Fredholm's third theorem for second order singular Dirichlet problem
Abstract
There are found conditions guarantee the validity of the third Fredholm's theorem for the second-order singular Dirichlet problem.
Full text
Lom a idze and Opluš il Bounda y Value P oblems 2014, 2014:59
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R E S E A R C H Open Access
F edholm’s hi d heo em o second-o de
singula Di ichle p oblem
Alexande Lom a idze1,2 and Zdenˇ
ek Opluš il2*
*Co espondence:
[email p o ec ed].cz
2Ins i u e o Ma hema ics, Facul y o
Mechanical Enginee ing, B no
Uni e si y o Technology,
Technická 2, B no, 616 69, Czech
Republic
Full lis o au ho in o ma ion is
a ailable a he end o he a icle
Abs ac
Conside he singula Di ichle p oblem
u =p( )u+q( );
u(a)=0, u(b)=0,
whe e p,q:]a,b[→Ra e locally Lebesgue in eg able unc ions. I is p o ed ha i
b
a
(s–a)(b–s)p(s)–ds <+∞and b
a
(s–a)(b–s)q(s)ds <+∞,
hen F edholm’s hi d heo em emains ue.
MSC: 34B05
Keywo ds: singula Di ichle p oblem; F edholm’s hi d heo em
1 In oduc ion
Conside he bounda y alue p oblem
u =p( )u+q( ), ()
u(a)=, u(b)=, ()
whe e p,q∈Lloc(]a,b[). We a e mainly in e es ed in he case, when he unc ions pand
qa e no in eg able on [a,b]. In his case, he p oblem (), ()issaid obesingula .I is
p o ed in [] ha i
b
a
(s–a)(b–s)p(s)–ds <+∞()
and
b
a
(s–a)(b–s)q(s)ds <+∞,()
hen, o he singula p oblem (), (), he F edholm al e na i e holds. Mo e p ecisely, he
ollowing heo em is ue.
©2014 Lom a idze and Opluš il; licensee Sp inge . This is an Open Access a icle dis ibu ed unde he e ms o he C ea i e Com-
mons A ibu ion License (h p://c ea i ecommons.o g/licenses/by/2.0), which pe mi s un es ic ed use, dis ibu ion, and ep o-
duc ion in any medium, p o ided he o iginal wo k is p ope ly ci ed.
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Theo em . ([, Theo em .]) Le () hold.Then he p oblem (), () is uniquely sol able
o any q sa is ying () iff he co esponding homogeneous equa ion
u =p( )u(a)
has no non i ial solu ion sa is ying ().
The aim o his pape is o show ha , unde he assump ion (), he F edholm’s hi d
heo em emains ue. Be o e o mula ion o he main esul s, we in oduce he ollowing
no a ion.
Ris he se o eal numbe s.
Fo x∈R,wepu [x]–=
(|x|–x).
C(I), whe e I⊂R, is a se o con inuous unc ions u:I→R.
Fo u∈C([α,β]), we pu u[α,β]=max{|u( )|: ∈[α,β]}.
AC
loc(]α,β[) is he se o unc ions u:]α,β[→R, which a e absolu ely con inuous o-
ge he wi h hei fi s de i a i e on e e y closed subin e al o ]α,β[.
Lloc(]α,β[) is he se o unc ions p:]α,β[→R,whicha eLebesguein eg ableone e y
closed subin e al o ]α,β[.
By (a)( esp., (b)) we deno e he igh ( esp., le ) limi o he unc ion :]a,b[→Ra
he poin a( esp., b).
By a solu ion o equa ion () we unde s and a unc ion u∈AC
loc(]a,b[), which sa isfies i
almos e e ywhe e in ]a,b[.Asolu iono equa ion()sa is ying() is said o be a solu ion
o he p oblem (), ().
Wewillsay ha ace ainp ope yholdsin]α,β[ i i akes place on e e y closed subin-
e al o ]α,β[.
Recall ha we conside he p oblem (), (), whe e p,q∈Lloc(]a,b[).
Theo em . Le () hold.Then he homogeneous p oblem (a), () hasnomo e hanone,
up o a cons an mul iple,non i ial solu ion.
Rema k . Belowwewillshow(seeP oposi ion.) ha i () holds and uis a non i ial
solu ion o (a), (), hen he e exis s >such ha
u( )≤ ( –a)(b– ) o ∈[a,b].
Theo em . Le () hold and he homogeneous p oblem (a), () ha e a non i ial so-
lu ion u.Then he p oblem (), (), whe e he unc ion q sa isfies (), is sol able iff he
condi ion
b
a
q(s)u(s)ds = ()
is ulfilled.
Rema k . In iew o Rema k . and condi ion (), he unc ion quis in eg able on
[a,b] and, he e o e, condi ion () is meaning ul.
2 Auxilia y s a emen s
Fi s o all, o con enience o e e ences, we ecall wo lemmas om [].
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Lemma . ([, Lemma .]) Le () and () hold.Then, o any α∈[a,b[and β∈]α,b],
e e y solu ion u o equa ion () sa is ying
u(α)=, u(β)=
admi s he es ima e
( –a)(b– )u( )≤u[α,β]b–a+b
a
(s–a)(b–s)p(s)–ds
+b
a
(s–a)(b–s)q(s)ds o ∈]α,β[.
Lemma . ([, Lemma .]) Le () hold.Then he e exis a∈]a,b[, b∈]a,b[, and
>such ha , o any α∈[a,a[, β∈]b,b], and q sa is ying (), e e y solu ion u o
equa ion () sa is ying
u(α)=
admi s he es ima e
u( )≤( –a)u[α,a]+
a
(s–a)q(s)ds +( –a)a
q(s)ds o ∈]α,a],
while e e y solu ion u o equa ion () sa is ying
u(β)=
admi s he es ima e
u( )≤(b– )u[b,β]+b
(b–s)q(s)ds +(b– )
bq(s)ds o ∈[b,β[.
Nex p oposi ion immedia ely ollows om Lemma ..
P oposi ion . Le () hold and ube a non i ial solu ion o he homogeneous p oblem
(a), (). Then he e exis s >such ha
u( )≤ ( –a)(b– ) o ∈[a,b].
P oposi ion . Le () hold and ube a non i ial solu ion o (a) sa is ying u(a)=
( espec i ely,u(b)=).Then he e exis s a∈]a,b[( espec i ely,b∈]a,b[) such ha
u( )= o ∈]a,a] espec i ely,u( )= o ∈[b,b[.()
P oo In iew o () he eexis sa∈]a,b[( espec i ely,b∈]a,b[) such ha
a
a
(s–a)p(s)–ds < espec i ely, b
b
(b–s)p(s)–ds <
.
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Hence, he inequali y
a
a
(s–a)(a–s)p(s)–ds <a–a
espec i ely, b
b
(s–b)(b–s)p(s)–ds <b–b
holds, as well. The la e inequali y, by i ue o [,Lemma.],implies ha o anya<
< <a( espec i ely, b< < <b), he p oblem
u =p( )u;u( )=, u( )=
has no non i ial solu ion.
Now suppose ha uis a non i ial solu ion o (a)sa is yingu(a)=( espec i ely,
u(b) = ). Then i ollows om he abo e ha ei he
u( )= o ∈]a,a] espec i ely, u( )= o ∈[b,b[,()
o he e is a ∈]a,a]( espec i ely, ∈[b,b[) such ha
u( )= o ∈]a, [, u( )=
espec i ely, u( )= o ∈] ,b[, u( )=
.
()
I is now clea ha ()holdswi ha=a( espec i ely, b=b)i ()holds,andwi h
a=a+
( espec i ely, b= +b
)i ()issa isfied.
Lemma . Le () and () hold.Le ,mo eo e ,u be a solu ion o he p oblem (), () and
ube a solu ion o he p oblem (a), (). Then
lim
→a+u( )u( )–u( )u
( )=, lim
→b–u( )u( )–u( )u
( )=. ()
P oo I is clea ha
u( )u( )–u( )u
( )=q( )u( ) o ∈]a,b[.
Hence,
u( )u( )–u( )u
( )=δ–c
q(s)u(s)ds o ∈]a,b[, ()
whe e
c=a+b
and δ=u(c)u(c)–u(c)u
(c).
By i ue o P oposi ion . and condi ion (), he unc ion quis in eg able on [a,b].
Thus, i ollows om () ha he e exis s a fini e limi
lim
→a+u( )u( )–u( )u
( )=ε.()
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Nowwewillshow ha ε= . Suppose he con a y, le
ε>. ()
Then he e is α∈]a,b[such ha
u( )u( )–u( )u
( )>ε
o ∈]a,α]. ()
On accoun o P oposi ion ., we can assume wi hou loss o gene ali y ha
u( )= o ∈]a,α]. ()
Then i ollows om () ha
u( )
u( )
>ε
u
( ) o ∈]a,α].
Hence
μu( )–u( )>ε
u( )α
ds
u
(s) o ∈]a,α], ()
whe e μ=u(α)
u(α).
Taking now in o accoun P oposi ion .,wege om() ha
μu( )–u( )>εu( )
–a–
α–a o ∈]a,α],
whe e ε=ε
(b–a). The la e inequali y, in iew o he condi ions u(a)=andu(a)=,
implies ha
lim
→a+
|u( )|
–a=. ()
On he o he hand, by i ue o Lemma ., he eisM>such ha
( –a)u( )≤M o ∈]a,α]. ()
In iew o ()and(), we ge
lim
→a+u( )u( )=lim
→a+( –a)u( )
|u( )|
–a=,
and he e o e, on accoun o (), we ob ain
lim
→a+u( )u
( )=ε.
Now, le α∈]a,α[besuch ha
u( )u
( )>ε
o ∈]a,α].
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Then i is clea ha
u[a,b]u
( )>ε
o ∈]a,α]
and consequen ly
u[a,b]u( )>ε
( –a) o ∈]a,α].
Howe e , he la e inequali y and ()yield ha ε≤, which con adic s (). The con-
adic ion ob ained p o es he fi s equali y in (). By he same a gumen s one can p o e
he second equali y in ().
We will need he nex lemma in he p oo o he sufficiency pa o Theo em . and
hus, we will suppose ha Theo em . and he necessi y pa o Theo em . a e ue.
Lemma . Le () hold and he homogeneous p oblem (a), () ha e a non i ial solu-
ion u.Then he e exis n∈Nand >such ha , o any q sa is ying () and () and
e e y n >n, he solu ion u o he p oblem
u =p( )+
np( )–u+q( ); u(a)=, u(b)=
admi s he es ima e
u( )≤ b
a
(s–a)(b–s)q(s)ds o ∈[a,b].
P oo Suppose he con a y, le he asse ion o he lemma be iola ed. Then, o any n∈
N, he eexis kn≥n,qn∈Lloc(]a,b[), and un∈AC
loc(]a,b[) such ha
b
a
(s–a)(b–s)qn(s)ds <+∞,b
a
qn(s)u(s)ds =,
u
n( )=p( )+
knp( )–un( )+qn( ) o ∈]a,b[,
un(a)=, un(b)=
and
un[a,b]>nb
a
(s–a)(b–s)qn(s)ds.
In oduce he no a ion
˜
un( )=
un[a,b]
un( ), ˜
qn( )=
un[a,b]
qn( ) o ∈]a,b[.
Then i is clea ha
˜
u
n( )=p( )+
knp( )–˜
un( )+˜
qn( ) o ∈]a,b[, ()
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˜
un(a)=, ˜
un(b)=,
˜
un[a,b]=, ()
b
a
(s–a)(b–s)˜
qn(s)ds <
n()
and
b
a
˜
qn(s)u(s)ds = . ()
By i ue o Lemma . (wi h q( )=
kn[p( )]–˜
un( )+˜
qn( )) and (), we ha e
( –a)(b– )˜
u
n( )≤b–a+n+
nb
a
(s–a)(b–s)p(s)–ds
+b
a
(s–a)(b–s)˜
qn(s)ds o ∈]a,b[, ()
while, by i ue o Lemma . (wi h q( )=
kn[p( )]–˜
un( )+˜
qn( )), he e exis a∈]a,b[,
b∈]a,b[, and >such ha
˜
un( )≤ –a+a
a
(s–a)
knp(s)–˜
un(s)+˜
qn(s)
ds o ∈]a,a],
˜
un( )≤b– +b
(b–s)
knp(s)–˜
un(s)+˜
qn(s)
ds o ∈[b,b[.
()
On accoun o ()and(), he sequence {un}+∞
n= is uni o mly bounded and equicon-
inuous in ]a,b[. Thus, by i ue o he A zelà-Ascoli lemma, we can assume wi hou loss
o gene ali y ha
lim
n→+∞˜
un( )= ( )uni o mlyin]a,b[, ()
whe e ∈C(]a,b[) and, mo eo e ,
lim
n→+∞˜
u
na+b
=c.()
In iew o ()i isclea ha
˜
un( )=˜
una+b
+ –a+b
˜
u
na+b
+
a+b
s
a+b
p(ξ)+
knp(ξ)–˜
un(ξ)+˜
qn(ξ)dξds o ∈]a,b[.
Hence, on accoun o (), (), (), and (), we ge
( )= a+b
+c –a+b
+
a+b
s
a+b
p(ξ) (ξ)dξds o ∈]a,b[.
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The e o e, ∈AC
loc(]a,b[) and is a solu ion o equa ion (a). On he o he hand, i
ollows om (), in iew o (), (), and (), ha
( )≤( –a) o ∈]a,a]and ( )≤(b– ) o ∈[b,b[,
and hus is a solu ion o he p oblem (a), ().
By i ue o ()and(), i is clea ha he e a e n∈N,a∈]a,a], and b∈[b,b[
such ha
˜
un( )< o ∈[a,a]∪[b,b],n>n.
The e o e, ˜
un[a,b]= o n>n. Taking now in o accoun (), we ge ha [a,b]=
and, he e o e, is a non i ial solu ion o he p oblem (a), ().
By i ue o Theo em ., he eisλ=such ha
( )=λu( ) o ∈[a,b]. ()
Mo eo e , in iew o he necessi y pa o Theo em . (wi h q( )=
kn[p( )]–˜
un( )+˜
qn( )),
(), (), (), and (), we ge
b
ap(s)–˜
un(s) (s)ds =. ()
Le now α∈]a,b[andβ∈]α,b[ be a bi a y. Then, in iew o (), we ha e
lim
n→+∞β
αp(s)–˜
un(s) (s)ds =β
αp(s)–
(s)ds.()
On accoun o (), (), and P oposi ion ., he unc ion [p]– is in eg able on [a,b].
Taking in o accoun (), we ge
α
ap(s)–˜
un(s) (s)ds
≤α
ap(s)– (s)ds
and
b
βp(s)–˜
un(s) (s)ds
≤b
βp(s)– (s)ds.
Hence, () implies he inequali y
β
αp(s)–˜
un(s) (s)ds ≤α
ap(s)– (s)ds +b
βp(s)– (s)ds,
which, oge he wi h (), esul s in
β
αp(s)–
(s)ds ≤α
ap(s)– (s)ds +b
βp(s)– (s)ds.
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Since αand βwe e a bi a y, we ge om he la e inequali y ha
b
ap(s)–
(s)ds =.
Taking now in o accoun ha ≡ , we ge [p]–≡, i.e.,p( )≥ o ∈]a,b[. Howe e , in
his case he p oblem (a), () has no non i ial solu ion, which con adic s he assump ion
o he lemma.
3P oo s
P oo o Theo em . Le uand be any non i ial solu ions o (a). By i ue o
Lemma . (wi h u≡ and q≡), we ge
lim
→a+u
( ) ( )–u( )
( )=.
On he o he hand, clea ly
u
( ) ( )–u( )
( )= o ∈]a,b[,
and, he e o e,
u
( ) ( )–u( )
( )= o ∈[a,b]. ()
Choose ∈]a,b[such ha
u
( )=.
I is clea ha u( )=sinceo he wiseu≡. Then i ollows om () ha
( )=
and as abo e ( )=.Pu λ=u( )
( )and
w( )=u( )–λ ( ) o ∈[a,b].
E iden ly, wis a solu ion o equa ion (a)andw( ) = . Howe e , i ollows om () ha
w( )=.Consequen ly,w≡and husu≡λ .
P oo o Theo em . Le ube a non i ial solu ion o (a), ()whileube a solu ion o
(), (). Pu
( )=u( )u( )–u( )u
( ) o ∈]a,b[.
I is clea ha
( )=q( )u( ) o ∈]a,b[.