Open Access

Nontrivial solutions to boundary value problem with general singular differential operator

Boundary Value Problems20152015:70

https://doi.org/10.1186/s13661-015-0331-7

Received: 7 October 2014

Accepted: 13 April 2015

Published: 25 April 2015

Abstract

In this paper, a class of boundary value problems with general singular differential operator is investigated. The nonlinear term in the boundary value problem is sign-changing and may be unbounded from below. By means of the topological degree of a completely continuous field, the existence of nontrivial solutions is obtained. Finally, an example is given to illustrate the application of our main result.

Keywords

singularitieseigenvaluetopological degreenontrivial solutions

MSC

34B16

1 Introduction

In [1], Han and Wu considered the singular boundary value problem
$$ \begin{aligned} &u''+h(t)f(u)=0, \quad t\in(0,1), \\ &u(0)=u(1)=0 \end{aligned} $$
(1.1)
under the following assumptions:
  1. (H1)
    There exist three constants \(b>0\), \(c>0\) and \(\alpha\in (0,1)\) such that
    $$f(u)\geq-b-c|u|^{\alpha}, \quad u\in\mathbb{R}. $$
     
  2. (H2)

    \(h\in C((0,1), [0,+\infty))\), \(h(t)\not\equiv0\) in \((0,1)\) and \(\int_{0}^{1}t(1-t)h(t)\,\mathrm{d}t<+\infty\).

     
  3. (H3)

    \(f:\mathbb{R}\rightarrow \mathbb{R}\) is continuous.

     
Denote by \(\tilde{\lambda}_{1}\) the first eigenvalue of the eigenvalue problem
$$ u''+\lambda h(t)u=0, \qquad u(0)=u(1)=0. $$
(1.2)
By computing the Leray-Schauder degree, they established the following result.

Theorem A

Assume that (H1)-(H3) hold. If
$$\liminf_{u\to+\infty}\frac{f(u)}{u}>\tilde{\lambda}_{1} \quad\textit{and} \quad \limsup_{u\to0} \biggl| \frac{f(u)}{u} \biggr|< \tilde{\lambda}_{1}, $$
then the singular boundary value problem (1.1) has at least one nontrivial solution.
This paper is mainly focused on the existence of nontrivial solutions to singular boundary value problem with general differential operator such as
$$ \begin{aligned} &u''+a(t)u'+b(t)u+ h(t)f(t,u)=0, \quad t\in(0,1), \\ &u(0)=u(1)=0. \end{aligned} $$
(1.3)
We will make the following assumptions:
  1. (A1)

    \(a\in C(0,1)\cap L^{1}(0,1)\); \(b\in C((0, 1),(-\infty,0))\) and \(\int_{0}^{1}s(1-s)|b(s)|\,\mathrm{d}s<+\infty\).

     
  2. (A2)

    \(h:(0, 1)\rightarrow[0, \infty)\) is continuous and does not vanish identically on any subinterval of \((0,1)\); furthermore, \(\int_{0}^{1} s(1-s)h(s)\,\mathrm{d}s<+\infty\).

     
  3. (A3)
    \(f: [0, 1]\times \mathbb{R}\rightarrow \mathbb{R}\) is continuous and there exist two nonnegative functions \(c, d\in C[0,1]\), \(d\not\equiv0\) and \(B\in C(\mathbb{R}, [0,+\infty))\) with
    $$\begin{aligned}& \lim_{|u|\to+\infty}\frac{B(u)}{|u|}=0 \end{aligned}$$
    (1.4)
    such that
    $$\begin{aligned}& f(t, u)\geq-c(t)-d(t) B(u), \quad t\in[0,1], u\in\mathbb{R}. \end{aligned}$$
    (1.5)
     
  4. (A4)

    \(\liminf_{u\to+\infty}\frac{f(t,u)}{u}>\lambda _{1}\) uniformly on \(t\in[0,1]\).

     
  5. (A5)

    \(\limsup_{u\to0} | \frac{f(t,u)}{u} |<\lambda_{1}\) uniformly on \(t\in[0,1]\).

     
Here \(\lambda_{1}\) is the first eigenvalue of the following linear eigenvalue problem corresponding to the boundary value problem (1.3):
$$ \begin{aligned} &u''+a(t)u'+b(t)u+ \lambda h(t)u=0 , \quad t\in(0,1), \\ &u(0)=u(1)=0. \end{aligned} $$
(1.6)

We state our main theorem as follows.

Theorem 1.1

Let (A1)-(A5) hold, then the singular boundary value problem (1.3) has at least one nontrivial solution.

Remark 1.1

(A1) and (A2) show that the functions a, b and h are all allowed to be singular at \(t=0,1\), which means that the differential operator in the equation of problem (1.3) is more general than the classic Sturm-Liouville differential operator where \(a, b\in C[0,1]\) are required. It is worth mentioning that the solvability of singular boundary value problems with general differential operators were also investigated in [2], where the results on the existence of solutions do not involve the eigenvalue of the corresponding linear eigenvalue problem, therefore the conditions laid on the functions a, b in [2] are more general than the conditions in this paper.

Remark 1.2

The existence of the first eigenvalue \(\lambda_{1}\) and the properties of its corresponding positive eigenfunction \(\varphi_{1} \) in (1.6) (see Lemma 2.4 in Section 2) play a very important role in the proof of our main theorem. The presence of \(a(t)u'+b(t)u\), however, brings difficulties when it comes to proving that \(\varphi_{1}\in C^{1}[0,1]\). We find a way to overcome this problem. For the special case of (1.6), i.e., the eigenvalue problem (1.2), the complete results on eigenvalues and eigenfunctions were obtained by Asakawa [3].

Remark 1.3

Choose \(B(u)=|u|^{\alpha}\), \(\alpha\in(0,1)\) and the functions c, d to be constants on \([0,1]\). Then (A3) can be reduced to (H1). Therefore, the main theorem in this paper generalizes Theorem A in [1].

We note that the results on positive solutions or nontrivial solutions for other kinds of boundary value problems have been considered in many publications such as [310] and the references therein. The existence of nontrivial solutions involving the relation between the principal eigenvalue and the growth of the nonlinearity has been investigated in [11].

The rest of the paper is arranged as follows. In Section 2, we state some notations and prove some preliminary results. In Section 3, we prove our main result. In Section 4, an example is given to illustrate the application of our main result.

2 Notations and preliminary results

In this section, we recall some notations, abstract theorems and auxiliary results, which are important for proving our main result.

We denote by \(C[0,1]\) the Banach space with the norm \(\|u\|=\max_{t\in[0,1]}|u(t)|\). Let
$$P=\bigl\{ u\in C[0,1] \mid u(t)\geq0, t\in[0,1]\bigr\} $$
be a positive cone in \(C[0,1]\). Denote by \(B_{r}=\{u \in C[0, 1] \mid \|u\|< r\}\) (\(r>0\)) the open ball of radius r. We denote by \(AC[0,1]\) the space of all absolute continuous functions on \([0,1]\). Let
$$\begin{aligned}& AC_{\mathrm{loc}}[0, 1)= \bigl\{ u \mid u|_{[0,d]}\in AC[0, d] \mbox{ for every compact interval } [0, d]\subseteq[0, 1) \bigr\} , \\& AC_{\mathrm{loc}}(0, 1]= \bigl\{ u \mid u|_{[d,1]}\in AC[d, 1] \mbox{ for every compact interval } [d, 1]\subseteq(0, 1] \bigr\} . \end{aligned}$$

Lemma 2.1

[4]

Suppose that (A1) holds. Then
  1. (i)
    the initial value problem
    $$ \begin{aligned} &u''+a(t)u'+b(t)u=0 , \quad t\in(0,1), \\ &u(0)=0, \qquad u'(0)=1 \end{aligned} $$
    (2.1)
    has a unique solution \(\alpha\in AC[0,1]\cap C^{1}[0,1)\) and \(\alpha '\in AC_{\mathrm{loc}}[0,1)\);
     
  2. (ii)
    the initial value problem
    $$ \begin{aligned} &u''+a(t)u'+b(t)u=0 , \quad t\in(0,1), \\ &u(1)=0, \qquad u'(1)=-1 \end{aligned} $$
    (2.2)
    has a unique solution \(\beta\in AC[0,1]\cap C^{1}(0,1]\) and \(\beta'\in AC_{\mathrm{loc}}(0,1]\);
     
  3. (iii)

    α is nondecreasing on \([0,1]\), β is nonincreasing on \([0,1]\).

     

We remark here that from the proof of Lemma 2.2 and Lemma 2.3 in [4], the existence and uniqueness of α and β for the initial value problem (2.1) and (2.2) have no relation with the sign of the functions a and b. The sign condition laid on b in (A1) is meant to obtain the monotonicity of α and β.

Let
$$\begin{aligned}& G(t,s)= \frac{1}{\rho} \left \{ \begin{array}{@{}l@{\quad}l} \alpha(s)\beta(t), & 0\leq s \leq t\leq1,\\ \alpha(t)\beta(s), & 0\leq t \leq s\leq1, \end{array} \right . \end{aligned}$$
(2.3)
where \(\rho=\alpha'(\frac{1}{2})\beta(\frac{1}{2})-\alpha(\frac{1}{2})\beta'(\frac{1}{2})\) is a positive constant.

We give the following remark. Although its proof is trivial, the consequences of the result are of major importance.

Remark 2.1

It follows from the above lemma that there exist positive constants \(c_{1}\), \(c_{2}\), \(\rho_{1}\) and \(\rho_{2}\) such that
$$c_{1}t\leq\alpha(t)\leq c_{2}t,\qquad \rho_{1}(1-t) \leq\beta(t)\leq\rho_{2}(1-t),\quad t\in[0,1]. $$
Furthermore, for \(t,s\in[0,1]\),
$$(\ast)\quad\rho G(t,s)\leq\rho G(s,s)=\alpha(s)\beta(s)\leq c_{2} \rho_{2}s(1-s). $$
From Lemma 2.6 in [4], the singular boundary value problem (1.3) can be converted into the equivalent integral equation
$$u(t)=\int_{0}^{1}G(t,s)q(s)h(s)f\bigl(s, u(s) \bigr)\,\mathrm{d}s, \quad t\in[0,1], $$
where \(q(s)=\exp(\int_{\frac{1}{2}}^{s}a(\tau)\,\mathrm{d}\tau)\). Define \(\tilde {h}(t)=q(t)h(t)\) for ease of notation. Combining (A2) and the definition of q, we see that \(\tilde{h}\) satisfies the following:
(\(\tilde{\mathrm{A}}2\)): 

\(\tilde{h}:(0, 1)\rightarrow[0, \infty)\) is continuous and does not vanish identically on any subinterval of \((0,1)\); furthermore, \(0<\int_{0}^{1} s(1-s)\tilde{h}(s)\,\mathrm{d}s<+\infty\).

For any \(u\in C[0,1]\), let
$$\begin{aligned}& (Au) (t)=\int_{0}^{1}G(t,s)\tilde{h}(s)f\bigl(s, u(s)\bigr)\,\mathrm{d}s, \end{aligned}$$
(2.4)
$$\begin{aligned}& (Tu) (t)=\int_{0}^{1}G(t,s)\tilde{h}(s)u(s)\, \mathrm{d}s. \end{aligned}$$
(2.5)
From Lemma 2.9 in [4], then \(A: C[0,1]\rightarrow C[0,1]\) and \(T: C[0,1]\rightarrow C[0,1]\) are completely continuous operators, respectively.

Lemma 2.2

Suppose that (A1) and (A2) hold. Let \(\sigma\in(0, 1)\), \(r\in C(0,1)\) be continuous and \(\int_{0}^{1}s(1-s)| r(s) |<+\infty\). If ω is a solution of
$$ \begin{aligned} &u''+a(t)u'+b(t)u+r(t)u=0 , \quad t\in(0,1), \\ &u(\sigma)= u'(\sigma)=0 \end{aligned} $$
(2.6)
such that \(\omega\in C[0, 1]\cap C^{1}(0,1)\) and \(\omega'\in AC_{\mathrm{loc}}(0, 1)\), then \(\omega\equiv0\).

Proof

Let ω be a solution of initial value problem (2.6). Multiplying both sides of the equation of (2.6) by q and integrating it from σ to t, we get
$$\omega'(t)=-\frac{1}{q(t)}\int_{\sigma}^{t}q(s) \bigl(b(s)+r(s)\bigr)\omega(s) \,\mathrm{d}s,\quad t\in(0,1). $$
Integrating the above equation from σ to t again, we have
$$ \omega(t)=-\int_{\sigma}^{t} \frac{1}{q(\tau)}\int _{\sigma}^{\tau}q(s) \bigl(b(s)+r(s)\bigr)\omega(s)\, \mathrm{d}s\,\mathrm{d}\tau, \quad t\in(0,1). $$
(2.7)
Let
$$\kappa_{1}=\min_{t\in[0,1]}q(t), \qquad \kappa_{2}=\max_{t\in[0,1]}q(t). $$
If \(t\in[\sigma, 1)\), then it follows from (2.7) that
$$\begin{aligned} \bigl|\omega(t)\bigr|&\leq\frac{\kappa_{2}}{\kappa_{1}}\int_{\sigma}^{t} \int_{\sigma}^{\tau}\bigl|b(s)+r(s)\bigr| \bigl|\omega(s)\bigr| \,\mathrm{d}s \,\mathrm{d}\tau \\ &=\frac{\kappa _{2}}{\kappa_{1}}\int_{\sigma}^{t}(t-s)\bigl|b(s)+r(s)\bigr| \bigl| \omega(s)\bigr| \,\mathrm{d}s \\ &\leq\frac{\kappa_{2}}{\kappa_{1}}\int_{\sigma}^{t}(1-s)\bigl|b(s)+r(s)\bigr| \bigl|\omega (s)\bigr| \,\mathrm{d}s. \end{aligned}$$
Since \(\int_{\sigma}^{1}(1-s)|b(s)+r(s)|\,\mathrm{d}s<+\infty\), we get \(\omega (t)\equiv0\) for \(t\in[\sigma,1]\) by Gronwall’s inequality. Using the same argument, we can easily get \(\omega(t)\equiv0\) for \(t\in[0, \sigma ]\). This completes the proof. □

We now turn to the eigenvalue problem (1.6). Before stating the results on the eigenvalue and eigenfunction, we give the following lemma, which is a direct result of the Krein-Rutman theorem.

Lemma 2.3

[12, 13]

Suppose that \(T: C[0, 1]\rightarrow C[0,1]\) is a completely continuous linear operator and \(T(P)\subset P\). If there exist \(\phi\in C[0, 1]\backslash(-P)\) and a constant \(c>0\) such that \(cT\phi\geq\phi\), then the spectral radius \(r(T )\neq0\) and T has a positive eigenfunction corresponding to its first eigenvalue \(\lambda _{1}=(r(T))^{-1}\).

We are now in a position to give the results on eigenvalue problem (1.6).

Lemma 2.4

Suppose that (A1) and (A2) are satisfied. Then T has a principle eigenvalue \(\lambda_{1}=(r(T))^{-1}\) and a positive eigenfunction \(\varphi_{1}\in P\) corresponding to \(\lambda_{1}\). Furthermore,
  1. (i)

    \(\varphi'_{1}(0)>0\), \(\varphi'_{1}(1)<0\) and there exist positive constants \(\nu_{1}\), \(\nu_{2}\) such that \(\nu_{1}t(1-t)\leq\varphi_{1}(t)\leq\nu _{2}t(1-t)\), \(t\in[0,1]\);

     
  2. (ii)

    there exist \(\delta_{1}, \delta_{2}>0\) such that \(\delta_{1} G(t,s)\leq\varphi_{1}(s)\leq\delta_{2} s(1-s)\), \(t,s\in[0,1]\).

     

Proof

It is obvious that there is \(t_{1}\in(0,1)\) such that \(G(t_{1}, t_{1})\tilde {h}(t_{1})>0\). Thus there exists \([a_{1}, b_{1}]\in(0, 1)\) such that \(t_{1}\in(a_{1}, b_{1})\) and \(G(t,s)\tilde {h}(s)>0\) for all \(t,s\in[a_{1}, b_{1}]\). Take \(\psi\in C[0, 1]\) such that \(\psi(t)\geq0\), \(t\in[0,1]\), \(\psi(t_{1})>0\) and \(\psi(t)= 0\), \(t\notin[a_{1}, b_{1}]\). Then, for \(t\in[a_{1}, b_{1}]\),
$$ (T\psi) (t)=\int_{0}^{1}G(t,s)\tilde{h}(s)\psi(s) \,\mathrm{d}s\geq\int_{a_{1}}^{b_{1}}G(t,s)\tilde{h}(s) \psi(s)\,\mathrm{d}s>0. $$
(2.8)
So there exists a constant \(c>0\) such that \(c(T \psi)(t)\geq\psi(t)\), \(t\in[0,1]\). From Lemma 2.3, we know that the spectral radius \(r(T) \neq0\) and T has a positive eigenfunction \(\varphi_{1}\) corresponding to its first eigenvalue \(\lambda_{1}=(r(T))^{-1}\).
If (A1) and (A2) hold, then the conclusions of Lemma 2.1 are still valid for the initial value problems
$$u''+a(t)u'+b(t)u+\lambda_{1} h(t)u=0, \qquad u(0)=0,\qquad u'(0)=1 $$
and
$$u''+a(t)u'+b(t)u+\lambda_{1} h(t)u=0, \qquad u(1)=0,\qquad u'(1)=-1. $$
We denote the unique solutions of the above two initial problems by ξ and ζ, respectively. Then Lemma 2.1 shows that \(\xi\in AC[0,1]\cap C^{1}[0,1)\), \(\xi'\in AC_{\mathrm{loc}}[0,1)\) and \(\zeta\in AC[0,1]\cap C^{1}(0,1]\), \(\zeta'\in AC_{\mathrm{loc}}(0,1]\).
(i) From the definition of eigenvalue, we know that \(\varphi_{1}\) satisfies
$$ \begin{aligned} &\varphi_{1}''+a(t) \varphi_{1}'+b(t)\varphi_{1}+ \lambda_{1}h(t)\varphi_{1}=0 , \quad t\in(0,1), \\ &\varphi_{1}(0)=\varphi_{1}(1)=0. \end{aligned} $$
(2.9)
Since \(\varphi_{1}(0)=\varphi_{1}(1)=0\), there exists \(\tau\in(0,1)\) such that \(\varphi_{1}(\tau)=\|\varphi_{1}\|>0\) and \(\varphi_{1}'(\tau)=0\). Let
$$ \chi(t)=q(t)\varphi_{1}'(t)\xi(t)-q(t) \varphi_{1}(t)\xi'(t), \quad t\in(0,1). $$
(2.10)
Then \(\chi\in C^{1}(0,1)\) and for \(t\in(0,1)\), it is easy to compute that
$$\chi'(t)=\bigl(q(t)\varphi_{1}'(t) \bigr)'\xi(t)-\bigl(q(t)\xi'(t)\bigr)' \varphi_{1}(t)=0. $$
For \(0< t<\tau\), integrating \(\chi'\) from t to τ and letting \(t\to0\), we have
$$ 0=\lim_{t\to0}\int_{t}^{\tau}\chi'(s)\,\mathrm{d}s=\chi(\tau)-\lim_{t\to 0}\chi(t). $$
(2.11)
By Lemma 2.3 in [4] and the fact that \(\xi\in C^{1}[0,1)\), \(\xi (0)=0\), we have
$$ \lim_{t\to0}q(t)\varphi_{1}'(t) \xi(t)=q(0)\lim_{t\to0}\varphi_{1}'(t) \xi(t)=0. $$
(2.12)
Combining (2.10), (2.11), (2.12) and
$$\lim_{t\to0}q(t)\varphi_{1}(t)\xi'(t)=q(0) \varphi_{1}(0)\xi '(0)=0, \qquad \chi(\tau)=-q(\tau) \varphi_{1}(\tau)\xi'(\tau), $$
we get
$$q(\tau)\varphi_{1}(\tau)\xi'(\tau)=0. $$
Since \(q(\tau)\neq0\) and \(\varphi_{1}(\tau)\neq0\), we have \(\xi'(\tau )=0\). Then \(\xi(\tau)\neq0\) due to Lemma 2.2. Let us define the function ϕ by
$$\phi(t)=\varphi_{1}(t)-\frac{\varphi_{1}(\tau)}{\xi(\tau)}\xi(t), \quad t\in[0,1], $$
it is easy to check that ϕ is the solution of the following problem:
$$\begin{aligned} &\phi''+a(t)\phi'+b(t)\phi+ \lambda_{1}h(t)\phi=0 , \quad t\in(0,1), \\ &\phi'(\tau)=0, \qquad \phi(\tau)=0. \end{aligned}$$
Then Lemma 2.2 yields \(\phi\equiv0\), that is, \(\varphi_{1}=[\varphi_{1}(\tau )/\xi(\tau)]\xi\). In particular, \(\varphi_{1}'(0)=\varphi_{1}(\tau)/\xi(\tau)\neq0\); furthermore, \(\varphi_{1}'(0)>0\) by positivity of \(\varphi_{1}\) on \((0,1)\). In the same manner, we can see that \(\varphi_{1}=[\varphi_{1}(\tau)/\zeta (\tau)]\zeta\) and \(\varphi_{1}'(1)=-\varphi_{1}(\tau)/\zeta(\tau)< 0\).
Define
$$ \Theta(t)=\left \{ \begin{array}{@{}l@{\quad}l} \varphi_{1}'(0), &t=0, \\ \frac{\varphi_{1}(t)}{t(1-t)}, &0< t<1, \\ -\varphi_{1}'(1), &t=1. \end{array} \right . $$
Then \(\Theta(\cdot)\) is continuous and \(\Theta(t)> 0\) on \([0,1]\), so there exist positive constants \(\nu_{1}\) and \(\nu_{2}\) such that \(\nu_{1}\leq \Theta(t)\leq\nu_{2}\) for \(t\in[0,1]\), i.e.,
$$(**)\quad\nu_{1}t(1-t)\leq\varphi_{1}(t)\leq \nu_{2}t(1-t),\quad t\in[0,1]. $$
(ii) From Lemma 2.1(iii), () and (), for \(t,s\in[0,1]\), we have
$$\rho G(t,s)\leq\rho G(s,s)=\alpha(s)\beta(s)\leq c_{2} \rho_{2}s(1-s)\leq \frac{c_{2}\rho_{2}}{\nu_{1}}\nu_{1}s(1-s)\leq \frac{c_{2}\rho_{2}}{\nu_{1}}\varphi_{1}(s). $$
Define
$$\delta_{1}:=\frac{\rho\nu_{1}}{c_{2}\rho_{2}},\qquad \delta_{2}:= \nu_{2}, $$
then
$$\delta_{1} G(t,s)\leq\varphi_{1}(s)\leq\delta_{2} s(1-s),\quad t,s\in[0,1]. $$

This completes the proof. □

Lemma 2.5

Suppose that (A1) and (A2) are satisfied. Let
$$P_{1}= \biggl\{ u\in P \Bigm| \int_{0}^{1} \varphi_{1}(t)\tilde{h}(t)u(t)\,\mathrm{d} t\geq\lambda_{1}^{-1} \delta_{1}\|u\| \biggr\} , $$
where \(\varphi_{1}\) and \(\delta_{1}\) are defined by Lemma  2.4. Then \(P_{1}\) is a cone in \(C[0,1]\) and \(T(P)\subset P_{1}\).

Proof

It follows from (\(\tilde{\mathrm{A}}2\)) and () that
$$\int_{0}^{1}\varphi_{1}(t) \tilde{h}(t)\bigl|u(t)\bigr|\,\mathrm{d}t\leq\delta_{2}\int_{0}^{1}t(1-t) \tilde{h}(t)\bigl|u(t)\bigr|\,\mathrm{d}t< +\infty $$
for any \(u\in C[0,1]\), then \(P_{1}\) is well defined. Furthermore, for any \(u\in P\), from Lemma 2.4, we have
$$\begin{aligned} \int_{0}^{1}\varphi_{1}(t)\tilde{h}(t) (Tu) (t)\,\mathrm{d}t &=\int_{0}^{1} \biggl[ \varphi _{1}(t)\tilde{h}(t)\int_{0}^{1} G(t,s)\tilde{h}(s) u(s)\,\mathrm{d}s \biggr]\,\mathrm{d}t \\ &=\int_{0}^{1}\int_{0}^{1} \varphi_{1}(t)\tilde{h}(t) G(t,s)\tilde{h}(s) u(s)\,\mathrm{d}s\, \mathrm{d}t \\ &=\int_{0}^{1}\int_{0}^{1} \varphi_{1}(t)\tilde{h}(t) G(t,s)\tilde{h}(s) u(s)\,\mathrm{d}t\, \mathrm{d}s \\ &=\int_{0}^{1} \biggl[\tilde{h}(s) u(s)\int _{0}^{1}G(t,s)\tilde{h}(t)\varphi _{1}(t) \,\mathrm{d}t \biggr]\,\mathrm{d}s \\ &=\lambda_{1}^{-1}\int_{0}^{1} \tilde{h}(s)\varphi_{1}(s)u(s)\,\mathrm{d}s \\ &\geq\lambda_{1}^{-1}\delta_{1}\int _{0}^{1}G(t,s)\tilde{h}(s)u(s)\,\mathrm{d}s = \lambda_{1}^{-1}\delta_{1}(Tu) (t), \end{aligned}$$
then \(\int_{0}^{1}\varphi_{1}(t)\tilde{h}(t)(Tu)(t)\,\mathrm{d}t\geq\lambda _{1}^{-1}\delta_{1}\|Tu\|\), i.e., \(T(P)\subset P_{1}\).

This completes the proof. □

3 Proof of the main result

Proof of Theorem 1.1

From (A4), for \(\varepsilon>0\), there exists \(L>0\) such that
$$ f(t,u)\geq\lambda_{1}(1+\varepsilon)u, \quad t\in[0,1], u\geq L. $$
(3.1)
Combining (A3) and (3.1), we can see that there exists a nonnegative function \(c_{1}\in C[0,1]\) such that
$$ f(t,u)\geq\lambda_{1}(1+\varepsilon)u-c_{1}(t)-d(t)B(u), \quad t\in [0,1], u\in\mathbb{R}. $$
(3.2)
In the following we shall prove
$$ u-Au\neq\mu\varphi_{1}, \quad \forall u\in C[0,1], \|u\|=R, \mu \geq0, $$
(3.3)
provided that R is large enough.
In fact, if (3.3) is not true, then there exist \(u_{0}\in C[0,1]\), \(\| u_{0}\|=R\) and \(\mu_{0}\geq0\) such that
$$ u_{0}-Au_{0}= \mu_{0} \varphi_{1}. $$
(3.4)
For \(t\in[0,1]\) and for any \(u\in C[0,1]\), we set
$$\begin{aligned}& \eta_{0}(t)=\int_{0}^{1}G(t,s) \tilde{h}(s)u_{0}(s)\,\mathrm{d}s, \qquad \eta _{1}(t)=\int _{0}^{1}G(t,s)\tilde{h}(s)c_{1}(s)\, \mathrm{d}s, \end{aligned}$$
(3.5)
$$\begin{aligned}& \eta_{2}(t)=\int_{0}^{1}G(t,s) \tilde{h}(s)d(s)B\bigl(u_{0}(s)\bigr)\,\mathrm{d}s, \end{aligned}$$
(3.6)
$$\begin{aligned}& g(u)=\int_{0}^{1}\varphi_{1}(t) \tilde{h}(t)u(t)\,\mathrm{d}t. \end{aligned}$$
(3.7)
Then from (\(\tilde{\mathrm{A}}2\)) and (), functional g is well defined. Thus we have
$$\begin{aligned}[b] &u_{0}(t)+\eta_{1}(t)+ \eta_{2}(t)\\ &\quad=(Au_{0}) (t)+ \mu_{0} \varphi_{1}(t)+\eta_{1}(t)+\eta_{2}(t) \\ &\quad=\int_{0}^{1}G(t,s)\tilde{h}(s) \bigl[f \bigl(s, u_{0}(s)\bigr)+c_{1}(s)+d(s)B\bigl(u_{0}(s) \bigr) \bigr]\,\mathrm{d}s+\mu_{0} \varphi_{1}(t). \end{aligned} $$
Recall that
$$\varphi_{1}(t)=\lambda_{1}^{-1}\int _{0}^{1}G(t,s)\tilde{h}(s)\varphi_{1}(s) \,\mathrm{d}s. $$
It follows from Lemma 2.5 and (3.2) that \(u_{0}+\eta_{1}+\eta_{2}\in P_{1}\).
From Lemma 2.4 and (3.5)-(3.7), we have
$$\begin{aligned} g(\eta_{0})&=\int_{0}^{1} \varphi_{1}(t)\tilde{h}(t)\eta_{0}(t)\,\mathrm{d}t =\int _{0}^{1} \biggl[\varphi_{1}(t) \tilde{h}(t) \int_{0}^{1}G(t,s) \tilde{h}(s)u_{0}(s)\,\mathrm{d}s \biggr]\,\mathrm{d}t \\ &=\int_{0}^{1} \biggl[\tilde{h}(s)u_{0}(s) \int_{0}^{1} G(t,s)\varphi_{1}(t) \tilde{h}(t)\,\mathrm{d}t \biggr]\,\mathrm{d}s \\ &= \lambda_{1}^{-1}\int_{0}^{1} \varphi_{1}(s)\tilde{h}(s)u_{0} (s)\,\mathrm{d}s= \lambda_{1}^{-1}g(u_{0}). \end{aligned}$$
(3.8)
By the same manner, we get
$$ \begin{aligned} &g(\eta_{1})= \lambda_{1}^{-1}\int_{0}^{1} \varphi_{1}(s)\tilde{h}(s)c_{1}(s) \,\mathrm{d}s, \\ &g(\eta_{2})= \lambda_{1}^{-1}\int _{0}^{1}\varphi_{1}(s)\tilde{h}(s)d(s)B \bigl(u_{0}(s)\bigr) \,\mathrm{d}s. \end{aligned} $$
(3.9)
By (3.2), (3.7) and (3.8), we have
$$\begin{aligned} g(Au_{0})&=\int_{0}^{1} \biggl[ \varphi_{1}(t) \tilde{h}(t)\int_{0}^{1}G(t,s) \tilde{h}(s)f\bigl(s,u_{0}(s)\bigr)\,\mathrm{d}s \biggr]\,\mathrm{d}t \\ &\geq\lambda_{1}(1+\varepsilon)g(\eta_{0})-g( \eta_{1})-g(\eta_{2}) \\ &= (1+\varepsilon)g(u_{0}) -g(\eta_{1})-g( \eta_{2}). \end{aligned}$$
Then, from the above inequality, (3.8), (3.9) and Lemma 2.5, we have
$$\begin{aligned} &g(Au_{0})-g(u_{0}) \\ &\quad\geq\varepsilon g(u_{0}) -g(\eta_{1})-g( \eta_{2}) \\ &\quad=\varepsilon g(u_{0}+\eta_{1}+\eta_{2}) -(1+\varepsilon)g(\eta_{1})-(1+\varepsilon)g(\eta_{2}) \\ &\quad=\varepsilon\int_{0}^{1} \varphi_{1}(s)\tilde{h}(s) \bigl[u_{0} (s)+ \eta_{1}(s)+\eta_{2}(s) \bigr]\,\mathrm{d}s- \lambda_{1}^{-1}(1+\varepsilon)\int_{0}^{1} \varphi_{1}(s)\tilde{h}(s)c_{1}(s) \,\mathrm{d}s \\ &\qquad{}-\lambda_{1}^{-1}(1+\varepsilon)\int _{0}^{1}\varphi_{1}(s)\tilde {h}(s)d(s)B \bigl(u_{0}(s)\bigr) \,\mathrm{d}s \\ &\quad\geq\varepsilon\lambda_{1}^{-1} \delta_{1}\|u_{0} +\eta_{1}+\eta_{2}\|-\lambda_{1}^{-1}(1+ \varepsilon)\int_{0}^{1}\varphi_{1}(s) \tilde {h}(s)c_{1}(s) \,\mathrm{d}s \\ &\qquad{}-\lambda_{1}^{-1}(1+\varepsilon)\int _{0}^{1}\varphi_{1}(s)\tilde {h}(s)d(s)B \bigl(u_{0}(s)\bigr) \,\mathrm{d}s \\ &\quad\geq\varepsilon\lambda_{1}^{-1} \delta_{1}\bigl(\|u_{0}\|- \|\eta_{1}\|-\|\eta_{2}\|\bigr)-\lambda_{1}^{-1}(1+ \varepsilon)\int_{0}^{1}\varphi _{1}(s) \tilde{h}(s)c_{1}(s) \,\mathrm{d}s \\ &\qquad{}-\lambda_{1}^{-1}(1+\varepsilon)\int _{0}^{1}\varphi_{1}(s)\tilde {h}(s)d(s)B \bigl(u_{0}(s)\bigr) \,\mathrm{d}s. \end{aligned}$$
(3.10)

We divide the proof into two cases as follows.

Case 1. \(B(\cdot)\) is bounded on \(\mathbb{R}\).

In this case, there exists \(M_{2}>0\), for any \(u\in\mathbb{R}\), \(B(u)\leq M_{2}\). Thus,
$$\begin{aligned} \|\eta_{2}\|&=\max_{t\in[0,1]}\int _{0}^{1}G(t,s)\tilde {h}(s)d(s)B \bigl(u_{0}(s)\bigr)\,\mathrm{d}s \\ &\leq M_{2}\int_{0}^{1}G(s,s) \tilde{h}(s)d(s)\,\mathrm{d}s. \end{aligned}$$
(3.11)
Therefore, from (3.10), (3.11) and the expression of \(\eta_{1}\), we get
$$\begin{aligned} g(Au_{0})-g(u_{0})\geq{}&\varepsilon\lambda_{1}^{-1} \delta_{1}\bigl(\| u_{0}\|-\|\eta _{1}\|-\| \eta_{2}\|\bigr)-\lambda_{1}^{-1}(1+\varepsilon)\int _{0}^{1}\varphi_{1}(s)\tilde {h}(s)c_{1}(s) \,\mathrm{d}s \\ &{}-\lambda_{1}^{-1}(1+\varepsilon)\int_{0}^{1} \varphi_{1}(s)\tilde {h}(s)d(s)B\bigl(u_{0}(s)\bigr) \, \mathrm{d}s \\ \geq{}&\varepsilon\lambda_{1}^{-1}\delta_{1} \biggl( \| u_{0}\|-\int_{0}^{1}G(s,s) \tilde{h}(s)c_{1}(s) \,\mathrm{d}s-M_{2}\int _{0}^{1}G(s,s)\tilde{h}(s)d(s)\,\mathrm{d}s \biggr) \\ &{}-\lambda_{1}^{-1} (1+\varepsilon)\int_{0}^{1} \varphi_{1}(s)\tilde{h}(s)c_{1}(s) \,\mathrm{d}s -\lambda_{1}^{-1}(1+\varepsilon) M_{2}\int _{0}^{1}\varphi_{1}(s)\tilde{h}(s)d(s) \, \mathrm{d}s \\ >{}&0 \end{aligned}$$
provided that we take
$$R>A+B+C+D:=R_{1}, $$
where
$$\begin{aligned} &A=\int_{0}^{1}G(s,s)\tilde{h}(s)c_{1}(s) \,\mathrm{d}s, \qquad B=M_{2}\int_{0}^{1}G(s,s) \tilde{h}(s)d(s)\,\mathrm{d}s, \\ &C=\frac{(1+\varepsilon)}{\varepsilon\delta_{1}}\int_{0}^{1} \varphi_{1}(s)\tilde {h}(s)c_{1}(s) \,\mathrm{d}s,\qquad D= \frac{(1+\varepsilon)}{\varepsilon\delta_{1}} M_{2}\int_{0}^{1} \varphi_{1}(s)\tilde{h}(s)d(s) \,\mathrm{d}s. \end{aligned}$$

Case 2. \(B(\cdot)\) is unbounded on \(\mathbb{R}\).

From (1.4), for \(\rho>0\), there exists a positive constant \(M_{1}\) such that
$$ B(u)< \rho |u|, \quad |u|\geq M_{1}. $$
(3.12)
In this case, there exists a positive constant \(M_{3}>M_{1}\) such that
$$ B(u)\leq B(M_{3}), \quad |u|\leq M_{3}. $$
(3.13)
Since \(\|u_{0}\|=R\), then for \(s\in[0,1]\), \(|u_{0}(s)|\leq R< M_{3}\). By (3.12) and (3.13), we have
$$\begin{aligned} \|\eta_{2}\|&=\max_{t\in[0,1]}\int _{0}^{1}G(t,s)\tilde {h}(s)d(s)B \bigl(u_{0}(s)\bigr)\,\mathrm{d}s \leq\int_{0}^{1}G(s,s) \tilde{h}(s)d(s) B(M_{3}) \,\mathrm{d}s \\ &\leq\int_{0}^{1}G(s,s)\tilde{h}(s)d(s) \rho M_{3} \,\mathrm{d}s= \rho M_{3}\int_{0}^{1}G(s,s) \tilde{h}(s)d(s) \,\mathrm{d}s. \end{aligned}$$
(3.14)
Therefore, from (3.10), (3.13) and (3.14), we have
$$\begin{aligned} g(Au_{0})-g(u_{0})\geq{}&\varepsilon\lambda_{1}^{-1} \delta_{1}\bigl(\| u_{0}\|-\|\eta _{1}\|-\| \eta_{2}\|\bigr)- \lambda_{1}^{-1}(1+\varepsilon) \int _{0}^{1}\varphi_{1}(s) \tilde{h}(s)c_{1}(s) \,\mathrm{d}s \\ &{} -\lambda_{1}^{-1} (1+\varepsilon)\int _{0}^{1}\varphi_{1}(s) \tilde{h}(s)d(s)B \bigl(u_{0}(s)\bigr) \,\mathrm{d}s \\ \geq{}&\varepsilon\lambda_{1}^{-1}\delta_{1} \biggl( \| u_{0}\|-\int_{0}^{1}G(s,s) \tilde{h}(s)c_{1}(s) \,\mathrm{d}s-\rho M_{3}\int _{0}^{1}G(s,s)\tilde{h}(s)d(s)\,\mathrm{d}s \biggr) \\ &{} -\lambda_{1}^{-1} (1+\varepsilon)\int _{0}^{1}\varphi_{1}(s) \tilde{h}(s)c_{1}(s) \,\mathrm{d}s -\lambda_{1}^{-1}(1+\varepsilon) \rho M_{3} \int_{0}^{1}\varphi_{1}(s) \tilde{h}(s)d(s) \,\mathrm{d}s \\ >{}&0 \end{aligned}$$
provided that we take
$$R>A+E+C+F:=R_{2}, $$
where
$$\begin{aligned} &E=\rho M_{3}\int_{0}^{1}G(s,s) \tilde{h}(s)d(s)\,\mathrm{d}s, \qquad F=\frac{(1+\varepsilon)}{\varepsilon\delta_{1}} \rho M_{3} \int_{0}^{1}\varphi_{1}(s) \tilde{h}(s)d(s) \,\mathrm{d}s. \end{aligned}$$
Thus, no matter which case happens, if we choose \(R>\max\{R_{1}, R_{2}\}\), we have
$$ g(Au_{0})-g(u_{0})> 0. $$
(3.15)
On the other hand, from (3.4), (\(\tilde{\mathrm{A}}2\)) and the fact that \(\varphi_{1}(t)>0\) for \(t\in(0,1)\) and \(\mu_{0}\geq 0\), we have
$$g(u_{0})-g(Au_{0})=g(\mu_{0} \varphi_{1})=\mu_{0}g(\varphi_{1})\geq0, $$
which contradicts with (3.15). Therefore, (3.3) is true and we have
$$ \operatorname{deg}(I-A, B_{R}, \theta)=0. $$
(3.16)
From (A5), for \(0<\varepsilon_{1}<\lambda_{1}\), there exists \(0< r<R\) such that
$$ \bigl| f(t,u) \bigr|\leq (\lambda_{1}-\varepsilon_{1}) | u |, \quad t \in[0,1], |u |< r. $$
(3.17)
In the following we will prove that
$$ u\neq\mu Au, \quad u\in\partial B_{r}, \mu\in[0,1]. $$
(3.18)
If (3.18) is not true, then there exist \(u_{1}\in\partial B_{r}\) and \(\mu _{1}\in(0,1]\) such that \(u_{1}= \mu_{1} Au_{1}\), and then by (3.17) we have
$$\begin{aligned} g\bigl(\bigl|u_{1}(t)\bigr|\bigr)&=g\bigl(\mu_{1} \bigl|(Au_{1}) (t)\bigr|\bigr)=\mu_{1}\int_{0}^{1} \biggl[ \varphi_{1}(t)\tilde{h}(t) \biggl|\int_{0}^{1}G(t,s) \tilde{h}(s)f\bigl(s, u_{1}(s)\bigr)\,\mathrm{d}s \biggr| \biggr]\,\mathrm{d}t \\ &\leq\mu_{1}(\lambda_{1}-\varepsilon_{1}) \int _{0}^{1} \biggl[\varphi_{1}(t) \tilde{h}(t) \int_{0}^{1}G(t,s) \tilde{h}(s)\bigl|u_{1}(s)\bigr|\,\mathrm{d}s \biggr]\,\mathrm{d}t \\ &=\mu_{1}(\lambda_{1}-\varepsilon_{1}) \int _{0}^{1} \biggl[\tilde{h}(s)\bigl|u_{1}(s)\bigr| \int_{0}^{1}G(t,s)\varphi_{1}(t) \tilde{h}(t )\,\mathrm{d}t \biggr]\,\mathrm{d}s \\ &=\mu_{1}\bigl(1-\lambda_{1}^{-1} \varepsilon_{1}\bigr) \int_{0}^{1} \varphi_{1}(s)\tilde{h}(s)\bigl|u_{1}(s)\bigr|\,\mathrm{d}s \\ &= \mu_{1}\bigl(1-\lambda_{1}^{-1} \varepsilon_{1}\bigr)g\bigl(\bigl|u_{1}(t)\bigr|\bigr), \end{aligned}$$
which implies \(g(|u_{1}(t)|)\leq0\).
On the other hand, from \(\|u_{1}\|=r>0\) and \(\varphi_{1}(t)>0\) for \(t\in (0,1)\), we have
$$g\bigl(\bigl|u_{1}(t)\bigr|\bigr)=\int_{0}^{1} \varphi_{1}(t)\tilde{h}(t)\bigl|u_{1}(t)\bigr|\,\mathrm{d}t>0,\quad t \in[0,1], $$
which is a contradiction. Thus (3.18) holds. According to the invariance property of the Leray-Schauder degree, we have
$$ \operatorname{deg}(I-A, B_{r}, \theta)=1. $$
(3.19)
By (3.16), (3.19) and the additivity of the Leray-Schauder degree, we obtain
$$\operatorname{deg}(I-A, B_{R}\backslash\bar{B}_{r}, \theta)=\operatorname{deg}(I-A, B_{R}, \theta)-\operatorname{deg}(I-A, B_{r}, \theta)=-1. $$
Therefore, A has at least one fixed point on \(B_{R}\backslash\bar {B}_{r}\), i.e., problem (1.3) has at least one nontrivial solution. This completes the proof. □

4 Example

In this section, an example is given to illustrate the application of our main result (Theorem 1.1). Consider the following boundary value problem:
$$ \begin{aligned} &u''- \frac{u}{t(1-t)}+\frac{1}{t(1-t)}f(t,u)=0, \quad t\in(0,1), \\ &u(0)=u(1)=0, \end{aligned} $$
(4.1)
where
$$\begin{aligned}& a\equiv0, \qquad b(t)=-\frac{1}{t(1-t)},\qquad h(t)=\frac{1}{t(1-t)}, \\& f(t,u)= \left \{ \begin{array}{@{}l@{\quad}l} -\sqrt{|u|}, & -\infty< u\leq-1,\\ -u^{2}, & -1<u<0,\\ u^{2}, & 0\leq u<\infty. \end{array} \right . \end{aligned}$$
Obviously, conditions (A1)-(A3) of Theorem 1.1 are satisfied. For the linear eigenvalue problem corresponding to (4.1), we compute that
$$\lambda_{1}=3, \qquad \varphi_{1}(t)=t(1-t). $$
Moreover,
$$\liminf_{u\to+\infty}\frac{f(t,u)}{u}=+\infty>\lambda _{1}=3>0= \limsup_{u\to0} \biggl| \frac{f(t,u)}{u} \biggr|, $$
which implies that (A4) and (A5) are satisfied. So, all of conditions in Theorem 1.1 are fulfilled. Therefore, the boundary value problem (4.1) has at least one nontrivial solution according to Theorem 1.1.

Declarations

Acknowledgements

The authors would like to thank the anonymous referees for their careful reading of the manuscript and for the constructive comments. This work is supported by the Natural Science Foundation of Gansu Province, China (1308RJZA113), the National Natural Science Foundation of China (11362008), by Youth Science Foundation of Lanzhou Jiaotong University (2012019) and the Fundamental Research Funds for the Universities of Gansu Province (212084).

Open Access This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.

Authors’ Affiliations

(1)
Department of Mathematics, Lanzhou Jiaotong University
(2)
School of Mathematics and Statistics, Zhejiang University of Finance and Economics

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© Li and Yan; licensee Springer. 2015