A Reliable Method for Solving Fractional Sturm–Liouville Problems

Abstract

In this paper, a reliable method for solving fractional Sturm–Liouville problem based on the operational matrix method is presented. Some of our numerical examples are presented.

1. Introduction

The Sturm–Liouville theory plays an important role for the development of spectral methods and the theory of self-adjoint operators [1]. Several applications on SLPs are studied as boundary-value problems [2]. The Sturm–Liouville eigenvalue problem has played an important role in modeling many physical problems. The theory of the problem is well developed and many results have been obtained concerning the eigenvalues and corresponding eigenfunctions. It should be noted that, since finding analytical solutions for this problem is an extremely difficult task, several numerical algorithms have been developed to seek approximate solutions. However, to date, mostly integer-order differential operators in SLPs have been used, and such operators do not include any fractional differential operators. Fractional calculus is a theory which unifies and generalizes the notions of integer-order differentiation and integration to any real order [3,4,5].

Recently, the fractional Sturm–Liouville problems were formulated in [6,7]. Authors in these papers considered several types of the fractional Sturm–Liouville equations and they investigated the eigenvalues and eigenfunctions properties of the fractional Sturm–Liouville operators.

Djrbashian [8] studied the existence of a solution to the fractional boundary value problem. In [9], authors discussed the aforementioned relation between eigenvalues and zeros of Mittag–Leffler function. In [10], they used the Homotopy Analysis method while, in [11], they used the fractional differential transform method to compute the eigenvalues. In [12], researchers used the Fourier series to solve this problem while, in [13,14], they used the method of Haar wavelet operational matrix. In [15,16,17,18,19], researchers extended the scope of some spectral properties of fractional Sturm–Liouville problems. Variational methods and Inverse Laplace transform method were applied in [20,21], respectively. Recently, in [22], authors constructed numerical schemes using radial basis functions while, in [23], they used Galerkin finite element method for such problems. Greenberg and Marletta [24,25] developed their own code using Theta Matrices (SLEUTH). In [26], researchers implemented the iterated variation method.

In this article, we present a numerical technique for solving class of FSLPs of the form

$$D^{\gamma }\left[f\left(t\right)D^{\gamma }y\left(t\right)\right]+\mu g\left(t\right)y\left(t\right)=h\left(t\right),0\le t\le 1,\frac{1}{2}<\gamma \le 1$$

(1)

subject to

$$\begin{array}{ccc}\hfill c_{0}y\left(0\right)+c_1{D}^{\gamma }y\left(0\right)& =& 0,c_0^2+c_1^2>0,\hfill \end{array}$$

(2)

$$\begin{array}{ccc}\hfill c_{2}y\left(1\right)+c_3{D}^{\gamma }y\left(1\right)& =& 0,c_2^2+c_3^2>0,\hfill \end{array}$$

(3)

where $c_0,c_1,c_{2}and{c}_3$ are constants such that $\det \left(\begin{array}{cc}{c}_0& c_1\\ c_2& c_3\end{array}\right)\ne 0$, $f\left(t\right),g\left(t\right),$ $h\left(t\right)$ are continuous functions with $f\left(t\right),g\left(t\right)>0$ for all $t\in [0,1]$, and $D^{\gamma }$ is the Caputo derivative.

Next, we present some results related to the Caputo fractional derivative, as well as the definition of the fractional-order functions.

Definition 1.

The Rimann–Liouville fractional integral operator $I^{\gamma }$ of order $\gamma >0$ on $L_1[0,1]$ is given by

$$\begin{array}{ccc}\hfill I^{\alpha }y\left(t\right)& =& \frac{1}{\mathrm{\Gamma }\left(\gamma \right)}\underset{0}{\overset{t}{\int }}\frac{y\left(s\right)}{{(t-s)}^{1-\gamma }}ds,\hfill \\ \hfill I^{0}y\left(t\right)& =& y\left(t\right),\hfill \end{array}$$

where $\mathrm{\Gamma }\left(\gamma \right)$ is the Euler Gamma functionv(see [5,27]).

For any $\gamma$, $\zeta \ge 0$, and $\zeta$ > $-1$, $I^{\alpha }$ existsfor any $t\in [0,1]$ and

$$I^{\gamma }{t}^{\zeta }=\frac{\mathrm{\Gamma }(\zeta +1)}{\mathrm{\Gamma }(\zeta +\gamma +1)}{t}^{\zeta +\gamma }.$$

(4)

Definition 2.

The Caputo fractional derivative of order γ is defined by

$$D^{\gamma }y\left(t\right)=I^{n-\gamma }{D}^{l}y\left(t\right)=\frac{1}{\mathrm{\Gamma }(l-\gamma )}\underset{0}{\overset{t}{\int }}\frac{y^{\left(l\right)}\left(s\right)}{{(t-s)}^{\gamma -l+1}}ds,$$

provided that the integral exists, where $l=\left[\gamma \right]+1,$ $\left[\gamma \right]$ is the integer part of the positive real number $\gamma ,t>0$.

For $y\in L_1[0,1]$ and $\gamma \ge 0$:

$$I^{\gamma }{D}^{\gamma }y\left(t\right)=y\left(t\right)-\sum _{r=0}^{l-1}{y}^{\left(r\right)}\left(0^{+}\right)\frac{t^r}{r!},$$

(5)

Let ${\Delta }_n$ be defined by

$${\Delta }_n=Span\{1,t^{\gamma },t^{2\gamma },\dots ,t^{n\gamma }\}.$$

The inner product on the set ${\Delta }_n$ is given by

$$\left(f\left(t\right),g\left(t\right)\right)={}_0^{1}f\left(t\right)g\left(t\right)dt.$$

Theorem 1.

The sequence of functions defined as follows are orthogonal:

$$y_i\left(t\right)=(t^{\gamma }-a_i)y_{i-1}\left(t\right)-b_i{y}_{i-2}\left(t\right),i=2,3,\dots$$

(6)

with $y_0\left(t\right)=1,y_1\left(t\right)=t^{\gamma }-a_1,$ and

$$a_i=\frac{\left(t^{\gamma }{y}_{i-1}\left(t\right),g\left(t\right)\right)}{\left(y_{i-1}\left(t\right),y_{i-1}\left(t\right)\right)},b_i=\frac{\left(t^{\gamma }{y}_{i-1}\left(t\right),y_{i-2}\left(t\right)\right)}{\left(y_{i-2}\left(t\right),y_{i-2}\left(t\right)\right)}.$$

Proof. 

For $i=1,$

$$\begin{array}{cc}\hfill \left(y_1\left(t\right),y_0\left(t\right)\right)& =\left(t^{\gamma }-a_1,y_0\left(t\right)\right)\hfill \\ & =\left(t^{\alpha },1\right)-\frac{\left(t^{\alpha },1\right)}{\left(1,1\right)}\left(1,1\right)=0.\hfill \end{array}$$

Assume the result of the theorem is true for $i>1$. Then, for any $j\in \{0,1,\dots ,i-2\},$ we have

$$\begin{array}{ll}\left(y_{i+1}\left(t\right),y_j\left(t\right)\right)\hfill & =\left((t^{\gamma }-a_{i+1})y_i\left(t\right)-b_{i+1}{y}_{i-1}\left(t\right),y_j\left(t\right)\right)\hfill \\ & =\left(t^{\gamma }{y}_i\left(t\right),y_j\left(t\right)\right)-a_{i+1}\left(y_i\left(t\right),y_j\left(t\right)\right)-b_{i+1}\left(y_{i-1}\left(t\right),y_j\left(t\right)\right)\hfill \\ & =\left(t^{\gamma }{y}_i\left(t\right),y_j\left(t\right)\right)\hfill \\ & =\left(y_i\left(t\right),t^{\gamma }{y}_j\left(t\right)\right)\hfill \\ & =\left(y_i\left(t\right),y_{j+1}\left(t\right)+a_{j+1}{y}_j\left(t\right)+b_{j+1}{y}_{j-1}\left(t\right)\right)\hfill \\ & =\left(y_i\left(t\right),y_{j+1}\left(t\right)\right)+a_{j+1}\left(y_i\left(t\right),y_j\left(t\right)\right)+b_{j+1}\left(y_i\left(t\right),y_{j-1}\left(t\right)\right)\hfill \\ & =0.\hfill \end{array}$$

For $j=i-1$,

$$\begin{array}{ll}\left(y_{i+1}\left(t\right),y_{i-1}\left(t\right)\right)\hfill & =\left((t^{\gamma }-a_{i+1})y_i\left(t\right)-b_{i+1}{y}_{i-1}\left(t\right),y_{i-1}\left(t\right)\right)\hfill \\ & =\left(t^{\gamma }{y}_i\left(t\right),y_{i-1}\left(t\right)\right)-a_{i+1}\left(y_i\left(t\right),y_{i-1}\left(t\right)\right)-b_{i+1}(y_{i-1}\left(t\right),y_{i-1}\left(t\right))\hfill \\ & =\left(t^{\gamma }{y}_i\left(t\right),y_{i-1}\left(t\right)\right)-b_{i+1}(y_{i-1}\left(t\right),y_{i-1}\left(t\right))\hfill \\ & =\left(t^{\gamma }{y}_i\left(t\right),y_{i-1}\left(t\right)\right)-\frac{\left(t^{\gamma }{y}_i\left(t\right),y_{i-1}\left(t\right)\right)}{\left(y_{i-1}\left(t\right),y_{i-1}\left(t\right)\right)}(y_{i-1}\left(t\right),y_{i-1}\left(t\right))\hfill \\ & =0.\hfill \end{array}$$

□

2. Operational Matrices of Fractional Integration

A set of l Block Pulse Functions (BPFs) in the interval $[0,1)$ are given by $\{b_0\left(t\right),b_1\left(t\right),\dots ,b_{l-1}\left(t\right)\}$ such that

$$b_i\left(t\right)=\left\{\begin{array}{cc}1,& \frac{i}{l}\le t<\frac{i+1}{l}\\ 0,& otherwise\end{array}\right\}$$

(7)

for $i=0,1,\dots ,l-1$. The following are some of the BPFs properties

$$b_i\left(t\right)b_j\left(t\right)=\left\{\begin{array}{ll}{b}_i\left(t\right),& i=j\\ 0,& i\ne j\end{array}\right\}$$

(8)

and

$${}_0^1{b}_i\left(t\right)b_j\left(t\right)dt=\left\{\begin{array}{cc}\frac{1}{l},& i=j\\ 0,& i\ne j\end{array}\right\}.$$

(9)

If $y\in L_2[0,1]$, then

$$y\left(t\right)=Y_{l-1}^T{B}_{l-1}\left(t\right)$$

(10)

where

$$Y_{l-1}=\left[\begin{array}{c}{y}_0\\ y_1\\ ⋮\\ y_{l-1}\end{array}\right],B_{l-1}\left(t\right)=\left[\begin{array}{c}{b}_0\left(t\right)\\ b_1\left(t\right)\\ ⋮\\ b_{l-1}\left(t\right)\end{array}\right],$$

and

$$y_i=l_{\frac{i}{l}}^{\frac{i+1}{l}}y\left(t\right)dt,i=0,1,\dots l-1.$$

(11)

Theorem 2.

Let $I^{\gamma }$ be the Rimann–Liouville functional operator. Then,

$$I^{\gamma }{B}_{l-1}\left(t\right)=P_l^{\gamma }{B}_{l-1}\left(t\right)$$

(12)

where

$$P_l^{\gamma }=\frac{1}{l^{\gamma }}\frac{1}{\mathrm{\Gamma }(\gamma +2)}\left[\begin{array}{ccccc}1& {\epsilon }_1& {\epsilon }_2& \dots & {\epsilon }_{l-1}\\ 0& 1& {{\epsilon }_1}_{.}& \cdots & {\epsilon }_{l-2}\\ 0& 0& 1& \cdots & {\epsilon }_{l-3}\\ 0& 0& 0& \ddots & ⋮\\ 0& 0& 0& 0& 1\end{array}\right]$$

and ${\epsilon }_r={(r+1)}^{\gamma +1}-{2r}^{\gamma +1}+{(r-1)}^{\gamma +1},r=1:l-1.$

Proof. 

For each $i=0,1,\dots ,l-1$, we can write $I^{\gamma }{b}_i$ as

$$I^{\gamma }{b}_i{=}_{j=0}^{l-1}{c}_{ij}{b}_j\left(t\right).$$

Multiply both sides by $b_r\left(t\right),$ for $0\le r\le l-1$, then integrate both sides to get

$$\begin{array}{cc}{c}_{ir}\hfill & =r_{\frac{r}{l}}^{\frac{r+1}{l}}{I}^{\gamma }{b}_i\left(t\right)dt={\frac{r}{\mathrm{\Gamma }\left(\gamma \right)}}_{\frac{r}{l}}^{\frac{r+1}{l}}\underset{0}{\overset{t}{\int }}\frac{b_i\left(t\right)}{{(t-t)}^{1-\gamma }}dtdt.\hfill \\ & =\left\{\begin{array}{cc}0,& i>r\ge 0\\ 1,& i=r\\ {(r+1)}^{\gamma +1}-{2r}^{\gamma +1}+{(r-1)}^{\gamma +1}& i<r\le l-1\end{array}\right\}.\hfill \end{array}$$

For more details, see [28,29]. □

Theorem 3.

Let $Y_{M-1}\left(t\right)=\left[\begin{array}{c}{y}_0\left(t\right)\\ y_1\left(t\right)\\ ⋮\\ y_{M-1}\left(t\right)\end{array}\right]$. Then, there exists an $M\times l$ matrix $Q^{\gamma }$ such that

$$Y_{M-1}\left(t\right)=Q_{M\times l}^{\gamma }{B}_{l-1}\left(t\right)$$

(13)

where

$${\left(Q_{M\times l}^{\gamma }\right)}_{i,k}=l_{\frac{k}{l}}^{\frac{k+1}{l}}{y}_i\left(t\right)dt$$

for $i=0:M-1$ and $k=0:l-1$.

Proof. 

It is easy to see that $y_i\left(t\right)\in L_2[0,1),$ for each $i=0:M-1$. Using Equations (10) and (11), we get

$$Y_{M-1}\left(t\right)=Q_{M\times l}^{\gamma }{B}_{l-1}\left(t\right)$$

where

$${\left(Q_{M\times l}^{\gamma }\right)}_{i,k}=l_{\frac{k}{l}}^{\frac{k+1}{l}}{y}_i\left(t\right)dt.$$

for $i=0:M-1$ and $k=0:l-1$ which ends the proof.

From now on, let $M=l$. □

Theorem 4.

If $0<\gamma <1$, then $Q_{l\times l}^{\gamma }$ is nonsingular matrix.

Proof. 

Theorem 3 implies that

$$Y_{l-1}\left(t\right)Y_{l-1}{\left(t\right)}^T=Q_{l\times l}^{\gamma }{B}_{l-1}\left(t\right)B_{l-1}{\left(t\right)}^T{Q}_{l\times l}^{\gamma T}.$$

Integrate both sides with respect to t on (0,1) to get

$${}_0^1{Y}_{l-1}\left(t\right)Y_{l-1}{\left(t\right)}^{T}dt=Q_{l\times l}^{\gamma }\left({}_0^1{B}_{l-1}\left(t\right)B_{l-1}{\left(t\right)}^{T}dt\right)Q_{l\times l}^{\gamma T}.$$

Theorem 1 and Equation (9) yield

$$D_1=Q_{l\times l}^{\gamma }{D}_2{Q}_{l\times l}^{\gamma T}$$

(14)

where

$$D_1=\left[\begin{array}{cccc}{}_0^1{y}_0\left(t\right)y_0\left(t\right)dt& 0& \cdots & 0\\ 0& {}_0^1{y}_1\left(t\right)y_1\left(t\right)dt& \ddots & ⋮\\ ⋮& \ddots & \ddots & 0\\ 0& \cdots & 0& {}_0^1{y}_{l-1}\left(t\right)y_{l-1}\left(t\right)dt\end{array}\right]$$

and

$$D_2=\frac{1}{\gamma }\left[\begin{array}{cccc}\frac{1}{l^{\gamma }}& 0& \cdots & 0\\ 0& \frac{2^{\gamma }-1}{l^{\gamma }}& \ddots & ⋮\\ ⋮& \ddots & \ddots & 0\\ 0& \cdots & 0& \frac{l^{\gamma }-{(l-1)}^{\gamma }}{l^{\gamma }}\end{array}\right].$$

Then, $\det \left(D_1\right)>0$ and $\det \left(D_2\right)>0$. Equation (14) gives

$${\left(\det \left(Q_{l\times l}^{\gamma }\right)\right)}^2=\frac{\det \left(D_1\right)}{\det \left(D_2\right)}>0.$$

Thus, $Q_{l\times l}^{\gamma }$ is nonsingular. □

Operational Matrix of Fractional Integration

If $y\in C^1[0,1],$ then

$$y\left(t\right){=}_{k=0}^{\infty }{u}_k{y}_k\left(t\right)$$

where

$$u_k=\frac{{}_0^{1}u\left(t\right)y_k\left(t\right)dt}{{}_0^1{f}_k\left(t\right)y_k\left(t\right)dt}.$$

Approximate the function $y\left(t\right)$ by

$$U_{l-1}\left(t\right){=}_{k=0}^{l-1}u{y}_k\left(t\right)=U^T{Y}_{l-1}\left(t\right),$$

(15)

where

$$U=\left[\begin{array}{c}{u}_0\\ u_1\\ ⋮\\ u_{l-1}\end{array}\right]\mathrm{and}{Y}_{l-1}\left(t\right)=\left[\begin{array}{c}{y}_0\left(t\right)\\ y_1\left(t\right)\\ ⋮\\ y_{l-1}\left(t\right)\end{array}\right].$$

(16)

Theorem 5.

$I^{\gamma }{Y}_{l-1}\left(t\right)=H_l^{\gamma }{Y}_{l-1}\left(t\right)$

where

$$H_l^{\gamma }=Q_{l\times l}^{\gamma }{P}_l^{\gamma }{\left(Q_{l\times l}^{\gamma }\right)}^{-1}$$

Proof. 

Let $H_l^{\gamma }$ be given by

$$I^{\gamma }{Y}_{l-1}\left(t\right)=H_l^{\gamma }{Y}_{l-1}\left(t\right).$$

(17)

From Equations (13) and (17), we get

$$I^{\gamma }{Y}_{l-1}\left(t\right)=H_l^{\gamma }{Y}_{l-1}\left(t\right)=H_l^{\gamma }{Q}_{l\times l}^{\gamma }{B}_{l-1}\left(t\right)$$

(18)

and

$$\begin{array}{cc}{I}^{\gamma }{Y}_{l-1}\left(t\right)\hfill & =I^{\gamma }{Q}_{l\times l}^{\gamma }{B}_{l-1}\left(t\right)\hfill \\ & =Q_{l\times l}^{\gamma }{I}^{\gamma }{B}_{l-1}\left(t\right)\hfill \\ & =Q_{l\times l}^{\gamma }{P}_l^{\gamma }{B}_{l-1}\left(t\right).\hfill \end{array}$$

(19)

Combining Equations (18) and (19), we get

$$H_l^{\gamma }{Q}_{l\times l}^{\gamma }{B}_{l-1}\left(t\right)=Q_{l\times l}^{\gamma }{P}_l^{\gamma }{B}_{l-1}\left(t\right).$$

(20)

Therefore,

$$H_l^{\gamma }=Q_{l\times l}^{\gamma }{P}_l^{\gamma }{\left(Q_{l\times l}^{\gamma }\right)}^{-1}.$$

□

3. Method of Solution

Using Equations (10) and (13), we get

$$D^{\gamma }\left[f\left(t\right)D^{\gamma }y\left(t\right)\right]=U^T{Y}_{l-1}\left(t\right)=U^T{Q}_{l\times l}^{\gamma }{B}_{l-1}\left(t\right).$$

Thus,

$$f\left(t\right)D^{\gamma }y\left(t\right)-f\left(0\right)\varpi =I^{\gamma }{U}^T{Y}_{l-1}\left(t\right)$$

where $\varpi =D^{\gamma }y\left(0\right)$. Theorem 5 and Equations (10) and (13) imply that

$$\begin{array}{cc}{D}^{\gamma }y\left(t\right)\hfill & =\frac{1}{f\left(t\right)}\left(U^T{I}^{\gamma }{Y}_{l-1}\left(t\right)+f\left(0\right)\varpi \right)\hfill \\ & =\frac{1}{f\left(t\right)}\left(U^T{H}_l^{\gamma }{Y}_{l-1}\left(t\right)+f\left(0\right)\varpi \right)\hfill \\ & =U^T{H}_l^{\gamma }{Q}_{l\times l}^{\gamma }\frac{B_{ml-1}\left(t\right)}{f\left(t\right)}+\frac{f\left(0\right)\varpi }{f\left(t\right)}\hfill \\ & =U^T{H}_l^{\gamma }{Q}_{l\times l}^{\gamma }\left[\begin{array}{c}{b}_0\left(t\right)/f\left(t\right)\\ b_1\left(t\right)/f\left(t\right)\\ ⋮\\ b_{l-1}\left(t\right)/f\left(t\right)\end{array}\right]+f\left(0\right)\varpi \left[\begin{array}{c}1/f\left(t\right)\\ 1/f\left(t\right)\\ ⋮\\ 1/f\left(t\right)\end{array}\right].\hfill \end{array}$$

Hence,

$$D^{\gamma }y\left(t\right)=\left(U^T{H}_l^{\gamma }{Q}_{l\times l}^{\gamma }P{F}_1+f\left(0\right)\varpi F_2\right)B_{l-1}\left(t\right)$$

Thus,

$$y\left(t\right)=\left(U^T{H}_l^{\gamma }{Q}_{l\times l}^{\gamma }{F}_1+f\left(0\right)\varpi F_2\right)I^{\gamma }{B}_{l-1}\left(t\right)+\psi$$

where $\psi =y\left(0\right)$. Therefore,

$$y\left(t\right)=\left(U^T{H}_l^{\gamma }{Q}_{l\times l}^{\gamma }{F}_1+f\left(0\right)\varpi F_2\right)Y_l^{\gamma }{B}_{l-1}\left(t\right)+\psi .$$

(21)

Hence,

$$U^T{Q}_{l\times l}^{\gamma }{B}_{m-1}\left(t\right)+\mu g\left(t\right)\left(\left(U^T{H}_l^{\gamma }{Q}_{l\times l}^{\gamma }{F}_1+f\left(0\right)\varpi F_2\right)F{Y}_l^{\gamma }{B}_{l-1}\left(t\right)+\psi \right)=h\left(t\right)$$

or

$$U^T(Q_{l\times l}^{\gamma }+\mu q\left(t\right)H_l^{\gamma }{Q}_{l\times l}^{\gamma }{F}_1{Y}_l^{\gamma })B_{l-1}\left(t\right)=h\left(t\right)-\mu g\left(t\right)f\left(0\right)\varpi F_2{Y}_l^{\gamma }{B}_{l-1}\left(t\right)-\mu \psi g\left(t\right).$$

(22)

Using the boundary conditions in Equations (2) and (3), we get the following cases

• if $c_0=0,\varpi =0,c_1\ne 0,c_2\ne 0$, and

  $$\psi =-c_2{U}^T{H}_l^{\gamma }{Q}_{l\times m}^{\gamma }{F}_1{Y}_l^{\gamma }{B}_{l-1}\left(1\right)-\frac{c_3}{c_2}{U}^T{H}_l^{\gamma }{Q}_{l\times l}^{\gamma }P{F}_1{B}_{l-1}\left(1\right)$$
• if $c_0\ne 0,\psi =-\frac{c_1}{c_0}\varpi$ and

  $$\varpi =\frac{-c_2{U}^T{H}_l^{\gamma }{Q}_{l\times l}^{\gamma }{F}_1{Y}_l^{\gamma }{B}_{l-1}\left(1\right)-c_3{U}^T{H}_l^{\gamma }{Q}_{l\times l}^{\gamma }{Y}_1{B}_{l-1}\left(1\right)}{f\left(0\right)F_2{Y}_l^{\gamma }{B}_{l-1}\left(1\right)-\frac{c_1{c}_2}{c_0}+c_{3}f\left(0\right)Y_2{B}_{l-1}\left(1\right)}.$$

Thus,

$$U^T\left(\begin{array}{c}{Q}_{l\times l}^{\gamma }+\mu g\left(t\right)H_l^{\gamma }{Q}_{l\times l}^{\gamma }{F}_1{Y}_l^{\gamma }{B}_{l-1}\left(t\right)+\\ \left(\mu g\left(t\right)f\left(0\right)F_2{Y}_l^{\gamma }{B}_{l-1}\left(t\right)-\frac{c_1}{c_0}\mu g\left(t\right)\right)\left(\frac{-c_2{H}_l^{\gamma }{Q}_{l\times l}^{\gamma }{F}_1{Y}_l^{\gamma }{B}_{l-1}\left(1\right)-c_3{H}_l^{\gamma }{Q}_{l\times l}^{\gamma }{F}_1{B}_{l-1}\left(1\right)}{f\left(0\right)F_2{F}_l^{\gamma }{B}_{l-1}\left(1\right)-\frac{c_1{c}_2}{c_0}+c_{3}f\left(0\right)F_2{B}_{l-1}\left(1\right)}\right)\end{array}\right)=h\left(t\right).$$

(23)

We use the collocation points

$$t_i=\frac{r+1}{l+1},i=0:l-1.$$

Substitute these values into Equation (23) and take the transpose of both sides to get a system of linear equations in terms of U of the form

$$G\left(\mu \right)U=R.$$

(24)

To have a nonzero solution to the system in Equation (24), $G\left(\mu \right)$ must be nonsingular. Thus,

$$\det \left(G\right(\mu \left)\right)=0.$$

(25)

Therefore, we find the eigenvalues from Equation (25) and we find the corresponding eigenfunctions from Equation (21).

4. Numerical Results

We present two examples for $l=16$. In this paper, we focus only on the eigenvalues.

Example 1.

Consider

$$D^{\gamma }\left[D^{\gamma }y\left(t\right)\right]+\mu y\left(t\right)=0,t\in [0,1],\gamma \in (0.5,1],$$

$$y\left(0\right)=0,y\left(1\right)=0.$$

Using the procedure described in the previous section, the generated eigenvalues are reported in

Table 1. Eigenvalues for different values of $\gamma$.

$\mathit{\gamma }=0.75$	$\mathit{\gamma }=0.95$	$\mathit{\gamma }=0.99$
8.7825905605957	8.2711826449023	9.6635258705797
14.0844503539395	58.990163159836	38.044080578817
	96.673652759078	84.971438095925
	148.295350243613	150.13372170205
	199.571402506686	233.59863572826
	277.107135647923	335.09777723091
	295.450149615306	454.76440655382
		590.93089519416

.

For $\gamma =1$, the exact eigenvalues are well-known and they are given by

$${\mu }_n=n^2{\pi }^2,n=1,2,3,\dots .$$

It is worth mentioning that the eigenvalues of the problem in this example approach to $n^2{\pi }^2$ when $\gamma$ approaches to 1. Let

$${\delta }_{i,j}=\left|{}_0^1{y}_i\left(t\right)y_j\left(t\right)dt\right|.$$

For $\gamma =0.75$, ${\delta }_{1,2}=5.7\times {10}^{-16}$. Sample of these values for $\gamma =0.95$ are given as

$${\delta }_{1,2}=5.7\times {10}^{-16},{\delta }_{4,6}=2.6\times {10}^{-16},{\delta }_{1,6}=8.3\times {10}^{-16}.$$

Similarly, for $\gamma =0.99,$

$${\delta }_{1,2}=3.1\times {10}^{-16},{\delta }_{4,6}=4.2\times {10}^{-16},{\delta }_{1,7}=2.0\times {10}^{-16}.$$

This means the orthogonality relation holds. We notice that the eigenvalues satisfy the increasing property.

Example 2.

Consider

$$D^{\alpha }\left[D^{\alpha }y\left(t\right)\right]+\lambda (1+t^{\alpha })y\left(t\right)=0,t\in [0,1],\gamma \in (0.5,1],$$

$$u\left(0\right)=0,u\left(1\right)=0.$$

Using the procedure described in the previous section, the generated eigenvalues are reported in

Table 2. Eigenvalues for different values of $\gamma$.

$\mathit{\gamma }=0.501$	$\mathit{\gamma }=0.75$	$\mathit{\gamma }=0.95$
3.74496847027704	4.90596508821842	5.82711926402061
5.59359607314814	9.95423834570763	21.8630977993855
25.47511927569108	14.2468657217155	100.868795211963
151.8458499360075	25.8797084072818	234.225682145921
	124.475138197374	439.200912754629
		721.009344587213
		984.124781340994

.

Let

$${\delta }_{i,j}=\left|{}_0^1{y}_i\left(t\right)y_j\left(t\right)g\left(t\right)dt\right|.$$

For $\gamma =0.502$, ${\delta }_{1,2}=3.3\times {10}^{-16}$ and ${\delta }_{2,4}=4.9\times {10}^{-16}$. Samples of these values for $\gamma =0.75$ are given as

$${\delta }_{1,2}=2.2\times {10}^{-16},{\delta }_{4,5}=4.1\times {10}^{-16},{\delta }_{1,5}=6.9\times {10}^{-16}.$$

Similarly, for $\gamma =0.95,$

$${\delta }_{1,2}=1.2\times {10}^{-16},{\delta }_{4,6}=2.1\times {10}^{-16},{\delta }_{1,7}=4.6\times {10}^{-16}.$$

This means the orthogonality relation holds. We notice that the eigenvalues satisfy the property

$${\mu }_1\le {\mu }_2\le \dots .$$

5. Conclusions

In this article, a reliable method for solving fractional Sturm–Liouville problem based on the operational matrix method is presented. Two of our numerical examples are presented. From the previous discussion, we notice the following.

• From previous section, we can find the eigenvalues with the following property

  $${\lambda }_1<{\lambda }_2<{\lambda }_3<\dots <{\lambda }_n<\dots .$$
• From previous section, the orthogonality property

  $${}_0^1{y}_i\left(t\right)y_j\left(t\right)q\left(t\right)=0,i\ne j$$

  holds.
• The proposed method can be generalized to other applications in Physics and Engineering.