In this paper, we were concerned with the global behavior of positive solutions for third-order semipositone problems with an integral boundary condition
y‴+βy″+αy′+λf(t,y)=0,t∈(0,1),y(0)=y′(0)=0,y(1)=χ∫10y(s)ds,
where α∈(0,∞) and β∈(−∞,∞) are two constants, λ,χ are two positive parameters, and f∈C([0,1]×[0,∞),R) with f(t,0)<0. Our analysis mainly relied on the bifurcation theory.
Citation: Zhonghua Bi, Sanyang Liu. Global structure of positive solutions for third-order semipositone integral boundary value problems[J]. AIMS Mathematics, 2024, 9(3): 7273-7292. doi: 10.3934/math.2024353
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In this paper, we were concerned with the global behavior of positive solutions for third-order semipositone problems with an integral boundary condition
y‴+βy″+αy′+λf(t,y)=0,t∈(0,1),y(0)=y′(0)=0,y(1)=χ∫10y(s)ds,
where α∈(0,∞) and β∈(−∞,∞) are two constants, λ,χ are two positive parameters, and f∈C([0,1]×[0,∞),R) with f(t,0)<0. Our analysis mainly relied on the bifurcation theory.
Third-order boundary value problems emerge in applied mathematics and physics, which have been investigated via various methods; see [1,2,3,4,5,6,7]. In [5], by using the disconjugacy theory [8], Ma and Lu obtained the optimal intervals to guarantee that the Green's functions corresponding to the third order linear problem are of one sign. Such results have been extended in [1].
Lemma 1.1. ([1,Theorem 2.1]) Suppose that there exist two constants α∈(0,∞) and β∈(−∞,∞), which satisfy
0<4α−β2<π2ande−β2√4α−β2+βsin(√4α−β22)≥√4α−β2cos(√4α−β22). | (1.1) |
The property of disconjugacy of y‴+βy″+αy′=0 is valid.
To our knowledge, the behavior of positive solutions for second-order boundary value problems under the semipositone condition have been discussed extensively; see [9,10,11,12,13,14,15]. It is worth noting that based upon the bifurcation theory, excluding Ambrosetti et al. [9] who explored semipositone elliptic problems under the linear boundary condition, recently, Ma and Wang [14,15] proved the solvability for second-order problems with a nonlinear boundary condition. In comparison to second-order cases, the challenge in studying third-order cases lies in that the third-order differential operator is non-self-adjoint.
Inspired by the aforementioned literature, we investigate the behavior of positive solutions for the following problems
y‴+βy″+αy′+λf(t,y)=0,t∈(0,1),y(0)=y′(0)=0,y(1)=χ∫10y(s)ds, | (1.2) |
where α∈(0,∞) and β∈(−∞,∞) are two constants, λ,χ are two positive parameters, and
f:[0,1]×[0,∞)→R |
is continuous and satisfies that the following assumption holds.
(F1) (semipositone) f(t,0)<0,∀t∈[0,1].
The integral boundary conditions can be decomposed into various different situations, such as multipoint and nonlocal boundary conditions. As far as we are concerned, few literatures on the solvability of third-order semipositone problems with an interval boundary condition since the maximum principle may fail. In order to overcome this difficulty, as a first step, Henderson [3] translated the third-order semipositone problems into the positone problems under the three-point boundary condition. Concurrently, we found that they only showed that the discussed problems have at least one positive solution. No detailed information about the global behavior of solutions was investigated since the spectral structure of third-order linear eigenvalue problems has not yet been found. Meanwhile, in order to use the fixed point index theory and bifurcation theory to show our results, we need to obtain a suitable cone utilizing the property of Green's function in a Banach space. Recently, Cabada used a new technique to construct a smaller positive cone (see [16,17]).
To sum up, we first obtain the relationship between the Green's function of (1.2) and the Green's function corresponding to (1,2), (2,1)-conjugate boundary conditions. As we will see, the expression of G(t,s) of (1.2) is so complex with two parameters that it is hard to obtain the suitable cone. Fortunately, we look for a condition showed in [18] to avoid the complex computations by the form of G(t,s) of (1.2). We will be concerned with the limits of G(t,s)/G(1,s) as s=0 and s=1.
(PG) There is a continuous function ϕ: [0,1]→(0,∞) and k1,k2∈C[0,1] such that 0<k1(s)<k2(s) satisfies
ϕ(t)k1(s)≤G(t,s)≤ϕ(t)k2(s),∀(t,s)∈[0,1]×[0,1]. |
Finally, relying on the linear behavior of f(t,y) as y→∞, we are employed to consider three cases. All conclusions are derived that a global branch of solutions for (1.2) exists and bifurcates from infinity. As λ nears to the bifurcation point, we demonstrate that the solutions of large norms are still positive, enabling the application of bifurcation theory or topological methods.
From Lemma 1.1, by means of the method introduced in [8,Page 105–106], we enunciate the following expression of the Green's function.
Theorem 2.1. Assume that Lemma 1.1 is satisfied, then,
y‴(t)+βy″(t)+αy′(t)=h1(t),t∈(0,1),y(0)=y′(0)=y(1)=0, | (2.1) |
has a unique solution given by
y(t)=∫10G1(t,s)h1(s)ds, |
where h1∈X and G1(t,s) is the Green's function of (2.1), that is,
G1(t,s)={−e−β(s−t)g(t)g(1−s)α√4α−β2g(1),0≤t≤s≤1,−e−β(s−t)(g(t)g(1−s)−g(1)g(t−s))α√4α−β2g(1),0≤s≤t≤1, | (2.2) |
where
g(ξ)=√4α−β2−√4α−β2e−βξ2cos(√4α−β22ξ)−βe−βξ2sin(√4α−β22ξ),ξ∈[0,1]. |
Corollary 2.1. The following properties of the function G1(t,s) are satisfied.
(1) G1(t,s) is negative on (t,s)∈(0,1)×(0,1).
(2) G1(0,s)=∂G1∂t(0,s)=G1(1,s)=0, ∀s∈(0,1).
(3) G1(t,1)=∂G1∂s(t,1)=G1(t,0)=0, ∀t∈(0,1).
(4) ∂2G1∂t2(0,s)<0<∂G1∂t(1,s), ∀s∈(0,1).
(5) ∂G1∂s(t,0)<0 and ∂2G1∂s2(t,1)<0, ∀t∈(0,1).
First of all, we point out that the following problem
y‴(t)+βy″(t)+αy′(t)=0,t∈(0,1),y(0)=y′(0)=0,y(1)=1, | (2.3) |
has no solution if, and only if, Lemma 1.1 is fulfilled.
In another case, (2.3) has a unique solution ω, that is,
ω(t)=g(t)g(1). | (2.4) |
Obviously, ω(t) is positive on t∈(0,1]. Moreover, by denoting
D:=∫10ω(τ)dτ>0, |
we have that it is given by the following expression
D=(2α−β)√4α−β2+2β√4α−β2e−β2cos(√4α−β22)−2(2α−β2)e−β2sin(√4α−β22)2αg(1). | (2.5) |
From the disconjugacy theory, if we investigate the following problem (h2∈X),
v‴+βv″+αv′=h2(t),t∈(0,1),v(0)=v(1)=v′(1)=0, | (2.6) |
one can know that (2.6) is the adjoint of (2.1) and the Green's function of (2.6) satisfies
G2(t,s)=−G1(s,t). |
Furthermore, we have
z(s)=∫10G2(s,η)dη=−∫10G1(η,s)dη | (2.7) |
as the unique solution satisfying the following problem
z‴(s)+βz″(s)+αz′(s)=1,t∈(0,1),z(0)=z(1)=z′(1)=0. | (2.8) |
Moreover, from Lemma 1.1 and [8,Theorem 11], z(s)>0 on s∈(0,1), z′(s)>0, and z″(1)>0.
Lemma 2.1. Suppose that Lemma 1.1 is satisfied. The problem
y‴(t)+βy″(t)+αy′(t)+ℏ(t)=0,t∈(0,1),y(0)=y′(0)=0,y(1)=χ∫10y(s)ds, | (2.9) |
is equivalent to the following integral equation if, and only if, χD≠1, where ℏ∈X, i.e.,
y(t)=∫10G(t,s)ℏ(s)ds, |
where G(t,s) represents the Green's function of (2.9), i.e.,
G(t,s)=−G1(t,s)−χω(t)1−λD∫10G1(η,s)dη, | (2.10) |
and ω(t) and D are given in (2.4) and (2.5), respectively.
Proof. Since (2.1) and (2.4) have the solution v and ω, respectively, then the unique solution of (2.9) is
y=v+χω∫10y(s)ds. |
Furthermore,
y=−∫10G1(t,s)ℏ(s)ds+χω(t)∫10y(s)ds. | (2.11) |
Put
A:=∫10y(s)ds, |
then integrating (2.11) on the interval (0,1), we get
A=−∫10∫10G1(η,s)ℏ(s)dsdη+χ∫10ω(η)∫10y(s)dsdη=−∫10∫10G1(η,s)ℏ(s)dsdη+χA∫10ω(η)dη, |
and, furthermore,
A=−∫10ℏ(s)∫10G1(η,s)dηds1−χ∫10ω(η)dη. |
Replacing A in (2.11), we arrive at
y=−∫10G1(t,s)ℏ(s)ds−χω(t)∫10ℏ(s)∫10G1(η,s)dηds1−χ∫10ω(η)dη. |
Theorem 2.2. From Lemma 1.1 and χ∈(0,1D), we derive that G(t,s)>0 on (t,s)∈[0,1]×[0,1].
Proof. From Corollary 2.1, G1(t,s) is negative. Next, by the property of (PG), we show that there is a positive constant R and a function l∈X satisfying l(t)>0 on t∈(0,1] and l(0)=0, for which the following result is satisfied, that is,
l(t)χ1−χDz(s)≤G(t,s)≤Rχ1−χDz(s),(t,s)∈[0,1]×[0,1]. | (2.12) |
Now, set
ψ(t,s):=G(t,s)G(1,s)=1−χDχG1(t,s)∫10G1(η,s)dη+ω(t). | (2.13) |
Obviously, ψ(t,s) is continuous and
G(1,s)=−χ1−χD∫10G1(η,s)dη. |
By Corollary 2.1, z(s) is the solution of (2.8). From the L'Hôptial rule, we derive
lims→0+G1(t,s)∫10G1(η,s)dη=lims→0+−G1(t,s)z(s)=lims→0+−∂G1∂s(t,s)z′(s)=−∂G1∂s(t,0)z′(0)>0. |
Thus,
lims→0+ψ(t,s)=1−χDχ(−∂G1∂s(t,0)z′(0))+ω(t):=ψ1(t)>0. |
Analogously,
lims→1−G1(t,s)∫10G1(η,s)dη=lims→1−−G1(t,s)z(s)=lims→1−−∂2G1∂s2(t,s)z″(s)=−∂2G1∂s2(t,1)z″(1)>0 |
and
lims→1−ψ(t,s)=1−χDχ(−∂2G1∂s2(t,1)z″(1))+ω(t):=ψ2(t)>0. |
Since the limits functions ψ1(t) and ψ2(t) exist and are finite, then at s=0,1, ψ(t,s) has removable discontinuities, so it can be extended to a function ˉψ∈C((0,1)×(0,1)). Therefore,
l(t)=mins∈[0,1]ˉψ(t,s) |
is continuous such that
l(0)=0,0<l(t)≤ˉψ(t,s)≤R:=max(t,s)∈[0,1]×[0,1]ˉψ(t,s). |
Corollary 2.2. From Lemma 1.1 and χ∈(0,1D), for all constant δ∈(0,1), there is a constant γ∈(0,1) depending on δ, such that the following result is satisfied, that is,
γχ1−χDz(s)≤G(t,s),(t,s)∈[δ,1]×[0,1]. | (2.14) |
Proof. The result is satisfied from the definition of the function l.
Define a cone
P:={y∈X|y(t)≥0,t∈[0,1]andmint∈[δ,1]y(t)≥γ‖y‖}, | (2.15) |
where δ satisfies Corollary 2.2.
Remark 2.1. In this paper, in view of the boundary conditions, the integral boundary condition is more widely used than the obtained third-order conjugate boundary condition in [1,5].
The work space is X=C[0,1] with the norm
‖y‖:=maxt∈[0,1]|y(t)|. |
We also set
Br:={y∈X:‖y‖<r} |
with r>0.
Next, by means of the Krein-Rutman theorem [19,Theorem 19.3 (a)], we discuss the existence of the principal eigenvalue for the linear eigenvalue problem as follows
−y‴−βy″−αy′=λb(t)y,t∈(0,1),y(0)=y′(0)=0,y(1)=χ∫10y(s)ds. | (3.1) |
Denote A: P→X as the map,
Ay(t):=λ∫10G(t,s)b(s)y(s)ds,t∈[0,1]. |
Lemma 3.1. Assume that χ∈(0,1D) and Lemma 1.1 are satisfied, (3.1) has a principal eigenvalue λ1, and the corresponding eigenfunction ϕ1(t) is positive.
Proof. It follows from (2.15), then P is normal and has nonempty interior, so X=¯P−P. From Theorem 2.2, A is a strong positive operator and A∈intP. By the Krein-Rutman theorem, the spectral radius r(A) is positive, and ϕ1∈X exists, satisfying ϕ1>0 and Aϕ1=r(A)ϕ1. Thus,
λ1=(r(A))−1>0. |
Let A∗ be the conjugate operator of A, then A∗ϕ2=r(A∗)ϕ2, where ϕ2∈X such that ϕ2>0 on (0,1), corresponding to λ1. Since
∫10(Aϕ1)ϕ2dt=λ1∫10ϕ1ϕ2dt=∫10ϕ1(A∗ϕ2)dt, |
then the algebraic multiplicity of λ1 is 1. Thus, λ1 is the principal eigenvalue of (3.1).
Denote a nonlinear operator K: X→X by
y:=Kℏ. |
From the above notation, it follows that (1.2) is equivalent to
y−λKf(⋅,y)=0,y∈X. | (3.2) |
Throughout the paper, we will use the same symbol to represent both the function and the corresponding Nemytskii operator.
We denote if there is a sequence (μn,yn) with μn→λ∞ and yn∈X, such that yn−μnKf(yn)=0 and ‖yn‖→∞, then λ∞ is a bifurcation from infinity for (3.2).
In some cases, such as what we will later discuss in detail, by application of an appropriate rescaling to look for bifurcation from infinity based on the Leray-Schauder topological degree, which represents deg(⋅,⋅,⋅), it is worth noting that K is continuous and compact. Thus it is reasonable to investigate the topological degree of I−λKf, where I is the identity map.
Theorem 3.1. Assume that χ∈(0,1D) and Lemma 1.1 are satisfied. We suppose that
f∈C([0,1]×[0,∞),R) |
satisfies (F1) and (F2).
(F2) There exists a positive function c∈X such that
limy→∞f(⋅,y)y=c. |
There is a positive constant ε such that (1.2) exist positive solutions if either
(Ⅰ) υ1>0 (possibly ∞) in [0,1] and λ∈[λ∞−ε,λ∞),
or
(Ⅱ) υ2<0 (possibly −∞) in [0,1] and λ∈(λ∞,λ∞+ε],
where
λ∞:=λ1c |
and
υ1(t):=lim infy→∞(f(t,y)−cy),υ2(t):=lim supy→∞(f(t,y)−cy). |
In order to show Theorem 3.1 is valid, we first extend f(t,⋅) to R and set
F(t,y):=f(t,|y|) | (3.3) |
and
Φ(λ,y):=y−λKF(t,y),y∈X. | (3.4) |
Obviously, the solution y>0 of Φ(λ,y)=0 is equivalent to a positive solution of (1.2).
Lemma 3.2. Assume that χ∈(0,1D) and Lemma 1.1 are satisfied. For every compact interval ℵ⊂[0,+∞)∖{λ∞}, there is a positive constant r such that, for all λ∈ℵ, ‖y‖≥r, Φ(λ,y)≠0. Moreover,
(1) if υ1>0, then ℵ=[λ∞,λ], forλ>λ∞;
(2) if υ2<0, then ℵ=[0,λ∞].
Proof. Let μn→μ≥0∈ℵ, that is, μ≠λ∞ and ‖yn‖→∞ be such that
yn=μnKF(t,yn). |
Setting
wn:=yn‖yn‖, |
it follows that
wn=μn‖yn‖−1KF(t,yn). |
From (F2) and (3.3), up to a subsequence in X, wn→w is fulfilled for which w satisfies the problem
−w‴−βw″−αw′=μc|w|,t∈(0,1),w(0)=w′(0)=0,w(1)=χ∫10w(s)ds, |
and ‖w‖=1. By Theorem 2.2, w≥0, then μc=λ1, i.e., μ=λ∞, which contradicts with the assumption of μ≠λ∞.
Next, we give a short illustration of Lemma 3.2 (1), and (2) follows similarly. Now, assume that there is a sequence (μn,yn)∈(0,∞)×X, where μn→λ∞, ‖yn‖→∞, and μn>λ∞, such that
Φ(μn,yn)=0. | (3.5) |
Note that yn∈X has a unique decomposition
yn=vn+snϕ1, | (3.6) |
where sn∈R, since yn>0, ϕ2>0, and
∫10vn(t)ϕ2(t)dt=0, |
and by (3.6), we obtain
sn=(∫10un(t)ϕ2(t)dt)(∫10ϕ1(t)ϕ2(t)dt)−1>0,n∈N. | (3.7) |
From (3.5), it follows that
∫10ϕ2(t)yn(t)dt=μn∫10ϕ2(t)KF(t,yn(t))dt. |
Since
∫10ϕ2(t)KF(t,yn(t))dt=−λ1∫10ϕ2(t)yn(t)dt, |
we obtain
−λ1∫10ϕ2(t)yn(t)dt=∫10(y‴n+βy″n+αy′n)ϕ2(t)dt=−∫10μnf(t,yn(t))ϕ2(t)dt=−∫10μn(f(t,yn(t))−cyn(t))ϕ2(t)dt−∫10μnyn(t)cϕ2(t)dt, |
then
(μnc−λ1)∫10ϕ2(t)yn(t)dt=−∫10μn(f(t,yn(t))−cyn(t))ϕ2(t)dt. |
For n large enough, μn>λ∞ and
∫10yn(t)ϕ2(t)dt>0 |
hold, then we infer that
∫10(f(t,yn(t))−cyn(t))ϕ2(t)dt<0 |
and from the Fatou lemma, we yield
0≥lim infn→∞∫10(f(t,yn)−cyn)ϕ2dt≥∫10υ1ϕ2dt, |
which contradicts with υ1>0.
Lemma 3.3. Assume that χ∈(0,1D) and Lemma 1.1 are satisfied. For λ∈(λ∞,∞), there is a positive constant r such that
Φ(λ,y)≠τϕ1,forallτ≥0,‖y‖≥r. |
Proof. If there is a sequence {yn}∈X with ‖yn‖→∞ and numbers τn≥0 satisfy Φ(λ,yn)=τnϕ1, then
−y‴n−βy″n−αy′n=λF(t,yn)+τnλ1ϕ1. |
Since F(t,y)≈c|y|→∞ and τnλ1ϕ1≥0, to the maximum principle, yn>0 for all t∈(0,1).
Choose ϵ>0 such that
λ∞<λ(1−ϵ). |
From condition (F2), there is a constant R0>0 such that
f(t,y)≥(1−ϵ)cy,∀y>R0,t∈(0,1). |
By ‖yn‖→∞, we know there is a positive constant N∗ such that
yn>R0,∀n≥N∗ |
and
f(t,yn)≥(1−ϵ)cyn. | (3.8) |
By (3.7) and (3.8), we derive
snλ1∫10ϕ1(t)ϕ2(t)dt=∫10−(y‴n+βy″n+αy′n)ϕ2(t)dt=λ∫10F(t,yn)ϕ2(t)dt+τλ1∫10ϕ1(t)ϕ2(t)dt≥λ∫10F(t,yn)ϕ2(t)dt≥λ∫10(1−ϵ)cyn(t)ϕ2(t)dt=λ(1−ϵ)csn∫10ϕ1(t)ϕ2(t)dt. |
Thus,
λ∞≥λ(1−ϵ), |
which is a contradiction.
For y≠0, we set
Z:=y‖y‖2. |
Set
Ψ(λ,Z):=Φ(λ,y)‖y‖2=y−λKF(t,y)‖y‖2=Z−λ‖Z‖2KF(t,Z‖Z‖2). |
λ∞ is a bifurcation from infinity for (3.4) if, and only if, λ∞ is a bifurcation from Z=0 for Ψ=0. From Lemma 3.2, for λ∈(0,λ∞), by homotopy, we have
deg(Ψ(λ,⋅),B1r,0)=deg(Ψ(0,⋅),B1r,0)=deg(I,B1r,0)=1. | (3.9) |
Similarly, from Lemma 3.2, for τ∈[0,1] and λ∈(λ∞,∞),
deg(Ψ(λ,⋅),B1r,0)=deg(Ψ(0,⋅)−τϕ2,B1r,0)=deg(Ψ(0,⋅)−ϕ1,B1r,0)=0. | (3.10) |
Set
Σ:={(λ,y)∈[0,∞)×X:y≠0,Φ(λ,y)=0}. |
It follows from (3.9) and (3.10) that the following lemma is fulfilled.
Lemma 3.4. λ∞ is a bifurcation from infinity for (3.4). To be precise, there is an unbounded, closed, connected set Σ∞⊂Σ which bifurcates from infinity. Furthermore, if υ1>0, Σ∞ bifurcates to the left (respectively, if υ2<0, Σ∞ bifurcates to the right).
Proof of Theorem 3.1. From Lemmas 3.2–3.4, it is sufficient to demonstrate that if μn→λ∞ and ‖yn‖→∞, for all t∈(0,1) and n large enough, yn is positive. Set
wn=yn‖yn‖. |
By utilizing the preceding results up to subsequence in X, wn→w and w=ϑϕ1 with ϑ>0 are satisfied, then, yn>0 for n large enough.
Theorem 3.2. Assume that χ∈(0,1D) and Lemma 1.1 are satisfied. We suppose that
f∈C([0,1]×[0,∞),R) |
satisfies (F1) and (F3).
(F3) There is a positive function c∈X such that
limy→∞f(⋅,y)yp=c,p∈(1,∞). |
There is a constant λ∗>0, such that for λ∈(0,λ∗], (1.2) exist positive solutions. More specifically, there exist a connected set of positive solutions for (1.2), which bifurcates from infinity at λ∞=0.
Set
F(t,y):=F(t,y)−c|y|p, |
where F(t,y) is denoted as (3.3).
Next, using the rescaling w=dy and λ=dp−1 with d>0, shows that λ∞=0 is a bifurcation from infinity for
y−λKF(t,y)=0, | (3.11) |
which is equivalent to (λ,y), which is a solution of (3.11) if, and only if,
w−KˆF(d,w)=0, | (3.12) |
where
ˆF(d,w):=c|w|p+dpF(d−1w). |
As d=0, we set
ˆF(0,w):=c|w|p. |
By (F3), we know that ˆF(d,w) is continuous for (d,w)∈[0,∞)×R. Let
˜S(d,w):=w−KˆF(d,w),d∈(0,∞). |
Thus, ˜S(d,⋅) is compact. For d=0, solution of ˜S(0,w)=0 are nothing but solutions of
−w‴−βw″−αw′=c|w|p,t∈(0,1),w(0)=w′(0)=0,w(1)=χ∫10w(s)ds. | (3.13) |
Now, we show that the following results are fulfilled:
˜S(0,w)≠0,for all‖w‖≥R2, | (3.14) |
˜S(0,w)≠0,for all0<‖w‖≤R1, | (3.15) |
deg(˜S(0,w),PR∖¯Pr,0)=−1,forr∈(0,R1],R∈[R2,∞), | (3.16) |
where R1,R2 are two constants with 0<R1<R2.
First, we claim that there exists R>0 such that for ‖w‖≥R, ˜S(0,w)≠0.
Assume that (3.13) has a sequence {wn} satisfying
limn→∞‖wn‖=∞, |
i.e.,
−w‴n−βw″n−αw′n=(c|wn|p−1)wn,t∈(0,1),wn(0)=w′n(0)=0,wn(1)=χ∫10wn(s)ds. |
Note that
limn→∞c|wn|p−1=∞,t∈(0,1). |
By the remarks in the final paragraph on [20,Page 56], wn must change its sign in (0,1), which is a contradiction.
Second, we prove that for 0<‖w‖≤R1, ˜S(0,w)≠0, where R1>0 is a constant.
On the contrary, if (3.15) does not hold, then (3.13) exists a sequence of solutions wn, which satisfies
‖wn‖→0,n→∞. | (3.17) |
Let
vn=wn‖wn‖. |
From (3.13), we have
−v‴n−βv″n−αv′n=c|wn|p‖wn‖,t∈(0,1),vn(0)=v′n(0)=0,vn(1)=χ∫10vn(s)ds. |
From (3.17), we have
limn→∞vn=0 |
uniformly in t∈(0,1).
According to the standard argument, after taking a subsequence and relabeling if necessary, v∗∈X exists and ‖v∗‖=1, satisfying
vn→v∗,n→∞ |
and
−v‴∗−βv″∗−αv′∗=0,t∈(0,1),v∗(0)=v′∗(0)=0,v∗(1)=χ∫10v∗(s)ds, |
which shows that v∗=0 holds. However, this is a contradiction. Hence, (3.15) holds.
In the end, we show (3.16) is valid. Denote
Pr:={y∈P:‖y‖<r}. |
Now, by (3.14) and (3.15), we derive that for all w∈∂PR and w∈∂Pr, ~S(0,w)≠0 is valid. So for all
w∈∂(PR∖¯Pr), ˜S(0,w)≠0 |
is still fulfilled. Hence, deg(˜S(0,w),PR∖¯Pr,0) is well defined.
Note that ˜f(w)=|w|p, so we claim that
deg(˜S(0,w),PR∖¯Pr,0)=−1. |
It is easy to verify the following conditions:
(H1) f0:=limw→0+˜f(w)w=0;
(H2) f∞:=limw→+∞˜f(w)w=∞.
Choose M1>0, which satisfies
M1γχ1−χD∫10z(s)c(s)ds>1. |
From (H2), there is a constant R2>0, such that ∀w≥R2, f(w)>M1w is satisfied. Choosing R>max{R1,R2}, we claim that
‖KˆF(0,w)‖>‖w‖ |
for w∈∂PR. In fact, for w∈∂PR,
(KˆF(0,w))(t)=∫10G(t,s)c(s)|w|pds≥M1γχ1−χD‖w‖∫10z(s)c(s)ds>‖w‖. |
Hence, it follows from the fixed point index theorem of [19] that
i(KˆF(0,⋅),PR,P)=0. | (3.18) |
From (H1), there exists a constant ι>0 such that w∈[0,ι], and
˜f(w)≤M2w, |
for which M2>0 satisfies
M2Rχ1−χD∫10z(s)c(s)ds≤1. |
Choose 0<r<min{ι,R2}, for w∈∂Pr,
‖KˆF(0,w)‖=maxt∈[δ,1]∫10G(t,s)c(s)|w|pds≤M2Rχ1−χD‖w‖∫1δz(s)c(s)ds≤‖w‖. |
Obviously, KˆF(0,w)≠w for w∈∂Pr. By the fixed point index theorem of [19],
i(KˆF(0,⋅),Pr,P)=1. | (3.19) |
From the additivity of the fixed point index, (3.18), and (3.19), we get
i(KˆF(0,⋅),PR∖¯Pr,P)=−1. | (3.20) |
From (3.20) and
˜S(0,w):X→PR∖¯Pr, |
we obtain
deg(˜S(0,w),PR∖¯Pr,0)=−1. |
Lemma 3.5. Assume that χ∈(0,1D) and Lemma 1.1 are fulfilled, then there is a constant d0>0 such that
(ⅰ) deg(˜S(d,⋅),PR∖¯Pr,0)=−1, ∀d∈[0,d0];
(ⅱ) if ˜S(d,w)=0,d∈[0,d0],‖w‖∈[r,R], then w>0.
Proof. On the contrary, there is a sequence (dn,wn), where dn→0, ‖wn‖∈{r,R}, and
wn=KˆF(dn,wn). |
Since the operator K is compact, up to a subsequence, wn→w and
˜S(0,w)=0,‖wn‖∈{r,R}, |
which contradicts with (3.14) and (3.15). Therefore, (ⅰ) holds.
In order to show (ⅱ) is valid, we once argue by means of contradiction. From the preceding results, we can present a sequence {wn}∈X with {t∈[0,1]:wn≤0}≠∅, which satisfies wn→w, ‖w‖∈{r,R} and ˜S(0,w)=0; that is, w is a solution of (3.13). From Theorem 2.2, w>0 holds. Therefore, we get a contradiction that for n large enough, wn is positive.
Proof of Theorem 3.2. From Lemma 2.1, ∀d∈[0,d0], and (3.12) exists a positive solution wd. Recalling for d>0 and λ=dp−1, y=wd is a solution (λ,yλ) of (3.11) for
0<λ<λ∗:=dp−10. |
For wd>0, (λ,yλ) is the positive solution of (1.2). So, ∀d∈[0,d0], ‖wd‖≥d implies that
‖yd‖=‖wd‖/d→∞ |
as d→0.
Theorem 3.3. Assume that χ∈(0,1D) and Lemma 1.1 are satisfied. We suppose that
f∈C([0,1]×[0,∞),R) |
satisfies (F1) and (F4).
(F4) There is a positive function c∈X, such that
limy→∞f(⋅,y)yq=c,q∈[0,1). |
There is a constant λ∗>0, such that for λ∈[λ∗,∞), positive solutions of (1.2) exist. To be precise, there is a connected set of positive solutions of (1.2), which bifurcates from infinity for λ∞=∞.
In this case, we will show that (1.2) exist positive solutions, which branch off from ∞ as λ∞=∞. As the same treatment as the superlinear problems, we again use w=dy, λ=dq−1 with q replacing p. In the case of superlinear problems, (λ,y) is the solution of (3.11) if (d,w) satisfies (3.12). Now, by q∈[0,1),
λ→∞⇔d→0. | (3.21) |
Lemma 3.6. Assume that χ∈(0,1D) and Lemma 1.1 are satisfied. For q∈(0,1), the problem
−w‴(t)−βw″(t)−αw′(t)=c(t)wq(t),t∈(0,1),w(0)=w′(0)=0,w(1)=χ∫10w(s)ds | (3.22) |
has a unique positive solution w0.
Proof. Assume that w1, w2 are two positive solutions of (3.22), i.e.,
−w‴1(t)−βw″1(t)−αw′1(t)=c(t)wq1,w1(0)=w′1(0)=0,w1(1)=χ∫10w1(s)ds,−w‴2(t)−βw″2(t)−αw′2(t)=c(t)wq2,w2(0)=w′2(0)=0,w2(1)=χ∫10w2(s)ds. |
We will show that w1≥w2 and w2≥w1. If w1⪈w2, we discuss that the element ˉw satisfies the following form, that is,
ˉw(t)=w1(t)−ϵw2(t),t∈[0,1], |
where ϵ∈(0,1). Let there exist a point ζ0∈(0,1) such that
ˉw(ζ0)=w1(ζ0)−ϵ0w2(ζ0)=0. | (3.23) |
On the other hand,
−ˉw‴(t)−βˉw″(t)−αˉw′(t)=−(w1(t)−ϵ0w2(t))‴−β(w1(t)−ϵ0w2(t))″−α(w1(t)−ϵ0w2(t))′=c(t)[wq1(t)−ϵ0wq2(t)]≥c(t)[ϵq0wq2(t)−ϵ0wq2(t)]>0,ˉw(0)=ˉw′(0)=0,ˉw(1)=χ∫10ˉw(s)ds. |
So, ˉw(t)>0, which contradicts with (3.23). Therefore, w1≥w2. By the same method, we may prove that w1≤w2.
As the same treatment as the superlinear case, we also can obtain that the following results are fulfilled:
˜S(0,w)≠0for all‖w‖≥R3, | (3.24) |
˜S(0,w)≠0for all0<‖w‖≤R4, | (3.25) |
and
deg(˜S(0,w),PR∖¯Pr,0)=1forr∈(0,R3],R∈[R4,∞), | (3.26) |
where R3 and R4 are two constants satisfying 0<R3<R4. Therefore, the following results can be obtained.
Lemma 3.7. Assume that χ∈(0,1D) and Lemma 1.1 are satisfied. There exists d0>0 such that
(ⅰ) deg(˜S(d,⋅),PR∖¯Pr,0)=1, ∀d∈[0,d0];
(ⅱ) If ˜S(d,w)=0,d∈[0,d0],‖w‖∈[r,R], then w>0.
Proof of Theorem 3.3. By the continuation, a connected subset Γ of solutions of ˜S(d,w)=0 satisfies (0,w0)∈Γ. From Lemma 3.7, we derive that there is a positive constant d0, such that for d<d0, these solutions are positive. Since λ=dq−1, u=wd, then a connected subset of solutions of (1.2) exists, which it can be transformed by Γ, so for
λ>λ∗:=dq−10, |
these solutions are positive. Furthermore, from (3.21), Σ∞ bifurcates from infinity for λ∞=+∞.
Remark 3.1. Comparing with [3], we no longer consider the nonlinear term as a translation and we obtain the optimal interval of the parameter λ under the three cases.
In order to illustrate that Theorems 3.1–3.3 are valid, we take α=1 and β=0, then Lemma 1.1 holds and
χ∈(0,1−cos(1)1−sin(1)). |
If
f(t,y)=2y˜p+tln(1+y)−t−1,˜p∈[0,∞), |
then f(t,0)=−t−1<0 is satisfied for t∈[0,1]. Let λ1 be the first positive eigenvalue corresponding to the linear problem (3.1) and ϕ1 be the positive eigenfunction corresponding to λ1. Next, we will check that all conditions in Theorems 3.1–3.3 are fulfilled. In fact,
c=limy→∞f(t,y)y˜p=2. |
In view of Theorem 3.1, λ∞=λ12, where ˜p=1,
υ1(t)=lim infy→∞(f(t,y)−cy)=lim infy→∞(tln(1+y)−t−1)>0,t∈[0,1]. |
Thus, Theorem 3.1 is valid. In the case of the superlinear problem,
λ∞:=λ12y1−˜p, |
where ˜p>1, so we derive that
limy→∞λ∞=λ12y1−˜p=0. |
Therefore, Theorem 3.2 is valid. Similarly, we also show that Theorem 3.3 is fulfilled.
By virtue of bifurcation theory or topological methods, we show the global behavior of positive solutions of (1.2) in the cases of asymptotically linear, superlinear, and sublinear as y→∞, and for λ near the bifurcation value, where the solutions norms are indeed positive. The domain of (1.2) offers potential for further studies. For example, in this kind of semipositone case, f(t,0)<0 is bounded. A natural question is whether or not there exist positive solutions for (1.2) if f(t,0)→−∞. Furthermore, the methods from this paper can be used for studying neutral-type and impulse differential equations as forms in [21,22,23,24,25]. We can also further discuss the stability analysis of the obtained positive solutions.
The authors declare they have not used Artificial Intelligence (AI) tools in the creation of this article.
This work was supported by the National Natural Science Foundation of China (Grant No. 12271419) and Natural Science Basic Research Program of Shaanxi Province of China (No. 2023-JC-QN-0081).
The authors declare that they have no competing interests in this paper.
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