Photocopying permitted bylicenseonly the Gordon and BreachScience Publishersimprint.
Printed inMalaysia.
The HELP Inequality for Hamiltonian
Systems
B.M. BROWN
a,
W.D. EVANSbandM.MARLETTAc’*aDepartmentofComputerScience, Universityof Wales,Cardiff,
PO Box916,CardiffCF23XF,UK;bSchool of Mathematics,University ofWales, Cardiff,Senghennydd Road, Cardiff, CF2 4YH, UK; DepartmentofMathematics andComputerScience,University of Leicester, University Road,
LeicesterLE17RHUK
(Received30 July 1998; Revised 15September1998)
Wc extend the Hardy-Evcritt-Littlcwood-Polya inequality, hitherto established for 2nth orderformally sclfadjoint ordinary differential equations,toa wideclass oflinear Hamiltonian systems. The method follows Dias (Ph.D. thesis, Cardiff: University of Wales,1994)butwithouttheHilbcrtspace settingwhichheuses.
Keywords: Hardy; Everitt; Littlewood; Polya; Hamiltonian system; Inequality AMS(MOS) SubjectClassifications: 26D10(26D1,34B, 34L,47F05)
1
INTRODUCTION
In
the seminalpaper10]
Hardyand Littlewoodintroducedthe inequality(/0 If’l
2dx<
4/0 Ifl
2dx/0 If"l
2dx(1)
which is requiredtoholdfor all
functionsfsuch
that theright-handsideof
(1)
is finite. Equalityisattainedwhen,for anyp>
0 and for 0<
x<
o,f(x)
Aexp(-px/2)
sin(px/27r/3).
* Corresponding author.
57
No lessthan threeproofs of
(1)
may be found inthe book ofHardyetal.
[11].
Everitt in
[9]
introducedthe inequality2
{p(x)lf’(x)[
2+ (q(x) TW(X))lf(x)[ }
dxfa 12fa
< K(r) w(x)lf(x)
dxw(x)lA/l[f](x rf(x)[
2dx(2)
where isthe second-order Sturm-Liouville operator
kd[f]
:=-1(-(pf)’ +
qf)
w
with b
>
a>
cx,w(x) >
0, p(x)>
0, q(x)E with 1/p, q locally integrableandrareal parameter.In[9]
itisassumed thatJ4isregular at a and singular at b and satisfies the so-called strong-limit-point condition.Underthese conditions Everitt showed that theexistence or otherwiseofK(r)
in(2)
dependsonthespectral propertiesofA[f]
and the existence criteria for(2)
can bedetermined intermsof the behaviour of theTitchmarsh-Weylm-function associated with.M.
Everitt’sproof in[9]
ismodelledonthecalculusof variationsproofgiven in 11 for(1).
Howeveranoperator theoreticproofisgiven by Evansand Zettl in
[8]
and bothproofsarereproducedin the article
[6]
ofEvansandEverittin which aneatcharacterisationofthe criteria foravalidinequalityisgiven in terms of theTitchmarsh-Weylm-function: it is thislatterapproach that will serveasamodel for the workreportedonin thispaper.The inequality(2)
has been extended[7]
tothecasewhen bis aregular pointor k4isinthe limit-circlecase atb.Thetheoryassociated with the existence ofananalogous inequality,inwhichA//isnowtheformally symmetric 2Nth order expressionN
:=
Wj=0
isproved byDiasin
[5] (see
also[2]).
Inthislatterinequalitythe criteria for the existenceofavalid inequalitycan again be determinedby the spectral properties of the self-adjoint operator generated byA
inL2w(a,b)
but this time it is formulated in terms of the associated Titchmarsh-Weyl Mmatrix.In examplesofthe inequality,theproblem ofprovingthe existence and also ofdeterminingthe bestconstant is oftenahardanalyticproblem, depending asitdoeson knowledge of the closed form expression for themfunctionorMmatrix.
A
complete catalogueof all knownexamples tothe present time istobefoundin[3]. In
view ofthe analytic difficulties inobtaininginformationabout the best constant in particularexamples, anumerical approachtotheproblemhas beenundertakenby Brownetal. This isreported in,forexample,
[4].
This paper reports onthe development ofa
HELP-type
inequality associated with thelinearHamiltonian system(AA + (3)
where Yis a 2n vector and
Jt
andB
are 2n x 2n real matrices with.A* Jt >
0 and/3*B.
Jisthe 2n x 2n matrix0 -i)
andIthenxnidentitymatrix.Itisshown in
[16]
that the 2nth orderformallysymmetric differential equationA//[f]
:=-(pj(x) y
Wj=0
may bewritten as a Hamiltoniansystem, andin
[1]
it is shown that the matrix-vectorSturm-Liouvilleproblem(P(x) Y’)’ + Q(x)
Y ,kYwhereP and
Q
are nxn matrices and Yis an n vectormay also be embeddedinaHamiltonian system. Thiscurrentworkis notjustasimple extension of the work of Dias[5],
since asA
is semi-definite andnot definite,weareunabletousetheHilbertspace settingfor theproblem that hewasabletoexploit andareforcedtoworkwithout thisabstract structure. Howeverthegeneralapproachof the secondproofin[6]
can be made toworktoyieldaHELP-type
theoryfor linear Hamiltonian systems. Also examples can be found that are not covered by any previous work.2
THE HAMILTONIAN SYSTEM AND THE DIRICHLET INTEGRAL Let C(x)
andB(x)
be real symmetricn xnmatrices, andA(x)
an nxn real matrix, such that the elements ofA,
BandCarelocally integrable over aninterval[a, o).
LetKbean nxndiagonalmatrixof the formK
diag(kl(x),..., km(x), O,
0,...,0), (4)
wherern
<
nandwhere thediagonalelementskj
havelocallyintegrable reciprocals,togetherwiththepropertiesess
inf(kj) > O,
j 1,...,m.We
denote by Kthe 2n x 2n diagonal matrix whose first rndiagonal elementsarethekj
and whoseremaining2n rndiagonal elementsareall zero;wedenotebyK and/
thepseudo-inversesofKand/respectively.
Thus,forexample,
K
diag(1/kl(x),..., 1/km(x), O,
0,...,0). (6)
For absolutely continuous n-vector functions u and v we define the operators
L1
andL2
by[u, v] -v’ + C(x)u (x)v, Z:[u, v]
:=’- (x)u (x)v.
(7)
For
2n-vector functions y, partitionedin anobviousnotation as Y--(UYvy )
wedefine
[, v])
z(y)
:=L:[y, Vy] (8)
Wedefinea bilinearform
(., .)
byIf, g) rg ax u}u
dx,(9)
forany2n-vector
functionsfand
g for which theintegralconverges.We
alsodefinethesetof
admissiblefunctions
4tobe thesetof absolutely continuous2n-vectorfunctionsfsuch
thatI(f,f)l < +c, (10)
L2[uf, vf]
OE]n, (11)
(I-- KK)L1 [uf, f]
0 ]1n. (12)
We
make thefollowing crucialassumption.ASSUMPTION
The bilinear form(., .)
ispositive definiteonthesetA
of admissible functions y which are solutions ofL(y) AKy
for any 6C.
FollowingReid
[15]
wedefine theDirichletform
associated withtwo n- vectorfunetionsfand
g:O(f, g):-- fa [vnvg-+- ufug]dx.
Now
supposethatf,
g 4. Wehave(f,/tL(g)) ug(IZl [Ug, llg])
dxu)Ll[Ug, Vg]
dx (using(12)
forg)Uf (--
Vg+ Cug _AT lg)
dx[-u}g] + (Auf + Bf)*vf + u}Cug u}Ag
dx(a [z, z] ol [-}1 + e(f,g.
Thuswehave provedthe integration-by-parts formula
(f,/tL(g)) [-uvg] + D(f, g) (13)
forf,
gA.
3
THE HINTON-SHAW-TITCHMARSH-WEYL M-MATRIX
In this sectionwe review very briefly someresults from the paper of Hinton and Shaw[12].
Thesewillbe very important in therestof the paper.Thepaperof Hinton andShaw dealswithsolutions of Hamiltonian systems of differential equations of the form
L(y) AK(x)y, A
E C.(14)
In general,notall solutions y of this equation will satisfy
(y, y) < + .
Thosewhichdowill lieintheset
.A
of admissible functions because the structureof ensures that(11)
and(12)
aresatisfied byUy,Vywhen y solves(14).
Inaddition toAssumption 1,HintonandShaw require the following.
ASSUMPTION 2 (’Limit-point assumption’) Supposef,g E
Jr.
Then(15)
Here denotes Hermitianconjugation.
THEOREM 3.1 (Hinton and
Shaw) Suppose
thatA-0,
and that Assumptions and 2 both hold. Then thedifferential
equation(14)
possessesnlinearly independent solutionsbl (x, A),..., bn(X, A)
such that(bj,j) < +o,
j 1,...,n.Moreover,
these solutionsmay be normalisedsothat/f(x, A)
isthe 2n xnmatrixwhose columnsare
b](x, A),..., bn(X, A)
then(a, A) (-M(A) )
Iwhere
M(A)
is an analytic n n matrixfunction of A (called
theTitehmarsh-Weyl matrix). Thismatrixhasthepropertythat
M(A)
is aNevanlinnafunetion,inthesensethat
is positive
definite for A >
0,M(A)
negativedefinite for < O. (16)
Moreover,
4
CRITERIA FOR A VALID HELP
INEQUALITYIn
this sectionwereproducetheresultsofDias[5]
onexistence ofaHELP
inequality, but in themoregeneralcontextofaHamiltoniansystemL(y) AK(x)y, (18)
orequivalently
L1 [u, v] AK(x)u, L2[u, v]
0.(19)
Weassumethat the followingstrong limitpointproperty holds.
ASSUMPTION
3 Forallfunctions f,
gEA,
lim
uf(x)*vg(X)
0 ]1n.
X----O
(20)
Notethat Assumption 3 impliesAssumption2.
We
definethespaceA0
toconsistofthose
functionsf A
such thatuf(a)
0n, vf(a)
0 ]tn. (21)
ForA #
+
iv,v 0,wedefinethedeficiencyspacesN+ (A)
andN_()
by
N+(A) {f AlL(f) ARf), N_(A) {f
EAlL(f) Rf}.
(22) As
the matricesA,
BandC occurringinthedefinitions ofL1
andL2
areallreal-valued,andas isreal-valued,itiseasytoseethat y E
N+(A) = 37 N_(A).
In
particular,therefore, dimN+(A)= dimN_(A). Recallthat the matrix(x; A)
introducedinSection 3hascolumns whichformabasisofN+(A).
Thus any function h in
N+(A)
maybeexpressedas alinear combination of the columns of0(.; A),
h
0(.; A)a,
whereaisann-vectorofconstantsdepending onlyon
A.
We shallestablish
HELP
inequalitiesfor functionsfin
theset/x .=zx0,
N/(a),
N_(23)
The first resultwerequireisthat, despitetheappearanceof
A
ontheright hand side of(23), A
doesnotdependonA.
LEMMA4.1 A=.,4.
Proof
It is clearthat A C_A
so wejust need to prove the oppositeinclusion.
Following the notation ofSection 3, we form the 2n xn matrix whose columns are solutions of
L(y)= AKy,
subject to the initial conditions(a;) (-M())I (24)
Itisnotdifficultto seethat the columns of
0(.; A)
spanN+(A)
and the columns of0(.; A)
spanN_(A).
FromTheorem 3.1 wealso knowthatM(A) M(A)
and henceThus
(x;
((a;A),(a;X)) -()I
i 0Hencethe 2n x 2n matrix
((a; A), (a; A))
isoffull rank if andonlyif3M(A)
isnonsingular.From
Hinton and Shaw[12]
weactually know thatM(A)
is positive definite forA >
0 and negative definite forA <
0,which certainlytherefore impliesthat((a; A), (a; A))
isof fullrank. Thusgivenanyfunction
fE A
wecanchoosea2n-vectore=(
cl)c2
suchthat
uf(a)) ((a; A) (a;/))c (a; A)Cl + (a; )c2.
f(a)=
vf(a)
Defining
g(x) =f(x) (x;/)Cl I/(x; )C2,
itisclearthat gis anelementof.A suchthat
ug(a) vg(a)
0EThus g E
A0.
Also,(.; )k)C N+(,)k)
and(x;/)C2 N_().
Hencef A,
whichcompletestheproof.Remark This result is essentially the
Von Neumann
decomposition formula see[13,
Lemma10.2.5].
Wehave been forcedtoproveitin this verydirectway bythe fact that there doesnotappearto be anatural Hilbertspace setting for theseproblems,exceptinspecialcases.Lemma4.1 establishes thedecomposition
e4
Ao
@N+ (/)
@N_(). (25)
Weshallusethisextensivelyinthesequel.
LnMMA4.2 Thedirect sum appearingin
(25)
isanorthogonalsum withrespecttothe sesquilinear
form
(f,g) (t (L(/) #f),t(L(g) #g)) + u2(f,g). (26) Proof Suppose thatf Ao
andg EN+(,k),
so thatL(g)
#Kg+
iuKg.Then
(f, g), (/t (L(f) lzf), ivg) +
u2(f, g)
iu
(L1 [uf, vf])*KtKKtKug -i#u u)KKtKKtKug
We
nowuse(12)
tosimplify the firstintegral,andtheresultKK
K K tosimplify thesecond.Theseyieldfa fa
(f g)
A it,,L [uf, vf]*Ug
iAuuKug. (27)
Using theintegration-by-parts formula
(13)
twice, togetherwith(20)
and(21)
toeliminate theboundaryterms,weobtain(28)
Now
L(g) AKg
implies thatLl[Ug Vg]--,,Kug;
substitutingthisback into(28)
givesLl [uf, vf]*Ug uKug. (29)
Substituting
(29)
into(27)
yields(f,g)
=0,asrequired.
Similarlyone mayprovethat
(f, g)=
0 forfE A0
and gEN_(A).
Itremains toshow that
(f, g)
0forf N+(A)
and gN_(A).
This case is easy:wehaveL(f) Afand L(g) A[fg,
sothat(f,g), (iutf, iutg) + v2 (f,g)
u2
u}KId KKIO
Ug+
u2u}Kug O,
since
KKtK=
K.Thiscompletestheproof.LEMMA4.3
<I,, ff) 2u@(M(A)). (30)
Proof
Using the definition of(., .)a
wehave(, ) (t(L() #),/t(L() #)) + v2(, ).
Nowwe usethe fact that
L(9) (# + iv)K9
toobtain(9, 9)
u(t/9)*KR(t/9) + u2(9, 9).
Fromtheidentity
RRtR R
thissimplifiestogive(9, 9),\
2v2(9, 9). (31)
Butnow
1
9"L(9)
- {u,(a)v,(a) + D(9, 9)},
the last identityfollowingfrom theintegration-by-parts formula
(13).
Fromthe initial conditionson9,wehave
u(a) M(A)
andv(a) I,
yielding1
{-M*(A) + D(9, 9)}
(,I,, ,I,)
Multiplying both sidesby
A
andtakingthe imaginary partsyieldsv(9, 9) -(M*(A)) .(M(A)), (32)
thelast equalityonthe rightfollowingfromthewell known fact thatMis a symmetric matrix
[12].
Combining(31)
and(32)
gives the required result.LEMMA 4.4 Let
f E.A
be a real-valuedfunction, and suppose thatf fo +
h+
h,wherefo Ao
and hN+(A),
h EN_(A). Suppose
thath 9a,
(33)
wherea
(a
l,an)
isaconstantvector. Then(f,f),x (J,J),x + 4va*(M(A))a. (34) Proof
From Lemma4.2weknow that thesum +
h+
hisorthogonalwithrespectto
(., "/.
Thuswehave(f,f)a (fo,fo) + (h,h) + (h,h) (J,3)a + 2(h,h). (35) From (33),
itisclearthata*( )aa.
(h,h)
Using
(30)
yields(h, h), 2ua*(M(A))a. (36)
Combining
(35)
and(36)
givesthe required result.For
real-valuedfE A
consider(f, f).
From(26)
wehave(f,f), ([(tL(f), [(tL(f)) 2#(RtRf, RtL(f))
+ #2 ([(t[(f, [(tiff) + u (f,f)
(gL(f), gL(f)) 2/z(f, gL(f)) + IAI
2(f,f),
wherewehave used the identity
*
severaltimes.Nowweapplytheintegration-by-partsformula
(13)
totheterm(f, L(f))
toobtain(f, f) ([tL(f), ’tL(f)) 2#D(f, f) + I 12 (f,f 2#uf(a)*vf(a).
(37) We
definetheformJ(f) <’tL(f), ilL(f)) 2#D(f, f) + IAI
2(f,f). (38)
Then
(37)
immediately yieldsJ(f) (f, f), + 2#uf(a)*vf(a). (39)
As f=fo +
h+
h withf0
EA0,Eq. (21)
yieldsuf(a)= 2N(uh(a))
andvf(a) 2N(vh(a)).
Equations(24)
and(33)
giveuh(a) -M(A)a, vh(a)
a.(40)
Substituting into
(39)
weobtainJ(f) (f, f)A 8#((a))X(M(A)a). (41)
Combiningthiswith
(34)
yieldsJ(f) (J, f0) + 4ua*.(M())a 8#((a))(M()a). (42) We
nowseparateM(A)
anda intoreal andimaginaryparts, denotedbyM(A) MR
q- iMi, a aR 4-iai.
Some simple algebrashows that
(42)
mayberearrangedintheform4 4
J(f) (fo, fo), +- (uai +/ZaR)TMI(uaI + paR)+-- at.(-A:ZM(/))aR.
(43)
In order to make further progress we must introduce a further assumptionabout theset4of admissible functions.
ASSUMPTION4 Forall
non-zerofE A, (f,f) >
O.Withthisassumptionwecanrewritethe expression
(38)
asaquadratic formin p :=1)1.
Let peisothat/z
pcos4.Thensomesimplealgebra shows thatJ(f) & (f)
:=(f, f) (P D(f, f) (f,f)
cos0’
2+ (L(f) L(f)) D(f, f)
cos2b
(f,f)
(44)
Similarly,whenever
A
pei(+r),
we obtainD(f,f)
cos)
2J(f) Jo+r(f)
:=(f,f)
P+ (f,f + (tL(f) tL(f)) D(f,f)cos
2b
(f,f)
(45)
Givenanynon-zero
fand
anyb
E[0, 7r/2)
onemayalways chooseD(f,f)
cos(46)
Thiseliminatesthefirst term onthe right hand side ofoneof
(44, 45).
The remainingtermwillb
non-negative if andonlyifD(f,f) <
sec2(f,f)(tL(f), tL(f)). (47)
Thisisclearlya
HELP
inequality,provided0< b < 7r/2.
Weare nowin apositiontoprovethe following theorem.
THEOREM 4.5 Let Sbetheset
of
all valuesof 4
E(0, 7r/2]
such thatthefollowingtwoconditionshold:
(-A2M(A))>0 (A=pe i0)
}
Vp>0.(48)
(A2M(A)) >_
0(A
pei(o+r))
Then
7r/2
$.Let00
inf(S). ThenaHELPinequalityD(f,f) <_ t(f,f)([ftL(f),tL(f)) (49)
holds
for
allf A if
andonlyif Oo < 7r/2.
Moreover,the bestconstant inthe inequalityisgivenbyt
sec20o.
Proof Suppose
that00 < 7r/2.
Chooseb=00.
Givenf,
choose p accordingto(46),
eliminating the firstterm onthe right handside of one of(44, 45).
The conditions(48)
thenimplythat(47)
holds, givingaHELP
inequalitywithsec200.
Thus the condition00 < r/2
is sufficient foraHELP
inequality.To
see that sec200
isthe best constant, supposewe look for a smallerconstant2 sec2q
forsomeb
E(0, 00). By
definition of00
therewill exist somep
>
0such thatatleast one of thefollowingconditions fails:(-)2M(,)) >_
0, ,k--pei, (50)
(A2M(A)) _>
0,A
pei(o+r). (51) Suppose
the first condition fails. Thenwe canfindarealvectoraRsuch thatWith #=pcos0, v=psin0, we choose a real vector
a =-/z/PaR.
Nextwedefine h
(aR +
iai)andf
h/hsothat from(43)
wehaveJ(f) <
0. From(44)
this shows that the HELP inequality fails with=sec2b.
This covers the case where(50)
does not hold.A
similarargument dealswith thecase where
(51)
doesnothold, so the choicesecZ0
is clearly best-possible. This also proves that the condition00 < 7r/2
isnecessary foraHELP
inequalitytohold.5
GENERALISATIONS
Itwould be nicetoremoveAssumption 4,and indeedwe cando thisforat least part of the resultinTheorem 4.5. If Assumption 4isremoved then the condition
00 < 7r/2
ofTheorem4.5 stillimpliesaHELP
inequalityfor
thosenonzero
functions f
E.At
suchthat(f,f) >
O.Whathappens
iffis
anonzerofunction such that(f,f)
0?In
thiscase(38)
loses thetermIAI
2(f,f)
ontherighthandside. Equation(43)
stillholds,andsothe condition
00 < 7r/2
impliespositive-definitenessofj(f) (t L(f), L(f)) 2/zD(f,f).
Wecan
take/z
tobe arbitrarilylargeandofeithersign in(38)
andso we deducethat,when00 < 7r/2,
(f,f)
0= D(f,f)
O.Thuswestillhavea
HELP
inequalityinthiscase,albeitatrivialone.6
EXAMPLES
We present three examples of
HELP
inequalities associated with Hamiltonian systems. Two of these may be derived fromHELP
inequalities for even-order differential operators; one, however, is completelynew.First,however,werequireatechnicallemma.LEMMA
6.1 (CompactnessLemma) Suppose
that there existsb
E(0, 7r/2)
such that the conditions(-A2M(A)) >
0,A
pei, (52)
(A2M(A)) >
0,A
-pei(53)
hold
for
all 0 [,7r/21, for
allsufficientlylarge p andfor
all sufficiently smallp. Then, in thenotationof
Theorem 4.5,00 < 7r/2,
andsoaHELP inequality holds.Proof
The Nevanlinna property ofM(A)
ensures that(52)
holdsfor 07r/2. Hence
bycontinuity, for each p>
0 there existsCp (0, 7r/2)
such that(52)
holds forany0E[bp, 7r/2].
Wecan chooseCp
as acontinuous functionofp.Thus given anyset[Pmin,Pmax],
we canchooseb. (0, 7r/2)
such that
(- A2M(A)) >
0 forA
pei0,
for 0[b., 7r/2]
and PE[Pmin,Pmax]. However,
by hypothesis,wecanchoosePminsufficiently small and Pmax sufficiently large to ensure that(52)
holds for all 0 [b,Tr/2],
for all p(O, Pmin]l,.J[Pmax,X3). Hence(52)
holds for all 0 [min(b,,), 7r/2]
for allp>
0.A
similarargument deals with(53).
6.1 Example
I
We
considerthe Hamiltonian system withn 2for which, in our earlier notation,(0 0) A=(0 1)
C= 0 0 0
(1 0) B__(2 0)
K= 0 0 0
(54)
Inordertodeterminethe square-integrablesolutionsofthis systemwe require theeigenvaluesand eigenvectors of the matrix
Theeigenvaluesarereadilyseentobe 4-#+,where
#+/-
(1 A
4-v/A - A) 1/2. (56)
Hereweadoptthe convention that the squareroothaspositivereal part.
Theeigenvectorassociated withaneigenvalue#is givenby
V--
(]Z(lZ
2-]-1),
]Z2t_ _]_1,-/]2,/)T. (57)
Given aneigenvalue #and eigenvectorvthe associated solution of the Hamiltonian systemis vexp(#x). Itiseasy tocheck that thepositive- definiteness condition inAssumption holds for anylinearcombination ofourfour solutions here.
For
A
in thefirstquadrantof thecomplex planewithIA[
large,itiseasy to seethat theeigenvalues
with negative realpartare -/+ and#_.The associatedsolutionsyl(x)andy2(x)of theHamiltoniansystemare givenby
#_
(#2_ + 1) exp(#_x)
(#2 + A + 1)exp(#_x)
(Yl (x)y2(x)) --A#
2exp(#_x)
A#_
exp(#_x)
_#+(#2+ + 1)exp(-#+x)
(#+ + A +1)exp(-#+x)
_p,2+ exp(-#+x)
-A#2+ exp(-/z+x)
From
Theorem 3.1weknow thatv
(yl (0)y2 (0))
i
forsomenon-singularn n matrixV.Thisallowsus todeduce that
{ (u-
-(1
-I-#_/z+)/
-(1 -+-/z_U+)/A
)
(U- -/z+)(A + 1)/A (58)
Putting
A
pexp(i0),0<
0< 7r/2,
p>
0,it is easytoseethat#+1/
and#_
ix/x/ exp(i0/2)
forlargep.This allowsustodeducethat(-,,2ml ()) -x/p
3/2cos(30/2), (59)
(-,2m22())) -xp
5/2cos(50/2), (60)
(_2m12() p3/2 cos(30/2). (61)
Itisthereforeeasytocheck that for allsufficiently largep,thereexistsa sectorsurrounding 0
7r/2
inwhich(-A2M(A))
ispositive definite.Nextwecheck thecasewhere issmall.Itiseasytocheck that when issmall,theeigenvaluesof thematrixSin
(55)
with negative real part are #+and # Also,#+ ,,-1 #_
2’
2The M-matrix is obtainedbyreplacing#_ by #_ in
(58),
and hence(-2 .2/2)
’k2M(A) ,k2/2-2,X
Putting pexp(i0)weget
_,k2M(A) ( p2/2
2psinOsin 20
\
_p2/2
sin20’
2psin 0
J
Againitis easytocheck that for all sufficientlysmall p, thereexists a sectorsurrounding 0
7r/2
in which this matrix ispositivedefinite.Wemust nowcheck the case where
A
pexp(i(0+ 7r)),
0<
0< 7r/2:
againwe mustconsiderseparatelythecasesoflargeandsmallp.Wehave
/z+
(1 +
pei4-V/p2e
2io+
peio) 1/2,
andit iseasytocheckthat,bothinthecaseof small p andinthecaseof largep, -#+ and -#_ arethe eigenvalues ofSwithnegativereal part.
Thus, changing#_ to -#_ in
(58),
+ v-)/.
M(,) -(1 -v+v-)/,
-( +-/)
’
-[( + )/](+ + -) )
For largepwehave
#+
x/v/-fiexp(i0),
whilefor small p
q- 4-
21-V/-pe i0/2.
Henceitiseasily shown that forlargep,
sin(30/2)
9(2M()) x/’p3/2
sin(30/2)
sin(30/2)
)
-psin(50/2)
Itiseasily checked that for all sufficientlylargep thereexists a sector surrounding0
7r/2
in whichthis ispositive definite.Forsmall pweobtain
M(,k)
2psin021
P cos 01/2PCOS0)
and for all sufficiently smallpthereexists a sectorsurrounding 0
7r/2
inwhich this is positive definite.
By
the compactness lemma(Lemma 6.1)
wetherefore haveaHELP
inequality for thisproblem,valid forallfunctionsffor
which(f,f) >
0.It is interesting to observe that for this Hamiltonian system, Assumption 4 fails. The function
0
f(x) _(1/2)e_X/C e-X/
(1/ x/’
e Xv5is anadmissible function with
(f,f)
0.6.2 Example 2
For
oursecondexampleweconsider the Hamiltonian system associated withthesecondordermatrix-vectorSturm-Liouville equation_y,,+(Xo -xO) Y=AY’ x(Oc)., (62)
Here
Yis a2-vector.Thesystemcanbecastin the Hamiltonian form-V’= AU+
0
U’=
V.ItiseasytocheckthatourDirichlet form
D(f,f)
isthesame astheusual Dirichlet integral for suchaproblem:D(f,f) Y’ll
2+Y*
x0 -x0 Y dx, wheref= y,
Itisalso easytocheck that Assumptions 1-4 all hold.For
fixedA
with(A) :
0 letyl(x)denoteanontrivialL2solutionofthescalar equation
y" +
xyAy
and letY2denoteanontrivialL2solution of the scalar equationy"
xyAy.
Then(y,)
0Y= (0)
yareL2solutions of
(62). We
formthe matricesU(x)= (yl(x)
0y2(x)
0) V(x)--(yx) y(x)
0)
and observe that
(W)
specifies an admissible matrix solution of the Hamiltonian system. TheM-matrixistherefore givenbyM(,X) v(o) r(o) -yl(O)/y[(O)
00
-y2(O)/yi(O
We
nowrecognise thatM(A)
diag(ml(A), m2(A))
where ml and m2 are, respectively, the scalar m-functions associated withtheSturm-Liouvilleequations
-y" +
xyAy, -y"-xy Ay,
The conditions for a
HELP
inequality reduce to the corresponding conditionsfor each of thesetwoscalar equations.For the first equation there is only point spectrum. The first few eigenvalues, subjecttotheNeumann boundarycondition
y’(O)=
0,are given approximatelybyA0
1.01879297,A1
10.5507875,A2
23.2333564.(These
were obtained using the code SL02F[14].)
The equationy" +
xyAy
hasanassociated ’shifted’HELP
inequalityoftheform(2)
for7- Ak,k 0,1,2, Forthe second equation, the whole real line consists of continuous spectrum, and there is aHELP
inequality [6, Example5]
of theform(2)
for any value of7-.Thusfor the equation(62)
thereis noHELP
inequality, but thereareHELP
inequalities associated witheach of the ’shifted’ equationsx(O, oo),
k=O,1,2,...As
anasideweobserve thatfor(62)
thereiscontinuous spectrumonthe whole real line with discontinuities of the spectral function at the eigenvalues of the equation-y"+
xyAy
with boundary conditiony’(0)
0.6.3 Example3
Weconsiderafourth-ordermatrix-vectorSturm-Liouvilleproblem
y(iv)+ (ql(x)
0q.(x)
0)
Y=AY,
x(0, o). (63)
We
convertthisinto aHamiltonian systemby defining(,) (-,,,,)
U
y
Vy.
sothat
A-q(x)
0 0 0 0 0 0 0-V’--
0A-qv.(x)
0 0U+
0 0 0 00 0 0 0 0 0 0
V,
0 0 0 0 0 0 0
0 0 0 0 0 0 0
U’
0 0 0U+
0 0 0 0 V.0 0 0 0 0 0 0
0 0 0 0 0 0 0
Once againit iseasytocheckAssumptions 1-4andit isalsoeasyto see that
D(f,f)
is equivalenttothe usual Dirichlet form associated with(63).
Clearlytheoriginalfourth order equation
(63)
canbedecoupledinto twoscalarfourth order equations. Under suitable limit-pointhypoth- eses, letylandZldenote linearlyindependent L2solutions ofy(iV) +
ql(x)y Ay,
and lety2andz2denote linearlyindependent L2solutionsof
y(iV) + q2(x)y Ay.
Wecanthen form four admissible solutions forourHamiltonian system:
f (x)a" (y (x). O. y (x). O. y’ (x). O. y (x). 0).
f2(x)
T(Zl (x),
0,zl(x),
0,z"(x), O,zt(x), 0),
f3 (x)T (0,
Y2(x),
0,Yl (x),
0,y’(x),
0,y (x)),
f4(x)
T(0, z2(x) O, zi(x O,-
zm[x,,
O,zo.x...
The 4 x 4 M-matrix is thus givenby
M()
ylO) zl(O)
0y2(O)
0z2(O)
0|y(O) z (0)
0 0\o
0y(O) (0)
-y]" (0) -z]" (0)
00 0
-y’(0)
y’(o) z’(o) o
0 0
y(O)
-1
-’(0)
0
(o)
Thisgives
M(A) P[
0()
\
where 0 denotes the 2 x 2zeromatrix,
M(A)
(j1,2)
denote the 2 x 2 Titchmarsh-WeylmatriceszZ l l)(
andPisthe 4 x 4permutationmatrix
0 0 0
0 0 0
0 0 0
0 0 0
The matrices
M1
andM2
areTitchmarsh-Weylmatricesfor the fourth orderproblemsy(iv) +
ql(x)y Ay, y(iV) + q2(x)y Ay. (64)
Thus, as in the case of Example 2, the conditions for a
HELP
inequality in Theorem 4.5 become equivalenttothe conditionsfor both ofthe scalar equations in(64)
tohave associatedHELP inequalities.The coefficientmatrix
Q(x) (
ql0(x) q2(x)
0)
in
(63)
canbereplaced by anymatrixof the formRQ(x)R
TwhereRis orthogonalandconstant,andthe same result will hold: the conditions for aHELP
inequality for the matrix-vector system reducetothe conditions forHELP
inequalitiesfor the associated scalarproblems.7
CONCLUDING REMARKS
In this paper we have generalised the work of Dias to Hamiltonian systems.
In
so doing, we have been forced to abandon the natural Hilbert-space setting which Diasusesfor the 2nth order scalarselfadjoint case; in particular, therefore, our formula for the so-calledVon
Neumann decomposition hashadtobeprovedinaverydirectway.References
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