numberical analysis of nonlinear equations
DESCRIPTION
Presented are lecture notes on numerical analysis of nonlinear equationsTRANSCRIPT
![Page 1: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/1.jpg)
Lecture Notes on
NUMERICAL ANALYSIS
of
OF NONLINEAR EQUATIONS
Eusebius Doedel
1
![Page 2: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/2.jpg)
Persistence of Solutions
We discuss the persistence of solutions to nonlinear equations.
2
![Page 3: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/3.jpg)
◦ Newton’s method for solving a nonlinear equation
G(u) = 0 , G(·) , u ∈ Rn ,
may not converge if the “initial guess” is not close to a solution.
◦ To alleviate this problem one can introduce an artificial “homotopy” pa-rameter in the equation.
◦ Actually, most equations already have parameters.
◦ We discuss the persistence of solutions to such parameter-dependent equa-tions.
3
![Page 4: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/4.jpg)
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 .
PROOF : See a good Analysis book · · ·
4
![Page 5: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/5.jpg)
EXAMPLE: (AUTO demo hom.)
Letg(u, λ) = (u2 − 1) (u2 − 4) + λ u2 ecu ,
where c is fixed, e.g., c = 0.1 .
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 .
5
![Page 6: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/6.jpg)
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.
Hence each of the four solutions when λ = 0 is isolated .
Thus each of these solutions persists as λ becomes nonzero,
( at least for “small” values of | λ | ).
6
![Page 7: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/7.jpg)
0.0 0.2 0.4 0.6 0.8 1.0
λ
−2
−1
0
1
2
u
Four solution families of g(u, λ) = 0 . Note the fold.
7
![Page 8: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/8.jpg)
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 two other homotopies meet at a fold .
◦ IFT condition (ii) is not satisfied at this fold. (Why not?)
8
![Page 9: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/9.jpg)
Consider the equation
G(u, λ) = 0 , u , G(·, ·) ∈ Rn , λ ∈ R .
Letx ≡ (u , λ) .
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., ifRank(G0
x) = n .
9
![Page 10: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/10.jpg)
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) .
Above,
N (G0u) denotes the null space of G0
u ,
and
R(G0u) denotes the range of G0
u ,
i.e., the linear space spanned by the n columns of G0u .
10
![Page 11: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/11.jpg)
THEOREM. Letx0 ≡ ( 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 ,
then either G0u is nonsingular and by the IFT we have
u = u(λ) near x0 ,
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 unique solution family passes through a regular solution. •
11
![Page 12: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/12.jpg)
NOTE:
◦ Such a solution family is sometimes also called a solution branch .
◦ Case (ii) above is that of a simple fold , to be discussed later.
◦ Thus even near a simple fold there is a unique solution family.
◦ However, near such a fold, the family can not be parametrized by λ.
12
![Page 13: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/13.jpg)
EXAMPLE: The A→ B → C reaction. (AUTO demo abc.)
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 = 1.21 is the heat transfer coefficient .
We compute stationary solutions for varying D.
13
![Page 14: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/14.jpg)
0.110 0.115 0.120 0.125 0.130 0.135 0.140 0.145 0.150
Damkohler number
0
1
2
3
4
5
6
7
8
L2-
norm
2
2
3
4
4
Stationary solution families of the A→ B → C reaction for β = 1.15.Solid/dashed lines denote stable/unstable solutions.
The red square denotes a Hopf bifurcation.(AUTO demo abc). Note the two folds .
14
![Page 15: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/15.jpg)
Examples of IFT Application
Here we give examples where the IFT shows that a given solution persists, atleast locally, when a problem parameter is changed. We also identify some caseswhere the conditions of the IFT are not satisfied.
15
![Page 16: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/16.jpg)
A Predator-Prey Model
(AUTO demo pp2.)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) .
16
![Page 17: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/17.jpg)
The Jacobian matrix is
Gu =
(3− 6u1 − u2 − 5λe
−5u1 −u1
3u2 −1 + 3u1
)= Gu(u1, u2;λ) .
Gu(0, 0; 0) =
(3 00 −1
); eigenvalues 3,-1 (unstable) .
Gu(1, 0; 0) =
( −3 −10 2
); eigenvalues -3,2 (unstable) .
Gu(1
3, 2; 0) =
( −1 −13
6 0
); eigenvalues
(−1− µ)(−µ) + 2 = 0µ2 + µ+ 2 = 0
µ± = −1±√−7
2
Re(µ±) < 0 (stable) .
All three Jacobians at λ = 0 are nonsingular.
Thus, by the IFT, all three stationary points persist for (small) λ 6= 0 .
17
![Page 18: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/18.jpg)
In this problem we can explicitly find all solutions (see Figure 1) :
Branch I :
(u1, u2) = (0, 0) .
Branch II :
u2 = 0 , λ =3u1(1− u1)
1− e−5u1.
(Note that limu1 → 0
λ = limu1 → 0
3(1− 2u1)
5e−5u1=
3
5.)
Branch III :
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 branch points, one of which is
(u1, u2, λ) = (0, 0, 3/5) .
18
![Page 19: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/19.jpg)
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
Figure 1: Stationary solution families of the predator-prey model. Solid/dashedlines denote stable/unstable solutions. Note the fold , the bifurcations (opensquares), and the Hopf bifurcation (red square).
19
![Page 20: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/20.jpg)
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
Figure 2: Stationary solution families of the predator-prey model, showing fishversus quota. Solid/dashed lines denote stable/unstable solutions.
20
![Page 21: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/21.jpg)
◦ Stability of branch I :
Gu((0, 0);λ) =
(3− 5λ 0
0 −1
); eigenvalues 3− 5λ, − 1 .
Hence the trivial solution is :
unstable if λ < 3/5 ,
and
stable if λ > 3/5 ,
as indicated in Figure 2.
◦ Stability of branch II :
This family has no stable positive solutions.
21
![Page 22: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/22.jpg)
◦ Stability of branch III :
At λH ≈ 0.67 ,
(the red square in Figure 2) the complex eigenvalues cross the imaginary axis.
This crossing is a Hopf bifurcation, a topic to be discussed later.
Beyond λH there are periodic solutions whose period T increases as λ in-creases. (See Figure 4 for some representative periodic orbits.)
The period becomes infinite at λ = λ∞ ≈ 0.70 .
This final orbit is called a heteroclinic cycle.
22
![Page 23: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/23.jpg)
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
max
fish
Figure 3: Stationary (blue) and periodic (red) solution families of the predator-prey model.
23
![Page 24: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/24.jpg)
0.1 0.2 0.3 0.4 0.5 0.6
fish
0.0
0.1
0.2
0.3
0.4
0.5
shar
ks
Figure 4: Some periodic solutions of the predator-prey model. The largest orbitsare very close to a heteroclinic cycle.
24
![Page 25: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/25.jpg)
From Figure 3 we can deduce the solution behavior for (slowly) increasing λ :
- Branch III is followed until λH ≈ 0.67 .
- Periodic solutions of increasing period until λ = λ∞ ≈ 0.70 .
- Collapse to trivial solution (Branch I).
EXERCISE.
Use AUTO to repeat the numerical calculations (demo pp2) .
Sketch phase plane diagrams for λ = 0, 0.5, 0.68, 0.70, 0.71 .
25
![Page 26: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/26.jpg)
The Gelfand-Bratu Problem
(AUTO demo exp.)
u′′(x) + λ eu(x) = 0 , ∀x ∈ [0, 1] ,
u(0) = u(1) = 0 .
If λ = 0 then u(x) ≡ 0 is a solution.
We’ll prove that this solution is isolated, so that there is a continuation
u = u(λ) , for |λ| small .
26
![Page 27: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/27.jpg)
Consider
u′′(x) + λ eu(x) = 0 ,
u(0) = 0 , u′(0) = p ,
⇒ u = u( x ; p, λ ) .
We want to solveu( 1 ; p, λ )︸ ︷︷ ︸≡G( p, λ )
= 0 ,
for |λ| small .
HereG( 0 , 0 ) = 0 .
27
![Page 28: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/28.jpg)
We must show (IFT) that
Gp( 0 , 0 ) ≡ up( 1 ; 0 , 0 ) 6= 0 :
u′′p(x) + λ0 e
u0(x) up(x) = 0 ,
up(0) = 0 , u′p(0) = 1 ,
where u0(x) ≡ 0 .
Now up( x ; 0 , 0 ) satisfiesu′′p(x) = 0 ,
up(0) = 0 , u′p(0) = 1 .
Henceup( x ; 0 , 0 ) = x , up( 1 ; 0 , 0 ) = 1 6= 0 .
EXERCISE. Compute the solution family of the Gelfand-Bratu problem asrepresented in Figures 5 and 6. (AUTO demo exp.)
28
![Page 29: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/29.jpg)
0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5
λ0.0
0.5
1.0
1.5
2.0
2.5
3.0
3.5
∫ 1 0u(x
)dx
1 2 3 45
67
8
9
10
11
1213
14
Figure 5: Bifurcation diagram of the Gelfand-Bratu equation.
29
![Page 30: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/30.jpg)
0.0 0.2 0.4 0.6 0.8 1.0
x0
1
2
3
4
5
6
u(x
)
1 2 3
4
5 6
789
10
11
12
13
14
Figure 6: Some solutions to the Gelfand-Bratu equation.
30
![Page 31: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/31.jpg)
A Nonlinear Eigenvalue Problem
(AUTO demo nev.)Consider the nonlinear boundary value problem
u′′ + λ (u + u2) = 0 ,
u(0) = u(1) = 0 ,
which has u(t) ≡ 0 as solution for all λ .
Equivalently, we want the solution
u = u( t ; p , λ) ,
of the initial value problemu′′ + λ (u + u2) = 0 ,
u(0) = 0 , u′(0) = p ,
that satisfiesG( p , λ ) ≡ u( 1 ; p , λ ) = 0 .
31
![Page 32: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/32.jpg)
HereG : R × R → R ,
withG( 0 , λ ) = 0 , for all λ .
Let
up( t ; p , λ ) =du
dp( t ; p , λ ) ,
Gp( p , λ ) = up( 1 ; p , λ ) .
32
![Page 33: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/33.jpg)
Then up ( = u0p ) satisfies
u′′p + λ (1 + 2u) up = 0 ,
up(0) = 0 , u′p(0) = 1 ,
which, about u ≡ 0 , givesu′′p + λ up = 0 ,
up(0) = 0 , u′p(0) = 1 .
By the variation of parameters formula
up(t; p, λ) =sin√λt√λ
, λ ≥ 0 , (independent of p).
Gp(0, λ) = up(1; p, λ) =sin(√λ)√λ
= 0 , if λ = λk ≡ (kπ)2 .
Thus the conditions of the IFT fail to be satisfied at λk = (kπ)2 .
(We will see that these solutions are branch points .)
33
![Page 34: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/34.jpg)
0 1 2 3 4 5
λ (scaled)
−10
−5
0
5
10
u′ a
tx=
0
1
23
4
5
Figure 7: Solution families to the nonlinear eigenvalue problem.
34
![Page 35: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/35.jpg)
0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0x
−1.0
−0.5
0.0
0.5
1.0
1.5
2.0
u
1
2
3
4
5
Figure 8: Some solutions to the nonlinear eigenvalue problem.
35
![Page 36: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/36.jpg)
Numerical Continuation
Here we discuss algorithms for computing families of solutions to nonlinear equa-tions. The IFT is important in the design of such continuation methods.
36
![Page 37: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/37.jpg)
◦ Newton’s method for solving a nonlinear equation
G(u) = 0 , G(·) , u ∈ Rn ,
may not converge if the “initial guess” is not close to a solution.
◦ To deal with this one can introduce a “homotopy parameter” .
◦ Most equations already naturally have parameters.
◦ We now discuss computing such families of solutions.
37
![Page 38: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/38.jpg)
Consider the equation
G(u, λ) = 0 , u , G(·, ·) ∈ Rn , λ ∈ R .
Letx ≡ (u , λ) .
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 n (rows) by n + 1 (columns)matrix
G0x ≡ Gx(x0) ,
has maximal rank, i.e., ifRank(G0
x) = n .
38
![Page 39: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/39.jpg)
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) .
Above,
N (G0u) denotes the null space of G0
u ,
and
R(G0u) denotes the range of G0
u ,
i.e., the linear space spanned by the n columns of G0u .
39
![Page 40: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/40.jpg)
THEOREM. Letx0 ≡ ( u0 , λ0 )
be a regular solution ofG(x) = 0 .
Then, near x0 , there exists a unique one-dimensional continuum of solutions
x(s) with x(0) = x0 .
PROOF. Since
Rank( G0x ) = Rank( G0
u | G0λ ) = n ,
then either G0u is nonsingular and by the IFT we have
u = u(λ) near x0 ,
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 unique solution family passes through a regular solution. •
40
![Page 41: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/41.jpg)
NOTE:
◦ Such a continuum of solutions is called a solution family or a solution branch.
◦ Case (ii) above, namely,
dim N (G0u) = 1 and G0
λ 6∈ R(G0u) ,
is that of a simple fold , to be discussed later.
41
![Page 42: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/42.jpg)
0.140 0.145 0.150 0.155 0.160 0.165 0.170 0.175
Damkohler number
1
2
3
4
5
6
7
L2-
norm
2929
30
31
32
33
33
34
Figure 9: A solution family with four folds. (AUTO demo abc with β = 1.25 .)
42
![Page 43: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/43.jpg)
Parameter Continuation
Here the continuation parameter is taken to be λ .
Suppose we have a solution (u0, λ0) of
G(u, λ) = 0 ,
as well as the direction vector u0 .
Here
u ≡ du
dλ.
We want to compute the solution u1 at λ1 ≡ λ0 + ∆λ .
43
![Page 44: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/44.jpg)
��
��
��
��
����������������������������������
������������
������������
"u"
u
u
λ λλ
u 1(0)
dud λ
at λ0 )u
10
1
0
∆λ
(=
Figure 10: Graphical interpretation of parameter-continuation.
44
![Page 45: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/45.jpg)
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 the Newton convergence theory guarantees thatthis iteration will converge.
45
![Page 46: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/46.jpg)
After convergence, the new direction vector u1 can be computed by solving
Gu(u1, λ1) u1 = −Gλ(u1, λ1) .
This equation follows from differentiating
G(u(λ), λ) = 0 ,
with respect to λ at λ = λ1 .
NOTE:
◦ u1 can be computed without another LU -factorization of Gu(u1, λ1) .
◦ Thus the extra work to find u1 is negligible.
46
![Page 47: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/47.jpg)
EXAMPLE: The Gelfand-Bratu problem (AUTO demo exp) :
u′′(x) + λ eu(x) = 0 for x ∈ [0, 1] , u(0) = 0 , u(1) = 0 .
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 .
47
![Page 48: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/48.jpg)
Let
u ≡
u1
u2
·uN−1
.
Then we can write the above as
G( u , λ ) = 0 ,
where
G : Rn × R → Rn , n ≡ N − 1 .
48
![Page 49: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/49.jpg)
Parameter-continuation : Suppose we have
λ0 , u0 , and u0 .
Setλ1 = 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 ,
ν = 0, 1, 2, · · · ,
withu
(0)1 = u0 + ∆λ u0 .
After convergence find u1 from
Gu(u1, λ1) u1 = −Gλ(u1, λ1) .
Repeat the above procedure to find u2 , u3 , · · · .
49
![Page 50: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/50.jpg)
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.
The solution family has a fold where the parameter-continuation method fails.(AUTO demo exp: See the earlier Figures 5 and 6).
50
![Page 51: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/51.jpg)
Keller’s Pseudo-Arclength Continuation
This method allows continuation of a solution family past a fold.
Suppose we have a solution (u0, λ0) of
G( u , λ ) = 0 ,
as well as the direction vector (u0, λ0) of the solution branch.
Pseudo-arclength continuation solves the following equations for (u1, λ1) :
G(u1, λ1) = 0 ,
(u1 − u0)∗ u0 + (λ1 − λ0) λ0 − ∆s = 0 .
See Figure 11 for a graphical interpretation.
51
![Page 52: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/52.jpg)
����u 0
u 0 ∆ s����
λ 0��
����������
����
"u"
λ λλ
u
10
1
u( ), λ 00
Figure 11: Graphical interpretation of pseudo-arclength continuation.
52
![Page 53: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/53.jpg)
Solve the equations
G(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
.
Next direction vector : G1u G1
λ
u∗0 λ0
( u1
λ1
)=
0
1
.
53
![Page 54: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/54.jpg)
NOTE:
◦ In practice (u1, λ1) can be computed with one extra backsubstitution.
◦ The orientation of the branch is preserved if ∆s is sufficiently small.
◦ The direction vector must be rescaled, so that indeed ‖ u1 ‖2 + λ21 = 1 .
54
![Page 55: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/55.jpg)
THEOREM.
The Jacobian of the pseudo-arclength system is nonsingular
at a regular solution point.
PROOF. Letx ≡ (u , λ) ∈ Rn+1 .
Then pseudo-arclength continuation can be written as
G(x1) = 0 ,
(x1 − x0)∗ x0 − ∆s = 0 , (‖ x0 ‖ = 1 ) .
(See Figure 12 for a graphical interpretation.)
55
![Page 56: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/56.jpg)
∆ s��
�� ����
1x
0x
"X-space"
0x
Figure 12: Parameter-independent pseudo-arclength continuation.
56
![Page 57: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/57.jpg)
The matrix in Newton’s method at ∆s = 0 is
(G0
x
x∗0
).
At a regular solution we have
N (G0x) = Span{x0} .
We must show that
(G0
x
x∗0
)
is nonsingular at a regular solution.
57
![Page 58: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/58.jpg)
If on the contrary (G0
x
x∗0
)is singular then
G0x z = 0 and x∗0 z = 0 ,
for some vector z 6= 0 .
Thus
z = c x0 , for some constant c .
But then0 = x∗0 z = c x∗0 x0 = c ‖ x0 ‖2 = c ,
so that z = 0 , which is a contradiction. •
58
![Page 59: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/59.jpg)
EXAMPLE:
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 :
◦ ◦ ◦◦ ◦ ◦ ◦◦ ◦ ◦ ◦◦ ◦ ◦ ◦◦ ◦ ◦ ◦◦ ◦ ◦ ◦◦ ◦ ◦ ◦◦ ◦ ◦
◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦
.
59
![Page 60: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/60.jpg)
Following Folds
When a parameter passes a fold, then the behavior of a system can changedrastically. Thus it is useful to determine how the location of a fold changeswhen a second parameter changes, i.e., we want the compute a “critical stabilitycurve”, or a “locus of fold points”, in 2-parameter space.
60
![Page 61: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/61.jpg)
Simple Folds
A regular solution x0 ≡ (u0, λ0) of G(u, λ) = 0 , is called a simple fold if
dim N (G0u) = 1 and G0
λ 6∈ R(G0u) .
From differentiatingG(u(s), λ(s)) = 0 ,
we haveGu(u(s), λ(s)) u(s) + Gλ(u(s), λ(s))λ(s) = 0 .
In particular,G0
u u0 = − λ0 G0λ .
At a fold we have G0λ 6∈ R(G0
u) . Thus
λ0 = 0 .
Hence G0uu0 = 0 . Thus, since dim N (G0
u) = 1 , we have
N (G0u) = Span{u0} .
61
![Page 62: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/62.jpg)
Differentiating again, we have
G0u u0 + G0
λ λ0 + G0uu u0 u0 + 2G0
uλ u0 λ0 + G0λλ λ0 λ0 = 0 .
At a simple fold (u0, λ0) let
N (G0u) = Span{φ} , (φ = u0) ,
andN ((G0
u)∗) = Span{ψ} .
Multiply by ψ∗ and use λ0 = 0 and ψ ⊥ R(G0u) to find
ψ∗G0λ λ0 + ψ∗G0
uu φ φ = 0 .
Here ψ∗G0λ 6= 0 , since G0
λ 6∈ R(G0u) . Thus
λ0 = −ψ∗G0uuφφ
ψ∗G0λ
.
If the curvature λ0 6= 0 then (u0, λ0) is called a simple quadratic fold.
62
![Page 63: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/63.jpg)
The Extended System
To continue a fold in two parameters we use the extended system
G(u, λ, µ) = 0 ,
Gu(u, λ, µ) φ = 0 ,
φ∗φ0 − 1 = 0 .
Here µ ∈ R is a second parameter in the equations.
The vector φ0 is from a “reference solution” (u0,φ0, λ0, µ0) .
(In practice this is the latest computed solution point on the branch.)
The above system has the form
F(U, µ) = 0 , U ≡ (u,φ, λ), F : R2n+1 × R→ R2n+1 ,
or, using the “parameter-free” formulation,
F(X) = 0 , X ≡ (U, µ) , F : R2n+2 → R2n+1 .
63
![Page 64: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/64.jpg)
Parameter Continuation
First consider continuing a solution
(u0 , φ0 , λ0) at µ = µ0 ,
in µ (although, in practice, we use pseudo-arclength continuation).
By the IFT there is a smooth solution family
U(µ) = ( u(µ) , φ(µ) , λ(µ) ) ,
if the Jacobian
F0U ≡
dF
dU(U0) =
G0
u O G0λ
G0uuφ0 G0
u G0uλφ0
0∗ φ∗0 0
,
is nonsingular.
64
![Page 65: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/65.jpg)
THEOREM.
A simple quadratic fold with respect to λ can be continued locally, using thesecond parameter µ as continuation parameter.
PROOF.
Suppose F0U is singular. Then
(i) G0ux + zG0
λ = 0 ,
(ii) G0uuφ0x + G0
uy + zG0uλφ0 = 0 ,
(iii) φ∗0y = 0 ,
for some
x,y ∈ Rn , z ∈ R .
65
![Page 66: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/66.jpg)
Since G0λ 6∈ R(G0
u) we have from (i) that z = 0 , and hence
x = c1 φ0 , for some c1 ∈ R .
Multiply (ii) on the left by ψ∗0 , to get
c1 ψ∗0G
0uuφ0φ0 = 0 .
Thus c1 = 0 , because by assumption ψ∗0G0uuφ0φ0 6= 0 .
Therefore x = 0 , and from (ii) we now have
G0uy = 0 , i .e. , y = c2 φ0 .
But then by (iii) c2 = 0 . Thus
x = y = 0 , z = 0 ,
and hence F0U is nonsingular. ◦
66
![Page 67: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/67.jpg)
NOTE:
◦ The zero eigenvalue of G0u need not be be algebraically simple.
◦ Thus, for example, G0u may have the form
G0u =
(0 10 0
),
provided the fold is simple and quadratic.
◦ Parameter-continuation fails at folds w.r.t µ on a solution family to
F(U , µ) = 0 .
◦ Such points represent cusps, branch points, or isola formation points.
67
![Page 68: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/68.jpg)
Pseudo-Arclength Continuation of Folds
Treat µ as one of the 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 ,
(1)
and the added pseudo-arclength equation
(u− u0)∗u0 + (φ− φ0)∗φ0 + (λ− λ0)λ0 + (µ− µ0)µ0 − ∆s = 0 . (2)
As before,( u0 , φ0 , λ0 , µ0 ) ,
is the direction of the branch at the current solution point
( u0 , φ0 , λ0 , µ0) .
68
![Page 69: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/69.jpg)
The Jacobian of F with respect to u , φ , λ , and µ , at
X0 = ( u0 , φ0 , λ0 , µ0 ) ,
is now
F0X ≡
dF
dX(X0) ≡
G0u O G0
λ G0µ
G0uuφ0 G0
u G0uλφ0 G0
uµφ0
0∗ φ∗0 0 0
.
For pseudo-arclength continuation we must check that F0X has full rank.
For a simple quadratic fold with respect to λ this follows from the Theorem.
Otherwise, ifψ∗0 G0
uu φ0 φ0 6= 0 ,
andG0λ ∈ R(G0
u) , G0µ 6∈ R(G0
u) ,
i.e., if we have a simple quadratic fold with respect to µ , then we can apply thetheorem to F0
X with the second last column struck out, to see that F0X still
has full rank.
69
![Page 70: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/70.jpg)
EXAMPLE: The A→ B → C reaction. (AUTO demo abc.)
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 .
We will compute solutions for varying D and β .
70
![Page 71: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/71.jpg)
0.110 0.115 0.120 0.125 0.130 0.135 0.140 0.145 0.150
Damkohler number
0
1
2
3
4
5
6
7
8
L2-
norm
2
2
3
4
4
Figure 13: A stationary solution family of demo abc; β = 1.15.
71
![Page 72: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/72.jpg)
0.10 0.12 0.14 0.16 0.18 0.20 0.22
Damkohler number
1.15
1.20
1.25
1.30
1.35
1.40
β
2
3
4
5
Figure 14: The locus of folds of demo abc.
72
![Page 73: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/73.jpg)
0.10 0.15 0.20 0.25 0.30 0.35 0.40
Damkohler number
0
1
2
3
4
5
6
7
L2-
norm
Figure 15: Stationary solution families for β = 1.15, 1.17, . . . , 1.39.
73
![Page 74: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/74.jpg)
Numerical Treatment ofBifurcations
Here we discuss branch switching, and the detection of branch points.
74
![Page 75: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/75.jpg)
Simple Singular Points
LetG : Rn+1 → Rn .
A solutionx0 ≡ x(s0) of G(x) = 0 ,
is called a simple singular point if
G0x ≡ Gx(x0) has rank n− 1 .
75
![Page 76: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/76.jpg)
In the parameter formulation, where
G0x = ( G0
u | G0λ ) ,
we have that
x0 = (u0, λ0) is a simple singular point
if and only if
(i) dim N (G0u) = 1 , G0
λ ∈ R(G0u) ,
or
(ii) dim N (G0u) = 2 , G0
λ 6∈ R(G0u) .
76
![Page 77: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/77.jpg)
−1.00 −0.75 −0.50 −0.25 0.00 0.25 0.50 0.75 1.00
λ
−1.00
−0.75
−0.50
−0.25
0.00
0.25
0.50
0.75
1.00
u
Figure 16: Solution curves of u (λ− u) = 0 , with simple singular point.
77
![Page 78: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/78.jpg)
An example of case (ii) is
G(u, λ) =(λ − u2
1 − u22
u1 u2
), at λ = 0 , u1 = u2 = 0 .
HereG0
u =(
0 00 0
), and G0
λ =(
10
).
u 2
−1.0
−0.5
0.0
0.5
1.0λ
0.00.2
0.40.6
0.81.0
u1
−1.0
−0.5
0.0
0.5
1.0
L
Figure 17: Solution curves of G(u, λ) = 0 , with simple singular point.
78
![Page 79: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/79.jpg)
Suppose we have a solution family x(s) of
G(x) = 0 ,
where s is some parametrization.
Let
x0 ≡ (u0, λ0) ,
be a simple singular point.
Thus, by definition of simple singular point,
N (G0x) = Span{φ1,φ2} , N (G0∗
x ) = Span{ψ} .
79
![Page 80: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/80.jpg)
We also have
G(x(s)) = 0 , G0 = G(x0) = 0 ,
Gx(x(s)) x(s) = 0 , G0x x0 = 0 ,
Gxx(x(s)) x(s) x(s) + Gx(x(s)) x(s) = 0 , G0xx x0 x0 + G0
x x0 = 0 .
Thus x0 = α φ1 + β φ2 , for some α, β ∈ R , and
ψ∗G0xx (α φ1 + β φ2) (α φ1 + β φ2) + ψ∗ G0
x︸ ︷︷ ︸=0
x0 = 0 ,
(ψ∗G0xxφ1φ1)︸ ︷︷ ︸c11
α2 + 2 (ψ∗G0xxφ1φ2)︸ ︷︷ ︸c12
αβ + (ψ∗G0xxφ2φ2)︸ ︷︷ ︸c22
β2 = 0 .
This is the Algebraic Bifurcation equation (ABE).
80
![Page 81: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/81.jpg)
We want solution pairs(α , β) ,
of the ABE, with not both α and β equal to zero.
If the discriminant∆ ≡ c2
12 − c11 c22 ,
satisfies∆ > 0 ,
then the ABE has two real, distinct (i.e., linearly independent) solutions,
(α1, β1) and (α2, β2) ,
which are unique up to scaling.
In this case we have a bifurcation, (or branch point), i.e., two distinct branchespass through x0 .
81
![Page 82: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/82.jpg)
Examples of Bifurcations
First we construct the Algebraic Bifurcation Equation (ABE) for our simplepredator-prey model, in order to illustrate the necessary algebraic manipulations.Thereafter we present a more elaborate application to a nonlinear eigenvalueproblem
82
![Page 83: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/83.jpg)
A Predator-Prey Model
In the 2-species predator-prey model
u′1 = 3u1(1− u1)− u1u2 − λ(1− e−5u1 ) ,
u′2 = −u2 + 3u1u2 .
we have
Gx = (Gu1|Gu2|Gλ) =
3− 6u1 − u2 − 5λe−5u1 −u1 −(1− e−5u1)
3u2 −1 + 3u1 0
,
and
Gxx =
((−6 + 25λe−5u1 ,−1,−5e−5u1) (−1, 0, 0) (−5e−5u1 , 0, 0)(0, 3, 0) (3, 0, 0) (0, 0, 0)
).
83
![Page 84: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/84.jpg)
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
Figure 18: Two branch points (Solutions 2 and 4) in AUTO demo pp2.
84
![Page 85: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/85.jpg)
At Solution 2 in Figure 18 we have u1 = u2 = 0 , λ = 3/5 , so that
x0 = ( 0 , 0 , 3/5 ) ,
G0x =
(0 0 00 −1 0
), G0∗
x =
0 00 −10 0
,
N (G0x) = Span
0
01
,
100
, N (G0∗
x ) = Span
{(10
)},
and
G0xx =
((9,−1,−5) (−1, 0, 0) (−5, 0, 0)(0, 3, 0) (3, 0, 0) (0, 0, 0)
).
G0xx φ1 =
( −5 0 00 0 0
), G0
xx φ2 =
(9 −1 −50 3 0
).
85
![Page 86: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/86.jpg)
Thus
ψ∗ G0xx φ1 φ1 = ψ∗
( −5 0 00 0 0
)φ1 = ψ∗
(00
)= 0 ,
ψ∗ G0xx φ1 φ2 = ψ∗
( −5 0 00 0 0
)φ2 = ψ∗
( −50
)= −5 ,
ψ∗ G0xx φ2 φ2 = ψ∗
(9 −1 −50 3 0
)φ2 = ψ∗
(90
)= 9 .
Therefore the ABE is−10 α β + 9 β2 = 0 .
which has two linearly independent solutions, namely,
(αβ
)=
(10
),(
910
).
86
![Page 87: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/87.jpg)
Thus the (non-normalized) directions of the two bifurcation families at x0 are
x0 = (1) φ1 + (0) φ2 =
001
=
u1
u2
λ
,
and
x0 = (9) φ1 + (10) φ2 =
1009
=
u1
u2
λ
.
NOTE:
◦ The first direction is that of the zero solution family.
◦ The second direction is that of the bifurcating nonzero solution family.
◦ Since λ 6= 0 for the second direction, this is a transcritical bifurcation .
◦ (The case where λ = 0 may correspond to a pitchfork bifurcation .)
87
![Page 88: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/88.jpg)
A Nonlinear Eigenvalue Problem
This example makes extensive use of the method of “Variation of Parameters”for solving linear differential equations. We first recall the use of this method.
Variation of Parameters
Ifv′(t) = A(t) v(t) + f(t) ,
then
v(t) = V (t) [v(0) +∫ t
0V (s)−1f(s) ds] ,
where V (t) is the solution (matrix) of
V ′(t) = A(t) V (t) ,
V (0) = I .
Here V (t) is called the fundamental solution matrix.
88
![Page 89: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/89.jpg)
EXAMPLE:
Apply Variation of Parameters to the equation
v′′ + λ v = f ,
rewritten as v′1 = v2 ,
v′2 = −λ v1 + f ,
or (v′1v′2
)=
(0 1−λ 0
) (v1
v2
)+
(0f
).
Then
V (t) =
(cos√λt sin
√λt/√λ
−√λ sin√λt cos
√λt
).
89
![Page 90: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/90.jpg)
We find that
V −1(t) =
(cos√λt − sin
√λt/√λ√
λ sin√λt cos
√λt
),
V −1(s)
(0f
)=
( − sin√λs f(s)/
√λ
cos√λs f(s)
),
so that
(v1(t)v2(t)
)=
cos√λt sin
√λt/√λ
−√λ sin√λt cos
√λt
v1(0)
v2(0)
+∫ t
0
− sin√λsf(s)/
√λ
cos√λsf(s)
ds
Hence
v1(t) = cos(√λt) v1(0) +
sin(√λt)√λ
v2(0)
− cos√λt
∫ t
0
sin√λs f(s)√λ
ds +sin√λt√λ
∫ t
0cos√λs f(s) ds .
90
![Page 91: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/91.jpg)
For the specific initial value problem
v′′ + λ v = f ,
v(0) = v′(0) = 0 ,
we have
v(t) = v1(t) =sin√λt√λ
∫ t
0cos√λs f(s) ds − cos
√λt√λ
∫ t
0sin√λs f(s) ds .
91
![Page 92: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/92.jpg)
Singular points
Consider the nonlinear boundary value problemu′′ + λ (u + u2) = 0 ,
u(0) = u(1) = 0 ,
which has u(t) ≡ 0 as solution for all λ .
Equivalently, we want the solution
u = u( t ; p , λ) ,
of the initial value problemu′′ + λ (u + u2) = 0 ,
u(0) = 0 , u′(0) = p ,
that satisfiesG( p , λ ) ≡ u( 1 ; p , λ ) = 0 .
92
![Page 93: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/93.jpg)
u′′ + λ (u + u2) = 0 , u(0) = 0 , u′(0) = p
G( p , λ ) ≡ u( 1 ; p , λ ) = 0
We have that
G : R × R → R ,
withG( 0 , λ ) = 0 , for all λ .
Let
up( t ; p , λ ) =du
dp( t ; p , λ ) ,
Gp( p , λ ) = up( 1 ; p , λ ) , etc.
93
![Page 94: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/94.jpg)
u′′ + λ (u + u2) = 0 u(0) = 0 , u′(0) = p
Then up ( = u0p ) satisfies
u′′p + λ (1 + 2u) up = 0 ,
up(0) = 0 , u′p(0) = 1 ,
which, about u ≡ 0 , givesu′′p + λ up = 0 ,
up(0) = 0 , u′p(0) = 1 .
By the variation of parameters formula
up(t; p, λ) =sin√λt√λ
, λ ≥ 0 , (independent of p).
Gp(0, λ) = up(1; p, λ) =sin(√λ)√λ
= 0 , if λ = λk ≡ (kπ)2 .
(We will see that the λk are branch points.)
94
![Page 95: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/95.jpg)
u′′ + λ (u + u2) = 0 u(0) = 0 , u′(0) = p
Next, uλ satisfiesu′′λ + u + u2 + λ (1 + 2u) uλ = 0 ,
uλ(0) = 0 , u′λ(1) = 0 ,
which, about u ≡ 0 , givesu′′λ + λ uλ = 0 ,
uλ(0) = 0 , u′λ(0) = 0 .
from which,
uλ( t ; p , λ ) ≡ 0 , Gλ(0, λ) = uλ(1; 0, λ) = 0 ,
which holds, in particular, at
λ = λk = (kπ)2 .
95
![Page 96: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/96.jpg)
Thus, so far we know that
G(0, λ) = 0 , for all λ ,
and ifλ = λk ≡ (kπ)2 ,
then, withx ≡ (p, λ) ,
we haveG0
x ≡ ( Gp(0, λk) | Gλ(0, λk) ) = (0 | 0) ,
N (G0x) = Span{
(10
),(
01
)} , (2D) ,
N (G0∗
x ) = Span{(1)} , (1D) .
Thus the solutionsp = 0 , λ = λk = (kπ)2 ,
correspond to simple singular points.
96
![Page 97: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/97.jpg)
Construction of the ABE
The ABE is
(ψ∗G0xxφ1φ1) α2 + 2(ψ∗G0
xxφ1φ2) α β + (ψ∗G0xxφ2φ2) β2 = 0 ,
whereψ = 1 ,
G0xx =
((G0
pp | G0pλ)
∣∣∣ (G0λp|G0
λλ)),
with
G0pp = Gpp(0, λk) = upp(1; 0, λk) , etc.
97
![Page 98: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/98.jpg)
If the ABE has 2 independent solutions then the bifurcation directions are
(pλ
)= α1
(10
)+ β1
(01
)=
(α1
β1
),
and (pλ
)= α2
(10
)+ β2
(01
)=
(α2
β2
).
Sincep = 0 ,
is a solution for all λ , one direction is
(pλ
)=
(01
).
98
![Page 99: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/99.jpg)
u′′ + λ (u + u2) = 0 u(0) = 0 , u′(0) = p
Now upp satisfiesu′′pp + 2λ u2
p + λ (1 + 2u) upp = 0 ,
upp(0) = u′pp(0) = 0 ,
which, about
u ≡ 0 , λ = λk , up(t; p, λ) =sin√λt√λ
,
gives u′′pp + λk upp = − 2λk u
2p = − 2 sin2
√λt ,
upp(0) = u′pp(0) = 0 .
that is, u′′pp + (kπ)2 upp = − 2 sin2(kπt) ,
upp(0) = u′pp(0) = 0 .
99
![Page 100: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/100.jpg)
By the variation of parameters formula
upp( t ; p , λ ) =
sin kπt
kπ
∫ t
0cos kπs(−2 sin2 kπs) ds − cos kπt
kπ
∫ t
0sin kπs(−2 sin2 kπs) ds
= − 2 sin kπt
kπ
∫ t
0sin2 kπs cos kπs ds +
2 cos kπt
kπ
∫ t
0sin3 kπs ds ,
and hence
Gpp(0, λk) = upp(1; 0, λk) =2
3(kπ)2[1− (−1)k] =
0 , k even ,
43(kπ)2
, k odd .
100
![Page 101: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/101.jpg)
u′′ + λ (u + u2) = 0 u(0) = 0 , u′(0) = p
Next, upλ satisfies
u′′pλ + (1 + 2u) up + 2λ up uλ + λ (1 + 2u) upλ = 0 ,
upλ(0) = u′pλ(0) = 0 .
which, about
u ≡ 0 , λk = (kπ)2 , up(t; p, λ) =sin√λt√λ
, uλ = 0 ,
gives u′′pλ + k2 π2 upλ = − up = − sin kπt
kπ,
upλ(0) = u′pλ(0) = 0 .
101
![Page 102: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/102.jpg)
Using the variation of parameters formula
upλ(t; p, λ) =
sin kπt
kπ
∫ t
0cos kπs(
− sin kπs
kπ) ds − cos kπt
kπ
∫ t
0sin kπs(
− sin kπs
kπ) ds
=sin kπt
(kπ)2
∫ t
0sin kπs cos kπs ds +
cos kπt
(kπ)2
∫ t
0sin2 kπs ds ,
and hence
Gpλ(0, λk) = upλ(1; 0, λk) =1
2(kπ)2(−1)k =
1
2(kπ)2, k even ,
−12(kπ)2
, k odd .
102
![Page 103: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/103.jpg)
u′′ + λ (u + u2) = 0 u(0) = 0 , u′(0) = p
Finally uλλ satisfiesu′′λλ + (1 + 2u) uλ + (1 + 2u) uλ + 2λ u2
λ + λ (1 + 2u) uλλ = 0 ,
uλλ(0) = u′λλ(0) = 0 ,
which, about
u = 0 , uλ = 0 , λ = λk = k2π2 ,
gives u′′λλ + k2 π2 uλλ = 0 ,
uλλ(0) = u′λλ(0) = 0 ,
so thatuλλ(t; p, λ) ≡ 0 , Gλλ(0, λk) = uλλ(1; 0, λk) = 0 .
103
![Page 104: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/104.jpg)
Thus we have found that
G0xx =
((G0
pp|G0pλ)∣∣∣(G0
λp|G0λλ)
)=
((0| 1
2(kπ)2)∣∣∣ ( 1
2(kπ)2|0))
, k even,
(( 4
3(kπ)2| −12(kπ)2
)∣∣∣ ( −1
2(kπ)2|0)), k odd.
The coefficients of the ABE are
ψ∗G0xxφ1φ1 =
0 , k even,
43(kπ)2
, k odd,ψ∗G0
xxφ1φ2 =
1
2(kπ)2, k even,
−12(kπ)2
, k odd,
ψ∗G0xxφ2φ2 =
0 , k even,
0 , k odd.
104
![Page 105: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/105.jpg)
Thus the ABE is
α β = 0 , k even, Roots : (α, β) = (1, 0) , (0, 1) ,
43α2 − 1
2α β = 0 , k odd, Roots : (α, β) = (0, 1) , (3, 8) .
The directions of the bifurcating families are:
x =(pλ
)= α φ1 + β φ2 = α
(10
)+ β
(01
),
where
(pλ
)=
(01
),(
10
), k even, (“pitch-fork bifurcation”) ,
(01
),(
38
), k odd, (“transcritical bifurcation”) .
105
![Page 106: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/106.jpg)
0 1 2 3 4 5
λ (scaled)
−10
−5
0
5
10
u′ a
tx=
0
1
23
4
5
Figure 19: The bifurcating families of the nonlinear eigenvalue problem.
106
![Page 107: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/107.jpg)
EXERCISE.
◦ Check the above calculations (!).
◦ Use the AUTO demo nev to compute some bifurcating families.
◦ Do the numerical results support the analytical results?
◦ Also carry out the above analysis and AUTO computations for
u′′ + λ (u + u3) = 0 ,
u(0) = u(1) = 0 .
107
![Page 108: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/108.jpg)
Branch Switching
◦ Along a solution family we may find branch points .
◦ We give two methods for switching branches .
◦ We also give a method to detect branch points.
108
![Page 109: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/109.jpg)
Computing the bifurcation direction
LetG : Rn+1 → Rn .
Suppose that we have a solution family x(s) of
G(x) = 0 ,
and thatx0 ≡ x(s0) ,
is a simple singular point , i.e., the n by n+ 1 matrix
G0x ≡ Gx(x0) has rank n− 1 ,
withN (G0
x) = Span{φ1,φ2} , N (G0∗
x ) = Span{ψ} .
109
![Page 110: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/110.jpg)
NOTATION.
◦ x0 = α1 φ1 + β1 φ2 denotes the direction of the ”given” family.
◦ x′0 = α2 φ1 + β2 φ2 denotes the direction of the bifurcating family.
The two coefficient vectors
(α1 , β2) and (α2 , β2) ,
correspond to two linearly independent solutions of the ABE
c11 α2 + 2 c12 α β + c22 β
2 = 0 .
Assume that the discriminant is positive:
∆ ≡ c212 − c11 c22 > 0 .
110
![Page 111: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/111.jpg)
Since along the ”given” family we have
G0x x0 = 0 ,
we can takeφ1 = x0 , ( x0 = α1 φ1 + β1 φ2 ) .
Thus(α1, β1) = (1, 0)
is a solution of the ABE
c11 α2 + 2 c12 α β + c22 β
2 = 0 .
Thusc11 = 0 , and (since c2
12 − c11c22 > 0 ) c12 6= 0 .
The second solution then satisfies
2 c12 α + c22 β = 0 ,
from which,
(α2, β2) = = (c22 ,−2c12) , (unique, up to scaling) .
111
![Page 112: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/112.jpg)
To evaluate c12 and c22 , we need the null vectors φ1 and φ2 of G0x .
φ1 : We already have chosen φ1 = x0 .
φ2 : Choose φ2 ⊥ φ1 . Then φ2 is a null vector of
F0x =
(G0
x
x∗0
)i .e., F 0
x φ2 =(
G0x
x∗0
)φ2 = 0 .
Note that F0x is the Jacobian of the pseudo-arclength system at x0 !
The null space of F0x is indeed one-dimensional. (Check!)
ψ : is the left null vector: (G0x)∗ ψ = 0 , so that also
(F0x)∗
(ψ0
)= ((G0
x)∗ | x0)(ψ0
)= 0 ,
i.e., ψ is also the left null vector of F0x .
112
![Page 113: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/113.jpg)
NOTE:
◦ Left and right null vectors of a matrix can be computed at little cost , oncethe matrix has been LU decomposed.
◦ After determining the coefficients α2 and β2 , scale the direction vector
x′0 ≡ α2 φ1 + β2 φ2 ,
of the bifurcating family so that
‖ x′0 ‖ = 1 .
113
![Page 114: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/114.jpg)
Switching branches
The first solution x1 on the bifurcating family can be computed from :
G(x1) = 0 ,
(x1 − x0)∗x′0 − ∆s = 0 ,(3)
wherex′0 is the direction of the bifurcating branch.
As initial approximation in Newton’s method take
x(0)1 = x0 + ∆s x′0 .
For a graphical interpretation see Figure 20.
NOTE: Computing x′0 requires evaluation of G0xx .
114
![Page 115: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/115.jpg)
1x
0x
∆ s
��������
0x��
��
x 0
Figure 20: Switching branches using the correct bifurcation direction.
115
![Page 116: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/116.jpg)
Simplified branch switching
Instead of Eqn. (3) for the first solution on the bifurcating branch, use :
G(x1) = 0 ,
(x1 − x0)∗φ2 − ∆s = 0 .
where φ2 is the second null vector of G0x , with, as before,
φ2 ⊥ φ1, (φ1 = x0) ,i.e., (
G0x
x∗0
)φ2 = 0 , ‖ φ2 ‖ = 1 .
As initial approximation, now use
x(0)1 = x0 + ∆s φ2 .
For a graphical interpretation see Figure 21.
116
![Page 117: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/117.jpg)
0x
∆ s
��������
x )0 1φ(=
1x
����
����
φ2
Figure 21: Switching branches using the orthogonal direction.
117
![Page 118: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/118.jpg)
NOTE:
◦ The simplified branch switching method may fail in some situations.
◦ The advantage is that it does not need second derivatives .
◦ The orthogonal direction φ2 can be computed at little cost .
◦ In fact, φ2 is the null vector of the pseudo-arclength Jacobian(G0
x
x∗0
)at the branch point.
118
![Page 119: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/119.jpg)
Detection of Branch Points
LetG : Rn+1 → Rn .
Recall that a solution
x0 ≡ x(s0) of G(x) = 0 ,
is a simple singular point if
G0x ≡ Gx(x0) has rank n− 1 .
Suppose that we have a solution family x(s) of
G(x) = 0 ,and that
x0 = x(0) ,
is a simple singular point.
Let x0 be the unit tangent to x(s) at x0 .
Assume that x(s) is parametrized by its projection onto x0 . (See Figure 22.)
119
![Page 120: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/120.jpg)
0x
s
x(s)
Figure 22: Parametrization of a solution family near a branch point.
120
![Page 121: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/121.jpg)
Consider the pseudo-arclength system
F( x ; s ) ≡ G(x)
(x− x0)∗x0 − s
. (4)
Then
Fx( x ; s ) = Fx(x) =(
Gx(x)x∗0
),
and
F0x ≡ Fx(x0) =
(G0
x
x∗0
).
NOTE: Fx does not explicitly depend on s .
121
![Page 122: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/122.jpg)
Takeφ1 = x0 ,
as the first null vector of G0x .
Thus
F0x =
(G0
x
φ∗1
).
Choose the second null vector φ2 of G0x such that
φ∗2 φ1 = 0 .
Then
F0x φ2 =
(G0
x
φ∗1
)φ2 = 0 ,
so that φ2 is also a null vector of F0x , while φ1 is not.
In fact, F0x has a one-dimensional nullspace .
122
![Page 123: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/123.jpg)
The null vector of
F0∗x =
(G0
x
φ∗1
)∗= ( (G0
x)∗ | φ1 ) ,
is given by
Ψ ≡(ψ0
),
where ψ is the null vector of (G0x)∗ .
NOTE: SinceG0
x has n rows and n+ 1 columns ,
andN (G0
x) = Span{φ1, φ2} is assumed two-dimensional ,
it follows that(G0
x)∗ has n+ 1 rows and n columns ,
andN (G0∗
x ) = Span{ψ} is one-dimensional .
123
![Page 124: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/124.jpg)
THEOREM. Letx0 = x(0) ,
be a simple singular point on a smooth solution family x(s) of
G(x) = 0 .
Let F(x; s) be as above, i.e., F(x; s) ≡(
G(x)(x− x0)∗x0 − s
).
Assume that
◦ the discriminant ∆ of the ABE is positive ,
◦ 0 is an algebraically simple eigenvalue of
F0x ≡
(G0
x
x∗0
).
Then
det Fx(x(s)) = det(
Gx(x(s))x∗0
),
changes sign at x0 .
124
![Page 125: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/125.jpg)
PROOF.
Consider the parametrized eigenvalue problem
Fx( x(s) ) φ(s) = κ(s) φ(s) ,
where κ(s) and φ(s) are smooth near s = 0 , with
κ(0) = 0 and φ(0) = φ2 ,
i.e., the eigen pair( κ(s) , φ(s) ) ,
is the continuation of (0,φ2) .
(This can be done because 0 is an algebraically simple eigenvalue.)
125
![Page 126: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/126.jpg)
Differentiating
Fx( x(s) ) φ(s) = κ(s) φ(s) ,
gives
Fxx(x(s)) x(s) φ(s) + Fx(x(s)) φ(s) = κ(s) φ(s) + κ(s) φ(s).
Evaluating at s = 0 , using
κ(0) = 0 ,
andx0 = φ1 and φ(0) = φ2 ,
gives
F0xx φ1 φ2 + F0
x φ(0) = κ0 φ2 .
126
![Page 127: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/127.jpg)
F0xx φ1 φ2 + F0
x φ(0) = κ0 φ2 .
Multiplying this on the left by Ψ∗ we find
κ0 =Ψ∗F0
xxφ1φ2
Ψ∗φ2
=(ψ∗, 0)
(G0
xx
0
)φ1φ2
Ψ∗φ2
=ψ∗G0
xxφ1φ2
Ψ∗φ2
.
The left and right null vectors ( Ψ and φ2 ) of F0x cannot be orthogonal
(because the eigenvalue 0 is assumed to be algebraically simple .
Thus
Ψ∗φ2 6= 0 .
127
![Page 128: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/128.jpg)
Note that
ψ∗G0xxφ1φ2 = c12 ,
is a coefficient of the ABE.
By assumption, the discriminant satisfies
∆ 6= 0 .
As before this implies that
c12 6= 0 ,
and hence
κ0 6= 0 •
128
![Page 129: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/129.jpg)
NOTE:
◦ The Theorem implies that det(
Gx(x(s))x∗0
)changes sign .
◦ Note that x0 is kept fixed.
◦ We haven’t proved that det Fx(x(s)) = det(
Gx(x(s))x(s)∗
)changes sign.
◦ The latter follows from a similar, but more elaborate argument.
◦ Detection of simple singular points is based upon this fact.
◦ During continuation we monitor the determinant of the matrix Fx .
◦ If a sign change is detected then an iterative method can be used toaccurately locate the singular point.
◦ For large systems a scaled determinant avoids overflow .
129
![Page 130: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/130.jpg)
The following theorem states that there must be a bifurcation at x0 .
(This result can be proven by degree theory .)
THEOREM.
Let x(s) be a smooth solution family of
F( x ; s ) = 0 ,
whereF : Rn+1 × R → Rn+1 is C1 ,
and assume that
det Fx(x(s); s) changes sign at s = 0 .
Then x(0) is a bifurcation point , i.e., every open neigborhood of x0 containsa solution of F(x; s) = 0 that does not lie on x(s) .
130
![Page 131: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/131.jpg)
Boundary Value Problems
131
![Page 132: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/132.jpg)
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
0q( u(s) , µ , λ ) ds = 0 , q(·) ∈ Rnq .
132
![Page 133: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/133.jpg)
This boundary value problem (BVP) is of the form
F( X ) = 0 ,
whereX = ( u , µ , λ ) ,
to which we add the pseudo-arclength equation
< X−X0 , X0 > − ∆s = 0 ,
where X0 represents the preceding solution on the branch.
In detail, the pseudo-arclength equation is
∫ 1
0(u(t)− u0(t))∗u0(t) dt + (µ− µ0)µ0
+ (λ− λ0)λ0 − ∆s = 0 .
133
![Page 134: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/134.jpg)
◦ We want to solve BVP for u(·) and µ .
◦ We can think of λ as the continuation parameter .
◦ (In pseudo-arclength continuation, we don’t distinguish µ and λ .)
◦ 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 .
134
![Page 135: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/135.jpg)
Discretization
Here we discuss the method of “orthogonal collocation with piecewise polyno-mials”, for solving boundary value problems. This method is very accurate, andallows adaptive mesh-selection.
135
![Page 136: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/136.jpg)
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 .
136
![Page 137: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/137.jpg)
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 points) .
See Figure 23 for a graphical interpretation.
137
![Page 138: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/138.jpg)
�� �� ��
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
Figure 23: The mesh {0 = t0 < t1 < · · · < tN = 1} . Collocation points and“extended-mesh points” are shown for the case m = 3, in the jth mesh interval.Also shown are two of the four local Lagrange basis polynomials.
138
![Page 139: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/139.jpg)
Since each local polynomial is determined by
(m+ 1) n ,
coefficients, the total number of degrees of freedom (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 ) .
139
![Page 140: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/140.jpg)
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 µ are also superconvergent.
140
![Page 141: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/141.jpg)
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.
141
![Page 142: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/142.jpg)
The collocation equations are
p′
j(zj,i) = f( pj(zj,i) , µ , λ ) , i = 1, · · · ,m, j = 1, · · · , N .
The discrete 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 .
142
![Page 143: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/143.jpg)
The pseudo-arclength 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 on the solution branch, and
( u0 , µ0 , λ0 ) ,
is the normalized direction of the branch at the previous solution.
The discretized pseudo-arclength equation is
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 .
143
![Page 144: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/144.jpg)
Numerical Linear Algebra
The complete discretization consists of
m n N + nb + nq + 1 ,
nonlinear equations, in the unknowns
{uj− im} ∈ RmnN+n , µ ∈ Rnµ , λ ∈ R .
These equations can be solved by a Newton-Chord iteration .
144
![Page 145: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/145.jpg)
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 pseudo-arclength 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 .
145
![Page 146: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/146.jpg)
u0 u 13
u 23
u1 u2 u3 uN µ λ
• • • • • • • • • •• • • • • • • • • •• • • • • • • • • •• • • • • • • • • •• • • • • • • • • •• • • • • • • • • •
• • • • • • • • • •• • • • • • • • • •• • • • • • • • • •• • • • • • • • • •• • • • • • • • • •• • • • • • • • • •
• • • • • • • • • •• • • • • • • • • •• • • • • • • • • •• • • • • • • • • •• • • • • • • • • •• • • • • • • • • •
• • • • • • • • • •• • • • • • • • • •• • • • • • • • • •• • • • • • • • • •• • • • • • • • • •• • • • • • • • • •
• • • • • •• • • • • •• • • • • • • • • • • • • • • • • • • • • • • • • • • •• • • • • • • • • • • • • • • • • • • • • • • • • • • •
Figure 24: The structure of the Jacobian
146
![Page 147: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/147.jpg)
u0 u 13
u 23
u1 u2 u3 uN µ λ
• • • • • • • • • •• • ◦ • • • • • • •• • ◦ ◦ • • • • • •• • ◦ ◦ ◦ • • • • •• • ◦ ◦ ◦ ◦ • • • •• • ◦ ◦ ◦ ◦ • • • •
• • • • • • • • • •• • ◦ • • • • • • •• • ◦ ◦ • • • • • •• • ◦ ◦ ◦ • • • • •• • ◦ ◦ ◦ ◦ • • • •• • ◦ ◦ ◦ ◦ • • • •
• • • • • • • • • •• • ◦ • • • • • • •• • ◦ ◦ • • • • • •• • ◦ ◦ ◦ • • • • •• • ◦ ◦ ◦ ◦ • • • •• • ◦ ◦ ◦ ◦ • • • •
• • • • • • • • • •• • ◦ • • • • • • •• • ◦ ◦ • • • • • •• • ◦ ◦ ◦ • • • • •• • ◦ ◦ ◦ ◦ • • • •• • ◦ ◦ ◦ ◦ • • • •
• • • • • •• • • • • •• • ◦ ◦ ◦ ◦ • • ◦ ◦ ◦ ◦ • • ◦ ◦ ◦ ◦ • • ◦ ◦ ◦ ◦ • • • •• • ◦ ◦ ◦ ◦ • • ◦ ◦ ◦ ◦ • • ◦ ◦ ◦ ◦ • • ◦ ◦ ◦ ◦ • • • •
Figure 25: The system after condensation of parameters.
147
![Page 148: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/148.jpg)
u0 u 13
u 23
u1 u2 u3 uN µ λ
• • • • • • • • • •• • ◦ • • • • • • •• • ◦ ◦ • • • • • •• • ◦ ◦ ◦ • • • • •? ? ◦ ◦ ◦ ◦ ? ? ? ?? ? ◦ ◦ ◦ ◦ ? ? ? ?
• • • • • • • • • •• • ◦ • • • • • • •• • ◦ ◦ • • • • • •• • ◦ ◦ ◦ • • • • •? ? ◦ ◦ ◦ ◦ ? ? ? ?? ? ◦ ◦ ◦ ◦ ? ? ? ?
• • • • • • • • • •• • ◦ • • • • • • •• • ◦ ◦ • • • • • •• • ◦ ◦ ◦ • • • • •? ? ◦ ◦ ◦ ◦ ? ? ? ?? ? ◦ ◦ ◦ ◦ ? ? ? ?
• • • • • • • • • •• • ◦ • • • • • • •• • ◦ ◦ • • • • • •• • ◦ ◦ ◦ • • • • •? ? ◦ ◦ ◦ ◦ ? ? ? ?? ? ◦ ◦ ◦ ◦ ? ? ? ?
? ? ? ? ? ?? ? ? ? ? ?? ? ◦ ◦ ◦ ◦ ? ? ◦ ◦ ◦ ◦ ? ? ◦ ◦ ◦ ◦ ? ? ◦ ◦ ◦ ◦ ? ? ? ?? ? ◦ ◦ ◦ ◦ ? ? ◦ ◦ ◦ ◦ ? ? ◦ ◦ ◦ ◦ ? ? ◦ ◦ ◦ ◦ ? ? ? ?
Figure 26: The preceding matrix, showing the decoupled ? sub-system.
148
![Page 149: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/149.jpg)
u0 u 13
u 23
u1 u2 u3 uN µ λ
• • • • • • • • • •• • ◦ • • • • • • •• • ◦ ◦ • • • • • •• • ◦ ◦ ◦ • • • • •? ? ◦ ◦ ◦ ◦ ? ? · · ? ?? ? ◦ ◦ ◦ ◦ ◦ ? · · ? ?
• • • • • • • • • •• • ◦ • • • • • • •• • ◦ ◦ • • • • • •• • ◦ ◦ ◦ • • • • •
? ? ◦ ◦ ◦ ◦ ◦ ◦ ? ? ? ?? ? ◦ ◦ ◦ ◦ ◦ ◦ ? ? ? ?
• • • • • • • • • •• • ◦ • • • • • • •• • ◦ ◦ • • • • • •• • ◦ ◦ ◦ • • • • •? ? ◦ ◦ ◦ ◦ ? ? · · ? ?? ? ◦ ◦ ◦ ◦ ◦ ? · · ? ?
• • • • • • • • • •• • ◦ • • • • • • •• • ◦ ◦ • • • • • •• • ◦ ◦ ◦ • • • • •
? ? ◦ ◦ ◦ ◦ ◦ ◦ ? ? ? ?? ? ◦ ◦ ◦ ◦ ◦ ◦ ? ? ? ?
? ? ? ? ? ?? ? ? ? ? ?? ? ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ? ? ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ? ? ? ?? ? ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ? ? ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ? ? ? ?
Figure 27: Stage 1 of the nested dissection to solve the decoupled ? system.
149
![Page 150: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/150.jpg)
u0 u 13
u 23
u1 u2 u3 uN µ λ
• • • • • • • • • •• • ◦ • • • • • • •• • ◦ ◦ • • • • • •• • ◦ ◦ ◦ • • • • •? ? ◦ ◦ ◦ ◦ ? ? · · ? ?? ? ◦ ◦ ◦ ◦ ◦ ? · · ? ?
• • • • • • • •• • ◦ • • • • • • •• • ◦ ◦ • • • • • •• • ◦ ◦ ◦ • • • • •
? ? ◦ ◦ ◦ ◦ ◦ ◦ ? ? · · ? ?? ? ◦ ◦ ◦ ◦ ◦ ◦ ◦ ? · · ? ?
• • • • • • • • • •• • ◦ • • • • • • •• • ◦ ◦ • • • • • •• • ◦ ◦ ◦ • • • • •? ? ◦ ◦ ◦ ◦ ? ? · · ? ?? ? ◦ ◦ ◦ ◦ ◦ ? · · ? ?
• • • • • • • • • •• • ◦ • • • • • • •• • ◦ ◦ • • • • • •• • ◦ ◦ ◦ • • • • •
? ? ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ? ? ? ?? ? ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ? ? ? ?? ? ? ? ? ?? ? ? ? ? ?? ? ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ? ? ? ?? ? ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ? ? ? ?
Figure 28: Stage 2 of the nested dissection to solve the decoupled ? system.
150
![Page 151: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/151.jpg)
u0 u 13
u 23
u1 u2 u3 uN µ λ
• • • • • • • • • •• • ◦ • • • • • • •• • ◦ ◦ • • • • • •• • ◦ ◦ ◦ • • • • •? ? ◦ ◦ ◦ ◦ ? ? · · ? ?? ? ◦ ◦ ◦ ◦ ◦ ? · · ? ?
• • • • • • • • • •• • ◦ • • • • • • •• • ◦ ◦ • • • • • •• • ◦ ◦ ◦ • • • • •
? ? ◦ ◦ ◦ ◦ ◦ ◦ ? ? · · ? ?? ? ◦ ◦ ◦ ◦ ◦ ◦ ◦ ? · · ? ?
• • • • • • • • • •• • ◦ • • • • • • •• • ◦ ◦ • • • • • •• • ◦ ◦ ◦ • • • • •? ? ◦ ◦ ◦ ◦ ? ? · · ? ?? ? ◦ ◦ ◦ ◦ ◦ ? · · ? ?
• • • • • • • • • •• • ◦ • • • • • • •• • ◦ ◦ • • • • • •• • ◦ ◦ ◦ • • • • •
+ + ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ + + + ++ + ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ + + + ++ + + + + ++ + + + + ++ + ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ + + + ++ + ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ + + + +
Figure 29: The preceding matrix showing the final decoupled + sub-system.
151
![Page 152: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/152.jpg)
u0 u 13
u 23
u1 u2 u3 uN µ λ
• • • • • • • • • •• • ◦ • • • • • • •• • ◦ ◦ • • • • • •• • ◦ ◦ ◦ • • • • •? ? ◦ ◦ ◦ ◦ ? ? · · ? ?? ? ◦ ◦ ◦ ◦ ◦ ? · · ? ?
• • • • • • • • • •• • ◦ • • • • • • •• • ◦ ◦ • • • • • •• • ◦ ◦ ◦ • • • • •
? ? ◦ ◦ ◦ ◦ ◦ ◦ ? ? · · ? ?? ? ◦ ◦ ◦ ◦ ◦ ◦ ◦ ? · · ? ?
• • • • • • • • • •• • ◦ • • • • • • •• • ◦ ◦ • • • • • •• • ◦ ◦ ◦ • • • • •? ? ◦ ◦ ◦ ◦ ? ? · · ? ?? ? ◦ ◦ ◦ ◦ ◦ ? · · ? ?
• • • • • • • • • •• • ◦ • • • • • • •• • ◦ ◦ • • • • • •• • ◦ ◦ ◦ • • • • •
A A ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ B B + +A A ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ B B + ++ + + + + ++ + + + + ++ + ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ + + + ++ + ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ + + + +
Figure 30: The Floquet Multipliers are the eigenvalues of −B−1A .
152
![Page 153: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/153.jpg)
Accuracy Test
The Table shows the location of the fold in the Gelfand-Bratu problem for
◦ 4 Gauss collocation points per mesh interval
◦ N mesh intervals
N Fold location2 3.51378975504 3.51383086018 3.5138307211
16 3.513830719132 3.5138307191
153
![Page 154: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/154.jpg)
A Singularly-Perturbed BVP
ε u′′(x) = u(x) u′(x) (u(x)2 − 1) + u(x) .
with boundary conditions
u(0) =3
2, u(1) = γ .
Computational formulation
u′1 = u2 ,
u′2 =λ
ε( u1 u2 (u2
1 − 1) + u1 ) ,
with boundary conditions
u1(0) = 3/2 , u1(1) = γ .
When λ = 0 an exact solution is
u1(x) =3
2+ (γ − 3
2) x , u2(x) = γ − 3
2.
154
![Page 155: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/155.jpg)
COMPUTATIONAL STEPS:
◦ λ is a homotopy parameter to locate a starting solution.
◦ In the first run λ varies from 0 to 1 .
◦ In the second run ε is decreased by continuation.
◦ In the third run ε = 10−3 , and the solution is continued in γ .
◦ This third run takes many continuation steps if ε is very small.
155
![Page 156: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/156.jpg)
−2.0 −1.5 −1.0 −0.5 0.0 0.5 1.0 1.5γ
13
14
15
16
17
18
19
20
21
22
norm
u
Figure 31: Bifurcation diagram of the singularly-perturbed BVP.
156
![Page 157: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/157.jpg)
0.0 0.2 0.4 0.6 0.8 1.0x
−2.5
−2.0
−1.5
−1.0
−0.5
0.0
0.5
1.0
1.5
u
Figure 32: Some solutions along the solution family.
157
![Page 158: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/158.jpg)
Hopf Bifurcation and PeriodicSolutions
We first introduce the concept of Hopf bifurcation for a “linear” problem. Thenwe state the Hopf Bifurcation Theorem (without proof), and we give examplesof periodic solutions emanating from Hopf bifurcation points.
158
![Page 159: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/159.jpg)
A Linear Example
The linear problem : u′1 = λu1 − u2 ,
u′2 = u1 ,(5)
which can also be written as(u′1u′2
)=
(λ −11 0
)(u1
u2
),
is of the form
u′(t) = A(λ) u(t) ≡ G(u, λ) ,
with stationary solutions,
u1 = u2 = 0 , for all λ .
159
![Page 160: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/160.jpg)
The eigenvalues µ of the Jacobian matrix
Gu(u, λ) = A(λ) =
(λ −11 0
)= Gu(0, λ) ,
satisfydet( A(λ) − µ I ) = µ2 − λ µ + 1 = 0 ,
from which
µ1,2 =λ±√λ2 − 4
2.
Consider the initial value problem
u′(t) = A(λ) u(t) ≡ G(u, λ) ,
with
u(0) =(u1(0)u2(0)
)=
(p1
p2
)≡ p .
160
![Page 161: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/161.jpg)
Then
u(t) = etA(λ) p = V (λ) etΛ(λ) V −1(λ) p
= V (λ)
(etµ1(λ) 0
0 etµ2(λ)
)V −1(λ) p ,
(6)
where
A(λ) V (λ) = V (λ) Λ(λ) , A(λ) = V (λ) Λ(λ) V −1(λ) ,
and
Λ(λ) =
(µ1(λ) 0
0 µ2(λ)
), V (λ) =
(v11(λ) v12(λ)v21(λ) v22(λ)
).
161
![Page 162: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/162.jpg)
Assume that−2 < λ < 2 ,
and recall that
u(t) = V (λ)
(etµ1(λ) 0
0 etµ2(λ)
)V −1(λ) p ,
and
µ1,2 =λ±√λ2 − 4
2.
Thus we see that
u(t) → 0 if λ < 0 ,
and
u(t) → ∞ if λ > 0 ,
i.e., the zero solution is stable if λ is negative, and unstable if λ is positive.
162
![Page 163: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/163.jpg)
However, if λ = 0 , then
A0 ≡ A(0) =
(0 −11 0
),
andµ1 = i, µ2 = − i ,
V0 ≡ V (0) =
(1 −i−i 1
),
V −10 =
1
2
(1 ii 1
),
so that
u(t) = V0 etΛ V −1
0 p =1
2
(1 −i−i 1
)(eit 00 e−it
)(1 ii 1
) (p1
p2
)
=1
2
(eit + e−it i(eit − e−it)−i(eit − e−it) eit + e−it
)(p1
p2
).
163
![Page 164: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/164.jpg)
Thus, if λ = 0 , then
(u1(t)u2(t)
)=
(cos(t) − sin(t)sin(t) cos(t)
)(p1
p2
)=
(p1 cos(t) − p2 sin(t)p1 sin(t) + p2 cos(t)
),
and we see that
◦ This solution is periodic, with period 2π , for any p1 , p2 .
◦ u1(t)2 + u2(t)2 = p21 + p2
2 , (The orbits are circles.) ,
◦ We can fix the phase by setting, for example, p2 = 0 .
◦ (Then u2(0) = 0 .)
◦ This leaves a one-parameter family of periodic solutions.
(See Figures 33 and 34.).
◦ For nonlinear problems the family is generally not “vertical”.
164
![Page 165: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/165.jpg)
EXERCISE. (Demo lhb .)
Use AUTO to compute the zero stationary solution family, a Hopf bifurcation,and the emanating family of periodic solutions, of the “linear” Hopf bifurcationproblem, i.e., of
u′1 = λu1 − u2 ,
u′2 = u1 .(7)
NOTE:
◦ The family of periodic solutions is “vertical”, i.e., λ = 0 along it.
◦ This is not “typical” (not “generic”) for Hopf bifurcation.
◦ For the numerically computed family, λ = 0 up to numerical accuracy.
◦ The period is constant, namely, 2π , along the family.
◦ This is also not generic for periodic solutions from a Hopf bifurcation.
◦ The numerical computation of periodic solutions will be considered later.
165
![Page 166: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/166.jpg)
−1.0 −0.5 0.0 0.5 1.0
λ
−0.5
0.0
0.5
1.0
1.5
2.0
2.5
norm
Figure 33: Bifurcation diagram of the “linear” Hopf bifurcation problem.
166
![Page 167: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/167.jpg)
−3 −2 −1 0 1 2 3u1
−3
−2
−1
0
1
2
3
u2
Figure 34: A phase plot of some periodic solutions.
167
![Page 168: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/168.jpg)
The Hopf Bifurcation Theorem
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, i.e., a family of periodic solutions bifurcatesfrom the stationary solution at (u0, λ0) . ◦
NOTE: The assumptions also imply that f0u is nonsingular, so that the station-
ary solution family can indeed be parametrized locally using λ .
168
![Page 169: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/169.jpg)
EXERCISE. (AUTO Demo vhb .)
Use AUTO to compute the zero stationary solution family, a Hopf bifurcation,and the emanating family of periodic solutions for the equation
u′1 = λu1 − u2 ,
u′2 = u1 (1− u1) .(8)
NOTE:
◦ u(t) ≡ 0 is a stationary solution for all λ .
◦ u(t) ≡(
1λ
)is another stationary solution .
169
![Page 170: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/170.jpg)
NOTE:
The Jacobian along the solution family u(t) ≡ 0 is
(λ −11 0
),
with eigenvaluesλ ± √λ2 − 4
2.
◦ The eigenvalues are complex for λ ∈ (−2, 2) .
◦ The eigenvalues cross the imaginary axis when λ passes through zero.
◦ Thus there is a Hopf bifurcation along u ≡ 0 at λ = 0 .
◦ A family of periodic solutions bifurcates from u = 0 at λ = 0 .
170
![Page 171: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/171.jpg)
−1.0 −0.5 0.0 0.5 1.0
λ
−0.2
0.0
0.2
0.4
0.6
0.8
1.0
1.2
norm
Figure 35: Bifurcation diagram for Equation (8).
171
![Page 172: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/172.jpg)
−0.4 −0.2 0.0 0.2 0.4 0.6 0.8 1.0u1
−0.6
−0.4
−0.2
0.0
0.2
0.4
0.6
u2
Figure 36: A phase plot of some periodic solutions to Equation (8).
172
![Page 173: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/173.jpg)
0.0 0.2 0.4 0.6 0.8 1.0
Scaled Time
−0.4
−0.2
0.0
0.2
0.4
0.6
0.8
1.0
u1
Figure 37: u1 as a function of the scaled time variable t for Equation (8).
173
![Page 174: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/174.jpg)
0.0 0.2 0.4 0.6 0.8 1.0
Scaled Time
−0.6
−0.4
−0.2
0.0
0.2
0.4
0.6
u2
Figure 38: u2 as a function of the scaled time variable t for Equation (8).
174
![Page 175: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/175.jpg)
NOTE:
◦ The family of periodic solutions is also “vertical” (non-generic).
◦ The period changes along this family; in fact, the period tends to infinity.
◦ The terminating infinite period orbit is an example of a homoclinic orbit.
◦ This homoclinic orbit contains the stationary point (u1, u2) = (1, 0) .
◦ In the solution diagrams, showing u1 and u2 versus time t , note howthe “peak” in the solution remains in the same location.
◦ This is a result of the numerical “phase-condition”, to be discussed later.
175
![Page 176: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/176.jpg)
EXERCISE. (Demo het .)
Use AUTO to compute the zero stationary solution family, a Hopf bifurcation,and the emanating family of periodic solutions, of the equation
u′1 = λu1 − u2 ,
u′2 = u1 (1− u21) .
(9)
NOTE:
◦ u(t) ≡ 0 is a stationary solution for all λ .
◦ There is a Hopf bifurcation along u ≡ 0 at λ = 0 .
◦ u(t) ≡(
1λ
),
(−1λ
)are two more stationary solutions .
176
![Page 177: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/177.jpg)
−1.0 −0.5 0.0 0.5 1.0
λ
−0.2
0.0
0.2
0.4
0.6
0.8
1.0
1.2
norm
Figure 39: Bifurcation diagram for Equation (9).
177
![Page 178: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/178.jpg)
−1.0 −0.5 0.0 0.5 1.0u1
−0.6
−0.4
−0.2
0.0
0.2
0.4
0.6
u2
Figure 40: A phase plot of some periodic solutions to Equation (9).
178
![Page 179: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/179.jpg)
0.0 0.2 0.4 0.6 0.8 1.0
Scaled Time
−1.0
−0.5
0.0
0.5
1.0
u1
Figure 41: u1 as a function of the scaled time variable t for Equation (9).
179
![Page 180: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/180.jpg)
0.0 0.2 0.4 0.6 0.8 1.0
Scaled Time
−0.6
−0.4
−0.2
0.0
0.2
0.4
0.6
u2
Figure 42: u2 as a function of the scaled time variable t for Equation (9).
180
![Page 181: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/181.jpg)
NOTE:
◦ This family of periodic solutions is also “vertical” (non-generic).
◦ The period along this family also tends to infinity.
◦ The terminating infinite period orbit is an example of a heteroclinic cycle.
◦ This heteroclinic cycle is made up of two heteroclinic orbits.
◦ The heteroclinic orbits contains the stationary points
(u1, u2) = (1, 0) and (u1, u2) = (−1, 0) .
181
![Page 182: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/182.jpg)
EXERCISE. (AUTO demo pp3 .)
Compute the families of periodic solutions that bifurcate from the four Hopfbifurcation points in the following system; taking p4 = 4 .
u′1(t) = u1 (1− u1) − p4 u1 u2 ,
u′2(t) = − 14u2 + p4 u1 u2 − 3 u2 u3 − p1 (1− e−5u2) ,
u′3(t) = − 12u3 + 3 u2 u3 .
This is a simple predator-prey model where, say,
u1 = plankton, u2 = fish, u3 = sharks, λ ≡ p1 = fishing quota.
The factor
(1 − e−5u2) ,
models that the quota cannot be met if the fish population is small.
182
![Page 183: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/183.jpg)
0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8
quota
0.0
0.2
0.4
0.6
0.8
1.0
1.2
max
plan
kton
Figure 43: A bifurcation diagram for the 3-species model; with p4 = 4 .
183
![Page 184: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/184.jpg)
NOTE:
◦ These periodic solution families are not “vertical”. (The generic case.)
◦ One family connects the two Hopf points along the stationary family alongwhich u1 is constant.
◦ The second family connects the two Hopf points along the stationary familyalong which u1 is not constant.
◦ These two families of periodic solutions “intersect” at a branch point ofperiodic solutions, at λ ≈ 0.3012 .
◦ At this point there is an “interchange of stability” between the families.
◦ Stable periodic orbits are denoted by solid circles in the diagram.
◦ Unstable periodic orbits are denoted by open circles in the diagram.
184
![Page 185: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/185.jpg)
0.0 0.1 0.2 0.3 0.4 0.5 0.6
plankton
0.0
0.1
0.2
0.3
0.4
0.5
fish
Figure 44: Part of the planar orbit family for the 3-species model; p4 = 4 .
185
![Page 186: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/186.jpg)
0.0 0.1 0.2 0.3 0.4 0.5 0.6
plankton
0.0
0.1
0.2
0.3
0.4
0.5
fish
Figure 45: The remainder of the planar orbit family; p4 = 4 .
186
![Page 187: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/187.jpg)
fish0.10.2
0.30.4
plankton0.1
0.20.3
0.40.5
0.6
sharks
−0.25
−0.20
−0.15
−0.10
−0.05
0.00
0.05
Figure 46: Part of the 3D orbit family for the 3-species model; p4 = 4 .
187
![Page 188: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/188.jpg)
fish0.10.2
0.30.4
plankton0.2
0.3
0.4
0.5
0.6
sharks
0.025
0.050
0.075
0.100
Figure 47: The remainder of the 3D orbit family; p4 = 4 .
188
![Page 189: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/189.jpg)
Computing Periodic Solutions
Periodic solutions can be computed very effectively using a boundary value ap-proach. This method also determines the period very accurately. Moreover,the technique allows asymptotically unstable periodic orbits to be computed aseasily as asymptotically stable ones.
189
![Page 190: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/190.jpg)
The BVP Approach.
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.
190
![Page 191: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/191.jpg)
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 ) .
Specifically, 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.
191
![Page 192: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/192.jpg)
An example is the Poincare orthogonality condition
(uk(0) − uk−1(0))∗ u′
k−1(0) = 0 .
(Below we derive a numerically more suitable phase condition.)
u k-1 (0)
��
��
uk-1 (0)
u (0)k
Figure 48: Graphical interpretation of the Poincare phase condition.
192
![Page 193: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/193.jpg)
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) ‖2
2 dt .
The optimal solutionuk(t+ σ) ,
must satisfy the necessary condition
D′(σ) = 0 .
193
![Page 194: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/194.jpg)
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
∫ 10 uk(t)
∗ u′k−1(t) dt = 0 .
This is the integral phase condition.
194
![Page 195: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/195.jpg)
Pseudo-Arclength Continuation
We use pseudo-arclength continuation to follow a family of periodic solutions.
This allows calculation past folds along a family of periodic solutions.
It also allows calculation of a “vertical family” of periodic solutions.
For periodic solutions the pseudo-arclength equation is
∫ 1
0(uk(t)− uk−1(t))∗uk−1(t) dt + (Tk − Tk−1)Tk−1 + (λk − λk−1)λk−1 = ∆s .
195
![Page 196: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/196.jpg)
In summary, we have the following equations for continuing periodic solutions:
u′k(t) = T f( uk(t) , λk ) ,
uk(0) = uk(1) ,
∫ 1
0uk(t)
∗ u′
k−1(t) dt = 0 ,
with pseudo-arclength continuation equation
∫ 1
0(uk(t)− uk−1(t))∗uk−1(t) dt + (Tk − Tk−1)Tk−1 + (λk − λk−1)λk−1 = ∆s .
Here
u(·) , f(·) ∈ Rn , λ , T ∈ R .
196
![Page 197: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/197.jpg)
Starting at a Hopf Bifurcation
Let(u0 , λ0) ,
be a Hopf bifurcation point, i.e.,
fu( u0 , λ0 ) ,
has a simple conjugate pair of purely imaginary eigenvalues
± i ω0 , ω0 6= 0 ,
and no other eigenvalues on the imaginary axis.
Also, the pair crosses the imaginary axis transversally with respect to λ .
By the Hopf Bifurcation Theorem, a family of periodic solutions bifurcates.
197
![Page 198: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/198.jpg)
Asymptotic estimates for periodic solutions near the Hopf bifurcation :
u( t ; ε ) = u0 + ε φ(t) + O(ε2) ,
T (ε) = T0 + O(ε2) ,
λ(ε) = λ0 + O(ε2) .
Here ε locally parametrizes the family of periodic solutions.
T (ε) denotes the period, and
T0 =2π
ω0
.
The function φ(t) is the normalized nonzero periodic solution of the linearized,constant coefficient problem
φ′(t) = fu(u0, λ0) φ(t) .
198
![Page 199: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/199.jpg)
To compute a first periodic solution
( u1(·) , T1 , λ1 ) ,
near a Hopf bifurcation (u0, λ0) , we still have
u′1(t) = T f( u1(t) , λ1 ) , (10)
u1(0) = u1(1) . (11)
Initial estimates for Newton’s method are
u(0)1 (t) = u0 + ∆s φ(t) , T
(0)1 = T0 , λ
(0)1 = λ0 .
199
![Page 200: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/200.jpg)
Above, φ(t) is a nonzero solution of the time-scaled, linearized equations
φ′(t) = T0 fu(u0, λ0) φ(t) , φ(0) = φ(1) ,
namely,
φ(t) = sin(2πt) ws + cos(2πt) wc ,
where( ws , wc ) ,
is a null vector in
( −ω0 I fu(u0, λ0)fu(u0, λ0) ω0 I
) (ws
wc
)=
(00
), ω0 =
2π
T0
.
The nullspace is generically two-dimensional since(−wc
ws
),
is also a null vector.
200
![Page 201: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/201.jpg)
For the phase equation we “align” u1 with φ(t) , i.e.,
∫ 10 u1(t)∗ φ′(t) dt = 0 .
Since
λ0 = T0 = 0 ,
the pseudo-arclength equation for the first step reduces to
∫ 10 ( u1(t)− u0(t) )∗ φ(t) dt = ∆s .
201
![Page 202: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/202.jpg)
Accuracy Test
EXERCISE.
A simple accuracy test is to treat the linear equation
u′1(t) = λ u1 − u2 ,
u′2(t) = u1 ,
as a bifurcation problem.
◦ It has a Hopf bifurcation point at λ = 0 from the zero solution family.
◦ The bifurcating family of periodic solutions is vertical.
◦ Along the family, the period remains constant, namely, T = 2π .
202
![Page 203: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/203.jpg)
For the above problem:
◦ Compute the family of periodic solutions, for different choices of the numberof mesh points and the number of collocation points.
◦ Determine the error in the period for the computed solutions.
Typical results are shown in Table 1.
ntst ncol = 2 ncol = 3 ncol = 44 0.47e-1 (5.5) 0.85e-3 (7.9) 0.85e-5 (8.9)8 0.32e-2 (4.0) 0.14e-4 (6.0) 0.35e-7 (8.0)
16 0.20e-3 (4.0) 0.22e-6 (6.0) 0.14e-9 (8.0)32 0.13e-4 0.35e-8 0.54e-11
Table 1: Accuracy of T for the linear problem
203
![Page 204: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/204.jpg)
EXAMPLE: (AUTO demo fhn.)
The FitzHugh-Nagumo model
u′1 = u1 − u31
3− u2 + I ,
u′2 = a( u1 + b − cu2 ) .
is a model of spike generation in squid giant axons, where
u1 is the membrane potential ,
u2 is a recovery variable ,
I is the stimulus current .
Take I as bifurcation parameter, and
a = 0.08 , b = 0.7 , c = 0.8 .
If I = 0 then (u1, u2) = (−1.19941 ,−0.62426) is a stationary solution.
204
![Page 205: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/205.jpg)
0.0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6
current
−2.0
−1.5
−1.0
−0.5
0.0
0.5
1.0
1.5
2.0
max
u1
Figure 49: Bifurcation diagram of the Fitzhugh-Nagumo equations.
205
![Page 206: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/206.jpg)
0.0 0.2 0.4 0.6 0.8 1.0
scaled time
−2.0
−1.5
−1.0
−0.5
0.0
0.5
1.0
1.5
2.0
u1
Figure 50: A stable periodic solutions .
206
![Page 207: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/207.jpg)
Periodically Forced Systems
Here we illustrate computing periodic solutions to a periodically forced system.
In AUTO this can be done by adding a nonlinear oscillator with the desiredperiodic forcing as one of its solution components.
An example of such an oscillator is
x′ = x + βy − x (x2 + y2) ,
y′ = −βx + y − y (x2 + y2) ,
which has the asymptotically stable solution
x = sin(βt) , y = cos(βt) .
207
![Page 208: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/208.jpg)
EXAMPLE. (AUTO demo ffn.)
Couple the oscillator
x′ = x + βy − x (x2 + y2) ,y′ = −βx + y − y (x2 + y2) ,
to the Fitzhugh-Nagumo equations :
v′ = c (v − v3
3+ w − r y) ,
w′ = −(v − a + b ∗ w)/c ,
whereb = 0.8 , c = 3 , and β = 10 .
Note that ifa = 0 and r = 0 ,
then a solution is
x = sin(βt) , y = cos(βt) , v(t) ≡ 0 , w(t) ≡ 0 ,
Continue this solution from r = 0 to, say, r = 10 .
208
![Page 209: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/209.jpg)
0 1 2 3 4 5 6 7 8 9 10
Forcing amplitude r
0.0
0.5
1.0
1.5
2.0
2.5
3.0
max
v
1
2
2
3
4
Figure 51: Continuation from r = 0 to r = 10 . Solution 1 is unstable. Solution2 corresponds to a torus bifurcation.
209
![Page 210: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/210.jpg)
0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0
Scaled time−3
−2
−1
0
1
2
3
v(t
) 1
2
34
Figure 52: Some solutions along the path from r = 0 to r = 10 .
210
![Page 211: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/211.jpg)
NOTE:
◦ The starting solution at r = 0, with v = w = 0, is unstable.
◦ The oscillation becomes stable when r passes the value rT ≈ 4.52 .
◦ At r = rT there is a torus bifurcation.
211
![Page 212: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/212.jpg)
General Non-Autonomous Systems
If the forcing is not periodic, or difficult to model by an autonomous oscillator,then the equations can be rewritten in autonomous form as follows:
u′(t) = f( t , u(t) ) , u(·) , f(·) ∈ Rn , t ∈ [0, 1] ,
b( u(0) , u(1) ) = 0 , b(·) ∈ Rn ,
can be transformed into
u′(t) = f( v(t) , u(t) ) ,
v′(t) = 1 , v(·) ∈ R ,
b( u(0) , u(1) ) = 0 ,
v(0) = 0 ,
which is autonomous, with n+ 1 ODEs and n+ 1 boundary conditions.
212
![Page 213: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/213.jpg)
Periodic Solutions of Conservative Systems
EXAMPLE:u′1 = − u2 ,
u′2 = u1 (1− u1) .
PROBLEM:
◦ This equation has a family of periodic solutions, but no parameter !
◦ This system has a constant of motion, namely the Hamiltonian
H(u1, u2) = − 1
2u2
1 −1
2u2
2 +1
3u3
1 .
213
![Page 214: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/214.jpg)
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 .
(This is AUTO demo lhb; see Figures 33 and 34.)
214
![Page 215: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/215.jpg)
−1.0 −0.5 0.0 0.5 1.0
λ
−0.2
0.0
0.2
0.4
0.6
0.8
1.0
1.2
norm
Figure 53: Bifurcation diagram of the “linear” Hopf bifurcation problem.
215
![Page 216: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/216.jpg)
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 software.
216
![Page 217: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/217.jpg)
EXAMPLE : The Circular Restricted 3-Body Problem (CR3BP).
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.
For the Earth-Moon system µ ≈ 0.01215 .
217
![Page 218: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/218.jpg)
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
,
where
r1 =√
(x+ µ)2 + y2 + z2 , r2 =√
(x− 1 + µ)2 + y2 + z2 .
218
![Page 219: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/219.jpg)
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 + pseudo-arclength equation.
219
![Page 220: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/220.jpg)
NOTE:
◦ One can use standard BVP continuation and bifurcation software.
◦ The “unfolding term” λ ∇v regularizes the continuation.
◦ λ will be ”zero”, once solved for.
◦ Other unfolding terms are possible.
220
![Page 221: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/221.jpg)
Families of Periodic Solutions of the Earth-Moon system.
221
![Page 222: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/222.jpg)
The planar Lyapunov family L1.
222
![Page 223: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/223.jpg)
The Halo family H1.
223
![Page 224: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/224.jpg)
The Halo family H1.
224
![Page 225: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/225.jpg)
The Vertical family V1.
225
![Page 226: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/226.jpg)
The Axial family A1.
226
![Page 227: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/227.jpg)
Following Periodic Orbit Folds
Fold-following algorithms also apply to folds along solution families of boundaryvalue problems and, in particular, folds along families of periodic solutions.
227
![Page 228: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/228.jpg)
EXAMPLE: The A→ B → C reaction. (AUTO demo abc.)
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 ,
with
α = 1 , σ = 0.04 , B = 8 .
We will compute solutions for varying D and β .
228
![Page 229: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/229.jpg)
0.20 0.22 0.24 0.26 0.28 0.30 0.32 0.34 0.36
Damkohler number
1
2
3
4
5
6
7
8
9
10
max
u3
22
3
34
4
5
5
77
88 9
1010
11
12
13
Figure 54: Stationary and periodic solutions of demo abc; β = 1.55.
229
![Page 230: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/230.jpg)
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
0u(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 ) .
230
![Page 231: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/231.jpg)
At a fold with respect to λ we have
FX( X , λ ) Φ = 0 ,
< Φ , Φ > = 1 .where
X = ( u , T ) , Φ = ( v , S ) ,
or, written in detail,
v′(t) − T fu( u(t) , λ ) v − S f( u(t) , λ ) = 0 ,
v(0) − v(1) = 0 ,
∫ 1
0v(t)∗u
′
0(t) dt = 0 ,
∫ 1
0v(t)∗v(t) dt + S2 = 1 .
231
![Page 232: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/232.jpg)
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 pseudo-arclength equation
< X−X0 , X0 > + < Φ−Φ0 , Φ0 > + (λ−λ0) λ0 + (µ−µ0) µ0 − ∆ s = 0 .
232
![Page 233: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/233.jpg)
In detail:u′(t) − T f( u(t) , λ , µ ) = 0 ,
u(0) − u(1) = 0 ,∫ 1
0u(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
0v(t)∗u
′
0(t) dt = 0 ,
with normalization ∫ 1
0v(t)∗v(t) dt + S2 − 1 = 0 ,
and pseudo-arclength 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 .
233
![Page 234: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/234.jpg)
0.21 0.22 0.23 0.24 0.25 0.26 0.27 0.28
Damkohler number
1.55
1.56
1.57
1.58
1.59
1.60
1.61
1.62
β
2
5
6 7
89
10
12
Figure 55: The locus of periodic solution folds of demo abc.
234
![Page 235: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/235.jpg)
0.20 0.22 0.24 0.26 0.28 0.30 0.32 0.34 0.36
Damkohler number
1
2
3
4
5
6
7
8
9
10
max
u3
22
3
34
4
5
5
77
88 9
1010
11
12
13
Figure 56: Stationary and periodic solutions of demo abc; β = 1.55.
235
![Page 236: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/236.jpg)
0.20 0.22 0.24 0.26 0.28 0.30 0.32 0.34 0.36
Damkohler number
1
2
3
4
5
6
7
8
9
10
max
u3
2
2
3
34
4
5
5
7 7
8
89
1010
1112
13
Figure 57: Stationary and periodic solutions of demo abc; β = 1.56.
236
![Page 237: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/237.jpg)
0.20 0.22 0.24 0.26 0.28 0.30 0.32 0.34 0.36
Damkohler number
1
2
3
4
5
6
7
8
9
10
max
u3
2
2
3
34
4
5
5
7
7
88
9
1010
1111
12
1313
14
15
Figure 58: Stationary and periodic solutions of demo abc; β = 1.57.
237
![Page 238: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/238.jpg)
0.20 0.22 0.24 0.26 0.28 0.30 0.32 0.34 0.36
Damkohler number
1
2
3
4
5
6
7
8
9
10
max
u3
2
2
3
3 4
4
5
5
78
99
10
11
1213
Figure 59: Stationary and periodic solutions of demo abc; β = 1.58.
238
![Page 239: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/239.jpg)
0.20 0.25 0.30 0.35 0.40
Damkohler number
1
2
3
4
5
6
7
8
9
10
max
u3
2
2
3
3
4
4
5
5
7
8
9 9
10
1112
13
Figure 60: Stationary and periodic solutions of demo abc; β = 1.61.
239
![Page 240: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/240.jpg)
0.20 0.25 0.30 0.35 0.40
Damkohler number
1
2
3
4
5
6
7
8
9
10
max
u3
2
23
34
4
5
5
7
8
99
10
11
Figure 61: Stationary and periodic solutions of demo abc; β = 1.62.
240
![Page 241: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/241.jpg)
Following Hopf Bifurcations
We consider the persistence of a Hopf bifurcation as a second parameter is varied,and we give an algorithm for computing a 2-parameter locus of Hopf bifurcationpoints.
241
![Page 242: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/242.jpg)
A Hopf bifurcation along a stationary solution family (u(λ), λ) , of
u′ = f(u, λ) ,
occurs when 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 , α(λ0) 6= 0 , and β0 = β(λ0) 6= 0 ,
also assuming there are no other eigenvalues on the imaginary axis.
The assumptions imply that f0u is nonsingular , so that stationary solution
family can indeed be parametrized locally using λ .
The right and left complex eigenvectors of f0u = fu(u(λ0), λ0) are defined by
f0u φ0 = i β0 φ0 , (f0
u)∗ ψ0 = − i β0 ψ0 .
242
![Page 243: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/243.jpg)
Transversality and Persistence
THEOREM. The eigenvalue crossing in the Hopf Bifurcation Theorem is transver-sal if
Re ( ψ∗0 [f0uu (f0
u)−1 f0λ − f0
uλ] φ0 ) 6= 0 .
PROOF. Since the eigenvaluei β0 ,
is algebraically simple, there is a smooth solution family (at least locally) to theparametrized right and left eigenvalue-eigenvector equations :
(a) fu(u(λ) , λ) φ(λ) = κ(λ) φ(λ) ,
(b) ψ(λ)∗fu(u(λ), λ) = κ(λ) ψ∗(λ) ,
(c) ψ(λ)∗φ(λ) = 1, (and also φ(λ)∗φ(λ) = 1) ,
withκ(λ0) = i β0 .
(Above, ∗ denotes conjugate transpose.)
243
![Page 244: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/244.jpg)
fu(u(λ) , λ) φ(λ) = κ(λ) φ(λ) , ψ(λ)∗fu(u(λ), λ) = κ(λ) ψ∗(λ) , ψ(λ)∗φ(λ) = 1
Differentiation with respect to λ gives
(a) fuuuφ + fuλφ + fuφ = κφ + κφ ,
(b) ψ∗fuuu + ψ∗fuλ + ψ∗fu = κψ∗ + κψ
∗,
(c) ψ∗φ + ψ∗φ = 0 .
Multiply(a) on the left by ψ∗ ,
and(b) on the right by φ ,
to get
(a) ψ∗fuuu φ + ψ∗fuλ φ + ψ∗fu φ = κ ψ∗φ + κ ψ∗φ ,
(b) ψ∗fuu u φ + ψ∗fuλ φ + ψ∗fu φ = κ ψ∗φ + κ ψ
∗φ .
244
![Page 245: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/245.jpg)
ψ∗fuuu φ + ψ∗fuλ φ + ψ∗fu φ = κ ψ∗φ + κ ψ∗φ
ψ∗fuu u φ + ψ∗fuλ φ + ψ∗fu φ = κ ψ∗φ + κ ψ
∗φ
Adding the above, and using
ψ∗fu︸ ︷︷ ︸=κψ∗
φ + ψ∗fuφ︸︷︷︸=κφ
= κ(ψ∗φ + ψ∗φ) = κ
d
dλ(ψ∗φ︸ ︷︷ ︸
=1
) = 0 ,
we find
κ = ψ∗[fuuu + fuλ] φ .
245
![Page 246: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/246.jpg)
κ = ψ∗[fuuu + fuλ] φ
From differentiating
f(u(λ), λ) = 0 ,
with respect to λ , we have
u = − (fu)−1 fλ ,
so thatκ = ψ∗[−fuu (fu)−1 fλ + fuλ] φ .
Thus the eigenvalue crossing is transversal if
α(0) = Re(κ0) = Re(ψ∗0[f0uu (f0
u)−1 f0λ − f0
uλ] φ0) 6= 0 . •
246
![Page 247: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/247.jpg)
NOTE:
The transversality condition of the Theorem, i.e.,
Re ( ψ∗0 [f0uu(f0
u)−1f0λ − f0
uλ] φ0 ) 6= 0 ,
is also needed for persistence of the Hopf bifurcation, as discussed below.
247
![Page 248: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/248.jpg)
The extended system for following Hopf bifurcations 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 typically is the latest computed solution point of a family.
248
![Page 249: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/249.jpg)
First consider parametrizing in the second parameter µ , i.e., we seek a family
( u(µ) , φ(µ) , β(µ) , λ(µ) ) .
(In practice, pseudo-arclength continuation is used.)
The derivative with respect to
( u , φ , β , λ ) ,
at the solution point( u0 , φ0 , β0 , λ0 , µ0 ) ,
is f0u O 0 f0
λ
f0uuφ0 f0
u − iβ0I −iφ0 f0uλφ0
0∗ φ∗0 0 0
. (12)
249
![Page 250: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/250.jpg)
The Jacobian is of the form
A O 0 c1
C D −iφ0 c2
0∗ φ∗0 0 0
,
where
A = f0u (nonsingular), C = f0
uu φ0 , D = f0u − i β0 I ,
andc1 = f0
λ , c2 = f0uλ φ0 ,
withN (D) = Span{φ0} , N (D∗) = Span{ψ0} ,
whereψ∗0 φ0 = 1 , φ∗0 φ0 = 1 .
250
![Page 251: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/251.jpg)
THEOREM.
If the eigenvalue crossing is transversal, i.e., if
Re(κ0) 6= 0 ,
then the Jacobian matrix (12) is nonsingular.
Hence there locally exists a solution family
( u(µ) , φ(µ) , β(µ) , λ(µ) )
to the extended system
F( u , φ , β , λ ; µ ) = 0 ,
i.e., the Hopf bifurcation persists under small perturbations of µ .
251
![Page 252: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/252.jpg)
PROOF. We prove this by constructing a solution
x ∈ Rn , y ∈ Cn , z1 , z2 ∈ R ,
to A O 0 c1
C D −iφ0 c2
0∗ φ∗0 0 0
xyz1
z2
=
fgh
,
wheref ∈ Rn , g ∈ Cn , h ∈ C .
From the first equationAx + z2 c1 = f ,
we havex = A−1f − z2 A
−1 c1 .
The second equation can then be written
C A−1f − z2 C A−1 c1 + D y − z1 i φ0 + z2 c2 = g .
252
![Page 253: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/253.jpg)
C A−1f − z2 C A−1 c1 + D y − z1 i φ0 + z2 c2 = g .
Multiply on the left by ψ∗0 to get
ψ∗0 C A−1 f − z2 ψ∗0 C A−1 c1 − z1 i ψ
∗0 φ0 + z2 ψ
∗0 c2 = ψ∗0 g .
Recall thatψ∗0 φ0 = 1 .
Definingf ≡ C A−1 f , c1 ≡ C A−1 c1 .
we havei z1 + ψ∗0 (c1 − c2) z2 = ψ∗0 (f − g) .
Computationally f and c1 are obtained from
A f = f , A c1 = c1 ,
f = C f , c1 = C c1 .
253
![Page 254: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/254.jpg)
i z1 + ψ∗0 (c1 − c2) z2 = ψ∗0 (f − g)
Separate real and imaginary part of this equation, and use the fact that
z1 and z2 are real ,
to getRe( ψ∗0 [c1 − c2] ) z2 = Re( ψ∗0 [f − g] ) ,
z1 + Im( ψ∗0 [c1 − c2] ) z2 = Im( ψ∗0 [f − g] ) ,
from which
z2 =Re( ψ∗0 [f − g] )
Re( ψ∗0 [c1 − c2] ),
z1 = − z2 Im( ψ∗0 [c1 − c2] ) + Im( ψ∗0 [f − g] ) .
254
![Page 255: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/255.jpg)
Now solve for x in
A x = f − z2 c1 ,
and compute a particular solution yp to
D y = g − C x + i z1 φ0 − z2 c2 .
Then
y = yp + α φ0 , α ∈ C .
The third equation is
φ∗0 y = φ∗0 yp + α φ∗0 φ0 = h ,
from which, using φ∗0 φ0 = 1 ,
α = h − φ∗0 yp .
255
![Page 256: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/256.jpg)
The above construction can be carried out if
Re( ψ∗0 [c1 − c2] ) 6= 0 .
However, using the definition of c1 and c2 , we have
Re( ψ∗0 [c1 − c2] ) = Re( ψ∗0 [C A−1 c1 − c2] )
= Re( ψ∗0 [f0uu φ0 (f0
u)−1f0λ − f0
uλ φ0] )
= Re( ψ∗0 [f0uu (f0
u)−1f0λ − f0
uλ] φ0 )
= Re(κ0) 6= 0 . •
256
![Page 257: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/257.jpg)
Practical Continuation of Hopf Bifurcations
Recall that the extended system for following Hopf bifurcations 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 ) , with u ∈ Rn, φ ∈ Cn, β, λ, µ ∈ R .
In practice, we treat µ as an unknown, and add the continuation equation
(u−u0)∗u0 + (φ−φ0)∗φ0 +(β−β0)β0 + (λ−λ0)λ0 + (µ−µ0)µ0 − ∆s = 0 .
257
![Page 258: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/258.jpg)
EXERCISE.
Investigate the Hopf bifurcations in the system
u′1(t) = u1 (1− u1) − p4 u1 u2 ,
u′2(t) = − 14u2 + p4 u1 u2 − 3 u2 u3 − p1 (1− e−5u2) ,
u′3(t) = − 12u3 + 3 u2 u3 .
This is the predator-prey model where,
u1 ∼ plankton, u2 ∼ fish, u3 ∼ sharks, λ ≡ p1 ∼ fishing quota.
(See also Figure 43.)
258
![Page 259: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/259.jpg)
0.0 0.1 0.2 0.3 0.4 0.5
quota
0.0
0.2
0.4
0.6
0.8
1.0
1.2
1.4
max
plan
kton
Figure 62: A bifurcation diagram for the 3-species model; with p4 = 3 .
259
![Page 260: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/260.jpg)
0.10 0.15 0.20 0.25 0.30 0.35 0.40 0.45 0.50 0.55
quota
2.75
3.00
3.25
3.50
3.75
4.00
p 4
Figure 63: Loci of Hopf bifurcations for the 3-species model.
260
![Page 261: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/261.jpg)
0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8
quota
0.0
0.2
0.4
0.6
0.8
1.0
1.2
max
plan
kton
Figure 64: A bifurcation diagram for the 3-species model; with p4 = 4 .
261
![Page 262: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/262.jpg)
EXERCISE. (AUTO demo abc-hb .)
Compute loci of Hopf bifurcation points for the A→ B → C reaction
u′1 = −u1 + D(1− u1)eu3 ,
u′2 = −u2 + D(1− u1)eu3 − Dσu2eu3 ,
u′3 = −u3 − βu3 + DB(1− u1)eu3 + DBασu2eu3 ,
with
α = 1 , σ = 0.04 , B = 8 ,
and varying D and β .
262
![Page 263: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/263.jpg)
0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8
Damkohler number
1.2
1.3
1.4
1.5
1.6
1.7
1.8
1.9
2.0
2.1
2.2
β
Figure 65: A locus of Hopf bifurcations.
263
![Page 264: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/264.jpg)
0.10 0.15 0.20 0.25 0.30 0.35 0.40
Damkohler number
0
1
2
3
4
5
6
7
L2-
norm
Figure 66: Diagrams for β = 1.2 , 1.3 , 1.4 , 1.5 , 1.6 , 1.7 , 1.8 .
264
![Page 265: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/265.jpg)
Stable and Unstable Manifolds
◦ One can also use continuation to compute solution families of IVP.
◦ In particular, one can compute stable and unstable manifolds .
265
![Page 266: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/266.jpg)
EXAMPLE: The Lorenz Equations. (AUTO demos lor, lrz, man.)
x′ = σ (y − x) ,
y′ = ρ x − y − x z ,
z′ = x y − β z ,
whereσ = 10 and β = 8/3 .
266
![Page 267: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/267.jpg)
Figure 67: Bifurcation diagram of the Lorenz equations.
267
![Page 268: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/268.jpg)
NOTE:
◦ The zero solution is unstable for ρ > 1 .
◦ Two nonzero stationary solutions bifurcate at ρ = 1 .
◦ The nonzero stationary solutions become unstable for ρ > ρH .
◦ At ρH ( ρH ≈ 24.7 ) there are Hopf bifurcations.
◦ Unstable periodic solutions emanate from each Hopf bifurcation.
◦ These families end in homoclinic orbits (infinite period) at ρ ≈ 13.9 .
◦ For ρ > ρH there is the famous Lorenz attractor.
268
![Page 269: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/269.jpg)
Figure 68: Unstable periodic orbits of the Lorenz equations.
269
![Page 270: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/270.jpg)
The Lorenz Manifold
◦ For ρ > 1 the origin is a saddle point .
◦ The Jacobian has two negative eigenvalues 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 Lorenz Manifold helps us understand the Lorenz attractor .
270
![Page 271: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/271.jpg)
-30 -20 -10 0 10 20 30X
0
25
50
75
100
125
150
175
Z
The Lorenz Equations: rho = 60
Figure 69: Three orbits whose initial conditions agree to >11 decimal places !
271
![Page 272: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/272.jpg)
Figure 70: A small portion of a Lorenz Manifold · · ·272
![Page 273: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/273.jpg)
Figure 71: Intersection of a Lorenz Manifold with a sphere (ρ = 35, R = 100).
273
![Page 274: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/274.jpg)
Figure 72: Intersection of a Lorenz Manifold with a sphere (ρ = 35, R = 100).
274
![Page 275: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/275.jpg)
Figure 73: Intersection of a Lorenz Manifold with a sphere (ρ = 35, R = 100).
275
![Page 276: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/276.jpg)
How was the Lorenz Manifold computed?
First compute an orbit u0(t) , for t from 0 to T0 (where T0 < 0) , with
u0(0) close to the origin 0 ,and
u0(0) in the stable eigenspace spanned by v1 and v2 ,
that is,
u0(0) = 0 + ε
(cos(2πθ)
|µ1| v1 − sin(2πθ)
|µ2| v2
),
for, say, θ = 0 .
276
![Page 277: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/277.jpg)
Scale time
t → t
T0
,
Then the initial orbit satisfies
u′0(t) = T0 f( u0(t) ) , for 0 ≤ t ≤ 1 ,
andu0(0) =
ε
|µ1| v1 .
The initial orbit has length
L = T0
∫ 1
0|| f(u0(s)) || ds .
277
![Page 278: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/278.jpg)
Thus the initial orbit corresponds to a solution X0 of the equation
F(X) = 0 ,
where
F(X) ≡
u′(t)− T f(u(t))
u(0) − ε(
cos(θ)|µ1| v1 − sin(θ)
|µ2| v2
)T∫ 1
0 || f(u) || ds − L
with X = ( u(·) , θ , T ) , (for given L and ε) ,
and
X0 = ( u0(·) , 0 , T0 ) .
278
![Page 279: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/279.jpg)
As before, the continuation system is
F(Xk) = 0 ,
< Xk −Xk−1 , Xk−1 > − ∆s = 0 , ( ‖ Xk−1 ‖ = 1 ) ,
and
X = ( u(·) , θ , T ) , (keeping L and ε fixed) ,
or
X = ( u(·) , θ , L ) , (keeping T and ε fixed) ,
or ( for computing the starting orbit u0(t) ) ,
X = ( u(·) , L, T ) , (keeping θ and ε fixed) .
Other variations are possible · · · .
279
![Page 280: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/280.jpg)
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 cannot suddenly change a lot in any continuation step.
◦ This allows the ”entire manifold ” to be computed.
280
![Page 281: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/281.jpg)
NOTE:
◦ Crossings of the Lorenz manifold with the plane z = ρ− 1 can be located.
◦ Connections between the origin and the nonzero equilibria can be located.
◦ There are subtle variations on the algorithm !
281
![Page 282: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/282.jpg)
Figure 74: Crossings of the Lorenz Manifold with the plane z = ρ− 1282
![Page 283: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/283.jpg)
Heteroclinic connections
◦ During the computation of the 2D stable manifold of the origin one canlocate heteroclinic orbits between the origin and the nonzero equilibria.
◦ The same heteroclinic orbits can be detected during the computation ofthe 2D unstable manifold of the nonzero equilibria.
283
![Page 284: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/284.jpg)
Representation of the orbit family in the stable manifold.
284
![Page 285: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/285.jpg)
A heteroclinic connection in the Lorenz equations.
285
![Page 286: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/286.jpg)
Another heteroclinic connection in the Lorenz equations.
286
![Page 287: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/287.jpg)
· · · and another · · ·287
![Page 288: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/288.jpg)
· · · and another · · ·288
![Page 289: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/289.jpg)
This continuation located 512 connections!
289
![Page 290: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/290.jpg)
NOTE:
◦ The heteroclinic connections have a combinatorial structure.
◦ We can also continue each heteroclinic connection as ρ varies.
◦ They spawns homoclinic orbits, having their own combinatorial structure.
◦ These results shed some light on the Lorenz attractor as ρ changes.
More details:
E. J. Doedel, B. Krauskopf, H. M. Osinga, Global bifurcations of the Lorenzmodel, Nonlinearity 19, 2006, 2947-2972.
290
![Page 291: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/291.jpg)
Example: The Circular Restricted 3-Body Problem
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( x , y , z) ,
is the position of the zero-mass body, and
r1 =√
(x + µ)2 + y2 + z2 , r2 =√
(x− 1 + µ)2 + y2 + z2 .
For the Earth-Moon system µ ≈ 0.01215 .
291
![Page 292: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/292.jpg)
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 .
292
![Page 293: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/293.jpg)
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.
293
![Page 294: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/294.jpg)
NOTE:
◦ The “unfolding term” λ ∇v regularizes the continuation.
◦ λ will be ”zero”, once solved for.
◦ Other unfolding terms are possible.
◦ The unfolding term allows using BVP continuation.
294
![Page 295: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/295.jpg)
Families of Periodic Solutions of the Earth-Moon system.
295
![Page 296: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/296.jpg)
A family of Halo orbits.
296
![Page 297: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/297.jpg)
NOTE:
• ”Small” Halo orbits have one real Floquet multiplier outside the unit circle.
• Such Halo orbits are unstable .
• They have a 2D unstable manifold .
297
![Page 298: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/298.jpg)
Continuation, keeping the endpoint x(1) fixed.
298
![Page 299: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/299.jpg)
NOTE:
• 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 .
299
![Page 300: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/300.jpg)
NEW1:Continuation, keeping the endpoint x(1) fixed.
300
![Page 301: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/301.jpg)
NEW1:Continuation, keeping the endpoint x(1) fixed.
301
![Page 302: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/302.jpg)
NEW1:Continuation, keeping the endpoint x(1) fixed.
302
![Page 303: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/303.jpg)
NEW1:Continuation, keeping the endpoint x(1) fixed.
303
![Page 304: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/304.jpg)
NEW1:Continuation, keeping the endpoint x(1) fixed.
304
![Page 305: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/305.jpg)
NEW1:Continuation, keeping the endpoint x(1) fixed.
305
![Page 306: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/306.jpg)
NEW1:Continuation, keeping the endpoint x(1) fixed.
306
![Page 307: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/307.jpg)
NEW1:Continuation, keeping the endpoint x(1) fixed.
307
![Page 308: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/308.jpg)
NEW1:Continuation, keeping the endpoint x(1) fixed.
308
![Page 309: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/309.jpg)
NEW1:Continuation, keeping the endpoint x(1) fixed.
309
![Page 310: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/310.jpg)
NEW1:Continuation, keeping the endpoint x(1) fixed.
310
![Page 311: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/311.jpg)
NEW1:Continuation, keeping the endpoint x(1) fixed.
311
![Page 312: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/312.jpg)
NEW1:Continuation, keeping the endpoint x(1) fixed.
312
![Page 313: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/313.jpg)
NEW1:Continuation, keeping the endpoint x(1) fixed.
313
![Page 314: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/314.jpg)
NEW1:Continuation, keeping the endpoint x(1) fixed.
314
![Page 315: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/315.jpg)
NEW1:Continuation, keeping the endpoint x(1) fixed.
315
![Page 316: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/316.jpg)
NEW1:Continuation, keeping the endpoint x(1) fixed.
316
![Page 317: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/317.jpg)
NEW1:Continuation, keeping the endpoint x(1) fixed.
317
![Page 318: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/318.jpg)
NEW1:Continuation, keeping the endpoint x(1) fixed.
318
![Page 319: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/319.jpg)
NEW1:Continuation, keeping the endpoint x(1) fixed.
319
![Page 320: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/320.jpg)
NEW1:Continuation, keeping the endpoint x(1) fixed.
320
![Page 321: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/321.jpg)
NEW1:Continuation, keeping the endpoint x(1) fixed.
321
![Page 322: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/322.jpg)
NEW1:Continuation, keeping the endpoint x(1) fixed.
322
![Page 323: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/323.jpg)
NEW1:Continuation, keeping the endpoint x(1) fixed.
323
![Page 324: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/324.jpg)
NEW1:Continuation, keeping the endpoint x(1) fixed.
324
![Page 325: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/325.jpg)
NEW1:Continuation, keeping the endpoint x(1) fixed.
325
![Page 326: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/326.jpg)
NEW1:Continuation, keeping the endpoint x(1) fixed.
326
![Page 327: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/327.jpg)
NEW1:Continuation, keeping the endpoint x(1) fixed.
327
![Page 328: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/328.jpg)
NEW1:Continuation, keeping the endpoint x(1) fixed.
328
![Page 329: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/329.jpg)
NEW1:Continuation, keeping the endpoint x(1) fixed.
329
![Page 330: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/330.jpg)
NEW1:Continuation, keeping the endpoint x(1) fixed.
330
![Page 331: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/331.jpg)
NEW1:Continuation, keeping the endpoint x(1) fixed.
331
![Page 332: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/332.jpg)
NEW1:Continuation, keeping the endpoint x(1) fixed.
332
![Page 333: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/333.jpg)
NEW1:Continuation, keeping the endpoint x(1) fixed.
333
![Page 334: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/334.jpg)
NEW1:Continuation, keeping the endpoint x(1) fixed.
334
![Page 335: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/335.jpg)
NEW1:Continuation, keeping the endpoint x(1) fixed.
335
![Page 336: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/336.jpg)
NEW2:Continuation, keeping the endpoint x(1) fixed.
336
![Page 337: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/337.jpg)
NEW2:Continuation, keeping the endpoint x(1) fixed.
337
![Page 338: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/338.jpg)
NEW2:Continuation, keeping the endpoint x(1) fixed.
338
![Page 339: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/339.jpg)
NEW2:Continuation, keeping the endpoint x(1) fixed.
339
![Page 340: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/340.jpg)
NEW2:Continuation, keeping the endpoint x(1) fixed.
340
![Page 341: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/341.jpg)
NEW2:Continuation, keeping the endpoint x(1) fixed.
341
![Page 342: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/342.jpg)
NEW2:Continuation, keeping the endpoint x(1) fixed.
342
![Page 343: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/343.jpg)
NEW2:Continuation, keeping the endpoint x(1) fixed.
343
![Page 344: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/344.jpg)
NEW2:Continuation, keeping the endpoint x(1) fixed.
344
![Page 345: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/345.jpg)
NEW2:Continuation, keeping the endpoint x(1) fixed.
345
![Page 346: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/346.jpg)
NEW2:Continuation, keeping the endpoint x(1) fixed.
346
![Page 347: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/347.jpg)
NEW2:Continuation, keeping the endpoint x(1) fixed.
347
![Page 348: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/348.jpg)
NEW2:Continuation, keeping the endpoint x(1) fixed.
348
![Page 349: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/349.jpg)
NEW2:Continuation, keeping the endpoint x(1) fixed.
349
![Page 350: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/350.jpg)
NOTE:
• Continuation with x(1) fixed can lead to a Halo-to-torus connection.
• This heteroclinic 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) .
350
![Page 351: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/351.jpg)
Traveling Waves
◦ One can also use continuation to compute traveling wave phenomena.
◦ We illustrate this for a particular model from Biology.
351
![Page 352: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/352.jpg)
Wave Phenomena in a Distributed System
◦ We want to find traveling waves in a parabolic PDE.
◦ The PDE has one space dimension.
◦ Traveling waves are periodic solutions of a ”reduced ” ODE system.
◦ Solitary waves correspond to homoclinic orbits in the reduced system.
◦ Moving fronts are heteroclinic orbits in the reduced system.
◦ Thus ODE continuation techniques can be used.
352
![Page 353: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/353.jpg)
An Enzyme Model
◦ We consider an enzyme catalyzed reaction involving two substrates.
◦ The reaction takes place inside a single compartment.
◦ A membrane separates the compartment from an outside reservoir.
◦ In the reservoir the substrates are kept at a constant level.
◦ Enzymes are embedded in the membrane.
◦ The enzymes activate the reaction.
353
![Page 354: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/354.jpg)
A simple model of such a reaction is
s′(t) = (s0 − s) − ρR(s, a) ,
a′((t) = α(a0 − a) − ρR(s, a) .
◦ s and a denote the concentrations of two chemical species.
◦ The reaction takes place inside a compartment .
◦ An excess of concentration of s inhibits the reaction.
◦ a always activates the reaction.
◦ The reaction rate is proportional to
R(s, a) =a
κ1 + a
s
1 + s+ κ2s2.
◦ s0 and a0 are the constant concentrations in the outside reservoir .
◦ The reaction is catalyzed by an enzyme.
354
![Page 355: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/355.jpg)
This equation has been used to model a reaction with
substrates Oxaloacetate and NADH ,and catalyzed by
Malate Deshydrogenase .
In this case appropriate parameter values are:
κ1 = 3.4 , κ2 = 0.023 , α = 0.2 .
Thus there are three parameters left, namely,
s0 , a0 , ρ .First we also fix
ρ = 210 ,and we use
a0 as bifurcation parameter ,for each of the values
s0 = 143.0 , s0 = 144.5 , s0 = 145.0 .
355
![Page 356: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/356.jpg)
590 600 610 620 630 640 650 660 670a0
0
50
100
150
200
L2-n
orm
Bifurcation diagram of the ODE for s0 = 143.0, ρ = 210.
Red: Stationary states ; Blue: Periodic orbits.
The periodic family connects two Hopf bifurcations .
Solid: Stable ; Dashed: Unstable.
356
![Page 357: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/357.jpg)
590 600 610 620 630 640 650 660 670a0
0
50
100
150
200L2
-nor
m
Bifurcation diagram of the ODE for s0 = 144.5, ρ = 210.
The periodic family ends in a saddle-node homoclinic orbit .
357
![Page 358: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/358.jpg)
590 600 610 620 630 640 650 660 670a0
0
50
100
150
200L2
-nor
m
Bifurcation diagram of the ODE for s0 = 145.0, ρ = 210.
The periodic family ends in a saddle homoclinic orbit .
358
![Page 359: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/359.jpg)
0 5 10 15 20 25 30 35 40 45s
0
20
40
60
80
100
120
140a
The saddle homoclinic orbit that terminates the periodic family.
(s0 = 145.0 ; The two stationary points are also indicated.)
359
![Page 360: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/360.jpg)
NOTE:
◦ For s0 = 143 a family of stable periodic orbits connects the Hopf points.
◦ For s0 = 144.5 there is only one Hopf bifurcation.
◦ At the other end the family ends in a saddle node homoclinic orbit .
◦ For s0 = 145 there is also only one Hopf bifurcation.
◦ At the other end the family ends in a saddle homoclinic orbit .
◦ The bifurcation diagrams are shown superimposed in the next Figure.
360
![Page 361: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/361.jpg)
590 600 610 620 630 640 650 660 670a0
0
50
100
150
200L2
-nor
m
The superimposed bifurcation diagrams.
361
![Page 362: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/362.jpg)
580 600 620 640 660 680 700 720a0
142
143
144
145
146
147
148s 0
s0 = 143
s0 = 144.5
s0 = 145
HBLP
LP
Hom
Loci of folds, Hopf bifurcations, and homoclinic orbits.
362
![Page 363: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/363.jpg)
NOTE:
◦ The preceding 2-parameter diagram shows loci of ”singular points ” .
◦ (Loci of folds , Hopf bifurcations , and homoclinic orbits .)
◦ There is a cusp on the locus of folds in the 2-parameter diagram.
◦ The Hopf bifurcation locus terminates on the fold locus near the cusp.
◦ At this end point the Hopf bifurcation has infinite period .
◦ (The steady state Jacobian has a double zero eigenvalue there.)
◦ (The geometric multiplicity of this eigenvalue is 1 .)
◦ This singular point is called a Takens-Bogdanov (TB) bifurcation.
363
![Page 364: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/364.jpg)
NOTE: continued · · ·
◦ The locus of homoclinic orbits also emanates from the TB point.
◦ Part of the locus of homoclinic orbits follows the fold locus.
◦ These homoclinic orbits are called saddle-node homoclinic orbits .
◦ The stationary point on these homoclinic orbits is a fold point.
◦ (Thus this stationary point has a zero eigenvalue .)
◦ Compare the 2-parameter diagram to the 1-parameter diagrams !
364
![Page 365: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/365.jpg)
The Enzyme Model with Diffusion
Now consider the s-a system with diffusion , namely, the PDE
st = sxx − λ[ρR(s, a)− (s0 − s)] ,
at = βaxx − λ[ρR(s, a)− α(a0 − a)],
with, as before, a0 as a free parameter , and
ρ = 210 , κ1 = 3.4 , κ2 = 0.023 ,and
s0 = 145 , β = 5 , λ = 3 .
Look for traveling waves:
s(x, t) = s(x− ct) , a(x, t) = a(x− ct) .
This reduces the PDE to two coupled ODEs:
s′′ = −cs′ + λ[ρR(s, a)− (s0 − s)] ,
a′′ = − cβa′ + λ
β[ρR(s, a)− α(a0 − a)] .
365
![Page 366: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/366.jpg)
Rewrite the reduced system
s′′ = −cs′ + λ[ρR(s, a)− (s0 − s)] ,
a′′ = − cβa′ + λ
β[ρR(s, a)− α(a0 − a)] ,
as a first order system
s′ = u ,
u′ = −cu + λ[ρR(s, a)− (s0 − s)] ,
a′ = v ,
v′ = − cβv + λ
β[ρR(s, a)− α(a0 − a)] .
The stationary states satisfy
u = v = 0 ,
ρR(s, a)− (s0 − s) = 0 ,
ρR(s, a)− α(a0 − a) = 0 .
366
![Page 367: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/367.jpg)
NOTE:
◦ The stationary states do not depend on the wave speed c.
◦ The stationary states are those of the system without diffusion .
◦ The Jacobian of the stationary states now depends on c.
◦ Thus the Hopf bifurcations need not be present.
◦ However, for large c there must be a Hopf bifurcation.
◦ The Hopf bifurcation approaches the ODE Hopf as c gets large.
◦ Thus there are PDE wave trains for large c , when s0 = 145. .
◦ The ODE homoclinic orbit implies PDE solitary waves for large c.
367
![Page 368: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/368.jpg)
Wave Trains and Solitary Waves
For the first bifurcation diagram for the reduced system we use c = 100 .
Indeed, we find that
◦ The stationary states are those of the system without diffusion .
◦ There is a Hopf bifurcation near the ODE Hopf bifurcation.
◦ The family of periodic orbits indeed ends in a homoclinic orbit .
368
![Page 369: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/369.jpg)
590 600 610 620 630 640 650 660 670a0
0
50
100
150
200L2
-nor
m
Bifurcation diagram of the reduced system for c = 100.
369
![Page 370: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/370.jpg)
From the diagram for c = 100 we can conclude that
◦ The PDE has wave trains of wave speed c = 100 .
◦ The PDE has a solitary wave of wave speed c = 100 .
Note that
◦ Stabilities are different from those for the system without diffusion.
◦ (The diagrams do not show stability now.)
Next:
◦ Are there low speed wave trains and low speed solitary waves ?
◦ To find out we compute the locus of homoclinics of the reduced system .
◦ As free parameters we use a0 and the wave speed c .
370
![Page 371: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/371.jpg)
600 610 620 630 640 650a0
0.0
0.2
0.4
0.6
0.8
1.0W
ave
Spe
ed
Wave speeds
c = 0.4
Wave speeds
c = 0.47
Wave speeds
c = 0.5
Wave speeds
c = 0.6
Wave speeds
c = 0.7
Wave speedsc = 0.747
The locus of solitary waves. (Shown for smaller values of the wave speed.)
371
![Page 372: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/372.jpg)
From the preceding diagram we can draw the following conclusions :
◦ For wave speeds between 0.48 and 0.77 there are three solitary waves .
◦ These are at at different a0-values, but have the same wave speed .
◦ Near a0 = 605 there is a solitary wave of wave speed zero .
◦ (This is a stationary wave .)
◦ The circled special point will be discussed later.
◦ We first show a 1-parameter diagram for c = 0.4 .
372
![Page 373: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/373.jpg)
590 600 610 620 630 640 650 660 670a0
0
50
100
150
200
L2-n
orm
Bifurcation diagram of the reduced system for c = 0.4.The blue branch represents wave trains.
Its homoclinic end point is a solitary wave.Stability is not shown in this diagram.
373
![Page 374: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/374.jpg)
For the diagram for c = 0.4 we note that
◦ There is a fold on the wave train family.
◦ We can compute the locus of such folds for varying a0 and c .
◦ We can also compute the locus of Hopf points for varying a0 and c .
◦ We add these loci to the diagram with the locus of solitary waves .
◦ We also show 1-parameter diagrams for more values of c .
374
![Page 375: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/375.jpg)
635 636 637 638 639 640 641 642 643a0
0.0
0.5
1.0
1.5
2.0W
ave
Spe
ed
PLP
Hom
HB Hom
Homoclinic orbits (solitary waves), Hopf bifurcations, and folds.
375
![Page 376: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/376.jpg)
590 600 610 620 630 640 650 660 670a0
0
50
100
150
200L2
-nor
m
Bifurcation diagram of the reduced system for c = 0.47.
376
![Page 377: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/377.jpg)
590 600 610 620 630 640 650 660 670a0
0
50
100
150
200L2
-nor
m
Bifurcation diagram of the reduced system for c = 0.5.
377
![Page 378: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/378.jpg)
590 600 610 620 630 640 650 660 670a0
0
50
100
150
200L2
-nor
m
Bifurcation diagram of the reduced system for c = 0.6.
378
![Page 379: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/379.jpg)
590 600 610 620 630 640 650 660 670a0
0
50
100
150
200L2
-nor
m
Bifurcation diagram of the reduced system for c = 0.7.
379
![Page 380: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/380.jpg)
590 600 610 620 630 640 650 660 670a0
0
50
100
150
200L2
-nor
m
1
23
Bifurcation diagram of the reduced system for c = 0.747.
380
![Page 381: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/381.jpg)
NOTE:
◦ There is a fold w.r.t. the wave speed c on the solitary wave locus.
◦ (This fold is near c = 0.45 , a0 = 637.2 .)
◦ Two new solitary waves appear at that point.
635 636 637 638 639 640 641 642 643a0
0.0
0.5
1.0
1.5
2.0
Wav
eS
peed
PLP
Hom
HB Hom
381
![Page 382: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/382.jpg)
Now consider the circled point near a0 = 638.5 , c = 0.749 :
◦ It then represents a transition from one solitary wave to another.
◦ In the diagram for c = 0.747 note the near-vertical family.
◦ It is a wave-train family connecting the solitary waves 1 and 3 .
◦ At the section through the circled point the family is exactly vertical .
635 636 637 638 639 640 641 642 643a0
0.0
0.5
1.0
1.5
2.0
Wav
eS
peed
PLP
Hom
HB Hom
382
![Page 383: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/383.jpg)
0.0 0.2 0.4 0.6 0.8 1.0x
0
20
40
60
80
100
120
140
160
a
12
3
Three periodic solutions that represent solitary waves.The independent variable has been scaled to the interval [0,1].
383
![Page 384: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/384.jpg)
Traveling Waves on a Ring
◦ We locate traveling wave solutions of given wave length L .
◦ If necessary we put two or more waves in series to get wave length L .
◦ For the choice L = 22 we found five distinct waves this way.
◦ We can continue these for varying a0 and c , with L = 22 fixed .
◦ This corresponds to computing traveling waves on a ring of size L = 22 .
◦ Results are shown projected onto the a0 − c-plane in the next diagram.
384
![Page 385: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/385.jpg)
600 605 610 615 620 625 630 635 640External Concentration a0
0.0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8W
ave
Spe
ed
7
11
Loci of traveling wave solutions on a ring of size L = 22.
385
![Page 386: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/386.jpg)
Concentration
s
10
20
30
40
50
A traveling wave on the ring. (The solution with label 7.)
386
![Page 387: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/387.jpg)
Stationary Waves on a Ring
◦ There are also families of stationary waves (”patterns ”) on the ring.
◦ Like traveling waves, these are not unique .
◦ (They can be phase shifted , i.e., rotated around the ring).
◦ Such patterns can be found by time-integrating unstable traveling waves.
◦ For example, traveling wave 11 is unstable .
◦ After time integration it approaches a stationary wave .
◦ This stationary wave is shown in the following diagram.
387
![Page 388: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/388.jpg)
Concentration
s
1520253035404550
A stationary wave on the ring, obtained after time integrationof the unstable traveling wave with label 11.
388
![Page 389: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/389.jpg)
NOTE:
◦ Stationary waves can be continued as traveling waves with c = 0 .
◦ A phase condition is necessary in this continuation.
◦ (Because stationary waves can be phase-shifted.)
◦ The next diagram shows a skeleton of the solution structure.
389
![Page 390: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/390.jpg)
MAX
a
25
50
75
100
125
150
175a0
590600
610620
630640
650660
Wave
Speed
0.1
0.2
0.3
0.4
Traveling waves, stationary waves and uniform states
390
![Page 391: Numberical Analysis of Nonlinear Equations](https://reader030.vdocument.in/reader030/viewer/2022020219/55cf9132550346f57b8b6963/html5/thumbnails/391.jpg)
NOTE:
◦ The S-shaped curve with c = 0 represents spatially uniform states .
◦ Other curves in the c = 0 plane represent stationary waves .
◦ Curves that rise above the c = 0 plane represent traveling waves .
◦ Time integration of unstable traveling waves gives other solutions .
◦ For example, waves bouncing off each other are found.
◦ Some of these are stable, while others are transient .
391