VOL. 21 NO. 3 (1998) 571-580
A SAMPLING THEOREM ASSOCIATED WITH BOUNDARY-VALUE PROBLEMS WITH NOT NECESSARILY SIMPLE EIGENVALUES
MAHMOUD H.ANNABY
Lehrstuhl
A
fiirMathematik,RWTH Aachen,
D-52056Aachen, Germany.
HASSAN A. HASSAN
Department
of Mathematics,Faculty of Science,Cairo University, Giza,Egypt.
(Received
July 10, 1996 andinrevised formMarch13, 1997)
ABSTRACT. We
use a new versionofKramer’s
theoremto deriveasamplingtheorem asso- ciated with secondorderboundary-value problems whose eigenvaluesare notnecessarily simple.KEY WORDS AND PHRASES: Kramer’s theorem, Lagrange
interpolations, eigenvaluc problems.1991
AMS SUBJECT CLASSIFICATION CODES: 34A05,
94A24.1.
INTRODUCTION.
In [3], Kramer
derivedasampling theoremwhichgeneralizes the Whittaker-Shannon-Kotel’nikov sampling theorem[6,
8, pp.16-17]. It
statesthatTHEOREM
1.1.Let I
beafiniteclosed interval.Let K(x,t) I C
Cbeafunctionsuch thatK(z,t) . L-(I),
Vt(C.Let {t}e z
beasequenceofreal numberssuchthat{g(x,t)}ez
isacomplete orthogonalset in
L2(I). Let
g_ L2(I)
and suppose thatI(t) [ g(, t)g()
d.J Then
where
f(t) f(t)S(t), a.z
S(t) f K(x,t)K(x,t)dx ilK(x,t)ll2
DEFINITION
1.1.A
functionK(z,t) I C C
is called aKramer-type
kernel ifK(x,t)
EL(I), Vt
EC
and thereexists a sequence{t}
CC
suchthat, {K(x,t)}
is a complete orthogonalset inL(I).
Thepointnowis that,where canonefind
Kramer-type
kernels?An
aswertothis question isgiven byKramer [3]
asfollows:Consider theself-adjoint boundary value problem
Ly EP,(x)y(’-’)(x) ty,
xe I [a,b], (1.1)
j 1,2,...,
(1.2)
Typeset
byfl..A-TEX
572 M. H. ANNABY AND H. A. HASSAN
Assume
thatu(x, t)
isasolution of(1.1)
such that the zeros,{t },
ofBj(u(x,t))
arethesameV3.Thus,
[3],
thezerosofBj(u(x,t))
aretheeigenvaluesof theproblem(1.1)-(1.2),
and{u(x,t)}
isacomplete orthogonalsetofeigenfunctions. Then
THEOREM
1.2.Let L(y)
ty,B(y) O,
j 1,..., n, beaself-adjoint boundaryvalue problem onI. Suppose
thatthere exists a solutionu(x,t)
of(1.1)
such that the set ofzerosE, {t}
oftS,(u(x,t))
isindependent ofi.Let
g_ n(I).If
f(t) [ ,(, t)g()
then,f
has the representationwhere
s(t) L (,)(,)
i1(, )11,
Kramer’s
theorem stated aboveisnotalwaystrue,since one canfindaboundary-valueprob- lemof thetype(1.1)-(1.2)
andasolutionu(x,t)
such thatBj(u(x,t))
has thesame zeros{t},
Vj, but neither
{t
isthe setofeigenvalues,nor{u(x, tk)}
isthecompleteset ofeigenfunctions.For
example,considertheboundaryvalueproblemB,(y) y(O) y(r) O, B(y) y’(0) y’(Tr)
0.(1.4) We
have,u(x, t)
cosv/
cosv/x
isasolutionof(1.3)
withB(u) cosv/g(1-
cosv/Tr), B(u) -vcos
sinv/r.
Obviously
B(u),B2(u)
have the same set ofzeros,{tk k2}’=o,
but neither{t}’=
0 is the sequence ofeigenvalues, nor{cos--coskx}’=o
is the complete set ofeigenfunctions.So
it is not practical to discuss theexistenceofKramer-type
kernels associated withproblemsoftype(1.3)-(1.4),
i. e., when theeigenvaluesare notnecessarilysimple. When theeigenvalues ofthe problemaresimple, manyKramer-type
expansions associated with theboundary-value problemswerederived
[1, 2, 9].
There are two ways introduced by Zayed
[7, 8]
to obtain sampling;series associated withproblem
(1.3)-(1.4).
The first one[8,
pp.50-52]
is given by taking the kernel ofthesampled
integral transformtobeTherefore,
ifforsome
F
EL=(0, 7r),
thenwhere
b(x, t) A
cosV
x+ B
sinV
x.l(t) F()(z, t)
dz,f(t) f(0_) sin(nv/) + B
2sin(v/)} v/ +a { a v/sin(Trv) (t-4k)
(t
4k) + b? A (-4k) sin(vq)
(t-4k -) B
(2k) sin(zrv/)
+ (t-4k)
J
a F(z)
coskzb _2 F(z)
sinkz dx.The last series is not asampling expansion of
f(t)
sincethe coefficientsak andbk
can not be uniquely expressed in terms ofthe sampled values off
at the eigenvalues. Ifwe denote the Hilberttransformoff
by,
whereweobtain
F(x) (A
sinvx B
cosv/x)
dx,wherer
B/A.
Thesecond wayisgivenbytakingthe kernel ofthesampled integral transformtobe
O(x,t) P(t)G(x,o,t),
where
G(x,, t)
istheGreen’s
function of(1.3)-(1.4), 0
ischosenin[0, r]
asin[7],
andP(t)
is thecanonicalproduct-(t)
P(t) = I t
4k2,
k1,2,....
Then,
forF e L2(0, 0,
we havef (t) F(x)4(x, t) dx,
f(t) f(t,) P(t)
(t- t,)P’(t,)
k--O
f(0)
2sin( v) 4(_l)kv sin(vf) -vq + /(t) .(t-4)"
As
wehaveenthereis noKramer-ty
reprentations ciated withproblem(1.3)-(1.4).
In
th article we u another version ofKramer’s
theorem,Lemma
3.1, so that we can obtainanewKramer-type
mplingreprentation ciated with second orderbounda-lue
problems whichmay have multiple eigenvalues.
2.
PRELIMINARIES.
Considerthesecond-order eigenvalueproblem
Ly y" q(x)y -Ay,
x6I [a,b], A
6C,
U,(y) a,,y(a) + a,2y’(a) + B,,y(b) + o.y’(b) O, 1,2, (2.2)
where
a,,, fl,
arerealconstants,
andq(x)
is a continuousreal-valuedfunctionon[a, hi.
(2.3)
574 M. H. ANNABY AND H. A. HASSAN
Let
u,v EC2(a, b).
ThentheGreen’s
formulafor this eigenvalueproblemisaz U,V, + UV + UV, + U,V,, (2.4)
where
Uj, _<
j_<
4,arelinearlyindependent linearformsofu(a), u’(a), u(b), u’(b),
andVj, _<
j
_<
4, are linearly independent linear forms ofv(a),v’(a),v(b), v’(b). Here V 0,
j 1,2,arethe adjointboundaryconditionsof
(2.2),
cf.[5]. Moreover
problem(2.1)-(2.2)
with(2.3)
is self-adjoint,[5],
and has at most countable set of realeigenvalueswith nofinite limit points.Let {,(z,A),(z,A)}
be thefundamental set of solutions of(2.1)
definedby,(a,A)
1,’,(a,A)
0,(2.5) (a,A)=O, (a,A)=l.
Any
solutionof(2.1)
canbe writtenas(, )
c,, (x, + c(z, ),
wherecl,c2arearbitrary constants. The function
(x, A)
is aneigenfunctionof theself-adjoint eigenvalueproblem(2.1)-(2.3)
ifitsatisfies(2.2),
i. e., when thesystemu(,) u() c
has anontrivial solution. Thishappenswhen
A(A) U’(I)U2(,) U()U’()[
=0.Thatisthe rootsof
A()
arethe eigenvalues of theproblem. Theeigenvaluesofproblem(2.1)- (2.3)
arenot necessarilysimple.Assume {A,,, A,,}, {A3,,}
arethe sequences of double and simpleeigenvalues respectively.Let X,(x, A),
1,2, 3,
be the functionsx,(x,) ,(z,), x(z,) (z,),) + c(),(z,),
and
,(,) (,) + u(,(,)) u,((,))
where
() f (, ), (, )
i1,(, )11
and a aconstant chon such that
X(x, A.,) 0,
Vk, k1,2,...
We
can see that{X,(x,A,.,),X(x,A.,)}
and{X(x,A.,)}
arethe quences oforthogonal eigenfunctions coespondingto{A, A,, }, {A.,
respectively. Thment
canbe eilyderived
usg
the factthataneigenvalueA* of problem(2.1)-(2.3)
issimpleif andonlyifoneoftheentriesof
A(A*)
does not vanh.Now
aumethatthezerosofA(A),
i.e. the eigenlues{A,.,}, 1,2,3,
have theymptotic behaviourA,., O(k )
k. The,
forexample, tes place
if the boundaryconditions areregular[5,
p.64].
Alsume that themultiplicities zerosofA(A)
areatmosttwo.3.
A SAMPLING THEOREM.
In
th section, we state and prove the main threm of the paper. Theorem 3.1 below asamplingtheorem ciated withacond-orderboundary-lueproblem
whoseeigenvaluesarenot necessarily simple.
We
startourstudy bythefollowingLemma,
takenfrom[1]. It
isanewversionof
Kramer’s
theorem.LEMMA
3.1.Let {A,. },=
besequencesof numbers.Let K,: In,
b xC C, 1,2,...,
n be n functions such thatK,(z,A)
EL2(a,b), VA
EC,
and thattJ,"= {K,(z,A,,)}
forms acomplete orthogonal set in
L2(a,b). Let H,
be the subspacegenerated
by{K,(z,A,.)},
1,2,... ,n. Then
L(a,b) E,=, BH,. Assume
thatf ,%, Bf, L(a,b), f, . H,,
andF(A) Z F,(A) f,(x)K,(x,A)d.x. (3.1)
,=1
Then
where
F(A) F, (A,,)S,(A), (3.2)
t----1 k---I
S..(,k) f K,(x,A)K,(x,A,.,.)
dx(3.3)
flK,(=,,,,,,)l=
andv, dim
H,.
THEOREM
3.1.Let H,
be the subspace generated by{X, (x, A,.)},
1,2, 3,
and letL(a,b) , BH,. Let f ,, Sf, L2(a,b), f, H,. Assume
thatF(A) F,(A) f,(x)x,(x,A)dx. (3.4)
,=i
Then
F
admitsthefollowingrepresentationF(A) F,($,.,)
(A- , ,)C’,.,(A,.,)
1=1 k=l
where
G,.,.(A) [X,(X,X),X,(x,A,..)],
i=1,2,3,
and
[u, v] uv’-u’v.
The threeseriesconverge uniformlyonanycompactsubsetof thecomplex plane.Moreover
Ca.:(A) G(A)
1(3.6)
I=1 k=l
ifzeroisnotaneigenvalue. These productsmustbemultiplied by$ifzeroisasimple eigenvalue, andby
X
2ifzeroisadoubleeigenvalue.PROOF.
SettingG,.(.X) [XI(x,A)Lx,(x,A,.,.) X,(x,A,..)Lx,(x,A)] dz, 1,2,3,
and integratingby parts,weobtain
C,.,(A) [xICz, A), X,(z,A,.,)] t’,,,
i= 1,2,3.(3.7)
On
theotherhand,
using(2.1),
onegetsC,.,(X) CA- A,..) X,(X,A)X,(X,A,.,)dx
andtherefore
(3.8)
(3.9)
576 M. H. ANNABY AND H. A. HAS SAN Since
f,
EH,,
i--1,2, 3, thenwhere
c,, f,(x)x,(x,A,,)
dxF,(A,,). (3.10)
Using Parseval’s equality,weobtain
F,() $,()x,(z, )
=,
IIx,(-, )ll
C,,(X)
1
2, 3, F,(A,.)
(A A..,)G’. (A..)
(cf. (3.8), (3.9),
and(3.10)).
The proof of theuniform convergence canbe established in[8,
pp.
4].
We
nowshowthatG3,($),
fork 1,2,..., hno zerosotherthan theeigenlues.We
use t’hesametechniqueof[2]. om (3.7)
it clear that each,.,
1,2, 3, k 1,2,..., azeroof
G3.(A). Suppo
"
is anotherzeroofG,(A). It
will be shown that"
aneigenvalue of(2.1)-(2.2). From (2.4)
and(3.7),
weobtainG
,XU V, + + U,, V
for all
A,
where theU3,
1<
j_<
4,arelinearformsinx3(a, A), X’:(a, A), xa(b, A), X’a(b, A),
andtheV,
1<
j<
4, arelinearforms inXa(tt,.a,t:),X’3(a, A3,k.),X.3(b,,,3,k),)(.t3(b, A3,#.).
Since)(3(X,,a,E)
isaneigenfunction,then
V,(x(=,:,,,,)) o,
j ,2.Obviously
Ul(X3(X, ))= {A()), U2(X3(x, ) A()k),
hence(3.11)
where
v(x(=,:,.)) [V,(x(=,,x.,,))+ V(x(=, .,,)).].
Since
A"
is a zero ofG3,k(A),
thenC3,k(A’) A(A)V(x:(x,A:,#.))
0.Now
assume thatV(X3(X, A3,#,))
0.SinceY(x3(x,A:,.))
isindependent ofA,
by(3.11), G3,,(A) =-
0. ThusG3,,(A)
isidentically zero, which contradictsthe fact that
G’3,,(A3,k) =/=
0(cf.(3.9)). So, V(X:(x,A,.))
isnotzeroforall eigenvalues. Thuswehave
A(A) 0,
andsoA"
isaneigenvalue of(2.1)
and(2.2).
Finallywe show that
G3.,(A)
may take the from(3.6).
Indeed by Hadamard’sfactorization[5, 55]
wetheorem
[4,
p.24]
forentirefunctions andbynoting thatG3,,(A)
isof order p. canwrite
c,,,(,)
(the
orderofG3, ).
ThusP(z) c(k)
where
P(z)
isapolynomial whosedegree
does not exceed]is aconstant depending onlyonk.The convergence ofevery
product
inG3,k(A)
isguaranteed sincethe eigenvalues behaves likeO(k-)
ask oc. ObviouslyC.,,(.) C(,)
c,.,, (,.,,) c,(,)
sowithoutlossofgeneralitywemay assumethat
Ga,(A) G(A).
Thiscompletesthe proofof theorem3.1.In
some cases thethree-series summation(3.5)
can bereduced, (see
examples 1, 2below)
into atwo-series one. Thefirst iswrittenin terms of
F(A,,k A2,k)
andthe secondin termsofFa(Aa,). In
suchcases wemay need to reformtheintegral transform(3.4)
into asuitableoneaswesee inthefollowingcorollary
COROLLARY
3.1.Assume
thatG,,.(A)
#(k)h(A),
where#isafunction
depends
onlyonk and his anentirefunctiondepends onlyonA. Let F=(A) f:(x)._(x, A) dx,
whr
.(=,,) h(,)X=(=,),). H,,
forF(A) F (A) + F=()) + F(A),
wehave
k=l k-I
PROOF.
InsteadofG.(A),
weconsiderG2,(A) (A- A2,.) .:(x,A)X2(x,A=,,)dx h(A)G2,(A),
and
G2,,(A2,, h(A2,,)G=,,(A=,,).
ThenC,() h()C,() (,) C,,()
,=, (=,) h(=. )C’=, (=,,) (=,) C’ ,, (=,,)
and
4.
EXAMPLES.
EXAMPLE
4.1. Consider theperiodic eigenvalue problem-V" Ay t2y,
0_<
z<_
r,(4.1)
578 M.H. ANNABY AND H. A. HASSAN
u, () (0) ()
0,v() ’(0) ’()
0.(.2)
This is a regular self-adjoint eigenvalue problem. The fundamental set ofsolutions of
(4.1)
subjecttoconditions
(2.5)
isThus
(,))
ot,e (, )
sintxA(A)
1 cosTrtsinrt 1 cosrt 4sin zrt
2"
The eigenvaluesare
A
4k,
k0,1, 2,...,
whereA
4k,
k 1,2,...
aredouble eigenlues and thecorresponding eigenfunctionsare{
cos2kx, sin2k2kx}
nd A
0istheonlysimpleeigenvaluewiththeeigenfunctionCa(x, 0),
whereHence 3(x, 0)
r.Now
t
COSX t
I
cos 7rtcostx
---" +
sinzrt 1 cosrtsin
t(r x)
sintxcost( x) + +
costx.So
G(,\)=tsinrt, G2(A)=
sinTrtG3(A)=47rsin
t 2a’,,,(,,) 7’ a’,,,(,,) g, a’,0(0) ,
where
A., ,k.,
4k. Let L*(a,b) ,
$H,, whereH, H, H
arethesubspaeesgener-ated by
{cos 2kx}=, {sin2kx}
2k= ,{}
respectively.
Let F(A) F, (A) + F(A) + F(A),
whereF,(A) /,(x),(x,A), f,
6H,,
i=1,2,3.
Then
2tsinrt
8k
sinrt 4sin*(t)
(4.3) F() ] F (4 ’),( 4’) + ] F’(4),t( 4’-) + (0) ,,t---.
k=l k=l
In
thefollowingwe seethat theform(4.3)
canbe reducedintoanotherformwhichissimilar to thoseresultinginthecaseofsimple eigenvaluesby redefiningthesampledintegraltransformas described intheabovecorollary.Indeed,
e’,*()G,()A h(A)
isentire.Let
F(,) F (A)+ F(,) + F3(A),
where
F() /(z);(, )
d.Then,
noting thatF3(0) F(0),
we have2tsinrt
F(A) l(4k )
k=l
r(A
+ P(o)
4i( t)
EXAMPLE
4.2. Consider theanti-periodiceigenvalueproblem-y" Ay t2y,
0<_
x<_
zf,(4.4) u, (v) v(0) + ()
0,(4.5) u(v) v’(o) + v’() o.
This isa regular self-adjoint eigenvalue
problem. For
the same fundamental solutions inthe previousexample,wehaveA(A)
4cos.
The eigenvaluesare{(2k- 1)2}=,
all of themaredouble,
theircorrespondingeigenfunctionsareNow
and
sinrt
G,,k(A)
-tsinnt,
t
G,,(Ax,k) , G2,k(A2,)= 2(2k- 1)
2.For
the corresponding integraltransform,F(A),
definedasinthetheorem,wegetthefollowing sampling representation-2tsinrt
F(A) F, ((2k 1) 2)
r(A (2k 1) 2)
k----1
-2(2k I)2
sin+ F2((2k 1) 2)
rt(A (2k "-" "
Alsowe have, G,,()a2,(4) i
h(A). Let
Hence
F(A) F(A) + F2(A), F(A) f()A(, A)
d.(,k)- ((gk- 1) 2)
--2tsinrt7[(- (2k- 1)2)
REMARK.
Unlike thecaseof simple eigenvalues,asthe above twoexamplesshow, we do nothavethe relationsG,,(A)= cG,(A),
1,2,wherea,(A)
ifzeroisnotaneigenvalue, ifzerois aneigenvalue,
and
c
is aconstantdependingonk.In
factit canbeeasilyseenthatG,, G2.,
intheprevious examples, havezeros morethan thoseofG,, G2.
ACKNOWLEDGMENT.
The authors wish to express theirgratitude to ProfessorM. A.
EI-Sayed, CairoUniversity for readingthe manuscript and forhis constructivecomments. The first author wishes to thank Alexander yon Humboldt Foundation for supporting hisstay in
Germany,
under the numberIV-1039259,
when heprepared
therevised versionof thepaper.580 M. H. ANNABY AND H. A. HASSAN
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Samplingtheorems associated withfourth
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159-177.[3] Kramer, H. P., A
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Math. Phys. 38(1959),
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Levin,B.,
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Naimark,M. A.,
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