In this section, we apply the previous abstract result on fixed points for discontinuous operators in order to look for \(W^{2,1}\)solutions for the following singular secondorder ODE with separated BCs:
$$ \left \{ \textstyle\begin{array}{@{}l} u''(t)+g(t)f(t,u(t))=0 \quad\mbox{for almost every (a.e.) } t \in I=[0,1], \\ \alpha u(0)  \beta u'(0)=0, \\ \gamma u(1) + \delta u'(1)=0, \end{array}\displaystyle \right . $$
(3.1)
where \(\alpha, \beta, \gamma, \delta\ge0\) and \(\Gamma=\gamma \beta+ \alpha\gamma+ \alpha\delta>0\).
This kind of secondorder BVPs have received a lot of attention in the literature. For example, in the monograph [7] the method of lower and upper solutions is used to look for \(\mathcal {C}^{2}\)solutions in the case of continuous nonlinearities and \(W^{2,1}\)solutions in the case of Carathéodory ones. This method is also applied in [8] to a continuous φLaplacian problem with separated BCs. On the other hand, a monotone method is applied in [9] in order to look for extremal solutions for a functional problem with derivative dependence in the nonlinearity. As a main novelty of the present work, we allow the nonlinearity f to have a countable number of discontinuities with respect to its spatial variable, and we require no monotonicity conditions. Moreover, the function g can be singular.
To apply our new fixed point theorem to the BVP (3.1), we recall that \(u \in W^{2,1}(I)\) is a solution of (3.1) if (and only if) u is a solution of the following Hammerstein integral equation:
$$ u(t)= \int_{0}^{1} k(t,s) g(s) f\bigl(s,u(s)\bigr)\,ds, $$
(3.2)
whenever the integral in (3.2) has sense and where k is the corresponding Green function, which is given by (see, for example, [10])
$$ k(t,s)=\frac{1}{\Gamma} \left \{ \textstyle\begin{array}{@{}l@{\quad}l} (\gamma+ \delta \gamma t)(\beta+ \alpha s) &\mbox{if }0 \le s \le t \le1, \\ (\beta+ \alpha t)(\gamma+ \delta \gamma s) &\mbox{if }0 \le t < s \le1. \end{array}\displaystyle \right . $$
(3.3)
It is known [10] that k is nonnegative. Furthermore, note that k is continuous (and therefore bounded) in the square \([0,1] \times [0,1]\) and that its partial derivatives \(\frac{\partial k}{\partial t}\) and \(\frac{\partial k}{\partial s}\) can be discontinuous on the diagonal \(t=s\). However, these partial derivatives are essentially bounded on the square.
In the sequel, we consider the Banach space \(X=\mathcal{C}^{1}(I)\) of continuously differentiable functions defined on I with the norm
$$\u\=\sup_{t \in I} \biglu(t)\bigr + \sup_{t \in I} \biglu'(t)\bigr. $$
Lemma 3.1
Assume that:
 (H_{1}):

\(g \in L^{1}(I)\);
 (H_{2}):

there exist
\(R >0\)
and
\(H_{R} \in L^{\infty}(I)\)
such that for a.e. \(t \in I\)
and all
\(u \in[R,R]\)
we have
\(f(t,u) \le H_{R}(t)\);
 (H_{3}):

the following estimate holds:
$$\H_{R}\_{L^{\infty}} (M_{1}+M_{2}) \le R, $$
where
$$ M_{1}=\sup_{t \in I} \int_{0}^{1} k(t,s) \biglg(s)\bigr\,ds,\qquad M_{2}= \sup _{t \in I} \int_{0}^{1} \biggl\vert \frac{\partial k}{\partial t}(t,s) g(s) \biggr\vert \,ds; $$
(3.4)
 (H_{4}):

for each
\(u \in\overline{B}_{R}=\{u \in X : \u\ \le R\}\)
the composition
\(t \in I \longmapsto f(t,u(t))\)
is a measurable function.
Then the operator
\(T: \overline{B}_{R} \longrightarrow X\)
given by
$$Tu(t)= \int_{0}^{1} k(t,s) g(s) f\bigl(s,u(s)\bigr)\,ds $$
is well defined and maps
\(\overline{B}_{R}\)
into itself.
Remark 3.2
Since k is the Green’s function related to a homogeneous secondorder BVP, \(Tu \in W^{2,1}(I)\) for all u, so, in particular, \((Tu)'\) is absolutely continuous (then \(Tu \in X\)), and \((Tu)''\) exists almost everywhere on I. This will be used later in our argumentations.
Proof of Lemma 3.1
Let \(R>0\) given by condition (H_{2}). First, note that the kernel k has the form (3.3). Therefore, for each \(t \in[0,1]\), \(k(t,\cdot)\) is a continuous function, and for \(s\neq t\), the function \(s \in[0,1] \to\frac {\partial k}{\partial t}(t,s)\) is well defined and integrable. Then, conditions (H_{1}), (H_{2}) and (H_{4}) imply that for \(u \in \overline{B}_{R}\) the function Tu is well defined.
On the other hand, for \(u \in\overline{B}_{R}\) we have
$$\begin{aligned} \Tu\ &\le\sup_{t \in I} \int_{0}^{1} k(t,s) \biglg(s)\bigr \biglf\bigl(s,u(s)\bigr)\bigr\,ds + \sup_{t \in I} \int_{0}^{1} \biggl\vert \frac{\partial k}{\partial t}(t,s) \biggr\vert \biglg(s)\bigr \biglf(s,u(s)\bigr\,ds \\ & \le\H_{R}\_{\infty} (M_{1}+M_{2}), \end{aligned}$$
and then condition (H_{3}) implies that \(\Tu\ \le R\). □
Lemma 3.3
Under the assumptions of Lemma
3.1, \(T(\overline{B}_{R})\)
is relatively compact in
X.
Proof
We have shown in Lemma 3.1 that \(T(\overline{B}_{R}) \subset \overline{B}_{R}\). Therefore, the set \(T(\overline{B}_{R})\) is totally bounded in X. Now, to see that \(T(\overline{B}_{R})\) is equicontinuous, we only have to notice that, for a.e. \(t \in I\) and every \(u \in\overline{B}_{R}\), we have
$$\bigl(Tu)''(t)\bigr \le\biglg(t)\bigr H_{R}(t), $$
which implies that
$$\bigl(Tu)'(t)(Tu)'(s)\bigr \le \int_{t}^{s} \bigl(Tu)''(r)\bigr\,dr \le \int_{t}^{s} \biglg(r)\bigr H_{R}(r)\,dr. $$
Then \(T(\overline{B}_{R})\) is relatively compact in X. □
In a similar way as in Definition 4.1 of [1], we introduce the admissible discontinuities for our nonlinearities.
Definition 3.4
We say that \(\gamma:[a,b] \subset I \longrightarrow \mathbb {R}\), \(\gamma\in W^{2,1}([a,b])\), is an admissible discontinuity curve for the differential equation \(u''(t)+g(t)f(t,u(t))=0\) if one of the following conditions holds:

(i)
\(\gamma''(t)=g(t) f(t,\gamma(t))\) for a.e. \(t \in[a,b]\);

(ii)
there exist \(\psi\in L^{1}([a,b])\), \(\psi>0\) almost everywhere, and \(\varepsilon>0\) such that
$$\begin{aligned}& \begin{aligned}[b] &\mbox{either }{}\gamma''(t) + \psi(t) < g(t)f(t,y)\\ &\quad\mbox{for a.e. }t \in[a,b]\mbox{ and all } y \in\bigl[\gamma (t)\varepsilon, \gamma(t)+\varepsilon\bigr], \end{aligned} \end{aligned}$$
(3.5)
$$\begin{aligned}& \mbox{or }{}\gamma''(t)  \psi(t) > g(t)f(t,y) \\& \quad\mbox{for a.e. }t \in[a,b]\mbox{ and all }y \in\bigl[\gamma (t)\varepsilon, \gamma(t)+\varepsilon\bigr]. \end{aligned}$$
(3.6)
If (i) holds, then we say that γ is viable for the differential equation; if (ii) holds, we say that γ is inviable.
The previous definition says, roughly speaking, that a timedependent discontinuity curve γ is admissible if one of the following holds: either γ solves the differential equation on its domain, or, if it does not, the solutions are pushed ‘far away’ from γ.
To prove our main result on the existence of solutions for problem (3.1) by using admissible discontinuity curves, we need some auxiliary theoretical results on integrable functions. The reader can see their proofs in [1].
Lemma 3.5
([1], Lemma 4.1)
Let
\(a,b \in \mathbb {R}\), \(a< b\), and let
\(g, h \in L^{1}(a,b)\), \(g \ge0\)
a.e., and
\(h>0\)
a.e. in
\((a,b)\).
For every measurable set
\(J \subset(a,b)\)
with
\(m(J)>0\), there is a measurable set
\(J_{0} \subset J\)
with
\(m(J \setminus J_{0})=0\)
such that, for every
\(\tau_{0} \in J_{0}\), we have
$$ \lim_{t \to\tau_{0}^{+}}\frac{\int_{[\tau_{0},t]\setminus J}g(s)\,ds}{\int_{\tau_{0}}^{t}{h(s)\,ds}}=0=\lim _{t \to\tau_{0}^{}}\frac {\int_{[t,\tau_{0}]\setminus J}g(s)\,ds}{\int_{t}^{\tau_{0}}{h(s)\,ds}}. $$
(3.7)
Corollary 3.6
([1], Corollary 4.2)
Let
\(a,b \in \mathbb {R}\), \(a< b\), and let
\(h \in L^{1}(a,b)\)
be such that
\(h>0\)
a.e. in
\((a,b)\).
For every measurable set
\(J \subset(a,b)\)
with
\(m(J)>0\), there is a measurable set
\(J_{0} \subset J\)
with
\(m(J \setminus J_{0})=0\)
such that, for all
\(\tau_{0} \in J_{0}\), we have
$$ \lim_{t \to\tau_{0}^{+}} \frac{\int_{[\tau_{0},t] \cap J}h(s)\,ds}{\int_{\tau_{0}}^{t}{h(s)\,ds}}=1=\lim _{t \to\tau_{0}^{}}\frac {\int_{[t,\tau_{0}]\cap J}h(s)\,ds}{\int_{t}^{\tau_{0}}{h(s)\,ds}}. $$
(3.8)
Corollary 3.7
([1], Corollary 4.3)
Let
\(a,b \in \mathbb {R}\), \(a< b\), and let
\(f, f_{n}:[a,b] \longrightarrow \mathbb {R}\)
be absolutely continuous functions on
\([a,b]\) (\(n \in \mathbb {N}\)) such that
\(f_{n} \to f\)
uniformly on
\([a,b]\)
and, for a measurable set
\(A \subset [a,b]\)
with
\(m(A)>0\), we have
$$\lim_{n \to\infty}f_{n}'(t)=g(t) \quad\textit{for a.a. }t \in A. $$
If there exists
\(M \in L^{1}(a,b)\)
such that
\(f'(t) \le M(t)\)
a.e. in
\([a,b]\)
and also
\(f_{n}'(t) \le M(t)\)
a.e. in
\([a,b]\) (\(n \in \mathbb {N}\)), then
\(f'(t)=g(t)\)
for a.a. \(t \in A\).
Now we can show the main result in this section.
Theorem 3.8
Let
f
and
g
satisfy (H_{1})(H_{4}) and the following:
 (H_{5}):

there exist admissible discontinuity curves
\(\gamma_{n}: I_{n} =[a_{n},b_{n}] \longrightarrow \mathbb {R}\), \(n \in \mathbb {N}\), such that for a.e. \(t \in I\)
the function
\(f(t,\cdot)\)
is continuous in
\([R,R] \setminus\bigcup_{n : t \in I_{n}} \{ \gamma_{n}(t)\}\).
Then problem (3.1) has at least one solution in
\(\overline{B}_{R}\).
Proof
We consider the multivalued operator \(\mathbb {T}\) associated to T as in (2.1). Therefore, \(\mathbb {T}\) is upper semicontinuous with nonempty, convex and compact values and, as T, maps \(\overline{B}_{R}\) into itself. Moreover, \(\mathbb {T}(\overline{B}_{R})\) is relatively compact in X by Lemma 3.3. Therefore, if we show that \(\{u\} \cap \mathbb {T}u \subset\{Tu\}\), then we obtain by Theorem 2.7 that T has a fixed point in \(\overline{B}_{R}\), which corresponds to a solution of the BVP (3.1). This part of the proof now follows the lines of [1], Theorem 4.4, but we include it for completeness and for highlighting the main differences between the two results. Thus, we fix \(u \in\overline{B}_{R}\) and consider three cases.
Case 1: \(m(\{t \in I_{n} : u(t)=\gamma_{n}(t)\} )=0\) for all \(n \in \mathbb {N}\).
Then we have that \(f(t,\cdot)\) is continuous for a.e. \(t \in I\), and therefore if \(u_{k} \to u\) in \(\overline{B}_{R}\) then we obtain \(f(t,u_{k}(t)) \to f(t,u(t))\) for a.e. \(t \in I\). This, together with (H_{2}) and (H_{3}), implies that \(Tu_{k}\) converges uniformly to Tu in X. Then, T is continuous at u, and therefore we obtain \(\mathbb {T}u= \{ Tu\}\).
Case 2: there exists \(n \in \mathbb {N}\) such that \(\gamma _{n}\) is inviable and \(m(\{t \in I_{n} : u(t)=\gamma_{n}(t)\})>0\). We will show that, in this case, \(u \notin \mathbb {T}u\).
To do this, assume that \(\gamma_{n}\) satisfies (3.6) (the other case is similar). Let \(\psi\in L^{1}(I)\) and \(\varepsilon>0\) given by (3.6) and set
$$J=\bigl\{ t \in I_{n} : u(t)=\gamma_{n}(t)\bigr\} ,\qquad M(t)=\biglg(t)\bigr H_{R}(t). $$
Notice that (H_{1}) and (H_{2}) imply that \(M \in L^{1}(I)\), and so we deduce from Lemma 3.5 that there is a measurable set \(J_{0} \subset J\) with \(m(J_{0})=m(J)>0\) such that, for all \(\tau_{0} \in J_{0}\), we have
$$ \lim_{t \to\tau_{0}^{+}}\frac{2\int_{[\tau_{0},t] \setminus J}M(s)\,ds}{(1/4)\int_{\tau_{0}}^{t}{\psi(s)\,ds}}=0=\lim _{t \to\tau _{0}^{}}\frac{2\int_{[t,\tau_{0}]\setminus J}M(s)\,ds}{(1/4)\int _{t}^{\tau_{0}}{\psi(s)\,ds}}. $$
(3.9)
By Corollary 3.6 there exists \(J_{1} \subset J_{0}\) with \(m(J_{0} \setminus J_{1})=0\) such that, for all \(\tau_{0} \in J_{1}\), we have
$$ \lim_{t \to\tau_{0}^{+}}\frac{\int_{[\tau_{0},t] \cap J_{0}}\psi(s)\,ds}{\int_{\tau_{0}}^{t}{\psi(s)\,ds}}=1=\lim _{t \to\tau _{0}^{}}\frac{\int_{[t,\tau_{0}]\cap J_{0}}\psi(s)\,ds}{\int_{t}^{\tau _{0}}{\psi(s)\,ds}}. $$
(3.10)
Let us now fix a point \(\tau_{0} \in J_{1}\). From (3.9) and (3.10) we deduce that there exist \(t_{}<\tau_{0}\) and \(t_{+}>\tau_{0}\), \(t_{\pm}\) sufficiently close to \(\tau_{0}\), such that the following inequalities are satisfied:
$$\begin{aligned}& 2\int_{[\tau_{0},t_{+}]\setminus J}M(s)\,ds < \frac{1}{4} \int_{\tau _{0}}^{t_{+}}{\psi(s)\,ds}, \end{aligned}$$
(3.11)
$$\begin{aligned}& \int_{[\tau_{0},t_{+}] \cap J}\psi(s)\,ds \ge \int_{[\tau_{0},t_{+}] \cap J_{0}}\psi(s)\,ds>\frac{1}{2} \int_{\tau_{0}}^{t_{+}}\psi(s)\,ds, \end{aligned}$$
(3.12)
$$\begin{aligned}& 2 \int_{[t_{},\tau_{0}]\setminus J}M(s)\,ds < \frac{1}{4} \int _{t_{}}^{\tau_{0}}{\psi(s)\,ds}, \end{aligned}$$
(3.13)
$$\begin{aligned}& \int_{[t_{},\tau_{0}] \cap J}\psi(s)\,ds>\frac{1}{2} \int_{t_{}}^{\tau_{0}}\psi(s)\,ds. \end{aligned}$$
(3.14)
Finally, we define the positive number
$$ \rho=\min \biggl\{ \frac{1}{4} \int_{t_{}}^{\tau_{0}}{\psi(s)\,ds}, \frac{1}{4} \int_{\tau_{0}}^{t_{+}}{\psi(s)\,ds} \biggr\} , $$
(3.15)
and we are now in a position to prove that \(u \notin{\mathbb {T}}u\). By Proposition 2.3 it suffices to prove the following claim.
Claim
Let
\(\varepsilon>0\)
be given by our assumptions over
\(\gamma_{n}\), and let
ρ
be as in (3.15). For every finite family
\(u_{i} \in B_{\varepsilon}(u)\cap\overline{B}_{R} \)
and
\(\lambda _{i} \in[0,1]\) (\(i=1,2,\dots,m\)) with
\(\sum\lambda_{i}=1\), we have
$$\biggl\ u\sum\lambda_{i} Tu_{i}\biggr\ _{{\mathcal {C}}^{1}} \ge\rho. $$
Let \(u_{i}\) and \(\lambda_{i}\) be as in the claim and, for simplicity, denote \(v=\sum\lambda_{i} Tu_{i}\). Then for a.a. \(t \in J=\{t \in I_{n} : u(t)=\gamma_{n}(t)\}\) we have
$$ v''(t)=\sum _{i=1}^{m} \lambda_{i} (Tu_{i})''(t)= \sum_{i=1}^{m}{\lambda_{i} g(t) f \bigl(t,u_{i}(t)\bigr)}. $$
(3.16)
On the other hand, for every \(i \in\{1,2,\dots,m\}\) and every \(t \in J\), we have
$$\biglu_{i}(t)\gamma_{n}(t)\bigr=\biglu_{i}(t)u(t)\bigr< \varepsilon, $$
and then the assumptions on \(\gamma_{n}\) ensure that, for a.a. \(t \in J\), we have
$$ v''(t)=\sum _{i=1}^{m}{\lambda_{i} g(t) f \bigl(t,u_{i}(t)\bigr)}< \sum_{i=1}^{m}{ \lambda_{i} \bigl(\gamma_{n}''(t) \psi(t)\bigr)}=u''(t)\psi(t). $$
(3.17)
Now we compute
$$\begin{aligned} v'(\tau_{0})v'(t_{})={}& \int_{t_{}}^{\tau_{0}}{v''(s)\,ds}= \int _{[t_{},\tau_{0}]\cap J}{v''(s)\,ds}+ \int_{[t_{},\tau_{0}]\setminus J}{v''(s)\,ds} \\ < {}& \int_{[t_{},\tau_{0}]\cap J}{u''(s)\,ds} \int_{[t_{},\tau_{0}]\cap J}{\psi(s)\,ds} \\ &{} + \int_{[t_{},\tau_{0}]\setminus J}{M(s)\,ds} \quad\bigl(\mbox{by (3.17), (3.16) and (H$_{2}$)}\bigr) \\ ={}&u'(\tau_{0})u'(t_{}) \int_{[t_{},\tau_{0}]\setminus J}{u''(s)\,ds} \int_{[t_{},\tau_{0}]\cap J}{\psi(s)\,ds} + \int_{[t_{},\tau_{0}]\setminus J}{M(s)\,ds} \\ \le{}& u'(\tau_{0})u'(t_{}) \int_{[t_{},\tau_{0}]\cap J}{\psi(s)\,ds}+2 \int_{[t_{},\tau_{0}]\setminus J}{M(s)\,ds} \\ < {}&u'(\tau_{0})u'(t_{})\frac{1}{4} \int_{t_{}}^{\tau_{0}}{\psi(s)\,ds} \quad\bigl(\mbox{by (3.13) and (3.14)}\bigr). \end{aligned}$$
Hence, \(\uv\_{{\mathcal {C}}^{1}} \ge v'(t_{})u'(t_{}) \ge\rho\), provided that \(v'(\tau_{0}) \ge u'(\tau_{0})\).
Similar computations with \(t_{+}\) instead of \(t_{}\) show that if \(v'(\tau _{0}) \le u'(\tau_{0})\), then we also have \(\uv\_{{\mathcal {C}}^{1}} \ge \rho\). The claim is proven.
Case 3: \(m(\{t \in I_{n} : u(t)=\gamma_{n}(t)\} )>0\) only for some of those \(n \in \mathbb {N}\) such that \(\gamma_{n}\) is viable. We will show that, in this case, \(u \in \mathbb {T}u\) implies \(u=Tu\).
To see that, we consider the subsequence of all viable admissible discontinuity curves in the conditions of Case 3, which we can denote without loss of generality by \(\{\gamma_{n}\}_{n \in{\scriptsize \mathbb {N}}}\). We have \(m(J_{n})>0\) for all \(n \in \mathbb {N}\), where
$$J_{n}=\bigl\{ t \in I_{n} : u(t)=\gamma_{n}(t) \bigr\} . $$
For each \(n \in \mathbb {N}\) and for a.a. \(t \in J_{n}\), we have
$$u''(t)=\gamma_{n}''(t)=g(t)f \bigl(t,\gamma_{n}(t)\bigr)=g(t)f\bigl(t,u(t)\bigr), $$
and therefore \(u''(t)=f(t,u(t))\) a.e. in \(J=\bigcup_{n \in \scriptsize{ \mathbb {N}}}J_{n}\).
Now we assume that \(u \in{\mathbb {T}}u\), and we prove that it also implies that \(u''(t)=g(t) f(t,u(t))\) a.e. in \(I \setminus J\), thus showing that \(u=Tu\).
Since \(u \in{\mathbb {T}}u\), for each \(k \in \mathbb {N}\), we can use Proposition 2.3 with \(\varepsilon=\rho=1/k\) to guarantee that we can find functions \(u_{k,i} \in B_{1/k}(u)\cap\overline{B}_{R} \) and coefficients \(\lambda_{k,i} \in[0,1]\) (\(i=1,2,\dots,m(k)\)) such that \(\sum\lambda_{k,i}=1\) and
$$\Biggl\Vert u\sum_{i=1}^{m(k)} \lambda_{k,i}Tu_{k,i}\Biggr\Vert _{{\mathcal {C}}^{1}}< \frac{1}{k}. $$
Let us denote \(v_{k}= {\sum_{i=1}^{m(k)}}\lambda _{k,i}Tu_{k,i}\). Notice that \(v_{k}' \to u'\) uniformly in I and
$$\u_{k,i}u\_{{\mathcal {C}}^{1}}\le\frac{1}{k} $$
for all \(k \in \mathbb {N}\) and all \(i \in\{1,2,\dots,m(k)\}\).
For a.a. \(t \in I\setminus J\), we have that \(g(t) f(t,\cdot)\) is continuous at \(u(t)\), so for any \(\varepsilon>0\), there is some \(k_{0}=k_{0}(t) \in \mathbb {N}\) such that, for all \(k \in \mathbb {N}\), \(k \ge k_{0}\), we have
$$\biglg(t) f\bigl(t,u_{k,i}(t)\bigr)g(t) f\bigl(t,u(t)\bigr)\bigr < \varepsilon \quad\mbox{for all }i \in\bigl\{ 1,2,\dots,m(k)\bigr\} , $$
and therefore
$$\biglv_{k}''(t)g(t) f\bigl(t,u(t)\bigr)\bigr\le \sum_{i=1}^{m(k)}\lambda_{k,i}\biglg(t) f \bigl(t,u_{k,i}(t)\bigr)g(t) f\bigl(t,u(t)\bigr)\bigr < \varepsilon. $$
Hence, \(v_{k}''(t) \to g(t) f(t,u(t))\) for a.a. \(t \in I \setminus J\), and then Corollary 3.7 guarantees now that \(u''(t)=g(t) f(t,u(t))\) for a.a. \(t \in I \setminus J\).
Then, we have proven that \(\{u\} \cap \mathbb {T}u \subset\{Tu\}\) for all \(u \in\overline{B}_{R}\). By application of Theorem 2.7 we obtain that T has at least one fixed point in \(\overline{B}_{R}\), which corresponds to a solution of the BVP (3.1) in \(\overline{B}_{R}\). □
Remark 3.9
Note that if \(g(t)f(t,0)=0\) for almost all \(t \in[0,1]\), then 0 is a solution of the BVP (3.1). Therefore, when \(g(t)f(t,0)\neq 0\) in a set of positive measure, then Theorem 3.8 provides the existence of a nontrivial solution. In this case, since the kernel k is nonnegative and if, moreover, \(g(t)f(t,u)\geq0\) almost everywhere, then we obtain the existence of a nonnegative solution with a nontrivial norm.
Remark 3.10
The improvement with respect to Theorem 4.4 of [1] relies not only on the fact that we can deal with a more general set of BCs but also on the fact that we do not require global
\(L^{1}\) estimates on f, allowing a more general class of nonlinearities. On the other hand, notice that our result can be extended to other type of BCs whenever condition (H_{3}) makes sense for the corresponding Green’s function.
Finally, we illustrate our results by an example.
Example 3.11
For \(n \in \mathbb {N}\), we denote by \(\phi(n)\) the function such that \(\phi (1)=2\) and, for \(n \ge2\), \(\phi(n)\) counts the number of divisors of n. Thus defined, \(\phi(n) \ge2\) for all \(n \in \mathbb {N}\), ϕ is not bounded, and, since there are infinitely many prime numbers, \({\liminf_{n \to\infty}} \phi(n)=2\). Now we define the function
$$ (t,u) \in(0,1] \times \mathbb {R}\longmapsto\tilde{f}(t,u)= \phi^{\lambda }\bigl(n(t,u)\bigr), \quad\lambda\in(0,1), $$
(3.18)
where
$$n(t,u):= \left \{ \textstyle\begin{array}{@{}l@{\quad}l} 1 & \mbox{if } u \in(\infty,t), \\ n & \mbox{if } \frac{t}{n} \le u < \frac{t}{n+1} \mbox{ and }t \le u < 0, \\ n & \mbox{if } (n1) \sqrt{t} \le u < n \sqrt{t} \mbox{ and } u \ge0. \end{array}\displaystyle \right . $$
We are concerned with the ODE
$$ u''(t)=\frac{\phi^{\lambda}(n(t,u))}{\sqrt{t}} \quad\mbox{for a.e. }t \in I=[0,1], $$
(3.19)
coupled with separated BCs.
We claim that this problem has at least one solution. In order to show this, note that we can rewrite the ODE (3.19) in the form \(u''(t)+g(t)f(t,u(t))=0\), where \(g(t)=\frac{1}{\sqrt{t}}\) and \(f=\tilde{f}\) with f̃ as in (3.18). We now show that the functions g and f satisfy conditions (H_{1})(H_{5}).
First, it is clear that \(g \in L^{1}(I)\), and so (H_{1}) holds. On the other hand, since for all \(n \in \mathbb {N}\) it is \(\phi(n) \le\max\{2,n\}\), we obtain that we have \(u \in[n,n] \Rightarrow f(t,u) \le\max\{2,n\}^{\lambda}\) for each \(n \in \mathbb {N}\). Then, if we take \(R \in \mathbb {N}\), \(R \ge2\), large enough such that \(M_{1}+M_{2} \le R^{1\lambda}\) (with \(M_{1}\), \(M_{2}\) as in (3.4)), then we can guarantee that (H_{2}) and (H_{3}) hold.
To check (H_{4}), note that for every continuous function u, we can write the composition \(t \in I \longmapsto f(t,u(t))\) as
$$ t \longmapsto f\bigl(t,u(t)\bigr)=\sum _{n=1}^{\infty} \phi^{\lambda}(n) \bigl( \chi_{I_{n}}(t)+\chi_{J_{n}}(t) \bigr)+\phi(1) \chi_{K}(t), $$
(3.20)
where χ denotes the characteristic function, and \(I_{n}\), \(J_{n}\), K are the following measurable sets:
$$\left \{ \textstyle\begin{array}{@{}l@{\quad}l} I_{n}=u^{1}([(n1) \sqrt{t},n\sqrt{t}) \cap[0,+\infty)),& n \in \mathbb {N},\\ J_{n}=u^{1} ( [\frac{1}{n}t,\frac{1}{n+1} t ) \cap [t,0) ),& n \in \mathbb {N}, \\ K=u^{1}((\infty,t)). \end{array}\displaystyle \right . $$
Then (3.20) is a measurable function, and therefore condition (H_{4}) is satisfied.
Finally, we check condition (H_{5}). For a.a. \(t \in I\), the function \(f(t,\cdot)\) has a countable number of discontinuities of the form \(\gamma_{k}(t)=k\sqrt{t}\), and \(\hat{\gamma}_{k}(t)=\frac{1}{k+1}t\), \(k \in N \subset \mathbb {N}\), but all these discontinuity curves are inviable for the differential equation. Indeed, notice that, for \(k \in N\) and \(t \in I\), we have \(\gamma_{k}''(t)=\frac{k}{4t^{3/2}}>0\), \(\hat {\gamma}_{k}''(t)=0\) and
$$g(t)f(t,y) \le\frac{2^{\lambda}}{\sqrt{t}} \le2^{\lambda}\le 1 \quad\mbox{for all }y\in \mathbb {R}, $$
taking into account that \(\phi(n) \ge2\) for all \(n \in \mathbb {N}\). Then, condition (H_{5}) holds (it suffices to take, for example, the same function \(\psi\equiv\frac{1}{2}\) for all discontinuity curves).
We can conclude that the differential equation (3.19), coupled with separated BCs, has at least one solution in \(\overline{B}_{R}\) provided that \(M_{1}+M_{2} \le R^{1\lambda}\). Note that the solution is nontrivial since the zero function does not satisfy the ODE.
In the special case of \(\alpha=\beta=\gamma=\delta=1\) and \(\lambda =1/3\), we obtain (rounded to the third decimal place) \(M_{1}+M_{2}=2{,}336\) and \(R=4\).