Critical Point Theory - Sandwich and Linking Systems

347 Pages·2020·3.409 MB·English

by  Schechter, Martin

Martin Schechter
Critical Point
Theory
Sandwich and Linking Systems

Martin Schechter
Critical Point Theory
Sandwich and Linking Systems
MartinSchechter
Brooklyn,NY,USA
ISBN978-3-030-45602-3 ISBN978-3-030-45603-0 (eBook)
https://doi.org/10.1007/978-3-030-45603-0
MathematicsSubjectClassification:35A15,35B38,49J40,58E05,70G75,35-02,47-02,49-02,58-02
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Preface
The Purposeofthe Book
Manyproblemsarisinginscienceandengineeringcallforthesolvingofnonlinear
partialdifferentialequationsorsystems.Theseequationsaredifficulttosolve,and
there are very few general techniques that can be applied to solve them. Different
equationsseemtorequiredifferentmethodsofsolution.(Ofcourse,somenonlinear
equationshavenosolutions.)However,itwasnoticedthatmanyoftheseequations
areEulerequationsoffunctionals(i.e.,functionsfromsomespacetoR).Thismeans
thatforeachequationthereisaC1 functionalG(u)suchthatanysolutionuofthe
equationsatisfies
G(cid:2)(u)=0. (1)
(Here C1 refers to continuously differentiable functions.) As an illustration, the
equation
−(cid:2)u(x)=f(x,u(x))
istheEulerequationofthefunctional
(cid:2)
1
G(u)= (cid:3)∇u(cid:3)2− F(x,u(x))dx
2
onanappropriatespaceS,where
(cid:2)
t
F(x,t)= f(x,s)ds, (2)
0
andthenormisthatofL2.(Here
(cid:3)n ∂2u(x ,x ,··· ,x )
(cid:2)u(x)= 1 2 n ,
∂2x
k=1 k
vii
viii Preface
∇u=(∂u(x)/∂x ,∂u(x)/∂x ,··· ,∂u(x)/∂x ),
1 2 n
and
(cid:2)
(cid:3)u(cid:3)2 = |u(x)|2.)
S
The solving of the Euler equation is tantamount to finding critical points of the
corresponding functional. This has motivated researchers to study critical points
of functionals in order to solve the corresponding Euler equations and systems. It
has led to the development of several techniques to find critical points. In many
cases,thesetechniquesleadtoresultssuperiortothoseobtainedbyothermethods
ofsolvingtheseequations.Thisvolumeisdedicatedtothelatestdevelopmentsand
applicationsofthesetechniques.
In order to explain the nature of the problem, we describe in detail the one-
dimensionalcase.
AOne-DimensionalProblem
Considertheproblemoffindingasolutionof
−u(cid:2)(cid:2)(x)+u(x)=f(x,u(x)), x ∈I =[0,2π], (3)
undertheconditions
u(0)=u(2π), u(cid:2)(0)=u(cid:2)(2π). (4)
We assume that the function f(x,t) is continuous in I ×R and is periodic in x
withperiod2π.Inordertosolveit,youmayfirstwanttosolvethelinearproblem
correspondingto(3),(4),namely
−u(cid:2)(cid:2)(x)+u(x)=f(x), x ∈I =[0,2π], (5)
undertheconditions(4),wherethefunctionf(x)iscontinuousinI andisperiodic
inx withperiod2π.Afterasubstantialcalculation,onefindsthatthereisaunique
solutiongivenby
(cid:2)
x
u(x)=Aex +Be−x + sinh(t −x)f(t)dt, (6)
0
where
(cid:2)
e2π 2π
2A= e−tf(t)dt (7)
e2π −1
0
Preface ix
and
(cid:2)
e−2π 2π
2B = etf(t)dt. (8)
1−e−2π
0
Canthissolutionbeusedtosolve(3),(4)?Itcaniff(x,t)isboundedforallx
andt.Forthen,wecandefine
(cid:2)
x
Tu(x)=A(u)ex +B(u)e−x + sinh(t −x)f(t,u(t))dt, (9)
0
where
(cid:2)
e2π 2π
2A(u)= e−tf(t,u(t))dt (10)
e2π −1
0
and
(cid:2)
e−2π 2π
2B(u)= etf(t,u(t))dt. (11)
1−e−2π
0
Thenasolutionof(3),(4)willexistifwecanfindafunctionu(x)suchthat
Tu(x)=u(x), x ∈I. (12)
SuchafunctioniscalledafixedpointoftheoperatorT.Allsolutionsof(3),(4)are
fixedpointsoftheoperatorT.Inthepresentcase,onecanshowthatthereisindeed
a fixed point for the operator T when f(x,t) is bounded. If more than one exists,
thereisnoorganizedwayoffindingallofthem.
However,thereisanotherwayofsolving(3),(4)calledthevariationalapproach.
Ifu(x)isasolution,thenwehave
(u(cid:2),v(cid:2))+(u,v)=(−u(cid:2)(cid:2)+u,v)=(f(·,u),v)
forallv ∈C1(I)satisfying(4).Here,
(cid:2)
2π
(u,v)= u(x)v(x)dx,
0
C1(I)isthesetofcontinuouslydifferentiableperiodicfunctionsonI,andweused
thefactthatnoboundarytermsariseintheintegrationbyparts.Theexpression
(u,v) =(u(cid:2),v(cid:2))+(u,v) (13)
H
isascalarproductcorrespondingtothenorm
(cid:3)u(cid:3) =((cid:3)u(cid:2)(cid:3)2+(cid:3)u(cid:3)2)1/2. (14)
H
x Preface
Thus,asolutionof(3),(4)satisfies
(u,v) =(f(·,u),v) (15)
H
forallv ∈ C1(I)satisfying(4).WeletH bethecompletionofperiodicfunctions
inC1(I)withrespecttothenorm(14).
In searching for solutions of (3) and (4), we look for functions u ∈ H which
satisfy(15).If{u }isaCauchysequenceinH offunctionsinC1(I),then
k
(cid:3)u −u (cid:3)→0, (cid:3)u(cid:2) −u(cid:2)(cid:3)→0.
j k j k
Thismeansthattherearefunctionsu,h∈L2(I)suchthat
u →u, u(cid:2) →h in L2(I). (16)
k k
Notethat
(u ,v(cid:2))=−(u(cid:2),v), v ∈C1(I), (17)
k k
byintegrationbyparts.Thusinthelimit,
(u,v(cid:2))=−(h,v), v ∈C1(I). (18)
Ifu∈C1(I), h∈L2(I),and(18)holds,thenh=u(cid:2)a.e.
Now, suppose that u ∈ L2(I) and there is an h ∈ L2(I) such that (18) holds.
Notethathisunique.EventhoughuisnotinC1(I)andwedonotknowwhether
ornotithasaderivativeatanypoint,wedefinethe“weak”derivativeofutobeh
(cid:2)
anddenoteitbyu.Itbehaveslikeaderivativewithrespecttointegrationbyparts.
Since H is the completion of C1(I) with respect to the norm given by (14), then
everyfunctioninH hasaweakderivativeinL2(I).Conversely,everyfunctionin
L2(I)thathasaweakderivativeinL2(I)isinH.
Although functions in H need not be in C1(I), they are in C(I), the set of
continuousfunctionsonI.Infact,thereisaconstantK suchthat
|u(x)|≤K(cid:3)u(cid:3) , x ∈I, u∈H. (19)
H
Moreover,
u(0)=u(2π). (20)
Moreprecisely,everyfunctionu ∈ H isalmosteverywhereequaltoafunctionin
C(I).Inequality(19)holdsforthecontinuousfunctionequaltoua.e.,i.e.,itholds
foruifweadjustitonasetofmeasurezerotomakeitcontinuous.Thesameistrue
of(20).Moreover,ifthesequence{u }convergesinH,thenitconvergesuniformly
k
onI.Wealsohave
Preface xi
Theorem1 Iff ∈L2(I), u∈H,and
(u,v) =(f,v), v ∈C1(I), (21)
H
thenu(cid:2) ∈H andu(cid:2)(cid:2) =(u(cid:2))(cid:2) =u−f.Inparticular,u(cid:2) iscontinuousinI andisthe
(cid:2)(cid:2)
derivativeofuintheusualsense.If,inaddition,f isinC(I),thenu iscontinuous
in I, and u(cid:2)(cid:2) = u−f in the usual sense. In particular, it is a solution of (3), (4).
Consequently,u∈H isasolutionof(3),(4)iffitsatisfies(15)forallv ∈H.
Assumethatf(x,t)isacontinuousfunctiononI ×Rsatisfying
|f(x,t)|≤C(|t|+1), x ∈I, t ∈R. (22)
Welet
(cid:2)
1 2π
G(u)= (cid:3)u(cid:3)2 − F(x,u(x))dx, (23)
2 H
0
where
(cid:2)
t
F(x,t)= f(x,s)ds. (24)
0
NotethatG(u)isdefinedonH.
Next,wecalculatethederivativeofG.Wehave
1
G(u+εv)−G(u)= ((cid:3)u+εv(cid:3)2 −(cid:3)u(cid:3)2 )
2 H H
(cid:2)
2π
− [F(x,u+εv)−F(x,u)]dx
0
1
= ((cid:3)u(cid:3)2 +2ε(u,v) +ε2(cid:3)v(cid:3)2 −(cid:3)u(cid:3)2 )
2 H H H H
(cid:2) (cid:2)
2π 1 d
− F(x,u+εθv)dθdx
dθ
0 0
(cid:2) (cid:2)
1 2π 1
=ε(u,v) +ε2 (cid:3)v(cid:3)2 −ε f(x,u+εθv)vdθdx.
H 2 H
0 0
Hence,
(cid:2)
2π
[G(u+εv)−G(u)]/ε−(u,v) + f(x,u)vdx
H
0
(cid:2) (cid:2)
ε 2π 1
= (cid:3)v(cid:3)2 − [f(x,u+εθv)−f(x,u)]vdθdx.
2 H
0 0