Applied Mathematical Sciences, Vol. 1, 216, no. 11, 543-548 HIKARI Ltd, www.m-hikari.com http://dx.doi.org/1.12988/ams.216.512743 On a Boundary-Value Problem for Third Order Operator-Differential Equations on a Finite Interval Sabir S. Mirzoyev 1,2 and Saadat B. Heydarova 2 1 Baku State University, Baku, Azerbaijan 2 Institute of Mathematics and Mechanics of NAS of Azerbaijan Baku, Azerbaijan Copyright c 216 Sabir S. Mirzoyev and Saadat B. Heydarova. This article is distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. Abstract The conditions which ensures the existence and uniqueness of a regular solution of a boundary value problem on a finite interval for operatordifferential equation in a Hilbert space are found. These conditions are expressed in terms of properties of operator coefficients. Thus, the estimations of the norms of the operators of intermediate derivatives on a finite interval are obtained and their connection with the solvability conditions of the given boundary value problem is shown. Mathematics Subject Classification: 39B42, 46C5, 34K1 Keywords: Hilbert space, operator-differential equation, boundary-value problem, linear operator 1 Introduction and preliminaries Let H γ be a separable Hilbert space and A be a positive definite self-adjoint operator with domain D(A). With respect to the scalar product (x, y) γ = (A γ x, A γ y), x, y D (A γ ) the linear set D (A γ ) domain of the operator A γ (γ ) becomes a Hilbert space H γ. When γ =, we consider H = H, (x, y) = (x, y).
544 Sabir S. Mirzoyev and Saadat B. Heydarova Let us consider the boundary value problem in H: u (t) A 3 u (t) + A 3 j u (j) (t) = f (t), t (, 1), (1) j= u () = u () =, u (1) =, (2) where f (t), u (t) are functions defined almost everywhere on an interval (, 1) with the values in H, the operator coefficients satisfy the following conditions: 1. A is a positive definite self-adjoint operator in H; 2. B j = A j A j ( j =, 3 ) are bounded operators in H. All derivatives here and below are understood in the sense of distributions [1]. We denote by L 2 ((, 1) : H) a Hilbert space of functions, determined on (, 1), almost everywhere with values from H, measurable, square-integrable in Bochner sense with the norm ( 1 1/2 f L2 ((,1):H) = f (t) dt) 2. Let us introduce the following Hilbert spaces [1] W 3 2 ((, 1) : H) = { u : u L 2 ((, 1) : H), A 3 u L 2 ((, 1) : H) }, W 3 2 ((, 1) : H) = { u : u W 3 2 ((, 1) : H), u () = u () =, u (1) = }, with a norm u W 3 2 ((,1):H) = ( u 2 L 2 ((,1):H) + A 3 u 2 L 2 ((,1):H)) 1/2. Definition. If for every f (t) L 2 ((, 1) : H) there exists a function u (t) W 2 3 ((, 1) ; H), such that satisfies the equation (1) almost everywhere in (, 1) and the estimation u W 3 2 ((,1):H) < const f L 2 ((,1):H), then the problem (1), (2) is called regularly solvable. In this paper we find the conditions on coefficients of the equation (1), which provide regular solvability of the problem (1), (2). It should be noted that the regular solvability of some boundary value problems for the equation (1) on semi-axis are investigated ( see, for example, [2-4]). In a bounded field the regular solvability of boundary value problems is studied relatively little (see, [5,6]). For the second-order equation similar research problems have been investigated in detail [7-9].
BVP for third order operator-differential equations 545 In the space W 3 2 ((, 1) : H) the following operators are defined P u = u (t) A 3 u (t), P 1 u = A 3 j u (j) (t), u W2 3 ((, 1) : H). j= From the conditions 1) and 2) it follows that the operators P and P 1 are boundedly acts from W 2 3 ((, 1) : H) to L 2 ((, 1) : H). The following lemma is true. Lemma 1. Let the condition 1) be fulfilled. Then for u W 2 3 ((, 1) : H) the following equality holds: P u 2 L 2 ((,1):H) = u 2 L 2 ((,1):H) + A 3 u 2 L 2 ((,1):H) + u (1) 2 3/2. (3) Proof. Let u W2 3 ((, 1) : H). Then we have: P u 2 L 2 ((,1):H) = u A 3 u 2 L 2 ((,1):H) = = u 2 (L 2 (,1):H) + A 3 u 2 (L 2 (,1):H) 2Re ( u, A 3 u ) (L 2 (,1):H). (4) Integrating by parts, we have ( u, A 3 u) L 2 ((,1):H) = ( A 1/2 u (t), A 5/2 u (t) ) 1 ( A 3/2 u (t), A 3/2 u (t) ) 1 + + ( A 5/2 u (t), A 1/2 u (t) ) 1 ( A 3 u, u ) L 2 ((,1):H) = = u (1) 2 3/2 ( A 3 u, u ) L 2 ((,1):H), i.e. 2Re (u, A 3 u) L2 ((,1):H) = u (1) 2 3/2. Taking into account this equality in (4) we obtain the validity of the equality (3). 2 Main results At first we will study the solvability of the equation P u = f. Theorem 1. Let the condition 1) be fulfilled. Then the operator P forms an isomorphism between the spaces W2 3 ((, 1) : H) and L 2 ((, 1) : H). Proof. By Banach s theorem on the inverse operator it is sufficient to prove that KerP = {} and ImP = L 2 ((, 1) : H). From the equality (3) it follows that KerP = {}. Let us show that ImP = L 2 ((, 1) : H). It is easy to see that the function u 1 (t) = 1 + ( (i 3 λ 3 E A 3) 1 ) 1 f (s) e iλs ds e iλt dλ, t (, 1), 2π
546 Sabir S. Mirzoyev and Saadat B. Heydarova belongs to W2 3 ((, 1) : H) and satisfies the equation u (t) A 3 u (t) = f (t) almost everywhere in the interval (, 1). The trace theorem [1] implies that u (j) 1 () H 3 j 1/2, u (j) 1 (1) H 3 j 1/2, j =, 1, 2. Now, the solution of the equation P u = f will be sought in the form u (t) = u 1 (t) + e ω1ta ϕ 1 + e ω2ta ϕ 2 + e ω3(t 1)A ϕ 3, t (, 1), ( ) where ϕ 1, ϕ 2, ϕ 3 H 5/2 are unknown vectors, ω 1 = 1 2 1 + i 3, ω2 = ( ) 1 2 1 i 3, ω3 = 1. Then, using lemma 1 and taking into account u 1 () H 5/2, u 1 () H 3/2 and u 1 (1) H 1/2, from the condition (2), we uniquely find the vectors ϕ 1, ϕ 2, ϕ 3 H 5/2. Consequently, u W2 3 ((, 1) : H) and P u = f. The theorem is proved. This theorem implies that norms u W 3 2 ((,1):H) and P u L2 ((,1):H) are equivalent in space W2 3 ((, 1) : H). Therefore norms are finite N j = sup A 3 j u (j) P L2 ((,1):H) u 1 L 2 ((,1):H), j =, 3. (5) u W2 3((,1):H) Theorem 2. Norms N j ( j =, 3 ) satisfy the following estimations : N j c j, where c = c 3 = 1, c 1 = 2 3 1/2, c 2 = 2 1/2 3 1/4. Proof. From the Lemma 1 it follows that N 1, N 3 1. For u W 3 2 ((, 1) : H) Au 2 L 2 ((,1):H) = (Au, Au ) L2 ((,1):H) = = ( A 3/2 u (t), A 1/2 u (t) ) 1 ( A 2 u, u ) L 2 ((,1.:H)) A 2 u L2 ((,1):H) u L2 ((,1):H). (6) Then, is easily verified that the equality is true for any ε > εau A 2 u + ε 1 A 3 u 2 L 2 ((,1):H) = ε2 Au 2 L 2 ((.,1):H) Au 2 L 2 ((,1):H) + +ε 2 A 3 u 2 L 2 ((,1):H) ε 1/2 A 3/2 u (1) ε 1/2 A 5/2 u (1) 2. Hence, we obtain that for any ε > : Au 2 L 2 ((,1):H) ε2 Au 2 L 2 ((,1):H) + ε 2 A 3 u 2 L 2 ((,1):H). Let ε = A 3 u 1/2 L 2 ((,1):H) Au 1/2 L 2 ((,1):H). Then we have Au 2 L 2 ((,1):H) 2 Au L2 (,1):H A 3 u L2 ((,1):H). (7)
BVP for third order operator-differential equations 547 Taking into account the inequality (6) in (7), we obtain: or Au 2 L 2 ((,1):H) 2 A 3 u L2 ((,1):H) Au 1/2 L 2 (,1):H u 1/2 L 2 ((,1):H) Au L2 ((,1):H) 22/3 A 3 u 2/3 u 1/3 L 2 ((,1):H). Then for any δ > the following inequality holds: Au 2 L 2 ((,1):H) (δ 24/3 A 3 u ( ) 2 2/3 1/3 1 L 2 ((,1):H)) δ 2 u 2 L 2 ((,1):H) ( 2 2 4/3 3 δ A 3 u 2 + 1 ) L 2 ((,1):H) 3δ 2 u 2 L 2 ((,1):H). Assuming δ = 2 1/3, we have: Au 2 L 2 ((,1):H) 4 3 u 2 W 3 2 (,1):H 4 3 P u 2 L 2 ((,1):H). Consequently, N 1 c 1 = 2 3 1/2. Then, from the inequality (6) it follows that N 2 c 2 = 2 1/2 3 1/4. The theorem is proved. Let us prove the following main Theorem 3. Let the conditions 1), 2) be fulfilled and the following inequality ( q = c j B 3 j < 1, Bj = A j A j, j =, 3 ), j= ( ) holds, where the numbers c j j =, 3 are defined from the theorem 2. Then the problem (1), (2) is regularly solvable. Proof. Let us write the problem (1), (2) in the equation form P u+p 1 u = f, f L 2 ((, 1) : H), u W 2 3 ((, 1) : H) By theorem 1, the operator P 1 : L 2 ((, 1) : H) W 2 3 ((, 1) : H) is an isomorphism. Assuming u = P 1 v, we have the equation v + P 1 P 1 v = f in L 2 ((, 1) : H). Taking into account the theorem 2, for any v L 2 ((, 1) : H) we obtain: P1 P 1 v = P L2 ((,1):H) 1u L2 ((,1):H) B 3 j A 3 j u (j) L2 ((,1):H) j= c j B 3 j P u L2 ((,1):H) = q v L 2 ((,1):H). j= As q < 1, then u = P 1 Theorem is proved. ( ) E + P1 P 1 1 f and u W 3 2 ((,1):H) const f L 2 (,1 :H).
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