On a nabla fractional boundary value problem with general boundary conditions

1.   Introduction

Let $a$, $b \in \mathbb{R}$ with $b - a \in \mathbb{N}_{1}$. Consider the homogeneous nabla fractional boundary value problem with general boundary conditions:

where $1 < \nu < 2$, $\alpha^{2} + \beta^{2} > 0$ and $\gamma^{2} + \delta^{2} > 0$. Brackins & Peterson ^[5] proved that the boundary value problem (1.1) has only the trivial solution if, and only if

In the following theorem, Brackins & Peterson ^[5] gave an explicit expression for its Green's function.

Theorem 1.1 (See ^[5]). Assume (1.2) holds. The Green's function for the boundary value problem (1.1) is given by

where

and

We show that this Green's function is nonnegative and obtain an upper bound for its maximum value. Using the Green's function, we will then develop a Lyapunov-type inequality for the nabla fractional boundary value problem

where $q : \mathbb{N}^{b}_{a + 1} \rightarrow \mathbb{R}$.

2.   Preliminaries

We shall use the following notations, definitions and known results of nabla fractional calculus throughout the article ^{[1,2,3,6,9,10,11,12,13]}. Denote by $\mathbb{N}_{a} : = \{a, a + 1, a + 2, \ldots\}$ and $\mathbb{N}^{b}_{a} : = \{a, a + 1, a + 2, \ldots, b\}$ for any $a$, $b \in \mathbb{R}$ such that $b - a \in \mathbb{N}_{1}$.

Definition 2.1(See ^[4]). The backward jump operator $\rho : \mathbb{N}_{a} \rightarrow \mathbb{N}_{a}$ is defined by

Definition 2.2 (See ^[14,15]). The Euler gamma function is defined by

Using its well-known reduction formula, the Euler gamma function can be extended to the half-plane $\Re(z) \leq 0$ except for $z \in \{\ldots, -2, -1, 0\}$.

Definition 2.3 (See ^[7]). For $t \in \mathbb{R} \setminus \{\ldots, -2, -1, 0\}$ and $r \in \mathbb{R}$ such that $(t + r) \in \mathbb{R} \setminus \{\ldots, -2, -1, 0\}$, the generalized rising function is defined by

Also, if $t \in \{\ldots, -2, -1, 0\}$ and $r \in \mathbb{R}$ such that $(t + r) \in \mathbb{R} \setminus \{\ldots, -2, -1, 0\}$, then we use the convention that $t^{\overline{r}} : = 0$.

Definition 2.4 (See ^[7]). Let $\mu \in \mathbb{R} \setminus \{\ldots, -2, -1\}$. Define the $\mu^{th}$-order nabla fractional Taylor monomial by

provided the right-hand side exists. Observe that $H_{\mu}(a, a) = 0$ and $H_{\mu}(t, a) : = 0$ for all $\mu \in \{\ldots, -2, -1\}$ and $t \in \mathbb{N}_{a}$.

Definition 2.5 (See ^[4]). Let $u: \mathbb{N}_{a} \rightarrow \mathbb{R}$ and $N \in \mathbb{N}_1$. The first order backward (nabla) difference of $u$ is defined by

and the $N^{th}$-order nabla difference of $u$ is defined recursively by

Definition 2.6 (See ^[7]). Let $u: \mathbb{N}_{a + 1} \rightarrow \mathbb{R}$ and $N \in \mathbb{N}_1$. The $N^{\text{th}}$-order nabla sum of $u$ based at $a$ is given by

where by convention $\big{(}\nabla ^{-N}_{a}u\big{)}(a) = 0$. We define $\big{(}\nabla ^{-0}_{a}u\big{)}(t) : = u(t)$ for all $t \in \mathbb{N}_{a + 1}$.

Definition 2.7 (See ^[7]). Let $u: \mathbb{N}_{a + 1} \rightarrow \mathbb{R}$ and $\nu > 0$. The $\nu^{\text{th}}$-order nabla sum of $u$ based at $a$ is given by

where by convention $\big{(}\nabla ^{-\nu}_{a}u\big{)}(a) = 0$.

Definition 2.8 (See ^[7]). Let $u: \mathbb{N}_{a + 1} \rightarrow \mathbb{R}$, $\nu > 0$ and choose $N \in \mathbb{N}_1$ such that $N - 1 < \nu \leq N$. The $\nu^{\text{th}}$-order nabla difference of $u$ is given by

The following properties of gamma function, generalized rising function, and fractional nabla Taylor monomial will be used in Section 3.

Proposition 1 (See ^[7]). Assume the following generalized rising functions and fractional nabla Taylor monomials are well defined.

1. $\Gamma(t) > 0$ for $t > 0$, and $\Gamma(t) < 0$ for $-1 < t < 0$.

2. $t^{\overline\nu}(t + \nu)^{\overline \mu} = t^{\overline {\nu + \mu}}$.

3. $\nabla(\nu + t)^{\overline {\mu}} = \mu(\nu + t)^{\overline {\mu - 1}}$.

4. $\nabla(\nu - t)^{\overline {\mu}} = -\mu(\nu - \rho(t))^{\overline {\mu - 1}}$.

5. $\nabla H_{\mu}(t, a) = H_{\mu - 1}(t, a)$.

6. $H_{\mu}(t, a) - H_{\mu - 1}(t, a) = H_{\mu}(t, a + 1)$.

7. $\sum^{t}_{s = a + 1}H_{\mu}(s, a) = H_{\mu + 1}(t, a)$.

8. $\sum^{t}_{s = a + 1}H_{\mu}(t, \rho(s)) = H_{\mu + 1}(t, a)$.

Proposition 2 (See ^[7]). Let $\nu \in \mathbb{R}^{+}$ and $\mu \in \mathbb{R}$ such that $\mu$, $\mu + \nu$ and $\mu - \nu$ are nonnegative integers. Then, for all $t \in \mathbb{N}_{a}$,

(i) $\nabla_{a}^{-\nu}(t - a)^{\overline{\mu}} = \frac{\Gamma(\mu + 1)}{\Gamma(\mu + \nu + 1)}(t - a)^{\overline{\mu + \nu}}$.

(ii) $\nabla_{a}^{\nu}(t - a)^{\overline{\mu}} = \frac{\Gamma(\mu + 1)}{\Gamma(\mu - \nu + 1)}(t - a)^{\overline{\mu - \nu}}$.

(iii) $\nabla_{a}^{-\nu} H_{\mu}(t, a) = H_{\mu + \nu}(t, a)$.

(iv) $\nabla_{a}^{\nu} H_{\mu}(t, a) = H_{\mu - \nu}(t, a)$.

Proposition 3 (See ^[8]). Let $\mu > -1$ and $s \in \mathbb{N}_{a}$. Then, the following hold:

(a) If $t \in \mathbb{N}_{\rho(s)}$, then $H_{\mu}(t, \rho(s)) \geq 0$, and if $t \in \mathbb{N}_{s}$, then $H_{\mu}(t, \rho(s)) > 0$.

(b) If $t \in \mathbb{N}_{\rho(s)}$ and $\mu > 0$, then $H_{\mu}(t, \rho(s))$ is a decreasing function of $s$.

(c) If $t \in \mathbb{N}_{s}$ and $-1 < \mu < 0$, then $H_{\mu}(t, \rho(s))$ is an increasing function of $s$.

(d) If $t \in \mathbb{N}_{\rho(s)}$ and $\mu \geq 0$, then $H_{\mu}(t, \rho(s))$ is a nondecreasing function of $t$.

(e) If $t \in \mathbb{N}_{s}$ and $\mu > 0$, then $H_{\mu}(t, \rho(s))$ is an increasing function of $t$.

(f) If $t \in \mathbb{N}_{s + 1}$ and $-1 < \mu < 0$, then $H_{\mu}(t, \rho(s))$ is a decreasing function of $t$.

Proposition 4 (See ^[8]). If $0 < \nu \leq \mu$, then $H_{\nu}(t, a) \le H_{\mu}(t, a)$, for each fixed $t \in \mathbb{N}_{a}$.

Proposition 5 (See ^[8]). Let $f$, $g$ be nonnegative real-valued functions on a set $S$. Moreover, assume $f$ and $g$ attain their maximum in $S$. Then, for each fixed $t \in S$,

Proposition 6. Let $\mu > -1$, $s \in \mathbb{N}_{a + 1}$, and $t \in \mathbb{N}_{s}$. Denote by

Then, the following hold:

(I) $0 < h_{\mu}(t, s)$.

(II) If $\mu > 0$, then $h_{\mu}(t, s) \leq 1$, and if $-1 < \mu < 0$, then $h_{\mu}(t, s) \geq 1$. In particular, $h_{0}(t, s) = 1$.

(III) If $\mu > 0$, then $h_{\mu}(t, s)$ is a nondecreasing function of $t$.

(IV) If $-1 < \mu < 0$, then $h_{\mu}(t, s)$ is a nonincreasing function of $t$.

Proof. (Ⅰ) First, consider

Since $\Gamma(t - a)$, $\Gamma(t - a + \mu)$, $\Gamma(t - s + 1)$, $\Gamma(t - s + \mu + 1) > 0$, it follows from (2.1) that $h_{\mu}(t, s) > 0$.

(Ⅱ) The proof of (Ⅱ) follows from the monotonicity of $H_{\mu}(t, \rho(s))$ with respect to $s$.

(Ⅲ) Next, consider

Since $\Gamma(t - a - 1)$, $\Gamma(t - a + \mu)$, $\Gamma(t - s + \mu)$, $\Gamma(t - s + 1) > 0$, and $(s - a - 1) \geq 0$, it follows from (2.2) that $\nabla h_{\mu}(t, s) \geq 0$, implying that (Ⅲ) holds.

(Ⅳ) Clearly, from (2.2), we have

Since $\Gamma(t - a - 1)$, $\Gamma(t - a - \mu)$, $\Gamma(t - s - \mu)$, $\Gamma(t - s + 1) > 0$, $(s - a - 1) \geq 0$, it follows from (2.3) that $\nabla h_{-\mu}(t, s) \leq 0$, implying that (IV) holds.

3.   Properties of Green's function

In this section, we obtain a few properties of $G(t, s)$ which we use in the later part of the article.

Lemma 1. Assume $\alpha$, $\beta$, $\gamma$, $\delta \geq 0$ and $\beta \geq \alpha$ such that (1.2) holds. Then,

1. $\xi > 0$ for all $t \in \mathbb{N}^{b}_{a}$.

2. $u(t, s) \geq 0$ for all $(t, s) \in \mathbb{N}^{b}_{a} \times \mathbb{N}^{b}_{a + 1}$ such that $t \leq s - 1$.

3. $v(t, s) \geq 0$ for all $(t, s) \in \mathbb{N}^{b}_{a} \times \mathbb{N}^{b}_{a + 1}$ such that $t \geq s$.

Proof. (1) From Proposition 3, we have $H_{\nu - 1}(b, a)$, $H_{\nu - 2}(b, a) > 0$ implying that

(2) From Proposition 3, we have $H_{\nu - 1}(b, \rho(s))$, $H_{\nu - 2}(b, \rho(s)) > 0$ for all $s \in \mathbb{N}^{b}_{a + 1}$, and $H_{\nu - 1}(t, a) \geq 0$ for all $t \in \mathbb{N}^{b}_{a}$. Also, from (1), we have $\xi > 0$ for all $t \in \mathbb{N}^{b}_{a}$. Thus, we obtain

for all $(t, s) \in \mathbb{N}^{b}_{a} \times \mathbb{N}^{b}_{a + 1}$ such that $t \leq s - 1$.

(3) Consider

where

We already know that $\xi > 0$ for all $t \in \mathbb{N}^{b}_{a}$. Now, we show that

From Proposition 3, we have $H_{\nu - 2}(b, \rho(s)) > 0$ for all $s \in \mathbb{N}^{b}_{a + 1}$. So, we obtain

Again, from Proposition 3, we have $H_{\nu - 1}(t, \rho(s)) \leq H_{\nu - 1}(b, \rho(s))$ for all $(t, s) \in \mathbb{N}^{b}_{a} \times \mathbb{N}^{b}_{a + 1}$ such that $t \geq s$, implying that

From Proposition 3, we have $H_{\nu - 1}(t, \rho(s)) \leq H_{\nu - 1}(t, a)$, $H_{\nu - 2}(b, a) \leq H_{\nu - 2}(b, \rho(s))$ for all $(t, s) \in \mathbb{N}^{b}_{a} \times \mathbb{N}^{b}_{a + 1}$ such that $t \geq s$, implying that

Now, consider

From Proposition 3, we have $H_{\nu - 1}(b, a)$, $H_{\nu - 1}(t, \rho(s)) > 0$, and $h_{\nu - 1}(b, s) \geq h_{\nu - 1}(t, s)$ for all $(t, s) \in \mathbb{N}^{b}_{a} \times \mathbb{N}^{b}_{a + 1}$ such that $t \geq s$, implying that

Therefore, we obtain $v(t, s) \geq 0$ for all $(t, s) \in \mathbb{N}^{b}_{a} \times \mathbb{N}^{b}_{a + 1}$ such that $t \geq s$. The proof is complete.

Theorem 3.1. Assume $\alpha$, $\beta$, $\gamma$, $\delta \geq 0$ and $\beta \geq \alpha$ such that (1.2) holds. Then, $G(t, s) \geq 0$ for all $(t, s) \in \mathbb{N}^{b}_{a} \times \mathbb{N}^{b}_{a + 1}$.

Proof. The proof follows from the preceding lemma.

Lemma 2. Assume $\alpha$, $\beta$, $\gamma$, $\delta \geq 0$ and $\beta \geq \alpha$ such that (1.2) holds. Then,

1. $u(t, s)$ is an increasing function of $t$ for all $(t, s) \in \mathbb{N}^{b}_{a} \times \mathbb{N}^{b}_{a + 1}$ such that $t \leq s - 1$.

2. $v(t, s)$ is a decreasing function of $t$ for all $(t, s) \in \mathbb{N}^{b}_{a} \times \mathbb{N}^{b}_{a + 1}$ such that $t \geq s$.

Proof. (1) Consider

From Proposition 3, we have $H_{\nu - 1}(b, \rho(s))$, $H_{\nu - 2}(b, \rho(s)) > 0$ for all $s \in \mathbb{N}^{b}_{a + 1}$, and $H_{\nu - 2}(t, a) > 0$ for all $t \in \mathbb{N}^{b}_{a + 1}$. Also, from (1), we have $\xi > 0$ for all $t \in \mathbb{N}^{b}_{a + 1}$. Thus, we obtain $\nabla_{t} u(t, s) > 0$, implying that (1) holds.

(2) From (3.2), we obtain

where

Clearly, $\xi > 0$ for all $t \in \mathbb{N}^{b}_{a + 1}$. Now, we show that

From Proposition 3, we have $H_{\nu - 2}(t, \rho(s)) > 0$ for all $(t, s) \in \mathbb{N}^{b}_{a} \times \mathbb{N}^{b}_{a + 1}$ such that $t \geq s$, implying that

From Proposition 3, we have $H_{\nu - 2}(t, \rho(s)) \geq H_{\nu - 2}(t, a)$, $H_{\nu - 1}(b, a) \geq H_{\nu - 1}(b, \rho(s))$ for all $(t, s) \in \mathbb{N}^{b}_{a} \times \mathbb{N}^{b}_{a + 1}$ such that $t \geq s$, implying that

Now, consider

From Proposition 3, we have $H_{\nu - 2}(b, a)$, $H_{\nu - 2}(t, \rho(s)) > 0$, and $h_{\nu - 2}(t, s) \geq h_{\nu - 2}(b, s)$ for all $(t, s) \in \mathbb{N}^{b}_{a} \times \mathbb{N}^{b}_{a + 1}$ such that $t \geq s$, implying that

Therefore, (2) holds. The proof is complete.

Theorem 3.2. Assume $\alpha$, $\beta$, $\gamma$, $\delta \geq 0$ and $\beta \geq \alpha$ such that (1.2) holds. The following inequality holds for the Green's function $G(t, s)$:

where

Proof. From Lemma 2, we have

Consider

Denote by

Then, by Lemma 1 and Proposition 3, we have

Now, consider

It follows from Lemma 1 that

Since

we have

Thus, by Proposition 3, (3.5), (3.7) and (3.8), we obtain

The proof is complete.

Theorem 3.3. Assume $\alpha$, $\beta$, $\gamma$, $\delta \geq 0$ and $\beta \geq \alpha$ such that (1.2) holds. The following inequality holds for the Green's function $G(t, s)$:

for all $(t, s) \in \mathbb{N}^{b}_{a} \times \mathbb{N}^{b}_{a + 1}$, where

Proof. Consider

Since $H_{\nu}(t, a) \geq 0$ for all $t \in \mathbb{N}^{b}_{a}$ and

we obtain (3.9). The proof is complete.

Theorem 3.4 (See ^[5]). Let $h : \mathbb{N}^{b}_{a + 1} \rightarrow \mathbb{R}$. If (1.1) has only the trivial solution, then the nonhomogeneous boundary value problem

has a unique solution given by

Now, we are able to establish a Lyapunov-type inequality for the nabla fractional boundary value problem (1.6).

Theorem 3.5. Assume $\alpha$, $\beta$, $\gamma$, $\delta \geq 0$ and $\beta \geq \alpha$ such that (1.2) holds. If the nabla fractional boundary value problem (1.6) has a nontrivial solution, then

Proof. Let $\mathcal{B}$ be the Banach space of functions endowed with norm

It follows from the above Theorem that a solution to (1.6) satisfies the equation

Hence

or, equivalently,

The proof is complete.

4.   Application

Here, we estimate a lower bound for the eigenvalues of the nabla fractional eigenvalue problem corresponding to the nabla fractional boundary value problem (1.6).

Theorem 4.1. Assume that $1 < \nu < 2$ and $u$ is a nontrivial solution of the nabla fractional eigenvalue problem

where $u(t) \neq 0$ for each $t \in \mathbb{N}^{b - 1}_{a + 1}$. Then,

Acknowledgments

We thank referees for helpful comments and suggestions on our article.

Conflict of interest

The author declares no conflicts of interest in this paper.