The infinite horizon positive LQ-problem for linear continuous time - - PowerPoint PPT Presentation

The infinite horizon positive LQ-problem for linear continuous time systems

Charlotte Beauthier1 Joint work with M. Laabissi2 and J. J. Winkin1

1University of Namur (FUNDP) – Belgium 2University Chouaib Doukkali – Morocco

CESAME Seminar UCL, Louvain-La-Neuve May 23, 2006

• Ch. Beauthier

The positive LQ-problem for linear continuous time systems

Motivations Problem Statement Stable Case Unstable Case Conclusion and future work

Outline

1

Motivations

2

Problem Statement Preliminary Concepts Problem

3

Stable Case Main Results Compartmental systems Design Methodology for Q and R

4

Unstable Case

5

Conclusion and future work

• Ch. Beauthier

The positive LQ-problem for linear continuous time systems

Motivations Problem Statement Stable Case Unstable Case Conclusion and future work

Outline

1

Motivations

2

Problem Statement

3

Stable Case

4

Unstable Case

5

Conclusion and future work

• Ch. Beauthier

The positive LQ-problem for linear continuous time systems

Motivations Problem Statement Stable Case Unstable Case Conclusion and future work

Motivations

◮ The class of positive linear time-invariant systems

→ Dynamical models where all the variables should remain

nonnegative.

→ Large class of applications : see [Farina, Rinaldi, 2000]

Economics models Chemical processes Compartmental systems Biological systems

• L. Farina, S. Rinaldi

Positive Linear Systems : Theory and applications Wiley, New York, 2000.

• Ch. Beauthier

The positive LQ-problem for linear continuous time systems

Motivations Problem Statement Stable Case Unstable Case Conclusion and future work

Motivations

◮ Feedback control law for a positive system

→ AIM : Keeping positivity for the closed-loop system. Stabilizability and holdability problem.

• Ch. Beauthier

The positive LQ-problem for linear continuous time systems

Motivations Problem Statement Stable Case Unstable Case Conclusion and future work

Motivations

• A. Berman, M. Neumann, R. J. Stern

Nonnegative matrices in dynamic systems John Wiley and Sons, 1989.

→ Existence of state feedback.

• S. Boyd, L. El Ghaoui, E. Feron, V. Balakrishnan

Linear matrix inequalities in system and control theory Society for industrial and applied mathematicals, Philadelphia, 1994.

→ Compute of state feedback LMI.

• Ch. Beauthier

The positive LQ-problem for linear continuous time systems

Motivations Problem Statement Stable Case Unstable Case Conclusion and future work

Motivations

◮ Feedback of LQ type

→ POSITIVE LQ-PROBLEM :

Given a positive system, ? Conditions ? such that the resulting LQ-optimal closed-loop system is positive ? Design methodology for choosing the weighting matrices in the quadratic cost ?

• Ch. Beauthier

The positive LQ-problem for linear continuous time systems

Motivations Problem Statement Stable Case Unstable Case Conclusion and future work

Motivations

◮ Positive optimal control

• W. P

. M. H. Heemels, S. J. L. Van Eijndhoven, A. A. Stoorvogel Linear quadratic regulator problem with positive controls

• Int. J. Control, Vol. 70, No. 4, 551-578, 1998.
• Ch. Beauthier

The positive LQ-problem for linear continuous time systems

Motivations Problem Statement Stable Case Unstable Case Conclusion and future work

Motivations

◮ Iterative scheme for the solution of the Riccati equation and unsigned matrices.

Chun-hua Guo and A. J. Laub On a Newton-like Method for Solving Algebraic Riccati Equations SIAM J. Matrix Anal. Appl., Vol. 21, No. 2, 2000. Chun-hua Guo and A. J. Laub On the iterative solution of a class of nonsymmetric algebraic Riccati equations SIAM J. Matrix Anal. Appl., Vol. 22, No. 2, 2000.

• Ch. Beauthier

The positive LQ-problem for linear continuous time systems

Motivations Problem Statement Stable Case Unstable Case Conclusion and future work

Motivations

◮ Nash Riccati equation and positive games theory.

• G. Jank, D. Kremer

Open loop Nash games and positive systems - Solvability conditions for non symmetric Riccati equations Lehrstuhl II für Mathematik, RWTH Aachen, Germany, May 2004.

• Ch. Beauthier

The positive LQ-problem for linear continuous time systems

Motivations Problem Statement Stable Case Unstable Case Conclusion and future work Preliminary Concepts Problem

Outline

1

Motivations

2

Problem Statement Preliminary Concepts Problem

3

Stable Case

4

Unstable Case

5

Conclusion and future work

• Ch. Beauthier

The positive LQ-problem for linear continuous time systems

Motivations Problem Statement Stable Case Unstable Case Conclusion and future work Preliminary Concepts Problem

Preliminary Concepts - Positive Systems

Consider the following linear time-invariant system description, denoted by R = [A, B ] : ˙ x(t) = A x(t) + B u(t), x(0) = x0 ◮ Definitions of positive system

• A linear system R = [A, B ] is said to be positive if

∀ x0 ≥ 0, ∀ u(t) ≥ 0, ∀ t ≥ 0 : x(t) ≥ 0

• A linear system ˙

x = A x(t) is said to be positive if ∀ x0 ≥ 0, ∀ t ≥ 0 : x(t) ≥ 0.

• Ch. Beauthier

The positive LQ-problem for linear continuous time systems

Motivations Problem Statement Stable Case Unstable Case Conclusion and future work Preliminary Concepts Problem

Preliminary Concepts - Positive Systems

◮ A well-known characterization of positive systems THM : A linear system R = [A, B ] is positive ⇔ A is a Metzler matrix and B ≥ 0. where

• A is a Metzler matrix

⇔ ∀ i = j : aij ≥ 0.

• B is a nonnegative matrix, denoted by B ≥ 0,

⇔ ∀ i, j : bij ≥ 0.

• Ch. Beauthier

The positive LQ-problem for linear continuous time systems

Motivations Problem Statement Stable Case Unstable Case Conclusion and future work Preliminary Concepts Problem

Preliminary Concepts - M-Matrix

• A is a Z-matrix if

∀ i = j : aij ≤ 0,

→

−A is a Metzler matrix.

→

A is a Z-matrix ⇔ ∃ s ∈ IR, B ≥ 0 such that A = s I − B.

See e.g. [Bapat, Raghavan, 1997].

• R. B. Bapat, T. E. S. Raghavan

Nonnegative matrices and applications Encyclopedia of mathematics and its applications 64, Cambridge university press, 1997.

• Ch. Beauthier

The positive LQ-problem for linear continuous time systems

Motivations Problem Statement Stable Case Unstable Case Conclusion and future work Preliminary Concepts Problem

Preliminary Concepts - M-Matrix

• A is a M-matrix if

∃ s ≥ ρ(B), B ≥ 0 such that A = s I − B.

• A is a nonsingular M-matrix if

∃ s > ρ(B), B ≥ 0 such that A = s I − B.

• Ch. Beauthier

The positive LQ-problem for linear continuous time systems

Motivations Problem Statement Stable Case Unstable Case Conclusion and future work Preliminary Concepts Problem

Preliminary Concepts - M-Matrix

Theorem Let A a Z-matrix, then the following assertions are equivalent :

1

A is nonsingular and A−1 ≥ 0.

2

There exists a vector x ≥ 0 such that A x >> 0.

3

There exists a vector x >> 0 such that A x >> 0.

4

All eigenvalues of A have positive real parts.

5

A is a nonsingular M-matrix.

• Ch. Beauthier

The positive LQ-problem for linear continuous time systems

Motivations Problem Statement Stable Case Unstable Case Conclusion and future work Preliminary Concepts Problem

Preliminary Concepts - Kronecker product

The Kronecker product of two matrices A = (aij) ∈ IRm×n and B = (bij) ∈ IRp×q is the matrix A ⊗ B given by : A ⊗ B :=    a11 B · · · a1n B . . . · · · . . . am1 B · · · amn B    ∈ IRmp×nq For each matrix A = (aij) ∈ IRm×n, we associate the vector vect(A) ∈ IRmn defined by : vect(A) := [a11, . . . , am1, . . . , a1n, . . . , amn ]T.

• Ch. Beauthier

The positive LQ-problem for linear continuous time systems

Motivations Problem Statement Stable Case Unstable Case Conclusion and future work Preliminary Concepts Problem

Preliminary Concepts - Kronecker product → Rewriting of some equations

e.g. the Sylvester equation : A X + X B = C ⇔ [I ⊗ A + BT ⊗ I] vect(X) = vect(C).

→ If A and B are M-matrices

Then [I ⊗ A + BT ⊗ I] is also a M-matrix.

σ([I ⊗ A + BT ⊗ I]) = σ(A) + σ(B)

P . Lancaster and M. Tismenetsky The theory of matrices 2nd ed., Academic Press, Orlando, Floride, 1985.

• Ch. Beauthier

The positive LQ-problem for linear continuous time systems

Motivations Problem Statement Stable Case Unstable Case Conclusion and future work Preliminary Concepts Problem

Problem Statement

Given a linear time-invariant system [A, B ]. Minimize J(x0, u) = ∞ [x(t)TQ x(t) + u(t)TR u(t)] dt, with R ∈ IRm×m positive definite (pd) symmetric matrix Q ∈ Rn×n positive semidefinite (psd) symmetric matrix

• Ch. Beauthier

The positive LQ-problem for linear continuous time systems

Motivations Problem Statement Stable Case Unstable Case Conclusion and future work Preliminary Concepts Problem

Well-known results

If    (A, B) is (exponentially) stabilizable and (Q, A) is (exponentially) detectable Then the algebraic Riccati equation (ARE) ATX + X A − X B R−1BTX = −Q has a unique stabilizing psd solution X. Moreover :

→

Optimal control : state feedback type : uopt(t) = Kopt x(t) = −R−1BTXx(t)

→

Optimal cost : J(x0, uopt) = xT

0 X x0.

• Ch. Beauthier

The positive LQ-problem for linear continuous time systems

Motivations Problem Statement Stable Case Unstable Case Conclusion and future work Preliminary Concepts Problem

Remark

Proposition If A is a stable Metzler matrix and the solution X of the ARE is such that (A − B R−1BTX) is a Metzler matrix, Then X ≥ 0 whenever Q ≥ 0. Proof :

• Rewriting of ARE

⇔ [I ⊗ (−AT) + (−A + B R−1BTX)T ⊗ I ] vect(X) = vect(Q).

• A is stable matrix ⇔ ∀ λ ∈ σ(A) : Re(λ) < 0
• Properties of M-matrices.
• Ch. Beauthier

The positive LQ-problem for linear continuous time systems

Motivations Problem Statement Stable Case Unstable Case Conclusion and future work Preliminary Concepts Problem

Main objective

Assume that the system [A, B ] is positive Main objective : ? Conditions on Q and R ? such that the closed-loop system is positive i.e. A − B R−1BTX is a Metzler matrix.

• Ch. Beauthier

The positive LQ-problem for linear continuous time systems

Motivations Problem Statement Stable Case Unstable Case Conclusion and future work Main Results Compartmental systems Design Methodology for Q and R

Outline

1

Motivations

2

Problem Statement

3

Stable Case Main Results Compartmental systems Design Methodology for Q and R

4

Unstable Case

5

Conclusion and future work

• Ch. Beauthier

The positive LQ-problem for linear continuous time systems

Motivations Problem Statement Stable Case Unstable Case Conclusion and future work Main Results Compartmental systems Design Methodology for Q and R

Stable Case - Assumptions

(H1) The matrix A is stable. (H2) Q ≫ 0 and B R−1BT≥ 0. (H3) −A + B R−1BTX1 is a Z-matrix, where X1 is solution of the Lyapunov equation ATX1 + X1 A = −Q.

• Ch. Beauthier

The positive LQ-problem for linear continuous time systems

Motivations Problem Statement Stable Case Unstable Case Conclusion and future work Main Results Compartmental systems Design Methodology for Q and R

Stable Case - Riccati equation & Iterative scheme

Observe that :

→ D := −A is a nonsingular M-matrix → Rewriting of ARE : DTX + X D + X B R−1BTX = Q.

Consider the ITERATIVE SCHEME :        X0 = 0 (DT + Xk B R−1BT) Xk+1 + Xk+1 (D + B R−1BTXk) = Xk B R−1BTXk + Q, ∀ k ≥ 0

• Ch. Beauthier

The positive LQ-problem for linear continuous time systems

Motivations Problem Statement Stable Case Unstable Case Conclusion and future work Main Results Compartmental systems Design Methodology for Q and R

Stable Case - Auxiliary result

Lemma Consider a linear time-invariant positive system [A, B ]. Assume that conditions (H1) − (H2) − (H3) hold. Then (Xk)k≥1, defined by the iterative scheme, is a psd matrix sequence, which is (elementwise) decreasing such that ∀ k ≥ 1 : 0 << Xk+1 ≤ Xk ≤ X1 and

• I ⊗ (DT + Xk B R−1BT) + (D + B R−1BTXk)T ⊗ I
• is a nonsingular M-matrix.
• Ch. Beauthier

The positive LQ-problem for linear continuous time systems

Motivations Problem Statement Stable Case Unstable Case Conclusion and future work Main Results Compartmental systems Design Methodology for Q and R

Stable Case - Auxiliary result

Proof : by induction and by using the iterative scheme the properties of M-matrices the Kronecker product

• Ch. Beauthier

The positive LQ-problem for linear continuous time systems

Motivations Problem Statement Stable Case Unstable Case Conclusion and future work Main Results Compartmental systems Design Methodology for Q and R

Stable Case - Main Results

Theorem Consider a linear time-invariant positive system [A, B ]. Assume that conditions (H1) − (H2) − (H3) hold. Then the stabilizing psd solution X of the ARE is such that the corresponding LQ-optimal closed-loop system is positive, i.e. (A − B R−1BTX) is a Metzler matrix.

• Ch. Beauthier

The positive LQ-problem for linear continuous time systems

Motivations Problem Statement Stable Case Unstable Case Conclusion and future work Main Results Compartmental systems Design Methodology for Q and R

Stable Case - Main Results

PROOF : Decrease of the sequence (Xk)k≥1 (see lemma)

→ the limit of this sequence X exists and is psd matrix.

One can take the limit in the iterative scheme

→ X is the stabilizing solution of ARE.

By (H3) and the monotonicity of the sequence, −A + B R−1BTXk is a Z-matrix

→ by limit : A − B R−1BTX is a Metzler matrix.

• Ch. Beauthier

The positive LQ-problem for linear continuous time systems

Motivations Problem Statement Stable Case Unstable Case Conclusion and future work Main Results Compartmental systems Design Methodology for Q and R

Stable Case - Main Results

The assumption (H3) −A + B R−1BTX1 is a Z-matrix, where X1(>> 0) is solution

• f the Lyapunov equation

ATX1 + X1 A = −Q. can be replaced by the following one : (H4) There exists a symmetric matrix Y(>> 0) such that ATY + Y A ≤ −Q and − A + B R−1BTY is a Z-matrix.

• Ch. Beauthier

The positive LQ-problem for linear continuous time systems

Motivations Problem Statement Stable Case Unstable Case Conclusion and future work Main Results Compartmental systems Design Methodology for Q and R

Stable Case - Compartmental systems

A matrix A = (aij) ∈ IRn×n is a compartmental matrix if

⋆

A is a Metzler matrix.

⋆

∀ j ∈ n

n

• i=1

aij ≤ 0.

The positive system [A, B ] is said to be a compartmental system if A is a compartmental matrix.

• Ch. Beauthier

The positive LQ-problem for linear continuous time systems

Motivations Problem Statement Stable Case Unstable Case Conclusion and future work Main Results Compartmental systems Design Methodology for Q and R

Stable Case - Compartmental systems

Here, we consider a compartmental matrix A such that : ∀ j ∈ n

n

• i=1

aij < 0. By [Van Den Hof, 1998], this property is a sufficient condition for the stability of an irreducible compartmental matrix A.

• J. M. Van Den Hof

Positive linear observers for linear compartmental systems SIAM J. Control Optimization, Vol. 36, No.2, p. 590-608, 1998.

• Ch. Beauthier

The positive LQ-problem for linear continuous time systems

Motivations Problem Statement Stable Case Unstable Case Conclusion and future work Main Results Compartmental systems Design Methodology for Q and R

Stable Case - Compartmental systems

Moreover, this condition implies that Z A << 0 and consequently ATZ + Z A << 0, where Z = (zij) ∈ IRn×n is given by ∀ i, j ∈ n zij = 1.

• Ch. Beauthier

The positive LQ-problem for linear continuous time systems

Motivations Problem Statement Stable Case Unstable Case Conclusion and future work Main Results Compartmental systems Design Methodology for Q and R

Stable Case - Compartmental systems

In this case, the theorem remains valid if the assumption (H2) Q ≫ 0 and B R−1BT≥ 0. is replaced by the weaker condition (H5) Q ≥ 0 and B R−1BT≥ 0. Proof : By using the iterative scheme where X1 is replaced by Z.

• Ch. Beauthier

The positive LQ-problem for linear continuous time systems

Motivations Problem Statement Stable Case Unstable Case Conclusion and future work Main Results Compartmental systems Design Methodology for Q and R

Stable Case - Design Methodology for Q and R

? Q and R such that (H2) and (H4) (or (H3)) hold ?

→

Find Q and R such that : ◮ Q >> 0 ◮ B R−1BT ≥ 0 ◮ There exists a symmetric matrix Y ≫ 0 such that ATY + Y A ≤ −Q and −A + B R−1BTY is a Z-matrix.

• Ch. Beauthier

The positive LQ-problem for linear continuous time systems

Motivations Problem Statement Stable Case Unstable Case Conclusion and future work Main Results Compartmental systems Design Methodology for Q and R

Stable Case - Design Methodology for Q and R

Choose Q = CTC such that C >> 0.

→ Q >> 0.

Choose R = s I − W such that W is a symmetric nonnegative matrix and s > ρ(W).

→ R nonsingular pd matrix

⇒ R−1 ≥ 0 and B R−1BT ≥ 0.

• Ch. Beauthier

The positive LQ-problem for linear continuous time systems

Motivations Problem Statement Stable Case Unstable Case Conclusion and future work Main Results Compartmental systems Design Methodology for Q and R

Stable Case - Design Methodology for Q and R

Let P a symmetric psd matrix such that P ≥ Q. Let Y the solution of the Lyapunov equation : ATY + Y A = −P Since A is stable, the solution is :

(see e.g. [Callier, Desoer, 1991])

Y = +∞ eAT tP eA t dt

• F. M. Callier, C. A. Desoer

Linear System Theory Springer-Verlag, New-York, 1991.

• Ch. Beauthier

The positive LQ-problem for linear continuous time systems

Motivations Problem Statement Stable Case Unstable Case Conclusion and future work Main Results Compartmental systems Design Methodology for Q and R

Stable Case - Design Methodology for Q and R → Y is a symmetric psd matrix and Y >> 0, since Q >> 0. → ATY + Y A ≤ − Q. → ? −A + B R−1BTY is a Z-matrix ?

Assume that α := min{aij : i = j} > 0 Since lim

s→+∞(sI − W)−1 = 0 and R = s I − W :

∃ s > ρ(W) such that −A + B R−1BTY is a Z-matrix

• Ch. Beauthier

The positive LQ-problem for linear continuous time systems

Motivations Problem Statement Stable Case Unstable Case Conclusion and future work Main Results Compartmental systems Design Methodology for Q and R

Stable Case - Example (1)

A = −2 1 1 −2

• B

= 1 1

• Q

=

• 4

0.01 0.01 6

• R

=

• 6

−2 −2 5

• → the closed-loop matrix is a Metzler matrix :

A + B K = A − B R−1BTX =

• −2.2692

0.7688 0.7789 −2.4100

• Ch. Beauthier

The positive LQ-problem for linear continuous time systems

Motivations Problem Statement Stable Case Unstable Case Conclusion and future work Main Results Compartmental systems Design Methodology for Q and R

Stable Case - Example (2)

Choose Q = CTC such that C >> 0

→ Take C = Q1/2 =

• 2

0.022 0.022 2.4495

• .

→ Q =

• 4

0.01 0.01 6

• >> 0
• Ch. Beauthier

The positive LQ-problem for linear continuous time systems

Motivations Problem Statement Stable Case Unstable Case Conclusion and future work Main Results Compartmental systems Design Methodology for Q and R

Stable Case - Example (3)

Choose R = s I − W with W ≥ 0, sym and s > ρ(W).

→ Take, e.g., W =

4 2 2 5

• with ρ(W) = 6.5616.

→ Take s the least integer which is larger than ρ(W) s = 7. → But s should be chosen sufficiently large for another

condition, see later. Here we take s = 7 + 3 = 10.

→ R =

• 6

−2 −2 5

• nonsingular and pd matrix such that :

B R−1BT =

• 0.1923

0.0769 0.0769 0.2308

• ≥ 0
• Ch. Beauthier

The positive LQ-problem for linear continuous time systems

Motivations Problem Statement Stable Case Unstable Case Conclusion and future work Main Results Compartmental systems Design Methodology for Q and R

Stable Case - Example (4)

Let P a symmetric psd matrix such that P ≥ Q.

→ Take P = Q + β I with e.g. β = 10 → P =

• 14

0.01 0.01 10

• .

Let Y the solution of the Lyapunov equation :

→ Y =

4.7517 2.5033 2.5033 5.2517

• >> 0 symmetric psd matrix.

min{aij : i = j} = 1 α > 0.

• Ch. Beauthier

The positive LQ-problem for linear continuous time systems

Motivations Problem Statement Stable Case Unstable Case Conclusion and future work Main Results Compartmental systems Design Methodology for Q and R

Stable Case - Example (5) → ∃ s > ρ(W) such that − A + B R−1BTY is a Z-matrix s must be sufficiently large ! (R = s I − W)

s −A + B R−1BTY Z-matrix A − B R−1BTX Metzler Y solution of Lyapunov X solution of Riccati 7 „ 9.2550 6.7550 7.5067 12.3808 « KO „ −2.9864 −0.2193 −0.0639 −3.7515 « KO 8 „ 4.4077 1.2517 1.4396 5.2517 « KO „ −2.4855 0.5060 0.5435 −2.7888 « OK 9 „ 3.5008 0.2823 0.3762 3.9541 « KO „ −2.3427 0.6851 0.7026 −2.5350 « OK 10 „ 3.1063 −0.1146 −0.0568 3.4045 « OK „ −2.2692 0.7688 0.7789 −2.4100 « OK

• Ch. Beauthier

The positive LQ-problem for linear continuous time systems

Motivations Problem Statement Stable Case Unstable Case Conclusion and future work Main Results Compartmental systems Design Methodology for Q and R

Stable Case - Example (5) → ∃ s > ρ(W) such that − A + B R−1BTY is a Z-matrix s must be sufficiently large ! (R = s I − W)

s −A + B R−1BTY Z-matrix A − B R−1BTX Metzler Y solution of Lyapunov X solution of Riccati 7 „ 9.2550 6.7550 7.5067 12.3808 « KO „ −2.9864 −0.2193 −0.0639 −3.7515 « KO 8 „ 4.4077 1.2517 1.4396 5.2517 « KO „ −2.4855 0.5060 0.5435 −2.7888 « OK 9 „ 3.5008 0.2823 0.3762 3.9541 « KO „ −2.3427 0.6851 0.7026 −2.5350 « OK 10 „ 3.1063 −0.1146 −0.0568 3.4045 « OK „ −2.2692 0.7688 0.7789 −2.4100 « OK

• Ch. Beauthier

The positive LQ-problem for linear continuous time systems

Motivations Problem Statement Stable Case Unstable Case Conclusion and future work Main Results Compartmental systems Design Methodology for Q and R

Stable Case - Example (6)

→ Nonnegative & stable optimal state trajectories

0.2 0.4 0.6 0.8 1 From: In(1) To: Out(1) 1 2 3 4 0.2 0.4 0.6 0.8 1 To: Out(2) From: In(2) 1 2 3 4 Optimal state trajectories x(t) Time (sec) Amplitude

• Ch. Beauthier

The positive LQ-problem for linear continuous time systems

Motivations Problem Statement Stable Case Unstable Case Conclusion and future work Main Results Compartmental systems Design Methodology for Q and R

Stable Case - Example (7)

→ Optimal control not nonnegative → But always nonpositive !! uopt(t) = −R−1BTX x(t) ≤ 0

−0.4 −0.3 −0.2 −0.1 From: In(1) To: Out(1) 1 2 3 4 −0.5 −0.4 −0.3 −0.2 −0.1 To: Out(2) From: In(2) 1 2 3 4 Optimal control u(t) Time (sec) Amplitude

• Ch. Beauthier

The positive LQ-problem for linear continuous time systems

Motivations Problem Statement Stable Case Unstable Case Conclusion and future work Main Results Compartmental systems Design Methodology for Q and R

Stable Case - Design Methodology for Q and R

Proposition If [A, B ] is a positive system such that A is stable and α := min{aij : i = j} > 0. Then there exist Q and R such that (H2) and (H4) are satisfied.

→

a design methodology for choosing Q and R

to get the positivity of the closed-loop system.

• Ch. Beauthier

The positive LQ-problem for linear continuous time systems

Motivations Problem Statement Stable Case Unstable Case Conclusion and future work Main Results Compartmental systems Design Methodology for Q and R

Stable Case - Design Parameters

Design Parameters ◮ Construction of Q → Choice of C and P ◮ Construction of R → Choice of s and W Assumption ◮ Solution of Lyapunov equation : Y

→

Choice of P

• Ch. Beauthier

The positive LQ-problem for linear continuous time systems

Motivations Problem Statement Stable Case Unstable Case Conclusion and future work Main Results Compartmental systems Design Methodology for Q and R

Stable Case - Design Parameters

Design Parameters ◮ Construction of Q → Choice of C and P ◮ Construction of R → Choice of s and W Assumption ◮ Solution of Lyapunov equation : Y

→

Choice of P

• Ch. Beauthier

The positive LQ-problem for linear continuous time systems

Motivations Problem Statement Stable Case Unstable Case Conclusion and future work

Outline

1

Motivations

2

Problem Statement

3

Stable Case

4

Unstable Case

5

Conclusion and future work

• Ch. Beauthier

The positive LQ-problem for linear continuous time systems

Motivations Problem Statement Stable Case Unstable Case Conclusion and future work

Unstable Case - Assumptions

We consider a positive system [A, B ] where A is not stable. ⇒ s(A) = sup

• Re(λ) : λ ∈ σ(A)
• ≥ 0.

◮ −A is not necessarily a nonsingular M-matrix. ◮ Take D = s I − A with s > s(A) ⇒ D is a nonsingular M-matrix.

(If A is a Metzler matrix then ∀ t ≥ 0 : eA t ≥ 0 ⇒ ∀ λ > s(A) (λ I − A)−1 = ∞ e−λ t eA t dt ≥ 0.)

• Ch. Beauthier

The positive LQ-problem for linear continuous time systems

Motivations Problem Statement Stable Case Unstable Case Conclusion and future work

Unstable Case - Assumptions → Riccati equation ARE, with D = s I − A :

DTX + X D + X B R−1BTX = Q + 2sX

→ Like the stable case, we make a similar assumption :

(H′

3) ∃ s > s(A) : −A + B R−1BTX1 is a Z-matrix

and B R−1BTX1 ≥ 2s I, where X1 is the solution of the Lyapunov equation : (A−s I)TX1 + X1 (A−s I) + Q = 0 and (H2) Q ≫ 0 and B R−1BT≥ 0.

• Ch. Beauthier

The positive LQ-problem for linear continuous time systems

Motivations Problem Statement Stable Case Unstable Case Conclusion and future work

Unstable Case - Assumptions

Consider the ITERATIVE SCHEME :        X0 = 0 (DT + Xk B R−1BT) Xk+1 + Xk+1 (D + B R−1BTXk) = Q + 2sXk + Xk B R−1BTXk, ∀ k ≥ 1

• Ch. Beauthier

The positive LQ-problem for linear continuous time systems

Motivations Problem Statement Stable Case Unstable Case Conclusion and future work

Unstable Case - Auxiliary Result

Lemma Consider a linear time-invariant positive system [A, B ]. Assume that conditions (H2) − (H′

3) hold.

Then (Xk)k≥1, defined by the iterative scheme, is a psd matrix sequence, which is (elementwise) decreasing such that ∀ k ≥ 1 : 0 << Xk+1 ≤ Xk ≤ X1 and

• I ⊗ (DT + Xk B R−1BT) + (D + B R−1BTXk)T ⊗ I
• is a nonsingular M-matrix.
• Ch. Beauthier

The positive LQ-problem for linear continuous time systems

Motivations Problem Statement Stable Case Unstable Case Conclusion and future work

Unstable Case - Main Results

Theorem Consider a linear time-invariant positive system [A, B ]. Assume that conditions (H2) − (H′

3) hold.

Then the psd solution X of the ARE is such that the corresponding LQ-optimal closed-loop system is positive, i.e. (A − B R−1BTX) is a Metzler matrix. PROOF : Similar to the proof for the stable case !

• Ch. Beauthier

The positive LQ-problem for linear continuous time systems

Motivations Problem Statement Stable Case Unstable Case Conclusion and future work

Outline

1

Motivations

2

Problem Statement

3

Stable Case

4

Unstable Case

5

Conclusion and future work

• Ch. Beauthier

The positive LQ-problem for linear continuous time systems

Motivations Problem Statement Stable Case Unstable Case Conclusion and future work

Summary

◮ Sufficient conditions for the existence of a solution of ARE corresponding to positive linear systems.

⋆

Construction of a decreasing sequence of positive matrices.

⋆

Convergence of this sequence to the solution of ARE.

◮ Weaker sufficient conditions for compartmental systems. ◮ Design method for the weighting matrices Q and R in the stable case.

• Ch. Beauthier

The positive LQ-problem for linear continuous time systems

Motivations Problem Statement Stable Case Unstable Case Conclusion and future work

Open Questions

Find necessary conditions for the positivity of the closed-loop system. Find a design methodology for Q and R in the unstable case. Applications. Inverse problem approach : LMI approach. Extension to infinite dimensional systems.

• Ch. Beauthier

The positive LQ-problem for linear continuous time systems

Motivations Problem Statement Stable Case Unstable Case Conclusion and future work

Thanks for your attention.

• Ch. Beauthier

The positive LQ-problem for linear continuous time systems