Consider the linear two-point boundary value problem

${\mu}_{2}(x){y}^{\u2033}+{\mu}_{1}(x){y}^{\prime}+\lambda {(x)}_{0}^{C}{D}_{x}^{\alpha}y+{\mu}_{0}(x)y=f(x),\phantom{\rule{1em}{0ex}}0<x<1,0<\alpha <1,$

(3.1)

with the boundary conditions

where ${}_{0}{}^{C}D_{x}$ is the Caputo fractional derivative operator.

An approximate solution for

$y(x)$ is represented by the formula

${y}_{n}(x)=\sum _{k=-M}^{M}{c}_{k}{S}_{k}(x),\phantom{\rule{1em}{0ex}}n=2M+1,$

(3.2)

where

${S}_{k}(x)$ is the function

$S(k,h)o\varphi (x)$ defined in (2.9) for some fixed step size

*h*. The unknown coefficients

${c}_{k}$ in (3.2) are determined by orthogonalizing the residual with respect to the basis functions,

*i.e.*,

$\u3008{\mu}_{2}(x){y}^{\u2033},{S}_{k}\u3009+\u3008{\mu}_{1}(x){y}^{\prime},{S}_{k}\u3009+\u3008\lambda {(x)}_{0}^{C}{D}_{x}^{\alpha}y,{S}_{k}\u3009+\u3008{\mu}_{0}(x)y,{S}_{k}\u3009=\u3008f(x),{S}_{k}\u3009.$

(3.3)

The inner product used for the sinc-Galerkin method is defined by

$\u3008f,\eta \u3009={\int}_{a}^{b}f(x)\eta (x)w(x)\phantom{\rule{0.2em}{0ex}}dx,$

where

$w(x)$ is a

*weight function* and it is convenient to take

$w(x)=\frac{1}{{\varphi}^{\prime}(x)}$

for the case of second-order problems.

A complete discussion on the choice of the weight function can be found in [17].

**Lemma 3.1** [19]

*Let* *ϕ* *be the conformal one*-

*to*-

*one mapping of the simply connected domain* ${D}_{E}$ *onto* ${D}_{S}$ *given by* (2.8).

*Then* $\begin{array}{c}{\delta}_{jk}^{(0)}=[S(j,h)o\varphi (x)]{|}_{x={x}_{k}}=\{\begin{array}{ll}1,& j=k,\\ 0,& j\ne k,\end{array}\hfill \\ {\delta}_{jk}^{(1)}=h\frac{d}{d\varphi}[S(j,h)o\varphi (x)]{|}_{x={x}_{k}}=\{\begin{array}{ll}0,& j=k,\\ \frac{{(-1)}^{k-j}}{k-j},& j\ne k,\end{array}\hfill \\ {\delta}_{jk}^{(2)}={h}^{2}\frac{{d}^{2}}{d{\varphi}^{2}}[S(j,h)o\varphi (x)]{|}_{x={x}_{k}}=\{\begin{array}{ll}-\frac{{\pi}^{2}}{3},& j=k,\\ \frac{-2{(-1)}^{k-j}}{{(k-j)}^{2}},& j\ne k.\end{array}\hfill \end{array}$

The method of approximating the integrals in (3.3) begins by integrating by parts to transfer all derivatives from *y* to ${S}_{k}$. The following theorems, which can easily be proved by using Lemma 3.1 and Definition 2.2, are used to solve equation (3.1).

**Theorem 3.2** [20]

*The following relations hold*:

$\u3008{\mu}_{2}(x){y}^{\u2033},{S}_{k}\u3009\cong h\sum _{k=-M}^{M}\sum _{i=0}^{2}\frac{y({x}_{k})}{{\varphi}^{\prime}({x}_{k}){h}^{i}}{\delta}_{jk}^{(i)}{g}_{2,i},$

(3.4)

$\u3008{\mu}_{1}(x){y}^{\prime},{S}_{k}\u3009\cong -h\sum _{k=-M}^{M}\sum _{i=0}^{1}\frac{y({x}_{k})}{{\varphi}^{\prime}({x}_{k}){h}^{i}}{\delta}_{jk}^{(i)}{g}_{1,i}$

(3.5)

*and*
$\u3008G,{S}_{k}\u3009\cong h\frac{G({x}_{k})w({x}_{k})}{{\varphi}^{\prime}({x}_{k})},$

(3.6)

*where*
$\begin{array}{r}{g}_{2,2}=({\mu}_{2}w){\left({\varphi}^{\prime}\right)}^{2},\phantom{\rule{2em}{0ex}}{g}_{2,1}=({\mu}_{2}w){\varphi}^{\u2033}+2{({\mu}_{2}w)}^{\prime}{\varphi}^{\prime},\phantom{\rule{2em}{0ex}}{g}_{2,0}={({\mu}_{2}w)}^{\u2033},\\ {g}_{1,1}=({\mu}_{1}w){\varphi}^{\prime},\phantom{\rule{2em}{0ex}}{g}_{1,0}={({\mu}_{1}w)}^{\prime}\end{array}$

*and*
$G={\mu}_{0}y\phantom{\rule{1em}{0ex}}\mathit{\text{or}}\phantom{\rule{1em}{0ex}}f(x).$

**Theorem 3.3** *For* $0<\alpha <1$,

*the following relation holds*:

$\begin{array}{c}\u3008\lambda {(x)}_{0}^{C}{D}_{x}^{\alpha}y(x),{S}_{k}\u3009\hfill \\ \phantom{\rule{1em}{0ex}}\cong -\frac{{h}_{M}}{\mathrm{\Gamma}(1-\alpha )}\sum _{k=-M}^{M}\frac{y({x}_{k})}{{\varphi}^{\prime}({x}_{k})}\frac{d}{dx}\left[{h}_{L}\sum _{r=-L}^{L}\frac{{({x}_{r}-x)}^{-\alpha}K({x}_{r})}{{\xi}^{\prime}({x}_{r})}\right]{|}_{x={x}_{k}},\phantom{\rule{1em}{0ex}}-M\le j\le M,\hfill \end{array}$

(3.7)

*where* $K(x)=\lambda (x){S}_{k}(x)w(x)$, $\xi (t)=ln(\frac{t-x}{1-t})$.

*Proof* The inner product with sinc basis element is given by

$\u3008\lambda {(x)}_{0}^{C}{D}_{x}^{\alpha}y(x),{S}_{k}\u3009={\int}_{0}^{1}{(\lambda (x){S}_{k}(x)w(x))}_{0}^{C}{D}_{x}^{\alpha}y(x)\phantom{\rule{0.2em}{0ex}}dx.$

Using Definition 2.2, we can write

${\int}_{0}^{1}{(\lambda (x){S}_{k}(x)w(x))}_{0}^{C}{D}_{x}^{\alpha}y(x)\phantom{\rule{0.2em}{0ex}}dx={\int}_{0}^{1}y{(x)}_{x}^{R}{D}_{1}^{\alpha}(K(x))\phantom{\rule{0.2em}{0ex}}dx,$

(3.8)

where

$K(x)=\lambda (x){S}_{k}(x)w(x)$. By the definition of the Riemann-Liouville fractional derivative given in (2.2), we have

${}_{x}{}^{R}D_{1}^{\alpha}(K(x))=-\frac{1}{\mathrm{\Gamma}(1-\alpha )}\frac{d}{dx}{\int}_{x}^{1}{(t-x)}^{-\alpha}K(t)\phantom{\rule{0.2em}{0ex}}dt.$

(3.9)

We will use the sinc quadrature rule given with equation (

2.13) to compute it because the integral given in (3.9) is divergent on the interval

$[x,1]$. For this purpose, a conformal map and its inverse image that denotes the sinc grid points are given by

$\xi (t)=ln\left(\frac{t-x}{1-t}\right)$

and

${x}_{r}={\xi}^{-1}(r{h}_{L})=\frac{{e}^{r{h}_{L}}+x}{1+{e}^{r{h}_{L}}},$

respectively. Then, according to equality (2.13), we write

$-\frac{1}{\mathrm{\Gamma}(1-\alpha )}\frac{d}{dx}{\int}_{x}^{1}{(t-x)}^{-\alpha}K(t)\phantom{\rule{0.2em}{0ex}}dt\cong -\frac{1}{\mathrm{\Gamma}(1-\alpha )}\frac{d}{dx}\left[{h}_{L}\sum _{r=-L}^{L}\frac{{({x}_{r}-x)}^{-\alpha}K({x}_{r})}{{\xi}^{\prime}({x}_{r})}\right],$

where

${h}_{L}=\pi /\sqrt{L}$. Thus, the right-hand side of (3.8) can be rewritten as follows:

${\int}_{0}^{1}y{(x)}_{x}^{R}{D}_{1}^{\alpha}(K(x))\phantom{\rule{0.2em}{0ex}}dx\cong -\frac{1}{\mathrm{\Gamma}(1-\alpha )}{\int}_{0}^{1}(y(x)\frac{d}{dx}\left[{h}_{L}\sum _{r=-L}^{L}\frac{{({x}_{r}-x)}^{-\alpha}K({x}_{r})}{{\xi}^{\prime}({x}_{r})}\right])\phantom{\rule{0.2em}{0ex}}dx.$

(3.10)

To apply the sinc quadrature rule given in (2.13) on the right-hand side of (3.10), a conformal map and its inverse image are given by

$\varphi (t)=ln\left(\frac{t}{1-t}\right)$

and

${x}_{k}={\varphi}^{-1}(kh)=\frac{{e}^{kh}}{1+{e}^{kh}},$

respectively. Consequently, when the rule is applied, it is obtained

$\begin{array}{c}\u3008\lambda {(x)}_{0}^{C}{D}_{x}^{\alpha}y(x),{S}_{k}\u3009\hfill \\ \phantom{\rule{1em}{0ex}}\cong -\frac{h}{\mathrm{\Gamma}(1-\alpha )}\sum _{k=-M}^{M}\frac{y({x}_{k})}{{\varphi}^{\prime}({x}_{k})}\frac{d}{dx}\left[{h}_{L}\sum _{r=-L}^{L}\frac{{({x}_{r}-x)}^{-\alpha}K({x}_{r})}{{\xi}^{\prime}({x}_{r})}\right]{|}_{x={x}_{k}},\phantom{\rule{1em}{0ex}}-M\le j\le M,\hfill \end{array}$

where $h=\pi /\sqrt{M}$. This completes the proof. □

Replacing each term of (3.3) with the approximation defined in (3.4)-(3.7), replacing $y({x}_{k})$ with ${c}_{k}$ and dividing by *h*, we obtain the following theorem.

**Theorem 3.4** *If the assumed approximate solution of boundary value problem* (3.1)

*is* (3.2),

*then the discrete sinc*-

*Galerkin system for the determination of the unknown coefficients* ${\{{c}_{k}\}}_{k=-M}^{M}$ *is given by* $\begin{array}{c}\sum _{k=-M}^{M}\{\sum _{i=0}^{2}\frac{1}{{h}^{i}}{\delta}_{jk}^{(i)}\frac{{g}_{2,i}({x}_{k})}{{\varphi}^{\prime}({x}_{k})}{c}_{k}-\sum _{i=0}^{1}\frac{1}{{h}^{i}}{\delta}_{jk}^{(i)}\frac{{g}_{1,i}({x}_{k})}{{\varphi}^{\prime}({x}_{k})}{c}_{k}\hfill \\ \phantom{\rule{2em}{0ex}}-\frac{1}{\mathrm{\Gamma}(1-\alpha )}\frac{{c}_{k}}{{\varphi}^{\prime}({x}_{k})}\frac{d}{dx}\left[{h}_{L}\sum _{r=-L}^{L}\frac{{({x}_{r}-x)}^{-\alpha}K({x}_{r})}{{\xi}^{\prime}({x}_{r})}\right]{|}_{x={x}_{k}}\}+\frac{{\mu}_{0}({x}_{j})w({x}_{j})}{{\varphi}^{\prime}({x}_{j})}{c}_{j}\hfill \\ \phantom{\rule{1em}{0ex}}=\frac{f({x}_{j})w({x}_{j})}{{\varphi}^{\prime}({x}_{j})},\phantom{\rule{1em}{0ex}}-M\le j\le M.\hfill \end{array}$