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
$$ G(t,s)= \textstyle\begin{cases} t(1s),&0 \leq t \leq s \leq 1, \\ s(1t),&0 \leq s \leq t \leq 1. \end{cases} $$
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
in \((0,1)\) for n large. □
Example 2.1
Let us consider the fourthorder 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 \(\lambda_{\infty }\).