Internat. J. Math. & Math. Sci.
Vol. 23, No. 3 (2000) 181–188
S0161171200002015
© Hindawi Publishing Corp.
DUALITY IN THE OPTIMAL CONTROL OF HYPERBOLIC
EQUATIONS WITH POSITIVE CONTROLS
JONG YEOUL PARK and MI JIN LEE
(Received 11 September 1998 and in revised form 21 October 1998)
Abstract. We study the duality theory for hyperbolic equations. Also, we consider distributed control systems with positive control and convex cost functionals.
Keywords and phrases. Hyperbolic equations, duality theory, optimal control, convex cost
function, positive control, distributed control systems.
2000 Mathematics Subject Classification. Primary 49K20, 49N15, 93C20.
1. Introduction. Recently Lions, motivated by various practical problems, started
a program of studying the optimal control distributed systems. The developments of
distributed systems are the establishing of the optimality systems which characterize
the optimal control [2, 3, 4, 5]. Duality theory for the corresponding parabolic equations with positive control has been given by Chan [1]. But, in this paper, we study
the duality theory for hyperbolic distributed control systems. In fact, we consider
distributed control systems with positive control and convex cost functionals. The
approach presented exploits the fundamental results of Lions [2] on the optimality
system which characterizes the optimal control. The method can be used to construct
dual optimal systems when the controls are positive.
2. Duality in the optimal control hyperbolic equations with second-order operator. Let Ω be a bounded open set in Rn with smooth boundary Γ and Q = Ω × (0, T ).
The norm on L2 (Q) is denoted by | · | and the corresponding inner product by (·, ·).
In Q, information on the state is given by
∂2y
+ ∆y = u,
∂t 2
in Q, u ∈ Uad , y ∈ L2 (Q),
y(0) ∈ K0 ,
y = 0, on
,
where
Uad = u | u ∈ L2 (Q), u ≥ 0 in Q ,
K0 = φ | φ ∈ H −1 (Ω), φ ≥ 0 in Ω ,
= Γ × (0, T ),
∆=
n
∂2
.
∂xi2
i=1
(2.1)
(2.2)
(2.3)
182
J. Y. PARK AND M. J. LEE
Given u ∈ Uad and y satisfying (2.1) we set
J(y, u) =
1
y − zd 2 + 1 Nu, u ,
2
2
(2.4)
where zd is given in L2 (Q) and N > 0. The problem of optimal control is find inf J(y,u),
u ∈ Uad , u and y being connected by (2.1). The above problem admits a unique solution {u, y} which is characterized by the solution {u, y, p} of the optimality system
∂2y
+ ∆y = u
∂t 2
y = 0 on
,
for u ≥ 0,
y(x, 0; u) = y0 (x),
u≥0
∂y
(x, 0; u) = y1 (x) in Ω,
∂t
(α1 )
a.e., in Q,
∂2p
+ ∆p = y − ᐆd
∂t 2
p = 0 on
,
p(T ) = 0,
p + Nu ≥ 0
in Q,
∂p
(T ) = 0
∂t
in Q,
(β1 )
in Ω,
p(0) ≥ 0,
u(p + Nu) = 0,
∂p(0)
y(0) = 0,
∂t
∂y(0)
p(0) = 0
∂t
in Ω.
(γ1 )
Next we prove the duality theorem.
Theorem 1. Let J = 1/2|y − ᐆd |2 + 1/2(Nu, u), K = −1/2|y|2 +1/2|ᐆd |2 − 1/2(Nu,
u). Assume y0 , u0 , p0 satisfy (α1 ), (β1 ), (γ1 ); y, u in J satisfy (α1 ); and y, u in K
satisfy (β1 ). Then
(2.5)
inf J = J y0 , u0 = K y0 , u0 = sup K.
(α1 )
(β1 )
Proof. (i) We begin by showing that J = K at y0 , u0 , p0 .
J y0 , u0 = J y0 , u0 − u0 , p0 − u0 , Nu0
(2.6)
∂ 2 p0
+
∆p
− u0 , Nu0 = K y0 , u0 .
= J y 0 , u0 − y 0 ,
0
2
∂t
(ii) To show inf (α1 )J = J y0 , u0 , we must check that J(y, u) ≥ J y0 , u0 , where
(y, u) satisfy (α1 ) and y0 , u0 , p0 satisfy (α1 ), (β1 ), (γ1 ). Now, we have
J(y, u) − J y0 , u0 ≥ y0 − zd , y − y0 + Nu0 , u − u0
2
∂ p0
+
∆p
,
y
−
y
+ Nu0 , u − u0
=
(2.7)
0
0
∂t 2
= p0 + Nu0 , u − u0 ≥ 0.
Thus inf J = J y0 , u0 .
183
DUALITY IN THE OPTIMAL CONTROL OF HYPERBOLIC EQUATIONS . . .
(iii) To show sup(β1 ) K = K y0 , u0 , we have to check that K(y, u) ≤ K y0 , u0 , where
(y, u) satisfy (β1 ). But we have
J y0 , u0 − J(y, u) ≥ y − zd , y0 − y + Nu, u0 − u =
∂2p
+ ∆p, y0 − y
∂t 2
2
∂ y0
+
∆y
−
u
,
p
−
p
+ Nu, u0 − u +
0
0
0
∂t 2
= − y − zd , y − (Nu, u) + y0 − zd , y0 + Nu0 , u0
+ u0 , Nu + p − u0 , p0 + Nu0
≥ − y − zd , y − (Nu, u) + y0 − zd , y0 + Nu0 , u0 .
(2.8)
Therefore,
J y0 , u0 − y0 − zd , y0 − Nu0 , u0 ≥ J(y, u) − y − zd , y − (Nu, u),
(2.9)
or
K y0 , u0 ≥ K(y, u).
(2.10)
This completes the proof.
Now, we define the cost functional as
J=
1
y(T ; u) − zd 22 + 1 (Nu, u).
L (Ω)
2
2
(2.11)
In Q = Ω × (0, T ), we consider the following system:
∂2y
+ ∆y = u,
∂t 2
y = 0, on
,
y(x, 0; u) = y0 ,
u≥0
(α2 )
∂y(x, 0; u)
= y1 ,
∂t
a.e., in Q,
∂2p
+ ∆p = 0 in Q,
∂t 2
p = 0 on
,
p(x, T ; u) = 0
for x ∈ Ω,
∂p(x, T ; u)
= y(x, T ; u) − zd
∂t
− p + Nu ≥ 0 in Q,
u(−p + Nu) = 0
∂p(0)
y(0) = 0,
∂t
(β2 )
for x ∈ Ω,
a.e., in Q,
∂y(0)
p(0) = 0
∂t
on Ω.
(γ2 )
184
J. Y. PARK AND M. J. LEE
Theorem 2. Let J = 1/2|y(T ; u)−zd |2L2 (Ω) +1/2(Nu, u), K = −1/2|y(T ; u)|2L2 (Ω) +
1/2|zd |2L2 (Ω) −1/2(Nu, u). Assume y0 , u0 , p0 satisfy (α2 ), (β2 ), (γ2 ); y, u in J satisfy (α2 );
and y, u in K satisfy (β2 ). Then
inf J = J y0 , u0 = K y0 , u0 = sup K.
(α2 )
(β2 )
(2.12)
Proof. (i) Now we prove that J = K at y0 , u0 , p0 . Then
J y0 , u0 = J y0 , u0 + p0 , u0 − Nu0 , u0
2
∂ y0
+
∆y
,
p
= J y 0 , u0 +
0
0 − Nu0 , u0
(2.13)
2
∂t
1
1
1
2
= − y0 x, T ; u L2 (Ω) + |zd |2L2 (Ω) − Nu0 , u0 = K y0 , u0 .
2
2
2
(ii) We show that J(y, u) ≥ J y0 , u0 , where (y, u) satisfy (α2 ) and y0 , u0 , p0 satisfy (α2 ), (β2 ), (γ2 ).
J(y, u) − J y0 , u0 ≥ y0 (T ; u) − zd , y(T ; u) − y0 (T ; u) L2 (Ω) + Nu0 , u − u0
= y0 (T ; u) − zd , y(T ; u) − y0 (T ; u) L2 (Ω) + Nu0 , u − u0
2
(2.14)
∂ p0
+
∆p
,
y(t;
u)
−
y
(t;
u)
−
0
0
∂t 2
= − p0 + Nu0 , u − u0 ≥ 0.
(iii) Now we claim that sup(β2 ) K = K y0 , u0 .
J y0 , u0 − J(y, u) ≥ y(T ; u) − zd , y0 (T ; u) − y(T , u) L2 (Ω) + Nu, u0 − u
= y(T ; u) − zd , y0 (T ; u) L2 (Ω) − y(T ; u) − zd , y(T ; u) L2 (Ω)
2
∂ y0
+ Nu, u0 − (Nu, u) −
+
∆y
−
u
,
p
−
p
0
0
0
∂t 2
= − y(T ; u) − zd , y(T ; u) L2 (Ω) − (Nu, u)
+ y0 (T ; u) − zd , y0 (T ; u) l2 (Ω) + Nu0 , u0
+ − p + Nu, u0 + p0 − Nu0 , u0 .
(2.15)
Therefore,
J y0 , u0 − y0 (T ; u) − zd , y0 (T ; u) L2 (Ω) − Nu0 , u0
≥J(y, u) − y(T ; u) − zd , y(T ; u) L2 (Ω) − (Nu, u),
(2.16)
and this implies
sup K(y, u) = K y0 , u0 .
(β2 )
Thus, the proof is complete.
(2.17)
DUALITY IN THE OPTIMAL CONTROL OF HYPERBOLIC EQUATIONS . . .
185
3. Duality in the optimal control of hyperbolic equation with fourth-order operator. Let us consider the fourth-order differential operator.
Uad = u | u ∈ L2 (Q), u ≥ 0 in Q ,
∆=
n
∂2
.
∂xi2
i=1
We consider a function a(x, t) such that
a ∈ C 1 ]0, T ]; L∞ (Ω) .
We introduce
V = φ | φ, ∆φ ∈ L2 (Ω) ,
and
a t; φ, ψ =
(3.1)
(3.2)
H = L2 (Ω)
(3.3)
Ω
a(x, t) ∆φ ∆ψ dx,
∀φ, ψ ∈ V ,
(3.4)
given u ∈ Uad and we set
J(y, u) =
1
y − zd 2 + 1 Nu, u ,
2
2
(3.5)
where zd ∈ L2 (Q), u ∈ Uad and N > 0.
∂2y
+ ∆(a∆y) = u,
∂t 2
∂∆
∆y = 0,
y = 0,
∂n
on
,
∆y(x, 0; u) = y0 (x),
∂y
(x, 0; u) = y1 (x),
∂t
u ≥ 0,
∂∆
y(0) ≥ 0.
∂n
∂y(0) ≥ 0,
∂p
+ ∆(a∆p) = y − zd in Q,
∂t 2
∂∆p
∆p = 0,
= 0 on
,
∂n
p(x, T ; u) = 0,
p + Nu ≥ 0
∂p(x, T ; u)
=0
∂t
x ∈ Ω,
(α3 )
(β3 )
on Ω,
in Q.
u(p + Nu) = 0,
∂p(0)
y(0) = 0,
∂t
∂y(0)
p(0) = 0
∂t
on Ω,
(γ3 )
Now we claim the following.
Theorem 3. Let J = (1/2)|y − zd |2 + (1/2)(Nu, u), K = −(1/2)|y|2 + (1/2)|zd |2 −
(1/2)(Nu, u). Assume y0 , u0 , p0 satisfy (α3 ), (β3 ), (γ3 ); y, u in J satisfy (α3 ) and y, u
in K satisfy (β3 ). Then
inf J = J y0 , u0 = K y0 , u0 = sup K.
(3.6)
(α3 )
(β3 )
186
J. Y. PARK AND M. J. LEE
Proof. (i) We begin by showing that J = K at y0 , u0 , p0 .
J y0 , u0 = J y0 , u0 − u0 , p0 − Nu0 , u0
2
∂ y0
+
∆
a∆y
,
p
= J y 0 , u0 −
0
0 − Nu0 , u0
∂t 2
= J y0 , u0 − y0 , y0 − zd − Nu0 , u0 = K y0 , u0 .
(3.7)
(ii) We show that J(y, u) ≥ J y0 , u0 .
J(y, u) − J y0 , u0 ≥ (y0 − zd , y − y0 ) + Nu0 , u − u0
2
∂ p0
+
∆
a∆p
,
y
−
y
=
0
0 + Nu0 , u − u0
2
∂t
+ Nu0 , u − u0 = p0 + Nu0 , u − u0 ≥ 0.
(3.8)
(iii) We prove that K(y, u) ≤ K y0 , u0 .
J y0 , u0 − J(y, u) ≥ y − zd , y0 − y + Nu, u0 − u
2
∂ p
+
∆(a∆P
),
y
−
y
+ Nu, u0 − u
=
0
∂t 2
2
∂ y0
+
∆
a∆y
,
p
−
p
(3.9)
−
u
+
0
0
0
∂t 2
= − y − zd , y + y0 − zd , y0 + p + Nu, u0 − Nu0 + p0 , u0
− (Nu, u) + Nu0 , u0 + (u0 , p) − y − ᐆd , y0
≥ − y − zd , y + y0 − z0 , y0 − (Nu, u) + Nu0 , u0 .
Therefore,
K y0 , u0 ≥ K(y, u).
(3.10)
Now, we set the following cost function:
J=
1
y(T ; u) − zd 22 + 1 Nu, u ,
L (Ω)
2
2
(3.11)
where zd ∈ L2 (Q), u ∈ Uad and N > 0 and associated following systems:
∂2y
+ ∆(a∆y) = u,
∂t 2
∂∆
∆y = 0,
y =0
∂n
on
,
∆y(x, 0; u) = y0 (x),
∂y
(x, 0; u) = y1 (x),
∂t
u ≥ 0,
∂∆
y(0) ≥ 0.
∂n
∂y(0) ≥ 0,
x ∈ Ω,
(α4 )
187
DUALITY IN THE OPTIMAL CONTROL OF HYPERBOLIC EQUATIONS . . .
∂2p
+ ∆(a∆p) = 0 in Q,
∂t 2
∂∆
∆p = 0,
p = 0 on
∂n
∂p
p(x, T ; u) = 0,
(x, T ; u) = y(x, T ; u) − zd
∂t
− p + Nu ≥ 0 in Q,
(β4 )
on Ω,
u(p + Nu) = 0,
∂p(0)
y(0) = 0,
∂t
∂y(0)
p(0) = 0
∂t
(γ4 )
on Ω.
we have the duality result.
Theorem 4. Let J=(1/2)|y(T ; u)−zd |2L2 (Ω)+(1/2)(Nu, u), K = −(1/2)|y(T ; u)|2L2 (Ω)
+(1/2)|zd |2L2 (Ω) −(1/2)(Nu, u). Assume y0 , u0 , p0 satisfy (α4 ), (β4 ), (γ4 ); y, u in J satisfy (α4 ) and y, u in K satisfy (β4 ). Then
inf J = J y0 , u0 = K y0 , u0 = sup K.
(α4 )
(β4 )
(3.12)
Proof. (i) We begin to prove that
J y0 , u0 = J y0 , u0 − u0 , p0 − Nu0 , u0
2
∂ y0
+
∆
a∆y
− Nu0 , u0
,
p
= J y 0 , u0 −
0
0
2
∂t
= J y0 , u0 − y0 (T ; u) − zd , y0 (T ; u) L2 (Ω)
∂ 2 p0
+ y0 ,
+
∆
a∆p
− Nu0 , u0
0
2
∂t
2
1 2
1
1
= − y0 (T ; u)L2 (Ω) + zd L2 (Ω) − Nu0 , u0 = K y0 , u0 .
2
2
2
(3.13)
(ii) We claim that J(y, u) ≥ J(y0 , u0 ).
J y, u − J y0 , u0 ≥ y0 (T ; u) − zd , y(T ; u) − y0 (T ; u) L2 (Ω) + Nu0 , u − u0
= y0 (T ; u) − zd , y(T ; u) − y0 (T ; u) L2 (Ω) + Nu0 , u − u0
2
(3.14)
∂ p0
+
∆
a∆p
(t;
u)
,
y(t;
u)
−
y
−
0
0
2
∂t
= − p + Nu0 , u − u0 ≥ 0.
(iii) Now, we verify that K(y, u) ≤ K y0 , u0 .
J y0 , u0 − J y, u ≥ y(T ; u) − zd , y0 (T ; u) − y(T ; u) L2 (Ω) + Nu, u0 − u
= y(T ; u) − zd , y0 (T ; u) L2 (Ω) − y(T ; u) − zd , y(T ; u) L2 (Ω)
188
J. Y. PARK AND M. J. LEE
2
∂ y0
+ Nu, u0 − (Nu, u) −
+
∆
a∆y
,
p
−
p
−
u
0
0
0
∂t 2
= − y(T ; u) − zd , y(T ; u) L2 (Ω) − (Nu, u)
+ y0 (T ; u) − zd , y0 (T ; u) L2 (Ω) + Nu0 , u0
+ − p + Nu, u0 + p0 − Nu0 , u0 .
(3.15)
This implies that
sup K = K y0 , u0 .
(β4 )
(3.16)
Acknowledgement. This work was supported by KOSEF, 1996.
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Park: Department of Mathematics, Pusan National University, Pusan 609-735, Korea
E-mail address: jyepark@hyowon.pusan.ac.kr
Lee: Department of Mathematics, Pusan National University, Pusan 609-735, Korea