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Novel results on conformable Bessel functions

Novel results on conformable Bessel functions 1IntroductionFractional calculus is theoretically a powerful analysis technique for investigating arbitrary order integrals and derivatives. At the beginning, this field of research has been only presented purely, and until very recently, researchers have realized the powerful applicability of this field in modeling many phenomena from natural sciences and engineering much better than using the ordinary usual calculus due to several properties in fractional calculus that can provide a good explanation of physical behavior of certain system (see refs. [1,2,3,4,5,6,7]). The applications of conformable and fractional calculus have been recently discussed in some research studies [8,9,10,11,12,13,14,15]. The analysis of conformable derivatives and integrals has been discussed in detail in earlier studies [16,17].Riemann–Liouville and Caputo fractional operators [18,19] have been initially introduced, and recently, various new or generalized definitions have been proposed by researchers. In 2014, Khalil et al. proposed a local definition of fractional derivative, known as conformable derivative [20]. This topic has been discussed in some research studies whether conformable derivatives are considered fractional derivatives or not. While Tarasov has discussed in an earlier study [21] that some recent definitions of fractional derivatives are not fractional derivatives, Almeida et al. have concluded in an earlier study [22] that conformable derivative is still interesting to study it further. A new generalized definition of conformable derivative that coincides with the definitions of Caputo and Riemann–Liouville fractional derivatives has been proposed in an earlier study [23]. Some mathematicians such as Zhao and Luo and Khalil et al. discussed the conformable derivative’s physical and geometrical meanings in an earlier studies [24,25], respectively. Therefore, the conformable derivative in this study will only be considered as a modified form of local usual derivative. The analysis of Bessel functions in the context of conformable derivative has been rarely discussed in any other research studies. Therefore, our results in this article are novel and worthy due to the importance of Bessel functions in various modeling scenarios in science and engineering.The article is outlined as follows: some essential notions of the conformable calculus and conformable Bessel functions are presented in Section 2. In Section 3, we propose and prove new properties of Bessel functions of the first kind involving their conformable derivatives or their zeros. We conclude with Section 4, where the orthogonality of such functions in the interval [0, 1] is studied.2PreliminariesDefinition 2.1[20] For a function f:[0,∞)→Rf:{[}0,\infty )\to R, the conformable derivative of order α\alpha is expressed as follows:(1)(Tαf)(t)=limε→0f(t+εt1−α)−f(t)ε.({T}_{\alpha }f)(t)=\mathop{\mathrm{lim}}\limits_{\varepsilon \to 0}\frac{f(t+\varepsilon {t}^{1-\alpha })-f(t)}{\varepsilon }.∀t>0\forall t\gt 0; 0<α≤10\lt \alpha \le 1. If ffis α\alpha -differentiable in some (0,a)(0,a), a>0a\gt 0, and limt→0+(Tαf)(t)\mathop{\mathrm{lim}}\limits_{t\to {0}^{+}}({T}_{\alpha }f)(t)exists, then it is defined as follows:(2)(Tαf)(0)=limt→0+(Tαf)(t).({T}_{\alpha }f)(0)=\mathop{\mathrm{lim}}\limits_{t\to {0}^{+}}({T}_{\alpha }f)(t).As a result, we have Theorem 2.1 as follows:Theorem 2.1[20]. If a function f:[0,∞)→Rf:{[}0,\infty )\to Ris α\alpha -differentiable at t0>0{t}_{0}\gt 0; 0<α≤10\lt \alpha \le 1, then ffis continuous at t0{t}_{0}.Hence, Tα{T}_{\alpha }satisfies some properties as follows:Theorem 2.2[20]. Suppose that 0<α≤10\lt \alpha \le 1and ff, ggare α\alpha -differentiable at a point t>0t\gt 0. Then, we get the following:(i)Tα(af + bg) = a(Tαf) + b(Tαg), for all a, b ∈ R.(ii)Tα(tp) = ptp−αfor all p ∈ R.(iii)Tα(λ) = 0, for all constant functions f(t) =λ.(iv)Tα(fg) = f(Tαg) + g(Tαf).(v)Tαfg=g(Tαf)−f(Tαg)g2;g≠0{T}_{\alpha }\left(\frac{f}{g}\right)=\frac{g({T}_{\alpha }f)-f({T}_{\alpha }g)}{{g}^{2}};g\ne 0.(vi)Additionally, if f is a differentiable function, then (Tαf)(t)=t1−αdfdt(t)({T}_{\alpha }f)(t)={t}^{1-\alpha }\frac{\text{d}f}{\text{d}t}(t).From the above definition, the conformable derivatives of certain functions are written as follows:(i)Tα(1) = 0.(ii)Tα(sin(at)) = at1−αcos(at), a ∈ R.(iii)Tα(cos(at)) = −at1−αsin(at), a ∈ R.(iv)Tα(eat) =aeat, a ∈ R.Theorem 2.3(Rolle’s theorem). [20]. Suppose that a>0a\gt 0; α∈(0,1]\alpha \in (0,1], and f:[a,∞)→Rf:{[}a,\infty )\to Ris a function that satisfies the following:–ffis continuous on a,ba,b.–ffis α\alpha -differentiable on (a,b)(a,b).–f(a)=f(b)f(a)=f(b).Then, ∃c∈(a,b)\exists \hspace{.25em}c\in (a,b)such that (Tαf)(c)({T}_{\alpha }f)(c)= 0.Theorem 2.4[26]. Suppose that a>0a\gt 0; α∈(0,1]\alpha \in (0,1], and f:[a,∞)→Rf:{[}a,\infty )\to Ris function that satisfies the following:–ffis continuous in [a,b]{[}a,b].–ffis α\alpha -differentiable on (a,b)(a,b).If (Tαf)(c)=0({T}_{\alpha }f)(c)=0∀t∈(a,b)\forall t\in (a,b), then ffis a constant on [a,b]{[}a,b].Definition 2.2The left-conformable derivative beginning from aaof a function f:[a,∞)→Rf:{[}a,\infty )\to Rof f\hspace{.25em}fof order 0<α≤10\lt \alpha \le 1, [27], is expressed as follows:(3)(Tαaf)(t)=limε→0f(t+ε(t−a)1−α)−f(t)ε.({T}_{\alpha }^{a}f)(t)=\mathop{\mathrm{lim}}\limits_{\varepsilon \to 0}\frac{f(t+\varepsilon {(t-a)}^{1-\alpha })-f(t)}{\varepsilon }.When a=0a=0, it is denoted as (Tαf)(t)({T}_{\alpha }f)(t). If ffis α\alpha -differentiable in some (a,b)(a,b), then we get the following equation:(4)(Tαaf)(a)=limt→a+(Tαaf)(t).({T}_{\alpha }^{a}f)(a)=\mathop{\mathrm{lim}}\limits_{t\to {a}^{+}}({T}_{\alpha }^{a}f)(t).It is noticeable that if ffis differentiable function, then (Tαaf)(a)=(t−a)1−αdfdt(t)({T}_{\alpha }^{a}f)(a)={(t-a)}^{1-\alpha }\frac{\text{d}f}{\text{d}t}(t). Theorem 2.2 holds for definition 2.2 when changing by (t−a)(t-a).Theorem 2.5(Chain rule). [27]. Assume f,g:(a,∞)→Rf,g:(a,\infty )\to Rbe left α-differentiable function functions, where 0<α≤10\lt \alpha \le 1. Suppose that h(t)=f(g(t))h(t)=f(g(t)). The h(t)h(t)is α-differentiable function ∀t≠a\forall t\ne aand g(t)≠0g(t)\ne 0, hence we get the following equation:(5)(Tαah)(t)=(Tαaf)(g(t))⋅(Tαag)(t).(g(t))α−1.({T}_{\alpha }^{a}h)(t)=({T}_{\alpha }^{a}f)(g(t))\cdot ({T}_{\alpha }^{a}g)(t).{(g(t))}^{\alpha -1}.If t=at=a, then we define as follows:(6)(Tαah)(a)=limt→a+(Tαaf)(g(t))⋅(Tαag)(t)⋅(g(t))α−1.({T}_{\alpha }^{a}h)(a)=\mathop{\mathrm{lim}}\limits_{t\to {a}^{+}}({T}_{\alpha }^{a}f)(g(t))\cdot ({T}_{\alpha }^{a}g)(t)\cdot {(g(t))}^{\alpha -1}.Remark 2.1In an earlier study [27], the left-conformable derivative at aafor some smooth functions has been investigated. Suppose that 0<α≤10\lt \alpha \le 1and n∈Z+n\in {Z}^{+}, then the left sequential conformable derivative of order nnis expressed as follows: Tαa(n)f(t)=TαaTαa…Tαaf(t)︸n-times{}^{(n)}T_{\alpha }^{a}f(t)=\mathop{\underbrace{{T}_{\alpha }^{a}{T}_{\alpha }^{a}\ldots {T}_{\alpha }^{a}f(t)}}\limits_{n\text{-times}}.We show via induction that if ffis continuously α\alpha -differentiable and 0<α≤1n0\lt \alpha \le \frac{1}{n}then, the nth order sequential conformable derivative is continuous and vanishes at the end point a.Theorem 2.6[27]. Suppose that f is infinitely α\alpha -differentiable function, for some 0<α≤10\lt \alpha \le 1at a neighborhood of a point t0{t}_{0}. Then, ffhas a conformable power series expansion as follows:(7)f(t)=∑k=0∞((k)Tαt0)(t0)αkk!(t−t0)kα,t0<t<t0+R1α.f(t)=\mathop{\sum }\limits_{k=0}^{\infty }\frac{((k){T}_{\alpha }^{{t}_{0}})({t}_{0})}{{\alpha }^{k}k\&#x0021;}{(t-{t}_{0})}^{k\alpha },\hspace{1em}{t}_{0}\lt t\lt {t}_{0}+{R}^{\frac{1}{\alpha }}.Here, ((k)Tαt0)(t0)((k){T}_{\alpha }^{{t}_{0}})({t}_{0})indicates that we are applying the conformable derivative kktimes.The α\alpha -integral of a function ffbeginning from a≥0a\ge 0is expressed as follows:Definition 2.3[20]. Iαa(f)(t)=∫atf(x)x1−α⋅dx{I}_{\alpha }^{a}(f)(t)={\int }_{a}^{t}\frac{f(x)}{{x}^{1-\alpha }}\cdot \text{d}x, where it is a usual Riemann improper integral, and α∈(0,1]\alpha \in (0,1].As a result, we obtain the following:Theorem 2.7TαaIαa(f)(t)=f(t){T}_{\alpha }^{a}{I}_{\alpha }^{a}(f)(t)=f(t), for t≥at\ge a, wheref\hspace{.25em}fis any continuous function in the domain of Iα.{I}_{\alpha }.Lemma 2.1[27]. Let f:(a,b)→Rf:(a,b)\to Rbe differentiable and α∈(0,1]\hspace{.25em}\alpha \in (0,1]. Then, ∀a>0\forall \hspace{.25em}a\gt 0, and we get the following:(9)IαaTαa(f)(t)=f(t)−f(a).{I}_{\alpha \hspace{.25em}}^{a}{T}_{\alpha }^{a}(f)(t)=f(t)-f(a).Theorem 2.8[27]. Let f:[a,b]→Rf:{[}a,b]\to Rbe two functions ∋fg\ni fgis differentiable. Then we have the following equation:(10)∫abf(x)Tαa(g)(x)dα(x)=fg]ab−∫abg(x)Tαa(f)(x)dα(x).\begin{array}{c}\underset{a}{\overset{b}{\int }}f(x){T}_{\alpha }^{a}(g)(x)\text{d}\alpha (x)={fg]}_{a}^{b}\\ \hspace{1em}-\underset{a}{\overset{b}{\int }}g(x){T}_{\alpha }^{a}(f)(x)\text{d}\alpha (x).\end{array}Now, consider the sequential conformable Bessel equation [28]:(11)t2αTαTαy+αtαTαy+α2(t2α−p2)y=0.{t}^{2\alpha }{T}_{\alpha }{T}_{\alpha }y+\alpha {t}^{\alpha }{T}_{\alpha }y+{\alpha }^{2}({t}^{2\alpha }-{p}^{2})y=0.where 0<α≤10\lt \alpha \le 1and ppis any real number. If α=1\alpha =1, then Eq. (1) is a usual Bessel equation [29]. t=0t=0is a α-\alpha \text{-}regular singular point for the equation. In this case, for t>0t\gt 0, the solution can be investigated via a conformable Fröbenius series as:y=trα∑n=0∞cntnα=∑n=0∞cnt(n+r)αy={t}^{r\alpha }\mathop{\sum }\limits_{n=0}^{\infty }{c}_{n}{t}^{n\alpha }=\mathop{\sum }\limits_{n=0}^{\infty }{c}_{n}{t}^{(n+r)\alpha }We let:Tαy=∑n=0∞α(n+r)cnt(n+r−1)α{T}_{\alpha }y=\mathop{\sum }\limits_{n=0}^{\infty }\alpha (n+r){c}_{n}{t}^{(n+r-1)\alpha }TαTαy=∑n=0∞α2(n+r)(n+r−1)cnt(n+r−2)α{T}_{\alpha }{T}_{\alpha }y=\mathop{\sum }\limits_{n=0}^{\infty }{\alpha }^{2}(n+r)(n+r-1){c}_{n}{t}^{(n+r-2)\alpha }Let substitute these expressions in Eq. (11) to get the following:l(r)c0trα+l(r+1)c1t(r+1)α+∑n=0∞[l(r+n)cn+α2cn−2]t(r+n)α=0\begin{array}{c}l(r){c}_{0}{t}^{r\alpha }+l(r+1){c}_{1}{t}^{(r+1)\alpha }\\ \hspace{1em}+\mathop{\sum }\limits_{n=0}^{\infty }{[}l(r+n){c}_{n}+{\alpha }^{2}{c}_{n-2}]{t}^{(r+n)\alpha }=0\end{array}where l(r)=r(r−1)α2+rα2−α2p2l(r)=r(r-1){\alpha }^{2}+r{\alpha }^{2}-{\alpha }^{2}{p}^{2}.Let us c0≠0{c}_{0}\ne 0, then we obtain the following:l(r)=0l(r)=0Since α2≠0{\alpha }^{2}\ne 0, the following can be written as follows:l(r)=0⇒r(r−1)α2+rα2−α2p2=0⇒r2−p2=0.l(r)=0\hspace{.25em}\Rightarrow \hspace{.25em}r(r-1){\alpha }^{2}+r{\alpha }^{2}-{\alpha }^{2}{p}^{2}=0\hspace{.25em}\Rightarrow \hspace{.25em}{r}^{2}-{p}^{2}=0.Hence, we find:r1=p,r2=−p.{r}_{1}=p\hspace{.25em},\hspace{.25em}{r}_{2}=-p.Now, for p>0p\gt 0, the solutions of the conformable Bessel equation of order ppare analyzed as follows:In this case, for r1=p{r}_{1}=p, we have:l(r1+1)c1=[α2p(p+1)+α2(p+1)−α2p2]c1=0.l({r}_{1}+1){c}_{1}={[}{\alpha }^{2}p(p+1)+{\alpha }^{2}(p+1)-{\alpha }^{2}{p}^{2}]{c}_{1}=0.(2p+1)c1=0.(2p+1){c}_{1}=0.Due to p>0p\gt 0, it follows that c1=0{c}_{1}=0. The recurrence relation is as follows:cn=−cn−2n(n+p).{c}_{n}=-\frac{{c}_{n-2}}{n(n+p)}.From c1=0{c}_{1}=0and last recurrence relation, we obtain the following:c3=c5=…=0.{c}_{3}={c}_{5}=\ldots =0.c2n=(−1)nc022nn!(p+1)(p+2)…(p+n),n≥1.{c}_{2n}=\frac{{(-1)}^{n}{c}_{0}}{{2}^{2n}n\&#x0021;(p+1)(p+2)\ldots (p+n)}\hspace{.25em},\hspace{.25em}n\ge 1.Let us c0=cΓ(p+1){c}_{0}=\frac{c}{\text{&#x0393;}(p+1)}. Thus, the first solution of the conformable Bessel equation of order pphas the following form:(12)y1(t)=c∑n=0∞(−1)nn!Γ(p+n+1)tα22n+p.{y}_{1}(t)=c\mathop{\sum }\limits_{n=0}^{\infty }\frac{{(-1)}^{n}}{n\&#x0021;\text{&#x0393;}(p+n+1)}{\left(\frac{{t}^{\alpha }}{2}\right)}^{2n+p}.Besides, the Bessel function of order ppis valid.(13)(Jα)p(t)=∑n=0∞(−1)nn!Γ(p+n+1)tα22n+p.{({J}_{\alpha })}_{p}(t)=\mathop{\sum }\limits_{n=0}^{\infty }\frac{{(-1)}^{n}}{n\&#x0021;\text{&#x0393;}(p+n+1)}{\left(\frac{{t}^{\alpha }}{2}\right)}^{2n+p}.For r2=−p{r}_{2}=-p, if r1−r2=2p{r}_{1}-{r}_{2}=2pis not a positive integer, then the second solution of the conformable Bessel equation of order ppis written as follows:(14)y2(t)=c∑n=0∞(−1)nn!Γ(−p+n+1)tα22n−p.{y}_{2}(t)=c\mathop{\sum }\limits_{n=0}^{\infty }\frac{{(-1)}^{n}}{n\&#x0021;\text{&#x0393;}(-p+n+1)}{\left(\frac{{t}^{\alpha }}{2}\right)}^{2n-p}.The second type Bessel functions of order pphas the following form:(15)(Jα)−p(t)=∑n=0∞(−1)nn!Γ(−p+n+1)tα22n−p.{({J}_{\alpha })}_{-p}(t)=\mathop{\sum }\limits_{n=0}^{\infty }\frac{{(-1)}^{n}}{n\&#x0021;\text{&#x0393;}(-p+n+1)}{\left(\frac{{t}^{\alpha }}{2}\right)}^{2n-p}.3Some basic properties of conformable Bessel functionsFirst, we will study the convergence of series (13) and (15).Theorem 3.1The series that defines the conformable Bessel functions of the first kind of order ppand −p-pare absolutely convergent for all t>0\hspace{.25em}t\gt 0.ProofThe radius of convergence of conformable series is as follows:∑n=0∞(−1)nn!Γ(±p+n+1)xnα\mathop{\sum }\limits_{n=0}^{\infty }\frac{{(-1)}^{n}}{n\&#x0021;\text{&#x0393;}(\pm p+n+1)}{x}^{n\alpha }which is easily found by ratio test [30],1R=limn→∞n!Γ(±p+n+1)(n+1)!Γ(±p+n+2)=limn→∞1(n+1)(±p+n+1)=0\begin{array}{c}\frac{1}{R}=\mathop{\mathrm{lim}}\limits_{n\to \infty }\frac{n\&#x0021;\Gamma (\pm p+n+1)}{(n+1)\&#x0021;\text{&#x0393;}(\pm p+n+2)}\\ \hspace{1em}=\mathop{\mathrm{lim}}\limits_{n\to \infty }\frac{1}{(n+1)(\pm p+n+1)}=0\end{array}which implies R=∞R=\infty .□Remark 3.1As series (13) and (15) are both convergent ∀t>0\forall t\gt 0, we may do a term-by-term differentiation for them [29].As (Jα)p{({J}_{\alpha })}_{p}and (Jα)−p{({J}_{\alpha })}_{-p}are the second order conformable linear differential equation’s solutions, our goal is to see that they are linearly independent, and thus they form a basis for the vector space of the solutions of Eq. (11).Theorem 3.2If 2p2pis not a positive integer, then the conformable Bessel functions of the first kind (Jα)p{({J}_{\alpha })}_{p}and (Jα)−p{({J}_{\alpha })}_{-p}, are linearly independent. In that case, for any solution yyof (11), ∃A,B∈R\exists A,B\in R(16)y(t)=A(Jα)p(t)+B(Jα)−p(t).y(t)=A{({J}_{\alpha })}_{p}(t)+B{({J}_{\alpha })}_{-p}(t).ProofIt is enough to see that the α-\alpha \text{-}Wronskian of (Jα)p(t){({J}_{\alpha })}_{p}(t)and (Jα)−p(t){({J}_{\alpha })}_{-p}(t)Wα((Jα)p(t),(Jα)−p(t))=(Jα)p(t)(Jα)−p(t)Tα(Jα)p(t)Tα(Jα)−p(t),{W}^{\alpha }({({J}_{\alpha })}_{p}(t),{({J}_{\alpha })}_{-p}(t))=\left|\begin{array}{cc}{({J}_{\alpha })}_{p}(t)& {({J}_{\alpha })}_{-p}(t)\\ {T}_{\alpha }{({J}_{\alpha })}_{p}(t)& {T}_{\alpha }{({J}_{\alpha })}_{-p}(t)\end{array}\right|,does not vanish at any point. Since (Jα)p{({J}_{\alpha })}_{p}and (Jα)−p(t){({J}_{\alpha })}_{-p}(t)satisfy Eq. (11)t2αTαTα(Jα)p(t)+αtαTα(Jα)p(t)+α2(t2α−p2)(Jα)p(t)=0,{t}^{2\alpha }{T}_{\alpha }{T}_{\alpha }{({J}_{\alpha })}_{p}(t)+\alpha {t}^{\alpha }{T}_{\alpha }{({J}_{\alpha })}_{p}(t)+{\alpha }^{2}({t}^{2\alpha }-{p}^{2}){({J}_{\alpha })}_{p}(t)=0,t2αTαTα(Jα)−p(t)+αtαTα(Jα)−p(t)+α2(t2α−p2)(Jα)−p(t)=0.{t}^{2\alpha }{T}_{\alpha }{T}_{\alpha }{({J}_{\alpha })}_{-p}(t)+\alpha {t}^{\alpha }{T}_{\alpha }{({J}_{\alpha })}_{-p}(t)+{\alpha }^{2}({t}^{2\alpha }-{p}^{2}){({J}_{\alpha })}_{-p}(t)=0.By the multiplication of the two equations by (Jα)−p(t){({J}_{\alpha })}_{-p}(t)and (Jα)p(t){({J}_{\alpha })}_{p}(t), respectively. Subtracting one from the other, and dividing by tα{t}^{\alpha }, we get the following:tα((Jα)p(t)TαTα(Jα)−p(t)−(Jα)−p(t)TαTα(Jα)p(t))−α((Jα)p(t)Tα(Jα)−p(t)−(Jα)−p(t)Tα(Jα)p(t))=0.\begin{array}{c}{t}^{\alpha }({({J}_{\alpha })}_{p}(t){T}_{\alpha }{T}_{\alpha }{({J}_{\alpha })}_{-p}(t)-{({J}_{\alpha })}_{-p}(t){T}_{\alpha }{T}_{\alpha }{({J}_{\alpha })}_{p}(t))\\ \hspace{1em}-\alpha ({({J}_{\alpha })}_{p}(t){T}_{\alpha }{({J}_{\alpha })}_{-p}(t)-{({J}_{\alpha })}_{-p}(t){T}_{\alpha }{({J}_{\alpha })}_{p}(t))=0.\end{array}This is equivalent toTα(tα((Jα)p(t)Tα(Jα)−p(t)−(Jα)−p(t)Tα(Jα)p(t)))=Tα(tαWα((Jα)p(t),(Jα)−p(t)))=0.\begin{array}{c}{T}_{\alpha }({t}^{\alpha }({({J}_{\alpha })}_{p}(t){T}_{\alpha }{({J}_{\alpha })}_{-p}(t)-{({J}_{\alpha })}_{-p}(t){T}_{\alpha }{({J}_{\alpha })}_{p}(t)))\\ \hspace{1em}={T}_{\alpha }({t}^{\alpha }{W}^{\alpha }({({J}_{\alpha })}_{p}(t),{({J}_{\alpha })}_{-p}(t)))=0.\end{array}This implies Wα((Jα)p(t),(Jα)−p(t))=Ctα{W}^{\alpha }({({J}_{\alpha })}_{p}(t),{({J}_{\alpha })}_{-p}(t))=\frac{C}{{t}^{\alpha }}, (see Theorem 2.4), CCis a constant that needs to be found. By considering the first term in the series (13), we have the following:(Jα)p(t)=tα2pΓ(p+1)(1+O(t2α)),{({J}_{\alpha })}_{p}(t)=\frac{{\left(\frac{{t}^{\alpha }}{2}\right)}^{p}}{\text{&#x0393;}(p+1)}(1+O({t}^{2\alpha })),Tα(Jα)p(t)=tα2p−12Γ(p)(1+O(t2α)).{T}_{\alpha }{({J}_{\alpha })}_{p}(t)=\frac{{\left(\frac{{t}^{\alpha }}{2}\right)}^{p-1}}{2\text{&#x0393;}(p)}(1+O({t}^{2\alpha })).The same applies to (Jα)−p(t){({J}_{\alpha })}_{-p}(t). Then, with the help of Euler’s reflection formula, Γ(z)Γ(1−p)=πsinπz,∀z∈C−Z\left(\Gamma (z)\Gamma (1-p)=\frac{\pi }{\sin \hspace{.25em}\pi z}\hspace{.25em},\hspace{.25em}\forall z\in C-Z\right)Wα((Jα)p(t),(Jα)−p(t))=1tα1Γ(p+1)Γ(−p)−1Γ(−p+1)Γ(p)+O(tα)=−2αsinpππtα+O(tα).\begin{array}{l}{W}^{\alpha }({({J}_{\alpha })}_{p}(t),{({J}_{\alpha })}_{-p}(t))\\ \hspace{1em}=\frac{1}{{t}^{\alpha }}\left(\frac{1}{\text{&#x0393;}(p+1)\text{&#x0393;}(-p)}-\frac{1}{\text{&#x0393;}(-p+1)\text{&#x0393;}(p)}\right)\\ \hspace{1em}+O({t}^{\alpha })=-\frac{2\alpha \hspace{.25em}\sin \hspace{.25em}p\pi }{\pi {t}^{\alpha }}+O({t}^{\alpha }).\end{array}However, Wα((Jα)p(t),(Jα)−p(t))=Ctα{W}^{\alpha }({({J}_{\alpha })}_{p}(t),{({J}_{\alpha })}_{-p}(t))=\frac{C}{{t}^{\alpha }}have been stated previously, so the last O(tα)O({t}^{\alpha })must be zero andWα((Jα)p(t),(Jα)−p(t))=−2αsinpππtα,{W}^{\alpha }({({J}_{\alpha })}_{p}(t),{({J}_{\alpha })}_{-p}(t))=-\frac{2\alpha \hspace{.25em}\sin \hspace{.25em}p\pi }{\pi {t}^{\alpha }},which only vanishes when ppis an integer. From the hypothesis, 2p2pis not an integer, so neither is ppnorWα((Jα)p(t),(Jα)−p(t))≠0,∀t>0.{W}^{\alpha }({({J}_{\alpha })}_{p}(t),{({J}_{\alpha })}_{-p}(t))\ne 0\hspace{.25em},\hspace{.25em}\forall t\gt 0.Therefore, (Jα)p(t){({J}_{\alpha })}_{p}(t)and (Jα)−p(t){({J}_{\alpha })}_{-p}(t)are linearly independent solutions of (Eq. (11)) which is a second-order conformable linear differential equation. Due to the fact that solutions constructing a two-dimensional vector space, and (Jα)p(t){({J}_{\alpha })}_{p}(t)and (Jα)−p(t){({J}_{\alpha })}_{-p}(t)being linearly independent, any solution can be expressed as a linear combination of them.We will derive some basic facts about the zeros of the conformable Bessel function: (Jα)p(t){({J}_{\alpha })}_{p}(t)and its α-\alpha \text{-}derivative Tα(Jα)p(t){T}_{\alpha }{({J}_{\alpha })}_{p}(t).□Remark 3.2As in the case of classical Bessel functions [30,31], the positive zeros of the conformable Bessel function: (Jα)p(t){({J}_{\alpha })}_{p}(t)\hspace{.25em}can be arranged as a sequence:(17)0<ap,1<ap,2<…<ap,n<…,limn→∞ap,n=∞.0\lt {a}_{p,1}\lt {a}_{p,2}\lt \ldots \lt {a}_{p,n}\lt \ldots ,\hspace{.25em}\mathop{\mathrm{lim}}\limits_{n\to \infty }{a}_{p,n}=\infty .Theorem 3.3All zeros of (Jα)p(t){({J}_{\alpha })}_{p}(t), except t=0t=0possibly, are simple.ProofIf t0≠0{t}_{0}\ne 0is a multiple zero of (Jα)p(t){({J}_{\alpha })}_{p}(t), then we have at least that (Jα)p(t0)=0{({J}_{\alpha })}_{p}({t}_{0})=0and Tα(Jα)p(t0)=0{T}_{\alpha }{({J}_{\alpha })}_{p}({t}_{0})=0. As t0≠0{t}_{0}\ne 0, it follows from the conformable differential Eq. (11) that also TαTα(Jα)p(t0)=0{T}_{\alpha }{T}_{\alpha }{({J}_{\alpha })}_{p}({t}_{0})=0. Iteration then leads to Tα(n)(Jα)p(t0)=0{}^{(n)}T_{\alpha }{({J}_{\alpha })}_{p}({t}_{0})=0for all n=0,1,2,…n=0,1,2,\ldots , which implies that is identically zero. This is a trivial contradiction.□Theorem 3.4Let p>0p\gt 0. All zeros of Tα(Jα)p(t){T}_{\alpha }{({J}_{\alpha })}_{p}(t), except t=0t=0or t=pt=ppossibly, are simple.ProofIf t0{t}_{0}is a multiple zero of Tα(Jα)p(t){T}_{\alpha }{({J}_{\alpha })}_{p}(t), then we have at least that Tα(Jα)p(t0)=0{T}_{\alpha }{({J}_{\alpha })}_{p}({t}_{0})=0and TαTα(Jα)p(t0)=0{T}_{\alpha }{T}_{\alpha }{({J}_{\alpha })}_{p}({t}_{0})=0. For t0≠0{t}_{0}\ne 0and t0≠p{t}_{0}\ne p, then it follows from the conformable differential Eq. (11) that also (Jα)p(t0)=0{({J}_{\alpha })}_{p}({t}_{0})=0. In addition, this leads to (Jα)p(t){({J}_{\alpha })}_{p}(t)being identically zero which is clearly not true.□Theorem 3.5Let ppis a nonnegative integer. It is verified that between any two consecutive zeros of (Jα)p(t){({J}_{\alpha })}_{p}(t), ∃\exists precisely one zero of (Jα)p−1(t){({J}_{\alpha })}_{p-1}(t)and precisely one zero of (Jα)p+1(t){({J}_{\alpha })}_{p+1}(t).ProofLet 0<a<b0\lt a\lt bbe two consecutive zeros of (Jα)p(t){({J}_{\alpha })}_{p}(t). Therefore, tpα(Jα)p(t){t}^{p\alpha }{({J}_{\alpha })}_{p}(t)vanishes at aaand bb. By theorem 2.3, we have:Tα(tpα(Jα)p(c))=0forsomec∈(a,b).{T}_{\alpha }({t}^{p\alpha }{({J}_{\alpha })}_{p}(c))=0\hspace{.5em}\text{for}\hspace{.5em}\text{some}\hspace{.5em}c\in (a,b).As given by Yazici and Gözütok [32],Tα(tpα(Jα)p(t))=αtpα(Jα)p−1(t).{T}_{\alpha }({t}^{p\alpha }{({J}_{\alpha })}_{p}(t))=\alpha {t}^{p\alpha }{({J}_{\alpha })}_{p-1}(t).Hence, we obtain (Jα)p−1(c)=0{({J}_{\alpha })}_{p-1}(c)=0.Repeating the above argument with the identity given by Uddin et al. [33], Tα(t−pα(Jα)p(t))=−αt−pα(Jα)p+1(t){T}_{\alpha }({t}^{-p\alpha }{({J}_{\alpha })}_{p}(t))=-\alpha {t}^{-p\alpha }{({J}_{\alpha })}_{p+1}(t), we get that (Jα)p+1(t){({J}_{\alpha })}_{p+1}(t)has a root in (a,b)(a,b).Thus, we have proved that both (Jα)p−1(t){({J}_{\alpha })}_{p-1}(t)and (Jα)p−1(t){({J}_{\alpha })}_{p-1}(t)have at least one root in (a,b)(a,b).If (Jα)p−1(t){({J}_{\alpha })}_{p-1}(t)has two roots in (a,b)(a,b), then from the above, we conclude that (Jα)p(t){({J}_{\alpha })}_{p}(t)would have a root in (a,b)(a,b). However, this contradicts the assumption that c and d are consecutive roots. Thus, (Jα)p−1(t){({J}_{\alpha })}_{p-1}(t)has exactly one root in (a,b)(a,b). Similarly, (Jα)p+1(t){({J}_{\alpha })}_{p+1}(t)has exactly one root in (a,b)(a,b).□4Orthogonality of conformable Bessel’s functionIn the classical sense, two functions f,gf,\hspace{.25em}gare orthogonal on an interval [a,b]{[}a,b]; if ∫abf(t)g(t)dt=0{\int }_{a}^{b}f(t)g(t)\text{d}t=0. For the case of Bessel functions, the interval that mathematicians consider is [0,1]{[}0,1]for physical applications [30,31]. Now, we want to study the orthogonality of conformable Bessel functions on such interval.Theorem 4.1Let ppis a nonnegative integer. If aaand bbbe roots of the equation (Jα)p(t)=0{({J}_{\alpha })}_{p}(t)=0, then we have the following:(18)∫01tα(Jα)p(at)(Jα)p(bt)dtt1−α=0,ifa≠b,α2(Jα)p+12(a),ifa=b.\begin{array}{c}\underset{0}{\overset{1}{\int }}{t}^{\alpha }{({J}_{\alpha })}_{p}(at){({J}_{\alpha })}_{p}(bt)\frac{\text{d}t}{{t}^{1-\alpha }}\\ \hspace{1em}=\left\{\begin{array}{ll}0,& \text{if}\hspace{.25em}a\ne b,\\ \frac{\alpha }{2}{({J}_{\alpha })}_{p+1}^{2}(a),& \text{if}\hspace{.25em}a=b.\end{array}\right.\end{array}ProofFirst, let us write the conformable Bessel Eq. (11) as follows:(19)xαTα(xαTα(Jα)p(x))−α2(x2α−p2)(Jα)p(x)=0.{x}^{\alpha }{T}_{\alpha }({x}^{\alpha }{T}_{\alpha }{({J}_{\alpha })}_{p}(x))-{\alpha }^{2}({x}^{2\alpha }-{p}^{2}){({J}_{\alpha })}_{p}(x)=0.The variable ppneeds to be a noninteger. It turns out to be useful to define a new variable ttby x=atx=at, where aais a constant, which we will take to be a zero of (Jα)p{({J}_{\alpha })}_{p}, that is, (Jα)p(a)=0{({J}_{\alpha })}_{p}(a)=0. Let us define:(20)u(t)=(Jα)p(at).u(t)={({J}_{\alpha })}_{p}(at).which implies u(1)=0u(1)=0, and by substituting into Eq. (18), the following is obtained:(21)tαTα(tα(Tαu)(t))−α2(a2αt2α−p2)u(t)=0.{t}^{\alpha }{T}_{\alpha }({t}^{\alpha }({T}_{\alpha }u)(t))-{\alpha }^{2}({a}^{2\alpha }{t}^{2\alpha }-{p}^{2})u(t)=0.We can also write down the equation obtained by picking another zero, bbsay. Let us define:(22)v(t)=(Jα)p(bt)sov(1)=0.v(t)={({J}_{\alpha })}_{p}(bt)\hspace{.5em}\text{so}\hspace{.5em}v(1)=0.We have:(23)tαTα(tα(Tαv)(t))−α2(b2αt2α−p2)v(t)=0.{t}^{\alpha }{T}_{\alpha }({t}^{\alpha }({T}_{\alpha }v)(t))-{\alpha }^{2}({b}^{2\alpha }{t}^{2\alpha }-{p}^{2})v(t)=0.To derive the orthogonality relation, we multiply Eq. (21) by vv, and Eq. (23) by uu. Subtracting and dividing by tα{t}^{\alpha }gives the following:(24)v(t)Tα(tα(Tαu)(t))−u(t)Tα(tα(Tαv)(t))+(a2α−b2α)tαu(t)v(t)=0.\begin{array}{c}v(t){T}_{\alpha }({t}^{\alpha }({T}_{\alpha }u)(t))-u(t){T}_{\alpha }({t}^{\alpha }({T}_{\alpha }v)(t))\\ \hspace{1em}+({a}^{2\alpha }-{b}^{2\alpha }){t}^{\alpha }u(t)v(t)=0.\end{array}The first two terms in Eq. (24) can be combined as follows:(25)Tα(tαv(t)(Tαu)(t)−tαu(t)(Tαv)(t)).{T}_{\alpha }({t}^{\alpha }v(t)({T}_{\alpha }u)(t)-{t}^{\alpha }u(t)({T}_{\alpha }v)(t)).As the extra terms are presented in Eq. (25), but not in Eq. (24), when the derivatives are expanded out, which are equal and opposite, and hence, they are cancelled. Hence, we get the following:Tα(tαv(t)(Tαu)(t)−tαu(t)(Tαv)(t))+(a2α−b2α)tαu(t)v(t)=0.\begin{array}{c}{T}_{\alpha }({t}^{\alpha }v(t)({T}_{\alpha }u)(t)-{t}^{\alpha }u(t)({T}_{\alpha }v)(t))\\ \hspace{1em}+({a}^{2\alpha }-{b}^{2\alpha }){t}^{\alpha }u(t)v(t)=0.\end{array}In addition, we α-\alpha \text{-}integrate this over the range of ttfrom 00to 11, which gives the following:(26)[tαv(t)(Tαu)(t)−tαu(t)(Tαv)(t)]01+(a2α−b2α)∫01tαu(t)v(t)dtt1−α=0.\begin{array}{c}{{[}{t}^{\alpha }v(t)({T}_{\alpha }u)(t)-{t}^{\alpha }u(t)({T}_{\alpha }v)(t)]}_{0}^{1}\\ \hspace{1em}+({a}^{2\alpha }-{b}^{2\alpha })\underset{0}{\overset{1}{\int }}{t}^{\alpha }u(t)v(t)\frac{\text{d}t}{{t}^{1-\alpha }}=0.\end{array}The integrated term vanishes at the lower limit because t=0t=0, and it also vanishes at the upper limit because u(1)=v(1)=0u(1)=v(1)=0. Hence, if a≠ba\ne b, Eq. (26) gives the following:(27)∫10tαu(t)v(t)dtt1−α=0.\underset{1}{\overset{0}{\displaystyle \int }}{t}^{\alpha }u(t)v(t)\frac{\text{d}t}{{t}^{1-\alpha }}=0.which by using Eqs. (20) and (21), the following can be written as follows:(28)∫01tα(Jα)p(at)(Jα)p(bt)dtt1−α=0.\underset{0}{\overset{1}{\int }}{t}^{\alpha }{({J}_{\alpha })}_{p}(at){({J}_{\alpha })}_{p}(bt)\frac{\text{d}t}{{t}^{1-\alpha }}=0.This is the desired orthogonality equation. It is good to remember that we require that aaand bbare distinct zeroes of (Jα)p{({J}_{\alpha })}_{p}; hence, both Bessel functions in Eq. (28) vanish at the upper limit.Now, we consider the case a=ba=b. First, Eq. (26) can be rewritten as follows:(29)(a2α−b2α)∫01tα(Jα)p(at)(Jα)p(bt)dtt1−α=bα(Jα)p(a)Tα(Jα)p(b)−aα(Jα)p(b)Tα(Jα)p(a).\begin{array}{c}({a}^{2\alpha }-{b}^{2\alpha })\underset{0}{\overset{1}{\int }}{t}^{\alpha }{({J}_{\alpha })}_{p}(at){({J}_{\alpha })}_{p}(bt)\frac{\text{d}t}{{t}^{1-\alpha }}\\ \hspace{1em}={b}^{\alpha }{({J}_{\alpha })}_{p}(a){T}_{\alpha }{({J}_{\alpha })}_{p}(b)\\ \hspace{1em}-{a}^{\alpha }{({J}_{\alpha })}_{p}(b){T}_{\alpha }{({J}_{\alpha })}_{p}(a).\end{array}Moreover, if a=ba=b, ∫01tα(Jα)p2(at)dtt1−α{\int }_{0}^{1}{t}^{\alpha }{({J}_{\alpha })}_{p}^{2}(at)\frac{\text{d}t}{{t}^{1-\alpha }}is 00\frac{0}{0}form. To overcome this challenge, let aabe a root of the equation (Jα)p(t)=0{({J}_{\alpha })}_{p}(t)=0, so that (Jα)p(a)=0{({J}_{\alpha })}_{p}(a)=0, let also b=a+εb=a+\varepsilon . By substituting (Jα)p(a)=0{({J}_{\alpha })}_{p}(a)=0, b=a+εb=a+\varepsilon , and taking limit ε→0\varepsilon \to 0in Eq. (29), we obtain the following:limε→0∫01tα(Jα)p(at)(Jα)p((a+ε)t)dtt1−α=limε→0−aα(Jα)p(a+ε)Tα(Jα)p(a)a2α−(a+ε)2α.\begin{array}{c}\mathop{\mathrm{lim}}\limits_{\varepsilon \to 0}\underset{0}{\overset{1}{\int }}{t}^{\alpha }{({J}_{\alpha })}_{p}(at){({J}_{\alpha })}_{p}((a+\varepsilon )t)\frac{\text{d}t}{{t}^{1-\alpha }}\\ \hspace{1em}=\mathop{\text{lim}}\limits_{\varepsilon \to 0}\frac{-{a}^{\alpha }{({J}_{\alpha })}_{p}(a+\varepsilon ){T}_{\alpha }{({J}_{\alpha })}_{p}(a)}{{a}^{2\alpha }-{(a+\varepsilon )}^{2\alpha }}.\end{array}It is still 00\frac{0}{0}form, so by applying L’Hopital’s rule on the right-hand side:limε→0∫01tα(Jα)p(at)(Jα)p((a+ε)t)dtt1−α=12α(Tα(Jα)p(a))2.\begin{array}{c}\mathop{\mathrm{lim}}\limits_{\varepsilon \to 0}\underset{0}{\overset{1}{\int }}{t}^{\alpha }{({J}_{\alpha })}_{p}(at){({J}_{\alpha })}_{p}((a+\varepsilon )t)\frac{\text{d}t}{{t}^{1-\alpha }}\\ \hspace{1em}=\frac{1}{2\alpha }{({T}_{\alpha }{({J}_{\alpha })}_{p}(a))}^{2}.\end{array}Finally, if a=ba=b, the corresponding integral is given by the following equation:(30)∫01tα(Jα)p2(at)dtt1−α=12α(Tα(Jα)p(a))2.□\hspace{2em}\underset{0}{\overset{1}{\int }}{t}^{\alpha }{({J}_{\alpha })}_{p}^{2}(at)\frac{\text{d}t}{{t}^{1-\alpha }}=\frac{1}{2\alpha }{({T}_{\alpha }{({J}_{\alpha })}_{p}(a))}^{2}.\hspace{4em}<mml:mpadded xmlns:ali="http://www.niso.org/schemas/ali/1.0/"xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance">\square </mml:mpadded>Remark 4.1From Property 3.1 (iii) and (iv) in an earlier study [29], the above equation can be written as follows:(31)∫01tα(Jα)p2(at)dtt1−α=12α(Tα(Jα)p(a))2=α2(Jα)p+12(a)α2=α2(Jα)p−12(a)2.\begin{array}{c}\underset{0}{\overset{1}{\int }}{t}^{\alpha }{({J}_{\alpha })}_{p}^{2}(at)\frac{\text{d}t}{{t}^{1-\alpha }}=\frac{1}{2\alpha }{({T}_{\alpha }{({J}_{\alpha })}_{p}(a))}^{2}\\ \hspace{1em}=\frac{\alpha }{2}{({J}_{\alpha })}_{p+1}^{2}(a)\frac{\alpha }{2}=\frac{\alpha }{2}{({J}_{\alpha })}_{p-1}^{2}{(a)}^{2}.\end{array}Remark 4.2Hence, we denote the positive zeros of (Jα)p(t){({J}_{\alpha })}_{p}(t)by λn{\lambda }_{n}, arranging them as in Eq. (19). Then, on the interval [0,1]{[}0,1], the conformable Bessel functions can be written as follows: (Jα)p(a1t),(Jα)p(a2t),…,(Jα)p(ant),…{({J}_{\alpha })}_{p}({a}_{1}t),\hspace{.25em}{({J}_{\alpha })}_{p}({a}_{2}t),\hspace{.25em}\ldots ,\hspace{.25em}{({J}_{\alpha })}_{p}({a}_{n}t),\ldots , which constitutes as an orthogonal system with weight function tα{t}^{\alpha }.5ConclusionNovel results on conformable Bessel functions have been successfully obtained in this study. Some essential properties of the Bessel functions of first order involving their conformable derivatives or their zeros have been accomplished. These functions’ orthogonality in [0, 1] has been introduced and proven systematically. This study’s outcomes are considered as an indication that the obtained results in the sense of conformable derivative coincide with the obtained classical integer order results. Our results can be further extended into applications of the obtained results and numerical analysis can be also studied in future research studies. http://www.deepdyve.com/assets/images/DeepDyve-Logo-lg.png Nonlinear Engineering de Gruyter

Novel results on conformable Bessel functions

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de Gruyter
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© 2022 Francisco Martínez et al., published by De Gruyter
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2192-8029
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2192-8029
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10.1515/nleng-2022-0002
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Abstract

1IntroductionFractional calculus is theoretically a powerful analysis technique for investigating arbitrary order integrals and derivatives. At the beginning, this field of research has been only presented purely, and until very recently, researchers have realized the powerful applicability of this field in modeling many phenomena from natural sciences and engineering much better than using the ordinary usual calculus due to several properties in fractional calculus that can provide a good explanation of physical behavior of certain system (see refs. [1,2,3,4,5,6,7]). The applications of conformable and fractional calculus have been recently discussed in some research studies [8,9,10,11,12,13,14,15]. The analysis of conformable derivatives and integrals has been discussed in detail in earlier studies [16,17].Riemann–Liouville and Caputo fractional operators [18,19] have been initially introduced, and recently, various new or generalized definitions have been proposed by researchers. In 2014, Khalil et al. proposed a local definition of fractional derivative, known as conformable derivative [20]. This topic has been discussed in some research studies whether conformable derivatives are considered fractional derivatives or not. While Tarasov has discussed in an earlier study [21] that some recent definitions of fractional derivatives are not fractional derivatives, Almeida et al. have concluded in an earlier study [22] that conformable derivative is still interesting to study it further. A new generalized definition of conformable derivative that coincides with the definitions of Caputo and Riemann–Liouville fractional derivatives has been proposed in an earlier study [23]. Some mathematicians such as Zhao and Luo and Khalil et al. discussed the conformable derivative’s physical and geometrical meanings in an earlier studies [24,25], respectively. Therefore, the conformable derivative in this study will only be considered as a modified form of local usual derivative. The analysis of Bessel functions in the context of conformable derivative has been rarely discussed in any other research studies. Therefore, our results in this article are novel and worthy due to the importance of Bessel functions in various modeling scenarios in science and engineering.The article is outlined as follows: some essential notions of the conformable calculus and conformable Bessel functions are presented in Section 2. In Section 3, we propose and prove new properties of Bessel functions of the first kind involving their conformable derivatives or their zeros. We conclude with Section 4, where the orthogonality of such functions in the interval [0, 1] is studied.2PreliminariesDefinition 2.1[20] For a function f:[0,∞)→Rf:{[}0,\infty )\to R, the conformable derivative of order α\alpha is expressed as follows:(1)(Tαf)(t)=limε→0f(t+εt1−α)−f(t)ε.({T}_{\alpha }f)(t)=\mathop{\mathrm{lim}}\limits_{\varepsilon \to 0}\frac{f(t+\varepsilon {t}^{1-\alpha })-f(t)}{\varepsilon }.∀t>0\forall t\gt 0; 0<α≤10\lt \alpha \le 1. If ffis α\alpha -differentiable in some (0,a)(0,a), a>0a\gt 0, and limt→0+(Tαf)(t)\mathop{\mathrm{lim}}\limits_{t\to {0}^{+}}({T}_{\alpha }f)(t)exists, then it is defined as follows:(2)(Tαf)(0)=limt→0+(Tαf)(t).({T}_{\alpha }f)(0)=\mathop{\mathrm{lim}}\limits_{t\to {0}^{+}}({T}_{\alpha }f)(t).As a result, we have Theorem 2.1 as follows:Theorem 2.1[20]. If a function f:[0,∞)→Rf:{[}0,\infty )\to Ris α\alpha -differentiable at t0>0{t}_{0}\gt 0; 0<α≤10\lt \alpha \le 1, then ffis continuous at t0{t}_{0}.Hence, Tα{T}_{\alpha }satisfies some properties as follows:Theorem 2.2[20]. Suppose that 0<α≤10\lt \alpha \le 1and ff, ggare α\alpha -differentiable at a point t>0t\gt 0. Then, we get the following:(i)Tα(af + bg) = a(Tαf) + b(Tαg), for all a, b ∈ R.(ii)Tα(tp) = ptp−αfor all p ∈ R.(iii)Tα(λ) = 0, for all constant functions f(t) =λ.(iv)Tα(fg) = f(Tαg) + g(Tαf).(v)Tαfg=g(Tαf)−f(Tαg)g2;g≠0{T}_{\alpha }\left(\frac{f}{g}\right)=\frac{g({T}_{\alpha }f)-f({T}_{\alpha }g)}{{g}^{2}};g\ne 0.(vi)Additionally, if f is a differentiable function, then (Tαf)(t)=t1−αdfdt(t)({T}_{\alpha }f)(t)={t}^{1-\alpha }\frac{\text{d}f}{\text{d}t}(t).From the above definition, the conformable derivatives of certain functions are written as follows:(i)Tα(1) = 0.(ii)Tα(sin(at)) = at1−αcos(at), a ∈ R.(iii)Tα(cos(at)) = −at1−αsin(at), a ∈ R.(iv)Tα(eat) =aeat, a ∈ R.Theorem 2.3(Rolle’s theorem). [20]. Suppose that a>0a\gt 0; α∈(0,1]\alpha \in (0,1], and f:[a,∞)→Rf:{[}a,\infty )\to Ris a function that satisfies the following:–ffis continuous on a,ba,b.–ffis α\alpha -differentiable on (a,b)(a,b).–f(a)=f(b)f(a)=f(b).Then, ∃c∈(a,b)\exists \hspace{.25em}c\in (a,b)such that (Tαf)(c)({T}_{\alpha }f)(c)= 0.Theorem 2.4[26]. Suppose that a>0a\gt 0; α∈(0,1]\alpha \in (0,1], and f:[a,∞)→Rf:{[}a,\infty )\to Ris function that satisfies the following:–ffis continuous in [a,b]{[}a,b].–ffis α\alpha -differentiable on (a,b)(a,b).If (Tαf)(c)=0({T}_{\alpha }f)(c)=0∀t∈(a,b)\forall t\in (a,b), then ffis a constant on [a,b]{[}a,b].Definition 2.2The left-conformable derivative beginning from aaof a function f:[a,∞)→Rf:{[}a,\infty )\to Rof f\hspace{.25em}fof order 0<α≤10\lt \alpha \le 1, [27], is expressed as follows:(3)(Tαaf)(t)=limε→0f(t+ε(t−a)1−α)−f(t)ε.({T}_{\alpha }^{a}f)(t)=\mathop{\mathrm{lim}}\limits_{\varepsilon \to 0}\frac{f(t+\varepsilon {(t-a)}^{1-\alpha })-f(t)}{\varepsilon }.When a=0a=0, it is denoted as (Tαf)(t)({T}_{\alpha }f)(t). If ffis α\alpha -differentiable in some (a,b)(a,b), then we get the following equation:(4)(Tαaf)(a)=limt→a+(Tαaf)(t).({T}_{\alpha }^{a}f)(a)=\mathop{\mathrm{lim}}\limits_{t\to {a}^{+}}({T}_{\alpha }^{a}f)(t).It is noticeable that if ffis differentiable function, then (Tαaf)(a)=(t−a)1−αdfdt(t)({T}_{\alpha }^{a}f)(a)={(t-a)}^{1-\alpha }\frac{\text{d}f}{\text{d}t}(t). Theorem 2.2 holds for definition 2.2 when changing by (t−a)(t-a).Theorem 2.5(Chain rule). [27]. Assume f,g:(a,∞)→Rf,g:(a,\infty )\to Rbe left α-differentiable function functions, where 0<α≤10\lt \alpha \le 1. Suppose that h(t)=f(g(t))h(t)=f(g(t)). The h(t)h(t)is α-differentiable function ∀t≠a\forall t\ne aand g(t)≠0g(t)\ne 0, hence we get the following equation:(5)(Tαah)(t)=(Tαaf)(g(t))⋅(Tαag)(t).(g(t))α−1.({T}_{\alpha }^{a}h)(t)=({T}_{\alpha }^{a}f)(g(t))\cdot ({T}_{\alpha }^{a}g)(t).{(g(t))}^{\alpha -1}.If t=at=a, then we define as follows:(6)(Tαah)(a)=limt→a+(Tαaf)(g(t))⋅(Tαag)(t)⋅(g(t))α−1.({T}_{\alpha }^{a}h)(a)=\mathop{\mathrm{lim}}\limits_{t\to {a}^{+}}({T}_{\alpha }^{a}f)(g(t))\cdot ({T}_{\alpha }^{a}g)(t)\cdot {(g(t))}^{\alpha -1}.Remark 2.1In an earlier study [27], the left-conformable derivative at aafor some smooth functions has been investigated. Suppose that 0<α≤10\lt \alpha \le 1and n∈Z+n\in {Z}^{+}, then the left sequential conformable derivative of order nnis expressed as follows: Tαa(n)f(t)=TαaTαa…Tαaf(t)︸n-times{}^{(n)}T_{\alpha }^{a}f(t)=\mathop{\underbrace{{T}_{\alpha }^{a}{T}_{\alpha }^{a}\ldots {T}_{\alpha }^{a}f(t)}}\limits_{n\text{-times}}.We show via induction that if ffis continuously α\alpha -differentiable and 0<α≤1n0\lt \alpha \le \frac{1}{n}then, the nth order sequential conformable derivative is continuous and vanishes at the end point a.Theorem 2.6[27]. Suppose that f is infinitely α\alpha -differentiable function, for some 0<α≤10\lt \alpha \le 1at a neighborhood of a point t0{t}_{0}. Then, ffhas a conformable power series expansion as follows:(7)f(t)=∑k=0∞((k)Tαt0)(t0)αkk!(t−t0)kα,t0<t<t0+R1α.f(t)=\mathop{\sum }\limits_{k=0}^{\infty }\frac{((k){T}_{\alpha }^{{t}_{0}})({t}_{0})}{{\alpha }^{k}k\&#x0021;}{(t-{t}_{0})}^{k\alpha },\hspace{1em}{t}_{0}\lt t\lt {t}_{0}+{R}^{\frac{1}{\alpha }}.Here, ((k)Tαt0)(t0)((k){T}_{\alpha }^{{t}_{0}})({t}_{0})indicates that we are applying the conformable derivative kktimes.The α\alpha -integral of a function ffbeginning from a≥0a\ge 0is expressed as follows:Definition 2.3[20]. Iαa(f)(t)=∫atf(x)x1−α⋅dx{I}_{\alpha }^{a}(f)(t)={\int }_{a}^{t}\frac{f(x)}{{x}^{1-\alpha }}\cdot \text{d}x, where it is a usual Riemann improper integral, and α∈(0,1]\alpha \in (0,1].As a result, we obtain the following:Theorem 2.7TαaIαa(f)(t)=f(t){T}_{\alpha }^{a}{I}_{\alpha }^{a}(f)(t)=f(t), for t≥at\ge a, wheref\hspace{.25em}fis any continuous function in the domain of Iα.{I}_{\alpha }.Lemma 2.1[27]. Let f:(a,b)→Rf:(a,b)\to Rbe differentiable and α∈(0,1]\hspace{.25em}\alpha \in (0,1]. Then, ∀a>0\forall \hspace{.25em}a\gt 0, and we get the following:(9)IαaTαa(f)(t)=f(t)−f(a).{I}_{\alpha \hspace{.25em}}^{a}{T}_{\alpha }^{a}(f)(t)=f(t)-f(a).Theorem 2.8[27]. Let f:[a,b]→Rf:{[}a,b]\to Rbe two functions ∋fg\ni fgis differentiable. Then we have the following equation:(10)∫abf(x)Tαa(g)(x)dα(x)=fg]ab−∫abg(x)Tαa(f)(x)dα(x).\begin{array}{c}\underset{a}{\overset{b}{\int }}f(x){T}_{\alpha }^{a}(g)(x)\text{d}\alpha (x)={fg]}_{a}^{b}\\ \hspace{1em}-\underset{a}{\overset{b}{\int }}g(x){T}_{\alpha }^{a}(f)(x)\text{d}\alpha (x).\end{array}Now, consider the sequential conformable Bessel equation [28]:(11)t2αTαTαy+αtαTαy+α2(t2α−p2)y=0.{t}^{2\alpha }{T}_{\alpha }{T}_{\alpha }y+\alpha {t}^{\alpha }{T}_{\alpha }y+{\alpha }^{2}({t}^{2\alpha }-{p}^{2})y=0.where 0<α≤10\lt \alpha \le 1and ppis any real number. If α=1\alpha =1, then Eq. (1) is a usual Bessel equation [29]. t=0t=0is a α-\alpha \text{-}regular singular point for the equation. In this case, for t>0t\gt 0, the solution can be investigated via a conformable Fröbenius series as:y=trα∑n=0∞cntnα=∑n=0∞cnt(n+r)αy={t}^{r\alpha }\mathop{\sum }\limits_{n=0}^{\infty }{c}_{n}{t}^{n\alpha }=\mathop{\sum }\limits_{n=0}^{\infty }{c}_{n}{t}^{(n+r)\alpha }We let:Tαy=∑n=0∞α(n+r)cnt(n+r−1)α{T}_{\alpha }y=\mathop{\sum }\limits_{n=0}^{\infty }\alpha (n+r){c}_{n}{t}^{(n+r-1)\alpha }TαTαy=∑n=0∞α2(n+r)(n+r−1)cnt(n+r−2)α{T}_{\alpha }{T}_{\alpha }y=\mathop{\sum }\limits_{n=0}^{\infty }{\alpha }^{2}(n+r)(n+r-1){c}_{n}{t}^{(n+r-2)\alpha }Let substitute these expressions in Eq. (11) to get the following:l(r)c0trα+l(r+1)c1t(r+1)α+∑n=0∞[l(r+n)cn+α2cn−2]t(r+n)α=0\begin{array}{c}l(r){c}_{0}{t}^{r\alpha }+l(r+1){c}_{1}{t}^{(r+1)\alpha }\\ \hspace{1em}+\mathop{\sum }\limits_{n=0}^{\infty }{[}l(r+n){c}_{n}+{\alpha }^{2}{c}_{n-2}]{t}^{(r+n)\alpha }=0\end{array}where l(r)=r(r−1)α2+rα2−α2p2l(r)=r(r-1){\alpha }^{2}+r{\alpha }^{2}-{\alpha }^{2}{p}^{2}.Let us c0≠0{c}_{0}\ne 0, then we obtain the following:l(r)=0l(r)=0Since α2≠0{\alpha }^{2}\ne 0, the following can be written as follows:l(r)=0⇒r(r−1)α2+rα2−α2p2=0⇒r2−p2=0.l(r)=0\hspace{.25em}\Rightarrow \hspace{.25em}r(r-1){\alpha }^{2}+r{\alpha }^{2}-{\alpha }^{2}{p}^{2}=0\hspace{.25em}\Rightarrow \hspace{.25em}{r}^{2}-{p}^{2}=0.Hence, we find:r1=p,r2=−p.{r}_{1}=p\hspace{.25em},\hspace{.25em}{r}_{2}=-p.Now, for p>0p\gt 0, the solutions of the conformable Bessel equation of order ppare analyzed as follows:In this case, for r1=p{r}_{1}=p, we have:l(r1+1)c1=[α2p(p+1)+α2(p+1)−α2p2]c1=0.l({r}_{1}+1){c}_{1}={[}{\alpha }^{2}p(p+1)+{\alpha }^{2}(p+1)-{\alpha }^{2}{p}^{2}]{c}_{1}=0.(2p+1)c1=0.(2p+1){c}_{1}=0.Due to p>0p\gt 0, it follows that c1=0{c}_{1}=0. The recurrence relation is as follows:cn=−cn−2n(n+p).{c}_{n}=-\frac{{c}_{n-2}}{n(n+p)}.From c1=0{c}_{1}=0and last recurrence relation, we obtain the following:c3=c5=…=0.{c}_{3}={c}_{5}=\ldots =0.c2n=(−1)nc022nn!(p+1)(p+2)…(p+n),n≥1.{c}_{2n}=\frac{{(-1)}^{n}{c}_{0}}{{2}^{2n}n\&#x0021;(p+1)(p+2)\ldots (p+n)}\hspace{.25em},\hspace{.25em}n\ge 1.Let us c0=cΓ(p+1){c}_{0}=\frac{c}{\text{&#x0393;}(p+1)}. Thus, the first solution of the conformable Bessel equation of order pphas the following form:(12)y1(t)=c∑n=0∞(−1)nn!Γ(p+n+1)tα22n+p.{y}_{1}(t)=c\mathop{\sum }\limits_{n=0}^{\infty }\frac{{(-1)}^{n}}{n\&#x0021;\text{&#x0393;}(p+n+1)}{\left(\frac{{t}^{\alpha }}{2}\right)}^{2n+p}.Besides, the Bessel function of order ppis valid.(13)(Jα)p(t)=∑n=0∞(−1)nn!Γ(p+n+1)tα22n+p.{({J}_{\alpha })}_{p}(t)=\mathop{\sum }\limits_{n=0}^{\infty }\frac{{(-1)}^{n}}{n\&#x0021;\text{&#x0393;}(p+n+1)}{\left(\frac{{t}^{\alpha }}{2}\right)}^{2n+p}.For r2=−p{r}_{2}=-p, if r1−r2=2p{r}_{1}-{r}_{2}=2pis not a positive integer, then the second solution of the conformable Bessel equation of order ppis written as follows:(14)y2(t)=c∑n=0∞(−1)nn!Γ(−p+n+1)tα22n−p.{y}_{2}(t)=c\mathop{\sum }\limits_{n=0}^{\infty }\frac{{(-1)}^{n}}{n\&#x0021;\text{&#x0393;}(-p+n+1)}{\left(\frac{{t}^{\alpha }}{2}\right)}^{2n-p}.The second type Bessel functions of order pphas the following form:(15)(Jα)−p(t)=∑n=0∞(−1)nn!Γ(−p+n+1)tα22n−p.{({J}_{\alpha })}_{-p}(t)=\mathop{\sum }\limits_{n=0}^{\infty }\frac{{(-1)}^{n}}{n\&#x0021;\text{&#x0393;}(-p+n+1)}{\left(\frac{{t}^{\alpha }}{2}\right)}^{2n-p}.3Some basic properties of conformable Bessel functionsFirst, we will study the convergence of series (13) and (15).Theorem 3.1The series that defines the conformable Bessel functions of the first kind of order ppand −p-pare absolutely convergent for all t>0\hspace{.25em}t\gt 0.ProofThe radius of convergence of conformable series is as follows:∑n=0∞(−1)nn!Γ(±p+n+1)xnα\mathop{\sum }\limits_{n=0}^{\infty }\frac{{(-1)}^{n}}{n\&#x0021;\text{&#x0393;}(\pm p+n+1)}{x}^{n\alpha }which is easily found by ratio test [30],1R=limn→∞n!Γ(±p+n+1)(n+1)!Γ(±p+n+2)=limn→∞1(n+1)(±p+n+1)=0\begin{array}{c}\frac{1}{R}=\mathop{\mathrm{lim}}\limits_{n\to \infty }\frac{n\&#x0021;\Gamma (\pm p+n+1)}{(n+1)\&#x0021;\text{&#x0393;}(\pm p+n+2)}\\ \hspace{1em}=\mathop{\mathrm{lim}}\limits_{n\to \infty }\frac{1}{(n+1)(\pm p+n+1)}=0\end{array}which implies R=∞R=\infty .□Remark 3.1As series (13) and (15) are both convergent ∀t>0\forall t\gt 0, we may do a term-by-term differentiation for them [29].As (Jα)p{({J}_{\alpha })}_{p}and (Jα)−p{({J}_{\alpha })}_{-p}are the second order conformable linear differential equation’s solutions, our goal is to see that they are linearly independent, and thus they form a basis for the vector space of the solutions of Eq. (11).Theorem 3.2If 2p2pis not a positive integer, then the conformable Bessel functions of the first kind (Jα)p{({J}_{\alpha })}_{p}and (Jα)−p{({J}_{\alpha })}_{-p}, are linearly independent. In that case, for any solution yyof (11), ∃A,B∈R\exists A,B\in R(16)y(t)=A(Jα)p(t)+B(Jα)−p(t).y(t)=A{({J}_{\alpha })}_{p}(t)+B{({J}_{\alpha })}_{-p}(t).ProofIt is enough to see that the α-\alpha \text{-}Wronskian of (Jα)p(t){({J}_{\alpha })}_{p}(t)and (Jα)−p(t){({J}_{\alpha })}_{-p}(t)Wα((Jα)p(t),(Jα)−p(t))=(Jα)p(t)(Jα)−p(t)Tα(Jα)p(t)Tα(Jα)−p(t),{W}^{\alpha }({({J}_{\alpha })}_{p}(t),{({J}_{\alpha })}_{-p}(t))=\left|\begin{array}{cc}{({J}_{\alpha })}_{p}(t)& {({J}_{\alpha })}_{-p}(t)\\ {T}_{\alpha }{({J}_{\alpha })}_{p}(t)& {T}_{\alpha }{({J}_{\alpha })}_{-p}(t)\end{array}\right|,does not vanish at any point. Since (Jα)p{({J}_{\alpha })}_{p}and (Jα)−p(t){({J}_{\alpha })}_{-p}(t)satisfy Eq. (11)t2αTαTα(Jα)p(t)+αtαTα(Jα)p(t)+α2(t2α−p2)(Jα)p(t)=0,{t}^{2\alpha }{T}_{\alpha }{T}_{\alpha }{({J}_{\alpha })}_{p}(t)+\alpha {t}^{\alpha }{T}_{\alpha }{({J}_{\alpha })}_{p}(t)+{\alpha }^{2}({t}^{2\alpha }-{p}^{2}){({J}_{\alpha })}_{p}(t)=0,t2αTαTα(Jα)−p(t)+αtαTα(Jα)−p(t)+α2(t2α−p2)(Jα)−p(t)=0.{t}^{2\alpha }{T}_{\alpha }{T}_{\alpha }{({J}_{\alpha })}_{-p}(t)+\alpha {t}^{\alpha }{T}_{\alpha }{({J}_{\alpha })}_{-p}(t)+{\alpha }^{2}({t}^{2\alpha }-{p}^{2}){({J}_{\alpha })}_{-p}(t)=0.By the multiplication of the two equations by (Jα)−p(t){({J}_{\alpha })}_{-p}(t)and (Jα)p(t){({J}_{\alpha })}_{p}(t), respectively. Subtracting one from the other, and dividing by tα{t}^{\alpha }, we get the following:tα((Jα)p(t)TαTα(Jα)−p(t)−(Jα)−p(t)TαTα(Jα)p(t))−α((Jα)p(t)Tα(Jα)−p(t)−(Jα)−p(t)Tα(Jα)p(t))=0.\begin{array}{c}{t}^{\alpha }({({J}_{\alpha })}_{p}(t){T}_{\alpha }{T}_{\alpha }{({J}_{\alpha })}_{-p}(t)-{({J}_{\alpha })}_{-p}(t){T}_{\alpha }{T}_{\alpha }{({J}_{\alpha })}_{p}(t))\\ \hspace{1em}-\alpha ({({J}_{\alpha })}_{p}(t){T}_{\alpha }{({J}_{\alpha })}_{-p}(t)-{({J}_{\alpha })}_{-p}(t){T}_{\alpha }{({J}_{\alpha })}_{p}(t))=0.\end{array}This is equivalent toTα(tα((Jα)p(t)Tα(Jα)−p(t)−(Jα)−p(t)Tα(Jα)p(t)))=Tα(tαWα((Jα)p(t),(Jα)−p(t)))=0.\begin{array}{c}{T}_{\alpha }({t}^{\alpha }({({J}_{\alpha })}_{p}(t){T}_{\alpha }{({J}_{\alpha })}_{-p}(t)-{({J}_{\alpha })}_{-p}(t){T}_{\alpha }{({J}_{\alpha })}_{p}(t)))\\ \hspace{1em}={T}_{\alpha }({t}^{\alpha }{W}^{\alpha }({({J}_{\alpha })}_{p}(t),{({J}_{\alpha })}_{-p}(t)))=0.\end{array}This implies Wα((Jα)p(t),(Jα)−p(t))=Ctα{W}^{\alpha }({({J}_{\alpha })}_{p}(t),{({J}_{\alpha })}_{-p}(t))=\frac{C}{{t}^{\alpha }}, (see Theorem 2.4), CCis a constant that needs to be found. By considering the first term in the series (13), we have the following:(Jα)p(t)=tα2pΓ(p+1)(1+O(t2α)),{({J}_{\alpha })}_{p}(t)=\frac{{\left(\frac{{t}^{\alpha }}{2}\right)}^{p}}{\text{&#x0393;}(p+1)}(1+O({t}^{2\alpha })),Tα(Jα)p(t)=tα2p−12Γ(p)(1+O(t2α)).{T}_{\alpha }{({J}_{\alpha })}_{p}(t)=\frac{{\left(\frac{{t}^{\alpha }}{2}\right)}^{p-1}}{2\text{&#x0393;}(p)}(1+O({t}^{2\alpha })).The same applies to (Jα)−p(t){({J}_{\alpha })}_{-p}(t). Then, with the help of Euler’s reflection formula, Γ(z)Γ(1−p)=πsinπz,∀z∈C−Z\left(\Gamma (z)\Gamma (1-p)=\frac{\pi }{\sin \hspace{.25em}\pi z}\hspace{.25em},\hspace{.25em}\forall z\in C-Z\right)Wα((Jα)p(t),(Jα)−p(t))=1tα1Γ(p+1)Γ(−p)−1Γ(−p+1)Γ(p)+O(tα)=−2αsinpππtα+O(tα).\begin{array}{l}{W}^{\alpha }({({J}_{\alpha })}_{p}(t),{({J}_{\alpha })}_{-p}(t))\\ \hspace{1em}=\frac{1}{{t}^{\alpha }}\left(\frac{1}{\text{&#x0393;}(p+1)\text{&#x0393;}(-p)}-\frac{1}{\text{&#x0393;}(-p+1)\text{&#x0393;}(p)}\right)\\ \hspace{1em}+O({t}^{\alpha })=-\frac{2\alpha \hspace{.25em}\sin \hspace{.25em}p\pi }{\pi {t}^{\alpha }}+O({t}^{\alpha }).\end{array}However, Wα((Jα)p(t),(Jα)−p(t))=Ctα{W}^{\alpha }({({J}_{\alpha })}_{p}(t),{({J}_{\alpha })}_{-p}(t))=\frac{C}{{t}^{\alpha }}have been stated previously, so the last O(tα)O({t}^{\alpha })must be zero andWα((Jα)p(t),(Jα)−p(t))=−2αsinpππtα,{W}^{\alpha }({({J}_{\alpha })}_{p}(t),{({J}_{\alpha })}_{-p}(t))=-\frac{2\alpha \hspace{.25em}\sin \hspace{.25em}p\pi }{\pi {t}^{\alpha }},which only vanishes when ppis an integer. From the hypothesis, 2p2pis not an integer, so neither is ppnorWα((Jα)p(t),(Jα)−p(t))≠0,∀t>0.{W}^{\alpha }({({J}_{\alpha })}_{p}(t),{({J}_{\alpha })}_{-p}(t))\ne 0\hspace{.25em},\hspace{.25em}\forall t\gt 0.Therefore, (Jα)p(t){({J}_{\alpha })}_{p}(t)and (Jα)−p(t){({J}_{\alpha })}_{-p}(t)are linearly independent solutions of (Eq. (11)) which is a second-order conformable linear differential equation. Due to the fact that solutions constructing a two-dimensional vector space, and (Jα)p(t){({J}_{\alpha })}_{p}(t)and (Jα)−p(t){({J}_{\alpha })}_{-p}(t)being linearly independent, any solution can be expressed as a linear combination of them.We will derive some basic facts about the zeros of the conformable Bessel function: (Jα)p(t){({J}_{\alpha })}_{p}(t)and its α-\alpha \text{-}derivative Tα(Jα)p(t){T}_{\alpha }{({J}_{\alpha })}_{p}(t).□Remark 3.2As in the case of classical Bessel functions [30,31], the positive zeros of the conformable Bessel function: (Jα)p(t){({J}_{\alpha })}_{p}(t)\hspace{.25em}can be arranged as a sequence:(17)0<ap,1<ap,2<…<ap,n<…,limn→∞ap,n=∞.0\lt {a}_{p,1}\lt {a}_{p,2}\lt \ldots \lt {a}_{p,n}\lt \ldots ,\hspace{.25em}\mathop{\mathrm{lim}}\limits_{n\to \infty }{a}_{p,n}=\infty .Theorem 3.3All zeros of (Jα)p(t){({J}_{\alpha })}_{p}(t), except t=0t=0possibly, are simple.ProofIf t0≠0{t}_{0}\ne 0is a multiple zero of (Jα)p(t){({J}_{\alpha })}_{p}(t), then we have at least that (Jα)p(t0)=0{({J}_{\alpha })}_{p}({t}_{0})=0and Tα(Jα)p(t0)=0{T}_{\alpha }{({J}_{\alpha })}_{p}({t}_{0})=0. As t0≠0{t}_{0}\ne 0, it follows from the conformable differential Eq. (11) that also TαTα(Jα)p(t0)=0{T}_{\alpha }{T}_{\alpha }{({J}_{\alpha })}_{p}({t}_{0})=0. Iteration then leads to Tα(n)(Jα)p(t0)=0{}^{(n)}T_{\alpha }{({J}_{\alpha })}_{p}({t}_{0})=0for all n=0,1,2,…n=0,1,2,\ldots , which implies that is identically zero. This is a trivial contradiction.□Theorem 3.4Let p>0p\gt 0. All zeros of Tα(Jα)p(t){T}_{\alpha }{({J}_{\alpha })}_{p}(t), except t=0t=0or t=pt=ppossibly, are simple.ProofIf t0{t}_{0}is a multiple zero of Tα(Jα)p(t){T}_{\alpha }{({J}_{\alpha })}_{p}(t), then we have at least that Tα(Jα)p(t0)=0{T}_{\alpha }{({J}_{\alpha })}_{p}({t}_{0})=0and TαTα(Jα)p(t0)=0{T}_{\alpha }{T}_{\alpha }{({J}_{\alpha })}_{p}({t}_{0})=0. For t0≠0{t}_{0}\ne 0and t0≠p{t}_{0}\ne p, then it follows from the conformable differential Eq. (11) that also (Jα)p(t0)=0{({J}_{\alpha })}_{p}({t}_{0})=0. In addition, this leads to (Jα)p(t){({J}_{\alpha })}_{p}(t)being identically zero which is clearly not true.□Theorem 3.5Let ppis a nonnegative integer. It is verified that between any two consecutive zeros of (Jα)p(t){({J}_{\alpha })}_{p}(t), ∃\exists precisely one zero of (Jα)p−1(t){({J}_{\alpha })}_{p-1}(t)and precisely one zero of (Jα)p+1(t){({J}_{\alpha })}_{p+1}(t).ProofLet 0<a<b0\lt a\lt bbe two consecutive zeros of (Jα)p(t){({J}_{\alpha })}_{p}(t). Therefore, tpα(Jα)p(t){t}^{p\alpha }{({J}_{\alpha })}_{p}(t)vanishes at aaand bb. By theorem 2.3, we have:Tα(tpα(Jα)p(c))=0forsomec∈(a,b).{T}_{\alpha }({t}^{p\alpha }{({J}_{\alpha })}_{p}(c))=0\hspace{.5em}\text{for}\hspace{.5em}\text{some}\hspace{.5em}c\in (a,b).As given by Yazici and Gözütok [32],Tα(tpα(Jα)p(t))=αtpα(Jα)p−1(t).{T}_{\alpha }({t}^{p\alpha }{({J}_{\alpha })}_{p}(t))=\alpha {t}^{p\alpha }{({J}_{\alpha })}_{p-1}(t).Hence, we obtain (Jα)p−1(c)=0{({J}_{\alpha })}_{p-1}(c)=0.Repeating the above argument with the identity given by Uddin et al. [33], Tα(t−pα(Jα)p(t))=−αt−pα(Jα)p+1(t){T}_{\alpha }({t}^{-p\alpha }{({J}_{\alpha })}_{p}(t))=-\alpha {t}^{-p\alpha }{({J}_{\alpha })}_{p+1}(t), we get that (Jα)p+1(t){({J}_{\alpha })}_{p+1}(t)has a root in (a,b)(a,b).Thus, we have proved that both (Jα)p−1(t){({J}_{\alpha })}_{p-1}(t)and (Jα)p−1(t){({J}_{\alpha })}_{p-1}(t)have at least one root in (a,b)(a,b).If (Jα)p−1(t){({J}_{\alpha })}_{p-1}(t)has two roots in (a,b)(a,b), then from the above, we conclude that (Jα)p(t){({J}_{\alpha })}_{p}(t)would have a root in (a,b)(a,b). However, this contradicts the assumption that c and d are consecutive roots. Thus, (Jα)p−1(t){({J}_{\alpha })}_{p-1}(t)has exactly one root in (a,b)(a,b). Similarly, (Jα)p+1(t){({J}_{\alpha })}_{p+1}(t)has exactly one root in (a,b)(a,b).□4Orthogonality of conformable Bessel’s functionIn the classical sense, two functions f,gf,\hspace{.25em}gare orthogonal on an interval [a,b]{[}a,b]; if ∫abf(t)g(t)dt=0{\int }_{a}^{b}f(t)g(t)\text{d}t=0. For the case of Bessel functions, the interval that mathematicians consider is [0,1]{[}0,1]for physical applications [30,31]. Now, we want to study the orthogonality of conformable Bessel functions on such interval.Theorem 4.1Let ppis a nonnegative integer. If aaand bbbe roots of the equation (Jα)p(t)=0{({J}_{\alpha })}_{p}(t)=0, then we have the following:(18)∫01tα(Jα)p(at)(Jα)p(bt)dtt1−α=0,ifa≠b,α2(Jα)p+12(a),ifa=b.\begin{array}{c}\underset{0}{\overset{1}{\int }}{t}^{\alpha }{({J}_{\alpha })}_{p}(at){({J}_{\alpha })}_{p}(bt)\frac{\text{d}t}{{t}^{1-\alpha }}\\ \hspace{1em}=\left\{\begin{array}{ll}0,& \text{if}\hspace{.25em}a\ne b,\\ \frac{\alpha }{2}{({J}_{\alpha })}_{p+1}^{2}(a),& \text{if}\hspace{.25em}a=b.\end{array}\right.\end{array}ProofFirst, let us write the conformable Bessel Eq. (11) as follows:(19)xαTα(xαTα(Jα)p(x))−α2(x2α−p2)(Jα)p(x)=0.{x}^{\alpha }{T}_{\alpha }({x}^{\alpha }{T}_{\alpha }{({J}_{\alpha })}_{p}(x))-{\alpha }^{2}({x}^{2\alpha }-{p}^{2}){({J}_{\alpha })}_{p}(x)=0.The variable ppneeds to be a noninteger. It turns out to be useful to define a new variable ttby x=atx=at, where aais a constant, which we will take to be a zero of (Jα)p{({J}_{\alpha })}_{p}, that is, (Jα)p(a)=0{({J}_{\alpha })}_{p}(a)=0. Let us define:(20)u(t)=(Jα)p(at).u(t)={({J}_{\alpha })}_{p}(at).which implies u(1)=0u(1)=0, and by substituting into Eq. (18), the following is obtained:(21)tαTα(tα(Tαu)(t))−α2(a2αt2α−p2)u(t)=0.{t}^{\alpha }{T}_{\alpha }({t}^{\alpha }({T}_{\alpha }u)(t))-{\alpha }^{2}({a}^{2\alpha }{t}^{2\alpha }-{p}^{2})u(t)=0.We can also write down the equation obtained by picking another zero, bbsay. Let us define:(22)v(t)=(Jα)p(bt)sov(1)=0.v(t)={({J}_{\alpha })}_{p}(bt)\hspace{.5em}\text{so}\hspace{.5em}v(1)=0.We have:(23)tαTα(tα(Tαv)(t))−α2(b2αt2α−p2)v(t)=0.{t}^{\alpha }{T}_{\alpha }({t}^{\alpha }({T}_{\alpha }v)(t))-{\alpha }^{2}({b}^{2\alpha }{t}^{2\alpha }-{p}^{2})v(t)=0.To derive the orthogonality relation, we multiply Eq. (21) by vv, and Eq. (23) by uu. Subtracting and dividing by tα{t}^{\alpha }gives the following:(24)v(t)Tα(tα(Tαu)(t))−u(t)Tα(tα(Tαv)(t))+(a2α−b2α)tαu(t)v(t)=0.\begin{array}{c}v(t){T}_{\alpha }({t}^{\alpha }({T}_{\alpha }u)(t))-u(t){T}_{\alpha }({t}^{\alpha }({T}_{\alpha }v)(t))\\ \hspace{1em}+({a}^{2\alpha }-{b}^{2\alpha }){t}^{\alpha }u(t)v(t)=0.\end{array}The first two terms in Eq. (24) can be combined as follows:(25)Tα(tαv(t)(Tαu)(t)−tαu(t)(Tαv)(t)).{T}_{\alpha }({t}^{\alpha }v(t)({T}_{\alpha }u)(t)-{t}^{\alpha }u(t)({T}_{\alpha }v)(t)).As the extra terms are presented in Eq. (25), but not in Eq. (24), when the derivatives are expanded out, which are equal and opposite, and hence, they are cancelled. Hence, we get the following:Tα(tαv(t)(Tαu)(t)−tαu(t)(Tαv)(t))+(a2α−b2α)tαu(t)v(t)=0.\begin{array}{c}{T}_{\alpha }({t}^{\alpha }v(t)({T}_{\alpha }u)(t)-{t}^{\alpha }u(t)({T}_{\alpha }v)(t))\\ \hspace{1em}+({a}^{2\alpha }-{b}^{2\alpha }){t}^{\alpha }u(t)v(t)=0.\end{array}In addition, we α-\alpha \text{-}integrate this over the range of ttfrom 00to 11, which gives the following:(26)[tαv(t)(Tαu)(t)−tαu(t)(Tαv)(t)]01+(a2α−b2α)∫01tαu(t)v(t)dtt1−α=0.\begin{array}{c}{{[}{t}^{\alpha }v(t)({T}_{\alpha }u)(t)-{t}^{\alpha }u(t)({T}_{\alpha }v)(t)]}_{0}^{1}\\ \hspace{1em}+({a}^{2\alpha }-{b}^{2\alpha })\underset{0}{\overset{1}{\int }}{t}^{\alpha }u(t)v(t)\frac{\text{d}t}{{t}^{1-\alpha }}=0.\end{array}The integrated term vanishes at the lower limit because t=0t=0, and it also vanishes at the upper limit because u(1)=v(1)=0u(1)=v(1)=0. Hence, if a≠ba\ne b, Eq. (26) gives the following:(27)∫10tαu(t)v(t)dtt1−α=0.\underset{1}{\overset{0}{\displaystyle \int }}{t}^{\alpha }u(t)v(t)\frac{\text{d}t}{{t}^{1-\alpha }}=0.which by using Eqs. (20) and (21), the following can be written as follows:(28)∫01tα(Jα)p(at)(Jα)p(bt)dtt1−α=0.\underset{0}{\overset{1}{\int }}{t}^{\alpha }{({J}_{\alpha })}_{p}(at){({J}_{\alpha })}_{p}(bt)\frac{\text{d}t}{{t}^{1-\alpha }}=0.This is the desired orthogonality equation. It is good to remember that we require that aaand bbare distinct zeroes of (Jα)p{({J}_{\alpha })}_{p}; hence, both Bessel functions in Eq. (28) vanish at the upper limit.Now, we consider the case a=ba=b. First, Eq. (26) can be rewritten as follows:(29)(a2α−b2α)∫01tα(Jα)p(at)(Jα)p(bt)dtt1−α=bα(Jα)p(a)Tα(Jα)p(b)−aα(Jα)p(b)Tα(Jα)p(a).\begin{array}{c}({a}^{2\alpha }-{b}^{2\alpha })\underset{0}{\overset{1}{\int }}{t}^{\alpha }{({J}_{\alpha })}_{p}(at){({J}_{\alpha })}_{p}(bt)\frac{\text{d}t}{{t}^{1-\alpha }}\\ \hspace{1em}={b}^{\alpha }{({J}_{\alpha })}_{p}(a){T}_{\alpha }{({J}_{\alpha })}_{p}(b)\\ \hspace{1em}-{a}^{\alpha }{({J}_{\alpha })}_{p}(b){T}_{\alpha }{({J}_{\alpha })}_{p}(a).\end{array}Moreover, if a=ba=b, ∫01tα(Jα)p2(at)dtt1−α{\int }_{0}^{1}{t}^{\alpha }{({J}_{\alpha })}_{p}^{2}(at)\frac{\text{d}t}{{t}^{1-\alpha }}is 00\frac{0}{0}form. To overcome this challenge, let aabe a root of the equation (Jα)p(t)=0{({J}_{\alpha })}_{p}(t)=0, so that (Jα)p(a)=0{({J}_{\alpha })}_{p}(a)=0, let also b=a+εb=a+\varepsilon . By substituting (Jα)p(a)=0{({J}_{\alpha })}_{p}(a)=0, b=a+εb=a+\varepsilon , and taking limit ε→0\varepsilon \to 0in Eq. (29), we obtain the following:limε→0∫01tα(Jα)p(at)(Jα)p((a+ε)t)dtt1−α=limε→0−aα(Jα)p(a+ε)Tα(Jα)p(a)a2α−(a+ε)2α.\begin{array}{c}\mathop{\mathrm{lim}}\limits_{\varepsilon \to 0}\underset{0}{\overset{1}{\int }}{t}^{\alpha }{({J}_{\alpha })}_{p}(at){({J}_{\alpha })}_{p}((a+\varepsilon )t)\frac{\text{d}t}{{t}^{1-\alpha }}\\ \hspace{1em}=\mathop{\text{lim}}\limits_{\varepsilon \to 0}\frac{-{a}^{\alpha }{({J}_{\alpha })}_{p}(a+\varepsilon ){T}_{\alpha }{({J}_{\alpha })}_{p}(a)}{{a}^{2\alpha }-{(a+\varepsilon )}^{2\alpha }}.\end{array}It is still 00\frac{0}{0}form, so by applying L’Hopital’s rule on the right-hand side:limε→0∫01tα(Jα)p(at)(Jα)p((a+ε)t)dtt1−α=12α(Tα(Jα)p(a))2.\begin{array}{c}\mathop{\mathrm{lim}}\limits_{\varepsilon \to 0}\underset{0}{\overset{1}{\int }}{t}^{\alpha }{({J}_{\alpha })}_{p}(at){({J}_{\alpha })}_{p}((a+\varepsilon )t)\frac{\text{d}t}{{t}^{1-\alpha }}\\ \hspace{1em}=\frac{1}{2\alpha }{({T}_{\alpha }{({J}_{\alpha })}_{p}(a))}^{2}.\end{array}Finally, if a=ba=b, the corresponding integral is given by the following equation:(30)∫01tα(Jα)p2(at)dtt1−α=12α(Tα(Jα)p(a))2.□\hspace{2em}\underset{0}{\overset{1}{\int }}{t}^{\alpha }{({J}_{\alpha })}_{p}^{2}(at)\frac{\text{d}t}{{t}^{1-\alpha }}=\frac{1}{2\alpha }{({T}_{\alpha }{({J}_{\alpha })}_{p}(a))}^{2}.\hspace{4em}<mml:mpadded xmlns:ali="http://www.niso.org/schemas/ali/1.0/"xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance">\square </mml:mpadded>Remark 4.1From Property 3.1 (iii) and (iv) in an earlier study [29], the above equation can be written as follows:(31)∫01tα(Jα)p2(at)dtt1−α=12α(Tα(Jα)p(a))2=α2(Jα)p+12(a)α2=α2(Jα)p−12(a)2.\begin{array}{c}\underset{0}{\overset{1}{\int }}{t}^{\alpha }{({J}_{\alpha })}_{p}^{2}(at)\frac{\text{d}t}{{t}^{1-\alpha }}=\frac{1}{2\alpha }{({T}_{\alpha }{({J}_{\alpha })}_{p}(a))}^{2}\\ \hspace{1em}=\frac{\alpha }{2}{({J}_{\alpha })}_{p+1}^{2}(a)\frac{\alpha }{2}=\frac{\alpha }{2}{({J}_{\alpha })}_{p-1}^{2}{(a)}^{2}.\end{array}Remark 4.2Hence, we denote the positive zeros of (Jα)p(t){({J}_{\alpha })}_{p}(t)by λn{\lambda }_{n}, arranging them as in Eq. (19). Then, on the interval [0,1]{[}0,1], the conformable Bessel functions can be written as follows: (Jα)p(a1t),(Jα)p(a2t),…,(Jα)p(ant),…{({J}_{\alpha })}_{p}({a}_{1}t),\hspace{.25em}{({J}_{\alpha })}_{p}({a}_{2}t),\hspace{.25em}\ldots ,\hspace{.25em}{({J}_{\alpha })}_{p}({a}_{n}t),\ldots , which constitutes as an orthogonal system with weight function tα{t}^{\alpha }.5ConclusionNovel results on conformable Bessel functions have been successfully obtained in this study. Some essential properties of the Bessel functions of first order involving their conformable derivatives or their zeros have been accomplished. These functions’ orthogonality in [0, 1] has been introduced and proven systematically. This study’s outcomes are considered as an indication that the obtained results in the sense of conformable derivative coincide with the obtained classical integer order results. Our results can be further extended into applications of the obtained results and numerical analysis can be also studied in future research studies.

Journal

Nonlinear Engineeringde Gruyter

Published: Jan 1, 2022

Keywords: conformable derivative; conformable Bessel functions; conformable Bessel equation

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