1. Introduction
The first and second initial boundary value problem for Schr¨odinger in conical
domains were researched in [2, 3]. The unique solvability of the general boundary
value problems for Schr¨odinger systems in domains with conical point is completed
in [4].
In this paper, we are concerned with the regularity with respect to time variables of solutions of the problem in [4].
This paper includes following sections: In the first section, we define the problem. The regularity with respect to time variable is dealt with in sections 2. Finally,
in section 3, we apply the obtained results to a problem of mathematical physics.

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JOURNAL OF SCIENCE OF HNUE
Natural Sci., 2011, Vol. 56, No. 3, pp. 3-12
ON THE REGULARITY OF SOLUTION OF THE INITIAL
BOUNDARY VALUE PROBLEM FOR SCHRO¨DINGER SYSTEMS
IN CONICAL DOMAINS
Nguyen Thi Lien
Hanoi National University of Education
E-mail: Lienhnue@gmail.com
Abstract. The purpose of this paper is to establish the regularity with
respect to time variable of solution of the initial boundary value problems
for Schro¨dinger systems in the cylinders with base containing the conical
point.
Keywords: Regularity, generalized solution, conical domain.
1. Introduction
The first and second initial boundary value problem for Schro¨dinger in conical
domains were researched in [2, 3]. The unique solvability of the general boundary
value problems for Schro¨dinger systems in domains with conical point is completed
in [4].
In this paper, we are concerned with the regularity with respect to time vari-
ables of solutions of the problem in [4].
This paper includes following sections: In the first section, we define the prob-
lem. The regularity with respect to time variable is dealt with in sections 2. Finally,
in section 3, we apply the obtained results to a problem of mathematical physics.
2. Notations and formulation of the problem
Let Ω be a bounded domain in Rn (n ≥ 2) with the boundary ∂Ω. We suppose
that S = ∂Ω \ {0} is a smooth manifold and Ω in a neighbourhood U of the origin
0 coincides with the cone K = {x : x/ | x |∈ G}, where G is a smooth domain on
the unit sphere Sn−1 in Rn.
Let T be a positive real number or T = ∞. Set Ωt = Ω× (0, t), St = S× (0, t).
We use notations and the functional spaces in [4].
Now we introduce a differential operator of order 2m
L(x, t,D) =
m∑
|p|,|q|=0
(−1)|p|Dp
(
apq(x, t)D
q
)
,
3
Nguyen Thi Lien
where apq are s × s matrices whose smooth elements in ΩT , apq = a
⋆
pq (a
∗
qp is the
transportated conjugate matrix to apq).
We introduce also a system of boundary operators
Bj = Bj(x, t,D) =
∑
|p|≤µj
bj,p(x, t)D
p, j = 1, ..., m,
on S. Suppose that bj,p(x, t) are s× s matrices whose smooth elements in ΩT and
ordBj = µj ≤ m− 1 for j = 1, ..., χ,
m ≤ ordBj = µj ≤ 2m− 1 for j = χ+ 1, ..., m.
Assume that coefficients ofBj are independent of t if ordBj < m and {Bj(x, t,D)}
m
j=1
is a normal system on S for all t ∈ [0, T ], i.e., the two following conditions are sat-
isfied:
(i) µj 6= µk for j 6= k,
(ii) Boj (x, t, ν(x)) 6= 0 for all (x, t) ∈ ST , j = 1, ..., m.
Here ν(x) is the unit outer normal to S at point x and Boj (x, t,D) is the
principal part of Bj(x, t,D),
Boj = B
o
j (x, t,D) =
∑
|p|=µj
bj,p(x, t)D
p, j = 1, ..., m.
Furthermore, Boj (0, t, ν(x)) 6= 0 for all x ∈ S closed enough to the origin 0.
We set
HmB (Ω) =
{
u ∈ Hm(Ω) : Bju = 0 on S for j = 1, .., χ
}
with the same norm in Hm(Ω) and
B(t, u, v) =
m∑
|p|,|q|=0
∫
Ω
apqD
quDpvdx, t ∈ [0, T ].
We assume that B(t, ., .) satisfies the following inequality:
(−1)mB(t, u, u) ≥ µ0‖u(x, t)‖
2
Hm(Ω) (2.1)
for all u ∈ HmB (Ω) and t ∈ [0, T ], where µ0 is a positive constant independen of u
and t.
4
On the regularity of solution of the initial boundary value problem...
Assume that it can be choose boundary operators B′j on ST , j = 1, ..., m
satisfying
B(t, u, v) =
∫
Ω
Luvdx+
χ∑
j=1
∫
S
B′juBjvds+
m∑
j=χ+1
∫
S
BjuB
′
jvds. (2.2)
Denote H−mB (Ω) the dual space to H
m
B (Ω). We write
〈
., .
〉
to stand for the
pairing between HmB (Ω) and H
−m
B (Ω), and (., .) to define the inner product in L2(Ω).
We then have the continuous imbeddings HmB (Ω) →֒ L2(Ω) →֒ H
−m
B (Ω) with the
equation 〈
f, v
〉
= (f, v) for f ∈ L2(Ω) ⊂ H
−m
B (Ω), v ∈ H
m
B (Ω).
We study the following problem in the cylinder ΩT :
(−1)m−1iL(x, t,D)u− ut = f(x, t) in ΩT , (2.3)
Bju = 0, on ST , j = 1, ..., m, (2.4)
u |t=0= φ, on Ω, (2.5)
where f ∈ L2((0, T );H
−m
B (Ω)) and φ ∈ L2(Ω) are given functions.
The function u ∈ H((0, T );HmB (Ω), H
−m
B (Ω)) is called a generalized solution of
the problem (2.3) - (2.5) iff u(., 0) = φ and the equality
(−1)m−1iB(t, u, v)−
〈
ut, v
〉
=
〈
f(t), v
〉
(2.6)
holds for a.e. t ∈ (0, T ) and all v ∈ HmB (Ω).
3. The regularity with respect to time variable
For k ∈ N, u, v ∈ Hm,0(ΩT ), t ∈ [0, T ] we set
Btk(t, u, v) =
m∑
|p|,|q|=0
∫
Ω
∂kapq
∂tk
DquDpvdx,
BTtk(u, v) =
T∫
0
Btk(t, u, v)dt,
BT (u, v) = BTt0(u, v).
Now we improve slightly the regularity of generalized solutionu by making the
initial data φ and the right-hand side f more regularity. We denote X = L2(ΩT ) or
X = H((0, T );H−mB (Ω)).
5
Nguyen Thi Lien
Lemma 3.1. Let φ ∈ HmB (Ω) and f ∈ X. Then the generalized solution of problem
(2.3) - (2.5) belongs to H((0, T );HmB (Ω), L2(Ω)) and satisfies the following estimate
‖u‖H((0,T );Hm
B
(Ω),L2(Ω)) ≤ C(‖φ‖
2
Hm
B
(Ω) + ‖f‖
2
X). (3.1)
Here the constant C is independent of g, f, u.
Proof. (i) Let us consider first the case f ∈ L2(ΩT ).
Let uN be the functions defined as in the proof of Theorem 3.2 in [4] with Ck =
(φ, ψk), k = 1, 2, ... replaced by Ck = ‖ψk‖
−2
Hm
B
(Ω)(φ, ψk)HmB (Ω), where (., .)HmB (Ω) de-
notes the inner product in HmB (Ω). Remember that in [4] we have
(−1)m−1i
m∑
|p|,|q|=0
∫
Ω
apqD
quNDpψldx−
∫
Ω
uNt ψldx =
〈
f, ψl
〉
, l = 1, ..., N.
Multiplying both sides of this equality by
dCNl
dt
, then taking sum with respect to l
from 1 to N , we arrive at
−‖uNt ‖
2
L2(ΩT )
+ (−1)m−1i
m∑
|p|,|q|=0
∫
Ω
apqD
quNDpuNt dxdt =
T∫
0
(f, uNt )dt.
Adding with its complex conjugate, we obtain
‖uNt ‖
2
L2(ΩT )
= −Re
T∫
0
(f, uNt )dt. (3.2)
Using Cauchy’s inequality, we get
| − Re
T∫
0
(f, uNt )dt| ≤ ǫ‖u
N
t ‖
2
L2(ΩT )
+
1
4ǫ
‖f‖2L2(ΩT ) (0 < ǫ < 1).
Hence
‖uNt ‖
2
L2(ΩT )
≤ C‖f‖2L2(ΩT ).
Letting T tends to ∞ yields
‖uNt ‖
2
L2(ΩT )
≤ C‖f‖2L2(ΩT ) (3.3)
In [4], we had the following estimate
‖uN‖2L2((0,T );HmB (Ω)) ≤ C
(
‖f‖2
L2((0,T );H
−m
B
(Ω))
+ ‖φ‖2L2(Ω)
)
. (3.4)
6
On the regularity of solution of the initial boundary value problem...
Combining (3.3) and (3.4) we have
‖uN‖H((0,T );Hm
B
(Ω),L2(Ω)) ≤ C(‖φ‖
2
Hm
B
(Ω) + ‖f‖
2
L2(ΩT )
). (3.5)
This implies that the sequence {uN} contains a subsequence which weakly converges
to a function v ∈ H((0, T );HmB (Ω), L2(Ω)). Passing to the limit of the subsequence,
we can see that v is a generalized solution of problem (2.3) - (2.5). Thus u = v ∈
H((0, T );HmB (Ω), L2(Ω)). The estimate (3.1) with X = L2(ΩT ) follows from (3.5).
(ii) Consider the second case f ∈ H((0, T );H−mB (Ω)).
Because of ft ∈ L2((0, T );H
−m
B (Ω)), f is continuous on [0, T ]. So we can represent
f(t) = f(s) +
t∫
s
ft(τ)dτ, ∀s, t ∈ [0, T ].
This implies
‖f(t)‖2
H−m
B
(Ω)
≤
(
‖f(s)‖H−m
B
(Ω) +
∫
J
‖ft(τ)‖H−m
B
(Ω)dτ
)2
≤ 2
(
‖f(s)‖2
H−m
B
(Ω)
+
∫
J
‖ft(τ)‖
2
H−m
B
(Ω)
dτ
)
, (3.6)
where J = [a, b] ⊂ [0, T ] such that a ≤ s, t ≤ b and b− a = 1. Integrating both sides
of (3.6) with respect to s on J , we obtain
‖f(t)‖2
H−m
B
(Ω)
≤ 2‖f‖2
H((0,T );H−m
B
(Ω))
, (t ∈ [0, T ]). (3.7)
By the same way to get (3.2), we have
‖uNt ‖
2
L2(ΩT )
= −Re
T∫
0
〈
f, uNt
〉
dt.
On the other hand
T∫
0
〈
f, uNt
〉
dt = −
T∫
0
〈
ft, u
N
〉
dt+
〈
f(., T ), uN(., T )
〉
−
〈
f(., 0), uN(., 0)
〉
. (3.8)
Noting that ‖ft‖
2
L2((0,T );H
−m
B
(Ω))
≤ ‖f‖2H((0,T );Hm
B
(Ω),L2(Ω))
, using (3.7) with t = 0 and
t = T , we get from (3.8) that
‖uNt ‖
2
L2(ΩT )
= −Re
T∫
0
〈
f, uNt
〉
dt
≤ C(ǫ)‖f‖2H((0,T );Hm
B
(Ω),L2(Ω))
+ ǫ
(
‖uN‖2L2((0,T );HmB (Ω)) + ‖u
N(T )‖2Hm(Ω) + ‖u
N(0)‖2Hm
B
(Ω)
)
.
7
Nguyen Thi Lien
Using (3.4) we obtain
‖uNt ‖
2
L2(ΩT )
≤ C
(
‖f‖2
H((0,T );H−m
B
(Ω))
+ ‖φ‖2L2(ΩT )
)
.
Letting T tends to ∞ yields
‖uNt ‖
2
L2(ΩT )
≤ C
(
‖f‖2
H((0,T );H−m
B
(Ω))
+ ‖φ‖2L2(ΩT )
)
. (3.9)
Combining (3.4) and (3.9), we get
‖uNt ‖
2
H((0,T );Hm
B
(Ω),L2(Ω))
≤ C
(
‖f‖2
H((0,T );H−m
B
(Ω))
+ ‖φ‖2L2(ΩT )
)
. (3.10)
By the similar argument to the part (i) above, we obtain the assertion of the Lemma
for the case f ∈ H((0, T );H−mB (Ω)). This completes the proof of the Lemma.
Remark 3.1. From the proof of above Lemma, we can see that if φ ∈ HmB (Ω) and
f = f1 + f2 where f1 ∈ L2(ΩT ) and f2 ∈ H((0, T );H
−m
B (Ω)) then the generalized
solution u of problem (2.3 - (2.5) belongs to H((0, T );HmB (Ω), L2(Ω)). The estimate
holds with ‖f‖2X replaced by ‖f1‖
2
L2(ΩT )
+ ‖f2‖
2
H((0,T );H−m
B
(Ω))
Now we investigate the regularity of the solution of problem (2.3) - (2.5). For h is
a non-negative integer, we denote
Ltk = Ltk(x, t,D) =
m∑
|p|,|q|=0
Dp(
∂kapq
∂tk
Dq),
φ0 = φ, φ1 := f(., 0)− L(x, 0, D)φ0, ...,
φh = fth−1(., 0)−
h−1∑
k=0
(
h− 1
k
)
Lth−1−k(x, 0, D)φk. (3.11)
We say that the hth-order compatibility conditions for problem (2.3) - (2.5) are
fulfilled if φ0, ..., φh−1 belong to H
2m(Ω) and
s∑
k=0
(
s
k
)
(Bj)ts−k(x, 0, D)φk |S= 0, s = 0, ..., h− 1, j = 1, ..., m. (3.12)
Theorem 3.1. Let h is a non-negative integer. Suppose that φ, f on L2(ΩT ) satis-
fying φk ∈ H
m(Ω), ftk ∈ L2(ΩT ) for k = 0, ..., h. Furthermore if h ≥ 1, the h
th-order
compatibility conditions for problem (2.3) - (2.5) are fulfilled. Then the generalized
solution u ∈ H((0, T );HmB (Ω), H
−m
B (Ω)) of problem (2.3) - (2.5) satisfies
utk ∈ H((0, T );H
m
B (Ω), L2(Ω)) for k = 0, ..., h (3.13)
and
h∑
k=0
‖utk‖
2
H((0,T );Hm
B
(Ω),L2(Ω))
≤ C
h∑
k=0
(
‖φk‖
2
Hm(Ω) + ‖ftk‖
2
L2(ΩT )
)
, (3.14)
where C is the constant independent of u, f, φ.
8
On the regularity of solution of the initial boundary value problem...
Proof. We will show by induction on h that not only the assertion (3.13), (3.14)
but also the following equalities hold
utk(0) = φk, k = 1, ..., h (3.15)
and
(−1)m−1i
h∑
k=0
(
h
k
)
Bth−k(t, utk , η)− (uth+1 , η) = (fth , η) , ∀η ∈ H
m
B (Ω). (3.16)
The case h = 0 is just proved by Lemma 3.1. Assuming now that they hold for h−1,
we will prove them for h (h ≥ 1). We consider first the following problem : find a
function v ∈ H((0, T );HmB (Ω), H
−m
B (Ω)) satisfying
v(0) = φk
and
(−1)m−1iB(t, v, η)−
〈
vt, η
〉
= (fth , η)− (−1)
m−1i
h−1∑
k=0
(
h
k
)
Bth−k(t, utk , η) (3.17)
for all η ∈ HmB (Ω) and a.e. t ∈ [0, T ]. Let F (t), t ∈ [0, T ], be function defined by
〈
F (t), η
〉
= (fth , η)− (−1)
m−1i
h−1∑
k=0
(
h
k
)
Bth−k(t, utk , η). (3.18)
By the induction hypothesis,
〈
F (t), η
〉
= (fth , η) − (fth−1 , η) + (uth, η). So F ∈
L2((0, T );H
−m
B (Ω)). According to Theorem 3.2 in [4], problem (3.17) has a solution
v ∈ H((0, T );HmB (Ω), H
−m
B (Ω)). We set
w(x, t) = φh−1(x) +
t∫
0
v(x, τ)dτ, x ∈ Ω, t ∈ [0, T ].
Then we have w(0) = φh−1, wt = v, wt(0) = φh.Using (3.17), noting that
∂
∂t
B(t, w, η) =
Bt(t, w, η) +B(t, wt, η) we obtain
(−1)m−1i
∂
∂t
B(t, w, η)−
〈
wtt, η
〉
= (fth , η) + (−1)
m−1i
∂
∂t
Bt(t, w − uth−1)
−(−1)m−1i
h−2∑
k=0
(
h− 1
k
)
Bth−1−k(t, utk , η) (3.19)
9
Nguyen Thi Lien
It follows from equality (2.2) that∫
Ω
Lψηdx = B(t, ψ, η) +
m∑
j=1
∫
S
BjψB′jηds,
for all ψ ∈ H2m(Ω), η ∈ HmB (Ω) and t ∈ [0, T ]. Derivativing (h− 1 − k) times both
sides of this equality with respect to t and taking ψ = φk (0 ≤ k ≤ h− 1), we have∫
Ω
Lth−1−kφkηdx =Bth−1−k(t, φk, η)
+
m∑
j=1
∫
S
h−1−k∑
s=0
(
h− 1− k
s
)
Bjφk(B′j)tsηds.
Multiplying both sides of this equality with
(
h−1
k
)
, taking sum in k from 0 to h− 1
and noting that
(
h−1
k
)(
h−1−k
s
)
=
(
h−1
s
)(
h−1−s
k
)
, we obtain
∫
Ω
h−1∑
k=0
(
h− 1
k
)
Lth−1−kφkηdx =
h−1∑
k=0
(
h− 1
k
)
Bth−1−k(t, φk, η)
+
m∑
j=1
h−1∑
s=0
(
h− 1
s
)∫
S
h−1−s∑
k=0
(
h− 1− s
k
)
(Bj)th−1−s−kφk(B
′
j)tsηds. (3.20)
From this equality taking t = 0 together with (3.11), (3.12) we can see that
(wt(0), η) = (φh, η) = (fth−1(0), η)−
h−1∑
k=0
(
h− 1
k
)
Bth−1−k(0, φk, η). (3.21)
Using (3.21) and integrating equality (3.19) with respect to t from 0 to t, we arrive
at
〈
wt, η
〉
+ (−1)m−1iB(t, w, η) = (fth−1 , η) + (−1)
m−1i
t∫
0
Bt(τ, w − uth−1 , η)dτ
−(−1)m−1i
h−1∑
k=0
(
h− 1
k
)
Bth−1−k(t, utk , η). (3.22)
Put z = w − uth−1 , then z(0) = 0. It follows from (3.22) and the induction (3.16)
with h replaced by h− 1 that
−
〈
zt, η
〉
+ (−1)m−1iB(t, z(t), η) = (−1)mi
t∫
0
Bτ (τ, z(., τ), η)dτ. (3.23)
10
On the regularity of solution of the initial boundary value problem...
Noting that |Bτ (t, u, u) ≤ C‖u‖
2
Hm
B
(Ω), doing the same in the Theorem 3.1 in [4],
we can show that z ≡ 0 on ΩT . So v = uth ∈ H((0, T );H
m
B (Ω), H
−m
B (Ω)) and
uth(0) = wt(0) = φh. Now we show v = uth ∈ H((0, T );H
m
B (Ω), L2(Ω)). We rewrite
(3.17) in the form
(−1)m−1iB(t, u, η)−
〈
vt, η
〉
= (fth , η) +
〈
F̂ (t), η
〉
, (3.24)
where F̂ (t), t ∈ [0, T ] on HmB (Ω) defined by
〈
F̂ (t), η
〉
= (−1)mi
h−1∑
k=0
(
h
k
)
Bth−k(t, utk , η), η ∈ H
m
B (Ω). (3.25)
Since utk ∈ L2((0, T );H
m
B (Ω)) for k = 1, .., h, we can see from (3.25) that F̂t ∈
L2((0, T );H
−m
B (Ω))). According to the remark below Lemma 3.1, we obtain from(3.24)
that uth ∈ H((0, T );H
m
B (Ω), L2(Ω)) and
‖u‖2H((0,T );Hm
B
(Ω),L2(Ω))
≤ C
(
‖φk‖
2
Hm
B
(Ω) + ‖fth‖
2
L2(ΩT )
+ ‖F̂‖2
H((0,T );H−m
B
(Ω))
)
. (3.26)
On the other hands, (3.25) follows that
‖F̂‖2
L2((0,T );H
−m
B
(Ω))
≤ C
( h−1∑
k=0
‖utk‖
2
L2((0,T );HmB (Ω))
)
(3.27)
and
〈
F̂t(t), η
〉
= (−1)mi
h−1∑
k=0
(
h+ 1
k
)
Bth+1−k(t, utk , η) + (−1)
mihBt(t, uth, η).
Thus
‖F̂t‖
2
L2((0,T );H
−m
B
(Ω))
≤ C
( h∑
k=0
‖utk‖
2
L2((0,T );HmB (Ω))
)
. (3.28)
Noting that
‖uth‖
2
L2((0,T );HmB (Ω))
≤ ‖uth−1‖
2
H((0,T );Hm
B
(Ω),L2(Ω))
So using induction hypothesis, (3.27) and (3.28), we arrive at
h∑
k=0
‖utk‖
2
H((0,T );Hm
B
(Ω),L2(Ω))
≤ C
h∑
k=0
(
‖φk‖
2
Hm(Ω) + ‖ftk‖
2
L2(ΩT )
)
.
The proof of this theorem is completed.
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Nguyen Thi Lien
4. An example
In this section we apply the previous results to the first boundary value prob-
lem for the Schro¨dinger equation. We consider the following problem:
∆u+ iut = f(x, t) in ΩT , (4.1)
u |t=0= 0, on Ω, (4.2)
u |ST= 0, (4.3)
where ∆ is the Laplace operator. By
◦
H1(Ω) we denote the completion of
◦
C∞ (Ω) in
the norm of the space H1(Ω). Then H((0, T );H1B(Ω), H
−1
B (Ω)) = H((0, T );
◦
H1(Ω)
,
◦
H−1(Ω)). From this fact and Theorem 3.1 we obtain the following result:
Theorem 4.1. Let h is a nonnegative integer. Suppose that f on L2(ΩT ) satisfying
ftk ∈ L2(ΩT ) for k = 0, ..., h. Furthermore if h ≥ 1, the h
th-order compatibility
conditions for problem (4.1) - (4.3) are fulfilled. Then the generalized solution u ∈
H((0, T );
◦
H1(Ω),
◦
H−1(Ω)) of problem (4.1) - (4.3) satisfies
utk ∈ H((0, T );
◦
H1(Ω), L2(Ω)) for k = 0, ..., h
and
h∑
k=0
‖utk‖
2
H((0,T );
◦
H1(Ω),L2(Ω))
≤ C
h∑
k=0
‖ftk‖
2
L2(ΩT )
,
where C is the constant independent of u, f .
Acknowledgement. This work was supported by National Foundation for
Science and Technology Development (NAFOSTED), Vietnam, under project No.
101.01.58.09.
REFERENCES
[1] R. A. Adams. Sobolev Spaces, 1975. Academic Press.
[2] Nguyen Manh Hung and Cung The Anh, 2010. Asymtotic expansions of solutions
of the first initial boundary value problem for the Schro¨dinger system near conical
points of the boundary. Differentsial’nye Uravneniya, Vol. 46, No. 2, pp. 285-289.
[3] Nguyen Manh Hung and Nguyen Thi Kim Son, 2009.On the regularity of solution
of the second initial boundary value problem for Schro¨dinger systems in domains
with conical points. Taiwanese journal of Mathematics. Vol. 13, No. 6, pp. 1885-
1907.
[4] Nguyen Thi Lien, 2010. On the solvability of the initial boundary value problem
for Schro¨dinger systems in conical domains. Journal of Science of Hanoi National
University of Education, Vol. 55, No. 6, pp. 82-89.
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