Now, let us establish Green’s function for the following two-point BVP

$$\begin{aligned}& ^{c}D_{+0}^{\zeta }u(t)+h(t)=0,\quad t \in [{c,d}], \end{aligned}$$

(3.1)

$$\begin{aligned}& u(c)=u'(c)=0,\quad u(d)=0 \end{aligned}$$

(3.2)

with \(c \neq \beta \), \(c\neq d\) and afterwards, supposing that the solution of the following three-point BVP

$$\begin{aligned}& ^{c}D_{+0}^{\zeta }y(t)+h(t)=0,\quad t \in [{c,d}], \end{aligned}$$

(3.3)

$$\begin{aligned}& y(c)=y'(c)=0,\qquad y(d)=ky(\beta ) \end{aligned}$$

(3.4)

can be stated as follows

$$\begin{aligned} y(t)=u(t)+\bigl(\lambda _{0}+\lambda _{1}t+\lambda _{2}t^{2}\bigr)u(\beta ), \end{aligned}$$

where \(\lambda _{0}\), \(\lambda _{1}\), and \(\lambda _{2}\) are constants that will be specified later. We will estimate Green’s function for (3.3) and (3.4), respectively.

### Proposition 3.1

*If* \(h:[{c,d}] \to \mathbb{R}\) *is continuous mapping*, *then BVP* (3.1) *and* (3.2) *has a unique solution given by*

$$\begin{aligned} u(t)= \int _{c}^{t}{ \biggl[ \frac{(c-t)^{2} (d-s)^{\zeta -1}}{\Gamma (\zeta )(c-d)^{2}}- \frac{(t-s)^{\zeta -1}}{\Gamma (\zeta )} \biggr]h(s)} \,ds + \int _{t}^{d}{ \biggl[ \frac{(c-t)^{2} (d-s)^{\zeta -1}}{\Gamma (\zeta )(c-d)^{2}} \biggr]h(s)} \,ds , \end{aligned}$$

*which can be expressed in compact form as*

$$\begin{aligned} u(t)= \int _{c}^{d}{R(t,s)h(s)\,ds }, \end{aligned}$$

*where*

$$ R(t,s)= \textstyle\begin{cases}\frac{(c-t)^{2} (d-s)^{\zeta -1}}{\Gamma (\zeta )(c-d)^{2}}- \frac{(t-s)^{\zeta -1}}{\Gamma (\zeta )},& {c\le s \le t \le d,} \\ \frac{(c-t)^{2} (d-s)^{\zeta -1}}{\Gamma (\zeta )(c-d)^{2}},& c \le t \le s \le d. \end{cases} $$

(3.5)

### Proof

It is well known that Problem (3.1) and (3.2) is similar to solving the integral equation

$$\begin{aligned} u(t)=c_{1}+c_{2}(t-c)+c_{3}(t-c)^{2}- \frac{1}{\Gamma (\zeta )} \int _{c}^{t}(t-s)^{\zeta -1}h(s)\,ds , \end{aligned}$$

where \(c_{1}\), \(c_{2}\), and \(c_{3}\) are some real constants. Using boundary conditions given in (1.4), we can obtain

$$\begin{aligned}& c_{1}=\frac{c^{2}}{\Gamma (\zeta )(c-d)^{2}} \int _{c}^{d} (d-s)^{ \zeta -1}h(s) \,ds , \\& c_{2}=-\frac{2c}{\Gamma (\zeta )(c-d)^{2}} \int _{c}^{d} (d-s)^{ \zeta -1}h(s)\,ds , \\& c_{3}=\frac{1}{\Gamma (\zeta )(c-d)^{2}} \int _{c}^{d} (d-s)^{\zeta -1}h(s) \,ds . \end{aligned}$$

Thus, we get

$$\begin{aligned} u(t) =& \int _{c}^{d} \frac{(c-t)^{2} (d-s)^{\zeta -1}}{\Gamma (\zeta )(c-d)^{2} } h(s)\,ds - \int _{c}^{t} \frac{(t-s)^{\zeta -1}}{\Gamma (\zeta ) } h(s)\,ds \\ =& \int _{c}^{t} \frac{(c-t)^{2} (d-s)^{\zeta -1}}{\Gamma (\zeta )(c-d)^{2} } h(s)\,ds + \int _{t}^{d} \frac{(c-t)^{2} (d-s)^{\zeta -1}}{\Gamma (\zeta )(c-d)^{2} } h(s)\,ds \\ &{} - \int _{c}^{t} \frac{(t-s)^{\zeta -1}}{\Gamma (\zeta ) }h(s)\,ds \\ =& \int _{c}^{t} \biggl[ \frac{(c-t)^{2} (d-s)^{\zeta -1}}{\Gamma (\zeta )(c-d)^{2}}- \frac{(t-s)^{\zeta -1}}{\Gamma (\zeta )} \biggr]h(s)\,ds \\ &{}+ \int _{t}^{d} \biggl[ \frac{(c-t)^{2} (d-s)^{\zeta -1}}{\Gamma (\zeta )(c-d)^{2}} \biggr]h(s)\,ds . \end{aligned}$$

(3.6)

□

The unique result exists from the assumption that the completely homogeneous BVP has only the trivial solution. So Proposition 3.1 has been proved.

### Proposition 3.2

*Let* \(h:[{c,d}] \to \mathbb{R}\) *be a continuous mapping*, *if* \(k(c-\beta )^{2} \neq (c-d)^{2}\) *and* \(c\neq \beta \), \(c\neq d\), *then BVP* (3.3) *and* (3.4) *has a unique solution given by*

$$\begin{aligned} y(t)=u(t)+\frac{k(c-t)^{2}}{(c-d)^{2}-k(c-\beta )^{2}}u(\beta ). \end{aligned}$$

*The solution can be written further as*

$$\begin{aligned} y(t)= \int _{c}^{d}G(t,s)h(s)\,ds , \end{aligned}$$

*where*

$$ G(t,s)=R(t,s)+\frac{k(c-t)^{2}}{(c-d)^{2}-k(c-\beta )^{2}}R(\beta ,s). $$

(3.7)

### Proof

Let

$$\begin{aligned} y(t)=u(t)+\bigl(\lambda _{0}+\lambda _{1}t+\lambda _{2}t^{2}\bigr)u(\beta ), \end{aligned}$$

where \(\lambda _{0}\), \(\lambda _{1}\), \(\lambda _{2}\) are constants that will be determined using boundary conditions given in (3.4) and

$$\begin{aligned} u(t)= \int _{c}^{d}R(t,s)h(s)\,ds . \end{aligned}$$

Therefore, to compute \(\lambda _{0}\), \(\lambda _{1}\), \(\lambda _{2}\), we proceed as

$$\begin{aligned} y(c) =&u(c)+\bigl(\lambda _{0}+\lambda _{1}c+\lambda _{2}c^{2}\bigr)u(\beta )=\bigl( \lambda _{0}+\lambda _{1}c+\lambda _{2}c^{2} \bigr)u(\beta ), \\ y'(c) =&u'(c)+(\lambda _{1}+2\lambda _{2}c)u(\beta )=(\lambda _{1}+2 \lambda _{2}c)u(\beta ), \\ y(d) =&u(d)+\bigl(\lambda _{0}+\lambda _{1}d+\lambda _{2}d^{2}\bigr)u(\beta )=\bigl( \lambda _{0}+\lambda _{1}d+\lambda _{2}d^{2} \bigr)u(\beta ), \\ y(\beta ) =&u(\beta )+\bigl(\lambda _{0}+\lambda _{1} \beta +\lambda _{2} \beta ^{2}\bigr)u(\beta )=u(\beta ) \bigl(\lambda _{0}+\lambda _{1}\beta + \lambda _{2}\beta ^{2}+1\bigr). \end{aligned}$$

We get

$$\begin{aligned} \bigl(\lambda _{0} &{}+\lambda _{1}c+\lambda _{2}c^{2}\bigr)u(\beta )=0, \\ (\lambda _{1} &{}+2\lambda _{2}c)u(\beta )=0, \\ \bigl(\lambda _{0} &{}+\lambda _{1}d+\lambda _{2}d^{2}\bigr)u(\beta )=ku(\beta ) \bigl( \lambda _{0}+\lambda _{1}\beta +\lambda _{2}\beta ^{2}+1\bigr),\vadjust{\goodbreak} \end{aligned}$$

or

$$\begin{aligned}& \lambda _{0}+\lambda _{1}c+\lambda _{2}c^{2}=0, \\& (\lambda _{1}+2\lambda _{2}c)=0, \\& (1-k)\lambda _{0}+(d-k\beta )\lambda _{1}+ \bigl(d^{2}-k\beta ^{2}\bigr) \lambda _{2}=k. \end{aligned}$$

Solving the system, we get the corresponding values as

$$\begin{aligned}& \lambda _{0}=\frac{c^{2} k}{(c-d)^{2}-k(c-\beta )^{2}},\qquad \lambda _{1}= \frac{-2c k}{(c-d)^{2}-k(c-\beta )^{2}}, \\& \lambda _{2}=\frac{k}{(c-d)^{2}-k(c-\beta )^{2}}. \end{aligned}$$

Therefore, the final solution becomes

$$\begin{aligned} y(t) =&u(t)+ \biggl(\frac{c^{2} k}{(c-d)^{2}-k(c-\beta )^{2}}- \frac{2c kt}{(c-d)^{2}-k(c-\beta )^{2}}+ \frac{k t^{2}}{(c-d)^{2}-k(c-\beta )^{2}} \biggr)u(\beta ) \\ =&u(t)+\frac{k(c-t)^{2}}{(c-d)^{2}-k(c-\beta )^{2}}u(\beta ). \end{aligned}$$

Now we derive the proof of the uniqueness. Let *z* be also a solution to (3.3) and (3.4), that is

$$\begin{aligned}& ^{c}D_{+0}^{\zeta }z(t)+h(t)=0,\quad t\in [{c,d}], \end{aligned}$$

(3.8)

$$\begin{aligned}& z(c)=z'(d)=0,\qquad z(d)=kz(\beta ). \end{aligned}$$

(3.9)

Let \(\Omega (t)=z(t)-y(t)\), \(t\in [{c, d}]\). Due to linearity property of the Caputo non-integer order derivative, we have

$$\begin{aligned} ^{c}D_{+0}^{\zeta }\Omega (t) =&^{c}D_{+0}^{\zeta }z(t)-^{c}D_{+0}^{ \zeta }y(t) =-h(t)+h(t)=0,\quad t\in [{c, d}]. \end{aligned}$$

Therefore, \(\Omega (t)=c_{1}+c_{2}t+c_{3}t^{2}\), where \(c_{1}\), \(c_{2}\), and \(c_{3}\) are constants that will be computed later. We have

$$\begin{aligned}& \Omega (c)=z(c)-y(c), \\& \Omega '(c)=z'(c)-y'(c), \\& \Omega (d)=z(d)-y(d)=kz(\beta )-ky(\beta )=k\bigl(z(\beta )-y(\beta )\bigr)=k \Omega (\beta ), \end{aligned}$$

or

$$\begin{aligned} \Omega (c) =&c_{1}+c_{2}c+c_{3}c^{2}, \\ \Omega '(c) =&c_{2}+2c_{3}c, \\ \Omega (d) =&c_{1}+c_{2}d+c_{3}d^{2}=k \bigl(c_{1}+c_{2}\beta +c_{3}\beta ^{2}\bigr)=k \Omega (\beta ). \end{aligned}$$

We get the following homogeneous system

$$\begin{aligned} \textstyle\begin{cases} {c_{1}+c_{2}a+c_{3}c^{2}=0,} \\ {c_{2}+2c_{3}c=0,} \\ {c_{1}(1-k)+c_{2}(d-k\beta )+c_{3} \bigl(d^{2}-k\beta ^{2}\bigr)=0,} \end{cases}\displaystyle \end{aligned}$$

with determinant

$$ \begin{vmatrix} 1 & c & c^{2} \\ 0 & 1 & 2c \\ 1-k & d-k\beta & d^{2}-k\beta ^{2} \end{vmatrix}=(c-d)^{2}-k(c-\beta )^{2}\neq 0.$$

Therefore, the homogeneous system contains only the trivial solution, and hence \(\Omega (t)\equiv 0\), \(t\in [{c,d}]\) or \(y(t)\equiv z(t)\), \(t\in [{c, d}]\). Thus, the proof is completed. □