A BOUNDARY VALUE PROBLEM FOR THE WAVE EQUATION

Traditionally, boundary value problems have been studied for elliptic differential equations. The mathematical systems described in these cases turn out to be “well posed”. However, it is also important, both mathematically and physically, to investigate the question of boundary value problems for hyperbolic partial differential equations. In this regard, prescribing data along characteristics as formulated by Kalmenov [5] is of special interest. The most recent works in this area have resulted in a number of interesting discoveries [3, 4, 5, 7, 8]. Our aim here is to extend some of these results to a more general domain which includes the characteristics of the underlying wave equation as a part of its boundary.


Introduction.
The well-known two point boundary value problem for the massspring system has an analog in the continuum case which was first formulated in [5,7] as follows Lu = u tt − u xx = F(x, t), (x, t) ∈ R, (1.1) u(x, 0) = 0, 0 ≤ x ≤ 2, (1.2) Here, R is a region bounded by the characteristics and the line segment t = 0, 0 ≤ x ≤ 2, as described below.R = (x, t) : t < x < 2 − t, 0 < t < 1 . (1.4) In [5], it is shown that a unique solution u ∈ W 1 2 (R) ∩ W 1 2 (∂R) ∩ C( R) [1] of (1.1), (1.2), and (1.3) can be constructed, and that L and L −1 are both self adjoint in L 2 (R).Furthermore, in the case where F(x,t) = λu, a complete set of eigenfunctions and corresponding eigenvalues are explicitly computed.In [7] on the other hand, where F(x,t) = λp(x, t)u, such explicit computations are not possible.The selfadjointness of L in L p 2 (R), the set of weighted L 2 functions in R with weight p, and thereby the existence of a complete set of eigenfunctions, is established by constructing a symmetric Hilbert Schmidt kernel [9] for L −1 .In addition it is demonstrated that one can replace the boundary conditions (1.2) and (1.3) with u t (x, 0) = 0, 0 ≤ x ≤ 2, u(1, 1) = 0, (1.5) and obtain self adjointness.In the subsequent papers [4,8], some generalizations of these results were considered.These generalizations were focused on the dimension of the space variable while preserving both the complete continuity [9] and self adjointness of the resulting integral operator.Our goal in this paper is to study equation (1.1) along with boundary data prescribed on the characteristics and noncharacteristic C 1 curve.The effect of this change in the boundary is that the symmetry of the kernel of the integral operator is lost.However, one can preserve the compactness of the integral operator and obtain information about the spectrum of the operator L using the theory of compact operators.

The boundary value problem
(2.1) The characteristic curves C 1 ,C 2 and noncharacteristic Γ are defined by ( Let τ ∈ [0, 1] be fixed.The line s = −t + f (τ) + τ meets Γ and C 1 at points S 1 and S 2 , respectively, and the line s = t + f (τ) − τ meets Γ and C 2 at S 1 , S 3 .Denote the point of intersection of C 1 and C 2 by S 4 .The coordinates of these points are as follows. (2.3) Let α ≠ 1, 0, be a constant in R. Consider the boundary value problem ) where ω is known.Our goal is to convert problem (2.4), (2.5), and (2.6) into an integral equation and study the kernel of the resulting integral operator.

The solution of the boundary value problem. Make the change of variables
Then, u(s, t) = v(x, y) and equation (2.4) has the simple form for v, where Ᏺ(x, y) = (1/4)F ((y + x)/2,(y − x)/2).Under the last transformation the region Ω is mapped into where g(x) is the solution of the equation for y, and M is the x-coordinate of the point of intersection of y = g(x) and y = 2.We note that therefore, the two lines do intersect.Let (σ , η) be an interior point of T .For a fixed σ ∈ [0,M], the boundary condition (2.5) in these coordinates will be In order to convert (3.2) and (3.6) into an integral equation, keeping in mind that v along the curve y = g(x) is given, we integrate (3.2) over the rectangle with vertices at (0, 2), (σ , 2), (0,η), and (σ , η).We have The boundary condition (3.6) and equality (3.8) imply If we substitute g −1 (η) for σ in (3.7) we have The substitution of g −1 (η) for σ in (3.9) yields In equations (3.10) and (3.11 Substitute v(0,η) from (3.12) and v(σ , 2) from (3.9) into (3.7) and solve for v(σ , η) to obtain If we combine the integrals in the above equation we can write (3.13) in the more compact form where G the Green's function in T × T is defined as follows.Let (σ , η) be a point in T .Define the sets T i ⊆ T , 1 ≤ i ≤ 6 by Then G, has values (3.16) Remark.We note that for α = −1 the function G is symmetric.Also, we had chosen the positive semi axis t = 0 instead of s = f (t), we would have had y = x as a part of the boundary of T .If, in addition, datum on t = 0 was chosen to be zero then the result of [7] would be applicable to v.
We have the following uniqueness result.Proof.Let v 1 and v 2 be two solutions.Then their difference V satisfies the following equations: The condition (3.18) is the simplified version of condition (3.6) due to the compatibility condition Choose a point (σ , η) in T , and integrate equation (3.17) over the rectangle in T with vertices at (0,η), (σ , η), (σ , 2), and (0, 2).We will have, Also, if we substitute g(σ ) for η in (3.21) and use the compatibility condition we have This equation together with the condition (3.18) imply that V (σ ,2) and V (0, g(σ )) are both zero.But these are the values of V along y = 2 and x = 0. Substituting zero for V (σ ,2) and V (0,η) in (3.21), we will have V (σ ,η) ≡ 0.
The case that the right-hand side of equation (3.2) depends on the unknown function v is treated differently.We will consider this situation in two separate settings next.

The eigenvalue problem
4.1.The datum along Γ is identically zero.Let λ be a parameter, and p a nonnegative measurable function in L ∞ (T ).Denote by B the Banach space of weighted square integrable functions, i.e., with norm in B defined by, In what follows we consider the equation and λ = 1/λ .For v ∈ B, (4.5) is an eigenvalue problem of functional analysis.The theory of the spectrum of a compact operator [2,9] allows us to investigate the solutions of this operator equation.Our task is to show that K is a compact (or completely continuous, e.g., [9]) operator in B. Here, for the sake of convenience, we include a few definitions that we will use in the sequel.
Definition 4.1.A set of functions {φ} is said to be uniformly bounded if there exists a constant c such that |φ| ≤ c, for all functions φ in the set.Definition 4.2.A set of continuous functions {φ} is said to be equicontinuous in a region Ω, if given , there exists δ( ) such that for any two points P 1 ,P 2 in Ω, for all functions φ in the set [2].According to Arzela's theorem [9], any uniformly bounded equicontinuous set {φ} of functions has a uniformly convergent subset.The first integral on the right-hand side can be made small as follows.Let P 1 and P 2 have coordinates (σ , η) and (σ ,η ), respectively.Assume without loss of generality that σ < σ and η < η.Define the rectangles R 1 -R 6 in T as follows (Figure 1) (4.12) We note that, since g is a continuous function, the area ∆R i of each rectangle is a function of .Let A( ) represent, A( ) = max ∆R i , i = 1,...,6 .The difference |G(P 1 ; x, y) − G(P 2 ; x, y)| takes on the following values in T .
which is independent of P 1 and P 2 .Then we have, The inequalities (4.10), (4.11), and (4.16) imply that Kφ(P 1 ) − Kφ(P 2 ) ≤ δ( ), (4.17) where δ( ) = 6L M 2 p ∞ A( ).This shows that the set {Kφ} is equicontinuous.Therefore, by Arzela's theorem {Kφ} has a uniformly convergent subset.But uniform convergence implies convergence in • p 2 norm.Hence, K maps a bounded subset of B to a compact set, i.e., K is a compact operator.Now, we are in a position to discuss the spectrum of K. We note first that, the conjugate operator K * is generated by the kernel G(x, y; σ ,η), and therefore is bounded.This suffices to imply that K * K is compact.Furthermore, if we equip the space B with the inner product φ, ψ = T φψp dx dy, (4.18) the operator K * K is also selfadjoint and positive over B. The eigenvalues of K * K are positive and countable.We let J be the index set of the eigenvalues of K * K and call µ j , the singular values of K. Also, ᏺ(K) will represent the space of functions in L p 2 (T ) that are mapped to zero by K [6], We have the following theorem [6,9] Theorem 4.2 (Singular Value Decomposition).Let µ 1 ≥ µ 2 ≥ µ 3 ••• > 0 be the ordered sequence of positive singular values of K, counted relative to its multiplicity.Then there exists orthonormal systems (φ j ) and (ψ j ) both subsets of L p

(T ) with the following properties:
Kφ j = µ j ψ j and K * ψ j = µ j φ j for all j ∈ J. ( The system (µ j ,φ j ,ψ j ) is called the singular system of K. Every φ in L p 2 (T ) possesses the singular value decomposition for some φ 0 ∈ ᏺ(K) and Kφ = j∈J µ j φ, φ j ψ j .( A necessary and sufficient condition for the existence of a solution of equation (4.5) is given by a theorem of Picard [6].
Theorem 4.3.Let (µ j ,v j ,u j ) be a singular system for the compact operator K.A necessary and sufficient condition for the equation (4.5) to be solvable is that

.24)
In this case is a solution of equation (4.5).

4.2.
The datum along Γ is not identically zero.We now turn to the more general case where the datum along the curve y = g(x) is not identically zero and the righthand side of (3.2) depends on v.The problem we are considering is The method of Section 3 shows that the solution of (

Higher dimensions.
The extension of the boundary value problem (1.1), (1.2), and (1.3) to analogous selfadjoint problems in higher dimensions is also of much interest.In [8], the equation (1.1) in n-dimensional space is considered.It is shown there that, in a characteristic cone, it is possible to construct symmetric Green functions which will convert the operator L of (5.1) to a selfadjoint integral operator.However, the boundary values that give rise to such selfadjoint problems are not known.In [4], a two space dimensional extension of (  (5.9) Upon further change of variable r (1−n)/2 u = U we have U tt − U r r = r (n−1)/2 F(r ,t).
(5.10) Equation (5.10) along with conditions (5.7) and (5.8) can be treated by the procedures of Sections 2 and 3.The only difference now is that when dealing with the case F(r ,t) = λp(r , t)u, the factor r n−1 appears on the right-hand side of (5.10), i.e., we will have U tt − U r r = λr n−1 p(r , t)U.
(5.11)However, since this factor is continuous over Ω, it does not pose any new difficulty.We can modify the weight function to P (r ,t) = r n−1 p(r , t), proceed as before and obtain the results of Section 4.

Figure 1 .
Figure 1.The regions R i .