IJMMS 2003:27, 1703–1714
PII. S0161171203209273
http://ijmms.hindawi.com
© Hindawi Publishing Corp.
DUALITY IN THE OPTIMAL CONTROL FOR DAMPED
HYPERBOLIC SYSTEMS WITH POSITIVE CONTROL
MI JIN LEE, JONG YEOUL PARK, and YOUNG CHEL KWON
Received 10 September 2002
We study the duality theory for damped hyperbolic equations. These systems have
positive controls and convex cost functionals. Our main results lie in the application of duality theorem, that is, inf J = sup K, on various cost functions.
2000 Mathematics Subject Classification: 49J20, 49N15, 93C20.
1. Introduction. Lions [2] introduced optimal control problems of the variety of distributed parameter systems, for example, elliptic, parabolic, and
hyperbolic. Here, we see that adjoint state systems are given cost functional
and distributed parameter systems. Duality theory is inf J = sup K, where J satisfies systems and K satisfies adjoint state systems. The duality theory for the
corresponding parabolic systems has been given by Chan [1] and Tanimoto [7].
Park and Lee [5, 6] studied the duality theory for hyperbolic systems. Also, they
[4] obtained same results for hyperbolic systems with damping terms. In this
paper, we study the duality theory for damped hyperbolic systems with positive controls. These systems have various convex cost functionals. The main
objective is to prove the duality theorem for damped hyperbolic systems with
positive controls and various cost functions. The main tools are integration by
parts and Green’s formula.
2. Preliminaries. Let X be a Hilbert space (·, ·) and let · X denote the
inner product and the induced norm on X; X denotes the dual space of X and
·, ·X ,X denotes the dual pairing between X and X. We introduce underlying
Hilbert spaces to describe damped second-order evolution equations. Let H
be a real pivot Hilbert space; its norm · H is simply denoted by | · |H . For
i = 1, 2, let Vi be a real separable Hilbert space. Assume that each pair (Vi , H)
is a Gelfand triple space with a notation Vi H ≡ H Vi . We suppose that
V1 is continuously embedded in V2 . Then we see that V1 V2 H ≡ H V2 V and the equalities φ, ψV1 ,V1 = φ, ψV2 ,V2 for φ ∈ V2 , ψ ∈ V , and
φ, ψV1 ,V1 = (φ, ψ)H for φ ∈ H, ψ ∈ V hold. Let T be a positive number. We
define a function space W (0, T ) by
W (0, T ) = y | y ∈ L2 0, T ; V1 , y ∈ L2 0, T ; V2 , y ∈ L2 0, T ; V1
(2.1)
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MI JIN LEE ET AL.
with an inner product
y1 , y2 W (0,T ) =
T 0
y1 (t), y2 (t) V1 + y1 (t), y2 (t) V2 + y1 (t), y2 (t) V dt.
1
(2.2)
This becomes a Hilbert space with norm
1/2
,
yW (0,T ) = y2L2 (0,T ;V ) + y 2L2 (0,T ;V ) + y 2L2 (0,T ;V )
1
2
1
(2.3)
where = d/dt and = d2 t/dt 2 .
From now on, we set V = V2 = H01 (Ω) and H = L2 (Ω), where Ω is a bounded
open set in Rn with smooth boundary Γ , and let Q = Ω ×(0, T ). We will give an
exact description of damped second-order evolution equations. We consider
the bilinear forms defined by
a1 (t; φ, ψ) =
n Ω
i,j=1
aij (t, x)
∂φ(x) ∂ψ(x)
dx
∂xi
∂xj
+
a2 (t; φ, ψ) =
Ω
a0 (t, x)φ(x)ψ(x)dx
n i,j=1
∀φ, ψ ∈ V ,
(2.4)
∂φ(x) ∂ψ(x)
bij (t, x)
dx
∂xi
∂xj
Ω
+
Ω
b0 (t, x)φ(x)ψ(x)dx
∀φ, ψ ∈ V ,
where aij , bij , a0 , and b0 are the functions satisfying the following properties:
(i) aij = aji and bij = bji ,
(ii) aij , bij , a0 , b0 ∈ C 1 ([0, T ]; L∞ (Ω)),
n
ai,j (t, x)ξi ξj ≥ c1 (ξ12 + · · · + ξn2 ), c1 > 0, ξi ∈ R,
(iii)
i,j=1
n
2
2
(iv)
i,j=1 bi,j (t, x)ξi ξj ≥ c2 (ξ1 + · · · + ξn ), c2 > 0, ξi ∈ R.
Using the above properties, we can show the coercivity condition of a1 and a2 .
Indeed, by (i) and (ii), there exists K > 0 such that |a0 (t, x)| ≤ K a.e., x ∈ Ω,
and for all t ∈ [0, T ]. The coercivity condition of a1 follows from
a1 (t; φ, ψ) ≥ c1
≥
n ∂φ(x) 2
φ(x)2 dx
∂x dx − K
Ω
Ω
i
i=1
(2.5)
c1 φ2H 1 (Ω) − K|φ|2L2 (Ω) .
0
Similarly, we can show the coercivity condition of a2 . Then we can define the
operator Ai (t) ∈ ᏸ(Vi , Vi ) for t ∈ [0, T ] deduced by the relation
ai (t; φ, ψ) = Ai (t)φ, ψ V ,Vi
i
∀φ, ψ ∈ Vi , i = 1, 2.
(2.6)
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DUALITY IN THE OPTIMAL CONTROL . . .
Let y0 ∈ H01 (Ω) and y1 ∈ L2 (Ω). Then by Nakagiri and Ha [3], there exists a
solution y ∈ W (0, T ) of
∂y(u; t)
∂ 2 y(u; t)
+ A1 (t)y(u; t) = u in (0, T ),
+ A2 (t)
∂t 2
∂t
y(u; 0) = y0 , y (u; 0) = y1 in Ω, y = 0 on Σ.
(2.7)
3. Duality. We choose a control variable space ᐁ = L2 (Q). Let u ∈ ᐁad =
{u | u ≥ 0 a.e. in Q} ⊂ ᐁ and y satisfying (2.7). we consider a cost functional
given by
J(y, u) =
1
2
T
0
y(u) − zd 2 dt + 1
2
T
0
(Nu, u)dt.
(3.1)
For our purpose, we consider the following systems:
y (u; t) + A2 (t)y (u; t) + A1 (t)y(u; t) = u in Q,
y (u; 0) = y1
y(u; 0) = y0 ,
u ≥ 0 a.e. in Q,
in Ω,
(3.2)
y(0) ≥ 0,
y (0) ≥ 0,
y =0
on Σ,
p (u; t) − A2 (t)p (u; t) + A1 (t) − A2 (t) p(u; t) = y − zd
p=0
on Σ,
p + Nu ≥ 0
p(u; T ) = 0,
in Q,
in Q,
p (u; T ) = 0
p(0) ≥ 0,
(3.3)
p (0) ≥ 0,
u(p + Nu) = 0,
p(0)y(0) = 0
y (0)p(0) = 0,
in Ω,
(3.4)
y(0)p (0) = 0.
T
T
Theorem 3.1. Let J = (1/2) 0 |y − zd |2 dt + (1/2) 0 (Nu, u)dt and K =
T
T
T
−(1/2) 0 |y|2 dt + (1/2) 0 |zd |2 dt − (1/2) 0 (Nu, u)dt. Assume that y0 , u0 ,
and p0 satisfy (3.2), (3.3), and (3.4), respectively, y and u in J satisfy (3.2), and
y and u in K satisfy (3.3). Then
inf J = J y0 , u0 = K y0 , u0 = sup K.
(3.2)
(3.5)
(3.3)
Proof. (i) We begin by showing that J = K at (y0 , u0 , p0 ).
J y0 , u 0 = J y 0 , u 0 −
= J y0 , u0 −
T
0
T
0
u0 , p0 dt −
T
0
u0 , Nu0 dt
y0 + A2 (t)y0 + A1 (t)y0 , p0 dt −
T
0
u0 , Nu0 dt
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MI JIN LEE ET AL.
= J y0 , u 0 −
−
T
0
T
0
y0 , p0 − A2 (t)p0 − A2 (t)p0 + A1 (t)p0 dt
u0 , Nu0 dt
= J y0 , u0 −
T
0
y0 , y0 − zd dt −
T
0
u0 , Nu0 dt.
(3.6)
(ii) To show inf (3.2) J = J(y0 , u0 ), we must check that J(y, u) ≥ J(y0 , u0 )
under (3.2) for (y, u, p) and under (3.2), (3.3), and (3.4) for (y0 , u0 , p0 ). Now,
we have
T
T
y0 − zd , y − y0 dt +
Nu0 , u − u0 dt
J(y, u) − y y0 , u0 ≥
0
=
T
0
+
=
0
+
=
p0 − A2 (t)p0 + A1 (t) − A2 (t) p0 , y − y0 dt
T
T
−
0
0
p0 , y + A2 (t)y + A1 (t)y dt
T
0
T
T
0
Nu0 , u − u0 dt
0
(3.7)
p0 , y0 + A2 (t)y0 + A1 (t)y0 dt
Nu0 , u − u0 dt
p0 , u − u0 dt +
T
0
Nu0 , u − u0 dt
≥ 0.
(iii) We claim that K(y, u) ≤ K(y0 , u0 ) under (3.3) for (y, u, p):
J y0 , u0 − J(y, u)
≥
=
T
0
T
0
+
T
0
Nu, u − u0 dt
p − A2 (t)p + A1 (t) − A2 (t) p, y0 − y dt +
T
0
=−
+
y − zd , y − y0 dt +
T
0
T
0
0
Nu, u0 − u dt
y0 + A2 (t)y0 + A1 (t)y0 − u0 , p0 − p dt
y − zd , y dt −
T
Nu0 , u0 dt +
T
0
T
0
(Nu, u)dt +
T
0
y0 , y0 − zd dt
Nu + p, u0 dt −
T
0
u0 , p0 + Nu0 dt
1707
DUALITY IN THE OPTIMAL CONTROL . . .
≥−
+
T
0
T
0
y − zd , y dt −
T
0
y0 , y0 − zd dt +
(Nu, u)dt
T
0
Nu0 , u0 dt.
(3.8)
Therefore
K y0 , u0 ≥ K(y, u).
(3.9)
This completes the proof.
Now, we observe the terminal value of y(u; t). Since the observation z(u)
is given by y(u; T ), the cost function is given as
J(y, u) =
1
y(T ) − zd 2 + 1
2
2
T
0
(Nu, u)dt.
(3.10)
We introduce the following systems:
y (u; t) + A2 (t)y (u; t) + A1 (t)y(u; t) = u in Q,
y (u; 0) = y1
y(u; 0) = y0 ,
u ≥ 0 a.e. in Q,
y(0) ≥ 0,
in Ω,
(3.11)
y (0) ≥ 0,
y =0
on Σ,
p (u; t) − A2 (t)p (u; t) + A1 (t) − A2 (t) p(u; t) = 0
p=0
in Q,
on Σ,
(3.12)
p (u; T ) = y(T ) − zd ,
p(u; T ) = 0,
− p + Nu ≥ 0
in Q,
p(0) ≥ 0,
p (0) ≥ 0,
u(−p + Nu) = 0,
p(0)y(0) = 0
in Ω,
y (0)p(0) = 0,
y(0)p (0) = 0.
(3.13)
T
Theorem 3.2. Let J = (1/2)|y(T ) − zd |2 + (1/2) 0 (Nu, u)dt and K =
T
−(1/2)|y(T )|2 + (1/2)|zd | − (1/2) 0 (Nu, u)dt. Assume that y0 , u0 , and p0
satisfy (3.11), (3.12), and (3.13), respectively, y and u in J satisfy (3.11), and y
and u in K satisfy (3.12). Then
inf J = J y0 , u0 = K y0 , u0 = sup K.
(3.11)
(3.12)
(3.14)
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MI JIN LEE ET AL.
Proof. (i) We now prove that J(y0 , u0 ) = K(y0 , u0 ):
J y0 , u 0 = J y 0 , u 0 +
= J y0 , u0 +
T
0
T
0
p0 , u0 dt −
T
0
Nu0 , u0 dt
y0 + A2 (t)y0 + A1 (t)y0 , p0 dt −
T
0
Nu0 , u0 dt
2 1 T
1
Nu0 , u0 dt
= y0 (T ) − zd − y0 (T ) − zd , y0 (T ) −
2
2 0
T
1
1
1
2
2
= − y0 (T ) + zd −
Nu0 , u0 dt
2
2
2 0
= K y0 , u0 .
(3.15)
(ii) We show that J(y, u) ≥ J(y0 , u0 ) under (3.11) for (y, u, p) and under
(3.11), (3.12), and (3.13) for (y0 , u0 , p0 ):
J(y, u) − J y0 − u0 ≥ y0 (T ) − zd , y(T ) − y0 (T ) +
= y0 (T ) − zd , y(T ) − y0 (T ) +
−
T
0
T
0
T
0
Nu0 , u − u0 dt
Nu0 , u − u0 dt
p − A2 (t)p + A1 (t) − A2 (t) p, y − y0 dt
= y0 (T ) − zd , y(T ) − y0 (T )
+
−
=
T
0
T
0
T
0
Nu0 , u − u0 dt − y0 (T ) − zd , y(T ) − y0 (T )
p0 , y − y0 + A2 (t)y − A2 (t)y0
+ A1 (t)y − A1 (t)y0 dt
− p0 + Nu0 , u − u0 dt
≥ 0.
(3.16)
(iii) We have to check that K(y, u) ≤ K(y0 , u0 ) under (3.12) for (y, u, p):
J y0 , u0 − J(y, u) ≥ y(T ) − zd , y0 (T ) − y(T ) +
= y(T ) − zd , y0 (T ) − y(T ) +
−
T
0
T
0
T
0
Nu, u0 − u dt
Nu, u0 − u dt
y0 + A2 (t)y0 + A1 (t)y0 − u0 , p0 − p dt
DUALITY IN THE OPTIMAL CONTROL . . .
= − y(T ) − zd , y(T ) −
+
+
T
T
Nu0 , u0 dt +
p0 − Nu0 , u0 dt
0
T
0
T
0
0
≥ − y(T ) − zd , y(T ) −
1709
(Nu, u) + y0 (T ) − zd , y0 (T ) dt
− p + Nu, u0 dt
T
0
+ y0 (T ) − zd , y0 (T ) +
(Nu, u)dt
T
Nu0 , u0 dt.
0
(3.17)
This shows that
K y0 , u0 ≥ K(y, u).
(3.18)
Therefore, Theorem 3.2 is proved.
When the observation z(u) is given by z(u) = y (u), the cost function is
defined as
1
2
J(y , u) =
T
0
y (u) − zd 2 dt + 1
2
T
0
(Nu, u)dt.
(3.19)
We will consider the following systems:
y (u; t) + A2 (t)y (u; t) + A1 (t)y(u; t) = u in (0, T ),
y(u; 0) = y0 ,
u≥0
y (u; 0) = y1
a.e. in Q,
y(0) ≥ 0,
in Ω,
p (u; t) − A2 (t)p (u; t) + A1 (t)p(u; t) +
= y (u; t) − zd
p=0
on Σ,
− p + Nu ≥ 0
y =0
T
t
on Σ,
A1 (σ )p(u; σ )dσ
in Q,
p(u; T ) = 0,
in Q,
(3.20)
y (0) ≥ 0,
(3.21)
p (u; T ) = 0,
p(0) ≥ 0,
p (0) ≥ 0,
u(−p + Nu) = 0
p(0)y(0) = 0, y (0)p(0) = 0, y(0)p (0) = 0,
T
y(0)
A (σ )p(σ )dσ = 0 in Ω.
(3.22)
0
T
T
Theorem 3.3. Let J = (1/2) 0 |y − zd |2 dt + 0 (1/2)(Nu, u)dt and K =
T 2
T
T
−(1/2) 0 |y | dt + (1/2) 0 |zd |2 dt − (1/2) 0 (Nu, u)dt. Assume that y0 , u0 ,
and p0 satisfy (3.20), (3.21), and (3.22), respectively, y and u in J satisfy (3.20),
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MI JIN LEE ET AL.
and y and u in K satisfy (3.21). Then
inf J = J y0 , u0 = K y0 , u0 = sup K.
(3.20)
(3.23)
(3.21)
Proof. (i) First, we claim that J(y0 , u0 ) = K(y0 , u0 ):
J y0 , u0 = J y0 , u0 +
= J y0 , u0 +
= J y0 , u0 +
T
0
T
0
p0 , u0 dt −
T 0
=−
1
2
T
0
T
0
0
Nu0 , u0 dt
p0 , y0 + A2 (t)y0 + A1 (t)y0 dt −
T
0
Nu0 , u0 dt
− p0 + A2 (t)p0 − A1 (t)p0
−
= J y0 , u0 −
T
T
t
A (σ )p(σ )dσ , y
y0 − zd dt −
2
y dt + 1
0
2
T
0
T
0
dt −
T
0
Nu0 , u0 dt
Nu0 , u0 dt
2
zd dt − 1
2
T
0
Nu0 , u0 dt.
(3.24)
(ii) Second, we must show that inf J = J(y0 , u0 ):
J(y , u) − J y0 , u0 ≥
T
0
y0 − zd , y − y0 dt +
T
0
Nu0 , u − u0 dt
T =
p0 − A2 (t)p0 + A1 (t)p0
0
+
+
=
T
0
+
=
T
0
t
A1 (σ )p0 (σ )dσ , y − y0
dt
Nu0 , u − u0 dt
p0 , − y − y0 − A2 (t) y − y0 − A1 (t) y − y0 dt
T
T
0
T
0
Nu0 , u − u0 dt
− p0 + Nu0 , u − u0 dt
≥ 0.
(3.25)
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DUALITY IN THE OPTIMAL CONTROL . . .
(iii) Finally, we check that sup K = K(y0 , u0 ):
J y0 , u0 − J(y , u)
≥
T
0
y − zd , y0 , y dt +
T
0
Nu, u0 − u dt
T
T A1 (σ )p(σ )dσ , y0 − y dt
=
p − A2 (t)p + A1 (t)p +
0
−
t
T
0
=−
T
0
y0 + A2 (t)y0 + A1 (t)y0 − u0 , p0 − p dt +
y − zd , y dt +
T
0
T
0
Nu, u0 − u dt
p , −y0 − A2 (t)y0 − A1 (t)y0 dt
T T
−
A (σ )p0 (σ )dσ dt
y0 , −p0 + A2 (t)p0 − A1 (t)p0 −
0
+
+
T
0
T
0
=−
+
−
0
T
0
T
0
≥−
+
T
T
0
T
0
t
Nu, u0 dt −
u0 , p0 dt −
T
0
T
0
(Nu, u)dt +
u0 , p dt +
T
0
T
0
0
u0 , p dt
u0 , p dt
y − zd , y dt −
Nu, u0 dt +
T
T
0
(Nu, u)dt +
T
0
y0 , y0 − zd dt
u0 , p0 dt
Nu0 , u0 dt −
y − zd , y dt −
y0 − zd , y0 dt +
T
T
0
T
0
0
Nu0 , u0 dt
(Nu, u)dt
Nu0 , u0 dt.
(3.26)
This implies that
K y0 , u0 ≥ K(y, u).
(3.27)
So, we claimed Theorem 3.3.
In this case, we observe the terminal value y (u; T ). Since the observation
z(u) is given by y (u; T ), the cost functional is given as
J(y , u) =
1
y (u; T ) − zd 2 + 1
2
2
T
0
(Nu, u)dt.
(3.28)
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MI JIN LEE ET AL.
We introduce the following systems:
y (u; t) + A2 (t)y (u; t) + A1 (t)y(u; t) = u in Q,
y (u; 0) = y1
y(u; 0) = y0 ,
u≥0
in Ω,
y(0) ≥ 0,
a.e. in Q,
(3.29)
y (0) ≥ 0
in Ω,
y =0
p (u; t) − A2 (t)p (u; t) + A1 (t) − A2 (t) p(u; t) = 0
on Σ,
in Q,
p = 0 on Σ,
p(u; T ) = y (T ) − zd ,
p + Nu ≥ 0
p(0) ≥ 0,
(3.30)
p (u; T ) = A2 (T )y (T ) − zd ,
in Q,
p (0) ≥ 0
in Ω,
u(p + Nu) = 0,
p(0)y(0) = 0,
(3.31)
y (0)p(0) = 0,
y(0)p (0) = 0
in Ω.
T
Theorem 3.4. Let J = (1/2)|y (T ) − zd |2 + (1/2) 0 (Nu, u)dt and K =
T
−(1/2)|y (T )|2 + (1/2)|zd | − (1/2) 0 (Nu, u)dt. Assume that y0 , u0 , and p0
satisfy (3.29), (3.30), and (3.31), respectively, y and u in J satisfy (3.29), and y
and u in K satisfy (3.30). Then
inf J = J y0 , u0 = K y0 , u0 = sup K.
(3.29)
(3.32)
(3.30)
Proof. (i) First, we show that J = K at (y0 , u0 , p0 ):
J y0 , u0 = J y0 , u0 −
= J y0 , u0 −
=
T
0
T
0
u0 , p0 dt −
−
0
0
u0 , Nu0 dt
y0 + A2 (t)y0 + A1 (t)y0 , p0 dt −
1
y (T ) − zd 2 + 1
0
2
2
T
T
T
0
T
0
Nu0 , u0 dt
Nu0 , u0 dt − y0 (T ) − zd , y0 (T )
Nu0 , u0 dt
2 1 2 1
1
= − y0 (T ) + zd −
2
2
2
= K y0 , u0 .
T
0
Nu0 , u0 dt
(3.33)
DUALITY IN THE OPTIMAL CONTROL . . .
1713
(ii) Second, we check that J(y , u) ≥ J(y0 , u0 ):
J(y , u) − J y0 , u0
≥ y0 (T ) − zd , y (T ) − y0 (T ) +
= y0 (T ) − zd , y (T ) − y0 (T ) +
+
=
T
0
+
=
T
0
0
0
T
0
Nu0 , u − u0 dt
Nu0 , u − u0 dt
p − A2 (t)p + A1 (t) − A2 (t) p, y − y0 dt
p, y − y0 + A2 (t)y − A2 (t)y0 + A1 (t)y − A1 (t)y0 dt
T
T
T
0
(3.34)
Nu0 , u − u0 dt
p + Nu0 , u − u0 dt
≥ 0.
(iii) Finally, we claim that K(y , u) ≤ K(y0 , u0 ):
J y0 , u0 − J(y , u)
≥ y0 (T ) − zd , y (T ) − y0 (T ) +
= y0 (T ) − zd , y (T ) − y0 (T ) +
+
T
0
T
0
Nu0 , u0 dt +
T
T
0
T
0
Nu0 , u − u0 dt
Nu0 , u − u0 dt
0
T
0
(Nu, u)dt + y0 (T ) − zd , y (T )
Nu + p, u0 dt −
≥ − y (T ) − zd , y (T ) −
+
0
y0 + A2 (t)y0 + A1 (t)y0 − u0 , p0 − p dt
= − y (T ) − zd , y (T ) −
+
T
T
0
T
0
(3.35)
p0 + Nu0 , u0 dt
(Nu, u)dt + y0 (T ) − zd , y (T )
Nu0 , u0 dt.
This implies that
K y0 , u0 ≥ K(y , u).
This completes the proof.
(3.36)
1714
MI JIN LEE ET AL.
Acknowledgment. This work was supported by the Korea Research Foundation Grant KRF-2001-037-DA0010.
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Mi Jin Lee: Department of Mathematics, Pusan National University (PNU), Pusan 609–
735, Korea
E-mail address: jin8719@hanmail.net
Jong Yeoul Park: Department of Mathematics, Pusan National University (PNU), Pusan
609–735, Korea
E-mail address: jyepark@pusan.ac.kr
Young Chel Kwon: Department of Mathematics, Dong-A University, Pusan 604–714,
Korea
E-mail address: yckwun@daunet.donga.ac.kr