**Oscillatory Behavior of a Type of Generalized Proportional Fractional Differential Equations with Forcing and Damping Terms**

**Jehad Alzabut 1,\* ,† , James Viji 2,†, Velu Muthulakshmi 2,† and Weerawat Sudsutad 3,†**


Received: 29 May 2020; Accepted: 23 June 2020; Published: 25 June 2020

**Abstract:** In this paper, we study the oscillatory behavior of solutions for a type of generalized proportional fractional differential equations with forcing and damping terms. Several oscillation criteria are established for the proposed equations in terms of Riemann-Liouville and Caputo settings. The results of this paper generalize some existing theorems in the literature. Indeed, it is shown that for particular choices of parameters, the obtained conditions in this paper reduce our theorems to some known results. Numerical examples are constructed to demonstrate the effectiveness of the our main theorems. Furthermore, we present and illustrate an example which does not satisfy the assumptions of our theorem and whose solution demonstrates nonoscillatory behavior.

**Keywords:** generalized proportional fractional operator; oscillation criteria; nonoscillatory behavior; damping and forcing terms

#### **1. Introduction**

Fractional calculus is a mathematical branch investigating the properties of derivatives and integrals of non-integer orders. The significance of this subject falls in the fact that the fractional derivative has the feature of nonlocal nature. This property makes these derivatives suitable to simulate more physical phenomena such as earthquake vibrations, polymers, and so forth; see, for example, References [1–10] and the references cited therein.

In recent years, there have appeared different types of fractional derivatives. However, it has been realized that most of these derivatives lose some of their basic properties that classical derivatives have such as the product rule and the chain rule. Fortunately, Khalil et al. [11] defined a new well-behaved fractional derivative, called the "conformable fractional derivative", which depends entirely on the classical limit definition of the derivative. Thereafter, researchers developed the conformable derivative and obtained different results exposing its features [12–14]. Recently, Jarad et al. [15] introduced the generalized proportional fractional (GPF) derivative of Caputo and Riemann-Liouville type involving exponential functions in their kernels. The GPF derivative not only preserves classical properties but also verifies semi group property and of nonlocal behavior. For recent results involving GPF derivative, one can refer to References [16–18].

In 2012, Grace et al. [19] initiated the study of oscillation theory for fractional differential equations. Thereafter, many researchers have investigated the oscillatory properties of fractional differential

equations; see for instance References [20–25]. In 2019, Aphithana et al. [24] studied forced oscillatory properties of solutions to the conformable initial value problem of the form

$$\begin{cases} \, \_aD^{1+a,\rho}x(t) + p(t)\_aD^{a,\rho}x(t) + q(t)f(x(t)) = g(t), \quad t > a, \\\lim\_{t \to a^+} \mathcal{J}^{j-a,\rho}x(t) = b\_j \quad (j = 1, 2, \dots, m), \end{cases}$$

where *<sup>m</sup>* <sup>=</sup> ⌈*α*⌉, 0 <sup>&</sup>lt; *<sup>ρ</sup>* <sup>≤</sup> 1,*p*, *<sup>g</sup>* <sup>∈</sup> <sup>C</sup>(R+, <sup>R</sup>), *<sup>q</sup>* <sup>∈</sup> <sup>C</sup>(R+, <sup>R</sup>+), *<sup>f</sup>* <sup>∈</sup> <sup>C</sup>(R, <sup>R</sup>) are continuous functions, *<sup>a</sup>Dα*,*<sup>ρ</sup>* is the left conformable derivative of order *<sup>α</sup>* <sup>∈</sup> <sup>C</sup> of *<sup>x</sup>*, *Re*(*α*) <sup>≥</sup> 0 in the Riemann-Liouville setting and *<sup>a</sup>*J *j*−*α*,*ρ* is the left conformable integral operator of order *<sup>j</sup>* <sup>−</sup> *<sup>α</sup>* <sup>∈</sup> <sup>C</sup>, *<sup>b</sup><sup>j</sup>* <sup>∈</sup> <sup>R</sup>, *<sup>j</sup>* <sup>=</sup> 1, 2, . . . , *<sup>m</sup>*.

They also studied the forced oscillation of conformable initial value problems in the Caputo setting of the form

$$\begin{cases} \ \_a^C D^{1+a,\rho} \mathbf{x}(t) + p(t) \mathbf{\bar{f}} \, \_a^D D^{a,\rho} \mathbf{x}(t) + q(t) f(\mathbf{x}(t)) = \mathbf{g}(t), \quad t > a, \\\\ \ \_a^k D^{\rho} \mathbf{x}(a) = b\_k \quad (k = 0, 1, \dots, m - 1), \end{cases}$$

where *<sup>m</sup>* <sup>=</sup> ⌈*α*⌉, 0 <sup>&</sup>lt; *<sup>ρ</sup>* <sup>≤</sup> 1, and *<sup>C</sup> <sup>a</sup> Dα*,*<sup>ρ</sup>* is the left conformable derivative of order *<sup>α</sup>* <sup>∈</sup> <sup>C</sup> of *<sup>x</sup>*, *Re*(*α*) <sup>≥</sup> <sup>0</sup> in the Caputo setting.

In 2020, Sudsutad et al. [26] established some oscillation criteria for the following generalized proportional fractional differential equation

$$\begin{cases} \, \_aD^{a,\rho}x(t) + \xi\_1(t, x(t)) = \mu(t) + \xi\_2(t, x(t)), \quad t > a \ge 0, \\\lim\_{t \to a^+} \, \_aI^{j-a,\rho}x(t) = b\_j \quad (j = 1, 2, \dots, n), \end{cases}$$

with *<sup>n</sup>* <sup>=</sup> ⌈*α*⌉, *<sup>a</sup>Dα*,*<sup>ρ</sup>* is the generalized proportional fractional derivative operator of order *<sup>α</sup>* <sup>∈</sup> <sup>C</sup>, *Re*(*α*) ≥ 0, 0 < *ρ* ≤ 1 in the Riemann-Liouville setting and *<sup>a</sup> I α*,*ρ* is the generalized proportional fractional integral operator.

In this paper, motivated by the above papers, we establish some sufficient conditions for forced oscillation criteria of all solutions of the generalized proportional fractional (GPF) initial value problem with damping term in the Riemann-Liouville type of the form:

$$\begin{cases} \, \_aD^{1+a,\rho}y(l) + p(l)\_aD^{a,\rho}y(l) + q(l)f(y(l)) = g(l), \quad l > a \ge 0, \\\lim\_{l \to a^+} \, \_al^{j-a,\rho}y(l) = b\_{\bar{j}} \quad (j = 1, 2, \dots, m), \end{cases} \tag{1}$$

where *<sup>m</sup>* <sup>=</sup> ⌈*α*⌉, 0 <sup>&</sup>lt; *<sup>ρ</sup>* <sup>≤</sup> 1, *<sup>a</sup>Dα*,*<sup>ρ</sup>* is the left GPF derivative of order *<sup>α</sup>* <sup>∈</sup> <sup>C</sup> of *<sup>y</sup>*, *Re*(*α*) <sup>≥</sup> 0 in the Riemann-Liouville setting and *<sup>a</sup> I j*−*α*,*ρ* is the left GPF integral of order *<sup>j</sup>* <sup>−</sup> *<sup>α</sup>* <sup>∈</sup> <sup>C</sup>, *Re*(*<sup>j</sup>* <sup>−</sup> *<sup>α</sup>*) <sup>&</sup>gt; 0, *<sup>b</sup><sup>j</sup>* <sup>∈</sup> <sup>R</sup>, *<sup>j</sup>* <sup>=</sup> 1, 2, . . . , *<sup>m</sup>* and *<sup>p</sup>*, *<sup>g</sup>* <sup>∈</sup> <sup>C</sup>(R+, <sup>R</sup>), *<sup>q</sup>* <sup>∈</sup> <sup>C</sup>(R+, <sup>R</sup>+), *<sup>f</sup>* <sup>∈</sup> <sup>C</sup>(R, <sup>R</sup>).

Moreover, we study the forced oscillation criteria of all solutions of the GPF initial value problem with damping term in the Caputo type of the form

$$\begin{aligned} \, ^C\_a D^{1+a,\rho} y(l) + p(l) \, ^C\_a D^{a,\rho} y(l) + q(l) f(y(l)) &= g(l), \quad l > a \ge 0, \\\, ^D\_c D^{k,\rho} y(a) = b\_k \quad (k = 0, 1, \dots, n-1), \end{aligned} \tag{2}$$

where *<sup>n</sup>* <sup>=</sup> ⌈*α*⌉, 0 <sup>&</sup>lt; *<sup>ρ</sup>* <sup>≤</sup> 1, *<sup>C</sup> <sup>a</sup> Dα*,*<sup>ρ</sup>* is the left GPF derivative of order *<sup>α</sup>* <sup>∈</sup> <sup>C</sup> of *<sup>y</sup>*, *Re*(*α*) <sup>≥</sup> 0 in the Caputo setting and *Dk*,*<sup>ρ</sup>* = *D <sup>ρ</sup>D ρ* · · · *D ρ* | {z } k times , and *D<sup>ρ</sup>* is the proportional derivative defined in Reference [13].

We claim that the results of this paper improve and generalize previously existing oscillation results in Reference [24].

**Definition 1.** *The solution y of problem* (1) *(respectively* (2)*) is called oscillatory if it has arbitrarily large zeros on* (0, ∞)*; otherwise, it is called nonoscillatory. An equation is called oscillatory if all its solutions are oscillatory.*

#### **2. Preliminaries**

In this section, we provide some basic definitions and results which will be used throughout this paper. For the justifications and proofs, the reader can consult References [13,15].

#### **Definition 2.** *[15] (Modified Conformable Derivatives).*

*For <sup>ρ</sup>* <sup>∈</sup> [0, 1]*, let the functions <sup>k</sup>*0, *<sup>k</sup>*<sup>1</sup> : [0, 1] <sup>×</sup> <sup>R</sup> <sup>→</sup> [0, <sup>∞</sup>) *be continuous such that for all <sup>l</sup>* <sup>∈</sup> <sup>R</sup> *we have*

$$\lim\_{\rho \to 0^+} k\_1(\rho, l) = 1, \quad \lim\_{\rho \to 0^+} k\_0(\rho, l) = 0, \quad \lim\_{\rho \to 1^-} k\_1(\rho, l) = 0, \quad \lim\_{\rho \to 1^-} k\_0(\rho, l) = 1,\tag{3}$$

*and k*1(*ρ*, *l*) 6= 0*, ρ* ∈ [0, 1)*, k*0(*ρ*, *l*) 6= 0*, ρ* ∈ (0, 1]*.*

Then, Anderson et al. [13] defined the modified conformable differential operator of order *ρ* by

$$D^{\rho}f(l) = k\_1(\rho, l)f(l) + k\_0(\rho, l)f'(l),\tag{4}$$

provided that the right-hand side exists at *<sup>l</sup>* <sup>∈</sup> <sup>R</sup> and *<sup>f</sup>* ′ (*l*) = *<sup>d</sup> dl f* . The derivative given in (4) is called a proportional derivative. For more details about the control theory of the proportional derivatives and its component functions *k<sup>0</sup>* and *k1*, we refer the reader to [27].

Of special interest, we shall restrict ourselves to the case when *k*1(*ρ*, *l*) = (1 − *ρ*) and *k*0(*ρ*, *l*) = *ρ*. Therefore, (4) becomes

$$D^{\rho}f(l) = (1 - \rho)f(l) + \rho f'(l). \tag{5}$$

Notice that lim *<sup>ρ</sup>*→0<sup>+</sup> *D<sup>ρ</sup> f*(*l*) = *f*(*l*) and lim *ρ*→1<sup>−</sup> *D<sup>ρ</sup> f*(*l*) = *f* ′ (*l*). It is clear that the derivative (5) is somehow more general than the conformable derivative which does not tend to the original function as *ρ* tends to 0.

To find the associated integral to the proportional derivative in (5), we solve the following equation

$$D^\rho g(l) = (1 - \rho)g(l) + \rho g'(l) = f(l), \quad l \ge a.$$

The above equation is a first order linear differential equation and its solution is given by

$$\mathcal{g}(l) = \frac{1}{\rho} \int\_a^l e^{\frac{\rho - 1}{\rho}(l - s)} f(s) ds.$$

Define the proportional integral associated to *D<sup>ρ</sup>* by

$$\,\_aI^{1,\rho}f(l) = \frac{1}{\rho} \int\_a^l e^{\frac{\rho - 1}{\rho}(l-s)} f(s)ds,\tag{6}$$

where we accept that *<sup>a</sup> I* 0,*ρ f*(*l*) = *f*(*l*).

**Lemma 1.** *[15] Let f be defined on* [*a*, <sup>∞</sup>) *and differentiable on* (*a*, <sup>∞</sup>) *and <sup>ρ</sup>* ∈ (0, 1]*. Then, we have*

$$
\partial\_a I^{1,\rho} D^{\rho} f(l) = f(l) - e^{\frac{\rho - 1}{\rho}(l - a)} f(a). \tag{7}
$$

**Definition 3.** *[15] For <sup>ρ</sup>* <sup>∈</sup> (0, 1] *and <sup>α</sup>* <sup>∈</sup> <sup>C</sup>*, Re*(*α*) <sup>&</sup>gt; 0, *we define the left GPF integral of f by*

$$(\,\_a\mathrm{I}^{a,\rho}f)(l) = \frac{1}{\rho^a \Gamma(a)} \int\_a^l e^{\frac{\rho-1}{\rho}(l-s)} (l-s)^{a-1} f(s) ds = \rho^{-a} \,\_a e^{\frac{\rho-1}{\rho}l} \,\_a \mathrm{I}^a \left( e^{\frac{1-\rho}{\rho}l} f(l) \right), \tag{8}$$

*where <sup>a</sup> I α is the left Riemann-Liouville fractional integral of order α.*

*The right GPF integral ending at b, however, can be defined by*

$$(I\_b^{a,\rho}f)(l) = \frac{1}{\rho^a \Gamma(a)} \int\_l^b e^{\frac{\rho - 1}{\rho}(s - l)} (s - l)^{a - 1} f(s) ds = \rho^{-a} e^{\frac{\rho - 1}{\rho}l} I\_b^a \left( e^{\frac{1 - \rho}{\rho}l} f(l) \right), \tag{9}$$

*where I<sup>α</sup> b is the right Riemann-Liouville fractional integral of order α.*

**Definition 4.** *[15] For <sup>ρ</sup>* <sup>∈</sup> (0, 1] *and <sup>α</sup>* <sup>∈</sup> <sup>C</sup>*, Re*(*α*) <sup>≥</sup> 0, *we define the left GPF derivative of f by*

$$\begin{aligned} (\,\_{a}D^{a,\rho}f)(l) &= \,\_{a}D^{a,\rho}\,\_{a}I^{n-a,\rho}f(l) \\ &= \,\_{\rho}\frac{D^{n,\rho}\_{l}}{\rho^{n-a}\Gamma(n-a)}\int\_{a}^{l}e^{\frac{\rho-1}{\rho}(l-s)}(l-s)^{n-a-1}f(s)ds.\end{aligned} \tag{10}$$

*The right GPF derivative ending at b is defined by*

$$\begin{aligned} (\boldsymbol{D}\_b^{\boldsymbol{a},\boldsymbol{\rho}}f)(l) &= \ \_ {\ominus} \boldsymbol{D}^{\boldsymbol{a},\boldsymbol{\rho}} \boldsymbol{I}\_b^{\boldsymbol{n}-\boldsymbol{a},\boldsymbol{\rho}} f(l) \\ &= \ \_ {\boldsymbol{\rho}} \boldsymbol{D}\_l^{\boldsymbol{n},\boldsymbol{\rho}} {}\_ {\boldsymbol{\rho}} \int\_l ^ {b} {}\_ {\boldsymbol{\rho}} {}^ {\boldsymbol{\rho}-1} (\boldsymbol{s}-l)^{\boldsymbol{n}-\boldsymbol{a}-1} f(\boldsymbol{s}) \boldsymbol{s} \, \end{aligned} \tag{11}$$

*where n* = [*Re*(*α*)] + 1*.*

If we let *ρ* = 1 in Definition 4 , then one can obtain the left and right Riemann-Liouville fractional derivatives as in [6]. Moreover, it is clear that

$$\lim\_{\alpha \to 0} D^{\alpha, \rho} f(l) = f(l) \text{ and } \lim\_{\alpha \to 1} D^{\alpha, \rho} f(l) = D^{\rho} f(l).$$

**Lemma 2.** *[15] Let Re*(*α*) > 0, *n* = −[−*Re*(*α*)], *f* ∈ *L*1(*a*, *b*) *and* (*<sup>a</sup> I α*,*ρ <sup>f</sup>*)(*l*) <sup>∈</sup> *AC<sup>n</sup>* [*a*, *b*]. *Then,*

$$({}\_a 1^{a, \rho} {}\_a \mathbf{D}^{a, \rho} f)(l) = f(l) - e^{\frac{\rho - 1}{\rho} (l - a)} \sum\_{j = 1}^n ({}\_a 1^{j - a, \rho} f)(a^+) \frac{(l - a)^{a - j}}{\rho^{a - j} \Gamma(a + 1 - j)}. \tag{12}$$

**Definition 5.** *[13] (Partial Conformable Derivatives). Let <sup>ρ</sup>* <sup>∈</sup> [0, 1]*, and let the functions <sup>k</sup>*0, *<sup>k</sup>*<sup>1</sup> : [0, 1] <sup>×</sup> <sup>R</sup> <sup>→</sup> [0, <sup>∞</sup>) *be continuous and satisfy* (3)*. Given a function <sup>f</sup>* : <sup>R</sup><sup>2</sup> <sup>→</sup> <sup>R</sup> *such that <sup>∂</sup> ∂l f*(*l*,*s*) *exists for each fixed <sup>s</sup>* <sup>∈</sup> <sup>R</sup>*, define the partial differential operator D<sup>ρ</sup> l via*

$$D\_{l}^{\rho}f(l,s) = k\_{1}(\rho,l)f(l,s) + k\_{0}(\rho,l)\frac{\partial}{\partial l}f(l,s). \tag{13}$$

**Definition 6.** *[13] (Conformable Exponential Function). Let <sup>ρ</sup>* <sup>∈</sup> (0, 1]*, the points <sup>s</sup>*, *<sup>l</sup>*, <sup>∈</sup> <sup>R</sup> *with <sup>s</sup>* <sup>≤</sup> *l, and let the function <sup>p</sup>* : [*s*, *<sup>l</sup>*] <sup>→</sup> <sup>R</sup> *be continuous. Let <sup>k</sup>*0, *<sup>k</sup>*<sup>1</sup> : [0, 1] <sup>×</sup> <sup>R</sup> <sup>→</sup> [0, <sup>∞</sup>) *be continuous and satisfy* (3)*, with p*/*k*<sup>0</sup> *and k*1/*k*<sup>0</sup> *Riemann integrable on* [*s*, *l*]*. Then the exponential function with respect to D<sup>ρ</sup> in* (4) *is defined to be*

$$e\_p(l,s) := e^{\int\_s^l \frac{p(\tau) - k\_1(\rho, \tau)}{k\_0(\rho, \tau)} d\tau}, \quad e\_0(l,s) := e^{-\int\_s^l \frac{k\_1(\rho, \tau)}{k\_0(\rho, \tau)} d\tau}. \tag{14}$$

*Using* (4) *and* (14)*, we have the following basic results.*

**Lemma 3.** *[13] (Basic Derivatives). Let the conformable differential operator D<sup>ρ</sup> be given as in* (4)*, where <sup>ρ</sup>* <sup>∈</sup> [0, 1]*. Let the function <sup>p</sup>* : [*s*, *<sup>l</sup>*] <sup>→</sup> <sup>R</sup> *be continuous. Let <sup>k</sup>*0, *<sup>k</sup>*<sup>1</sup> : [0, 1] <sup>×</sup> <sup>R</sup> <sup>→</sup> [0, <sup>∞</sup>) *be continuous and satisfy* (3)*, with p*/*k*<sup>0</sup> *and k*1/*k*<sup>0</sup> *Riemann integrable on* [*s*, *l*]*. Assume the functions f and g are differentiable as needed. Then*

(*i*) *D<sup>ρ</sup>* [*a f* + *bg*] = *aD<sup>ρ</sup>* [ *f* ] + *bD<sup>ρ</sup>* [*g*] *for all a*, *<sup>b</sup>* <sup>∈</sup> <sup>R</sup>*;*


$$D\_l^\rho[e\_p(l,s)] = p(l)e\_p(l,s)$$

*for ep*(*l*,*s*) *given in* (14)*;*

(*vi*) *for ρ* ∈ (0, 1] *and for the exponential function e*<sup>0</sup> *given in* (14)*, we have*

$$D^{\rho} \left[ \int\_{a}^{l} \frac{f(s)e\_{0}(l,s)}{k\_{0}(\rho,s)} ds \right] = f(l).$$

**Definition 7.** *[15] For <sup>ρ</sup>* <sup>∈</sup> (0, 1] *and <sup>α</sup>* <sup>∈</sup> <sup>C</sup> *with Re*(*α*) <sup>≥</sup> <sup>0</sup>*, we define the left GPF derivative of Caputo type starting at a by*

$$\begin{aligned} (^{\mathbb{C}}\_{d}D^{a,\rho}f)(l) &= \,\_{d}I^{n-a,\rho}(D^{a,\rho}f)(l) \\ &= \,\_{\rho} \frac{1}{\rho^{n-a}\Gamma(n-a)} \int\_{a}^{l} e^{\frac{\rho-1}{\rho}(l-s)} (l-s)^{n-a-1} (D^{a,\rho}f)(s) ds. \end{aligned} \tag{15}$$

*The right GPF derivative of Caputo ending at b is defined by*

$$\begin{aligned} (\prescript{\mathsf{C}}{}{\mathcal{D}}\_{b}^{a,\rho}f)(l) &= \prescript{\mathsf{I}}{}{\mathcal{I}}\_{b}^{n-a,\rho} \left( {}\_{\ominus} \prescript{\mathsf{I}}{}{\mathcal{D}}\_{f}^{a,\rho}f \right)(l) \\ &= \frac{1}{\rho^{n-a} \Gamma(n-a)} \int\_{l}^{b} e^{\frac{\rho - 1}{\rho}(s-l)} (s-l)^{n-a-1} (\prescript{\mathsf{I}}{}{\mathcal{D}}^{a,\rho}f)(s) ds, \end{aligned} \tag{16}$$

*where n* = [*Re*(*α*)] + 1*.*

**Lemma 4.** *[15] For ρ* ∈ (0, 1] *and n* = [*Re*(*α*)] + 1*, we have*

$$\,\_aI^{a,\rho}(\,\_a^C D^{a,\rho} f)(l) = f(l) - \sum\_{k=0}^{n-1} \frac{(D^{k,\rho} f)(a)}{\rho^k k!} (l-a)^k e^{\frac{\rho - 1}{\rho}} (l-a). \tag{17}$$

**Proposition 1.** *[15] Let <sup>α</sup>*, *<sup>β</sup>* <sup>∈</sup> <sup>C</sup> *be such that Re*(*α*) <sup>≥</sup> <sup>0</sup> *and Re*(*β*) <sup>&</sup>gt; <sup>0</sup>*. Then, for any* <sup>0</sup> <sup>&</sup>lt; *<sup>ρ</sup>* <sup>≤</sup> <sup>1</sup> *and n* = [*Re*(*α*)] + 1*, we have*

$$\text{(i)}\quad \left( {}\_{a}I^{a}{}^{\rho}e^{\frac{\rho-1}{\rho}l}(l-a)^{\beta-1} \right) \left( y \right) = \frac{\Gamma(\beta)}{\Gamma(\beta+a)\rho^{a}} e^{\frac{\rho-1}{\rho}y} (y-a)^{\beta+a-1}, \quad \text{Re}(a) > 0.$$

$$\text{2(ii)}\quad \left(\_{a}D^{a,\rho}e^{\frac{\rho-1}{\rho}l}(l-a)^{\beta-1}\right)(y) = \frac{\rho^a \Gamma(\beta)}{\Gamma(\beta-a)}e^{\frac{\rho-1}{\rho}y}(y-a)^{\beta-a-1}, \quad \text{Re}(a) \ge 0.$$

$$\text{(iii)} \quad \left( {}^{\mathbb{C}}\_{a}D^{a,\rho}e^{\frac{\rho-1}{\rho}l}(l-a)^{\beta-1} \right)(y) = \frac{\rho^{a}\Gamma(\beta)}{\Gamma(\beta-a)}e^{\frac{\rho-1}{\rho}y}(y-a)^{\beta-a-1}, \quad \text{Re}(a) > n.$$

#### **3. Oscillation Results via Riemann-Liouville Operator**

In this section, we establish the oscillation criteria for the GPF initial value problem (1). We prove our results under the following assumption:

 $\text{If } (H) \text{ } p \in \mathbb{C}(\mathbb{R}^+, \mathbb{R}), q \in \mathbb{C}(\mathbb{R}^+, \mathbb{R}^+), \text{ $g \in \mathbb{C}(\mathbb{R}^+, \mathbb{R})$ ,  $f \in \mathbb{C}(\mathbb{R}, \mathbb{R})$  with  $\frac{f(u)}{u} > 0$  for all  $u \neq 0$ .}$ 

For our convenience, we set the following notations:

$$\Phi(l) \quad := \quad \Gamma(a)e^{\frac{\rho-1}{\rho}(l-a)}\sum\_{j=1}^{m} \frac{\rho^j b\_j(l-a)^{a-j}}{\Gamma(a+1-j)},\tag{18}$$

$$\Lambda(l,L) \quad := \int\_{a}^{L} e^{\frac{\rho-1}{\rho}(l-s)} (l-s)^{a-1} \left( \frac{e^{\frac{\rho-1}{\rho}(s-l\_{l})}M + l\_{l\_{l}}I^{1,\rho}\left(\rho\lg(s)V(s)\right)}{V(s)} \right) ds,\tag{19}$$

$$V(l) \quad := \quad \exp \int\_{l\_1}^{l} \frac{\rho p(\tau) - (1 - \rho)}{\rho} d\tau,\tag{20}$$

$$M \quad := \quad {}\_{l\_1}D^{a,\rho}y(l\_1)V(l\_1), \quad M \text{ is an arbitrary constant.} \tag{21}$$

**Theorem 1.** *Assume that* (*H*) *holds. If*

$$\lim\_{l\to\infty} \inf l^{1-\alpha} \int\_{L}^{l} e^{\frac{\rho-1}{\rho}(l-s)} (l-s)^{\alpha-1} \left[ \frac{e^{\frac{\rho-1}{\rho}(s-l\_1)}M + \,\_{l\_1}I^{1,\rho} \left(\rho g(s)V(s)\right)}{V(s)} \right] ds = -\infty \tag{22}$$

*and*

$$\lim\_{l\to\infty} \sup l^{1-a} \int\_{L}^{l} e^{\frac{\rho-1}{\rho}(l-s)} (l-s)^{a-1} \left[ \frac{e^{\frac{\rho-1}{\rho}(s-l\_1)} M + \,\_1I^{1,\rho} \left( \rho g(s) V(s) \right)}{V(s)} \right] ds = \infty,\tag{23}$$

*for every sufficiently large L, where V*(*l*) *and M are defined as in* (20) *and* (21) *respectively, then every solution of problem* (1) *is oscillatory.*

**Proof.** Suppose that *y*(*l*) is a nonoscillatory solution of problem (1). Without loss of generality, let *L* > *a* be large enough and *l*<sup>1</sup> ≥ *L* such that *y*(*l*) > 0 for all *l* ≥ *l*1. Using Lemma 3 (iii), Equations (5) and (13), we have

$$\begin{split} \mathcal{D}^{\rho} \left[ {}\_{a} \mathcal{D}^{\mu,\rho} \mathcal{y}(l) V(l) \right] &= \ {}\_{a} \mathcal{D}^{\mu,\rho} \mathcal{y}(l) D^{\rho} V(l) + V(l) D^{\rho} \left( {}\_{a} \mathcal{D}^{\mu,\rho} \mathcal{y}(l) \right) - (1 - \rho) {}\_{a} \mathcal{D}^{\mu,\rho} \mathcal{y}(l) V(l) \\ &= \ {}\_{a} \mathcal{D}^{\mu,\rho} \mathcal{y}(l) D^{\rho} V(l) + V(l) \left[ (1 - \rho) {}\_{a} D^{\mu,\rho} \mathcal{y}(l) + \rho \frac{d}{dl} \left( {}\_{a} D^{\mu,\rho} \mathcal{y}(l) \right) \right] \\ &- (1 - \rho) {}\_{a} D^{\mu,\rho} \mathcal{y}(l) V(l) \\ &= \ \rho \left[ {}\_{a} \mathcal{D}^{1 + \mu,\rho} \mathcal{y}(l) + \mathcal{p}(l)\_{a} D^{\mu,\rho} \mathcal{y}(l) \right] V(l) \\ &= \ \rho \left[ -\mathcal{q}(l) f(\mathcal{y}(l)) + \mathcal{g}(l) \right] V(l) \\ &< \rho g(l) V(l). \end{split}$$

Taking the proportional integral operator *<sup>l</sup>*<sup>1</sup> *I* 1,*<sup>ρ</sup>* on both sides to the above inequality, we obtain

$$\leq\_{l\_1} I^{1,\rho} \left( D^{\rho} \left[ \_a D^{a,\rho} y(l) V(l) \right] \right) < \,\_1 I^{1,\rho} \left( \rho g(l) V(l) \right) \,. \tag{24}$$

Using Lemma 1 on the L.H.S of (24), we have

$$\,\_{a}D^{\alpha}\!\_{\ell}\mathcal{Y}(l) < \frac{e^{\frac{\rho-1}{\ell}(l-l\_{1})}\mathcal{M} + \,\_{l\_{1}}I^{1,\rho}\left(\rho\mathcal{g}(l)V(l)\right)}{V(l)}.$$

Taking the left GPF integral operator *<sup>a</sup> I <sup>α</sup>*,*ρ*on both sides to the above inequality, we get

$$\delta\_{a^{\;l}}I^{a,\rho}\left(\_{a}D^{a,\rho}y(l)\right) < \,\_{a}I^{a,\rho}\left[\frac{e^{\frac{\rho-1}{\rho}(l-l\_{1})}M + \_{l\_{1}}I^{1,\rho}\left(\rho g(l)V(l)\right)}{V(l)}\right].\tag{25}$$

Using Lemma 2 on the L.H.S of (25), we have

$$y(l) - e^{\frac{\rho - 1}{\rho}(l - a)} \sum\_{j=1}^{m} \frac{b\_j(l - a)^{a - j}}{\rho^{a - j} \Gamma(a + 1 - j)} < \,\_4I^{\mathbb{1}, \rho} \left[ \frac{e^{\frac{\rho - 1}{\rho}(l - l\_1)} M + \,\_1I^{1, \rho} \left( \rho g(l) V(l) \right)}{V(l)} \right]. \tag{26}$$

Applying the left GPF integral formula on the R.H.S of (26) , we have

$$\begin{aligned} \left(y(l) \right) &< \left. e^{\frac{\rho - 1}{\rho}(l - a)} \sum\_{j=1}^{m} \frac{b\_j(l - a)^{a - j}}{\rho^{a - j} \Gamma(a + 1 - j)} \\ &+ \frac{1}{\rho^a \Gamma(a)} \int\_a^l e^{\frac{\rho - 1}{\rho}(l - s)} (l - s)^{a - 1} \left[ \frac{e^{\frac{\rho - 1}{\rho}(s - l\_1)} M + \iota\_1 I^{1, \rho} \left(\rho \operatorname{g}(s) V(s)\right)}{V(s)} \right] ds, \end{aligned}$$

for every sufficiently large *L*. If we multiply the above inequality by *ρ <sup>α</sup>*Γ(*α*), we get

$$\begin{split} \rho^{a}\Gamma(a)y(l) &< \quad \Gamma(a)e^{\frac{\rho-1}{\rho}(l-a)} \sum\_{j=1}^{m} \frac{\rho^{l}b\_{j}(l-a)^{a-j}}{\Gamma(a+1-j)} \\ &+ \int\_{a}^{l} \frac{e^{\frac{\rho-1}{\rho}(l-s)}}{e^{\rho}}(l-s)^{a-1} \left[ \frac{e^{\frac{\rho-1}{\rho}(s-l\_{1})} \, M + \, l\_{1}}{V(s)} \frac{\rho(\rho(s)V(s))}{\,} \right] ds \\ &+ \int\_{L}^{l} e^{\frac{\rho-1}{\rho}(l-s)}(l-s)^{a-1} \left[ \frac{e^{\frac{\rho-1}{\rho}(s-l\_{1})} \, M + \, l\_{1} \, l^{1,\rho} \left(\rho g(s)V(s)\right)}{V(s)} \right] ds \\ &= \quad \Phi(l) + \Lambda(l,L) + \int\_{L}^{l} \frac{e^{\frac{\rho-1}{\rho}(l-s)}}{e^{\rho}}(l-s)^{a-1} \left[ \frac{e^{\frac{\rho-1}{\rho}(s-l\_{1})} \, M + \, l\_{1} \, l^{1,\rho} \left(\rho g(s)V(s)\right)}{V(s)} \right] ds, \end{split} \tag{27}$$

where Φ(*l*) and Λ(*l*, *L*) are defined in (18) and (19), respectively.

Multiplying (27) by *l* 1−*α* , we get

$$\begin{split} 0 &< \quad l^{1-a} \rho^a \Gamma(a) y(l) \\ &< \quad l^{1-a} \Phi(l) + l^{1-a} \Lambda(l, L) \\ &+ l^{1-a} \int\_L e^{\frac{\rho-1}{\rho}(l-s)} (l-s)^{a-1} \left[ \frac{e^{\frac{\rho-1}{\rho}(s-l\_1)} M + \,\_1I\_1^{1, \rho} \left( \rho g(s) V(s) \right)}{V(s)} \right] ds. \end{split} \tag{28}$$

Let us consider the following two cases for *L*<sup>1</sup> ≥ *L*. Case(i): Let 0 < *α* ≤ 1. Then *m* = 1. Since *e ρ*−1 *ρ* (*l*−*a*) <sup>≤</sup> 1 and the function *<sup>h</sup>*1(*l*) = *l*−*a l α*−<sup>1</sup> is decreasing for *ρ* > 0 , 0 < *α* < 1 , we get for *l* ≥ *L*1,

$$\left| l^{1-a} \Phi(l) \right| = \left| e^{\frac{\rho - 1}{\rho} (l - a)} \rho b\_1 \left( \frac{l - a}{l} \right)^{a - 1} \right| \le \rho |b\_1| \left( \frac{L\_1 - a}{L\_1} \right)^{a - 1} := \mathbb{C}\_1(L\_1), \tag{29}$$

and

$$\begin{split} \left| l^{1-\mathfrak{a}} \Lambda(l,L) \right| &= \left| l^{1-\mathfrak{a}} \int\_{a}^{L} e^{\frac{\rho-1}{\rho}(l-s)} (l-s)^{\mathfrak{a}-1} \left( \frac{e^{\frac{\rho-1}{\rho}(s-l\_{l})} M +\_{l\_{l}} \mathfrak{l}^{1,\mathfrak{a}} \left( \rho \mathfrak{g}(s) V(s) \right)}{V(s)} \right) ds \right| \\ &\leq \int\_{a}^{L} \left| e^{\frac{\rho-1}{\rho}(l-s)} \right| \left( \frac{l-a}{l} \right)^{a-1} \left| \frac{e^{\frac{\rho-1}{\rho}(s-l\_{l})} M +\_{l\_{l}} \mathfrak{l}^{1,\mathfrak{a}} \left( \rho \mathfrak{g}(s) V(s) \right)}{V(s)} \right| ds \\ &\leq \int\_{a}^{L} \left( \frac{L\_{1} - a}{L\_{1}} \right)^{a-1} \left| \frac{e^{\frac{\rho-1}{\rho}(s-l\_{l})} M +\_{l\_{l}} \mathfrak{l}^{1,\mathfrak{a}} \left( \rho \mathfrak{g}(s) V(s) \right)}{V(s)} \right| ds \\ &:=: \quad \mathcal{C}\_{2}(L, L\_{1}). \end{split} \tag{30}$$

From (28), (29) and (30), we get, for *l* ≥ *L*1,

$$\left[I^{1-a}\int\_{L}^{l}e^{\frac{\rho-1}{\rho}(l-s)}(l-s)^{a-1}\left[\frac{e^{\frac{\rho-1}{\rho}(s-l\_1)}M+l\_1I^{1,\rho}\left(\rho g(s)V(s)\right)}{V(s)}\right]ds \geq -[\mathbb{C}\_1(L\_1)+\mathbb{C}\_2(L\_1L\_1)].$$

Since the R.H.S of the above inequality is a negative constant, it follows that

$$\lim\_{l\to\infty} \inf l^{1-\alpha} \int\_L^l e^{\frac{\rho-1}{\rho}(l-s)} (l-s)^{\alpha-1} \left[ \frac{e^{\frac{\rho-1}{\rho}(s-l\_1)}M + \_{l\_1}I^{1,\rho} \left(\rho g(s)V(s)\right)}{V(s)} \right] ds > -\infty\rho$$

which leads to a contradiction with (22).

Case(ii): Let *<sup>α</sup>* <sup>&</sup>gt; 1. Then *<sup>m</sup>* <sup>≥</sup> 2 and *l*−*a l α*−<sup>1</sup> < 1 for *α* > 1 and *ρ* > 0. Since *e ρ*−1 *ρ* (*l*−*a*) <sup>≤</sup> 1 and the function *h*2(*l*) = (*l* − *a*) 1−*j* is decreasing for *j* > 1 and *ρ* > 0, for *l* ≥ *L*1, we have

$$\begin{split} \left| l^{1-a} \Phi(l) \right| &= \left| l^{1-a} \Gamma(a) e^{\frac{\rho - 1}{\rho} (l - a)} \sum\_{j=1}^{m} \frac{\rho^{j} b\_{j} (l - a)^{a - j}}{\Gamma(a + 1 - j)} \right| \\ &\leq \quad \Gamma(a) \left( \frac{l - a}{l} \right)^{a - 1} \sum\_{j=1}^{m} \frac{\rho^{j} |b\_{j}| (l - a)^{1 - j}}{\Gamma(a + 1 - j)} \\ &\leq \quad \Gamma(a) \sum\_{j=1}^{m} \frac{\rho^{j} |b\_{j}| (L\_{1} - a)^{1 - j}}{\Gamma(a + 1 - j)} \\ &:= \quad \mathcal{C}\_{3}(L\_{1}), \end{split} \tag{31}$$

and

$$\begin{split} \left| \left| I^{1-\mathfrak{a}} \Lambda(l,L) \right| \right| &= \left| \int\_{a}^{L} e^{\frac{\rho - 1}{\rho}(l-s)} \left( \frac{l-s}{l} \right)^{a-1} \left( \frac{e^{\frac{\rho - 1}{\rho}(s-l\_{l})} M + l\_{l} I^{1,\rho} \left( \rho g(s) V(s) \right)}{V(s)} \right) ds \right| \\ &\leq \int\_{a}^{L} \left| \frac{e^{\frac{\rho - 1}{\rho}(s-l\_{l})} M + l\_{l} I^{1,\rho} \left( \rho g(s) V(s) \right)}{V(s)} \right| ds \\ &:= \quad \mathsf{C}\_{4}(L). \end{split} \tag{32}$$

From (28), (31) and (32), we conclude that for *l* ≥ *L*1,

$$I^{1-a} \int\_{L}^{l} e^{\frac{\rho-1}{\rho}(l-s)} (l-s)^{a-1} \left[ \frac{e^{\frac{\rho-1}{\rho}(s-l\_1)}M + \,\_{l\_1}I^{1,\rho} \left(\rho g(s)V(s)\right)}{V(s)} \right] ds \ge -[\mathbb{C}\_3(L\_1) + \mathbb{C}\_4(L)].$$

Since, the R.H.S of the above inequality is a negative constant, it follows that

$$\lim\_{l\to\infty} \inf l^{1-\mathfrak{a}} \int\_L e^{\frac{\rho-1}{\rho}(l-s)} (l-s)^{\mathfrak{a}-1} \left[ \frac{e^{\frac{\rho-1}{\rho}(s-l\_1)} M + \mathfrak{l}\_1 I^{1,\mathfrak{a}} \left( \rho \mathfrak{g}(s) V(s) \right)}{V(s)} \right] ds > -\infty \,\,\mu$$

which is a contradiction to (22).

Therefore, *y*(*l*) is oscillatory. If *y*(*l*) is eventually negative, by a similar argument, we get a contradiction with condition (23). Hence the theorem.

#### **4. Oscillation Results via Caputo Operator**

In this section, we establish the oscillation criteria for the GPF initial value problem (2) under the assumption (H):

We set

$$\Psi(l) \quad := \quad \Gamma(a) \sum\_{k=0}^{m-1} \frac{b\_k}{\rho^{k-a} k!} (l-a)^k e^{\frac{\rho - 1}{\rho} (l-a)},\tag{33}$$

$$\Omega(l,L) \quad := \int\_{a}^{L} e^{\frac{\rho - 1}{\rho}(l - s)} (l - s)^{a - 1} \left( \frac{e^{\frac{\rho - 1}{\rho}(s - l\_1)} M^\* + \,\_1I^{1,\rho} \left( \rho \mathbf{g}(s) V(s) \right)}{V(s)} \right) ds,\tag{34}$$

$$M^\* \quad := \quad \, ^C\_a D^{\mu,\rho} y(a) V(a), \quad M^\* \text{ is an arbitrary constant.} \tag{35}$$

**Theorem 2.** *Assume that* (*H*) *holds. If*

$$\lim\_{l\to\infty} \inf l^{1-n} \int\_{L}^{l} e^{\frac{\rho-1}{\rho}(l-s)} (l-s)^{a-1} \left[ \frac{e^{\frac{\rho-1}{\rho}(s-l\_1)} M^\* + l\_1 I^{1,\rho} \left( \rho g(s) V(s) \right)}{V(s)} \right] ds = -\infty \tag{36}$$

*and*

$$\lim\_{l\to\infty} \sup l^{1-n} \int\_{L}^{l} e^{\frac{\rho-1}{\rho}(l-s)} (l-s)^{a-1} \left[ \frac{e^{\frac{\rho-1}{\rho}(s-l\_1)} M^\* + \_1 I^{1,\rho} \left( \rho \mathbf{g}(s) V(s) \right)}{V(s)} \right] ds = \infty,\tag{37}$$

*for every sufficiently large L, where V*(*l*) *and M*∗ *are defined as in* (20) *and* (35)*, respectively, then every solution of problem* (2) *is oscillatory.*

**Proof.** Suppose that *y*(*l*) is a nonoscillatory solution of problem (2). Without loss of generality, let *L* > *a* be large enough and *l*<sup>1</sup> ≥ *L* such that *y*(*l*) > 0 for *l* ≥ *l*1. Using Lemma 3 (iii), Equations (5) and (13), we have

$$\begin{split} \left[D^{\theta}\right]\_{a}^{\mathbb{C}}D^{a,\rho}\mathcal{Y}(l)V(l) &= \, \, \_{a}^{\mathbb{C}}D^{a,\rho}\mathcal{Y}(l)D^{\rho}V(l) + V(l)D^{\rho}\left(\, \_{a}^{\mathbb{C}}D^{a,\rho}\mathcal{Y}(l)\right) - (1-\rho)^{\mathbb{C}}\_{a}D^{a,\rho}\mathcal{Y}(l)V(l) \\ &= \, \, \_{a}^{\mathbb{C}}D^{a,\rho}\mathcal{Y}(l)D^{\rho}V(l) + V(l)\left[(1-\rho)^{\mathbb{C}}\_{a}D^{a,\rho}\mathcal{Y}(l) + \rho\frac{d}{dl}\left(\, \_{a}^{\mathbb{C}}D^{a,\rho}\mathcal{Y}(l)\right)\right] \\ &\quad - (1-\rho)^{\mathbb{C}}\_{a}D^{a,\rho}\mathcal{Y}(l)V(l) \\ &= \, \, \_{\rho}\left[\, \_{a}^{\mathbb{C}}D^{1+a,\rho}\mathcal{Y}(l) + p(l)\left(\, \_{a}^{\mathbb{C}}D^{a,\rho}\mathcal{Y}(l)\right)V(l) \\ &= \, \, \, \rho\left[-q(l)f(y(l)) + g(l)\right]V(l) \\ &< \, \, \rho g(l)V(l). \end{split}$$

Taking the proportional integral operator *<sup>l</sup>*<sup>1</sup> *I* 1,*<sup>ρ</sup>* on both sides to the above inequality, we obtain

$$\left(\_{l\_1}I^{1,\rho}\left(\mathbf{D}^{\rho}\left[\prescript{\mathbf{C}}{a}{\mathbf{D}}^{a,\rho}\mathscr{Y}(l)V(l)\right]\right)<\_{l\_1}I^{1,\rho}\left(\rho\mathbf{g}(l)V(l)\right).\tag{38}$$

Using Lemma (1) on the L.H.S of (38) , we have

$$\, \_a^C D^{\alpha, \rho} y(l) < \frac{e^{\frac{\rho - 1}{\rho}(l - l\_1)} M^\* + \,\_1I^{1, \rho} \left(\rho g(l) V(l)\right)}{V(l)}.$$

Applying the left GPF integral operator *<sup>a</sup> I <sup>α</sup>*,*ρ*on both sides to the above inequality, we get

$$\,\_aI^{a,\rho}\left( \,\_a^C D^{a,\rho} \mathcal{Y}(l) \right) < \,\_aI^{a,\rho} \left[ \frac{e^{\frac{\rho-1}{\rho}(l-l\_1)} M^\* + \,\_1I^{1,\rho} \left( \rho \mathcal{g}(l) V(l) \right)}{V(l)} \right]. \tag{39}$$

Using Lemma 4 on the L.H.S of (39), we have

$$(y(l) - \sum\_{k=0}^{n-1} \frac{(D^k \varrho)(a)}{\rho^k k!} \ (l - a)^k e^{\frac{\rho - 1}{\rho}(l - a)} < \,\_4I^{a, \rho} \left[ \frac{e^{\frac{\rho - 1}{\rho}(l - l\_1)} M^\* + \,\_1I^{1, \rho} \left( \rho g(l) V(l) \right)}{V(l)} \right]. \tag{40}$$

Applying the left GPF integral formula on the R.H.S of (40), we have

$$\begin{aligned} \mathcal{Y}(l) &< \sum\_{k=0}^{n-1} \frac{b\_k}{\rho^k k!} (l-a)^k e^{\frac{\rho - 1}{\rho} (l-a)} \\ &+ \frac{1}{\rho^a \Gamma(a)} \int\_a^l e^{\frac{\rho - 1}{\rho} (l-s)} (l-s)^{a-1} \left[ \frac{e^{\frac{\rho - 1}{\rho} (s - l\_1)} M^\* +\_{l\_1} I^{1, \rho} \left( \rho g(s) V(s) \right)}{V(s)} \right] ds, \end{aligned}$$

for every sufficiently large *L*. If we multiply the above inequality by *ρ <sup>α</sup>*Γ(*α*), we get

$$\begin{split} \rho^{a}\Gamma(a)y(l) &< \quad \rho^{a}\Gamma(a) \sum\_{k=0}^{n-1} \frac{\frac{\rho\_{k}}{\rho^{a}k!} (l-a)^{k}}{\rho^{a}k!} (l-a)^{k} \, e^{\frac{\rho-1}{\rho}(l-a)} \\ &+ \int\_{a}^{L} e^{\frac{\rho-1}{\rho}(l-s)} (l-s)^{a-1} \left[ \frac{e^{\frac{\rho-1}{\rho}(s-l\_{l})} M^{\*} +\_{l\_{l}} I^{1,\rho} \left( \rho g(s) V(s) \right)}{V(s)} \right] ds \\ &+ \int\_{L}^{l} e^{\frac{\rho-1}{\rho}(l-s)} (l-s)^{a-1} \left[ \frac{e^{\frac{\rho-1}{\rho}(s-l\_{l})} M^{\*} +\_{l\_{l}} I^{1,\rho} \left( \rho g(s) V(s) \right)}{V(s)} \right] ds \\ &= \quad \Psi(l) + \Omega(l,L) \\ &+ \int\_{L}^{l} e^{\frac{\rho-1}{\rho}(l-s)} (l-s)^{a-1} \left[ \frac{e^{\frac{\rho-1}{\rho}(s-l\_{l})} M^{\*} +\_{l\_{l}} I^{1,\rho} \left( \rho g(s) V(s) \right)}{V(s)} \right] ds, \end{split} \tag{41}$$

where Ψ(*l*) and Ω(*l*, *L*) are defined in (33) and (34), respectively. Multiplying (41) by *l* 1−*n* , we get

$$\begin{aligned} 0 &< \quad l^{1-n} \Gamma(\mathfrak{a}) y(l) \\ &< \quad l^{1-n} \Psi(l) + l^{1-n} \Omega(l, L) \\ &+ l^{1-n} \int\_L e^{\frac{\rho-1}{\rho}(l-s)} (l-s)^{\mathfrak{a}-1} \left[ \frac{e^{\frac{\rho-1}{\rho}(s-l\_1)} M^{\mathfrak{a}} + \,\_l I^{1, \rho} \left( \rho \mathfrak{g}(s) V(s) \right)}{V(s)} \right] ds. \end{aligned} \tag{42}$$

Let us consider the following two cases for *L*<sup>1</sup> ≥ *L*.

Case(i): Let 0 < *α* ≤ 1. Then *n* = 1. Since *e ρ*−1 *ρ* (*l*−*a*) ≤ 1 and the function *h*3(*l*) = (*l* − *s*) *α*−1 is decreasing for 0 < *α* < 1 , we get for *l* ≥ *L*<sup>1</sup> ,

$$\left| l^{1-n} \Psi(l) \right| = \left| l^{1-n} \Gamma(a) \sum\_{k=0}^{n-1} \frac{b\_k}{\rho^{k-a} k!} (l-a)^k e^{\frac{\rho - 1}{\rho} (l-a)} \right| \le \rho^a \Gamma(a) |b\_0| := \mathbb{C}\_5(L), \tag{43}$$

and

$$\begin{split} \left| l^{1-n} \Omega(l, L) \right| &= \left| l^{1-n} \int\_{a}^{L} e^{\frac{\rho - 1}{\rho}(l - s)} (l - s)^{a - 1} \left( \frac{e^{\frac{\rho - 1}{\rho}(s - l\_{l})} M^{s} + \_{l\_{l}} I^{1, \rho}(\rho g(s) V(s))}{V(s)} \right) ds \right| \\ &\leq \int\_{a}^{L} (l - s)^{a - 1} \left| \frac{e^{\frac{\rho - 1}{\rho}(s - l\_{l})} M^{s} + \_{l\_{l}} I^{1, \rho}(\rho g(s) V(s))}{V(s)} \right| ds \\ &\leq \int\_{a}^{L} (L\_{1} - s)^{a - 1} \left| \frac{e^{\frac{\rho - 1}{\rho}(s - l\_{l})} M^{s} + \_{l\_{l}} I^{1, \rho}(\rho g(s) V(s))}{V(s)} \right| ds \\ &:= \quad \mathscr{C}\_{6}(L, L\_{1}). \end{split} \tag{44}$$

Then, from (42) and *l* ≥ *L*1, we get

$$I^{1-n} \int\_{L}^{l} e^{\frac{\rho-1}{\rho}(l-s)} (l-s)^{a-1} \left[ \frac{e^{\frac{\rho-1}{\rho}(s-l\_1)} M^\* + \,\_{l\_1} I^{1,\rho} \left( \rho \text{g}(s) V(s) \right)}{V(s)} \right] ds \ge -[\mathbb{C}\_5(L) + \mathbb{C}\_6(L, L\_1)].$$

Since, the R.H.S of the above inequality is a negative constant, it follows that

$$\lim\_{l\to\infty} \inf l^{1-n} \int\_L e^{\frac{\rho-1}{\rho}(l-s)} (l-s)^{a-1} \left[ \frac{e^{\frac{\rho-1}{\rho}(s-l\_1)} M^\* + \_{l\_1} I^{1,\rho} \left( \rho g(s) V(s) \right)}{V(s)} \right] ds > -\infty \,\mu$$

which leads to a contradiction with the condition (36).

Case(ii): Let *<sup>α</sup>* <sup>&</sup>gt; 1. Then *<sup>n</sup>* <sup>≥</sup> 2 and *l*−*a l n*−<sup>1</sup> < 1 for *n* ≥ 2 and *α* > 1. Since *e ρ*−1 *ρ* (*l*−*a*) ≤ 1 and the function *h*4(*l*) = (*l* − *a*) *k*−*n*+1 is decreasing for *k* > *n* − 1 and for *l* ≥ *L*1, we have

$$\begin{split} \left| l^{1-n} \Psi(l) \right| &= \left| l^{1-n} \Gamma(a) \sum\_{k=0}^{n-1} \frac{b\_k}{\rho^{k-a} k!} (l-a)^k e^{\frac{\rho-1}{\rho} (l-a)} \right| \\ &= \left| \rho^n \Gamma(a) \left( \frac{l-a}{l} \right)^{n-1} \sum\_{k=0}^{n-1} \frac{b\_k}{\rho^k k!} (l-a)^{k-n+1} e^{\frac{\rho-1}{\rho} (l-a)} \right| \\ &\leq \left| \rho^n \Gamma(a) \sum\_{k=0}^{n-1} \frac{|b\_k|}{\rho^k k!} (l-a)^{k-n+1} \\ &\leq \left| \rho^n \Gamma(a) \sum\_{k=0}^{n-1} \frac{|b\_k| (L\_1 - a)^{k-n+1}}{\rho^k k!} \right| \\ &:= \, \, \, \_C\!\!/ (L\_1) . \end{split} \tag{45}$$

and

$$\begin{split} \left| l^{1-n} \Omega(l, L) \right| &= \left| l^{1-n} \int\_{a}^{L} e^{\frac{\rho-1}{\rho}(l-s)} (l-s)^{a-1} \left( \frac{e^{\frac{\rho-1}{\rho}(s-l\_{l})} M^{s} + \iota\_{l\_{l}} l^{1, \rho}}{V(s)} \right) ds \right| \\ &= \left| l^{a-n} \int\_{a}^{L} e^{\frac{\rho-1}{\rho}(l-s)} \left( \frac{l-s}{l} \right)^{a-1} \left( \frac{e^{\frac{\rho-1}{\rho}(s-l\_{l})} M^{s} + \iota\_{l\_{l}} l^{1, \rho} \left( \rho g(s) V(s) \right)}{V(s)} \right) ds \right| \\ &\leq \int\_{a}^{L} \left| \frac{e^{\frac{\rho-1}{\rho}(s-l\_{l})} M^{s} + \iota\_{l\_{l}} l^{1, \rho} \left( \rho g(s) V(s) \right)}{V(s)} \right| ds \\ &:= \, \, \mathcal{C}\_{8}(L) . \end{split} \tag{46}$$

From Equations (42), (45) and (46), we conclude that for *l* ≥ *L*1,

$$I^{1-n} \int\_{L}^{l} e^{\frac{\rho-1}{\rho}(l-s)} (l-s)^{\alpha-1} \left[ \frac{e^{\frac{\rho-1}{\rho}(s-l\_1)} M^\* + \,\_l I^{1,\rho} \left(\rho \text{g}(s) V(s)\right)}{V(s)} \right] ds \ge -[\mathbb{C}\_7(L\_1) + \mathbb{C}\_8(L)].$$

Since, the R.H.S of the above inequality is a negative constant, it follows that

$$\lim\_{l\to\infty} \inf l^{1-n} \int\_L^l e^{\frac{\rho-1}{\rho}(l-s)} (l-s)^{a-1} \left[ \frac{e^{\frac{\rho-1}{\rho}(s-l\_1)} M^\* + \,\_l I^{1,\rho} \left( \rho \gsim (s) V(s) \right)}{V(s)} \right] ds > -\infty \,\_r$$

which contradicts the (36).

Therefore, *y*(*l*) is oscillatory. If *y*(*l*) is eventually negative, by a similar argument, we get a contradiction with condition (37). Hence the theorem.

**Remark 1.** *If we put ρ* = 1 *in Theorem* (1) *and Theorem* (2)*, then they reduced to Theorem 3.1 and Theorem 4.1, respectively, of [24].*

#### **5. Examples**

This section include some examples for the illustration of our main results.

**Example 1.** *Consider the following GPF initial value problem*

$$\begin{cases} \ \_0D^{\frac{2}{2},1}y(l) - \_0D^{\frac{1}{2},1}y(l) + (l+7)^2(y+3)e^{\cos 2y} = e^{2l}\sin l, & l > 0, \\\\ \lim\_{l \to 0^+} \ \_0I^{\frac{1}{2},1}y(l) = b\_1. \end{cases} \tag{47}$$

Setting *α* = <sup>1</sup> 2 , *ρ* = 1, *a* = 0, *p*(*l*) = −1, *q*(*l*) = (*l* + 7) 2 , *f*(*y*) = (*y* + 3)*e* cos 2*y* , *g*(*l*) = *e* 2*l* sin *l* and *V*(*l*) = *e l*1−*l* . The assumption (*H*) is satisfied if *y*(*l*) > 0. Then,

$$\begin{aligned} \,\_1I \, ^{1,\phi} \left( \rho g(s) V(s) \right) &= \,\_1\frac{1}{\rho} \int\_{l\_1}^s e^{\frac{\rho - 1}{\rho}(s - \tau)} \rho g(\tau) V(\tau) d\tau \\ &= \int\_{l\_1}^s e^{l\_1 + \tau} \sin \tau d\tau \\ &= \,\_2\frac{e^{l\_1 + s}}{2} \left( \sin s - \cos s \right) - \frac{e^{2l\_1}}{2} \left( \sin l\_1 - \cos l\_1 \right) \\ &= \,\_1\frac{\sqrt{2}e^{l\_1 + s}}{2} \sin \left( s - \frac{\pi}{4} \right) - \frac{e^{2l\_1}}{2} \left( \sin l\_1 - \cos l\_1 \right) \end{aligned}$$

Set a point *l*<sup>1</sup> = *<sup>π</sup>* 2 . Hence, we compute that

$$\begin{aligned} &l^{1-\alpha} \int\_0^l e^{\frac{\rho-1}{\rho}(l-s)} (l-s)^{\alpha-1} \left[ \frac{e^{\frac{\rho-1}{\rho}(s-l\_1)} M + \_{l\_1} I^{1,\rho} \left( \rho g(s) V(s) \right)}{V(s)} \right] ds \\ &= \quad l^{\frac{1}{2}} \int\_0^l (l-s)^{-\frac{1}{2}} e^{s - \frac{\pi}{2}} \left[ \left( M - \frac{e^{\pi}}{2} \right) + \frac{\sqrt{2}}{2} e^{\frac{\pi}{2} + s} \sin \left( s - \frac{\pi}{4} \right) \right] ds. \end{aligned}$$

By setting *l* − *s* = *τ* 2 , we can get the above integral as

$$\begin{split} &\quad l^{\frac{1}{2}}\int\_{0}^{l}(l-s)^{-\frac{1}{2}}e^{s-\frac{\pi}{2}}\left[\left(M-\frac{e^{\pi}}{2}\right)+\frac{\sqrt{2}}{2}e^{\frac{\pi}{2}+s}\sin\left(s-\frac{\pi}{4}\right)\right]ds\\ &=\quad l^{\frac{1}{2}}\int\_{\sqrt{l}}^{0}\frac{1}{\pi}e^{l-\tau^{2}-\frac{\pi}{2}}\left[\left(\frac{2M-e^{\pi}}{2}\right)+\frac{\sqrt{2}}{2}e^{\frac{\pi}{2}+l-\tau^{2}}\sin\left(l-\tau^{2}-\frac{\pi}{4}\right)\right](-2\tau)d\tau\\ &=\quad \left(2M-e^{\pi}\right)l^{\frac{1}{2}}e^{l-\frac{\pi}{2}}\int\_{0}^{\sqrt{l}}e^{-\tau^{2}}d\tau+\sqrt{2}l^{\frac{1}{2}}e^{2l}\int\_{0}^{\sqrt{l}}e^{-2\tau^{2}}\sin\left(l-\tau^{2}-\frac{\pi}{4}\right)d\tau\\ &=\quad \left(2M-e^{\pi}\right)l^{\frac{1}{2}}e^{l-\frac{\pi}{2}}\int\_{0}^{\sqrt{l}}e^{-\tau^{2}}d\tau+\sqrt{2}l^{\frac{1}{2}}e^{2l}\sin\left(l-\frac{\pi}{4}\right)\int\_{0}^{\sqrt{l}}e^{-2\tau^{2}}\cos\tau^{2}d\tau\\ &\quad -\sqrt{2}l^{\frac{1}{2}}e^{2l}\cos\left(l-\frac{\pi}{4}\right)\int\_{0}^{\sqrt{l}}e^{-2\tau^{2}}\sin\tau^{2}d\tau. \end{split}$$

Let *<sup>l</sup>* → +<sup>∞</sup> as the result of |*<sup>e</sup>* −2*τ* 2 cos *τ* 2 | ≤ *e* −2*τ* 2 , |*e* −2*τ* 2 sin *τ* 2 | ≤ *e* −2*τ* 2 and lim*l*→+<sup>∞</sup> R √ *l* 0 *e* −2*τ* 2 *dτ* = √ 2*π* 4 . Thus, we know that lim*l*→+<sup>∞</sup> R √ *l* 0 *e* −2*τ* 2 cos *τ* <sup>2</sup>*dτ* and lim*l*→+<sup>∞</sup> R √ *l* 0 *e* −2*τ* 2 sin *τ* <sup>2</sup>*dτ* are convergent.

Hence, we set lim*l*→+<sup>∞</sup> R √ *l* 0 *e* −2*τ* 2 cos *τ* <sup>2</sup>*d<sup>τ</sup>* <sup>=</sup> *<sup>A</sup>* and lim*l*→+<sup>∞</sup> R √ *l* 0 *e* −2*τ* 2 sin *τ* <sup>2</sup>*dτ* = *B*. Select the sequence {*lk*} = n 3*π* <sup>2</sup> + *<sup>π</sup>* <sup>4</sup> <sup>+</sup> <sup>2</sup>*k<sup>π</sup>* <sup>−</sup> arctan − *B A* o, lim*l*→+<sup>∞</sup> *<sup>l</sup><sup>k</sup>* <sup>=</sup> <sup>∞</sup>, then

$$\lim\_{k \to +\infty} \left\{ l\_k^{\frac{1}{2}} e^{l\_k} \left[ \left( 2M - e^{\pi \tau} \right) e^{-\frac{\pi}{2}} \int\_0^{\sqrt{l\_k}} e^{-\tau^2} d\tau + \sqrt{2} e^{l\_k} \left( \sin \left( l\_k - \frac{\pi}{4} \right) \int\_0^{\sqrt{l\_k}} e^{-2\tau^2} \cos \tau^2 d\tau \right. \right. \right. \\ \left. \left. + \left. \frac{\pi}{4} \right|\_0^{\sqrt{l\_k}} e^{-2\tau^2} \sin \tau^2 d\tau \right] \right\}. \tag{48}$$
 
$$- \cos \left( l\_k - \frac{\pi}{4} \right) \int\_0^{\sqrt{l\_k}} e^{-2\tau^2} \sin \tau^2 d\tau \right) \Big]. \tag{49}$$

Firstly, we consider the following limit:

$$\begin{split} &\lim\_{k\to+\infty} \left\{ \sin\left(l\_k - \frac{\pi}{4}\right) \int\_0^{\sqrt{l\_k}} e^{-2\tau^2} \cos\tau^2 d\tau - \cos\left(l\_k - \frac{\pi}{4}\right) \int\_0^{\sqrt{l\_k}} e^{-2\tau^2} \sin\tau^2 d\tau \right\} \\ &= \quad A \cdot \lim\_{k\to+\infty} \sin\left(\frac{3\pi}{2} + 2k\pi - \arctan\left(-\frac{B}{A}\right)\right) - B \cdot \lim\_{k\to+\infty} \cos\left(\frac{3\pi}{2} + 2k\pi - \arctan\left(-\frac{B}{A}\right)\right) \\ &= \quad A \cdot \sin\left(\frac{3\pi}{2} - \arctan\left(-\frac{B}{A}\right)\right) - B \cdot \cos\left(\frac{3\pi}{2} - \arctan\left(-\frac{B}{A}\right)\right) \\ &= \quad - \sqrt{A^2 + B^2} .\end{split}$$

Secondly, we know that lim*k*→+<sup>∞</sup> *e <sup>l</sup><sup>k</sup>* = +<sup>∞</sup> and lim*k*→+<sup>∞</sup> <sup>2</sup>*Mel<sup>k</sup> <sup>e</sup>* − *π* 2 R √ *lk* 0 *e* −*τ* 2 *dτ* = 2*Me*<sup>−</sup> *<sup>π</sup>* 2 √ *π* <sup>2</sup> <sup>=</sup> <sup>√</sup> *πMe*<sup>−</sup> *<sup>π</sup>* <sup>2</sup> . Hence, for (48), we have

$$\begin{split} & \lim\_{k \to +\infty} \left\{ l\_k^{\frac{1}{k}} e^{l\_k} \left[ \left( 2M - e^{\pi} \right) e^{-\frac{\pi}{2}} \int\_0^{\sqrt{l\_k}} e^{-\tau^2} d\tau + \sqrt{2} e^{l\_k} \left( \sin \left( l\_k - \frac{\pi}{4} \right) \int\_0^{\sqrt{l\_k}} e^{-2\tau^2} \cos \tau^2 d\tau \right) \right. \\ & \left. - \cos \left( l\_k - \frac{\pi}{4} \right) \int\_0^{\sqrt{l\_k}} e^{-2\tau^2} \sin \tau^2 d\tau \right) \right] \Big\} \\ & = \left[ \sqrt{\pi} M e^{-\frac{\pi}{2}} + ( + \infty ) \left( -\sqrt{A^2 + B^2} \right) \right] \\ & = -\infty. \end{split}$$

Then, we obtain

$$\lim\_{l\to\infty} \inf l^{1-\alpha} \int\_0^l e^{\frac{\rho-1}{\rho}(l-s)} (l-s)^{\alpha-1} \left[ \frac{e^{\frac{\rho-1}{\rho}(s-l\_1)} M + \_{l\_1} I^{1,\rho} \left( \rho \varrho(s) V(s) \right)}{V(s)} \right] ds = -\infty < 0.$$

Similarly, selecting the sequence {*lr*} = n 3*π* <sup>2</sup> + *<sup>π</sup>* <sup>4</sup> <sup>+</sup> <sup>2</sup>*r<sup>π</sup>* <sup>−</sup> arctan − *B A* o, we can obtain

$$\lim\_{l\to\infty} \sup l^{1-a} \int\_0^l e^{\frac{\rho-1}{\rho}(l-s)} (l-s)^{a-1} \left[ \frac{e^{\frac{\rho-1}{\rho}(s-l\_1)}M + \iota\_1 I^{1,\rho} \left(\rho \mathcal{g}(s)V(s)\right)}{V(s)} \right] ds = +\infty > 0.$$

Therefore, by Theorem 1 all solutions of the problem (47) are oscillatory.

**Example 2.** *Consider the following GPF Caputo initial value problem*

$$\begin{cases} \ ^{\mathbb{C}}D^{\frac{3}{2},1}y(l) - \ ^{\mathbb{C}}D^{\frac{1}{2},1}y(l) + e^{(l+1)^2}\ln(y^2+e) = e^{2l}\cos l, \quad l > 0, \\\\ y(0) = b\_0. \end{cases} \tag{49}$$

Setting *α* = <sup>1</sup> 2 , *ρ* = 1, *a* = 0, *p*(*l*) = −1, *q*(*l*) = *e* (*l*+1) 2 , *f*(*y*) = ln(*y* <sup>2</sup> + *e*), *g*(*l*) = *e* 2*l* cos *l* and *V*(*l*) = *e l*1−*l* . The assumption (*H*) is satisfied if *y*(*l*) > 0. Then, we get

$$\begin{aligned} \,\_1I\_1^{1,\varrho} \left( \rho g(s) V(s) \right) &= \,\_1\frac{1}{\rho} \int\_{l\_1}^s e^{\frac{\rho - 1}{\rho}(s - \tau)} \rho g(\tau) V(\tau) d\tau \\ &= \int\_{l\_1}^s e^{l\_1 + \tau} \cos \tau d\tau \\ &= \,\_1\frac{e^{l\_1 + s}}{2} \left( \sin s + \cos s \right) - \frac{e^{2l\_1}}{2} \left( \sin l\_1 + \cos l\_1 \right) \\ &= \,\_1\frac{\sqrt{2}e^{l\_1 + s}}{2} \sin \left( s + \frac{\pi}{4} \right) - \frac{e^{2l\_1}}{2} \left( \sin l\_1 + \cos l\_1 \right) \end{aligned}$$

Set *l*<sup>1</sup> = *<sup>π</sup>* <sup>2</sup> with *n* = 1. Hence, we can compute that

$$\begin{aligned} &l^{1-n}\int\_0^l e^{\frac{\rho-1}{\rho}(l-s)}(l-s)^{a-1} \left[\frac{e^{\frac{\rho-1}{\rho}(s-l\_1)}M^\* + \_{l\_1}I^{1,\rho}\left(\rho g(s)V(s)\right)}{V(s)}\right]ds\\ &=&\int\_0^l (l-s)^{-\frac{1}{2}}e^{s-\frac{\pi}{2}} \left[\left(M^\* - \frac{e^{\pi}}{2}\right) + \frac{\sqrt{2}}{2}e^{\frac{\pi}{2}+s}\sin\left(s + \frac{\pi}{4}\right)\right]ds.\end{aligned}$$

By setting *l* − *s* = *τ* 2 , we can get the above integral as

$$\begin{split} &\int\_{0}^{l} (l-s)^{-\frac{1}{2}} e^{s-\frac{\pi}{2}} \left[ \left( M^\* - \frac{e^{\pi}}{2} \right) + \frac{\sqrt{2}}{2} e^{\frac{\pi}{2} + s} \sin \left( s + \frac{\pi}{4} \right) \right] ds \\ &= \quad \int\_{\sqrt{l}}^{0} \frac{1}{\tau} e^{l-\tau^2 - \frac{\pi}{2}} \left[ \left( \frac{2M^\* - e^{\pi}}{2} \right) + \frac{\sqrt{2}}{2} e^{\frac{\pi}{2} + l - \tau^2} \sin \left( l - \tau^2 + \frac{\pi}{4} \right) \right] (-2\tau) d\tau \\ &= \quad \left( 2M^\* - e^{\pi} \right) e^{l - \frac{\pi}{2}} \int\_{0}^{\sqrt{l}} e^{-\tau^2} d\tau + \sqrt{2} e^{2l} \int\_{0}^{\sqrt{l}} e^{-2\tau^2} \sin \left( l - \tau^2 + \frac{\pi}{4} \right) d\tau \\ &= \quad \left( 2M^\* - e^{\pi} \right) e^{l - \frac{\pi}{2}} \int\_{0}^{\sqrt{l}} e^{-\tau^2} d\tau + \sqrt{2} e^{2l} \sin \left( l + \frac{\pi}{4} \right) \int\_{0}^{\sqrt{l}} e^{-2\tau^2} \cos \tau^2 d\tau \\ & \quad - \sqrt{2} e^{2l} \cos \left( l + \frac{\pi}{4} \right) \int\_{0}^{\sqrt{l}} e^{-2\tau^2} \sin \tau^2 d\tau. \end{split}$$

Let *<sup>l</sup>* → +<sup>∞</sup> as the result of |*<sup>e</sup>* −2*τ* 2 cos *τ* 2 | ≤ *e* −2*τ* 2 , |*e* −2*τ* 2 sin *τ* 2 | ≤ *e* −2*τ* 2 and lim*l*→+<sup>∞</sup> R √ *l* 0 *e* −2*τ* 2 *dτ* = √ 2*π* 4 . Thus, we know that lim*l*→+<sup>∞</sup> R √ *l* 0 *e* −2*τ* 2 cos *τ* <sup>2</sup>*dτ* and lim*l*→+<sup>∞</sup> R √ *l* 0 *e* −2*τ* 2 sin *τ* <sup>2</sup>*dτ* are convergent.

Hence, we can set lim*l*→+<sup>∞</sup> R √ *l* 0 *e* −2*τ* 2 cos *τ* <sup>2</sup>*d<sup>τ</sup>* <sup>=</sup> *<sup>A</sup>* and lim*l*→+<sup>∞</sup> R √ *l* 0 *e* −2*τ* 2 sin *τ* <sup>2</sup>*dτ* = *B*. Select the sequence {*lk*} = n 7*π* <sup>2</sup> <sup>−</sup> *<sup>π</sup>* <sup>4</sup> <sup>+</sup> <sup>2</sup>*k<sup>π</sup>* <sup>−</sup> arctan − *B A* o, lim*l*→+<sup>∞</sup> *<sup>l</sup><sup>k</sup>* <sup>=</sup> <sup>∞</sup>, then we compute the following term:

$$\lim\_{k \to +\infty} \left\{ e^{l\_k} \left[ \left( 2M^\* - e^{\pi i} \right) e^{-\frac{\pi}{2}} \int\_0^{\sqrt{l\_k}} e^{-\tau^2} d\tau + \sqrt{2} e^{l\_k} \left( \sin \left( l\_k + \frac{\pi}{4} \right) \int\_0^{\sqrt{l\_k}} e^{-2\tau^2} \cos \tau^2 d\tau \right. \right. \right. \\ \left. \left. + \frac{\pi}{4} \left( \cos \left( l\_k - \frac{\pi}{4} \right) \int\_0^{\sqrt{l\_k}} e^{-2\tau^2} \sin \tau^2 d\tau \right. \right) \right] \\ \left. - \frac{\pi}{4} \left( \cos \left( l\_k + \frac{\pi}{4} \right) \int\_0^{\sqrt{l\_k}} e^{-2\tau^2} \sin \tau^2 d\tau \right) \right] \\ \left. - \frac{\pi}{4} \left( \cos \left( l\_k - \frac{\pi}{4} \right) \int\_0^{\sqrt{l\_k}} e^{-2\tau^2} \sin \tau^2 d\tau \right) \right] \\ \left. + \frac{\pi}{4} \left( \cos \left( l\_k + \frac{\pi}{4} \right) \int\_0^{\sqrt{l\_k}} e^{-2\tau^2} \sin \tau^2 d\tau \right) \right] \\ \left. + \frac{\pi}{4} \left( \cos \left( l\_k - \frac{\pi}{4} \right) \int\_0^{\sqrt{l\_k}} e^{-2\tau^2} \sin \tau^2 d\tau \right) \right] \end{aligned} \tag{50}$$

Firstly, we consider the following limit:

$$\begin{split} &\lim\_{k\to+\infty} \left\{ \sin\left(l\_k + \frac{\pi}{4}\right) \int\_0^{\sqrt{l\_k}} e^{-2\tau^2} \cos\tau^2 d\tau - \cos\left(l\_k + \frac{\pi}{4}\right) \int\_0^{\sqrt{l\_k}} e^{-2\tau^2} \sin\tau^2 d\tau \right\} \\ &= \quad A \cdot \lim\_{k\to+\infty} \sin\left(\frac{7\pi}{2} + 2k\pi - \arctan\left(-\frac{B}{A}\right)\right) - B \cdot \lim\_{k\to+\infty} \cos\left(\frac{7\pi}{2} + 2k\pi - \arctan\left(-\frac{B}{A}\right)\right) \\ &= \quad A \cdot \sin\left(\frac{7\pi}{2} - \arctan\left(-\frac{B}{A}\right)\right) - B \cdot \cos\left(\frac{7\pi}{2} - \arctan\left(-\frac{B}{A}\right)\right) \\ &= \quad - \sqrt{A^2 + B^2}. \end{split}$$

Secondly, we know that lim*k*→+<sup>∞</sup> *e <sup>l</sup><sup>k</sup>* = +<sup>∞</sup> and lim*k*→+<sup>∞</sup> <sup>2</sup>*M*<sup>∗</sup> *e lk e* − *π* 2 R √ *lk* 0 *e* −*τ* 2 *dτ* = 2*M*∗ *e* − *π* 2 √ *π* <sup>2</sup> = √ *πM*∗ *e* − *π* <sup>2</sup> . Hence, for (50), we have

$$\begin{split} & \lim\_{k \to +\infty} \left\{ e^{l\_k} \left[ \left( 2M^\* - e^{\pi} \right) e^{-\frac{\pi}{2}} \int\_0^{\sqrt{l\_k}} e^{-\tau^2} d\tau + \sqrt{2} e^{l\_k} \left( \sin \left( l\_k + \frac{\pi}{4} \right) \int\_0^{\sqrt{l\_k}} e^{-2\tau^2} \cos \tau^2 d\tau \right) \right. \\ & \left. - \cos \left( l\_k + \frac{\pi}{4} \right) \int\_0^{\sqrt{l\_k}} e^{-2\tau^2} \sin \tau^2 d\tau \right) \right] \right\} \\ & = \quad \left[ \sqrt{\pi} M^\* e^{-\frac{\pi}{2}} + (+\infty) \left( -\sqrt{A^2 + B^2} \right) \right] \\ & = \quad - \infty. \end{split}$$

Then, we obtain

$$\lim\_{l\to\infty} \inf l^{1-n} \int\_0^l e^{\frac{\rho-1}{\rho}(l-s)} (l-s)^{a-1} \left[ \frac{e^{\frac{\rho-1}{\rho}(s-l\_1)} M^\* + l\_1 I^{1,\rho} \left( \rho \mathcal{g}(s) V(s) \right)}{V(s)} \right] ds = -\infty < 0.$$

Similarly, selecting the sequence {*lr*} = n *π* <sup>2</sup> <sup>−</sup> *<sup>π</sup>* <sup>4</sup> <sup>+</sup> <sup>2</sup>*r<sup>π</sup>* <sup>−</sup> arctan − *B A* o, we can obtain

$$\lim\_{l\to\infty} \sup l^{1-n} \int\_0^l e^{\frac{\rho-1}{\rho}(l-s)} (l-s)^{\alpha-1} \left[ \frac{e^{\frac{\rho-1}{\rho}(s-l\_1)} M^\* + \_{l\_1} I^{1,\rho} \left( \rho \mathcal{g}(s) V(s) \right)}{V(s)} \right] ds = +\infty > 0.$$

Therefore, by Theorem 2 all solutions of the problem (49) are oscillatory.

**Example 3.** *Consider the following GPF Riemann-Liouville initial value problem*

$$\begin{cases} \ \_0D^{\frac{2}{2},1}y(l) + \sqrt{l}\left(\frac{4}{\sqrt{\pi}} + \frac{\epsilon^{3}\sqrt{-9}}{(-y)^{\frac{1}{4}}}\right) = \mathcal{e}^{3l}, \quad l > 0, \\\ \lim\_{l \to 0^{+}} \mathcal{I}^{\frac{1}{2},1}y(l) = 0. \end{cases} \tag{51}$$

Setting *α* = <sup>1</sup> 2 , *<sup>ρ</sup>* <sup>=</sup> 1, *<sup>a</sup>* <sup>=</sup> *<sup>p</sup>*(*l*) = 0, *<sup>q</sup>*(*l*) = <sup>√</sup> *l*, *f*(*y*) = √ 4 *π* + *<sup>e</sup>* 3 √ −*y* (−*y*) 1 4 , *g*(*l*) = *e* <sup>3</sup>*<sup>l</sup>* and *V*(*l*) = 1. The assumption (*H*) is satisfied if *y*(*l*) > 0. Then,

$$\partial\_{l\_1} I^{1, \rho} \left( \rho g(\mathbf{s}) V(\mathbf{s}) \right) = \frac{1}{\rho} \int\_{l\_1}^{\mathbf{s}} e^{\frac{\rho - 1}{\rho} (\mathbf{s} - \tau)} \rho g(\tau) V(\tau) d\tau = \int\_{l\_1}^{\mathbf{s}} e^{\mathbf{3} \tau} d\tau = \frac{e^{\mathbf{3} \mathbf{s}}}{\mathbf{3}} - \frac{e^{\mathbf{3} \mathbf{l}\_1}}{\mathbf{3}} = \frac{1}{\mathbf{3}} \left( e^{\mathbf{3} \mathbf{s}} - e^{\mathbf{3} \mathbf{l}\_1} \right).$$

By setting *l*<sup>1</sup> = <sup>1</sup> 3 and *l* − *s* = *τ* 2 , it follows that

$$\begin{split} &l^{1-\alpha} \int\_{0}^{l} e^{\frac{\rho-1}{\rho}(l-s)} (l-s)^{\alpha-1} \left[ \frac{e^{\frac{\rho-1}{\rho}(s-l\_{1})} M + \,\_{l\_{1}} I^{1,\rho} \left( \rho g(s) V(s) \right)}{V(s)} \right] ds \\ &= \quad l^{\frac{1}{2}} \int\_{0}^{l} (l-s)^{-\frac{1}{2}} \left[ M + \frac{1}{3} \left( e^{3s} - \varepsilon \right) \right] ds. \\ &= \quad l^{\frac{1}{2}} \left[ \left( M - \frac{e}{3} \right) \int\_{0}^{l} (l-s)^{-\frac{1}{2}} ds + \frac{1}{3} \int\_{0}^{l} (l-s)^{-\frac{1}{2}} e^{3s} ds \right] \\ &= \quad \quad \quad \quad \quad \quad 2l^{\frac{1}{2}} \left[ \left( M - \frac{e}{3} \right) \sqrt{l} + \frac{1}{3} e^{3l} \int\_{0}^{\sqrt{l}} e^{-3\tau} d\tau \right]. \end{split}$$

However, the condition (22) does not holds since

$$\begin{split} &\lim\_{l\to\infty}\inf\_{l}l^{1-\alpha}\int\_{0}^{l}e^{\frac{\rho-1}{\rho}(l-s)}(l-s)^{\alpha-1}\left[\frac{e^{\frac{\rho-1}{\rho}(s-a)}\,M+\_{l\_{1}}I^{1,\rho}\left(\rho g(s)V(s)\right)}{V(s)}\right]ds \\ &=\lim\_{l\to\infty}\inf\left\{2l^{\frac{1}{2}}\left[\left(M-\frac{\varrho}{3}\right)\sqrt{l}+\frac{1}{3}e^{3l}\int\_{0}^{\sqrt{l}}e^{-3\tau}d\tau\right]\right\} \\ &=\quad\left[\left(M-\frac{\varrho}{3}\right)(+\infty)+(+\infty)\frac{\sqrt{\pi}}{2}\right]=\infty. \end{split}$$

Using Proposition 1 (ii) with *α* = <sup>3</sup> 2 , *β* = 3 and *ρ* = 1, we get *<sup>a</sup>D α*,*ρ <sup>y</sup>*(*l*) = <sup>−</sup><sup>4</sup> √ √ *l π* , it is easy to verify that *y*(*l*) = −*l* 2 is a nonoscillatory solution of (51). Figure 1 demonstrates the solution *y*(*l*) = −*l* 2 .

**Figure 1.** The nonoscillatory behavior of the solution *y*(*l*) = −*l* 2 .

#### **6. Conclusions**

In this paper, the oscillatory behavior of solutions of generalized proportional fractional initial value problem is studied. Forced and damped oscillation results are obtained via GPF operators in the frame of Riemann-Liouville and Caputo settings. The main theorems of this paper improve and generalize some existing oscillation theorems reported in the literature. In particular, for the choice of *ρ* = 1, our contributions obtained using GPF operators cover the results discussed in Reference [24] which are obtained via conformable operators. At the end, we presented some numerical examples with particular values of parameters to illustrate the validity of the proposed results. Interestingly, we provided an example demonstrating that the failure of any condition forces the existence of a nonoscillatory solution. This justifies the advantage of our findings.

We believe that the results of this paper are of great importance for the audience of interested researchers. Several types of oscillation conditions could be generalized by considering respective equations within GPF derivatives.

**Author Contributions:** All authors contributed equally and significantly to this paper. All authors have read and approved the final version of the manuscript.

**Funding:** This research received no external funding.

**Acknowledgments:** J. Alzabut would like to thank Prince Sultan University for funding this work through research group Nonlinear Analysis Methods in Applied Mathematics (NAMAM) group number RG-DES-2017-01-17. The second author was supported by DST-INSPIRE Scheme (No.DST/INSPIRE Fellowship/2018/IF180260) New Delhi, India. The third author was supported by DST-FIST Scheme (Grant No. SR/FST/MSI-115/2016), New Delhi, India. The fourth author was partially supported by Navamindradhiraj University Research Fund (NURS), Navamindradhiraj University, Thailand.

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

#### **References**


© 2020 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 (http://creativecommons.org/licenses/by/4.0/).

### *Article* **The Analytical Analysis of Time-Fractional Fornberg–Whitham Equations**

#### **A. A. Alderremy <sup>1</sup> , Hassan Khan 2,\*, Rasool Shah <sup>2</sup> , Shaban Aly 1,3 and Dumitru Baleanu 4,5**


Received: 17 April 2020; Accepted: 11 June 2020; Published: 16 June 2020

**Abstract:** This article is dealing with the analytical solution of Fornberg–Whitham equations in fractional view of Caputo operator. The effective method among the analytical techniques, natural transform decomposition method, is implemented to handle the solutions of the proposed problems. The approximate analytical solutions of nonlinear numerical problems are determined to confirm the validity of the suggested technique. The solution of the fractional-order problems are investigated for the suggested mathematical models. The solutions-graphs are then plotted to understand the effectiveness of fractional-order mathematical modeling over integer-order modeling. It is observed that the derived solutions have a closed resemblance with the actual solutions. Moreover, using fractional-order modeling various dynamics can be analyzed which can provide sophisticated information about physical phenomena. The simple and straight-forward procedure of the suggested technique is the preferable point and thus can be used to solve other nonlinear fractional problems.

**Keywords:** Adomian decomposition method; Caputo operator; Natural transform; Fornberg–Whitham equations

#### **1. Introduction**

It is well known that in many fields of physics, the studies of non-linear wave problems and their effects are of wide significance. Traveling wave solutions are a significant kind of result for the non-linear partial differential condition and numerous non-linear fractional differential equations (FDEs) have been shown to an assortment of traveling wave results. Although water wave are among the extremely important of all-natural phenomena, they have an extraordinarily rich mathematical structure. Water waves are one of the most complicated fields in wave dynamics, including the study in non-linear, electromagnetic waves in 1 space and 3-time dimensions [1–5]. For illustration, the well-known Korteweg–de Vries equation

$$D\_t \mu - 6\mu D\_x \mu + \mu D\_{\text{xxx}} \mu = 0\,\text{s}$$

has a simple solitary-wave solution [6]. Camassa-Holm equation

$$D\_t \mu - D\_{\text{xxt}} \mu + \mathfrak{Z} \mu D\_{\text{x}} \mu = 2 D\_{\text{x}} \mu D\_{\text{x} \text{x}} \mu + \mu D\_{\text{xxx}} \mu\_{\text{xx}}$$

a model approximation for symmetric non-linear dispersive waves in shallow water, was suggested by Camassa and Holm [7]. Due to its useful mathematical proprieties, this scenario has attracted much attention during the past decade. It has been found that the Camassa Holm equation includes poles, composite wave, stumpons, and cuspons solutions [8]. The specific Camassa–Holm equation solutions were studied by Vakhnenko and Parkes [9]. In mathematical physics the Fornberg-Whitham (FW) model is a significant mathematical equation. The FWE [10,11] is expressed as

$$D\_t \mu - D\_{\text{xxt}} \mu + D\_{\text{x}} \mu = \mu D\_{\text{xxx}} \mu - \mu D\_{\text{x}} \mu + \Im D\_{\text{x}} \mu D\_{\text{xx}} \mu\_{\text{x}}$$

where *µ*(*x*, *t*) is the fluid velocity, *x* is the spatial co-ordinate and *t* is the time. In 1978 Fornberg and Whitham derived a *µ*(*x*, *t*) = *Ce <sup>x</sup>* <sup>2</sup> <sup>−</sup> <sup>4</sup>*<sup>t</sup>* <sup>3</sup> peaked solution with an arbitrary constant of *C* [12]. This algorithm was developed to analyze the breakup of dispersive nonlinear water waves. The FWE has been found to require peakon results as a simulation for limiting wave heights as well as the frequency of wave breaks. In fractional calculus (FC) has gained considerable significance and popularity, primarily because of its well-shown applications in a wide range of apparently disparate areas of engineering and science [13]. Many scholars, such as Singh et al. [14], Merdan et al. [15], Saker et al. [16], Gupta and Singh [17] etc., have therefore researched the fractional extensions of the FW model for the Caputo fractional-order derivative [18].

The existence, uniqueness and stability are the important ingredient to show for any mathematical problems in science and engineering. In this connection Li et al. have determine the existance and unique of the solutions for some nonlinear fractional differential equations [19]. Becani et al. have discussed the theory of existence and uniqueness for some singular PDEs [20]. The generalized theorem of existence and uniqueness for nth order fractional DEs was analyzed by Dannan et al. in [21]. Similarly the stability of solutions for the Fornberg-Whitham equation was investigated by Xiujuan Gao et al. in [22]. Shan et al. have discussed the optimal control of the Fornberg-Whitham equation [23].

Recently, the researchers have taken greater interest in FC, i.e., the study of integrals and derivatives of fractional-order non-integer. Major importance have been demonstrated in the analysis of the FC and its various implementations in the field engineering [24–27]. FDEs are widely utilized to model in a variety of fields of study, including an analysis of fractional random walking, kinetic control schemes theory, signal processing, electrical networks, reaction and diffusion procedure [28,29]. FD provides a splendid method for characterizing the memories and genetic properties of different procedures [30,31].

Over the last few years, FDEs have become the subject of several studies owing to their frequent use in numerous implementations in viscoelasticity, biology, fluid mechanics, physics, dynamical schemes, electrical network, physics, signal and optics process, as they can be modelled by linear and nonlinear FDEs [32–36]. FD offer an outstanding method for explaining the memories and inherited properties of specific materials and processes. Fractional-order integrals and derivatives have proven more effective in formulating such electrical and chemical problems than the standard models. Non-linear FPDEs have many applications in various areas of engineering such as heat and mass transfer, thermodynamics and micro-electro mechanics scheme [37–39].

The technique of natural decomposition (NDM) was initially developed by Rawashdeh and Maitama in 2014 [40–42], to solve ODEs and PDEs that appear in different fields of mathematics. The suggested technique is mixing of the Adomian technique (ADM) and natural transformation. The key benefit of this suggested technique is the potential to integrate two important methods of achieving fast convergent series for PDEs. Many scholars have recently solved different types of fractional-order PDEs, for example heat and wave equations [43], coupled Burger equations [44], hyperbolic telegraph equation [45], Harry Dym equation [46] and diffusion equations [47].

#### **2. Preliminaries**

**Definition 1.** *Let g* ∈ *C<sup>β</sup> and β* ≥ −1*, then the Riemann–Liouville integral of order γ, γ* > 0 > *is given by [48–50]:*

$$\int\_{t}^{\gamma} \mathbf{g}(\mathbf{x}, t) = \frac{1}{\Gamma(\gamma)} \int\_{0}^{t} (t - \theta)^{\gamma - 1} \mathbf{g}(\mathbf{x}, \theta) d\theta, \qquad t > 0. \tag{1}$$

**Definition 2.** *Let <sup>g</sup>* <sup>∈</sup> <sup>C</sup>*<sup>t</sup> and <sup>t</sup>* ≥ −1, *then Caputo definition of fractional derivative of order <sup>γ</sup> if <sup>m</sup>* <sup>−</sup> <sup>1</sup> <sup>&</sup>lt; *<sup>γ</sup>* <sup>≤</sup> *m with m* <sup>∈</sup> <sup>N</sup> *is describe as [48–50]*

$$D\_t^\gamma g(t) = \begin{cases} \frac{d^m g(t)}{dt^m}, & \gamma = m \in N, \\\frac{1}{\Gamma(m-\gamma)} \int\_0^t (t-\theta)^{m-\gamma-1} g^{(m)}(\theta) d\theta, & m-1 < \gamma < m, \quad m \in N, \end{cases} \tag{2}$$

**Remark 1.** *Some basic properties are below [48–50]*

$$\begin{aligned} D\_{\mathbf{x}}^{\gamma} I\_{\mathbf{x}}^{\gamma} g(\mathbf{x}) &= g(\mathbf{x}), \\ I^{\gamma} \mathbf{x}^{\lambda} &= \frac{\Gamma(\lambda + 1)}{\Gamma(\gamma + \lambda + 1)} \mathbf{x}^{\gamma + \lambda}, \qquad \gamma > 0, \lambda > -1, \quad \mathbf{x} > 0, \\ D\_{\mathbf{x}}^{\gamma} I\_{\mathbf{x}}^{\gamma} g(\mathbf{x}) &= g(\mathbf{x}) - \sum\_{k=0}^{m} g^{(k)}(0^{+}) \frac{\mathbf{x}^{k}}{k!}, \qquad \text{for} \quad \mathbf{x} > 0. \end{aligned}$$

**Definition 3.** *The natural transform of the function g*(*t*) *is expressed by N*[*g*(*t*)] *for t* ∈ *R and is given by [40–42,51]*

$$N\left[\mathcal{g}(t)\right] = \mathcal{G}(s,\omega) = \int\_{-\infty}^{\infty} e^{-st} \mathcal{g}(\omega t) dt; \quad s, \omega \in (-\infty, \infty), a.s.$$

*where s and ω are the NT variables. If g*(*t*)*H*(*t*) *is defined for positive real numbers, then NT can be presented as [40–42,51]*

$$N[\mathcal{g}(t)\mathcal{Q}(t)] = N^+[\mathcal{g}(t)] = \mathcal{G}^+(s,\omega) = \int\_0^\infty e^{-st}\mathcal{g}(\omega t)dt; \quad s,\omega \in (0,\infty), \quad and \quad t \in \mathbb{R}.\tag{3}$$

*where Q*(*t*) *denotes the Heaviside function.*

**Theorem 1.** *The NT of the Caputo derivative of fractional order of any function g*(*t*) *is defined as [40–42,51]*

$$N^{+}[^{c}D^{\gamma}g(t)] = \frac{s^{\gamma}}{\omega^{\gamma}}\mathcal{G}(s,\omega) - \sum\_{k=0}^{m-1} \frac{s^{\gamma-(k+1)}}{\omega^{\gamma-k}}[D^{k}g(t)]\_{t=0\prime} \quad m-1 \le \gamma < m. \tag{4}$$

*where m is the natural number and γ represent the order of the derivative with fractional order.*

**Remark 2.** *Some basic NT properties are listed below [40–42,51]*

$$\begin{aligned} N^+[1] &= \frac{1}{s'} \\ N^+[t^\gamma] &= \frac{\Gamma(\gamma+1)\omega^\gamma}{s^{\gamma+1}} , \\ N^+[\mathcal{g}^{(m)}(t)] &= \frac{s^m}{\omega^m} \mathcal{R}(s,\omega) - \sum\_{k=0}^{m-1} \frac{s^{m-(k+1)}}{\omega^{m-k}} \frac{\Gamma(\gamma+1)\omega^\gamma}{s^{\gamma+1}} .\end{aligned}$$

#### **3. NDM Procedure**

In this section, NDM procedure is introduced to solve general FPDEs of the form [41,42]

$$D^{\gamma}\mu(\mathbf{x},t) + L\mu(\mathbf{x},t) + N\mu(\mathbf{x},t) = P(\mathbf{x},t), \quad \mathbf{x}, t \ge 0, \quad \ell - 1 < \gamma < \ell,\tag{5}$$

The fractional derivative in Equation (5) is represented by Caputo operator. The linear and nonlinear terms are denoted by *L* and *N* respectively and the source term is *P*(*x*, *t*).

The solution at *t* = 0 is

$$
\mu(\mathbf{x}, \mathbf{0}) = h(\mathbf{x}).\tag{6}
$$

Using NT to Equation (5), we get [41,42]

$$N^{+}\left[D^{\gamma}\mu(\mathbf{x},t)\right] + N^{+}\left[L\mu(\mathbf{x},t) + N\mu(\mathbf{x},t)\right] = N^{+}\left[P(\mathbf{x},t)\right],\tag{7}$$

Applying the differential property of NT [41,42]

$$\frac{s^{\gamma}}{\omega^{\gamma\gamma}}N^{+}\left[\mu(\mathbf{x},t)\right]-\frac{s^{\gamma-1}}{\omega^{\gamma\gamma}}\mu(\mathbf{x},0) = N^{+}\left[P(\mathbf{x},t)\right]-N^{+}\left[L\mu(\mathbf{x},t)+N\mu(\mathbf{x},t)\right],$$

$$\begin{split}N^{+}\left[\mu(\mathbf{x},t)\right]&=\frac{1}{s}\mu(\mathbf{x},0)+\frac{\omega^{\gamma}}{s^{\gamma}}N^{+}\left[P(\mathbf{x},t)\right]-\frac{\omega^{\gamma}}{s^{\gamma}}N^{+}\left[L\mu(\mathbf{x},t)+N\mu(\mathbf{x},t)\right].\\N^{-}\left[\mu(\mathbf{x},t)\right]&=\frac{\omega^{\gamma}}{\omega^{\gamma}}N^{+}\left[\mu(\mathbf{x},t)\right].\end{split}$$

Now *µ*(*x*, 0) = *k*(*x*),

$$N^{+}\left[\mu(\mathbf{x},t)\right] = \frac{h(\mathbf{x})}{s} + \frac{\omega^{\gamma}}{s^{\gamma}}N^{+}\left[P(\mathbf{x},t)\right] - \frac{\omega^{\gamma}}{s^{\gamma}}N^{+}\left[L\mu(\mathbf{x},t) + N\mu(\mathbf{x},t)\right].\tag{8}$$

The infinite series of NDM *µ*(*x*, *t*) is shown by

$$
\mu(\mathbf{x}, t) = \sum\_{\ell=0}^{\infty} \mu\_{\ell}(\mathbf{x}, t). \tag{9}
$$

Adomian polynomial for nonlinear terms is

$$N\mu(\mathbf{x},t) = \sum\_{\ell=0}^{\infty} A\_{\ell\ell} \tag{10}$$

$$A\_{\ell} = \frac{1}{\ell!} \left[ \frac{d^{\ell}}{d\lambda^{\ell}} \left[ N \sum\_{\ell=0}^{\infty} (\lambda^{\ell} \mu\_{\ell}) \right] \right]\_{\lambda=0} , \qquad \ell = 0, 1, 2 \cdots \tag{11}$$

putting Equations (9) and (11) into Equation (8), we have

$$N^{+}\left[\sum\_{\ell=0}^{\infty}\mu\_{\ell}(\mathbf{x},t)\right] = \frac{h(\mathbf{x})}{\mathbf{s}} + \frac{\omega^{\gamma}}{\mathbf{s}^{\gamma}}N^{+}\left[P(\mathbf{x},t)\right] - \frac{\omega^{\gamma}}{\mathbf{s}^{\gamma}}N^{+}\left[L\sum\_{\ell=0}^{\infty}\mu\_{\ell}(\mathbf{x},t) + \sum\_{\ell=0}^{\infty}A\_{\ell}\right].\tag{12}$$

$$N^{+}\left[\mu\_{0}(\mathbf{x},t)\right] = \frac{h(\mathbf{x})}{s} + \frac{\omega^{\gamma}}{s^{\gamma}} N^{+}\left[P(\mathbf{x},t)\right],\tag{13}$$

$$N^{+}\left[\mu\_{1}(\mathbf{x},t)\right] = -\frac{\omega^{\gamma}}{s^{\gamma}}N^{+}\left[L\mu\_{0}(\mathbf{x},t) + A\_{0}\right].$$

We will usually compose

$$N^{+}\left[\mu\_{\ell+1}(\mathbf{x},t)\right] = -\frac{\omega^{\gamma}}{s^{\gamma}}N^{+}\left[L\mu\_{\ell}(\mathbf{x},t) + A\_{\ell}\right], \quad \ell \ge 1. \tag{14}$$

Using the inverse NT to Equations (13) and (14) [41,42].

$$
\mu\_0(\mathbf{x}, t) = h(\mathbf{x}) + N^- \left[ \frac{\omega^\gamma}{s^\gamma} N^+ \left[ P(\mathbf{x}, t) \right] \right],
$$

$$
\upsilon\_{\ell+1}(\mathbf{x}, t) = -N^- \left[ \frac{\omega^\gamma}{s^\gamma} N^+ \left[ L\mu\_\ell(\mathbf{x}, t) + A\_\ell \right] \right]. \tag{15}
$$

#### **4. NDM Implementation**

**Example 1.** *The following nonlinear Fornberg-Whitham with fractional derivative is considered [14]*

$$D\_t^\gamma \mu - D\_{\text{xxx}} \mu + D\_\mathbf{x} \mu = \mu D\_{\text{xxx}} \mu - \mu D\_\mathbf{x} \mu + \mathfrak{H}\_\mathbf{x} \mu D\_\mathbf{x} \mu, \quad t > 0, \quad 0 < \gamma \le 1,\tag{16}$$

*having initial solution as*

$$
\mu(\mathbf{x},0) = \exp\left(\frac{\mathbf{x}}{2}\right). \tag{17}
$$

*Applying NT to Equation (16), we have*

$$\frac{\mathcal{S}^{\gamma}}{\omega^{\gamma}}N^{+}\left[\mu(\mathbf{x},t)\right]-\frac{\mathcal{S}^{\gamma-1}}{\omega^{\gamma}}\mu(\mathbf{x},0) = N^{+}\left[D\_{\text{xx}t}\mu - D\_{\text{x}}\mu + \mu D\_{\text{xxx}}\mu - \mu D\_{\text{x}}\mu + 3D\_{\text{x}}\mu D\_{\text{xx}}\mu\right].$$

$$N^{+}\left[\mu(\mathbf{x},t)\right]-\frac{1}{s}\mu(\mathbf{x},0) = \frac{\omega^{\gamma}}{s^{\gamma}}N^{+}\left[D\_{\text{xxx}t}\mu - D\_{\text{x}}\mu + \mu D\_{\text{xxx}}\mu - \mu D\_{\text{x}}\mu + 3D\_{\text{x}}\mu D\_{\text{xx}}\mu\right].$$

*Using inverse natural transformation*

$$\mu(\mathbf{x},t) = \mathbf{N}^- \left[ \frac{\mu(\mathbf{x},0)}{s} - \frac{\omega^\gamma}{s^\gamma} \mathbf{N}^+ \left[ \mathbf{D}\_{\mathbf{x}\mathbf{x}t}\mu - \mathbf{D}\_\mathbf{x}\mu + \mu\mathbf{D}\_{\mathbf{x}\mathbf{x}\mathbf{x}}\mu - \mu\mathbf{D}\_\mathbf{x}\mu + \mathbf{3}\mathbf{D}\_\mathbf{x}\mu \mathbf{D}\_{\mathbf{x}\mathbf{x}}\mu \right] \right].$$

*Applying the ADM process, we have*

*µ*0(*x*, *t*) = *N* − *µ*(*x*, 0) *s* = *N* − " exp *x* 2 *s* # , *<sup>µ</sup>*0(*x*, *<sup>t</sup>*) = exp *<sup>x</sup>* 2 , (18) ∞ ∑ ℓ=0 *µ*ℓ+<sup>1</sup> (*x*, *t*) = *N* − " *ω<sup>γ</sup> s γ N* + " ∞ ∑ ℓ=0 (*Dxxtµ*)<sup>ℓ</sup> − ∞ ∑ ℓ=0 (*Dxµ*)<sup>ℓ</sup> + ∞ ∑ ℓ=0 *A*<sup>ℓ</sup> − ∞ ∑ ℓ=0 *B*<sup>ℓ</sup> + 3 ∞ ∑ ℓ=0 *C*ℓ ## , ℓ = 0, 1, 2, · · · *A*0(*µDxxxµ*) = *µ*0*Dxxxµ*0, *A*1(*µDxxxµ*) = *µ*0*Dxxxµ*<sup>1</sup> + *µ*1*Dxxxµ*0, *A*2(*µDxxxµ*) = *µ*1*Dxxxµ*<sup>2</sup> + *µ*1*Dxxxµ*<sup>1</sup> + *µ*2*Dxxxµ*0, *B*0(*µDxµ*) = *µ*0*Dxµ*0, *B*1(*µDxµ*) = *µ*0*Dxµ*<sup>1</sup> + *µ*1*Dxµ*0, *B*2(*µDxµ*) = *µ*1*Dxµ*<sup>2</sup> + *µ*1*Dxµ*<sup>1</sup> + *µ*2*Dxµ*0, *C*0(*DxµDxxµ*) = *Dxµ*0*Dxxµ*0, *C*1(*DxµDxxµ*) = *Dxµ*0*Dxxµ*<sup>1</sup> + *Dxµ*1*Dxxµ*0, *C*2(*DxµDxxµ*) = *Dxµ*1*Dxxµ*<sup>2</sup> + *Dxµ*1*Dxxµ*<sup>1</sup> + *Dxµ*2*Dxxµ*0,

*for* ℓ = 1

$$\begin{split} \mu\_{1}(\mathbf{x},t) &= N^{-} \left[ \frac{\omega^{\gamma}}{s^{\gamma}} N^{+} \left[ (D\_{\mathbf{x}\mathbf{x}t}\mu)\_{0} - (D\_{\mathbf{x}}\mu)\_{0} + A\_{0} - B\_{0} + \mathfrak{X}\_{0} \right] \right], \\ \mu\_{1}(\mathbf{x},t) &= -\frac{1}{2} N^{-} \left[ \frac{\omega^{\gamma} \exp(\frac{\mathbf{x}}{2})}{s^{\gamma+1}} \right] = -\frac{1}{2} \exp\left(\frac{\mathbf{x}}{2}\right) \frac{t^{\gamma}}{\Gamma(\gamma+1)}. \end{split} \tag{19}$$

*for* ℓ = 2

$$\begin{split} \mu\_{2}(\mathbf{x},t) &= N^{-} \left[ \frac{\omega^{\gamma}}{s^{\gamma}} N^{+} \left[ (D\_{\text{xxt}}\mu)\_{1} - (D\_{\text{x}}\mu)\_{1} + A\_{1} - B\_{1} + 3\mathbb{C}\_{1} \right] \right], \\ \mu\_{2}(\mathbf{x},t) &= -\frac{1}{8} \exp\left(\frac{\chi}{2}\right) \frac{t^{2\gamma-1}}{\Gamma(2\gamma)} + \frac{1}{4} \exp\left(\frac{\chi}{2}\right) \frac{t^{2\gamma}}{\Gamma(2\gamma+1)}, \end{split} \tag{20}$$

*for* ℓ = 3

$$\begin{split} \mu\_{3}(\mathbf{x},t) &= N^{-} \left[ \frac{\omega^{\gamma}}{s^{\gamma}} N^{+} \left[ (D\_{\text{xxt}}\mu)\_{2} - (D\_{\text{x}}\mu)\_{2} + A\_{2} - B\_{2} + 3\mathfrak{C}\_{2} \right] \right], \\ \mu\_{3}(\mathbf{x},t) &= -\frac{1}{32} \exp\left(\frac{\mathbf{x}}{2}\right) \frac{t^{3\gamma-2}}{\Gamma(3\gamma-1)} + \frac{1}{8} \exp\left(\frac{\mathbf{x}}{2}\right) \frac{\gamma^{3\gamma-1}}{\Gamma(3\gamma)} - \frac{1}{8} \exp\left(\frac{\mathbf{x}}{2}\right) \frac{t^{3\gamma}}{\Gamma(3\gamma+1)}, \end{split} \tag{21}$$

*The NDM solution for problem (16) is*

$$
\mu(\mathbf{x},t) = \mu\_0(\mathbf{x},t) + \mu\_1(\mathbf{x},t) + \mu\_2(\mathbf{x},t) + \mu\_3(\mathbf{x},t) + \mu\_4(\mathbf{x},t) \cdots \ .
$$

$$
\begin{split}
\mu(\mathbf{x},t) &= \exp\left(\frac{\mathbf{x}}{2}\right) - \frac{1}{2}\exp\left(\frac{\mathbf{x}}{2}\right)\frac{t^{\gamma}}{\Gamma(\gamma+1)} \\
&- \frac{1}{8}\exp\left(\frac{\mathbf{x}}{2}\right)\frac{t^{2\gamma-1}}{\Gamma(2\gamma)} + \frac{1}{4}\exp\left(\frac{\mathbf{x}}{2}\right)\frac{t^{2\gamma}}{\Gamma(2\gamma+1)} - \frac{1}{32}\exp\left(\frac{\mathbf{x}}{2}\right) \\
&\frac{t^{3\gamma-2}}{\Gamma(3\gamma-1)} + \frac{1}{8}\exp\left(\frac{\mathbf{x}}{2}\right)\frac{\gamma^{3\gamma-1}}{\Gamma(3\gamma)} - \frac{1}{8}\exp\left(\frac{\mathbf{x}}{2}\right)\frac{t^{3\gamma}}{\Gamma(3\gamma+1)} - \cdots .
\end{split}
\tag{22}
$$

*The simplification of Equation (22);*

$$\mu(\mathbf{x},t) = \exp\left(\frac{\mathbf{x}}{2}\right) \left[1 - \frac{t^{\gamma}}{2\Gamma(\gamma+1)} - \frac{1}{8} \frac{t^{2\gamma-1}}{\Gamma(2\gamma)} + \frac{1}{4} \frac{t^{2\gamma}}{\Gamma(2\gamma+1)} - \frac{1}{32} \frac{t^{3\gamma-2}}{\Gamma(3\gamma-1)} + \frac{1}{8} \frac{t^{3\gamma-1}}{\Gamma(3\gamma)} - \frac{1}{8} \frac{t^{3\gamma}}{\Gamma(3\gamma+1)} + \dotsb\right].\tag{23}$$

*The exact result of Example 1*

$$\mu(\mathbf{x},t) = \exp\left(\frac{\mathbf{x}}{2} - \frac{2t}{3}\right) \Big|\_{\mathbf{x}} \tag{24}$$

In Table 1, the NDM-solutions at different fractional-order derivatives, *γ* = 0.5, 0.7 and 1 are shown. The NDM-solutions at various time level, *t* = 0.2, 0.4 and *t* = 1 are determined. The absolute error of the proposed method at *γ* = 1 is also displayed. From Table 1, it is investigated that suggested method has the desire rate of convergence and considered to be the best tool for the analytical solution of FPDEs. In Table 2, the NDM and LADM solutions are compared at various fractional-order of the derivatives. It is observed that the NDM has the higher degree of accuracy as compared to LDM. The comparison has been done at *γ* = 0.5, 0.7 and 0.9. It is also investigated that the fractional-order solutions of NDM have the higher accuracy as compared LDM.


**Table 1.** The NDM solutions and absolute error of Example 1 at *γ* = 0.5, 0.7 and 1.

**Table 2.** Two terms comparison of NDM and LDM [16] of different fractional-order at *γ* = 0.5, 0.7 and 0.9 of Example 1.


In Figures 1 and 2, the NDM and actual solution of Example 1 are plotted. It is observed that NDM solutions are in closed contact with the exact solutions of Example 1. In Figures 3 and 4, the solutions of Example 1 at various fractional-order of the derivatives are plotted. The graphical representation has shown the convergence phenomena of fractional-order solution towards the solution at integer order of Example 1.

**Figure 1.** Exact and NDM solutions at *γ* = 1 of Example 1.

**Figure 2.** Exact and NDM solutions at *γ* = 1 of Example 1.

**Figure 3.** The NDM solutions of different valve of *γ* of Example 1.

**Figure 4.** Solution graph of Example 1, at various value of *γ*.

**Example 2.** *Consider the following nonlinear time-fractional Fornberg–Whitham equation [18]*

$$D\_t^\gamma \mu - D\_{\text{xxx}} \mu + D\_\mathbf{x} \mu = \mu D\_{\text{xxx}} \mu - \mu D\_\mathbf{x} \mu + \mathfrak{D}\_\mathbf{x} \mu D\_\mathbf{x} \mu, \quad t > 0, \quad 0 < \gamma \le 1,\tag{25}$$

*with initial condition*

$$
\mu(\mathbf{x},0) = \cosh^2\left(\frac{\mathbf{x}}{4}\right),
\tag{26}
$$

*Applying natural transformation of Equation (25),*

$$\frac{\mathcal{S}^{\gamma}}{\omega^{\gamma}} N^{+} \left[ \mu(\mathbf{x}, t) \right] - \frac{\mathcal{S}^{\gamma - 1}}{\omega^{\gamma}} \mu(\mathbf{x}, 0) = N^{+} \left[ D\_{\text{xx}t} \mu - D\_{\text{x}} \mu + \mu D\_{\text{xxx}} \mu - \mu D\_{\text{x}} \mu + 3 D\_{\text{x}} \mu D\_{\text{xx}} \mu \right].$$

$$N^{+} \left[ \mu(\mathbf{x}, t) \right] - \frac{1}{s} \mu(\mathbf{x}, 0) = \frac{\omega^{\gamma}}{s^{\gamma}} N^{+} \left[ D\_{\text{xx}t} \mu - D\_{\text{x}} \mu + \mu D\_{\text{xxx}} \mu - \mu D\_{\text{x}} \mu + 3 D\_{\text{x}} \mu D\_{\text{xx}} \mu \right].$$

*Using inverse natural transformation*

$$\mu(\mathbf{x},t) = N^- \left[ \frac{\mu(\mathbf{x},0)}{s} - \frac{\omega^\gamma}{s^\gamma} N^+ \left[ D\_{\mathbf{x}\mathbf{x}t} \mu - D\_\mathbf{x} \mu + \mu D\_{\mathbf{x}\mathbf{x}\mathbf{x}} \mu - \mu D\_\mathbf{x} \mu + 3D\_\mathbf{x} \mu D\_{\mathbf{x}\mathbf{x}} \mu \right] \right].$$

*Applying the ADM process, we have*

$$
\mu\_0(\mathbf{x}, t) = N^- \left[ \frac{\mu(\mathbf{x}, 0)}{\mathbf{s}} \right] = N^- \left[ \frac{\cosh^2 \left( \frac{\mathbf{x}}{4} \right)}{\mathbf{s}} \right],
$$

$$
\mu\_0(\mathbf{x}, t) = \cosh^2 \left( \frac{\mathbf{x}}{4} \right),
\tag{27}
$$

$$
\sum\_{\ell=0}^\infty \mu\_{\ell+1}(\mathbf{x}, t) = N^- \left[ \frac{\omega^\gamma}{s^\gamma} N^+ \left[ \sum\_{\ell=0}^\infty (D\_{\text{xxt}} \mu)\_\ell - \sum\_{\ell=0}^\infty (D\_{\text{x}} \mu)\_\ell + \sum\_{\ell=0}^\infty A\_\ell - \sum\_{\ell=0}^\infty B\_\ell + 3 \sum\_{\ell=0}^\infty C\_\ell \right] \right], \quad \ell = 0, 1, 2, \cdots
$$

$$
A\_0(\mu D\_{\text{xxt}} \mu) = \mu\_0 D\_{\text{xxt}} \mu\_0.
$$

$$\begin{aligned} A\_0(\mu D\_{\text{xxx}} \mu) &= \mu\_0 D\_{\text{xxx}} \mu\_0, \\ A\_1(\mu D\_{\text{xxx}} \mu) &= \mu\_0 D\_{\text{xxx}} \mu\_1 + \mu\_1 D\_{\text{xxx}} \mu\_0, \\ A\_2(\mu D\_{\text{xxx}} \mu) &= \mu\_1 D\_{\text{xxx}} \mu\_2 + \mu\_1 D\_{\text{xxx}} \mu\_1 + \mu\_2 D\_{\text{xxx}} \mu\_0, \\ B\_0(\mu D\_{\text{x}} \mu) &= \mu\_0 D\_{\text{x}} \mu\_0, \\ B\_1(\mu D\_{\text{x}} \mu) &= \mu\_0 D\_{\text{x}} \mu\_1 + \mu\_1 D\_{\text{x}} \mu\_0, \\ B\_2(\mu D\_{\text{x}} \mu) &= \mu\_1 D\_{\text{x}} \mu\_2 + \mu\_1 D\_{\text{x}} \mu\_1 + \mu\_2 D\_{\text{x}} \mu\_0. \end{aligned}$$

$$\mathbb{C}\_0(D\_{\text{x}} \mu D\_{\text{xx}} \mu) = D\_{\text{x}} \mu\_0 D\_{\text{xx}} \mu\_0.$$

$$\begin{aligned} \mathbf{C}\_{0}(\boldsymbol{\Sigma}\_{\mathbf{x}}\mu \mathbf{D}\_{\mathbf{x}\mathbf{x}}\mu) &= \mathbf{D}\_{\mathbf{x}\boldsymbol{\mu}}\mu \mathbf{D}\_{\mathbf{x}\mathbf{x}}\mu\_{0}\nu \\ \mathbf{C}\_{1}(D\_{\mathbf{x}}\mu D\_{\mathbf{x}\mathbf{x}}\mu) &= D\_{\mathbf{x}}\mu\_{0}D\_{\mathbf{x}\mathbf{x}}\mu\_{1} + D\_{\mathbf{x}}\mu\_{1}D\_{\mathbf{x}\mathbf{x}}\mu\_{0}\nu \\ \mathbf{C}\_{2}(D\_{\mathbf{x}}\mu D\_{\mathbf{x}\mathbf{x}}\mu) &= D\_{\mathbf{x}}\mu\_{1}D\_{\mathbf{x}\mathbf{x}}\mu\_{2} + D\_{\mathbf{x}}\mu\_{1}D\_{\mathbf{x}\mathbf{x}}\mu\_{1} + D\_{\mathbf{x}}\mu\_{2}D\_{\mathbf{x}\mathbf{x}}\mu\_{0}\nu \end{aligned}$$

*for* ℓ = 1

$$\begin{split} \mu\_{1}(\mathbf{x},t) &= N^{-} \left[ \frac{\omega^{\gamma}}{s^{\gamma}} N^{+} \left[ (D\_{\text{xxt}}\mu)\_{0} - (D\_{\text{x}}\mu)\_{0} + A\_{0} - B\_{0} + 3\mathbf{C}\_{0} \right] \right], \\ \mu\_{1}(\mathbf{x},t) &= -\frac{11}{32} N^{-} \left[ \frac{\omega^{\gamma}\sinh\left(\frac{\chi}{2}\right)}{s^{\gamma+1}} \right] = -0.3437 \sinh\left(\frac{\chi}{4}\right) \frac{t^{\gamma}}{\Gamma(\gamma+1)}, \end{split} \tag{28}$$

*for* ℓ = 2

$$\begin{aligned} \mu\_2(\mathbf{x}, t) &= N^- \left[ \frac{\omega^\gamma}{s^\gamma} N^+ \left[ (D\_{\text{x1}t} \mu)\_1 - (D\_x \mu)\_1 + A\_1 - B\_1 + 3 \mathbf{C}\_1 \right] \right], \\ \mu\_2(\mathbf{x}, t) &= -0.08593 \sinh\left(\frac{\chi}{4}\right) \frac{t^\gamma}{\Gamma(\gamma + 1)} + 0.11816 \cosh\left(\frac{\chi}{4}\right) \frac{t^{2\gamma}}{\Gamma(2\gamma + 1)}, \end{aligned} \tag{29}$$

*for* ℓ = 3

$$\begin{aligned} \mu\_{3}(\mathbf{x},t) &= N^{-} \left[ \frac{\omega^{\gamma}}{s^{\gamma}} N^{+} \left[ (D\_{\text{xx}}\mu)\_{2} - (D\_{\text{x}}\mu)\_{2} + A\_{2} - B\_{2} + 3C\_{2} \right] \right], \\ \mu\_{3}(\mathbf{x},t) &= -0.08593 \sinh\left(\frac{\mathbf{x}}{4}\right) \frac{t^{\gamma}}{\Gamma(\gamma+1)} + 0.11816 \cosh\left(\frac{\mathbf{x}}{4}\right) \frac{t^{2\gamma}}{\Gamma(2\gamma+1)} - 0.02707 \sinh\left(\frac{\mathbf{x}}{4}\right) \frac{t^{3\gamma}}{\Gamma(3\gamma+1)}. \end{aligned} \tag{30}$$

*The NDM result for problem 2 is*

$$\mu(\mathbf{x},t) = \mu\_0(\mathbf{x},t) + \mu\_1(\mathbf{x},t) + \mu\_2(\mathbf{x},t) + \mu\_3(\mathbf{x},t) + \mu\_4(\mathbf{x},t) \dots$$

$$\mu(\mathbf{x},t) = \cosh^2\left(\frac{\mathbf{x}}{4}\right) - 0.3437\sinh\left(\frac{\mathbf{x}}{4}\right)\frac{t^{\gamma}}{\Gamma(\gamma+1)} - 0.08593\sinh\left(\frac{\mathbf{x}}{4}\right)\frac{t^{\gamma}}{\Gamma(\gamma+1)}$$

$$\begin{split} &+ 0.11816\cosh\left(\frac{\mathbf{x}}{4}\right)\frac{t^{2\gamma}}{\Gamma(2\gamma+1)} - 0.08593\sinh\left(\frac{\mathbf{x}}{4}\right)\frac{t^{\gamma}}{\Gamma(\gamma+1)} \\ &+ 0.11816\cosh\left(\frac{\mathbf{x}}{4}\right)\frac{t^{2\gamma}}{\Gamma(2\gamma+1)} - 0.02707\sinh\left(\frac{\mathbf{x}}{4}\right)\frac{t^{3\gamma}}{\Gamma(3\gamma+1)} - \cdots \end{split}$$

*The exact result is;*

$$
\mu(\mathbf{x}, t) = \cosh^2\left(\frac{\mathbf{x}}{4} - \frac{11t}{24}\right).
$$

*In Figures 5 and 6, the solution graph of exact and NDM of Example 2 at integer-order are plotted. The closed relation is observed between NDM and exact solution of Example 2. In Figures 7 and 8, the fractional-order solutions of Example 2 are presented. The graphical representation have confirmed the different dynamics of Example 2, which are correlated with each other.*

**Figure 5.** The graph of exact and approximate solution of Example 2.

**Figure 6.** The graph of exact and approximate solution of Example 2.

**Figure 7.** The NDM solutions of different valve of *γ* of Example 2.

**Figure 8.** The graph of Example 2, for different value of *γ*.

#### **5. Conclusions**

In the current work. an innovative technique is used to find the solution of fractional Fornberg-Whitham equations. The fractional-derivatives are discussed within Caputo operator. The solutions are determined for fractional-order problems and an aesthetically a strong relation is found. The fractional models have shown convergence to the ordinary model as the order of the derivative tends towards to an integer. The graphical representation has provided similar behavior of actual and derived results. It is also noted the current method needs small calculation and higher convergence to achieve the solution of the targeted problems.

**Author Contributions:** Conceptualization, R.S. and H.K.; Methodology, R.S.; Software, A.A.A.; Validation, D.B. and S.A; Formal Analysis, H.K.; Investigation, R.S. and A.A.A.; Resources, H.K. and R.S.; Writing—Original Draft Preparation, R.S.; Writing—Review and Editing, H.K., D.B. and S.A.; Visualization, H.K.; Supervision, H.K., H.K.; Project Administration, D.B.; Funding Acquisition, A.A.A. and S.A. All authors have read and agreed to the published version of the manuscript.

**Funding:** The authors extend their appreciation to the Deanship of Scientific Research at King Khalid University, Saudi Arabia for founding this work through Research Groups program under grant number (R.G.P2./99/41).

**Acknowledgments:** The authors extend their appreciation to the Deanship of Scientific Research at King Khalid University, Saudi Arabia for founding this work through Research Groups program under grant number (R.G.P2./99/41).

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

#### **References**


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