On existence of certain delta fractional difference models

Definition 2.1

Definition 2.1 see Brackins (2014)

For the homogeneous FDE $$\left({{}_{p_0+1}{}^{\mathrm{RL}}\Delta }^{\alpha }\left(z\nabla y\right)\right)\left(x\right)=0,$$ we say $\phi (x,\bar{\sigma }\left(r\right))$ , where $\phi :{\mathbb{N}}_{p_0+1}\times {\mathbb{N}}_{p_0+1}\to \mathbb{R}$ , as a Cauchy function such that $\phi (\cdot ,\bar{\sigma }\left(r\right))$ is the unique solution of

$$\left\{\begin{array}{c}\left({{}_r{}^{\mathrm{RL}}\Delta }^{\alpha }\left(z\nabla \phi \right)\right)\left(x\right)=0,x\in {\mathbb{N}}_{r+1},\hfill \\ \phi (r-1)=0,\left(\nabla \phi \right)\left(r\right)=\frac{1}{z\left(r\right)},\hfill \end{array}\right.$$ for each fixed $r\in {\mathbb{N}}_{p_0+1}$ . Also, it is expressed by

(2.6)

$$\phi (x,\bar{\sigma }\left(r\right))=\sum _{s=r}^x\frac{{(s+\alpha -\sigma \left(r\right))}^{\underline{\alpha -1}}}{\Gamma \left(\alpha \right)z\left(s\right)},x\in {\mathbb{N}}_{p_0+1},$$ where $\sigma \left(\tau \right)=\tau +1$ and $\bar{\sigma }\left(\tau \right)=\tau -1$ .

Lemma 2.1

Let $0<\theta <1$ , $A_0\in \mathbb{R}$ , and $h,g$ be defined on ${\mathbb{N}}_{p_0}$ . Then the fractional IVP

$$\left\{\begin{array}{cc}\left({{}_{p_0+1}{}^{\mathrm{RL}}\Delta }^{\theta }y\right)\left(x\right)=g(x+\theta ),\hfill & x\in {\mathbb{N}}_{p_0+1},\hfill \\ y(p_0+1)=A_0,\hfill & \hfill \end{array}\right.$$ has the unique solution $$y\left(x\right)=\left({{}_{p_0+1}{}^{\mathrm{RL}}\Delta }^{-\theta }g\right)(x+\theta )+\left[A_0-g(p_0+1)\right]\frac{{(x-p_0-2+\theta )}^{\underline{\theta -1}}}{\Gamma \left(\theta \right)},s.t.x\in {\mathbb{N}}_{p_0+2}.$$

Proof

By taking ${{}_{p_0+2-\theta }{}^{\mathrm{RL}}\Delta }^{-\theta }$ on both sides of (2.7), we have

$$\left({{}_{p_0+2-\theta }{}^{\mathrm{RL}}\Delta }^{-\theta }\left({{}_{a+1}{}^{\mathrm{RL}}\Delta }^{\theta }y\right)\right)\left(x\right)=\left({{}_{p_0+2-\theta }{}^{\mathrm{RL}}\Delta }^{-\theta }g(x+\theta )\right)\left(x\right).$$

Computing the left side of (2.8), we see that

$$\left({{}_{p_0+2-\theta }{}^{\mathrm{RL}}\Delta }^{-\theta }\left({{}_{a+1}{}^{\mathrm{RL}}\Delta }^{\theta }y\right)\right)\left(x\right)≔\left({{}_{p_0+2-\theta }{}^{\mathrm{RL}}\Delta }^{-\theta }h\right)\left(x\right)=\frac{1}{\Gamma \left(\theta \right)}\sum _{r=p_0+2-\theta }^{x-\theta }{\left(x-\sigma \left(r\right)\right)}^{\underline{\theta -1}}h\left(r\right)=\frac{1}{\Gamma \left(\theta \right)}\sum _{r=p_0+1-\theta }^{x-\theta }{\left(x-\sigma \left(r\right)\right)}^{\underline{\theta -1}}h\left(r\right)-\frac{1}{\Gamma \left(\theta \right)}{\left(x-(p_0+1-\theta )-1\right)}^{\underline{\theta -1}}h(p_0+1-\theta )=\left({{}_{p_0+1-\theta }{}^{\mathrm{RL}}\Delta }^{-\theta }{{}_{p_0+1}{}^{\mathrm{RL}}\Delta }^{\theta }y\right)\left(x\right)-\frac{1}{\Gamma \left(\theta \right)}{\left(x-p_0-2+\theta \right)}^{\underline{\theta -1}}h(p_0+1-\theta )=y\left(x\right)-\frac{1}{\Gamma \left(\theta \right)}{(x-p_0-2+\theta )}^{\underline{\theta -1}}{A}_0,$$ where it is used that $$h(p_0+1-\theta )=\frac{1}{\Gamma (-\theta )}\sum _{r=p_0+1}^{p_0+1}{(p_0-\theta -r)}^{\underline{\theta -1}}y\left(r\right)=\frac{1}{\Gamma (-\theta )}{(-\theta -1)}^{\underline{-\theta -1}}y(p_0+1)=y(p_0+1)=A_0.$$

Next, by computing the right side of (2.8), we have

$$\left({{}_{p_0+2-\theta }{}^{\mathrm{RL}}\Delta }^{-\theta }g(x+\theta )\right)\left(x\right)=\frac{1}{\Gamma \left(\theta \right)}\sum _{r=p_0+2-\theta }^{x-\theta }{\left(x-\sigma \left(r\right)\right)}^{\underline{\theta -1}}g(r+\theta )=\frac{1}{\Gamma \left(\theta \right)}\sum _{r=p_0+1-\theta }^{x-\theta }{\left(x-\sigma \left(r\right)\right)}^{\underline{\theta -1}}g(r+\theta )-\frac{1}{\Gamma \left(\theta \right)}{\left(x-(p_0+1-\theta )-1\right)}^{\underline{\theta -1}}g(p_0+1)=\left({{}_{p_0+1}{}^{\mathrm{RL}}\Delta }^{-\theta }g\right)\left(x+\theta \right)-\frac{1}{\Gamma \left(\theta \right)}{(x-p_0-2+\theta )}^{\underline{\theta -1}}g(p_0+1).$$ Having the left and right sides of (2.8) as in (2.9) and (2.10), respectively, can give the desired result. □

Example 2.1

Consider the fractional IVP $$\left({{}_1{}^{\mathrm{RL}}\Delta }^{\mu }y\right)\left(x\right)=2,$$ $$y\left(1\right)=\pi .$$ By considering Lemma 2.1, we have $$y\left(x\right)=\left({{}_{p_0+1}{}^{\mathrm{RL}}\Delta }^{-\mu }g\right)(x+\mu )+(\pi -2)\frac{{(x-p_0+\mu -2)}^{(\mu -1)}}{\Gamma \left(\mu \right)}=\frac{2}{\Gamma (\mu +1)}{(x+\mu -p_0-1)}^{\left(\mu \right)}+(\pi -2)\frac{{(x-p_0+\mu -2)}^{(\mu -1)}}{\Gamma \left(\mu \right)}=\frac{2}{\Gamma (\mu +1)}\frac{\Gamma (x+\mu -p_0)}{\Gamma (x-p_0)}+\frac{\pi -2}{\Gamma \left(\mu \right)}\frac{\Gamma (x-p_0+\mu -1)}{\Gamma (x-p_0)},$$ for $x\in {\mathbb{N}}_2$ . In addition, we have represented the values of $y\left(x\right)$ for $x\in \{2,3,\dots ,10\}$ and different values of $\mu$ in Fig. 1.

Theorem 3.1

Let $0<\mu <1$ , $z:{\mathbb{N}}_{p_0+1}\to (0,\infty )$ , and $g_1,g_2:{\mathbb{N}}_{p_0+2}\to \mathbb{R}$ . Then, the fractional IVP

$$\left\{\begin{array}{c}\left({{}_{a+1}{}^{\mathrm{RL}}\Delta }^{\mu }\left(z\nabla y\right)\right)\left(x\right)+g_1(x+\mu )y(x+\mu -1)=g_2(x+\mu ),x\in {\mathbb{N}}_{p_0+2},\hfill \\ y\left(p_0\right)=A_0,y(p_0+1)=B_0,\hfill \end{array}\right.$$ has a unique solution for $y:{\mathbb{N}}_{p_0}\to \mathbb{R}$ .

Proof

Rewriting (3.1) by using (2.2), we have

$$\frac{1}{\Gamma (-\mu )}\sum _{r=p_0+1}^{x+\mu }{\left(x-r-1\right)}^{\underline{-\mu -1}}\left[z\nabla y\right]\left(r\right)+g_1(x+\mu )y(x+\mu -1)=g_2(x+\mu ).$$ With $x=p_0+2-\mu$ , the above equality becomes $$g_2(p_0+2)=\frac{1}{\Gamma (-\mu )}\sum _{r=p_0+1}^{p_0+2}{\left(p_0+1-\mu -r\right)}^{\underline{-\mu -1}}\left[z\nabla y\right]\left(r\right)+g_1\left(p_0+2\right)y\left(p_0+1\right)=\frac{1}{\Gamma (-\mu )}\left[{(-\mu )}^{\underline{-\mu -1}}\left[z\nabla y\right](p_0+1)+{(-\mu -1)}^{\underline{-\mu -1}}\left[z\Delta y\right](p_0+2)\right]+g_1(p_0+2)y(p_0+1)=z(p_0+2)y(p_0+2)+A_0\mu z(p_0+1)+B_0\left[g_1(p_0+2)-\mu z(p_0+1)-z(p_0+2)\right].$$ We solve this equation for $y(p_0+2)$ to get $$y(p_0+2)=\frac{1}{z(p_0+2)}\left[g_2(p_0+2)-A_0\mu z(p_0+1)-B_0\left(g_1(p_0+2)-\mu z(p_0+1)-z(p_0+2)\right)\right].$$ This implies that $y(p_0+2)$ can be determined uniquely by considering $y\left(p_0\right)=A_0$ and $y(p_0+1)=B_0$ and the given values of $z,g_1$ , and $g_2$ .

We are continuing by using induction and we have to demonstrate that $y\left(x\right)$ is uniquely determined on ${\mathbb{N}}_{p_0}$ . For this, let $y\left(x\right)$ be the unique solution to (3.1), for $x\in {\mathbb{N}}_{p_0}^{x_0}$ and $x_0\in {\mathbb{N}}_{p_0+2}$ . Then, we have to prove that $y(x_0+1)$ is also the unique solution of (3.1).

To do this, we use $x=x_0+1-\mu$ in (3.2), we get $$g_2(x_0+1)=\sum _{r=p_0+1}^{x_0+1}\frac{{\left(x_0-\mu -r\right)}^{\underline{-\mu -1}}}{\Gamma (-\mu )}\left[z\nabla y\right]\left(r\right)+g_1\left(x_0+1\right)y\left(x_0\right)=\sum _{r=p_0+1}^{x_0}\frac{{\left(x_0-\mu -r\right)}^{\underline{-\mu -1}}}{\Gamma (-\mu )}\left[z\nabla y\right]\left(r\right)+g_1\left(x_0+1\right)y\left(x_0\right)+z(x_0+1)y(x_0+1)-z(x_0+1)y\left(x_0\right).$$ This can be solved for $y(x_0+1)$ , $$y(x_0+1)=\frac{1}{z(x_0+1)}\left[g_2(x_0+1)-\sum _{r=p_0+1}^{x_0}\frac{{(x_0-\mu -r)}^{\underline{-\mu -1}}}{\Gamma (-\mu )}\left[z\nabla y\right]\left(r\right)-g_1(x_0+1)y\left(x_0\right)+z(x_0+1)y\left(x_0\right)\right].$$ Thus, by considering the hypothesis, each $y\left(x\right)$ , for $x$ in ${\mathbb{N}}_{p_0}^{x_0}$ , are known. Therefore, we can say that $y(x_0+1)$ is also the unique solution of (3.1). Consequently, $y\left(x\right)$ is the unique solution of (3.1) on ${\mathbb{N}}_{p_0}^{x_0+1}$ . This give us our proof. □

Theorem 3.2

The solution to (3.3) can be expressed by $$y\left(x\right)=\sum _{r=p_0+2}^x\phi (x,\bar{\sigma }\left(r\right))f\left(r\right),$$ where $\phi (x,\bar{\sigma }\left(r\right))$ is as defined in Definition 2.1.

Proof

Assume that $h\left(x\right)≔\left(z\nabla y\right)\left(x\right)$ and $y\left(x\right)$ is a solution of the fractional IVP (3.3). Therefore, $h\left(x\right)$ is a solution of $$\left({{}_{p_0}{}^{\mathrm{RL}}\Delta }^{\mu }h\right)\left(x\right)=f(x+\mu ),h(p_0+1)=z(p_0+1)\nabla y(p_0+1)=0.$$ Moreover, by Lemma 2.1, its solution can be represented as $$h\left(x\right)=\left({{}_{p_0+1}{}^{\mathrm{RL}}\Delta }^{-v}f\right)(x+\mu )-\frac{{(x-p_0-2+\mu )}^{\underline{\mu -1}}}{\Gamma \left(\mu \right)}f(p_0+1)=\sum _{r=p_0+1}^x\left[\frac{{\left[x-\sigma \left(r\right)+\mu \right]}^{\underline{\mu -1}}}{\Gamma \left(\mu \right)}f\left(r\right)\right]-\frac{{(x-p_0-2+\mu )}^{\underline{\mu -1}}}{\Gamma \mu }f(p_0+1).$$ It can be divided both sides by $z\left(x\right)$ to get $$\left(\nabla y\right)\left(x\right)=\sum _{r=p_0+1}^x\frac{{(x+\mu -\sigma \left(r\right))}^{\underline{\mu -1}}}{\Gamma \left(\mu \right)z\left(x\right)}f\left(r\right).$$ By summing both sides ${\sum }_{s=p_0+2}^x$ , we have

$$y\left(x\right)-y(p_0+1)=\sum _{s=p_0+2}^x\left[\sum _{r=p_0+2}^s\frac{{(s+\mu -\sigma \left(r\right))}^{\underline{\mu -1}}}{\Gamma \left(\mu \right)z\left(s\right)}f\left(r\right)\right],$$ where we have used that $$\sum _{s=p_0+2}^x\left(\nabla y\right)\left(x\right)=y\left(x\right)-y(p_0+1).$$ Now, we should interchange the order of the sums in (3.4) and use $y(p_0+1)=0$ to obtain $$y\left(x\right)=\sum _{r=p_0+2}^x\left[\sum _{s=r}^x\frac{{(s+\mu -\sigma \left(r\right))}^{\underline{\mu -1}}}{\Gamma \left(\mu \right)z\left(s\right)}f\left(r\right)\right]=\sum _{r=p_0+2}^x\left[\sum _{s=r}^x\frac{{(s+\mu -\sigma \left(r\right))}^{\underline{\mu -1}}}{\Gamma \left(\mu \right)z\left(s\right)}\right]f\left(r\right)=\sum _{r=p_0+2}^x\phi (x,\bar{\sigma }\left(r\right))f\left(r\right).$$ This ends our proof. □

Corollary 3.1

Let $z:{\mathbb{N}}_{p_0+1}\to (0,\infty )$ and $w_1\left(x\right),w_2\left(x\right)$ can satisfy $$\left({{}_{p_0+1}{}^{\mathrm{RL}}\Delta }^{\mu }\left(z\nabla w_1\right)\right)\left(x\right)\ge \left({{}_{p_0+1}{}^{\mathrm{RL}}\Delta }^{\mu }\left(z\nabla w_2\right)\right)\left(x\right),x\in {\mathbb{N}}_{p_0+2},$$ $$w_1\left(p_0\right)=w_2\left(p_0\right),w_1(p_0+1)=w_2(p_0+1).$$ Then, $w_1\left(x\right)\ge w_2\left(x\right)$ , for $x\in {\mathbb{N}}_{p_0}$ .

Proof

Let us set $w\left(x\right)=w_1\left(x\right)-w_2\left(x\right)$ , and $$g\left(x\right)≔\left({{}_{p_0+1}{}^{\mathrm{RL}}\Delta }^{\mu }\left(z\nabla w\right)\right)\left(x\right)=\left({{}_{p_0+1}{}^{\mathrm{RL}}\Delta }^{\mu }\left(z\nabla w_1\right)\right)\left(x\right)-\left({{}_{p_0+1}{}^{\mathrm{RL}}\Delta }^{\mu }\left(z\nabla w_2\right)\right)\left(x\right)\ge 0,x\in {\mathbb{N}}_{p_0+2}.$$ Therefore, $w$ solves the IVP $$\left({{}_{p_0+1}{}^{\mathrm{RL}}\Delta }^{\mu }\left(z\nabla w\right)\right)\left(x\right)=g\left(x\right),x\in {\mathbb{N}}_{p_0+2},$$ $$w(p_0+1)=0,\left(\nabla w\right)(p_0+1)=0.$$ Thus, by using Theorem 3.2, we see that $$w\left(x\right)=\sum _{r=p_0+2}^{x}g\left(r\right)\phi (x,\bar{\sigma }\left(r\right))=\sum _{r=p_0+2}^{x}g\left(r\right)\sum _{s=r}^x\frac{{(s-\sigma \left(r\right))}^{\underline{\mu -1}}}{\Gamma \left(\mu \right)z\left(s\right)}\ge 0,$$ and this implies that $w\left(x\right)=w_1\left(x\right)-w_2\left(x\right)\ge 0$ . Thus, the proof has been done. □

Theorem 4.1

Let $p_0,m$ be two real numbers such that $m-p_0\in {\mathbb{N}}_1$ , $h,z:{\mathbb{N}}_{p_0+1}^m\to \mathbb{R}$ , and $z\left(x\right)>0$ . Then, the fractional BVP

$$\left\{\begin{array}{c}-\left({{}_{p_0+1}{}^{\mathrm{RL}}\Delta }^{\mu }\left(z\nabla y\right)\right)\left(x\right)=h(x+\mu ),x\in {\mathbb{N}}_{p_0+1}^m,\hfill \\ y\left(p_0\right)=0,y\left(m\right)=0,\hfill \end{array}\right.$$ has the unique solution

(4.2)

$$y\left(x\right)=\sum _{x=p_0+1}^{m}G(x,r)h\left(r\right),$$ where

(4.3)

$$G(x,r)=\left\{\begin{array}{c}\frac{\phi (m,\bar{\sigma }\left(r\right))}{\phi (m,p_0)}\phi (x,p_0),x\le r-1,\hfill \\ \frac{\phi (m,\bar{\sigma }\left(r\right))}{\phi (m,p_0)}\phi (x,p_0)-\phi (x,\bar{\sigma }\left(r\right)),x\ge r,\hfill \end{array}\right.$$ and $\phi (\cdot ,\cdot )$ is as defined in Definition 2.1.

Proof

Let $\phi \left(x\right)=\left(z\nabla y\right)\left(x\right)$ , and let $A_0≔\phi (p_0+1)=z(p_0+1)\left(\nabla y\right)(p_0+1)$ . Then, by using Lemma 2.1, the solution $\phi \left(x\right)$ of the fractional IVP $$\left\{\begin{array}{c}-\left({{}_{p_0+1}{}^{\mathrm{RL}}\Delta }^{\mu }\phi \right)\left(x\right)=h(x+\mu ),\hfill \\ \phi (p_0+1)=A_0,\hfill \end{array}\right.$$ is given by $$\phi \left(x\right)=-\left({{}_{p_0+1}{}^{\mathrm{RL}}\nabla }^{-\mu }h\right)(x+\mu )-\left[A_0-h(p_0+1)\right]\frac{{(x-p_0-2+\mu )}^{\underline{\mu -1}}}{\Gamma \left(\mu \right)}.$$ By setting $k_0≔A_0-h(p_0+1)$ and using $\left(\nabla y\right)\left(x\right)=\phi \left(x\right)/z\left(x\right)$ , we have $$\left(\nabla y\right)\left(x\right)=\frac{-1}{z\left(x\right)}\left[\left({{}_{p_0+1}{}^{\mathrm{RL}}\Delta }^{-\mu }h\right)(x+\mu )-k_0\frac{{(x-p_0-2+\mu )}^{\underline{\mu -1}}}{\Gamma \left(\mu \right)}\right]=\frac{-1}{z\left(x\right)}\left[\frac{1}{\Gamma \left(\mu \right)}\sum _{r=p_0+1}^x{(x+\mu -\sigma \left(r\right))}^{\mu -1}h\left(r\right)+k_0\frac{{(x-p_0-2+\mu )}^{\underline{\mu -1}}}{\Gamma \left(\mu \right)}\right].$$ By summing both sides ${\sum }_{s=p_0+1}^x$ we have $$y\left(x\right)=-\sum _{s=p_0+1}^x\left[\sum _{r=p_0+1}^s\frac{{(s+\mu -\sigma \left(r\right))}^{\underline{\mu -1}}}{\Gamma \left(\mu \right)}h\left(r\right)+k_0\frac{{(s-p_0-2+\mu )}^{\underline{\mu -1}}}{\Gamma \left(\mu \right)z\left(s\right)}\right].$$ We change the order of sums to have

$$y\left(x\right)=-\sum _{r=p_0+1}^{x}h\left(x\right)\sum _{s=r}^x\frac{{(s+\mu -\sigma \left(r\right))}^{\underline{\mu -1}}}{\Gamma \left(\mu \right)z\left(s\right)}-k_0\sum _{s=p_0+1}^x\frac{{(s-p_0-2+\mu )}^{\underline{\mu -1}}}{\Gamma \left(\mu \right)z\left(s\right)}=-\sum _{r=p_0+1}^{x}h\left(r\right)\phi (x,\bar{\sigma }\left(r\right))-k_0\phi (x,p_0).$$ If we let $x=m$ and solve it for $k_0$ , then we have $$k_0=\frac{-\sum _{r=p_0+1}^{m}h\left(r\right)\phi (m,\bar{\sigma }\left(r\right))}{\phi (m,p_0)}.$$ Substituting $k_0$ into (4.4) to obtain $$y\left(x\right)=-\sum _{r=p_0+1}^{x}h\left(r\right)\phi (x,\bar{\sigma }\left(r\right))+\frac{\phi (x,p_0)}{\phi (m,p_0)}\sum _{r=p_0+1}^{m}h\left(r\right)\phi (m,\bar{\sigma }\left(r\right))=-\sum _{r=p_0+1}^{x}h\left(r\right)\phi (x,\bar{\sigma }\left(r\right))+\frac{\phi (x,p_0)}{\phi (m,p_0)}\sum _{r=p_0+1}^{x}h\left(r\right)\phi (m,\bar{\sigma }\left(r\right))+\frac{\phi (x,p_0)}{\phi (m,p_0)}\sum _{r=x+1}^{m}h\left(r\right)\phi (m,\bar{\sigma }\left(r\right))=\sum _{r=p_0+1}^{x}h\left(r\right)\left[\frac{\phi (m,\bar{\sigma }\left(r\right))}{\phi (m,p_0)}\phi (x,p_0)-\phi (x,\bar{\sigma }\left(r\right))\right]+\sum _{r=x+1}^{m}h\left(r\right)\left[\frac{\phi (m,\bar{\sigma }\left(r\right))}{\phi (m,p_0)}\phi (x,p_0)\right]=\sum _{r=p_0+1}^{m}h\left(r\right)G(x,r),$$ where $$G(x,r)=\left\{\begin{array}{c}\frac{\phi (m,\bar{\sigma }\left(r\right))}{\phi (m,p_0)}\phi (x,p_0),x\le r-1,\hfill \\ \frac{\phi (m,\bar{\sigma }\left(r\right))}{\phi (m,p_0)}\phi (x,p_0)-\phi (x,\bar{\sigma }\left(r\right)),x\ge r.\hfill \end{array}\right.$$ Consequently, any solution $y$ of the BVP (4.1) is necessarily given by (4.2) as we just established. Furthermore, the uniqueness of $y$ can follow from Theorem 3.1. □

Lemma 4.1

The fractional self-adjoint BVP (4.5) has only the trivial solution iff $$\Lambda =\frac{{\xi }_1{\xi }_2}{z(p_0+1)}+{\xi }_0{\xi }_2\sum _{s=p_0+2}^m\frac{{(s-p_0-2+\mu )}^{\underline{\mu -1}}}{\Gamma \left(\mu \right)z\left(s\right)}+\frac{{\xi }_0{\xi }_3{(m-p_0-2+\mu )}^{\underline{\mu -1}}}{\Gamma \left(\mu \right)z\left(m\right)}\ne 0.$$

Proof

If we consider $-\left({{}_{p_0+1}{}^{\mathrm{RL}}\Delta }^{{\xi }_0}\left(z\nabla y\right)\right)\left(x\right)=0$ , then it follows from Lemma 2.1 that $$\left(z\nabla y\right)\left(x\right)=k_0\frac{{(x-p_0-2+\mu )}^{\underline{\mu -1}}}{\Gamma \left(\mu \right)},$$ that is,

$$\left(\nabla y\right)\left(x\right)=k_0\frac{{(x-p_0-2+\mu )}^{\underline{\mu -1}}}{\Gamma \left(\mu \right)z\left(x\right)}.$$ It can be rewritten as follows $$y\left(x\right)-y\left(p_0\right)=\sum _{s=p_0+1}^x{k}_0\frac{{(s-p_0-2+\mu )}^{\underline{\mu -1}}}{\Gamma \left(\mu \right)z\left(s\right)}.$$ Let us set $y\left(p_0\right)=k_1$ to have

(4.7)

$$y\left(x\right)=k_0\sum _{s=p_0+1}^x\frac{{(s-p_0-2+\mu )}^{\underline{\mu -1}}}{\Gamma \left(\mu \right)z\left(s\right)}+k_1.$$ We can find the values of $k_0$ and $k_1$ by considering BCs. By computing both Eqs. (4.6) and (4.7) at $x=p_0+1$ , we have $$y(p_0+1)=\frac{k_0}{z(p_0+1)}+k_1,\left(\nabla y\right)(p_0+1)=\frac{k_0}{z(p_0+1)}.$$ Thus, the BC ${\xi }_{0}y(p_0+1)-{\xi }_1\nabla y(p_0+1)=0$ gives

(4.8)

$$k_0\left(\frac{{\xi }_0-{\xi }_1}{z(p_0+1)}\right)+k_1{\xi }_0=0.$$ Again, by evaluating Eqs. (4.6) and (4.7) at $x=m$ , we see that $$y\left(m\right)=\sum _{s=p_0+1}^m{k}_0\frac{{(s-p_0-2+\mu )}^{\underline{\mu -1}}}{\Gamma \left(\mu \right)z\left(s\right)}+k_1,\nabla y\left(m\right)=k_0\frac{{(m-p_0-2+\mu )}^{\mu -1}}{\Gamma \left(\mu \right)z\left(m\right)}.$$ Therefore, the BC ${\xi }_{2}y\left(m\right)+{\xi }_3\nabla y\left(m\right)=0$ becomes

(4.9)

$$k_0\left({\xi }_2\sum _{s=p_0+1}^m\frac{{(s-p_0-2+\mu )}^{\underline{\mu -1}}}{\Gamma \left(\mu \right)z\left(s\right)}+{\xi }_3\frac{{(m-p_0-2+\mu )}^{\underline{\mu -1}}}{\Gamma \left(\mu \right)z\left(m\right)}\right)+k_1{\xi }_2=0.$$ It is known that the system of Eqs. (4.8) and (4.9) (for $k_0$ and $k_1$ ) has only the trivial solution iff the determinant of the system $$\Lambda ≔-\left|\begin{array}{cc}\frac{{\xi }_0-{\xi }_1}{z(p_0+1)}\hfill & \hfill {\xi }_0\hfill \\ {\xi }_2\sum _{s=p_0+1}^m\frac{{(s-p_0-2+\mu )}^{\underline{\mu -1}}}{\Gamma \left(\mu \right)z\left(s\right)}+{\xi }_3\frac{{(m-p_0-2+\mu )}^{\underline{\mu -1}}}{\Gamma \left(\mu \right)z\left(s\right)}\hfill & \hfill {\xi }_2\hfill \end{array}\right|,$$ is not equal to $0$ . This equals to $$\Lambda =\frac{{\xi }_0{\xi }_2}{z(p_0+1)}-\frac{{\xi }_1{\xi }_2}{z(p_0+1)}-{\xi }_0{\xi }_2\sum _{s=p_0+1}^m\frac{{(s-p_0-2+\mu )}^{\underline{\mu -1}}}{\Gamma \left(\mu \right)z\left(s\right)}-{\xi }_0{\xi }_2\frac{{(m-p_0-2+\mu )}^{\underline{\mu -1}}}{\Gamma \left(\mu \right)z\left(m\right)}=-\frac{{\xi }_1{\xi }_2}{z(p_0+1)}-{\xi }_0{\xi }_2\sum _{s=p_0+2}^m\frac{{(s-p_0-2+\mu )}^{\underline{\mu -1}}}{\Gamma \left(\mu \right)z\left(s\right)}-\frac{{\xi }_0{\xi }_3{(m-p_0-2+\mu )}^{\underline{\mu -1}}}{\Gamma \left(\mu \right)z\left(m\right)}.$$ Consequently, we found that $k_0$ and $k_1$ are not both equal to $0$ . That is, $y$ is nontrivial iff $\Lambda \ne 0$ . The proof is done. □