Combining local and non-local e ms in a nonlinea ellip ic
p oblem1
F ancisco Julio S.A. Co ˆ
ea1and An onio Su´
a ez2
1. Uni e sidade Fede al de Campina G ande
Cen o de Ciˆencias e Tecnologia
Unidade Acadˆemica de Ma em´a ica e Es a ´ıs ica
CEP:58.109-970, Campina G ande - PB - B azil
2. Dp o. de Ecuaciones Di e enciales y An´alisis Num´e ico
Fac. de Ma em´a icas, Uni . de Se illa
Calle Ta ia s/n, 41012-Se illa, Spain
E-mail add esses: [email p o ec ed], [email p o ec ed],
Abs ac
In his pape we s udy he exis ence, uniqueness, mul iplici y and s abili y o pos-
i i e solu ion o a non-linea ellip ic p oblem ha combines local and non-local e ms
aking he o m o an in eg al in space. The p oo s a e mainly based on ixed poin
heo ems, bi u ca ion echniques, sub-supe solu ions and con inua ion a gumen s.
AMS Classi ica ion. ??.
Keywo ds. Non-local ellip ic equa ions, bi u ca ion echniques, a p io i bounds.
1 In oduc ion
Th oughou his wo k we conside he ollowing p oblem



−∆u=λup+ZΩ
uβin Ω,
u= 0 on ∂Ω,
(1.1)
whe e Ω ⊂IRNis a bounded egula domain, λ∈IR and p, β > 0.
Du ing ecen yea s he so called non-local ellip ic p oblems ha e a ac ed he a -
en ion o a lo o esea che s due wo main aspec s: Fi s ly due o hei ma hema ical
impo ance. The p esence o non-local e ms p o okes some di icul ies which, some imes,
do no appea in he local ones. So, he beha iou o hese p oblems may be, in gene al,
dis inc o hei local coun e pa . Secondly, hese p oblems a ise om p ac ical mo i-
a ions om Biology, Physics, Hea T ans e , Mechanics and so on, which makes hei
s udies pa icula ly in e es ing. See, o ins ance, he e iew pape [8].
In pa icula , in p oblem (1.1) he e exis s a combina ion o a local and a non-local
e ms in addi i e way. Obse e ha while o λ= 0 equa ion (1.1) is a non-local ellip ic
equa ion, when λ < 0 he e is a compe i ion be ween bo h e ms. I is in e es ing o s udy
1AS ha e been suppo ed by he Spanish Minis y o Science and Technology unde G an MTM2009-
12367.
Augus 14, 2010 F. J. S. A. Co ˆea and A. Su´a ez
he beha iou o he se o posi i e solu ions o (1.1) depending o he size o pand βand
o cou se o he sign o λ.
P oblem (1.1) has been p e iously analyzed in [14] and [12], a leas o ou knowledge,
only he case λ≤0, β > 1 and p≥1. In bo h wo ks, he pa abolic p oblem ela ed o
(1.1) was s udied. In pa icula , bo h wo ks showed he alue p=β ep esen s a c i ical
blow-up exponen . Indeed, hey p o ed ha i β > p o β=pand λ > −|Ω|, he blow-up
can occu in ini e ime. Howe e , when β < p o β=pand λ≤ −|Ω|all he solu ions
a e global and bounded.
Wi h espec o he ellip ic p oblem (1.1), he au ho s p o ed he exis ence o posi i e
solu ion o λsmall in he pa icula case λ < 0, p > β > 1. In his pape , we comple e
his s udy, and gi e esul s o all he alues o pand β.
Be o e p oceeding o he s a emen o he main esul s, we need o in oduce some
no a ion. Gi en egula , non-nega i e and non- i ial unc ions a, b and m, we deno e by
λ1(−∆ + m;a, b) he p incipal eigen alue o he ollowing in eg o-di e en ial eigen alue
p oblem
−∆u+m(x)u−a(x)ZΩ
b(x)u=λu in Ω, u= 0 on ∂Ω, (1.2)
(see Sec ion 2 o a de ailed s udy o his p oblem). Deno e also
λ1:= λ1(−∆; 0,0) and σ1:= λ1(−∆; 1,1).
We use he p incipal eigen alues o (1.2) o cha ac e ize he s abili y o he solu ions wi h
espec o he pa abolic coun e pa p oblem. We say ha a posi i e solu ion u0o (1.1)
is s able ( esp. uns able) i he p incipal eigen alue o he linea iza ion o (1.1) a ound u0
is posi i e ( esp. nega i e), i.e.,
λ1(−∆−λpup−1
0;β;uβ−1
0)>0 ( esp. <0.)
We also say ha u0is neu ally s able i i is ze o. Obse e ha pand βcan be less han
one, and so he eigen alue p oblem (1.2) can ha e singula e ms.
We can now s a e ou main esul s, which depend on he size o pand β.
Fi s , i is clea ha i (p, β) = (1,1) hen (1.1) is an eigen alue p oblem and i
possesses posi i e solu ion i λ=σ1. So, we assume ha (p, β)6= (1,1).
In he case p= 1 we can ob ain:
Theo em 1.1. Assume ha p= 1. Then, he e exis s a unique posi i e solu ion o (1.1)
o λ<λ1, and no posi i e solu ions o λ≥λ1. Mo eo e , he solu ion is s able o β < 1
and uns able o β > 1. Finally,
lim
λ→λ1
kuk∞=(0when β > 1,
+∞when β < 1.and lim
λ→−∞ kuk∞=(+∞when β > 1,
0when β < 1.
In Figu e 1 we ha e ep esen ed he bi u ca ion diag ams co esponding o he case
p= 1. Case 1 ep esen s he solu ions o (1.1) when β < 1 and and Case 2 shows he case
β > 1. Obse e ha we ha e a bi u ca ion om ze o when β > 1 and a bi u ca ion om
in ini y when β < 1 a λ=λ1.
In he case p < 1, we ge :
Theo em 1.2. Assume ha p < 1.
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Combining local and non-local e ms Augus 14, 2010
uu
1 1
Case Case
Figu e 1: Bi u ca ion diag ams o equa ion (1.1) o p= 1.
a) Assume also ha β= 1.
(a) I σ1>0 he e exis s a posi i e solu ion o (1.1) i , and only i , λ > 0. The
solu ion is unique and s able. Mo eo e ,
lim
λ→0kuλk∞= 0,lim
λ→∞ kuλk∞=∞.
(b) I σ1= 0 he e exis s a posi i e solu ion o (1.1) i , and only i , λ= 0. The e
a e in ini e posi i e solu ions and hey a e neu ally s able.
(c) I σ1<0 he e exis s a posi i e solu ion o (1.1) i , and only i , λ < 0. The
solu ion is unique and uns able. Mo eo e ,
lim
λ→0kuλk∞= 0,lim
λ→−∞ kuλk∞=∞.
b) Assume also ha β > 1. The e exis s a alue λ > 0such ha he e exis s a posi i e
solu ion o (1.1) i and only i λ≤λ. The e is a unique and uns able posi i e solu ion
o λ≤0and a leas wo solu ions, uλ
1< uλ
2, o λ > 0and small, uλ
1is s able and
uλ
2uns able. Mo eo e ,
lim
λ→−∞ kuλk∞=∞and lim
λ→0kuλ
1k∞= 0.
c) Assume now ha β < 1.
(a) I β < p he e exis s a posi i e solu ion o (1.1) o all λ∈IR. The solu ion is
unique and s able. Mo eo e ,
lim
λ→−∞ kuλk∞= 0 and lim
λ→∞ kuλk∞=∞.
(b) I β=p he e exis s λ0<0such ha he e exis s posi i e solu ion i and only i
λ>λ0. In ac , λ0∈(−|Ω|,−RΩϕp
1), being ϕ1>0 he eigen unc ion associa ed
o λ1such ha kϕ1k∞= 1. Fu he mo e, he solu ion is unique and s able and
lim
λ↓λ0
kuλk∞= 0 and lim
λ→∞ kuλk∞=∞.
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Augus 14, 2010 F. J. S. A. Co ˆea and A. Su´a ez
(c) I p<β he e exis s λ0<0such ha he e exis s posi i e solu ion i and only
i λ≥λ0. Mo eo e , o λ≥0 he solu ion is unique and s able, and o λ
nega i e and small he e exis a leas wo posi i e solu ions, uλ
1< uλ
2,uλ
1is
uns able and uλ
2s able. Mo eo e ,
lim
λ→0kuλ
2k∞= 0 and lim
λ→∞ kuλk∞=∞.
In Figu e 2 we ha e d awn he bi u ca ion diag ams o (1.1) co esponding o he case
p < 1. Cases 1, 2 and 3 ep esen he solu ions when β= 1 and σ1>0, σ1= 0 and
σ1<0, espec i ely. Case 4 shows he case β > 1, and when β < 1 we ha e he Cases 5,
6 and 7 when β > p,β=pand β < p, espec i ely.
u
u
u
u
u
Case 1 Case 2 Case 3
Case 4
u
Case 6
Case 5
u
Case 7
Figu e 2: Bi u ca ion diag ams o equa ion (1.1) o p < 1.
Le us compa e some o ou esul s wi h he well-known ones o he local equa ion
−∆u=λup+uβ.
In he case p= 1 he exis ence esul s a e a he simila o he local case. Howe e , o he
case β > 1 in he non-local case we do no need impose he condi ion β < (N+2)/(N−2)
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Combining local and non-local e ms Augus 14, 2010
o ob ain he exis ence o a p io i bounds. Mo eo e , in his case we show ha he solu ion
is uns able (simila o he local case) bu he solu ion is unique, unlike he local case.
Wi h espec o he case p < 1 we would like o poin ou ha in he non-local case
any non-nega i e and non- i ial solu ion is posi i e in all Ω. This con as s wi h he
local case in which o λnega i e could exis non-nega i e and non- i ial solu ions ha
anishes in a pa o Ω, he dead co e.
Obse e ha in he case p < 1< β he esul ob ained is a he simila o he case o
he local equa ion s udied in [1]. Howe e , again in ou case we do no need o impose he
condi ion β < (N+ 2)/(N−2). Also, he esul ob ained in he case p < β < 1 is simila
o he local equa ion analyzed in [7].
Le us ema k ha o ob ain he exis ence esul s in he p e ious esul s, we can no
use he a ia ional me hods due o he equa ion (1.1) has no a a ia ional s uc u e. In
ac , we ha e used basically a ixed poin a gumen and he sub-supe solu ion me hod o
ob ain abo e esul s.
Fo he case p > 1 we a e no able o use he ixed poin a gumen . So, we ha e
in oduced ou equa ion (1.1) in a mo e gene al equa ion, see equa ion (4.18), and use
bi u ca ion me hods and classical esul s om [2]. Fo ha , we need o ob ain a p io i
bounds o posi i e solu ions o (1.1). This is no a i ial p oblem. We dis inguish wo
cases. When λ < 0 we ob ain a p io i bounds excep in he case β=p≥1. The case
λ > 0 is ha de . Basically, we ha e used o di e en a gumen s: boo -s apping and
blow-up a gumen s o ob ain he esul s. Fo he case λ > 0 we ha e p o ed ha i
p < 1,∀β > 0 o p= 1, β > 1,(1.3)
o ,
1<p<(N+ 2)/(N−2),∀β > 0,(1.4)
o
p≥(N+ 2)/(N−2),and β > (N/2)(p−1),(1.5)
hen he e exis a p io i bounds o (1.1). Obse e ha (1.4) is he classical es ic ion in
he local case. On he o he hand, (1.3) means ha when pis small, we ob ain a p io i
bounds o all he alues o β; while (1.5) gi es a p io i bounds when βis la ge, e en when
pis g ea e ha c i ical exponen (N+ 2)/(N−2).
Mo eo e , hese esul s a e op imal in some way, because o λnega i e and β=p >
and o λposi i e and β < 1 = pwe p o e ha he e exis a bi u ca ion om in ini y
o some λ, and so a p io i bounds do no exis . Mo eo e , we show ha o p=β >
(N+ 2)/(N−2) he e is no posi i e solu ion o λla ge.
Theo em 1.3. Assume ha p > 1.
a) Assume ha β= 1.
(a) Suppose ha σ1>0. I he e exis s a posi i e solu ion hen λ > 0. I λ > 0and
p < (N+ 2)/(N−2) hen he e exis s a leas a posi i e solu ion. The solu ion
is uns able and
lim
λ→0kuλk∞=∞and lim
λ→∞ kuλk∞= 0.(1.6)
(b) I σ1= 0 he e exis s a posi i e solu ion i , and only i , λ= 0. The e a e
in ini e posi i e solu ions and hey a e neu ally s able.
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Augus 14, 2010 F. J. S. A. Co ˆea and A. Su´a ez
(c) I σ1<0 he e exis s a posi i e solu ion i , and only i , λ < 0. The solu ion is
unique and s able and and
lim
λ→−∞ kuλk∞= 0,lim
λ→0kuλk∞=∞.(1.7)
b) Assume also ha β > 1.
(a) I β > p he e exis s a unique and uns able posi i e solu ion o λ≤0. I ,
mo eo e , he e exis a p io i bounds, he e exis s posi i e solu ions o λ > 0
and i is uns able. Mo eo e ,
lim
λ→−∞ kuλk∞=∞.
(b) I β=p, he e exis s λ0<0such ha (1.1) possesses posi i e solu ion o
λ∈(λ0,0] and
lim
λ→λ0
kuk∞= +∞.(1.8)
Mo eo e , his solu ion is unique and uns able.
I , mo eo e , he e exis a p io i bounds, he e exis s posi i e solu ions o λ > 0
and i is uns able.
(c) I β < p, he e exis s λ0<0such ha (1.1) possesses posi i e solu ion o
λ∈[λ0,0]. Mo eo e , i λis small and nega i e, he e exis a leas wo posi i e
solu ions, uλ
1< uλ
2,uλ
1is uns able and uλ
2s able and
lim
λ↑0kuλ
2k∞= +∞.
I , mo eo e , he e exis a p io i bounds, he e exis s posi i e solu ions o λ > 0
and i is uns able.
c) Assume now ha β < 1. The e exis s a unique and s able posi i e solu ion o (1.1)
o λ≤0. Assume now he exis ence o a a p io i bounds. The e exis s λ > 0such
ha he e exis s a posi i e solu ion i , and only i , λ≤λ. Mo eo e , λ > 0and small
he e exis a leas wo posi i e solu ions uλ
1, uλ
2,uλ
1is s able and
lim
λ→−∞ kuλk∞= +0,lim
λ↑0kuλ
2k∞= +∞.
In Figu e 3 we ha e ep esen ed he bi u ca ion diag ams o (1.1) co esponding o he
case p > 1. Cases 1, 2 and 3 ep esen he solu ions when β= 1 and σ1>0, σ1= 0 and
σ1<0, espec i ely. Cases 4, 5 and 6 show he cases β > p,β=pand β < p, espec i ely.
Finally, Case 7 ep esen s β < 1.
An ou line o he pape is: in Sec ion 2 we s udy he eigen alue p oblem and some
p elimina ies esul s; Sec ion 3 is de o ed o ob ain a p io i bounds o posi i e solu ions
o (1.1) and in he las Sec ion we p o e Theo ems 1.1, 1.2 and 1.3.
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Combining local and non-local e ms Augus 14, 2010
u
u
u
u
u
Case 1 Case 2 Case 3
Case 4
u
Case 6
Case 5
u
Case 7
Figu e 3: Bi u ca ion diag ams o equa ion (1.1) o p > 1.
2 The eigen alue p oblem and p elimina ies esul s
In his sec ion we s udy a non-local and singula eigen alue p oblem, which appea s
when one linea izes a ound a posi i e solu ion o (1.1). Speci ically, we s udy he ollowing
p oblem



−∆u+m(x)u−a(x)ZΩ
b(x)u=σu in Ω,
u= 0 on ∂Ω,
(2.1)
whe e m∈C1(Ω), a∈C(Ω) and b∈C1(Ω) and e i y: o some α∈(−1,1) and γ < 1
(Hm)|∂im|d(x, ∂Ω)2−αa e bounded o all x∈Ω and i= 1, ..., N;
(Hb) he e exis s K > 0 such ha b(x)≤Kd(x, ∂Ω)−γ,
whe e d(x, ∂Ω) := dis (x, ∂Ω) The nex esul was p o ed in [4]:
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Augus 14, 2010 F. J. S. A. Co ˆea and A. Su´a ez
Theo em 2.1. Assume ha m e i ies (Hm),a∈C1(Ω) ∩C(Ω), is a non-nega i e and
non- i ial unc ion, b∈C1(Ω) is a non-nega i e and non- i ial unc ion and i e i ies
(Hb). Then, he e exis s a p incipal eigen alue o (2.1), deno ed by λ1(−∆ + m;a;b),
which has an associa ed posi i e eigen unc ion ϕ1∈C2(Ω) ∩C1,δ
0(Ω) o some δ∈(0,1),
and ∂ϕ1
∂n <0on ∂Ω,(2.2)
whe e ndeno es he ou wa d uni no mal ec o . Mo eo e , λ1(−∆ + m;a;b)is simple,
and i is he unique eigen alue ha ing an associa ed eigen unc ion wi hou change o sign.
In he ollowing esul we gi e a c i e ia o asce ain he sign o λ1(−∆ + m;a;b), see
also [4]:
P oposi ion 2.2. a) Assume ha he e exis s a posi i e unc ion u∈C2(Ω)∩C1,δ
0(Ω),
δ∈(0,1), such ha
−∆u+m(x)u−a(x)ZΩ
b(x)u > 0in Ω.
Then,
λ1(−∆ + m;a;b)>0.
b) Assume ha he e exis s a posi i e unc ion u∈C2(Ω) ∩C1,δ
0(Ω),δ∈(0,1), such
ha
−∆u+m(x)u−a(x)ZΩ
b(x)u < 0in Ω.
Then,
λ1(−∆ + m;a;b)<0.
Along he pape , we a e going o deno e by
λ1:= λ1(−∆; 0; 0) and σ1:= λ1(−∆; 1; 1).
The nex esul cha ac e izes he sign o λ1(−∆ + m;a;b) on e ms o he solu ion o he
p oblem
(−∆ζ+m(x)ζ=b(x) in Ω,
ζ= 0 on ∂Ω. (2.3)
Thanks o P oposi ion 2.5 in [11] i λ1(−∆ + m; 0; 0) >0 he e exis s a unique posi i e
solu ion ζ∈C2(Ω) ∩C1,δ
0(Ω), δ∈(0,1), o (2.3).
Lemma 2.3. Assume ha λ1(−∆ + m; 0; 0) >0. Then,
sgn (λ1(−∆ + m;a;b)) = sgn 1−ZΩ
a(x)ζ.
P oo . Deno e by ϕ1a posi i e eigen unc ion associa ed o λ1(−∆ + m;a;b). Mul iplying
(2.3) by ϕ1, and in eg a ing we ob ain ha
1−ZΩ
a(x)ζZΩ
b(x)ϕ1=λ1(−∆ + m;a;b)ZΩ
ϕ1ζ.
This concludes he esul .
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Combining local and non-local e ms Augus 14, 2010
The nex esul shows he mono ony o he p incipal eigen alue wi h espec o he
domain.
Lemma 2.4. Conside a sub-domain Ω0⊂Ω, and ha he unc ions a, b and m e i y he
condi ions o Theo em 2.1. Deno e by λ0and λ1 he p incipal eigen alues λ1(−∆+m;a;b)
in Ω0and Ω, espec i ely. Then, λ1< λ0.
P oo . Conside ϕ∗
0 he adjoin posi i e eigen unc ion associa ed o λ0, ha is
−∆ϕ∗
0+m(x)ϕ∗
0−b(x)ZΩ0
a(x)ϕ∗
0=λ0ϕ∗
0in Ω0,ϕ∗
0= 0 on ∂Ω0.
Then, p olonging ϕ∗
0by ze o a Ω, and mul iplying by ϕ1, a posi i e eigen unc ion associ-
a ed o λ1, we ge
ZΩ0
a(x)ϕ∗
0ZΩ0
b(x)ϕ1−ZΩ
b(x)ϕ1+Z∂Ω0
∂ϕ∗
0
∂n ϕ1= (λ1−λ0)ZΩ0
ϕ∗
0ϕ1,
whence, using (2.2), we deduce ha λ1< λ0.
Wi h espec o he mono ony on he po en ials, we ha e:
Lemma 2.5. Assume ha m1≤m2,a1≥a2and b1≥b2. Then,
λ1(−∆ + m1;a1;b1)≤λ1(−∆ + m2;a2;b2).
P oo . Le ϕ > 0 an eigen unc ion associa ed o λ1(−∆ + m1;a1;b1). Then
−∆ϕ+m2ϕ−λ1(−∆+m1;a1;b1)ϕ−a2ZΩ
b2ϕ= (m2−m1)ϕ+a1ZΩ
b1ϕ−a2ZΩ
b2ϕ≥0,
and so, by P oposi ion 2.2, λ1(−∆ + m2−λ1(−∆ + m1;a1;b1); a2;b2)≥0, ha is,
λ1(−∆ + m2;a2;b2)≥λ1(−∆ + m1;a1;b1).
Finally, he ollowing esul will be e y use ul du ing he wo k:
Lemma 2.6. Assume a > 0in Ω. I holds ha
lim
λ→+∞λ1(−∆ + m;λa;b) = −∞.
P oo . Conside a ball B⊂Ω such ha b≥b0>0 in Band such ha λB
1(−∆+m; 0; 0) >
0. By Lemma 2.4
λΩ
1(−∆ + m;λa;b)< λB
1(−∆ + m;λa;b).
We a e going o p o e ha λB
1(−∆+m;λa;b)→ −∞ as λ→+∞. Indeed, since λB
1(−∆+
m; 0; 0) >0 he e exis s a unique posi i e solu ion, deno ed by e, o he equa ion
−∆e+m(x)e=b(x) in B,e= 0 on ∂B.
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Augus 14, 2010 F. J. S. A. Co ˆea and A. Su´a ez
P oposi ion 3.6. Assume ha 0<β<p<(N+ 2)/(N−2),p > 1, and λ∈Λ, wi h
Λ⊂IR+compac such ha 0/∈Λ. Then, he e exis s a p io i bound o posi i e solu ions
o (1.1).
P oo . We use again a Gidas-Sp uck a gumen . Wi h he same no a ion ha P oposi ion
3.4 we ge
−∆wn=λwp
n+M−p
nZΩ
uβ
nin Ωn.
Bu ,
M−p
nZΩ
uβ
n≤M−p+β
n|Ω| → 0,
and so passing o he limi we again ob ain
−∆w=λwpin IRNo IRN
+.
Finally, we analyze he case p≤1. Obse e ha when p= 1 > β we will show ha
he e exis s bi u ca ion om in ini y a λ=λ1>0 (see Theo em 1.1). So, we s udy he
case p < 1 and β≤1.
P oposi ion 3.7. Assume ha p < 1and β≤1, and λ∈Λ, wi h Λ⊂IR+compac such
ha 0/∈Λ. Then, he e exis s a p io i bound o posi i e solu ions o (1.1).
P oo . Assume ha he e exis s a sequence λn→λ0>0 and posi i e solu ions uno (1.1)
such ha kunk∞→ ∞. Deno e by
wn:= un
kunk∞
.
I is clea ha wn e i ies
−∆wn=λnwp
nkunkp−1
∞+kunkβ−1
∞ZΩ
wβ
nin Ω, wn= 0 on ∂Ω.
Then, wn→win C2(Ω) being wa solu ion o
−∆w= 0 i β < 1−∆w−ZΩ
w= 0 i β= 1.
In he i s case, i is clea ha w≡0. In he second one, since he e exis s posi i e solu ion
o λn>0 we ge ha σ1>0, and hen w≡0. In bo h cases, we a i e a con adic ion
because kwk∞= 1.
In he ollowing esul , we show ha (1.1) does no possess classical posi i e solu ions
o β=p > (N+ 2)/(N−2) and λla ge.
P oposi ion 3.8. Assume ha Ωis bounded and s a shaped wi h espec o some poin
x0∈Ω,β=p > (N+ 2)/(N−2) and λ > C(N), o some posi i e cons an depending
on N. Then, (1.1) does no possess posi i e solu ion.
16

Combining local and non-local e ms Augus 14, 2010
P oo . We a e going o use a Pohozae ’s a gumen , see o ins ance Chap e 1.5 in [12].
Mul iplying (1.1) by x· ∇uwe ge
N−2
2ZΩ
|∇u|2+1
2Z∂Ω
∂u
∂n
2
x·n=Nλ
p+ 1 ZΩ
up+1 +NZΩ
uβZΩ
u,
and hen
0<1
2Z∂Ω
∂u
∂n
2
x·n=λN
p+ 1 −N−2
2ZΩ
up+1 +N+ 2
2ZΩ
uβZΩ
u.
By H¨olde inequali y, we ge (using β=p)
0<ZΩ
uβZΩ
u≤C(Ω) ZΩ
up+1.
Hence
0<1
2Z∂Ω
∂u
∂n
2
x·n≤λN
p+ 1 −N−2
2+C(Ω)N+ 2
2ZΩ
up+1,
an absu dum o λla ge.
Wi h a comple ely analogous a gumen , we can p o e:
Co olla y 3.9. Assume ha Ωis bounded and s a shaped wi h espec o some poin
x0∈Ω,β=p < (N+ 2)/(N−2) and λ < −C(N), o some posi i e cons an depending
on N. Then, (1.1) does no possess posi i e solu ion.
4 P oo o he main esul s
In his sec ion we p o e he main esul s o he pape s a ed in Sec ion 1. Fi s ly,
obse e ha i (p, β) = (1,1) hen (1.1) is an eigen alue p oblem, and so he e exis
posi i e solu ions i , and only i ,
λ=σ1.
Recall ha sgn(σ1) = sgn(1 −RΩe) whe e eis de ined in (3.1).
So, om now on we assume ha (p, β)6= (1,1).
Also, o λ= 0 and β6= 1 he e exis s a unique posi i e solu ion
u=eZΩ
uβ=⇒u=eZΩ
eβ1/(1−β)
.
( he case β= 1 is an eigen alue p oblem). Mo eo e , by P oposi ion 2.9, uis s able o
β < 1 and uns able o β > 1. So, we assume λ6= 0.
4.1 Some use ul esul s
A i s a emp o s udy (1.1) is conside
R=ZΩ
uβ,
17
Augus 14, 2010 F. J. S. A. Co ˆea and A. Su´a ez
and hen we ha e o s udy he equa ion
(−∆u=λup+Rin Ω,
u= 0 on ∂Ω,(4.1)
and a e ha , o ind a poin ixed o
R=ZΩ
uβ
R⇐⇒ 1 = ZΩ
wRβ≡h(R) (4.2)
being uRa posi i e solu ion o (4.1) and wR=uR/R1/β and so posi i e solu ion o
(−∆w=λR(p−1)/βwp+R(β−1)/β in Ω,
w= 0 on ∂Ω.(4.3)
In he ollowing esul we s udy in de ail he map R7→ h(R).
P oposi ion 4.1. Assume R > 0.
a) Assume p= 1. Then (4.3) possesses a posi i e solu ion, deno ed by wR, i , and only
i , λ<λ1. The solu ion is unique. Mo eo e ,
wR=eλRβ−1
β,(4.4)
being eλ he unique posi i e solu ion o
((−∆−λ)eλ= 1 in Ω,
eλ= 0 on ∂Ω.(4.5)
b) Assume p < 1. Then (4.3) possesses a unique posi i e solu ion, deno ed by wR, o
all λ∈IR. Mo eo e , he map R∈(0,∞)7→ h(R)is con inuous and de i able. Fo
λ > 0
lim
R→0h(R) = ∞.
(a) When β= 1.
i. I λ > 0,R7→ h(R)is dec easing and
lim
R→∞ h(R) = ZΩ
e.
ii. I λ < 0,R7→ h(R)is inc easing and
lim
R→0h(R) = 0,lim
R→∞ h(R) = ZΩ
e.
(b) When β > 1,
i. I λ > 0
lim
R→∞ h(R) = ∞.
18
Combining local and non-local e ms Augus 14, 2010
ii. I λ < 0,R7→ h(R)is inc easing and
lim
R→0h(R)=0,lim
R→∞ h(R) = ∞.
(c) When β < 1and λ > 0,R7→ h(R)is dec easing, and o all λ
lim
R→∞ h(R)=0.
Mo eo e ,
i. I β < p and λ < 0,R7→ h(R)is dec easing, and o all λ
lim
R→0h(R) = ∞.
ii. I β=p, he map R7→ h(R)is dec easing and
lim
R→0h(R) = (∞i λ > 0,
ρ0(λ)i λ < 0,
whe e
ρ0(λ)∈(−1/λ)ZΩ
ϕβ
1,(−1/λ)|Ω|,(4.6)
ϕ1is he posi i e eigen unc ion associa ed o λ1such ha kϕ1k∞= 1, and
ρ0(λ)is a non-dec easing unc ion in λ o λ < 0.
iii. I β > p,
lim
R→0h(R) = (∞i λ > 0,
0i λ < 0.
c) Assume p > 1and λ < 0, hen he e exis s a unique posi i e solu ion, deno ed by
wR, o (4.3). Mo eo e , he map R∈(0,∞)7→ h(R)is con inuous and de i able.
(a) When β= 1. The map R7→ h(R)is dec easing and
lim
R→0h(R) = ZΩ
e, lim
R→∞ h(R)=0.
(b) When β > 1.
i. I β > p,R7→ h(R)is inc easing and
lim
R→0h(R)=0,lim
R→∞ h(R)=+∞.
ii. I β=p, he map R7→ h(R)is inc easing
lim
R→0h(R)=0,lim
R→∞ h(R) = ρ0(λ),
wi h ρ0(λ)as in (4.6).
iii. I β < p
lim
R→0h(R) = 0,lim
R→∞ h(R) = 0.
19
Augus 14, 2010 F. J. S. A. Co ˆea and A. Su´a ez
(c) When β < 1. The map R7→ h(R)is dec easing and
lim
R→0h(R)=+∞lim
R→∞ h(R)=0.
P oo . a) Assume ha p= 1, hen wR e i ies
(−∆−λ)wR=R(β−1)/β
and he esul o pa ag aph a) is ob ained easily.
Fo he o he cases, i is clea ha (w,w) = (0, Ke) is a pai o sub-supe solu ion o
(4.3) o K e i ying
K≥λKpepR(p−1)/β +R(β−1)/β.(4.7)
I is enough o ake Kla ge in any case.
On he o he hand, o λ≥0 and p < 1 he uniqueness ollows by [3] and o λ < 0
hanks o λR(p−1)/βwpis a dec easing map in R.
The con inui y and de i abili y o he map R7→ wRis s anda d.
I λ≥0 we ha e ha −∆w≥R(β−1)/β and so
wR≥R(β−1)/βe. (4.8)
Analogously, i λ≤0 we ha e ha
wR≤R(β−1)/βe. (4.9)
Mo eo e , i λ≥0 and p < 1 we ha e ha −∆w≥λwpR(p−1)/β and hen
wR≥R−1/βλ1/(1−p)w1,(4.10)
whe e w1is he unique posi i e solu ion o (2.6).
Also, by he maximum p inciple o λ < 0 we ge
kwRkp
∞≤R(β−p)/β
−λ.(4.11)
Finally, εϕ1is subsolu ion o (4.3), kϕ1k∞= 1, i
ελ1≤λεpϕp
1R(p−1)/β +R(β−1)/β.(4.12)
b) Assume ha p < 1. Then, i is clea by (4.10) ha o λ > 0
lim
R→0h(R)=+∞,
and o β > 1 by (4.8)
lim
R→∞ h(R)=+∞.
Take now λ≤0, hen i is clea by (4.9) ha
lim
R→0h(R) = 0 i β > 1, lim
R→∞ h(R) = 0 i β < 1,
and by (4.11)
lim
R→0h(R) = 0 i β > p, lim
R→∞ h(R) = 0 i β < p.
20
Combining local and non-local e ms Augus 14, 2010
Now, conside β > 1 and λ < 0. Obse e ha he map R7→ λR(p−1)/β +R(β−1)/β is
inc easing, and so R7→ wRalso. In his case, o (4.12) is enough
ελ1−λεpR(p−1)/β =R(β−1)/β.(4.13)
F om his equali y we deduce ha ε(R)→ ∞ as R→ ∞, and hen h(R)→ ∞.
Fo β < 1 we ha e o dis inguish se e al cases. I β < p and λ < 0 hen again by
(4.13), we ge ha ε(R)→ ∞ as R→0. Fo he case β=pwe ha e ha
εp(R)→ −1
λas R→0.
Mo eo e , by (4.11) we deduce ha h(R)≤−1
λ|Ω|.
Fo λ > 0, o (4.7) is enough
K−λkekp
∞KpR(p−1)/β =R(β−1)/β.(4.14)
I p, β < 1 i is clea ha K(R)→0 as R→ ∞.
Finally, o β= 1 and λ > 0 obse e ha
e≤wR≤K(R)e
and K(R)→1 as R→ ∞ by (4.14).
Fo λ < 0, wR≤eand ε(R)eis subsolu ion i
ε−λεpCRp−1= 1.(4.15)
I is clea ha ε(R)→1 as R→ ∞.
Obse e also ha in he pa icula case β=p,wR e i ies
−∆w=R(β−1)/β(λwp+ 1)
and hen, by he maximum p inciple
λwp(x)+1≥0 o all x∈Ω. (4.16)
Indeed, i λ≥0 hen (4.16) is clea . I λ < 0 obse e ha
λwp(x)+1≥λkwkp
∞+ 1 ≥0.
Then, i R1< R2and β < 1, we ge ha wR2is sub-solu ion o (4.3) o R=R1, and
hen wR2< wR1. This p o es ha R7→ h(R) is dec easing.
Hence, he e exis s he ollowing limi
lim
R↓0h(R) := ρ0(λ).
Mo eo e , i λ1< λ2<0, wλ1,R is subsolu ion o he equa ion (4.3) wi h λ=λ2, and so
wλ1,R < wλ2,R. Taking limi we ha e ha
ρ0(λ1)≤ρ0(λ2).
21

Augus 14, 2010 F. J. S. A. Co ˆea and A. Su´a ez
Finally, assume ha β≤p. Obse e ha wR e i ies
−∆w=R(β−1)/β(λR(p−β)/βwp+ 1).
Obse e ha by he maximum p inciple and since λ < 0 we ge
λR(p−β)/βwp(x)+1≥λR(p−β)/βkwkp
∞(x)+1≥0.
Take R1< R2, hen
R(β−1)/β
1(λR(p−β)/β
1wp
R1+ 1) ≥R(β−1)/β
2(λR(β−p)/β
2wp
R1+ 1),
and hen wR1is a supe solu ion o he equa ion (4.3) wi h R=R2. We conclude ha
wR1≥wR2.
c) Assume ha p > 1 and λ < 0. F om (4.9) we ha e ha h(R)→0 as R→0 i β > 1
and h(R)→0 as R→ ∞ i β < 1.
In his case i β < 1 i is clea ha ε(R)→ ∞ as R→0 om (4.13).
Now, assume β > 1. I p<β hen ε(R)→ ∞ as R→ ∞, i p=β,εp(R)→ −1/λ and
o p>βwe ha e ha K(R)→0.
Finally, o β= 1, and using again (4.15), we ha e ha h(R)→RΩei R→0.
In he ollowing esul we p o e a s abili y esul o a posi i e solu ion u0o (1.1),
ob ained such ha u0=uR0 o some R0>0, in unc ion on he map hde ined in (4.2).
P oposi ion 4.2. Le u0be a posi i e solu ion o (1.1) ob ained such ha u0=uR0 o
some R0>0. Then, i
h0(R0)<0( esp. h0(R0)>0) hen u0is s able ( esp. u0is uns able).
P oo . Le u0=uR0a posi i e solu ion o (1.1). Assume ha h0(R0)<0, we wan o
show ha
λ1(−∆−λpup−1
0;β;uβ−1
0)>0,(4.17)
(analogous a gumen in he case h0(R0)>0).
Fi s , obse e ha he map R7→ uRis inc easing (uRde ined in (4.1)), and so i s
de i a i e u0
R>0 in Ω, being u0
R he unique solu ion o
−∆u0
R=λpup−1
Ru0
R+ 1 in Ω, u0
R= 0 on ∂Ω.
On he o he hand, obse e ha since h0(R0)<0 and using ha h(R) = (1/R)ZΩ
uβ
R, we
ge
βZΩ
uβ−1
R0u0
R0<ZΩ
uβ
R0
R0
=h(R0)=1.
To p o e (4.17) we use P oposi ion 2.2 wi h u=u0
R0>0. Indeed, obse e ha
−∆u0
R0−λpup−1
0u0
R0−βZΩ
uβ−1
0u0
R0= 1 −βZΩ
uβ−1
0u0
R0>0,
and hen he s abili y ollows.
22
Combining local and non-local e ms Augus 14, 2010
Fo he case p > 1 and λ > 0 we wo k wi h he o iginal equa ion. In ac , assume
β≥1 and conside he ollowing auxilia p oblem:



−∆u=µu +λup+ZΩ
uβin Ω,
u= 0 on ∂Ω.
(4.18)
Lemma 4.3. Assume p > 1,β≥1and λ > 0. Then, o µ=λ1when β > 1and
µ=σ1when β= 1 bi u ca es om he i ial solu ion a non-bounded con inuum Co
posi i e solu ions o (4.18). Mo eo e , assuming he exis ence o a p io i bound o (4.18)
o µ∈Λ,Λa compac subse o IR, he e exis s a posi i e solu ion i , and only i , µ<λ1
i β > 1and µ<σ1 o β= 1.
P oo . Fi s , obse e ha i uis a posi i e solu ion o (4.18) we ha e
µ=λ1(−∆−λup−1; 1; uβ−1)< λ1(−∆; 1; uβ−1),
ha is, µ<λ1i β > 1 and µ<σ1in he case β= 1.
Tha µ=λ1 o β > 1 and µ=σ1 o β= 1 is a bi u ca ion poin om he i ial
solu ion is consequence o he C andall-Rabinowi z Theo em [5], see also [6].
The exis ence o an unbounded con inuum C ollows by he classical Rabinowi z The-
o em [13].
4.2 P oo o Theo em 1.1
Assume ha p= 1. I is clea ha (1.1) does no possess posi i e solu ion o λ≥λ1.
By P oposi ion 4.1 a) he e exis s a unique posi i e solu ion o λ<λ1. The s abili y
esul s ollow by P oposi ion 2.9 a) and b).
We s udy now he beha iou wi h espec o λ. Obse e ha i uis a posi i e solu ion
o (1.1) we ha e
(−∆−λ)u=ZΩ
uβ,
and so,
u=eλZΩ
eβ
λ1/(1−β)
,
and so aking in o accoun ha o ϕ1>0 wi h kϕ1k∞= 1, ϕ1eigen unc ion associa ed
o λ1,1
λ1−λϕ1≤eλin Ω,
we ge ha o β < 1,
u≥1
(λ1−λ)1/(1−β)ϕ1ZΩ
ϕβ
11/(1−β)
and so kuk∞→ ∞ as λ→λ1.
Assume ha β > 1 and conside a sequence λn< λ1,λn→λ1and un he posi i e
solu ion o (4.18) o λ=λn. We know by P oposi ion 3.3 ha kunk∞is bounded, and so
passing o he limi we ge ha un→u0in C2(Ω) as λn→λ1, wi h u0posi i e solu ion
o λ=λ1. Then, u0≡0.
Finally, he beha iou as λ→ −∞ ollows by Lemma 2.8.
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Augus 14, 2010 F. J. S. A. Co ˆea and A. Su´a ez
4.3 P oo o Theo em 1.2
a) Assume p < 1 = β. Assume ha σ1>0, hen i is clea ha λ > 0. Obse e ha
since σ1>0, applying Lemma 2.3 wi h a≡b≡1 and m≡0 we ge ha RΩe < 1.
Now, he exis ence and uniqueness ollow by P oposi ion 4.1 b). The s abili y ollows by
P oposi ion 2.9. Finally, obse e ha (see (2.7))
uλ=λ1/(1−p)u1,
being u1 he unique solu ion o (1.1) o λ= 1. F om he e, we can deduce he beha iou
as λ→0 and λ→ ∞. The o he cases can be ea ed simila ly.
b) Assume p < 1< β. The exis ence and uniqueness in he case λ < 0 ollow by
P oposi ion 4.1. Also, he s abili y ollows by P oposi ion 2.9. Now conside λ > 0.
Deno e
eR:= R(β−1)/βe.
Take R0>0 small such ha
ZΩ
eβ
R0=Rβ−1
0ZΩ
eβ<1.
Fix such R0>0. Now, i is clea wR0→eR0in L∞(Ω) as λ→0, hence h(R0)<1 o
λ≤λ0, wi h λ0small. So, since
lim
R→0h(R) = lim
R→∞ h(R) = +∞,
he e exis a leas wo posi i e alues R1
0< R0< R2
0such ha h(Ri
0) = 1, i= 1,2, and
so wo posi i e solu ions uλ
i=uRi
0o (1.1) o λ≤λ0wi h u1< u2, and
h0(R1
0)<0< h0(R2
0).
Thank o P oposi ion 4.2 we ha e ha uλ
1is s able and uλ
2uns able.
Now, we show ha he e does no exis posi i e solu ions o (1.1) o λla ge. Obse e
ha
u≥λ1/(1−p)w1,(4.19)
whe e w1is de ined in (2.6), and hen
−∆u≥λ1/(1−p)wp
1+λ(β−1)/(1−p)ZΩ
wβ−1
1u
and so
λ1(−∆; λ(β−1)/(1−p);wβ−1
1)>0.
This is an absu dum because by Lemma 2.6
λ1(−∆; λ(β−1)/(1−p);wβ−1
1)→ −∞ as λ→ ∞.
Then, we can de ine
Λ := {λ∈IR : he e exis s a leas a posi i e solu ion o (1.1)}.
24
Combining local and non-local e ms Augus 14, 2010
We ha e p o ed ha 0 < λ := sup Λ <∞. Thanks o he bounds by P oposi ion 3.3,
he e exis s posi i e solu ion o λ=λ. Now, i is clea ha i λ∈(0, λ) hen (εw1, uλ)
is a pai o sub-supe solu ion o (1.1) wi h εsmall and uλa posi i e solu ion o (1.1) o
λ=λ. Obse e ha his me hod wo ks o non-local equa ion, see o ins ance [9].
On he o he hand, conside uλ
1 o λ∈(0, λ0). Since o λ= 0 he solu ion is uns able,
we can assu e ha
lim
λ→0kuλ
1k∞= 1.
Finally, he beha iou o he solu ion as λ→ −∞ ollows by Lemma 2.8.
c) Assume p, β < 1. In his case, i is clea by P oposi ions 4.1 and 4.2 he exis ence,
uniqueness and s abili y o λ > 0.
(a) Suppose ha β < p. In his case we ha e again by P oposi ion 4.1 he exis ence and
uniqueness o all λ∈IR and he s abili y ollows by P oposi ion 2.9.
(b) Suppose β=p. Obse e ha since he map h(R) is dec easing, in case o exis ence o
posi i e solu ion, i is unique. Mo eo e , by (4.6) and since ρ0(λ) is non-dec easing, he e
exis s a unique alue λ0<0 such ha
ρ0(λ)≤1 o λ≤λ0and ρ0(λ)>1 o λ∈(λ0,0).
Hence, he e exis s a posi i e solu ion o (1.1) i , and only i , λ>λ0. Then,
lim
R→0h(R)≤1 o λ<λ0, lim
R→0h(R)>1 o λ>λ0,
and
lim
λ→λ0
kuλk∞= 0.
Again, hanks o P oposi ion 4.2 we know ha he solu ion is s able.
(c) Suppose β > p. Wi h a simila a gumen o he used in he pa ag aph b) we can show
he exis ence o wo posi i e solu ions o λnega i e and small. Indeed, in his case
lim
R→0h(R) = lim
R→∞ h(R)=0,
and he e exis a leas wo posi i e alues R1
0< R0< R2
0such ha h(Ri
0) = 1, i= 1,2,
and so wo posi i e solu ions uλ
i=uRi
0o (1.1) o λ≤λ0wi h uλ
1< uλ
2, and
h0(R1
0)>0> h0(R2
0),
and hen uλ
1is uns able and uλ
2s able.
We p o e now he non-exis ence o posi i e solu ions o λ e y nega i e. Indeed,
obse e ha by (2.5) we ha e
−λ≤Ckukβ−p
∞≤C(4.20)
his las inequali y by Lemma 3.1. Then, i he e exis s a posi i e solu ion o (1.1) we ge
ha λ≥ −C.
Again, we can de ine
Λ := {λ∈IR : he e exis s a leas a posi i e solu ion o (1.1)}.
We know ha λ:= in Λ >−∞ and λ < 0, and using as sub-supe solu ion he pai
(uλ, Ke) o Kla ge, we p o e he exis ence o posi i e solu ion o all λ∈(λ, 0). Finally,
by (4.20) i can no occu ha o a sequence (λn, un) we ha e λn→λ0<0 and kunk∞→
0. Mo eo e , since he solu ion o λ= 0 we can conclude ha
lim
λ→0kuλ
1k∞= 0.
25