R E S E A R C H
Open Access
Existence of positive solutions of elliptic
mixed boundary value problem
Guofa Li
**Correspondence:
[email protected] Department of Mathematics and Information Science, Qujing Normal University, Qujing, 655011, P.R. China
Abstract
In this paper, we use variational methods to prove two existence of positive solutions of the following mixed boundary value problem:
⎧ ⎪ ⎨ ⎪ ⎩
–
u=f(x,u), x∈
,
u= 0, x∈
σ
,∂u
∂ν=g(x,u), x∈
.
One deals with the asymptotic behaviors off(x,u) near zero and infinity and the other deals with superlinear off(x,u) at infinity.
MSC: 35M12; 35D30
Keywords: elliptic mixed boundary value problem; positive solutions; mountain pass theorem; Sobolev embedding theorem
1 Introduction and preliminaries
This paper is concerned with the existence of positive solutions of the following elliptic mixed boundary value problem:
⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩
–u=f(x,u), x∈,
u= , x∈σ,
∂u
∂ν =g(x,u), x∈,
()
whereis a bounded domain inRnwith Lipschitz boundary∂,σ∪=∂,σ∩=Ø,
is a sufficiently smooth (n– )-dimensional manifold, andνis the outward normal vector on∂. We assumef :×R→R,g:×R→Rare continuous and satisfy
(S) f(x,t)≥,∀t≥,x∈,f(x, ) = .f(x,t)≡,∀t< ,x∈. (S) For almost everyx∈, f(xt,t)is nondecreasing with respect tot> . (S) limt→f(xt,t)=p(x),limt→+∞f(xt,t)=q(x)≡uniformly in a.e.x∈, where
p(x)∞<λ,λis the first eigenvalue of (),≤p(x),q(x)∈L∞().
(S) There existsc,c> such that|f(x,t)| ≤c+c|t|p–for somep∈(,n–n)asn≥
andp∈(, +∞)asn= , .
The eigenvalue problem of () is studied by Liu and Su in []
⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩
–u=λu in,
u= onσ,
∂u
∂ν =λu on.
()
There exists a set of eigenvalues {λk} and corresponding eigenfunctions {uk} which
solve problem (), where ≤λ ≤λ ≤ · · · ≤λk ≤ · · ·, λk → ∞ as k → ∞, λ = inf=u∈V
|∇u|dx
|u|dx+
|u|ds.
There have been many papers concerned with similar problems at resonance under the boundary condition; see [–]. Moreover, some multiplicity theorems are obtained by the topological degree technique and variational methods; interested readers can see [–]. Problem () is different from the classical ones, such as those with Dirichlet, Neuman, Robin, No-flux, or Steklov boundary conditions.
In this paper, we assumeV:={v∈H() :v|
σ = }is a closed subspace ofH(). We
define the norm in V asu=|∇u|dx+|γu|ds, · Lp() is the Lp() norm,
· Lp() is theLp() norm,γ :V →L() is the trace operator with γu=u for all
u∈H(), that is continuous and compact (see []). Furthermore, we define g=γf, ≤g(x,t)≤ |γf(x,t)|fort> (see []). Then, by (S), we obtain
lim
t→+∞ g(x,t)
t ≤t→lim+∞
|γf(x,t)|
t =q(x)≡, a.e.x∈. ()
Letbe a bounded domain with a Lipschitz boundary; there is a continuous embedding
V →Ly() fory∈[, n
n–] whenn≥, andy∈[, +∞) whenn= , . Then there exists γy> , such that
uLy()≤γyu, ∀u∈V. ()
Moreover, there is a continuous boundary trace embeddingV→Lz() forz∈[,(nn––)] whenn≥, andz∈[, +∞) whenn= , . Then there existskz> , such that
uLz()≤kzu, ∀u∈V. ()
It is well known that to seek a nontrivial weak solution of problem () is equivalent to finding a nonzero critical value of theCfunctional
J(u) =
|∇u|dx–
F(x,u)dx–
G(s,u)ds, ()
whereu∈V,F(x,u) =uf(x,t)dt,G(x,u) =ug(x,t)dt. Moreover, by (S) and the Strong maximum principle, a nonzero critical point ofJis in fact a positive solution of (). In order to find critical points of the functional (), one often requires the technique condition, that is, for someμ> ,∀|u| ≥M> ,x∈,
<μF(x,u)≤uf(x,u), F(x,u) =
u
It is easy to see that the condition (AR) implies thatlimu→+∞F(ux,u)= +∞, that is,f(x,u)
must be superlinear with respect touat infinity. In the present paper, motivated by [] and [], we study the existence and nonexistence of positive solutions for problem () with the asymptotic behavior assumptions (S) offat zero and infinity. Moreover, we also study superlinear off at infinity withq(x)≡+∞in (S), which is weaker than the (AR) condition, that is the (AR) condition does not hold.
In order to get our conclusion, we define the minimization problem
=inf
|∇u|dx:u∈V,
q(x)udx+
q(s)uds= , ()
then> , which is achieved by someϕ∈V withϕ(x) > a.e. in; see Lemma .
We denote byc,c,cuniversal constants unless specified otherwise. Our main results are as follows.
Theorem Let conditions (S) to (S) hold, then:
(i) If> , then the problem () has no any positive solution inV. (ii) If< , then the problem () has at least one positive solution inV.
(iii) If= , then the problem () has one positive solutionu(x)∈Vif and only if there exists a constantc> such thatu(x) =cϕ(x)andf(x,u) =q(x)u(x),
g(x,u) =q(x)u(x)a.e.x∈, whereϕ(x) > is the function which achieves.
Corollary Let conditions (S) to (S) with q(x)≡l> hold, then:
(i) Ifl<λ, then the problem () has no any positive solution inV.
(ii) Ifλ<l< +∞, then the problem () has at least one positive solution inV. (iii) Ifl=λ, then the problem () has one positive solutionu(x)∈Vif and only if there
exists a constantc> such thatu(x) =cϕ(x)andf(x,u) =λu(x),g(x,u) =λu(x) a.e.x∈, whereϕ(x) > is the eigenfunction of theλ.
Theorem Let conditions (S) to (S) with q(x)≡+∞hold, then the problem () has at least one positive solution in V .
2 Some lemmas
We need the following lemmas.
Lemma If q(x)∈L∞(), q(x)≥, q(x)≡, then> and there existsϕ(x)∈V such
that=|∇ϕ|dx and
q(x)ϕ
dx+
q(s)ϕ
ds= . Moreover,ϕ(x) > a.e. in V .
Proof By the Sobolev embedding functionV →L() and Fatou’s lemma, it is easy to know that > and there existsϕ(x)∈V, which satisfies, that is,
q(x)ϕ
dx+
q(s)ϕds= . Furthermore, we assumeϕ(x)≥, thenϕ(x) could replace by|ϕ(x)|.
By the Strong maximum principle, we knowϕ(x) > a.e. inV.
Lemma If conditions (S) to (S) hold, then there existsβ,ρ> such that J|∂Bρ()≥β, ∀u∈V ,u=ρ.
Proof By condition (S), there existsδ> ,ε> such that f(xu,u)≤λ–ε, g(xu,u)≤ γf(ux,u) ≤ λ–εas <|u| ≤δ. Which implies thatF(x,u)≤(λ–ε)u+c|u|y,G(x,u)≤(λ–ε)u+
By () and (), we obtain
J(u) =
∇u
L()–
F(x,u)dx–
G(s,u)ds
≥
∇u
L()+
γu
L()–
γu
L()–
(λ–ε)u
L()
–cuyLy()–
(λ–ε)u
L()–cuzLz()
≥
u –
(λ–ε) λ
u–cγyyuy–
(λ–ε+ ) λ+
u–ckzzuz
=
ε(λ+ ) λ(λ+ )
–
u–cγyyuy–ckz zuz.
Hence,y,z> ; we takeεwhich satisfies ε(λ+) λ(λ+)–
> , that is,ε> λ(λ+)
λ+ . Then we
take a positive constantβsuch thatJ|∂Bρ()≥βasu=ρ, and is small enough.
Lemma If conditions (S) to (S) hold,< ,ϕ(x) > is defined by Lemma , then
J(tϕ(x))→–∞as t→+∞.
Proof If< ,ϕ(x) > is defined by Lemma , by Fatou’s lemma, and (S), we have
lim
t→+∞
J(tϕ(x))
t
=
∇ϕ(x)
dx– lim
t→+∞
F(x,tϕ(x))dx
t –t→lim+∞
G(s,tϕ(s))ds
t
≤
∇ϕ(x)
dx–
lim
t→+∞
F(x,tϕ(x))
tϕ (x)
ϕ(x)dx
–
lim
t→+∞
G(s,tϕ(s))
tϕ (s)
ϕ(s)ds
=
∇ϕ(x)dx–
f(x,tϕ(x))
tϕ(x)
ϕ(x)dx–
g(s,tϕ(s))
tϕ(s)
ϕ(s)ds
=
∇ϕ(x)
dx–
q(x)ϕ(x)dx+
q(s)ϕ(s)ds
=
(– )
∇ϕ(x)
dx
< .
So,J(tϕ(x))→–∞ast→+∞.
Lemma Let conditions (S) and (S) hold. If a sequence{un} ⊂V satisfiesJ(un),un →
as n→+∞, then there exists a subsequence of{un}, still denoted by{un}such that J(tun)≤
+t
n +J(un)for all t> , n≥.
Proof SinceJ(un),un → asn→+∞, for a subsequence, we may assume that
–
n<
J(un),un
=∇unL()–
f(x,un)undx–
g(s,un)unds<
For any fixedx∈andn≥, set
ψ(t) =
t
f(x,un)un–F(x,tun), ψ(t) =
t
g(s,un)un–G(s,tun).
Then (S) implies that
ψ(t) =tf(x,un)un–f(x,tun)un
=tun
f(x,un) –
f(x,tun)
t = ⎧ ⎨ ⎩
≥, <t≤;
≤, t> .
It implies thatψ(t)≤ψ(),∀t> . Following the same procedures, we obtainψ(t)≤ ψ(),∀t> .
For allt> and positive integern, by (), we have
J(tun) =
t
∇un
L()–
F(x,tun)dx–
G(s,tun)ds
≤ t
n+
f(x,un)undx+
g(s,un)unds
–
F(x,tun)dx–
G(s,tun)ds
≤ t
n+
f(x,un)un–F(x,un)
dx+
g(s,un)un–G(s,un)
ds. ()
On the other hand, by (), one has
J(un) =
∇un
L()–
F(x,un)dx–
G(s,un)ds
≥ – n+
f(x,un)undx+
g(s,un)unds
–
F(x,un)dx–
G(s,un)ds
= – n+
f(x,un)un–F(x,un)
dx+
g(s,un)un–G(s,un)
ds. One has
f(x,un)un–F(x,un)
dx+
g(s,un)un–G(s,un)
ds≤J(un) +
n. ()
Combining () and (), we haveJ(tun)≤+t
n +J(un).
Lemma (see []) Suppose E is a real Banach space, J∈C(E,R)satisfies the following
geometrical conditions:
(ii) There existse∈E\Bρ()such thatJ(e)≤. Letbe the set of all continuous paths joiningande:
=
h∈C[, ],E|h() = ,h() =e,
and
c= inf
h∈ max
t∈[,]J
h(t).
Then there exists a sequence {un} ⊂ E such that J(un)→c≥β and ( +un)×
J(u
n)E∗→.
3 Proofs of main results
Proof of Theorem (i) Ifu∈Vis one positive solution of problem (), by (), one has
=J(u),u=
|∇u|dx–
f(x,u)u dx–
g(s,u)u ds.
That is,
|∇u|dx=
f(x,u)u dx+
g(s,u)u ds
≤
q(x)udx+
q(s)uds= .
It implies that≤. This completes the proof of Theorem (i).
(ii) By Lemma , there existsβ,ρ> such thatJ|∂Bρ()≥βwithu=ρ. By Lemma ,
we obtainJ(tϕ(x)) < ast→+∞. Define
=h∈C[, ],V|h() = ,h() =tϕ(x)
, ()
c= inf
h∈tmax∈[,]J
h(t), ()
whereϕ(x) > is given by Lemma . Thenc≥β> and by Lemma , there exists{un} ⊂
V such that
J(un) =
∇un
L()–
F(x,un)dx–
G(s,un)ds=c+o(), ()
+unJ(un)V∗→. ()
() implies that
J(un),un
=∇unL()–
f(x,un)undx–
g(s,un)unds=o(). ()
If{un}is bounded inV, whenis bounded andf(x,u),g(x,u) are subcritical, we can get
{un}has a subsequence strong convergence to a critical value ofJ, and our proof is
com-plete. So, to prove the theorem, we only need show that{un}is bounded inV. Supposing
that{un}is unbounded, that is,un →+∞asn→+∞. We order
tn=
√c un
, wn=tnun=
√cun
un
. ()
Then{wn}is bounded inV. By extracting a subsequence, we supposewn→wis a strong
convergence inL(),w
n→wis a convergence a.e.x∈,wnwis a weak convergence
inV.
We claim thatw= . In fact, by (S) and (S), we know∀x∈,un≥, and there exists
M,M> such that|f(xu,unn)| ≤M,|g(xu,unn)| ≤M. Ifw= ,wn→ is a strong convergence
inL(), and by () and () we know
c = tnun=tn
∇unL()+γunL()
= tn
f(x,un)undx+tn
g(s,un)unds+tnγunL()+o()
=
f(x,un)
un
wndx+
g(s,un)
un
wnds+tnunL()+o()
≤ M
wndx+M
wnds+wnL()+o()
→.
It is contradiction withc> , sow= . As follows, we provew= satisfies
∇ϕ(x)∇w(x)dx–
q(x)ϕ(x)w(x)dx–
q(s)ϕ(s)w(s)ds= .
We order
pn(x) =
⎧ ⎨ ⎩
f(x,un)/un, un≥,x∈,
, un< ,x∈,
qn(x) =
⎧ ⎨ ⎩
g(x,un)/un, un≥,x∈,
, un< ,x∈.
By (S) and (S), there existsM> such that ≤pn(x)≤M, ≤qn(x)≤M,∀x∈. We select a suitable subsequence and there existsh(x)∈L(),h(x)∈L() such that
pn(x)→h(x) is a strong convergence inL(),qn(x)→h(x) is a strong convergence in
L(), and ≤h
(x)≤M, ≤h(x)≤M,∀x∈.
It follows fromwn→wis a strong convergence inL() that
pn(x)wn(x)ϕ(x)dx=
pn(x)w+n(x)ϕ(x)dx→
h(x)w+(x)ϕ(x)dx,
qn(s)wn(s)ϕ(s)ds=
qn(s)w+n(s)ϕ(s)ds→
Hence,{pn(x)wn(x)}is bounded inL(),pn(x)wn(x)h(x)w+(x) inL();{qn(x)wn(x)}
is bounded inL(),qn(x)wn(x)h(x)w+(x) inL(). By (), we have
∇wn(x)∇ϕ(x)dx–
pn(x)wn(x)ϕ(x)dx–
qn(s)wn(s)ϕ(s)ds
=
∇tnun(x)
∇ϕ(x)dx–
pn(x)tnun(x)ϕ(x)dx–
qn(s)tnun(s)ϕ(s)ds
=
√ c un
∇un(x)∇ϕ(x)dx–
pn(x)un(x)ϕ(x)dx–
qn(s)un(s)ϕ(s)ds
→.
Sincewnwis a weak convergence inV, we obtain
∇ϕ(x)∇w(x)dx–
h(x)ϕ(x)w+(x)dx–
h(s)ϕ(s)w+(s)ds= , ϕ∈V.
We orderϕ=w–; this yieldsw–= , sow=w+≥. By the Strong maximum principle, we knoww> a.e. in, soun→ ∞a.e. in. Combining (S) and (), we obtain
∇ϕ(x)∇w(x)dx–
q(x)ϕ(x)w(x)dx–
q(s)ϕ(s)w(s)ds= , ∀ϕ∈V.
This is a contradiction with< . This completes the proof of Theorem (ii). (iii) If= , by Lemma , there exists someϕ(x) > , such that
∇v(x)∇ϕ(x)dx=
q(x)v(x)ϕ(x)dx+
q(s)v(s)ϕ(s)ds. ()
Ifuis a positive solution of (), for the aboveϕ(x), we have
∇u(x)∇ϕ(x)dx=
fx,u(x)ϕ(x)dx+
gs,u(s)ϕ(s)ds. ()
We orderv=uin (), and it follows from () that
∇u(x)∇ϕ(x)dx=
q(x)u(x)ϕ(x)dx+
q(s)u(s)ϕ(s)ds
=
fx,u(x)ϕ(x)dx+
gs,u(s)ϕ(s)ds
≤
q(x)u(x)ϕ(x)dx+
q(s)u(s)ϕ(s)ds,
which implies that(f(x,u) –q(x)u(x))ϕ(x)dx+
(g(s,u) –q(s)u(s))ϕ(s)ds= .
Whenϕ(x) > a.e. in, combining (S), (S), and (), we obtain
f(x,u)≤q(x)u(x), g(x,u)≤q(x)u(x).
Then we must havef(x,u) =q(x)u(x),g(x,u) =q(x)u(x) a.e. in,u(x) > also achieves (= ). Whenu=cϕ,c> , we have
|∇ϕ|dx=
q(x)ϕdx+
q(s)ϕds, which
On the other hand, if for somec> ,u(x) =cϕ(x) andf(x,cϕ(x)) =cq(x)ϕ(x),g(x,u) =
cq(x)ϕ(x) a.e.x∈, sincecϕ(x) also achieves. This meansu(x) =cϕ(x) is a solution
of problem () as= . This completes the proof of Theorem (iii).
Proof of Corollary Note that whenq(x)≡l, then=λ
l. The conclusion follows from
Theorem .
Proof of Theorem Whenq(x)≡+∞, we can replaceϕbyϕin () and definecas in (), then following the same procedures as in the proof of Theorem (ii), we need to show only that{un}is bounded inV. For this purpose, let{wn}be defined as in (). If{wn}is
bounded inV, we knowwn→wis a strong convergence inL(),wn→wis convergence
a.e.x∈,wnwis a weak convergence inV, andw∈V.
Ifun →+∞, thentn→ andw(x)≡. We set={x∈:w(x) = },={x∈:
w(x)= }. Obviously, by (),|un| →+∞a.e. in. Whenq(x)≡+∞in (S), there exists
K,K> andnlarge enough we have|f(xu,unn)| ≥K,|g(xu,nun)| ≥K uniformly in x∈. Hence, by () and (), we obtain
c= lim
n→+∞t
nun
= lim
n→+∞t
n
∇unL()+γunL()
= lim
n→+∞t
n
f(x,un)undx+
g(s,un)unds+γunL()
= lim
n→+∞
f(x,un)
un
wndx+
g(s,un)
un
wnds+tnγunL()
≥K
wdx+K
wds+wL().
Noticing thatw(x)= inandK,K can be chosen large enough, som≡ and thenw(x)≡ in.
Then we knowlimn→+∞
F(x,wn)dx+limn→+∞
G(s,wn)ds= , and consequently,
J(wn) =
∇wn
L()+o()
= wn
– wn
L()+o()
≥
– λ+
wn+o()
= c
– λ+
+o(). ()
Byun →+∞,tn→ asn→+∞, then it follows Lemma and (), we obtain
J(wn) =J(tnun)≤
+tn
n ≤c. ()
Obviously, () and () are contradictory. So{un}is bounded inV. This completes the
4 Example
In this section, we give two examples onf(x,u): One satisfies (S) to (S) withq(x)≡+∞, but does not satisfy the (AR) condition; the other illustrates how the assumptions on the boundary are not trivial and compatible with the inner assumptions in.
Example Set:
f(x,t) =
⎧ ⎨ ⎩
, t≤;
tln( +t), t> .
Then it is easy to verify thatf(x,t) satisfies (S) to (S) withp(x) = ast→ andq(x) = +∞ast→+∞. In addition,
F(x,t) = t
ln( +t) – t
+ t–
ln( +t).
So, for someμ> ,μF(x,t) =tln( +t)(μ –lnμ(+t)+tlnμ(+t)–μt) >tln( +t), for allt
large.
This meansf(x,t) does not satisfy the (AR) condition.
Example Consider the following problem:
⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩
–u(x) =αu(x), <x<l,
u() = ,
u(l) =αu(l),
()
whereα> is a constant. It is obvious thatg=γf asf(x,u) =αu(x). Problem () is a case of (); we can obtain the nontrivial solution:u(x) =Csin√αx,C= .
Competing interests
The author declares that he has no competing interests.
Author’s contributions
Li G carried out all studies in this article.
Acknowledgements
The author would like to thank the referees for carefully reading this article and making valuable comments and suggestions.
Received: 19 January 2012 Accepted: 6 August 2012 Published: 16 August 2012 References
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