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

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. 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. 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