Cauchy-Newmann problem for schrodinger equation regularity of weak solution of Cauchy-Newmann problem for schrodinger equation in infinite nonsmooth cylinders

JOURNAL OF SCIENCE OF HNUE Natural Sci., 2008, Vol. 53, N ◦ . 5, pp. 31-40 CAUCHY-NEWMANN PROBLEM FOR SCHR  ODINGER EQUATION REGULARITY OF WEAK SOLUTION OF CAUCHY-NEWMANN PROBLEM FOR SCHR  ODINGER EQUATION IN INFINITE NONSMOOTH CYLINDERS Nguyen Manh Hung and Nguyen Kim Son Hanoi National University of Education Abstract. The main goal of this paper is to obtain the regularity of weak solution of Cauchy-Newmann problem for classical Schrodinger equation in infinite cylinders with the bases containing conical points. Keywords: The Newmann problem, Schrodinger equation, weak solution, smoothness, regularity. 1. Introduction The Schrodinger equation received a great deal of attention from mathemati- cians, in particular because of its applications to quantum mechanics and optics. Indeed, some simplified models lead to certain Schrodinger equation, see [1,5] for example. In this paper we consider the Cauchy-Newmann problem for classical Schroding- er equation in infinite nonsmooth domains. The Cauchy-Dirichlet problem for gen- eral Schrodinger systems in domains containing conical points has been investigated in [3,4,5]. The Cauchy-Newmann problems have been dealt with for the class heat equations (see [10]) and for general second order parabolic equations in [2] in do- mains with edges. This problem in domains with conical points has been investi- gated for general hyperbolic system in [7]. Here in our paper, that problem for class Schrodinger equations will be dealt with in infinite cylinders Q∞ = Ω × (0,+∞) where Ω is a domain containing conical points. Our main purpose is to study the regularity of the weak solution of the problem. The paper is organized as follows. In Section 2, we introduce some notations and functional spaces that we use throughout the text. We also introduce the formu- lation of the problem and recall some results of the unique existence and smoothness with the time variable of the weak solution of the problem. Our main result is stated in Section 3 with Theorem 3.1. The proof of this theorem is given in Section 4 with some auxiliary lemmas and propositions. 31 Nguyen Manh Hung and Nguyen Kim Son 2. Preliminaries Let Ω be a bounded domain in Rn, n > 2; Ω and ∂Ω denote the closure and the boundary of Ω in Rn. We suppose that Γ = ∂Ω\{0} is a infinitely differentiable surface everywhere except the origin and Ω coincides with the cone K = {x : x |x| ∈ G} in a neighborhood of the origin 0, where G is a smooth domain on the unit sphere Sn−1 in Rn. We begin by introducing some notations and functional spaces which are used fluently in the rest. Denote Q∞ = Ω × (0,+∞), S∞ = Γ × (0,+∞), x = (x1, ..., xn) ∈ Ω, ∂xj = ∂/∂xj , uxj = ∂xju, utk = ∂ ku/∂tk, r = |x| = √ x21 + ...+ x 2 n. For each multi-index α = (α1, ..., αn)(αi ∈ N, i = 1, ..., n), set |α| = α1 + · · ·+ αn, ∂α = ∂αx = ∂ α1 x1 · · ·∂ αn xn . In this paper we will use usual functional spaces: C∞(Ω), L2(Ω), H m(Ω), where m, l, k ∈ N (see [4,5] for the precise definitions). Denoted by H lβ(Ω) the space of all measurable complex functions u(x, t) that satisfy ‖u‖Hl β (Ω) = (∑ |α|6l ∫ Ω r2(β+|α|−l)|∂αu|2dx ) 1 2 < +∞. Hm,l(e−γt, Q∞) (γ > 0) the space of all measurable complex functions u(x, t) that have generalized derivatives up to order m with respect to x and up to order l with respect to t with the norm ‖u‖Hm,l(e−γt,Q∞) = ( ∫ Q∞ [ ∑ |α|6m |∂αu|2 + ∑ j6l |utj | 2 ] e−2γtdxdt ) 1 2 < +∞. H l,kβ (e −γt, Q∞)- the space of all measurable complex functions u(x, t) with the norm ‖u‖Hl,kβ (e−γt,Q∞) = [ ∫ Q∞ ( l∑ |α|=0 r2(β+|α|−l)|∂αu|2 + k∑ j=1 |utj | 2 ) e−2γtdxdt ] 1 2 < +∞. H lβ(e −γt, Q∞)-the weighted space with the norm ‖u‖Hl β (e−γt,Q∞) = ( ∑ |α|+j6l ∫ Q∞ r2(β+|α|+j−l)|∂αutj | 2e−2γtdxdt ) 1 2 < +∞. Let X be a Banach space. Denoted by L∞(0,∞;X) the space of all measurable functions u : (0,∞) −→ X, t 7−→ u(t) with the norm ‖u‖L∞(0,∞;X) = ess sup 0<t<+∞ ‖u(t)‖X < +∞. 32 Cauchy-Newmann problem for Schrodinger equation regularity of weak solution... In cylinder Q∞ we consider the following problem i∆u − ut = f in Q∞, (2.1) u(x, 0) = 0 on Ω, (2.2) ∂u ∂ν = n∑ k=1 ∂u ∂xk cos(xk, ν) = 0 on S∞, (2.3) where ν is the unit exterior normal to S∞. The function u(x, t) is called weak solution in space H1,0(e−γt, Q∞) of the problem (2.1)−(2.3) if u(x, t) ∈ H1,0(e−γt, Q∞), satisfying for each T ∈ (0,+∞) −i n∑ k=1 ∫ Q∞ ∂u ∂xk (x, t) ∂η ∂xk (x, t)dxdt+ ∫ Q∞ u(x, t)ηt(x, t)dxdt= ∫ Q∞ f(x, t)η(x, t)dxdt (2.4) for all test functions η(x, t) ∈ H1,1(e−γt, Q∞), η(x, t) = 0 for t ∈ [T,+∞). By applying theorems 3.1, 3.2 and 4.1 in [6] for the problem (2.1)− (2.3), we have the following lemmas. Lemma 2.1. (The solvability of the problem) Let f, ft ∈ L ∞(0,∞, L2(Ω)). Then for every γ > 0, the Cauchy-Newmann problem (2.1)−(2.3) has exactly one weak solution u(x, t) in H1,0(e−γt, Q∞), that satisfies ‖u‖2H1,0(e−γt,Q∞) 6 C(‖f‖ 2 L∞(0,∞,L2(Ω)) + ‖ft‖ 2 L∞(0,∞,L2(Ω)) ), where the constant C does not depend on u, f . Lemma 2.2. (The regularity with respect to time variable) Let h be a nonnegative integer. Suppose that ftk ∈ L ∞(0,∞, L2(Ω)) for all k 6 h + 1, f(x, 0) = 0 and if h > 2 then ftk(x, 0) = 0 for all k 6 h−1, for all x ∈ Ω. Then the weak solution u(x, t) of the problem (2.1)− (2.3) has generalized derivatives with respect to time variable up to order h, which belongs to H1,0(e−γt, Q∞), with γ > 0 arbitrary, moreover ‖uts‖ 2 H1,0(e−γt,Q∞) 6 C h+1∑ k=0 ‖ftk‖ 2 L∞(0,∞,L2(Ω)) , s = 0, 1, ..., h, where C is a constant independent of u, f . By the same arguments as in [4,5] and using the lemmas 2.1, 2.2, we can prove the following lemma. 33 Nguyen Manh Hung and Nguyen Kim Son Lemma 2.3. Assume that u(x, t) is a weak solution of the problem (2.1) − (2.3) in the space H1,0(e−γt, Q∞) and f, ft, ftt ∈ L ∞(0,∞, L2(Ω)), f(x, 0) = 0. Then for almost all t ∈ (0,+∞) the equation n∑ k=1 ∫ Ω ∂u ∂xk ∂χ ∂xk dx = −i ∫ Ω [ ut + f ] χdx holds for all functions χ = χ(x) ∈ H1(Ω). 3. Formulation of the main result Let ω = (ω1, ..., ωn−1) be a local coordinate system on S n−1 and r = |x|. Then the Laplace operator can write in the form: ∆u(r, ω) = 1 rn−1 ∂ ∂r (rn−1 ∂ ∂r )u(r, ω) + 1 r2 ∆ωu(r, ω), where ∆ω is Laplace-Beltrami operator on S n−1. Now we consider the Newmann problem for the following equation ∆ωv + [(iλ) 2 + i(2− n)λ]v = 0, ω ∈ G. (3.1) Because ∆ω is a strong elliptic operator that satisfies strictly Garding inequality and self-adjoint, then for every t ∈ (0,+∞) the spectrum of this problem is discrete and has an enumerable set of eigenvalues. We give here the main result of this paper. Theorem 3.1. Let l be a nonnegative integer. Assume that u(x, t) be a weak solution in the space H1,0(e−γt, Q∞) of the problem (2.1)− (2.3) and ftk ∈ L ∞(0,∞, H l0(Ω)) if k 6 3 and ftk(x, 0) = 0 if k 6 l + 1. Assume further that the strip 1− n 2 6 Imλ 6 l + 2− n 2 does not contain any eigenvalue of the Newmann problem for equation (3.1) for all t ∈ [0,+∞). Then u ∈ H l+20 (e −γt, Q∞) and the following estimate holds ‖u‖2 Hl+20 (e −γt,Q∞) 6 C l+3∑ k=0 ‖ftk‖ 2 L∞(0,∞,Hl0(Ω)) , where C is the constant independent of u, f . 34 Cauchy-Newmann problem for Schrodinger equation regularity of weak solution... 4. Proof of Theorem 3.1 Lemma 4.1. Let u(x, t) be a weak solution in H1,0(e−γt, Q∞) of the problem (2.1)− (2.3) such that u(x, t) = 0 when |x| > R =constant. Moreover, we assume that f, ft, ftt ∈ L ∞(0,∞, L2(K)), f(x, 0) = 0. Then for almost all t ∈ [0,+∞), we have (i) if n > 3 then u ∈ H21 (K), (ii) if n = 2 then u ∈ H21+ε(K), where ε > 0 is arbitrary. Proof. Because f, ft, ftt ∈ L ∞(0,∞, L2(K)), f(x, 0) = 0, from Lemma 2.2 we have ut ∈H 1,0(e−γt, Q∞). Following Lemma 2.3, u(x, t) is a solution of the Newmann problem for the elliptic equation ∆u = F, where F = −i(ut + f) ∈ L2(K) for almost t ∈ (0,+∞). Denote Ωk = {x ∈ Ω : 2−k 6 |x| 6 2−k+1}, k = 1, 2, ... Let k0 be large enough such that 2−k0+2 < R. From the theory of the regular of solution of the boundary value problem for elliptic systems in smooth domains and near the piece smooth boundary of domain (see [9]), we have u ∈ H2(Ωk0) for almost t ∈ (0,+∞) and the following inequality holds∫ Ωk0 |∂αu(x, t)|2dx 6 C [ ∫ Ωk0−1∪Ωk0∪Ωk0+1 |F (x, t)|2dx+ ∫ Ωk0−1∪Ωk0∪Ωk0+1 |u(x, t)|2dx ] , |α| 6 2, where C is a positive constant. By choosing k1 > k0 and setting x = (2 k0/2k1)x′, one has ∫ Ωk0 |∂αu(x′, t)|2dx′ 6 C ∫ Ωk0−1∪Ωk0∪Ωk0+1 [ |F (x′, t)|2 ( 2k0 2k1 )4 + |u(x′, t)|2 ] dx′. Return to the variable x, we get( 2k0 2k1 )2|α|∫ Ωk1 |∂αu(x, t)|2dx 6 C ∫ Ωk1−1∪Ωk1∪Ωk1+1 [ |F (x, t)|2 ( 2k0 2k1 )4 + |u(x, t)|2 ] dx. (4.1) a) Case 1: n > 3. Then ∫ K r−2|u|2dx 6 C ∫ K r−n|u|2dx < +∞. (4.2) It follows from (4.1) that∫ Ωk1 r2(|α|−1)|∂αu|2dx 6 C ∫ Ωk1−1∪Ωk1∪Ωk1+1 [ |F (x, t)|2r2 + r−2|u|2 ] dx, 35 Nguyen Manh Hung and Nguyen Kim Son where C is a constant independent of u, f, k1. Taking sum with respect to k1 > k0, one has ∑ k1>k0 ∫ Ωk1 r2(|α|−1)|∂αu|2dx 6 C ∑ k1>k0 ∫ Ωk1 [ |F (x, t)|2r2 + r−2|u|2 ] dx. This implies ∫ ⋃ k>k0 Ωk r2(|α|−1)|∂αu|2dx 6 C ∫ ⋃ k>k0 Ωk [|F (x, t)|2r2 + r−2|u|2]dx. Because out of a neighborhood of conical point K is a smooth domain, so we have∫ K r2(|α|−1)|∂αu|2dx 6 C ∫ K [|F (x, t)|2+r−2|u|2]dx (4.3) for all |α| 6 2, almost all t ∈ (0,+∞). From (4.2), (4.3) and F ∈ L2(K) we receive u ∈ H21 (K) for almost all t ∈ (0,+∞). b) Case 2: n = 2. Since u ∈ H1,0(e−γt, Q∞) so for almost all t ∈ (0,+∞) one has ∫ K r0|∂βu|2dx < ∞, |β| = 1. This implies ∫ K rδ|∂βu|2dx 6 Rδ ∫ K |∂βu|2dx < +∞, where δ > 0 is arbitrary. For all δ > 0 we have δ > 0 = 1− n 2 , so it follows from Lemma 7.1.1, page 268 in [8] that∫ K r2(δ/2−1)|u|2dx 6 C ∑ |β|=1 ∫ K rδ|∂βu|2dx 6 C ∑ |β|=1 ∫ K |∂βu|2dx < +∞. (4.4) From the inequality (4.1), for all |α| 6 2 one gets ∫ Ωk1 r 2(|α|−1+ δ 2 ) |∂αu|2dx 6 C ∫ Ωk1−1∪Ωk1∪Ωk1+1 [ |F (x, t)|2r2+δ + r−2+δ|u|2 ] dx, (4.5) where C is a constant independent of u, f, k1. By using analogous arguments used in the proof of part a, from (4.4), (4.5) we have ∫ K r2(1+ δ 2 +|α|−2)|∂αu|2dx 6 C ∫ K [ |F (x, t)|2+ ∑ |β|=1 |∂βu|2 ] dx < +∞ for all |α| 6 2, almost all t ∈ (0,+∞). That is u ∈ H2 1+ δ 2 (K). The lemma is proved. 36 Cauchy-Newmann problem for Schrodinger equation regularity of weak solution... Proposition 4.1. Let ftk ∈ L ∞(0,∞, L2(K)), k 6 3, and f(x, 0) = ft(x, 0) = 0 for x ∈ K. Assume that u(x, t) is a weak solution in H1,0(e−γt, Q∞) of the problem (2.1)− (2.3) such that u ≡ 0 when |x| > R = constant. In addition, suppose that the strip 1− n 2 6 Imλ 6 2− n 2 does not contain any point of the spectrum of Newmann problem for the system (3.1) for all t ∈ [0,+∞). Then u ∈ H20 (e −γt, Q∞). Proof. We need to prove∑ |α|+k62 ∫ K∞ r2(|α|+k−2)|∂αutk | 2e−2γtdxdt < +∞. (4.6) It follows from Lemma 2.2 that ∫ K∞ |ut2| 2e−2γtdxdt < +∞. Hence, in order to prove (4.6), we just need to prove that both u and ut are in the space H 2,0 0 (e −γt, Q∞). Indeed, if n > 3 then by applying Lemma 4.1 and Lemma 2.2 we have u ∈ H21 (K), F = i(ut+f) ∈ L2(K) for almost t ∈ (0,+∞). Since in the strip 1− n 2 6 Imλ 6 2− n 2 there is no spectral point of Newmann problem for (3.1) for all t ∈ [0,+∞), then following Theorem 3.2 page 37 in [11], one gets u ∈ H20 (K) and satisfies ‖u‖2H20 (K) 6 C [ ‖F‖2L2(K) + ‖u‖ 2 H21(K) ] , for almost t ∈ (0,+∞), where C is a positive constant. Using the arguments as used in the proof of Lemma 4.1, we have ‖u‖2H20 (K) 6 C [ ‖ut‖ 2 L2(K) + ‖f‖2L2(K) + ∫ K r−n|u|2dx ] for almost t ∈ (0,+∞), where C = constant > 0. Multiplying this inequality with e−2γt, then integrating with respect to t from 0 to +∞, we obtain ‖u‖2 H2,00 (e −γt,K∞) 6 C [ 2∑ k=0 ‖ftk‖ 2 L∞(0,∞,L2(K)) + ∫ K r−n|u|2dx ] < +∞. Then u is a function in the space H2,00 (e −γt, Q∞). And if n = 2 then following Lemma 4.1 we have u ∈ H21+ε(K) for almost t ∈ (0,+∞). Because the straight line Imλ = 1− n 2 does not contain any point from the spectrum of Newmann problem for (3.1) for all t ∈ (0,+∞) then for each t ∈ (0,+∞) there 37 Nguyen Manh Hung and Nguyen Kim Son exists ε(t) > 0 such that the strip 1 − ε(t) − n 2 6 Imλ 6 1 − n 2 does not contain any spectral point of Newmann problem for (3.1). In another way, we also have F ∈ H01 (K). This implies from Theorem 3.2 of [11] that u ∈ H 2 1 (K) satisfying the equality ‖u‖2 H21 (K) 6 C [ ‖F‖2 H01 (K) +‖u‖2 H21+ε ] . Repeating the proof for the case n > 3 we achieve u ∈ H2,00 (e −γt, K∞), too. Now differentiating (2.1) with respect to t, we have ∆v = −i(vt + ft), where v = ut ∈ H 1,0(e−γt, Q∞). Repeating arguments used for function u we receive v ∈ H2,00 (e −γt, K∞) or ut ∈ H 2,0 0 (e −γt, K∞). So (4.6) is proved. This completes the proof of this proposition. Proposition 4.2. Let ftk ∈ L ∞(0,∞, H l0(K)), k 6 3, and ftk(x, 0) = 0 for k 6 l+1, x ∈ K. Assume that u(x, t) is a weak solution in H1,0(e−γt, Q∞) of the problem (2.1)− (2.3) such that u ≡ 0 when |x| > R = constant. In addition, suppose that the strip 1− n 2 6 Imλ 6 l + 2− n 2 does not contain any point of the spectrum of Newmann problem for system (3.1) for all t ∈ [0,+∞). Then u ∈ H l+20 (e −γt, Q∞), satisfies ‖u‖2 Hl+20 (e −γt,Q∞) 6 C l+3∑ k=0 ‖ftk‖ 2 L∞(0,∞,Hl0(K)) , (4.7) where C is a constant independent of u, f. Proof. We use induction by l. For l = 0 then we had Proposition 4.1. Assume that this theorem holds up to l− 1, we need to prove that the theorem holds for l. That is we have to prove the following inequality ‖uts‖ 2 Hl+2−s0 (e −γt,Q∞) 6 C l+3∑ k=0 ‖ftk‖ 2 L∞(0,∞,Hl0(K)) (4.8) for s = l, l − 1, .., 0, where C=constant>0. Since ftk ∈ L ∞(0,∞, H l0(K)) for k 6 3, so ftk ∈ L ∞(0,∞, L2(K)) for k 6 l+3. In another way, ftk(x, 0) = 0 for k 6 l + 1. Then from Lemma 2.2 we have utl+1 ∈ H1,0(e−γt, Q∞), uts ∈ H 1,0(e−γt, Q∞) for s 6 l. Hence, by using similar arguments used in the proof of Proposition 4.1 we get utl ∈ H 2 0 (e −γt, Q∞). So (4.8) holds for s = l. Assume that (4.8) holds for s = l, l−1, ..., j+1. Put v = utj ∈ H l−j+1 0 (e −γt, Q∞) (by inductive hypothesis) and differentiate (2.1) j-times with respect to t, we have ∆v = −i(vt+ftj ). Following assumptions of induction of s one has vt ∈ H l−j+1 0 (e −γt , Q∞), ftj ∈ H l−j 0 (e −γt, Q∞). Hence Fj = −i(vt + ftj ) ∈ H l−j 0 (e −γt, Q∞). In another 38 Cauchy-Newmann problem for Schrodinger equation regularity of weak solution... way sinceH l−j0 (e −γt, Q∞) ⊆ H l−j−1,0 −1 (e −γt, Q∞), so Fj ∈ H l−j−1 −1 (e −γt, Q∞) for almost t ∈ (0,+∞). Since in the strip l + 1 − j − n 2 6 Imλ 6 l + 2 − j − n 2 there is no spectral point of Newmann problem for (3.1) for all t ∈ [0,+∞), then following Theorem 3.2 page 37 in [11], one gets v ∈ H l+1−j−1 (K). This implies v ∈ H l+1−j,0 −1 (e −γt, Q∞). Note that Fj ∈ H l−j,0 0 (e −γt, Q∞) then applying Theorem 7.3.2 in [8] one gets v ∈ H l+2−j,00 (e −γt, Q∞) satisfying ‖v‖2 Hl+2−j,00 (e −γt,Q∞) 6 C 3∑ k=0 ‖ftk‖ 2 Hl0(e −γt,Q∞) , where C =constant > 0. In another way, it is easy to see that ‖utj‖ 2 Hl+2−j0 (e −γt,Q∞) 6 ‖utj+1‖ 2 Hl+2−j−10 (e −γt,Q∞) + ‖utj‖ 2 Hl+2−j,00 (e −γt,Q∞) . Hence from the inductive assumptions we receive ‖utj‖ 2 Hl+2−j0 (e −γt,Q∞) 6 C 3∑ k=0 ‖ftk‖ 2 Hl0(e −γt,Q∞) , where C is a positive constant. It means that (4.8) is proved. Let s = 0 in (4.8) the proposition is proved. Proof of Theorem 3.1 Proof. Denote U0 be a neighborhood of conical point with a sufficiently small diam- eter so that the intersection of Ω and U0 coincides with the cone K. Take a function u0 = ϕ0u, where ϕ0 ∈ o C∞(U0) and ϕ0 ≡ 1 in a neighborhood of the origin. The function u0 satisfies i∆u0 − (u0)t = ϕ0f + Lu, where Lu = (i∆ϕ0 − (ϕ0)t)u+ igrap ϕ0grap u is a linear differential operator order 1. Coefficients of this operator depend on the choice of the function ϕ0 and equal to 0 outside U0. Denote u1 = ϕ1u = (1− ϕ0)u. It is easy to see that u1 is equal to 0 in a neighborhood of conical point. This follows that we can apply the theorem on the smoothness of a solution of elliptic problem in a smooth domain to this function to conclude that u1 = ϕ1u ∈ H l+2 0 (K) for almost t ∈ [0,+∞). By applying Lemma 2.2 we receive u1 ∈ H l+2 0 (e −γt, Q∞) and ‖u1‖ 2 Hl+20 (e −γt,Q∞) 6 C l+3∑ k=0 ‖ftk‖ 2 Hl0(e −γt,Q∞) , C = const > 0. (4.9) 39 Nguyen Manh Hung and Nguyen Kim Son Let us prove Theorem 3.1 by induction by l. When l = 0 then the function u0, f̂ = ϕ0f + Lu satisfy the hypotheses of Proposition 4.1. So u0 ∈ H 2 0 (e −γt, Q∞). It follows that u = u0+u1 is inH 2 0 (e −γt, Q∞). Assume that the theorem holds up to l−1. Then we have u ∈ H l+10 (e −γt, Q∞). By using analogous arguments used in the proof of Proposition 4.2, with note that f̂tj ∈ H l−j 0 (e −γt, Q∞) (from the hypothesis of induction), we can prove that u0 ∈ H l+2 0 (e −γt, Q∞). So u ∈ H l+2 0 (e −γt, Q∞). The inequality in Theorem 3.1 can derive from inequality (4.7) (for u0) and inequality (4.9). This completes the proof of Theorem 3.1. REFERENCES [1] Dautray R. and Lions J.L., 1990. Mathematical analysis and numerical methods for science and technology. Vol. 1-3-5, Springer - Verlag, Berlin - New York. [2] Frolova E.V., 1994. An initial boundary-value problem with a non coercive bound- ary condition in domains with edges. Zap. Nauchn. Semis. 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