- Trang Chủ
- Lịch sử - Văn hoá
- On the regularity of solution of the initial boundary value problem for schrodinger systems in conical domains
Xem mẫu
- 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 SCHRODINGER 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 Schr¨
odinger 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 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 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 S n−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
m
X
L(x, t, D) = (−1)|p| D p apq (x, t)D q ,
|p|,|q|=0
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
X
Bj = Bj (x, t, D) = bj,p (x, t)D p , j = 1, ..., m,
|p|≤µj
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 of Bj 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) Bjo (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 Bjo (x, t, D) is the
principal part of Bj (x, t, D),
X
Bjo = Bjo (x, t, D) = bj,p (x, t)D p , j = 1, ..., m.
|p|=µj
Furthermore, Bjo (0, t, ν(x)) 6= 0 for all x ∈ S closed enough to the origin 0.
We set
HBm (Ω) = u ∈ H m (Ω) : Bj u = 0 on S for j = 1, .., χ
with the same norm in H m (Ω) and
m
X Z
B(t, u, v) = apq D q uD p vdx, t ∈ [0, T ].
|p|,|q|=0 Ω
We assume that B(t, ., .) satisfies the following inequality:
(−1)m B(t, u, u) ≥ µ0 ku(x, t)k2H m (Ω) (2.1)
for all u ∈ HBm (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 Bj′ on ST , j = 1, ..., m
satisfying
Z χ Z
X m Z
X
B(t, u, v) = Luvdx + Bj′ uBj vds + Bj uBj′ vds. (2.2)
Ω j=1 S j=χ+1 S
Denote HB−m (Ω) the dual space to HBm (Ω). We write ., . to stand for the
pairing between HBm (Ω) and HB−m (Ω), and (., .) to define the inner product in L2 (Ω).
We then have the continuous imbeddings HBm (Ω) ֒→ L2 (Ω) ֒→ HB−m (Ω) with the
equation
f, v = (f, v) for f ∈ L2 (Ω) ⊂ HB−m (Ω), v ∈ HBm (Ω).
We study the following problem in the cylinder ΩT :
(−1)m−1 iL(x, t, D)u − ut = f (x, t) in ΩT , (2.3)
Bj u = 0, on ST , j = 1, ..., m, (2.4)
u |t=0 = φ, on Ω, (2.5)
where f ∈ L2 ((0, T ); HB−m(Ω)) and φ ∈ L2 (Ω) are given functions.
The function u ∈ H((0, T ); HBm(Ω), HB−m (Ω)) is called a generalized solution of
the problem (2.3) - (2.5) iff u(., 0) = φ and the equality
(−1)m−1 iB(t, u, v) − ut , v = f (t), v (2.6)
holds for a.e. t ∈ (0, T ) and all v ∈ HBm (Ω).
3. The regularity with respect to time variable
For k ∈ N, u, v ∈ H m,0 (ΩT ), t ∈ [0, T ] we set
m
X Z k
∂ apq q p
Btk (t, u, v) = D uD vdx,
∂tk
|p|,|q|=0 Ω
ZT
BtTk (u, v) = Btk (t, u, v)dt,
0
B (u, v) = BtT0 (u, v).
T
Now we improve slightly the regularity of generalized solution u by making the
initial data φ and the right-hand side f more regularity. We denote X = L2 (ΩT ) or
X = H((0, T ); HB−m(Ω)).
5
- Nguyen Thi Lien
Lemma 3.1. Let φ ∈ HBm (Ω) and f ∈ X. Then the generalized solution of problem
(2.3) - (2.5) belongs to H((0, T ); HBm(Ω), L2 (Ω)) and satisfies the following estimate
kukH((0,T );HBm (Ω),L2 (Ω)) ≤ C(kφk2HBm (Ω) + kf k2X ). (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 = kψk k−2
m (Ω) (φ, ψk )H m (Ω) , where (., .)H m (Ω) de-
HB B B
m
notes the inner product in HB (Ω). Remember that in [4] we have
m
X Z Z
m−1 q N
(−1) i apq D u D p ψl dx − uN
t ψl dx = f, ψl , l = 1, ..., N.
|p|,|q|=0 Ω Ω
dClN
Multiplying both sides of this equality by , then taking sum with respect to l
dt
from 1 to N, we arrive at
m
X Z ZT
2
−kuN
t kL2 (ΩT ) + (−1) m−1
i q N
apq D u D p uN
t dxdt = (f, uN
t )dt.
|p|,|q|=0 Ω 0
Adding with its complex conjugate, we obtain
ZT
2
kuN
t kL2 (ΩT ) = −Re (f, uN
t )dt. (3.2)
0
Using Cauchy’s inequality, we get
ZT
N 2 1
| − Re (f, uN
t )dt| ≤ ǫkut kL2 (ΩT ) + kf k2L2(ΩT ) (0 < ǫ < 1).
4ǫ
0
Hence
2 2
kuN
t kL2 (ΩT ) ≤ Ckf kL2 (ΩT ) .
Letting T tends to ∞ yields
2 2
kuN
t kL2 (ΩT ) ≤ Ckf kL2 (ΩT ) (3.3)
In [4], we had the following estimate
kuN k2L2 ((0,T );HBm (Ω)) ≤ C kf k2L2 ((0,T );H −m (Ω)) + kφk2L2 (Ω) . (3.4)
B
6
- On the regularity of solution of the initial boundary value problem...
Combining (3.3) and (3.4) we have
kuN kH((0,T );HBm (Ω),L2 (Ω)) ≤ C(kφk2HBm (Ω) + kf k2L2(ΩT ) ). (3.5)
This implies that the sequence {uN } contains a subsequence which weakly converges
to a function v ∈ H((0, T ); HBm(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 ); HBm(Ω), L2 (Ω)). The estimate (3.1) with X = L2 (ΩT ) follows from (3.5).
(ii) Consider the second case f ∈ H((0, T ); HB−m(Ω)).
Because of ft ∈ L2 ((0, T ); HB−m(Ω)), f is continuous on [0, T ]. So we can represent
Zt
f (t) = f (s) + ft (τ )dτ, ∀s, t ∈ [0, T ].
s
This implies
Z
2
kf (t)k2H −m (Ω) ≤ kf (s)kH −m (Ω) + kft (τ )kH −m (Ω) dτ
B B B
J
Z
≤2 kf (s)k2H −m (Ω) + kft (τ )k2H −m (Ω) dτ , (3.6)
B B
J
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
kf (t)k2H −m (Ω) ≤ 2kf k2H((0,T );H −m (Ω)) , (t ∈ [0, T ]). (3.7)
B B
By the same way to get (3.2), we have
ZT
2
kuN
t kL2 (ΩT ) = −Re f, uN
t dt.
0
On the other hand
ZT ZT
f, uN
t dt = − ft , uN dt + f (., T ), uN (., T ) − f (., 0), uN (., 0) . (3.8)
0 0
Noting that kft k2L ((0,T );H −m (Ω)) ≤ kf k2H((0,T );H m (Ω),L2 (Ω)) , using (3.7) with t = 0 and
2 B B
t = T , we get from (3.8) that
ZT
2
kuN
t kL2 (ΩT ) = −Re f, uN
t dt
0
2
≤ C(ǫ)kf kH((0,T );HBm (Ω),L2 (Ω)) + ǫ kuN k2L2 ((0,T );HBm (Ω)) + kuN (T )k2H m (Ω) + kuN (0)k2HBm (Ω) .
7
- Nguyen Thi Lien
Using (3.4) we obtain
2 2 2
kuN
t kL2 (ΩT ) ≤ C kf kH((0,T );H −m (Ω)) + kφkL2 (ΩT ) .
B
Letting T tends to ∞ yields
2 2 2
kuN
t kL2 (ΩT ) ≤ C kf kH((0,T );H −m (Ω)) + kφkL2 (ΩT ) . (3.9)
B
Combining (3.4) and (3.9), we get
2 2 2
kuN k m
t H((0,T );HB (Ω),L2 (Ω)) ≤ C kf k H((0,T );H −m
(Ω))
+ kφk L2 (ΩT ) . (3.10)
B
By the similar argument to the part (i) above, we obtain the assertion of the Lemma
for the case f ∈ H((0, T ); HB−m(Ω)). This completes the proof of the Lemma.
Remark 3.1. From the proof of above Lemma, we can see that if φ ∈ HBm (Ω) and
f = f1 + f2 where f1 ∈ L2 (ΩT ) and f2 ∈ H((0, T ); HB−m(Ω)) then the generalized
solution u of problem (2.3 - (2.5) belongs to H((0, T ); HBm(Ω), L2 (Ω)). The estimate
holds with kf k2X replaced by kf1 k2L2 (ΩT ) + kf2 k2H((0,T );H −m (Ω))
B
Now we investigate the regularity of the solution of problem (2.3) - (2.5). For h is
a non-negative integer, we denote
m
X ∂ k apq q
Ltk = Ltk (x, t, D) = Dp( D ),
∂tk
|p|,|q|=0
φ0 = φ, φ1 := f (., 0) − L(x, 0, D)φ0 , ...,
h−1
X
h−1
φh = fth−1 (., 0) − Lth−1−k (x, 0, D)φk . (3.11)
k=0
k
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
Xs
s
(Bj )ts−k (x, 0, D)φk |S = 0, s = 0, ..., h − 1, j = 1, ..., m. (3.12)
k=0
k
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 hth -order
compatibility conditions for problem (2.3) - (2.5) are fulfilled. Then the generalized
solution u ∈ H((0, T ); HBm(Ω), HB−m (Ω)) of problem (2.3) - (2.5) satisfies
utk ∈ H((0, T ); HBm(Ω), L2 (Ω)) for k = 0, ..., h (3.13)
and
h
X h
X
kutk k2H((0,T );HBm (Ω),L2 (Ω)) ≤ C kφk k2H m (Ω) + kftk k2L2 (ΩT ) , (3.14)
k=0 k=0
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
Xh
m−1 h
(−1) i Bth−k (t, utk , η) − (uth+1 , η) = (fth , η) , ∀η ∈ HBm (Ω). (3.16)
k=0
k
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 ); HBm(Ω), HB−m (Ω)) satisfying
v(0) = φk
and
h−1
X
m−1
m−1 h
(−1) iB(t, v, η) − vt , η = (fth , η) − (−1) i Bth−k (t, utk , η) (3.17)
k=0
k
for all η ∈ HBm (Ω) and a.e. t ∈ [0, T ]. Let F (t), t ∈ [0, T ], be function defined by
h−1
X
m−1 h
F (t), η = (fth , η) − (−1) i Bth−k (t, utk , η). (3.18)
k=0
k
By the induction hypothesis, F (t), η = (fth , η) − (fth−1 , η) + (uth , η). So F ∈
L2 ((0, T ); HB−m(Ω)). According to Theorem 3.2 in [4], problem (3.17) has a solution
v ∈ H((0, T ); HBm(Ω), HB−m (Ω)). We set
Zt
w(x, t) = φh−1 (x) + v(x, τ )dτ, x ∈ Ω, t ∈ [0, T ].
0
∂
Then we have w(0) = φh−1 , wt = v, wt (0) = φh . Using (3.17), noting that B(t, w, η) =
∂t
Bt (t, w, η) + B(t, wt , η) we obtain
∂
∂
(−1)m−1 i B(t, w, η) − wtt , η = (fth , η) + (−1)m−1 i Bt (t, w − uth−1 )
∂t ∂t
h−2
X
m−1 h−1
−(−1) i Bth−1−k (t, utk , η) (3.19)
k
k=0
9
- Nguyen Thi Lien
It follows from equality (2.2) that
Z m Z
X
Lψηdx = B(t, ψ, η) + Bj ψBj′ ηds,
Ω j=1 S
for all ψ ∈ H 2m (Ω), η ∈ HBm (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
Z
Lth−1−k φk ηdx =Bth−1−k (t, φk , η)
Ω
m Z
X X
h−1−k
h−1−k
+ Bj φk (Bj′ )ts ηds.
j=1 S s=0
s
Multiplying both sides of this equality with h−1 , taking sum in k from 0 to h − 1
h−1 h−1−k
h−1 h−1−s
k
and noting that k s
= s k
, we obtain
h−1
Z X h−1
X
h−1 h−1
Lth−1−k φk ηdx = Bth−1−k (t, φk , η)
k k
Ω k=0 k=0
Xm Xh−1 X h − 1 − s
Z h−1−s
h−1
+ (Bj )th−1−s−k φk (Bj′ )ts ηds. (3.20)
j=1 s=0
s k=0
k
S
From this equality taking t = 0 together with (3.11), (3.12) we can see that
h−1
X
h−1
(wt (0), η) = (φh , η) = (fth−1 (0), η) − Bth−1−k (0, φk , η). (3.21)
k=0
k
Using (3.21) and integrating equality (3.19) with respect to t from 0 to t, we arrive
at
Zt
wt , η + (−1)m−1 iB(t, w, η) = (fth−1 , η) + (−1)m−1 i Bt (τ, w − uth−1 , η)dτ
0
h−1
X
m−1 h−1
−(−1) i Bth−1−k (t, utk , η). (3.22)
k
k=0
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
− zt , η + (−1)m−1 iB(t, z(t), η) = (−1)m i Bτ (τ, z(., τ ), η)dτ. (3.23)
0
10
- On the regularity of solution of the initial boundary value problem...
Noting that |Bτ (t, u, u) ≤ Ckuk2H m (Ω) , doing the same in the Theorem 3.1 in [4],
B
we can show that z ≡ 0 on ΩT . So v = uth ∈ H((0, T ); HBm(Ω), HB−m (Ω)) and
uth (0) = wt (0) = φh . Now we show v = uth ∈ H((0, T ); HBm(Ω), L2 (Ω)). We rewrite
(3.17) in the form
(−1)m−1 iB(t, u, η) − vt , η = (fth , η) + Fb(t), η , (3.24)
where Fb(t), t ∈ [0, T ] on HBm (Ω) defined by
h−1
X
h
b m
F (t), η = (−1) i Bth−k (t, utk , η), η ∈ HBm (Ω). (3.25)
k=0
k
Since utk ∈ L2 ((0, T ); HBm(Ω)) for k = 1, .., h, we can see from (3.25) that Fbt ∈
L2 ((0, T ); HB−m(Ω))). According to the remark below Lemma 3.1, we obtain from(3.24)
that uth ∈ H((0, T ); HBm(Ω), L2 (Ω)) and
kuk2H((0,T );HBm (Ω),L2 (Ω)) ≤ C kφk k2HBm (Ω) + kfth k2L2 (ΩT ) + kFbk2H((0,T );H −m (Ω)) . (3.26)
B
On the other hands, (3.25) follows that
h−1
X
kFbk2L2 ((0,T );H −m (Ω)) ≤ C kutk k2L2 ((0,T );HBm (Ω)) (3.27)
B
k=0
and
h−1
X
h + 1
Fbt (t), η = (−1) i
m
Bth+1−k (t, utk , η) + (−1)m ihBt (t, uth , η).
k=0
k
Thus
h
X
kFbt k2L2 ((0,T );H −m (Ω)) ≤ C kutk k2L2 ((0,T );HBm (Ω)) . (3.28)
B
k=0
Noting that
kuth k2L2 ((0,T );HBm (Ω)) ≤ kuth−1 k2H((0,T );HBm (Ω),L2 (Ω))
So using induction hypothesis, (3.27) and (3.28), we arrive at
h
X h
X
kutk k2H((0,T );HBm (Ω),L2 (Ω)) ≤ C kφk k2H m (Ω) + kftk k2L2 (ΩT ) .
k=0 k=0
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 Schr¨odinger 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 H 1 (Ω) we denote the completion of C ∞ (Ω) in
◦
the norm of the space H 1 (Ω). Then H((0, T ); HB1 (Ω), HB−1 (Ω)) = H((0, T ); H 1 (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 hth -order compatibility
conditions for problem (4.1) - (4.3) are fulfilled. Then the generalized solution u ∈
◦ ◦
H((0, T ); H 1 (Ω), H −1 (Ω)) of problem (4.1) - (4.3) satisfies
◦
utk ∈ H((0, T ); H 1 (Ω), L2 (Ω)) for k = 0, ..., h
and
h
X h
X
2
kutk k ◦ ≤C kftk k2L2 (ΩT ) ,
H((0,T );H 1 (Ω),L2 (Ω))
k=0 k=0
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.
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