Global behavior of positive solutions for some semipositone fourth-order problems

Abstract

In this paper, we study the global behavior of positive solutions of fourth-order boundary value problems

$$ \textstyle\begin{cases} u''''=\lambda f(x,u), \quad x\in (0,1), \\ u(0)=u(1)=u''(0)=u''(1)=0, \end{cases} $$

where $f: [0,1]\times \mathbb{R^{+}} \to \mathbb{R}$ is a continuous function with $f(x,0)<0$ in $(0, 1)$, and $\lambda >0$. The proof of our main results are based upon bifurcation techniques.

Introduction

In this paper, we study the global behavior of positive solutions of fourth-order boundary value problems

$$ \textstyle\begin{cases} u''''=\lambda f(x,u),\quad x\in (0,1), \\ u(0)=u(1)=u''(0)=u''(1)=0, \end{cases} $$

(1.1)

where $\lambda > 0$ and $f:[0,1]\times \mathbb{R^{+}}\rightarrow \mathbb{R}$. If $f(x, 0)\geq 0$, then (1.1) is called a positone problem.

On the contrary, here we deal with the so-called semipositone problem when f is such that

$(f_{1})$ :

$f(x, 0) < 0$ $\forall x \in (0,1)$.

The existence of positive solutions of second-order positone problems have been extensively studied via the Leray–Schauder degree theory, fixed point theorem on a cone, and the method of lower and upper solutions; see [1,2,3] and the references therein.

Ambrosetti [4] studied the existence of positive solutions for semipositone elliptic problems via bifurcation theory. Recently, Hai and Shivaji [5] obtained the existence of positive solutions for second-order semipositone problems

$$ \textstyle\begin{cases} -u''=\lambda h(t)f(u), & t\in(0,1), \\ u(0)=0,&u'(1)+c (u(1) )u(1)=0 \end{cases} $$

via a Krasnosel’skii fixed-point-type theorem in a Banach space.

The existence and multiplicity of positive solutions of fourth-order positone problems have been studied by several authors; see [6,7,8,9,10,11] along this line. However, there are few results for fourth-order semipositone problems; see [12]. Ma [12] used the fixed point theorem in cones to show that the problem

$$ \textstyle\begin{cases} u''''=\lambda\tilde{f} (x,u(x),u'(x) ), \quad t\in (0,1), \\ u(0)=u'(0)=u''(1)=u'''(1)=0 \end{cases} $$

has a positive solution if $\lambda >0$ is small enough, where $\tilde{f}(x, u, p)\geq -M$ for some positive constant M, and

$$ \lim_{p\rightarrow \infty }\frac{f(x,u,p)}{p}=\infty . $$

There is a big difference in the study of fourth- and second-order problems. For example:

• 1. Spectrum theory for singular second-order linear eigenvalue problems has been established via Prüfer transform in [13]. However, the spectrum structure of singular fourth-order linear eigenvalue problems is not established so far.

• 2. The uniqueness of solutions of second-order problems

  $$ \textstyle\begin{cases} -u''=\lambda u^{q}, &x\in (a,b), \\ u>0,&x\in (a,b), \\ u(a)=u(b)=0 \end{cases} $$

  has been obtained in [14]. However, the uniqueness of solution of

  $$ \textstyle\begin{cases} w''''=b\vert w \vert ^{\alpha }, \quad x\in (0,1), \\ w(0)=w(1)=w''(0)=w''(1)=0 \end{cases} $$

  is not obtained so far.

• 3. It is well known that, for a second-order differential equation with periodic, Neumann, or Dirichlet boundary conditions, the existence of a well-ordered pair of lower and upper solutions $\alpha \leq \beta$ is sufficient to ensure the existence of a solution in the sector enclosed by them. However, this result it is not true for fourth-order differential equations; see Remark 3.1 in [15].

Motivated by Ambrosetti [4], we investigate the global behavior of positive solutions of the fourth-order boundary value problem (1.1). Depending on the behavior of $f = f(x, s)$ as $s\rightarrow +\infty$, we handle both asymptotically linear, superlinear, and sublinear problems. All results are obtained by showing that there exists a global branch of solutions of (1.1) “emanating from infinity” and proving that for λ near the bifurcation value, solutions of large norms are indeed positive to which bifurcation theory or topological methods apply in a classical fashion. Since there are a lot of differences between second- and fourth-order cases, we have to overcome several new difficulties in the proof of our main results.

We deal in Sect. 2 with asymptotically linear problems. In Sect. 3, we discuss superlinear problems, and we show that (1.1) possesses positive solutions for $0 < \lambda < \lambda^{*}$. Similar arguments can be used in the sublinear case, discussed in Sect. 4, to show that (1.1) has positive solutions provided that λ is large enough.

Asymptotically linear problems

For Lebesgue spaces, we use standard notation. We work in $X=C[0,1]$. The usual norm in such spaces is denoted by $\Vert u \Vert =\max_{t \in [0,1]}\vert u(t) \vert$, and we set $B_{r}=\{u\in X: \Vert u \Vert \leq r\}$. The first eigenvalue of $u''''$ with boundary conditions $u(0)=u(1)=u''(0)=u''(1)=0$ is denoted by $\lambda_{1}$; $\phi_{1}$ is the corresponding eigenfunction such that $\phi_{1}>0$ in $(0,1)$. We also set $\mathbb{R^{+}}=[0, \infty )$.

We define $K: X\rightarrow X$ by

$$ Ku(t):= \int_{0}^{1} \int_{0}^{1}G(t,s)G(s,\tau )f \bigl(\tau ,u(\tau ) \bigr)\,d\tau \,ds $$

and

$$ G(t,s)= \textstyle\begin{cases} t(1-s),&0 \leq t \leq s \leq 1, \\ s(1-t),&0 \leq s \leq t \leq 1. \end{cases} $$

We write $u=Kv$ if

$$ \textstyle\begin{cases} u''''=v,\quad x\in (0,1), \\ u(0)=u(1)=u''(0)=u''(1)=0. \end{cases} $$

With this notation, problem (1.1) is equivalent to

$$ u-\lambda Kf(u)=0,\quad u\in X. $$

(2.1)

Hereafter we will use the same symbol to denote both the function and the associated Nemitski operator.

We say that $\lambda_{\infty }$ is a bifurcation from infinity for (2.1) if there exist $\mu_{n}\rightarrow \lambda_{\infty }$ and $u_{n} \in X$ such that $u_{n}-\mu_{n}Kf(u_{n})=0$ and $\Vert u_{n} \Vert \rightarrow \infty$.

In some situations, like the specific ones we will discuss later, an appropriate rescaling allows us to find bifurcation from infinity by means of the Leray–Schauder topological degree, denoted by $\deg (\cdot , \cdot ,\cdot )$. Recall that $K: X\rightarrow X$ is (continuous and) compact, and hence it makes sense to consider the topological degree of $I-\lambda Kf$, where I is the identity map.

We suppose that $f\in C([0,1]\times \mathbb{R^{+}}, \mathbb{R})$ satisfies $(f_{1})$ and

$(f_{2})$ :

there is $m>0$ such that

$$ \lim_{u\rightarrow +\infty }\frac{f(x,u)}{u}=m. $$

Let $\lambda_{\infty }=\frac{\lambda_{1}}{m}$ and define

$$ a(x)=\liminf_{u\rightarrow +\infty } \bigl(f(x,u)-mu \bigr),\qquad A(x)= \limsup _{u\rightarrow +\infty } \bigl(f(x,u)-mu \bigr). $$

Theorem 2.1

Suppose that f satisfies $(f_{1})$ and $(f_{2})$. Then there exists $\epsilon >0$ such that (1.1) has positive solutions, provided that either

• (i) $a>0$ (possibly +∞) in $[0,1]$, and $\lambda \in ( \lambda_{\infty }-\epsilon ,\lambda_{\infty })$; or

• (ii) $A <0$ (possibly −∞) in $[0,1]$, and $\lambda \in ( \lambda_{\infty },\lambda_{\infty }+\epsilon )$.

The proof of Theorem 2.1 will be carried out in several steps. First of all, we extend $f(x,\cdot )$ to the whole $\mathbb{R}$ by setting

$$ F(x, u)=f \bigl(x,\vert u \vert \bigr). $$

For $u \in X$,

$$ \varPhi (\lambda ,u):=u-\lambda KF(u). $$

Clearly, any $u>0$ such that $\varPhi (\lambda ,u)=0$ is a positive solution of (1.1).

Lemma 2.1

For every compact interval $\varLambda \subset \mathbb{R^{+}}\backslash \{\lambda_{\infty }\}$, there exists $r>0$ such that

$$ \varPhi (\lambda ,u)\neq 0\quad \forall \Vert u \Vert \geq r. $$

Moreover,

• (i) if $a>0$, then we can also take $\varLambda =[\lambda_{\infty }, \lambda ]$ for all $\lambda > \lambda_{\infty }$, and

• (ii) if $A<0$, then we can also take $\varLambda =[0,\lambda_{\infty }]$.

Proof

Suppose on the contrary that there exists a sequence $\{(\mu _{n}, u_{n})\}$ satisfying

$$ \mu _{n}\in \varLambda ; \qquad \Vert u_{n} \Vert \geq n \quad \text{for } n\in \mathbb{N}; \qquad u_{n}=\mu _{n}KF(u_{n}). $$

Obviously, $\Vert u_{n} \Vert \geq n$ implies that $u_{n}(x)\not \equiv 0$. We may assume that $\mu _{n}\rightarrow \mu$ for some $\mu \neq \lambda _{\infty }$.

Setting $w_{n}=u_{n}\Vert u_{n} \Vert ^{-1}$, we find

$$ w_{n}=\mu_{n}\Vert u_{n} \Vert ^{-1}KF(u_{n}). $$

Since $w_{n}$ is bounded in X, after taking a subsequence if necessary, we have that $w_{n}\rightarrow w$ in X, where w is such that $\Vert w \Vert =1$ and satisfies

$$ \textstyle\begin{cases} w''''=\mu m\vert w \vert , \quad x\in (0,1), \\ w(0)=w(1)=w''(0)=w''(1). \end{cases} $$

By the maximum principle it follows that $w\geq 0$. Since $\Vert w \Vert =1$, we infer that $\mu m=\lambda_{1}$, namely $\mu =\lambda_{\infty }$, a contradiction that proves the first statement.

We will give a short sketch of (i). Taking $\mu_{n}\downarrow \lambda_{\infty }$, it follows that $w\geq 0$ satisfies

$$ \textstyle\begin{cases} w''''=\lambda_{1}w,\quad x\in (0,1), \\ w(0)=w(1)=w''(0)=w''(1)=0, \end{cases} $$

(2.2)

and hence there exists $\beta >0$ such that $w=\beta \phi_{1}$. Then we have $u_{n}=\Vert u_{n} \Vert w_{n}\rightarrow +\infty$ and $F(u_{n})=f(u_{n})$ for n large.

From $\varPhi (\lambda_{n},u_{n})=0$ it follows that

$$ \lambda_{1} \int_{0}^{1}u_{n}\phi_{1}\,dx= \mu_{n} \int_{0}^{1} \bigl(f(u_{n})-mu _{n} \bigr)\phi_{1}\,dx+\mu_{n}m \int_{0}^{1}u_{n}\phi_{1}\,dx. $$

(2.3)

Since $\mu_{n}>\lambda_{\infty }$ and $\int_{0}^{1}u_{n}\phi_{1}\,dx>0$ for n large, we infer that $\int_{0}^{1}(f(u_{n})-mu_{n})\phi_{1}\,dx<0$ for n large, and the Fatou lemma yields

$$\begin{aligned} 0&\geq \liminf\int_{0}^{1} \bigl(f(u_{n})-mu_{n} \bigr)\phi_{1}\,dx \\ &\geq \int_{0} ^{1}a\phi_{1}\,dx, \end{aligned}$$

a contradiction if $a>0$.

We prove statement (ii) similarly to (i). Taking $\mu_{n}\uparrow \lambda_{\infty }$, it follows that $w\geq 0$ satisfies (2.2), and hence there exists $\beta >0$ such that $w=\beta \phi_{1}$. Then we have $u_{n}=\Vert u_{n} \Vert w_{n}\rightarrow +\infty$ and $F(u_{n})=f(u_{n})$ for n large.

From $\varPhi (\lambda_{n},u_{n})=0$ we have (2.3); since $\mu_{n}< \lambda_{\infty }$ and $\int_{0}^{1}u_{n}\phi_{1}\,dx>0$ for n large, we infer that $\int_{0}^{1}(f(u_{n})-mu_{n})\phi_{1}\,dx>0$ for n large, and the Fatou lemma yields

$$ 0\leq \liminf \int_{0}^{1} \bigl(f(u_{n})-mu_{n} \bigr)\phi_{1}\,dx\leq \int_{0} ^{1}A\phi_{1}\,dx, $$

a contradiction if $A<0$. □

Lemma 2.2

If $\lambda >\lambda_{\infty }$, then there exists $r>0$ such that

$$ \varPhi (\lambda ,u)\neq t\phi_{1}\quad \forall t\geq 0,\Vert u \Vert \geq r. $$

Proof

Taking into account that $F(x, u)\simeq m\vert u \vert$ as $\vert u \vert \rightarrow \infty$, we can repeat the arguments of Lemma 3.3 of [16] with some minor changes. □

For $u\neq 0$, we set $z=u\Vert u \Vert ^{-2}$. Letting

$$\begin{aligned} \varPsi (\lambda ,z)&=\Vert u \Vert ^{2}\varPhi (\lambda ,u) =z-\lambda \Vert z \Vert ^{2}KF \biggl(\frac{z}{ \Vert z \Vert ^{2}} \biggr), \end{aligned}$$

we have that $\lambda_{\infty }$ is a bifurcation from infinity for (2.1) if and only if it is a bifurcation from the trivial solution $z=0$ for $\varPsi =0$. From Lemma 2.1 by homotopy it follows that

$$\begin{aligned} \deg \bigl(\varPsi (\lambda ,\cdot ),B_{1/r},0 \bigr)&=\deg \bigl(\varPsi (0, \cdot ),B_{1/r},0 \bigr) \\ &= \deg (I,B_{1/r},0)=1\quad \forall \lambda < \lambda_{\infty }. \end{aligned}$$

(2.4)

Similarly, by Lemma 2.2 we infer that, for all $\tau \in [0,1]$ and $\lambda >\lambda_{\infty }$,

$$\begin{aligned} \deg \bigl(\varPsi (\lambda ,\cdot ),B_{1/r},0 \bigr)&=\deg \bigl(\varPsi ( \lambda ,\cdot )- \tau \phi_{1},B_{1/r},0 \bigr) \\ &=\deg \bigl(\varPsi (\lambda ,\cdot )-\phi_{1},B_{1/r},0 \bigr)=0\quad \forall \lambda < \lambda_{\infty }. \end{aligned}$$

(2.5)

Let us set

$$ \varSigma = \bigl\{ (\lambda ,u)\in \mathbb{R^{+}}\times X: u\neq 0,\varPhi ( \lambda ,u)=0 \bigr\} . $$

From (2.4) and (2.5) and the preceding discussion we deduce the following:

Lemma 2.3

$\lambda_{\infty }$ is a bifurcation from infinity for (2.1). More precisely, there exists an unbounded closed connected set $\varSigma_{\infty }\subset \varSigma$ that bifurcates from infinity. Moreover, $\varSigma_{\infty }$ bifurcates to the left (to the right), provided that $a>0$ (respectively, $A<0$).

Proof of Theorem 2.1

By the previous lemmas it suffices to show that if $\mu_{n}\rightarrow \lambda_{\infty }$ and $\Vert u_{n} \Vert \rightarrow \infty$, then $u_{n}>0$ in $[0,1]$ for n large. Setting

$$ w_{n}=u_{n}\Vert u_{n} \Vert ^{-1} $$

and using the preceding arguments, we find that, up to subsequence, $w_{n}\rightarrow w$ in X and $w=\beta \phi_{1}$, $\beta >0$. Then it follows that

$$ u_{n}>0 $$

in $(0,1)$ for n large. □

Example 2.1

Let us consider the fourth-order semipositone boundary value problem

$$ \textstyle\begin{cases} x''''(t)=\lambda f(t,x),\quad t\in (0,1), \\ x(0)=x(1)=x''(0)=x''(1)=0, \end{cases} $$

(2.6)

where $\lambda >0$ and $f(t,x)=10x+t\ln (1+x)-t$.

Obviously,

$$\begin{aligned}& f(t, 0)< 0,\quad t\in (0,1); \\& \lim_{x\to \infty }\frac{f(t,x)}{x}=10=:m; \\& a(t)=\liminf_{x\rightarrow +\infty } \bigl(f(t,x)-mx \bigr)=\liminf _{x\rightarrow +\infty } \bigl(t\ln (1+x)-t \bigr)>0, \quad t\in (0,1). \end{aligned}$$

Notice that $\lambda_{1}=\pi^{4}$ and $\lambda_{\infty }=\frac{\pi ^{4}}{10}$. Thus by Theorem 2.1 there exists $\epsilon >0$ such that (2.6) has positive solutions, provided that $\lambda \in ( \lambda_{\infty }-\epsilon , \lambda_{\infty })$. Moreover, Lemma 2.3 guarantees that there exists an unbounded closed connected set of positive solutions $\varSigma_{\infty }\subset \varSigma$ that bifurcates from infinity and bifurcates to the left of $\lambda_{\infty }$.

Superlinear problems

We study the existence of positive solutions of problem (1.1) when $f(x,\cdot )$ is superlinear. Precisely, we suppose that $f\in C([0,1] \times \mathbb{R^{+}},\mathbb{R})$ satisfies $(f_{1})$ and

$(f_{3})$ :

there is $b\in C([0,1]),b>0$, such that $\lim_{u\rightarrow \infty }u^{-p}f(x,u) =b$ uniformly in $x\in [0,1]$ with $p>1$.

Lemma 3.1

([6])

Let X be a Banach space, and let $\varOmega \subset X$ be a cone in X. For $p>0$, define $\varOmega_{p}=\{x\in\varOmega \mid \vert x \vert < p\}$. Assume that $F:\varOmega_{p}\rightarrow \varOmega$ is completely continuous such that

$$ Fx\neq x, \quad x\in \partial \varOmega_{p}= \bigl\{ x\in \varOmega \mid \vert x \vert =p \bigr\} . $$

• (1) If $\Vert Fx \Vert \leq \Vert x \Vert$ for $x\in \partial \varOmega_{p}$, then $i(F,\varOmega_{p},\varOmega )=1$.

• (2) If $\Vert Fx \Vert \geq \Vert x \Vert$ for $x\in \partial \varOmega_{p}$, then $i(F,\varOmega_{p},\varOmega )=0$.

Our main result is the following:

Theorem 3.1

Let $f \in C([0,1]\times \mathbb{R^{+}},\mathbb{R})$ satisfy $(f_{1})$ and $(f_{3})$. Then there exists $\lambda_{*}>0$ such that (1.1) has positive solutions for all $0<\lambda \leq \lambda_{*}$. More precisely, there exists a connected set of positive solutions of (1.1) bifurcating from infinity at $\lambda_{\infty }=0$.

Proof

As before, we set

$$ F(x,u)=f \bigl(x,\vert u \vert \bigr) $$

and let

$$ G(x,u)=F(x,u)-b\vert u \vert ^{p}. $$

For the remainder of the proof, we omit the dependence with respect to $x\in [0,1]$.

To prove that $\lambda_{\infty }=0$ is a bifurcation from infinity for

$$ u-\lambda KF(u)=0, $$

(3.1)

we use the rescaling $w=\gamma u, \lambda =\gamma^{p-1}, \gamma >0$. A direct calculation shows that $(\lambda ,u)$, $\lambda >0$, is a solution of (3.1) if and only if

$$ w-K\tilde{F}(\gamma ,w)=0, $$

(3.2)

where

$$ \tilde{F}(\gamma ,w):=b\vert w \vert ^{p}+\gamma^{p}G \bigl(\gamma^{-1}w \bigr). $$

(3.3)

We can extend F̃ to $\gamma =0$ by setting

$$ \tilde{F}(0,w)=b\vert w \vert ^{p}, $$

and by $(f_{3})$ such an extension is continuous. We set

$$ S(\gamma ,w)=w-K\tilde{F}(\gamma ,w),\quad \gamma \in \mathbb{R^{+}}. $$

Let us point out explicitly that $S(\gamma ,\cdot )=I-K$ with compact K. For $\gamma =0$, solutions of $S_{0}(w):=S(0,w)=0$ are nothing but solutions of

$$ \textstyle\begin{cases} w''''=b\vert w \vert ^{p},\quad x\in (0,1), \\ w(0)=w(1)=w''(0)=w''(1)=0. \end{cases} $$

(3.4)

We claim that there exist two constants $R>r>0$ such that

$$\begin{aligned}& S_{0}(w)\neq 0 \quad \forall \Vert w \Vert \geq R, \end{aligned}$$

(3.5)

$$\begin{aligned}& S_{0}(w)\neq 0 \quad \forall \Vert w \Vert \leq r. \end{aligned}$$

(3.6)

Assume on the contrary that (3.5) is not true. Then there exists a sequence $\{w_{n}\}$ of solutions of (3.4) satisfying

$$ \Vert w_{n} \Vert \to \infty , n\rightarrow \infty . $$

(3.7)

In fact, we have from (3.4) that

$$ \textstyle\begin{cases} w_{n}''''= (b\vert w_{n} \vert ^{p-1} )w_{n}, \quad x\in (0,1), \\ w_{n}(0)=w_{n}(1)=w_{n}''(0)=w_{n}''(1)=0, \end{cases} $$

since

$$ \lim_{n\rightarrow \infty } \bigl(b\vert w_{n} \vert ^{p-1} \bigr)=\infty \quad \text{uniformly in } x\in [1/4,3/4], $$

which means that $w_{n}$ must change its sign in $[1/4,3/4]$. However, this is a contradiction. Therefore (3.5) is valid.

Assume on the contrary that (3.6) is not true. Then there exists a sequence ${w_{n}}$ of solutions of (3.4) satisfying

$$ \Vert w_{n} \Vert >0 \quad \forall {n\in \mathbb{N}}; \qquad \Vert w_{n} \Vert \rightarrow 0, \quad n\rightarrow \infty . $$

(3.8)

Let $v_{n}:= w_{n}/\Vert w_{n} \Vert$. From (3.4) we have

$$ \textstyle\begin{cases} v_{n}''''=(b\vert w_{n} \vert ^{p-1})v_{n},\quad x\in (0,1), \\ v_{n}(0)=v_{n}(1)=v_{n}''(0)=v_{n}''=0. \end{cases} $$

(3.9)

By the standard argument, after taking a subsequence and relabeling if necessary, it follows that

$$ \lim_{n\rightarrow \infty } \bigl(b\vert w_{n} \vert ^{p-1} \bigr)=0\quad \text{uniformly in } x\in [0,1], $$

and there exists $v_{*}\in X$ with $\Vert v_{*} \Vert =1$ such that

$$ v_{n}\rightarrow v_{*}, \quad n\rightarrow \infty , $$

and

$$ \textstyle\begin{cases} v_{*}''''=0,\quad x\in (0,1), \\ v_{*}(0)=v_{*}(1)=v_{*}''(0)=v_{*}''(1)=0, \end{cases} $$

which implies that $v_{*}=0$. However, this is a contradiction, Therefore (3.6) is valid.

Now, from (3.5) and (3.6), we deduce

$$ S_{0}(w)\neq 0\quad \forall w\in \partial \varOmega_{R},\qquad S_{0}(w) \neq 0 \quad \forall w\in \partial \varOmega_{r}. $$

This implies

$$ S_{0}(w)\neq 0\quad \forall w\in \partial (\bar{\varOmega }_{R}\setminus \varOmega_{r}). $$

Thus the degree $\deg (S_{0}, \varOmega_{R}\setminus \varOmega_{r},0)$ is well defined.

Next, we show that

$$ \deg (S_{0}, \varOmega_{R}\setminus \bar{\varOmega }_{r},0)=-1. $$

To this end, let us define

$$ \varOmega = \bigl\{ u\in C[0,1]:u(t)\geq 0 \text{ for } t\in [0,1] \bigr\} $$

and

$$ \varOmega_{\rho }= \bigl\{ u\in \varOmega : \bigl\Vert u(t) \bigr\Vert < \rho \bigr\} . $$

Using Lemma 3.1 and an argument similar to that in the proof of [6], Theorem 3, we deduce

$$ \mathrm{i} \bigl(K\tilde{F}(0,\cdot ),\varOmega_{r},\varOmega \bigr)=1, \qquad \mathrm{i} \bigl(K\tilde{F}(0,\cdot ),\varOmega_{R},\varOmega \bigr)=0. $$

(3.10)

By the excision and the additivity properties of the degree it follows that

$$ \mathrm{i} \bigl(K\tilde{F}(0,\cdot ),\varOmega_{R}\setminus \bar{\varOmega }_{r}, \varOmega \bigr)+\mathrm{i} \bigl(K\tilde{F}(0,\cdot ),\varOmega_{r},\varOmega \bigr)= \mathrm{i} \bigl(K\tilde{F}(0,\cdot ),\varOmega_{R},\varOmega \bigr), $$

(3.11)

and accordingly,

$$ \mathrm{i} \bigl(K\tilde{F}(0,\cdot ),\varOmega_{R}\setminus \bar{\varOmega }_{r}, \varOmega \bigr)=\mathrm{i} \bigl(K\tilde{F}(0,\cdot ),\varOmega_{R},\varOmega \bigr) - \mathrm{i} \bigl(K\tilde{F}(0,\cdot ),\varOmega_{r},\varOmega \bigr)=-1, $$

(3.12)

that is,

$$ \deg (S_{0},\varOmega_{R}\setminus \bar{\varOmega }_{r},0)=-1. $$

Lemma 3.2

There exists $\gamma >0$ such that

• (i) $\deg (S(\gamma , \cdot ), \varOmega_{R}\setminus \bar{\varOmega }_{r}, 0)=-1$ $\forall 0 \leq \gamma \leq \gamma_{0}$;

• (ii) if $S(\gamma ,w)=0$, $\gamma \in [0,\gamma_{0}]$, $r \leq \Vert w \Vert \leq R$, then $w>0$ in $(0,1)$.

Proof

Clearly, (i) follows if we show that

$$ S(\gamma ,w)\neq 0, \quad 0 \leq \gamma \leq \gamma_{0}, \Vert w\Vert \in \{r,R\}. $$

Otherwise, there exists a sequence $(\gamma_{n}, w_{n})$ with $\gamma_{n}\rightarrow 0$, $\Vert w_{n} \Vert \in \{r,R\}$, and $w_{n}=K \tilde{F}(\gamma_{n}, w_{n})$. Since K is compact, then, up to a subsequence, $w_{n}\rightarrow w$, and

$$ S_{0}(w)=0,\quad \Vert w \Vert \in \{r,R\}, $$

a contradiction with (3.5) and (3.6).

Thus, by (3.7) and homotopy we get that

$$ \deg \bigl(S(\gamma ,\cdot ), \varOmega_{R}\setminus \varOmega_{r},0 \bigr)=-1. $$

To prove (ii), we argue again by contradiction. As in the preceding argument, we find a sequence $w_{n}\in X$ with $\{x\in [0,1]: w_{n}(x) \leq 0\}\neq \varnothing$ such that $w_{n}\rightarrow w$, $\Vert w \Vert \in [r,R]$, and $S_{0}(w)=0$; namely, w solves (3.4). By the maximum principle, $w>0$ on (0,1) and X. Moreover, without relabeling, $w_{n}\rightarrow w$ in X. Therefore

$$ w_{n}>0,\quad x\in (0,1), $$

for n large, a contradiction. □

Proof of Theorem 3.1 completed

By Lemma 3.2 problem (3.2) has a positive solution $w_{\gamma }$ for all $0\leq \gamma \leq \gamma_{0}$. As remarked before, for $\gamma >0$, the rescaling $\lambda =\gamma^{p-1}, u=w/\gamma$ gives a solution $(\lambda , u_{\lambda })$ of (3.1) for all $0<\lambda <\lambda_{*}:=\gamma_{0}^{p-1}$. Since $w_{\gamma }>0$, $(\lambda , u_{\lambda })$ is a positive solution of (1.1). Finally, $\Vert w_{\gamma } \Vert \geq r$ for all $\gamma \in [0,\gamma_{0}]$ implies that

$$ \Vert u_{\lambda } \Vert =\Vert w \Vert _{\gamma }/\gamma \rightarrow \infty \quad \text{as }\gamma \rightarrow 0. $$

This completes the proof. □

Sublinear problems

In this final section, we deal with sublinear f, namely with $f\in C([0,1]\times \mathbb{R^{+}},\mathbb{R})$ that satisfy $(f_{1})$ and

$(f_{4})$ :

$\exists b\in C([0,1]), b>0$, such that $\lim_{u\rightarrow \infty }u^{-q}f(x,u)= b$ uniformly in $x\in [0,1]$ with $0\leq q<1$.

We will show that in this case positive solutions of (1.1) branch off from ∞ for $\lambda_{\infty }=+\infty$. First, some preliminaries are in order. It is convenient to work on X. Following the same procedure as for the superlinear case, we employ the rescaling $w=\gamma u,\lambda =\gamma^{q-1}$ and use the same notation with q instead of p. As before, $(\lambda , u)$ solves (3.1) if and only if $(\gamma ,w)$ satisfies (3.2). Note that now, since $0\leq q<1$, we have that

$$ \lambda \rightarrow +\infty \quad \Leftrightarrow \quad \gamma \rightarrow 0. $$

(4.1)

For future reference, note that by Lemma 3.1

$$ \textstyle\begin{cases} u''''=bw^{q},\quad x\in (0,1), \\ u(0)=u(1)=u''(0)=u''(1)=0 \end{cases} $$

(4.2)

has a unique positive solution $w_{0}$.

We claim that there exist two constants $R>r>0$ such that

$$\begin{aligned}& S_{0}(w)\neq 0 \quad \forall \Vert w \Vert \leq R; \end{aligned}$$

(4.3)

$$\begin{aligned}& S_{0}(w)\neq 0 \quad \forall \Vert w \Vert \geq r; \end{aligned}$$

(4.4)

$$\begin{aligned}& \deg (S_{0}, O_{R}\setminus O_{r},0)=1. \end{aligned}$$

(4.5)

Assume on the contrary that (4.3) is not true. Then there exists a sequence ${w_{n}}$ of solutions of (4.4) satisfying

$$ \Vert w_{n} \Vert \rightarrow 0, \quad n\rightarrow \infty , $$

(4.6)

then $w_{n}\equiv 0$ in $[0,1]$ for n large.

Let $v_{n}:= w_{n}/\Vert w_{n} \Vert$. From (3.4) we have

$$ \textstyle\begin{cases} v_{n}''''=(b\vert w_{n} \vert ^{q-1})v_{n},\quad x\in (0,1),\\ v_{n}(0)=v_{n}(1)=v_{n}''(0)=v_{n}''(1)=0. \end{cases} $$

(4.7)

By the standard argument, after taking a subsequence and relabeling if necessary, it follows that

$$ \lim_{n\rightarrow \infty } \bigl(b\vert w_{n} \vert ^{q-1} \bigr)=0\quad \text{uniformly in } x\in [0,1], $$

and there exists $v_{*}\in X$ with $\Vert v_{*} \Vert =1$ such that

$$ v_{n}\rightarrow v_{*}, \quad n\rightarrow \infty , $$

and

$$ \textstyle\begin{cases} v_{*}''''=0,\quad x\in (0,1), \\ v_{*}(0)=v_{*}(1)=v_{*}''(0)=v_{*}''(1)=0, \end{cases} $$

which implies that $v_{*}=0$. However, this is a contradiction, Therefore (4.3) is valid.

Assume on the contrary that (4.4) is not true. Then there exists a sequence $\{w_{n}\}$ of solutions of (4.4) satisfying

$$ \Vert w_{n} \Vert \to \infty , \quad n\rightarrow \infty . $$

(4.8)

In fact, we have from (3.4) that

$$ \textstyle\begin{cases} w_{n}''''= (b\vert w_{n} \vert ^{q-1} )w_{n}, \quad x\in (0,1), \\ w_{n}(0)=w_{n}(1)=w_{n}''(0)=w_{n}''(1)=0, \end{cases} $$

since

$$ \lim_{n\rightarrow \infty } \bigl(b\vert w_{n} \vert ^{q-1} \bigr)=\infty \quad \text{uniformly in } x\in [1/4,3/4], $$

which shows that $w_{n}$ must change its sign in $[1/4,3/4]$. However, this is a contradiction. Therefore (4.4) is valid.

Now, from (4.3) and (4.4) we deduce

$$ S_{0}(w)\neq 0 \quad \forall w\in \partial O_{R},\qquad S_{0}(w)\neq 0 \quad \forall w\in \partial O_{r}. $$

This implies that

$$ S_{0}(w)\neq 0\quad \forall w\in \partial (\bar{O}_{R} \setminus O _{r}). $$

Thus, the degree $\deg (S_{0}, O_{R}\setminus \bar{O}_{r},0)$ is well defined.

Next, we show that

$$ \deg (S_{0}, O_{R}\setminus \bar{O}_{r},0)=1. $$

To this end, let us define

$$ O= \bigl\{ u\in C[0,1]:u(t)\geq 0 \text{ for } t\in [0,1] \bigr\} $$

and

$$ O_{\rho }= \bigl\{ u\in O:\Vert u \Vert < \rho \bigr\} . $$

Using Lemma 3.1 and an argument similar to that in the proof of [6], Theorem 3, we deduce

$$ \mathrm{i} \bigl(K\tilde{F}(0,\cdot ),O_{r},O \bigr)=0,\qquad \mathrm{i} \bigl(K \tilde{F}(0,\cdot ),O_{R},O \bigr)=1. $$

By the excision and the additivity properties of the degree it follows that

$$ \mathrm{i} \bigl(K\tilde{F}(0,\cdot ),O_{R}\setminus \bar{O}_{r},O \bigr)+ \mathrm{i} \bigl(K\tilde{F}(0,\cdot ),O_{r},O \bigr)=\mathrm{i} \bigl(K\tilde{F}(0, \cdot ),O_{R},O \bigr) , $$

and accordingly,

$$ \mathrm{i} \bigl(K\tilde{F}(0,\cdot ),O_{R}\setminus \bar{O}_{r},O \bigr)= \mathrm{i} \bigl(K\tilde{F}(0,\cdot ),O_{R},O \bigr) -\mathrm{i} \bigl(K\tilde{F}(0, \cdot ),O_{r},O \bigr)=1, $$

that is,

$$ \deg (S_{0},O_{R}\setminus \bar{O}_{r},0)=1. $$

Lemma 4.1

There exists $\gamma >0$ such that

• (i) $\deg (S(\gamma , \cdot ), O_{R}\setminus \bar{O}_{r}, 0)=1$ $\forall 0 \leq \gamma \leq \gamma_{0}$;

• (ii) if $S(\gamma ,w)=0$, $\gamma \in [0,\gamma_{0}]$, $r \leq \Vert w \Vert \leq R$, then $w>0$ in $(0,1)$.

Proof

Clearly, (i) follows if we show that

$$ S(\gamma ,w)\neq 0, \quad 0 \leq \gamma \leq \gamma_{0}, \Vert w \Vert \in \{r,R\}. $$

Otherwise, there exists a sequence $(\gamma_{n}, w_{n})$ with $\gamma_{n}\rightarrow 0$, $\Vert w_{n} \Vert \in \{r,R\}$, and $w_{n}=K \tilde{F}(\gamma_{n}, w_{n})$. Since K is compact, then, up to a subsequence, $w_{n}\rightarrow w$, and

$$ S_{0}(w)=0, \quad \Vert w \Vert \in \{r,R\}, $$

a contradiction with (4.3) and (4.4).

To prove (ii), we argue again by contradiction. As in the preceding argument, we find a sequence $w_{n}\in X$ with $\{x\in [0,1]: w_{n}(x) \leq 0\}\neq \varnothing$ such that $w_{n}\rightarrow w,\Vert w \Vert \in [r,R]$, and $S_{0}(w)=0$; namely, w solves (3.2). By the maximum principle, $w>0$ on (0,1) and X. Moreover, without relabeling, $w_{n}\rightarrow w$ in X. Therefore

$$ w_{n}>0, \quad x\in (0,1), $$

for n large, a contradiction. □

Theorem 4.1

Let $f\in C([0,1]\times \mathbb{R^{+}}, \mathbb{R})$ satisfy $(f_{1})$ and $(f_{4})$. Then there is $\lambda^{*}>0$ such that (1.1) has positive solutions for all $\lambda \geq \lambda^{*}$. More precisely, there exists a connected set of positive solutions of (1.1) bifurcating from infinity for $\lambda_{\infty }=+\infty$.

Proof of Theorem 4.1

By Lemma 4.1 problem (3.2) has a positive solution $w_{\gamma }$ for all $0\leq \gamma \leq \gamma_{0}$. As remarked before, for $\gamma >0$, the rescaling

$$ \lambda =\gamma^{q-1},\quad u=w/\gamma $$

gives a solution $(\lambda , u_{\lambda })$ of (3.1) for all $\lambda \geq \lambda^{*}:=\gamma_{0}^{q-1}$. Since $w_{\gamma }>0$, $(\lambda , u_{\lambda })$ is a positive solution of (1.1). Finally, $\Vert w_{\gamma } \Vert \geq r$ for all $\gamma \in [0,\gamma_{0}]$ implies that

$$ \Vert u_{\lambda } \Vert =\Vert w \Vert _{\gamma }/\gamma\rightarrow \infty \quad \text{as } \gamma \rightarrow 0. $$

This completes the proof. □