Recall that \(E=\{\lambda>0 \mid \mbox{(1.1) has a positive solution}\}\). In this section, we shall establish criteria for E to contain an interval (Theorem 3.5), and for E to be an interval (Corollary 3.7), which may either be bounded or unbounded (Theorem 3.9).
To begin, we consider the initial value problem
$$ \left \{ \begin{array}{@{}l} y^{(q)}(t)=x(t),\quad t\in(0,1),\\ y(0)=y'(0)=y''(0)=\cdots=y^{(q-1)}(0)=0. \end{array} \right . $$
(3.1)
Noting the initial conditions in (3.1), we have by repeated integration
$$ y^{(k)}(t)=\int_{0}^{t}\int _{0}^{s_{1}}\int_{0}^{s_{2}} \cdots\int_{0}^{s_{q-k-1}}x (s_{q-k} )\,ds_{q-k}\cdots \,ds_{1},\quad 0\leq k\leq q-1. $$
(3.2)
Denote the integral
$$ J^{k}x(t)=\int_{0}^{t}\int _{0}^{s_{1}}\int_{0}^{s_{2}} \cdots\int_{0}^{s_{k-1}}x (s_{k} )\,ds_{k}\cdots \,ds_{1},\quad k\geq1. $$
(3.3)
Then (3.2) is simply
$$ y^{(k)}(t)= J^{q-k}x(t),\quad 0\leq k\leq q-1. $$
(3.4)
In view of (3.1) and (3.4), we can rewrite (1.1) as the following \((m-q)\)
th order Sturm-Liouville boundary value problem:
$$ \left \{ \begin{array}{@{}l} x^{(m-q)}(t)+\lambda F (t,\tilde{J}x(t))=0,\quad t\in(0,1), \\ x^{(k)}(0)=0,\quad 0\leq k\leq m-q-3, \\ \zeta x^{(m-q-2)}(0)-\theta x^{(m-q-1)}(0)=0,\qquad \rho x^{(m-q-2)}(1)+\delta x^{(m-q-1)}(1)=0, \end{array} \right . $$
(3.5)
where we denote
$$\tilde{J}x(t)= \bigl(J^{q}x(t),J^{q-1}x(t),\ldots,Jx(t),x(t) \bigr). $$
If (3.5) has a solution \(x^{*}\), then the boundary value problem (1.1) has a solution \(y^{*}\) given by
$$ y^{*}(t)=J^{q}x^{*}(t)=\int_{0}^{t}\int _{0}^{s_{1}}\int_{0}^{s_{2}} \cdots\int_{0}^{s_{q-1}}x^{*} (s_{q} )\,ds_{q}\cdots \,ds_{1}. $$
(3.6)
Hence, the existence of a solution of (1.1) follows from the existence of a solution of (3.5). Further, it is obvious from (3.6) that \(y^{*}\) is positive if \(x^{*}\) is. An eigenvalue of (3.5) is thus also an eigenvalue of (1.1), i.e.,
$$E= \bigl\{ \lambda>0 \mid (1.1)\mbox{ has a positive solution} \bigr\} = \bigl\{ \lambda>0 \mid (3.5) \mbox{ has a positive solution} \bigr\} . $$
We shall study the eigenvalue problem (1.1) via (3.5) and a new technique will be developed to handle the nonlinear term F.
For easy reference, the conditions that will be mentioned later are listed below.
-
(C1)
There exist continuous functions \(f:(0,\infty)^{q+1} \to(0,\infty)\) and \(a,b:(0,1)\to[0,\infty)\) such that for \(t\in(0,1)\) and \(u_{j}\in(0,\infty)\), \(1\leq j\leq q+1\),
$$a(t)f(u_{1},\ldots,u_{q+1})\leq F(t,u_{1}, \ldots,u_{q+1})\leq b(t)f(u_{1},\ldots,u_{q+1}). $$
-
(C2)
\(a(t)\) is not identically zero on any nondegenerate subinterval of \((0,1)\) and there exists \(r\in (0,1]\) such that \(a(t)\geq rb(t)\) for all \(t\in(0,1)\).
-
(C3)
\(0<\int_{0}^{1}(\theta+\zeta t)[\delta +\rho(1-t)]b(t)\,dt<\infty\).
-
(C4)
f is nondecreasing in each of its arguments, i.e., for \(u_{j},v,w\in(0,\infty)\), \(1\leq j\leq q+1\) with \(v\leq w\), we have
$$f(u_{1},\ldots,u_{i-1},v,u_{i+1}, \ldots,u_{q+1})\leq f(u_{1},\ldots ,u_{i-1},w,u_{i+1}, \ldots,u_{q+1}),\quad 1\leq i\leq q+1. $$
-
(C5)
For \(t\in(0,1)\) and \(u_{j},v,w\in(0,\infty)\), \(1\leq j\leq q+1\) with \(v\leq w\), we have
$$\begin{aligned} F(t,u_{1},\ldots,u_{i-1},v,u_{i+1}, \ldots,u_{q+1})\leq F(t,u_{1},\ldots ,u_{i-1},w,u_{i+1}, \ldots,u_{q+1}),& \\ 1\leq i\leq q+1.& \end{aligned}$$
Let the Banach space
$$B= \bigl\{ x\in C^{(m-q)}(0,1)\cap C^{(m-q-1)}[0,1] \mid x^{(k)}(0)=0, 0\leq k\leq m-q-3 \bigr\} $$
be equipped with the norm
$$\| x\|=\sup_{t\in[0,1]}\bigl| x^{(m-q-2)}(t)\bigr|. $$
Throughout the paper, let \(\eta\in (0,\frac{1}{2} )\) be fixed. Define the cone C in B by
$$ C= \Bigl\{ x\in B \bigm| x^{(m-q-2)}(t)\geq0, t\in [0,1]; \min _{t\in[\eta,1-\eta]} x^{(m-q-2)}(t)\geq\gamma\|x\| \Bigr\} , $$
(3.7)
where \(\gamma=rK_{\eta}/L\). For a constant \(M>0\), let
$$C(M)=\bigl\{ x\in C \mid \|x\|\leq M\bigr\} . $$
Lemma 3.1
[36, 37]
Let
\(x\in B\). For
\(0\leq i\leq m-q-2\), we have
$$ \bigl|x^{(i)}(t)\bigr|\leq\frac{t^{m-q-2-i}}{(m-q-2-i)!} \|x\|,\quad t\in[0,1]. $$
(3.8)
In particular,
$$ \bigl|x(t)\bigr|\leq\frac{1}{(m-q-2)!} \|x\|,\quad t\in[0,1]. $$
(3.9)
Lemma 3.2
[36, 37]
Let
\(x\in C\). For
\(0\leq i\leq m-q-2\), we have
$$ x^{(i)}(t)\geq0,\quad t\in[0,1], $$
(3.10)
and
$$ x^{(i)}(t)\geq(t-\eta)^{m-q-2-i}\frac{\gamma}{(m-q-2-i)!} \|x\|,\quad t \in[ \eta,1-\eta]. $$
(3.11)
In particular, we have, for fixed
\(z\in(\eta,1-\eta)\),
$$ x(t)\geq(z-\eta)^{m-q-2}\frac{\gamma}{(m-q-2)!} \|x\|,\quad t\in [z,1-\eta]. $$
(3.12)
Remark 3.1
If \(x^{*}\in C\) is a nontrivial solution of (3.5), then (3.10) and (3.12) imply that \(x^{*}\) is a positive solution of (3.5). As noted earlier a positive solution \(y^{*}\) of (1.1) can be obtained via (3.6).
The next result is useful in handling the nonlinear term F.
Lemma 3.3
Let
\(x\in C\)
and let
\(z\in(\eta,1-\eta)\)
be fixed. Then we have, for
\(1\leq k\leq q\),
$$ J^{k} x(t)\leq\frac{1}{(m-q-2)!} \|x\|,\quad t\in[0,1], $$
(3.13)
and
$$ J^{k} x(t)\geq(z-\eta)^{m-q-2+k}\frac{\gamma}{(m-q-2+k)!} \|x\| ,\quad t \in[z,1-\eta]. $$
(3.14)
Proof
Since \(x\in C\subset B\), using (3.9) we obtain for \(1\leq k\leq q\) and \(t\in[0,1]\),
$$\begin{aligned} J^{k}x(t) =& \int_{0}^{t}\int _{0}^{s_{1}}\int_{0}^{s_{2}} \cdots\int_{0}^{s_{k-1}}x (s_{k} )\,ds_{k}\cdots \,ds_{1} \\ \leq& \int_{0}^{1}\int_{0}^{1} \int_{0}^{1}\cdots\int_{0}^{1} \frac{\|x\|}{(m-q-2)!}\,ds_{k}\cdots \,ds_{1} = \frac{\|x\|}{(m-q-2)!}. \end{aligned}$$
Next, since \(x\in C\), it follows from (3.11) that
$$ x(t)\geq(t-\eta)^{m-q-2}\frac{\gamma\|x\|}{(m-q-2)!},\quad t\in[\eta,1-\eta]. $$
(3.15)
Let \(t\in[z,1-\eta]\). Using (3.15) we find, for \(1\leq k\leq q\),
$$\begin{aligned} J^{k} x(t) =&\int_{0}^{t}\int _{0}^{s_{1}}\int_{0}^{s_{2}} \cdots\int_{0}^{s_{k-1}}x (s_{k} )\,ds_{k}\cdots \,ds_{1} \geq\int_{\eta}^{z}\int_{\eta}^{s_{1}} \int_{\eta}^{s_{2}}\cdots\int_{\eta}^{s_{k-1}}x (s_{k} )\,ds_{k}\cdots \,ds_{1} \\ \geq&\int_{\eta}^{z}\int_{\eta}^{s_{1}} \int_{\eta}^{s_{2}}\cdots\int_{\eta}^{s_{k-1}}(s_{k}- \eta)^{m-q-2} \frac{\gamma\|x\|}{(m-q-2)!}\,ds_{k}\cdots \,ds_{1} \\ =& (z-\eta)^{m-q-2+k}\frac{\gamma\|x\|}{(m-q-2+k)!}. \end{aligned}$$
□
In view of Remark 3.1, to obtain a positive solution of (3.5), we shall seek a fixed point of the operator S in the cone C, where \(S:C\rightarrow B\) is defined by
$$ Sx(t)= \lambda\int_{0}^{1} g_{m-q}(t,s)F \bigl(s,\tilde{J}x(s) \bigr)\,ds,\quad t\in[0,1]. $$
(3.16)
Recall that \(g_{m-q}(t,s)\) (see Lemma 2.2(d)) is the Green’s function of ((2.3)
n
)\(_{m-q}\), thus (3.16) is equivalent to
$$ (Sx)^{(m-q-2)}(t)= \lambda\int_{0}^{1} G(t,s)F \bigl(s,\tilde{J}x(s) \bigr)\,ds,\quad t\in[0,1], $$
(3.17)
where \(G(t,s)\) is the Green’s function of (2.1).
We further define the operators \(U,V:C\to B\) by
$$Ux(t)=\lambda\int_{0}^{1}g_{m-q}(t,s)a(s)f \bigl(\tilde{J}x(s) \bigr)\,ds \quad\mbox{and}\quad Vx(t)=\lambda\int _{0}^{1}g_{m-q}(t,s)b(s)f \bigl(\tilde{J}x(s) \bigr)\,ds. $$
As in (3.17), differentiating gives
$$(Ux)^{(m-q-2)}(t)=\lambda\int_{0}^{1}G(t,s)a(s)f \bigl(\tilde{J}x(s) \bigr)\,ds $$
and
$$(Vx)^{(m-q-2)}(t)=\lambda\int_{0}^{1}G(t,s)b(s)f \bigl(\tilde{J}x(s) \bigr)\,ds. $$
If (C1) is satisfied, then it is clear that
$$ Ux(t)\leq Sx(t)\leq Vx(t),\quad t\in[0,1], $$
(3.18)
and
$$ (Ux)^{(m-q-2)}(t)\leq(Sx)^{(m-q-2)}(t)\leq(Vx)^{(m-q-2)}(t),\quad t\in [0,1]. $$
(3.19)
Lemma 3.4
Let (C1)-(C4) hold. Then the operator
S
is compact on the cone
C.
Proof
Let us consider the case when \(a(t)\) is unbounded in a deleted right neighborhood of 0 and also in a deleted left neighborhood of 1. Clearly, \(b(t)\) is also unbounded near 0 and 1. For \(n\in\{1,2,3,\ldots\}\), let \(a_{n}, b_{n}:[0,1]\to[0,\infty)\) be defined by
$$a_{n}(t)= \left \{ \begin{array}{@{}l@{\quad}l} a ( \frac{1}{n+1} ),& 0\leq t\leq \frac{1}{n+1} ,\\ a(t), & \frac{1}{n+1}\leq t\leq \frac{n}{n+1}, \\ a ( \frac{n}{n+1} ),& \frac{n}{n+1}\leq t\leq1, \end{array} \right . $$
and
$$b_{n}(t)= \left \{ \begin{array}{@{}l@{\quad}l} b ( \frac{1}{n+1} ), & 0\leq t\leq \frac{1}{n+1}, \\ b(t), & \frac{1}{n+1}\leq t\leq \frac{n}{n+1},\\ b ( \frac{n}{n+1} ), & \frac {n}{n+1}\leq t\leq1. \end{array} \right . $$
Also, we define the operators \(U_{n},V_{n}:C\to B\) by
$$U_{n}x(t)=\lambda\int_{0}^{1}g_{m-q}(t,s)a_{n}(s)f \bigl(\tilde{J}x(s) \bigr)\,ds $$
and
$$V_{n}x(t)=\lambda\int_{0}^{1}g_{m-q}(t,s)b_{n}(s)f \bigl(\tilde{J}x(s) \bigr)\,ds. $$
It is standard that, for each n, both \(U_{n}\) and \(V_{n}\) are compact operators on C. Let \(M>0\) and \(x\in C(M)\). For \(t\in [0,1]\), we obtain
$$\begin{aligned} \bigl\vert V_{n}x(t)-Vx(t)\bigr\vert \leq& {\lambda\int _{0}^{1}g_{m-q}(t,s) \bigl|b_{n}(s)-b(s)\bigr|f \bigl(\tilde{J}x(s) \bigr)\,ds} \\ =& \lambda\int_{0}^{\frac{1}{n+1}}g_{m-q}(t,s) \bigl|b_{n}(s)-b(s)\bigr|f \bigl(\tilde{J}x(s) \bigr)\,ds \\ &{} + \lambda\int^{1}_{\frac{n}{n+1}} g_{m-q}(t,s)\bigl|b_{n}(s)-b(s)\bigr|f \bigl(\tilde{J}x(s) \bigr)\,ds \\ =& \lambda\int_{0}^{\frac{1}{n+1}}g_{m-q}(t,s) \biggl\vert b_{n} \biggl(\frac{1}{n+1} \biggr)-b(s)\biggr\vert f \bigl(\tilde{J}x(s) \bigr)\,ds \\ &{} + \lambda\int^{1}_{\frac{n}{n+1}} g_{m-q}(t,s) \biggl\vert b_{n} \biggl(\frac{n}{n+1} \biggr)-b(s)\biggr\vert f \bigl(\tilde{J}x(s) \bigr)\,ds. \end{aligned}$$
Since \(\|x\|\leq M\), from (3.13) we get
$$J^{k} x(s)\leq\frac{M}{(m-q-2)!},\quad s\in[0,1], 1\leq k\leq q. $$
Hence, together with (3.9), it follows from the monotonicity of f (condition (C4)) that
$$\begin{aligned} f \bigl(\tilde{J}x(s) \bigr) =& f \bigl(J^{q}x(s),J^{q-1}x(s), \ldots,Jx(s),x(s) \bigr) \\ \leq& f \biggl(\frac{M}{(m-q-2)!},\frac{M}{(m-q-2)!},\ldots, \frac{M}{(m-q-2)!} \biggr) \equiv \bar{f}_{M}. \end{aligned}$$
(3.20)
Applying (3.20) and Lemma 2.2(e), we obtain
$$\begin{aligned} \bigl\vert V_{n}x(t)-Vx(t)\bigr\vert \leq& \lambda\bar{f}_{M} \biggl[\int_{0} ^{\frac{1}{n+1}} \frac{LG(s,s)}{(m-q-2)!}\biggl\vert b \biggl(\frac {1}{n+1} \biggr)-b(s)\biggr\vert \,ds \\ &{}+ \int^{1}_{\frac{n}{n+1}} \frac{LG(s,s)}{(m-q-2)!}\biggl\vert b \biggl(\frac{n}{n+1} \biggr)-b(s)\biggr\vert \,ds \biggr]. \end{aligned}$$
The integrability of \(G(s,s)b(s)\) (which is simply (C3)) ensures that \(V_{n}\) converges uniformly to V on \(C(M)\). Hence, V is compact on C. By a similar argument, we see that \(U_{n}\) converges uniformly to U on \(C(M)\) and therefore U is also compact on C. It follows immediately from inequality (3.18) that the operator S is compact on C. □
Remark 3.2
From the proof of Lemma 3.4, we see that if the functions a and b are continuous on the close interval \([0,1]\), then the conditions (C3) and (C4) are not needed in Lemma 3.4.
The first main result shows that E contains an interval.
Theorem 3.5
Let (C1)-(C4) hold. Then there exists
\(\ell>0\)
such that the interval
\((0,\ell] \subseteq E\).
Proof
For a given \(M>0\), we define
$$ \ell=M \biggl[\bar{f}_{M} L\int_{0}^{1}G(s,s)b(s)\,ds \biggr]^{-1}. $$
(3.21)
Let \(\lambda\in(0, \ell]\). We shall prove that \(S (C(M))\subseteq C(M)\). For this, let \(x\in C(M)\). First, we shall show that \(Sx\in C\). From (3.19), it is clear that
$$ (Sx)^{(m-q-2)}(t)\geq\lambda\int_{0}^{1}G(t,s)a(s) f \bigl(\tilde{J}x(s) \bigr)\,ds\geq0,\quad t\in[0,1]. $$
(3.22)
Also, (3.19) and Lemma 2.2(b) provide
$$(Sx)^{(m-q-2)}(t) \leq\lambda\int_{0}^{1}G(t,s)b(s) f \bigl(\tilde{J}x(s) \bigr)\,ds \leq \lambda\int_{0}^{1}LG(s,s) b(s)f \bigl( \tilde{J}x(s) \bigr)\,ds,\quad t\in[0,1], $$
which immediately implies
$$ \|Sx\|\leq\lambda\int_{0}^{1}LG(s,s) b(s)f \bigl(\tilde{J}x(s) \bigr)\,ds. $$
(3.23)
Further, using (3.19), Lemma 2.2(c), (C2), and (3.23) successively, we find, for \(t\in [\eta,1-\eta]\),
$$\begin{aligned} (Sx)^{(m-q-2)}(t) \geq& \lambda\int_{0}^{1}G(t,s)a(s)f \bigl(\tilde{J}x(s) \bigr)\,ds \\ \geq& {\lambda\int_{0}^{1}K_{\eta}G(s,s)a(s) f \bigl(\tilde{J}x(s) \bigr)\,ds} \\ \geq& {\lambda\int_{0}^{1}K_{\eta}G(s,s) rb(s) f \bigl(\tilde{J}x(s) \bigr)\,ds} \\ \geq& \frac{K_{\eta}}{L} r\|Sx\| = \gamma\|Sx\|. \end{aligned}$$
Hence,
$$ \min_{t\in[\eta,1-\eta]} (Sx)^{(m-q-2)}(t)\geq\gamma\| Sx\|. $$
(3.24)
Inequalities (3.22) and (3.24) show that \(Sx\in C\).
Next, we shall prove that \(\|Sx\|\leq M\). Noting that \(\|x\|\leq M\), we use (3.20), Lemma 2.2(b), and (3.21) to get
$$\begin{aligned} (Sx)^{(m-q-2)}(t) \leq& \ell \int_{0}^{1}G(t,s)b(s) f \bigl(\tilde{J}x(s) \bigr)\,ds \\ \leq& {\ell\bar{f}_{M}\int_{0}^{1} LG(s,s)b(s)\,ds} = M, \quad t\in[0,1], \end{aligned}$$
or equivalently
Hence, \(S(C(M))\subseteq C(M)\). Also, the standard arguments yield that S is completely continuous. By Schauder’s fixed point theorem, S has a fixed point in \(C(M)\). Clearly, this fixed point is a positive solution of (3.5) and therefore λ is an eigenvalue of (3.5). Noting that \(\lambda\in(0, \ell]\) is arbitrary, it follows immediately that the interval \((0,\ell] \subseteq E\). □
Remark 3.3
From the proof of Theorem 3.5, it is clear that conditions (C1) and (C2) ensure that \(S:C\to C\).
Theorem 3.6
Let (C1)-(C5) hold. Suppose that
\(\lambda^{*}\in E\). For any
\(\lambda\in(0,\lambda^{*})\), we have
\(\lambda\in E\), i.e., \((0,\lambda^{*}]\subseteq E\).
Proof
Let \(x^{*}\) be the eigenfunction corresponding to the eigenvalue \(\lambda^{*}\). Then we have
$$ x^{*}(t)= Sx^{*}(t) = \lambda^{*}\int_{0}^{1} g_{m-q}(t,s)F \bigl(s,\tilde{J}x^{*}(s) \bigr)\,ds,\quad t\in[0,1]. $$
(3.25)
Define
$$A= \bigl\{ x\in B \mid 0\leq x(t)\leq x^{*}(t), t\in[0,1] \bigr\} . $$
Let \(\lambda\in (0,\lambda^{*})\) and \(x\in A\). It is obvious from definition (3.3) that
$$J^{k}x(t)\leq J^{k}x^{*}(t),\quad t\in[0,1], 1\leq k\leq q. $$
Now, applying (C5) and noting (3.25), we obtain
$$\begin{aligned} 0 \leq& Sx(t)= \lambda\int_{0}^{1} g_{m-q}(t,s)F \bigl(s,\tilde{J}x(s) \bigr)\,ds \\ \leq&\lambda^{*}\int_{0}^{1} g_{m-q}(t,s)F \bigl(s,\tilde{J}x^{*}(s) \bigr)\,ds \\ =& Sx^{*}(t),\quad t\in[0,1]. \end{aligned}$$
This shows that the operator S maps A into A. Moreover, the operator S is continuous and completely continuous. Schauder’s fixed point theorem guarantees that S has a fixed point in A which is a positive solution of (3.5). Hence, λ is an eigenvalue of (3.5), i.e., \(\lambda\in E\). □
The next result states that E is itself an interval.
Corollary 3.7
Let (C1)-(C5) hold. If
\(E\neq\emptyset\), then
E
is an interval.
Proof
Suppose E is not an interval. Then there exist \(\lambda_{0}, \lambda_{1}\in E\) (\(\lambda_{0}<\lambda_{1}\)), and \(\tau\in (\lambda_{0}, \lambda_{1})\) with \(\tau\notin E\). However, this is not possible as Theorem 3.6 guarantees that \(\tau\in E\). Hence, E is an interval. □
The next result gives upper and lower bounds of an eigenvalue.
Theorem 3.8
Let (C1)-(C4) hold. Let
λ
be an eigenvalue of (3.5) and
\(x\in C\)
be a corresponding eigenfunction. Further, let
\(\|x\|=p\)
and
\(z\in(\eta,1-\eta)\)
be fixed. Then
$$ \lambda\geq\frac{p}{f (\frac{p}{(m-q-2)!},\frac {p}{(m-q-2)!},\ldots,\frac{p}{(m-q-2)!} )} \biggl[\int_{0}^{1} LG(s,s)b(s)\,ds \biggr]^{-1}, $$
(3.26)
and
$$\begin{aligned} \lambda\leq{}&\frac{p}{f (\frac{\gamma p}{(m-2)!}(z-\eta)^{m-2},\frac{\gamma p}{(m-3)!}(z-\eta)^{m-3},\ldots,\frac{\gamma p}{(m-q-1)!}(z-\eta)^{m-q-1},\frac{\gamma p}{(m-q-2)!}(z-\eta)^{m-q-2} )} \\ &{}\times \biggl[\int _{z}^{1-\eta} G(t_{0},s)a(s)\,ds \biggr]^{-1}, \end{aligned}$$
(3.27)
where
\(t_{0}\)
is any number in
\((0,1)\)
such that
\(x^{(m-q-2)}(t_{0})\neq0\).
Proof
Let \(t_{1}\in[0,1]\) be such that
$$p=\|x\|=x^{(m-q-2)}(t_{1}). $$
Then, applying (3.19), Lemma 2.2(b), (3.20), and (C4), we find
$$\begin{aligned} p =& x^{(m-q-2)}(t_{1})= (Sx)^{(m-q-2)}(t_{1}) \\ \leq& \lambda \int_{0}^{1}G(t_{1},s)b(s)f \bigl(\tilde{J}x(s) \bigr)\,ds \\ \leq& {\lambda\int_{0}^{1}LG(s,s) b(s) \bar{f}_{p} \,ds,} \end{aligned}$$
which gives (3.26) immediately.
Next, noting (3.19), (3.14), (3.12), and (C4), we get
$$\begin{aligned} p \geq& x^{(m-q-2)}(t_{0}) \\ \geq& \lambda {\int_{0}^{1}G(t_{0},s)a(s) f \bigl(\tilde{J}x(s) \bigr)\,ds} \\ \geq& \lambda {\int_{z}^{1-\eta} G(t_{0},s)a(s) f \bigl(\tilde{J}x(s) \bigr)\,ds} \\ \geq& \lambda{ \int_{z}^{1-\eta}} G(t_{0},s)a(s)\,ds \\ &{} \times f \biggl(\frac{\gamma p}{(m-2)!}(z-\eta)^{m-2}, \frac {\gamma p}{(m-3)!}(z-\eta)^{m-3},\ldots,\frac{\gamma p}{(m-q-1)!}(z- \eta)^{m-q-1},\\ &{}\frac{\gamma p}{(m-q-2)!}(z-\eta)^{m-q-2} \biggr), \end{aligned}$$
from which (3.27) is immediate. □
The next result gives the criteria for E to be a bounded/unbounded interval.
Theorem 3.9
Define
$$\begin{aligned}& P_{B}= \biggl\{ f \Bigm| \frac {u}{f(u,u,\ldots,u)} \textit{ is bounded for }u\in(0, \infty) \biggr\} , \\& P_{0}= \biggl\{ f \Bigm| {\lim_{u\to\infty}\frac{u}{ f(u,u,\ldots,u)}=0} \biggr\} , \\& P_{\infty}= \biggl\{ f \Bigm| {\lim_{u\to\infty }\frac{u}{ f(u,u,\ldots,u)}= \infty} \biggr\} . \end{aligned}$$
-
(a)
Let (C1)-(C5) hold. If
\(f\in P_{B}\), then
\(E=(0,\ell)\)
or
\((0,\ell]\)
for some
\(\ell\in(0, \infty)\).
-
(b)
Let (C1)-(C5) hold. If
\(f\in P_{0}\), then
\(E=(0,\ell]\)
for some
\(\ell\in(0,\infty)\).
-
(c)
Let (C1)-(C4) hold. If
\(f\in P_{\infty}\), then
\(E=(0,\infty)\).
Proof
(a) This follows from (3.27) and Corollary 3.7.
(b) Since \(P_{0}\subseteq P_{B}\), we have from Case (a) that \(E= (0,\ell)\) or \((0,\ell]\) for some \(\ell\in(0,\infty)\). In particular,
Let \(\{\lambda_{n}\}_{n=1}^{\infty}\) be a monotonically increasing sequence in E which converges to ℓ, and let \(\{x_{n}\}_{n=1}^{\infty}\) be a corresponding sequence of eigenfunctions in the context of (3.5). Further, let \(p_{n}= \| x_{n}\|\). Then (3.27) together with \(f\in P_{0}\) implies that no subsequence of \(\{p_{n}\}_{n=1}^{\infty}\) can diverge to infinity. Thus, there exists \(M>0\) such that \(p_{n}\leq M\) for all n. So \(\{x_{n}\}_{n=1}^{\infty}\) is uniformly bounded. This implies that there is a subsequence of \(\{x_{n}\}_{n=1}^{\infty}\), relabeled as the original sequence, which converges uniformly to some x, where \(x(t) \geq0\) for \(t\in[0,1]\). Clearly, we have \(Sx_{n}=x_{n}\), i.e.,
$$ x_{n}(t)=\lambda_{n}\int_{0}^{1} g_{m-q}(t,s)F \bigl(s,\tilde{J}x_{n}(s) \bigr)\,ds,\quad t \in[0,1]. $$
(3.28)
Since \(x_{n}\) converges to x and \(\lambda_{n}\) converges to ℓ, letting \(n\to\infty\) in (3.28) leads to
$$x(t)=\ell\int_{0}^{1} g_{m-q}(t,s)F \bigl(s,\tilde{J}x(s) \bigr)\,ds,\quad t\in[0,1]. $$
Hence, ℓ is an eigenvalue with corresponding eigenfunction x, i.e., \(\ell=\sup E\in E\). This completes the proof for Case (b).
(c) Let \(\lambda>0\) be fixed. Choose \(\epsilon>0\) so that
$$ \lambda \frac{L}{(m-q-2)!}\int_{0}^{1} G(s,s)b(s)\,ds\leq\frac {1}{\epsilon}. $$
(3.29)
If \(f\in P_{\infty}\), then there exists \(M=M(\epsilon)>0\) such that
$$ f(u,u,\ldots,u)<\epsilon u, \quad u\geq\frac{M}{(m-q-2)!}. $$
(3.30)
We shall show that \(S(C(M))\subseteq C(M)\). Let \(x\in C(M)\). From the proof of Theorem 3.5, we have (3.22) and (3.24) and so \(Sx\in C\). It remains to show that \(\|Sx\|\leq M\). Applying (3.19), Lemma 2.2(b), (3.20), (3.30), and (3.29), we find, for \(t\in[0,1]\),
$$\begin{aligned} (Sx)^{(m-q-2)}(t) \leq& \lambda \int_{0}^{1}G(t,s)b(s)f \bigl(\tilde{J}x(s) \bigr)\,ds \\ \leq& {\lambda\int_{0}^{1}LG(s,s)b(s) f \biggl( \frac{M}{(m-q-2)!},\frac{M}{(m-q-2)!},\ldots, \frac{M}{(m-q-2)!} \biggr)\,ds} \\ \leq& {\lambda\epsilon \frac{M}{(m-q-2)!}\int_{0}^{1} LG(s,s)b(s)\,ds} \leq M. \end{aligned}$$
It follows that \(\|Sx\|\leq M\) and hence \(S(C(M))\subseteq C(M)\). Also, S is continuous and completely continuous. Schauder’s fixed point theorem guarantees that S has a fixed point in \(C(M)\). Clearly, this fixed point is a positive solution of (3.5) and therefore λ is an eigenvalue of (3.5). Since \(\lambda>0\) is arbitrary, it shows that \(E=(0,\infty)\). □
Example 3.1
Consider the Sturm-Liouville boundary value problem
$$ \left \{ \begin{array}{@{}l} y^{(5)}(t)+\lambda F (t,y(t),y'(t),y''(t),y'''(t) )=0, \quad t\in(0,1),\\ y(0)=y'(0)=y''(0)=0,\qquad 2y^{(3)}(0)-y^{(4)}(0)=0,\qquad -y^{(3)}(1)+3y^{(4)}(1)=0, \end{array} \right . $$
(3.31)
where \(\lambda>0\) and
$$\begin{aligned} F\bigl(t,y,y',y'',y''' \bigr) =& \biggl[\frac{300+900t+270t^{2}+90t^{3}+10t^{4}-3t^{5}}{200}+2 \biggr]^{-1/2} \\ &{}\times \biggl(\frac{y+y'+y''+y'''}{4} +2 \biggr)^{1/2}. \end{aligned}$$
Here, \(m=5\), \(q=3\), \(\zeta=2\), \(\theta=1\), \(\rho=-1\), and \(\delta=3\). Clearly, (C1)-(C5) are satisfied with
$$a(t)=b(t)= \biggl[\frac{300+900t+270t^{2}+90t^{3}+10t^{4}-3t^{5}}{200}+2 \biggr]^{-1/2} $$
and
$$f(u_{1},u_{2},u_{3}, u_{4})= \biggl( \frac{u_{1}+u_{2}+u_{3}+u_{4}}{4}+2 \biggr)^{1/2}. $$
It is obvious that \(f\in P_{\infty}\). Hence, by Theorem 3.9(c) we have \(E=(0,\infty)\). In fact, when \(\lambda=\frac{36}{5}\in(0,\infty)\), (3.31) has a positive solution \(y(t)=t^{3}+\frac{1}{2}t^{4}-\frac{3}{50}t^{5}\).