On the regularity of solution of the initial boundary value problem for schrodinger systems in conical domains

Chia sẻ: Minh Minh | Ngày: | Loại File: PDF | Số trang:10

Thêm vào BST

Báo xấu

13
lượt xem 1
download

Download Vui lòng tải xuống để xem tài liệu đầy đủ

The purpose of this paper is to establish the regularity with respect to time variable of solution of the initial boundary value problems for Schrodinger systems in the cylinders with base containing the conical point.

Chủ đề:

Bình luận(0) Đăng nhập để gửi bình luận!

Lưu

Nội dung Text: On the regularity of solution of the initial boundary value problem for schrodinger systems in conical domains

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. 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 Schr¨odinger 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 Schr¨odinger 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 Schr¨odinger systems in conical domains. Journal of Science of Hanoi National University of Education, Vol. 55, No. 6, pp. 82-89. 12