# Positive Solutions of Two-Point Boundary Value Problems for Monge-Ampère Equations

Authors: Baoqiang Yan; Meng Zhang
Journal: Journal of Function Spaces
 (2015)

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## Abstract

This paper considers the following boundary value problem:((-u'(t))n)'=ntn-1f(u(t)),0<t<1,u'(0)=0,u(1)=0, where n>1 is odd. We establish the method of lower and upper solutions for some boundary value problems which generalizes the above equations and using this method we present a necessary and sufficient condition for the existence of positive solutions to the above boundary value problem and some sufficient conditions for the existence of positive solutions.


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## Body

## 1. Introduction

In [1], Kutev obtained the existence of a unique nontrivial solution of the following boundary value problem (BVP) (1)-u′tn′=ntn-1fut,0<t<1,u′0=0,u1=0,where n>1 and f(u)=up, ∀0<p≠n, which is related to the following Dirichlet problem of the Monge-Ampère equations in Rn(2)det⁡D2u=λf-uinB,u=0on∂B,where B=x∈Rn:x<1 is the unit ball in Rn and D2u=∂2u/∂xi∂xj is the Hessian of u.After Kutev’s works, the Monge-Ampère equation has attracted a growing attention in recent years because of its important role in several areas of applied mathematics. For instance, in [2], Hu and Wang obtained some sufficient conditions for the existence and multiplicity of the positive solutions for problem (1) and Dai [3] and Wang [4] discussed the unilateral global bifurcation results for the problem with f(u)=un+g(u). And there are some results on the existence and multiplicity and nonexistence of nontrivial radial convex solutions of systems of Monge-Ampère equations (see [5–8]) also. We notice that the nonlinearity f(u) is continuous at u=0 in the above works. For the case that f(u) is singular at u=0, there are fewer results for BVP (1). But some interesting results are presented for BVP (2) in [9–12] where f(u) is singular at u=0. We also refer to [7, 8, 13, 14] and references therein for further discussions regarding solutions of the Monge-Ampère equations.One goal in this paper is to consider the existence of positive solutions under the conditions thatn>1 and f(u) is singular at u=0. Compared to the results in [10], we do not need the monotonicity of f.Our paper is organized as follows. Section2 lists some lemmas. In Section 3, we obtain a theorem on the upper and lower solutions which is an extension of this method to the class of problem we consider. Section 4 is devoted to our main results on some sufficient conditions for positive solutions of the BVP (1).


## 2. Preliminaries

LetX be the Banach space C[0,1] with v=maxt∈[0,1]⁡vt.Define a setK⊆X by (3)
K
=
v
∈
X
:
v
t
is  nonincreasing  in
t
,
v
1
=
0
,
v
t
≥
v
1
-
t
∀
t
∈
0,1
.It can be easily verified that K is indeed a normal cone in X (see [2]).Lemma 1 (see [15, 16]).
LetE be a Banach space, and let D⊆E be a bounded, closed, and convex set. Assume that F:D→D is a continuous compact operator. Then, F has at least one fixed point in D.Remark 2.
Obviously, assume thatE is a Banach space, D⊆E is a bounded, closed, and convex set and let F:E→D be a continuous compact operator. Then, F has at least one fixed point in D also.Lemma 3.
Assume that functionv∈C[0,1] with v(t)≥0, v(0)=v=maxt∈0,1vt, and v′(t) decreasing in [0,1]. Then, (4)vt≥1-tv,t∈0,1.Proof.
Sincev′(t) is decreasing in [0,1], we have (5)vt-v0=∫0tv′sds≥tv′t,∀t∈0,1,v1-vt=∫t1v′sds≤1-tv′t,∀t∈0,1,which implies (6)1-tvt-v0=1-t∫0tv′sds≥1-ttv′t,∀t∈0,1,tv1-vt=t∫t1v′sds≤t1-tv′t,∀t∈0,1.Then, for t∈[0,1], we have (7)1-tvt-v0≥tv1-vt.That is, (8)vt≥1-tv0+tv1.Since v(0)=maxt∈[0,1]⁡vt and v(1)≥0, we have (9)vt≥1-tv.The proof is complete.Remark 4.
The idea of the proof of Lemma3 comes from Lemma 2.2 in [2].Now, we list some conditions for convenience:(C1)(10)f:0,∞⟶-∞,∞is  continuous;(C2) n is odd.Throughout this paper, we always assume that condition(C2) holds; that is to say, n is odd.


## 3. Upper and Lower Solutions

Consider the following two-boundary value problem:(11)-v′tn′=ntn-1fvt,0<t<1,v′0=a,v1=b,where f:D→R is continuous and D⊆R.Now, we give following definitions.Definition 5.
Ifα∈C[0,1]∩C10,1∩C2(0,1) and (12)-α′tn′≤ntn-1fαt,0<t<1,α′0=a,α1≤b,then α(t) is called a lower solution of the BVP (11).Similarly, we can define an upper solution of BVP (11).Definition 6.
Ifβ∈C[0,1]∩C10,1∩C2(0,1) and (13)-β′tn′≥ntn-1fβt,0<t<1,β′0=a,β1≥b,then β(t) is called an upper solution of BVP (11).Ifα(t)≤β(t), define (14)Dαβt≔x∈R:αt≤x≤βt∀t∈0,1.Theorem 7.
Letα,β∈C[0,1]∩C10,1∩C20,1, respectively, be lower and upper solutions of BVP (11) with α(t)≤β(t) on [0,1] and Dαβ(t)⊆D for all t∈(0,1) and (15)there  existsh∈L10,1with∫01ntn-1htdt<∞,such  thatsupx∈Dαβt⁡fx≤htfor  a.e.t∈0,1holds. Then, BVP (11) has at least one solution v∈C[0,1]∩C10,1∩C2(0,1) and (16)αt≤vt≤βt,t∈0,1.Proof.
Consider the modified boundary value problem of the form(17)-v′tn′=ntn-1Fvt,0<t<1,v′0=a,v1=b,where the function (18)Fvt=fαt+min⁡1,αt-vt,vt<αt,fvt,αt≤vt≤βt,fβt+min⁡1,vt-βt,vt>βt.
Obviously, if the problem (17) has a solution v(t) with α(t)≤v(t)≤β(t), t∈[0,1], then problem (11) has a solution v(t), t∈[0,1].
Forv∈X=C[0,1], define an operator (19)Avt=b+t-1a+∫t1∫0snτn-1Fvτdτ1/nds,0≤t≤1.A standard argument shows that A:X→{v:v≤a+b+(∫01‍nτn-1[h(τ)+1]dτ)1/n} is continuous and compact. Lemma 1 guarantees that A has a fixed point v∈C[0,1]∩C10,1∩C2(0,1); that is BVP (17) has a solution v(t).
Now, we claim that(20)αt≤vt≤βt,t∈0,1,which guarantees that v∈C[0,1]∩C10,1∩C2(0,1) is a solution of BVP (11).
Letx(t)=α(t)-v(t), t∈[0,1].
First, we show that(21)x0≤0,that  is,α0≤v0.
Suppose thatx(0)>0. Then, ∃t0>0 such that x(t)>0, t∈[0,t0]. Let t∗=sup{t:x(s)>0,s∈0,t}. Obviously, t∗≤1 and (22)αt>vt,t∈0,t∗,xt∗=0.Let y(t)=(-α′(t))n-(-v′(t))n, t∈(0,t∗). Then (23)y′t=-α′tn′--v′tn′≤ntn-1fαt-ntn-1fαt+min⁡1,αt-vt<0,t∈0,t∗,which implies that y(t) is decreasing on [0,t∗). And so y(t)<y(0) for all t∈(0,t∗); that is, (24)-α′tn--v′tn<-α′0n--v′0n=0.That is,(25)-α′tn<-v′tn,t∈0,t∗.From (C2), we have (26)-α′t<-v′t.That is, (27)x′t=α′t-v′t>0,which means that x(t) is increasing on [0,t∗). Then, x(t)>x(0)>0, t∈(0,t∗). The continuity of x(t) at t=t∗ means that x(t∗)≥x(0)>0, which contradicts (22).
Secondly, we prove that(28)αt≤vt,∀t∈0,1.
Assume thatv(t)<α(t) for some t∈(0,1). Since α(0)≤v(0) and α(1)≤v(1), there exists a t~∈(0,1) such that x(t~)=maxt∈[0,1]⁡xt=maxt∈[0,1]⁡αt-vt>0 with x′(t~)=0 (note α′(t~)=v′(t~)). Then, (29)-v′t~n′=nt~n-1Fvt~=nt~n-1fαt~+nt~n-1min⁡1,αt~-vt~>nt~n-1fαt~≥-α′t~n′.From the continuity of -v′tn′ and -α′tn′ in (0,1), there exists a δ>0 such that (30)-v′tn′>-α′tn′,t∈t~-δ,t~+δ⊆0,1.Integrating both sides of the above inequality from t~ to t∈(t~,t~+δ), we get (31)-v′tn--v′t~n>-α′tn--α′t~n,t∈t~,t~+δ.That is, (32)-v′tn>-α′tn,t∈t~,t~+δ.From (C2), we have (33)-v′t>-α′t,t∈t~,t~+δ.And so (34)α′t-v′t>0,t∈t~,t~+δ.Then, (35)x′t>0,t∈t~,t~+δ,which contradicts the fact that x(t~)=maxt∈[0,1]x(t). Hence, (28) is true.
The same argument shows that(36)vt≤βt,∀t∈0,1.
Consequently, (28) and (36) show that (20) holds. Then, v(t) is a solution to BVP (11).Remark 8.
The idea of the proof comes from [17, 18].


## 4. Main Results

Suppose thatK is defined in Section 2. Now, we list the following condition for convenience:(H1)  f:(0,+∞)→(0,+∞), continuous and ∃λ,μ,δ,(-∞<λ<0<μ<1,0<δ≤1), ∀v∈(0,+∞) such that (37)c0μfv≤fc0v≤c0λfv,0≤c0≤δ,c0λfv≤fc0v≤c0μfv,c0≥1δ.Remark 9.
Assume that(38)fv=v-α1+v-α2+vβ1+vβ2,v∈0,+∞,where α1,α2∈(0,+∞), β1,β2∈(0,1) are constants. Let δ=1, λ=min{-α1,-α2}, μ=max{β1,β2}. Then, if 0≤c0≤δ, we have (39)fc0v=c0v-α1+c0v-α2+c0vβ1+c0vβ2=c0-α1v-α1+c0-α2v-α2+c0β1vβ1+c0β2vβ2≤c0λv-α1+v-α2+vβ1+vβ2=cλfv,fc0v=c0v-α1+c0v-α2+c0vβ1+c0vβ2=c0-α1v-α1+c0-α2v-α2+c0β1vβ1+c0β2vβ2≥c0μv-α1+v-α2+vβ1+vβ2=cμfv.A similar argument shows that if c0≥1/δ, we have (40)c0λfv≤fc0v≤c0μfv.We have the following theorems.Theorem 10.
Suppose that(H1) holds and 2(μ-λ)/(1-μ)<1. Then, the necessary and sufficient condition of the existence of the positive solution in C1[0,1]∩C2(0,1) for BVP (1) is (41)0<∫01tn-1f1-tdt<∞.Proof.
Necessity. Suppose that v∈C1[0,1]∩C2(0,1) is a positive solution. We have (42)n-v′tn-1-v′′t=ntn-1fvt>0,t∈0,1,-∞<v′t=-∫0tnτn-1fvτdτ1/n<0,t∈0,1,vt=∫t1∫0snτn-1fvτdτ1/nds≤1-t∫01nτn-1fvτdτ1/n.Let (43)Mv=∫01nτn-1fvτdτ1/n.From v(t)>0 and v(1)=0, Lemma 3 implies that v∈K and (44)v1-t≤vt≤Mv1-t.Let c1>0 be a constant such that c1Mv≤δ, c1≤δ. By (37) and (44), we have (45)fvt=f1c1c1vt1-t1-t≥1c1λfc1vt1-t1-t≥c1μ-λvt1-tμf1-t≥c1μ-λvμf1-t,t∈0,1.That is, (46)ntn-1fvt≥nc1μ-λvμtn-1f1-t,t∈0,1.Since (47)-v′1n=∫01nsn-1fvsds<+∞we have from (46) (48)0<∫01ntn-1f1-tdt≤∫01nc1λ-μv-μtn-1fvtdt=c1λ-μv-μ-v′1n<+∞.Hence, (41) holds.
Sufficiency. Since (41) holds, set (49)ht≔∫t1∫0snτn-1f1-τdτ1/nds,0≤t≤1.Obviously h(t)≥0 and h′(t) are decreasing in [0,1]. Then, Lemma 3 means that h(t)≥h(1-t). From (41), we know that there exists N>0 such that ∫01‍ntn-1f(1-t)dt≤N. Consequently, (50)a11-t≤ht≤a21-t,where a1=min⁡1,h, a2=max⁡{1,N1/n}.
Letα(t)≔k1h(t) and βt≔k2h(t), where c2≥1/a1δ is big enough such that k1=(1/2)(c2λ-μa1λ)1/(1-μ)≤1/2, k2=2(c2μ-λa1μ)1/(1-μ)≥2. Obviously, (51)0≤αt≤βt,t∈0,1.Since (note (50)) (52)fc2ht=fc2ht1-t1-t≥c2λht1-tλf1-t≥c2λa1λf1-t,t∈0,1,we have (note k1/c2≤δ) (53)fαt=fk1ht≥k1c2μfc2ht≥k1μc2λ-μa1λf1-t.From (54)k1=12c2λ-μa1λ1/1-μ,we can obtain (55)k11-μ=121-μc2λ-μa1λ.That is, (56)c2λ-μa1λ=k11-μ12μ-1.And so (53) is (57)fαt≥12μ-1k11-μk1μf1-t≥k11-μk1μf1-t=k1f1-t,t∈0,1.Similarly, we have (58)fβt=fk2ht≤1c2λfc2k2ht≤k2f1-t.Differentiating α(t) on (0,1), we have (59)α′t=-k1∫0tnτn-1f1-τdτ1/n,-α′tn=k1n∫0tnτn-1f1-τdτ,-α′tn′=k1nntn-1f1-t≤k1ntn-1f1-t≤ntn-1fαt,with α′(0)=0, α(1)=0, which implies that α(t) is a lower solution to BVP (1).
Similarly, we have(60)-β′tn′≥ntn-1fβt,with β′(0)=0, β(1)=0. Hence, β(t) is an upper solution to BVP (1).
Theorem7 guarantees that BVP (1) has at least one positive solution in C1[0,1]∩C2(0,1).Remark 11.
Idea of the proof comes from [19].Theorem 12.
Assume that(C1) holds and the following conditions are true:(1)
there existL>0 and ɛ>0 such that (61)fx>L,∀x∈0,ɛ;(2)
there existsM>0 such that (62)supx≥M⁡fx<+∞.Then, BVP (1) has at least one positive solution v∈C[0,1]∩C1[0,1)∩C2(0,1). If, moreover, f is decreasing, then the solution is unique.Proof.
By assumption (1), there isɛm>0 such that (63)fx>L,∀x∈0,ɛm.Without restriction, we can assume that {ɛm} is a decreasing sequence and limm→+∞⁡ɛm=0 with ɛm<min⁡{1,L} for all m∈N.
Let(64)γt≔∫t1∫0snτn-11-τdτ1/nds,t∈0,1.It is easy to see that (65)0<γt,t∈0,1,γ∈C20,1.Let (66)k0≔12+γɛ1,which guarantees that (67)0≤k0γt<ɛ1,t∈0,1.
Set(68)α1t:≡ɛ1,t∈0,1.From (63), we have (69)-α1′tn′=0<ntn-1L<ntn-1fɛ1=ntn-1fα1t,t∈0,1,α1′0=0,α11=ɛ1.From (C1) and condition (2), we choose an M′>max{ɛ1,M} such that (70)supx∈M′,+∞⁡fx=R<+∞.Let (71)β1t≔M′+∫t1∫0snτn-1M′+R+1dτ1/nds,t∈0,1.Equation (70) guarantees that (72)fβ1t≤R,∀t∈0,1,-β1′tn′=ntn-1M′+R+1≥ntn-1fβ1t,t∈0,1,β1′0=0,β11=M′≥ɛ1.Now, Theorem 7 implies that there exists u1∈C[0,1]∩C1[0,1)∩C2(0,1) such that (73)-u1′tn′=ntn-1fu1t,t∈0,1,u1′0=0,u11=ɛ1.Let α2(t)≔ɛ2, t∈[0,1], and β2(t)≔u1(t), t∈[0,1]. It is easy to prove that (74)-α2′tn′=0<ntn-1L<ntn-1fɛ2=ntn-1fα2t,t∈0,1,α2′0=0,α21=ɛ2,-β2′tn′=ntn-1fβ2t,t∈0,1,β2′0=0,β21=ɛ1>ɛ2.Theorem 7 guarantees that there exists u2∈C[0,1]∩C1[0,1)∩C2(0,1) such that (75)-u2′tn′=ntn-1fu2t,t∈0,1,u2′0=0,u21=ɛ2.Proceeding the above proof, we can get a sequence {um(·)} such that (76)ɛm≤umt≤um-1t,t∈0,1,-um′tn′=ntn-1fumt,t∈0,1,um′0=0,um1=ɛm.
Now, we claim that(77)umt≥k0γt,t∈0,1,m∈N.
First, we show that(78)um0≥k0γ0,um1≥k0γ1,m∈N.
Obviously,um(1)=ɛm>0=k0γ(1). Suppose um(0)<k0γ(0). Let h(t)≔um(t)-k0γ(t), t∈[0,1], and t′=sup{t:h(s)<0,s∈[0,t)}. Obviously, t′<1 (note um(1)≥k0γ(1)) and h(t′)=0, h(t)<0, for all t∈[0,t′), which implies that (79)ɛm≤umt<k0γt<ɛ1,t∈0,t′.For t∈(0,t′), condition (1) guarantees that (80)-um′tn′=ntn-1fumt>ntn-1L>ntn-1k01-t=-k0γ′tn′.Integrate (80) from 0 to t∈(0,t′) to obtain (81)-um′tn>-k0γ′tn,t∈0,t′.From (C2), we have (82)-um′t>-k0γ′t,t∈0,t′.That is, (83)h′t=um′t-k0γ′t<0,t∈0,t′,which means that h(t) is decreasing on (0,t′]. The continuity of h(t) at t′ means that h(t′)<0. This contradicts that h(t′)=0. Hence, (78) is true.
Secondly, we show that(84)umt≥k0γt,∀t∈0,1,m∈N.
Suppose that there is at∈(0,1) such that h(t)<0. There exists a t0∈(0,1) with h(t0)=mint∈[0,1]h(t)<0 and h′(t0)=0; that is, um′(t0)=k0γ′(t0). Now, h(t0)<0 implies that ɛm≤um(t0)<k0γ(t0)<ɛ1, which together with condition (1) guarantees that (85)-um′t0n′=nt0n-1fumt0>nt0n-1L>nt0n-1k01-t0=-k0γ′t0n′.The continuity of -um′tn′ and -k0γ′tn′ at t=t0 guarantees that there exists a δ>0 such that (86)-um′tn′>-k0γ′tn′,t∈t0-δ,t0+δ.Integrate (86) from t0 to t∈(t0,t0+δ) to obtain (87)-um′tn>-k0γ′tn,t∈t0,t0+δ.From (C2), we have (88)-um′t>-k0γ′t,t∈t0,t0+δ.That is, (89)h′t=um′t-k0γ′t<0,t∈t0,t0+δ,which means that h(t) is decreasing on (t0,t0+δ). This contradicts that h(t0)=mint∈[0,1]⁡h(t). Consequently, (84) is true.
Consider now the pointwise limit as follows:(90)u∗t≔limm→+∞⁡umt,t∈0,1.It is clear that (91)k0γt≤u∗t≤umt,∀t∈0,1,which means that (92)u∗1=0.For any 1>ϵ>0, from (91) and (93)-um′tn′=ntn-1fumt,∀t∈0,1,we have (94)sup⁡ntn-1fx:t∈0,1-ϵ,k0γt≤x≤u1t<+∞.That is, ∀t∈[0,1-ϵ](95)
-
u
m
′
t
n
′
≤
sup
⁡
n
t
n
-
1
f
x
:
t
∈
0,1
-
ϵ
,
k
0
γ
t
≤
x
≤
u
1
t
<
+
∞
.The Ascoli-Arzela theorem guarantees that {um(t)} converges uniformly u∗(t) on [0,1-ϵ] and {um′(t)} converges uniformly u∗′(t) on [0,1-ϵ]. Since ϵ>0 is arbitrary, we find that (96)u∗∈C10,1,-u∗′tn′=ntn-1fu∗t,t∈0,1,u∗′0=0.It remains only to prove the continuity of u∗(t) at t=1.
Letɛ>0 be given. Take umɛ(t) with umɛ(0)<ɛ. Since umɛ(t) is continuous at t=1, we can find δ>0 such that (97)0<umɛt<ɛ,∀t∈1-δ,1.From (91), we obtain that (98)0<u∗t<ɛ,t∈1-δ,1,which together with (92) means that u∗(t) is continuous at t=1. From (92) and (96), u∗ is positive solution to BVP (1) in C1[0,1)∩C2(0,1).
Now, we show that the uniqueness ofu∗ follows when f is decreasing on (0,+∞).
Suppose that, by contradiction,v1 and v2 are two different solutions. Obviously (99)v11=v21=0.
First, we show that(100)v10=v20.
In fact, ifv1(0)<v2(0), let t∗=sup{t:v1(s)<v2(s),s∈(0,t)}. Obviously 1≥t∗>0, v1(t)<v2(t) for all t∈[0,t∗] and v1(t∗)=v2(t∗) (notice v1(1)=v2(1)=0). The monotonicity of f implies that (101)-v1′tn′=ntn-1fv1t>ntn-1fv2t=-v2′tn′,∀t∈0,t∗.Integrate (101) from 0 to t to obtain (102)-v1′tn>-v2′tn,∀t∈0,t∗.From (C2), we have (103)-v1′t>-v2′t,∀t∈0,t∗.That is, (104)v1t-v2t′<0,∀t∈0,t∗.Then, v1(t)-v2(t) is decreasing on (0,t∗), which together with v1(0)-v2(0)<0 means that v1(t∗)-v2(t∗)<0. This is a contradiction. Hence, (100) is true.
Secondly, we show that(105)v1t≡v2t,t∈0,1.
Suppose that there existst0∈(0,1) with v1(t0)≠v2(t0). Without loss of generality, we suppose that v1(t0)<v2(t0). Let t′=mint∈[0,1](v1(t)-v2(t)). Equation (100) guarantees that t′∈(0,1), v1(t′)<v2(t′), and v1′(t′)=v2′(t′). The monotonicity of f implies that (106)-v1′t′n′=nt′n-1fv1t′>nt′n-1fv2t′=-v2′t′n′.The continuity of -v1′tn′ and -v2′tn′ at t=t′ implies that there exists a δ>0 such that (107)-v1′tn′>-v2′tn′,∀t∈t′-δ,t′+δ.Integrate (107) from t′ to t∈(t′,t′+δ) to obtain (108)-v1′tn>-v2′tn,∀t∈t′,t′+δ.Now, (C2) implies that (109)-v1′t>-v2′t,∀t∈t′,t′+δ.That is, (110)v1t-v2t′<0,∀t∈t′,t′+δ.Then, v1(t)-v2(t) is decreasing on t∈(t′,t′+δ), which contradicts t′=mint∈[0,1]⁡(v1(t)-v2(t)). Hence, (105) is true.



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*Source: 620251-2015-08-20.xml*