16mars2006 NicolasDelanoue,LucJaulin,BertrandCottenceau Stabililityanalysisofanonlinearsystemusingintervalanalysis.

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Tools Lyapunov theory Algorithm to prove stability Future work

Stabilility analysis of a nonlinear system using interval analysis.

Groupe de Travail - M´ethode ensembliste

Nicolas Delanoue, Luc Jaulin, Bertrand Cottenceau

Universit´ e d’Angers - LISA

16 mars 2006

Nicolas Delanoue, Luc Jaulin, Bertrand Cottenceau Stabilility analysis of a nonlinear system using interval analysis.

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Let us consider the following dynamical system : x ˙ = f (x)

x ∈ R ⁿ where f ∈ C ^∞ ( R ⁿ , R ⁿ ). (1)

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x ˙ = f (x)

x ∈ R ⁿ where f ∈ C ^∞ ( R ⁿ , R ⁿ ).

Notation

φ ^. (x) : R → R ⁿ

t 7→ φ ^t (x) denotes the solution of (1) satisfying φ ⁰ (x) = x.

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With [x 0 ] ⊂ R ⁿ ,

Find [x ∞ ] ⊂ [x] ⊂ [x 0 ] such that ∃x _∞ ∈ [x ∞ ] :

t→+∞ ∞

[ x _∞ ] ⊂ [ x ] ⊂ [ x 0 ]

(5)

With [x 0 ] ⊂ R ⁿ ,

Find [x ∞ ] ⊂ [x] ⊂ [x 0 ] such that ∃x _∞ ∈ [x ∞ ] : f (x ∞ ) = 0.

∀x ∈ [x], ∀t ∈ R ⁺ , φ ^t (x) ∈ [x ₀ ].

∀x ∈ [x], φ ^t (x) →

t→+∞ x ∞ .

[ x _∞ ] ⊂ [ x ] ⊂ [ x 0 ]

(6)

With [x 0 ] ⊂ R ⁿ ,

Find [x ∞ ] ⊂ [x] ⊂ [x 0 ] such that ∃x _∞ ∈ [x ∞ ] : f (x ∞ ) = 0.

∀x ∈ [x], ∀t ∈ R ⁺ , φ ^t (x) ∈ [x ₀ ].

[ x _∞ ] ⊂ [ x ] ⊂ [ x 0 ]

(7)

With [x 0 ] ⊂ R ⁿ ,

Find [x ∞ ] ⊂ [x] ⊂ [x 0 ] such that ∃x _∞ ∈ [x ∞ ] : f (x ∞ ) = 0.

∀x ∈ [x], ∀t ∈ R ⁺ , φ ^t (x) ∈ [x ₀ ].

∀x ∈ [x], φ ^t (x) →

t→+∞ x ∞ .

[ x _∞ ] ⊂ [ x ] ⊂ [ x 0 ]

(8)

With [x 0 ] ⊂ R ⁿ ,

Find [x ∞ ] ⊂ [x] ⊂ [x 0 ] such that ∃x _∞ ∈ [x ∞ ] : f (x ∞ ) = 0.

∀x ∈ [x], ∀t ∈ R ⁺ , φ ^t (x) ∈ [x ₀ ] ⇔ φ ^R

⁺

([x]) ⊂ [x ₀ ]

∀x ∈ [x], φ ^t (x) →

t→+∞ x ∞ . ⇔ φ ^∞ ([x]) = {x _∞ }

[ x _∞ ] ⊂ [ x ] ⊂ [ x 0 ]

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With [x 0 ] ⊂ R ⁿ ,

Find [x ∞ ] ⊂ [x] ⊂ [x 0 ] such that ∃x _∞ ∈ [x ∞ ] :

f (x ∞ ) = 0 (Uniqueness - Interval Newton algorithm) φ ^R

⁺

([x]) ⊂ [x ₀ ] (Stability - Lyapunov)

φ ^∞ ([x]) = {x _∞ } (Convergence - Lyapunov)

[ x _∞ ] ⊂ [ x ] ⊂ [ x ₀ ]

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With [x 0 ] ⊂ R ⁿ ,

Find [x ∞ ] ⊂ [x] ⊂ [x 0 ] such that ∃x _∞ ∈ [x ∞ ] :

f (x ∞ ) = 0 (Uniqueness - Interval Newton algorithm)

[ x _∞ ] ⊂ [ x ] ⊂ [ x ₀ ]

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With [x 0 ] ⊂ R ⁿ ,

Find [x ∞ ] ⊂ [x] ⊂ [x 0 ] such that ∃x _∞ ∈ [x ∞ ] :

f (x ∞ ) = 0 (Uniqueness - Interval Newton algorithm) φ ^R

⁺

([x]) ⊂ [x ₀ ] (Stability - Lyapunov)

φ ^∞ ([x]) = {x _∞ } (Convergence - Lyapunov)

[ x _∞ ] ⊂ [ x ] ⊂ [ x ₀ ]

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With [x 0 ] ⊂ R ⁿ ,

Find [x ∞ ] ⊂ [x] ⊂ [x 0 ] such that ∃x _∞ ∈ [x ∞ ] :

f (x ∞ ) = 0 (Uniqueness - Interval Newton algorithm) φ ^R

⁺

([x]) ⊂ [x ₀ ] (Stability - Lyapunov)

φ ^∞ ([x]) = {x _∞ } (Convergence - Lyapunov)

[ x _∞ ] ⊂ [ x ] ⊂ [ x ₀ ]

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Outline

1 Tools

Interval Newton method Positivity

2 Lyapunov theory

Definitions of stability Lyapunov function The linear case

3 Algorithm to prove stability Algorithm

Example

4 Future work

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Interval Newton method Positivity

Newton-Raphson Method

Output

An approximation of a solution of f (x) = 0 where x ∈ [x] where f ∈ C ^∞ (R ⁿ , R ⁿ ).

n+1 n n n

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Newton-Raphson Method

Output

An approximation of a solution of f (x) = 0 where x ∈ [x] where f ∈ C ^∞ (R ⁿ , R ⁿ ).

Algorithm

Initialization x ₀

x n+1 = x n − Df ⁻¹ (x n )f (x n )

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Interval Newton Method

Output

A box enclosing the solution of f (x) = 0.

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Interval Newton Method

Output

A box enclosing the solution of f (x) = 0.

A flag indicating the uniqueness and the existence of the solution.

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Algorithm [x] ₀ = [x]

[x] n+1 = [x] n ∩ ρ x

1

([x] n )

where ρ _x

₁

: [x] → R ⁿ with ρ _x

₁

(x) = x ₁ − Df ⁻¹ (x)f (x ₁ ) and x 1 ∈ [x] n

[x]

₀

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Algorithm [x] ₀ = [x]

[x] n+1 = [x] n ∩ ρ x

1

([x] n )

where ρ _x

₁

: [x] → R ⁿ with ρ _x

₁

(x) = x ₁ − Df ⁻¹ (x)f (x ₁ ) and x 1 ∈ [x] n

[x]

₀

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Algorithm [x] ₀ = [x]

[x] n+1 = [x] n ∩ ρ x

1

([x] n )

where ρ _x

₁

: [x] → R ⁿ with ρ _x

₁

(x) = x ₁ − Df ⁻¹ (x)f (x ₁ ) and x 1 ∈ [x] n

[x]

₀

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Algorithm [x] ₀ = [x]

[x] n+1 = [x] n ∩ ρ x

1

([x] n )

where ρ _x

₁

: [x] → R ⁿ with ρ _x

₁

(x) = x ₁ − Df ⁻¹ (x)f (x ₁ ) and x 1 ∈ [x] n

[x]

₀

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Algorithm [x] ₀ = [x]

[x] n+1 = [x] n ∩ ρ x

1

([x] n )

where ρ _x

₁

: [x] → R ⁿ with ρ _x

₁

(x) = x ₁ − Df ⁻¹ (x)f (x ₁ ) and x 1 ∈ [x] n

[x]

₀

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Algorithm [x] ₀ = [x]

[x] n+1 = [x] n ∩ ρ x

1

([x] n )

where ρ _x

₁

: [x] → R ⁿ with ρ _x

₁

(x) = x ₁ − Df ⁻¹ (x)f (x ₁ ) and x 1 ∈ [x] n

[x]

₁

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Properties

Let f ∈ C ^∞ ( R ⁿ , R ⁿ ), x ₁ ∈ [x]. If 0 6∈ det(Df ([x])) then

1

x ^∗ ∈ [x], f (x ^∗ ) = 0 ⇒ x ^∗ ∈ ρ _x

₁

([x])

2

ρ _x

₁

([x]) ⊂ [x] ⇒ ∃!x ^∗ ∈ [x], f (x ^∗ ) = 0

[x]

₁

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Positivity

A flag indicating that f ([x]) ≥ 0.

3 cases :

Interval analysis is good for ∀x ∈ [x], f (x) > 0 Algebra calculus is good for ∀x ∈ [x], f (x) = 0. Otherwise.

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Positivity

A flag indicating that f ([x]) ≥ 0.

3 cases :

Otherwise.

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Positivity

A flag indicating that f ([x]) ≥ 0.

3 cases :

Interval analysis is good for ∀x ∈ [x], f (x) > 0

Algebra calculus is good for ∀x ∈ [x], f (x) = 0. Otherwise.

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Positivity

A flag indicating that f ([x]) ≥ 0.

3 cases :

Interval analysis is good for ∀x ∈ [x], f (x) > 0

Algebra calculus is good for ∀x ∈ [x], f (x) = 0.

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Positivity

A flag indicating that f ([x]) ≥ 0.

3 cases :

Interval analysis is good for ∀x ∈ [x], f (x) > 0 Algebra calculus is good for ∀x ∈ [x], f (x) = 0.

Otherwise.

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x

y

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Algebra calculus is not enough ...

Minimazing polynomial function.

Interval analysis is not enough ... In general, one only has :

f ([x]) & [f ]([x]).

multiple occurrence of variables. outward rounding.

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Algebra calculus is not enough ...

Minimazing polynomial function.

Interval analysis is not enough ...

In general, one only has :

f ([x]) & [f ]([x]).

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Algebra calculus is not enough ...

Minimazing polynomial function.

Interval analysis is not enough ...

In general, one only has :

f ([x]) & [f ]([x]).

multiple occurrence of variables.

outward rounding.

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Algebra calculus is not enough ...

Minimazing polynomial function.

Interval analysis is not enough ...

In general, one only has :

f ([x]) & [f ]([x]).

multiple occurrence of variables.

outward rounding.

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x y

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x y

Interval analysis

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x y

Algebra calculus and Interval Analysis

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Theorem

Let x ₀ ∈ E where E is a convex set of R ⁿ , and f ∈ C ² ( R ⁿ , R ). We have the following implication :

1

∃x ₀ such that f (x ₀ ) = 0 and Df (x ₀ ) = 0.

2

∀x ∈ E , D ² f (x ) > 0.

then ∀x ∈ E , f (x) ≥ 0.

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Example

Let us prove that f (x) ≥ 0, ∀x ∈ [−1/2, 1/2] ² where f : R ² → R is defined by

f (x, y) = − cos(x ² + √

2 sin ² y) + x ² + y ² + 1.

0 2 4 6 8 10 12 14 16 18 20

Z

−2 −3 0 −1 2 1

4 3 X

−3

−2−1

0 1

2 3

Y 4

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Example

Let f : R ² → R be the function defined by : f (x, y) = − cos(x ² + √

2 sin ² y) + x ² + y ² + 1.

1

One has : f (0, 0) = 0 and ∇f (0, 0) = 0

∇f (x, y) =

2x(sin(x ² + √

2 sin ² y ) + 1) 2 √

2 cos y sin y sin √

2 sin ² y + x ² + 2y

.

0 2 4 6 8 10 12 14 16 18 20

Z

−2 −3 0 −1 2 1

4 3 X

−3−2

−1 01

2

34

Y

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∇ ² f =

a _1,1 a _1,2 a 2,1 a 2,2

=

∂

²

f

∂x

²

∂

²

f

∂x∂y

∂

²

f

∂y∂x

∂

²

f

∂y

²

!

a 1,1 = 2 sin √

2 sin ² y + x ² +4x ² cos √

2 sin ² y + x ² + 2.

a _2,2 = −2 √

2 sin ² y sin( √

2 sin ² y + x ² ) +2 √

2 cos ² y sin( √

2 sin ² y + x ² ) +8 cos ² y sin ² y cos( √

2 sin ² y + x ² ) + 2.

a _1,2 = a _2,1 = = 4 √

2x cos y sin y cos √

2 sin y ² + x ² .

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Interval analysis : ∀x ∈ [−1/2, 1/2] ² , ∇ ² f (x) ⊂ [A]

[A] =

[1.9, 4.1] [−1.3, 1.4]

[−1.3, 1.4] [1.9, 5.4]

. It remains to check : ∀A ∈ [A], A is positive.

∀x ∈ R ⁿ − {0}, x ^T Ax > 0

The set of positive definite symmetric n × n matrices is denoted by

S ⁿ⁺ .

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Interval analysis : ∀x ∈ [−1/2, 1/2] ² , ∇ ² f (x) ⊂ [A]

[A] =

[1.9, 4.1] [−1.3, 1.4]

[−1.3, 1.4] [1.9, 5.4]

.

It remains to check : ∀A ∈ [A], A is positive.

Definition

A symmetric matrix A is positive definite if

∀x ∈ R ⁿ − {0}, x ^T Ax > 0

The set of positive definite symmetric n × n matrices is denoted by S ⁿ⁺ .

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Definition

A set of symmetric matrices [A] is an interval symmetric matrix if : [A] = {(a _ij ) ij , a ij = a ji , a ij ∈ [a] ij }

i.e.

[A, A] =

A symmetric, A ≤ A ≤ A .

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Example

Working in R ² , a symmetric matrix A A =

a _1,1 a _1,2 a _1,2 a _2,2

a

₁₁

a

₂₂

a

₁₂

A A A

c

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Remark - Rohn

Let V ([A]) denotes the finite set of corners of [A]. S ⁿ⁺ and [A] are convex subsets of S ⁿ :

[A] ⊂ S ⁿ⁺ ⇔ V ([A]) ⊂ S ⁿ⁺

S ⁿ is a vector space of dimension ⁿ⁽ⁿ⁺¹⁾ ₂ . #V ([A]) = 2

ⁿ⁽ⁿ⁺¹⁾²

.

a_1,1axis

a_2,2axis a_1,2axis

[A]

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Theorem - Adefeld

Let [A] be an interval symmetric matrix.

and C = {z ∈ R ⁿ tel que |z _i | = 1}

If ∀z ∈ C , A z = A c + Diag(z)∆Diag(z ) is positive definite.

then [A] is positive definite.

a_1,1axis

a_2,2axis a₁,2axis

[A]

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Let us consider the following dynamical system : x ˙ = f (x)

x ∈ R ⁿ where f ∈ C ^∞ ( R ⁿ , R ⁿ ).

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Definitions of stability Lyapunov function The linear case

Definition

Let x ∈ R , x is an equilibrium state if : f (x) = 0

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Definition

A set D is stable if :

φ ^R

⁺

(D) ⊂ D

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Definition

A set D is stable if :

φ ^R

⁺

(D) ⊂ D

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Non stable example

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Definition

An equilibrium state x ∞ is asymptotically (D, D 0 )-stable if

φ ^R

⁺

(D) ⊂ D 0

φ ^∞ (D) = {x _∞ }

D ⊂ D ⁰

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Definition

An equilibrium state x ∞ is asymptotically (D, D 0 )-stable if

φ ^R

⁺

(D) ⊂ D 0

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Definition

An equilibrium state x ∞ is asymptotically (D, D 0 )-stable if φ ^R

⁺

(D) ⊂ D 0

φ ^∞ (D) = {x _∞ }

D ⊂ D ⁰

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Definition

One says that a function L : R ⁿ → R is Lyapunov for (1) if :

h∇L(x ), f (x)i < 0, ∀x ∈ D − {x _∞ } With V : t 7→ L(x(t)), one has :

d

dt V (t) = _dt ^d (L(x(t)))

= _dx ^d L · _dt ^d x(t)

= h∇L(x), f (x(t))i < 0

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Definition

One says that a function L : R ⁿ → R is Lyapunov for (1) if :

1

L(x ) = 0 ⇔ x = x ∞

2

x ∈ D − {x _∞ } ⇒ L(x) > 0

3

h∇L(x ), f (x)i < 0, ∀x ∈ D − {x _∞ } With V : t 7→ L(x(t)), one has :

d

dt V (t) = _dt ^d (L(x(t)))

= _dx ^d L · _dt ^d x(t)

= h∇L(x), f (x(t))i < 0

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Definition

One says that a function L : R ⁿ → R is Lyapunov for (1) if :

1

L(x ) = 0 ⇔ x = x ∞

2

x ∈ D − {x _∞ } ⇒ L(x) > 0

d

dt V (t) = _dt ^d (L(x(t)))

= _dx ^d L · _dt ^d x(t)

= h∇L(x), f (x(t))i < 0

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Definition

One says that a function L : R ⁿ → R is Lyapunov for (1) if :

1

L(x ) = 0 ⇔ x = x ∞

2

x ∈ D − {x _∞ } ⇒ L(x) > 0

3

h∇L(x ), f (x)i < 0, ∀x ∈ D − {x _∞ }

With V : t 7→ L(x(t)), one has :

d

dt V (t) = _dt ^d (L(x(t)))

= _dx ^d L · _dt ^d x(t)

= h∇L(x), f (x(t))i < 0

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Definition

One says that a function L : R ⁿ → R is Lyapunov for (1) if :

1

L(x ) = 0 ⇔ x = x ∞

2

x ∈ D − {x _∞ } ⇒ L(x) > 0

3

h∇L(x ), f (x)i < 0, ∀x ∈ D − {x _∞ } With V : t 7→ L(x(t)), one has :

d

dt V (t) = _dt ^d (L(x(t)))

= _dx ^d L · _dt ^d x(t)

= h∇L(x), f (x(t))i < 0

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Lyapunov’s Theorem

Let D ⁰ be a compact subset of R ⁿ and x ∞ in the interior of D ⁰ . If L : D ⁰ → R is Lyapunov for (1) then

there exists a subset D of D ⁰ such that the equilibrium state x ∞ is

asymptotically D, D ⁰ -stable.

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For linear systems :

˙

x = Ax (2)

One usually takes L = x ^T Wx where W ∈ S ⁿ . and h∇L(x), f (x)i = x ^T (A ^T W + WA)x.

Lyapunov conditions translate into

1

W ∈ S ⁿ⁺ .

2

−(A ^T W + WA) ∈ S ⁿ⁺ .

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For linear systems :

˙

x = Ax (2)

One usually takes L = x ^T Wx where W ∈ S ⁿ . and h∇L(x), f (x)i = x ^T (A ^T W + WA)x.

Lyapunov conditions translate into

1

W ∈ S ⁿ⁺ .

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For linear systems :

˙

x = Ax (2)

One usually takes L = x ^T Wx where W ∈ S ⁿ . and h∇L(x), f (x)i = x ^T (A ^T W + WA)x.

Lyapunov conditions translate into

1

W ∈ S ⁿ⁺ .

2

−(A ^T W + WA) ∈ S ⁿ⁺ .

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For ˙ x = Ax , to find a Lyapunov function for 2, one solves the Lyapunov equation of unknown

A ^T W + WA = −I

and we check that W ∈ S ⁿ⁺ .

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Theorem

The system ˙ x = Ax is asymptotically stable is equivalent to for all Q ∈ S ⁿ⁺ , the matrix W solution of

A ^T W + WA = −Q is positive definite.

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Example The system

x ˙

˙ y

=

−1 0

1 −1

x y

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Algorithm Example

Algorithm

1

Prove that [x 0 ] contains an unique equilibrium state x ∞ .

2

Find [x ∞ ] ⊂ [x 0 ] which contains x ∞ .

3

Linearize the system around an approximation ˜ x ∞ .

4

Find a Lyapunov function L _x

_∞

for the linearized system.

5

Check that L _x

_∞

is also Lyapunov for ˙ x = f (x).

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Algorithm Example

Algorithm

1

Prove that [x 0 ] contains an unique equilibrium state x ∞ .

2

Find [x ∞ ] ⊂ [x 0 ] which contains x ∞ .

x

∞

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Algorithm Example

Algorithm

1

Prove that [x 0 ] contains an unique equilibrium state x ∞ .

2

Find [x ∞ ] ⊂ [x 0 ] which contains x ∞ .

3

Linearize the system around an approximation ˜ x ∞ .

4

Find a Lyapunov function L _x

_∞

for the linearized system.

5

Check that L _x

_∞

is also Lyapunov for ˙ x = f (x).

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Algorithm Example

Algorithm

1

Prove that [x 0 ] contains an unique equilibrium state x ∞ .

2

Find [x ∞ ] ⊂ [x 0 ] which contains x ∞ .

3

Linearize the system around an approximation ˜ x ∞ .

4

Find a Lyapunov function L _x

_∞

for the linearized system.

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Algorithm Example

Algorithm

1

Prove that [x 0 ] contains an unique equilibrium state x ∞ .

2

Find [x ∞ ] ⊂ [x 0 ] which contains x ∞ .

3

Linearize the system around an approximation ˜ x ∞ .

4

Find a Lyapunov function L _x

_∞

for the linearized system.

5

Check that L _x

_∞

is also Lyapunov for ˙ x = f (x).

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Algorithm Example

Explanations

Step 4 : L x

∞

(x) = (x − x ∞ ) ^T W x ˜

∞

(x − x ∞ ) Step 5 : It remains to check :

g x

∞

(x) = −h∇L _x

_∞

(x), f (x)i ≥ 0

x

∞

According to theorem of positivity, one only has to check that

∇ ² g x

∞

([x 0 ]) ⊂ S ⁿ⁺

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Algorithm Example

Explanations

Step 4 : L x

∞

(x) = (x − x ∞ ) ^T W x ˜

∞

(x − x ∞ ) Step 5 : It remains to check :

g x

∞

(x) = −h∇L _x

_∞

(x), f (x)i ≥ 0 One has :

g (x ∞ ) = 0

∇g _x

_∞

(x ∞ ) = 0

According to theorem of positivity, one only has to check that

∇ ² g x

∞

([x 0 ]) ⊂ S ⁿ⁺

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x ˙ 1

˙ x ₂

=

−x ₂ x ₁ − (1 − x ₁ ² )x ₂

where [x ₀ ] = [−0.6, 0.6] ² .

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Future work

To combine : These results.

Guaranteed integration of ODE.

Graph theory.

to compute a guaranteed approximation of the attraction domain of x

_∞

16mars2006 NicolasDelanoue,LucJaulin,BertrandCottenceau Stabililityanalysisofanonlinearsystemusingintervalanalysis.

Stabilility analysis of a nonlinear system using interval analysis.

Nicolas Delanoue, Luc Jaulin, Bertrand Cottenceau

Universit´ e d’Angers - LISA

16 mars 2006

Let us consider the following dynamical system : x ˙ = f (x)

x ∈ R n where f ∈ C ∞ ( R n , R n ). (1)

x ˙ = f (x)

x ∈ R n where f ∈ C ∞ ( R n , R n ).

Notation

φ . (x) : R → R n

t 7→ φ t (x) denotes the solution of (1) satisfying φ 0 (x) = x.

With [x 0 ] ⊂ R n ,

Find [x ∞ ] ⊂ [x] ⊂ [x 0 ] such that ∃x ∞ ∈ [x ∞ ] :

t→+∞ ∞

[ x ∞ ] ⊂ [ x ] ⊂ [ x 0 ]

With [x 0 ] ⊂ R n ,

Find [x ∞ ] ⊂ [x] ⊂ [x 0 ] such that ∃x ∞ ∈ [x ∞ ] : f (x ∞ ) = 0.

∀x ∈ [x], ∀t ∈ R + , φ t (x) ∈ [x 0 ].

∀x ∈ [x], φ t (x) →

t→+∞ x ∞ .

[ x ∞ ] ⊂ [ x ] ⊂ [ x 0 ]

With [x 0 ] ⊂ R n ,

Find [x ∞ ] ⊂ [x] ⊂ [x 0 ] such that ∃x ∞ ∈ [x ∞ ] : f (x ∞ ) = 0.

∀x ∈ [x], ∀t ∈ R + , φ t (x) ∈ [x 0 ].

[ x ∞ ] ⊂ [ x ] ⊂ [ x 0 ]

With [x 0 ] ⊂ R n ,

Find [x ∞ ] ⊂ [x] ⊂ [x 0 ] such that ∃x ∞ ∈ [x ∞ ] : f (x ∞ ) = 0.

∀x ∈ [x], ∀t ∈ R + , φ t (x) ∈ [x 0 ].

∀x ∈ [x], φ t (x) →

t→+∞ x ∞ .

[ x ∞ ] ⊂ [ x ] ⊂ [ x 0 ]

With [x 0 ] ⊂ R n ,

Find [x ∞ ] ⊂ [x] ⊂ [x 0 ] such that ∃x ∞ ∈ [x ∞ ] : f (x ∞ ) = 0.

∀x ∈ [x], ∀t ∈ R + , φ t (x) ∈ [x 0 ] ⇔ φ R

([x]) ⊂ [x 0 ]

∀x ∈ [x], φ t (x) →

t→+∞ x ∞ . ⇔ φ ∞ ([x]) = {x ∞ }

[ x ∞ ] ⊂ [ x ] ⊂ [ x 0 ]

With [x 0 ] ⊂ R n ,

Find [x ∞ ] ⊂ [x] ⊂ [x 0 ] such that ∃x ∞ ∈ [x ∞ ] :

f (x ∞ ) = 0 (Uniqueness - Interval Newton algorithm) φ R

([x]) ⊂ [x 0 ] (Stability - Lyapunov)

φ ∞ ([x]) = {x ∞ } (Convergence - Lyapunov)

[ x ∞ ] ⊂ [ x ] ⊂ [ x 0 ]

With [x 0 ] ⊂ R n ,

Find [x ∞ ] ⊂ [x] ⊂ [x 0 ] such that ∃x ∞ ∈ [x ∞ ] :

f (x ∞ ) = 0 (Uniqueness - Interval Newton algorithm)

[ x ∞ ] ⊂ [ x ] ⊂ [ x 0 ]

With [x 0 ] ⊂ R n ,

Find [x ∞ ] ⊂ [x] ⊂ [x 0 ] such that ∃x ∞ ∈ [x ∞ ] :

f (x ∞ ) = 0 (Uniqueness - Interval Newton algorithm) φ R

([x]) ⊂ [x 0 ] (Stability - Lyapunov)

φ ∞ ([x]) = {x ∞ } (Convergence - Lyapunov)

[ x ∞ ] ⊂ [ x ] ⊂ [ x 0 ]

With [x 0 ] ⊂ R n ,

Find [x ∞ ] ⊂ [x] ⊂ [x 0 ] such that ∃x ∞ ∈ [x ∞ ] :

f (x ∞ ) = 0 (Uniqueness - Interval Newton algorithm) φ R

([x]) ⊂ [x 0 ] (Stability - Lyapunov)

φ ∞ ([x]) = {x ∞ } (Convergence - Lyapunov)

[ x ∞ ] ⊂ [ x ] ⊂ [ x 0 ]

Outline

1 Tools

Interval Newton method Positivity

2 Lyapunov theory

Definitions of stability Lyapunov function The linear case

3 Algorithm to prove stability Algorithm

Example

4 Future work

Newton-Raphson Method

Output

An approximation of a solution of f (x) = 0 where x ∈ [x] where f ∈ C ∞ (R n , R n ).

n+1 n n n

Newton-Raphson Method

Output

An approximation of a solution of f (x) = 0 where x ∈ [x] where f ∈ C ∞ (R n , R n ).

Algorithm

Initialization x 0

x n+1 = x n − Df −1 (x n )f (x n )

Interval Newton Method

x ∈ R ⁿ where f ∈ C ^∞ ( R ⁿ , R ⁿ ). (1)

x ∈ R ⁿ where f ∈ C ^∞ ( R ⁿ , R ⁿ ).

φ ^. (x) : R → R ⁿ

t 7→ φ ^t (x) denotes the solution of (1) satisfying φ ⁰ (x) = x.

With [x 0 ] ⊂ R ⁿ ,

Find [x ∞ ] ⊂ [x] ⊂ [x 0 ] such that ∃x _∞ ∈ [x ∞ ] :

[ x _∞ ] ⊂ [ x ] ⊂ [ x 0 ]

With [x 0 ] ⊂ R ⁿ ,

Find [x ∞ ] ⊂ [x] ⊂ [x 0 ] such that ∃x _∞ ∈ [x ∞ ] : f (x ∞ ) = 0.

∀x ∈ [x], ∀t ∈ R ⁺ , φ ^t (x) ∈ [x ₀ ].

∀x ∈ [x], φ ^t (x) →

[ x _∞ ] ⊂ [ x ] ⊂ [ x 0 ]

With [x 0 ] ⊂ R ⁿ ,

Find [x ∞ ] ⊂ [x] ⊂ [x 0 ] such that ∃x _∞ ∈ [x ∞ ] : f (x ∞ ) = 0.

∀x ∈ [x], ∀t ∈ R ⁺ , φ ^t (x) ∈ [x ₀ ].

[ x _∞ ] ⊂ [ x ] ⊂ [ x 0 ]

With [x 0 ] ⊂ R ⁿ ,

Find [x ∞ ] ⊂ [x] ⊂ [x 0 ] such that ∃x _∞ ∈ [x ∞ ] : f (x ∞ ) = 0.

∀x ∈ [x], ∀t ∈ R ⁺ , φ ^t (x) ∈ [x ₀ ].

∀x ∈ [x], φ ^t (x) →

[ x _∞ ] ⊂ [ x ] ⊂ [ x 0 ]

With [x 0 ] ⊂ R ⁿ ,

Find [x ∞ ] ⊂ [x] ⊂ [x 0 ] such that ∃x _∞ ∈ [x ∞ ] : f (x ∞ ) = 0.

∀x ∈ [x], ∀t ∈ R ⁺ , φ ^t (x) ∈ [x ₀ ] ⇔ φ ^R

([x]) ⊂ [x ₀ ]

∀x ∈ [x], φ ^t (x) →

t→+∞ x ∞ . ⇔ φ ^∞ ([x]) = {x _∞ }

[ x _∞ ] ⊂ [ x ] ⊂ [ x 0 ]

With [x 0 ] ⊂ R ⁿ ,

Find [x ∞ ] ⊂ [x] ⊂ [x 0 ] such that ∃x _∞ ∈ [x ∞ ] :

f (x ∞ ) = 0 (Uniqueness - Interval Newton algorithm) φ ^R

([x]) ⊂ [x ₀ ] (Stability - Lyapunov)

φ ^∞ ([x]) = {x _∞ } (Convergence - Lyapunov)

[ x _∞ ] ⊂ [ x ] ⊂ [ x ₀ ]

With [x 0 ] ⊂ R ⁿ ,

Find [x ∞ ] ⊂ [x] ⊂ [x 0 ] such that ∃x _∞ ∈ [x ∞ ] :

[ x _∞ ] ⊂ [ x ] ⊂ [ x ₀ ]

With [x 0 ] ⊂ R ⁿ ,

Find [x ∞ ] ⊂ [x] ⊂ [x 0 ] such that ∃x _∞ ∈ [x ∞ ] :

f (x ∞ ) = 0 (Uniqueness - Interval Newton algorithm) φ ^R

([x]) ⊂ [x ₀ ] (Stability - Lyapunov)

φ ^∞ ([x]) = {x _∞ } (Convergence - Lyapunov)

[ x _∞ ] ⊂ [ x ] ⊂ [ x ₀ ]

With [x 0 ] ⊂ R ⁿ ,

Find [x ∞ ] ⊂ [x] ⊂ [x 0 ] such that ∃x _∞ ∈ [x ∞ ] :

f (x ∞ ) = 0 (Uniqueness - Interval Newton algorithm) φ ^R

([x]) ⊂ [x ₀ ] (Stability - Lyapunov)

φ ^∞ ([x]) = {x _∞ } (Convergence - Lyapunov)

[ x _∞ ] ⊂ [ x ] ⊂ [ x ₀ ]

An approximation of a solution of f (x) = 0 where x ∈ [x] where f ∈ C ^∞ (R ⁿ , R ⁿ ).

An approximation of a solution of f (x) = 0 where x ∈ [x] where f ∈ C ^∞ (R ⁿ , R ⁿ ).

Initialization x ₀

x n+1 = x n − Df ⁻¹ (x n )f (x n )

Algorithm [x] ₀ = [x]

where ρ _x

: [x] → R ⁿ with ρ _x

(x) = x ₁ − Df ⁻¹ (x)f (x ₁ ) and x 1 ∈ [x] n

Algorithm [x] ₀ = [x]

where ρ _x

: [x] → R ⁿ with ρ _x

(x) = x ₁ − Df ⁻¹ (x)f (x ₁ ) and x 1 ∈ [x] n

Algorithm [x] ₀ = [x]

where ρ _x

: [x] → R ⁿ with ρ _x

(x) = x ₁ − Df ⁻¹ (x)f (x ₁ ) and x 1 ∈ [x] n

Algorithm [x] ₀ = [x]

where ρ _x

: [x] → R ⁿ with ρ _x

(x) = x ₁ − Df ⁻¹ (x)f (x ₁ ) and x 1 ∈ [x] n

Algorithm [x] ₀ = [x]

where ρ _x

: [x] → R ⁿ with ρ _x

(x) = x ₁ − Df ⁻¹ (x)f (x ₁ ) and x 1 ∈ [x] n

Algorithm [x] ₀ = [x]

where ρ _x

: [x] → R ⁿ with ρ _x

(x) = x ₁ − Df ⁻¹ (x)f (x ₁ ) and x 1 ∈ [x] n

Let f ∈ C ^∞ ( R ⁿ , R ⁿ ), x ₁ ∈ [x]. If 0 6∈ det(Df ([x])) then

x ^∗ ∈ [x], f (x ^∗ ) = 0 ⇒ x ^∗ ∈ ρ _x

ρ _x