*Article* **Fractional Sturm–Liouville Eigenvalue Problems, II**

**Mohammad Dehghan \* and Angelo B. Mingarelli**

School of Mathematics and Statistics, Carleton University, Ottawa, ON K1S 5B6, Canada **\*** Correspondence: dehghan@math.carleton.ca

**Abstract:** We continue the study of a non-self-adjoint fractional three-term Sturm–Liouville boundary value problem (with a potential term) formed by the composition of a left Caputo and left Riemann– Liouville fractional integral under *Dirichlet type* boundary conditions. We study the existence and asymptotic behavior of the real eigenvalues and show that for certain values of the fractional differentiation parameter *α*, 0 < *α* < 1, there is a finite set of real eigenvalues and that, for *α* near 1/2, there may be none at all. As *α* → 1<sup>−</sup> we show that their number becomes infinite and that the problem then approaches a standard Dirichlet Sturm–Liouville problem with the composition of the operators becoming the operator of second order differentiation.

**Keywords:** Fractional Sturm–Liouville; fractional calculus; Laplace transform; Mittag-Lefflerfunction; eigenvalues; asymptotics

### **1. Introduction**

This is a continuation of [1] where the results therein are extended to three-term Fractional Sturm–Liouville operators (with a potential term) formed by the composition of a left Caputo and left-Riemann–Liouville fractional integral. Similar kinds of spectral problems have been considered in [2–11]. Specifically, the boundary value problem considered here is of the form,

$$-\,^c\mathcal{D}\_{0^+}^a \circ \mathcal{D}\_{0^+}^a y(t) + q(t)y(t) = \lambda y(t), \qquad 1/2 < a < 1, \quad 0 \le t \le 1,\tag{1}$$

with boundary conditions

$$\mathcal{L}\_{0^{+}}^{1-a}y(t)|\_{t=0} = c\_1, \quad \text{and} \quad \mathcal{L}\_{0^{+}}^{1-a}y(t)|\_{t=1} = c\_2. \tag{2}$$

where *<sup>c</sup>*1, *<sup>c</sup>*<sup>2</sup> are real constants and the real valued unspecified potential function, *<sup>q</sup>* <sup>∈</sup> *<sup>L</sup>*∞[0, 1]. We note that these are not self-adjoint problems and so there may be a non-real spectrum, in general. A well-known property of the Riemann–Liouville integral gives that if the solutions are continuous on [0, 1] then the boundary conditions (2) reduce to the usual fixed-end boundary conditions, *y*(0) = *y*(1) = 0, as *α* → 1.

For the analogue of the Dirichlet problem described above we study the existence and asymptotic behaviour of the real eigenvalues and show that for each *α*, 0 < *α* < 1, there is a finite set of real eigenvalues and that, for *α* near 1/2, there may be none at all. As *α* → 1<sup>−</sup> we show that their number becomes infinite and that the problem then approaches a standard Dirichlet Sturm–Liouville problem with the composition of the operators becoming the operator of second order differentiation acting on a suitable function space.

Our approach is different from most in this area. Specifically, we start with the existence and uniqueness of solutions of the Equation (1) along with the initial conditions (2), then we formulate the boundary value problem as an integral equation, after which we show that the solution of this integral equation as a function of *λ* is an entire function of *λ* of order of at most 1/2*α*. Since *α* is between 1/2 and 1, this entire function is of fractional order and therefore must have an infinite number of zeros, some of which may be complex. These zeros are the eigenvalues of our problem and therefore we get their existence. Using

**Citation:** Dehghan, M.; Mingarelli, A.B. Fractional Sturm–Liouville Eigenvalue Problems, II. *Fractal Fract.* **2022**, *6*, 487. https://doi.org/ 10.3390/fractalfract6090487

Academic Editor: António Lopes

Received: 29 July 2022 Accepted: 25 August 2022 Published: 30 August 2022

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**Copyright:** © 2022 by the authors. Licensee MDPI, Basel, Switzerland. This article is an open access article distributed under the terms and conditions of the Creative Commons Attribution (CC BY) license (https:// creativecommons.org/licenses/by/ 4.0/).

asymptotic methods and hard analysis, we proved our bounds on the real eigenvalues of which there must be a finite number for each *α*. Finally, we show that as *α* tends to 1, the number of real eigenvalues becomes infinite and the original problem approaches the standard Sturm–Liouville eigenvalue problem.

#### **2. Preliminaries**

We recall some definitions from fractional calculus and refer the reader to our previous paper [1] for further details.

**Definition 1.** *The left and the right Riemann–Liouville fractional integrals* <sup>I</sup>*<sup>α</sup> <sup>a</sup>*<sup>+</sup> *and* <sup>I</sup>*<sup>α</sup> <sup>b</sup>*<sup>−</sup> *of order <sup>α</sup>* <sup>∈</sup> <sup>R</sup><sup>+</sup> *are defined by*

$$\mathcal{Z}\_{a^{+}}^{a}f(t) := \frac{1}{\Gamma(a)} \int\_{a}^{t} \frac{f(s)}{(t-s)^{1-a}} ds, \quad t \in (a,b],\tag{3}$$

*and*

$$\mathcal{X}\_{b-}^{a}f(t) := \frac{1}{\Gamma(a)} \int\_{t}^{b} \frac{f(s)}{(s-t)^{1-a}} ds, \quad t \in [a,b), \tag{4}$$

*respectively. Here* Γ(*α*) *denotes Euler's Gamma function. The following property is easily verified.*

**Property 1.** For a constant *<sup>C</sup>*, we have <sup>I</sup>*<sup>α</sup> <sup>a</sup>*<sup>+</sup> *<sup>C</sup>* <sup>=</sup> (*t*−*a*)*<sup>α</sup>* <sup>Γ</sup>(*α*+1) · *C*.

The proof is by direct calculation.

**Definition 2.** *The left and the right Caputo fractional derivatives <sup>c</sup>*D*<sup>α</sup> <sup>a</sup>*<sup>+</sup> *and <sup>c</sup>*D*<sup>α</sup> <sup>b</sup>*<sup>−</sup> *are defined by*

$$\prescript{c}{}{\mathcal{D}}\_{a^{+}}^{a}f(t) := \mathcal{Z}\_{a^{+}}^{n-a} \circ \mathcal{D}^{n}f(t) = \frac{1}{\Gamma(n-a)} \int\_{a}^{t} \frac{f^{(n)}(s)}{(t-s)^{a-n+1}} ds, \quad t > a,\tag{5}$$

*and*

$$\, ^c \mathcal{D}\_{b^-}^a f(t) := (-1)^n \mathcal{Z}\_{b^-}^{n-a} \circ \mathcal{D}^a f(t) = \frac{(-1)^n}{\Gamma(n-a)} \int\_t^b \frac{f^{(n)}(s)}{(s-t)^{a-n+1}} ds, \quad t < b,\tag{6}$$

*respectively, where f is sufficiently differentiable and n* − 1 ≤ *α* < *n.*

**Definition 3.** *Similarly, the left and the right Riemann–Liouville fractional derivatives* <sup>D</sup>*<sup>α</sup> <sup>a</sup>*<sup>+</sup> *and* D*α <sup>b</sup>*<sup>−</sup> *are defined by*

$$\mathcal{D}\_{a^{+}}^{a}f(t) := \mathcal{D}^{a} \circ \mathcal{Z}\_{a^{+}}^{n-a}f(t) = \frac{1}{\Gamma(n-a)} \frac{d^{n}}{dt^{n}} \int\_{a}^{t} \frac{f(s)}{(t-s)^{a-n+1}} ds, \quad t > a,\tag{7}$$

*and*

$$\mathcal{D}\_{b^{-}}^{a}f(t) := (-1)^{n} \mathcal{D}^{n} \circ \mathcal{Z}\_{b^{-}}^{n-a} f(t) = \frac{(-1)^{n}}{\Gamma(n-a)} \frac{d^{n}}{dt^{n}} \int\_{t}^{b} \frac{f(s)}{(s-t)^{a-n+1}} ds, \quad t < b,\tag{8}$$

*respectively, where f is sufficiently differentiable and n* − 1 ≤ *α* < *n.*

**Property 2.** *For* (*ν*) > −1*,* 0 < *α* < 1*, and t* > 0*, we have*

$$D\_{0^{+}}^{\\\alpha}(t^{\nu}) = \frac{\Gamma(1+\nu)}{\Gamma(1+\nu-\alpha)}t^{\nu-\alpha}$$

**Property 3.** *For* (*ν*) > 0*,* 0 < *α* < 1*, and t* > 0*, we have*

$$^cD\_{0^+}^\alpha(t^\nu) = \frac{\Gamma(1+\nu)}{\Gamma(1+\nu-\alpha)}t^{\nu-\alpha}$$

**Property 4.** *If y*(*t*) <sup>∈</sup> *<sup>L</sup>*1(*a*, *<sup>b</sup>*) *and* <sup>I</sup>1−*<sup>α</sup> <sup>a</sup>*<sup>+</sup> *<sup>y</sup>*, <sup>I</sup>1−*<sup>α</sup> <sup>b</sup>*<sup>−</sup> *y* ∈ *AC*[*a*, *b*]*, then*

$$\begin{aligned} \mathcal{Z}\_{a^{+}}^{\alpha} \mathcal{D}\_{a^{+}}^{\alpha} \boldsymbol{y}(t) &= \boldsymbol{y}(t) - \frac{(t-a)^{\alpha-1}}{\Gamma(\alpha)} \mathcal{Z}\_{a^{+}}^{1-\alpha} \boldsymbol{y}(a), \\ \mathcal{Z}\_{b^{-}}^{\alpha} \mathcal{D}\_{b^{-}}^{\alpha} \boldsymbol{y}(t) &= \boldsymbol{y}(t) - \frac{(b-t)^{\alpha-1}}{\Gamma(\alpha)} \mathcal{Z}\_{b^{-}}^{1-\alpha} \boldsymbol{y}(b). \end{aligned}$$

**Property 5.** *If y*(*t*) ∈ *AC*[*a*, *b*] *and* 0 < *α* ≤ 1*, then*

$$\begin{aligned} T\_{a^+}^{a^-c} \mathcal{D}\_{a^+}^{\alpha} y(t) &= y(t) - y(a), \\ T\_{b^-}^{a^-c} \mathcal{D}\_{b^-}^{\alpha} y(t) &= y(t) - y(b). \end{aligned}$$

**Property 6.** *For* 0 < *α* < 1 *we have*

$$
\mathcal{D}\_{a^{+}}^{a}f(t) = \frac{f(a)}{\Gamma(1-\alpha)}(t-a)^{-\alpha} + ^c\mathcal{D}\_{a^{+}}^{a}f(t).
$$

$$The \text{ } Mittag\text{-}L\text{-}fhler \text{ } Function.$$

The function *Eδ*(*z*) defined by

$$E\_{\delta}(z) := \sum\_{k=0}^{\infty} \frac{z^{\delta}}{\Gamma(\delta k + 1)}, \quad (z \in \mathbb{C}, \Re(\delta) > 0), \tag{9}$$

was introduced by Mittag-Leffler [12]. In particular, when *δ* = 1 and *δ* = 2, we have

$$E\_1(z) = e^z, \qquad E\_2(z) = \cosh(\sqrt{z}). \tag{10}$$

The generalized Mittag-Leffler function *Eδ*,*<sup>θ</sup>* (*z*) is defined by

$$E\_{\delta,\theta}(z) = \sum\_{k=0}^{\infty} \frac{z^k}{\Gamma(\delta k + \theta)},\tag{11}$$

where *z*, *θ* ∈ C and Re (*δ*) > 0. When *θ* = 1, *Eδ*,*<sup>θ</sup>* (*z*) coincides with the Mittag-Leffler function (9):

$$E\_{\delta,1}(z) = E\_{\delta}(z). \tag{12}$$

Two other particular cases of (11) are as follows:

$$E\_{1,2}(z) = \frac{e^z - 1}{z}, \quad E\_{2,2}(z) = \frac{\sinh(\sqrt{z})}{\sqrt{z}}.\tag{13}$$

**Property 7.** *For any δ with* (*δ*) > 0 *and for any z* = 0 *we have*

$$E\_{\delta,\delta}(z) = \frac{1}{z} E\_{\delta,0}(z)$$

Further properties of this special function may be found in [13].

**Property 8** (See [14], p. 43)**.** *If* <sup>0</sup> <sup>&</sup>lt; *<sup>δ</sup>* <sup>&</sup>lt; <sup>2</sup> *and <sup>μ</sup>* <sup>∈</sup> ( *δπ* <sup>2</sup> , min(*π*, *δπ*))*, then the function Eδ*,*<sup>θ</sup>* (*z*) *has the following exponential expansion as* |*z*| → ∞

$$E\_{\delta, \theta}(z) = \begin{cases} \frac{1}{\delta} z^{\frac{1-\theta}{\delta}} \exp(z^{\frac{1}{\theta}}) - \sum\_{k=1}^{N} \frac{1}{\Gamma(\theta - \delta k)} \frac{1}{z^k} + O(\frac{1}{z^{N+1}}), & |\arg(z)| \le \mu, \\\\ -\sum\_{k=1}^{N} \frac{1}{\Gamma(\theta - \delta k)} \frac{1}{z^k} + O(\frac{1}{z^{N+1}}), & \mu \le |\arg(z)| \le \pi. \end{cases} \tag{14}$$

#### **3. Existence and Uniqueness of the Solution of SLPs**

In this section we convert (1) and (2) to an integral equation and prove that it has a solution that satisfies the relevant equations and initial conditions. First, we proceed formally. Separating terms in (1), we get

$$\mathbf{^ctD\_{0^+}^\mathbf{x}} \circ \mathcal{D}\_{0^+}^\mathbf{x} y(t) = (q(t) - \lambda)y(t), \qquad 1/2 < \mathfrak{a} < 1, \quad 0 \le t \le 1.$$

Taking the left Riemann–Liouville fractional integrals <sup>I</sup>*<sup>α</sup> <sup>a</sup>*<sup>+</sup> on both sides of the above equation and using Property 5, we have

$$
\mathcal{D}\_{0^+}^{\alpha}y(t) - \mathcal{D}\_{0^+}^{\alpha}y(t)|\_{t=0} = \mathcal{T}\_{0^+}^{\alpha} \left( (q(t) - \lambda)y(t) \right).
$$

Taking the left Riemann–Liouville fractional integrals <sup>I</sup>*<sup>α</sup> <sup>a</sup>*<sup>+</sup> from both sides of the above equation once again and using Property 4, we get

$$y(t) - \frac{t^{n-1}}{\Gamma(n)} \mathcal{Z}\_{0^{+}}^{1-a} y(t)|\_{t=0} - \mathcal{Z}\_{0^{+}}^{a} \left(\mathcal{D}\_{0^{+}}^{a} y(t)|\_{t=0}\right) = \mathcal{Z}\_{0^{+}}^{a} \left(\mathcal{Z}\_{0^{+}}^{a} \left((q(t) - \lambda)y(t)\right)\right)$$

Using Property 1, we can write

$$y(t) = c\_1 \frac{t^{\alpha - 1}}{\Gamma(\alpha)} + c\_2 \frac{t^{\alpha}}{\Gamma(\alpha + 1)} + \mathcal{Z}\_{0^+}^{\alpha} \left( \mathcal{Z}\_{0^+}^{\alpha} \left( (q(t) - \lambda) y(t) \right) \right),$$

in which

$$c\_1 = \mathcal{I}\_{0^+}^{1-\alpha} y(t)|\_{t=0\prime} \qquad c\_2 = \mathcal{D}\_{0^+}^{\alpha} y(t)|\_{t=0\prime}$$

We obtain, through the double fractional integral in the above equation, the following:

$$y(t) = c\_1 \frac{t^{a-1}}{\Gamma(a)} + c\_2 \frac{t^a}{\Gamma(a+1)} + \frac{1}{\Gamma^2(a)} \int\_0^t (t-s)^{a-1} \left( \int\_0^s \frac{(q(r) - \lambda) y(r)}{(s-r)^{1-a}} dr \right) ds.$$

By changing the order of integrals in the above equation we get

$$y(t) = c\_1 \frac{t^{\kappa - 1}}{\Gamma(\kappa)} + c\_2 \frac{t^{\kappa}}{\Gamma(\kappa + 1)} + \frac{1}{\Gamma^2(\kappa)} \int\_0^t (q(r) - \lambda) y(r) \left( \int\_r^t (t - s)^{\kappa - 1} (s - r)^{\kappa - 1} ds \right) dr$$

Solving the inner integral gives us

$$y(t,\lambda) = c\_1 \frac{t^{a-1}}{\Gamma(a)} + c\_2 \frac{t^a}{\Gamma(a+1)} + \frac{1}{\Gamma(2a)} \int\_0^t (q(s) - \lambda) y(s,\lambda) (t-s)^{2a-1} ds.\tag{15}$$

We will now show that (15) has a solution that exists in a neighbourhood of *t* = 0 and is unique there. Working backwards will then provide us with a unique solution to (1) and (2). Although this result already appears in [15], we give a shorter proof part of which will be required later.

To this end, let *t* > 0. Define

$$y\_n(t, \lambda) = y\_0(t, \lambda) + \frac{1}{\Gamma(2a)} \int\_0^t (t - s)^{2a - 1} (q(s) - \lambda) y\_{n - 1}(s, \lambda) ds,\tag{16}$$

where

$$y\_0(t, \lambda) = c\_1 \frac{t^{\alpha - 1}}{\Gamma(\alpha)} + c\_2 \frac{t^{\alpha}}{\Gamma(\alpha + 1)}.\tag{17}$$

Let *λ* ∈ **C**, |*λ*| < Λ, where Λ > 0 is arbitrary but fixed. Then,

$$\begin{split} |y\_1(t, \lambda) - y\_0(t, \lambda)| &\le \frac{1}{\Gamma(2a)} \int\_0^t (t - s)^{2a - 1} |q(s) - \lambda| |y\_0(s, \lambda)| ds \\ &\le \frac{||q||\_\infty + \Lambda}{\Gamma(2a)} \int\_0^t (t - s)^{2a - 1} |y\_0(s, \lambda)| ds, \end{split} \tag{18}$$

in which ||*q*||<sup>∞</sup> <sup>=</sup> sup*t*∈[0,1] <sup>|</sup>*q*(*t*)|. Substituting (17) in (18) and using the fact that,

$$\int\_{a}^{t} (t-s)^{\alpha-1} (s-a)^{\beta-1} ds = \frac{(t-a)^{\alpha+\beta-1} \Gamma(\alpha) \Gamma(\beta)}{\Gamma(\alpha+\beta)} \lambda$$

we have

$$|y\_1(t\_\prime\lambda) - y\_0(t\_\prime\lambda)| \le (||q||\_\infty + \Lambda) \left(\frac{c\_1}{\Gamma(3\alpha)} t^{3\alpha - 1} + c\_2 \frac{t^{3\alpha}}{\Gamma(3\alpha + 1)}\right) \lambda$$

Now, for *n* = 2 in (16) we get

$$\begin{split} |y\_2(t,\lambda) - y\_1(t,\lambda)| &\leq \frac{1}{\Gamma(2\mathfrak{a})} \int\_0^t (t-s)^{2\mathfrak{a}-1} |q(s) - \lambda| |y\_1(s,\lambda) - y\_0(s,\lambda)| ds \\ &\leq \frac{1}{\Gamma(2\mathfrak{a})} \int\_0^t (t-s)^{2\mathfrak{a}-1} |q(s) - \lambda| \left( (||q||\_{\infty} + \Lambda) \frac{c\_1}{\Gamma(3\mathfrak{a})} s^{3\mathfrak{a}-1} + c\_2 \frac{s^{3\mathfrak{a}}}{\Gamma(3\mathfrak{a}+1)} \right) ds \\ &\leq (||q||\_{\infty} + \Lambda)^2 \left( \frac{c\_1}{\Gamma(5\mathfrak{a})} t^{5\mathfrak{a}-1} + \frac{c\_2}{\Gamma(5\mathfrak{a}+1)} t^{5\mathfrak{a}} \right). \end{split} \tag{19}$$

Continuing in this way we get that the series

$$y(t,\lambda) := y\_0(t,\lambda) + \sum\_{n=1}^{\infty} \left( y\_n(t,\lambda) - y\_{n-1}(t,\lambda) \right) \tag{20}$$

where

$$\sum\_{n=1}^{\infty} |y\_n(t, \lambda) - y\_{n-1}(t, \lambda)| \le c\_1 t^{-1} \sum\_{n=1}^{\infty} \frac{(||q||\_{\infty} + \Lambda)^n}{\Gamma(2na + a)} t^{2na + a} + c\_2 \sum\_{n=1}^{\infty} \frac{(||q||\_{\infty} + \Lambda)^n}{\Gamma(2na + a + 1)} t^{2na + a} \tag{21}$$

converges uniformly on compact subsets of (0, 1]. Denote the sum of the infinite series in (20) by *y*(*t*, *λ*). So, by virtue of (17) and (21), (20) gives us,

$$\begin{split} &|y(t,\lambda)| \leq |y\_0(t,\lambda)| + \sum\_{n=1}^{\infty} |y\_n(t,\lambda) - y\_{n-1}(t,\lambda)| \\ &\leq \frac{c\_1}{\Gamma(a)} t^{a-1} + \frac{c\_2}{\Gamma(a+1)} t^a + c\_1 t^{-1} \sum\_{n=1}^{\infty} \frac{(||q||\_{\infty} + \Lambda)^n}{\Gamma(2na+a)} t^{2na+a} + c\_2 \sum\_{n=1}^{\infty} \frac{(||q||\_{\infty} + \Lambda)^n}{\Gamma(2na+a+1)} t^{2na+a} \\ &= c\_1 t^{-1} \sum\_{n=0}^{\infty} \frac{(||q||\_{\infty} + \Lambda)^n}{\Gamma(2na+a)} t^{2na+a} + c\_2 \sum\_{n=0}^{\infty} \frac{(||q||\_{\infty} + \Lambda)^n}{\Gamma(2na+a+1)} t^{2na+a} \\ &= c\_1 t^{a-1} E\_{2a,a}((||q||\_{\infty} + \Lambda) t^{2a}) + c\_2 t^a E\_{2a,a+1}((||q||\_{\infty} + \Lambda) t^{2a}). \end{split}$$

Note that for a solution *y*(*t*, *λ*) of (15) to be *C*([0, 1]), it is necessary and sufficient that *<sup>c</sup>*<sup>1</sup> <sup>=</sup> 0, i.e., <sup>I</sup>1−*<sup>α</sup>* <sup>0</sup><sup>+</sup> *y*(*t*)|*t*=<sup>0</sup> = 0. This then proves the global existence of a solution of (15) on [*δ*, 1], *<sup>δ</sup>* <sup>&</sup>gt; 0, since *<sup>q</sup>* <sup>∈</sup> *<sup>L</sup>*∞[0, 1] for given *<sup>c</sup>*<sup>1</sup> and *<sup>c</sup>*2, as defined in (2).

From the proof comes the following a priori estimate when *c*<sup>1</sup> = 0, that is,

$$\begin{aligned} |y(t,\lambda)| &\leq c\_2 t^{\alpha} \left( \frac{1}{\Gamma(\alpha+1)} + \left| E\_{2\alpha,\alpha+1}((||q||\_{\infty} + \Lambda)t^{2\alpha}) \right| \right) \\ &\leq c\_2 \left( \frac{1}{\Gamma(\alpha+1)} + \left| E\_{2\alpha,\alpha+1}((||q||\_{\infty} + \Lambda)t^{2\alpha}) \right| \right) \end{aligned}$$

valid for each *t* ∈ [0, 1] and all |*λ*| < Λ.

The previous bound can be made into an absolute constant by taking the sup over all *t* and |*λ*| < Λ. Of course, the bound goes to infinity as |*λ*| → ∞ over non-real values, as it must. Thus,

$$|y(t\_\prime \lambda)| \le c\_2 \left( \frac{1}{\Gamma(\alpha + 1)} + \sup\_{|\lambda| < \Lambda, t \in [0, 1]} \left| E\_{2a, a+1} ((||q||\_\infty + \Lambda)t^{2a}) \right| \right)$$

$$= c\_2 \left( \frac{1}{\Gamma(a + 1)} + |E\_{2a, a+1} ((||q||\_\infty + \Lambda))| \right) := c\_3. \tag{22}$$

for all |*λ*| < Λ, *t* ∈ [0, 1]. Uniqueness follows easily by means of Gronwall's inequality, as usual. Let *<sup>ε</sup>* <sup>&</sup>gt; 0. Assume that (15) has two solutions *<sup>y</sup>*(*t*, *<sup>λ</sup>*), *<sup>z</sup>*(*t*, *<sup>λ</sup>*). Since *<sup>q</sup>* <sup>∈</sup> *<sup>L</sup>*∞[0, 1] and |*λ*| < Λ we can derive that,

$$|y(t, \lambda) - z(t, \lambda)| \le \mathfrak{e} \mathfrak{e}^{\frac{1}{\Gamma(2\alpha)} (||q||\_{\infty} + \Lambda)\frac{t^{2\alpha}}{2\alpha}}.$$

and since *t* ∈ [0, 1], we get

$$|y(t, \lambda) - z(t, \lambda)| \le O(\varepsilon)$$

where the *O*-term can be made independent of both *t*, *λ*. Letting *ε* → 0 yields uniqueness for *t* ∈ [0, 1] and |*λ*| < Λ.

#### **4. Another Integral Equation**

In the previous section we showed that (15) has a solution that, for each *λ* ∈ **C**, exists on [0, 1], is unique, and is continuous there if and only if *c*<sup>1</sup> = 0. On the other hand, if *c*<sup>1</sup> = 0 then the solution is merely continuous on all compact subsets of (0, 1]. In this section we find another expression for the integral equation which is equivalent to both (15) and the problem (1) with boundary conditions (2).

**Lemma 1.** *For* 0 < *α* < 1 *and* 0 < *t* < 1*, we have*

$$- {}^{c} \mathcal{D}\_{0^{+}} {}^{\alpha} \mathcal{D}\_{0^{+}} \left( t^{a-1} E\_{2a,a} (-\lambda t^{2a}) \right) = \lambda t^{a-1} E\_{2a,a} (-\lambda t^{2a})$$

**Proof.** Using properties of the Mittag-Leffler function we can write

$$\begin{split} \mathcal{D}\_{0^{+}}^{\kappa} \left( t^{a-1} E\_{2a,\kappa} (-\lambda t^{2a}) \right) &= \mathcal{D}\_{0^{+}}^{a} \left( \sum\_{k=0}^{\infty} \frac{(-\lambda)^{k} t^{2ak+a-1}}{\Gamma(2ak+a)} \right) \\ &= \sum\_{k=0}^{\infty} \frac{(-\lambda)^{k} \mathcal{D}\_{0^{+}}^{\kappa} \left( t^{2ak+a-1} \right)}{\Gamma(2ak+a)} \\ &= \sum\_{k=0}^{\infty} \frac{(-\lambda)^{k} t^{2ak-1}}{\Gamma(2ak)} \\ &= t^{-1} E\_{2a,0} (\lambda t^{2a}) \\ &= -\lambda t^{2a-1} E\_{2a,2a} (-\lambda t^{2a}), \end{split} \tag{23}$$

in which the third and the last equalities come from Property 2 and Property 7, respectively. Now, taking the left Caputo fractional derivative of both sides of (23) we get

$$\begin{split} -\,^c\mathcal{D}\_{0^+}^{\mathfrak{a}}\cdot\mathcal{D}\_{0^+}^{\mathfrak{a}}\left(t^{a-1}E\_{2\mathfrak{a},\mathfrak{a}}\left(-\lambda t^{2\mathfrak{a}}\right)\right) &= \,^c\mathcal{D}\_{0^+}^{\mathfrak{a}}\left(\lambda t^{2a-1}E\_{2\mathfrak{a},2\mathfrak{a}}\left(-\lambda t^{2\mathfrak{a}}\right)\right) \\ &= \,^c\mathcal{D}\_{0^+}^{\mathfrak{a}}\left(\sum\_{k=0}^\infty \frac{(-\lambda)^k t^{2ak+2a-1}}{\Gamma(2ak+2a)}\right) \\ &= \lambda \left(\sum\_{k=0}^\infty \frac{(-\lambda)^k t^{2a}\mathcal{D}\_{0^+}^{\mathfrak{a}}\left(t^{2ak+2a-1}\right)}{\Gamma(2ak+2a)}\right) \\ &= \lambda t^{a-1}\sum\_{k=0}^\infty \frac{(-\lambda)^k t^{2ak}}{\Gamma(2ak+a)} \\ &= \lambda t^{a-1}E\_{2a,\mathfrak{a}}(-\lambda t^{2\mathfrak{a}}) \end{split}$$

as required.

**Lemma 2.** *For* 0 < *α* < 1 *and* 0 < *t* < 1*, we have*

$$- \, ^c \mathcal{D}\_{0^+}^{\mathfrak{a}} \mathcal{D}\_{0^+}^{\mathfrak{a}} \left( t^a E\_{2a, a+1} (- \lambda t^{2a}) \right) = \lambda t^a E\_{2a, a+1} (- \lambda t^{2a})$$

**Proof.** Once again, using the properties of the Mittag-Leffler function we can write

$$\begin{split} \mathcal{D}\_{0^{+}}^{\alpha} \left( t^{\mathbf{z}} E\_{2\kappa,\alpha+1} (-\lambda t^{2\alpha}) \right) &= \mathcal{D}\_{0^{+}}^{\alpha} \left( \sum\_{k=0}^{\infty} \frac{(-\lambda)^{k} t^{2ak+\alpha}}{\Gamma(2ak+\alpha+1)} \right) \\ &= \sum\_{k=0}^{\infty} \frac{(-\lambda)^{k} \mathcal{D}\_{0^{+}}^{\alpha} \left( t^{2ak+\alpha} \right)}{\Gamma(2ak+\alpha+1)} \\ &= \sum\_{k=0}^{\infty} \frac{(-\lambda)^{k} t^{2ak}}{\Gamma(2ak+1)} \\ &= E\_{2a,1} (-\lambda t^{2a}). \end{split} \tag{24}$$

in which the third equality comes from Property 3. Now, taking the left Caputo fractional derivative of both sides of (24) we get

$$\begin{split} \,^c\mathcal{D}\_{0^+}^{\kappa}\mathcal{D}\_{0^+}^{\kappa} \left( \,^{\mu}E\_{2a,a+1}(-\lambda t^{2\kappa}) \right) &= \,^c\mathcal{D}\_{0^+}^{\kappa} \left( E\_{2a,1}(\lambda t^{2\kappa}) \right) \\ &= \,^c\mathcal{D}\_{0^+}^{\kappa} \left( \sum\_{k=0}^{\infty} \frac{(-\lambda)^k t^{2ak}}{\Gamma(2ak+1)} \right) \\ &= \sum\_{k=0}^{\infty} \frac{(-\lambda)^{kc} \mathcal{D}\_{0^+}^{\kappa} \left( t^{2ak} \right)}{\Gamma(2ak+1)} \\ &= \sum\_{k=1}^{\infty} \frac{(-\lambda)^k t^{\kappa(2k-1)}}{\Gamma(1+\kappa(2k-1))} \\ &= \sum\_{k=0}^{\infty} \frac{(-\lambda)^{k+1} t^{\kappa(2k+1)}}{\Gamma(1+\kappa(2k+1))} \\ &= -\lambda t^a \sum\_{k=0}^{\infty} \frac{(-\lambda)^k t^{2ak}}{\Gamma(2ka+a+1)} \\ &= -\lambda t^a E\_{2a,a+1}(-\lambda t^{2a}) \end{split}$$

as desired.

**Lemma 3.** *For* 0 < *α* < 1 *and* 0 < *t* < 1*, we have*

$$-\, ^c \mathcal{D}\_{0^+}^{\mathbf{z}} \mathcal{D}\_{0^+}^{\mathbf{z}} \left( \int\_0^t (t-s)^{2\alpha-1} E\_{2\mathbf{z},2\mathbf{z}}(-\lambda(t-s)^{2\mathbf{z}}) q(s) y(s) ds \right) = -q(t)y(t) + \lambda \int\_0^t (t-s)^{2\alpha-1} E\_{2\mathbf{z},2\mathbf{z}}(-\lambda(t-s)^{2\mathbf{z}}) q(s) y(s) ds$$

**Proof.** Let *c*<sup>4</sup> = 1/Γ(1 − *α*). Observe that,

$$\begin{split} \mathbf{I}\_{0}^{1-a} \int\_{0}^{t} (t-s)^{2a-1} \mathbf{E}\_{2a,2a} (-\lambda (t-s)^{2a}) q(s) y(s) ds &= c\_{4} \int\_{0}^{t} \frac{\int\_{0}^{r} (r-s)^{2a-1} \mathbf{E}\_{2a,2a} (-\lambda (r-s)^{2a}) q(s) y(s) ds}{(t-r)^{a}} dr \\ &= c\_{4} \int\_{0}^{t} q(s) y(s) \left( \int\_{s}^{t} \frac{(r-s)^{2a-1}}{(t-r)^{a}} \mathbf{E}\_{2a,2a} (-\lambda (r-s)^{2a}) dr \right) ds \\ &= c\_{4} \int\_{0}^{t} q(s) y(s) \left( \sum\_{k=0}^{\infty} \frac{(-\lambda)^{k}}{\Gamma(2ak+2\pi)} \int\_{s}^{t} \frac{(r-s)^{2a-1+2ak}}{(t-r)^{a}} dr \right) ds \\ &= \int\_{0}^{t} q(s) y(s) \left( \sum\_{k=0}^{\infty} \frac{(-\lambda)^{k} (t-s)^{2a+k}}{\Gamma(2ak+a+1)} \right) ds \\ &= \int\_{0}^{t} q(s) y(s) (t-s)^{a} E\_{2a,2+1} (-\lambda (t-s)^{2a}) ds. \end{split} \tag{25}$$

Next, differentiating both sides of (25) with respect to *<sup>t</sup>* and noting that <sup>D</sup>*<sup>α</sup>* <sup>0</sup><sup>+</sup> <sup>=</sup> *<sup>D</sup>*(I1−*<sup>α</sup>* <sup>0</sup><sup>+</sup> ) we find,

$$\mathcal{D}\_{0^{+}}^{a}\left(\int\_{0}^{t}(t-s)^{2\kappa-1}E\_{2a,2a}(-\lambda(t-s)^{2a})q(s)y(s)ds\right) = \int\_{0}^{t}(t-s)^{\kappa-1}E\_{2a,a}(-\lambda(t-s)^{2a})q(s)y(s)ds.\tag{26}$$

as *<sup>d</sup> dt*(*t <sup>α</sup>E*2*α*,*α*+1(−*λ<sup>t</sup>* <sup>2</sup>*α*)) = *t <sup>α</sup>*−1*E*2*α*,*α*(−*λ<sup>t</sup>* <sup>2</sup>*α*). Next, we are going to take the left Caputo fractional derivative of both sides of (26). However, since the right hand side of (26) as a function of *t* is zero at *t* = 0, we can use Property 6 and replace the Caputo fractional derivative *<sup>c</sup>*D*<sup>α</sup>* <sup>0</sup><sup>+</sup> by the Riemann–Liouville one <sup>D</sup>*<sup>α</sup>* <sup>0</sup><sup>+</sup> . In order to do so, first we need to apply <sup>I</sup>1−*<sup>α</sup>* <sup>0</sup><sup>+</sup> followed by the classical derivative of the right hand side of (26) as follows,

$$\begin{split} \mathbb{E}\_{0^{+}}^{1-\mathfrak{a}} \int\_{0}^{t} (t-s)^{\mathfrak{a}-1} \mathbb{E}\_{2\mathfrak{a},\mathfrak{a}} (-\lambda (t-s)^{2\mathfrak{a}}) q(s) y(s) ds &= c\_{4} \int\_{0}^{t} \frac{\int\_{0}^{r} (r-s)^{\mathfrak{a}-1} \mathbb{E}\_{2\mathfrak{a},\mathfrak{a}} (-\lambda (r-s)^{2\mathfrak{a}}) q(s) y(s) ds}{(t-r)^{\mathfrak{a}}} dr \\ &= c\_{4} \int\_{0}^{t} q(s) y(s) \left( \int\_{s}^{t} \frac{(r-s)^{\mathfrak{a}-1}}{(t-r)^{\mathfrak{a}}} \, ^{2} \mathrm{E}\_{2\mathfrak{a},\mathfrak{a}} (-\lambda (r-s)^{2\mathfrak{a}}) dr \right) ds \\ &= c\_{4} \int\_{0}^{t} q(s) y(s) \left( \sum\_{k=0}^{\infty} \frac{(-\lambda)^{k}}{(2\lambda k + a)} \int\_{s}^{t} \frac{(r-s)^{\mathfrak{a}-1+2\mathfrak{a}k}}{(t-r)^{\mathfrak{a}}} dr \right) ds \\ &= \int\_{0}^{t} q(s) y(s) \left( \sum\_{k=0}^{\infty} \frac{(-\lambda)^{k} (t-s)^{2\mathfrak{a}k}}{\Gamma(2\mathfrak{a}k + 1)} \right) ds \\ &= \int\_{0}^{t} q(s) y(s) \mathrm{E}\_{2\mathfrak{a},\mathfrak{a}} (-\lambda (t-s)^{2\mathfrak{a}}) ds. \end{split}$$

Taking the derivative of the previous equation and using the fact stated in the previous paragraph, we get

$$\begin{split} \,^t \mathcal{D}\_{0^+}^t \int\_0^t (t-s)^{\alpha-1} \mathcal{E}\_{2a,a}(-\lambda(t-s)^{2a}) q(s) y(s) ds &= q(t) y(t) + \int\_0^t q(s) y(s) (t-s)^{-1} \mathcal{E}\_{2a,0}(-\lambda(t-s)^{2a}) ds \\ &= q(t) y(t) - \lambda \int\_0^t (t-s)^{2a-1} \mathcal{E}\_{2a,2a}(-\lambda(t-s)^{2a}) q(s) y(s) ds. \end{split} \tag{27}$$

where we used Property 7 to arrive at the second equality above. Combining (26) and (27) completes the proof.

**Theorem 1.** *For* 1/2 < *α* < 1*, the integral equation*

$$y(t,\lambda) = c\_1 t^{\mu-1} E\_{2a,a}(-\lambda t^{2a}) + c\_2 t^a E\_{2a,a+1}(-\lambda t^{2a}) + \int\_0^t (t-s)^{2a-1} E\_{2a,2a}(-\lambda (t-s)^{2a}) q(s) \, y(s,\lambda) \, ds \tag{28}$$

*satisfies* (1) *with initial conditions* <sup>I</sup>1−*<sup>α</sup>* <sup>0</sup><sup>+</sup> *<sup>y</sup>*(*t*)|*t*=<sup>0</sup> <sup>=</sup> *<sup>c</sup>*<sup>1</sup> *and* <sup>D</sup>*<sup>α</sup>* <sup>0</sup><sup>+</sup> *y*(*t*)|*t*=<sup>0</sup> = *c*<sup>2</sup> *in which c*<sup>1</sup> *and c*<sup>2</sup> *are given constants, and that this solution is unique.*

**Proof.** We apply <sup>−</sup>*c*D*<sup>α</sup>* <sup>0</sup><sup>+</sup> <sup>D</sup>*<sup>α</sup>* <sup>0</sup><sup>+</sup> on both sides of (28) to find,

$$\begin{split}-\mathscr{C}\mathcal{D}^{\mathfrak{a}}\_{0^{+}}\mathcal{D}^{\mathfrak{a}}\_{0^{+}}(y(t,\lambda)) &= -\,^{c}\mathscr{D}^{\mathfrak{a}}\_{0^{+}}\mathcal{D}^{\mathfrak{a}}\_{0^{+}}\Big(c\_{1}t^{a-1}\,^{E}\_{2a,\mathfrak{a}}(-\lambda t^{2a}) + c\_{2}t^{a}\,^{E}\_{2a,a+1}(-\lambda t^{2a})\Big) + \\ &- \,^{c}\mathscr{D}^{\mathfrak{a}}\_{0^{+}}\mathcal{D}^{\mathfrak{a}}\_{0^{+}}\Big(\int\_{0}^{t}(t-s)^{2a-1}\,^{E}\_{2a,2a}(-\lambda(t-s)^{2a})q(s)\,y(s,\lambda)\,ds\Big) \\ &= \lambda c\_{1}t^{a-1}\,^{E}\_{2a,a}(-\lambda t^{2a}) + \lambda c\_{2}t^{a}\,^{E}\_{2a,a+1}(-\lambda t^{2a}) - q(t)y(t) + \\ &\lambda\int\_{0}^{t}(t-s)^{2a-1}\,^{E}\_{2a,2a}(-\lambda(t-s)^{2a})q(s)\,y(s,\lambda)\,ds \\ &= -q(t)y(t) + \lambda\Big(c\_{1}t^{a-1}\,^{E}\_{2a,a}(-\lambda t^{2a}) + c\_{2}t^{a}\,^{E}\_{2a,a+1}(-\lambda t^{2a})\Big) + \\ &\lambda\left(\int\_{0}^{t}(t-s)^{2a-1}\,^{E}\_{2a,2a}(-\lambda(t-s)^{2a})q(s)\,y(s,\lambda)\,ds\right) \\ &= -q(t)y(t) + \lambda(y(t,\lambda)), \end{split} \tag{29}$$

in which second equality come from Lemmas 1–3. We verify the initial conditions. Taking <sup>I</sup>1−*<sup>α</sup>* <sup>0</sup><sup>+</sup> of both sides (28), we get,

$$\begin{split} \mathcal{I}\_{0^{+}}^{1-a}(y(t,\lambda)) &= \mathcal{I}\_{0^{+}}^{1-a} \Big( c\_{1}t^{a-1} \operatorname{E}\_{2a,a}(-\lambda t^{2a}) + c\_{2}t^{a} \operatorname{E}\_{2a,a+1}(-\lambda t^{2a}) \Big) + \\ &\quad \mathcal{I}\_{0^{+}}^{1-a} \Big( \int\_{0}^{t} (t-s)^{2a-1} \operatorname{E}\_{2a,2a}(-\lambda (t-s)^{2a}) q(s) \, y(s,\lambda) \, ds \Big) \\ &= c\_{1} \operatorname{E}\_{2a,1}(-\lambda t^{2a}) + c\_{2}t \operatorname{E}\_{2a,2}(-\lambda t^{2a}) + \int\_{0}^{t} (t-s)^{a} \operatorname{E}\_{2a,2a+1}(-\lambda (t-s)^{2a}) q(s) \, y(s,\lambda) \, ds, \end{split} \tag{30}$$

where the third term of the second equality comes from (25). Since *E*2*α*,1(−*λt* <sup>2</sup>*α*)|*t*=<sup>0</sup> <sup>=</sup> <sup>1</sup> and the other two terms of the above equality vanish when *t* = 0, we have verified the first initial condition. Again Taking <sup>D</sup>*<sup>α</sup>* <sup>0</sup><sup>+</sup> on both sides (28), we can find,

$$\begin{split} \mathcal{D}^{a}\_{0+}(y(t,\lambda)) &= \mathcal{D}^{a}\_{0+} \left( c\_{1}t^{a-1}E\_{2a,a}(-\lambda t^{2a}) + c\_{2}t^{a}E\_{2a,a+1}(-\lambda t^{2a}) \right) + \\ &\quad \mathcal{D}^{a}\_{0+} \left( \int\_{0}^{t} (t-s)^{2a-1}E\_{2a,2a}(-\lambda (t-s)^{2a}) q(s) \, y(s,\lambda) \, ds \right) \\ &= -c\_{1}\lambda t^{2a-1}E\_{2a,2a}(-\lambda t^{2a}) + c\_{2}E\_{2a,1}(-\lambda t^{2a}) + \int\_{0}^{t} (t-s)^{a-1}E\_{2a,a}(-\lambda (t-s)^{2a}) q(s) \, y(s,\lambda) \, ds, \end{split} \tag{31}$$

where the second equality above comes from (23), (24) and (26). The second initial condition can readily be obtained by substituting *t* = 0 in (31).

#### **5. Analyticity of Solutions with Respect to the Parameter** *λ*

In this section we show that the solutions (15) or (28) are, generally speaking, entire functions of the parameter *λ* for each *t* under consideration and *λ* ∈ **C**. First, we show continuity with respect to said parameter. Consider the case where *c*<sup>1</sup> = 0, i.e., *y* ∈ *C*[0, 1].

**Lemma 4.** *Let y* ∈ *C*[0, 1]*, λ* ∈ **C***. Then, for each fixed t* ∈ [0, 1]*, y*(*t*, *λ*) *is continuous with respect to λ .*

**Proof.** Let Λ > 0 be arbitrary but fixed, and let |*λ*|, |*λ*0| < Λ. Using (28),

$$\begin{split} y(t,\lambda) - y(t,\lambda\_0) &= \frac{1}{\Gamma(2a)} \int\_0^t (t-s)^{2a-1} ( (q(s)-\lambda)y(s,\lambda) - (q(s)-\lambda\_0)y(s,\lambda\_0) ) ds \\ &= \frac{1}{\Gamma(2a)} \int\_0^t (t-s)^{2a-1} ( (\lambda\_0-\lambda)y(s,\lambda) + (q(s)-\lambda\_0)(y(s,\lambda)-y(s,\lambda\_0)) ) ds. \end{split}$$
 
$$\text{So,}$$

$$y(t,\lambda) - y(t,\lambda\_0) = -(\lambda - \lambda\_0) \frac{1}{\Gamma(2a)} \int\_0^t (t-s)^{2a-1} y(s,\lambda) ds + \frac{1}{\Gamma(2a)} \int\_0^t (t-s)^{2a-1} (q(s) - \lambda\_0) (y(s,\lambda) - y(s,\lambda\_0)) ds. \tag{32}$$

Now, let  > 0 and |*λ* − *λ*0| < *δ* where *δ* > 0 is to be chosen later. Then,

$$|y(t,\lambda) - y(t,\lambda\_0)| \le \delta \frac{1}{\Gamma(2a)} \int\_0^t (t-s)^{2a-1} |y(s,\lambda)| ds + \frac{1}{\Gamma(2a)} \int\_0^t (t-s)^{2a-1} |q(s) - \lambda\_0||y(s,\lambda) - y(s,\lambda\_0)| ds.$$
 
$$\text{Using (22) and } C \text{ norm We in recall } \omega \text{ on } \omega$$

Using (22) and Gronwall's inequality, we get

$$\begin{split} |y(t,\lambda) - y(t,\lambda\_0)| &\leq \frac{\delta c\_3 t^{2\alpha}}{2\alpha \,\Gamma(2\alpha)} + \frac{1}{\Gamma(2\alpha)} \int\_0^t (t-s)^{2\alpha - 1} |q(s) - \lambda\_0||y(s,\lambda) - y(s,\lambda\_0)| ds \\ &\leq \frac{\delta c\_3}{\Gamma(2\alpha + 1)} e^{\frac{1}{\Gamma(2\alpha)} \int\_0^1 (t-s)^{2\alpha - 1} |q(s) - \lambda\_0| ds} \\ &\leq \frac{\delta c\_3}{\Gamma(2\alpha + 1)} e^{\frac{1}{\Gamma(2\alpha)} \int\_0^1 (1-s)^{2\alpha - 1} |q(s) - \lambda\_0| ds} :=: \mathcal{C}\delta \end{split}$$

where

$$\mathcal{C} = \frac{\mathcal{C}\_3}{\Gamma(2\alpha + 1)} \ e^{\frac{1}{\Gamma(2\alpha)} \int\_0^1 (1 - s)^{2\alpha - 1} |q(s) - \lambda\_0| \, ds}$$

is a function of *<sup>α</sup>* and *<sup>λ</sup>*<sup>0</sup> only as *<sup>q</sup>* <sup>∈</sup> *<sup>L</sup>*∞(0, 1). Thus, for any *<sup>t</sup>* <sup>∈</sup> [0, 1], the continuity of *y*(*t*, *λ*) follows by choosing *δ* < *<sup>ε</sup> <sup>C</sup>* . It also follows from this that,

$$\sup\_{t \in [0,1]} |y(t, \lambda) - y(t, \lambda\_0)| < \varepsilon, \quad |\lambda - \lambda\_0| < \delta. \tag{33}$$

Next, we consider the differentiability of *y*(*t*, *λ*) with respect to *λ*.

**Lemma 5.** *Let y* ∈ *C*[0, 1]*, λ* ∈ **C***. Then, for each fixed t* ∈ [0, 1]*, y*(*t*, *λ*) *is differentiable with respect to λ.*

**Proof.** As before let |*λ*| < Λ, *t* ∈ [0, 1]. Equation (32) can be rewritten as

$$\frac{y(t,\lambda)-y(t,\lambda\_0)}{\lambda-\lambda\_0} = -\frac{1}{\Gamma(2a)} \int\_0^t (t-s)^{2a-1} y(s,\lambda) \, ds + \frac{1}{\Gamma(2a)} \int\_0^t (t-s)^{2a-1} (q(s)-\lambda\_0) \frac{y(s,\lambda)-y(s,\lambda\_0)}{\lambda-\lambda\_0} \, ds.$$

As *y*(*t*, *λ*0) is given, we define *h*(*t*, *λ*0) to be the unique solution of the Volterra integral equation of the second kind,

$$h(t, \lambda\_0) = -\frac{1}{\Gamma(2a)} \int\_0^t (t - s)^{2a - 1} y(s, \lambda\_0) \, ds + \frac{1}{\Gamma(2a)} \int\_0^t (t - s)^{2a - 1} (q(s) - \lambda\_0) h(s, \lambda\_0) \, ds.$$
 
$$\text{So,}$$

$$\begin{split} \left| \frac{y(t,\lambda) - y(t,\lambda\_0)}{\lambda - \lambda\_0} - h(t,\lambda\_0) \right| &\leq \frac{1}{\Gamma(2a)} \int\_0^t (t-s)^{2a-1} |y(s,\lambda) - y(s,\lambda\_0)| ds \\ &+ \frac{1}{\Gamma(2a)} \int\_0^t (t-s)^{2a-1} |q(s) - \lambda\_0| \left| \frac{y(s,\lambda) - y(s,\lambda\_0)}{\lambda - \lambda\_0} - h(s,\lambda\_0) \right| ds. \end{split}$$

Let *ε* > 0 and choose *δ* > 0 as in (33). Using Gronwall's inequality and (33) we get, for *t* ∈ [0, 1],

$$\begin{split} \left| \frac{y(t,\lambda) - y(t,\lambda\_0)}{\lambda - \lambda\_0} - h(t,\lambda\_0) \right| &\leq \frac{\varepsilon}{2a\Gamma(2a)} + \frac{1}{\Gamma(2a)} \int\_0^t (t-s)^{2a-1} |q(s) - \lambda\_0| \left| \frac{y(s,\lambda) - y(s,\lambda\_0)}{\lambda - \lambda\_0} - h(s,\lambda\_0) \right| ds, \\ &\leq \frac{\varepsilon}{\Gamma(2a+1)} e^{\frac{1}{\Gamma(2a)} \int\_0^1 (t-s)^{2a-1} |q(s) - \lambda\_0| ds} = O(\varepsilon). \end{split} \tag{34}$$

for *λ* near *λ*<sup>0</sup> since, for *t* ∈ [0, 1], 1 <sup>0</sup> (*<sup>t</sup>* <sup>−</sup> *<sup>s</sup>*)2*α*−1|*q*(*s*) <sup>−</sup> *<sup>λ</sup>*0<sup>|</sup> *ds* <sup>=</sup> *<sup>O</sup>*(1). Thus,

$$\frac{\partial y(t,\lambda)}{\partial \lambda}|\_{\lambda=\lambda\_0} := \lim\_{\lambda \to \lambda\_0} \frac{y(t,\lambda) - y(t,\lambda\_0)}{\lambda - \lambda\_0} = h(t,\lambda\_0),$$

exists at *λ*0. Since *λ*<sup>0</sup> is arbitrary *yλ*(*t*, *λ*) exists for all *λ* with |*λ*| < Λ, real or complex and the result follows.

**Theorem 2.** *For each t* ∈ [0, 1]*, y*(*t*, *λ*) *is an entire function of λ.*

**Proof.** This follows from Lemma 5 since *λ* ∈ **C** and |*λ*| < Λ where Λ > 0 is arbitrary.

#### **6. A Dirichlet-Type Problem**

Let *y* ∈ *C*[0, 1], *λ* ∈ **C** be fixed. In this case we note that the first of the boundary conditions (2) is equivalent to the usual fixed end (Dirichlet) boundary conditions, that is,

$$y \in \mathbb{C}[0,1] \quad \iff \mathcal{I}\_{0^+}^{1-a} y(t,\lambda)|\_{t=0} = 0 \quad \Longleftrightarrow \quad y(0,\lambda) = 0.$$

The continuity assumption implies that there is a number *M* such that |*y*(*t*, *λ*)| ≤ *M*, for all *t* ∈ [0, 1]. Thus,

$$|\mathcal{Z}\_{0^{+}}^{1-\alpha}y(t,\lambda)| \leq \frac{M}{\Gamma(\alpha)} \int\_{0}^{t} (t-s)^{-\alpha} \, ds = \frac{M \, t^{1-\alpha}}{(1-\alpha)\Gamma(\alpha)},$$

and so <sup>I</sup>1−*<sup>α</sup>* <sup>0</sup><sup>+</sup> *y*(*t*, *λ*)|*t*=<sup>0</sup> = 0. On the other hand (15) now implies that *c*<sup>1</sup> = 0, i.e., *y*(0, *λ*) = 0, so that *y* ∈ *C*[0, 1]. However, the condition *y*(1, *λ*) = 0 is independent of the statement that <sup>I</sup>1−*<sup>α</sup>* <sup>0</sup><sup>+</sup> *y*(*t*, *λ*)|*t*=<sup>1</sup> = 0.

Since, for any *z* = 0, the Mittag-Leffler functions satisfy

$$E\_{\delta,\delta}(z) = \frac{1}{z} E\_{\delta,0}(z),$$

we get

$$
\hbar^{2a-1} E\_{2a,2a}(-\lambda t^{2a}) = -\frac{1}{\lambda t} E\_{2a,0}(-\lambda t^{2a}).\tag{35}
$$

Hence, using (28) and (35) we get

$$y(t,\lambda) = \varepsilon\_1 t^{a-1} E\_{2a,a}(-\lambda t^{2a}) + \varepsilon\_2 t^a E\_{2a,a+1}(-\lambda t^{2a}) - \int\_0^t \frac{E\_{2a,0}(-\lambda(t-s)^{2a})}{\lambda(t-s)} q(s)y(s,\lambda) \, ds. \tag{36}$$

**Remark 1.** *When α* → 1*, the integral Equation (36) becomes*

$$y(t,\lambda) = y(0,\lambda)\cos(\sqrt{\lambda}t) + y'(0,\lambda)\frac{\sin(\sqrt{\lambda}t)}{\sqrt{\lambda}} + \int\_0^t \frac{\sin(\sqrt{\lambda}(t-s))}{\sqrt{\lambda}} q(s)y(s,\lambda)ds,\tag{37}$$

*which is exactly the integral equation equivalent of the classical Sturm–Liouville equation* −*y* + *q*(*t*)*y* = *λy for λ* > 0*.*

**Remark 2.** *Observe that, for each α,*

$$\lim\_{s \to t^{-}} -\frac{E\_{2a,0}(-\lambda(t-s)^{2\alpha})}{\lambda(t-s)} = \begin{cases} \begin{array}{c} 0, \quad \text{if} \quad \alpha \in (1/2, 1],\\ 1, \quad \text{if} \quad \alpha = 1/2. \end{array} \end{cases}$$

*and so, for each* 1/2 < *α* < 1*, the kernel appearing in* (36) *is uniformly bounded on* [0, 1]*. This agrees with the equivalent result for the classical case* (37)*.*

#### **7. Existence and Asymptotic Distribution of the Eigenvalues**

Without loss of generality we may assume that *c*<sup>2</sup> = 1 in (36) and *y*(*t*, *λ*) is the corresponding solution. In the sequel we always assume that 1/2 < *α* < 1.

**Lemma 6.** *For each t* ∈ [0, 1]*,* 1/2 < *α* < 1*, and* | arg(−*λ*)| ≤ *μ where μ* ∈ (*απ*, *π*)*, we have t <sup>α</sup>E*2*α*,*α*+1(−*λ<sup>t</sup>* <sup>2</sup>*α*) → 0 *as* |*λ*| → ∞*.*

**Proof.** By (14) we can write

$$\begin{split} t^{a}E\_{2a,a+1}(-\lambda t^{2a}) &= t^{a} \left( \frac{1}{2a} (-\lambda t^{2a})^{\frac{1-(a+1)}{2a}} \right) \exp\left\{ (-\lambda t^{2a})^{\frac{1}{2a}} \right\} + \mathbf{O}\left(\frac{1}{\lambda}\right) \\ &= -\frac{i}{2a\sqrt{\lambda}} \exp\left\{ (-\lambda)^{\frac{1}{2a}}t \right\} + \mathbf{O}\left(\frac{1}{\lambda}\right) \\ &= -\frac{i}{2a\sqrt{\lambda}} \exp\left\{ |\lambda|^{\frac{1}{\pi a}} \left( \cos(\frac{\arg(-\lambda)}{2a}) + i\sin(\frac{\arg(-\lambda)}{2a}) \right) t \right\} + \mathbf{O}\left(\frac{1}{\lambda}\right). \end{split}$$

Therefore,

$$\left| t^{a} E\_{2a,a+1}(-\lambda t^{2a}) \right| = \frac{1}{2a\sqrt{\lambda}} \exp\left\{ |\lambda|^{\frac{1}{2a}} \cos(\frac{\arg(-\lambda)}{2a}) t \right\}.$$

Regarding the assumption on arg(−*λ*), we have cos( arg(−*λ*) <sup>2</sup>*<sup>α</sup>* ) < 0 and it completes the proof.

**Lemma 7.** *For each t* ∈ [0, 1]*, s* ∈ [0, *t*]*,* 1/2 < *α* < 1*, and* | arg(−*λ*)| ≤ *μ where μ* ∈ (*απ*, *π*)*, we have <sup>E</sup>*2*α*,0(−*λ*(*t*−*s*)2*α*) *λ*(*t*−*s*) <sup>→</sup> <sup>0</sup> *as* <sup>|</sup>*λ*| → <sup>∞</sup>*.*

**Proof.** By (14) we can write

$$\begin{split} \frac{E\_{2a,0}(-\lambda(t-s)^{2\alpha})}{\lambda(t-s)} &= \frac{\left(\frac{1}{2\pi}(-\lambda(t-s)^{2\alpha})^{\frac{1}{2\pi}}\right) \exp\left\{(-\lambda(t-s)^{2\alpha})^{\frac{1}{2\pi}}\right\} + \mathbf{O}\left(\frac{1}{\lambda}\right)}{\lambda(t-s)} \\ &= \frac{1}{2\alpha} \frac{(-\lambda)^{\frac{1}{2\pi}}}{\lambda} \exp\left\{(-\lambda)^{\frac{1}{2\pi}}(t-s)\right\} + \mathbf{O}\left(\frac{1}{\lambda^2}\right). \end{split}$$

Then,

$$\left|\frac{E\_{2a,0}(-\lambda(t-s)^{2a})}{\lambda(t-s)}\right| = \frac{1}{2a|\lambda|^{(2a-1)/2a}} \exp\left\{ (t-s)|\lambda|^{1/2a} \cos\left(\frac{\arg(-\lambda)}{2a}\right) \right\} + O\left(\frac{1}{|\lambda|^2}\right).$$

Arguing as in the previous lemma we reach the desired conclusion.

**Lemma 8.** *For each t* ∈ [0, 1]*, s* ∈ [0, *t*]*,* 1/2 < *α* < 1*, and* | arg(−*λ*)| ≤ *μ where μ* ∈ (*απ*, *π*)*, we have <sup>E</sup>*2*α*,1−*α*(−*λ*(*t*−*s*)2*α*) *<sup>λ</sup>*(*t*−*s*)*<sup>α</sup>* <sup>→</sup> <sup>0</sup> *as* <sup>|</sup>*λ*| → <sup>∞</sup>

**Proof.** By (14) we can write

$$\begin{split} \frac{E\_{2a,1-a}(-\lambda(t-s)^{2a})}{\lambda(t-s)^{a}} &= \frac{\left(\frac{1}{2\pi}(-\lambda(t-s)^{2a})^{\frac{1-(1-a)}{2a}}\right)\exp\left\{(-\lambda(t-s)^{2a})^{\frac{1}{2\pi}}\right\}+\mathbf{O}\left(\frac{1}{\lambda}\right)}{\lambda(t-s)^{a}} \\ &= \frac{i}{\sqrt{\lambda}2a}\exp\left\{(-\lambda)^{\frac{1}{2\pi}}(t-s)\right\}+\mathbf{O}\left(\frac{1}{\lambda^{2}}\right). \end{split}$$

Then,

$$\left|\frac{E\_{2a,1-a}\left(-\lambda\left(t-s\right)^{2a}\right)}{\lambda\left(t-s\right)^{a}}\right| = \frac{1}{2a\sqrt{|\lambda|}}\exp\left\{\left(t-s\right)|\lambda|^{1/2a}\cos\left(\frac{\arg\left(-\lambda\right)}{2a}\right)\right\}.\tag{38}$$

The result follows since the exponential term is uniformly bounded.

**Lemma 9.** *For each t* ∈ [0, 1]*, and* 1/2 < *α* < 1*, the solution y*(*t*, *λ*) *is an entire function of λ of order at most* 1/2*α.*

**Proof.** Let *λ* ∈ **C**. Define *f* by

$$y(t,\lambda) = \exp\left\{t|\lambda|^{1/2a}\cos\left(\frac{\arg(-\lambda)}{2a}\right)\right\}f(t). \tag{39}$$

Then, using (36),

$$f(t) = t^{\mathbb{1}} E\_{2a, a+1}(-\lambda t^{2a}) \exp\left\{-t|\lambda|^{1/2a} \cos\left(\frac{\arg(-\lambda)}{2a}\right)\right\}$$

$$= -\int\_0^t \frac{E\_{2a, 0}(-\lambda(t-s)^{2a})}{\lambda(t-s)} \exp\left\{- (t-s)|\lambda|^{1/2a} \cos\left(\frac{\arg(-\lambda)}{2a}\right)\right\} q(s) f(s) \, ds$$

Applying Lemma <sup>6</sup> there exists <sup>Λ</sup> <sup>∈</sup> **<sup>R</sup>**<sup>+</sup> such that for all <sup>|</sup>*λ*<sup>|</sup> <sup>&</sup>gt; <sup>Λ</sup> we have

$$|f(t)| \le 1 + \frac{1}{2a|\lambda|^{(2a-1)/2a}} \int\_0^t |q(s)| \, |f(s)| \, ds$$

which, on account of Gronwall's inequality, gives us

$$|f(t)| \le \exp\left\{ \frac{1}{2a|\lambda|^{(2a-1)/2a}} \int\_0^1 |q(s)| \, ds \right\} \tag{40}$$

for all sufficiently large <sup>|</sup>*λ*|. Thus, *<sup>f</sup>* <sup>∈</sup> *<sup>L</sup>*∞[0, 1] so that (39) yields, for some *<sup>M</sup>*,

$$|y(t, \lambda)| \le M \exp\left(|\lambda|^{1/2a}\right),$$

and the order claim is verified.

**Lemma 10.** *For each t* <sup>∈</sup> [0, 1]*,* <sup>I</sup>1−*<sup>α</sup>* <sup>0</sup><sup>+</sup> *y*(*t*, *λ*) *is an entire function of λ of order at most* 2*α.* **Proof.** This is clear from the definition, the possible values of *α*, and since *y*(*t*, *λ*) is itself entire and of order at most 1/2*α*, from Lemma 9.

**Lemma 11.** *The boundary value problem* (1) *and* (2) *has infinitely many complex eigenvalues (real eigenvalues are not to be excluded here).*

**Proof.** By Lemma 10, we know that <sup>I</sup>1−*<sup>α</sup>* <sup>0</sup><sup>+</sup> *y*(*t*, *λ*) is entire for each *t* ∈ [0, 1], and 1/2 < *α* < 1 as well. So, the eigenvalues of our problem are given by the zeros of <sup>I</sup>1−*<sup>α</sup>* <sup>0</sup><sup>+</sup> *y*(1, *λ*), which must be countably infinite in number since the latter function is of fractional order 1/2*α* (on account of the restriction on *α*). This gives us the existence of infinitely many eigenvalues, generally in **C**.

Next, we give the asymptotic distribution of these eigenvalues when *α* is either very close to 1/2 from the right or very close to 1 from the left. Recall (36) with *c*<sup>2</sup> = 1, so that

$$y(t,\lambda) = t^a E\_{2a,a+1}(-\lambda t^{2a}) - \int\_0^t \frac{E\_{2a,0}(-\lambda(t-s)^{2a})}{\lambda(t-s)} q(s) y(s,\lambda) \, ds. \tag{41}$$

An iterative method for solving for approximate solutions of (41) maybe found in [16]. Keeping in mind the boundary condition (2) at *<sup>t</sup>* <sup>=</sup> 1, we calculate <sup>I</sup>1−*<sup>α</sup>* <sup>0</sup><sup>+</sup> *y*(*t*, *λ*) and then evaluate this at *t* = 1 in order to find the dispersion relation for the eigenvalues. However, our derivation is theoretical in nature. A straightforward though lengthy calculation using (41) and the definition of the Mittag-Leffler functions show that

$$\begin{split} \mathcal{I}\_{0^{+}}^{1-a}y(t,\lambda) &= \mathcal{I}\_{0^{+}}^{1-a} \left\{ t^{a} E\_{2a,a+1}(-\lambda t^{2a}) \right\} + \mathcal{I}\_{0^{+}}^{1-a} \left( \int\_{0}^{t} \frac{E\_{2a,0}(-\lambda(t-s)^{2a})}{\lambda(t-s)} q(s) y(s,\lambda) \, ds \right) \\ &= t E\_{2a,2}(-\lambda t^{2a}) + \frac{1}{\lambda} \int\_{0}^{t} \frac{E\_{2a,1-a}(-\lambda(t-s)^{2a})}{(t-s)^{a}} q(s) y(s,\lambda) \, ds \end{split} \tag{42}$$

so that the eigenvalues of (1) and (2) are given by those *λ* ∈ **C** such that

$$E\_{2a,2}(-\lambda) + \frac{1}{\lambda} \int\_0^1 \frac{E\_{2a,1-a}(-\lambda(1-s)^{2a})}{(1-s)^a} q(s)y(s,\lambda) \, ds = 0. \tag{43}$$

Let us consider first the case where *λ* ∈ **R**. Lemma 8 implies that the right side of (38) tends to 0 as *λ* → ∞. Indeed this, combined with (39), implies that

$$\left| \frac{E\_{2a, 1-a}(-\lambda(t-s)^{2a})}{\lambda(t-s)^a} y(s, \lambda) \right| = \mathcal{O}\left(\frac{1}{\sqrt{|\lambda|}}\right).$$

for all sufficiently large *λ*.

Thus, the real eigenvalues of the problem (1) and (2) become the zeros of a transcendental equation of the form,

$$E\_{2\alpha,2}(-\lambda) + O\left(\frac{1}{\sqrt{\lambda}}\right) = 0.$$

We are concerned with the asymptotic behaviour of these real zeros. Recall the distribution of the real zeros of *E*2*α*,2(−*λ*) in [1]. There we showed that, for each *n* = 0, 1, 2, . . . , *N*<sup>∗</sup> − 1, where *N*<sup>∗</sup> depends on *α*, the interval

$$I\_n(a) := \left( \left( \frac{(2n + \frac{1}{2} + \frac{1}{2a})\pi}{\sin\left(\frac{\pi}{2a}\right)} \right)^{2a}, \quad \left( \frac{(2n + \frac{3}{2} + \frac{1}{2a})\pi}{\sin\left(\frac{\pi}{2a}\right)} \right)^{2a} \right),\tag{44}$$

always contains at least two real zeros of *E*2*α*,2(−*λ*). For *α* → 1, these intervals approach the intervals

$$\left(\left(2n+1\right)^2\pi^2, \left(2n+2\right)^2\pi^2\right),$$

whose end-points are each eigenvalues of the Dirichlet problem for the classical equation −*y* = *λ y* on [0, 1]. Since each interval *In* contains two zeros we can denote the first of these two zeros by *λ*2*n*(*α*). Equation (44) now gives the *a-priori* estimate

$$\left(\frac{(2n+\frac{1}{2}+\frac{1}{2n})\pi}{\sin(\frac{\pi}{2n})}\right)^{2n} \le \lambda\_{2n}(\alpha) \le \left(\frac{(2n+\frac{3}{2}+\frac{1}{2n})\pi}{\sin(\frac{\pi}{2n})}\right)^{2n}.\tag{45}$$

For each *α* < 1, and close to 1, and for large *λ*, the real zeros of the preceding equation approach those of *E*2*α*,2(−*λ*) and spread out towards the end-points of intervals of the form (44). For *α* close to 1/2 there are no zeros, the first two zeros appearing only when *α* ≈ 0.7325. For *α* larger than this critical value, the zeros appear in pairs and in intervals of the form (44).

Next, recall that for *α* < 1 there are only *finitely many* such real zeros, (see [1]) their number growing without bound as *α* → 1. It also follows from Lemma 11 that, for each *α*, the remaining infinitely many eigenvalues must be non-real. As *α* → 1<sup>−</sup> these non-real eigenvalues tend to the real axis thereby forming more and more real eigenvalues until the spectrum is totally real when *α* = 1 and the problem then reduces to a (classical) regular Sturm–Liouville problem.

Finally, for *α* close to 1, (45) leads to the approximation,

$$
\lambda\_{2n}(\alpha) \approx \left(\frac{(2n+2)\pi}{\sin\left(\frac{\pi}{2\alpha}\right)}\right)^{2\kappa},
$$

from which this, in conjunction with (44) and *α* → 1, we can derive the classical eigenvalue asymptotics, *<sup>λ</sup><sup>n</sup>* <sup>∼</sup> *<sup>n</sup>*2*π*<sup>2</sup> as *<sup>n</sup>* <sup>→</sup> <sup>∞</sup>.

#### **8. Conclusions**

We consider the fractional eigenvalue problem,

$$-^c \mathcal{D}\_{0^+}^\alpha \circ \mathcal{D}\_{0^+}^\alpha y(t) + q(t)y(t) = \lambda y(t), \quad 0 \le t \le 1,$$

where *α* is a real parameter, 1/2 < *α* < 1, *λ* is a generally unspecified complex parameter, with mixed Caputo and Riemann–Liouville derivatives and *q* an essentially bounded function, subject to the following boundary conditions involving the Riemann–Liouville integrals,

$$
\mathcal{T}\_{0^+}^{1-\alpha}y(t)|\_{t=0} = 0,\quad \text{and}\quad \mathcal{T}\_{0^+}^{1-\alpha}y(t)|\_{t=1} = 0.
$$

We show that this problem admits, for each *α* under consideration, and for eigenfunctions that are in *C*[0, 1], a finite number of real eigenvalues and an infinite number of non-real eigenvalues. The real eigenvalues, though finite in number for each *α*, are approximated by (44) and (45), which as *<sup>α</sup>* <sup>→</sup> 1 gives the classical asymptotic relation *<sup>λ</sup><sup>n</sup>* <sup>∼</sup> *<sup>n</sup>*2*π*<sup>2</sup> as *n* → ∞.

As *α* → 1<sup>−</sup> we observe that the spectrum obtained approaches the Sturm–Liouville spectrum of the classical problem

$$-y'' + q(t)y = \lambda y, \qquad y(0) = y(1) = 0.$$

The same results hold if the eigenfunctions are merely *C*(0, 1] (i.e., *c*<sup>1</sup> = 0) except that now the latter have an infinite discontinuity at *t* = 0 for each *α*. The proofs are identical and are therefore omitted.

**Author Contributions:** Conceptualization, M.D. and A.B.M.; methodology, M.D. and A.B.M.; software, M.D. and A.B.M.; validation, M.D. and A.B.M.; formal analysis, M.D. and A.B.M.; investigation, M.D. and A.B.M.; resources, M.D. and A.B.M.; data curation, Not applicable; writing—original draft preparation, M.D. and A.B.M.; writing—review and editing, M.D. and A.B.M.; visualization, Not applicable; supervision, M.D. and A.B.M.; project administration, M.D. and A.B.M.; funding acquisition, M.D. and A.B.M. All authors have read and agreed to the published version of the manuscript.

**Funding:** This research was partially supported by a grant from the Office of the Dean of Science, Carleton University.

**Institutional Review Board Statement:** Not applicable.

**Informed Consent Statement:** Not applicable.

**Data Availability Statement:** No data were used to support this study.

**Conflicts of Interest:** The authors declare no conflict of interest.

#### **References**

