Abstract
Consider positive solutions and multiple positive solutions for a discrete nonlinear third-order boundary value problem
which has the sign-changing Green’s function. Here \(T>8\) is a positive integer, \([1,T-1]_{\mathbb{Z}}=\{1,2,\dots ,T-2\}\), \(\alpha \in [0, \frac{1}{T-1})\), \(a:[0,T-2]_{\mathbb{Z}}\to (0,+\infty )\), \(f:[1,T-2]_{ \mathbb{Z}}\times [0,\infty )\to [0,\infty )\) is continuous.
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1 Introduction
Multi-point boundary value problems for differential equations have a wide application in computational physics, economics, and modern biological fields [1]. In 1992, Gupta [2] studied solvability of differential equation three-point boundary value problem. Soon afterwards, there arose many results on multi-point nonlinear boundary value problems [3,4,5,6]. In 1999, Ma [7] studied the existence of positive solution for a second-order differential equation three-point boundary value problem. Thereafter, many results for the existence of positive solutions on multi-point boundary value problems have been studied [7,8,9,10,11,12,13,14,15,16,17,18,19]. With the development of the computing science and the computer simulation, multi-point boundary value problems should be discretized, so we need to study corresponding difference equation [20,21,22,23,24,25,26]. In 1998, by using Krasnosel’skii’s fixed point theorem, Agarwal and Henderson [24] studied the discrete problem
They obtained the existence of positive solutions in two cases for \(\lambda =1\) and \(\lambda \neq 1\). Later, there were many interesting results on the positive solutions to the discrete boundary value problems, see, for instance, [23,24,25,26] and the references therein. It is noted that Green’s functions are positive in most of these results. However, when the Green’s function is sign-changing, could we also obtain the existence of positive solutions to these kinds of problems?
In 2015, by using the Guo–Krasnosel’skii fixed point theorem, Wang and Gao [25] studied the existence of positive solutions to the discrete third-order three-point boundary value problem
where \(\eta \in [1,[\frac{3T^{2}-3T-2}{6T+3}]]_{\mathbb{Z}}\) and the Green’s function is sign-changing. In 2016, Geng and Gao [26], by using the Guo–Krasnosel’skii fixed point theorem, studied the discrete third-order three-point boundary value problem
when f satisfies some superlinear and sublinear condition on 0 and ∞. For the continuous case, which has the sign-changing Green’s function, one can see [27,28,29] and the references therein. Inspired by the above works, in this paper, we consider the existence and multiple positive solutions to the following discrete nonlinear third-order three-point BVP:
where \(T>8\) is a positive integer, \(\alpha \in [0,\frac{1}{T-1})\), \(a:[1,T-2]_{\mathbb{Z}}\to (0,+\infty )\), \(f:[1,T-2]_{\mathbb{Z}} \times [0,\infty )\to [0,\infty )\) is continuous, η satisfies
- \((H_{0})\) :
-
\(\eta \in [\lfloor \frac{T-2}{2}\rfloor +1,T-2]_{ \mathbb{Z}} \).
Under assumption \((H_{0})\), the Green’s function of (1.1) changes its sign. The proof of our main results is based upon the following well-known Guo–Krasnoselskii fixed point theorems [30].
Theorem 1.1
Let E be a Banach space and \(K\subset E \) be a cone. Assume that \(\varOmega _{1}\), \(\varOmega _{2}\) are open bounded subsets of E with \(0\in \varOmega _{1}\), \(\overline{\varOmega }_{1}\subset \varOmega _{2}\). If
is a completely continuous operator such that
-
(i)
\(\|Au\|\leq \|u\|\), \(u\in K\cap \partial \overline{\varOmega }_{1}\), and \(\|Au\|\geq \|u\| \), \(u\in K\cap \partial \overline{\varOmega }_{2}\),
or
-
(ii)
\(\|Au\|\geq \|u\|\), \(u\in K\cap \partial \overline{\varOmega }_{1}\), and \(\|Au\|\leq \|u\| \), \(u\in K\cap \partial \overline{\varOmega }_{2}\),
then A has a fixed point in \(K\cap (\overline{\varOmega }_{2}\setminus \varOmega _{1})\).
Theorem 1.2
Let E be a Banach space and K be a cone in E. For some \(p>0\), define \(K_{p}=\{x\in K | \|x\|\leq p\}\). Assume that \(A:K_{p}\to K\) is a compact operator; if \(x\in \partial K_{p}=\{x \in k | \|x\|=p\}\), \(Ax\neq x\), we have
-
(i)
For \(\forall x\in \partial K_{p}\), if \(\|Ax\|\geq \|x\|\), then \(i(A,K_{p},K)=0\),
-
(ii)
For \(\forall x\in \partial K_{p}\), if \(\|Ax\|\leq \|x\|\), then \(i(A,K_{p},K)=1\).
2 Preliminaries
Lemma 2.1
Suppose that \((H_{0})\) holds. Then the linear problem
has a unique solution
where
and
Proof
Summing both sides of equation (2.1) from \(s=1\) to \(s=t-1\), we get
and
Then summing both sides from \(\tau =1\) to \(\tau =t\), we get
From boundary conditions \(\Delta u(0)=u(T)=0\), \(\Delta ^{2} u(\eta )- \alpha \Delta u(T-1)=0\), we have
Furthermore, we get
Then we have
□
Lemma 2.2
Suppose that (\(H_{0}\)) holds. Then the Green’s function \(G(t,s)\) changes its sign on \([0,T]_{\mathbb{Z}}\times [0,T]_{ \mathbb{Z}}\). More precisely,
-
(i)
If \(s\in [1,\eta ]_{\mathbb{Z}}\), then \(\Delta _{t} G(t,s)\leq 0\), \(G(t,s)\geq 0\) for \(\forall t\in [0,T]_{\mathbb{Z}}\). Furthermore,
$$\begin{aligned}& \min_{t\in [0,T]_{\mathbb{Z}}}G(t,s)=G(T,s)=0, \\& \max_{t\in [0,T]_{\mathbb{Z}}}G(t,s)=G(0,s)\leq \frac{T(T-1)(1+\alpha \eta )}{1-\alpha (T-1)}. \end{aligned}$$ -
(ii)
If \(s\in [\eta +1,T-2]_{\mathbb{Z}}\), then \(\Delta _{t}G(t,s) \geq 0\), \(G(t,s)\leq 0\) for \(\forall t\in [0,T]_{\mathbb{Z}}\). Furthermore,
$$\begin{aligned}& \max_{t\in [0,T]_{\mathbb{Z}}}G(t,s)=G(T,s)=0, \\& \min_{t\in [0,T]_{\mathbb{Z}}}G(t,s)=G(0,s)\geq \frac{-T(T-1)(1+ \alpha \eta )}{1-\alpha (T-1)}. \end{aligned}$$
Proof
(i) If \(s\in [1,\eta ]_{\mathbb{Z}}\), we have
If \(s\leq t-2\), since \(\alpha (T-1)-1<0\), we get
If \(t-2\leq s\), since \(\alpha <\frac{1}{T-1}\), so \(2-2\alpha (T-1)>0\), \(\alpha (T-s-1)-1<0\), we have
For \(\forall t\in [0,T-1]_{\mathbb{Z}}\), \(\Delta _{t}G(t,s)\leq 0\). If \(s\in [1,\eta ]_{\mathbb{Z}}\), \(G(t,s)\geq 0\). So
(ii) If \(s\in [\eta +1,T]_{\mathbb{Z}}\), we have
If \(t-2\leq s\), since \(\alpha <\frac{1}{T-1}\), \(2-2\alpha (T-1)>0\), we have
If \(s\leq t-2\), then \(t-s>0\), so
If \(s\in [\eta +1,T]_{\mathbb{Z}}\), for \(\forall t\in [0,T]_{ \mathbb{Z}}\), we have
In conclusion, if \(s\in [1,\eta ]_{\mathbb{Z}}\), \(G(t,s)\) is decreasing, then \(G(t,s)\geq 0\), since \(\min_{t\in [0,T]_{\mathbb{Z}}}G(t,s)=0\); if \(s\in [\eta +1,T]_{ \mathbb{Z}}\), \(G(t,s)\) is increasing, then \(G(t,s)\leq 0\) since \(\max_{t\in [0,T]_{\mathbb{Z}}}G(t,s)=0\). So, the Green’s function \(G(t,s)\) is a sign-changing function on \([0,T]_{\mathbb{Z}} \times [0,T]_{\mathbb{Z}}\). □
Remark
Now, we give a brief explanation for the reason why we choose
Consider the problem
From Lemma 2.1, we get
where
Clearly, \(u(t)\geq 0\) is equivalent to \(\phi (t)\geq 0\), and
Let \(\Delta \phi (t)=0\). Then \(t=0\) or \(t=1+\frac{2\eta -\alpha (T-1)(T-2)}{1- \alpha (T-1)}\). Therefore,
Now, we claim that if (2.3) holds, then \(\phi (t)\) is a positive solution of (2.4). In fact, if (2.3) holds, then \(\frac{2\eta +1- \alpha (T-1)^{2}}{1-\alpha (T-1)}\geq T-1\) in this case, this implies that
More precisely, \(\Delta \phi (0)=0\), \(\Delta \phi (t)<0\) for \(t\in [1,T-2]_{\mathbb{Z}}\) and \(\Delta \phi (T-1)<0\) for \(\frac{2 \eta +1-\alpha (T-1)^{2}}{1-\alpha (T-1)}> T-1\) and \(\Delta \phi (T-1)=0\) for \(\frac{2\eta +1-\alpha (T-1)^{2}}{1-\alpha (T-1)}= T-1\). Therefore, \(\phi (t)\) is a positive solution of the linear problem (2.4) since \(\phi (T)=0\).
Let
Then E is a Banach space with norm \(\|u\|=\max_{t\in [0,T]_{ \mathbb{Z}}}|u(t)|\).
Let
Then \(K_{0}\) is a cone in E.
Lemma 2.3
Let (\(H_{0}\)) hold. If \(y\in K_{0}\), then the solution \(u(t)\) of problem (2.1) belongs to \(K_{0}\), i.e., \(u\in K _{0}\). Furthermore, \(u(t)\) is concave on \([0,\eta ]_{\mathbb{Z}}\).
Proof
Firstly, if \(0\leq t-2\leq \eta \), then
Then we obtain that
Since \(y\in {K_{0}}\), we know that y is decreasing on \([0,T]_{ \mathbb{Z}}\). That is to say, \(y(t)\geq y(\eta )\) for \(t\leq \eta \) and \(y(t)\leq y(\eta )\) for \(t\geq \eta \). Therefore, if \(t-1\leq \eta \), then
If \(t-1>\eta \), then \(y(t-1)\leq y(\eta )\). Therefore, by (2.5), no matter \(t-1\leq \eta \) or \(t-1>\eta \), we always have
Combining this with \(\eta \geq \frac{T-2}{2}\), we have
Furthermore, we have
Now, if \(t-1\leq \eta \) and \(t\leq \eta \), then
Combining this with the monotonicity of y, we get that
For other cases, \(t-1\leq \eta < t\) and \(t-1> \eta \), we could also obtain \(\Delta ^{2}u(t-1)\leq 0\) for \(0\leq t-2\leq \eta \).
Secondly, if \(\eta < t-2\leq T-2\), then
and
Combining this with the fact that \(y\in K_{0}\) and the monotonicity of y, we get that
Since \(\eta \geq \frac{T-2}{2}\), then we get that
So, for \(\forall t\in [0,T-1]_{\mathbb{Z}}\), \(\Delta u(t)\leq 0\), which implies that \(u(t)\) is decreasing. Since \(u(T)=0\), for \(\forall t \in [0,T]_{\mathbb{Z}}\), we have \(u(t)\geq 0\) and \(u\in K_{0}\). For \(\forall t\in [1,\eta +1]_{\mathbb{Z}}\), \(\Delta ^{2}u(t-1)\leq 0\), we get that \(u(t)\) is concave on \([0,\eta +2]_{\mathbb{Z}}\). □
Lemma 2.4
Let (\(H_{0}\)) hold. If \(y\in K_{0}\), then the solution \(u(t)\) of (2.1) satisfies
where \(\theta ^{\ast }=\frac{\eta +2-\theta }{\eta +2}\), \(\theta \in [ \lfloor \frac{T}{2} \rfloor +1,\eta +2 ] _{\mathbb{Z}}\).
Proof
From Lemma 2.3, \(u(t)\) is concave on \([0,\eta +2]_{ \mathbb{Z}}\). So, \(u(t)\) satisfies
Meanwhile, from Lemma 2.3, \(u(t)\) is non-increasing on \([0,T]_{ \mathbb{Z}}\), which implies that \(u(0)=\|u\|\). Therefore,
□
3 Main results
- (\(H_{1}\)):
-
\(f:[1,T-2]_{\mathbb{Z}}\times [0,\infty )\to [0,\infty )\) is continuous. For \(u\in [0,\infty )\), \(f(t,u)\) is a decreasing function with respect to t, and for \(t\in [1,T-2]_{\mathbb{Z}}\), \(f(t,u)\) is an increasing function with respect to u.
- (\(H_{2}\)):
-
\(a:[1,T-2]_{\mathbb{Z}}\to [0,\infty )\) is a decreasing function.
Let
Then K is a cone in E. Define the operator \(S:K\to E\) as
Lemma 3.1
\(S:K\to K\) is completely continuous.
Proof
It is obvious that \(S:K\to E\) is completely continuous since the Banach space E is finite dimensional. Now, let us prove that \(S:K\to K\), that is to say, for any \(u\in K\), \(Su\in K\).
Let \(u\in K\). Then \(u\in K_{0}\), which implies that \(\Delta u(t) \leq 0\) and u is decreasing on t. Therefore, by \((H_{1})\), \(f(t,u(t))\) is a decreasing function of t. Let \(y(t):=a(t)f(t,u(t))\). Then, from (\(H_{1}\)) and (\(H_{2}\)), we obtain that \(y(t)\geq 0\) and y is also a decreasing function of t. Thus, \(y\in K_{0}\). Furthermore, by (3.1), we know that
and
Therefore, Su satisfies problem (2.1). Now, similar to the proof of Lemma 2.3, and using the fact \(y\in K_{0}\), we obtain that \(Su\in K _{0}\) and Su is concave on \([0,\eta ]_{\mathbb{Z}}\). Furthermore, by Lemma 2.4 and the fact \(Su\in K_{0}\), we know that
Therefore, \(Su\in K\) and \(S:K\to K\) is completely continuous. □
From (3.1) and Lemma 3.1, we know that if u is a fixed point of S in K, then u is a positive solution of (1.1). In the rest of this paper, we try to prove S has at least one or two fixed point(s) in K by using Theorem 1.1 and Theorem 1.2.
Let
Theorem 3.1
Suppose that (\(H_{0}\)), (\(H_{1}\)), and (\(H_{2}\)) hold. If there exist two constants r and R (\(r\neq R\)) such that
- (A1):
-
\(f(t,u)\leq \frac{r}{A}\), \((t,u)\in [1,T-2]_{\mathbb{Z}} \times [0,r]\);
- (A2):
-
\(f(t,u)\geq \frac{R}{B}\), \((t,u)\in [1,T-2]_{\mathbb{Z}} \times [\theta ^{\ast }R,R]\),
then problem (1.1) has at least one positive solution \(u\in K\) with \(\min \{r,R\}\leq \|u\|\leq \max \{r,R\}\).
Proof
Without loss of generality, suppose that \(r< R\), the other case could be treated similarly. Let \(\varOmega _{1}=\{u\in E:\|u\|< r \}\). From Lemma 2.2, \(G(t,s)\leq 0\) for \(s\in [\eta +1,T-2]_{ \mathbb{Z}}\); \(G(t,s)\geq 0\) for \(s\in [1,\eta ]_{\mathbb{Z}}\). Since (A1), we get, for \(\forall u\in K\cap \partial \varOmega _{1}\),
So, for \(u\in K\cap \partial \varOmega _{1} \),
Let \(\varOmega _{2}=\{u\in E:\|u\|< R\} \). Then, for \(u\in K \cap \partial {\varOmega _{2}}\),
In fact, by Lemma 2.2,
Furthermore,
Let
and
Then
and
Therefore, (3.5) holds. This implies that
So, for \(\forall u\in K\cap \partial \varOmega _{2}\),
Then, by Theorem 1.1, S has at least one fixed point \(u\in K\) and u will be a positive solution of problem (1.1). □
Theorem 3.2
Suppose that (\(H_{0}\)), (\(H_{1}\)), and (\(H_{2}\)) hold. If one of the following conditions holds:
or
then (1.1) has at least one positive solution.
Proof
Firstly, we prove the case that \((B_{1}) \) holds. Since \(f^{0}=0\), there exists a constant \(R_{1}>0\) such that
Since \(f_{\infty }=\infty \), there exists a constant \(R_{2}>R_{1}\) such that
From Theorem 3.1, problem (1.1) has at least one positive solution \(u\in K \).
Secondly, suppose that \((B_{2}) \) holds. Since \(f_{0}=\infty \), there exists a constant \(r_{1}>0\) such that
Let \(\varOmega _{1}=\{u\in E:\|u\|< r_{1} \}\). If \(u\in K\cap \partial \varOmega _{1}\), we have
Therefore, similar to the proof of (3.6), we have
On the other hand, since \(f^{\infty }=0\), there exists a constant \(r_{2}>0\) such that
If f is bounded, then there exists a constant \(N>0\) such that \(f\leq N\). So, we choose \(R'=\max \{2r_{1},NA\}\). If f is unbounded, then let \(R'>\max \{2r_{1},r_{2}\}\) such that \(f(t,u)\leq f(t,R_{2})\). Let \(\varOmega _{2}=\{u\in K:\|u\|< R'\}\) for \(\forall (t,u)\in [1,T-2]_{ \mathbb{Z}}\times [0,R_{2}]\). Similar to the proof of Theorem 3.1, we get, for \(\forall u\in k\cap \partial \varOmega _{2}\),
Thus, by Theorem 1.1, S has at least one fixed point \(u\in K\cap \overline{ \varOmega _{2}}\setminus \varOmega _{1}\), which is a positive solution of problem (1.1). □
Theorem 3.3
Assume that (\(H_{0}\)), (\(H_{1}\)), and (\(H_{2}\)) hold. If
- \((C_{1})\) :
-
\(f_{0}:=\lim_{u\to 0^{+}}\min_{t\in [1,T-2]_{\mathbb{Z}}}= \frac{f(t,u)}{u}=+\infty \), \(f_{\infty }:=\lim_{u\to \infty } \min_{t\in [1,T-2]_{\mathbb{Z}}}=\frac{f(t,u)}{u}=+\infty \),
and
- (\(C_{2}\)):
-
There exists a constant \(p>0\) such that \(f(t,u)<\gamma p\) for \(0\leq u\leq p\) and \(t\in [1,T-2]_{\mathbb{Z}} \), where
$$ \gamma = \Biggl( \frac{T(T-1)(1+\alpha \eta )}{1-\alpha (T-1)}\sum_{s=1} ^{T-2}a(s) \Biggr)^{-1}, $$
then problem (1.1) has at least two positive solutions \(u_{1}\) and \(u_{2}\) with \(0\leq \|u_{1}\|\leq p\leq \|u_{2}\|\).
Proof
Choose \(M>0\) such that
Since \(f_{0}=+\infty \), there exists a constant r with \(0< r< p\) such that \(f(t,u)\ge Mu\) for \(0\le u\le r\). Then, for \(\forall u\in \partial K_{r}\), we have
From Theorem 1.2, we get
Since \(f_{\infty }=+\infty \), there exists a constant \(R_{1}>0\) such that \(f(t,u)\ge Mu\) for \(\forall u\ge R_{1}\). Choose \(R>\max \{p, \frac{R _{1}}{\theta ^{*}}\}\), then for \(\forall u\in \partial K_{R}\), \(\min_{t\in [T-\theta ,\theta ]_{\mathbb{Z}}}u(t)\ge \theta ^{*}\|u\|>R_{1}\). Similar to the above proof, we have
Therefore,
From (\(C_{2}\)), for \(\forall u\in \partial K_{p}\),
Therefore, for \(\forall u\in \partial K_{p}\), \(\|Tu\|\le \|u\|\). By Theorem 1.2,
Thus,
So, S has a fixed point \(u_{1}\) in \(K_{p}\setminus \mathring{K}_{r}\) and another fixed point \(u_{2}\) in \(K_{R}\setminus \mathring{K}_{p}\). So, problem (1.1) has at least two positive solutions \(u_{1}\) and \(u_{2}\) with \(0\le \|u_{1}\|\le p\le \|u_{2}\|\). □
Theorem 3.4
Assume that \((H_{0})\), \((H_{1})\), and \((H_{2})\) hold. If
- \((D_{1})\) :
-
\(f^{0}:=\lim_{u\to 0^{+}}\max_{t\in [1,T-2]_{\mathbb{Z}}} \frac{f(t,u)}{u}=0\), \(f^{\infty }:=\lim_{u\to \infty } \max_{t\in [1,T-2]_{\mathbb{Z}}}\frac{f(t,u)}{u}=0\),
and
- \((D_{2})\) :
-
there exists a constant \(p>0\) such that \(f(t,u)>\beta p\) for \(\theta ^{*}p\le u\le p\) and \(t\in [1,T-2]_{\mathbb{Z}}\), where
$$ \beta = \Biggl( \theta ^{\ast }\frac{\theta (T-\theta )[2-\alpha (\theta -1)]}{2-2\alpha (T-1)}\sum _{s=T-\theta }^{\theta }a(s) \Biggr)^{-1}, $$
then (1.1) has at least two positive solutions \(u_{1}\) and \(u_{2}\) with \(0\le \|u_{1}\|\le p\le \|u_{2}\|\).
Proof
From (\(D_{1}\)), for \(\forall \epsilon >0\), there exists \(M_{1}>0\), if \(u>0\), \(t\in [0,T]_{\mathbb{Z}}\), we have \(f(t,u)\leq M _{1}+\epsilon u\). Then, for \(\forall u\in K\),
Choose \(\epsilon >0\) sufficiently small and \(R>p\) sufficiently large, then for \(\forall u\in \partial K_{R}\), \(\|Su\|\le \|u\|\), from Theorem 1.2, we have
In a similar way, if \(0< r< p\),
From (\(D_{2}\)), for \(\forall u\in \partial K_{p}\),
Then, for \(\forall u\in \partial K_{p}\), \(\|Su\|\geq \|u\|\). From Theorem 1.2, we have
Then, problem (1.1) has at least two solutions \(u_{1}\) and \(u_{2}\) with \(0\le \|u_{1}\|\le p\le \|u_{2}\|\). □
4 Example
Example 4.1
Consider the discrete three-point boundary problem
where \(a(t)=\frac{9-t}{10}\), and
Since \(T=9\) and \(\alpha =\frac{1}{9}\), \(\eta \in [4,7]_{\mathbb{Z}}\). Without loss of generality, let \(\eta =4\). Then, by the direct calculation, we get \(\theta \in [5,6]_{\mathbb{Z}}\). Choose \(\theta =5\), then \(\theta ^{\ast }=\frac{\eta +2-\theta }{\eta +2}= \frac{1}{6}\). So,
If we choose \(R=330\), \(r=1{,}000{,}000\), from Theorem 3.1, problem (4.1) has at least one positive solution.
Example 4.2
In this example, we continue to consider problem (4.1) with
Continue to take \(\alpha =\frac{1}{9}\), \(\eta =4\), \(\theta =5\), and \(\theta ^{\ast }=\frac{1}{6}\). Then
Furthermore, if we choose \(p=1\), then for \(\theta ^{\ast }p\leq u \leq p\), \(f(t,u)\geq \beta p=\frac{3}{68}\). From Theorem 3.4, problem (4.1) has at least two positive solutions \(u_{1}\) and \(u_{2}\) with \(0\le \|u_{1}\|\le p\le \|u_{2}\|\).
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The authors express their gratitude to the referee for reading the paper carefully and making several corrections and remarks.
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The datasets used and analysed during the current study are available from the corresponding author on reasonable request.
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This work was supported by the National Natural Science Foundation of China (No. 71561024). The key projects of Natural Science Foundation of Gansu province of China (No. 18JR3RA084).
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Xu, Y., Tian, W. & Gao, C. Existence of positive solutions of discrete third-order three-point BVP with sign-changing Green’s function. Adv Differ Equ 2019, 206 (2019). https://doi.org/10.1186/s13662-019-2145-x
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DOI: https://doi.org/10.1186/s13662-019-2145-x