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Nontrivial solutions of inverse discrete problems with signchanging nonlinearities
Advances in Difference Equations volume 2019, Article number: 450 (2019)
Abstract
This paper is concerned with the existence of solutions of an inverse discrete problem with signchanging nonlinearity. This kind of problems includes, as a particular case, nth order difference equations coupled with suitable conditions on the boundary of the interval of definition. It would be valid for the case in which the related Green’s function is positive on a subset of its rectangle of definition.
The existence results follow from spectral theory, as an application of the Krein–Rutman theorem and by means of degree theory.
Introduction
During the last years, many authors discussed the existence of solutions for boundary value problems by using various topological methods. However, there are just a few results with signchanging nonlinearities. Usually, in order to obtain positive solutions of semipositone problems for ordinary differential or difference equations, by using fixed point methods, the nonlinearity terms need to be bounded from below and ultimately positive. For example, in [2] the authors studied the following problem:
where \(g: [ a,b ] \times \mathbb{R}^{+}\rightarrow \mathbb{R}\) is continuous, bounded from below (i.e., \(g ( t,z ) +M>0\) for some \(M>0 \)) and \(\lim_{z\rightarrow \infty }\frac{f ( t,z ) }{z}=\infty \) uniformly for \(t\in [ \alpha ,\beta ] \subset ( a,b ) \). Later, Bai and Xu [4] discussed its discrete analog and it also required boundedness from below on the nonlinearity term, as well as a superlinear condition at ∞. We also refer the reader to [9, 11, 12], where the authors used topological methods to deduce, under a similar hypothesis, the existence results to a discrete fractional semipositone boundary value problem.
By using Guo–Krasnosel’skii’s fixed point theorem, Bai, Henderson, and Zeng [3] obtained the existence of positive solutions of the discrete Neumann boundary value problem
where the nonlinearity term \(f: [ 1,T ] _{\mathbb{Z}} \times \mathbb{R}^{+}\rightarrow \mathbb{R}\) is a continuous signchanging function, for which there exist a function \(h: [ 1,T ] _{\mathbb{Z}}\rightarrow \mathbb{R}^{+}\), with \(h\not \equiv 0\) on \([ 1,T ] _{\mathbb{Z}}\), and a constant \(L>0\) such that \(f ( t,z ) +Lz+h ( t ) \geq 0\), \(( t,z ) \in [ 1,T ] _{\mathbb{Z}}\times \mathbb{R}^{+}\).
In [17], using Krein–Rutman theorem, Zhang studied the fourth order singular boundary value problem
under some suitable conditions concerning the first characteristic value corresponding to the relevant linear operator. Here, h is allowed to be singular at both \(t=0\) and \(t=1\). In particular, \(f:\mathbb{R} \rightarrow \mathbb{R}\) may be a signchanging and unbounded function from below, and the existence of a control from below function related to f is not assumed. The existence results of nontrivial solutions and positivenegative solutions are given by the topological degree theory and the fixed point index theory, respectively.
A similar idea can be found in a very recent paper [18], where under suitable conditions concerning the first eigenvalue corresponding to the relevant linear problem, the authors established the existence of nontrivial solutions for boundary value problems of the following fourth order difference equation with a signchanging nonlinearity:
where \(T\geq 5\) is an integer and \(f: [ 2,T ] _{\mathbb{Z}} \times \mathbb{R}\rightarrow \mathbb{R}\) is a continuous function. The results are based on the topological degree theory and generalize some previous results obtained for this problem. It is important to point out that our results generalize some ones given in that reference, in fact, as we will see in Example 9, Theorems 3.2 and 3.1 in [18] are particular cases (respectively) of Theorems 5 and 6 in this work.
Our aim is to extend these results as we study the following nth order boundary value problem:
Here, \(f:I \times \mathbb{R} \to \mathbb{R}\) is a continuous function, \(ba\geq 2\), and
are linear operators, for which the following conditions for the related Green’s function \(G(k,s)\) are fulfilled:
 \((G1)\) :

\(G(k,s)\ge 0\) for all \(k\in I\) and \(s\in I\).
 \((G2)\) :

\(G(k,s)\) is symmetric for all \(k\in I\) and \(s\in I\), i.e., \(G(k,s)=G(s,k) \).
 \((G3)\) :

There is \(k_{0} \in I\) such that \(G(k_{0},s) >0\) for all \(s \in I\).
Remark 1
Notice that condition \((G2)\) implies that all the eigenvalues of the matrix \((G(k,s))_{k, s \in I}\) are real and that its associated Jordan matrix is diagonal. As a consequence, all its eigenvalues are zero if and only if G is identically zero which, from \((G3)\), is not true. So we deduce that the spectral radius is strictly positive.
There are several papers in the literature, where the related Green’s function of the studied difference equation verifies previous properties for a suitable value of parameters, see, for instance, [1, 6, 7] (and in [5] for order four), where the nth order Problem \((P)\) with periodic conditions, i.e.,
has been studied.
The Neumann conditions were considered in [8].
So, with this idea in mind, we study a more general problem that includes the difference equations as particular cases. That is, we consider the general inverse discrete problem
where G satisfies conditions \((G1)\)–\((G3)\), but it is not necessarily the Green’s function of a related linear operator.
To deduce the existence results of equation (2), we will work with spectral and degree theory. To this end, define the space E as the collection of all maps from I to \(\mathbb{R}\) equipped with the norm \(\Vert u \Vert =\max_{k\in I} \vert u(k) \vert \). Clearly, E is a Banach space.
Now, we recall some definitions that will be useful in the sequel. Let X be a Banach space. We say that \(K\subset X\) is a cone if K is a closed convex set such that \(\lambda K\subset K\) for all \(\lambda \geq 0\) and \(K\cap ( K ) = \{ 0 \} \). If \(\overline{KK}=X\), i.e., the set \(\{ uv\mid u,v\in K \} \) is dense in X, then K is called a total cone. If \(KK=X\), K is called a reproducing cone. If a cone has nonempty interior \(K^{0}\), then it is called a solid cone. Any solid cone has the property that \(KK=X\); in particular, it is total.
Let us recall Krein–Rutman theorem.
Theorem 2
(Krein–Rutman, [10, Theorem 19.2])
Let X be a Banach space, \(K\subset X\) be a total cone, and \(T:X\rightarrow X\) be a compact linear operator that is positive (i.e., \(T ( K ) \subset K \)) with positive spectral radius \(r ( T )\). Then \(r ( T ) \) is an eigenvalue with an eigenvector \(u\in K\backslash \{ 0 \} \), i.e., \(Tu=r(T)u\).
Now we introduce two useful results using the properties of the topological degree, which can be found in [13].
Lemma 3
Let X be a Banach space and Ω be a bounded open set in X. Suppose that \(A:\varOmega \rightarrow X\) is a continuous compact operator. If there exists \(u_{0}\in X\backslash \{ 0 \} \) such that
then the topological degree \(\deg (IA,\varOmega ,0)=0\).
Lemma 4
Let X be a Banach space and Ω be a bounded open set in X with \(0\in \varOmega \). Suppose that \(A:\varOmega \rightarrow X\) is a continuous compact operator. If
then the topological degree \(\deg (IA,\varOmega ,0)=1\).
The paper is scheduled as follows: In the next section we deduce the main properties of the kernel G. Section 3 is devoted to nonlinear problem (2). In it, under suitable assumptions on the nonlinearity f, which is allowed to change its sign, related to the first eigenvalue of the linear problem, the existence of a nontrivial solution is proved. The paper ends with some examples where the applicability of the results is pointed out.
Kernel properties
In this section we present some additional assumptions on the kernel G, and we deduce some suitable properties.
Define the cone K on E as follows:
To the end of the paper, let \(B_{\rho }= \{ u\in E: \Vert u \Vert <\rho \} \) for \(\rho >0\), \(\partial B_{\rho }= \{ u\in E: \Vert u \Vert =\rho \} \), and \(\overline{B_{\rho }}= \{ u\in E: \Vert u \Vert \leq \rho \} \).
Define the operator \(T:E\rightarrow E\) as follows:
By definition, we have that the solutions of problem (2) coincide with the fixed points of operator T.
Now, define the operator
From Remark 1, we have that \(r(L)\), the spectral radius of operator L, is such that \(r(L)>0\). Thus, since the compactness and the continuity properties are equivalent on finite dimensional spaces, we can apply Krein–Rutman theorem to operator L. So, it follows that
where \(v(k)\geq 0\) on I, \(v\not \equiv 0\) on I, is the corresponding eigenfunction.
Notice that if \(v(k_{0})=0\), from condition \((G3)\) we arrive at the following contradiction:
Moreover, since G is symmetric, we have
In the sequel, we introduce the following functions:
and
It is clear, from \((G1)\), that \(l_{1}\ge 0\) on I and, by condition \((G3)\), that \(l_{1}(k_{0})> 0\). From \((G1)\) and \((G3)\) we have that \(\varphi (s)>0\) for all \(s \in I\). Moreover, the following inequalities hold:
Nonlinear problem
This section is devoted to proving the existence of a nontrivial solution of Problem (2) which, as we have noted in previous sections, is equivalent to finding a fixed point of operator T defined on (3). The proofs follow similar arguments to the ones developed by Zhang, O’Regan, and Fu in [18] for equation (1).
So, we present some assumptions about the nonlinearity f:
 \((H0)\) :

\(f: I \times \mathbb{R}\to \mathbb{R}\) is a continuous function on \(I \times \mathbb{R}\).
 \((H1)\) :

\(\lim_{u\rightarrow 0^{+}}\inf \frac{f(k,u)}{u}> \frac{1}{r(L)}\), \(\lim_{u\rightarrow 0^{}}\sup \frac{f(k,u)}{u}<\frac{1}{r(L)}\) for all \(k\in I\).
 \((H2)\) :

\(\lim_{ \vert u \vert \rightarrow \infty } \sup \frac{ \vert f(k,u) \vert }{ \vert u \vert }<\frac{1}{r(L)}\) for all \(k\in I\).
Our first main result is as follows.
Theorem 5
Suppose that \((H0)\)–\((H2)\) hold. Then Problem (2) has at least one nontrivial solution.
Proof
From \((H1)\) there exist \(\varepsilon _{0}\in ( 0, \frac{1}{r(L)} ) \) and \(r>0\) such that
and
Using the above two inequalities, one can obtain that
and
Now, taking into account that
we introduce the following cone:
We claim that \(L ( K ) \subset K_{1}\). Indeed, using the definition of L, for all \(u\in K\), \(L u \ge 0\) on I. So, for any \(\sigma \in I\), we have that
Thus,
Notice that, since \(v=\frac{Lv}{r(L)}\) it follows that \(v\in K_{1}\).
Now, we claim that \(uTu\neq \mu v\) for all \(u\in \partial B_{r}\) and \(\mu \geq 0\). Assume, on the contrary, that there exist \(u_{1}\in \partial B_{r}\) and \(\mu _{1}\geq 0\) such that
Using (10) and (12), we have that
Multiplying both sides of the above inequality by \(v(k)\), summing from a to b, and using (5) and (6), we obtain that
whence \(\sum_{k=a}^{b}u_{1}(k)v(k)\leq 0\).
On the other hand, from (12) we have, for any \(k \in I\),
Now, using the latter together with \(L ( K ) \subset K_{1}\), \(v\in K_{1}\), and (11), gives us that \(u_{1} ( \frac{1}{r(L)}\varepsilon _{0} ) Lu_{1}\in K_{1}\).
Hence, from (5) and (6), we deduce
The above inequality implies that
which, together with \(( \frac{1}{r(L)}\varepsilon _{0} ) ^{1}>r(L)\), implies that \(u_{1}=0\), which contradicts the fact that \(u_{1}\in \partial B_{r}\).
Thus, (12) is false and, from Lemma 3, it follows that
From \((H2)\) there exist \(\varepsilon _{1}\in ( 0,\frac{1}{r(L)} ) \) and \(c_{1}>0\) such that
Let
Now, we will show that X is bounded in E.
To this end, as a direct consequence of [16, (57d)], we have that for any \(0<\alpha <\frac{1}{r(L)}\), its corresponding Neumann series converges in the operator norm:
In addition, since it is a sum of positive operators, we have that \(( I\alpha L ) ^{1}\) is a positive operator on E.
On the other hand, using (14), we have that the following inequalities are fulfilled for all \(u\in X\):
Clearly, the previous inequality is of the form \(\vert u \vert \leq \alpha L \vert u \vert +v\), with \(0<\alpha =\frac{1}{r(L)} \varepsilon _{1}<\frac{1}{r(L)}\) and \(v=\sum_{s=a}^{b}G(k,s)c _{1} \).
As a consequence, from the positiveness of \(( I\alpha L ) ^{1}\) on E, we deduce that \(\vert u \vert \leq ( I \alpha L ) ^{1}v\), u is bounded, i.e., X is bounded.
Now, one can choose \(R>\max \{ \sup_{u\in X} \Vert u \Vert ,r \} \) for r defined by (10). Then, \(\lambda Tu\neq u\) for all \(u\in \partial B_{R}\) and \(\lambda \in [ 0,1 ]\). From Lemma 4 it follows that \(\deg ( IT,B_{R},0 ) =1\).
Using the latter one, together with \(\deg ( IT,B_{r},0 ) =0\), gives us
Therefore, operator T has at least one fixed point in \(B_{R}\backslash \overline{B_{r}}\), which is a nontrivial solution of (2). □
Before formulating our second main result, let us introduce the following conditions about the nonlinearity:
 \((H3)\) :

There exist two constants \(a_{1}>0\), \(b_{1}>0\) and a function \(\varPhi \in C(\mathbb{R},\mathbb{R}^{+})\) such that
$$ f(k,u)\geq a_{1}b_{1}\varPhi (u)\quad \text{for all }u\in \mathbb{R}\text{ and }k\in I. $$  \((H4)\) :

\(\lim_{ \vert u \vert \rightarrow +\infty }\frac{ \varPhi (u)}{ \vert u \vert }=0\).
 \((H5)\) :

\(\lim_{ \vert u \vert \rightarrow +\infty } \inf \frac{f(k,u)}{ \vert u \vert }>\frac{1}{r(L)}\) for all \(k\in I\).
 \((H6)\) :

\(\lim_{ \vert u \vert \rightarrow 0}\sup \frac{ \vert f(k,u) \vert }{ \vert u \vert }<\frac{1}{r(L)}\) for all \(k\in I\).
The result is the following.
Theorem 6
Suppose that \((H0)\) and \((H3)\)–\((H6)\) hold. Then Problem (2) has at least one nontrivial solution.
Proof
From \((H5)\) there exist \(\varepsilon _{2}>0\) and \(N_{1}>0\) such that
For any given \(\varepsilon _{2} >b_{1}\varepsilon _{3}\), using \((H4)\), there exists \(N_{2}>N_{1}\) such that
From \((H3)\), since \(a_{1}>0\), \(b_{1}>0\) and Φ is a nonnegative function, we have
for all u with \(\vert u \vert >N_{2}\).
Let us denote \(c_{2}= ( \frac{1}{r(L)}+\varepsilon _{2}b_{1} \varepsilon _{3} ) N_{2}+\max_{k\in I, \vert u \vert \leq N_{2}} \vert f(k,u) \vert \), \(c_{3}=c_{2}+a_{1}\), and \(\varPhi ^{\ast }=\max_{ \vert u \vert \leq N_{2}}\varPhi (u)\). Then
Note that \(\varepsilon _{3}\) can be chosen arbitrarily small, and let us set
and
Now, we will prove that
Notice that, if the equality holds for \(\mu =0\), we have proved the existence of the fixed point of T.
Define the cone \(K_{2}\) as follows:
Using (9), we have that \(L ( K ) \subset K_{2}\). Moreover, \(v= \frac{1}{r(L)}Lv\in K_{2}\).
Let \(\widetilde{u}(k)=\sum_{s=a}^{b}G(k,s)(a_{1}+b_{1}\varPhi ( u_{2} ) +c_{2})\). Then, for all \(k \in I\), the following inequalities hold:
Therefore,
Note that from the definition of R, since \(R>R_{1}\), we have \(\Vert \widetilde{u} \Vert < R\).
Assume, on the contrary, that there exist \(u_{2}\in \partial B_{R}\) and \(\mu _{2}>0\) such that
From
using \((H3)\), \(L(K)\subset K_{2}\), and \(v\in K_{2}\), we deduce that \(u_{2}+\widetilde{u}\in K_{2}\).
As a result, we obtain that, for all \(k \in I\),
Now, we claim that
which is equivalent to
Since \(u_{2}+\widetilde{u}\in K_{2}\), we have that \(u_{2}(k)+ \widetilde{u}(k)\geq l_{1} ( k ) \Vert u_{2}+ \widetilde{u} \Vert \geq l_{1} ( k ) ( \Vert u_{2} \Vert  \Vert \widetilde{u} \Vert )\). Then
From (17) and from the definition of R, since \(R>R_{2}\), we deduce
Finally, from (16) we obtain
which proves our claim.
Now, using (20) it follows that
Then (18) gives us
Define \(\mu ^{\ast }:=\sup \{ \mu >0\mid u_{2}+ \widetilde{u}\geq \mu v \} \). Note that \(\mu _{2}\in \{ \mu >0\mid u_{2}+\widetilde{u}\geq \mu v \} \). Thus \(\mu ^{\ast }\geq \mu _{2}\) and \(u_{2}+\widetilde{u}\geq \mu ^{\ast }v\). Using (5) we deduce that
whence
which contradicts the definition of \(\mu ^{\ast }\).
Then (15) holds and Lemma 3 gives us that \(\deg ( IT,B_{R},0 ) =0\).
On the other hand, from \((H6)\), there exist \(\varepsilon _{4}\in ( 0,\frac{1}{r(L)} ) \) and \(r\in ( 0,R ) \) such that
We will prove that
Assume on the contrary that there exist \(u_{3}\in B_{r}\) and \(\mu _{3}\geq 1\) such that \(Tu_{3}=\mu _{3}u_{3}\). Then
and \(( I ( \frac{1}{r(L)}\varepsilon _{4} ) L ) \vert u_{3} \vert \leq 0\).
So, arguing as in the proof of Theorem 5, with \(\alpha = \frac{1}{r(L)}\varepsilon _{4}\), we deduce that \(u_{3}(k)\equiv 0\), in contradiction with \(u_{3}\in \partial B_{r}\). Hence (22) holds and from Lemma 4 we deduce
Thus,
Therefore, operator T has at least one fixed point in \(B_{R}\backslash \overline{B_{r}}\), which is a nontrivial solution of (2). □
Examples
In this section we introduce a few examples where the applicability of the existence results proved in the previous section is pointed out.
Example 7
Consider the second order problem with Dirichlet conditions
We have that this problem is equivalent to the operator equation
with
Clearly, \(G ( 0,s ) =0=G ( N,s ) \) for all \(s\in J_{1}\) and \(G ( k,s ) =G ( s,k ) >0\) for all \(( k,s ) \in J_{1}\times J_{1}\).
If we restrict the equation to the interval \(J_{1}\) and set the operator
we have that the Green’s function \(G ( k,s ) \) satisfies conditions \((G1)\)–\((G3)\) on \(J_{1}\times J_{1}\).
Moreover, the smallest eigenvalue of the eigenproblem
is \(\lambda _{1}=4\sin ^{2}\frac{\pi }{2N}\) and \(r_{1} ( L ) =\frac{1}{\lambda _{1}}=\frac{1}{4\sin ^{2}\frac{\pi }{2N}}\).
Similar results for the same second order problem with periodic conditions \(u ( 0 ) =u ( N ) \), \(u ( 1 ) =u ( N+1 ) \) or with Neumann conditions \(u ( 0 ) =u ( N ) \), \(\Delta u ( 0 ) =\Delta u ( N ) \) might be obtained.
Example 8
Now, consider the fourth order Lidstone problem
It is shown in [14] that the corresponding Green’s function \(G ( k,s ) \) is strictly positive and symmetric on \(J_{2}\times J_{2}\) and it also satisfies conditions \((G1)\)–\((G3)\).
One can compute that in this case
whence
Example 9
Consider now the fourth order Lidstone problem
One can verify (see [15, 18]) that the corresponding Green’s function \(G (k,s ) \) is strictly positive and symmetric on \(J_{3}\times J_{3}\) and it also satisfies conditions \((G1)\)–\((G3)\). Moreover,
As a direct consequence, we have that the enunciations of Theorems 5 and 6 for this problem coincide, respectively, with Theorems 3.2 and 3.1 in [18].
Example 10
Consider the inverse difference equation
which corresponds to the kernel
It is immediate to verify that the eigenvalues of the matrix function \(G(k,s) \) are \(N1\) and 0. So \(r_{4}(L)=N1\).
Example 11
Let \(N=101\) and consider the inverse difference equation
which corresponds to the kernel
By numerical approach, one may verify that the nonzero eigenvalues of G are given by \(1.45456\times 10^{6}\), \(107{,}101\), and 5940.08. In consequence,
For any of the previous examples, we can take
with \(p>\frac{1}{r_{i} ( L ) }> \vert ptq \vert \) for \(i=1,2,3,4\) and \(t \in I\).
Since
it is clear that conditions \((H0)\) and \((H3)\)–\((H6)\) hold.
We notice that a necessary condition on p and q is
From Theorem 6 it follows that all of the considered examples have at least one nontrivial solution.
Moreover, choose
with
Due to the fact that
we deduce that conditions \((H0)\)–\((H2)\) hold and, from Theorem 5, we have that all of the considered examples have at least one nontrivial solution.
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First author is partially supported by Xunta de Galicia (Spain), project EM2014/032 and AIE, Spain and FEDER, grant MTM201675140P. The second author is supported by the Bulgarian National Science Foundation under Project DN 12/4 “Advanced analytical and numerical methods for nonlinear differential equations with applications in finance and environmental pollution”.
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Cabada, A., Dimitrov, N.D. Nontrivial solutions of inverse discrete problems with signchanging nonlinearities. Adv Differ Equ 2019, 450 (2019). https://doi.org/10.1186/s136620192383y
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MSC
 39A10
 39A06
 39A70
Keywords
 Inverse discrete problem
 Difference equation
 Green’s function
 Spectral radius