Electronic Journal of Differential Equations, Vol. 2020 (2020), No. 28, pp. 1–20. ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu NONLOCAL PROBLEMS FOR HYPERBOLIC EQUATIONS FROM THE VIEWPOINT OF STRONGLY REGULAR BOUNDARY CONDITIONS LUDMILA S. PULKINA Abstract. In this article, we consider a nonlocal problem for hyperbolic equa- tion with integral conditions and show their close connection with the notion of strongly regular boundary conditions. This has an important bearing on the method of the study of solvability. We propose also a new approach which enables us to prove a unique solvability of the nonlocal problem with integral condition. 1. Introduction In this article, we consider the nonlocal problem for hyperbolic equations Lu ≡ utt − (a(x, t)ux)x + c(x, t)u = f(x, t). (1.1) The question is to find a solution of (1.1) in QT = (0, l) × (0, T ), with l, T < ∞, satisfying the initial conditions u(x, 0) = 0, ut(x, 0) = 0 (1.2) and nonlocal conditions ∫ l 0 Ki(x)u(x, t)dx = 0, i = 1, 2. (1.3) Various phenomena of modern natural science often lead to nonlocal problems on mathematical modeling, and nonlocal models turn out to be often more precise that local conditions; see [5]. Nonlocal problems form a relatively new division of differential equations theory and generate a need in developing some new methods of research [30]. Nowadays various nonlocal problems for partial differential equations are actively studied and one can find a lot of papers dealing with them; see [2, 9, 14, 13, 18] and references therein. We focus our attention on nonlocal problems with integral conditions for hyperbolic equations which have been studied in [1, 3, 4, 6, 12, 9, 25, 17, 19, 23, 27, 28]. Systematic studies of nonlocal problems with integral conditions originated with the papers by Cannon [10] and Kamynin [16]. These and further investigations of nonlocal problems show that classical methods most widely used to prove solvability of initial-boundary problems break down when applied to 2010 Mathematics Subject Classification. 35L10, 35L20, 35L99, 35D30, 34B10. Key words and phrases. Nonlocal problem; hyperbolic equation; nonlocal integral conditions; dynamical boundary conditions; strongly regular boundary conditions; weak solution. c©2020 Texas State University. Submitted December 17, 2019. Published April 6, 2020. 1 2 L. S. PULKINA EJDE-2020/28 nonlocal problems. Nowadays several methods have been devised for overcoming the difficulties arising because of nonlocal conditions. It appears that conditions for the existence and uniqueness of a solution to the nonlocal problem are closely related to the notion of regular boundary conditions [7, 8, 32]. It is known that the system of root functions of an ordinary differential operator with strongly regular boundary conditions form a Riesz basis in L2(0, 1). This property is particularly useful for obtaining results on solvability of bound- ary problems. For convenience we state here a criterium for strong regularity of boundary-value conditions for n = 2 in an easy-to-use form [24, pp. 72-73]. Sturm-Liuville problem: Consider y′′ + λy = 0 with the conditions a1y ′(0) + b1y ′(l) + a0y(0) + b0y(l) = 0, c1y ′(0) + d1y ′(l) + c0y(0) + d0y(l) = 0. (1.4) If the coefficients in (1.4) satisfy one of the following sets of conditions (I) a1d1 − b1c1 6= 0; (II) a1d1 − b1c1 = 0, |a1|+ |b1| > 0, b1c0 + a1d0 6= 0; (III) a1 = b1 = c1 = d1 = 0, a0d0 − b0c0 6= 0, then (1.4) are strongly regular. Before we return to the main problem (1.1)–(1.3), we mention one of initial works dealing with nonlocal problems. In 1897 Steclov [31] considered the problem for the heat equation with nonlocal boundary conditions a1ux(0, t) + b1ux(l, t) + a0u(0, t) + b0u(l, t) = 0, c1ux(0, t) + d1ux(l, t) + c0u(0, t) + d0u(l, t) = 0. (1.5) It is obvious that separating of variables in (1.5) leads to (1.4). Much later non- local problems with conditions of the form (1.5) were studied in [15, 14, 22] and other papers. A feature of the problems with nonlocal conditions is that operator generated by such conditions, in particular (1.5), is not self-adjoint. But if nonlocal conditions of the form (1.4) (as a result of the separation of variables in (1.5)) are strongly regular then there exists a unique solution to the nonlocal problem (see [15]). Now let us return to problem (1.1)–(1.3). Note that (1.3) are first-kind integral conditions as both of them include only integral terms. (The kind of a nonlocal integral condition depends on presence or lack of terms containing a trace of the required solution or its derivative outside the integral). Such conditions cause a considerable difficulties when we try to show that (1.1)–(1.3) is solvable. One method has been advanced for overcoming this difficulty [27]. Its essential idea is as follows. We reduce the first-kind integral conditions to the second-kind ones. To do this, we suppose that u(x, t) is a solution to (1.1)–(1.3), multiply (1.1) by Ki(x) EJDE-2020/28 NONLOCAL PROBLEMS FOR HYPERBOLIC EQUATIONS 3 and integrate over (0, l). As a result we obtain Ki(0)a(0, t)ux(0, t)−Ki(l)a(l, t)ux(l, t)−K ′i(0, )a(0, t)u(0, t) +K ′i(l)a(l, t)u(l, t)− ∫ l 0 [(K ′i(x)a(x, t))x −Ki(x)c(x, t)]u(x, t)dx = ∫ l 0 Ki(x)f(x, t)dx. (1.6) Let us denote a1(t) = K1(0)a(0, t), b1(t) = −K1(l)a(l, t), a0(t) = −K ′1(0)a(0, t), b0(t) = K ′1(l)a(l, t), c1(t) = K2(0)a(0, t), d1(t) = −K2(l)a(l, t), c0(t) = −K ′2(0)a(0, t), d0(t) = K ′2(l)a(l, t), Hi(x, t) = (K ′i(x)a(x, t))x −Ki(x)c(x, t), gi(t) = ∫ l 0 Ki(x)f(x, t)dx and write now (1.6) (omitting the arguments of coefficients) as a1ux(0, t) + b1ux(l, t) + a0u(0, t) + b0u(l, t)− ∫ l 0 H1(x, t)u(x, t)dx = g1(t), c1ux(0, t) + d1ux(l, t) + c0u(0, t) + d0u(l, t)− ∫ l 0 H2(x, t)u(x, t)dx = g2(t). (1.7) This system may be interpreted as perturbed Steclov conditions (1.5) (see [31, 29]). Thus, we establish certain formal connections between (1.7) and (1.4). We will consider it essentially in the next section and show that the nonlocal problem has a unique solution if coefficients of non-perturbed part satisfy one of conditions (I)– (III). The choice of a method depends on a particular criterion. 2. Solvability of nonlocal problems 2.1. Formulation of the problem. It was mentioned in the introduction that integral conditions (1.3) can be reduced to second-kind integral conditions (1.7). As problems (1.1)–(1.3) and (1.1)–(1.2), and (1.7) are equivalent [27], we will con- sider the problem with integral conditions (1.7). For convenience we formulate this problem here with a new indexing: find in QT a solution of the hyperbolic equation utt − (a(x, t)ux)x + c(x, t)u = f(x, t) (2.1) satisfying the initial conditions u(x, 0) = 0, ut(x, 0) = 0, x ∈ [0, l] (2.2) and nonlocal conditions (t ∈ [0, T ]), a1ux(0, t) + b1ux(l, t) + a0u(0, t) + b0u(l, t)− ∫ l 0 H1(x, t)u(x, t)dx = g1(t), c1ux(0, t) + d1ux(l, t) + c0u(0, t) + d0u(l, t)− ∫ l 0 H2(x, t)u(x, t)dx = g2(t). (2.3) Note that there is no loss of generality in supposing that initial conditions are homogeneous. 4 L. S. PULKINA EJDE-2020/28 2.2. Criterium I. For all t ∈ [0, T ], ∆1 ≡ a1d1 − b1c1 6= 0. (2.4) Solving (2.3) as a system with respect to ux(0, t), ux(l, t), we obtain a(0, t)ux(0, t) + α11(t)u(0, t) + α12(t)u(l, t) + ∫ l 0 H11(x, t)u(x, t)dx = g11(t), a(l, t)ux(l, t) + α21(t)u(0, t) + α22(t)u(l, t) + ∫ l 0 H12(x, t)u(x, t)dx = g12(t), (2.5) α11(t) = a0d1 − c0b1 ∆1 a(0, t), α12(t) = b0d1 − d0b1 ∆1 a(0, t), α21(t) = a0c1 − c0a1 ∆1 a(l, t), α22(t) = b0c1 − d0a1 ∆1 a(l, t), H11(x, t) = d1H1(x, t)− b1H2(x, t) ∆1 a(0, t), H12(x, t) = c1H1(x, t)− a1H2(x, t) ∆1 a(l, t), g11(t) = (d1g1(t)− b1g2(t))a(0, t) ∆1 , g12(t) = (c1g1(t)− a1g2(t))a(l, t) ∆1 . This form of integral conditions enables to apply, with only little modifications, a well-known method for boundary-value problem [21], based on a priori estimates. In our view, this approach is effective for studying nonlocal problems with conditions of the form (2.5). It was used for some particular cases [6, 27] so we will not demonstrate it here in detail. Problem 1. Find a solution u(x, t) to (2.1) satisfying (2.2) and (2.5). We consider the Sobolev space W 1 2 (QT ) and denote Ŵ 1 2 (QT ) = {v(x, t) : v ∈W 1 2 (QT ), v(x, T ) = 0}. Let u(x, t) be a solution to the Problem I and v ∈ Ŵ 1 2 (QT ). Using a standard method [21, p. 92] and taking into account (2.5) and ut(x, 0) = 0 we derive the equality∫ T 0 ∫ l 0 (−utvt + auxvx + cuv) dx dt − ∫ T 0 v(0, t)[α11u(0, t) + α12u(l, t)] dt+ ∫ T 0 v(l, t)[α21u(0, t) + α22u(l, t)] dt − ∫ T 0 v(0, t) ∫ l 0 H11(x, t)u(x, t) dx dt+ ∫ T 0 v(l, t) ∫ l 0 H12(x, t)u(x, t) dx dt = ∫ T 0 ∫ l 0 fv dx dt+ ∫ T 0 v(0, t)g11(t) dt− ∫ T 0 v(l, t)g12(t) dt. (2.6) A function u ∈W 1 2 (QT ) is said to be a weak solution to the Problem I if u(x, 0) = 0 and for every v ∈ Ŵ (QT ) the identity (2.6) holds. Theorem 2.1. Assume that (i) a ∈ C(Q̄T ) ∩ C1(QT ), c ∈ C(Q̄T ), a(x, t) > 0 for all (x, t) ∈ Q̄T ; EJDE-2020/28 NONLOCAL PROBLEMS FOR HYPERBOLIC EQUATIONS 5 (ii) H1i, H1it ∈ C(Q̄T ), f ∈ L2(QT ), g1i ∈W 1 2 (0, T ), i = 1, 2; (iii) α12 + α21 = 0, α11ξ 2 + 2α12ξη − α22η 2 ≤ 0. Then there exists a unique weak solution to Problem I. The proof of this theorem for a(x, t) = 1 one can find in [27]. It is not too difficult to show this theorem for a(x, t) not constant. 2.3. Criterium II. Now let ∆ ≡ a1d1 − b1c1 = 0 and |a1|+ |b1| > 0. We will not loss too much generality if suppose that the coefficients in (2.1) do not depend on t. This assumption simplifies the computational work. Then (2.3) can be written as a1ux(0, t) + b1ux(l, t) + a0u(0, t) + b0(t)u(l, t)− ∫ l 0 H1(x)u(x, t)dx = g1(t), c0u(0, t) + d0u(l, t)− ∫ l 0 H2(x)u(x, t)dx = g2(t). (2.7) Problem 2. Find a solution u(x, t) to (2.1) satisfying initial conditions u(x, 0) = 0, ut(x, 0) = 0 and nonlocal conditions (2.7). We can not give at once a definition of a weak solution to this problem as for Problem 1. In response to this, the following can be done. Let u(x, t) be a solution to the Problem 2. Differentiating the second relation of (2.7) with respect to t twice we obtain: c0utt(0, t) + d0utt(l, t) + ∫ l 0 H2(x)uttdx = g′′2 (t). As u(x, t) satisfies (2.1), we have∫ l 0 H2(x)utt(x, t)dx = ∫ l 0 H2(x)[(aux)x − cu+ f ]dx. After some manipulation,∫ l 0 H2(x)(aux)xdx = H2(l)a(l)ux(l, t)−H2(0)a(0)ux(0, t)−H ′2(l)a(l)u(l, t) +H ′2(0)a(0)u(0, t) + ∫ l 0 (H ′2(x)a(x))xu(x, t)dx. Then the second relation in (2.7) becomes c0utt(0, t) + d0utt(l, t)−H2(l)a(l)ux(l, t) +H2(0)a(0)ux(0, t) +H ′2(l)a(l)u(l, t) −H ′2(0)a(0)u(0, t)− ∫ l 0 [(H ′2(x)a(x))x −H2(x)c(x)]u(x, t)dx = g22(t) 6 L. S. PULKINA EJDE-2020/28 where g22(t) = g′′2 (t) + ∫ l 0 H2(x)f(x, t)dx. Equation (2.7) can be written as a1ux(0, t) + b1ux(l, t) + a0u(0, t) + b0(t)u(l, t)− ∫ l 0 H1u dx = g1(t), −H2(0)a(0)ux(0, t) +H2(l)a(l)ux(l, t) +H ′2(0)a(0)u(0, t)−H ′2(l)a(l)u(l, t) + c0utt(0, t) + d0utt(l, t)− ∫ l 0 [(H ′2(x)a(x))x −H2(x)c(x)]u(x, t)dx = g22(t). (2.8) If ∆2 = a1H2(l)a(l)+b1H2(0)a(0) 6= 0, then we can solve system (2.8) with respect to ux(0, t) and ux(l, t): a(0)ux(0, t) = α11u(0, t) + α12u(l, t) + β11utt(0, t) + β12utt + ∫ l 0 P1u dx+G1(t), a(l)ux(l, t) = α21u(0, t) + α22u(l, t) + β21utt(0, t) + β22utt + ∫ l 0 P2u dx+G2(t), (2.9) where αij , βij , Pi(x), Gi(t) i, j = 1, 2 can be find easily, for example, α11 = H ′2(0)b1a(0)−H2(l)a0a(l) ∆2 a(0), β11 = c0b1 ∆2 a(0), P1(x) = H1(x)H2(l)a(l) + (H ′2(x)a(x))xb1 −H2(x)c(x)b1 ∆2 a(0). (We do not cite all formulas because of their length). Conditions (2.9)) are known as dynamical conditions [11, 20, 33]. Thus if (2.7) holds then (2.9) also holds. The converse is also true if g2(0) = g′2(0) = 0. Indeed, let u(x, t) be a solution of (2.1) and let (2.9) hold. Then (2.8) holds. Integrating ∫ l 0 (H ′2(x)a(x))xu(x, t)dx by parts and taking into account that u(x, t) is a solution to (2.1) we easily arrive to d2 dt2 [c0u(0, t) + d0u(l, t) + ∫ l 0 H2(x)u(x, t)dx− g2(t)] = 0. Integrating this equality with respect to t twice, taking into account homoge- neous initial data c0u(0, 0) + d0u(l, 0) + ∫ l 0 H2(x)u(x, 0)dx− g2(0) = 0, c0ut(0, 0) + d0ut(l, 0) + ∫ l 0 H2(x)ut(x, 0)dx− g′2(0) = 0 we obtain (2.7). Thus the nonlocal conditions (2.7) and (2.9) are equivalent, so we will consider the Problem 2 as follows: find a solution u(x, t) to (2.1) satisfying (2.2) and (2.9). EJDE-2020/28 NONLOCAL PROBLEMS FOR HYPERBOLIC EQUATIONS 7 This form of nonlocal conditions enables us to introduce a notation of a weak solution. Following [21, p. 92], we obtain∫ T 0 ∫ l 0 (−utvt + auxvx + cuv) dx dt+ ∫ T 0 v(0, t)[α11u(0, t) + α12u(l, t)] dt − ∫ T 0 vt(0, t)[β11ut(0, t) + β12ut(l, t)] dt+ ∫ T 0 v(0, t) ∫ l 0 P1(x)u(x, t) dx dt − ∫ T 0 v(l, t)[α21u(0, t) + α22u(l, t)] dt + ∫ T 0 vt(l, t)[β21ut(0, t) + β22ut(l, t)] dt− ∫ T 0 v(l, t) ∫ l 0 P2(x)u(x, t) dx dt = ∫ T 0 ∫ l 0 f(x, t)v(x, t) dx dt− ∫ T 0 v(0, t)G1(t) dt+ ∫ T 0 v(l, t)G2(t) dt. (2.10) Let us denote Γ0 = {(x, t) : x = 0, t ∈ [0, T ]}, Γl = {(x, t) : x = l, t ∈ [0, T ]}, Γ = Γ0 ∪ Γl, W (QT ) = {u : u ∈W 1 2 (QT ), ut ∈ L2(Γ)}, Ŵ (QT ) = {v(x, t) : v(x, t) ∈W (QT ), v(x, T ) = 0}. A function u ∈W (QT ) is said to be a weak solution to the Problem 2 if u(x, 0) = 0 and for every v ∈ Ŵ (QT ) the (2.10) holds. Theorem 2.2. Assume the following conditions (i) a ∈ C(Q̄T ), a(x, t) > 0 for all (x, t) ∈ Q̄T , c ∈ C(Q̄T ); (ii) Pi ∈ C(Q̄T ), f ∈ L2(QT ), ft ∈ L2(QT ), Gi ∈ C[0, T ] ∩ C1(0, T ); (iii) β11ξ 2 + 2β21ξη − β22η 2 ≥ 0; (iv) α12 + α21 = 0, β12 + β21 = 0; p(v) β11 > 0, β22 < 0, β11 − |β21| > 0, −β22 − |β21| > 0. Then there exists a unique weak solution to Problem 2. Proof. Uniqueness. Suppose that u1 and u2 are two solutions to Problem 2. Then u = u1 − u2 satisfies initial condition u(x, 0) = 0, and the equation∫ T 0 ∫ l 0 (−utvt + auxvx + cuv) dx dt+ ∫ T 0 v(0, t)[α11u(0, t) + α12u(l, t)] dt − ∫ T 0 vt(0, t)[β11ut(0, t) + β12ut(l, t)] dt+ ∫ T 0 v(0, t) ∫ l 0 P1(x)u(x, t) dx dt − ∫ T 0 v(l, t)[α21u(0, t) + α22u(l, t)] dt + ∫ T 0 vt(l, t)[β21ut(0, t) + β22ut(l, t)] dt− ∫ T 0 v(l, t) ∫ l 0 P2(x)u(x, t) dx dt = 0. Setting v(x, t) = {∫ t τ u(x, η)dη, 0 ≤ t ≤ τ, 0, τ ≤ t ≤ T, 8 L. S. PULKINA EJDE-2020/28 where τ ∈ [0, T ] is arbitrary, and after some manipulation we obtain∫ l 0 [u2(x, τ) + a(x)v2 x(x, 0)]dx = 2 ∫ τ 0 ∫ l 0 cuv dx dt− β11u 2(0, τ) + 2β21u(0, τ)u(l, τ) + β22u 2(l, τ) + α22v 2(l, 0) + 2α21v(0, 0)v(l, 0)− α11v 2(0, 0) + 2 ∫ τ 0 (α12 + α21)v(0, t)vt(l, t) dt− 2 ∫ τ 0 (β12 + β21)u(0, t)ut(l, t) dt + ∫ τ 0 v(0, t) ∫ l 0 P1(x)u(x, t) dx dt− ∫ τ 0 v(l, t) ∫ l 0 P2(x)u(x, t) dx dt. (2.11) Taking into account condition (iii), namelly β11ξ 2 + 2β21ξη − β22η 2 ≥ 0, and (iiii) of Theorem 2.2 we obtain∫ l 0 [u2(x, τ) + a(x)v2 x(x, 0)]dx ≤ ∣∣∣2 ∫ τ 0 ∫ l 0 c(x)u(x, t)v(x, t) dx dt+ a22v 2(l, 0) + 2α21v(0, 0)v(l, 0) − α11v 2(0, 0) + ∫ τ 0 v(0, t) ∫ l 0 P1(x)u(x, t) dx dt − ∫ τ 0 v(l, t) ∫ l 0 P2(x)u(x, t) dx dt ∣∣∣. (2.12) Note that under conditions of Theorem 2.2 there exist positive numbers a0, c0, p such that max [0,l] |c(x)| ≤ c0, a(x) ≥ a0, max i ∫ l 0 P 2 i (x)dx ≤ p. Let us denote A = maxij |αij |. Now we estimate right-hand side of (2.12). Firstly, we use Cauchy and Cauchy-Bunyakovskii-Schwartz inequalities to obtain∫ l 0 [u2(x, τ) + a0v 2 x(x, 0)]dx ≤ ∫ l 0 [u2(x, τ) + a(x)v2 x(x, 0)]dx ≤ c0 ∫ τ 0 ∫ l 0 [u2(x, t) + v2(x, t)] dx dt+A[v2(0, 0) + v2(l, 0)] + ∫ τ 0 [v2(0, τ) + v2(l, τ)] dt+ 2p ∫ τ 0 ∫ l 0 u2(x, t) dx dt. Using trace inequalities, v2(zi, t) ≤ 2l ∫ l 0 v2 x(x, t)dx+ 2 l ∫ l 0 v2(x, t)dx, z1 = 0, z2 = l, (both are derived from v(zi, t) = ∫ zi x vξ(ξ, t)dξ + v(x, t)), we obtain∫ τ 0 [v2(0, τ) + v2(l, τ)] dt ≤ 4l ∫ τ 0 ∫ l 0 v2 x(x, t) dx dt+ 4 l ∫ τ 0 ∫ l 0 v2(x, t) dx dt. EJDE-2020/28 NONLOCAL PROBLEMS FOR HYPERBOLIC EQUATIONS 9 To estimate A[v2(0, 0)+v2(l, 0)] we use the following inequalities (a partial case for n = 1 in [21, p.77]): v2(zi, t) ≤ ε ∫ l 0 v2 x(x, t)dx+ c(ε) ∫ l 0 v2(x, t)dx, where z1 = 0, z2 = l and t ∈ [0, τ ]. Then we obtain v2(0, 0) ≤ ε ∫ l 0 v2 x(x, 0)dx+ c(ε) ∫ l 0 v2(x, 0)dx, v2(l, 0) ≤ ε ∫ l 0 v2 x(x, 0)dx+ c(ε) ∫ l 0 v2(x, 0)dx, where c(ε) = (l + ε)/lε. We note also that from representation of v(x, t) it follows that v2(x, t) ≤ τ ∫ τ 0 u2(x, t) dt. Hence, ∫ τ 0 ∫ l 0 v2(x, t) dx dt ≤ τ2 ∫ τ 0 ∫ l 0 u2(x, t) dx dt,∫ l 0 v2(x, 0)dx ≤ τ ∫ τ 0 ∫ l 0 u2(x, t) dx dt. Choosing ε with due care (ε = a0/4, then a0 − 2ε > 0) we obtain∫ l 0 [u2(x, τ) + a0 2 v2 x(x, 0)]dx ≤M ∫ τ 0 ∫ l 0 (u2(x, t) + v2 x(x, t)) dx dt, (2.13) where M = max{c0 + 2p, (c0 + 4 l )τ 2, A, 4l}. Let w(x, t) = ∫ t 0 ux(x, η)dη. Then vx(x, t) = w(x, t)− w(x, τ), vx(x, 0) = −w(x, τ). With the aid of these equalities we obtain∫ l 0 [u2(x, τ) + a0 2 w2(x, τ)]dx ≤ 2M ∫ τ 0 ∫ l 0 [u2 + w2] dx dt+ 2Mτ ∫ l 0 w2(x, τ) dx dt. As τ is arbitrary we choose it so that a0−4Mτ > 0. To be specific, let a0−4Mτ ≥ a0 2 . Then for all τ ∈ [0, a08M ] m1 ∫ l 0 [u2(x, τ) + w2(x, τ)]dx ≤ 2M ∫ τ 0 ∫ l 0 (u2 + w2) dx dt, where m1 = min{1, a0/4}. From Gronwall’s lemma it follows that ∫ l 0 [u2(x, τ) + w2(x, τ)]dx = 0. Hence u(x, τ) = 0 for all τ ∈ [0, a08M ]. Using the same reasoning as in [21, p.212], we obtain u(x, t) = 0 in QT . It means that there cannot be more than one weak solution to the Problem II. Existence. First, we construct approximations of the weak solution by the Faedo- Galerkin method. Let wk(x) ∈ C2[0, l] be a basis in W 1 2 (0, l). We define the approximations as follows um(x, t) = m∑ k=1 ck(t)wk(x) (2.14) 10 L. S. PULKINA EJDE-2020/28 and shall seek ck(t) from the equations∫ l 0 (umttwj + aumx w ′ j + cumwj)dx+ wk(0) [ α11u m(0, t) + α12u m(l, t) + β11u m tt (0, t) + β12u m(l, t) + ∫ l 0 P1(x)u(x, t)dx ] − wk(l) [ α21u m(0, t) + α22u m(l, t) + β21u m tt (0, t) + β22u m(l, t) + ∫ l 0 P2(x)u(x, t)dx ] = ∫ l 0 f(x, t)wj(x)dx− wj(0)G1(t) + wj(l)G2(t). (2.15) For every m, (2.15) represents a system of second-order ODE’s with respect to ck(t), m∑ k=1 Akjc ′′ k(t) + m∑ k=1 Bkjck(t) = fj(t), (2.16) where Akj = ∫ l 0 wk(x)wj(x)dx+ β11wk(0)wj(0) + β12wk(l)wj(0) − β21wk(0)wj(l)− β22wk(l)wj(l), Bkj = ∫ l 0 (a(x)w′k(x)w′j(x) + c(x)wk(x)wj(x))dx+ wk(0) ∫ l 0 P1(x)wj(x)dx − wk(l) ∫ l 0 P2(x)wj(x)dx+ α11wk(0)wj(0) + α12wk(l)wj(0) − α21wk(0)wj(l)− α22wk(0)wj(0), fj(t) = ∫ l 0 f(x, t)wj(x)dx− wj(0)G1(t) + wj(l)G2(t). Firstly we prove that this system is solvable with respect to c′′k(t). To this end consider the matrix {Aij} and show that it is positive definite. Consider the quadratic form q = ∑m i,j=1Akjξkξj and denote z(x) = ∑m k=1 ξkwk(x). After substituting Akj in q we obtain q = ∫ l 0 |z(x)|2dx+ β11|z(0)|2 + 2β12|z(0)‖z(l)| − β22|z(l)|2. From (iii) of Theorem 2.2, q ≥ 0. As q = 0 if and only if z = 0 and {wk} is linearly independent then ξk = 0 for k = 1, . . . ,m, therefore q is positive definite. Hence (2.16) is solvable with respect to c′′k(t). Thus we can state that under conditions of Theorem 2.2 Cauchy problem for (2.16) with initial conditions ck(0) = 0, c′k(0) = 0 has a solution for every m and {um} is constructed. To derive the estimate we multiply (2.15) by c′j(t), sum over j = 1, . . . ,m and integrate over (0, τ), where τ ∈ [0, T ] is arbitrary:∫ τ 0 ∫ l 0 (umttu m t + aumx u m xt + cumumt ) dx dt EJDE-2020/28 NONLOCAL PROBLEMS FOR HYPERBOLIC EQUATIONS 11 + ∫ τ 0 umt (0, t)[α11u m(0, t) + α12u m(l, t) + β11u m tt (0, t) + β12u m tt (l, t)] dt + ∫ τ 0 umt (0, t) ∫ l 0 P1(x)um(x, t) dx dt − ∫ τ 0 umt (l, t)[α21u m(0, t) + α22u m(l, t) + β21u m tt (0, t) + β22u m tt (l, t)] dt − ∫ τ 0 umt (l, t) ∫ l 0 P2(x)um(x, t) dx dt = ∫ τ 0 ∫ l 0 f(x, t)umt (x, t) dx dt− ∫ τ 0 umt (0, t)G1(t) dt+ ∫ τ 0 umt (l, t)G2(t) dt. Integration by parts and condition (iiii) lead to ∫ l 0 [(umt (x, τ))2 + a(x)(umx (x, τ))2]dx+ β11(umt (0, τ))2 − β22(umt (l, τ))2 = 2β21u m t (0, τ)umt (l, τ)− [α11(um(0, τ))2 + 2α21u m(0, τ)um(l, τ) − α22(um(l, τ))2]− 2 ∫ τ 0 ∫ l 0 cumumt dx dt + 2 ∫ τ 0 um(0, t) ∫ l 0 P1(x)umt dx dt− 2um(0, τ) ∫ l 0 P1(x)um(x, τ)dx − 2 ∫ τ 0 um(l, t) ∫ l 0 P2(x)umt dx dt+ 2um(l, τ) ∫ l 0 P2(x)um(x, τ)dx + 2 ∫ τ 0 ∫ l 0 fumt dx dt+ 2 ∫ τ 0 um(0, t)G1t(t) dt− 2 ∫ τ 0 um(l, t)G2t(t) dt + 2um(0, τ)G1(τ)− 2um(l, τ)G2(τ). (2.17) As β11 > 0 and β22 < 0, By (iiii) of Theorem 2.2, the left-hand side of (2.17) is positive. To estimate right-hand side of (2.17) we use the same technique as in the subsection for uniqueness. Therefore we demonstrate this procedure briefly. Using Cauchy and Cauchy-Bunyakovskii-Schwartz inequalities we obtain ∫ l 0 [(umt (x, τ))2 + a0(umx (x, τ))2]dx+ β11(umt (0, τ))2 − β22(umt (l, τ))2 ≤ C1 ∫ τ 0 ∫ l 0 [(um(x, t))2 + (umt (x, t))2] dx dt+ 2p ∫ l 0 (um(x, τ))2dx + 2 ∫ τ 0 [(um(0, t))2 + (um(l, t))2] dt + (2 + √ |a21|)[(um(0, τ))2 + (um(l, τ))2] + √ |b21|[(umt (0, τ))2 + (umt (l, τ))2] + ∫ τ 0 ∫ l 0 f2(x, t) dx dt + ∫ τ 0 [(G1t(t)) 2 + (G2t(t)) 2] dt+G2 1(τ) +G2 2(τ). (2.18) 12 L. S. PULKINA EJDE-2020/28 where C1 depends on a0, c0, p, l, T and do not depend on m. Using the inequality v2(zi, t) ≤ ε ∫ l 0 v2 x(x, t)dx+ c(ε) ∫ l 0 v2(x, t)dx, z1 = 0, z2 = l we obtain (um(0, τ))2 + (um(l, τ))2 ≤ 2ε ∫ l 0 (umx (x, τ))2dx+ 2c(ε) ∫ l 0 (um(x, τ))2dx,∫ τ 0 [(um(0, τ))2 + (um(l, τ))2] dt ≤ 2ε ∫ τ 0 ∫ l 0 (umx (x, τ))2 dx dt+ 2c(ε) ∫ τ 0 ∫ l 0 (um(x, τ))2 dx dt. Taking into account (iv) in Theorem 2.2, (um(x, τ))2 ≤ τ ∫ τ 0 (umt (x, t))2dt, and adding the inequality ∫ l 0 (um(x, τ))2dx ≤ τ ∫ τ 0 ∫ l 0 (umt (x, t))2 dx dt to the both sides of (2.18), we obtain∫ l 0 [(um(x, τ))2 + (umt (x, τ))2 + a0(umx (x, τ))2]dx + (β11 − √ |b21|)(umt (0, τ))2 + (−β22 − √ |b21|)(umt (l, τ))2 ≤ C2 ∫ τ 0 ∫ l 0 [(um(x, t))2 + (umt (x, t))2 + (umx (x, t))2] dx dt + 2 √ |a21|ε ∫ l 0 (umx (x, τ))2dx+ ∫ τ 0 ∫ l 0 f2(x, t) dx dt + ∫ τ 0 [(G1t(t)) 2 + (G2t(t)) 2] dt+G2 1(τ) +G2 2(τ). (2.19) Choosing ε such that ν = a0−2 √ |a21|ε > 0, we can carry 2 √ |a21|ε ∫ l 0 (umx (x, τ))2dx to the left-hand side of (2.19). Consequently,∫ l 0 [(um(x, τ))2 + (umt (x, τ))2 + (umx (x, τ))2]dx+ [(umt (0, τ))2 + (umt (l, τ))2] ≤ C3 ∫ τ 0 ∫ l 0 [(um(x, t))2 + (umt (x, t))2 + (umx (x, t))2] dx dt + C4 (∫ τ 0 ∫ l 0 f2(x, t) dx dt+ ∫ τ 0 [(G1t(t)) 2 + (G2t(t)) 2] dt +G2 1(τ) +G2 2(τ) ) . (2.20) In particular,∫ l 0 [(um(x, τ))2 + (umt (x, τ))2 + (umx (x, τ))2]dx ≤ C3 ∫ τ 0 ∫ l 0 [(um(x, t))2 + (umt (x, t))2 + (umx (x, t))2] dx dt + C4 (∫ τ 0 ∫ l 0 f2(x, t) dx dt+ ∫ τ 0 [(G1t(t)) 2 + (G2t(t)) 2] dt EJDE-2020/28 NONLOCAL PROBLEMS FOR HYPERBOLIC EQUATIONS 13 +G2 1(τ) +G2 2(τ) ) . Applying Gronwall’s lemma to the above inequality, after integrating over (0, T ), we obtain ‖um‖W 1 2 (QT ) ≤ r1, r1 = C4Te C3T (‖f‖2L2(QT ) + ‖G1‖2W 1 2 (0,T ) + ‖G2‖2W 1 2 (0,T )). Moreover, it follows also from (2.20) that (umt (0, τ))2 + (umt (l, τ))2 ≤ C3‖u‖2W 1 2 (QT ) + C4 ∫ τ 0 ∫ l 0 f2(x, t) dx dt + C4( ∫ τ 0 [(G1t(t)) 2 + (G2t(t)) 2] dt+G2 1(τ) +G2 2(τ)). Then ‖um‖L2(Γ) ≤ r2, r2 = TC3r1 + TC4(‖f‖2L2(QT ) + ‖G1‖2W 1 2 (0,T ) + ‖G2‖2W 1 2 (0,T )). Thus we have a priori estimate, ‖um‖W (QT ) ≤ R, R = max i {ri}, i = 1, 2. (2.21) Because of (2.21) we can extract a subsequence {uµ} from {um} such that as µ → ∞ {uµ} converges weakly to u ∈ W (QT ). This enables us to use standard technique [21, pp. 214-215] and show that the limit of {uµ} is the required weak solution to Problem 2. � 2.4. Criterium III. Let a1 = b1 = c1 = d1 = 0, ∆3 = a0d0− b0c0 6= 0. Then (2.3) can be write as u(0, t) + ∫ l 0 S1(x, t)u(x, t)dx = g31(t), u(l, t) + ∫ l 0 S2(x, t)u(x, t)dx = g32(t), (2.22) where S1(x, t) = b0H2(x, t)− d0H1(x, t) ∆3 , S2(x, t) = c0H1(x, t)− a0H2(x, t) ∆3 , g31(t) = d0g1(t)− b0g2(t) ∆3 , g32(t) = a0g2(t)− c0g1(t) ∆3 . Problem 3. Find a solution u(x, t) to equation (2.1) satisfying (2.2) and (2.22). We can use at least two methods to show the solvability of the Problem 3. One of them may be considered as a particular case of the method used in earlier section. Namely, we differentiate both equations in (2.22) with respect to t twice and arrive at dynamic nonlocal conditions. This method works for for a partial case: Si does not depend on t [28]. However, this method is not effective when Si depend on t also. The second method we can apply is a particular case of the technique initiated in [17]. The main idea of this method is the following: we introduce a new unknown function v(x, t) and arrive at the boundary-value problem for a loaded equation with respect to v(x, t) and can use the technique represented in [17]. 14 L. S. PULKINA EJDE-2020/28 Here we propose a third way. Using the idea in [17] to form a new unknown function. We suggest a different method. A function u(x, t) is said to be the solution to the Problem 3 if it satisfies equation (2.1) for almost all (x, t) ∈ QT , the initial condition (2.2), and conditions (2.22) in the L2(0, T ) sense. Theorem 2.3. Assume that: a, at, ax, att, c, ct ∈ C(Q̄T ), a0, b0, c0, d0 ∈ C2[0, T ], Si, Sit ∈ C2(Q̄T ), Si(0, t) = Si(l, t) = 0, 2l 3 ∫ l 0 (S1 + S2)2dξ < 1 for all t ∈ [0, T ], g3i ∈ C3[0, T ], g3i(x, 0) = g′3i(x, 0) = 0, andf, ft ∈ L2(QT ). Then there exists a unique solution u(x, t) to the problem 3. The proof is rather long, so we break it up into 3 steps. Step 1. Reduction to a problem for a loaded equation. Let u(x, t) be a solution to the Problem 3. We introduce a new function v(x, t) = u(x, t) + ∫ l 0 S̃(x, ξ, t)u(ξ, t)dξ − g̃(x, t) where S̃(x, ξ, t) = l − x l S1(ξ, t) + x l S2(ξ, t), g̃(x, t) = l − x l g31(t) + x l g32(t). Then v(x, t) satisfies the equation vtt − (avx)x + cv − ∫ l 0 (S̃u)ttdξ + (a(x, t) ∫ l 0 S̃x(x, ξ, t)u(ξ, t))xdξ − c(x, t) ∫ l 0 S̃(x, ξ, t)u(ξ, t)dξ = f + g̃tt + cg̃ + axg̃x. As u(x, t) satisfies (2.1), then∫ l 0 (S̃u)ttdξ = ∫ l 0 S̃ttdξ + 2 ∫ l 0 S̃tutdξ + ∫ l 0 S̃[(auξ)ξ − c(ξ, t)u+ f ]dξ. After little manipulations and taking into account the assumptions of Theorem 2.3 we obtain vtt − (avx)x + cv = ∫ l 0 M(x, ξ, t)u(ξ, t)dξ + 2 ∫ l 0 S̃tutdξ + S̃ξ(x, 0, t)a(0, t)u(0, t− S̃ξ(x, l, t)a(l, t)u(l, t) +G(x, t), (2.23) where M(x, ξ, t) = S̃tt(x, ξ, t) + (a(ξ, t)S̃ξ(x, ξ, t))ξ − (a(x, t)S̃x(x, ξ, t))x + [c(x, t)− c(ξ, t)]S̃(x, ξ, t), G(x, t) = f(x, t) + g̃tt(x, t) + c(x, t)g̃(x, t) + ax(x, t)g̃x(x, t) + ∫ l 0 S̃(x, ξ, t)f(ξ, t)dξ. EJDE-2020/28 NONLOCAL PROBLEMS FOR HYPERBOLIC EQUATIONS 15 It is easy to see that v(x, 0) = vt(x, 0) = 0, v(0, t) = v(l, t) = 0 (2.24) and we arrive to the next problem. Problem 4. Find a solution to equation (2.23) satisfying (2.24) Note that we are required to find not only v(x, t), but also u(x, t). Let us denote W 1 2,0(QT ) = {v(x, t) : v ∈W 1 2 (QT ), v(0, t) = v(l, t) = 0}, Ŵ 1 2,0(QT ) = {η(x, t) : η ∈W 1 2,0(QT ), η(x, T ) = 0}. A pair (u, v) is said to be a weak solution to Problem 4 if u ∈ W 1 2 (QT ), v ∈ W 1 2,0(QT ), v(x, 0) = 0, for every η ∈ Ŵ 1 2,0(QT ):∫ T 0 ∫ l 0 (−vtηt + avxηx + cvη) dx dt = ∫ T 0 ∫ l 0 η(x, t) ∫ l 0 M(x, ξ, t)u(ξ, t)dξ dx dt + ∫ T 0 ∫ l 0 η(x, t)[S̃ξ(x, 0, t)a(0, t)u(0, t)− S̃ξ(x, l, t)a(l, t)u(l, t)] dx dt + 2 ∫ T 0 ∫ l 0 η(x, t) ∫ l 0 S̃tutdξ dx dt+ ∫ T 0 ∫ l 0 G(x, t)η(x, t) dx dt (2.25) and u, v are related by v(x, t) = u(x, t) + ∫ l 0 S̃(x, ξ, t)u(ξ, t)dξ − g̃(x, t). (2.26) Step 2. Solvability of Problem 4. Theorem 2.4. Under the assumptions of Theorem 2.3 there exists a unique weak solution (u, v) to Problem 4. Proof. We approximate our weak solution as follows. Let u0 = 0 and define (un, vn) by ∫ T 0 ∫ l 0 (−vnt ηt + avnxηx + cvnη) dx dt = ∫ T 0 ∫ l 0 η(x, t) ∫ l 0 M(x, ξ, t)un−1(ξ, t)dξ dx dt + ∫ T 0 ∫ l 0 η(x, t)[S̃ξ(x, 0, t)a(0, t)un−1(0, t)− S̃ξ(x, l, t)a(l, t)un−1(l, t)] dx dt + 2 ∫ T 0 ∫ l 0 η(x, t) ∫ l 0 S̃tu n−1 t dξ dx dt+ ∫ T 0 ∫ l 0 G(x, t)η(x, t) dx dt, (2.27) vn(x, t) = un(x, t) + ∫ l 0 S̃(x, ξ, t)un(ξ, t)dξ − g̃(x, t). (2.28) 16 L. S. PULKINA EJDE-2020/28 As u0 = 0, then for v1 we have∫ T 0 ∫ l 0 (−v1 t ηt + av1 xηx + cv1η) dx dt = ∫ T 0 ∫ l 0 Gη dx dt. This means that v1(x, t) is a weak solution of the first initial boundary problem for vtt − (avx)x + cv = G(x, t). (2.29) It is known [21, pp. 213-215] that this solution is unique and ‖v1‖W 1 2,0 (QT ) ≤ C‖G‖L2(QT ). Moreover, as Gt ∈ L2(QT ) and a, at, att, ct are bounded then v1 ∈ W 2 2 (QT ) [21, pp. 216-219]. Now we can find u1(x, t) from (2.28) as under assumptions of Theorem 2.4 (2.28) is a second kind Fredholm integral equation with ‖S̃‖ < 1. Then we find v2(x, t) as a solution of the first initial boundary problem for the equation of the form (2.29) with G2(x, t) = ∫ l 0 M(x, ξ, t)u1(ξ, t)dξ + [ S̃ξ(x, 0, t)a(0, t)u1(0, t) − S̃ξ(x, l, t)a(l, t)u1(l, t) ] + 2 ∫ l 0 S̃t(x, ξ, t)u 1 t (ξ, t)dξ +G(x, t). Proceeding as above we obtain un(x, t) and vn(x, t). The conditions of Theorem 2.4 provide that for every n, Gn, Gnt ∈ L2(QT ). So un, vn ∈ W 2 2 (QT ) and the sequence of pairs (un, vn) is well defined. Now we show that this sequence converges as n→∞ in W 1 2,0 and the limit pair (u, v) is the weak solution of the Problem 4. Let zn = vn+1 − vn, rn = un+1 − un. From (2.27) and (2.28) we have∫ T 0 ∫ l 0 (−znt ηt + aznxηx + cznη) dx dt = ∫ T 0 ∫ l 0 η(x, t) ∫ l 0 M(x, ξ, t)rn−1(ξ, t)dξ dx dt + ∫ T 0 ∫ l 0 η(x, t)[S̃ξ(x, 0, t)a(0, t)rn−1(0, t)− S̃ξ(x, l, t)a(l, t)rn−1(l, t)] dx dt + 2 ∫ T 0 ∫ l 0 η(x, t) ∫ l 0 S̃tr n−1 t dξ dx dt, (2.30) zn(x, t) = rn(x, t) + ∫ l 0 S̃(x, ξ, t)rn(ξ, t)dξ. (2.31) The assumptions of Theorem 2.4 provide the existence of positive number s0 such that max Q̄T { ∫ l 0 S̃2dξ, ∫ l 0 S̃2 t dξ, ∫ l 0 S̃2 ξdξ} ≤ s0. Then for 1− 2s0l > 0 from (2.31), we obtain ‖rn‖2L2(QT ) ≤ 2 1− 2s0l ‖zn‖2L2(QT ). (2.32) EJDE-2020/28 NONLOCAL PROBLEMS FOR HYPERBOLIC EQUATIONS 17 From znt = rnt + ∫ l 0 S̃tr ndξ + ∫ l 0 S̃rnt dξ and znx = rnx + ∫ l 0 S̃xr ndξ, for 1 − 3s0l > 0, we obtain ‖rnt ‖2L2(QT ) ≤ 3 1− s0l ‖znt ‖2L2(QT ) + 6s0l (1− 2s0l)(1− 3s0l) ‖zn‖2L2(QT ), (2.33) ‖rnx‖2L2(QT ) ≤ ‖z n t ‖2L2(QT ) + 4s0l (1− 2s0l)2 ‖zn‖2L2(QT ). (2.34) Then from (2.32)–(2.34), ‖rn‖2W 1 2 (QT ) ≤ A‖z n‖2W 1 2 (QT ) (2.35) where A depends on a0, a1, c0, s0, l, T . To proceed further we prove the following statement. If v ∈W 2 2 (QT ) is a solution to the first initial-boundary problem in (2.29), then for almost all τ ∈ [0, T ],∫ l 0 [(v2(x, τ))2 + a(x, τ)v2 x(x, τ)]dx = ∫ τ 0 ∫ l 0 atv 2 x dx dt− 2 ∫ τ 0 ∫ l 0 cvvt dx dt+ 2 ∫ τ 0 ∫ l 0 Gvt dx dt. (2.36) We obtain (2.36) after integrating the equality (vtt−(avx)x+cv)vt = G(x, t)vt over Qτ = (0, l)× (0, τ). Note that we seek zn as a solution of (2.29) with G(x, t) = ∫ l 0 M(x, ξ, t)rn−1(ξ, t)dξ + [ S̃ξ(x, 0, t)a(0, t)rn−1(0, t) − S̃ξ(x, l, t)a(l, t)rn−1(l, t) ] + 2 ∫ l 0 S̃t(x, ξ, t)r n−1 t (ξ, t)dξ. Under the assumptions of Theorem 2.4 there exist positive numbers a0, a1, c0, s1 such that a(x, t) ≥ a0, |c(x, t)| ≤ c0, |a(x, t), ax(x, t), at(x, t)| ≤ a1, s1 = max { max Q̄T ∫ l 0 (S̃tt) 2dξ, max Q̄T ∫ l 0 (S̃ξξ) 2dξ } . From Theorem 2.1 and some manipulations (integrating and using Cauchy and Cauchy “with ε” inequalities) we obtain∫ l 0 [(zn(x, τ))2 + a(x, τ)(znx (x, τ))2]dx ≤ A1 ∫ τ 0 ∫ l 0 [(zn)2 + (znt )2 + (xnx)2] dx dt + c(ε) ∫ τ 0 ∫ l 0 (znt )2 dx dt+ ε ∫ τ 0 ∫ l 0 G2(x, t) dx dt. Estimating the last term with the help of Cauchy-Bunyakovskii-Schwartz inequality, and the trace inequalities (rn−1(zi, t)) 2 ≤ 2l ∫ l 0 (rn−1 x (x, t))2dx+ 2 l ∫ l 0 (rn−1(x, t))2dx, z1 = 0, z2 = l, 18 L. S. PULKINA EJDE-2020/28 we obtain ∫ l 0 [(zn(x, τ))2 + a(x, τ)(znx (x, τ))2]dx ≤ A2 ∫ τ 0 ∫ l 0 [(zn)2 + (znt )2 + (xnx)2] dx dt + εA3 ∫ τ 0 ∫ l 0 [(rn−1)2 + (rn−1 t )2 + (rn−1 x )2] dx dt. From Gronwall’s lemma and integrating over (0, T ) it follows that ‖zn‖2W 1 2 (QT ) ≤ εB‖r n−1‖2W 1 2 (QT ), (2.37) where B = TA3e A2T , Ai, i = 1, 2, 3, depend only on a0, a1, c0, s1, l, T . From (2.35) and (2.37) it follows that ‖zn‖2w1 2(QT ) ≤ ABε‖z n−1‖2w1 2(QT ), ‖rn‖2w1 2(QT ) ≤ ABε‖r n−1‖2w1 2(QT ). (2.38) We select a small ε such that 0 < εAB < 1. Hence, {un, vn} is a Cauchy sequence in W 1 2 (QT ). Thus, there exists a unique pair (u, v) ∈ W 1 2 (QT ) such that un → u, vn → v. Let n→∞ in (2.27), (2.28). From the converges of {un, vn} we see that (u, v) is the required solution of the Problem 4. � Step 3. Solvability of the Problem 3. As we noted above, under assumptions of Theorem 2.4, the weak solution (u, v) belongs to W 2 2 (QT ). That is why we can rewrite (2.25) as∫ T 0 ∫ l 0 (vtt − (avx)x + cv)η dx dt = ∫ T 0 ∫ l 0 η(x, t) ∫ l 0 M(x, ξ, t)u(ξ, t)dξ dx dt + ∫ T 0 ∫ l 0 η(x, t)[S̃ξ(x, 0, t)a(0, t)u(0, t)− S̃ξ(x, l, t)a(l, t)u(l, t)] dx dt + 2 ∫ T 0 ∫ l 0 η(x, t) ∫ l 0 S̃tutdξ dx dt+ ∫ T 0 ∫ l 0 G(x, t)η(x, t) dx dt. (2.39) Substituting v(x, t) represented by (2.26) into (2.39), after some manipulations, we obtain ∫ T 0 ∫ l 0 (utt − (aux)x + cu)η(x, t) dx dt = ∫ T 0 ∫ l 0 f(x, t)η(x, t) dx dt for all η ∈ W̊ 2 2 (QT ). So, u(x, t) is the solution to the (2.1). Obviously, the conditions (2.22) are fulfilled. This completes the proof of Theorem 2.3. Acknowledgments. I want to thank the anonymous referees for his/her comments that improved this article, and Professor Julio G. 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