Definition 2.1
[13, 14]
Let E be a real Banach space and \(A:E\rightarrow E\) be a nonlinear operator. A nonzero solution to the equation \(x=\lambda Ax\) is called an eigenvector of the nonlinear operator A; the corresponding number λ is called a characteristic value of A, and \(\lambda^{-1}\) is called a eigenvalue of A.
Definition 2.2
[14, 16]
Let \(E_{1}\), \(E_{2}\) be real Banach spaces and \(D\subset E_{1}\) contain the outside of a ball \(\{x\vert \Vert x\Vert \leq r\}\), \(A:D\rightarrow E_{2}\). The operator A is called asymptotically linear, if there is a continuous linear operator \(B:E_{1}\rightarrow E_{2}\) such that
$$\lim_{ \Vert x\Vert \rightarrow\infty}\frac{\Vert Ax-Bx\Vert }{\Vert x\Vert }=0. $$
The operator B involved in the definition of an asymptotically linear operator A is uniquely determined, it is called the derivative of A at infinity and is denoted by \(A'(\infty)\).
Lemma 2.1
[16]
Let
E
be a Banach space and
P
be a total cone in
E. Suppose
\(T :P\rightarrow P\)
is a bounded linear operator (therefore, T
can be uniquely extended to a bounded linear operator on
\(\overline{P-P}=E\), and the extended operator is denoted by
T
again) with the spectral radius
\(r(T)<1\). If
\(w,w_{0}\in E\)
such that
\(w\leq Tw+w_{0}\), then
$$w\leq(I-T)^{-1}w_{0}, $$
where
\((I-T)^{-1}\)
is the inverse operator of the operator
\(I-T\).
Lemma 2.2
[13, 16]
Suppose that
E
is a Banach space, \(A:E\rightarrow E\)
is a completely continuous and asymptotically linear operator. If 1 is not the eigenvalue of the linear operator
\(A'(\infty)\), then there exists
\(R_{0}>0\)
such that
$$\operatorname{deg}(I-A, B_{R}, \theta)=(-1)^{\gamma} $$
for any
\(R\geq R_{0}\), where
\(B_{R}=\{x\in E\vert \Vert x\Vert < R\} \), γ
is the sum of the algebraic multiplicities of the real eigenvalues of
\(A'(\infty)\)
in
\((1,+\infty)\).
Lemma 2.3
[16]
Let
E
be a Banach space and Ω be a bounded open set in
E
with
\(\theta\in\Omega\). Suppose that
\(A:\overline{\Omega}\rightarrow E\)
is a completely continuous operator. If
$$Au\neq\mu u,\quad \forall u\in\partial\Omega, \mu\geq1, $$
then the topological degree
\(\operatorname{deg}(I-A,\Omega,\theta)=1\).
Lemma 2.4
[14]
Let
E
be a Banach space, and
P
be a cone in
E, and Ω be a bounded open set in
E. Suppose that
\(A:P\cap\overline{\Omega}\rightarrow P\)
is a completely continuous operator. If
$$Au\neq\mu u,\quad \forall u\in P\cap\partial\Omega, \mu\geq1, $$
then the fixed point index
\(i(A,P\cap\Omega,P)=1\).
Remark 2.1
Let E be a Banach space and Ω be a bounded open set in E with \(\theta\in\Omega\). Suppose that \(A:\overline{\Omega}\rightarrow E\) is a completely continuous operator. If
$$\Vert Au\Vert < \Vert u\Vert ,\quad \forall u\in\partial\Omega, $$
then the topological degree \(\operatorname{deg}(I-A,\Omega,\theta)=1\). Furthermore, suppose that P is a cone in E and satisfies
$$\Vert Au\Vert < \Vert u\Vert ,\quad \forall u\in\pm P\cap\partial \Omega, $$
then the fixed point index \(i(A,\pm P\cap\Omega,\pm P)=1\).
Given \(h\in C[0,1]\), it follows from [8] that the unique solution of the problem
$$\textstyle\begin{cases} D^{\alpha}_{0^{+}}u(t)+h(t)=0,\quad 0< t< 1, \\ u(0)=u(1)=u'(0)=u'(1)=0, \end{cases} $$
can be expressed uniquely by \(u(t)=\int_{0}^{1}G(t,s)h(s)\,ds\), where
$$ G(t,s)=\frac{1}{\Gamma(\alpha)} \textstyle\begin{cases} (t-s)^{\alpha -1}+t^{\alpha-2}(1-s)^{\alpha-2} \bigl[(s-t)+(\alpha-2) (1-t)s \bigr],\\ \quad 0\leq s\leq t\leq1, \\ t^{\alpha-2}(1-s)^{\alpha-2} \bigl[(s-t)+(\alpha-2) (1-t)s \bigr],\\ \quad 0 \leq t\leq s\leq1. \end{cases} $$
(2.1)
It is easy to verify that \(G(t,s)>0\) for \(t,s\in(0,1)\) and
$$ (\alpha-2)t^{\alpha-2}(1-t)^{2}s^{2}(1-s)^{\alpha-2} \leq\Gamma( \alpha)G(t,s)\leq M_{0}t^{\alpha-2}(1-t)^{2},\quad \forall t,s\in[0,1], $$
(2.2)
where \(M_{0}=\max\{\alpha-1,(\alpha-2)^{2}\}\).
In the following, we introduce some notations. Let \(X=C[0,1]\) and
$$P=\bigl\{ x\in X\vert x(t)\geq0,t\in[0,1]\bigr\} , $$
then P is a cone in X. Set \(e(t)=t^{\alpha-2}(1-t)^{2}\),
$$\begin{aligned}& X_{e}=\{x \in X\vert \mbox{ there exists } \lambda>0\mbox{ such that }{-} \lambda e\leq x\leq\lambda e\}, \\& \Vert x\Vert _{e}=\inf\{\lambda>0\vert-\lambda e\leq x\leq \lambda e\},\quad \forall x\in E_{e}, \end{aligned}$$
then \(\Vert x\Vert _{e}\) is called the e-norm of the element \(x\in X_{e}\). It is easy to see that \(X_{e}\) becomes a Banach space under the norm \(\Vert \cdot \Vert _{e}\) and \(P_{e}=P\cap X_{e}\) is a normal solid cone in \(X_{e}\) (\(e\in\stackrel{\circ}{ P_{e}}\)).
Before proving the following lemmas, we need the following conditions:
- (H1):
-
\(f:[0,1]\times(-\infty,+\infty)\rightarrow(-\infty,+ \infty)\) is continuous and \(f(t,x)x>0\) for all \(x\in R\setminus \{0 \}\) and \(t\in[0,1]\).
- (H2):
-
\(\lim_{x\rightarrow\infty}\frac{f(t,x)}{x}= \beta_{\infty}(t)\) uniformly with respect to \(t\in[0,1]\).
Evidently, \(\beta_{\infty}(t)\in C[0,1]\) and \(\beta_{\infty}(t) \geq0\) for \(t\in[0,1]\). Define the operators K, F such that
$$\begin{aligned}& Ku(t)= \int^{1}_{0}G(t,s)u(s)\,ds,\quad u\in X, t\in[0,1], \end{aligned}$$
(2.3)
$$\begin{aligned}& (Fu) (t)=f \bigl(t,u(t) \bigr),\quad t\in[0,1], u\in X, \end{aligned}$$
(2.4)
and \(A=KF\).
Lemma 2.5
The operator
K
defined by (2.3) satisfies
\(K: X\rightarrow X_{e}\)
and
\(K: P\setminus\{\theta\}\rightarrow\stackrel{ \circ}{ P_{e}}\), where
\(\stackrel{\circ}{P_{e}}=\{x \in X\vert \textit{ there exist } \tilde{\alpha}>0, \tilde{\beta}>0 \textit{ such that } \tilde{\alpha} e\leq x\leq\tilde{\beta} e\}\).
Proof
For any \(u\in X\), by (2.2), (2.3) we have
$$\begin{aligned}& Ku(t)= \int_{0}^{1}G(t,s)u(s)\,ds \leq\frac{M_{0}}{\Gamma(\alpha)} \int_{0}^{1} \bigl\vert u(s) \bigr\vert \,ds\cdot e(t)=\tilde{\lambda}e(t),\quad t\in[0,1], \end{aligned}$$
(2.5)
$$\begin{aligned}& Ku(t)= \int_{0}^{1}G(t,s)u(s)\,ds \geq \int_{0}^{1}G(t,s) \bigl(- \bigl\vert u(s) \bigr\vert \bigr)\,ds \geq-\tilde{\lambda}e(t),\quad t\in[0,1], \end{aligned}$$
(2.6)
where \(\tilde{\lambda}=\frac{M_{0}}{\Gamma(\alpha)}\int_{0}^{1}\vert u(s)\vert \,ds\), i.e., \(K: X\rightarrow X_{e}\).
Moreover, for \(u\in P\setminus\{\theta\}\), by (2.2), one can get
$$ Ku(t)\geq\frac{\alpha-2}{\Gamma(\alpha)} \int_{0}^{1}s^{2}(1-s)^{ \alpha-2}u(s)\,ds \cdot e(t), \quad t\in[0,1], $$
(2.7)
then it follows from (2.5), (2.7) that \(\tilde{\alpha}e(t)\leq Ku(t) \leq\tilde{\beta}e(t)\), where
$$\tilde{\alpha}=\frac{\alpha-2}{\Gamma(\alpha)}\int_{0}^{1}s^{2}(1-s)^{ \alpha-2}u(s)\,ds,\qquad \tilde{\beta}=\frac{M_{0}}{\Gamma(\alpha)}\int_{0} ^{1}u(s)\,ds, $$
i.e., \(Ku\in\mathbin{\stackrel{\circ}{ P_{e}}}\), and so we show that \(K: P_{e}\setminus\{\theta\}\rightarrow\stackrel{\circ}{ P_{e}}\). □
Lemma 2.6
Suppose that (H2) holds, then the operator
\(A: X_{e}\rightarrow X_{e}\)
is asymptotically linear, and the derivative of
A
at infinity
\(A'(\infty)=B\), where
$$Bu(t)= \int_{0}^{1}G(t,s)\beta_{\infty}(s)u(s)\,ds,\quad u \in X_{e}. $$
Proof
It follows from Lemma 2.5 and the definitions of A, B that \(A,B: X_{e}\rightarrow X_{e}\). By (H2) we know that, for any \(\varepsilon>0\), there exists \(l>0\) such that
$$\begin{aligned} \biggl\vert \frac{f(t,x)}{x}-\beta_{\infty}(t) \biggr\vert < \varepsilon \end{aligned}$$
(2.8)
for any x with \(\vert x\vert \geq l\) and \(t\in[0,1]\).
Set \(T_{l}=\{x\in X_{e}\vert \Vert x\Vert _{e}\leq l\}\), and let
$$\begin{aligned} M=\sup_{ x\in T_{l}} \bigl\{ \Vert Ax\Vert _{e},\Vert Bx\Vert _{e} \bigr\} . \end{aligned}$$
(2.9)
For \(u\in X_{e}\), define
$$\begin{aligned}& \phi_{1}\bigl(u(t)\bigr)=\textstyle\begin{cases} u(t), & \vert u(t)\vert \leq l, \\ l, & \vert u(t)\vert \geq l, \end{cases}\displaystyle \\& \phi_{2}\bigl(u(t)\bigr)= \textstyle\begin{cases} l, & \vert u(t)\vert \leq l, \\ u(t), & \vert u(t)\vert \geq l. \end{cases}\displaystyle \end{aligned}$$
Then
$$\begin{aligned}& Au(t)=A\phi_{1} \bigl(u(t) \bigr)+A\phi_{2} \bigl(u(t) \bigr)-Al(t),\quad t\in[0,1],u\in X_{e}, \end{aligned}$$
(2.10)
$$\begin{aligned}& Bu(t)=B\phi_{1} \bigl(u(t) \bigr)+B\phi_{2} \bigl(u(t) \bigr)-Bl(t),\quad t\in[0,1],u\in X _{e}, \end{aligned}$$
(2.11)
where \(l(t)\equiv l\). It follows from (2.9) that
$$\begin{aligned} \bigl\Vert A\phi_{1}(u) \bigr\Vert _{e}\leq M,\qquad \Vert Al\Vert _{e}\leq M,\qquad \bigl\Vert B\phi_{1}(u) \bigr\Vert _{e}\leq M,\qquad \Vert Bl\Vert _{e}\leq M. \end{aligned}$$
(2.12)
Therefore, by (2.10), (2.11) and (2.12), we have
$$ \Vert Au-Bu\Vert _{e}\leq 4M+ \bigl\Vert (A \phi_{2})u-(B \phi_{2})u \bigr\Vert _{e}. $$
(2.13)
From (2.8), we get
$$\begin{aligned}& \bigl\vert (A\phi_{2})u(t)-(B\phi_{2})u(t) \bigr\vert \\& \quad = \biggl\vert \int_{0}^{1}G(t,s) \bigl[f \bigl(s, \phi_{2} \bigl(u(s) \bigr) \bigr)-\beta_{\infty}(s) \phi_{2} \bigl(u(s) \bigr) \bigr]\,ds \biggr\vert \\& \quad \leq \int_{0}^{1}G(t,s) \bigl\vert f \bigl(s, \phi_{2} \bigl(u(s) \bigr) \bigr)-\beta_{\infty}(s) \phi_{2} \bigl(u(s) \bigr) \bigr\vert \,ds \\& \quad \leq\frac{M_{0}}{\Gamma(\alpha)}\varepsilon \Vert u\Vert _{e} e(t) \end{aligned}$$
(2.14)
for any \(u\in X_{e}\), from which one deduces that
$$\bigl\Vert (A\phi_{2})u-(B\phi_{2})u\bigr\Vert \leq \frac{M_{0}}{\Gamma (\alpha)} \varepsilon \Vert u\Vert _{e} e(t), $$
and therefore from (2.13), one can get
$$\lim_{ \Vert u\Vert _{e}\rightarrow\infty}\frac{\Vert Au-Bu\Vert _{e}}{\Vert u\Vert _{e}}=0, $$
i.e., the operator A is asymptotically linear, and \(A'(\infty)=B\).
In this paper, we always denote by \(\Omega_{r}=\{u\in X:\Vert u\Vert _{e}< r\} \) (\(r>0\)) the open ball of radius r and by θ the zero function in \(X_{e}\). For the concepts and properties on the cone and the topological degree, one can refer to [12, 14, 16]. □