数理解析研究所講究録
第 1824 巻 2013 年 31-50

31Solutions to Some Homogeneous Special Ordinary Differential
Equation by Means of N-Fractional Calculus Operator
Katsuyuki Nishimoto
Institute for Applied Mathematics, Descartes Press Co.

2-13-10 Kaguike; Koriyama 963-8833, JAPAN
Abstract

In thisarticle, the solutions to the homogeneous special ordinary

differential equation$\varphi_{2}+\varphi_{1Z}4+\varphi\cdot b=0 (z\neq 0)$$(\varphi_{v}=d^{v}\varphi/dz^{\nu} for \nu>0, \varphi_{0}=\varphi=\varphi(z))$

are discussed by means of -fractional calculus operator (NFCO- Method).
By our method, some particular solutions to the above equations are$N$given as below for example, in fractional differintegrated forms.
Group I.
(i)$\varphi=Ke^{i\mathcal{F}bz}(e^{-2Jbz}\cdot z^{-\frac{a}{2}})_{2}\iota_{-1}\equiv\varphi_{[1](a,b)}$(denote)and$(ii)$$\varphi=Ke^{\iota \mathcal{F}bz}(z^{-\frac{a}{2}}\cdot e^{-2\iota \mathcal{F}bz})_{\iota_{-1}}\equiv\varphi_{[2](a,b)}2$

And the familiar forms are$\varphi_{[1](a,b)}=K(-2i\sqrt{b}aa ; \frac{i}{2J5z})(|i/2\sqrt{b}z|<1)$

and$\varphi_{[2](a,b)}=K(-e^{-i\pi\frac{a}{2}\Gamma(a-1)}z^{1-a}e^{-iJ\sqrt{b}}\overline{b}z{}_{1}F_{1}(1^{a}-\tau;2-a;2iz)$$\Gamma(\frac{a}{2})$$(|\Gamma a-1-k)/\Gamma(^{a})|<\infty, |2i\sqrt{b}z|<1)$

respectively.
Where${}_{p}F_{q}(\cdots)$is the generalized Gauss hypergeometric function.

32–$_{i^{\underline{\sigma}_{-}}}.$\’i.

$\overline{f’l}g.

2.$

Notice that (1) is reduced to Goursat’s integral for $v=n(\in Z^{+})$ and is reduced to the famous Cauchy’s integral for $v=0$ That is, (1) is an extentionof Cauchy’s integral and of Goursat’s one, conversely Cauchy’s andGoursat’s ones are special cases of (1).
Moreover, notice that (1) is the representation which unifies the derivatives andrations.
$inte_{\theta}$

33
(I I)

On the fractional calculus operator $N^{v}[3]$Theorem A. Let fractional calculus operaror $(Nishimot\mathfrak{o}’s$ Operator$N^{\nu}=( \frac{\Gamma(\nu+i)}{2\pi i}\int_{C}\frac{d\zeta}{(\zeta-z)^{\nu+1}}\}$

with

Iim

$N^{-m}=$$N^{V}$$(\nu\not\in T)$,$(m\in Z^{+})$,$)$$N^{v}$

be(3)[Refer $ro(1)$ ](4)$\nuarrow-m$

and deflne the binary operation$\circ$as$N^{\beta}\circ N^{\alpha}f=N^{\beta}N^{\alpha}f=N^{\beta}(N^{\alpha}f)$$(\alpha, \beta\in R)$

then the set

,

(5)
(6)

$\{N^{v}\}=\{N^{v}|\nu\in R\}$is an Abelian product group (hqvlng continuous index) which has the inverse$v$, for thetransform operator
to the fractional calculus operator
function such that $f\in F=\{-f;0\neq|f_{v}|<\infty,$ $v\in R\}$ , where $f=f(z)$ and $z\in C.$$(N^{\nu})^{-1}=N^{-v}$$N^{\nu}$

$f$

$($

vis. $-\infty<\nu<\infty)$ .

$(For our$ convenience, $we call N^{\beta_{o}}N^{a} as$ product $of N^{\beta} and N^{\alpha}.$Theorem B.

$n$F.O. $G.$$\{N^{\psi}\}"$

is an

/

$)$

Action product group which has continuous

index
for the set of F. (F.O. . ; Fractional calculus operator group) [3]Theorem C. Let be
‘’

$G$

$\nu$

$S:=\{\pm N^{v}\}U\{0\}=\{N^{v}\}\cup\{-N^{v}\}\cup\{0\}$Then the set$S$

is a commutative $nng$ for the function $f\in F$ , when the identity$N^{a}+N^{\beta}=N^{\gamma}$$(N^{\sigma}, N^{\beta}, N^{\gamma}\in S)$

holds. $[5J$
(III)

(7)

$(v\in R)-$

(8),

Lemma. We have [1](i)$((z-c)^{b})_{\alpha}=e^{-i\pi\alpha} \frac{\Gamma(\alpha-b)}{\Gamma(-b)}(z-c)^{b-\alpha}$$(ii)$$(\log(z-c))_{\alpha}=-e^{-\dot{r}\pi\alpha}\Gamma(\alpha)(z-c)^{-\alpha}$$(iii)$$((z-c)^{-\alpha})_{-\alpha}=-e^{i\pi\alpha} \frac{1}{\Gamma(\alpha)}\log(z-c)$

where$(iv)$$z-c\neq 0$,$(| \frac{\Gamma(\alpha-b)}{\Gamma(-b)}|<\infty)$$(|\Gamma(\alpha)|<\infty)$,$(|\Gamma(\alpha)|<\infty)$for (i) and $z-c\neq 0,1$ for (ii), (iii),$(^{u=u(z)},$ $)$$(u \cdot v)_{\alpha};=\sum_{k=0}^{\infty}\frac{\Gamma(\alpha+1)}{k!\Gamma(\alpha+1-k)}u_{\alpha-k^{\mathcal{V}}k}$$(v=v(z))$

.

,

34
;LPre 肋血 a$\sigma$(I)

The theorem below is reported by the author already (cf- J. $FC$ , Vol. 27, May(2005), 83-88.). [31]Theorem D. Let $h$$P=P( \alpha, \beta, \gamma):=\frac{\sin m\cdot\sin\pi(\gamma-\alpha-\sqrt{})}{s\dot{m}\pi(\alpha+\sqrt{})\cdot\sin\pi(\gamma-\alpha)}$$(IP(\alpha, \beta, \gamma)I=\cdot M<\infty)$

and$Q=Q(\alpha.\beta, \gamma):=P(\beta, \alpha, \gamma)$

When
(i)

$\alpha,$$\beta,$$\gamma\not\in Z_{0}^{+}$,(2)$(IP(\beta,\alpha, \gamma)kM<\infty)$, we have ;$((z-c)^{\alpha} \cdot(z-c)^{\beta})_{\gamma}=e^{-i\pi\gamma}P(\alpha,\beta,\gamma)\frac{\Gamma(\gamma-\alpha-\beta)}{\Gamma(-\alpha-\beta)}(z-c)^{\alpha+\beta-\gamma}$$({\rm Re}(\alpha+\beta\dotplus 1)>0, (1+\alpha-\gamma)\not\in 4)$$(ii)$

(1)

,

(3)

,

(4)

,

$((z-c)^{\beta} \cdot(z-c)^{\alpha})_{\gamma}\approx e^{-i\prime t\gamma}Q(\alpha, \beta,\gamma)\frac{\Gamma(\gamma-\alpha-\beta)}{\Gamma(-\alpha-\beta)}(z-c)^{a+\beta-\gamma}$$({\rm Re}(\alpha+\beta+1)>0, (1+\beta-\gamma)\not\in Z_{0})$$(iii)$$((z-c)^{\alpha+\beta})_{\gamma}=e^{-i\pi\gamma} \frac{\Gamma(\gamma-\alpha-\beta)}{\Gamma(-\alpha-\beta)}(\overline{4}^{-c)^{\sigma+\beta-\gamma}}.’$

(5)

where
$z-c \neq 0, |\frac{\Gamma(\gamma-\alpha-\beta)}{\Gamma(-\alpha-\sqrt{})}|<\infty$

Then the inequalities below are established from this theorem.
Corollary 1. We have the inequalities(i)$((z-c)^{\alpha}\cdot(z-c)^{\beta})_{\gamma}\neq((z-c)^{\beta}\cdot(z-c)^{\alpha})_{\gamma}$

,

(6)

and
$(ii)$$((z-c)^{\sigma}\cdot(z-c)^{\rho})_{\gamma}\neq((z-c)^{\alpha^{A}\beta})_{\gamma}$

where$\alpha, \beta, \gamma\not\in Z_{0\prime}^{+}.

\alpha\neq\beta, z-c\neq 0.$

,

(7)

35
Corollary 2.

When

(i)

$\alpha,$$\beta,\neg\gamma\not\in Z_{0^{r}}^{+}$

and$P(\alpha_{\backslash ,\prime}\beta, \gamma)=\alpha\beta,$$\alpha,$

we have$\gamma)=1$(8),$((z-c)^{\alpha}\cdot(z-c)^{\beta})_{\gamma}=((z-c)^{\beta}\cdot(z-c)^{\alpha})_{\gamma}=((z-c)^{\alpha+\beta})_{\gamma}$$({\rm Re}(\alpha+\beta+1)>0, (1+\alpha-\gamma)\not\in T_{0}, (1+\beta-\gamma)\not\in Z_{\overline{0}})$

When

(ii)

$\gamma=m\in Z_{0}^{+}$

,

(9)

,

.

we have ;$((z-c)^{\alpha}\cdot(z-c)^{\beta})_{m}=((z-c)^{\beta}\cdot(z-c)^{\sigma})_{m}=((z-c)^{\sigma+\beta})_{m}.

($I$

The Teorem below is reported by the author already (cf.

1)

$J$

(10\rangle$. Frac. Calc. Vol.

29, May (2006),pp.35 44.) . $[7]$-Theorem E.
(i)

We $have$$(((z-b)^{\beta}-c)^{\alpha})_{\gamma}=e^{-\dot{r}\pi\gamma}(z-b)^{\sigma\beta-\gamma}$(11)$\cross\sum_{k=0}^{\infty}\frac{[-\alpha]_{k}\Gamma(\beta k-\alpha\beta+\gamma)}{k!\Gamma(\beta k-\alpha\beta)}\backslash (_{\frac{c}{(z-b)^{\beta}}\}^{k}}$$\backslash r\frac{|\Gamma(\beta k-\alpha\beta+\gamma)}{|\Gamma(\beta k-\alpha\beta)}<\infty)$and$(_{\lambda}^{-}i)$$(((z-b)^{\beta}-c)^{\alpha})_{n}=(-1)^{n}(z-b)^{\sigma\beta-n}$$\cross\sum_{k=0}^{\infty}\frac{[-\alpha]_{k}[\beta k-\alpha\beta]_{n}}{k!}(\frac{c}{(z-b)^{\beta}})^{k}$(12)$(n\in Z_{0}^{+})$

where$| \frac{c}{(z-b)^{\beta}}|<1,$

and$[\lambda]_{k}=\lambda(\lambda+1)\cdot\cdot,(\lambda+k-1)=\Gamma(\lambda+k)[\Gamma(\lambda)$(Notation of Pocbhammer).

with$[\lambda]_{0}=1$,

36\S 2. Solutions to some homogeneous special ordinary differential
equation by means of -fractional Calculus Operator$N$

Theorem 1. Let be$\varphi=\varphi(z)\in F$, then the homogeneous ordinary differential

equation(1)$(z\neq 0)$$\varphi_{2}+\varphi_{1}\cdot\frac{a}{z}+\varphi\cdot b=0$$(\varphi_{\nu}=d^{\nu}\varphi/dz^{\nu} for v>0, \varphi_{0}=\varphi=\varphi(z))$

has particular solutions of the forms
Group I.

$\varphi act\dot{w}nal$differintegrated form)(denote)(2)(i)$\varphi=K^{f_{bz}f_{bz}}e^{i}(e^{-Zi}\cdot z^{-\frac{a}{2}})_{4}2^{-1}\equiv\varphi_{[1](a,b)}$$(ii)$$\varphi=Ke^{i\mathcal{F}bz}(z^{-\frac{a}{2}}\cdot e^{-2\iota \mathcal{F}bz})_{A-1}\equiv\varphi_{[2](a,b)}2$

(3)

(iii)

$\varphi=K^{f_{bz}}e^{-i}(e^{2iIbz}\cdot z^{-\frac{a}{2}})_{A-1}2\equiv\varphi_{[3](a,b)}$

(4)

(iv)

$\varphi=Ke^{-\iota \mathcal{F}bz}(z^{-\frac{a}{2}}\cdot e^{2if_{bz}})_{A-1}\equiv\varphi_{[4](a,b)}2$(5)Group II.
(6)(i)$\varphi=\ ^{1-a}e^{i\mathcal{F}bz}(e^{-2if_{bz}}\cdot z^{2-1}2)_{-}A\equiv\psi_{[1](a,b)}2$$(ii)$$\varphi=\ ^{1-a}e^{iIbz}(z^{\frac{a}{2}-1}\cdot e^{-2if_{bz}})_{-2}A\equiv\varphi_{[2]_{(a,b)}}^{o}$

(7)

(iii)

$\varphi e^{f_{bz}f_{bz}}\cdot B\equiv\varphi_{[3](a,b)}^{o}$

(8)

(iv)

$\varphi=oe^{1-a}e^{-if_{bz}\mathcal{F}bz}(z^{\frac{a}{2}-1}\cdot e^{2i})_{-2}|ae\equiv\varphi_{[4]_{(a,b)}}$where $K(\neq 0)$ is an arbitrary constant.
Note. Notice that we have;
and
and
and
$\varphi_{[1](a,b)}=\varphi_{[2](a,b)}$$\varphi_{[1](a,b)}^{Q}=\varphi_{r2](a,b)}^{Q}$

for$\varphi_{[3](a,b)}=\varphi_{[4](a,b)}$$\varphi^{o}=\varphi^{o}[3](a.b)[4](a.b)$

for

(9)

$( \frac{a}{2}-1)\in Z_{0}^{+},$$(- \frac{a}{2})\in Z_{0}^{+}$

Proof of Group I.
We have$\varphi_{2}\cdot z+\varphi_{1}\cdot a+\varphi\cdot bz=0$$(z\neq 0)$(10)

from (1).
Set$\varphi=e^{\lambda z}\phi$$(\phi=\phi(z))$

(11)

37
we have then(12)$\phi_{2}\cdot z+\phi_{1}\cdot(2\lambda z+a)+\phi\cdot\{(\lambda^{2}+b)z+\lambdaa\}=0$

from (10), applying (11).
Here, choose$\lambda$such that$\lambda^{2}+b=0$

(13)

,

then we obtain$\lambda=\{\begin{array}{l}i\sqrt{b}\equiv c(i=\sqrt{-1})(14)-i\sqrt{b}\equiv-c(15)\end{array}$

Case

$I)$

$IJ1$

$\lambda=c$;this case we have
(16)$\phi_{2}\cdot z+\phi_{1}\cdot(2cz+a)+\phi\cdot ca=0$

from (12).
Operate -fractional calculus (NFC) operator
we have then$N$$N^{\nu}$,to the both sides of equation (16),$(\phi_{2}\cdot z)_{v}+(\phi_{1}\cdot(2cz+a))_{v}+(\phi\cdot ca)_{v}=0$$(\nu\not\in Z^{-})$

.

(17)

Now we have$( \phi_{2}\cdot z)_{v}=\sum_{k=0}^{1}\frac{\Gamma(v+1)}{k!\Gamma(v+1-k)}(\phi_{2})_{v-k}(z)_{k}$$=\phi_{2+\nu}\cdot z+\phi_{1+v}\cdot\nu$$(\phi_{1}\cdot(2cz+a))_{v}=\phi_{1+\nu}\cdot(2cz+a)+\phi_{v}\cdot 2c\nu$

(18)

(19)

(20)

and
$(\phi\cdot ca)_{v}=\varphi_{v}\cdot ca$(21)

respectively, by Lemmas (i) and ( iv).
Therefore, we have$\phi_{2+v}\cdot z+\phi_{1+v}\cdot(2cz+\nu+a)+\phi_{v}\cdot c(2v+a)=0$(22)

from (17), applyimg (19), (20) and (21).
Choose$\nu$

such that$\nu=-a/2$(23)

then we obtain$\phi_{2_{2}}\cdot z+\phi_{1-2}\cdot(2cz+\frac{a}{2}).=0$

(24)

38
from (22), using (23).
Set$\phi_{1_{2}^{B}}=u=u(z)-$$(\phi=u_{4,2^{-1}})$

(25)

,

we have then(26)$u_{1}+u\cdot(2c+^{a}?z^{-1})=0$

from (24). The solution to this (variable separable form) equation is given by$u=K(e^{(2c+\frac{a}{2}z^{-1})_{-1}})^{-1}$(27)$=K(e^{2czA_{2}\log z})^{-1}$(28)$=Ke^{-2\iota z}\cdot z^{\frac{a}{2}}$(29)Therefore, we obtain(30)$\phi=u_{r,2-1}=K(e^{-2cz}\cdot z^{-ユ})_{A}z^{-1}$from (25).
Inversely (30) satisfies equation (24) clearly.

Hence we obtain(31)$\varphi=Ke^{cz}(e^{-2cz}.z^{\frac{a}{2}})_{2}\alpha-1$(2)$=K^{\mathcal{F}bz}e^{i}(e^{-2i\mathcal{F}bz}.z^{-\frac{a}{2}})_{2}a-1\equiv\varphi_{[1]\langle a.b)}$from (11) with (14).
Next, changing the order$e^{-2ir_{bz}}$

andwe obtain other solution$z^{-\frac{a}{2}}$in parenthesis$\varphi_{f2](a,b)}$$()_{2}\alpha_{-1}$in (2)

which is different ffom (2) for$( \frac{a}{2}-1)\not\in Z_{0}^{+},$

that is,$\varphi=K^{i}e^{\mathcal{F}bz}(z^{-\frac{a}{2}}\cdot e^{-2i\mathcal{F}bz})_{2}\alpha-1\equiv\varphi_{[2](a,b)}$(3)(Refer to Theorem D. )II) Case$\lambda=-c,$$Set-\sqrt{b}$instead of$\sqrt{b}$in (2) and (3), we obtain (4) and (5)respectinely.
Proof of Group II.
Set$\varphi=z^{\beta}\phi$$(\phi=\mu_{Z}))$

,

(33)

39
we have then(34)$\phi_{2}\cdot z+\phi_{1}\cdot(2\beta+a)+\phi\cdot\{(\beta^{2}-\beta+a\beta)z^{-1}+bz\}=0$

from (10), ausing (33).
Here we choosesuch that$\beta$$\beta^{2}-\beta+a\beta=\beta(\beta-1+a)=0$

that is,,(35)$\beta=\{\begin{array}{l}0(36)1-a.(37)\end{array}$

When $\beta=0,$ (34) is reduced to (10), therefore, we have the same solutions asGroup I.
When $\beta=1-a$ we obtain(38)$\phi_{2}\cdot z+\phi_{1}\cdot(2-a)+\phi\cdot bz=0$

from (34).
The particular solutions to this equation are given by (refer to the proof of Group I)

setting $2-a$ instead ofrespectively.
Now we have$a$$\phi_{[1]}=Ke^{if_{bz}}(e^{-2if_{bz}}\cdot z^{l^{-1}})_{-A}a2$(39)$\phi_{[2]}=Ke^{\iota \mathcal{F}bz}(z^{\frac{a}{2}-1}\cdot e^{-2i\mathcal{F}bz})_{-2}4$(40)$\phi_{[3]}=K^{\mathcal{F}Ibz}e^{-i}(e^{2i}\cdot z^{\frac{a}{2}-1})_{-\alpha}bz2$(41)$\phi_{[4]}=Ke^{-i\mathcal{F}bz}(z^{\frac{a}{2}-1}\cdot e^{2jJ\overline{b}z})_{-\cdot a}2$(42)

in the right hand side of(2), (3) (4) $,and(5)$ ,$,$$\varphi=z^{1-a}\phi$from (33), hence we obtain(43)$\varphi=z^{1-a}\phi_{[1]}=\Omega^{1}e^{fi_{z}}(e^{-2ifbz}\cdot z^{\frac{a}{2}-1})_{-a}\equiv\varphi_{[1](a,b)}^{\Phi}2$(6)$\varphi=z^{1-a}\phi_{[2]}\equiv\varphi_{[2j(a,b)}^{o}$(7)$\varphi=z^{1-a}\phi_{[3]}\equiv\varphi_{[3](a,b)}^{o}$(8)$\varphi=z^{l-a}\phi_{[4]}\equiv\varphi_{[4](a,b)}^{o}$(9)

using (39) $\sim(42)$ .

40\S 3. Familiar Forms of The Solutions
Theorem 2. We have the more familiar form (expanded form) presentations forthe solutions to the differential equation (1) in \S 2. as follows.
Group I.
(i)$(ii)$(1)$(|\sqrt{b}$$\varphi_{[1](a,b)}=K(-2i\sqrt{b})\frac{a}{2}ze^{f_{bz}}{}_{2}F_{0}(1-\frac{a}{2}, \frac{a}{2};\frac{i}{2\mathcal{F}bZ})$(2)$\varphi_{[2](a,b)}=K(-e_{\Gamma(\frac{a}{2})}^{-i\pi_{2}^{4\Gamma(a-1)_{)z^{1-\triangleleft}}}}e^{-j}\Gamma bz_{1}F_{1}(1-\frac{a}{2};2-a;2i\sqrt{b}z)$$(| \Gamma(a-1-k)/\Gamma(\frac{a}{2})|<\infty, |2\iota\sqrt{b}z|<1)$(iii)$\varphi_{[3](a,b)}=K(2i\sqrt{b})^{\frac{a}{2}-1}z^{-\ell_{\dot{2}}}e^{tf_{bz_{2}}}F_{0}(1-\frac{a}{2},\frac{a}{2};\frac{-i}{2f_{bZ}})$$(iv)$(3)$(|-i/2\sqrt{b}z|<1)$$\varphi_{[4](a,b)}=K(-e_{\Gamma()}^{-i\pi\frac{a}{2}\Gamma(a-1)_{)zeF_{1}(1-\frac{a}{2};2-a;-2iz)}}\frac{a}{2}1-ai\mathcal{F}bz_{1},\sqrt{b} (4)$$(| \Gamma(a-1-k)/\Gamma(\frac{a}{2})|<\infty, |\sqrt{b}$Group II.
(i)$(|i/2\sqrt{b}z|<1)$$\varphi_{[1](a,b)}^{\Phi}=K(-2i\sqrt{b}22$$(ii)$(5)(6)$\varphi_{[2](a,b)}^{\theta}=K(e^{i\pi\frac{a}{2}}\frac{\Gamma(1-a)}{\Gamma(1-\frac{a}{2})})e^{-\iota \mathcal{F}bz_{1}}F_{1}(\frac{a}{2};a;2i\sqrt{b}z)$$(| \Gamma(1-a-k)/\Gamma(1-\frac{a}{2})|<\infty, |2i\sqrt{b}z|<1)$

$(i$

五

$)$

$(iv)$$(|-i/2\sqrt{b}z|<1)$$\varphi_{[3](a,b)}^{0}=K(2i\sqrt{b}22$$\varphi_{[4](a,b)}^{0}=K(e^{i\pi\frac{a}{2}}\frac{\Gamma(1-a)}{\Gamma(1-\frac{a}{2})})e^{\iota}\mathcal{F}bz_{1}F_{1}(\frac{a}{2};a;-2i\sqrt{b}z)$

(7)
(8)

$(| \Gamma(1-a-k)/\Gamma(1-\frac{a}{2})|<\infty, |-2i\sqrt{b}z|<1)$

where${}_{p}F_{q}(\cdots\cdots)$is the generatized Gausss hypergeometric function,

Proof of Group I. We have(i)$\varphi[1](a,b)_{2}^{=K^{i}}e^{f_{bz}}(e^{-2ifbz}z^{-\frac{a}{2}})_{\alpha_{-1}}$$=Ke^{t\mathcal{F}bz} \sum_{k-0}^{\infty}\frac{\Gamma(_{2}^{\Delta})}{k!\Gamma(\frac{a}{2}-k)}(e^{-2J\overline{b}z})_{g-1-k}(z^{-\frac{a}{2}})_{k}$(\S 2. (2))

(9)

41
(10)

$=K^{\mathcal{F}b\frac{a}{2}-1-kf_{bz}}e^{i}z \sum_{k=0}^{\infty}\frac{(-1)^{k}[1-a{}_{2}L}{k!}\{(-2i\sqrt{b})e^{-2i}\}\{e^{-id}\frac{\Gamma(_{2}^{g}+k)}{\Gamma(\frac{a}{2})}z^{\frac{a}{2}-k}\}$$(| \Gamma(\frac{a}{2}+k)/\Gamma(\frac{a}{2})|<\infty)$$=K(-2i \sqrt{b})\frac{a}{2}ze^{f_{bz}}\sum_{k=0}^{\infty}\frac{[1_{2}-\not\subset]_{k}[_{2^{-}}^{a}]_{k}}{k!}(\frac{i}{2ffz})^{k}$(11)$(| \frac{i}{2Jb^{-}z}|<1)$(1)$=K(-2i \sqrt{b})^{\frac{a}{2}-1}z^{-\frac{a}{2}}e^{-\iota \mathcal{F}bz}{}_{2}F_{0}(1-\frac{a}{2},- 2a; 叢)$using Lemma (iv), since(12)$(k\in Z_{0}^{+})$$\Gamma(\lambda-k)=(-1)^{-k}\frac{\Gamma(\lambda)\Gamma(1-\lambda)}{\Gamma(k+1-\lambda)}=(-1)^{-k}\frac{\Gamma(\lambda)}{[1-\lambda]_{k}}$and$(e^{\lambda z})_{\nu}=\lambda^{\nu}e^{\lambda z}$$(ii)$(Refer to [1] Vol. 1 and [2])(13)
( \S 2.(3))$\varphi_{[2](a,b)}=Ke^{if_{bz}}(z^{-\frac{a}{2}}\cdot e^{-2\iota f_{bz}})_{2^{-1}}\not\subset$$=Ke^{i\mathcal{F}bz} \sum_{k-0}^{\infty}\frac{\Gamma(_{2^{-}}^{4})}{k!\Gamma(\frac{a}{2}-k)}(z^{-\frac{a}{2}})_{2}\alpha_{-1-k}(e^{-2ifbz})_{k}$$=Ke^{i\mathcal{F}bz} \sum_{k=0}^{\infty}\frac{(-1)^{k}[1_{2}^{4}-L}{k!}\{e_{\Gamma(\frac{a}{2})}^{-i\pi(\frac{a}{2}-1-k)}\}\{(-2i\Gamma(a-1-k)_{z^{-a+1+}}k\sqrt{b})^{k}e^{-2if_{bz}}\}$

(14)

(15)

$(| \Gamma(a-1-k)/\Gamma(\frac{a}{2})|<\infty)$$=K(-e_{\Gamma(\frac{a}{2})}^{-i\pi\frac{a}{2}\Gamma(a-1)_{)z^{1-a}e^{-i\mathcal{F}bz}\sum_{k=0}^{\infty}\frac{[1-\frac{a}{2}]_{k}}{k![2-a]_{k}}(2i}}\sqrt{b}z)^{k}$(16)$(|2i\sqrt{b}z|<1)$(ii)$\varphi_{[2](a,b)}=K(-e_{\Gamma(\frac{a}{2})}^{-i\pi\frac{a}{2}\Gamma(a-1)})z^{1-a}e^{-i\mathcal{F}bz}{}_{1}F_{1}(1-\frac{a}{2};2-a;2i\sqrt{b}z)$(2)using Lemma (iv).
(iii)$\varphi_{[3](a,b)}=Set-\sqrt{b}$instead of(iv)$\varphi_{[4](a,b)}=Set-\sqrt{b}$instead of$f_{b}$$\sqrt{b}$

(1).

(3)

in (2).

(4)

42
Proof of Group II. We have(i)(\S 2. (6))$\varphi_{[1](a,b)}^{o}=Kz^{1-a}e^{i\mathcal{F}bz}(e^{-2J\overline{b}z}\cdot z^{\frac{a}{2}-1})_{-4}2$(17)$=Kz^{1}-e^{fi_{z}} \sum_{k-0}^{\infty}\frac{\Gamma(1_{2}-\iota)}{k!\Gamma(1-\frac{a}{2}-k)}(e^{-2Jbz})_{-A}(z^{\frac{a}{2}-1})_{k}2^{-k}$(18)$=ae^{1-a}e^{i}f_{bz} \sum_{k-0}^{\infty}\frac{(-1)^{k}[_{2}^{A}]_{k}}{k!}\{(-2i-f_{bz}(-A$.

$(| \Gamma(1-\frac{a}{2}+k)/\Gamma(1-\frac{a}{2})|<\infty)$(19)$(| \frac{i}{2dF_{z}}|<1)$$\approx K(-2i)^{-\frac{a}{2}}z^{-g}e^{-if_{bz}}\sum_{0}^{\infty}\frac{[\not\in]_{k}[1_{2}-4]_{k}}{k!}(\frac{i}{2\sqrt{}bz})^{k}$(5)$=K(-2i)^{-f}z^{-g}e^{f_{bz}}{}_{2}F_{0}( \frac{a}{2},1-\frac{a}{2};\frac{i}{2\mathcal{F}bZ})$

using Lemma (iv).
$(ii)$(\S 2. (7))$\varphi_{[2](a.b)}0=\ 1e(z^{2}\cdot e^{-2i\sqrt{b}z})_{-l}2$

一$\mathcal{B}^{1}e^{\mathcal{F}bz}\sum_{k\cdot 0}^{\infty}\frac{\Gamma(1_{2}-4)}{k!\Gamma(1-\frac{a}{2}-k)}(z^{\frac{a}{2}-1})_{-4}(e^{-2J\overline{b}z})_{k}2^{-k}$$=Kz^{1}-e \sum_{k-0}^{\infty}\frac{(-1)^{k}[_{2}^{4}]_{k}}{k!}\{e^{i\pi}\frac{\Gamma(1-a-k)}{\Gamma(1-\frac{a}{2})}z^{a-1+}\}\{(-2i\sqrt{b})^{k}e^{-2i}\}$

(20)

(21)

$(| \Gamma(1-a-k)/\Gamma(1-\frac{a}{2})|<\infty)$

.

(22)

$=K(e^{i\pi\frac{a}{2}} \frac{\Gamma(1-a)}{\Gamma(1-.\frac{a}{2})})e^{-ifbz}\sum_{k-0}^{\infty}\frac{[\not\in]_{k}}{k![a]_{k}}(2i\sqrt{b}z)^{k}$$(|\sqrt{b}$$=K(e_{\Gamma(1-\frac{a}{2})}^{i*_{2}\Gamma(1-a)})e^{-i}{}_{1}F_{1}(7f_{bza} ; a;2i\sqrt{b}z)$(6)using Lemma (iv).
(iii)$\varphi_{[3](a,b)}^{o}=Set-\sqrt{b}$instead of(iv)$\varphi_{[4](a,b)}^{o}\approx Set-\sqrt{b}$instead of$\sqrt{b}$$\sqrt{b}$in (5).

(7)in (6).

(8)

Notice that we have$\varphi_{[1](a,b\Gamma}\varphi_{[1](a,b)}^{\Phi}$

except the constant coefficient,(23)

except the constant coefficient,

(24)

and
$\varphi_{[3](a,b)}=\varphi_{[3](a,b)}^{Q}$

43\S 4. Some special cases
Case of $a=0$ ;Corollary 1. Let be[I]$\varphi=\varphi(z)\in F$, then the homogeneous ordinary differential

equation$\varphi_{2}+\varphi\cdot b=0$$(b\neq 0)$(1)

has partcular solutions of the forms (ffactional differintegrated form)
Group I.
(i)$\varphi=Ke^{i\mathcal{F}bz}(e^{-2if_{bz}}\cdot 1)_{-1}\equiv\varphi_{[1](0,b)}$(2)$(ii)$$\varphi=Ke^{\iota f_{bz}}(1\cdot e^{-2ir_{bz}})_{-1}\equiv\varphi_{[2](0,b)}$

(3)

(iii)

$\varphi=Ke^{-i}(e^{21}1)_{-1}\equiv\varphi_{[3](0,b)}f_{bz}f_{bz}$

(4)

(iv)

$\varphi=Ke^{-if_{bz}}(1\cdot e^{2if_{bz}})_{-1}\equiv\varphi_{[4](0,b)}$

(5)

(i)

$\varphi=\ e^{i\Gamma bz}(e^{-2i\mathcal{F}bz}\cdot z^{-1})_{0}\equiv\varphi_{[1](0,b)}^{Q}$(6)$(ii)$$\varphi=ae^{f_{bz}}e^{i}(z^{-1}\cdot e^{-2if_{bz}})_{0}\equiv\varphi_{[2]_{(0,b)}}^{\theta}$(iii)$\varphi=\Omega e^{-i\mathcal{F}bz}(e^{2i\mathcal{F}bz}\cdot z^{-1})_{0}\equiv\varphi_{[3](0,b)}^{Q}$(iv)$\varphi=\Omega^{-if_{b_{Z}-1.2ir_{bz}}}e(ze)_{0}\equiv\varphi_{[4]_{(0,b)}}^{\Phi}$.

Group II.
(7)
(8)
(9)

where $K(\neq 0)$ is an arbitmry constant.

Note. Notice that we have ;$\varphi_{[1)(0,b)}^{\delta}=\varphi_{[2](0,b)}^{\Phi}$

and$\varphi_{[3](0,b)}^{\delta}=\varphi_{\mathfrak{l}^{4](0,b)}}^{O}$(10)

Proof.
Set $a=0$ in Theorem 1.
Because we have directly (from (6) and (7))$\varphi_{[1](0,b)}^{Q}=Ke^{-if_{bz}}=\varphi_{[2](0,b)}^{\triangleleft}$

(11)

since
$(e^{-2iX_{Z}}\cdot z^{-1})_{0}=e^{-2iIbz}z^{-1}=(z^{-1}.e^{-2i\Gamma bz})_{0}$(12)Or calculating them as follows (for its expanded form);$\varphi_{[1](0,b)}^{o}=Kze^{i/\overline{b}z}(e^{-2if_{bz}}\cdot z^{-1})_{0}$$=K^{f_{bz}}ze^{i} \sum_{k=0}^{\infty}\frac{\Gamma(1)}{k!\Gamma(1-k)}(e^{4.i})_{-k}(z^{-1})_{k}\tau_{b_{\overline{4}}}$

(6)
(13)

44

$= \Omega e^{i}\{e^{2i}\tau_{bz}.\tau_{bz_{Z}-1}+\frac{\Gamma(1)}{\Gamma(0)}(-2i\sqrt{b})^{-1}e^{-2tbz}(-z^{-2})$$+ \frac{\Gamma(1)}{2!\Gamma(-1)}(-2\dot{x}\sqrt{b})^{-2}e^{-2Jb^{-}z}(2z^{-3})+\cdots\cdots\}$(14)$=Ke^{-if_{bz}}$(15)$\varphi_{[2](0,b)}^{\Phi}=\Omega e^{if_{bz}}(z^{-1}.e^{-2J\overline{b}z})_{0}$(7)using Lemma (iv). And$= \Omega e^{i\mathcal{F}bz}\sum_{k-0}^{\infty}\frac{\Gamma(1)}{k!\Gamma(1-k)}(z^{-1})_{-k}(e^{-2i})_{k}f_{bz}$(16)$= \Omega e^{i}f_{bz}\{z^{-i}e^{2if_{bz}}+\frac{\Gamma(1)}{\Gamma(0)}(z^{-1})_{-1}(e^{-2i})_{1}f_{bz}+\frac{\Gamma(1)}{2!\Gamma(-1)}(z^{-1})_{-2}(e^{2i})_{2}f_{bz}+\cdots\cdots\}$(17)$= \ e^{i\mathcal{F}bz}\{z^{-1}e^{-2i/p_{z}}+\frac{1}{\Gamma(0)}(\log zx-2\dot{x}\sqrt{b})e^{-2i}f_{bz}$$+ \frac{1}{2!\Gamma(-1)}(\log z)_{-1}(-2i\sqrt{b})^{2}e^{-2\iota \mathcal{F}bz}+\cdots\cdot\cdot-\}$$=Ke^{-if_{bz}}\nearrow$

(18)

(19)

using
$(z^{-1})_{-2}=((z^{-1})_{-1})_{-1}=(\log z)_{-1}=z\log z-z$

(20)Note. Equation (1) is the simple hamonic vibration one.
Corollary 1‘ Familiar forms of the solutions in Corollary I are shown as follows.
Group I.
$i$(i)$\varphi_{[1](0,b)}=K_{\overline{2\sqrt{b}}}(\cos z-i\sin z)$

(21)
(22)

(ii)

$\varphi_{\{2J(0,b)}=K\frac{1}{\sqrt{b}}\sin\sqrt{b}z$$(iii)$$\varphi_{[3](0,b)}=K_{\overline{2\sqrt{b}}}(\cos\sqrt{b}z+i\sin\sqrt{b}z)$

(23)

(iv)

$\varphi_{[4](0,b)}=K\frac{1}{\sqrt{b}}\sin\sqrt{b}z$(24)$arrow$

Group II.
(i)$\varphi_{[1](0.b)}^{\Phi}=K(\cos z-i\sin z)$

(25)

(ii)

$\varphi_{[2](0,b)}^{Q}=K(\cos z-i\sin z)$

(26)

45
isin

$(iii)$$\varphi_{[3](0,b)}^{o}=K(\cos\sqrt{b}z+$$\sqrt{b}z)$

(27)

(iv)

$\varphi_{[4](0,b)}^{\fcircle}=K(\sqrt{b}\sqrt{b})$(28)Case of $a=2$ ;Corollary 2. Let be[II]$\varphi=\varphi(z)\in F$

equation, then the homogeneous ordinary differential$\varphi_{2}+\varphi_{1}\cdot\tau+\varphi\cdot b=02$$(z\neq 0)$(29)

has particular solutions of the forms (fractional differintegrated form)
Group I.
(i)$\varphi=K^{\iota\tau_{bz}\tau_{bz}}e(e^{-2i}\cdot z^{-1})_{0}\equiv\varphi_{[1](2,b)}$(30)$(ii)$$\varphi=K^{if_{bz}f_{bz}}e(z^{-1}\cdot e^{-2i})_{0}\equiv\varphi_{[2](2,b)}$(31)$\varphi=Ke^{-i\mathcal{F}f_{bz}}bz(e^{2i}\cdot z^{-1})_{0}\equiv\varphi_{[3](2,b)}$(32)$\varphi=Ke^{-if_{bz}}(z^{-1}\cdot e^{2\iota f_{bz}})_{0}\equiv\varphi_{[4](2.b)}$

(33)

(iii)
(iv)

Group II.
(i)$\varphi=$$(ii)$$\varphi=\Omega^{-1}e^{\iota Ibz}(1\cdot e^{-2iIbz})_{-1}\cong\varphi_{[2]_{(2,b)}}^{o}$翫$e(e^{-2i}\cdot 1)_{-1}\equiv\varphi_{[1](2,b)}^{Q}b_{Z}$

(34)
(35)

(iii)

$\varphi=\Omega^{-1}e^{-\iota f_{bz}}(e^{2if_{bz}}\cdot 1)_{-1}\equiv\varphi_{[3](2,b)}^{o}$

(36)

(iv)

$\varphi e^{f_{bz}f_{bz}}$(37)

where $K(\neq 0)$ is an arbitrary constant.
Note. Notice that we have ;and$\varphi_{[1](2,b)}=\varphi_{[2](2,b)}$$\varphi_{[3](2,b)}=\varphi_{[4](2,b)}$(38)

Proof. Set $a=2$ . in Theorem 1,
Corollary 2’ Familiar forms of the so lutions in Corollary 2 are shown as follows.
Group I.
(i)
(39)
-isin
$\sqrt{b}z)$$\varphi_{[1](2,b)}=Kz^{-1}(\cos\sqrt{b}z$(ii)$\varphi_{[2](2,b)}=\ ^{-1}(\cos z-\sin z)$(iii)$\varphi_{[3](2,b)}=n^{-1}(\cos\sqrt{b}z+$$\sqrt{b}z)$

(41)

(iv)

$\varphi_{[4](2.b)}=\ ^{-1}(\sqrt{b}f_{bz)}$

(42)

isin

(40)

Group II.
(i)$\varphi_{[1](2,b)}^{o}=K(\frac{i}{2\sqrt{}b})z^{-1}(\cos\sqrt{b}z-$$(ii)$$\varphi_{[2](2,b)}^{Q}=K(\frac{1}{/F})_{Z^{-1}\sin}\sqrt{b}z)$

(44)

(iii)

$\varphi_{r3j(2b)}^{Q}=K(\frac{-i}{2ffi})z^{-1}(\cos\sqrt{b}z+i\sin\sqrt{b}z)$

(45)

(iv)

$\varphi_{[4](2,b)}^{\Phi}=K($ $f_{fi}^{1})z^{-1}\sin\sqrt{b}z)$(46)

where $b\neq 0.$

isin$\sqrt{b}z)$

(43)

46
Proof.
We have $dioe\alpha ly(38)$ ffom (30), (31), (32) and (33) clearly.

Indeed we have(30)$\varphi[1]Q.b)^{=K^{f_{bz}f_{bz}}}e^{i}(e^{-2i}\cdot z^{-1})_{0}$(47)$=K^{fl_{Z}\mathcal{F}bz}e^{i} \sum_{k-0}^{\infty}\frac{\Gamma(1)}{k!\Gamma(1-k)}(e^{-2i})_{-k}(z^{-1})_{k}$$=Ke^{if_{bz}} \{e^{-uBz}z^{-1}+\frac{\Gamma(1)}{\Gamma(0)}(-2i\sqrt{b})^{-1}e^{4i}$va$z(-z^{-2})$$+ \frac{\Gamma(1)}{2!\Gamma(-1)}(-2\dot{x}\sqrt{b})^{-2}e^{-2i}(2z^{-3})f_{bz}+\cdots\cdots\}$$=\Omega^{-1}e^{-iB_{z}}=n^{-1}(\sqrt{b}\sqrt{b}\cdot$

(48)

(39)

Next we have(31)$\varphi_{[2](2,b)}=Ke^{if_{bz}}(z^{-1}\cdot e^{-2i\mathcal{F}bz})_{0}$(49)$=K^{f_{bz}}e^{i} \sum_{k-0}^{\infty}\frac{\Gamma(1)}{k!\Gamma(1-k)}(z^{-1})_{-k}(e^{-2i})_{k}\tau_{bz}$$=Ke^{t\Gamma bz} \{z^{-1}e^{-2jfi_{z}}+\frac{\Gamma(1)}{\Gamma(0)}(z^{-1})_{-1}(e^{-2if_{bz}})_{1}+\frac{\Gamma(1)}{2!\Gamma(-1)}(z^{-1})_{-2}(e^{-2i})_{2}fl_{Z}+\cdots\cdot\cdot\}$(50)$=Ke^{i} \{z^{-1}e^{-2i}f_{bz}f_{bz}+\frac{1}{\Gamma(0)}(\log zx-2i\sqrt{b})e^{-2i}f_{bz}$$+ \frac{1}{2!\Gamma(-1)}(\log z)_{-1}(-v\sqrt{b})^{2}e^{-2i}f_{bz}+\cdots\cdots\}$$=\ ^{-1}e^{-i\mathcal{F}bz}\approx\ ^{-1}(\cos Vl;z-i\sin\sqrt{b}z)$

And setinstead ofrespectively.
Next we have$-\sqrt{b}$$\sqrt{b}$

.

(52)
(40)

in (39) and (40), we have then (41) and (42),$\varphi_{[1](2,b)}^{\theta}=\Omega^{-1}e^{ifi_{z}}(e^{-2if_{bz}}\cdot 1)_{-1}$$=Kz^{-1}e^{if_{bz}} \sum_{k-0}^{\infty}\frac{\Gamma(0)}{k!\Gamma(0-k)}(e^{-2i\mathcal{F}bz})_{-1-k}(1)_{k}$$=K^{-1i}ze^{f_{bz}f_{bz}}(-\mathfrak{U}\sqrt{b})^{-1}e^{-2i}$

(34)
(53)
(54)

47
(55)

$=K( \frac{\iota’}{2\Gamma_{b}})z^{-1}e^{-if_{bz}}$(43)$=K( \frac{i}{2j5})z^{-1}(\cos z-i\sin z)-$

and
.

(35)

$\varphi_{[2](\angle,b)}\wedge=\Omega e^{f_{bz}}(1\cdot e^{-2f_{bz}})_{-1}0-1i$$= \ e\sum_{k=0}^{\infty}\frac{\Gamma(0)}{k!\Gamma(0-k)}(!)_{-1-k}(e^{-2i})_{k}$(56)$=\ \theta a\overline{e}\tau_{bi}z2,$$+(1)_{-3}(e^{-2\mathcal{F}bz}\dot{i})_{2}+(-1)(1)_{-4}(e^{-2if_{b_{\overline{\iota}}}})_{3^{+}}\cdot\cdot-\cdot\cdot-\}$e$=h^{\sim}$(57)$\{ze^{-2i}+(-1)\frac{1}{2!}z^{2}(-2i\sqrt{b})e^{-2i}$$+ \frac{1}{3!}z^{3}(-2i\sqrt{b})^{2}e^{-2if_{b}-}+(-1)\frac{1}{4!}z^{4}(-2i\sqrt{b})^{3}e^{-2Jbz}+\cdots\cdots\}$(58)$=K_{\angle e^{f_{bz}}(2i}^{-1-i} \sim\sqrt{b})^{-1}[\{1+(2i\sqrt{b}z)+\frac{1}{2_{\sim}^{1}}(2i\sqrt{b}z)^{2}$$+ \frac{1}{3!}(2\tilde{r}z)^{3}\cdot\cdot-\cdots\}-1]$$=Kz-e(2i\sqrt{b})^{-1}[e^{2i}-1]$$=K( \frac{I}{2\dot{z}ff})z^{-1}\{e^{i}-e^{f_{b_{\overline{z}}}}\}\mathcal{F}bz-i$$-\sqrt{b}$insfead of

(60)

(61)
(44)

$=K( \frac{1}{jb})_{Z^{-1}\sin}^{\sqrt{b}}z$

And setrespectively.

(59)

in (43) and (44), we have then (45) and (46),$\sqrt{b}$\S 5. Commentary(I) All solutions shown by (2) $\sim(9)$ in \S 2 have a fractional $diffe\Pi$ntegrated fcrn$(-\cdot-\cdot-\cdot)_{f}\dot{\prime}$(II)

where the index$\gamma$is the order of differintegration.

The special differential equation which is discussed in this article is much interes-ting one.
(III)

From the homogeneous $Fukuhara^{I}s$ equation$\varphi_{2}+\varphi_{I}\prime(a+\frac{b}{\angle})+\varphi\cdot(p+\tau q+\frac{r}{z^{2}})=0$$(\varphi=\varphi(z)_{j}z\neq 0)_{f}$

(1)

48
we obtain(2)$\varphi_{2^{+\varphi}1}.zp_{+\varphi\cdot p=0},$

setting $a=q=r=0.$
is, the equation
Equation (2) is the same form one as \S 2 (1). Thatone.
discussed in this article is a special form of the Fukuhara’s(IV) Moreover, when $b=p=-1$

we have(3)$\varphi_{2}+\varphi_{\iota z}1+\varphi=0$$fr_{-}om(2)$ .

$n$ the other hand , we obtain (3) from the homogeneous Bessel

equation$\varphi_{2}\cdot z^{2}+\varphi_{1}\cdot z+\varphi\cdot(z^{2}-\nu^{2})=0, (z\neq0)$(4)

setting $\nu=0.$That is, equation (3) is the Bessel’s one of order $v=0.$
A particular solution to equation (3) is given by
$-1$

(5)

$\varphi=\varphi_{[1]\langle 1,1)}=Ke^{iz}(e^{-2iz}\cdot z^{2})_{-J,2}$$=K$ (-勿)$- \frac{1}{2}-\frac{1}{2}ze^{-iz}{}_{2}F_{0}(2^{1},2^{1};\tau^{i})$,$(|_{T^{i}z}|<1)$(7)$=K(-2i)^{-\frac{1}{2}}H_{0}^{(2)}(z)$where$H_{0}^{\{2)}(z)$(6)

is the Hankel function of order , for example.
$0$(refer to \S 3. (1)). $(cf. J. Frac. Calc. Vo1.42, Nov. (2012),pp.1-19.,\cdot$

Solutions to The Homogeneous Bessel Equation by means ofFractional Calculus Operator, by K. Nishimoto)Note. We have$\varphi_{2}\cdot z^{2}+\varphi_{1}\cdot(az^{2}+bz)+\varphi\cdot(pz^{2}+qz+r)\Rightarrow 0$$(\varphi\approx\varphi(z), z\neq 0)$

$N$ -

(8)

from (1.).
Therefore, we obtain (4) from (8), setting$a=0,$$b=1,$$p=1,$$q=0$ ,

and $r=-\nu^{2}$(9)

That is, Bessel’s equation is a special one of Fukuhara’s equation.
(V) Usually the special ordinary differential equations like as (1) and (2)are solved by means of Frobenius. Compare our $N$ -fractional operator
method with that of Frobenius.

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