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on the occasion of his 70th birthday

NON-HOMOGENEOUS LINEAR FEEDBACK FOR OPTIMAL STABILIZATION BY BOUNDARY

CONTROL OF p-CALORIC EQUATION, p > 1

MARIA B ˘ATINET¸ U–GIURGIU

A boundary control problem forp-caloric equation,p >1, is considered. Only part of boundary conditions is controlled by a linear device. A feedback solution is obtained that stabilizes the evolution by minimizing a quadratic performance index.

AMS 2000 Subject Classification: 93C20, 93D15.

Key words: p-caloric equation, boundary control, non-homogeneous feedback, Lurie’s equations.

1. INTRODUCTION

Yakuboviˇc–Kalman–Popov’s lemma gives a criterion of absolute stability for nonlinear systems of differential equations. This lemma may be also used in problems of optimal control. Several proofs of this lemma were given in the finite dimensional case e.g. [4], [7], [9], [10]. A proof of this lemma in the case of Hilbert spaces and of linear bounded operators was given by Yakuboviˇc in [8].

In this paper, using the ideas from [8], [1] and [2] the optimal stabilization by boundary control of the p-caloric equation, p > 1, is studied. This is not a direct application of Yakuboviˇc’s result because here the operators are not bounded.

In [1] the optimal stabilization by boundary control of p-caloric equa- tion, p > 1, was studied. All the boundary conditions were controlled by a linear device and a feedback solution that stabilized the evolution minimizing a quadratic performance index was obtained. The optimal control appeared under the feedback form

MATH. REPORTS9(59),1 (2007), 3–20

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(∗) u(t) =−N−1(0 B)H





∩yy . . . .

p−1y z





zϕ0

(t),

whereH is the solution of the Lurie’ equations. We point out the connection between the problem of the infinite interval and that on the finite interval, considered in [2]. The connection consists of the fact that the solution of the Lurie’s equations (for the problem on the infinite interval) is a particular solu- tion of a Riccati type system that gave the optimal control on finite intervals.

This particular solution it is the stationary one.

In the problem considered in [1], all the boundary data were controlled.

But from the physical point of view it is more realistic to consider the case when only part of the boundary conditions is controlled.

This is the case considered in [2], where a boundary control problem with quadratic cost functional for thep-caloric equation,p≥1, with only part of the boundary conditions given by the solution of a differential equation involving control is approached and studied.

In [2], for the optimal control, a characterization under the feedback form was also got, but under a non-homogeneous form. By this we mean a representation

(∗∗) u(t) = p−1

i=0 γ2(t)

γ1(t) πi(t, x)iy(t, x)dx+πp(t)z(t) +πp+1(t),

where (πj(·,·))j=0,p are some of the components of the solution of a Riccati- type system andπp+1(·) is the solution of a linear differential equation.

In this paper, using the ideas from [8], [1–2], the optimal stabilization by boundary control ofp-caloric equation,p >1, is studied, in the case that only part of the boundary data is controlled.

The main result consists of the representation of the optimal control under the non-homogeneous feedback form

(∗ ∗ ∗) u(t) =−N−1(0 B 0)H







∩yy . . .

p−1y z α







ϕ z0

α

(t),

(3)

with H being the solution of the Lurie’s system. Also, the solution of the Lurie’s equations is the stationary solution of the Riccati-type system studied in [3].

2. STATEMENT OF THE PROBLEM

LetA, B,C be constant m×m,m×k, r×m matrices, with A stable, G2 a constant semipositive definite m×m matrix; N a constant, strictly positive definitek×kmatrix; U =L2([0,∞;Rk),ϕ(·)∈ C([0,1];Rp),z0Rm, K: [0,1]×[0,1]Rintegrable enjoying the property of positivity

1 0

1

0 K(x, ζ)χ(x)χ(ζ)dxdζ 0, ∀χ(·)∈ C([0,1];R);

D={(t, x)|t∈[0,∞), x∈[0,1]}.

Let us consider the boundary value problem

(2.1)















































py(t, x) = 0,

p = 2

∂x2

∂t (p)

iy(t,0) =α2i+1(t), iy(t,1) =α2(i+1)(t), i= 0, p−1

α(t) =Cz(t), α(t) =



 αj1(t) αj2(t)

... αjr(t)



, ∀t∈[0,∞),

{jk}k=1,r ⊂ {1, . . . ,2p}with j1 < j2 <· · ·< jr, dz

dt =Az+Bu(t), u∈ U z(0) =z0 Rm (given),

y(0, x),∩y(0, x), . . . ,∩p−1y(0, x)

=ϕ(x), ∀x∈[0,1]. Remark 2.1. Under the hypothesis that A is stable and using a Fourier transform argument, one sees thatu∈L2([0,∞);Rk) implieszu∈L2([0,∞);Rm), henceyu ∈L2([0,∞);R), forα∈L2([0,∞);R2p−r) withα(t) =

αm1(t)αm2(t) . . . αm2p−r(t)

,∀t∈[0,∞) and{m1, . . . , m2p−r}={1, . . . ,2p}\{j1, j2, . . . , jr}.

Denote by yu

zu

the solution of problem (2.1) corresponding to the control u(·). Associate with problem (2.1) the quadratic cost functional

(2.2) J

ϕ z0

α

(u) =

0 F

yu zu

, u

(t)dt,

(4)

where

(2.3) F

yu zu

, u

(t) = 1

0 1

0 K(x, ζ)yu(t, x)yu(t, ζ)dxdζ+

+ G2zu(t), zu(t)+ Nu(t), u(t).

The optimal control problem consists of finding a control functionu∈ U such that

(2.4) J

ϕz0

α

(u)≤J

ϕz0

α

(u), ∀u∈ U.

A straightforward calculation leads to the representation

(2.5) J

ϕ z0

α

(u) = u, Ru+

r

ϕ z0

α

, u

+

+

u, r

ϕ z0

α

+ρ

ϕ z0 α

of the cost functional, where R : U → U is a selfadjoint operator and r, ρ satisfy

r

 0 0 0

= 0, ρ

 0 0 0

= 0.

Remark 2.2. It follows from the hypothesis of positivity imposed onG2, N,K that

(2.6) J

0 00

(u) = u, Ru ≥δ||u||2, δ >0.

It follows from (2.6) by using a theorem of minimizing coercive forms that there uniquely exists an optimal control u

ϕz0

α

(·), for problem (2.1),

(2.4) that admits the representation

(2.7) u

ϕ z0

α

(t) =−R−1r

ϕ z0

α

(t).

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3. THE CHARACTERIZATION OF THE OPTIMAL CONTROL;

THE FEEDBACK FORM

Let the space

X = f

f(·)∈ C([0,1];Rp), z∈Rm, a column vector, β(·)∈L2

[0,∞);R2p−r that we endow with the scalar product

f1 z1

β1

,

f2 z2

β2

=

1

0 f1(x), f2(x)Rpdx+ z1, z2Rm+ +

0 β1(t), β2(t)R2p−rdt=

=

1

0 f2(x)f1(x)dx+ z1, z2Rm+

0 β2(t)β1(t)dt.

Theorem 3.1. Let V

ϕ z0

α

be the optimal value of the cost func- tional,

V

ϕ z0

α

=J

ϕ z0

α

u

ϕ z0

α

.

Then there exists a selfadjoint operator H:X → X such that

V

ϕ z0 α

=

ϕ z0 α

, H

ϕ z0 α

.

Proof. The result follows from a straightforward calculation using the representation formulae of the solution of problem (2.1), namely

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(3.1)





































yu(t, x) = p

i=1 1

0 i−1y(0, ξ)Gi(t, x; 0, ξ)dξ+ +

t 0

zu(τ), CG1(t, x;τ) dτ+

t 0

α(τ ),G2(t, x;τ) dτ=

=

1 0

ϕ(ξ),G(t, x; 0, ξ)

dξ+ +

t 0

zu(τ), CG1(t, x;τ)

dτ+

t 0

α(τ),G2(t, x;τ)

dτ zu(t) = eAtz0+

t

0 eA(t−s)Bu(s)ds,

whereGi(t, x;λ, µ) is the Green function of order idefined by Gi(t, x;λ, µ) = 1

2√π

Gi−1(t, x;u, v)G(u, v;λ, µ)dudv;

G(t, x;τ, ξ) is the Green function corresponding to the heat operator and the domain ∆;γis the functionγ :N→ {0,1}defined byγ(n) =

0, n= 2k 1, n= 2k+ 1 (Nis the set of all positive integers); G(·,·; 0) is the column vector

G(·,·; 0) =



G(·,·; 0) G2(·,·; 0)

. . . . Gp(·,·; 0)



; G1(t, x;τ) is the column vector

G1(t, x;τ) =







(1)j1−1∂G

∂ζ

j1 2

+γ(j1+1)

(t, x;τ, γ(j1+ 1)) . . . . (1)jr−1∂G

∂ζ

[jr2]+γ(jr+1)

(t, x;τ, γ(jr+ 1))







for any (t, x;τ)[0,∞)×[0,1]×[0,∞);G2(t, x;τ) is the column vector

G2(t, x;τ) =







(1)m1−1∂G

∂ζ

[m21]+γ(m1+1)

(t, x;τ, γ(m1+ 1)) . . . . (−1)m2p−r−1∂G

∂ζ

m2p−r 2

+γ(m2p−r+1)

(t, x;τ, γ(m2p−r+ 1))







 for any (t, x;τ)[0,∞)×[0,1]×[0,∞).

(7)

Denoting C1(s, λ) =

s

1

0 1 0

t

s BeA(θ−s)CG1(t, ζ;θ)dθ

×

×K(x, ζ)G(t, x; 0, λ)dxdζ

dt, C2(s) =

s

1 0

1

0 K(x, ζ)

t

s BeA(θ−s)CG1(t, ζ;θ)dθ

×

× 1

0

G1(t, x;τ)Cedτ

dxdζ

dt+

s BeA(t−s)G2eAtdt, C3(s, τ) =

s

1 0

1 0K(x, ζ)

t

sBeA(θ−s)CG1(t, x,;θ)dθ

G2(t, x;τ)dxdζ

dt, C4(s, τ) =

τ

1 0

1 0K(x, ζ)

t

sBeA(θ−s)CG1(t, x;θ)dθ

G2(t, x;τ)dxdζ

dt, C3(s) =

s

0 C3(s, τ)α(τ)dτ+

s C4(s, τ)α(τ)dτ, C5(s, τ) =

C3(s, τ), τ [0, s]

C4(s, τ), τ [s,∞), we have

(3.2)



















































r

ϕ z0 α

(s) =

1

0C1(s, λ)ϕ(λ)dλ+C2(s)z0+

0 C5(s, τ)α(τ)dτ =

=

1

0 C1(s, λ)ϕ(λ)dλ+C2(s)z0+C3(s), ρ

ϕ z0 α

=

0

1 0

1

0 K(x, ξ)

1

0 ϕ(λ)G(t, x; 0, λ)dλ+ +

t 0

G1(t, x;τ)Cez0dτ+

t 0

G2(t, x;τ)α(τ )dτ

×

× 1

0 ϕ(µ)G(t, ξ; 0, µ)dµ+

t 0

G1(t, ξ;θ)Cez0dθ+ +

t 0

G2(t, ξ;θ)α(θ)dθ

dxdξ+

G2eAtz0,eAtz0 dt.

(8)

Using (2.7) and the fact that R =R, we get J

ϕz0

α

u

ϕz0

α

=V

ϕ z0

α

=

r

ϕ z0 α

,−R−1r

ϕ z0 α

+

+ρ

ϕ z0

α

=

ϕ z0

α

, H

ϕ z0

α

,

a quadratic form in ϕ(·), z0 and α(·), where H : X → X is a selfadjoint

operator of the formH =

H11 H12 H13 H21 H22 H23 H31 H32 H33

 that is defined by (3.3)

H

ϕ z0

α

(β, τ) =









1

0 H11(λ, β)ϕ(λ)dλ+H12(β)z0+

0 H13(τ, β)α(τ)dτ

1

0 H21(λ)ϕ(λ)dλ+H22z0+

0 H23(τ)α(τ)dτ

1

0 H31(τ, λ)ϕ(λ)dλ+H32(τ)z0+

0 H33(θ, τ)α(θ)dθ









with

H11: [0,1]×[0,1]→ L(Rp,Rp), H11(α, β) =

0

−C1(s, α)R−1C1(s, β)+

+

1 0

1

0 K(x, ξ)G(s, x; 0, α)G(s, ξ; 0, β)dxdξ

ds, H12: [0,1]→ L(Rp,Rm),

H12(α) =

0

−C1(s, α)R−1C2(s)+

+

1 0

1

0 K(x, ξ)G(s, x; 0, α)

s 0

G1(s, ξ;θ)Cedθ

dxdξ

ds, H13: [0,∞)×[0,1]→ L(Rp,R2p−r),

H13(τ, λ) =

0

!−C1(s, λ)R−1C5(s, τ)"

ds, H21: [0,1]→ L(Rm,Rp), H21(α) =H12 (α),

(9)

H22is a constant m×m matrix defined by H22=

0

−C2(s)R−1C2(s) + eAsG2eAs+ +

1 0

1 0 K(x, ξ)

s

0eAθCG1(s, ξ;θ)dθ

× s

0

G1(s, x;τ)Cedτ

dxdξ

ds, H23: [0,∞)→ L(Rm,R2p−r), H23(τ) =

0 C2(s)R−1C5(s, τ)ds+ +

τ

1 0

1

0 K(x, ξ)

s

0 eAθCG1(s, x;θ)dθ

G2(s, ξ;τ)dxdξ

ds, H31: [0,∞)×[0,1]→ L(R2p−r;Rp), H31(τ, λ) =H13 (τ, λ),

H32: [0,∞)→ L(R2p−r;Rm), H32(τ) =H23 (τ), H33: [0,∞)×[0,∞)→ L

R2p−r,R2p−r , H33(θ, τ) =

0 C5(s, θ)R−1C5(s, τ)ds+H33(τ, θ), where

H33(τ, θ) =











τ

1 0

1

0 K(x, ξ)G2(s, x;τ)G2(s, ξ;θ)dxdξ

ds, θ [0, τ]

θ

1 0

1

0 K(x, ξ)G2(s, x;τ)G2(s, ξ;θ)dxdξ

ds, θ [τ,∞).

It follows from the properties of the Green function that

(3.4)











H11(0) =H11(1) =H11(·,0) =H11(·,1) = 0, H12(0) =H12(1) = 0, H21(0) =H21(1) = 0, H13(0) =H13(1) = 0, withH13(β) =

0 H13(τ, β)α(τ)dτ.

In order to get the feedback form of the optimal control we shall use the method of dynamic programming. For that we associate to problem (2.1),

(10)

(2.4) the family of problems

(3.5)











































py(t, x) = 0

iy(t,0) =α2i+1(t), iy(t,1) =α2(i+1)(t), i= 0, p−1 α(t) =Cz(t), α(t) =



αj1(t) αj2(t) . . . . αjr(t)



 for anyt∈[s,∞), {jk}k=1,r ⊂ {1, . . . ,2p} with j1< j2 <· · ·< jr,

dz

dt =Az+Bu(t), u∈ U,

y(s, x),∩y(s, x), . . . ,∩p−1y(s, x)

=ϕs(x), z(s) =z0s for anys >0 and∀x∈[0,1].

Define

(3.6) Js,

ϕs

zs0 α

(u) =

s F

yu zu

, u

(t)dt,

where

yu(·,·) zu(·)

is the solution of problem (3.5). Denote by (3.5)s the prob- lem of the family (3.5) which is concerned with the interval [s,∞) and with initial data

y(s, x), . . . , p−1y(s, x)

=ϕs(x), z(s) = zs0. Fix s > 0, ϕs(·), zs0 and consider problem (3.5)s. Consider also problem (2.1) with initial data y(0, x), . . . ,∩p−1y(0, x)

= ϕs(x), x [0,1] and z(0) = zs0. With this problem we associate the functional J

ϕs

zs0 α

(u). According to the previous

developments, there existu ϕ

z0ss

(·) and y

z

ϕs

z0s α

, the optimal control for

problem (2.1), (2.4) and, respectively, the solution of problem (2.1) corre- sponding to the optimal control with initial data

ϕs zs0 α

.

We give next without proof a few results (since they are analogous to the ones given in [3]).

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Lemma 3.1. Let s > 0, ϕs(·), zs0 be fixed. Let Js,

ϕs

zs0 α

(·) be given by

(3.6),and let u ϕ

s

zs0

, y

z

ϕs

z0s α

be the optimal control and the optimal solu-

tion of problem (2.1), (2.4) with initial data ϕs

zs0

. Denote by us, ϕ

s

z0s

(t) =

=u ϕ

s

zs0

(t−s),

y(t,·) z(t)

s,

ϕs

z0s

α

=

y(t−s,·)

z(t−s)

ϕs

z0s α

. Then y

z

s,

ϕs

zs0 α

is

the solution of problem (3.5)s corresponding to the control function

us, ϕ

s

zs0

(·), and us, ϕ

s

z0s

(·) is the optimal control of problem (3.5)s, (3.6).

Proof. See Lemma 2.1 from [3].

Proposition 3.1. Let H be given by Theorem 3.1. Let G y

z

, u

(·) be given by

G y

z

, u

(t) =F y

z

, u

(t)+

+











y(t,·)

∩y(t,·) . . . .

p−1y(t,·) z(t) α(t)









 , H











2y

∂x2(t, x) . . . .

2p−1y

∂x2 (t, x) Az(t) +Bu(t)

˙ α(t)









 (·)

+

+

H











2y

∂x2(t, x) . . . .

2p−1y

∂x2 (t, x) Az(t) +Bu(t)

˙ α(t)











 (·),











y(t,·)

∩y(t,·) . . . .

p−1y(t,·) z(t) α(t)











.

(12)

Then for all t for which





∩yy . . . .

p−1y z





ϕz0

α

is differentiable, we have

G y

z

ϕ z0

α

, v

(t)0 for any v ∈ U and G y

z

ϕ z0

α

,u

ϕ z0

α

(t) = 0.

Proof. See the proof of Proposition 2.1 from [3] and use the formula

(3.7) Gi(t+s, x; 0, ξ) =

1

0 G(t, x; 0, α)Gi(s, α; 0, ξ)dα

for the Green function of orderi= 1, p. This formula leads to the conclusion that

y z

ϕz0

α

(t+s,·) is the solution of the controlled system with initial

condition y

z

ϕz0

α

(s,·) and control function u

ϕz0

α

(t +s).

Proposition 3.2. LetWbe given by W=−N−1(0B 0)H. Then the optimal control u

ϕz0

α

(·)admits the representation under the feedback from

(3.8) u

ϕ z0

α

(t) =W







 y(t,·)

∩y(t, ·) . . . .

p−1y(t,·) z(t) α(t)









ϕ z0

α

.

Proof. See Proposition 2.2 from [3].

(13)

4. LURIE’S EQUATIONS

Lurie’s equations are the analog of the Riccati type system for the prob- lem on a finite interval. Set

h ϕ

z0

(t) = (0B0)H







y(t,·)

∩y(t, ·) . . . .

p−1y(t, ·) z(t) α(t)







ϕ z0

and

χ ϕ

z0

(t) =

1 0

1

0 K(x, ξ)y(t, x)y(t, ξ)dxdξ+ G2z(t),z(t)+

+







y(t,·)

y(t,·) . . . .

p−1y(t,·) z(t) α(t)







ϕz0

, H

 0 0 0

0 A 0

0 0 0







y(t, x)

∩y(t, x) . . . .

p−1y(t, x) z(t) α(t)







ϕz0

(·)

+

+

H

 0 0 0

0 A 0

0 0 0







y(t, ξ)

∩y(t, ξ) . . . .

p−1y(t, ξ) z(t) α(t)







ϕz0

(·),







y(t,·)

y(t,·) . . . .

p−1y(t,·) z(t) α(t)







ϕz0

+

+

H













2y

∂x2(t, x)

2 y

∂x2 (t, x) . . . .

2p−1y

∂x2 (t, x) 0 ˙ α(t)













ϕz0

(·),







y(t,·)

∩y(t,·) . . . .

p−1y(t,·) z(t) α(t)







ϕz0

+

(14)

+







y(t,·)

∩y(t,·) . . . .

p−1y(t,·) z(t) α(t)







ϕz0

, H













2y

∂x2(t, x)

2∩y

∂x2 (t, x) . . . .

2p−1y

∂x2 (t, x) 0 ˙ α(t)













ϕz0

(·)

.

According to the proof of Proposition 3.2, we have

(4.1) X ϕ

z0

(t) h ϕ

z0

(t), N−1h ϕ

z0

(t) 0.

By the definition ofH, taking into account that (Hij)i,j=1,3 satisfy conditions (3.4) and that the initial data y(0), . . . ,∩p−1y(0), z(0) are arbitrary, a straightforward calculation leads to the Lurie system (written forH11(x, ξ) = (Aij(x, ξ))i,j=0,p−1, H12(ξ) = (C0(ξ). . . Cp−1(ξ)), H21(x) =

B0(x)B1(x). . . Bp−1(x)

,H22=M(p+1)2,H23=#

0 H23(τ)α(τ)dτ =L1 ∈ L(Rm,R)):

(4.2)













































































2A00

∂x2 (x, ξ) +2A00

∂ξ2 (x, ξ) +C0(ξ)BN−1BB0(x) =K(x, ξ),

2A0k

∂x2 (x, ξ) +2A0k

∂ξ2 (x, ξ) +Ck(ξ)BN−1BB0(x)

−A0k−1(x, ξ) = 0, k= 1, p−1,

2Ak0

∂x2 (x, ξ) +2Ak0

∂ξ2 (x, ξ) +C0(ξ)BN−1BBk(x)

−Ak−10(x, ξ) = 0, k= 1, p1,

2Aij

∂x2 (x, ξ) +2Aij

∂ξ2 (x, ξ) +Cj(ξ)BN−1BBi(x)−Aij−1(x, ξ) =

= 0, i, j = 1, p−1,

2B0

∂x2 (x) +AB0(x) +M(p+1)2BN−1BB0(x)−

−C







(1)j1

∂A0 j1+1 2

−1

∂ξ (x, γ(j1)) ...

(1)jr

∂A0[jr+12 ]−1

∂ξ (x, γ(jr))







= 0,

(15)

























































































































































2Bk

∂x2 (x) +ABk(x) +M(p+1)2BN−1BBk(x)−

−C







 (1)j1

∂Ak j1+1 2

−1

∂ξ (x, γ(j1)) ...

(1)jr

∂Ak[jr+12 ]−1

∂ξ (x, γ(jr))







−Bk−1(x) = 0, k= 1, p−1,

2C0

∂ξ2 (ξ) +C0(ξ)A+C0(ξ)BN−1BM(p+1)2







 (1)j1

∂A j1+1 2

−1 0

∂x (γ(j1), ξ) ...

(1)jr∂A[jr+12 ]−1 0

∂x (γ(jr), ξ)







C = 0,

2Ck

∂ξ2 (ξ)+Ck(ξ)A+Ck(ξ)BN−1BM(p+1)2







 (1)j1

∂A j1+1 2

−1k

∂x (γ(j1), ξ) ...

(1)jr

∂A[jr+12 ]−1k

∂x (γ(jr), ξ)







C−Ck−1(ξ) = 0, k= 1, p−1,

M(p+1)2A+AM(p+1)2 +M(p+1)2BN−1BM(p+1)2







(1)j1

∂Bj1+1

2

−1

∂x (γ(j1)) ...

(1)jr

∂B[jr+12 ]−1

∂x (γ(jr))







C−

−C







(−1)j1

∂C j1+1 2

−1

∂x (γ(j1)) ...

(1)jr

∂C[jr+12 ]−1

∂x (γ(jr))







=G2,

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