u0 .
Here
u ≡ du
dλ.
We want to compute the solution u1 at λ1 ≡ λ0 + ∆λ .
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��
��
��
��
����������������������������������
������������
������������
"u"
u
u
λ λλ
u 1(0)
dud λ
at λ0 )u
10
1
0
∆λ
(=
Graphical interpretation of parameter-continuation.
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To solve the equationG(u1 , λ1) = 0 ,
for u1 (with λ = λ1 fixed) we use Newton’s method
Gu(u(ν)1 , λ1) ∆u
(ν)1 = − G(u
(ν)1 , λ1) ,
u(ν+1)1 = u
(ν)1 + ∆u
(ν)1 .
ν = 0, 1, 2, · · · .
As initial approximation use
u(0)1 = u0 + ∆λ u0 .
IfGu(u1, λ1) is nonsingular ,
and ∆λ sufficiently small then this iteration will converge [B55].
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After convergence, the new derivative u1 is computed by solving
Gu(u1, λ1) u1 = −Gλ(u1, λ1) .
This equation is obtained by differentiating
G(u(λ), λ) = 0 ,
with respect to λ at λ = λ1 .
Repeat the procedure to find u2 , u3 , · · · .
NOTE :
• u1 can be computed without another LU -factorization of Gu(u1, λ1) .
• Thus the extra work to compute u1 is negligible .
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EXAMPLE : The Gelfand-Bratu Problem .
u′′(x) + λ eu(x) = 0 for x ∈ [0, 1] , u(0) = 0 , u(1) = 0 .
We know that if λ = 0 then u(x) ≡ 0 is an isolated solution .
Discretize by introducing a mesh ,
0 = x0 < x1 < · · · < xN = 1 ,
xj − xj−1 = h , (1 ≤ j ≤ N) , h = 1/N .
The discrete equations are :
uj+1 − 2uj + uj−1
h2+ λ euj = 0 , j = 1, · · · , N − 1 ,
with u0 = uN = 0 .
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Let
u ≡
u1
u2
·uN−1
.
Then we can write the discrete equations as
G( u , λ ) = 0 ,
where
G : RN−1 × R → RN−1 .
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Parameter-continuation :
Suppose we have λ0 , u0 , and u0 . Set λ1 = λ0 + ∆λ .
Newton’s method :
Gu(u(ν)1 , λ1) ∆u
(ν)1 = −G(u
(ν)1 , λ1) ,
u(ν+1)1 = u
(ν)1 + ∆u
(ν)1 ,
for ν = 0, 1, 2, · · · , with
u(0)1 = u0 + ∆λ u0 .
After convergence compute u1 from
Gu(u1, λ1) u1 = −Gλ(u1, λ1) .
Repeat the procedure to find u2 , u3 , · · · .
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Here
Gu( u , λ ) =
− 2h2 + λeu1 1
h2
1h2 − 2
h2 + λeu2 1h2
. . .. . .
1h2 − 2
h2 + λeuN−1
.
Thus we must solve a tridiagonal system for each Newton iteration.
NOTE :
• The solution family has a fold where parameter-continuation fails !
• A better continuation method is “pseudo-arclength continuation ”.
• There are also better discretizations, namely collocation , as used in AUTO .
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Pseudo-Arclength Continuation
This method allows continuation of a solution family past a fold .
It was introduced by H. B. Keller (1925-2008) in 1977.
Suppose we have a solution (u0, λ0) of
G( u , λ ) = 0 ,
as well as the normalized direction vector (u0, λ0) of the solution family.
Pseudo-arclength continuation consists of solving these equations for (u1, λ1) :
G(u1, λ1) = 0 ,
〈 u1 − u0 , u0 〉 + (λ1 − λ0) λ0 − ∆s = 0 .
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����u 0
u 0 ∆ s����
λ 0��
����������
����
"u"
λ λλ
u
10
1
u( ), λ 00
Graphical interpretation of pseudo-arclength continuation.
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Solve the equationsG(u1, λ1) = 0 ,
〈 u1 − u0 , u0 〉 + (λ1 − λ0) λ0 − ∆s = 0 .
for (u1, λ1) by Newton’s method :
(G1u)(ν) (G1
λ)(ν)
u∗0 λ0
(∆u(ν)1
∆λ(ν)1
)= −
G(u(ν)1 , λ
(ν)1 )
〈 u(ν)1 − u0 , u0 〉+ (λ
(ν)1 − λ0)λ0 −∆s
.
Compute the next direction vector by solvingG1u G1
λ
u∗0 λ0
( u1
λ1
)=
0
1
,
and normalize it.
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NOTE :
• We can compute (u1, λ1) with only one extra backsubstitution .
• The orientation of the family is preserved if ∆s is sufficiently small.
• Rescale the direction vector so that indeed ‖ u1 ‖2 + λ21 = 1 .
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FACT : The Jacobian is nonsingular at a regular solution.
PROOF : Let x ≡(
uλ
)∈ Rn+1 .
Then pseudo-arclength continuation can be simply written as
G(x1) = 0 ,
〈 x1 − x0 , x0 〉 − ∆s = 0 , ( ‖ x0 ‖ = 1 ) .
∆ s��
�� ����
1x
0x
0x
Pseudo-arclength continuation.
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The pseudo-arclength equations are
G(x1) = 0 ,
〈 x1 − x0 , x0 〉 − ∆s = 0 , ( ‖ x0 ‖ = 1 ) .
The Jacobian matrix in Newton’s method at ∆s = 0 is(G0
x
x∗0
).
At a regular solution N (G0x) = Span{x0} .
We must show that
(G0
x
x∗0
)is nonsingular at a regular solution.
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If on the contrary
(G0
x
x∗0
)is singular then for some vector z 6= 0 we have :
G0x z = 0 ,
〈 x0 , z 〉 = 0 ,
Since by assumption N (G0x) = Span{x0} , we have
z = c x0 , for some constant c .
But then
0 = 〈 x0 , z 〉 = c 〈 x0 , x0 〉 = c ‖ x0 ‖2 = c ,
so that z = 0 , which is a contradiction . QED !
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EXAMPLE : The Gelfand-Bratu Problem .
Use pseudo-arclength continuation for the discretized Gelfand-Bratu problem.
Then the matrix (Gx
x∗
)=
(Gu Gλ
u∗ λ
),
in Newton’s method is a bordered tridiagonal matrix :
? ? ?? ? ? ?
? ? ? ?? ? ? ?
? ? ? ?? ? ? ?
? ? ? ?? ? ?
? ? ? ? ? ? ? ? ?
.
which can be decomposed very efficiently .
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Following Folds and Hopf Bifurcations
• At a fold the the behavior of a system can change drastically.
• How does the fold location change when a second parameter varies ?
• Thus we want the compute a locus of folds in 2 parameters.
• We also want to compute loci of Hopf bifurcations in 2 parameters.
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Following Folds
Treat both parameters λ and µ as unknowns , and compute a solution family
X(s) ≡ ( u(s) , φ(s) , λ(s) , µ(s) ) ,
to
F(X) ≡
G(u, λ, µ) = 0 ,
Gu(u, λ, µ) φ = 0 ,
〈 φ , φ0 〉 − 1 = 0 ,
and the added continuation equation
〈 u− u0 , u0 〉 + 〈 φ− φ0 , φ0 〉 + (λ− λ0)λ0 + (µ− µ0)µ0 − ∆s = 0 .
As before,( u0 , φ0 , λ0 , µ0 ) ,
is the direction of the family at the current solution point
( u0 , φ0 , λ0 , µ0) .
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EXAMPLE : The A→ B → C Reaction .
( Course demo : Chemical-Reactions/ABC-Reaction/Folds-SS )
The equations are
u′1 = −u1 + D(1− u1)eu3 ,
u′2 = −u2 + D(1− u1)eu3 − Dσu2eu3 ,
u′3 = −u3 − βu3 + DB(1− u1)eu3 + DBασu2eu3 ,
where
1− u1 is the concentration of A , u2 is the concentration of B ,
u3 is the temperature , α = 1 , σ = 0.04 , B = 8 ,
D is the Damkohler number , β is the heat transfer coefficient .
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A stationary solution family for β = 1.20.
Note the two folds and the Hopf bifurcation .
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0.00 0.05 0.10 0.15 0.20D
0.0
0.2
0.4
0.6
0.8
1.0
1.2
1.4
β
0.14 0.15 0.16 0.17 0.18 0.19 0.20D
1.20
1.25
1.30
1.35
1.40
β
A locus of folds (with blow-up) for the A→ B → C reaction.
Notice the two cusp singularities along the 2-parameter locus.
( There is a swallowtail singularity in nearby 3-parameter space. )
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0.12 0.14 0.16 0.18 0.20 0.22D
1
2
3
4
5
6
7
||u||
Stationary solution families for β = 1.20, 1.21, · · · , 1.42.( Open diamonds mark folds, solid red squares mark Hopf points. )
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Following Hopf Bifurcations
The extended system is
F(u,φ, β, λ;µ) ≡
f(u, λ, µ) = 0 ,
fu(u, λ, µ) φ − i β φ = 0 ,
〈 φ , φ0 〉 − 1 = 0 ,
whereF : Rn × Cn × R2 × R → Rn × Cn × C ,
and to which we want to compute a solution family
( u , φ , β , λ , µ ) ,
withu ∈ Rn, φ ∈ Cn, β, λ, µ ∈ R .
Above φ0 belongs to a “reference solution ”
( u0 , φ0 , β0 , λ0 , µ0 ) ,
which normally is the latest computed solution along a family.
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EXAMPLE : The A→ B → C Reaction .
( Course demo : Chemical-Reactions/ABC-Reaction/Hopf )
The stationary family with Hopf bifurcation for β = 1.20 .
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0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8D
0.5
1.0
1.5
2.0
β
The locus of Hopf bifurcations for the A→ B → C reaction.
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0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8D
0
1
2
3
4
5
6
7
||u||
Stationary solution families for β = 1.20, 1.20, 1.25, 1.30, · · · , 2.30,with Hopf bifurcations (the red squares) .
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Boundary Value Problems
Consider the first order system of ordinary differential equations
u′(t) − f( u(t) , µ , λ ) = 0 , t ∈ [0, 1] ,
where
u(·) , f(·) ∈ Rn , λ ∈ R, µ ∈ Rnµ ,
subject to boundary conditions
b( u(0) , u(1) , µ , λ ) = 0 , b(·) ∈ Rnb ,
and integral constraints∫ 1
0
q( u(s) , µ , λ ) ds = 0 , q(·) ∈ Rnq .
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This boundary value problem (BVP) is of the form
F( X ) = 0 ,where
X = ( u , µ , λ ) ,
to which we add the continuation equation
〈 X−X0 , X0 〉 − ∆s = 0 ,
where X0 represents the latest solution computed along the family.
In detail , the continuation equation is∫ 1
0
〈 u(t)− u0(t) , u0(t) 〉 dt + 〈 µ− µ0 , µ0 〉
+ (λ− λ0)λ0 − ∆s = 0 .
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NOTE :
• In the context of continuation we solve this BVP for (u(·) , λ , µ) .
• In order for problem to be formally well-posed we must have
nµ = nb + nq − n ≥ 0 .
• A simple case is
nq = 0 , nb = n , for which nµ = 0 .
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Discretization: Orthogonal Collocation
Introduce a mesh
{ 0 = t0 < t1 < · · · < tN = 1 } ,where
hj ≡ tj − tj−1 , (1 ≤ j ≤ N) ,
Define the space of (vector) piecewise polynomials Pmh as
Pmh ≡ { ph ∈ C[0, 1] : ph
∣∣∣[tj−1,tj ]
∈ Pm } ,
where Pm is the space of (vector) polynomials of degree ≤ m .
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The collocation method consists of finding
ph ∈ Pmh , µ ∈ Rnµ ,
such that the following collocation equations are satisfied :
p′h(zj,i) = f( ph(zj,i) , µ, λ ) , j = 1, · · · , N , i = 1, · · · ,m ,
and such that
ph satisfies the boundary and integral conditions .
The collocation points zj,i in each subinterval
[ tj−1 , tj ] ,
are the (scaled) roots of the mth-degree orthogonal polynomial (Gauss points3).
3 See Pages 261, 287 of the Background Notes on Elementary Numerical Methods.
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�� �� ��
0 1t t t t
t
0 1 2 N
z j,1 z j,2 z j,3t
tt j-1
j-1 j
j
l lj,3 j,1(t) (t)
j-1/3tj-2/3t
The mesh {0 = t0 < t1 < · · · < tN = 1} , with
collocation points and extended-mesh points shown for m = 3 .
Also shown are two of the four local Lagrange basis polynomials .
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Since each local polynomial is determined by
(m+ 1) n ,
coefficients, the total number of unknowns (considering λ as fixed) is
(m+ 1) n N + nµ .
This is matched by the total number of equations :
collocation : m n N ,
continuity : (N − 1) n ,
constraints : nb + nq ( = n + nµ ) .
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Assume that the solution u(t) of the BVP is sufficiently smooth.
Then the order of accuracy of the orthogonal collocation method is m , i.e.,
‖ ph − u ‖∞ = O(hm) .
At the main meshpoints tj we have superconvergence :
maxj | ph(tj)− u(tj) | = O(h2m) .
The scalar variables λ and µ are also superconvergent .
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Implementation
For each subinterval [ tj−1 , tj ] , introduce the Lagrange basis polynomials
{ `j,i(t) } , j = 1, · · · , N , i = 0, 1, · · · ,m ,
defined by
`j,i(t) =m∏
k=0,k 6=i
t− tj− km
tj− im− tj− k
m
,
where
tj− im≡ tj − i
mhj .
The local polynomials can then be written
pj(t) =m∑i=0
`j,i(t) uj− im.
With the above choice of basis
uj ∼ u(tj) and uj− im∼ u(tj− i
m) ,
where u(t) is the solution of the continuous problem.
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The collocation equations are
p′
j(zj,i) = f( pj(zj,i) , µ , λ ) , i = 1, · · · ,m, j = 1, · · · , N .
The boundary conditions are
bi( u0 , uN , µ , λ ) = 0 , i = 1, · · · , nb .
The integral constraints can be discretized as
N∑j=1
m∑i=0
ωj,i qk( uj− im, µ , λ) = 0 , k = 1, · · · , nq ,
where the ωj,i are the Lagrange quadrature weights .
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The continuation equation is∫ 1
0
〈 u(t) − u0(t) , u0(t) 〉 dt + 〈 µ − µ0 , µ0 〉 + (λ − λ0) λ0 − ∆s = 0 ,
where ( u0 , µ0 , λ0 ) ,
is the previous solution along the solution family, and
( u0 , µ0 , λ0 ) ,
is the normalized direction of the family at the previous solution .
The discretized continuation equation is of the form
N∑j=1
m∑i=0
ωj,i 〈 uj− im− (u0)j− i
m, (u0)j− i
m〉
+ 〈 µ − µ0 , µ0 〉 + (λ − λ0) λ0 − ∆s = 0 .
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Numerical Linear Algebra
The complete discretization consists of
m n N + nb + nq + 1 ,
nonlinear equations , with unknowns
{uj− im} ∈ RmnN+n , µ ∈ Rnµ , λ ∈ R .
These equations are solved by a Newton-Chord iteration .
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We illustrate the numerical linear algebra for the case
n = 2 ODEs , N = 4 mesh intervals , m = 3 collocation points ,
nb = 2 boundary conditions , nq = 1 integral constraint ,
and the continuation equation.
• The operations are also done on the right hand side , which is not shown.
• Entries marked “◦” have been eliminated by Gauss elimination.
• Entries marked “·” denote fill-in due to pivoting .
• Most of the operations can be done in parallel .
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u0 u 13
u 23
u1 u2 u3 uN µ λ
• • • • • • • • • •• • • • • • • • • •• • • • • • • • • •• • • • • • • • • •• • • • • • • • • •• • • • • • • • • •
• • • • • • • • • •• • • • • • • • • •• • • • • • • • • •• • • • • • • • • •• • • • • • • • • •• • • • • • • • • •
• • • • • • • • • •• • • • • • • • • •• • • • • • • • • •• • • • • • • • • •• • • • • • • • • •• • • • • • • • • •
• • • • • • • • • •• • • • • • • • • •• • • • • • • • • •• • • • • • • • • •• • • • • • • • • •• • • • • • • • • •
• • • • • •• • • • • •• • • • • • • • • • • • • • • • • • • • • • • • • • • •• • • • • • • • • • • • • • • • • • • • • • • • • • • •
The structure of the Jacobian .
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u0 u 13
u 23
u1 u2 u3 uN µ λ
• • • • • • • • • •• • ◦ • • • • • • •• • ◦ ◦ • • • • • •• • ◦ ◦ ◦ • • • • •• • ◦ ◦ ◦ ◦ • • • •• • ◦ ◦ ◦ ◦ • • • •
• • • • • • • • • •• • ◦ • • • • • • •• • ◦ ◦ • • • • • •• • ◦ ◦ ◦ • • • • •• • ◦ ◦ ◦ ◦ • • • •• • ◦ ◦ ◦ ◦ • • • •
• • • • • • • • • •• • ◦ • • • • • • •• • ◦ ◦ • • • • • •• • ◦ ◦ ◦ • • • • •• • ◦ ◦ ◦ ◦ • • • •• • ◦ ◦ ◦ ◦ • • • •
• • • • • • • • • •• • ◦ • • • • • • •• • ◦ ◦ • • • • • •• • ◦ ◦ ◦ • • • • •• • ◦ ◦ ◦ ◦ • • • •• • ◦ ◦ ◦ ◦ • • • •
• • • • • •• • • • • •• • ◦ ◦ ◦ ◦ • • ◦ ◦ ◦ ◦ • • ◦ ◦ ◦ ◦ • • ◦ ◦ ◦ ◦ • • • •• • ◦ ◦ ◦ ◦ • • ◦ ◦ ◦ ◦ • • ◦ ◦ ◦ ◦ • • ◦ ◦ ◦ ◦ • • • •
The system after condensation of parameters, which can be done in parallel .
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u0 u 13
u 23
u1 u2 u3 uN µ λ
• • • • • • • • • •• • ◦ • • • • • • •• • ◦ ◦ • • • • • •• • ◦ ◦ ◦ • • • • •• • ◦ ◦ ◦ ◦ • • • •• • ◦ ◦ ◦ ◦ • • • •
• • • • • • • • • •• • ◦ • • • • • • •• • ◦ ◦ • • • • • •• • ◦ ◦ ◦ • • • • •• • ◦ ◦ ◦ ◦ • • • •• • ◦ ◦ ◦ ◦ • • • •
• • • • • • • • • •• • ◦ • • • • • • •• • ◦ ◦ • • • • • •• • ◦ ◦ ◦ • • • • •• • ◦ ◦ ◦ ◦ • • • •• • ◦ ◦ ◦ ◦ • • • •
• • • • • • • • • •• • ◦ • • • • • • •• • ◦ ◦ • • • • • •• • ◦ ◦ ◦ • • • • •• • ◦ ◦ ◦ ◦ • • • •• • ◦ ◦ ◦ ◦ • • • •
• • • • • •• • • • • •• • ◦ ◦ ◦ ◦ • • ◦ ◦ ◦ ◦ • • ◦ ◦ ◦ ◦ • • ◦ ◦ ◦ ◦ • • • •• • ◦ ◦ ◦ ◦ • • ◦ ◦ ◦ ◦ • • ◦ ◦ ◦ ◦ • • ◦ ◦ ◦ ◦ • • • •
The preceding matrix, showing the decoupled • subsystem .
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u0 u 13
u 23
u1 u2 u3 uN µ λ
• • • • • • • • • •• • ◦ • • • • • • •• • ◦ ◦ • • • • • •• • ◦ ◦ ◦ • • • • •• • ◦ ◦ ◦ ◦ • • · · • •• • ◦ ◦ ◦ ◦ ◦ • · · • •
• • • • • • • • • •• • ◦ • • • • • • •• • ◦ ◦ • • • • • •• • ◦ ◦ ◦ • • • • •
• • ◦ ◦ ◦ ◦ ◦ ◦ • • • •• • ◦ ◦ ◦ ◦ ◦ ◦ • • • •
• • • • • • • • • •• • ◦ • • • • • • •• • ◦ ◦ • • • • • •• • ◦ ◦ ◦ • • • • •• • ◦ ◦ ◦ ◦ • • · · • •• • ◦ ◦ ◦ ◦ ◦ • · · • •
• • • • • • • • • •• • ◦ • • • • • • •• • ◦ ◦ • • • • • •• • ◦ ◦ ◦ • • • • •
• • ◦ ◦ ◦ ◦ ◦ ◦ • • • •• • ◦ ◦ ◦ ◦ ◦ ◦ • • • •
• • • • • •• • • • • •• • ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ • • ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ • • • •• • ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ • • ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ • • • •
Stage 1 of the nested dissection to solve the decoupled • subsystem.
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u0 u 13
u 23
u1 u2 u3 uN µ λ
• • • • • • • • • •• • ◦ • • • • • • •• • ◦ ◦ • • • • • •• • ◦ ◦ ◦ • • • • •• • ◦ ◦ ◦ ◦ • • · · • •• • ◦ ◦ ◦ ◦ ◦ • · · • •
• • • • • • • • • •• • ◦ • • • • • • •• • ◦ ◦ • • • • • •• • ◦ ◦ ◦ • • • • •
• • ◦ ◦ ◦ ◦ ◦ ◦ • • · · • •• • ◦ ◦ ◦ ◦ ◦ ◦ ◦ • · · • •
• • • • • • • • • •• • ◦ • • • • • • •• • ◦ ◦ • • • • • •• • ◦ ◦ ◦ • • • • •• • ◦ ◦ ◦ ◦ • • · · • •• • ◦ ◦ ◦ ◦ ◦ • · · • •
• • • • • • • • • •• • ◦ • • • • • • •• • ◦ ◦ • • • • • •• • ◦ ◦ ◦ • • • • •
• • ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ • • • •• • ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ • • • •• • • • • •• • • • • •• • ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ • • • •• • ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ • • • •
Stage 2 of the nested dissection to solve the decoupled • subsystem.
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u0 u 13
u 23
u1 u2 u3 uN µ λ
• • • • • • • • • •• • ◦ • • • • • • •• • ◦ ◦ • • • • • •• • ◦ ◦ ◦ • • • • •• • ◦ ◦ ◦ ◦ • • · · • •• • ◦ ◦ ◦ ◦ ◦ • · · • •
• • • • • • • • • •• • ◦ • • • • • • •• • ◦ ◦ • • • • • •• • ◦ ◦ ◦ • • • • •
• • ◦ ◦ ◦ ◦ ◦ ◦ • • · · • •• • ◦ ◦ ◦ ◦ ◦ ◦ ◦ • · · • •
• • • • • • • • • •• • ◦ • • • • • • •• • ◦ ◦ • • • • • •• • ◦ ◦ ◦ • • • • •• • ◦ ◦ ◦ ◦ • • · · • •• • ◦ ◦ ◦ ◦ ◦ • · · • •
• • • • • • • • • •• • ◦ • • • • • • •• • ◦ ◦ • • • • • •• • ◦ ◦ ◦ • • • • •
• • ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ • • • •• • ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ • • • •• • • • • •• • • • • •• • ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ • • • •• • ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ • • • •
The preceding matrix showing the final decoupled • subsystem .
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u0 u 13
u 23
u1 u2 u3 uN µ λ
• • • • • • • • • •• • ◦ • • • • • • •• • ◦ ◦ • • • • • •• • ◦ ◦ ◦ • • • • •• • ◦ ◦ ◦ ◦ • • · · • •• • ◦ ◦ ◦ ◦ ◦ • · · • •
• • • • • • • • • •• • ◦ • • • • • • •• • ◦ ◦ • • • • • •• • ◦ ◦ ◦ • • • • •
• • ◦ ◦ ◦ ◦ ◦ ◦ • • · · • •• • ◦ ◦ ◦ ◦ ◦ ◦ ◦ • · · • •
• • • • • • • • • •• • ◦ • • • • • • •• • ◦ ◦ • • • • • •• • ◦ ◦ ◦ • • • • •• • ◦ ◦ ◦ ◦ • • · · • •• • ◦ ◦ ◦ ◦ ◦ • · · • •
• • • • • • • • • •• • ◦ • • • • • • •• • ◦ ◦ • • • • • •• • ◦ ◦ ◦ • • • • •
A A ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ B B • •A A ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ B B • •• • • • • •• • • • • •• • ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ • • • •• • ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ • • • •
The approximate Floquet multipliers are the eigenvalues of M ≡ −B−1A .
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Accuracy Test
The Table shows the location of the fold in the Gelfand-Bratu problem,
for 4 Gauss collocation points per mesh interval, and N mesh intervals .
N Fold location2 3.51378975504 3.51383086018 3.5138307211
16 3.513830719132 3.5138307191
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Periodic Solutions
• Periodic solutions can be computed efficiently using a BVP approach.
• This method also determines the period very accurately.
• Moreover, the technique can compute unstable periodic orbits.
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Consider
u′(t) = f( u(t) , λ ) , u(·) , f(·) ∈ Rn , λ ∈ R .
Fix the interval of periodicity by the transformation
t → t
T.
Then the equation becomes
u′(t) = T f( u(t) , λ ) , u(·) , f(·) ∈ Rn , T , λ ∈ R .
and we seek solutions of period 1 , i.e.,
u(0) = u(1) .
Note that the period T is one of the unknowns .
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The above equations do not uniquely specify u and T :
Assume that we have computed
( uk−1(·) , Tk−1 , λk−1 ) ,
and we want to compute the next solution
( uk(·) , Tk , λk ) .
Then uk(t) can be translated freely in time :
If uk(t) is a periodic solution, then so is uk(t+ σ) , for any σ .
Thus, a phase condition is needed.
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An example is the Poincare phase condition
〈 uk(0) − uk−1(0) , u′
k−1(0) 〉 = 0 .
( But we will derive a numerically more suitable integral phase condition . )
u k-1 (0)
��
��
uk-1 (0)
u (0)k
Graphical interpretation of the Poincare phase condition.
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An Integral Phase Condition
If uk(t) is a solution then so is
uk(t+ σ) ,for any σ .
We want the solution that minimizes
D(σ) ≡∫ 1
0
‖ uk(t+ σ) − uk−1(t) ‖22 dt .
The optimal solutionuk(t+ σ) ,
must satisfy the necessary condition
D′(σ) = 0 .
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Differentiation gives the necessary condition
∫ 1
0
〈 uk(t+ σ) − uk−1(t) , u′k(t+ σ 〉 dt = 0 .
Writinguk(t) ≡ uk(t+ σ) ,
gives ∫ 1
0
〈 uk(t) − uk−1(t) , u′k(t) 〉 dt = 0 .
Integration by parts, using periodicity, gives
∫ 1
0〈 uk(t) , u
′
k−1(t) 〉 dt = 0 .
This is the integral phase condition.
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Continuation of Periodic Solutions
• Pseudo-arclength continuation is used to follow periodic solutions .
• It allows computation past folds along a family of periodic solutions.
• It also allows calculation of a “vertical family ” of periodic solutions.
For periodic solutions the continuation equation is
∫ 1
0
〈 uk(t)−uk−1(t) , uk−1(t) 〉 dt + (Tk−Tk−1)Tk−1 + (λk−λk−1)λk−1 = ∆s .
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SUMMARY :
We have the following equations for periodic solutions :
u′k(t) = T f( uk(t) , λk ) ,
uk(0) = uk(1) ,∫ 1
0
〈 uk(t) , u′
k−1(t) 〉 dt = 0 ,
with continuation equation
∫ 1
0
〈 uk(t)−uk−1(t) , uk−1(t) 〉 dt + (Tk−Tk−1)Tk−1 + (λk−λk−1)λk−1 = ∆s ,
where
u(·) , f(·) ∈ Rn , λ , T ∈ R .
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Stability of Periodic Solutions
In our continuation context, a periodic solution of period T satisfies
u′(t) = T f( u(t)) , for t ∈ [0, 1] ,
u(0) = u(1) ,
(for given value of the continuation parameter λ).
A small perturbation in the initial condition
u(0) + ε v(0) , ε small ,
leads to the linearized equation
v′(t) = T fu( u(t) ) v(t) , for t ∈ [0, 1] ,
which induces a linear map
v(0) → v(1) ,
represented byv(1) = M v(0) .
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v(1) = M v(0)
The eigenvalues of M are the Floquet multipliers that determine stability.
M always has a multiplier µ = 1 , since differentiating
u′(t) = T f( u(t)) ,
gives
v′(t) = T fu( u(t) ) v(t) ,
where
v(t) = u′(t) , with v(0) = v(1) .
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v(1) = M v(0)
• If M has a Floquet multiplier µ with | µ | > 1 then u(t) is unstable .
• If all multipliers (other than µ = 1) satisfy | µ | < 1 then u(t) is stable .
• At folds and branch points there are two multipliers µ = 1 .
• At a period-doubling bifurcation there is a real multiplier µ = −1 .
• At a torus bifurcation there is a complex pair on the unit circle.
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EXAMPLE : The Lorenz Equations .
( Course demo : Lorenz )
These equations were introduced in 1963 by Edward Lorenz (1917-2008)
as a simple model of atmospheric convection :
x′ = σ (y − x) ,
y′ = ρ x − y − x z ,
z′ = x y − β z ,
where (often)
σ = 10 , β = 8/3 , ρ = 28 .
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Course demo : Lorenz/Basic
Bifurcation diagram of the Lorenz equations for σ = 10 and β = 8/3 .
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Course demo : Lorenz/Basic
Unstable periodic orbits of the Lorenz equations.
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In the Lorenz Equations :
• The zero stationary solution is unstable for ρ > 1 .
• Two nonzero stationary families bifurcate at ρ = 1 .
• The nonzero stationary solutions are unstable for ρ > ρH .
• At ρH ≈ 24.7 there are Hopf bifurcations .
• Unstable periodic solution families emanate from the Hopf bifurcations.
• These families end in homoclinic orbits (infinite period) at ρ ≈ 13.9 .
• At ρ = 28 (and a range of other values) there is the Lorenz attractor .
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EXAMPLE : The A→ B → C Reaction .
( Course demo : Chemical-Reactions/ABC-Reaction/Periodic )
Stationary and periodic solution families of the A→ B → C reaction: β = 1.55 .
Note the coexistence of stable solutions, for example, solutions 1 and 2 .
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0.20 0.25 0.30 0.35D
2
3
4
5
6
7
8
9
max
u3
0.20 0.25 0.30 0.35D
2
3
4
5
6
7
8
9
max
u3
0.20 0.25 0.30 0.35D
2
3
4
5
6
7
8
9
max
u3
0.20 0.25 0.30 0.35D
2
3
4
5
6
7
8
9
max
u3
Top left: β = 1.55, right: β = 1.56, Bottom left: β = 1.57, right: β = 1.58.
( QUESTION : Is something missing somewhere ? )
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Following Folds for Periodic Solutions
Recall that periodic orbits families can be computed using the equations
u′(t) − T f( u(t) , λ ) = 0 ,
u(0) − u(1) = 0 ,∫ 1
0
〈 u(t) , u′
0(t) 〉 dt = 0 ,
where u0 is a reference orbit , typically the latest computed orbit.
The above boundary value problem is of the form
F( X , λ ) = 0 ,
whereX = ( u , T ) .
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At a fold with respect to λ we have
FX( X , λ ) Φ = 0 ,
〈 Φ , Φ 〉 = 1 ,
whereX = ( u , T ) , Φ = ( v , S ) ,
i.e., the linearized equations about X have a nonzero solution Φ .
In detail : v′(t) − T fu( u(t) , λ ) v − S f( u(t) , λ ) = 0 ,
v(0) − v(1) = 0 ,∫ 1
0
〈 v(t) , u′
0(t) 〉 dt = 0 ,
∫ 1
0
〈 v(t) , v(t) 〉 dt + S2 = 1 .
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The complete extended system to follow a fold is
F( X , λ , µ ) = 0 ,
FX( X , λ , µ ) Φ = 0 ,
〈 Φ , Φ 〉 − 1 = 0 ,
with two free problem parameters λ and µ .
To the above we add the continuation equation
〈 X−X0 , X0 〉 + 〈 Φ−Φ0 , Φ0 〉 + (λ−λ0) λ0 + (µ−µ0) µ0 − ∆s = 0 .
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In detail : u′(t) − T f( u(t) , λ , µ ) = 0 ,
u(0) − u(1) = 0 ,∫ 1
0
〈 u(t) , u′
0(t) 〉 dt = 0 ,
v′(t) − T fu( u(t) , λ , µ ) v − S f( u(t) , λ , µ ) = 0 ,
v(0) − v(1) = 0 ,∫ 1
0
〈 v(t) , u′
0(t) 〉 dt = 0 ,
with normalization∫ 1
0
〈 v(t) , v(t) 〉 dt + S2 − 1 = 0 ,
and continuation equation∫ 1
0
〈 u(t)−u0(t) , u0(t) 〉 dt +
∫ 1
0
〈 v(t)−v0(t) , v0(t) 〉 dt +
+ (T0 − T )T0 + (S0 − S)S0 + (λ− λ0)λ0 + (µ− µ0)µ0 − ∆s = 0 .
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EXAMPLE : The A→ B → C Reaction .
( Course demo : Chemical-Reactions/ABC-Reaction/Folds-PS )
Stationary and periodic solution families of the A→ B → C reaction .
(with blow-up) for β = 1.55 . Note the three folds , labeled 1, 2, 3 .
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0.21 0.22 0.23 0.24 0.25 0.26 0.27 0.28D
1.54
1.55
1.56
1.57
1.58
1.59
1.60
1.61
1.62
β
Loci of folds along periodic solution families for the A→ B → C reaction.
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Stationary and periodic solution families of the A→ B → C reaction: β = 1.56 .
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0.15 0.20 0.25 0.30 0.35 0.40D
1
2
3
4
5
6
7
8
9
max
u3
Stationary and periodic solution families of the A→ B → C reaction: β = 1.57 .
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Stationary and periodic solution families of the A→ B → C reaction: β = 1.58 .
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Stationary and periodic solution families of the A→ B → C reaction: β = 1.61 .
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Stationary and periodic solution families of the A→ B → C reaction: β = 1.62 .
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Periodic solutions along the isola for β = 1.58 .(Stable solutions are blue, unstable solutions are red.)
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Following Period-doubling Bifurcations
Let ( u(t) , T ) be a periodic solution , i.e., a solution of
u′(t) − T f( u(t) , λ ) = 0 ,
u(0) − u(1) = 0 ,∫ 1
0
〈 u(t) , u′
0(t) 〉 dt = 0 ,
where u0 is a reference orbit .
A necessary condition for a period-doubling bifurcation is that the followinglinearized system have a nonzero solution v(t) :
v′(t) − T fu( u(t) , λ ) v(t) = 0 ,
v(0) + v(1) = 0 ,∫ 1
0
〈 v(t) , v(t) 〉 dt = 1 .
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The complete extended system to follow a period-doubling bifurcation is
u′(t) − T f( u(t) , λ , µ ) = 0 ,
u(0) − u(1) = 0 ,∫ 1
0
〈 u(t) , u′
0(t) 〉 dt = 0 ,
v′(t) − T fu( u(t) , λ ) v(t) = 0 ,
v(0) + v(1) = 0 ,∫ 1
0
〈 v(t) , v(t) 〉 dt − 1 = 0 ,
and continuation equation∫ 1
0
〈 u(t)−u0(t) , u0(t) 〉 dt +
∫ 1
0
〈 v(t)−v0(t) , v0(t) 〉 dt +
+ (T0 − T )T0 + (λ− λ0)λ0 + (µ− µ0)µ0 − ∆s = 0 .
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EXAMPLE : Period-Doubling Bifurcations in the Lorenz Equations .
( Course demo : Lorenz/Period-Doubling )
• The Lorenz equations also have period-doubling bifurcations .
• In fact, there is a period-doubling cascade for large ρ .
• We start from numerical data .
• (Such data may be from simulation , i.e., initial value integration .)
• We also want to compute loci of period-doubling bifurcations .
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( Course demo : Lorenz/Period-Doubling )
Left panel : Solution families of the Lorenz equations.
The open diamonds denote period-doubling bifurcations .
Right panel : Solution 1 was found by initial value integration .
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( Course demo : Lorenz/Period-Doubling )
Left panel : A primary period-doubled solution.
Right panel : A secondary period-doubled solution.
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( Course demo : Lorenz/Period-Doubling )
Loci of period-doubling bifurcations for the Lorenz equations (with blow-up) .
Black: primary, Red: secondary, Blue: tertiary period-doublings .
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Periodic Solutions of Conservative Systems
EXAMPLE : A Model Conservative System .
( Course demo : Vertical-HB )
u′1 = − u2 ,
u′2 = u1 (1− u1) .
PROBLEM :
• This system has a family of periodic solutions, but no parameter !
• The system has a constant of motion , namely the Hamiltonian
H(u1, u2) = − 1
2u2
1 −1
2u2
2 +1
3u3
1 .
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REMEDY :
Introduce an unfolding term with unfolding parameter λ :
u′1 = λ u1 − u2 ,
u′2 = u1 (1− u1) .
Then there is a vertical Hopf bifurcation from the trivial solution at λ = 0 .
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Bifurcation diagram of the vertical Hopf bifurcation problem.( Course demo : Vertical-HB )
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NOTE :
• The family of periodic solutions is vertical .
• The parameter λ is solved for in each continuation step.
• Upon solving, λ is found to be zero , up to numerical precision.
• One can use standard BVP continuation and bifurcation algorithms.
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A phase plot of some periodic solutions.
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EXAMPLE : The Circular Restricted 3-Body Problem (CR3BP) .
( Course demo : Restricted-3Body/Earth-Moon/Orbits )
x′′ = 2y′ + x − (1− µ) (x+ µ)
r31
− µ (x− 1 + µ)
r32
,
y′′ = −2x′ + y − (1− µ) y
r31
− µ y
r32
,
z′′ = −(1− µ) z
r31
− µ z
r32
,
where
r1 =√
(x + µ)2 + y2 + z2 , r2 =√
(x− 1 + µ)2 + y2 + z2 .
and( x , y , z ) denotes the position of the zero-mass body .
NOTE : For the Earth-Moon system µ ≈ 0.01215 .
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The CR3BP has one integral of motion , namely, the “Jacobi-constant” :
J =x′2 + y′2 + z′2
2− U(x, y, z) − µ
1− µ2
,
where
U =1
2(x2 + y2) +
1− µr1
+µ
r2
,
and
r1 =√
(x+ µ)2 + y2 + z2 , r2 =√
(x− 1 + µ)2 + y2 + z2 .
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Boundary value formulation :
x′ = T vx
y′ = T vy
z′ = T vz
v′x = T [ 2vy + x − (1− µ)(x+ µ)r−31 − µ(x− 1 + µ)r−3
2 + λ vx ]
v′y = T [ − 2vx + y − (1− µ)yr−31 − µyr−3
2 + λ vy ]
v′z = T [ − (1− µ)zr−31 − µzr−3
2 + λ vz ]
with periodicity boundary conditions
x(1) = x(0) , y(1) = y(0) , z(1) = z(0) ,
vx(1) = vx(0) , vy(1) = vy(0) , vz(1) = vz(0) ,
+ phase constraint + continuation equation .
Here T is the period of the orbit.
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NOTE :
• One can use BVP continuation and bifurcation algorithms.
• The unfolding term λ ∇v regularizes the continuation.
• λ will be zero , once solved for.
• Other unfolding terms are possible.
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Schematic bifurcation diagram of periodic solution families of the Earth-Moon system .
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The planar Lyapunov family L1.
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The Halo family H1.
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The Halo family H1.
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The Vertical family V1.
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The Axial family A1.
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Stable and Unstable Manifolds
EXAMPLE : Phase-plane orbits: Fixed length .
These can be computed by orbit continuation .
Model equations are
x′ = εx − y3 ,
y′ = y + x3 .
where ε > 0 is small.
• There is only one equilibrium , namely, (x, y) = (0, 0) .
• This equilibrium has eigenvalues ε and 1 ; it is a source .
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For the computations :
• The time variable t is scaled to [0, 1].
• The actual integration time T is then an explicit parameter :
x′ = T ( εx − y3 ) ,
y′ = T ( y + x3 ) .
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These constraints are used :
• To put the initial point on a small circle around the origin :
x(0)− r cos(2πθ) = 0 ,
y(0)− r sin(2πθ) = 0 .
• To keep track of the end points :
x(1)− x1 = 0 ,
y(1)− y1 = 0 .
• To keep track of the length of the orbits∫ 1
0
√x′(t)2 + y′(t)2 dt− L = 0 .
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The computations are done in 3 stages :
• In the first run an orbit is grown by continuation :
- The free parameters are T, L, x1, y1 .
- The starting point is on the small circle of radius r.
- The starting point is in the strongly unstable direction .
- The value of ε is 0.5 in the first run.
• In the second run the value of ε is decreased to 0.01 :
- The free parameters are ε, T, x1, y1 .
• In the third run the initial point is free to move around the circle :
- The free parameters are θ, T, x1, y1 .
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( Course demo : Basic-Manifolds/2D-ODE/Fixed-Length )
Unstable Manifolds in the Plane: Orbits of Fixed Length .(The right-hand panel is a blow-up, and also shows fewer orbits.)
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EXAMPLE : Phase-plane orbits: Variable length .
These can also be computed by orbit continuation .
Model equations are
x′ = εx − y2 ,
y′ = y + x2 .
• The origin (x, y) = (0, 0) is an equilibrium .
• The origin has eigenvalues ε and 1 ; it is a source .
• Thus the origin has a 2-dimensional unstable manifold .
• We compute this stable manifold using continuation .
• (The equations are 2D; so we actually compute a phase portrait.)
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For the computations :
• The time variable t is scaled to [0, 1].
• The actual integration time T is then an explicit parameter :
x′ = T (εx− y2) ,
y′ = T (y + x2) .
NOTE :
• There is also a nonzero equilibrium
(x, y) = (ε13 ,−ε 2
3 ) .
• It is a saddle (1 positive, 1 negative eigenvalue).
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These constraints are used :
• To put the initial point on a small circle at the origin :
x(0)− r cos(2πθ) = 0 ,
y(0)− r sin(2πθ) = 0 .
• To keep track of the end points :
x(1)− x1 = 0 ,
y(1)− y1 = 0 .
• To keep track of the length of the orbits we add an integral constraint :∫ 1
0
√x′(t)2 + y′(t)2 dt− L = 0 .
• To allow the length L to contract :
(Tmax − T )(Lmax − L)− c = 0 .
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Again the computations are done in 3 stages :
• In the first run an orbit is grown by continuation :
- The free parameters are T, L, x1, y1, c .
- The starting point is on a small circle of radius r.
- The starting point is in the strongly unstable direction .
- In this first run ε = 0.5 .
• In the second run the value of ε is decreased to 0.05 :
- The free parameters are ε, T, L, x1, y1 .
• In the third run the initial point is free to move around the circle :
- The free parameters are θ, T, L, x1, y1 .
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( Course demo : Basic-Manifolds/2D-ODE/Variable-Length )
Unstable Manifolds in the Plane: Orbits of Variable Length .
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( Course demo : Basic-Manifolds/2D-ODE/Variable-Length )
Unstable Manifolds in the Plane: Orbits of Variable Length (Blow-up).
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EXAMPLE : A 2D unstable manifold in R3 .
This can also be computed by orbit continuation. The model equations are
x′ = εx− z3 ,
y′ = y − x3 ,
z′ = −z + x2 + y2 .
• We take ε = 0.05.
• The origin is a saddle with eigenvalues ε, 1, and −1.
• Thus the origin has a 2-dimensional unstable manifold .
• The initial point moves around a circle in the 2D unstable eigenspace .
• The equations are 3D; so we will compute a 2D manifold in R3 .
• There is also a nonzero saddle , so we use retraction .
• The set-up is similar to the 2D phase-portrait example.
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( Course demo : Basic-Manifolds/3D-ODE/Variable-Length )
Unstable Manifolds in R3: Orbits of Variable Length .
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EXAMPLE : Another 2D unstable manifold in R3 .
The model equations are
x′ = εx− y3 + z3 ,
y′ = y + x3 + z3 ,
z′ = −z − x2 + y2 .
• We take ε = 0.05.
• The origin is a saddle with eigenvalues ε, 1, and −1.
• Thus the origin has a 2-dimensional unstable manifold .
• The initial point moves around a circle in the 2D unstable eigenspace .
• The equations are 3D; so we will compute a 2D manifold in R3 .
• No retraction is needed, so we choose to compute orbits of fixed length .
• The set-up is similar to the 2D phase-portrait example.
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( Course demo : Basic-Manifolds/3D-ODE/Fixed-Length )
Unstable Manifolds in R3: Orbits of Fixed Length .
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The Lorenz Manifold
• For ρ > 1 the origin is a saddle point .
• The Jacobian has two negative and one positive eigenvalue .
• The two negative eigenvalues give rise to a 2D stable manifold .
• This manifold is known as as the Lorenz Manifold .
• The set-up is as for the earlier 3D model, using fixed length .
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Course demo : Lorenz/Manifolds/Origin/Fixed-Length
Part of the Lorenz Manifold (with blow-up). Orbits have fixed length L = 60 .
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Course demo : Lorenz/Manifolds/Origin/Fixed-Length
Part of the Lorenz Manifold. Orbits have fixed length L = 200 .
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Heteroclinic Connections.
• The Lorenz Manifold helps understand the Lorenz attractor .
• Many orbits in the manifold depend sensitively on initial conditions .
• During the manifold computation one can locate heteroclinic orbits .
• These are also in the 2D unstable manifold of the nonzero equilibria.
• The heteroclinic orbits have a combinatorial structure 4.
• One can also continue heteroclinic orbits as ρ varies.
4 Nonlinearity 19, 2006, 2947-2972.
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Course demo : Lorenz/Heteroclinics
Four heteroclinic orbits with very close initial conditions
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One can also determine the intersection of the Lorenz manifold with a sphere .
The set-up is as follows :
x′ = T σ (y − x) ,
y′ = T (ρ x − y − x z) ,
z′ = T (x y − β z) ,
which is of the form
u′(t) = T f( u(t) ) , for 0 ≤ t ≤ 1 ,
where
• T is the actual integration time , which is negative !
To this we add boundary and integral constraints .
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The complete set-up consists of the ODE
u′(t) = T f( u(t) ) , for 0 ≤ t ≤ 1 ,
subject to the following constraints :
u(0)− ε (cos(θ) v1 − sin(θ) v2) = 0 u(0) is on a small circle
u(1) − u1 = 0 to keep track of the end point u(1)
‖ u1 ‖ − R = 0 distance of u1 to the origin
〈 u1/ ‖ u1 ‖ , f(u1)/ ‖ f(u1) ‖ 〉 − τ = 0 to locate tangencies, where τ = 0
T∫ 1
0‖ f(u) ‖ ds − L = 0 to keep track of the orbit length
(T − Tn) (L− Ln) − c = 0 . allows retraction into the sphere
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The continuation system has the form
F(Xk) = 0 , where X = ( u(·) , Λ ) .
with continuation equation
〈 Xk −Xk−1 , Xk−1 〉 − ∆s = 0 , ( ‖ Xk−1 ‖= 1 ) .
The computations are done in 2 stages :
• In the first run an orbit is grown by continuation :
- The starting point is on the small circle of radius ε.
- The starting point is in the strongly stable direction .
- The free parameters are Λ = (T , L , c , τ , R ,u1) .
• In the second run the orbit sweeps the stable manifold.
- The initial point is free to move around the circle :
- The free parameters are Λ = (T , L , θ , τ , R , u1) .
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Course demo : Lorenz/Manifolds/Origin/Sphere
Intersection of the Lorenz Manifold with a sphere
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NOTE :
• We do not just change the initial point (i.e., θ) and integrate !
• Every continuation step requires solving a boundary value problem .
• The continuation stepsize ∆s controls the change in X .
• X can only change a little in any continuation step.
• This way the entire manifold (up to a given length L) is computed.
• The retraction constraint allows the orbits to retract into the sphere.
• This is necessary when heteroclinic connections are encountered.
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EXAMPLE : Unstable Manifolds of a Periodic Orbit .
( Course demo : Lorenz/Manifolds/Orbits/Rho24.3 )
Left: Bifurcation diagram of the Lorenz equations. Right: Labeled solutions.
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Both sides of the unstable manifold of periodic orbit 3 at ρ = 24.3 .
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EXAMPLE : Unstable Manifolds in the CR3BP .
( Course demo : Restricted-3Body/Earth-Moon/Manifolds/H1 )
• ”Small” Halo orbits have one real Floquet multiplier outside the unit circle.
• Such Halo orbits are unstable .
• They have a 2D unstable manifold .
• The unstable manifold can be computed by continuation .
• First compute a starting orbit in the manifold.
• Then continue the orbit keeping, for example, x(1) fixed .
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Part of the unstable manifold of three Earth-Moon L1-Halo orbits.
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• The initial orbit can be taken to be much longer · · ·• Continuation with x(1) fixed can lead to a Halo-to-torus connection!
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• The Halo-to-torus connection can be continued as a solution to
F( Xk ) = 0 ,
< Xk −Xk−1 , Xk−1 > − ∆s = 0 .
where
X = ( Halo orbit , Floquet function , connecting orbit) .
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In detail , the continuation system is
du
dτ− Tuf(u(τ), µ, l) = 0 ,
u(1)− u(0) = 0 ,∫ 1
0
〈 u(τ) , u0(τ) 〉 dτ = 0 ,
dv
dτ− TuDuf(u(τ), µ, l)v(τ) + λuv(τ) = 0 ,
v(1)− sv(0) = 0 (s = ±1) ,
〈 v(0) , v(0) 〉 − 1 = 0 ,
dw
dτ− Twf(w(τ), µ, 0) = 0 ,
w(0)− (u(0) + εv(0)) = 0 ,
w(1)x − xΣ = 0 .
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The system has
18 ODEs , 20 boundary conditions , 1 integral constraint .
We need
20 + 1 + 1 - 18 = 4 free parameters .
Parameters :
• An orbit in the unstable manifold: Tw , l , Tu , xΣ
• Compute the unstable manifold: Tw , l , Tu , ε
• Follow a connecting orbit: λu , l , Tu , ε
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The Solar Sail Equations
The equations in Course demo : Solar-Sail/Equations/equations.f90 :
x′′ = 2y′ + x− (1− µ)(x+ µ)
d3S
− µ(x− 1 + µ)
d3P
+β(1− µ)D2Nx
d2S
y′′ = − 2x′ + y − (1− µ)y
d3S
− µy
d3P
+β(1− µ)D2Ny
d2S
z′′ = − (1− µ)z
d3S
− µz
d3P
+β(1− µ)D2Nz
d2S
where
dS =√
(x+ µ)2 + y2 + z2 , dP =√
(x− 1 + µ)2 + y2 + z2 , r =√
(x+ µ)2 + y2
Nx = [cos(α)(x+ µ)− sin(α)y] [cos(δ)− sin(δ)z
r]/dS
Ny = [cos(α)y + sin(α)(x+ µ)] [cos(δ)− sin(δ)z
r]/dS
Nz = [cos(δ)z + sin(δ)r]/dS , D =x+ µ
dSNx +
y
dSNy +
z
dSNz
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Course demo : Solar-Sail/Sun-Jupiter/Libration/Points
Sun-Jupiter libration points, for β = 0, α = 0, δ = 0.
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Course demo : Solar-Sail/Sun-Jupiter/Libration/Points
Sun-Jupiter libration points, for β = 0.02, α = 0.02, δ = 0.
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Course demo : Solar-Sail/Sun-Jupiter/Libration/Loci
Sun-Jupiter libration points, with δ ∈ [−π2, π
2], for various β, with α = 0.
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Course demo : Solar-Sail/Sun-Jupiter/Libration/Loci
Sun-Jupiter libration points, with δ ∈ [−π2, π
2], for various α, with β = 0.15.
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Course demo : Solar-Sail/Sun-Jupiter/Libration/Homoclinic
Sun-Jupiter: detection of a homoclinic orbit at β = 0.050698, with α = 0, δ = 0.
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Course demo : Solar-Sail/Sun-Jupiter/Libration/Manifolds
Sun-Jupiter: unstable manifold orbits for δ ∈ [−π2, π
2], with β = 0.05, α = 0.1.
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Course demo : Solar-Sail/Sun-Jupiter/Libration/Manifolds
0.926 0.927 0.928 0.929 0.930 0.931 0.932 0.933x
−0.0075
−0.0050
−0.0025
0.0000
0.0025
0.0050
0.0075
z
−1.05 −1.00 −0.95 −0.90 −0.85 −0.80 −0.75 −0.70 −0.65y(T )
−0.04
−0.03
−0.02
−0.01
0.00
0.01
0.02
0.03
0.04
z(T
)
The libration points The end points
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Course demo : Solar-Sail/Sun-Jupiter/Libration/Manifolds
Some connecting orbits for α = 0.07 and varying β and δ.
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Course demo : Solar-Sail/Sun-Jupiter/Orbits
V1-orbits with β = 0.15, T = 6.27141, δ ∈ [0 , 0.6415] .
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