Abstract
In this paper, we give improved results on the existence of positive solutions for the following one-dimensional p-Laplacian equation with nonlinear boundary conditions:
where \(\phi_{p} ( s ) = | s | ^{ p-2 } s\), \(p >1 \). Constructing an available integral operator and combining fixed point index theory, we establish some optimal criteria for the existence of bounded positive solutions. The interesting point of the results is that the term \(b ( t ) \) may be singular at \(t=0\) and/or \(t=1\). Moreover, the nonlinear term \(g(t, y)\) is also allowed to have singularity at \(y=0\). In particular, our results extend and unify some known results.
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1 Introduction
In this paper, we investigate the existence of positive solutions for the following one-dimensional singular p-Laplacian equation with nonlinear boundary conditions:
where \(\phi_{p} ( s ) = | s |^{p-2}s\), \(p > 1 \), \(\phi_{q} = (\phi _{p} ) ^{-1} \), \(\frac{1}{p} + \frac{1}{q} =1\), \(\lambda_{1}, \lambda_{2} > 0\), \(\beta _{1}, \beta_{2}\geq0 \), \(b \in C (0, 1)\), \(b(t)\) may be singular at \(t = 0\) and/or \(t = 1\). \(g \in C ([0, 1]\times(0, +\infty))\), and \(g(t, y)\) may be singular at \(y=0\).
Third-order p-Laplacian equations with nonlinear boundary conditions play an important role in both theory and application. They have been paid much more attention to over the years, see [1–9] and the references therein. They are often used to model various phenomena in physics, biology, chemistry, and infections diseases in the positive energy problem. However, in various situations, including the cases just mentioned above, only positive solutions are meaningful; one refers the reader to see [2–23] for some references along this line. That is why people are particularly interested in studying positive solutions. The existence of positive solutions for p-Laplacian equation boundary value problems has been studied by many authors applying various nice methods such as topological degree, the Leray-Schauder continuation theorem, coincidence degree theory and so on (see [2, 4, 5, 8–24]).
In [3], He and Ge studied the following nonlinear boundary value problem:
Their main tool was the fixed point theorem in cones due to Krasnoselskii.
In [2], He studied the existence of double positive solutions for the following nonlinear three-point boundary value problems:
and
He employed a three-functional fixed point theorem due to Avery and Henderson.
Applying the fixed point theorem of cone expansion and compression of norm type, Su et al. [8] presented the existence of multiple positive solutions of the following nonlinear two-point boundary value problem:
Gupta and Trofimchuk [4] established prior bounds and the existence of positive solutions for the following boundary value problem:
where \(a_{i}\in R\), \(0 < \xi_{1} < \xi_{2} < \cdots< \xi_{m-2} < 1\), all \(a_{i}\) having the same sign, \(\alpha=\sum_{i=1}^{m-1}a_{i} > 1\), \(\sum_{i=1}^{m-2}a_{i}\xi_{i} \neq1\).
Feng and Webb [11] considered the following boundary value problem:
They presented the existence results with assumptions of nonlinear growth imposed on the nonlinear term f.
Motivated by the results mentioned above, in this paper we study the existence of positive solutions for problem (1.1). We should also assert here that our results are new and generalize the results in [1–3, 9–11].
The rest of the paper is organized as follows. In Section 2, we state some preliminaries and several lemmas in this work. In Section 3, we give the main results as well as some of their proofs. The existence of multiple positive solutions is obtained in Section 4. The existence of infinite positive solutions is presented in Section 5.
2 Preliminaries
Definition 2.1
Let X be a real Banach space. A nonempty closed convex set \(P\subset X\) is called a cone provided that
-
(i)
\(y \in P\), \(\lambda\geq0 \) implies \(\lambda y \in P\);
-
(i)
\(y\in P\), \(-y\in P\) implies \(y = 0\).
Definition 2.2
Let X be a real Banach space and P be a cone in X. A mapping α is called the nonnegative continuous concave functional on P if \(\alpha: P\longrightarrow[0,+\infty)\) is continuous and
Definition 2.3
A nonzero solution is said to be a \(C[0,1]\) solution of problem (1.1). We indicate a function \(y\in C [ 0, 1 ] \cap C ^{3} ( 0, 1 ) \) satisfying problem (1.1) with \(y(t)\) not identically zero on \((0,1)\). \(y(t)\) is said to be a \(C^{1}[0,1]\) solution, we indicate that \(y'\in C [ 0, 1 ]\). \(y(t)\) is called a positive solution of problem (1.1) if \(y(t)\) is a solution of problem (1.1) and \(y(t) > 0\) for each \(t\in(0,1)\).
Let \(X=C[0,1]\) be a Banach space with the norm \(\| y \|=\sup_{0\leq t\leq1}| y(t)|\), and let \(K=\{y\in X: y(t)\geq0, 0\leq t \leq1\}\). Then K is a positive cone in X. Throughout the paper, the partial ordering is always given by K. For the concepts and properties of Krein-Kutmann theorems and fixed point index theory, one refers the reader to see [13] and [24]. For \(\nu\in(0, \frac{1}{2})\), let
Obviously, \(P\subset K \subset X\). Denote \(P_{r} = \{ y\in P : \| y \| < r \}\), \(\partial P_{r} = \{ y\in P: \| y \| = r \}\), \(\overline{P}_{r,R} = \{ y\in P : r \leq\| y \| \leq R \}\) for any \(0 < r < R < +\infty\).
Throughout this paper, we suppose that the following conditions hold:
- (A1):
-
\(b \in C( ( 0, 1 ), [ 0, +\infty) )\), \(b(t)\) may be singular at \(t=0\) and/or \(t=1\), \(b(t)\not\equiv 0\) and
$$ 0 < \int_{0}^{1} b(s)\,ds < +\infty. $$ - (A2):
-
\(g(t, y)\in C([ 0,1]\times(0, +\infty); [0,+\infty))\), \(g(t, y)\) may be singular at \(y=0\) and for any \(0 < r < R < +\infty\) such that
$$ \lim_{n \rightarrow+ \infty}\sup_{y\in\overline{P}_{r, R}} \int_{E(n)} b(s)g\bigl(s, y(s)\bigr)\,ds = 0, $$where \(E(n) = [ 0, \frac{1}{n} ]\cup[ \frac{n-1}{n}, 1 ]\).
Remark 2.1
It follows from (A2) that
Remark 2.2
It is easy to see \(\phi_{q}(s)=|s|^{q-2}s\). In fact, from \(\frac{1}{p}+\frac{1}{q}=1\), we can obtain
Remark 2.3
By (A1), there exists \(t_{0}\in (0, 1)\) such that \(b(t_{0})>0\). Obviously, if \(g(t, y)\) is nonsingular at \(y=0\), that is, \(g\in C([0, 1]\times(0, +\infty), [0, +\infty))\), then (A2) is satisfied.
Denote
Lemma 2.1
Suppose that condition (A1) holds. Then there exists a constant \(\nu\in(0,\frac{1}{2})\) satisfying
Define a function \(f(t)\) on \([\nu, 1-\nu]\) given by
Obviously, \(f(t)\) is a continuous and positive function on \([\nu, 1-\nu]\) and has its minimum and maximum on \([\nu, 1-\nu]\). Therefore, there exist positive constants \(m > M > 0\) such that \(m \leq f(t) \leq M\), \(t\in[\nu, 1-\nu]\).
Lemma 2.2
[13]
Let X be a real Banach space and P be a cone in X, with \(\Omega(P)\) being a bounded open set in P, \(\theta\in\Omega(P)\). Suppose that \(T: \overline{\Omega }(P)\rightarrow P\) is a completely continuous operator. If there exists \(u_{0}\in P\setminus\{\theta\}\) such that \(u - Tu \neq\mu u_{0}\), \(\forall u\in\partial\Omega(P)\), \(\mu\geq0\). Then the fixed point index \(i(T, \Omega(P), P)=0\).
Lemma 2.3
[13]
Let X be a Banach space and P be a cone in X with \(\Omega(P)\) being a bounded open set in P, \(\theta\in\Omega(P)\). Assume that \(T: \overline{\Omega}(P)\to P\) is a completely continuous operator. If \(Tu\neq\eta u\), \(\forall u\in\partial\Omega(P)\), \(\eta\geq1\), then the fixed point index \(i(T, \Omega(P), P)=1\).
Lemma 2.4
[13]
Let X be a real Banach space and P be a positive cone in X. Let \(\Omega(P)\) be a bounded open set in P with \(\theta\in\Omega(P)\). Suppose that \(T: \overline {\Omega}(P)\to P\) is completely continuous and satisfies the following conditions:
-
(i)
If \(\|Tu\| > \|u\|\), \(\forall u\in \partial\Omega(P)\), then the fixed point index \(i(T, \Omega(P), P)=0\);
-
(ii)
If \(\|Tu\| < \|u\|\), \(\theta\in\Omega(P)\), \(\forall u\in \partial\Omega(P)\), then the fixed point index \(i(T, \Omega(P), P)=1\).
Lemma 2.5
Let \(a\in L(0, 1)\) be positive. Then the following boundary value problem
has a unique positive solution which is given explicitly by
Proof
It follows from \(y'(t)=\int^{\sigma^{*}}_{t}\phi_{q}(\int^{r}_{0}a(x)\,dx)\,dr\) and boundary condition (2.2) that \(y' ( 0 ) > 0 \) and \(y(0)= \phi_{q} (\frac{\beta_{1}}{\lambda_{1}}\phi_{p}( y' ( 0 ) ) ) > 0\), \(y'(1) < 0 \), with \(y(1)=-\phi_{q} (\frac{\beta_{2}}{\lambda_{2}}\phi _{p}(y'(1)) ) < 0\). Then there exists a constant \(\sigma^{*}\in[0,1]\) such that \(y'(\sigma^{*})=0\).
Integrating (2.1) from 0 to t, we get
By making use of \(y''(0)=0\), we obtain
Integrating (2.3) from t to \(\sigma^{*}\), we have
Since \(y'(\sigma^{*})=0\), we know that
Thus
Integrating (2.4) from 0 to t, we obtain
In view of (2.5) and (2.6) together with boundary conditions (2.2), we have
Therefore
Again, integrating (2.3) from \(\sigma^{*}\) to t, we see
From \(y'(\sigma^{*})=0\), we have
Then
Integrating (2.7) from t to 1, we get
Thus
Using (2.9) and (2.10) together with boundary conditions (2.2), we obtain
From (2.7) and (2.11) we see that the results of Lemma 2.5 hold. So the proof is complete. □
For \(y\in P\), we now define the integral operator \(T: P\setminus\{ \theta\} \to P\) by
Lemma 2.6
Suppose that conditions (A1) and (A2) hold. Then \(T: \overline{P}_{r,R} \to P\) is completely continuous, and the nonzero fixed point \(y\in \overline{P}_{r,R}\) of T is a positive solution of problem (1.1).
Proof
Firstly, we will show that \(T: \overline{P}_{r,R} \to P\). By a simple computation, for any \(y\in P_{r, R}\), we have
that is, \((Ty)(t)\) is a nonnegative concave function. We suppose that \(\tau=\inf\{\zeta\in[0,1]: \sup_{0\leq t\leq1} y (t) = y(\zeta )\}\). Now we shall make the following discussions.
Case (1) \(\tau\in[0, \nu]\). It follows from the concavity of \(y(t)\) that each point on chord between \((\tau, y(\tau))\) and \((1, y(1))\) is below the graph of \(y(t)\). Thus, for any \(t\in[\nu , 1-\nu]\), we get
which implies \(y(t)\geq\nu\| y\|\) for \(t\in[\nu, 1-\nu]\).
Case (2) \(\tau\in[\nu, 1-\nu]\). If \(t\in[\nu, \tau ]\), we obtain
which implies \(y(t)\geq\nu\| y\|\) for \(t\in[\nu, \tau]\).
If \(t\in[\tau, 1-\nu]\), we get
which implies \(y(t)\geq\nu\| y\|\) for \(t\in[\tau, 1-\nu]\). Therefore, \(y(t)\geq\nu\| y\|\) for \(t\in[\nu, 1-\nu]\).
Case (3) \(\tau\in[1-\nu, 1]\). By the same argument, for \(t\in[\nu, 1-\nu]\), we have
which implies \(y(t)\geq\nu\| y\|\) for \(t\in[\nu, 1-\nu]\). Using the above discussions, one can get \(y(t)\geq\nu\| y\|\) for \(t\in[\nu, 1-\nu]\). Thus \(TP_{r, R}\subset P\).
Next, for any \(0 < r < R < +\infty\), we will show
which implies that \(T: P\setminus\{0\}\to P\) is well defined.
From (A2) and Remark 2.1, for any \(0 < r < R < +\infty \), there exists a natural number k such that
Thus
For any \(y\in\partial P_{r}\), let \(y(t_{0})=\max_{t\in[0,1]}|y(t)|=r\), \(t_{0}\in[0,1]\). Denote
is the eigenvalue function of the set \(E[a, b]=\{t: a\leq t \leq b\}\). It follows from the concavity of \(y(t)\) on \([0,1]\) that
Consequently, from (2.14), we have
Denote
It follows from (A1) and (A2) with (2.13)-(2.16) that
Thus (2.12) holds. It also implies that \(T: \overline {P}_{r, R}\to P\) is well defined and \(T(Q)\) is uniformly bounded for any bounded set \(Q\subset\overline{P}_{r,R}\).
Now, we prove that for any \(0 < r < R < +\infty\), \(T(\overline {P}_{r,R})\) is equicontinuous. In fact, from (A2), for any \(\varepsilon>0\), there exists a natural number \(k_{0}\) such that
Let
Then
For the above \(\varepsilon> 0 \), take \(0 < \delta< \frac{\varepsilon}{2\widehat{G}+\varepsilon}\) such that for any \(t', t'' \in[0,1]\) satisfying \(|t'-t''| < \delta\) and for any \(s\in [\frac{1}{k_{0}}, \frac{k_{0}-1}{k_{0}} ]\), we have
This implies that \(T(\overline{P}_{r,R})\) is equicontinuous. Thus, by the Ascoli-Arzela theorem, we know that \(T: \overline{P}_{r,R}\to P\) is compact.
Finally we show that \(T: \overline{P}_{r,R}\to P\) is continuous. Suppose that \(y_{n}, y_{0}\in\overline{P}_{r,R}\) and \(\|y_{n}-y_{0}\|\to0\) (\(n\to \infty\)). Then \(r\leq\|y_{n}\|\leq R \) and \(r\leq\|y_{0}\|\leq R\). For any \(\varepsilon> 0\), by making use of (A2), there exists a natural number \(n_{0} > 0\) such that
On the other hand, from (2.15), we have
Since \(g(t, y)\) is uniformly continuous on \([\frac{1}{n_{0}}, \frac{n_{0}-1}{n_{0}} ]\times [\frac{r}{n_{0}}, R ]\), we have
holds uniformly on \(s\in[\frac{1}{n_{0}}, \frac{n_{0}-1}{n_{0}}]\). Then the Lebesgue dominated convergence theorem yields that
Thus, for the above \(\varepsilon>0\), there exists a natural number N, for any \(n > N\), we have
It follows from (2.18) and (2.19) that for any \(n > N\), we see
This implies that \(T: \overline{P}_{r,R}\to P\) is continuous. Thus \(T: \overline{P}_{r,R}\to P\) is completely continuous.
Obviously, if T has a nonzero fixed point \(y\neq0\), then \(y\in C[0,1]\cap C^{3}(0,1)\) and satisfies problem (1.1). On the other hand, by the maximum principle, we see that \(y(t)>0\), \(t\in (0,1)\). Hence y is a positive solution of problem (1.1). This completes the proof. □
3 The existence of at least one positive solution
For convenience, we let \(\omega^{*}\) and \(\omega_{*}\) be positive constants satisfying \(\omega^{*} > \frac{1}{m}\) and
Thus, for any positive constants l, L, r̂, R̂ satisfying \(\omega^{*} < l < +\infty\), \(0 < L < \omega_{*}\) and \(0 < \hat{r} < \widehat{R} < +\infty\), we have the following theorems.
Theorem 3.1
Suppose that conditions (A1) and (A2) hold. In addition, assume that
Then boundary value problem (1.1) has at least one positive solution ŷ satisfying \(\hat{r} \leq\hat{y} \leq\widehat{R}\).
Proof
From the first part of (3.1), \(g(t, y) \geq(l\hat{r})^{p-1}\), for \((t, y) \in[0, 1]\times[\nu\hat{r}, \hat{r}]\), let \(\psi\equiv1\). We prove that
If not, there exist \(y_{0}\in\partial P_{\hat{r}}\) and \(\mu_{0}\geq0\) such that \(y_{0} = Ty_{0} + \mu_{0} \psi\). Let \(\mu^{*}=\min_{t\in[0, 1]}y_{0}(t)\). Then, for \(t\in(0, \sigma^{*})\), we have
On the other hand, for \(t\in[\sigma^{*}, 1]\), by the same argument, we have
From the above discussions, we see \(y_{0} > \mu^{*}\), which implies \(\mu^{*} > \mu^{*}\), a contradiction with the definition of \(\mu^{*}\). Thus, we know that (3.2) holds. It follows from Lemma 2.2 that
Without loss of generality, we may assume that T has no fixed point on \(\partial P_{\hat{r}}\) and \(\partial P_{\widehat{R}}\). Now we prove that
In fact, if not, there exist \(y_{1}\in\partial P_{\widehat{R}}\) and \(\eta_{1}\geq1\) such that \(T y_{1} = \eta_{1} y_{1}\). Let \(y^{*}_{1}(t)=\min\{y_{1}(t), \widehat{R}\}\), thus \(y^{*}_{1}\in\partial P_{\widehat{R}}\). Thus, for \(t\in(0, \sigma^{*})\), we have
On the other hand, for \(t\in[\sigma^{*}, 1]\), by the same argument, we have
The above discussions imply that \(\|y^{*}_{1}\| < \|y^{*}_{1}\|\), a contradiction with the definition of \(y^{*}_{1}\). Therefore (3.4) holds. It follows from Lemma 2.3 that
From (3.3) and (3.5) together with the properties of fixed point index, we have
Thus T has at least one positive fixed point ŷ in \(P_{\hat{r},\widehat{R}}\), which means that problem (1.1) has at least one positive solution ŷ satisfying \(\hat{r} \leq\hat{y} \leq\widehat{R}\). This completes the proof. □
Theorem 3.2
Suppose that conditions (A1) and (A2) hold. In addition, assume that the following conditions hold:
- (H1):
-
\(0 < g^{0} < (\frac{\omega_{*}}{2})^{p-1}\),
- (H2):
-
\((\frac{\omega^{*}}{\nu})^{p-1} < g_{\infty} < +\infty\).
Then problem (1.1) has at least one positive solution.
Proof
From (H1), for any \(0 < \varepsilon< (\frac{\omega_{*}}{2})^{p-1} - g^{0}\), there exists a sufficiently small positive constant \(r_{1}\) such that \(0 < y \leq r_{1}\), for any \(t\in[0, 1]\), we have
Since \(y'(\sigma^{*})=0\), we find that
Thus, from (3.6) and (3.7), we obtain
Indeed, in view of Lemma 2.4, we deduce
Next from (H2), for any \(0 < \varepsilon< g_{\infty}-(\frac{\omega^{*}}{\nu})^{p-1} \), there exists a sufficiently large positive constant \(R^{*}\) such that for any \(y\geq R^{*}\), we have
Choose \(R_{1}=\max\{2r_{1}, \frac{R^{*}}{\nu}\}\), and let \(\varphi(t) \equiv1\), \(t\in[0, 1]\). Then \(R_{1} > r_{1}\) and \(\varphi \in\partial P _{1}\). Obviously, \(\varphi(t) \in P_{R_{1}}\setminus\{\theta\}\).
In the following we prove that
Otherwise, there exist \(y_{0}\in\partial P_{R_{1}}\) and \(\mu_{0} > 0\) such that \(y_{0} = Ty_{0} + \mu_{0} \varphi\). Let \(\xi=\min\{y_{0}(t): t\in[\nu, 1-\nu]\}\) and notice that for any \(t\in[\nu, 1-\nu]\), we have \(\min_{t\in[\nu, 1-\nu]}y(t)\geq\nu\|y_{0}\|=\nu R_{1}\geq R^{*}\).
Consequently, from (3.9), for any \(t\in[\nu, 1-\nu]\) and \(0 < t < \sigma^{*}\), we have
On the other hand, by the same argument, for \(t\in[\sigma^{*}, 1]\) and \(t\in[\nu, 1-\nu]\), from (3.9) and (3.7), we have
From (3.11) and (3.12), we easily obtain \(\xi> \xi\), a contradiction with the definition of ξ. From the above discussion we see that (3.10) holds. It follows from Lemma 2.2 that
From (3.8) and (3.13) together with the properties of fixed point index, we have
Thus T has at least one positive fixed point \(y^{*}\) in \(P_{r_{1}, R_{1}}\), which means that problem (1.1) has at least one positive solution \(y^{*}\) satisfying \(r_{1} \leq y^{*} \leq R_{1}\). This completes the proof. □
Theorem 3.3
Suppose that conditions (A1) and (A2) hold. In addition, assume that the following conditions hold:
- (H3):
-
\((\frac{\omega^{*}}{\nu})^{p-1} < g_{0} < +\infty \),
- (H4):
-
\(0 < g^{\infty} < (\frac{\omega_{*}}{2})^{p-1}\).
Then problem (1.1) has at least one positive solution.
Proof
From (H3), for any \(0 < \varepsilon< g_{0}-(\frac{\omega^{*}}{\nu})^{p-1} \), there exists a sufficiently large positive constant \(r_{2}\) such that for any \(t\in[0, 1]\) and \(0 < y \leq r_{2}\), we have
Thus, for any \(y\in\partial P_{r_{2}}\), we get
Let \(\varphi\equiv1\). Now we prove that
Otherwise, there exist \(y_{0}\in\partial P_{r_{2}}\) and \(\mu_{0} > 0\) such that \(y_{0} = Ty_{0} + \mu_{0} \varphi\). Let \(\zeta=\min\{y_{0}(t): t\in[\nu, 1-\nu]\}\), then for any \(t\in[\nu, 1-\nu]\) and \(0 < t < \sigma^{*}\), we have
On the other hand, by the same argument, for \(t\in[\sigma^{*}, 1]\) and \(t\in[\nu, 1-\nu]\), from (3.9), we have
From (3.15) and (3.16), we easily obtain \(\zeta> \zeta\), a contradiction with the definition of ζ. From the above discussion we see that (3.14) holds. It follows from Lemma 2.2 that
Next from (H4), for any \(0 < \varepsilon_{1} < (\frac{\omega_{*}}{2})^{p-1} - g^{\infty }\), there exists a sufficiently small positive constant \(R_{*}\) such that \(y \geq R_{*}\), \(t\in[0, 1]\), we have
Let \(M_{0}=\sup_{y\in\partial P_{R_{*}}}\int_{0}^{1}b(s)g(s, y(s))\,ds\). Then \(M_{0} < +\infty\) by (2.12). Take
Notice \(y\in\partial P _{R_{*}}\) implies that \(y(t) \leq\|y\| \leq R_{*}\). In addition, for any \(y \in\partial P_{R_{2}}\), let \(D[y]=\{t\in[0, 1]: y\geq R_{*}\}\), then for any \(t\in D[y]\), clearly, \(R_{*}\leq y \leq\|y\| = R_{2}\). Let \(y_{1}(t) = \min\{y(t), R_{*}\}\), then \(y_{1}\in\partial P_{R_{*}}\). Thus, for any \(y \in\partial P_{R_{2}}\), from (3.7) and (3.18), we have
Therefore \(\|Ty\| < \|y\|\) for any \(y \in\partial P_{R_{2}}\). It follows from Lemma 2.4 that
From (3.17) and (3.19) together with the properties of fixed point index, we have
Thus T has at least one positive fixed point \(y^{**}\) in \(P_{r_{2}, R_{2}}\) with \(r_{2} \leq y^{**} \leq R_{2}\). Therefore \(y^{**}\) is a positive solution of problem (1.1). This completes the proof. □
4 The existence of multiple positive solutions
Theorem 4.1
Suppose that conditions (A1) and (A2) hold. In addition, (H1) and (H4) hold, and there exists a positive constant \(r_{0}^{*} > 0\) such that
Then problem (1.1) has at least two positive solutions.
Proof
It follows from (H1) that there exists a positive constant \(r_{1}^{*}\) satisfying \(0 < r_{1}^{*} \leq r_{0}^{*} \) such that
By making use of (H4), there exists a positive constant \(r_{2}^{*}\) satisfying \(r_{2}^{*} > r_{0}^{*} >0 \) such that
Without loss of generality, we may assume that T has no fixed point on \(\partial P_{r^{*}_{1}}\) and \(\partial P_{r^{*}_{2}}\). From (4.2)-(4.3) with the corresponding proofs in Theorems 3.2 and 3.3, respectively, together with the permanence property of fixed point index, we have
and
For any \(y\in\partial P_{r^{*}_{0}}\), by the concavity of function \(y(t)\) for \(t\in[0, 1]\), we easily get \(0 < \nu\|y\| \leq y(t)\leq\|y\| =r_{0}^{*}\), \(t\in[0, 1]\). Thus, for any \(y\in\partial P_{r^{*}_{0}}\), in view of (4.1) and the corresponding proof in Theorem 3.1 together with the permanence property of fixed point index, we have
From (4.4) and (4.6) together with the properties of fixed point index, we obtain
Hence, T has at least two positive fixed points in \(P_{r^{*}_{1}, r^{*}_{0}}\) and \(P_{r^{*}_{0}, r^{*}_{2}}\), respectively. Therefore, problem (1.1) has at least two positive solutions. This completes the proof. □
Theorem 4.2
Suppose that conditions (A1) and (A2) hold. In addition, (H2) and (H3) hold, and there exists a positive constant \(\tilde{r}_{0}^{*} > \sqrt{\Gamma _{1}}\) such that
where \(\Gamma_{1}=\max \{ (1+\phi_{q} (\frac{\beta _{1}}{\lambda_{1}} ) ) \phi_{q} (2\int_{0}^{1}b(r)\,dr ), (1+\phi_{q} (\frac {\beta_{2}}{\lambda_{2}} ) ) \phi_{q} (2\int_{0}^{1}b(r)\,dr ) \}\).
Then problem (1.1) has at least two positive solutions.
Proof
It follows from (H3) that there exists a positive constant \(\tilde{r}_{1}^{*} >0 \) satisfying \(0 < \tilde{r}_{1}^{*} < \sqrt{\Gamma_{1}} \) such that
By making use of (H2), there exists a positive constant \(\tilde{r}_{3}^{*} > \tilde{r}_{0}^{*} > 0\) such that
Then, for any \(y\in\partial P_{\tilde{r}_{3}^{*}}\), we have
It follows from Lemma 2.6 that \(T:\overline{P}_{\tilde{r}_{1}^{*}, \tilde{r}_{3}^{*}}\longrightarrow P \) is a completely continuous operator. Now we extend the operator T, still denoted by T, then \(T: \overline{P}_{\tilde{r}_{3}^{*}}\to P\) is completely continuous. Without loss of generality, we may assume that T has no fixed point on \(\partial P_{\tilde{r}_{1}^{*}}\) and \(\partial P_{\tilde{r}_{3}^{*}}\). From (4.8) and (4.9), the corresponding proofs in Theorems 3.2 and 3.3, respectively, together with the permanence property of fixed point index, we have
and
Choose \(\sqrt{\Gamma_{1}} < \tilde{r}_{2}^{*} \leq\tilde{r}_{0}^{*}\). For all \(y\in\partial P_{\tilde{r}_{2}^{*}}\), by the concavity of function \(y(t)\) on \((0, 1)\), we see \(y(t)\geq\nu\|y\|\) for \(t\in(0, 1)\) and \(0 < y(t)\leq\tilde{r}_{2}^{*} \leq\tilde{r}_{0}^{*}\) for \(t\in(0, 1)\). It follows from (4.7) and \(y\in\partial P_{\tilde{r}_{2}^{*}}\) that
Therefore, for any \(y\in\partial P_{\tilde{r}_{2}^{*}}\), we obtain
Thus \(\|Ty\| < \|y\|\) for any \(y\in\partial P_{\tilde{r}_{2}^{*}}\). From Lemma 2.5 we have
By making use of (4.10) and (4.11) with (4.12), combining the properties of fixed point index, we have
Hence, T has at least two positive fixed points in \(P_{\tilde {r}^{*}_{1},\tilde{ r}^{*}_{2}}\) and \(P_{\tilde{r}_{2}^{*}, \tilde{r}_{3}^{*}}\), respectively. Therefore, problem (1.1) has at least two positive solutions. This completes the proof. □
5 The existence of infinite positive solutions
Now we shall discuss the existence of infinitely many positive solutions. For convenience, we make the following assumptions:
- (\(\mathrm{A}'_{1}\)):
-
There exists a nonincreasing sequence \(\{t_{n}\} _{n=1}^{\infty}\) such that \(0 < t_{n+1} < t_{n}, t_{1} < \frac{1}{2}\), \(\lim_{n\rightarrow \infty}t_{n}=t_{0} \geq0\), with \(\lim_{t\rightarrow t_{n}}b(t)= \infty\), \(n=1, 2, \ldots\) , and \(0 < \int^{1}_{0}b(t)\,dt< +\infty\). Moreover, \(b(t)\) does not vanish identical on any subinterval of \([0,1]\).
Theorem 5.1
Suppose that conditions (\(\mathrm{A}'_{1}\)) and (A2) hold. Let \(\{\nu_{k}\}_{k=1}^{\infty}\) be such that \(\nu_{k}\in (t_{k+1}, t_{k})\) (\(k=1, 2, \ldots\)). Let \(\{r_{k}\}_{k=1}^{\infty}\) and \(\{R_{k}\}_{k=1}^{\infty}\) be such that \(\nu_{k}r_{k} < r_{k} < R_{k}\), \(k=1, 2,\ldots \) . For each natural number k, we assume that g satisfies
- (E1):
-
\(g(t, y) \geq(lr_{k})^{p-1}\), \((t, y) \in[0, 1]\times[\nu_{k} r_{k}, R_{k}]\);
- (E2):
-
\(g(t, y)\leq(LR_{k})^{p-1}\), \((t, y)\in[0, 1]\times (0, R_{k}]\).
Then boundary value problem (1.1) has infinitely many positive solutions \(\{y_{k}\}_{k=1}^{\infty}\) such that \(\|y_{k}\|\) is between \(r_{k}\) and \(R_{k}\), \(k=1, 2,\ldots\) .
Proof
From (\(\mathrm{A}'_{1}\)), we know that \(t_{0} < t_{k+1} < \nu_{k} < t_{k} < \frac{1}{2}\), \(k=1, 2,\ldots \) , then for any \(k\in N\), \(y\in P\), we have
We easily know that \(\{P_{r_{k}}\}_{k=1}^{\infty}\) and \(\{P_{R_{k}}\} _{k=1}^{\infty}\) are subsets of P. For a fixed k and for any \(y\in P_{r_{k}}\), from (5.1) we have \(y(t) \geq\nu_{k} \|y\|=\nu_{k} r_{k}\), \(t\in[\nu_{k}, 1-\nu_{k}]\). For \([t_{1}, 1-t_{1}]\subset[\nu_{k}, 1-\nu_{k}]\subset[0, 1]\), let \(\psi\equiv1\). We prove that
If not, there exist \(\tilde{y}_{0}\in\partial P_{r_{k}}\) and \(\mu _{0}\geq0\) such that \(\tilde{y}_{0} = T\tilde{y}_{0} + \mu_{0} \psi\). Let \(\mu^{*}=\min_{t\in[0, 1]}\tilde{y}_{0}(t)\). Then, for \(t\in(0, \sigma^{*})\), we have
On the other hand, for \(t\in[\sigma^{*}, 1]\), by the same argument, we have
From the above discussions, we see \(\tilde{y}_{0} > \mu^{*}\), which implies \(\mu^{*} > \mu^{*}\), a contradiction with the definition of \(\mu^{*}\). Thus, we know that (5.2) holds. It follows from Lemma 2.2 that
Without loss of generality, we may assume that T has no fixed point on \(\partial P_{r_{k}}\) and \(\partial P_{R_{k}}\). Now we prove that
In fact, if not, there exist \(\tilde{y}_{1}\in\partial P_{R_{k}}\) and \(\tilde{\eta}_{1}\geq1\) such that \(T \tilde{y}_{1} = \tilde{\eta}_{1} \tilde{y}_{1}\). Let \(\tilde{y}^{*}_{1}(t)=\min\{\tilde{y}_{1}(t), R_{k}\}\), thus \(\tilde{y}^{*}_{1}\in\partial P_{R_{k}}\). Thus, for \(t\in(0, \sigma^{*})\), we have
On the other hand, for \(t\in[\sigma^{*}, 1]\), by the same argument, we have
The above discussions imply that \(\|\tilde{y}^{*}_{1}\| < \|\tilde {y}^{*}_{1}\|\), a contradiction with the definition of \(\tilde{y}^{*}_{1}\). Therefore (5.4) holds. It follows from Lemma 2.3 that
From (5.3) and (5.5) together with the properties of fixed point index, we have
Thus T has at least one positive fixed point \(y_{k}\) in \(P_{r_{k}, R_{k}}\), which means that problem (1.1) has at least one positive solution \(y_{k}\) satisfying \(r_{k} \leq y_{k} \leq R_{k}\). From the randomness of k we see that Theorem 5.1 holds. This completes the proof. □
Theorem 5.2
Suppose that conditions (\(\mathrm{A}'_{1}\)) and (A2) hold. Let \(\{\nu_{k}\}_{k=1}^{\infty}\) be such that \(\nu_{k}\in (t_{k+1}, t_{k})\) (\(k=1, 2, \ldots\)). Let \(\{r_{k}\}_{k=1}^{\infty}\) and \(\{R_{k}\}_{k=1}^{\infty}\) be such that \(\nu_{k}r_{k} < r_{k} < R_{k}\), \(k=1, 2,\ldots \) . For each natural number k, we assume that g satisfies
- (E3):
-
\(0 < g^{0} < (\frac{\omega_{*}}{2})^{p-1}\),
- (E4):
-
\((\frac{\omega^{*}}{\nu_{k}})^{p-1} < g_{\infty} < +\infty\).
Then boundary value problem (1.1) has infinitely many positive solutions \(\{y_{k}\}_{k=1}^{\infty}\) such that \(\|y_{k}\|\) is between \(r_{k}\) and \(R_{k}, k=1, 2,\ldots\) .
Theorem 5.3
Suppose that conditions (\(\mathrm{A}'_{1}\)) and (A2) hold. Let \(\{\nu_{k}\}_{k=1}^{\infty}\) be such that \(\nu_{k}\in (t_{k+1}, t_{k})\) (\(k=1, 2, \ldots\)). Let \(\{r_{k}\}_{k=1}^{\infty}\) and \(\{R_{k}\}_{k=1}^{\infty}\) be such that \(\nu_{k}r_{k} < r_{k} < R_{k}\), \(k=1, 2,\ldots \) . For each natural number k, we assume that g satisfies
- (E5):
-
\((\frac{\omega^{*}}{\nu_{k}})^{p-1} < g_{0} < +\infty\),
- (E6):
-
\(0 < g^{\infty} < (\frac{\omega_{*}}{2})^{p-1}\).
Then boundary value problem (1.1) has infinitely many positive solutions \(\{y_{k}\}_{k=1}^{\infty}\) such that \(\|y_{k}\|\) is between \(r_{k}\) and \(R_{k}\), \(k=1, 2,\ldots\) .
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Acknowledgements
The author is very grateful to the reviewers for their thorough reading of the manuscript and for their valuable comments and excellent suggestions. She would like to express her gratitude to Professor Lishan Liu and Professor RP Agarwal for their many valuable comments. The author was supported financially by the Foundation of Shanghai Natural Science (13ZR1430100) and the NNSF of P.R. China (No. 11371095, No. 2013M541455).
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Sun, Y. Positive solutions for one-dimensional third-order p-Laplacian boundary value problems. Adv Differ Equ 2017, 95 (2017). https://doi.org/10.1186/s13662-017-1153-y
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DOI: https://doi.org/10.1186/s13662-017-1153-y