This paper considers the existence of positive solutions of the boundary value problems v′′+λ(vr+vp-vq)=0 and v(-1)=v(1)=0, where p>r>q>-1 and λ is a positive parameter. Using a time-map approach, we obtain the exact number of positive solutions in different cases.

1. Introduction and Main Results

The study of multiplicity results to boundary value problem
(1)v′′+λf(v)=0,t∈(a,b),v(a)=v(b)=0,
where λ>0 is a positive parameter, is very interesting because of its applications. As we know, when f(v)=vp(x)+vq(x), the boundary value problem
(2)-v′′(x)=μvp(x)+vq(x),a≤x≤b,v(x)>0,x∈(a,b),v(a)=v(b)=0,
where 0<q<1<p and k≥0 are fixed given numbers and μ>0 is a parameter, comes from the elliptic problem
(3)-Δu=λuq+up,x∈Ω,u>0,x∈Ω,u=0,x∈∂Ω,
with 0<q<1<p, which was raised by Ambrosetti et al. in [1].

Under different assumptions on f, there are many results for the above problems and elliptic equations (see [2–8]).

In [9, 10], Liu considered the case of f(v)=vp+(1/λ)vq and f(v)=vp+vq+kv and gave the exact number of solutions and many interesting properties of the solutions.

Cheng [11] investigated the following two-point boundary value problem:
(4)-y′′=λ(yp-yq),t∈(-1,1),y(-1)=y(1)=0,
where λ>0 is a positive parameter and p>q>-1 and got the exact number of positive solutions.

Now, in this paper we consider the more general case
(5)-v′′=λ(vr+vp-vq),t∈(-1,1),v(-1)=v(1)=0,
where λ>0 is a positive parameter, p>r>q>-1, and (1/(r+1))+(1/(p+1))-(1/(q+1))<0.

Define β, where β satisfies
(6)βrr+1+βpp+1-βqq+1=0.

For p>r>q and -1<q<1, let λ1 be given by(7)λ1=r+12β1-rvv×(∫01dttq+1[(1-tr-q)+((r+1)/(p+1))βp-r(1-tp-q)])2.The main results of this paper are as follows.

Theorem 1.

If p>r>q≥1, (5) has exactly one positive solution for any λ>0.

Theorem 2.

If p>r≥1>q>-1, (5) has exactly one positive solution for λ∈(0,λ1] and none for λ>λ1.

Theorem 3.

If p>1>r>q≥0, (5) has exactly one positive solution for λ∈(0,λ1] and none for λ>λ1.

Theorem 4.

If (1/3)≥p>r>q≥0, (5) has exactly one positive solution for λ∈(λ1,+∞) and none for λ<λ1.

Theorem 5.

Assume that 1>p>r>0>q>-1. Define
(8)θ(r,p,q)=∫011-r[tq+1(1-m1βr-qtr-q-m2βp-qtp-q)]1/2dt-(r-q)∫011-tq+1[tq+1(1-m1βr-qtr-q-m2βp-qtp-q)]3/2dt-(p-r)(q+1)p+1βp-q×∫011-tp+1[tq+1(1-m1βr-qtr-q-m2βp-qtp-q)]3/2dt,
where m1=(q+1)/(r+1) and m2=(q+1)/(p+1). One has the following.

If θ(r,p,q)≥0, (5) has exactly one positive solution for λ∈(λ1,+∞) and none for λ∈(0,λ1).

If θ(r,p,q)<0, there exists λ0∈(0,λ1) such that (5) has exactly two positive solutions for λ∈(λ0,λ1], exactly one for λ∈(λ1,+∞) or λ=λ0, and none for λ∈(0,λ0).

2. The Proofs of Theorems <xref ref-type="statement" rid="thm1.1">1</xref>–<xref ref-type="statement" rid="thm1.4">4</xref>

We assume throughout this section that p>r>q>-1 and
(9)1r+1+1p+1-1q+1<0.
Denote that E=(β,+∞), where β is given by (6). For a≥β and t∈(0,1), let
(10)P(a,t)=∫ata(zr+zp)dz,Q(a,t)=∫atazqdz.
Define a function F:E→(0,+∞) as
(11)F(a)=∫0a[2∫va(zr+zp-zq)dz]-1/2dvfora∈E.
It is clear that ∫0β(zr+zp-zq)dz=0. Now we claim that ar+ap-aq>0 for a≥β and ∫0a[2∫va(zr+zp-zq)dz]-1/2dv<∞ if and only if a∈E.

Let g1(z)=zp and g2(z)=zr-zq; then f(z)=g1(z)+g2(z). Immediately, we get that g2′(x)=rzr-1-qzq-1=zq-1(rzr-q-q). In order to judge the sign of g2′(x), we just judge the sign of T(z)=rzr-q-q. Since T(0)=-q, T(1)=r-q, and T′(z)=r(r-q)zr-q-1, we can get the following two results.

For r>q>0, the function g2(z) has a stable point z0=(q/r)1/(r-q). When z∈(0,z0), we have g2′(z)<0, and when z∈(z0,+∞), we have g2′(z)>0.

For r>0>q, we have g2′(z)>0 on z∈(0,+∞). Combining g1′(x)=pzp-1 and g1′′(x)=p(p-1)zp-2 with the above two results, we obtain the monotony of f(z).

For p≥1>r>q>0, the function f(z) has a stable point z* on (0,1). When z∈(0,z*), we have f′(z)<0 and when z∈(0,z*), we have f′(z)>0.

For 1>p>r>q>0, we have that f′(z)>0 on (0,+∞).

For p>r>0>q or p>0>r>q, we have that f′(z)>0 on (0,+∞).

Then from ∫0β(zr+zp-zq)dz=0, we infer that ar+ap-aq>0 for a≥β.

We consider the integration ∫0a[2∫va(zr+zp-zq)dz]-1/2dv. It is clear that v=a is a flaw point. Since ∫0a[2∫va(zr+zp-zq)dz]-1/2dv<∫0a[2∫va(zp-zq)dz]-1/2dv on E, we consider the integration ∫0a[2∫va(zp-zq)dz]-1/2dv.

Using Lagrange theorem, we obtain that
(12)∫0a[2∫va(zp-zq)dz]-1/2dv=∫0a[2(ap+1p+1-vp+1p+1)-2(aq+1q+1-vq+1q+1)]-1/2dv=∫0a{2[v+θ1(a-v)]p(a-v)-2[v+θ2(a-v)]q(a-v)}-1/2dv=∫0a[2(a-v)]-1/2{[v+θ1(a-v)]p-[v+θ2(a-v)]q}-1/2dv<∞,
where θ1,θ2∈(0,1) are constants.

The following Lemma 6 is listed to show that to study the number of positive solutions of (5) is equivalent to study the shape of the map F(a) on E. Lemmas 7–9 show some properties of F(a) on E.

Lemma 6.

Let u(a,t) be the unique solution of the problem
(13)0≤u(t)≤a,∫u(t)a[2∫va(zr+zp-zq)dz]-1/2dv=F(a)t,t∈[0,1],
where a∈E. One has the following.

If λ>0 and v is a positive solution of (5), v(0)∈E, F(v(0))=λ, and v(t)=u(v(0),|t|) for t∈[-1,1].

If a∈E and F(a)=λ, v(t)=u(v(0),|t|) is a positive solution of (5) with v(0)=a.

Proof.

(1) Assume that λ>0 and v is a positive solution of (5). Let τ∈[-1,1] satisfy v(τ)=maxt∈[-1,1]v(t)=a. It follows from v′(τ)=0 that
(14)[v′(s)]2=∫τs2v′′(η)v′(η)dη=2λ∫v(s)a(zr+zp-zq)dz,s∈(-1,1).
This implies that a≥β and v′(s)≠0 if v(s)<a. And combine v′′(τ)=-λ(ar+ap-aq)<0 to obtain
(15)v′(s)>0fors∈(-1,τ),v′(s)<0fors∈(τ,1).
Then, we have that
(16)v′(s)[2∫v(s)a(zr+zp-zq)dz]-1/2=λ,s∈(-1,τ),-v′(s)[2∫v(s)a(zr+zp-zq)dz]-1/2=λ,s∈(τ,1).
It follows that
(17)∫v(t)a[2∫va(zr+zp-zq)dz]-1/2dv=(τ-t)λ,t∈[-1,τ],(18)∫v(t)a[2∫va(zr+zp-zq)dz]-1/2dv=(t-τ)λ,t∈[τ,1],(19)(1-τ)λ=∫0a[2∫va(zr+zp-zq)dz]-1/2dv=(τ+1)λ.
From (19) and a≥β we have that τ=0. With (17) and (18) we obtain the result (1) of this theorem.

(2) Since u(a,0)=a>0 and u(a,t) is a positive solution of the boundary value problem
(20)u′′+[F(a)]2(ur+up-uq)=0,t∈(0,1),u′(0)=0,u(1)=0,
we have that v(t)=u(a,|t|) is a positive solution of (5).

Lemma 7.

F is differentiable on (β,∞), and
(21)F(a)=142∫01H0(a,t)dt,a>β,(22)F′(a)=142∫01H1(a,t)dt,a>β,
where
(23)H0(a,t)=4a[P(a,t)-Q(a,t)]-1/2,H1(a,t)=2[P(a,t)-Q(a,t)]-3/2×[1-rr+1ar+1(1-tr+1)+1-pp+1ap+1(1-tp+1)-1-qq+1aq+1(1-tq+1)].

Proof.

Equation (21) can be obtained by (11), immediately. From
(24)∂P(a,t)∂a=∂∂a×[1-rr+1ar+1(1-tr+1)+1-pp+1ap+1(1-tp+1)]=ar(1-tr+1)+ap(1-tp+1),∂Q(a,t)∂a=∂∂a[1q+1aq+1(1-tq+1)]=aq(1-tq+1),
we have that
(25)∂H0(a,t)∂a=2[P(a,t)-Q(a,t)]-3/2×[1-rr+1ar+1(1-tr+1)+1-pp+1ap+1(1-tp+1)-1-qq+1aq+1(1-tq+1)]=H1(a,t).
It follows from (21) that F is differentiable on (β,∞) and (22) is true.

Lemma 8.

Consider the following:(26)lima→βF(a)={+∞,q≥1,λ1,q<1,
where λ1 is given by (7).

Proof.

From
(27)lima→β∫βadv2∫va(zr+zp-zq)dz=0
and the Lebesgue theorem, we have that
(28)lima→βF(a)=lima→β∫0βdv2∫va(zr+zp-zq)dz=∫0βdv2∫vβ(zr+zp-zq)dz.
On the other hand, from ∫0β(zr+zp-zq)dz=0, we can obtain that
(29)∫0βdv2∫vβ(zr+zp-zq)dz=∫0βdv2∫0v(zq-zr-zp)dz=∫01(βdt)×(2[(βt)q+1q+1-(βt)r+1r+1-(βt)p+1p+1])-1=∫01(βdt)×(2[βr+1r+1tq+1+βp+1p+1tq+100000000000-(βt)r+1r+1-(βt)p+1p+1])-1/2=r+12β(1-r)/2×∫01dttq+1[(1-tr-q)+((r+1)/(p+1))βp-r(1-tp-q)].
This completes the proof of Lemma 8.

Lemma 9.

Consider the following:(30)lima→+∞F(a)={0,p>1,r≥1orp>1,r<1,π22,p=1,r=1,+∞,p<1,r<1.

Proof.

From
(31)lima→+∞Q(a,t)P(a,t)=lima→+∞aq+1(1-tq+1)/(q+1)ar+1(1-tr+1)/(r+1)+ap+1(1-tp+1)/(p+1)=lima→+∞((1-tq+1)(r+1)(p+1))000000000×((q+1)(p+1)ar-q(1-tr+1)0000000000000+(q+1)(r+1)ap-q(1-tp+1))-1=0,
we have that(32)lima→+∞aP(a,t)-Q(a,t)=lima→+∞aP(a,t)=lima→+∞1(1/(r+1))ar-1(1-tr+1)+(1/(p+1))ap-1(1-tp+1).From Lemma 7 and the Lebesgue theorem, we can obtain the results of this lemma.

In the following section, we give the proofs of Theorems 1–5. For convenience, we denote that
(33)I(a,t)=1-rr+1ar+1(1-tr+1)+1-pp+1ap+1(1-tp+1)-1-qq+1aq+1(1-tq+1).
Hence from Lemma 7, we have that
(34)F′(a)=142∫012(P-Q)-3/2I(a,t)dt,fora>β.

Proof of Theorem <xref ref-type="statement" rid="thm1.1">1</xref>.

From p>r>q≥1, we obtain that
(35)I(a,t)<1-rr+1ar+1(1-tr+1)+1-pp+1ap+1(1-tp+1)-1-qq+1ap+1(1-tp+1)=1-rr+1ar+1(1-tr+1)+2(q-p)(p+1)(q+1)ap+1×(1-tp+1)<0,
for a>β, t∈(0,1). Thus F′(a)<0 for a>β. By Lemmas 6, 8, and 9, we have the results of this theorem.

Proof of Theorem <xref ref-type="statement" rid="thm1.2">2</xref>.

From p>r≥1>q>-1, (33), and (34), we have that F′(a)<0 for a>β. It follows from Lemmas 6, 8, and 9 that the results of Theorem 2 hold.

Proof of Theorem <xref ref-type="statement" rid="thm1.3">3</xref>.

Conditions p>1>r>q≥0 and (1/(r+1))+(1/(p+1))-(1/(q+1))<0 imply that rp-1>q(r+p+2)>0. With (33), we obtain that
(36)I(a,t)<1-rr+1ap+1(1-tp+1)+1-pp+1ap+1(1-tp+1)-1-qq+1aq+1(1-tq+1)<2(1-rp)(p+1)(r+1)ap+1(1-tp+1)-1-qq+1aq+1×(1-tq+1)<0,
for a>β, t∈(0,1), which means that F′(a)<0 for a>β. Thus by Lemmas 6, 8, and 9, we have the results of this theorem.

Proof of Theorem <xref ref-type="statement" rid="thm1.4">4</xref>.

From (1/3)≥p>r>q≥0 and (33), we have that
(37)I(a,t)>1-rr+1aq+1(1-tq+1)+1-pp+1aq+1(1-tq+1)-1-qq+1aq+1(1-tq+1)>(21-pp+1-1-qq+1)aq+1(1-tq+1)=q(3-p)+(1-3p)(p+1)(q+1)aq+1(1-tq+1)>0,
for a>β, t∈(0,1). Hence F′(a)>0 for a>β. By Lemmas 6, 8, and 9, we have the results of this theorem.

3. The Proof of Theorem <xref ref-type="statement" rid="thm1.5">5</xref>

In this section we always assume that 1>p>r>0>q>-1. Denote that
(38)S(a,t)=P(a,t)Q(a,t),a>β,t∈(0,1),(39)h1(s)=(s-1)(s-n),s∈(-∞,∞),(40)h2(s)=-3(s-m)(s-n)+2(s-1)(s-l),s∈(-∞,∞),
where P(a,t), Q(a,t), and β are given by (10), (6), and
(41)m=m(t)=q+1r+1-(p-r)(q+1)(p+1)(r+1)ap-q1-tp+11-tq+1,a>β,t∈(0,1),n=n(t)=1-q1-r+(p-r)(q+1)(p+1)(1-r)ap-q1-tp+11-tq+1,a>β,t∈(0,1),l=l(t)=1-q21-r2+(p2-r2)(q+1)(p+1)(1-r2)ap-q1-tp+11-tq+1,a>β,t∈(0,1).

Remark 10.

From 1>p>r>0≥q>-1, (6), and (41), it is obvious that m<1, n>1, and l<n.

Lemma 11.

For a>β and t∈(0,1),
(42)q+1r+1ar-q+q+1p+1ap-q<P(a,t)Q(a,t)<ar-q+ap-q.

Proof.

Let
(43)K(t)=arr+1[(r-q)tr+q+1+(q+1)tq-(r+1)tr]+app+1[(p-q)tp+q+1+(q+1)tq-(p+1)tp].
Condition 1>p>r>0>q>-1 implies that
(44)K′(t)=arr+1[tr-1(r-q)(r+q+1)tr+q+(q+1)qtq-1-(r+1)rtr-1]+app+1[tp-1(p-q)(p+q+1)tp+q+(q+1)qtq-1-(p+1)ptp-1]<arr+1[(r-q)(r+q+1)+(q+1)q-(r+1)r]tr+q+app+1[(p-q)(p+q+1)+(q+1)q-(p+1)p]tp+q=0,t∈(0,1).
With K(1)=0 and (44), we have that K(t)>0 for t∈(0,1).

It follows that
(45)∂∂tP(a,t)Q(a,t)=q+1(1-tq+1)2a-qK(t)>0,t∈(0,1).
Combining
(46)limt→0P(a,t)Q(a,t)=q+1r+1ar-q+q+1p+1ap-q,limt→1P(a,t)Q(a,t)=ar-q+ap-q,
we have the results of this lemma.

Lemma 12.

For 1>p>r>0>q>-1,
(47)lima→βF′(a)=122(q+1βq+1)1/2θ(r,p,q).

Proof.

From Lemma 7 we have that
(48)22F′(a)=∫01((1-rr+1ar+1(1-tr+1)+1-pp+1ap+1(1-tp+1)0000000000000-1-qq+1aq+1(1-tq+1))000000000×([P(a,t)-Q(a,t)]3/2)-11-pp+1)dt=∫011-r[P(a,t)-Q(a,t)]1/2dt-r-qq+1aq+1∫011-tq+1[P(a,t)-Q(a,t)]3/2dt-p-rp+1ap+1∫011-tp+1[P(a,t)-Q(a,t)]3/2dt.
It follows from Lebesgue theorem that
(49)lima→β∫011-r[P(a,t)-Q(a,t)]1/2dt=∫011-r[P(β,t)-Q(β,t)]1/2dt,lima→βr-qq+1aq+1∫011-tq+1[P(a,t)-Q(a,t)]3/2dt+lima→βp-rp+1ap+1∫011-tp+1[P(a,t)-Q(a,t)]3/2dt=r-qq+1βq+1∫011-tq+1[P(β,t)-Q(β,t)]3/2dt+p-rp+1βp+1∫011-tp+1[P(β,t)-Q(β,t)]3/2dt.
Finally, ∫0β(zr+zp-zq)dz=0 implies that
(50)P(β,t)-Q(β,t)=∫0βt(zq-zr-zp)dz=(βt)q+1q+1(1-m1βr-qtr-q-m2βp-qtp-q).
Combing (49) and (50), we complete the proof of this lemma.

Lemma 13.

F(a) has continuous derivatives up to second order on (β,∞) as follows:
(51)F′(a)=1-r22∫01G(a,t)h1(S(a,t))dt,a>β,(52)F′′(a)=1-r24a2∫01G(a,t)h2(S(a,t))dt,a>β,
where G(a,t)=[S(a,t)-1]-5/2[Q(a,t)]-1/2.

Proof.

Equation (51) can be obtained by
(53)H1(a,t)=2[P(a,t)-Q(a,t)]-3/2×[p-rp+1(1-r)P(a,t)-(1-q)Q(a,t)-p-rp+1ap+1(1-tp+1)]=2[P(a,t)Q(a,t)-1]-5/2[Q(a,t)]-1/2×(1-r)[P(a,t)Q(a,t)-1-q1-r-(p-r)(q+1)(p+1)(1-r)×ap-q1-tp+11-tq+1][P(a,t)Q(a,t)-1]=2(1-r)G(a,t)h1(S(a,t))
and Lemma 7, immediately. From (24) we have that
(54)a∂H1(a,t)∂a=-3(P-Q)-5/2×[p-rp+1(r+1)P-(q+1)Q+p-rp+1ap+1(1-tp+1)]×[(1-r)P-(1-q)Q-p-rp+1ap+1×(1-tp+1)p-rp+1]+2(P-Q)-3/2×[p2-r2p+1(1-r2)P-(1-q2)Q-p2-r2p+1ap+1(1-tp+1)]=(PQ-1)-3/2(1-r2)×{-3[P-q+1r+1Q+(p-r)(p+1)(r+1)×ap+1(1-tp+1)q+1r+1]×[P-1-q1-rQ-(p-r)(p+1)(1-r)ap+1×(1-tp+1)1-q1-r]+2(P-Q)×[P-1-q21-r2Q-(p2-r2)(p+1)(1-r2)×ap+1(1-tp+1)1-q21-r2]}=(1-r2)G(a,t)h2(S(a,t)),
and so (52) holds.

Lemma 14.

There exist δ(r,p,q)>0 and η(r,p,q)>0 such that
(55)F′′(a)+δ(r,p,q)F′(a)1+r2a>η(r,p,q)(ar-q+ap-q)-5/2a-(3+q)/2fora>β.

Proof.

From (41) and 1>p>r>0>q>-1, we have that
(56)mins∈[1,n]h2(s)=min{h2(1),h2(n)}=(n-1)min{3(1-m),2(n-l)}≥2(n-1)min{(1-m),(n-l)}>2(n-1)min{r-qr+1,(1-q)(r-q)1-r2}=2(n-1)r-qr+1.
And since h1((1+n)/2)=-((n-1)2/4)<0, let δ(r,p,q)>0 satisfy
(57)mins∈[1,n][h2(s)+δ(r,p,q)h1(s)]=2(n-1)r-qr+1.
Set
(58)h(s)=h2(s)+δ(r,p,q)h1(s),s∈(-∞,∞).
From h(s)=(δ-1)s2+(3m+3n-2l-2-δ-δn)s-3mn+2l+δn, we have that δ(r,p,q)-1>0 and
(59)h(s)≥mins∈[1,n]h(s)=2(n-1)r-qr+1,s∈(-∞,∞).
On the other hand, from (10), (38), and Lemma 11, we have that
(60)0<Q(a,t)<1q+1aq+1,0<q+1aq+1∫0a(zr+zp-zq)dz=q+1r+1ar-q+q+1p+1ap-q-1<S(a,t)-1<S(a,t)<ar-q+ap-q,
for a>β, t∈(0,1).

This means that
(61)G(a,t)=[S(a,t)-1]-5/2[Q(a,t)]-1/2>(ar-q+ap-q)-5/2a-(1+q)/21+q,
for a>β, t∈(0,1). It follows from Lemma 13 and (58) that
(62)F′′(a)+δ(r,p,q)F′(a)1+r2a=1-r24a2∫01G(a,t)h(S(a,t))dt.
Now, from (59)–(62) we have (55), where
(63)η(r,p,q)=(1-r)(r-q)1+q22∫01[n(t)-1]dt.

Lemma 15.

If θ(r,p,q)≥0, then F′(a)>0 for a>β. If θ(r,p,q)<0, then there exists a*>β, such that F′(a)<0 for a∈(β,a*) and F′(a)>0 for a>a*.

Proof.

It follows from Lemma 14 that if
(64)F′(a)=0,thenF′′(a)>0,
and then
(65)F′(a)has at most one zero in(β,∞).
By (40), (42), and (51) we can obtain that
(66)F′(a)>0forP(a,t)Q(a,t)>n(t).
With (10) and (41), (66) implies that
(67)F′(a)>0forq+1r+1ar+(q+1p+1-p-r1-r)ap>1-q1-raq.
If θ(r,p,q)≠0, then from Lemma 12 and (64)–(67) we have the results of this lemma, immediately. If θ(r,p,q)=0, then by Lemmas 12 and 14 we have that F′′(a)>0 for a near β, and so F′(a)>0 for a near β. Thus, it follows from (64)–(67) that F′(a)>0 for a>β.

Proof of Theorem <xref ref-type="statement" rid="thm1.5">5</xref>.

From Lemmas 6, 8, 9, and 15, we can obtain the results of Theorem 5.

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