ON A CLASS OF SINGULAR HYPERBOLIC EQUATION WITH A WEIGHTED INTEGRAL CONDITION

In this paper, we study a mixed problem with a nonlocal condition for a class of second order singular hyperbolic equations. We prove the existence and uniqueness of a strong solution. The proof is based on a priori estimate and on the density of the range of the operator generated by the studied problem.


Position of the problem.
In the domain Q = (0,R) × (0,T ), with 0 < R < ∞ and 0 < T < ∞, we consider a second order hyperbolic equation with the Bessel operator ᏸu = ν tt − 1 r ν r − ν r r = f(r , t), (r , t) ∈ Q. (1.1) We adjoin to the above equation the initial conditions 1 ν = ν(r , 0) = Φ(r ), r ∈ (0,R), 2 ν = ν t (r , 0) = Ψ (r ), r ∈ (0,R), (1.2) and the boundary conditions where f (r ,t), Φ(r ), Ψ (r ), m(t), and µ(t) are known functions. We assume that the data satisfies the following compatibility conditions: 3) can be viewed as a nonlocal boundary problem for a singular hyperbolic equation. This problem has not been studied previously. In the case when in equation (1.1), instead of the Bessel operator, we have the operator a(r , t)ν r r , with the Neumann condition and a linear constrain defined by 1 0 ν(r , t)dr = 0, we refer the reader to Bouziani [2]. For other problems with integral conditions, we turn back to Bouziani [3,1,4] and references therein.
In this paper, we prove the existence, uniqueness and continuous dependence upon the data of a strong solution of problem (1.1), (1.2), and (1.3). For this, we transform problem (1.1), (1.2), and (1.3) with inhomogeneous boundary conditions (1.3) to an equivalent problem with homogeneous conditions by introducing a new unknown function u defined as follows: Then, the problem (1.1), (1.2), and (1.3) can be formulated in the following way.
Instead of searching for the function ν, we search for the function u. So, the solution of problem (1.1), (1.2), and (1.3) is given by ν(r , t) = u(r , t) + U(r ,t).

Function spaces.
For the investigation of the posed problem, we need some function spaces. Let L 2 ρ (Q) be the weighted L 2 -space with finite norm The scalar product in L 2 ρ (Q) is defined by and with associated norm Weighted function spaces on the interval (0,R), such as L 2 ρ (0,R) and V 1 ρ (0,R), are used. Their definitions are analogous to those defined on Q.
The problem (1.8), (1.9), and (1.10) can be written in the following operator form: The domain of definition D(L) of an operator L is the set of all functions u ∈ L 2 (Q) for which u t , u tt , u r , u tr , u r r ∈ L 2 (Q) and satisfying conditions (1.10).
Let L be the closure of an operator L with domain of definition D(L).
Definition. We call a strong solution of the problem (1.8), (1.9), and (1.10), the solution of the operator equation (2.8)

Theorem 1. For any function u ∈ D(L), we have the energy inequality
where c is a positive constant independent of the solution u.
Proof. We consider the scalar product in L 2 (Q τ ) of the equation (1.8) and the operator Using conditions (1.9) and (1.10), and integrating by parts each integral term of the right-hand side of (3.3) gives Substituting (3.4), (3.5), (3.6), and (3.7) into (3.3), we obtain (3.8) Using the Cauchy inequality, the last three terms on the right-hand side of (3.8) can be estimated as follows: Substitution of (3.9), (3.10), and (3.11) in (3.8) gives the following inequality: (3.12) By virtue of the elementary inequality, and (3.12), we have where and Applying [2, Lem. 1], to the above inequality, we get (3.17) Since the first term on the left-hand side of (3.17) is positive, we have The right-hand side of the above inequality does not depend on τ. By taking the supremum with respect to τ over 0 to T , we get the desired inequality (3.1) with c = c  By virtue of the uniqueness of the limit in Ᏸ (Q), we conclude that f ≡ 0. According to (3.20), we also conclude that and that the canonical injection from V 1 By virtue of the uniqueness of the limit in Ᏸ (0,R), we conclude, from (3.26) and (3.29), that ϕ ≡ 0. Using the same procedure, we can show that ψ ≡ 0. This proves Proposition 1.
The inequality (3.1) can be extended to strong solutions after passing to limit, that is we have (3.30) The above inequality leads to the following results:

Existence of the solution
where c is a positive constant independent of the solution u.
Proof. From the inequality (3.1), it follows that the operator L has an inverse and, from Corollary 2, we deduce that the range R(L) of the operator L is closed. Hence, it suffices to prove the density of the set R(L) in F , that is R(L) = F . First, we need to prove the following proposition. Proposition 2. If, for any ω ∈ L 2 (Q) and for all u ∈ D 0 (L) = {u | u ∈ D(L) : then ω vanishes almost everywhere in Q.
Proof of the proposition. Relation (4.2) is given for any function u ∈ D 0 (L), so we can express it in the following particular form: Let u tt be the solution of the equation where Ψ (r , t) = T t ω(r , τ)dτ. (4.4) And let the function u be defined by Proof. For the proof, the reader should refer to [4].
To complete the proof of Proposition 2, we replace ω in (4.2) by its representation (4.6). We have − u tt ,u ttt L 2 ρ (Qs ) + u tt , 2 r ρu ttt L 2 ρ (Qs ) + u r ,u ttt L 2 (Qs ) − u r , 2 r ρu ttt L 2 (Qs ) + u r r ,u ttt L 2 ρ (Qs ) − u r r , 2 r ρu ttt L 2 ρ (Qs ) = 0. Using the Cauchy inequality, the right member of (4.13) can be bounded and results (4.14) We observe that the integrand in the second member of (4.14) is independent of s while in the first member depends on it. In order to avoid this difficulty, we introduce a new function ϑ defined by the formula ϑ(r , t) = T t u ττ dτ. Thus, inequality (4.14) can be written as  It follows from inequality (4.22) that ω ≡ 0 almost everywhere on Q T −s 0 . Since the length s is independent of the origin, we use the same procedure a finite number of times to show that ω ≡ 0 in Q. This completes the proof of Proposition 2.