A SAMPLING THEOREM ASSOCIATED WITH BOUNDARY-VALUE PROBLEMS WITH NOT NECESSARILY SIMPLE EIGENVALUES

We use a new version of Kramer's theorem to derive a sampling theorem associated 
with second order boundary-value problems whose eigenvalues are not necessarily simple.

In [3], Kramer derived a sampling theorem which generalizes the Whittaker-Shannon-Kotel'nikov sampling theorem [6, 8,  C is called a Kramer-type kernel if K(x,t) E L(I), Vt E C and there exists a sequence {t} C C such that, {K(x,t)} is a complete orthogonal set in L(I).
The point now is that, where can one find Kramer-type kernels?An aswer to this question is given by Kramer [3] as follows: Consider the self-adjoint boundary value problem Ly EP,(x)y('-')(x) ty, x e I [a,b], (1.1) j 1,2,..., (1.2) Typeset by fl..A-TEX Assume that u(x, t) is a solution of (1.1) such that the zeros, {t }, of Bj(u(x,t)) are the same V 3.
Thus, [3], the zeros of Bj(u(x,t)) are the eigenvalues of the problem (1.1)-(1.2),and {u(x,t)} is a complete orthogonal set of eigenfunctions.Then THEOREM 1.2.Let L(y) ty, B(y) O, j 1,..., n, be a self-adjoint boundary value problem on I. Suppose that there exists a solution u(x,t) of (1.1) such that the set of zeros E, {t} of tS,(u(x,t)) is independent of i.Let g _ n(I).If f(t) [,(, t)g() then, f has the representation where s(t) L (,) (,)   i1(, )11, Kramer's theorem stated above is not always true, since one can find a boundary-value prob- lem of the type (1.1)-(1.2) and a solution u(x,t) such that Bj(u(x,t)) has the same zeros {t}, Vj, but neither {t is the set of eigenvalues, nor {u(x, tk)} is the complete set of eigenfunctions.(1.4) We have, u(x, t) cos v/ cos v/x is a solution of (1.3) with B(u) cosv/g(1cos v/Tr), B(u) -vcos sin v/r.
Obviously B(u),B2(u) have the same set of zeros, {tk k2}'=o, but neither {t}'= 0 is the sequence of eigenvalues, nor {cos--coskx}'=o is the complete set of eigenfunctions.So it is not practical to discuss the existence of Kramer-type kernels associated with problems of type (1.3)-(1.4),i. e., when the eigenvalues are not necessarily simple.When the eigenvalues of the problem are simple, many Kramer-type expansions associated with the boundary-value problems were derived [1, 2, 9].There are two ways introduced by Zayed [7,8] to obtain sampling; series associated with problem (1.3)- (1.4).The first one [8, pp.50-52] is given by taking the kernel of the sampled integral transform to be Therefore, if for some F E L=(0, 7r), then where b(x, t) A cos Vx + B sin Vx.
The last series is not a sampling expansion of f(t) since the coefficients ak and bk can not be uniquely expressed in terms of the sampled values of f at the eigenvalues.If we denote the Hilbert transform of f by , where we obtain F(x) (A sin vx B cos v/x) dx, where r B/A.
The second way is given by taking the kernel of the sampled integral transform to be O(x,t) P(t)G(x,o,t), where G(x,, t) is the Green's function of (1.3)-(1.4),0 is chosen in [0, r] as in [7], and P(t) is the canonical product -(t)
In th article we u another version of Kramer's theorem, Lemma 3.1, so that we can obtain a new Kramer-type mpling reprentation ciated with second order bounda-lue problems which may have multiple eigenvalues.

PRELIMINARIES.
Consider the second-order eigenvalue problem Ly y" q(x)y -Ay, where a,,, fl, are real constants, and q(x) is a continuous real-valued function on [a, hi. (2.3) Let u, v E C2(a, b).Then the Green's formula for this eigenvalue problem is az U,V, + UV + UV, U,V,, (
In th section, we state and prove the main threm of the paper.Theorem 3.1 below a sampling theorem ciated with a cond-order boundary-lue problem whose eigenvalues are not necessarily simple.We start our study by the following Lemma, taken from [1].It is a new version of Kramer's theorem.
k=l k=l In the following we see that the form (4.3) can be reduced into another form which is similar to those resulting in the case of simple eigenvalues by redefining the sampled integral transform as described in the above corollary.Indeed, e',*() A h(A) is entire.Let G,() F(,) F (A)+ F(,) + F3(A), u, (v) v(0) + () 0, (4.5) This is a regular self-adjoint eigenvalue problem.For the same fundamental solutions in the previous example, we have A(A) 4 cos .The eigenvalues are {(2k-1)2}=, all of them are double, their corresponding eigenfunctions are Now and sin rt G,,k(A) -t sin nt, t G,,(Ax,k) , G2,k(A2,)= 2(2k-1) 2.
(,k)-((gk-1)2) --2t sin rt 7[(-(2k-1)2) REMARK.Unlike the case of simple eigenvalues, as the above two examples show, we do not have the relations G,,(A)= cG,(A), 1,2, where a,(A) if zero is not an eigenvalue, if zero is an eigenvalue, and c is a constant depending on k.In fact it can be easily seen that G,, G2., in the previous examples, have zeros more than those of G,, G2.
Let K(x,t) I C C be a function such that K(z,t) .L -(I), Vt ( C. Let {t}ez be a sequence of real numbers such that {g(x,t)}ez