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.
10 trang |
Chia sẻ: thanhle95 | Lượt xem: 382 | Lượt tải: 0
Bạn đang xem nội dung tài liệu On the regularity of solution of the initial boundary value problem for Schrodinger systems in conical domains, để tải tài liệu về máy bạn click vào nút DOWNLOAD ở trên
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.
11
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.
12