\documentclass[reqno]{amsart}

\AtBeginDocument{{\noindent\small
{\em Electronic Journal of Differential Equations},
Vol. 2003(2003), No. 87, pp. 1--64.\newline
ISSN: 1072-6691. URL: http://ejde.math.swt.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.swt.edu  (login: ftp)}
\thanks{\copyright 2003 Southwest Texas State University.}
\vspace{9mm}}

\begin{document}

\title[\hfilneg EJDE--2003/87\hfil The generalized Airy diffusion equation]
{The generalized Airy diffusion equation}

\author[F. M. Cholewinski \& J. A. Reneke\hfil EJDE--2003/87\hfilneg]
{Frank M. Cholewinski \& James A. Reneke}

\address{Frank M. Cholewinski\newline
Department of Mthematical Sciences\\
Clemson University \\
Clemson, SC 29634, USA}
\email{cholewa@innova.net}

\address{James A. Reneke\newline
Department of Mathematical Sciences\\
Clemson University \\
Clemson, SC 29634, USA}
\email{reneke@ces.clemson.edu}

\date{}
\thanks{Submitted February 3, 2003. Published August 22, 2003.}
\subjclass[2000]{33C10, 35C05, 44A15}
\keywords{Diffusion polynomial solutions, Bessel calculus, \hfill\break\indent
 positive definite kernels}


\begin{abstract}
 Solutions of a generalized Airy diffusion equation and an
 associated nonlinear partial differential equation are obtained.
 Trigonometric type functions are derived for a third order
 generalized radial Euler type operator. An associated complex
 variable theory and generalized Cauchy-Euler equations are
 obtained. Further, it is shown that the Airy expansions can be
 mapped onto the Bessel Calculus of Bochner, Cholewinski and
 Haimo.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{corollary}[theorem]{Corollary}
\newtheorem{proposition}[theorem]{Proposition}

\section{Introduction}

The Airy diffusion equation arises from the radial part of a third
order Laplace type operator on n-dimensional Euclidean space. In the one-dimensional
case, the Airy Diffusion equation of Widder \cite{Widd79} is obtained. The difficult
problem of representation of solutions encountered by Widder persists in the generalized
Airy equation case.

In this paper we obtain a sequence of polynomial solutions of the Airy diffusion equation,
which are analogous to the heat polynomials of Widder or the heat polynomials associated
with the generalized heat polynomials of Cholewinski and Haimo \cite{CholH68} or of
L. R. Bragg \cite{Brag65}. In the classical cases the heat polynomials are modified
Hermite polynomials and therefore have an orthogonality relation with respect to a positive
measure. The diffusion polynomial solutions obtained in this paper are 3-parity polynomials
and therefore by a result of Daboul and Rathie \cite{DaboR96} they can not be orthogonal in the
usual sense.

We also relate the solutions of the generalized Airy equation to solutions of a nonlinear
diffusion type partial differential equation.  The diffusion polynomial solutions lead to
dispersive waves which vanish at infinity.

Let $F(x_{1}, x_{2}, \dots, x_{n}) = F(r)$ be a radial function on
n-dimensional Euclidean space, where $r = (x_{1}^2 + x_{2}^2 +
\dots + x_{n}^2)^{1/2}$. Then a calculation shows that
\begin{equation}
\sum_{k = 1}^n\frac{r}{x_{k}}\frac{\partial ^3F}{\partial x_{k}^3}
= \frac{\partial ^3F}{\partial r^3} + \frac{3(n-1)}{r}\,\frac{\partial ^2F}{\partial r^2} - \frac{3(n -
1)}{r^2}\,\frac{\partial F}{\partial r}
\end{equation}
Let $\nu$ be a fixed nonnegative number, we define a linear third
order differential operator $\vartheta_{\nu}$ by
\begin{equation}
\vartheta_{\nu} = \frac{d^3}{dz^3} +
\frac{3\,\nu}{z}\,\frac{d^2}{dz^2} -
\frac{3\,\nu}{z}\,\frac{d}{dz}
\end{equation}
The generalized Airy diffusion equation is defined by
\begin{equation}
\vartheta_{\nu}\,u(x, t) = \frac{\partial u(x, t)}{\partial t}
\label{diffusion}
\end{equation}
If $\nu = 0$, we have the Airy equation of Widder \cite{Widd79}, and if $\nu = n - 1$ we
have a radial diffusion on $n$-dimensional Euclidean space. If $\nu$ is not an integer,
we have an analogue of the situation encountered by Bochner \cite{Boch55}, Weinstein
\cite{Wein53,Wein55} and others in the Bessel function case. The operator $\vartheta _{\nu}$
can be factored as
\begin{equation}
\vartheta _{\nu}  =  \frac{d}{dz}\,(z^{-3\nu}\,\frac{d}{dz}\,z^{3\nu}\,\frac{d}{dx})
 =  \frac{d}{dz}\,\Delta_{z}(\frac{2}{3}\,\nu)
\end{equation}
where
\begin{equation}
\Delta _{z}(\mu) = \frac{d^2}{dz^2} + \frac{2
\mu}{z}\,\frac{d}{dz}
\end{equation}
is the radial part of the Laplace operator on $n$-dimensional
Euclidean space $R^n$ with $\mu = \frac{n - 1}{2}$.

A number of solutions of the third order radial diffusion are shown to be related to
solutions of the radial heat equation. In fact we show that the source solution of the
generalized Airy diffusion equation is mapped onto the source solution of the
radial heat equation. In the case that $\nu = 0$, that is the one dimensional radial
diffusion, the source kernel is mapped onto the normal distribution function.

\section{Preliminary Results}

Let $\nu$ be a fixed nonnegative number. A simple
calculation shows that
\begin{equation}
\vartheta _{\nu}x^{3n} = 3^3n(n + \nu - 1/3)(n - 2/3)x^{3(n -1)} \label{SimpCal}
\end{equation}
and therefore $\vartheta _{\nu}$ acts as a delta operator on the ``basic'' sequence
$\{x^{3n}\}_{n = 0}^{\infty}$. By iteration we find that
\begin{equation}
\vartheta _{\nu}^{k}x^{3n} = 3^{3k}\,
\frac{\Gamma(n + 1)\Gamma(n + \nu + 2/3)\Gamma(n + 1/3)}{\Gamma(n - k + 1)
\Gamma(n - k + \nu + 2/3)\Gamma(n - k + 1/3)}x^{3(n - k)}
\label{iteration}
\end{equation}
For $k = n$, we get
\begin{equation}
\begin{aligned}
\vartheta _{\nu}^nx^{3n} & =  3^3n!\,\frac{\Gamma(n + 1/3)}{\Gamma(1/3)}\,
\frac{\Gamma(n + \nu 2/3)}{\Gamma(\nu + 2/3)}  \\
 & =  3^{3n}(1)_{n}(1/3)_{n}(\nu + 2/3)_{n}  \\
 & =  \alpha(3n, \nu) := \alpha_{3n}(\nu),
\end{aligned} \label{SpeCase}
\end{equation}
where
\begin{equation}
(a)_{n} = a(a + 1)(a + 2) \dots (a + n - 1) = \frac{\Gamma(a + n)}{\Gamma(a)}
\end{equation}
is the Pockhammer rising factorial function.

A Humbert type of Bessel function $G_{\nu}(z)$ is defined as the hypergeometric function
\begin{equation}
\begin{aligned}
G_{\nu}(z) & =  _{0}F_{2}(1/3, \nu + 2/3\,|\,(z/3)^3)  \\
 & = \sum_{n = 0}^{\infty} \frac{\Gamma(1/3)\Gamma(\nu + 2/3)z^{3n}}
 {3^{3n}\,n!\,\Gamma(n + 1/3)\Gamma(n + \nu + 2/3)}
\end{aligned}
\end{equation}
for $z \in \mathbb{C}$, the complex numbers. A calculation using Stirling's formula, shows
that
\[
\frac{\log \alpha_{3n}(\nu)}{3n\log 3n} \to  1 \quad \mbox{as } n\to \infty\]
and
\[\frac{3n}{(\alpha_{3n}(\nu))^{\frac{1}{3n}}} \to  e \quad\mbox{as } n\to \infty\]
It follows that $G_{\nu}(z)$ is an entire function of order one and of exponential
type 1.
By Marichev \cite{Mari83}, we also have the asymptotic estimate
\begin{equation}
G_{\nu}(z)\sim  \frac{\Gamma(1/3)\Gamma(\nu + 2/3)3^{\nu}}{\sqrt(3)\,2\pi}\,
\frac{e^{z}}{z^{\nu}} \quad\mbox{as } z\to \infty,\;
|\arg z| < \pi
\label{ser_exp}
\end{equation}
Furthermore, for ${\mathcal{G}}_{\nu}(z)$ defined by ${\mathcal{G}}_{\nu}(z) = G_{\nu}(-z)$
we have
\begin{equation}
\begin{aligned}
{\mathcal{G}}_{\nu}(z)
& =  G_{\nu}(-z) = _{0}F_{2}[1/3, \nu + 2/3\,|\,-(z/3)^3]  \\
 & \sim  \frac{\Gamma(1/3)\Gamma(\nu + 2/3)}{\sqrt{3}\,\pi}\,\frac{1}{(z/3)^{\nu}}\,
 e^{z/2}\cos (z\,\frac{\sqrt{3}}{2} - \frac{\nu\,\pi}{9})
\end{aligned}
\end{equation}
with $\arg \,z = 0$.
Using the estimate in \cite[p.~47]{Bate53a},
\[\frac{\Gamma(\nu + 2/3)}{\Gamma(\nu + 2/3 + n)}\sim (\nu + 2/3)^{-n}\;
\sim\;\nu^{-n}\quad\mbox{as } \nu\to \infty\]
and the Lebesgue Dominated Convergence Theorem, it follows that
\begin{equation}
G_{\nu}(3\,\nu^{1/3}\,(\frac{x}{2})^{2/3})\to 2^{-2/3}\,
\Gamma(1/3)x^{2/3}\,I_{-2/3}(x)\quad\mbox{as } \nu\to
\infty
\end{equation}
where $I_{\nu}(x)$ is a modified Bessel function of the first kind. Thus for large $\nu$,
behaves as an entire modified Airy Bessel function.
Recall that
\begin{equation}
\Gamma(\nu + 1)\,(z/2)^{-\nu}\,I_{\nu}(z) = \sum_{m = 0}^{\infty}\frac{z^{2m}}
{2^{2m}\,m!\,(\nu + 1)_{m}}
\end{equation}
is an entire function for $\nu > -1$.

In the Widder case of $\nu = 0$, we get upon application of the Gauss cubic factorial
equation that
\begin{equation}
G_{0}(z) = \sum_{k = 0}^{\infty} \frac{z^{3k}}{(3k)!} = \frac{1}{3}\,(e^z +
2\,e^{-z/2}\cos \frac{\sqrt{3}}{2}\,z)
\end{equation}
Using \eqref{SimpCal} it readily follows that
\begin{equation}
\vartheta_{\nu}\,G_{\nu}(xy) = \vartheta_{x}^{\nu}\,G_{\nu}(xy) =
y^3\,G_{\nu}(xy)
\end{equation}
Thus $G_{\nu}$ should play the role of the exponential function in a calculus associated
with the $\vartheta_{\nu}$ operator.

Next we define a generalized addition formula associated with the $\vartheta_{\nu}$
operator. This addition in terms of hypergeometric functions is an analogue of the
addition for Bessel functions presented by Bochner \cite{Boch55}. If $x$ and $y$ are
arbitrary complex numbers and $n$ is a nonnegative integer, we define
\begin{equation}
\begin{aligned}
&(x \oplus_{\nu}y)^{3n} \\
& =  x^{3n}\,_{3}F_{2}\left[
\begin{array}{c}
-n, -n + 2/3, -n + 1/3 - \nu \\[2pt]
1/3, \nu + 2/3
\end{array} \bigg|-(\frac{y}{x})^3\right] \\
& =  \sum_{k = 0}^n\binom{n}{k}
 \frac{\Gamma(\nu + 2/3)\Gamma(1/3)\Gamma(n + \nu + 2/3)\,y^{3k}x^{3(n - k)}}
 {\Gamma(k + 1/3)\Gamma(n - k + 1/3)\Gamma(n - k + \nu +2/3)\Gamma(k + \nu + 2/3)}
 \end{aligned}
 \label{plus}
\end{equation}
It readily follows that $(x \oplus_{\nu} y)^{3n}$ is a solution of the partial differential
equation
\begin{equation}
\frac{\partial ^3\,u}{\partial x^3} +
\frac{3\nu}{x}\,\frac{\partial ^2\,u} {\partial x^2} -
\frac{3\nu}{x^2}\,\frac{\partial u}{\partial x} = \frac{\partial
^3\,u}{\partial y^3} + \frac{3\nu}{y}\,\frac{\partial ^2\,u}
{\partial y^2} - \frac{3\nu}{y^2}\,\frac{\partial u}{\partial y}
\label{firstPDE}
\end{equation}
which satisfies the boundary conditions $u(x, 0) = x^{3n}$ and $u(0, y) = y^{3n}$.
\eqref{firstPDE} can be considered as a third order Euler-Poisson-Darboux equation,
see \cite{Cops58} or Weinstein \cite{Wein55}. The ordinary wave equation
\begin{equation}
\frac{\partial ^2\,u}{\partial x^2} = \frac{\partial
^2\,u}{\partial y^2}
\end{equation}
has solutions $f(x + y)$ corresponding to a boundary value function $f(x)$, whereas
\eqref{firstPDE} has in general solutions $f(x \oplus_{\nu} y)$ corresponding to
boundary value functions with 3rd order symmetry. Using Whipple's equation, see Henrici
\cite{Henr74}, p. 43, we also have that
\begin{equation}
(x \oplus_{\nu}y)^{3n} =  (x^3 + y^3)^n\,_{3}F_{2}\left[
\begin{array}{c}
-n/2, \frac{1 - n}{2}, -n + \nu \\
1/3, \nu + 2/3
\end{array} \bigg|\frac{4(xy)^3}{(x^3 + y^3)^2}\right]
\end{equation}
a result that relates the cubic addition associated with
$\vartheta_{\nu}$ to ordinary addition. Thus $(x \oplus_{\nu}
y)^{3n}\sim (x^3 + y^3)^n$ as $x\to \infty$ for $y$
fixed, for the $_{3}F_{2}$ polynomial goes to its constant term as
$x$ goes to infinity.

Furthermore, a calculation employing Equations \ref{iteration} and \ref{SpeCase} yields the
operational equation
\begin{equation}
G_{\nu}(y\,\vartheta_{x}(\nu)^{1/3})x^{3n} = (x \oplus_{\nu} y)^{3n}
\end{equation}
This equation is the 3rd order analogue of the extremely important equation
\begin{equation}
e^{yD}x^n = (x + y)^n
\end{equation}
where $D = \frac{d}{dx}$ is the derivative operator.
In the case of $\nu = 0$, we get the binomial formula
\begin{equation}
(1 \oplus_{0} x)^{3n} = \sum_{k = 0}^n \binom{3n}{3k}x^{3k}
\end{equation}
In the particular case of $x = 1$, we get the evaluation
\begin{equation}
(1 \oplus_{0} 1)^{3n} = \sum_{k = 0}^n
\binom{3n}{3k}  = \frac{1}{3}\,(2^{3n} + (-1)^n2)
\end{equation}
See \cite[p.~3]{GradR80}.
Furthermore, a series multiplication yields
\begin{equation}
G_{\nu}(x)\,G_{\nu}(y) = G_{\nu}(x \oplus_{\nu} y)
\end{equation}
which is the 3rd order analogue of the fundamental relation $e^{x}\,e^{y} = e^{x + y}$.
Using Stirling's formula, we also obtain the third order binomial limit
\begin{equation}
(1 \oplus_{\nu} \frac{x}{n})^{3n}\to G_{\nu}(x) \quad\mbox{as }n\to \infty
\end{equation}
Once again this is the analogue of the elementary calculus result
\begin{equation}
(1 + \frac{x}{n})^n\to e^{x} \quad\mbox{as }n\to \infty
\end{equation}
The function
\[
{\mathcal{G}}_{\nu}(z)  =  G_{\nu}(-z) = G_{\nu}(\beta_{3}z) \\
 =   _{0}F_{2}\big[\frac{1}{3}, \nu + 2/3|-(\frac{z}{3})^3\big]
\]
where $\beta_{3} = e^{i\pi/3}$ is a primitive root of $-1$, which plays the role of
$e^{-x}$ in many
calculations. However, it is not the multiplicative inverse of $G_{\nu}(z)$ as is seen
in the following development.  By Erd\'{e}lyi \cite[Volume 1, p.~186]{ErdeMOT53},
 we have
\[
_{0}F_{2}(a, b|z)\,_{0}F_{2}(a, b|-z)
  = \;_{3}F_{8}\left[
 \begin{array}{c}
 \frac{1}{3}\,(a + b - 1), \frac{1}{3}\,(a + b + 1) \\
  a, b, \frac{a}{2}, \frac{a + 1}{2}, \frac{b}{2}, \frac{b + 1}{2},
  \frac{a + b -1}{2},
 \frac{a + b}{2}
 \end{array}\bigg|-\frac{27}{64}\,z^2\right]
 \]
Taking $a = 1/3$ and $b = \nu + 2/3$, we get
\begin{equation}
\begin{aligned}
&G_{\nu}(z)\,{\mathcal{G}}_{\nu}(z) \\
& =  _{3}F_{8}\left[
 \begin{array}{c}
 \nu/3, \frac{\nu + 1}{3}\,\frac{\nu + 2}{3} \\
 1/3, \nu + 2/3, 1/6, 2/3, \frac{1}{2}\,(\nu + 2/3), \frac{1}{2}\,(\nu + 5/3),
 \frac{\nu}{2},
 \frac{\nu + 1}{2}
 \end{array}\bigg|-\frac{z^6}{12^3}\right]  \\
 & =  G_{\nu}((1 \oplus_{\nu}(-1))z)
\end{aligned}
\end{equation}
Since
\[
{\mathcal{G}}_{0}(x) = \frac{1}{3}(e^{-x} + 2\,e^{x/2}\cos\frac{\sqrt{3}}{2}x)
\]
we find that
\[G_{0}(x)\,{\mathcal{G}}_{0}(x) = \frac{1}{9}\{3 + 4\cosh(\frac{3}{2}x)\cos(\frac
{\sqrt{3}}{2}x) + 2\cos (\sqrt{3}x)\}\]
The generalized translation of a function $f(x) \in C^{\infty}$ is defined by
\[G_{\nu}(y\,\vartheta_{\nu}^{1/3})\,f(x) =
\sum_{n =
0}^{\infty}\frac{y^{3n}}{\alpha_{3n}}\,\vartheta_{\nu}^n\,f(x) =
f(x \oplus_{\nu} y)
\]
provided that the infinite series converges
locally uniformly in $x$ and $y$. In Section \ref{Source} we show
that if $f(x)$ is an entire function then $f(x \oplus_{\nu} y)$ is
also an entire function in the variables $x$ and $y$. The
translation operator can also be defined for formal power series.
Next we let $\mathcal{M}_{\nu}(R^{+})$ be the collection of positive
measures on $R^{+}$ such that the integral
\(\int_{0}^{\infty}e^{\tau\,y^2}\,d\gamma(y)\) is finite for $\tau
\geq \tau_{0} \geq 0$ and $d\gamma(y) \in \mathcal{M}_{\nu}$. If
$\gamma(y)$ is an increasing function on $R^{+}$ with compact
spectrum, that is the set of points of increase are contained in a
compact set, then the Stieltjes measure $d\gamma(y)$ is in
$\mathcal{M}_{\nu}(R^{+})$. Let $d\gamma(y)$ be an element of
$\mathcal{M}_{\nu}(R^{+})$ and let \(f(x) = \int_{0}^{\infty}
{\mathcal{G}}_{\nu}(xy)\,d\gamma(y)\). Since $G_{\nu}(x)$ is an entire
function of order one and type one, the integral converges locally
uniformly in $x$.

By the uniform convergence, we also get
\[\vartheta_{\nu}^n\,f(x) = \int_{0}^{\infty}(-y)^{3n}\,{\mathcal{G}}_{\nu}(xy)
\,d\gamma(y)\]
which is also uniformly convergent. Interchanging the summation and integration
we get
\[G_{\nu}(z\,\vartheta_{\nu}^{1/3})f(x) = \int_{0}^{\infty}{\mathcal{G}}_{\nu}(zy)\,
{\mathcal{G}}_{\nu}(xy)\,d\gamma(y) = f(x \oplus_{\nu} z)
\]
The uniform convergence justifies the interchange of summation and integration.
For $x$ and $z$ in a compact set; we have
\begin{align*}
\Big|\int_{0}^{\infty}G_{\nu}(zy)\,G_{\nu}(xy)\,d\gamma(y)\Big| &
\leq  \int_{0}^{\infty} G_{\nu}(Ay)\,G_{\nu}(By)\,d\gamma(y)  \\
 & \leq  M\,\int_{0}^{\infty}e^{(A + B)y}\,d\gamma(y) < \infty
\end{align*}
The functions $G_{\nu}(z)$ and ${\mathcal{G}}_{\nu}(z)$ have Poisson type integral
representations for $\nu \geq 1$ or $\mathop{\rm Re}\nu \geq 1$.

\begin{theorem} \label{thm1}
Let $\nu \geq 1$, then
\begin{equation}
G_{\nu}(z) = \frac{\Gamma(\nu +
2/3)}{3\,\Gamma(2/3)\,\Gamma(\nu)}\,\int_{0}^{1} \tau\,(1 +
\tau^3)^{\nu - 1}\,(e^{z\tau} + 2\,e^{-\frac{1}{2}\,z\tau}\cos
\frac{\sqrt{3}}{2}\,z\tau)\,d\tau \label{1st_result}
\end{equation}
and
\begin{equation}
{\mathcal{G}}_{\nu}(z) = \frac{\Gamma(\nu +
2/3)}{3\,\Gamma(2/3)\,\Gamma(\nu)}\,\int_{0}^{1} \tau\,(1 +
\tau^3)^{\nu - 1}\,(e^{-z\tau} + 2\,e^{\frac{1}{2}\,z\tau}\cos
\frac{\sqrt{3}}{2}\,z\tau)\,d\tau \label{2nd_result}
\end{equation}
\end{theorem}

\begin{proof}
 Using Gauss' cubic equation $(3n)! = \frac{3^{3n + 1/2}}{2\pi}\,n!
\,\Gamma(n + 1/3)\,\Gamma(n + 2/3)$ and the integral representation of the Beta function,
the general term of the $G_{\nu}(z)$ can be written as
\begin{align*}
&\frac{\Gamma(1/3)\,\Gamma(\nu + 2/3)\,z^{3n}}{3^{3n}\,n!\,\Gamma(n + 1/3)\,
\Gamma(n + \nu + 2/3)} \\
&= \frac{\sqrt{3}}{2\pi}\,\frac{\Gamma(1/3)\,\Gamma(\nu + 2/3)}{\Gamma(\nu)}\,
\frac{z^{3n}}{(3n)!}\,\int_{0}^{1}t^{\nu - 1}\,(1 - t)^{n - 1/3}\,dt
\end{align*}
Now the series $\sum_{n = 1}^{\infty}\frac{z^{3n}}{(3n)!}t^{\nu - 1}\,(1 - t)^{n - 1/3}$
converges uniformly with respect to $t$ in the interval $[0, 1]$ for $\nu \geq 1$.  Since
the term with $n = 0$ converges, we can interchange the summation.  Hence
\begin{align*}
G_{\nu}(z) & =  \frac{\sqrt{3}}{2\pi}\,\frac{\Gamma(1/3)\,\Gamma(\nu + 2/3)}{\Gamma(\nu)}\,
\int_{0}^{1}t^{\nu - 1}\,(1 - t)^{-1/3}\,
\sum_{n = 1}^{\infty}\frac{z^{3n}\,(1 - t)^n}{(3n)!}\,dt \\
 & =  \frac{\Gamma(\nu + 2/3)}{\Gamma(2/3)\,\Gamma(\nu)}\,\int_{0}^{1}
 t^{\nu - 1}\,(1 - t)^{-1/3}\,G_{0}((1 - t)^{1/3}\,z)\,dt
\end{align*}
With the change of variable $\tau = (1 - t)^{1/3}$, we get
\[G_{\nu}(z) = \frac{\Gamma(\nu + 2/3)}{\Gamma(2/3)\,\Gamma(\nu)}\,\int_{0}^{1}
\tau\,(1 - \tau^3)^{\nu - 1}\,(e^{z\tau} +
2\,e^{-\frac{1}{2}\,z\tau}
\cos\frac{\sqrt{3}}{2}\,z\tau)\,d\tau\] \eqref{2nd_result}
also follows from this result.
\end{proof}

The integral representations (\ref{1st_result}) and (\ref{2nd_result}), yield the
inequalities
\[|G_{\nu}(x)|
= G_{\nu}(x) \leq \frac{\Gamma(\nu + 2/3)}{\Gamma(2/3)\,\Gamma(\nu)}\,3\,e^{x}\]
and
\[|{\mathcal{G}}_{\nu}(x)|
\leq \frac{\Gamma(\nu + 2/3)}{\Gamma(2/3)\,\Gamma(\nu)}\,3\,e^{x/2}\]
for $x \geq 0$ and $\nu \geq 1$.

Next we introduce the entire Bessel functions
\begin{gather}
{\bf J}(z) = 2^{\nu - 1/2}\,\Gamma(\nu + 1/2)\,z^{1/2 - \nu}\,J_{\nu - 1/2}(z),\\
{\bf I}(z) = 2^{\nu - 1/2}\,\Gamma(\nu + 1/2)\,z^{1/2 - \nu}\,I_{\nu - 1/2}(z)
\end{gather}
where $J_{\nu - 1/2}(z)$ is the ordinary Bessel function of order $\nu - 1/2$
and $I_{\nu - 1/2}(z)$ is the Bessel function of imaginary argument. We have
the series expansion
\begin{equation}
{\bf J}_{\nu}(z) = \sum_{n =
0}^{\infty}\frac{(-1)^n\,z^{2n}}{b_{2n}(\nu)}
\end{equation}
where \(b_{2n}(\nu) = 2^{2n}\,n!\,(\nu + 1/2)_{n} = 2^{2n}\,n!\,
\frac{\Gamma(\nu + 1/2 + n)}{\Gamma(\nu + 1/2)}\).
The Bessel function $J_{\mu}(z)$ has the asymptotic expansion
\begin{equation}
J_{\mu}(z) \simeq (\frac{2}{\pi\,z})^{1/2}\cos (z - \frac{1}{2}\,\mu\,\pi -
\pi/4), \; -\pi < \arg \,z < \pi
\label{asym_exp}
\end{equation}
see Erdelyi \cite[Volume 2, p.~85]{Edre49}. In the next section it is shown that
the source  solution for the third
order diffusion is an ordinary Bessel function.

\section{The Source Function}\label{Source}

The formal Dirac delta function associated with our third order
calculus is
\begin{equation}
{\mathcal{D}}_{\nu}(x) =
\int_{0}^{\infty}{\mathcal{G}}_{\nu}(xy)\,d\eta_{\nu}(y)
\end{equation}
where $d\eta_{\nu}(y) = \frac{y^{3\nu + 1}\,dy}{3^{\nu - 1/3}\,\Gamma(\nu + 2/3)}$. This is
analogous to the classical representation
\begin{equation}
\delta(x) = \frac{1}{2\pi}\,\int_{-\infty}^{\infty}e^{ixy}\,dy
\end{equation}
or the Bessel representation
\begin{equation}
D_{\nu}(x) = \int_{0}^{\infty}{\bf J}_{\nu}(xy)\,d\mu_{\nu}(y)
\end{equation}
see \cite{CholH68}.
In general, solutions of the third order diffusion equation
\begin{equation}
\frac{\partial}{\partial t}\,u(x, t) = \frac{\partial^3u}{\partial
x^3} + \frac{3 \nu}{x}\, \frac{\partial^2u}{\partial x^2} -
\frac{3 \nu}{x^2}\,\frac{\partial \nu}{\partial x} =
\vartheta_{\nu}\,u(x, t) \label{diffeq}
\end{equation}
are formally given by the semigroup operation
\begin{equation}
u(x, t) = e^{t\,\vartheta_{\nu}}\,f(x)
\end{equation}
with $u(x, 0) = f(x)$.

We say that a function $u(x, t)$ in $C^3([0, a])$ in $x$ and
$C^{1}(0 \leq t \leq \sigma)$ in $t$ is in ${\mathcal{H}}_{\nu}([0, a]
\times [0, \sigma])$ if
\[\frac{\partial}{\partial t}\,u(x, t) = \vartheta_{\nu}\,u(x, t) \]
in the set $[0, a] \times [0, \sigma]$. We call a function in the class ${\mathcal{H}}_{\nu}$
an Airy or $\nu$-diffusion.
Taking $f(x) = {\mathcal{D}}_{\nu}(x)$, we get
\begin{equation}
\begin{aligned}
{\mathcal{K}}_{\nu}(x, t) =
e^{t\,\vartheta_{\nu}}\,{\mathcal{D}}_{\nu}(x) & =
e^{t\,\vartheta_{\nu}}\,\int_{0}^{\infty}{\mathcal{G}}_{\nu}(xy)\,d\eta_{\nu}(y)  \\
 & =  \int_{0}^{\infty}e^{-tx^3}\,{\mathcal{G}}_{\nu}(xy)\,d\eta_{\nu}(y)
\end{aligned}
\end{equation}
Using the series expansion \eqref{ser_exp}, term by term integration, and a change
of variables, we get
\begin{equation}
{\mathcal{K}}_{\nu}(x, t) = \frac{1}{(3t)^{\nu + 2/3}}\,\sum_{n=
0}^{\infty}\frac{(-1)^n\, x^{3n}t^{-n}}{3^{3n}\,n!\,(1/3)_{n}}
\end{equation}

Since ${\mathcal{G}}$ is an entire function of order 1, the term by term integration is valid based
on the absolutely and uniform convergence of the integral. Further, we have the representations
\begin{equation}
\begin{aligned}
{\mathcal{K}}_{\nu}(x, t) & =  \frac{1}{(3t)^{\nu +
2/3}}\,_{0}F_{1}(1/3\,|\,\frac{-x^3}{27t}) \\
 & =  \frac{\Gamma(4/3)}{3^{\nu + 2/3}}t^{\nu + 1}x\,J_{-2/3}(\frac{2x^{3/2}}
 {\sqrt{27t}}) \\
 & =  \frac{1}{(3t)^{\nu + 3/2}}\,{\bf J}_{-1/6}(\frac{2x^{3/2}}
 {\sqrt{27t}})
\end{aligned}
\end{equation}
Based on the asymptotic expansion (\ref{asym_exp}), we see that
\begin{equation}
{\mathcal{K}}_{\nu}(x, t) \sim \frac{\Gamma(2/3)}{\pi}\,3^{\nu + 3/2}t^{\nu + 19/12}\,
(\frac{x}{3})^{1/4}\cos (\frac{2x^{3/2}}{\sqrt{27t}} + \frac{\pi}{12})
\end{equation}
as $x \to \infty$.  In the classical case for the ordinary heat equation and for the
radial heat equation the source solutions are Gaussian type functions and go to zero at
infinity of order $e^{-cx^2}$.  In contrast, the source solution ${\mathcal{K}}_{\nu}(x, t)$
oscillates between infinite values for large values of $x$.

Applying the translation operator to the third order delta function we formally get
\begin{equation}
\begin{aligned}
G_{\nu}(z\vartheta_{\nu})\,{\mathcal{D}}_{\nu}(x) & =
{\mathcal{D}}_{\nu}(x \oplus_{\nu} z) \\
 & =  \int_{0}^{\infty}G_{\nu}(z\vartheta_{x})\,{\mathcal{G}}_{\nu}(xy)\,d\eta_{\nu}(y)
 \\
 & =  \int_{0}^{\infty}{\mathcal{G}}_{\nu}(x \oplus_{\nu} z)\,d\eta_{\nu}(y) \\
 & =  \int_{0}^{\infty}{\mathcal{G}}_{\nu}(xy)\,{\mathcal{G}}_{\nu}(zy)
 \,d\eta_{\nu}(z)
\end{aligned}
\end{equation}
Hence the Poisson type integral gives solutions of the third order diffusion
equation (\ref{diffeq}) with $u(x, 0) = f(x)$ has the formal solution
\begin{equation}
u(x, t) = e^{t\vartheta_{\nu}}\,f(x) = \int_{0}^{\infty}{\mathcal{K}}_{\nu}(x \oplus_{\nu} z,
t)\,f(z)\,d\eta_{\nu}(z)
\label{sol}
\end{equation}
Indeed the Poisson type integral gives solutions of the third order diffusion equation
(\ref{diffeq}) provided the function has suitable behavior at infinity.  It is easy to
see that if $f$ is a $C^{\infty}$ function with compact support in $[0, \infty)$, then
$u(x, t)$ given by (\ref{sol}) is a solution of the third order diffusion equation. The
case $\nu = 0$ is associated with Widder's Airy transform \cite{Widd79}.

To study the integral transform (\ref{sol}) we need to have order estimates on the
translated kernel ${\mathcal{K}}_{\nu}(x \oplus_{\nu} z, t)$. Let
\[B(x, y) = \frac{\Gamma(x)\,\Gamma(y)}{\Gamma(x + y)} = \int_{0}^{1}t^{x}\,
(1 - t)^{y - 1}\,dt\]
denote the beta function. From the integral representation for the beta function, it
readily follows that if $0 < x \leq x_{1}$ and $0 < y \leq y_{1}$, then
\[B(x_{1}, y_{1}) \leq B(x, y)\]
or equivalently
\begin{equation}
\frac{\Gamma(x + y)}{\Gamma(x)\,\Gamma(y)} \leq \frac{\Gamma(x_{1} + y_{1})}
{\Gamma(x_{1})\,\Gamma(y_{1})}
\end{equation}

\begin{theorem} \label{thm2}
If $\nu \geq 0$ and $x,y \geq 0$, then
\begin{equation}
(x \oplus_{\nu} y)^{3n} \leq \Gamma(1/3)\,\Gamma(\nu + 2/3)\,(n + 2\{\nu\} + 1)^
{3(\{\nu\} + 1)}\,(x + y)^{3(n + \{\nu\})}
\end{equation}
where $\{\nu\} = \mathop{\rm ceil}(\nu)$.
\end{theorem}

\begin{proof}. For $k \neq 0$ or $n$, using (\ref{sol}) we see that
\begin{equation}
\begin{aligned}
&\frac{(\nu + 2/3)_{n}}{(\nu + 2/3)_{n - k}\,(\nu + 2/3)_{k}} \\
& = \frac{\Gamma(\nu + 2/3)\,\Gamma(n + \nu + 2/3)}{\Gamma(n + 2\nu + 4/3)}\,
\frac{\Gamma(n + 2\nu + 4/3)}{\Gamma(n - k + \nu + 2/3)\,\Gamma(k + \nu + 2/3)}\\
 & \leq  \frac{\Gamma(\nu + 2/3)}{(n + \nu + 2/3)_{[\nu + 2/3]}}\,
 \frac{\Gamma(n + 2\{\nu\} + 2)}{\Gamma(n - k + \{\nu\} + 1)\,\Gamma(k + \{\nu\} + 1)}
 \\
 & \leq  \frac{\Gamma(\nu + 2/3)}{(n + \nu + 2/3)_{[\nu + 2/3]}}\,
 \frac{\Gamma(n + 2\{\nu\} + 2)}{\Gamma(n + \{\nu\} + 1)}\,
 \binom{n + \{\nu\}}{k} \\
 & \leq  \Gamma(\nu + 2/3)\,\frac{(n + \{\nu\} + 1)_{\{\nu\} + 1}}
 {(n + \nu + 2/3)_{[\nu + 2/3]}}\,
 \binom{n + \{\nu\}}{k}
\end{aligned}
\end{equation}
where $[x]$ is the greatest integer less than or equal to $x$.
Employing the asymptotic expansion for the quotients of Gamma functions, it follows that
\[\frac{(n + \{\nu\} + 1)_{\{\nu\} + 1}}{(n + \nu + 2/3)_{[\nu + 2/3]}} \sim
n^{\{\nu\} + 1 - [\nu + 2/3]} = O(n^2)
\]
as $n \to \infty$. In the same manner, we get
\[
\frac{(1/3)_{n}}{(1/3)_{n - k}\,(1/3)_{k}} \leq \frac{\Gamma(1/3)\,\Gamma(n + 1/3)}
{\Gamma(n + 2/3)}\,(n + 1)\,\binom{n}{k}
\leq \Gamma(1/3)\,(n + 1)\,\binom{n}{k}
\]
Note that $\Gamma(n + 1/3)/\Gamma(n + 2/3) \sim n^{-1/3}$ as $n \to \infty$.
Next we obtain an estimate on the binomial.  We have
\begin{equation}
\begin{aligned}
\binom{n}{k} & =  \frac{\Gamma(n + 1)}{\Gamma(n + 2)}\,
\frac{\Gamma(n + 2)}{\Gamma(n - k + 1)\,\Gamma(k + 1)} \\
& \leq  \frac{1}{n + 1}\,\frac{\Gamma(n + 2\{\nu\} + 2)}
 {\Gamma(n - k + \{\nu\} + 1)\,\Gamma(k + \{\nu\} + 1)} \\
& \leq  \frac{1}{n + 1}\,\frac{\Gamma(n + 2\{\nu\} + 2)}
 {\Gamma(n + \{\nu\} + 1)}\,\frac{\Gamma(n + \{\nu\} + 1)}
 {\Gamma(n - k + \{\nu\} + 1)\,k!}  \\
& =  \frac{1}{n + 1}\,(n + \{\nu\} + 1)_{\{\nu\} + 1}
 \binom{n + \{\nu\}}{k}
\end{aligned}
\end{equation}
It follows that
\begin{equation}
\begin{aligned}
\left[\begin{matrix}
\alpha_{3n} \\
\alpha_{3k}\end{matrix}\right]
&=\binom{n}{k}\,\frac{(1/3)_{n}}{(1/3)_{n - k}\,(1/3)_{k}}\,
\frac{(\nu + 2/3)_{k}}{(\nu + 2/3)_{n - k}\,(\nu + 2/3)_{k}}  \\
& \leq  \frac{\Gamma(\nu + 2/3)\,\Gamma(1/3)}{(1 + n)}\,
 \frac{\Gamma(n + 1/3)}{\Gamma(n + 2/3)}\,
 \frac{(n + \{\nu\} + 1)_{\{\nu\} + 1}^3}{(n + \nu + 2/3)_{[\nu + 2/3]}}\,
 \binom{n + \{\nu\}}{k}^3  \\
& \leq  \Gamma(\nu + 2/3)\,\Gamma(1/3)\,(n + 2\{\nu\} + 1)^{3(\{\nu\} + 1)}\,
 \binom{n + \{\nu\}}{k}^3
\end{aligned}
\end{equation}
Thus we find that
\begin{equation}
\begin{aligned}
&(x \oplus_{\nu} y)^{3n} \\
& =  x^{3n} + y^{3n} +
 \sum_{k = 1}^{n - 1}
 \binom{\alpha_{3n}(\nu)}{\alpha_{3k}(\nu)}
 x^{3k}\,y^{3(n - k)}  \\
& \leq  \Gamma(1/3)\,\Gamma(\nu + 1/3)\,(n + 2\{\nu\} + 1)^{3(\{\nu\} + 1)}\,
 \sum_{k = 0}^n
 \binom{n + \{\nu\}}{k}^3x^{3k}\,y^{3(n - k)} \\
 & \leq   \Gamma(1/3)\,\Gamma(\nu + 1/3)\,(n + 2\{\nu\} + 1)^{3(\{\nu\} + 1)}\,
 (x + y)^{3(n + \{\nu\})}
\end{aligned}
\end{equation}
Let $f(z) = \sum_{n = 0}^{\infty}a_{n}\,z^{3n}$ be an entire function, then by the
Cauchy-Hadamard formula
\[
\rho_{r} = [\limsup_{n \to \infty} |a_{n}|^{1/3n}]^{-1} = \infty
\]
Consider $f(z \oplus w) = \sum_{n = 0}^{\infty}a_{n}\,(z \oplus_{\nu} w)^{3n}$, we have
\[
|f(z \oplus w)| \leq \Gamma(1/3)\,\Gamma(\nu + 2/3)\,(|z| + |w|)^{3\{\nu\}}\,
\sum_{n = 0}^{\infty}|a_{n}|\,(n + 2\{\nu\} + 1)^{3(\{\nu\} + 1)}\,
(|z| + |w|)^{3n}
\]
Since $\limsup (n + 2\{\nu\} + 1)^{\frac{3(\{\nu\} +1)}{3n}} = 1$, it follows that
\[
[\limsup |a_{n}|^{\frac{1}{3n}}\,(n + 2\{\nu\} + 1)^{\frac{3(\{\nu\} +1)}{3n}}]^{-1}
= [\limsup |a_{n}|^{1/3n}]^{-1} = \infty
\]
Thus $f(z \oplus w)$ is an entire function in the $z$ and $w$ variables.

Further, let $\nu_{0} = \limsup_{n \to \infty} 3n\,|a_{n}|^{\frac{\rho}{3n}}$. Recall
that if $0 < \nu_{0} < \infty$ then $f(z)$ is of order $\rho$ and type $\tau$ if and only if
$\nu = e\tau\rho$. If $\nu_{0} = 0$ then $f$ is of growth $(\rho, 0)$ and if $\nu_{0} = \infty$
then $f$ is of growth not less than $(\rho, \infty)$.

If $f(z)$ is of growth $(\rho, \tau)$, we find that
\[
\limsup_{n \to \infty} 3n\,|a_{n}(n + 2\{\nu\} +1)^{3(\{\nu\} + 1)}|^{\frac{\rho}
{3n}} = \limsup_{n \to \infty} 3n\,|a_{n}|^{\frac{\rho}{3n}} = e\tau\rho
\]
Thus for fixed complex $w$, $f(z \oplus w)$ is also of growth $(\rho, \tau)$. The translated
functions occur as kernels in various integral transforms. Thus the convergence of the
transforms can be related to the growth properties of non-translated functions $f(z)$.

Applying the above results to ${\mathcal{K}}_{\nu}(x, t)$ and using Stirling's formula, it
follows that
\[
\rho = \limsup_{n \to \infty} \frac{3n\log 3^n}{log(\frac{1}{3^{3n}\,
t^n\,(1/3)_{n}\,n!})} = 3/2
\]
and
\[
\nu_{0} = \limsup_{n \to \infty}\frac{3n}{|3^{3n}t^n\,(1/3)_{n}\,n!|^{\frac{\rho}
{3n}}} = \frac{e}{(3t)^{1/2}} = e\tau\,\frac{3}{2}
\]
It follows that ${\mathcal{K}}_{\nu}(x, t)$ is of growth $(3/2,\frac{2}{\sqrt{27t}})$ for
$t > 0$. Since the derivative of a function$f$ of growth $(\rho, \tau)$ is also of growth
$(\rho, \tau)$, see \cite{Boas54}, p. 13, it follows that all the derivatives of
${\mathcal{K}}_{\nu}(x, t)$ are also of growth $(3/2,\frac{2}{\sqrt{27t}})$ for
$t > 0$.
\end{proof}

Now
\[\frac{\partial}{\partial t}\,{\mathcal{K}}_{\nu}(x, t) = \frac{1}{3^{\nu + 2/3}\,
t^{\nu + 5/3}}\,\sum_{n = 0}^{\infty}\frac{(-1)^{n + 1}\,(n + \nu + 2/3)\,
x^{3n}t^{-n}}{3^{3n}\,n!\,(1/3)_{n}}
\]
and a simple calculation shows that $\frac{\partial}{\partial t}\,{\mathcal{K}}_{\nu}(x, t)$
is of order 3/2. Since $\lim_{n \to \infty} (n + \nu + 2/3)^{\frac{1}{2n}} = 1$,
it also follows that $\frac{\partial}{\partial t}\,{\mathcal{K}}_{\nu}$ is also of type
$2/\sqrt{27t}$ for $t > 0$. This result is used to confirm local uniform convergence in
the following theorem.

\begin{theorem} \label{thm3}
 Let $f(x)$ be a function on $R^{+} = [0, \infty)$ such that $f(x) =
O(e^{-cx^{\rho}})$ with $\rho > 3/2$ and $c > 0$. Then
\begin{equation}
u(x, t) = \int_{0}^{\infty}{\mathcal{K}}_{\nu}(x \oplus_{\nu} y, t)\,f(y)\,d\eta_{\nu}(y)
\label{integral}
\end{equation}
is a solution of the Airy diffusion equation, that is, $u(x, t) \in {\mathcal{H}}_{\nu}$ for
$t > 0$.
\end{theorem}

\begin{proof}
 Based on the estimates for ${\mathcal{K}}_{\nu}(x
\oplus_{\nu} y, t)$ and its derivatives it follows that the
integral in \eqref{integral} and the integrals with kernels
$\frac{\partial^n}{\partial x^n}\,{\mathcal{K}}_{\nu} (x \oplus_{\nu}
y, t)$ locally converge uniformly, thus the operations of
differentiation and integration can be exchanged. Thus
\begin{equation}
\begin{aligned}
\vartheta_{\nu}\,u(x, t) & =  \int_{0}^{\infty}\vartheta_{\nu}\,
{\mathcal{K}}_{\nu}(x \oplus_{\nu} y, t)\,f(y)\,d\eta_{\nu}(y)  \\
 & =  \int_{0}^{\infty}\frac{\partial}{\partial t}\,{\mathcal{K}}_{\nu}(x \oplus_{\nu} y, t)\,
 f(y)\,d\eta_{\nu}(y)  \\
 & =  \frac{\partial}{\partial t}\,u(x, t)
\end{aligned}
\end{equation}
Since the function ${\mathcal{K}}_{\nu}(x \oplus_{\nu} y, t)$ is not an approximate identity
kernel, the boundary value $f(x)$ of $u(x, t)$ given by \eqref{integral} is only
recaptured in a formal way.
\end{proof}

Next we present an example of a $\nu$-Airy heat function which is not an entire function
of the space variable for a fixed $t$. We let
\[H_{1}(t) = \frac{2}{3^{(3\nu + 5)/2}\,\Gamma(\nu + 2/3)}t^{\frac{\nu}{2} +
\frac{5}{6}}\,K_{\nu - \frac{1}{3}}(2 \sqrt{\frac{t}{27}})\]
where $K_{\nu - \frac{1}{3}}$ is a modified Bessel function. By Erd\'{e}lyi \cite{ErdeMOT53}
we have the Mellin transform
\[\int_{0}^{\infty}t^{s - 1}H_{1}(t)\,dt = 3^{3s}\,\frac{\Gamma (s + 1) \Gamma (s + \nu + 2/3)}
{\Gamma (\nu + 2/3)}\]
for $\mathop{\rm Re}s > -1$. The function $H_{2}(t) = x^{1/3}e^{-x}/\Gamma (1/3)$ has the Mellin
transform
\[
\int_{0}^{\infty}t^{s - 1}H_{2}(t)\,dt = \frac{\Gamma (s + 1/3)}{\Gamma (1/3)}
\]
for $\mathop{\rm Re}s > -1/3$. Thus the mellin convolution of $H_{1}(x)$ and $H_{2}(x)$ is given by
\begin{align*}
{\mathcal{O}}_{\nu}(x) & =  \int_{0}^{\infty}H_{2}(x/t)H_{1}(t)\,\frac{dt}{t} \\
 & =  \frac{2x^{1/3}}{3^{(3\nu + 5)/2}\,\Gamma(1/3)\Gamma(\nu + 2/3)}\,
 \int_{0}^{\infty}t^{\frac{1}{2}\,(\nu - 1)}\,e^{-\frac{x}{t}}\,K_{\nu - 1/3}
 (2\sqrt{\frac{t}{27}})\,dt
\end{align*}
for $x \ge 0$.
By the Mellin convolution theorem \cite{Titc37} it follows that
\[
\int_{0}^{\infty}x^{s - 1}{\mathcal{O}}_{\nu}(x)\,dx  =  3^{3s}\,
\frac{\Gamma(s + 1)\Gamma(s + 1/3)\Gamma(s + \nu + 2/3)}{\Gamma(1/3)
\Gamma(\nu + 2/3)}
  =  \alpha _{3s}(\nu)
\]
for $\mathop{\rm Re}s > -1/3$. Letting
\[\phi(t) = \int_{0}^{\infty}e^{-tx}{\mathcal{O}}_{\nu}(x)x^{-1}\,dx\]
it follows that
\[
\phi^{(n)}(t) = (-1)^n\int_{0}^{\infty}e^{-tx}x^{n - 1}{\mathcal{O}}_{\nu}(x)\,dx\]
for $t \ge 0$ and $n = 0,1,2,\dots$. Thus $\phi(t) \in C^{\infty}(R^{+})$ and
$|\phi ^{(n)}(t)| \leq \alpha _{3n}(\nu)$ for $t \geq 0$. Hence
\[u(x,t) = \sum_{n = 0}^{\infty} \phi ^{(n)}(t)\,\frac{x^{3n}}{\alpha _{3n}(\nu)}\]
is a $\nu$-Airy diffusion for $-1 < x < 1$, $t > 0$. Further, we have
\[
u(x,0)  =  \sum_{n = 0}^{\infty} \phi ^{(n)}(0)\,\frac{x^{3n}}{\alpha _{3n}(\nu)}
 =  \sum_{n = 0}^{\infty}(-1)^nx^{3n}
 =  \frac{1}{1 + x^3}
\]
for $-1 < x < 1$. Hence the analytic extension $u(z,0)$ has singularities on the unit
circle at the cube roots of -1. Thus $u(z,0)$ can not be extended to an entire function.
It is holomorphic in the unit circle.

\section{Time Series Solutions}

Solutions for the partial differential equation $\vartheta_{\nu}\,u = u_{t}$ are obtained
for the Cauchy data given on the $t$-axis, that is,
\[u(0, t) = g(t) \quad\mbox{and}\quad u_{x}(0, t) = 0
\]
Let $D_{t}$ stand for differentiation with respect to $t$.
The function
\begin{equation}
u(x, t) = G_{\nu}(x\,D_{t}^{1/3})\,g(t)
\end{equation}
Gives a formal solution of the $\nu$-diffusion equation we have
\begin{align*}
\vartheta_{\nu}\,u(x, t) & =  \vartheta_{\nu}\,G_{\nu}(x\,D_{t}^{1/3})\,g(t) \\
 & =  D_{t}\,G_{\nu}(x\,D_{t}^{1/3})\,g(t) \\
 & =  D_{t}\,u(x, t)
\end{align*}
Using the power series expansion of $G_{\nu}(x)$, we define
\begin{equation}
G_{\nu}(x\,D_{t}^{1/3})\,g(t) = \sum_{n = 0}^{\infty}\frac{x^{3n}}{\alpha_{3n}}\,
g^{(n)}(t)
\end{equation}
provided the series converges locally uniformly for $g$ a $C^{\infty}$ function. Term by
term $x$-differentiation by the $\vartheta_{\nu}$ operator gives
\begin{equation}
\sum_{n = 1}^{\infty}\frac{x^{3(n - 1)}}{\alpha_{3(n - 1)}}\,g^{(n)}(t) =
\sum_{n = 0}^{\infty}\frac{x^{3n}}{\alpha_{3n}}\,g^{(n + 1)}(t)
\end{equation}
Thus the series gives the formal solution.

Let $C_{00}^{\infty}(0, \infty)$ be the class of functions $f(x)$ infinitely
differentiable and vanishing outside a compact subset of $0 < x < \infty$.

\begin{theorem} \label{cont_func}
If $f \in C_{00}^{\infty}(0, \infty)$ then the function
\begin{equation}
\hat{f}(t) = \int_{0}^{\infty}f(x)\,{\mathcal{G}}_{\nu}(xt)\,d\eta_{\nu}(x)
\end{equation}
is a continuous function in ${\mathcal{L}}^{1}((0, \infty), d\eta_{\nu}(x))$.
\end{theorem}

\begin{proof} Clearly the integral converges locally
uniformly and defines a continuous function. Since
$\vartheta_{\nu}\,G_{\nu}(xt) = -t^3\,G_{\nu}(xt)$, we find by
integration by parts that
\begin{align*}
t^3\,\hat{f}(t) & =
-\int_{0}^{\infty}f(x)\,\vartheta_{\nu}\,{\mathcal{G}}_{\nu}(xt)\,
d\eta_{\nu}(x) \\
& =  \int_{0}^{\infty}\left(\frac{\partial}{\partial x}x^{3\nu}\,\frac{\partial}
 {\partial x}x^{-3\nu}\,\frac{\partial}{\partial x}x^{3\nu + 1}\,f(x)\right)\,
 G_{\nu}(xt)\,\frac{dx}{c_{\nu}} \\
& =  \frac{1}{c_{\nu}}\,\int_{0}^{\infty}\vartheta_{-\nu}x^{3\nu + 1}\,f(x)\,
 {\mathcal{G}}_{\nu}(xt)\,dx
\end{align*}
Clearly $\vartheta_{-\nu}x^{3\nu + 1}\,f(x)$ is in $C_{00}^{\infty}(0, \infty)$.
Repeated applications yield
\[
t^{3n}\,\hat{f}(t) = \frac{1}{c_{\nu}}\,\int_{0}^{\infty}\vartheta_{-\nu}^n\,
(x^{3\nu + 1}\,f(x))\,{\mathcal{G}}_{\nu}(xt)\,dx
\]
Hence for $n$ sufficiently large, we have
\[\int_{1}^{\infty}|\hat{f}(t)|\,\frac{t^{3\nu + 1}}{c_{\nu}}\,dt =
\int_{1}^{\infty}t^{3n}\,|\hat{f}(t)|\,\frac{t^{-3n + 3\nu + 1}}{c_{\nu}}\,dt <
M_{2} < \infty
\]
and it follows that
\begin{align*}
\int_{0}^{\infty}|\hat{f}(t)|\,d\eta_{\nu}(t)
&= \int_{0}^{1}|\hat{f}(t)|t^{3\nu + 1}\,
\frac{dt}{c_{\nu}}
+ \int_{1}^{\infty}t^{3n}\,|\hat{f}(t)|\,
\frac{x^{-3n + 3\nu + 1}}{c_{\nu}}\,dt\\
& <  M_{1} + M_{2} < \infty\,.
\end{align*}
Next we let $g(t) = e^{-ty^3}$, then
\begin{equation}
G_{\nu}(x\,D_{t}^{1/3})\,e^{-ty^3}  =  \sum_{n =
0}^{\infty}\frac{x^{3n}}
{\alpha_{3n}}\,(-y)^{3n}\,e^{-ty^3}
 =  {\mathcal{G}}_{\nu}(xy)\,e^{-ty^3}
\end{equation}
\end{proof}

\begin{theorem} \label{thm5}
Let $\phi(x)$ be an entire function of growth $(2, \tau)$ and let
\begin{equation}
g(t) = \int_{0}^{\infty}e^{-ty^3}\,\phi(y)\,d\eta_{\nu}(y)
\label{phi_integral}
\end{equation}
then
\begin{equation}
G_{\nu}(x\,D_{t}^{1/3})\,g(t)  =
\int_{0}^{\infty}e^{-ty^3}\,\phi(y)\,
{\mathcal{G}}_{\nu}(xy)\,d\eta_{\nu}(y)
 =  u(x, t)
\end{equation}
\end{theorem}

\begin{proof}  Clearly the integral (\ref{phi_integral}) and all of its derivatives
converge uniformly for $t > 0$. Thus
\begin{align*}
G_{\nu}(x\,D_{t}^{1/3})\,g(t)
& = \sum_{n = 0}^{\infty}\frac{x^{3n}}{\alpha_{3n}}\,
\int_{0}^{\infty}(-y)^{3n}\,e^{-ty^3}\,\phi(y)\,d\eta_{\nu}(y) \\
 & =  \int_{0}^{\infty}e^{-ty^3}\,\phi(y)\,{\mathcal{G}}_{\nu}(xy)\,d\eta_{\nu}(y)
\end{align*}
Since ${\mathcal{G}}_{\nu}(z)$ is an entire function of order one, the uniform convergence
allows the interchange of summation and integration.
\end{proof}

\begin{corollary} \label{cor1}
If $\phi(x) \in C_{00}^{\infty}(0, \infty)$ and if
\[
g(t) = \int_{0}^{\infty}e^{-ty^3}\,\phi(y)\,d\eta_{\nu}(y)
\]
then
\begin{equation}
G_{\nu}(x\,D_{t}^{1/3})\,g(t) =
\int_{0}^{\infty}e^{-ty^3}\,\phi(y)\,
{\mathcal{G}}_{\nu}(xy)\,d\eta_{\nu}(y)
 =  u(x, t)
\end{equation}
is a $\nu$-diffusion for $t> 0$. Moreover, $u(x, t)$ is in ${\mathcal{L}}^{1}(R^{+},
d\eta_{\nu})$ and $u(x, 0) = \hat{\phi}(x)$.
\end{corollary}

\begin{proof} The integrability follows from Theorem \ref{cont_func}.
Next we define $g_{a}(t)$ for $a > 1/2$ by
\begin{equation}
g_{a}(t) =
\begin{cases}
\exp (-t^{-a}) & \mbox{for }  t > 0 \\
0 & \mbox{for }   t \leq 0
\end{cases}
\end{equation}
Then by an application of Cauchy's integral formula, there is a $\theta = \theta(a)$
such that
\begin{equation}
|g_{a}^{(k)}(t)| = \frac{k!}{(\theta t)^{k}}\exp (\frac{1}{2}t^{-a})
\end{equation}
see for example Widder \cite[p.~46]{Widd75}.  Let $u_{a}(x, t) = G_{\nu}(x\,D_{t}^{1/3})\,
g(t)$. Since $\frac{k!}{(3k)!} \leq \frac{1}{(2k)!}$, we find using Gauss' cubic equation
that for $t > 0$ and arbitrary complex $x$
\begin{equation}
\begin{aligned}
|u_{a}(x, t)|  & \leq  \sum_{k = 0}^{\infty}\frac{|x|^{3k}\,|g^{(k)}(t)|}
{\alpha_{3k}}  \\
 & \leq  \sum_{k = 0}^{\infty}\frac{k!}{\alpha_{3k}}\,\frac{|x|^{3k}}{(\theta t)^{k}}
 \exp(-\frac{1}{2}t^{-a})  \\
 & \leq  \sum_{k = 0}^{\infty}\frac{k!}{(3k)!}\,\frac{(2/3)_{k}}{(\nu + 2/3)_{k}}\,
 \frac{|x|^{3k}}{(\theta t)^{k}}\exp (-\frac{1}{2}t^{-a})  \\
 & \leq  \sum_{k = 0}^{\infty}\frac{1}{(2k)!}\,\frac{|x^{3/2}|^{2k}}{(\theta t)^{k}}
 \exp(-\frac{1}{2}t^{-a}) \\
 & = \cosh(\frac{|x^{3/2}|}{(\theta t)^{1/2}})\exp (-\frac{1}{2}t^{-a})  \\
 & \leq  \exp(\frac{|x^{3/2}|}{(\theta t)^{1/2}} - \frac{1}{2}t^{1/2 - a})  \\
 & = \exp(\frac{1}{t^{1/2}}\,\{\frac{|x|^{3/2}}{\theta^{1/2}}
 - \frac{1}{2}t^{1/2 - a}\})
\end{aligned}
\end{equation}
or
\begin{equation}
|u_{a}(x, t)|  \leq \exp\Big(\frac{1}{t^{1/2}}\,\{\frac{|x|^{3/2}}{\theta^{1/2}} -
\frac{1}{2}t^{1/2 - a}\}\Big)
\label{bound}
\end{equation}
Hence the series defining $u_{a}(x, t)$ converges uniformly and absolutely for $t > 0$
and $x$ in an arbitrary compact set.  The previous inequality shows that
\[\lim_{t \to 0^{+}} u_{a}(x, t)= 0\]
locally uniformly. It follows that
\begin{equation}
u_{a}(x, t) = \sum_{n = 0}^{\infty}\frac{x^{3n}}{\alpha_{3n}(\nu)}\,|g_{a}^{(n)}(t)|
\end{equation}
is a nontrivial solution of the $\nu$-diffusion equation with boundary values $u_{a}(x, 0) =
0$ for every $ a > 1$. Therefore the solutions of the $\nu$-diffusion boundary value
problem are not necessarily unique.

Inequality (\ref{bound}) shows that $u_{a}(x, t)$ is an entire function of growth $(3/2, 1/t)$
for $t > 0$. However, $u_{a}$ is not analytic in $t$ since $u_{a}(0, t)$ vanishes for all
$t \leq 0$ and $u_{a}(0, t) = g_{a}(t) > 0$ for $t > 0$. The $u_{a}(x, t)$ are
Tychonoff type solutions of the $\nu$-diffusion equation.
\end{proof}

\section{$\vartheta_{\nu}$-Trigonometric Functions}
\label{trig}

The ordinary trigonometric functions play an important role in the theory of the heat
and wave equations on Euclidean spaces. In order to study solutions of equations connected
with the $\vartheta_{\nu}$ operator, it is necessary to develop an associated
trigonometric theory.

We define the $\nu$-hyperbolic functions by the series
\begin{equation}
\label{eq66}
G_{\nu}(z)  =  \sum_{n = 0}^{\infty}\frac{z^{3n}}{\alpha_{3n}} =
\sum_{k = 0}^{\infty}\frac{z^{6k}}{\alpha_{6k}} +
\sum_{k = 0}^{\infty}\frac{z^{6k + 3}}{\alpha_{6k + 3}}
 :=  \cosh_{\nu}(z) + \sinh_{\nu}(z)
\end{equation}
It follows that $\cosh_{\nu}(z)$ and $\sinh_{\nu}(z)$ are entire functions of
order one.
Since
\begin{equation}
\alpha_{6k}(\nu) = 6^{6k}\,k!\,(1/2)_{k}\,(1/6)_{k}\,(2/3)_{k}\,(1/2 + 1/3)_{k}\,
(\nu/2 + 5/6)_{k}
\end{equation}
and
\begin{equation}
\alpha_{6k + 3}(\nu) = 9(\nu + 2/3)\,6^{6k}\,k!\,(3/2)_{k}\,(2/3)_{k}\,(7/6)_{k}\,
(\nu/2 + 5/6)_{k}\,(\nu/2 + 4/3)_{k}
\end{equation}
we obtain the hypergeometric representations
\begin{equation}
\cosh_{\nu}(z) := \sum_{k = 0}^{\infty}\frac{z^{6k}}{\alpha_{6k}}
 = _{0}F_{5}(1/2, 1/6, 2/3, \nu/2 + 1/3, \nu/2 + 5/6\,|\, \big(\frac{x}{6})^6)
\end{equation}
and
\begin{equation}
\begin{aligned}
\sinh_{\nu}(z) & :=  \sum_{k = 0}^{\infty}\frac{z^{6k + 3}}{\alpha_{6k + 3}} \\
 & =  \frac{x^3}{9(\nu + 2/3)}\,_{0}F_{5}(3/2, 2/3, 7/6, \nu/2 + 5/6, \nu/2 + 4/3\,
 |\big(\frac{x}{6}\big)^6)
\end{aligned}
\end{equation}
In particular for $\nu = 0$, a calculation shows that
\begin{gather}
\cosh_{0}(z) = \frac{1}{3}\cosh z + \frac{2}{3}
\cos (\frac{\sqrt{3}}{2}\,z)\cosh(z/2)\,, \\
\sinh_{0}(z) = \frac{1}{3}\sinh z + \frac{2}{3}\cos (\frac{\sqrt{3}}{2}\,z)
\sinh(z/2)
\end{gather}
Since
\begin{equation}
\cosh_{\nu}(ax) = \frac{1}{2}\,\{G_{\nu}(ax) + G_{\nu}(-ax)\}
\end{equation}
we get
\begin{gather}
\vartheta_{\nu}\cos h_{\nu}(ax) = a^3\sinh_{\nu}(ax)\,, \\
\vartheta_{\nu}\sinh_{\nu}(ax) = a^3\cos h_{\nu}(ax)
\end{gather}
Thus $\cosh_{\nu}(ax)$ and $\sinh_{\nu}(ax)$ are solutions of the harmonic type
equation
\begin{equation}
(\vartheta_{\nu}^2 - a^6)\,y(x) = 0
\end{equation}
We note that $\cosh_{\nu}$ is an even function and $\sinh_{\nu}$ is an odd function.
Since ${\mathcal{G}}_{\nu}(x) = G_{\nu}(-x) = \cosh_{\nu}(x) - \sinh_{\nu}(x)$,
we get
\begin{equation}
G_{\nu}(x)\,G_{\nu}(-x)  =  \cosh_{\nu}^2(x) - \sinh_{\nu}^2(x)
 =  G_{\nu}((1 \oplus_{\nu} (-1))x)
\end{equation}
See Equation (\ref{SpeCase}) for  a hypergeometric representation.

In terms of the umbral $\nu$-translation, we obtain a number of identities. We define an
umbral multiplication in terms of repeated addition.  We have
\begin{equation}
(2 \odot_{\nu} x)^{3n} = (x \oplus_{\nu} x)^{3n} = x^{3n}\,\sum_{k
= 0}^n \binom{\alpha_{3n}(\nu)}{\alpha_{3k}(\nu)}
\end{equation}
and, for $m \geq 2$, we have the generalized $\nu$-multinomial
\begin{equation}
\begin{aligned}
(m \odot_{\nu} x)^{3n} &= (x \oplus_{\nu} \dots \oplus_{\nu} x)^{3n}\\
& = x^{3n}\cdot
\sum_{\begin{array}{c}
\ell_{1} + \ell_{2} + \dots + \ell_{m} = n \\
\ell_{j} \geq 0
\end{array}}\frac{\alpha_{3n}(\nu)}{\alpha_{3\ell_{1}}\,\alpha_{3\ell_{2}}\dots\,
\alpha_{3\ell_{m}}}
\end{aligned}
\end{equation}
Now
\begin{equation}
\begin{aligned}
G_{\nu}(x)\,G_{\nu}(\pm y)
& =  G_{\nu}(x \oplus_{\nu} (\pm y))  \\
& = \cosh_{\nu}(x \oplus_{\nu} (\pm y)) + \sinh_{\nu}(x \oplus_{\nu} (\pm y))  \\
& = \cosh_{\nu}(x)\cos h_{\nu}(y) \pm sinh_{\nu}(x)\sinh_{\nu}(y)  \\
 &\quad + \sinh_{\nu}(x)\cos h_{\nu}(y) \pm \cosh_{\nu}(x)\sinh_{\nu}(y)\,.
\end{aligned}
\end{equation}
Equating the even and odd parts, we obtain the angle-sum and angle-difference
$\nu$-relations,
\begin{gather}
\cosh_{\nu}(x \oplus_{\nu} (\pm y)) =\ cosh_{\nu}(x)\cos h_{\nu}(y)
\pm \sinh_{\nu}(x) \sinh_{\nu}(y)\,,
\\
\sinh_{\nu}(x \oplus_{\nu} (\pm y)) = \sinh_{\nu}(x)\cos h_{\nu}(y)
\pm \cosh_{\nu}(x) \sinh_{\nu}(y)\,.
\end{gather}
Thus we also get the multiple angle relations
\begin{gather}
\sinh_{\nu}(2 \odot_{\nu} x) = 2\sinh_{\nu}(x)\cos h_{\nu}(x)
\label{sinh_double} \\
\cosh_{\nu}(2 \odot_{\nu} x) = \cosh_{\nu}^2(x) + \sinh_{\nu}^2(x)\,.
\label{cosh_double}
\end{gather}
Using the binomial theorem, we obtain the general multiple angle relations
\begin{gather}
\cosh_{\nu}(n \odot_{\nu} x) = \sum_{\ell = 0}^{[n/2]}
\binom{n}{2\ell}\sinh_{\nu}^{2\ell}(x)\cos h_{\nu}^{n - 2\ell}(x)
\label{cosh}\\
\sinh_{\nu}(n \odot_{\nu} x) = \sum_{\ell = 0}^{[n/2] - 1}
\binom{n}{2\ell + 1}\sinh_{\nu}^{2\ell + 1}(x)\cosh_{\nu}^{n - 2\ell - 1}(x)
\end{gather}
In terms of hypergeometric functions, using umbral variables, Equation (\ref{cosh})
yields the identity
\begin{equation}
\begin{aligned}
&_{0}F_{5}\big(1/2, 1/6, 2/3, \nu/2 + 1/3, \nu/2 + 5/6\,|\,
 \big(\frac{n \odot_{\nu} x}{6}\big)^6\big)  \\
& =  \sum_{\ell = 0}^{[n/2]}\binom{n}{2\ell}
\frac{x^{6\ell}}{9^{2\ell}\,(\nu + 2/3)^{2\ell}}\,
_{0}F_{5}(3/2, 2/3, 7/6, \nu/2 + 5/6, \nu/2 + 4/3\,|\, (\frac{x}{6})^6)^{2\ell}
\\
&\quad \times {} _{0}F_{5}(1/2, 1/6, 2/3, \nu/2 + 1/3, \nu/2 + 5/6\,|\,
 (\frac{x}{6})^6)^{n - 2\ell}\,.
\end{aligned}
\end{equation}
Also from \eqref{eq66} we get the hypergeometric identity
\begin{equation}
\begin{aligned}
&_{0}F_{2}\big(1/3, \nu + 2/3\,|\,\frac{(n \odot_{\nu} x)^3}{9}\big) \\
& =  \sum_{\ell = 0}^n\binom{n}{\ell}
 \frac{x^{3\ell}}{9^{\ell}\,(\nu + 2/3)^{\ell}}\,
_{0}F_{5}(3/2, 2/3, 7/6, \nu/2 + 5/6, \nu/2 + 4/3\,|\, (\frac{x}{6})^6)^{\ell}\\
&\quad\times {} _{0}F_{5}(1/2, 1/6, 2/3, \nu/2 + 1/3, \nu/2 + 5/6\,|\,
 (\frac{x}{6})^6)^{n - \ell}\,.
\end{aligned}
\end{equation}

Next we introduce an umbral subtraction for $\nu$-translation. The generalized subtraction
associated with the Bessel functions was developed by Cholewinski \cite{Chol88}. Since
$G_{\nu}(0) = 1$, we can define a ``$\nu$-Mobius'' sequence $\{\gamma_{3k}(\nu)\}_{k = 0}
^{\infty}$ by the equation
\begin{equation}
G_{\nu}(\gamma_{u}x) = \frac{1}{G_{\nu}(x)}\quad\mbox{or}\quad
G_{\nu}(\gamma_{u}x) G_{\nu}(x) = 1\,.
\end{equation}
By multiplication of series, we must have
\begin{equation}
\sum_{k = 0}^n\binom{\alpha_{3n}}{\alpha_{3k}}
\gamma_{3k} = \delta_{n, 0},
\end{equation}
the Kronecker delta.
For $n = 0$, we get $\gamma_{0} = 1$. Hence the sequence  $\{\gamma_{3k}(\nu)\}$
is obtained inductively by the equation
\begin{equation}
\gamma_{3n}(\nu) = -\sum_{k = 0}^{n - 1}\binom{\alpha_{3n}}{\alpha_{3k}}
\gamma_{3k}(\nu)
\end{equation}
The first four values are  $\gamma_{0} = 1$,
$\gamma_{3}(\nu) = -1$, $\gamma_{6}(\nu) = -1 + 8(\nu + 5/3)/(\nu
+ 2/3)$, and $\gamma_{9}(\nu) = -1 -126(\nu + 8/3)(\nu + 2)/(\nu +
2/3)^2$.
Further, we get
\begin{equation}
G_{\nu}(x)\,G_{\nu}(\gamma_{u}y)  =  G_{\nu}(x \oplus_{\nu} \gamma_{u}y)
 = \cosh_{\nu}(x \oplus_{\nu} \gamma_{u}y) +
 \sinh_{\nu}(x \oplus_{\nu} \gamma_{u}y)\,.
\end{equation}
Multiplying the series on the left and equating even and odd parts, it follows
that
\begin{gather}
\cosh_{\nu}(x \oplus_{\nu} \gamma_{u}y) = \cosh_{\nu}(x)\cosh_{\nu}(\gamma_{u}y)
+ \sinh_{\nu}(x)\sinh_{\nu}(\gamma_{u}y) = 1
\label{cosh_addition}\\
\sinh_{\nu}(x \oplus_{\nu} \gamma_{u}y) = \sinh_{\nu}(x)\cosh_{\nu}(\gamma_{u}y)
+ \cosh_{\nu}(x)\sinh_{\nu}(\gamma_{u}y) = 0
\end{gather}
The $\cosh_{\nu}$ identity above is an analogue of the familiar
result $\cosh^2x - \sinh^2x = 1$. Of course Equation
(\ref{cosh_addition}) can be expressed as a complicated
hypergeometric identity. With the obvious notation we get the
$\nu$-hyperbolic tangent identity
\[
1 + tanh_{\nu}(x)tanh_{\nu}(\gamma_{u}x) = sech_{\nu}(x)\,
sech_{\nu}(\gamma_{u}x)
\]
Using Equations (\ref{sinh_double}) and (\ref{cosh_double}), we obtain the
multiple angle relation
\begin{equation}
\tanh_{\nu}(2 \odot_{\nu} x) = \frac{2\,\tanh_{\nu}(x)}{1 +
\tanh_{\nu}^2(x)}\,.
\end{equation}
In the case $\nu = 0$, the $\nu$-hyperbolic tangent is
\begin{equation*}
\tanh_{0}(x) =  \frac{\sinh x - 2/3\cos \frac{\sqrt{3}}{2}x\sinh \frac{x}{2}}
{\cosh x + 2/3\cos \frac{\sqrt{3}}{2}x\cosh \frac{x}{2}}
 = \tanh(x)\frac{\Big(1 - \frac{\cos\frac{\sqrt{3}}{2}x}{3\cosh \frac{x}{2}}
 \Big)}{\Big(1 + (2/3)\frac{\cos\frac{\sqrt{3}}{2}x\cosh \frac{x}{2}}
 {\cosh x}\Big)}
\end{equation*}
Next
\begin{equation}
\tanh_{\nu}(x) = \frac{1 - G_{\nu}(-x)\,G_{\nu}(\gamma_{u}x)}{1 +
G_{\nu}(-x)\,G_{\nu}(\gamma_{u}x)}
\end{equation}
which implies that
\begin{equation}
\tanh_{\nu}(x) = 1 + 2\,\sum_{n =
1}^{\infty}(-1)^n\,\frac{G_{\nu}(-x)^n} {G_{\nu}(x)^n}
\label{tanh}
\end{equation}
which converges for $x > 0$, since $|G_{\nu}(-x)| < G_{\nu}(x)$ for $x > 0$. In umbral
notation Equation (\ref{tanh}) can be written as
\begin{equation}
\tanh_{\nu}(x) = 1 + 2\,\sum_{n = 1}^{\infty}(-1)^n\,G_{\nu}(n
\odot_{\nu} (x( -1 \oplus_{\nu} \gamma_{u})))\,.
\end{equation}
This is the analogue of the elementary formula
\begin{equation}
\tanh(x) = 1 + 2\,\sum_{n = 1}^{\infty}(-1)^n\,e^{-2nx},
\mbox{\hspace{.25in}for $\mathop{\rm Re}x > 0$}
\end{equation}

A complex $\nu$-exponential function $E_{\nu}(x)$ is defined by the equation
\begin{equation}
E_{\nu}(x) = G_{\nu}(\omega_{6}x) = \sum_{n = 0}^{\infty}
\frac{i^nx^{3n}}{\alpha_{3n}(\nu)}
\end{equation}
where $\omega_{6} = e^{\frac{\pi}{6}\,i} = \frac{\sqrt{3}}{2} + \frac{1}{2}\,i$. The
function $E_{\nu}(x)$ is the $e^{ix}$ in the the $\nu$-calculus associated with
$\vartheta_{\nu}$, it is an entire function of order one. We define generalized sine and
cosine functions by the equation
\begin{equation}
E_{\nu}(x)  =  \sum_{k = 0}^{\infty}\frac{(-1)^{k}x^{6k}}{\alpha_{6k}(\nu)}
+ i\,\sum_{k = 0}^{\infty}\frac{(-1)^{k}x^{6k+ 3}}{\alpha_{6k + 3}(\nu)}
 = \cos_{\nu}(x) + i\sin_{\nu}(x)\,.
\end{equation}
 From our previous results, we have
\begin{gather}
\cos_{\nu}(x) = \cosh_{\nu}(\omega_{6}x)\,, \\
\sinh_{\nu}(\omega_{6}x) = i\sin_{\nu}(x)\,.
\end{gather}
The hypergeometric representations are
\begin{gather}
\cos_{\nu}(x) = _{0}F_{5}\big(1/2, 1/6, 2/3, \nu/2 + 1/3, \nu/2 + 5/6\,|\,
-(\frac{x}{6})^6\big) \,,\\
\sin_{\nu}(x) = \frac{x^3}{9(\nu + 2/3)}\,_{0}F_{5}\big(3/2, 2/3,
7/6, \nu/2 + 5/6, \nu/2 +
4/3\,|\,-(\frac{x}{6})^6\big)\,.
\end{gather}
The trigonometric identities for the $\nu$-hyperbolic functions are easily converted
to identities for the $\nu$-sine and cosine functions. We have
\begin{gather}
\cos_{\nu}(x \oplus_{\nu} y) = \cos_{\nu}(x)\cos _{\nu}(y) -
\sin_{\nu}(x)\sin_{\nu}(y)\,,\\
\sin_{\nu}(x \oplus_{\nu} y) = \sin_{\nu}(x)\cos _{\nu}(y) +
\sin_{\nu}(x)\cos _{\nu}(y)\,.
\label{sin_addition}
\end{gather}
In particular,
\begin{gather}
\cos_{\nu}(2 \odot_{\nu} x) = \cos_{\nu}^2(x) -\sin_{\nu}^2(x)\,,\\
\sin_{\nu}(2 \odot_{\nu} x) = 2\sin_{\nu}(x)\cos _{\nu}(x)\,.
\end{gather}
In terms of hypergeometric functions, \eqref{sin_addition} yields the
identity
\begin{equation}
\begin{aligned}
&_{0}F_{5}\Big(1/2, 1/6, 2/3, \nu/2 + 1/3, \nu/2 + 5/6\,|\,
 -\big(\frac{x \oplus_{\nu} y}{6}\big)^6\Big)   \\
& = _{0}F_{5}\Big(1/2, 1/6, 2/3, \nu/2 + 1/3, \nu/2 + 5/6\,|\,
-(\frac{x}{6})^6\Big) \\
 & \quad\times{} _{0}F_{5}\Big(1/2, 1/6, 2/3,
\nu/2 + 1/3, \nu/2 + 5/6\,|\, -(\frac{y}{6})^6\Big)  \\
  & \quad -\frac{(xy)^6}{9^2\,(\nu + 2/3)}\,_{0}F_{5}\Big(
  3/2, 2/3, 7/6, \nu/2 + 5/6, \nu/2 + 4/3\,|\,
 -(\frac{x}{6})^6\Big) \\
  & \quad\times{} _{0}F_{5}\Big(3/2, 2/3, 7/6,
 \nu/2 + 5/6, \nu/2 + 4/3\,|\, -(\frac{y}{6})^6\Big)\,.
\end{aligned}
\end{equation}
In the case $\nu = 0$, the equations are
\begin{gather}
\sin_{0}(x) = -\frac{1}{3}\sin x + \frac{2}{3}\sin \frac{x}{2}\cosh
\frac{\sqrt{3}}{2}x\,,\\
\cos_{0}(x) = -\frac{1}{3}\cos x + \frac{2}{3}\cos \frac{x}{2}\cosh
\frac{\sqrt{3}}{2}x\,.
\end{gather}
The $\nu$-derivatives are
\begin{gather}
\vartheta_{\nu}\sin_{\nu}(ax) = a^3\cos _{\nu}(ax)\,,
\label{sin_deriv}\\
\vartheta_{\nu}\cos _{\nu}(ax) = -a^3\sin_{\nu}(ax)\,.
\label{cos_deriv}
\end{gather}
Thus $\sin_{\nu}(ax)$ and $\cos_{\nu}(ax)$ are solutions of the differential
equation
\[\vartheta_{\nu}\,y(x) + a^6\,y(x) = 0\,.
\]
Complex variable type addition equations are
\begin{gather*}
\cos_{\nu}(x \oplus_{\nu} \omega_{6}\,y) = \cos_{\nu}(x)cosh_{\nu}(y)
-i\sin_{\nu}^2(x)\sinh_{\nu}(y)\,,\\
\sin_{\nu}(x \oplus_{\nu} \omega_{6}\,y) = \sin_{\nu}(x)\cosh_{\nu}(y)
-i\cos _{\nu}(x)\sinh_{\nu}(y)
\end{gather*}
which are analogues of the complex variable formulas for $\sin(z)$ and $\cos(z)$
with $z = x + i\,y$.

\begin{theorem}[A $\nu$-DeMoivre's Formula] \label{thm6}
Let $\nu >0$ and $n$ be a positive integer. Then
\[
(\cos_{\nu}(x) + i\sin_{\nu}(x))^n = \cos_{\nu}(n \odot_{\nu} x) +
i\sin_{\nu}(n \odot_{\nu} x)
\]
\end{theorem}

\begin{proof} We have $E_{\nu}(x)^n = E_{\nu}(n \odot_{\nu}
x)$ and the equation follows.\end{proof}

Since $E_{\nu}(\gamma_{u}z)\,E_{\nu}(z) = G_{\nu}(\gamma_{u}\,\omega_{6}\,z)\,
G_{\nu}(\omega_{6}\,z) = 1$, it follows that
\begin{gather*}
\cos_{\nu}(z)\cos _{\nu}(\gamma_{u}z) - \sin_{\nu}(z)\sin_{\nu}(\gamma_{u}z)=1\,,\\
\sin_{\nu}(z)\cos _{\nu}(\gamma_{u}z) + \sin_{\nu}(\gamma_{u}z)\cos _{\nu}(z)
=0\,.
\end{gather*}
In the case that $\nu = 0$,
\[E_{0}(z) = \frac{1}{3}\,(e^{-iz} + 2\,e^{\frac{i\,z}{2}}\cosh(\frac{\sqrt{3}}
{2}\,z)\,.
\]
Obviously, the functions $G_{0}(x)$ and $E_{0}(x)$ are not periodic in the ordinary
sense. We will obtain umbral periods associated with the $\nu$-translation.

\begin{theorem} \label{thm7}
 For $\nu \geq 0$, the function ${\mathcal{G}}_{\nu}$ takes on the values +1
and -1 for infinitely many real values of $x$.
\end{theorem}

\begin{proof} The theorem easily follows from the asymptotic expansion of
${\mathcal{G}}_{\nu}$. We have
\begin{equation}
\begin{aligned}
{\mathcal{G}}_{\nu}(x) & =  _{0}F_{2}\left(1/3, \nu +
2/3\,|(-\frac{x}{3})^3 \right) \\
 & \sim  \frac{\Gamma(1/3)\Gamma(\nu + 2/3)}{\pi\,\sqrt{3}}\,
 (\frac{x}{3})^{-\nu}\,e^{x/2}\cos \big(\frac{\sqrt{3}}{2}x -
 \frac{\nu\pi}{2}\big)
\end{aligned}
\end{equation}
as $x \to \infty$, see Marichev \cite{Mari83}, for example. Hence the theorem
follows. \end{proof}

Since the zeros of an entire function cannot have a finite limit point, there is a
smallest $x_{2\pi}(\nu) > 0$ such that ${\mathcal{G}}_{\nu}(x_{2\pi}(\nu)) = 1$ and,
likewise, a smallest $x_{\pi}(\nu) > 0$ such that ${\mathcal{G}}_{\nu}(x_{\pi}(\nu)) = -1$.
Next we let $P_{2\pi}(\nu) = -x_{2\pi}(\nu)\,e^{-\frac{i\,\pi}{6}}e^{\frac{2\pi\,i}{3}}
\dots = -x_{2\pi}(\nu)\,i$, then
\[E_{\nu}(P_{2\pi}(\nu)) = G_{\nu}(-x_{2\pi}(\nu)\,e^{-\frac{i\,\pi}{6}}\,\omega_{6})
={\mathcal{G}}_{\nu}(x_{2\pi}(\nu)) = 1\,.
\]
Hence
\[
E_{\nu}(z \oplus_{\nu} P_{2\pi}(\nu))  =  E_{\nu}(z)\,E_{\nu}(P_{2\pi}(\nu))
=  E_{\nu}(z)
\]
and therefore $P_{2\pi}(\nu)$ is an umbral period of $E_{\nu}$.
In the case $\nu = 0$, $P_{2\pi}(0)$ is approximately -5.549831$i$.

By Theorem \ref{1st_result}, we obtain Poisson type representations for the trigonometric
functions of this section. For $\nu \geq 1$, we have
\begin{equation}
\label{Poisson} E_{\nu}(x) = \frac{\Gamma(\nu +
2/3)}{\Gamma(2/3)\,\Gamma(\nu)}\, \int_{0}^{1}\tau\,(1 -
\tau^3)^{\nu - 1}\,(e^{-ix\,\tau} + 2e^{\frac{ix\,\tau}{2}}
cosh\,\frac{\sqrt{3}}{2}x\,\tau)\,d\tau\,,
\end{equation}
\begin{gather*}
\cosh_{\nu}(x) = \frac{\Gamma(\nu + 2/3)}{\Gamma(2/3)\,\Gamma(\nu)}\,
\int_{0}^{1}\tau\,(1 - \tau^3)^{\nu - 1}\,(\cosh(x\,\tau) +
2\cos (\frac{\sqrt{3}}{2}\,
x\,\tau)\cosh(\frac{x\,\tau}{2})\,d\tau\,,\\
\sinh_{\nu}(x) = \frac{\Gamma(\nu + 2/3)}{\Gamma(2/3)\,\Gamma(\nu)}\,
\int_{0}^{1}\tau\,(1 - \tau^3)^{\nu - 1}(\sinh(x\,\tau) -
2\cos (\frac{\sqrt{3}}{2}\,
x\,\tau)\sinh(\frac{x\,\tau}{2})\,d\tau\,,\\
\cos_{\nu}(x) = \frac{\Gamma(\nu + 2/3)}{\Gamma(2/3)\,\Gamma(\nu)}\,
\int_{0}^{1}\tau\,(1 - \tau^3)^{\nu - 1}\,(\cos(x\,\tau) +
2\cos (\frac{\sqrt{3}}{2}\,
x\,\tau)\cos h(\frac{x\,\tau}{2})\,d\tau\,,\\
\sin_{\nu}(x) = \frac{\Gamma(\nu + 2/3)}{\Gamma(2/3)\,\Gamma(\nu)}\,
\int_{0}^{1}\tau\,(1 - \tau^3)^{\nu - 1}(-sin(x\,\tau) +
2\sin(\frac{\sqrt{3}}{2}
x\tau)\cosh(\frac{x\,\tau}{2})\,d\tau\,.
\end{gather*}
 From these equations, one can obtain order estimates for the individual functions,
for example
\[
E_{\nu}(x) = O(e^{\sqrt{3}x/2}) \quad \mbox{as } x \to \infty
\]
by \eqref{Poisson}.

In the following example we need to know the smallest real zero of
$\cos_{\nu}(2x)$. We have
\begin{align*}
\cos_{\nu}(2x) & = _{0}F_{5}\big(1/2, 1/6, 2/3, \nu/2 + 1/3, \nu/2 + 5/6\,|\,
-(\frac x3)^6\big) \\
 & \sim  \frac{\Gamma(1/2, 1/6, 2/3, \nu/2 + 1/3, \nu/2 + 5/6)}{\sqrt{6}\,(2\pi)^{5/2}}\,
 2\,(\frac{x}{2})^{-3}\,e^{\sqrt{3}x}\cos (x - \frac{\nu\pi}{6})
\end{align*}
as $x \to \infty$, where
\[
\Gamma(1/2, 1/6, 2/3, \nu/2 + 1/3, \nu/2 + 5/6) =
\Gamma(1/2)\,\Gamma(1/6)\,\Gamma(2/3)\,\Gamma(\frac{\nu}{2} + 1/3)\,
\Gamma(\frac{\nu}{2} + 5/6)\,.
\]
It follows from the asymptotic expansion that $\cos_{\nu}(2x)$ has infinitely
many real zeros. Since $\cos_{\nu}(z)$ is entire and $\cos_{\nu}(0) = 1$,
it follows that there is a smallest positive value $x_{0}$
such that $\cos_{\nu}(2x_{0}) = 0$.


Finally, we present a nontrivial application of the $\nu$-cosine function. Consider the
sixth order diffusion equation given by
\begin{equation}
\vartheta_{\nu}^2\,u(x, t) = \frac{\partial}{\partial t}\,u(x, t)\,.
\label{2nd_order_diff}
\end{equation}
Using \eqref{sin_deriv} and \eqref{cos_deriv}, it follows that the function
\[u(x, t) = e^{-t}\cos _{\nu}(x) - e^{-64t}\cos _{\nu}(2x)
\]
is a solution of the previous equations such that $u(x, 0) = \cos_{\nu}(x) -
\cos_{\nu}(2x)$ and $u_{t}(x, 0) = -\cos_{\nu}(x) + 64\cos _{\nu}(2x)$,
both being entire functions.

Let $x_{0}$ be the first positive root of $cos_{\nu}(2x)$, then for $-x_{0} < x < x_{0}$,
we have that $u(x, t) = 0$ on the curve
\[t = \frac{1}{63}\log \frac{cos_{\nu}(2x)}{cos_{\nu}(x)}
\]
Thus the uniqueness of solutions for the diffusion equation \eqref{2nd_order_diff}
 fails.

\section{$\nu$-Complex Functions}

The trigonometric identities of Section \ref{trig} involving the
$\cos_{\nu}(x \oplus_{\nu}\omega_{0}\,y)$ and
$\sin_{\nu}(x \oplus_{\nu} \omega_{0}\,y)$ suggest the possibility of
an analytic function theory associated with the $\vartheta_{\nu}$ operator. The
rudiments of such a theory is obtained in this section.

The operator limit
\begin{equation}
\frac{G_{\nu}(h\,\vartheta^{1/3}) -
1}{\frac{h^3}{\alpha_{3}(\nu)}} \; \to \; \vartheta_{x}
\end{equation}
gives a second way to compute $\vartheta_{x}\,f(x)$ for suitable
$f(x)$. We know that $\vartheta_{x}x^{3n} = 3^3\,n(n + \nu -
1/3)(n - 2/3)x^{3(n - 1)} = d_{3n}(\nu)\, x^{3(n - 1)}$.
Consider the quotient
\begin{align*}
\frac{(x \oplus_{\nu} h)^{3n} - x^{3n}}{\frac{h^3}{\alpha_{3}}} &
=  \sum_{k = 1}^n\alpha_{3}\, \binom{\alpha_{3n}}{\alpha_{3k}}
x^{3(n - k)}\,h^{3(k -1)} \\
 & \to  \alpha_{3}\, \binom{\alpha_{3n}}{\alpha_{3}}
x^{3(n - 1)}  =  d_{3n}(\nu)x^{3(n - 1)}
\end{align*}
as $h \to 0$. Thus
\begin{equation}
\lim_{h \to 0}\frac{(x \oplus_{\nu} h)^{3n} -
x^{3n}}{\frac{h^3}{\alpha_{3}}} = d_{3n}(\nu)x^{3(n - 1)}\,.
\end{equation}
Hence we define
\begin{equation}
\vartheta_{x}\,f(x) = \lim_{h \to 0}\, \frac{f(x
\oplus_{\nu} h) - f(x)} {\frac{h^3}{\alpha_{3}}} \label{vartheta}
\end{equation}
provided the limit exists for suitable functions. It is clear that
 \eqref{vartheta} is valid for polynomials and entire functions
of the variable $x^3$.

Next we introduce an umbral complex variable
$z_{\nu} \simeq (x \oplus_{\nu} \omega_{6} y)$,
where $\omega_{6} = e^{i \pi/6}$. We define
\[z_{\nu}^{3n} = (x \oplus_{\nu} \omega_{6} y)^{3n}\,.
\]
Further, we let $\Delta z_{\nu} \simeq (\Delta x \oplus_{\nu} \omega_{6} \Delta y)$, and we
define the $\nu$-derivative of a suitable function $f$ at a $z_{\nu}$ by
\begin{equation}
\begin{aligned}
\vartheta_{x}\,f(z_{\nu}) & =  \lim_{\Delta z \to 0} \frac{f(x \oplus_{\nu}
\omega_{6} y) \oplus_{\nu}(\Delta x \oplus_{\nu} \omega_{6} \Delta y)) -
f(x \oplus_{\nu} \omega_{6} y)}
{\frac{(\Delta x \oplus_{\nu} \omega_{6} \Delta y)^3}{\alpha_{3}(\nu)}} \\
 & =  \lim_{\Delta z \to 0 \mbox{ or }
 \Delta x, \Delta y \to 0}\, \frac{f(z_{\nu} \oplus_{\nu} \Delta z_{\nu}) -
 f(z_{\nu})}
{\frac{(\Delta z_{\nu})^3}{\alpha_{3}(\nu)}} \label{derivative}
\end{aligned}
\end{equation}
If the function $f$ has a $\nu$-derivative in a neighborhood of a point $(x, y)$ associated
with $z_{\nu}$, we say that $f$ is $\nu$-analytic at $z_{\nu}$.

\subsection*{Example}
Consider $f(z_{\nu}) = z_{\nu}^{3n} = (x \oplus_{\nu}
\omega_{6} y)^{3n}$. Then
\[
\frac{f(z_{\nu} \oplus_{\nu} \Delta z_{\nu}) - f(z_{\nu})}
{\frac{(\Delta z_{\nu})^3}{\alpha_{3}(\nu)}}  =  \sum_{k = 1}^n
\alpha_{3}\binom{\alpha_{3n}}{\alpha_{3k}}
z_{\nu}^{3(n - k)}\,(\Delta z_{\nu})^{3(k - 1)}
 \to  \alpha_{3n}(\nu)\,z_{\nu}^{3(n - 1)},
\]
as $\Delta x. \Delta y \to 0$\,.
Thus
\[\vartheta_{z}\,z_{\nu}^{3n} = d_{3n}(\nu)\,z_{\nu}^{3(n - 1)}
\]
It follows that polynomials $p(z_{\nu}) = \sum_{k =
0}^na_{k}\,z_{\nu}^{3k}$ and 3-parity analytic functions lead to
$\nu$-holomorphic functions. Moreover, if $f$ has a
$\nu$-derivative at $z_{\nu}$ then $f$ is continuous at $z_{\nu}$,
i.~e., $\lim_{\Delta z_{\nu} \to 0}f(z_{\nu} \oplus_{\nu}
\Delta z_{\nu}) = f(z_{\nu})$.

The complex $\nu$-derivative is also a linear operator. If $f$ and $g$ have
$\nu$-derivatives at $z_{\nu}$ and $a$ and $b$ are arbitrary complex numbers then
\[
\vartheta_{z}\,(af(z_{\nu}) + bg(z_{\nu})) = a\,\vartheta_{z}\,f(z_{\nu}) +
b\,\vartheta_{z}\,g(z_{\nu})
\]

\begin{theorem}[$\nu$-Cauchy-Riemann Equations] \label{thm8}
Let $f(z_{\nu}) = u(z_{\nu}) + iv(z_{\nu}) =
u(x, y) + iv(x, y)$ with $f$ differentiable at $z_{\nu}$ and let $u$ and $v$ be real
valued functions, then
\begin{equation}
\label{eq122}
\vartheta_{z}\,f(z_{\nu}) = \vartheta_{x}\,f(z_{\nu}) = -i\vartheta_{y}\,f(z_{\nu}),
\quad\mbox{complex form},
\end{equation}
or
\[
\vartheta_{x}\,u = \vartheta_{y}\,v \quad\mbox{and}\quad
\vartheta_{x}\,v = -\vartheta_{y}\,u, \quad\mbox{real form}
\]
\end{theorem}

\begin{proof}
Taking $\Delta y = 0$ in \eqref{derivative}, we get
\[\vartheta_{z}\,f(z_{\nu}) = \lim_{\Delta x \to 0} \frac{f(z_{\nu}
\oplus_{\nu} x) - f(z_{\nu})}{\frac{(\Delta x)^3}{\alpha_{3}}} =
\vartheta_{x}\,f(z_{\nu})\] Likewise, taking $\Delta x = 0$ in
\eqref{derivative}, we get
\[
\vartheta_{z}\,f(z_{\nu}) = \frac{1}{i} \lim_{\Delta y \to 0}
\frac{f(z_{\nu} \oplus_{\nu} \omega_{6}y) -
f(z_{\nu})}{\frac{(\Delta y)^3}{\alpha_{3}}} =
\frac{1}{i}\,\vartheta_{y}\, f(z_{\nu})
\]
Therefore, the complex Equations \eqref{eq122} hold. Clearly the $\nu$-derivatives
$\vartheta_{x}\,u$, $\vartheta_{x}\,v$, $\vartheta_{y}\,u$, and
$\vartheta_{y}\,v$ exist at $(x, y)$. Hence we find that
\begin{align*}
\vartheta_{z}\,f(z_{\nu})
 & =  \vartheta_{x}\,u(x, y) + i\vartheta_{x}\,v(x, y) \\
 & =  -i\vartheta_{y}\,f(z_{\nu}) \\
 & =  -i\vartheta_{y}\,u(x, y) + \vartheta_{y}\,v(x, y)\,.
\end{align*}
Consequently, $\vartheta_{x}\,u = \vartheta_{y}\,v$ and $\vartheta_{x}\,v =
-\vartheta_{y}\,u$.
\end{proof}

\begin{corollary} \label{cor2}
If $f$ has a second $\nu$-derivative at $z_{\nu}$, then
\begin{equation}
\vartheta_{x}^2\,f(z_{\nu}) + \vartheta_{y}^2\,f(z_{\nu}) = 0
\label{Laplace}
\end{equation}
\end{corollary}

\begin{proof} From the $\nu$-Cauchy-Riemann equations, it
follows that $\vartheta_{x}^2\,f(z_{\nu}) =
\vartheta_{z}^2\,f(z_{\nu})$ and $-\vartheta_{z}^2\,f(z_{\nu}) =
\vartheta_{y}^2\,f(z_{\nu})$. Thus
\[\vartheta_{x}^2\,f(z_{\nu}) + \vartheta_{y}^2\,f(z_{\nu}) =
\vartheta_{z}^2\,f(z_{\nu}) - \vartheta_{z}^2\,f(z_{\nu}) = 0\]
\end{proof}

 From \eqref{Laplace}, it follows that $\vartheta_{x}^2\,u(x, y) +
\vartheta_{y}^2\,u(x, y) = 0$ and $\vartheta_{x}^2\,vu(x, y) +
\vartheta_{y}^2\,v(x, y) = 0$. The operator $\Box_{\nu} =
\vartheta_{x}^2 + \vartheta_{y}^2$ is called the $\nu$-Laplacian.
Functions satisfying the $\nu$-Laplace equation $\Box_{\nu}\,u(x,
y) = 0$ in a domain are said to be $\nu$-harmonic in that domain.

\subsection*{Example}
 Consider $z_{\nu}^{3n} = (x \oplus_{\nu} \omega_{6} y)^{3n}$. Then
\begin{align*}
z_{\nu}^{3n} & =  \sum_{k = 0}^n \binom{\alpha_{3n}}{\alpha_{3k}}
  x^{3(n - k)}\,(\omega_{6} y)^{3k} \\
& =  \sum_{k = 0}^n \binom{\alpha_{3n}}{\alpha_{3k}}
  i^{k}\,y^{3k}x^{3(n - k)} \\
& =  \sum_{\ell = 0}^{[\frac{n}{2}]} \binom{\alpha_{3n}}{\alpha_{6\ell}}
 (-1)^{\ell}x^{3(n - 2\ell)}y^{6\ell}
+ i\,\sum_{\ell = 0}^{[\frac{n - 1}{2}]} \binom{\alpha_{3n}}{\alpha_{6\ell + 3}}
(-1)^{\ell}x^{3(n - 2\ell - 1)}y^{6\ell + 3} \\
 & = u_{3n}(x, y) + i\,v_{3n}(x, y)\,.
\end{align*}
Then $u_{3n}(x, y)$ and $v_{3n}(x, y$ are $\nu$-harmonic functions for all $x$, $y$. In
the particular case $n = 2$, we have
\[z_{\nu}^6 = x^6 - y^6 + i \binom{\alpha_{6}}{\alpha_{3}}
x^3\,y^3 = u_{6}(x, y) + i\,v_{6}(x, y)
\]
and
\[
\vartheta_{x}\,u_{6} = d_{6}x^3, \quad
\vartheta_{y}\,u_{6} = - d_{6}y^3\]
\[\vartheta_{x}\,v_{6} = d_{6}y^3, \quad\mbox{and \hspace{.15in}}
\vartheta_{y}\,v_{6} = d_{6}x^3 \quad\mbox{since }
\binom{\alpha_{6}}{\alpha_{3}}
 = \frac{d_{6}}{\alpha_{3}}
\]
Hence
\begin{gather*}
\vartheta_{x}^2\,u_{6} + \vartheta_{y}^2\,u_{6}
= \alpha_{6}(\nu) - \alpha_{6}(\nu)= 0\,\\
\vartheta_{x}^2\,v_{6} + \vartheta_{y}^2\,v_{6} = 0 + 0 = 0
\end{gather*}

\subsection*{Example}
It is clear that the Humbert function ${\mathcal{G}}_{\nu}(z_{\nu})$ is
$\nu$-analytic for all $z_{\nu}$. Further, we have
\[
{\mathcal{G}}_{\nu}(z_{\nu}) = {\mathcal{G}}_{\nu}(x \oplus_{\nu} \omega_{6} y)
 = {\mathcal{G}}_{\nu}(x){\mathcal{G}}_{\nu}(\omega_{6}y)
 =  {\mathcal{G}}_{\nu}(x)(cos_{\nu}(y) + i\sin_{\nu}(y))
\]
the $\nu$-analogue of Euler's equation. Therefore,
\[
{\mathcal{G}}_{\nu}(z_{\nu}) = {\mathcal{G}}_{\nu}(x)\cos _{\nu}(y) + i\,
{\mathcal{G}}_{\nu}(x)\sin_{\nu}(y) =u + iv
\]
and
\begin{gather*}
\vartheta_{x}u = {\mathcal{G}}_{\nu}(x)\cos _{\nu}(y), \quad
\vartheta_{y}v = - {\mathcal{G}}_{\nu}(x)\cos _{\nu}(y), \\
\vartheta_{xx}u = {\mathcal{G}}_{\nu}(x)\cos _{\nu}(y), \quad
\vartheta_{yy}v = - {\mathcal{G}}_{\nu}(x)\cos _{\nu}(y), \\
\vartheta_{y}u = - {\mathcal{G}}_{\nu}(x)\sin_{\nu}(y), \quad
\vartheta_{x}v = {\mathcal{G}}_{\nu}(x)\sin_{\nu}(y), \\
\vartheta_{yy}u = - {\mathcal{G}}_{\nu}(x)\cos _{\nu}(y),\quad
\vartheta_{xx}v = {\mathcal{G}}_{\nu}(x)\sin_{\nu}(y)\,.
\end{gather*}
Thus the real form of the $\nu$-Cauchy-Riemann equations hold and
${\mathcal{G}}_{\nu}(x)\cos _{\nu}(y)$ and ${\mathcal{G}}_{\nu}(x)\sin_{\nu}(y)$ are
$\nu$-harmonic functions.

One can verify that the following functions (1) $E_{\nu}(z_{\nu})$, (2)
$\cos_{\nu}(z_{\nu})$,
(3) $\sin_{\nu}(z_{\nu})$, (4) $\cosh_{\nu}(z_{\nu})$, and (5)
$\sinh_{\nu}(z_{\nu})$ are $\nu$-analytic functions for all $z_{\nu}$.

The generalized diffusion equation
\[
\Box_{\nu}\,u(x, y, t) = \frac{\partial}{\partial t}\,u(x, y, t)
\]
has a solution
$u(x, y, t) = \exp(-2\kappa^6t)\cos _{\nu}(\kappa x)\cos_{\nu}(\kappa y) +$
an arbitrary $\nu$-harmonic function. With suitable coefficients $\{a_{n,m}\}$
for uniform convergence,
the generalized wave equation
\[\Box_{\nu}\,u(x, y, t) = \vartheta_{t}^2\,u(x, y, t)
\]
has solutions
\[u(x, y, t) = \sum_{n,m = 0}^{\infty} a_{n,m}\sin_{\nu}(nx)\sin_{\nu}(my)\,
\cos(\lambda_{n, m}t) + h_{\nu}(x, y)
\]
where $\lambda_{n, m} =(n^2 + m^6)^{1/2}$ and $h_{\nu}(x, y)$ is a $\nu$-harmonic
function.
Also for suitable coefficients $\{a_{n,m}\}$,
the generalized wave equation
\begin{equation}
\label{eq2} \Box_{\nu}\,u(x, y, t) = \frac{\partial^2}{\partial
t^2}\,u(x, y, t)
\end{equation}
has solutions
\[
u(x, y, t) = \sum_{n,m = 0}^{\infty} a_{n,m}\sin_{\nu}(nx)\sin_{\nu}(my)\,
cos(\lambda_{n, m}t) + h_{\nu}(x, y)
\]
where $\lambda_{n, m} =(n^2 + m^6)^{1/2}$ and $h_{\nu}(x, y)$ is a $\nu$-harmonic
function. Two other forms for real solutions of \eqref{eq2}
are given by
\begin{gather*}
u(x, y, t) = \sum_{n,m = 0}^{\infty} a_{n,m}\cos _{\nu}(nx)\cos _{\nu}(my)
\cos(\lambda_{n, m}t) + h_{\nu}(x, y)\,\\
u(x, y, t) = \sum_{n,m = 0}^{\infty} a_{n,m}\sin_{\nu}(nx)\cos _{\nu}(my)
\cos(\lambda_{n, m}t) + h_{\nu}(x, y)\,.
\end{gather*}
Certainly the solution forms are valid for compact support sequences $\{a_{n,m}\}$.
Finally, we note that $\vartheta_{x}^2$ is the sixth order
operator given by
\[
\vartheta_{x}^2 = D^6 + \frac{6\nu}{x}\,D^{5} + \frac{3\nu(3\nu - 5)}{x^2}\,D^{4} +
\frac{36\nu(1 -\nu)}{x^3}\,D^3 + \frac{72\nu(\nu - 1)}{x^{4}}\,D^2
- \frac{72\nu(\nu - 1)}{x^{5}}\,D\] where $D =
\frac{\partial}{\partial x}$. Thus the techniques introduced in
this paper can solve some exceedingly difficult partial
differential equations.

\section{$\nu$-Diffusion Polynomials}\label{section7}

The simple set of polynomial solutions in ${\mathcal{H}}_{\nu}$ are
associated with the initial condition $u(x, 0) = x^{3n}$. In a
certain sense they are generalized Hermite polynomials. The
function $e^{tz^3}\,G_{\nu}(xz)$ is a $\nu$-diffusion. Multiplying
and rearranging the infinite series, we get a sequence of
polynomials given by
\begin{equation}
e^{tz^3}\,G_{\nu}(xz) = \sum_{n = 0}^{\infty} \frac{p_{n}^{\nu}(x,
t)\,z^{3n}} {\alpha_{3n}(\nu)}\,. \label{exponential}
\end{equation}
Thus
\begin{equation}
\begin{aligned}
p_{n}^{\nu}(x, t) & =  \sum_{k = 0}^n
\frac{\alpha_{3n}(\nu)}{\alpha_{3(n - k)}(\nu)}\,
\frac{x^{3(n - k)}t^{k}}{k!}  \\
 & =  (1/3)_{n}(\nu + 2/3)_{n}\,3^{3n}t^n\sum_{k = 0}^{\infty}
 \frac{(-1)^{k}(-n)_{k}}{(1/3)(\nu + 2/3)_{k}}\,
 \frac{x^{3k}t^{-k}}{k!\,3^{3k}}
\end{aligned} \label{polynomial}
\end{equation}
Therefore, the 3rd order polynomials $p_{n}^{\nu}(x, t)$ can be written as
hypergeometric functions given by
\begin{equation}
\begin{aligned}
p_{n}^{\nu}(x, t) & =  (1/3)_{n}(\nu +
2/3)_{n}\,3^{3n}t^n\,_{1}F_{2}\left[
\begin{matrix}
- n \\
1/3, \nu + 2/3
\end{matrix} \bigg| - \frac{x^3}{27t} \right]  \\
 & =  \frac{\alpha_{3n}(\nu)}{n!}t^n\,_{1}F_{2}\left[
\begin{matrix}
- n \\
1/3, \nu + 2/3
\end{matrix} \bigg| - \frac{x^3}{27t} \right]
\end{aligned}\label{polys}
\end{equation}
It is easy to show that $e^{t\vartheta_{\nu}}x^{3n} = p_{n}^{\nu}(x, t)$.
Thus the
$p_{n}^{\nu}(x, t)$ are in ${\mathcal{H}}_{\nu}$ and $p_{n}^{\nu}(x, 0) = x^{3n}$

For the ordinary diffusion equation $u_{xx} = u_{t}$, Widder
\cite{Widd75} has established a series expansion of solutions in
terms of heat polynomials which are essentially modified Hermite
polynomials. Cholewinski and Haimo \cite{CholH68} have presented a
similar development associated with Bessel functions and the Euler
operator $\Delta_{x} = D_{x}^2 + \frac{2\nu}{x}\,D_{x}$.
Expansions in terms of the $\nu$-diffusion polynomials
$p_{n}^{\nu}(x, t)$ does not yield as rich a theory as that of
Widder.

\begin{theorem} \label{thm9}
Let $\nu \geq 0$ and let $s$ and $t$ be arbitrary complex numbers, then
\begin{gather}
\vartheta_{x}\,p_{n}^{\nu}(x, t)  =  3^3n(n + \nu - 1/3)(n -
2/3)p_{n - 1}^{\nu}(x, t)
 =  d_{3n}(\nu)\,p_{n - 1}^{\nu}(x, t)\,,\\
p_{n}^{\nu}(x \oplus_{\nu} y, t) = \sum_{k = 0}^n
\binom{\alpha_{3n}}{\alpha_{3k}} p_{k}^{\nu}(x, t)\,y^{3(n - k)}\,,\\
p_{n}^{\nu}(x \oplus_{\nu} y, t + s) = \sum_{k = 0}^n
\binom{\alpha_{3n}}{\alpha_{3k}} p_{k}^{\nu}(x, t)\,p_{n - k}^{\nu}(x, s)\,,\\
(x \oplus_{\nu} y)^{3n} = \sum_{k = 0}^n
\binom{\alpha_{3n}}{\alpha_{3k}} p_{k}^{\nu}(x, t)\,p_{n - k}^{\nu}(y, - t)\,,\\
x^{3n} = \sum_{k = 0}^n
\frac{\alpha_{3n}}{\alpha_{3k}}\,\frac{(-1)^{n - k}} {(n -
k)!}t^{n - k}\,p_{k}^{\nu}(x, t) \label{x3n}\,,\\
p_{n}^{\nu}(p_{u/3}^{\nu}(x, s), t) = p_{n}^{\nu}(x, s + t), \quad\mbox{Huygens'
property}
\end{gather}
\end{theorem}

\begin{proof}
Since $\vartheta_{\nu}$ commutes with $e^{t\vartheta_{\nu}}$, we get
\begin{align*}
\vartheta_{\nu}\,p_{n}^{\nu}(x, t)
 & =  \vartheta_{\nu}\,e^{t\vartheta_{\nu}}x^{3n} \\
 & =  e^{t\vartheta_{\nu}}\,\vartheta_{\nu}x^{3n} \\
 & =  d_{3n}(\nu)\,e^{t\vartheta_{\nu}}x^{3(n - 1)} \\
 & =  d_{3n}(\nu)\,p_{n}^{\nu}(x, t)\,.
\end{align*}
Thus the $\{p_{n}^{\nu}(x, t)\}_{0}^{\infty}$ is a basic sequence for the delta operator
$\vartheta_{\nu}$. From \eqref{exponential}, we obtain by Cauchy multiplication
\begin{align*}
G_{\nu}(xz)
& =  \sum_{n = 0}^{\infty}\frac{p_{n}^{\nu}(x, t)}{\alpha_{3n}(\nu)}\,
z^{3n}\,\sum_{n = 0}^{\infty}\frac{(-1)^nt^nz^{3n}}{n!} \\
 & =  \sum_{n = 0}^{\infty}\frac{z^{3n}}{\alpha_{3n}(\nu)}\,
 \sum_{k = 0}^n\frac{\alpha_{3n}(\nu)}{\alpha_{3k}(\nu)}\,
 \frac{(-1)^{n - k}}{(n - k)!}t^{n - k}p_{k}^{\nu}(x, t) \\
 & =  \sum_{n = 0}^{\infty}\frac{x^{3n}z^{3n}}{\alpha_{3n}(\nu)}
\end{align*}
Therefore, comparing coefficients, it follows that
\[
x^{3n} = \sum_{k = 0}^n\frac{\alpha_{3n}(\nu)}{\alpha_{3k}(\nu)}\,
 \frac{(-1)^{n - k}}{(n - k)!}t^{n - k}p_{k}^{\nu}(x, t) \]
 which is \eqref{x3n}. Using \eqref{polynomial}, we get the equivalent
 composition property
 \[x^{3n} = p_{n}^{\nu}(p_{u/3}^{\nu}(x, t), -t)\,.
 \]
Since $G_{\nu}(xz)G_{\nu}(zy) = G_{\nu}(z(x \oplus_{\nu} y)))$
 we get
 \begin{align*}
 e^{tz^3}G_{\nu}(z(x \oplus_{\nu} y)))
 & =  e^{tz^3} G_{\nu}(xz)G_{\nu}(zy) \\
 & =  \sum_{n = 0}^{\infty}\frac{p_{n}^{\nu}(x \oplus_{\nu} y, t)}{\alpha_{3n}}\,
   z^{3n} \\
 & =  \sum_{n = 0}^{\infty} \frac{(yz)^{3n}}{\alpha_{3n}}\,
  \sum_{n = 0}^{\infty}\frac{p_{n}^{\nu}(x, t)}{\alpha_{3n}}\, z^{3n} \\
 & =  \sum_{n = 0}^{\infty}\frac{z^{3n}}{\alpha_{3n}}\,
  \sum_{k = 0}^{\infty} \binom{\alpha_{3n}}{\alpha_{3k}}
  p_{k}^{\nu}(x, t)\,y^{3(n - k)}\,.
 \end{align*}
Comparing coefficients, we get the generalized binomial property
\[
p_{n}^{\nu}(x \oplus_{\nu} y,t)  =  \sum_{k = 0}^{\infty}
\binom{\alpha_{3n}}{\alpha_{3k}}
p_{k}^{\nu}(x, t)\,y^{3(n - k)}
 =  (p_{u/3}^{\nu}(x, t) \oplus_{\nu} y)^{3n}
\]
where $\nu$ is an umbral variable, in this equation
\[
p_{u/3}^{\nu}(x, t)^{3k} = p_{3k/3}^{\nu}(x, t) = p_{k}^{\nu}(x, t)\,.
\]
In the same manner, we also obtain the Huygens type property
\begin{align*}
p_{n}^{\nu}(x \oplus_{\nu} y,t + s)
&=  \sum_{k = 0}^{\infty} \binom{\alpha_{3n}}{\alpha_{3k}}
p_{k}^{\nu}(x, t)\,p_{n - k}^{\nu}(y, s) \\
 & =  (p_{u/3}^{\nu}(x, t) \oplus_{\nu} p_{u/3}^{\nu}(y, s))^{3n}
\end{align*}
Since $p_{n}^{\nu}(x, 0) = x^{3n}$, we get
\[(x \oplus_{\nu} y)^{3n} = \sum_{k = 0}^{\infty}
 \binom{\alpha_{3n}}{\alpha_{3k}}
p_{k}^{\nu}(x, t)\,p_{n - k}^{\nu}(y, - t)
\]
Finally,
\begin{align*}
e^{(s + t)z^3}\,G_{\nu}(xz)
& =  e^{tz^3}\,\sum_{n = 0}^{\infty} \frac{p_{n}^{\nu}(x, s)}{\alpha_{3n}}\,z^{3n} \\
& =  \sum_{n = 0}^{\infty} \frac{t^nz^{3n}}{n!}\,\sum_{n = 0}^{\infty}
 \frac{p_{n}^{\nu}(x, s)}{\alpha_{3n}}\,z^{3n} \\
& =  \sum_{n = 0}^{\infty} \frac{z^{3n}}{\alpha_{3n}}\,\sum_{k = 0}^n
 \frac{\alpha_{3n}}{\alpha_{3k}}\,\frac{t^{n - k}}{(n - k)!}\,p_{k}^{\nu}(x, s) \\
& =  \sum_{n = 0}^{\infty}\frac{p_{n}^{\nu}(p_{\nu/3}^{\nu}(x, s),t)}{\alpha_{3n}}\,
 z^{3n} \\
& =  \sum_{n = 0}^{\infty} \frac{p_{n}^{\nu}(x, s + t)}{\alpha_{3n}}\,z^{3n}
\end{align*}
and, upon comparing coefficients, we get
\[p_{n}^{\nu}(p_{u/3}^{\nu}(x, s), t) = p_{n}^{\nu}(x, s + t)\]
\end{proof}

Next we introduce a class of generalized Laguerre polynomials associated with the
$\vartheta_{\nu}$ operator. For the purposes of this paper we are interested in the
$\nu$-diffusion counterparts.
We define the $n$th $\nu$-Laguerre polynomial by
\[
{\mathcal{L}}_{3n}^{\alpha,\nu}(x) = {\mathcal{L}}_{3n}^{\alpha}(x) =
(-1)^n(1 - \vartheta_{\nu})^{\alpha + n - 1}x^{3n},
\quad n = 0, 1, \dots
\]
Expanding the binomial, we get the representation
\begin{equation}
{\mathcal{L}}_{3n}^{\alpha}(x) = \sum_{k = 0}^n
 \binom{\alpha + n - 1}{n - k} \frac{\alpha_{3n}(\nu)}{\alpha_{3k}(\nu)}\,
(-1)^{k}x^{3k}
\label{Laguerre}
\end{equation}
Next we have
\begin{align*}
\frac{\vartheta_{x}}{\vartheta_{x} -1}\,{\mathcal{L}}_{3n}^{\alpha}(x)
& = (-1)^n\,\frac{\vartheta_{\nu}}{\vartheta_{\nu} - 1}\,(1 -
\vartheta_{\nu})^{\alpha
+n - 1}x^{3n} \\
& =  (-1)^{n + 1}d_{3n}(\nu)(1 - \vartheta_{\nu})^{\alpha
+ (n - 1) -1}x^{3(n - 1)} \\
& =  d_{3n}{\mathcal{L}}_{3(n - 1)}^{\alpha}(x)\,.
\end{align*}
Thus the sequence $\{{\mathcal{L}}_{3n}^{\alpha}(x)\}$ is the basic sequence
of 3-parity polynomials associated with the Laguerre type delta operator
$\frac{\vartheta_{x}}{\vartheta_{x} - 1}$. The $\{{\mathcal{L}}_{3n}^{\alpha}(x)$'s are
hypergeometric functions. From \eqref{Laguerre}, we have
\begin{align*}
{\mathcal{L}}_{3n}^{\alpha}(x)
& =  \sum_{k = 0}^n
\frac{(-1)^{k}\Gamma(\alpha + n)}{\Gamma(\alpha + k)}\,
\frac{\alpha_{3n}}{\alpha_{3k}}\, \frac{x^{3k}}{(n - k)!} \\
 & =  \frac{(\alpha)_{n}\alpha_{3n}(\nu)}{n!}\,\sum_{k = 0}^n
 \frac{(-n)_{k}x^{3k}}{3^{3k}\,k!\,(1/3)_{k}(\nu + 2/3)_{k}\,(\alpha)_{k}} \\
 & =  \frac{(\alpha)_{n}\,\alpha_{3n}(\nu)}{n!}\,_{1}F_{3}\,
 \left [ \begin{matrix}
 -n \\
 1/3, \nu + 2/3, \alpha
 \end{matrix} \bigg| \frac{x^3}{27} \right]
\end{align*}
In the case that $\alpha = 0$, we get
\[{\mathcal{L}}_{3n}(x) = \frac{\alpha_{3n}(\nu)}{n!}\,_{1}F_{3}\,
 \left [ \begin{matrix}
 -n \\
 1/3, \nu + 2/3, n
 \end{matrix} \bigg| \frac{x^3}{27} \right]
 \]
 Consider the series expansion
 \begin{align*}
 \frac{1}{(1 - z^3)^{\alpha}}\,G_{\nu}(x\,\frac{z}{(z^3 - 1)^{1/3}}) & =
 \sum_{k = 0}^{\infty}\frac{(-1)^{k}x^{3k}}{\alpha_{3k}}\,
 \frac{z^{3k}}{(1 - z^3)^{k + 1}} \\
  & =  \sum_{k = 0}^{\infty}\frac{(-1)^{k}x^{3k}z^{3k}}{\alpha_{3k}}\,
  \sum_{m = 0}^{\infty}\frac{(\alpha + k)_{m}}{m!}\,z^{3m} \\
  & =  \sum_{k = 0}^{\infty}\sum_{m = 0}^{\infty}\frac{(-1)^{k}x^{3k}(\alpha + k)_{m}
  z^{3(m + k)}}{\alpha_{3k}\,m!}\,.
 \end{align*}
 Since the resulting double series is absolutely convergent for $|z| < 1$, we may
 collect terms in $z^{3n}$ and we get
 \begin{equation}
 \begin{aligned}
 \frac{1}{(1 - z^3)^{\alpha}}\,G_{\nu}(x\,\frac{z}{(z^3 - 1)^{1/3}})
& =  \sum_{n = 0}^{\infty}\frac{z^{3n}}{\alpha_{3n}}\,
 \sum_{k = 0}{n}(-1)^{k}\binom{\alpha + n - 1 }{n - k}
\frac{\alpha_{3n}(\nu)}{\alpha_{3k}(\nu)}\, x^{3k}\\
 & =  \sum_{n = 0}^{\infty}\frac{{\mathcal{L}}_{3n}^{\alpha}(x)}{\alpha_{3n}}
 \,z^{3n}\,.
\end{aligned} \label{dblseries}
 \end{equation}
 From the generating function \eqref{dblseries}, it follows that
 \[
 {\mathcal{L}}_{3n}^{\alpha + \beta}(x \oplus_{\nu} y) =
 \sum_{k = 0}^n\binom{\alpha_{3n}}{\alpha_{3k}}
{\mathcal{L}}_{3k}^{\alpha}(x){\mathcal{L}}_{3(n - k)}^{\beta}(y)
\]
 It is also a consequence of the idempotent composition $\frac{\vartheta}{(\vartheta - 1)}
 \circ \frac{\vartheta}{(\vartheta - 1)} = \vartheta$ that
 \[{\mathcal{L}}_{3n}^{\alpha}({\mathcal{L}}_{\mu}^{\alpha}(x)) = x^{3n}
\]

 In quantum mechanics the associated Laguerre polynomials $L_{n}^{k}(x) = \frac{\partial^{k}}
 {\partial x^{k}}\,L_{n}(x)$ appear in various wave functions. We introduce $\nu$-associated
 Laguerre polynomials, given by
 \[\vartheta_{x}^n\,{\mathcal{L}}_{3k}^{\alpha}(x)\]

 \begin{theorem} \label{thm10}
 We have
 \[\vartheta_{x}^n\,{\mathcal{L}}_{3k}^{\alpha}(x) =
 (-1)^n\,\frac{\alpha_{3k}}{\alpha_{3(k - n)}}\,{\mathcal{L}}_{3(k - n)}^{\alpha + n}(x)\]
 \end{theorem}

\begin{proof} Consider
 \begin{align*}
 \vartheta_{x}^n \frac{1}{(1 - z)^{\alpha}}\,G_{\nu}(x\,\frac
  {z}{(z^3 - 1)^{1/3}})
& =  (-1)^n\,\frac{z^{3n}}{(1 - z^3)^{\alpha + n}}\,
  G_{\nu}(x\,\frac{z}{(z^3 - 1)^{1/3}}) \\
& =  \sum_{m = 0}^{\infty}(-1)^n\frac{{\mathcal{L}}_{3m}^{\alpha + n}(x)}
  {\alpha_{3m}}\,z^{3(m + n)} \\
& =  \sum_{k = n}^{\infty}(-1)^n\,\frac{\alpha_{3n}}{\alpha_{3(k - n)}}\,
  {\mathcal{L}}_{3(k - n)}^{\alpha + n}(x)\,\frac{z^{3k}}{\alpha_{3m}} \\
& =  \sum_{k = n}^{\infty}\frac{\vartheta_{x}^n\,{\mathcal{L}}_{3k}^{\alpha}(x)}
  {\alpha_{3k}}\,z^{3k}\,.
\end{align*}
 This theorem follows by comparing coefficients.
 By commutativity and the binomial theorem, it follows that
 \[{\mathcal{L}}_{3(k - n)}^{\alpha + n}(x) = \sum_{\ell = 0}^n
 \binom{n}{\ell}
\frac{\alpha_{3(k - n)}}{\alpha_{3(k - \ell - n)}}\,
 {\mathcal{L}}_{3(k - \ell - n)}^{\alpha + \ell}(x)\,.
\]
\end{proof}
 The polynomials $e^{t\vartheta_{x}}\,{\mathcal{L}}_{3n}^{\alpha}(x) =
 {\mathcal{L}}_{3n}^{\alpha}(x, t)$ are $\nu$-diffusion polynomial solutions of the initial
 value problem $\vartheta_{x}\,u(x, t) = \frac{\partial}{\partial t}\,u(x, t)$ with
 $u(x, 0) = {\mathcal{L}}_{3n}^{\alpha}(x, t)$. A simple calculation gives the composition
 equation ${\mathcal{L}}_{3n}^{\alpha}(x, t) = {\mathcal{L}}_{3n}^{\alpha}(p_{u/3}^{\nu}(x, t))$.
 The generating function of the sequence $\{{\mathcal{L}}_{3n}^{\alpha}\}_{0}^{\infty}$ is
 given by
\begin{equation}
\frac{1}{(1 - z^3)^{\alpha}}\,e^{t\,\frac{z^3}{z^3 - 1}}\,
G_{\nu}(x\,\frac{z}{(z^3 -1)^{1/3}}) = \sum_{n = 0}^{\infty}
\frac{{\mathcal{L}}_{3n}^{\alpha}(x, t)}{\alpha_{3n}}\,z^{3n}
\label{genfunc}
\end{equation}
  Moreover, a calculation using commutativity yields
\[
\frac{\vartheta_{x}}{\vartheta_{x} -1}\,{\mathcal{L}}_{3n}^{\alpha}(x, t) = d_{3n}(\nu)
  {\mathcal{L}}_{3(n - 1)}^{\alpha}(x, t)
\]
  Multiplying \eqref{genfunc} by $G_{\nu}(y\,\frac{z}{(z^3 - 1)^{1/3}})$
  yields the addition formula
  \[
  {\mathcal{L}}_{3n}^{\alpha}(x \oplus_{\nu} y, t) = \sum_{k = 0}^n
  \binom{\alpha_{3n}}{\alpha_{3k}}
  {\mathcal{L}}_{3k}^{\alpha}(x, t){\mathcal{L}}_{3(n - k)}^{\alpha}(y)
\]
Likewise, suitable multiplication of generating functions of the type
\eqref{genfunc}
gives the Huygens equation
\[
{\mathcal{L}}_{3n}^{\alpha}(x \oplus_{\nu} y, t + s) = \sum_{k = 0}^{\infty}
\binom{\alpha_{3n}}{\alpha_{3k}}
{\mathcal{L}}_{ek}^{\alpha}(x, t){\mathcal{L}}_{3(n - k)}^{\alpha}(y, s)
\]
which is a $\nu$-diffusion polynomial.

If we multiply the power series in $z^3$, we have upon
rearrangement
\[G_{\nu}(-z^2)G_{\nu}(xz) = \sum_{n = 0}^{\infty}
\frac{{\mathcal{H}}_{3n}^{\nu}(x)}{\alpha_{3n}}\,z^{3n}\,,
\]
where
\[
{\mathcal{H}}_{3n}^{\nu}(x) = \sum_{k = 0}^{[n/2]}
\frac{\alpha_{n}\,(-1)^{k}}{\alpha_{3k}\alpha_{3(n - 2k)}}x^{3(n - 2k)}\,.
\]
The ${\mathcal{H}}_{3n}^{\nu}(x)$'s are called $\nu$-Hermite polynomials. A simple
calculation shows that
\[
\vartheta_{\nu}\,{\mathcal{H}}_{3n}^{\nu}(x)
= d_{3n}(\nu){\mathcal{H}}_{3(n - 1)}^{\nu}(x)
\]
thus the ${\mathcal{H}}_{3n}^{\nu}(x)$'s form a basic set of polynomials associated with
$\vartheta_{\nu}$. Further, we have
\begin{gather*}
{\mathcal{H}}_{3n}^{\nu}(-x) = (-1)^n{\mathcal{H}}_{3n}^{\nu}(x)\,,\\
{\mathcal{H}}_{6\ell}^{\nu}(0) = (-1)^{\ell}\,\frac{\alpha_{6\ell}}{\alpha_{3\ell}},
\quad {\mathcal{H}}_{6\ell + 3}^{\nu}(0) = 0
\end{gather*}
A manipulation of the Pockhammer functions yields the identity
\begin{align*}
&\alpha_{3(n - 2k)} =\\
& \frac{\alpha_{3n}\,3^{-6k}}
{2^{6k}(-n/2)_{k}(-n/2 + 1/2)_{k}(1/3 - n/2)_{k}(2/3 - n/2)_{k}(1/6 - \frac{\nu + n}{2})_{k}
(1/6 - \frac{\nu + n - 1}{2})_{k}}
\end{align*}
Therefore, the $\nu$-Hermite polynomial can be represented as a hypergeometric function,
namely,
\begin{align*}
&{\mathcal{H}}_{3n}^{\nu}(x)=\\
& x^{3n}\,_{6}F_{2}\left [\begin{matrix}
 -n/2, n/2 + 1/2, 1/3 - n/2, 2/3 - n/2, 1/6 - \frac{\nu + n}{2}, 1/6 -
 \frac{\nu + n - 1}{2} \\
 1/3, \nu + 2/3 \end{matrix}
 \bigg| - \frac{12^3}{x^6}\right].
\end{align*}
Corresponding to the initial value problem
$\vartheta_{\nu}\,u(x, t) = u_{t}(x, t)$ with
$u(x, 0) = {\mathcal{H}}_{3n}^{\nu}(x)$ we get the polynomial solutions with the
generating functions
\[
e^{tz^3}\,G_{\nu}(-z^2)G_{\nu}(xz) = \sum_{n = 0}^{\infty}
\frac{{\mathcal{H}}_{3n}^{\nu}(x, t)}{\alpha_{3n}}\,z^{3n}
\]
It follows that $\vartheta_{\nu}\,{\mathcal{H}}_{3n}^{\nu}(x, t) = d_{3n}(\nu)
{\mathcal{H}}_{3(n - 1)}^{\nu}(x, t)$ and
\[{\mathcal{H}}_{3n}^{\nu}(x, t) = \sum_{k = 0}^{[n/2]}
\frac{\alpha_{3n}\,(-1)^{k}}{\alpha_{3k}\alpha_{3(n - 2k)}}\,
p_{n - 2k}^{\nu}(x, t)\]
The ${\mathcal{H}}_{3n}^{\nu}(x, t)$'s are called the $\nu$-Hermite diffusion
polynomials.

Various forms of the $\nu$-Hermite diffusion polynomials have appeared in the work of
Bateman \cite{Bate36} and Langer \cite{Lang55}. Bateman's polynomials are given by
\[
J_{n}^{\nu, \sigma}(x) =
 \frac{\Gamma((1/2)\nu + \sigma + n + 1)}{\Gamma(\sigma + (1/2)\nu + 1)n!}\,
 \frac{x^{\nu}}{\Gamma(\nu + 1)}\,_{1}F_{2}\left[ \begin{matrix}
 -n \\
 \nu + 1, (1/2)\nu + \sigma + 1 \end{matrix} \bigg| x^2 \right ]\,.
\]
Setting $\nu = -2/3$ and then $\sigma := \nu$, we get
\[
J_{n}^{-2/3, \mu}(x) =
\frac{(\nu + 2/3)_{n}}{n!\,\Gamma(1/3)}\,
 x^{-2/3}\,_{1}F_{2}\left[ \begin{matrix}  -n \\
 1/3, \nu + 2/3 \end{matrix} \bigg| x^2 \right ]\,.
\]
Letting $x^2 := z^3/(27t)$ and using \eqref{polys}, we
obtain
\begin{align*}
J_{n}^{-2/3, \mu}(\frac{z^{3/2}}{\sqrt{27t}})
& = \frac{(\nu + 2/3)_{n}}{n!\,\Gamma(1/3)}\,
 \frac{3t^{1/3}}{z}\,_{1}F_{2}\left[ \begin{matrix}  -n \\
 1/3, \nu + 2/3 \end{matrix} \bigg| \frac{z^3}{27t} \right ]\\
 & =  \frac{(-1)^n}{3^{3n}z}\,\frac{3t^{1/3}t^{-n}}{\Gamma(n + 1/3)\,n!}\,
 p_{n}^{\nu}(z, -t)\,.
\end{align*}
Thus
\[
p_{n}^{\nu}(z, -t) = (-1)^n\,3^{3n - 1}\,\Gamma(n + 1/3)\,n!t^{n - 1/3}\,z\,
J_{n}^{-2/3, \mu}\big(\sqrt{\frac{z^3}{27t}}\big)
\]
Bateman's generating function \cite[p.~574]{Bate36} is given by
\[
\sum_{n = 0}^{\infty}J_{n}^{(\nu, \sigma)}(x)t^n = t^{-\nu/2}\,
(1 - t)^{- \sigma - 1}\,J_{\nu}\big(\frac{2x\sqrt{t}}{\sqrt{1 - t}}\big),
\quad|t| < 1\,.
\]
Substituting ,we get the generating function
\[
\sum_{n = 0}^{\infty}\frac{(-1)^nt^{-n}p_{n}^{\nu}(z, -t)y^n}
{3^{3n}\Gamma(n + 1/3)\,n!} = \frac{(y/t)^{1/3}z}{3(1 - y)^{\nu + 1}}\,
J_{-2/3}\big(\frac{2z^{3/2}\sqrt{y}}{\sqrt{27t(1 - y)}}\big),
\]
$|y| < 1$ and $t > 0$ which is a source solution type generating function.
In terms of the source solutions
this is written as
\[
\sum_{n = 0}^{\infty}\frac{(-1)^nt^{-n}p_{n}^{\nu}(z, -t)y^n}
{3^{3n}(1/3)_{n}\,n!} = \big(\frac{3t}{y}\big)^{\nu + 2/3}\,
{\mathcal{K}}_{\nu}\big(x,
t\,\frac{1 - y}{y}\big)
\]
A recurrence formula for the functions $f_{n}(x) = _{1}F_{2}(-n; 1 + \alpha,
1 + \beta; \gamma)$ which are modified Bateman polynomials is given by
\begin{align*}
(\alpha + n)(\beta + n)f_{n}(z) - (3n^2 - 3n + 1 + (2n -
1)(\alpha + \beta) +
\alpha\beta - z)f_{n - 1}(z) & \\
 + (n - 1)(3n - 3 + \alpha + \beta)f_{n - 2} - (n - 1)(n - 2)f_{n - 3}(z) &= 0\,;
\end{align*}
see Rainville \cite{Rain45}. Letting $\alpha = -2/3$, $\beta = \nu
- 1/3$, and $z = - x^3/(27t)$, we obtain the recurrence relation
\begin{align*}
&p_{n}^{\nu}(x, t)\\
& = [3n^2 - 3n + 1 + (2n - 1)(\nu - 1) -(2/3)(\nu - 1/3) +
x^3/(27t)]27tp_{n - 1}^{\nu}(x, t) \\
 &\quad  - (n - 1)(3n - 3 + \nu - 1)(n - 5/3)(n + \nu
 - 4/3)(27t)^2p_{n - 2}^{\nu}(x, t) \\
 &\quad  -(n - 1)(n - 2)(n - 5/3)(n - 8/3)(\nu + n - 4/3)(\nu + n - 7/3)(27t)^3
 p_{n - 3}^{\nu}(x, t)
\end{align*}
with $p_{k}^{\nu}(x, t) = 0$, whenever the subscript $k$ is negative.

The first four $\nu$-diffusion polynomials are
\begin{gather*}
p_{0}^{\nu}(x, t) = 1, \quad
p_{1}^{\nu}(x, t) = x^3 + 9(\nu + 2/3)t\,,\\
p_{2}^{\nu}(x, t) = x^6 + 72(\nu + 5/3)x^3t + 324(\nu + 2/3)(\nu + 5/3)t^2\,,\\
\begin{aligned}
p_{3}^{\nu}(x, t) =& x^{9} + 63(3\nu + 8)x^6t + 3756(3\nu + 8)(3\nu + 5)x^3t^2 \\
&+756(3\nu + 2)(3\nu + 5)(3\nu + 8)t^2
\end{aligned}
\end{gather*}

\section{$\nu$-Airy Bernoulli Polynomials and Associated Airy Diffusions}

Corresponding to the classical Bernoulli polynomials we obtain $\nu$-Airy Bernoulli
and their associated $\nu$-diffusion polynomials. The properties of these polynomials
can be established using the Rota operator calculus \cite{Rota75}. However, the
basic properties used in this section follow by manipulation of infinite series.

We define the $\nu$-Airy Bernoulli polynomials by the generating function
\begin{equation}\label{Bernoulli1}
\frac{t^{3n}}{[G_{\nu}(t) - 1]^n}\,G_{\nu}(xt) = \sum_{k =
0}^{\infty}\frac{{\mathcal{B}}_{3k}^{n,\nu}(x)}
{\alpha_{3k}(\nu)}t^{3k}
\end{equation}
for $n = 0,\pm 1,\pm 2,\dots $ . For $n = 0$, we get ${\mathcal{B}}_{3k}^{0,\nu}(x) = x^{3k}$.
Setting $x = 0$, we obtain the $\nu$-Airy Bernoulli numbers given by
\begin{equation}\label{Bernoulli2}
\frac{t^{3n}}{[G_{\nu}(t) - 1]^n}\,G_{\nu}(xt) = \sum_{k =
0}^{\infty}\frac{{\mathcal{B}}_{3k}^{n,\nu}(0)}
{\alpha_{3k}(\nu)}t^{3k}\,.
\end{equation}
Let $\delta_{\nu} = G_{\nu}(\vartheta_{x}^{1/3}) - 1$ be a $\nu$-difference operator. In a suitable
Rota calculus, we have
\begin{equation}\label{Bernoulli3}
{\mathcal{B}}_{3k}^{n,\nu}(x) = \left [
\frac{\vartheta_{x}}{\delta_{\nu}} \right ]^nx^{3k}\,.
\end{equation}
This is easy to verify directly. Furthermore, from Equations \ref{Bernoulli1} and \ref{Bernoulli2}, we
get the $\nu$-binomial representation
\begin{equation}\label{Bernoulli4}
{\mathcal{B}}_{3k}^{n,\nu}(x)  =  \sum_{\ell = 0}^{k}
\binom{\alpha_{3k}}{\alpha_{3\ell}}
{\mathcal{B}}_{3(k - \ell)}^{n,\nu}(x^{3\ell})
  =  ({\mathcal{B}}_{\nu}^{n,\nu} \oplus_{\nu} x)^{3k}
\end{equation}
using umbral notation. It also follows from \eqref{Bernoulli1} that
\begin{equation}\label{Bernoulli5}
\begin{aligned}
\vartheta_{x}\,{\mathcal{B}}_{3k}^{n,\nu}(x)
& =  3^3k(k + \nu -1/3)(k - 2/3) {\mathcal{B}}_{3(k - 1)}^{n,\nu}(x)  \\
& =  d_{3k}(\nu){\mathcal{B}}_{3(k - 1)}^{n,\nu}(x)\,.
\end{aligned}
\end{equation}
Using the binomial theorem, we get
\begin{equation}\label{Bernoulli6}
\begin{aligned}
\delta_{\nu}^n\,G_{\nu}(xt)
& =  [G_{\nu}(\vartheta_{x}^{1/3}) - 1]^n\,G(xt)  \\
& = \sum_{k = 0}^n \binom{n}{k}
 G_{\nu}(\vartheta_{x}^{1/3})^{k}(-1)^{n - k} G_{nu}(xt)  \\
& =  \sum_{k = 0}^n (-1)^{n - k} \binom{n}{k}
 G_{\nu}(t)^{k}G_{nu}(xt)  \\
& =  [G_{\nu}(t) - 1]^n\,G_{\nu}(xt)\,.
\end{aligned}
\end{equation}
Applying $\delta_{\nu}$ to \eqref{Bernoulli1}, we get
\begin{equation}\label{Bernoulli7}
\begin{aligned}
\frac{t^{3n}}{[G_{\nu}(t) - 1]^n}\,G_{\nu}(xt)
& =  \sum_{k = 0}^{\infty}\frac{\delta_{\nu}{\mathcal{B}}_{3k}^{n,\nu}(x)}
{\alpha_{3k}(\nu)}t^{3k}  \\
 & =  \sum_{k = 0}^{\infty}\frac{{\mathcal{B}}_{3k}^{n - 1,\nu}(x)}
 {\alpha_{3k}(\nu)}t^{3(k + 1)} \\
 & =  \sum_{k = 0}^{\infty}d_{3k}(\nu)\frac{{\mathcal{B}}_{3k}^{n - 1,\nu}(x)}
 {\alpha_{3k}(\nu)}t^{3k}\,.
\end{aligned}
\end{equation}
Comparing coefficients, we get
\begin{equation}\label{Bernoulli8}
\delta_{\nu}{\mathcal{B}}_{3k}^{n,\nu}(x)
 =  {\mathcal{B}}_{3k}^{n,\nu}(x \oplus_{\nu} 1) -
{\mathcal{B}}_{3k}^{n,\nu}(x)  \\
 =  d_{3k}(\nu){\mathcal{B}}_{3(k - 1)}^{n - 1,\nu}(x)\,.
\end{equation}
This is an analogue of the basic difference equation for the classical Bernoulli
polynomials.

The equation
\begin{equation}\label{Bernoulli9}
\frac{t^{3(n + m)}}{[G_{\nu}(t) - 1]^{n + m}}\,G_{\nu}(t(x \oplus_{\nu} y))
 =  \sum_{k = 0}^{\infty}
\frac{{\mathcal{B}}_{3k}^{n + m,\nu}(x \oplus_{\nu} y)}{\alpha_{3k}(\nu)}t^{3k}
\end{equation}
yields the identity
\begin{equation}\label{Bernoulli10}
\begin{aligned}
{\mathcal{B}}_{3k}^{n + m,\nu}(x \oplus_{\nu} y)
& =  \sum_{\ell = 0}^{k}\binom{\alpha_{3k}}{\alpha_{3\ell}}
{\mathcal{B}}_{3\ell}^{n,\nu}(x)
{\mathcal{B}}_{3(k - \ell)}^{m,\nu}(y) \\
 & =  ({\mathcal{B}}_{\nu}^{n,\nu}(x) \oplus_{\nu}
 {\mathcal{B}}_{\nu}^{m,\nu}(y))^{3k}
\end{aligned}
\end{equation}
a binomial of  Rota type. Taking $m = -n$, we get the hypergeometric result
\begin{equation}\label{Bernoulli11}
(x \oplus_{\nu} y)^{3k} = \sum_{\ell = 0}^{k}
\binom{\alpha_{3k}}{\alpha_{3\ell}}
{\mathcal{B}}_{3\ell}^{n,\nu}(x)
{\mathcal{B}}_{3(k - \ell)}^{m,\nu}(y)
\end{equation}
Next we obtain solutions of the $\nu$-Airy heat equation corresponding to the
initial value polynomials ${\mathcal{B}}_{3k}^{n,\nu}(x)$. They are given by
\begin{equation}\label{Bernoulli12}
\begin{aligned}
e^{t\vartheta_{\nu}}{\mathcal{B}}_{3k}^{n,\nu}(x)
& =  {\mathcal{B}}_{3k}^{n,\nu}(x, t) \\
& =  \sum_{\ell = 0}^{k}\frac{t^{\ell}}{\ell !}\frac{\alpha_{3k}(\nu)}
{\alpha_{3(k - 1)}(\nu)} {\mathcal{B}}_{3(k - \ell)}^{n,\nu}(x) \\
& =  p_{3k}^{\nu}({\mathcal{B}}_{\nu}^{n,\nu}(x),t)
\end{aligned}
\end{equation}
where the $p_{ek}^{\nu}(x,t)$ are the $\nu$-diffusion polynomials of
Section \ref{section7}.
The $\nu$-Bernoulli heat polynomials are also given by the generating functions
\begin{equation}\label{Bernoulli13}
\frac{t^{3n}}{[G_{\nu}(t) - 1]^n}\,e^{ty^3}G_{\nu}(xy) = \sum_{k =
0}^{\infty}\frac{{\mathcal{B}}_{3k}
^{n,\nu}(x,t)}{\alpha_{3k}(\nu)}\,y^{3k}\,.
\end{equation}
The generalized Huygens property for the $\nu$-Bernoulli heat polynomials 
follows from \eqref{Bernoulli13}, adn we get
\begin{equation}\label{Bernoulli14}
\begin{aligned}
{\mathcal{B}}_{3k}^{n + m,\nu}(x \oplus_{\nu} y,t + s)
& =  \sum_{\ell = 0}^{k} \binom{\alpha_{3k}}{\alpha_{3\ell}}
{\mathcal{B}}_{3\ell}^{n,\nu}(x,t)
{\mathcal{B}}_{3(k - \ell)}^{m,\nu}(y,s) \\
& = ({\mathcal{B}}_{\nu}^{n,\nu}(x,t) \oplus_{\nu}{\mathcal{B}}_{\nu}
^{m,\nu}(y,s))^{3k}
\end{aligned}
\end{equation}
Taking $m = -n$, we get the relation
\begin{equation}\label{Bernoulli15}
p_{3k}^{\nu}(x \oplus_{\nu} y,t + s) =
\sum_{\ell = 0}^{k} \binom{\alpha_{3k}}{\alpha_{3\ell}}
{\mathcal{B}}_{3\ell}^{n,\nu}(x,t)
{\mathcal{B}}_{3(k - \ell)}^{-n,\nu}(y,s)\,.
\end{equation}
Letting $s = -t$, it follows that
\begin{equation}\label{Bernoulli16}
(x \oplus_{\nu} y)^{3k} = \sum_{\ell = 0}^{k}
\binom{\alpha_{3k}}{\alpha_{3\ell}}
{\mathcal{B}}_{3\ell}^{n,\nu}(x,t)
{\mathcal{B}}_{3(k - \ell)}^{-n,\nu}(y,-t)\,.
\end{equation}
Since $\delta_{\nu}$ and $e^{t\vartheta_{x}}$ commute, it follows from \eqref{Bernoulli8} that
\begin{equation}\label{Bernoulli17}
\delta_{\nu}^n{\mathcal{B}}_{3k}^{n,\nu}(x,t) = d_{3k}(\nu)
{\mathcal{B}}_{3(k - 1)}^{n - 1,\nu}(x,t)\,.
\end{equation}
For $k \geq n \geq 0$, it follows that
\begin{equation}\label{Bernoulli18}
\delta_{\nu}^n{\mathcal{B}}_{3k}^{n,\nu}(x,t) =
\frac{\alpha_{3k}(\nu)} {\alpha_{3(k - n)(\nu)}}\,p_{3(k -
n)}^{\nu}(x,t)\,.
\end{equation}
The $\nu$-Bernoulli heat polynomials can be expressed in terms of $\nu$-diffusion
polynomials. From \eqref{Bernoulli13} or \ref{Bernoulli14}, we get
\begin{equation}\label{Bernoulli19}
{\mathcal{B}}_{3k}^{n,\nu}(x,t) = \sum_{\ell = 0}^{k}
\binom{\alpha_{3k}}{\alpha_{3\ell}}
{\mathcal{B}}_{3(k - \ell)}^{n,\nu}(x,t)
p_{3\ell}^{\nu}(x,t)
\end{equation}
which expresses ${\mathcal{B}}_{3k}^{n,\nu}(x,t)$ as a sum of hypergeometric functions.

In order to calculate ${\mathcal{B}}_{3k}^{n,\nu}(x,t)$, we have to be able to calculate
the $\nu$-Airy Bernoulli numbers. The elementary relation
\begin{equation}\label{Bernoulli20}
\frac{t^{3n}}{[G_{\nu}(t) - 1]^n}\,\frac{t^{-3n}} {[G_{\nu}(t) -
1]^{-n}} = 1
\end{equation}
implies that
\begin{equation}\label{Bernoulli21}
({\mathcal{B}}_{\nu}^{n,\nu}(x,t) \oplus_{\nu}{\mathcal{B}}_{\nu}
^{m,\nu}(y,s))^{3k} = \delta_{0,k},k = 0,1,\dots
\end{equation}
where $\delta_{0,k}$ is the Kronecker delta function. In multinomial form
\eqref{Bernoulli21} can also be written as
\begin{equation}\label{Bernoulli22}
\sum_{\underline{\ell} = (\ell_{0},\ell_{1},\dots,\ell_{n}),
\underline{\ell} \in N^{n + 1},|\underline{\ell}| = k}
\binom{\alpha_{3k}}{\alpha_{3\underline{\ell}}}
{\mathcal{B}}_{3\ell_{0}}^{-1,\nu}
{\mathcal{B}}_{3\ell_{1}}^{-1,\nu}\cdots
{\mathcal{B}}_{3\ell_{n - 1}}^{n,\nu} = \delta_{0,k}\,.
\end{equation}
Since ${\mathcal{B}}_{3k}^{-1,\nu}(0) = d_{3k}(\nu)$, \eqref{Bernoulli22}
yields the recursion formula
\begin{equation}\label{Bernoulli23}
{\mathcal{B}}_{3k}^{n,\nu} = -\,\frac{\alpha_{3}^n(\nu)\alpha_{3k}}
{\alpha_{3(k + n)}}\,\sum \binom{\alpha_{3(k + n)}}{\alpha_{3\underline{\ell}}}
{\mathcal{B}}_{3\ell_{n}}^{n,\nu}\,,
\end{equation}
where the summation is taken over $|\underline{\ell}| = k + n$,
$0 \leq \ell_{n} < k$, $\ell_{i} > 0$, $i = 0,1,\dots,n -1$,
$\underline{\ell} = (\ell_{0},\ell_{1},\dots,\ell_{n})
\in N_{+}^{n + 1}$.

\section{A Primitive Integral for $\vartheta_{\nu}$}

Associated with the differential operator $\vartheta_{\nu}$, we introduce a formal
or primitive indefinite integral that commutes  with $\vartheta_{\nu}$ up to constants.
Recall that $d_{\alpha}(\nu) = \alpha(\alpha + 3\nu -1)(\alpha -2)$ and $d_{\alpha + 3}(\nu)
= (\alpha + 3)(\alpha + 3\nu + 2)(\alpha + 1)$. We define
\begin{equation}
\begin{aligned}
\int x^{\alpha}\,\partial_{\nu}(x)
& =  \frac{x^{\alpha +3}}{d_{\alpha + 3}(\nu)} + c_{1}x^2 + c_{2}x^{1 - 3\nu} +
c_{3},\quad\mbox{if }\alpha \neq -1, -3,
-(3\nu + 2)  \\
 & =  \frac{x^{\alpha + 3}}{(\alpha + 1)(\alpha + 3)^2} + c_{1}x^2 + c_{2}\ln x +
 c_{3}, \quad\mbox{for }\nu = 1/3
\end{aligned} \label{constants}
\end{equation}
\begin{align*}
\int x^{-3}\,\partial_{\nu}(x) & =  \frac{lnx}{2(1 - 3\nu)} +
c_{1}x^2 + c_{2}x^{1 - 3\nu} + c_{3},\quad \nu \neq 1/3 \\
 & =  -\frac{1}{4}\,(lnx)^2  + c_{2}x^2 + c_{2}\,lnx + c_{3},
 \quad\mbox{for }\nu = 1/3
\end{align*}
\begin{align*}
\int x^{-1}\,\partial_{\nu}(x)
& =  \frac{x^2\,lnx}{2(3\nu +
1)} + c_{1}x^2 +
c_{2}x^{1 - 3\nu} + c_{3},\quad \nu \neq -1/3 \\
 & =  -\frac{1}{4}x^2\,lnx + c_{1}x + c_{2}\,lnx + c_{3},\quad
 \nu = 1/3
\end{align*}
and
\[
\int x^{-(3(n + 2)}\,\partial_{\nu}(x) = \frac{x^{1 - 3\nu}}{(1 +
3\nu)(3\nu - 1)} + c_{1}x^2 + c_{2}x^{1 - 3\nu} + c_{3},
\quad \nu \neq 1/3\,.
\]
In each of the above, we have $\vartheta_{\nu}\,\int
x^{\alpha}\,\partial_{\nu}(x) = x^{\alpha}$. The indefinite
integral is extended to 3-parity polynomials and formal power
series in $x^3$. In which case we take $c_{1}$ and $c_{2}$ to be
zero. We are primarily interested in the case of $\alpha$ a
nonnegative integer divisible by three. If $p(x) = \sum_{k = 0}^n
a_{k}x^{3k}$, then a simple calculation shows that
\[
p(x) + c = \int \vartheta_{x}\,p(x)\,\partial_{\nu}(x) = \vartheta_{x}\,\int
p(x)\,\partial_{\nu}(x) + c
\]
In general, as in elementary calculus, if $\vartheta\,f(x) = F(x)$, we take
$\int F(x)\,\partial_{\nu}(x) = f(x) + c$.

The functions $G_{\nu}(x)$ and ${\mathcal{G}}_{\nu}(x)$ play the roles of $e^{x}$ and
$e^{-x}$ for the primitive indefinite integral.

\begin{theorem} \label{thm11}
For $\nu \geq 0$, we have
\begin{gather*}
\int G_{\nu}(x)\,\partial_{\nu}(x) = G_{\nu}(x) + c\,,\\
\int {\mathcal{G}}_{\nu}(x)\,\partial_{\nu}(x) = -{\mathcal{G}}_{\nu}(x) + c\,.
\end{gather*}
\end{theorem}

\begin{proof}
We have $\vartheta_{\nu}\,G_{\nu}(x) = G_{\nu}(x)$, or working with the
power,
\[
\int \sum_{n = 0}^{\infty}\frac{x^{3n}}{\alpha_{3n}}\,\partial_{\nu}(x)
 = \sum_{n = 0}^{\infty}\frac{x^{3n + 3}}{\alpha_{3n}d_{3(n + 1)}} + c \\
 = \sum_{n = 0}^{\infty}\frac{x^{3n}}{\alpha_{3n}} + c \\
 = G_{\nu}(x) + c
\]
since $d_{3(n + 1)}\alpha_{3n} = \alpha_{3(n + 1)}(\nu)$.
\end{proof}

\noindent $\nu$-Primitive Integrals:
\begin{gather*}
\int \cosh_{\nu}(x)\,\partial_{\nu}(x) = \sinh_{\nu}(x) + c\,,\\
\int \sinh_{\nu}(x)\,\partial_{\nu}(x) = \cosh_{\nu}(x) + c\,, \\
\int E_{\nu}(x)\,\partial_{\nu}(x) = -i\,E_{\nu}(x) + c\,, \\
\int \sin_{\nu}(x)\,\partial_{\nu}(x) = -\ cos_{\nu}(x) + c\,, \\
\int \cos_{\nu}(x)\,\partial_{\nu}(x) = \sin_{\nu}(x) + c \,,\\
\int p_{n}^{\nu}{\nu}^(x, t)\,\partial_{\nu}(x) = \frac{p_{n + 1}^{\nu}(x, t)}
d_{3(n + 1)} + c\,,\\
\int {\mathcal{L}}(x)_{3(k - n)}^{\alpha + n}\,\partial_{\nu}(x) =
(-1)^n\,\frac{\alpha_{3(k - n)}}{\alpha_{3k}}\,{\mathcal{L}}(x)_{3k}^{\alpha} + c\,, \\
\int {\mathcal{H}}_{3n}^{\nu}(x)\,\partial_{\nu}(x) = \frac
{{\mathcal{H}}_{3(n + 1)}^{\nu}(x)}{d_{3(n + 1)}} + c\,.
\end{gather*}
Actually, the constant $c$ can be replaced by the general solution
of the differential equation $\vartheta_{\nu}\,y(x) = 0$. In
general this is given by $y(x) = c_{1}x^2 + c_{2}x^{1 - 3\nu} +
c_{3}$, depending on the parameters as in \eqref{constants}.

A generalized $\nu$-definite integral associated with $\vartheta_{\nu}$ is defined by
\begin{align*}
\int_{y \oplus a}^{x \oplus b}z^{3n}\,\partial_{\nu}(x)
& =  \frac{z^{3(n + 1)}}{d_{3(n + 1)}}\Big|_{y \oplus a}^{x \oplus b} \\
& =  \frac{(x \oplus b)^{3(n + 1)} - (y \oplus a)^{3(n + 1)}}{d_{3(n + 1)}(\nu)}\,.
\end{align*}
The definite integral is extended by linearity to ${\mathcal{P}}_{3}$
the polynomials is $x^3$ and also to the formal power series
${\mathcal{F}}_{3}$ in $x^3$. The $\nu$-definite integral $\int_{y
\oplus a}^{x + \oplus b}$ commutes with $\vartheta_{\nu}$ on
${\mathcal{P}}_{3}$ (or ${\mathcal{F}}_{3}$). We have
\begin{align*}
\vartheta_{x}\,\int_{x \oplus a}^{x \oplus b}z^{3n}\,\partial_{\nu}(z)
& = \vartheta_{x}\,\frac{(x \oplus a)^{3(n + 1)} -
(x \oplus b)^{3(n + 1)}}{\alpha_{3(n + 1)}} \\
 & =  (x \oplus a)^{3n} - (x \oplus b)^{3n} \\
 & =  \int_{x \oplus a}^{x \oplus b}\vartheta_{z}\,z^{3n}\,\partial_{\nu}(z)\,.
\end{align*}
The result extends to ${\mathcal{P}}_{3}$ by linearity.
It is easy to show that
\begin{equation}
\int_{a}^{b}(z \oplus b)^{3n}\,\partial_{\nu}(z) = \int_{a \oplus b}^{x \oplus b}
z^{3n}\,\partial_{\nu}(z)
\label{chngVar}
\end{equation}
and therefore by linearity $\int_{a}^{b}p(z \oplus b)\,\partial_{\nu}(z) =
\int_{a \oplus b}^{x \oplus b}p(z)\,\partial_{\nu}(z)$.
Note that \eqref{chngVar} is the $\nu$-analogue of the elementary change of variable
formula
\[\int_{a}^{x}f(z + b)\,dz = \int_{a + b}^{x + b}f(z)\,dz\,.\]

\section{An Orthogonality Relation}

Since the $\nu$-diffusion polynomials $p_{n}(x, t)$ are three-parity polynomials they cannot
be orthogonal in the usual sense with respect to a measure, see \cite{DaboR96}. In this
section we obtain a multiple integral orthogonality relation.

By Erd\'{e}lyi,  \cite[p.~218,]{ErdeMOT53}, we have
\begin{equation}
\int_{0}^{\infty}e^{-pt}t^{\beta - 1}\,_{1}F_{2}(-n; \alpha + 1, \beta | \lambda t)\,dt
= n!\,\frac{\Gamma(\beta)}{(\alpha + 1)_{n}}\,p^{-\beta}\,L_{n}^{\alpha}(\lambda/p)
\label{intTrans}
\end{equation}
where $\mathop{\rm Re}p > 0$ and $L_{n}^{\alpha}$ is a Laguerre polynomial. Since
\[
p_{n}^{\nu}(x, t) = \frac{\alpha_{3n}(\nu)}{n!}t^n\,_{1}F_{2}(-n; 1/3, \nu + 2/3 |
- \frac{x^3}{27t})
\]
a change of variables in \eqref{intTrans} yields the following result.

\begin{theorem} \label{Thm9.1}
Let $\nu > 0$. Then
\[
\frac{(-1)^n}{3^{3n}t^{n + 1/3}\Gamma(n + 1/3)\,n!}\,\int_{0}^{\infty}
e^{- \frac{z^3}{27t\tau}}\,p_{n}^{\nu}(z, -t)\,dz =
\tau^{1/3}L_{n}^{\nu - 1/3}(\tau), \quad n = 0, 1,
\dots
\]
\end{theorem}

By the usual orthogonality of the Laguerre polynomials, we obtain
\begin{equation}
\int_{0}^{\infty}L_{m}^{\nu + 1/3}(\tau)L_{n}^{\nu - 1/3}(\tau)e^{-\tau}\,\tau^{\nu - 1/3}\,
d\tau = \frac{\Gamma(\nu + n + 2/3)}{n!}\,\delta_{n, m}
\label{Eq9.3}
\end{equation}
where $\delta_{n, m}$ is the Kronecker delta and $Re(\nu - 1/3) > -1$.
Thus \eqref{Eq9.3} is valid for $\nu \geq 0$.

\begin{theorem} \label{thm13}
For $\nu \geq 0$, we have
\begin{equation}\label{eq9.4}
\begin{aligned}
&\frac{1}{\sigma_{n}(t)}\,\int_{0}^{\infty}\int_{0}^{\infty}\int_{0}^{\infty}
\exp(- \frac{z^3 + y^3}{27t\tau})e^{-\tau}p_{n}^{\nu}(z,
-t)p_{m}^{\nu}(y, -t) \tau^{\nu -1}\,d\tau\,dz\,dy \\
&= \alpha_{3n}(\nu)\delta_{n, m}
\end{aligned}
\end{equation}
$n = 0, 1, 2, \dots$,
where $\sigma_{n}(t) = 3^{3n}t^{2n + 2/3}\Gamma(n + 1/3)\Gamma(1/3)\Gamma(\nu + 2/3)$.
\end{theorem}

\begin{proof} Using Theorem \ref{Thm9.1} and Fubini's Theorem, we get
\begin{align*}
&\int_{0}^{\infty}\tau^{1/3}L_{m}^{\nu - 1/3}(\tau)L_{n}^{\nu - 1/3}(\tau)e^{-\tau}
\tau^{\nu - 1}\,d\tau \\
& = \frac{1}{(3^{3n}t^{n + 1/3}\Gamma(n + 1/3)\,n!)^2}\\
&\quad\times \int_{0}^{\infty}e^{\tau}t^{\nu - 1}\,d\tau\,\int_{0}^{\infty}e^{-
\frac{z^3}{27t\tau}}
p_{n}^{\nu}(z, -t)\,dz\,\int_{0}^{\infty}e^{- \frac{y^3}{27t\tau}}p_{m}^{\nu}(y,
 -\tau)\,dy \\
& =  \frac{1}{(3^{3n}t^{n + 1/3}\Gamma(n + 1/3)\,n!)^2}\\
&\quad\times  \int_{0}^{\infty}\,\int_{0}^{\infty}\,\int_{0}^{\infty}e^{- \frac{z^3
 + y^3}{27t\tau}}
 e^{-\tau}p_{m}^{\nu}(z, -t)p_{n}^{\nu}(y, -t)\tau_{\nu - 1}\,d\tau\,dz\,dy \\
& =  \frac{\Gamma(\nu + n + 2/3)}{n!}\,\delta_{n, m}\,.
\end{align*}
Adjusting the constants yields the theorem.
\end{proof}

 By Erdelyi,  \cite[p. 149]{ErdeMOT53}, we have the relation
 \[\int_{0}^{\infty}e^{-t}t^{\nu - 1}e^{- \frac{\alpha}{4t}}\,dt =
 z(\frac{\alpha}{4})^{\nu/2}K_{\nu}(\alpha^{1/2})
 \]
 where $K_{\nu}$ is a modified Bessel function. With a suitable change of variables, we
 obtain
 \[
 \int_{0}^{\infty}e^{- \frac{z^3 + y^3}{27t\tau}}\tau^{\nu - 1}e^{-\tau}\,d\tau
 = 2\big[\frac{z^3 + y^3}{27t}\big]^{\nu/2}K_{\nu}\Big(2\big(\frac{z^3 + y^3}
 {27t}\big)^{1/2}\Big)\,.
 \]
 Substitution into \eqref{eq9.4}, gives the equation
 \begin{align*}
& \frac{2}{\sigma_{n}(t)}\,\int_{0}^{\infty}\int_{0}^{\infty}\Big[
 \frac{z^3 + y^3}{27t}\Big]^{\nu/2}K_{\nu}\Big[2\big(\frac{z^3 + y^3}
 {27t}\big)^{1/2}\Big]\,p_{n}^{\nu}(z, -t)p_{m}^{\nu}(y, -t)\,dz\,dy\\
& = \alpha_{3n}(\nu)\delta_{n, m}
\end{align*}
for $n = 0, 1, 2, \dots$.
Under suitable conditions on the sequence $\{a_{3n}\}_{0}^{\infty}$, the function
\[
u(x, t) = \sum_{m = 0}^{\infty}a_{3m}p_{m}^{\nu}(x, t)
\]
is in ${\mathcal{H}}_{\nu}(0, \infty)$ for $t > 0$. In the usual calculus manner the
coefficients are formally determined by
\[
a_{3n} = \frac{1}{\alpha_{3n}}\,\int_{0}^{\infty}\int_{0}^{\infty}
\Big[\frac{x^3 + y^3}{27t}\Big]^{\nu/2}K_{\nu}\Big[2\big(\frac{z^3 +
y^3} {27t}\big)^{1/2}\Big]u(z, -t)p_{n}^{\nu}(y, -t)\,dz\,dy
\]

We will consider convergence criteria in the next section.

\section{$\nu$-Diffusion Polynomial Expansions}

The diffusion polynomials $p_{n}^{\nu}(x, t)$ satisfy the $\nu$-Airy diffusion
equation \\
$\vartheta_{\nu}u(x, t) = D_{t}u(x, t)$. Since the partial differential equation
is linear, finite linear combinations of the diffusion polynomials are also solutions
in ${\mathcal{H}}_{\nu}(0 \leq x < \infty)$, for all $t$. Hence we expect to obtain infinite
series expansions
\[u(x, t) = \sum_{m = 0}^{\infty}a_{3m}p_{m}^{\nu}(x, t)\]
with possible convergence in a strip $|t| < \sigma$.

\begin{theorem} \label{thm14}
Let $\nu \geq 0$, then
\begin{equation}
|p_{n}^{\nu}(x, t)| \leq M\,\alpha_{3n}e^{|x| + |t|}
\label{Ineq10.2}
\end{equation}
\label{Thm10.1}
\end{theorem}

\begin{proof} Since
\[
e^{- tz^3}G_{\nu}(xz) = \sum_{n = 0}^{\infty}\frac{p_{n}^{\nu}(x, t)}
{\alpha_{3n}(\nu)}z^{3n}
\]
is an entire function, we find by Cauchy's Theorem that
\[
p_{n}^{\nu}(x, t) = \frac{\alpha_{3n}}{2\pi i}\,\int_{\Gamma}\frac
{e^{-tz^3}G_{\nu}(xz)}{z^{3n + 1}}\,dz\,,
\]
where $\Gamma$ is the
unit circle, $|z| = 1$. Since $G_{\nu}(z)$ is an entire function
of order one, we have $|G_{\nu}(xz)| \leq Me^{|xz|}$. Therefore,
\[
|p_{n}^{\nu}(x, t)|
\leq  M\,\frac{\alpha_{3n}}{2\pi}\,\int_{\Gamma} e^{|xz| - |t||z|^3}\,|dz|
 \leq M\alpha_{3n}(\nu)e^{|x| + |t|}
\]
Since $|G_{\nu}(xz)| \leq G_{\nu}(|x||z|) = G_{\nu}(|x|)$ on $|z| = 1$, we also get
the estimate
\[
|p_{n}^{\nu}(x, t)| \leq \alpha_{3n}G_{\nu}(|x|)e^{|t|}
\]
\end{proof}

\begin{theorem} \label{thm15}
Suppose $\sum_{n = 0}^{\infty}|a_{n}| < \infty$, then the series
\[
\sum_{n = 0}^{\infty}a_{n}\,\frac{p_{n}^{\nu}(x, t)}{\alpha_{3n}}
\]
converges absolutely and locally uniformly for $|x| < \infty$ and $|t| < \infty$.
\end{theorem}

\begin{proof}
 By the estimate  \eqref{Ineq10.2}, it follows that
\[
\Big|\sum_{n = 0}^{\infty}a_{n}\,\frac{p_{n}^{\nu}(x, t)}{\alpha_{3n}}\Big|
\leq Me^{|x| + |t|}\,\sum_{n = 0}^{\infty}|a_{n}| < \infty
\]
Thus the series converges absolutely and locally uniformly by the Weierstrass M-test.
Since $\frac{\vartheta_{\nu}}{\alpha_{3n}}\,p_{n}^{\nu}(x, t) = \frac{p_{n - 1}^{\nu}(x, t)}
{\alpha_{3(n - 1)}}$, it also follows that the differentiated series converges locally
uniformly and therefore the series represents a function in ${\mathcal{H}}_{\nu}$ for
$|x| < \infty$ and $|t| < \infty$.
\end{proof}

Let
\[{\mathcal{R}}_{\nu}(z)
= \sum_{n = 0}^{\infty}\frac{z^{3n}}{3^{3n}(1/3)_{n}(\nu + 2/3)_{n}}
= _{1}F_{2}\left(\begin{matrix}
1 \\
1/3, \nu + 2/3 \end{matrix}\bigg|\frac{z^3}{27t}\right)\,.
\]
Then using Stirling's formula we get
\[\limsup_{n \to \infty}\,\frac{3n\log 3n}{\log\,(3^{3n}(1/3)_{n}(\nu + 2/3)_{n})}
= 3/2\]
Therefore, ${\mathcal{R}}_{\nu}$ is an entire function of order $3/2$. Furthermore,
\[
\limsup_{n \to \infty} 3n\,\Big|\,\frac{1}{3^{3n}(1/3)_{n}(\nu + 2/3)_{n}} \Big|
^{\frac{3/2}{2n}} = \frac{e}{\sqrt{3}}\,.
\]
Therefore, ${\mathcal{R}}_{\nu}(x)$ is of type $\frac{2}{3\sqrt{3}}$.

\begin{lemma} \label{Lem10.3}
Let $\delta > 0$. Then
\[\frac{p_{n}^{\nu}(|x|, |t|)}{\alpha_{3n}(\nu)} \leq \frac{\delta^n}{n!}\,
(1 +
\frac{|t|}{\delta})^n{\mathcal{R}}_{\nu}\big(\frac{|x|}{\delta^{1/3}}\big)\]
\end{lemma}
\begin{proof} We have
\begin{align*}
\frac{p_{n}^{\nu}(|x|, |t|)}{\alpha_{3n}}
& \leq \frac{\delta^n}{n!}\, \sum_{n =
0}^{\infty}\,\left(\frac{|t|}{\delta}\right)^{n - k}\,
\frac{\frac{|x|^{3k}}{\delta^{k}}}{3^{3k}(1/3)_{k}(\nu + 2/3)_{k}} \\
 & \leq  {\mathcal{R}}_{\nu}(\frac{|x|}{\delta^{1/3}})\,\frac{\delta^n}{n!}
 \sum_{n = 0}^{\infty} \binom{n}{k}
 \left(\frac{|t|}{\delta}\right)^{n - k} \\
 & =  \frac{\delta^n}{n!}\,(1 + \frac{|t|}{\delta})^n{\mathcal{R}}_{\nu}
 (\frac{|x|}{\delta^{1/3}})
\end{align*}
since
\[\frac{\frac{|x|^{3k}}{\delta^{k}}}{3^{3k}(1/3)_{k}(\nu + 2/3)_{k}} < {\mathcal{R}}_{\nu}
(\frac{|x|}{\delta^{1/3}})\]
\end{proof}

\begin{lemma} \label{Lem10.4}
\[
p_{n}^{\nu}(x, t) \geq \frac{\alpha_{3n}}{n!}t^n,\quad\mbox{for }t, x > 0
\]
\end{lemma}

\begin{proof} Since the coefficients of $p_{n}^{\nu}$ are
positive, it follows that $p_{n}^{\nu}(x, t) \geq p_{n}^{\nu}(0,
t) = \frac{\alpha_{3n}}{n!}t^n$.
\end{proof}

\begin{theorem} \label{thm16}
If the series $\sum_{n = 0}^{\infty}a_{n}p_{n}^{\nu}(x_{0}, t_{0})$ converges for
$t_{0} > 0$ and $x_{0} > 0$, then the series
\[
\sum_{n = 0}^{\infty}a_{n}p_{n}^{\nu}(x, t) \quad\mbox{and }
\sum_{n = 0}^{\infty}a_{n}d_{3n}(\nu)p_{n - 1}^{\nu}(x, t)
\]
converge absolutely and locally uniformly in the strip $|t| < t_{0}$ and
$\sum_{n = 0}^{\infty}a_{n}p_{n}^{\nu}(x, t)$ is in ${\mathcal{H}}_{\nu}(R_{+})$ for
$|t| < t_{0}$.
\end{theorem}

\begin{proof} Since the general term of a convergent series must go to zero,
\[\lim_{n \to \infty}a_{n}p_{n}^{\nu}(x_{0}, t_{0}) = 0\,.
\] By Lemma \ref{Lem10.4},
it therefore follows that
\[a_{n} = O\big(\frac{n!}{\alpha_{3n}t_{0}^n}\big)\,.
\]
Using Lemma \ref{Lem10.3}, we get for $\delta > 0$
\begin{align*}
&\sum_{n = 0}^{\infty}a_{n}d_{3n}p_{n - 1}^{\nu}(x, t)  \ell
M\sum_{n = 1}^{infty}\frac{n!}
{\alpha_{3n}t_{0}^n}\,\frac{\alpha_{3n}}{n!}\,(\delta +
|t|)^n{\mathcal{R}}_{\nu}
(\frac{|x|}{\delta^{1/3}}) \\
 & \leq  M{\mathcal{R}}_{\nu}(\frac{|x|}{\delta^{1/3}})\sum_{n = 0}^{\infty}\big(
 \frac{\delta + |t|}{t_{0}}\big)^n
\end{align*}
which converges for $\delta + |t| < t_{0}$. Since $\delta > 0$ is arbitrary it converges for
$(\delta + |t|) < t_{0}$, and as before for $|t| < t_{0}$. The Weierstrass M-test provides
the local uniform convergence.

Let $f(z) = \sum_{n = 0}^{\infty}a_{n}z^n$ be an entire function
of order $\rho$, $\rho > 0$, and of type $0 < \sigma < \infty$.
The type is determined by
\[\limsup_{n \to \infty} \frac{3n}{e\rho}\,|a_{n}|^{\frac{\rho}{3n}} = \sigma\]
see for example Boas,  \cite[p.~11]{Boas54}.
Therefore,
\begin{equation}
|a_{n}| \leq M\left(\frac{e\sigma\rho}{3n}\right)^{3n/\rho}
\label{Eq10.9}
\end{equation}
\end{proof}

\begin{theorem}
If $f(z) = \sum_{n = 0}^{\infty} a_{n}z^{3n}$ is an entire function of order $\rho$ with
$0 < \rho < 3/2$ and of type $\sigma$, $0 < \sigma < \infty$, then
\begin{equation}
u(x, t) = \sum_{n = 0}^{\infty} a_{n}p_{n}^{\nu}(x, t)
\label{Eq10.10}
\end{equation}
is in ${\mathcal{H}}_{\nu}(R)$ in the strip $|t| < 1/(\sigma \rho)^{3/\rho}$ and $u(x, 0) =
f(x)$.
\end{theorem}

\begin{proof} Using \eqref{Eq10.9} and Lemma \ref{Lem10.3}, for $\delta> 0$
we obtain
\begin{equation}
\sum_{n = 0}^{\infty} a_{n}p_{n}^{\nu}(x, t) \ell M\sum_{n =
0}^{\infty}
\left(\frac{e\sigma\rho}{3n}\right)^{3n/\rho} \frac{\alpha_{3n}}{n!}\,
(\delta + |t|)^n{\mathcal{R}}_{\nu}(\frac{|x|}{\delta^{1/3}}
\label{Eq10.11}
\end{equation}
Using Stirling's formula, we get the estimate
\[
\left(\frac{e\sigma\rho}{3n}\right)^{3n/\rho}3^{3n}(1/3)_{n}(\nu + 2/3)_{n}
\sim \Big[\frac{e^{1 - \frac{2}{3}\rho}3^{\rho -1}}{n^{1 -
\frac{2}{3}\rho - \frac
{\nu\rho}{3^n}}}\Big]^{3n/\rho}\,\frac{2\pi(\sigma\rho)^{3n/\rho}}
{\Gamma(1/3)\Gamma(\nu + 2/3)}
\]
Now
\[
\Big[\frac{e^{1 -\frac{2}{3}\rho}3^{\rho -1}}{n^{1 - \frac{2}{3}\rho - \frac
{\nu\rho}{3^n}}}\Big]^{3n/\rho}\,\frac{2\pi(\sigma\rho)^{3n/\rho}}
{\Gamma(1/3)\Gamma(\nu + 2/3)} = O(1)
\]
for $0 < \rho < 3/2$. Thus the series in \eqref{Eq10.11} is dominated by
\[
M_{t}{\mathcal{R}}_{\nu}(\frac{|x|}{\delta^{1/3}})\sum_{n = 0}^{\infty}
\{(\sigma\rho)^{3/\rho}(\delta + |t|)\}^n
\]
which converges for $(\sigma\rho)^{3/\rho}(\delta + |t|) < 1$.
Since $\delta > 0$ is arbitrary we get absolute and local uniform convergence
for $|t| < \frac{1}{(\sigma\rho)^{3/\rho}}$, by the Weierstrass M-test. Since
the order and type of an entire functions is not changed by taking
derivatives, a similar type argument shows that the derived series
\[
\sum_{n = 1}^{\infty}a_{n}d_{2n}p_{n - 1}^{\nu}(x, t)
\]
also converges absolutely and locally uniformly for $|t| < \frac{1}{(\sigma\rho)^{3/\rho}}$.
It follows that $u(x, t)$ given by \eqref{Eq10.10} is in ${\mathcal{H}}_{\nu}$ in the
stated strip.
\end{proof}

In the classical case developed by Widder \cite{Widd79} or in the Bessel function case treated
by Bragg \cite{Brag65} and Cholewinski and Haimo \cite{CholH68}, a series expansion of the type
given by \eqref{Eq10.10}, leads to an integral representation of $u(x, t)$. In both
of those cases the representation depends on the fact that the diffusion polynomials can be
represented by Gaussian type integrals in terms of the source function. In the ${\mathcal{H}}_{\nu}$
case the corresponding integrals diverge.

The theorem does not imply that the given strip is the best possible. For example with
$G_{\nu}(x) = \sum_{n = 0}^{\infty}\frac{x^{3n}}{\alpha_{3n}}$, an entire function of
order one and type one, we get
\[u(x, t) = e^{t}G_{\nu}(x) = \sum_{n = 0}^{\infty}\frac{p_{n}^{\nu}(x, t)}{\alpha_{3n}}\]
which converges for all $x$ and $t$. The theorem gives the strip $|t| < 1$.

Next we consider a sequence of complex numbers $\{a_{n}\}_{0}^{\infty}$ for which
$|a_{n}|\alpha_{3n}(\nu) = O(1)$. Using Stirling's formula, we find that
\[
\limsup_{n \to \infty}\frac{3n\log 3n}{\log \frac{1}{|a_{n}|}} \leq 1
\]
Thus the function $f(z) = \sum_{n = 0}^{\infty}a_{n}z^n$ is an
entire function of order less than or equal to one. Likewise
another calculation using Stirling's formula yields
\[
\limsup_{n \to \infty}\frac{3n}{e}\,|a_{n}|^{1/3n} \leq 1
\]
Thus $f(z)$ is of growth $\leq \{1, 1\}$. We let $f_{r}(z) = \sum_{n = 0}^{\infty}
a_{n}r^{3n}z^{3n}$ with $0 < r < 1$ be the ``Abel means'' of $f$.

\begin{theorem}
If $|a_{n}|\alpha_{3n} = O(1)$, then
\[
u_{r}(x, t) = \sum_{n = 0}^{\infty}a_{n}r^{3n}p_{n}^{\nu}(x, t)
\]
is an entire function in the variables $x$ and $t$ and it is in ${\mathcal{H}}_{\nu}$ for all
$x$ and $t$.
\end{theorem}

\begin{proof}
By Theorem \ref{Thm10.1}, we get the domination
\begin{align*}
\sum_{n = 0}^{\infty}|a_{n}|r^{3n}|p_{n}^{\nu}(x, t)|
& \leq Me^{|x| + |t|}\sum_{n = 0}^{\infty}|a_{n}|\alpha_{3n}r^{3n} \\
& \leq  Me^{|x| + |t|}\sum_{n = 0}^{\infty}r^{3n} \\
& =  Me^{|x| + |t|}\,\frac{1}{1 - r^3} < \infty
\end{align*}
for $0 < r < 1$. Once again the Weierstrass M-test gives absolute and locally uniform
convergence for $x$ and $t$ in $C$, the complex numbers.

In the case $a_{n} = \frac{1}{\alpha_{3n}(\nu)}$, we get $G_{\nu}(rz) = \sum_{n = 0}^{\infty}
\frac{r^{3n}z^{3n}}{\alpha_{3n}}$ and we recover the generating function
\[
u_{r}(x, t) = e^{tr^3}G_{\nu}(rx) = \sum_{n = 0}^{\infty} \frac{p_{n}^{\nu}(x, t)}
{\alpha_{3n}}\,r^{3n}\,.
\]
\end{proof}

\section{Associated Functions}

A sequence of $\nu$-associated functions is defined as
\[
Q_{n}^{\nu}(x, t) = (- \vartheta_{\nu})^n\,{\mathcal{K}}_{\nu}(x, t)
= (-1)^n\,\frac{\partial^n}
{\partial t^n}{\mathcal{K}}_{\nu}(x, t)\,,
\]
where ${\mathcal{K}}_{\nu}(x,t)$ is the source solution of
$\vartheta_{\nu}\,u = u_{t}$.
Expansions in terms of the $Q_{n}^{\nu}(x, t)$ are related to
Laurent expansions for analytic functions, see Widder
\cite{Widd75} for the ordinary heat equation or Cholewinski and
Haimo \cite{CholH68} for the Bessel function case. Since $-
\vartheta_{\nu}\,Q_{n}^{\nu}(x, t) = (- \vartheta_{\nu})^{n  +
1}\, {\mathcal{K}}_{\nu}(x, t) = Q_{n + 1}^{\nu}(x, t)$,  and
$\frac{\partial}{\partial t}\,Q_{n}^{\nu}(x, t) = - Q_{n +
1}^{\nu}(x, t)$, it follows that $\vartheta_{\nu}\,Q_{n}^{\nu}(x,
t) = \frac{\partial}{\partial t}\,Q_{n}^{\nu}(x, t)$ for $t > 0$.

By our previous integral representation,
\[{\mathcal{K}}_{\nu}(x, t) = \int_{0}^{\infty}e^{-ty^3}{\mathcal{G}}_{\nu}(xy)\,d\eta_{\nu}(y),
\quad t > 0\]

Since ${\mathcal{G}}_{\nu}(xy)$ and its derivatives are entire functions of growth $\{1, 1\}$, we
obtain and integral representation for $Q_{n}^{\nu}(x, t)$, namely,
\begin{equation}
Q_{n}^{\nu}(x, t) =
\int_{0}^{\infty}y^{3n}e^{-ty^3}{\mathcal{G}}_{\nu}(xy)\,d\eta_{\nu}(y)
\label{Eq11.3}
\end{equation}
The growth condition yields the necessary domination integrals for the absolute and local uniform
convergence. Further, Fubini's Theorem yields the series expansion
\begin{equation}
\begin{aligned}
Q_{n}^{\nu}(x, t)
& =  \sum_{m = 0}^{\infty} \frac{(-1)^{m}x^{3m}}{\alpha_{3m}}\,\int_{0}^{\infty}
 e^{- ty^3}y^{3n + 3m + 3\nu + 1}\,\frac{dy}{c_{\nu}}  \\
& =  \sum_{m = 0}^{\infty}\frac{(-1)^{m}\Gamma(n + m + \nu + 2/3)}{\alpha_{3m}(\nu)
 3^{\nu + 2/3}\Gamma(\nu + 2/3)}\,\frac{x^{m}}{t^{n + m + \nu + 2/3}}  \\
& =  \frac{(\nu + 2/3)_{n}t^{- n}}{(3t)^{\nu + 2/3}}\,_{1}F_{2}
\left[\begin{matrix}  n + \nu + 2/3 \\
 1/3, \nu + 2/3\end{matrix}\bigg|- \frac{x^3}{27t}\right]\,.
\end{aligned} \label{Eq11.4}
\end{equation}
A direct calculation employing the coefficients of $Q_{n}^{\nu}(x, t)$ shows that $Q_{n}^{\nu}$ is an
entire function of growth $\{3/2, \frac{2}{3\sqrt{3}|t|^{1/2}}\}$, for $t \neq 0$, which is the
growth of ${\mathcal{K}}_{\nu}(x, t)$.

Next we obtain an upper bound of $|Q_{n}^{\nu}(x, t)|$ that applies to the variables $\nu$, $n$,
$x$, and $t$.

\begin{lemma} \label{lm3}
Let $\nu \geq 0$. Then
\begin{equation}
|Q_{n}^{\nu}(x, t)| \leq M(\nu + 2/3)_{n}\,\frac{e^{x^2/2}}{t^{n +
\nu + 2/3}},\quad  t > 0 \label{Eq11.5}
\end{equation}
\end{lemma}

\begin{proof}
 Since ${\mathcal{G}}_{\nu}(x)$ is of growth $\{1, 1\}$, we have
\[
|{\mathcal{G}}_{\nu}(xy)| \leq Me^{|x||y|} \leq Me^{\frac{|x|^2 + |y|^2}{2}}\,.
\]
Let $\epsilon > 0$. Since $y^2/2 \leq \epsilon y^3$, we get
\begin{align*}
|Q_{n}^{\nu}(x, t)|
& \leq Me^{|x|^2/2}\int_{0}^{\infty}y^{3n}e^{- ty^3}e^{y^2/2}\,
  d\eta_{\nu}(y) \\
& \leq  M\,\frac{e^{|x|^2/2}}{\Gamma(\nu + 3/2)}\,\int_{0}^{\infty}e^{- (t - \epsilon)y^3}
  y^{3n + 3\nu + 1}\,dy \\
& =  Me^{|x|^2/2}\,\frac{\Gamma(n + \nu + 2/3)}{(t - \epsilon)^{n + \nu + 2/3}},
 \quad\mbox{for }t > \epsilon > 0\,.
\end{align*}
Since $\epsilon > 0$ is arbitrary, we obtain inequality \ref{Eq11.5}.
\end{proof}

The integral \eqref{Eq11.3} leads to a number of generating functions involving the
associated functions $Q_{n}^{\nu}$. We have for $t > 0$
\[
\sum_{n = 0}^{\infty}\,\frac{Q_{n}^{\nu}(x, t)}{n!}\,z^n
= {\mathcal{K}}_{\nu}(x, - z + t)
\quad\mbox{with } z < t
\]
for
\[
\sum_{n = 0}^{\infty}\,\frac{Q_{n}^{\nu}(x, t)}{n!}\,z^n
= \int_{0}^{\infty}e^{zy^3 - ty^3}
{\mathcal{G}}_{\nu}(xy)\,d\eta_{\nu}(y)
\]
The integral is dominated by
\[
M\int_{0}^{\infty}e^{(z - t)y^2}e^{|xy|}\,d\eta_{\nu}(y)
\quad\mbox{for } z - t < 0\,.
\]
Therefore, the interchange of summation and integration is valid for
$z - t < 0$.
In a similar fashion, it follows that
\[
\sum_{n = 0}^{\infty} \frac{(-1)^nQ_{n}^{\nu}(x,t)}{\alpha_{3n}} z^{3n}
 =  \int_{0}^{\infty}
e^{- ty^3}{\mathcal{G}}_{\nu}(xy){\mathcal{G}}_{\nu}(yz) \eta_{\nu}(y)\,.
 =  {\mathcal{K}}(x \oplus_{\nu} z, t)
\]
This series is also the symbolic time series solution given by
${\mathcal{G}}_{\nu}(zD_{t}^{1/3}){\mathcal{K}}_{\nu}(x, t)$.

Let $\phi(z) = \sum_{n = 0}^{\infty}a_{n}z^n$ be an entire
function of growth $\{3, \sigma\}$,that is for $\epsilon > 0$
\[|\phi(z)| \leq Me^{(\sigma + \epsilon)|z|^3}\,.
\]

\begin{theorem} \label{thm19}
If $\phi$ is of growth $\{3, \sigma\}$, then
\[u(x,t) = \int_{0}^{\infty}e^{-ty^3}{\mathcal{G}}_{\nu}(xy)\phi(y)\,d\eta_{\nu}(y)\]
is in ${\mathcal{H}}_{\nu}$ for $t > \sigma \geq 0$.
\end{theorem}

\begin{proof} Since ${\mathcal{G}}_{\nu}$ is of growth $\{1, 1\}$, we find that
\begin{equation}
\begin{aligned}
\Big|\int_{0}^{\infty}{\mathcal{G}}_{\nu}(xy)e^{-
ty^2}\phi(y)\,d\eta_{\nu}(y)\Big|
 & \leq  M\int_{0}^{\infty}e^{xy}e^{- ty^3}e^{(\sigma + \epsilon)}y^3\,
 d\eta_{\nu}(y) \\
 & <  \infty, \quad\mbox{for\hspace{.15in}}t > \sigma + \epsilon\,.
\end{aligned} \label{Eq11.9}
\end{equation}
Since $\epsilon > 0$ is arbitrary, the integral converges absolutely and locally uniformly.
Since derivatives of ${\mathcal{G}}_{\nu}$ are also of growth $\{1, 1\}$, similar domination
integrals allow the interchange of integration and differentiation. It readily follows
that $u$ is in ${\mathcal{H}}_{\nu}$ for $t > \sigma \geq 0$.
\end{proof}

The integral representation also leads to an infinite series for $u(x, t)$.
 By \eqref{Eq11.9}, we also have
\[
\Big|\int_{0}^{\infty}e^{- ty^3}{\mathcal{G}}_{\nu}(xy)\sum_{n =
0}^{\infty}a_{n}y^3\, d\eta_{\nu}(y)\Big|
 \leq  M\int_{0}^{\infty}e^{xy}e^{- ty^3}e^{(\sigma + \epsilon)}y^3\,
 d\eta_{\nu}(y)
 <  \infty\,.
\]
Thus by Fubini's Theorem and domination for the differentiate integrals we get
\begin{align*}
u(x, t) & =  \sum_{n =0}^{\infty}a{n}\int_{0}^{\infty}{\mathcal{G}}_{\nu}
  (xy)y^{3n}e^{- ty^3}\, d\eta_{\nu}(y)\\
 & =  \sum_{n = 0}^{\infty}(-1)^na_{n}\vartheta_{x}^n\,\int_{0}^{\infty}
 {\mathcal{G}}_{\nu}(xy)e^{- ty^3}\,d\eta_{\nu}(y) \\
 & =  \sum_{n = 0}^{\infty}a_{n}(- \vartheta_{\nu})^n{\mathcal{K}}_{\nu}(x, t) \\
 & =  \sum_{n = 0}^{\infty}a_{n}Q_{n}^{\nu}(x, t)
\end{align*}
which converges of all $x$ and $t > \sigma$.

\begin{theorem} \label{thm20}
If the series
\[\sum_{n = 0}^{\infty}b_{n}Q_{n}^{\nu}(x, t)\]
converges absolutely for $t > \sigma$, then $\phi(y) = \sum_{n = 0}^{\infty}b_{n}y^{3n}$
is of growth $\{3, \sigma\}$.
\end{theorem}

\begin{proof} By the alternating series test
\[1 - \frac{(xy)^3}{\alpha_{3}} \leq {\mathcal{G}}_{\nu}(xy),
\quad x,y \geq 0\,.
\]
Hence we get
\[\int_{0}^{\infty}(1 - \frac{(xy)^3}{3})y^{3n}e^{- ty^3}\,d\eta_{\nu}(y)
\leq Q_{n}^{\nu}(x, t)\,.\]
Next
\[\int_{0}^{\infty}e^{- ty^3}y^{3n}\,d\eta_{\nu}(y) = \frac{(\nu + 2/3)_{n}}
{3^{\nu + 2/3}t^{n + \nu +2/3}}\,.\]
Therefore,
\[\frac{1}{(3t)^{\nu + 2/3}}\,\sum_{n = 0}^{\infty}\,\frac{|b_{n}|}{t^n}\,((\nu + 2/3)_{n}
- \frac{x^3}{\alpha_{3}}\,\frac{(\nu + 2/3)_{n + 1}}{t}) \leq
\sum_{n = 0}^{\infty}|b_{n}||Q_{n}^{\nu}(x, t)|\] and $\lim_{n
\to \infty}\,\frac{|b{n}|}{t^n}\,(\nu + 2/3)_{n + 1}^{1/n}
= 0$. By Stirling's formula $(\nu + 2/3)_{n}^{1/n}$ and $(\nu +
2/3)_{n + 1}^{1/n} \sim ne^{-1}$. It follows that $|b_{n}|^{1/n}
\leq \frac{(\sigma + \epsilon)}{\nu + 2/3)_{n}^{1/n}} \leq
\frac{e}{n}\,(\sigma + \epsilon)$ since the series converges for
$t = \sigma + \epsilon$. Consequently
\[
\limsup_{n \to \infty} 3n|b_{n}|^{\frac{3}{3n}} \leq \sigma + \epsilon
\]
and therefore $\epsilon > 0$ arbitrary implies that $\phi(y)$ is of growth type
$\{3, \sigma\}$.
\end{proof}

\subsection*{Example}
Let $\phi(x) = e^{ax^3}$, which is of growth
$\{3, |a|\}$. Applying the bound \eqref{Eq11.5}, Stirling's formula
and the root test to the series
\[
\sum_{n = 0}^{\infty}\,\frac{a^n}{n!}\,Q_{n}^{\nu}(x, t)
\]
we get
\[
\limsup_{n \to \infty}\Big\{\frac{|a|^n(\nu + 2/3)_{n}}{n!t^n}
\Big\}^{1/n} = \frac{|a|}{t}\,.
\]
Thus the convergence follows for $0 < |a| < t$.

\section{Bessel Calculus Connections}

The elements of the calculus associated with $\vartheta_{\nu}$ can be associated with the
elements of the calculus associated with the Bessel calculus. Formally the associations
can be obtained through the use of a generalized Hadamard product. However, we present
most of the results as integral representations between the elements of the respective
calculi.

Let $\nu \geq 0$ and $\mu = \nu + 1/6$, then the Bessel
coefficient is $b_{2n}(\mu) = b_{2n}(\nu + 1/6) = 2^{2n}\,n!\,(\nu
+ 2/3)_{n}$. Thus we have the coefficient relation
$\alpha_{3n}(\nu) = (\frac{27}{4})^n(1/3)_{n}b_{2n}(\nu)$. We
associated with a $\nu$-exponential power series
\[f(z) = \sum_{n = 0}^{\infty} \,\frac{a_{n}}{\alpha_{3n}(\nu)}\,z^{3n}\]
the power series
\[f_{\#}(z) = \sum_{n = 0}^{\infty} \,\frac{a_{n}}{b_{2n}(\mu)}\,z^{2n}\,.
\]

\begin{theorem} \label{thm21}
Let $\nu \geq 0$ and $f(z) = \sum_{n = 0}^{\infty} \,\frac{a_{n}}{\alpha_{3n}(\nu)}\,
z^{3n}$ be an entire function of order $\rho_{f}$ with $0 \leq \rho_{f} < 3$. then
$f_{\#}(z) = \sum_{n = 0}^{\infty} \,\frac{a_{n}}{b_{2n}(\mu)}\,z^{2n}$ is an entire
function of order $\rho_{f_{\#}} = 2\rho_{f}/(3 - \rho_{f})$.
\end{theorem}

\begin{proof}
 By Stirling's formula it follows that
\[\frac{log\,\frac{\alpha_{3n}(\nu)}{b_{2n}(\mu)}}{3n\log 3n} \to 1/3
\quad\mbox{as }n \to \infty\,.
\]
Since $\limsup_{n \to \infty}\,\frac{3n\log 3n}{\log\,|\frac{\alpha_{3n}}
{a_{n}}|} = \rho_{f}$, we find that
\[
\frac{3n\log 3n}{\log |\frac{\alpha_{3n}}{b_{2n}}\,\frac{b_{2n}}{a_{n}}|}
 = \frac{1}{\frac{\log \left|\frac{\alpha_{3n}}{b_{2n}}\right|}{3n\log 3n} +
\frac{2}{3}\,\frac{log\,\left|\frac{b_{2n}}{a_{n}}\right|}
{2n\log(\frac{e}{2}\,2n)}}
 \simeq  \frac{1}{1/3 + \frac{2}{3}\,\frac{1}{\rho_{f_{\#}}}} = \rho_{f}\,.
\]
Solving for $\rho_{f_{\#}}$, we get
\[\rho_{f_{\#}} = \frac{2\rho_{f}}{3 - \rho_{f}}\,.\]
\end{proof}

\subsection*{Example} consider the entire function $f(z) =
\exp(z^3)$ of order 3. We have
\[
f(z) = \sum_{n = 0}^{\infty}\frac{(-1)^n}{n!}\,z^{3n} = \sum_{n = 0}^{\infty}
\frac{(-1)^n3^{3n}(1/3)_{n}(\nu +
2/3)_{n}}{\alpha_{3n}(\nu)}\,z^{3n}
\] and
\[
f_{\#}(z)  =  \sum_{n = 0}^{\infty}
\frac{(-1)^n3^{3n}(1/3)_{n}(\nu + 2/3)_{n}}{b_{2n}(\mu)}\,z^{2n}
 =  _{1}F_{0}(1/3 | - \frac{27}{4}\,z^2)\,.
\]
Since $_{1}F_{0}(1/3 | z)$ converges for $|z| < 1$, $f_{\#}$ is holomorphic in
the disk $|z| < 2/3$. In this case $f_{\#}$ is not an  entire function.

\subsection*{Example}
 The function $G_{\nu}(z) = \sum_{n = 0}^{\infty}\,\frac{z^{3n}}
{\alpha_{3n}}$ is an entire function of order 1. We have
\[G_{\nu\,\#}(z) = \sum_{n = 0}^{\infty}\,\frac{z^{2n}}{b_{2n}(\mu)} = {\bf I}_{\nu +
1/6}(z)\]
is also a known entire function of order 1. The functions $G_{\nu}(z)$ and ${\bf I}_{\mu}(z)$
are zeta functions in their respective calculi. We also have ${\mathcal{G}}_{\nu\,\#}(z)
= {\bf J}_{\mu}(z)$.

\begin{proposition} \label{prop1}
Let $\nu \geq 0$ and $f(z) = \sum_{n = 0}^{\infty} \,\frac{a_{n}}{\alpha_{3n}(\nu)}\,
z^{3n}$ be an entire function of order $\{3, \sigma\}$ with $\sigma > 0$, then the
integral
\begin{equation}
\int_{0}^{\infty} e^{- \frac{4zy^3}{27x^2}}\,f(zy)\,dy
\label{Eq12.8}
\end{equation}
converges absolutely and locally uniformly for $0 < (xz)^2 <
\frac{4}{27\sigma}$ with $z > 0$.
\end{proposition}

\begin{proof}
For $\epsilon > 0$, we have the domination
 \[
 \Big|\int_{0}^{\infty} e^{- \frac{4zy^3}{27x^2}}\,f(zy)\,dy\Big|
\leq  M\,\int_{0}^{\infty} e^{- \frac{4zy^3}{27x^2}}\,e^{(\sigma
+ \epsilon)z^3y^3}\,dy
\]
 and the theorem follows.
\end{proof}

 If $f(z)$ is an entire function of order $0 \leq \rho_{f} < 3$, then it readily
 follows
 that the integral \eqref{Eq12.8} converges absolutely and locally uniformly for
$z, x > 0$. Thus in this case summations can be interchanged with the integration.

 \begin{proposition} \label{prop2}
 \[\int_{0}^{\infty} e^{- \frac{4zy^3}{27x^2}}\,y^{3n}\,dy =
 \frac{3^{3n}x^{2n + 2/3}}{2^{2n + 2/3}z^{n + 1/3}}\,\Gamma(n + 1/3)\]
 \label{Prop12.5}
 \end{proposition}

 \begin{proof} With the change of variables $s = y^3$, the integral reverts to a
 gamma function integral representation.
 \end{proof}

 \begin{theorem} \label{Thm12.6}
 Let $f(z) = \sum_{n = 0}^{\infty} \,\frac{a_{n}}{\alpha_{3n}(\nu)}\,
z^{3n}$ be an entire function of order $\leq 3$, then
\begin{equation}
\frac{z^{1/3}2^{2/3}}{x^{2/3}\Gamma(1/3)}\,\int_{0}^{\infty} e^{-
\frac{4zy^3} {27x^2}}\,f(zy)\,dy = \sum_{n =
0}^{\infty}\frac{a_{n}}{b_{2n}(\mu)}\,(xz)^{2n} = f_{\#}(zx)
\label{Eq12.11}
\end{equation}
for $x, z > 0$.
 \end{theorem}

\begin{proof}
Since $f$ is of order less than three, we can invert the integration and summation in the
integral \ref{Eq12.11}. the result the follows by a term for term application of Proposition
\ref{Prop12.5}. \end{proof}

The next result shows that the source solution in the $\nu$-calculus is related to the
Gaussian source solution of the radial heat equation
\begin{equation}
\Delta_{x}(u)\,u(x, t) = \frac{\partial^2 u}{\partial x^2} +
\frac{2\mu}{x}\, \frac{\partial u}{\partial x} = \frac{\partial
u}{\partial t}, \quad \mu = \nu + 1/6
\label{Eq12.12}
\end{equation}
see Cholewinski and Haimo \cite{CholH68}.

\begin{theorem} \label{Thm12.7} %\label{thm23}
Let $\nu \geq 0$. Then
\[
\frac{3^{\nu + 2/3}x^{- 2/3}}{2^{\nu}\Gamma(1/3)}\,\int_{0}^{\infty}
2^{- \frac{4y^3}{27x^2}}\,{\mathcal{K}}_{\nu}(y, t)\,dy =
\frac{1}{(2t)^{(\nu + 1/6) + 1/2}} \,e^{- \frac{x^2}{4t}}
\]
for $t> 0$ and $x \neq 0$.
\end{theorem}

\begin{proof}
 Since ${\mathcal{K}}_{\nu}(y, t)$ is an entire function of order 3/2, we
can interchange the summations and integrations in the following equations.
Using Proposition \ref{Prop12.5}, we obtain
\begin{align*}
\int_{0}^{\infty} e^{- \frac{4y^3}{27x^2}}\,{\mathcal{K}}_{\nu}(y,t)\,dy
& =  \frac{1}{(3t)^{\nu + 2/3}}\, \sum_{n =
0}^{\infty}\,\frac{(-1)^nt^{-n}}{3^{3n}\,n!\,(1/3)_{n}}\,
\int_{0}^{\infty}e^{- \frac{4y^3}{27x^2}}\,y^{3n}\,dy \\
 & =  \frac{\Gamma(1/3)x^{2/3}}{(3t)^{\nu + 2/3}2^{2/3}}\,\sum_{n = 0}^{\infty}\,\frac
 {(-1)^nx^{2n}t^{-n}}{2^{2n}\,n!} \\
 & =  \frac{\Gamma(1/3)x^{2/3}2^{\nu}}{3^{\nu + 2/3}}\,\frac{1}{(2t)^{\nu + 2/3}}\,
 e^{- \frac{x^2}{4t}}
\end{align*}
\end{proof}

Next we let $P_{n, m}(x, t)$ denote the radial heat polynomials of Cholewinski
and Haimo \cite{CholH68} and of Bragg \cite{Brag65}. We have
\[P_{n, \mu}(x, t) = \sum_{k = 0}^n2^{2k}
\binom nk \frac{\Gamma(\mu + 1/2 + n)}{\Gamma(\mu + 1/2 + n - k)}\,
x^{2(n - k)}t^{k}\,.\]

\begin{theorem} \label{Thm12.8}
Let $p_{n}^{\nu}(x, t)$ denote the $\nu$- diffusion polynomials
associated with $\vartheta_{\nu}$. Then
\[
\frac{2^{2n + 2/3}x^{- 2/3}}{3^{3n}\Gamma(n + 1/3)}\,\int_{0}^{\infty}
e^{-\frac{4y^3}{27x^2}}\, p_{n}^{\nu}(y, t)\,dy = P_{n, \nu +
1/6}(x, t)
\]
\end{theorem}

The proof follows by using \eqref{polys} and Proposition \ref{Prop12.5}, the result
follows by interchanging the finite summation and integration.

Theorems \ref{Thm12.7} and \ref{Thm12.8} show that $\nu$-diffusion polynomials
and radial heat functions can in some cases be related by an integral equation.

Let $W_{n, \nu}(x, t)$ denote the Appell transform of $P_{n, \mu}(x, t)$ given by
\begin{align*}
W_{n, \nu}(x, t)
& =  \frac{1}{(2t)^{\nu + 1/2}}e^{-\frac{x^2}{4t}} P_{n,\mu}(x/t, - 1/t) \\
& =  t^{-2n}G_{\nu}(x, t)P_{n, \mu}(x, -t) \\
& =  \frac{(-1)^n2^{2n}(\nu + 1/2)_{n}}{2^{\nu + 1/2}t^{n + \nu + 1/2}}\,
 _{1}F_{1}\left[\begin{matrix}
 n + \nu + 1/2 \\
 \nu + 1/2\end{matrix}\bigg|-\frac{x^2}{4t}\right]\,;
\end{align*}
see Cholewinski and Haimo \cite{CholH68}. $W_{n, \mu}(x, t)$ is a solution of the
radial heat equation, \eqref{Eq12.12}, and plays the role of $z^{- (n + 1)}$ in
radial
heat expansions, see for example Widder \cite{Widd75} for the classical heat theory.

\begin{theorem} \label{thm25}
Let $Q_{n}^{\nu}(x, t)$ be the $\nu$-associated function given by \eqref{Eq11.4},
then
\begin{align*}
&\frac{(-1)^n3^{\nu + 2/3}2^{2n -
\nu}}{\Gamma(1/3)x^{2/3}}\,\int_{0}^{\infty}
e^{-\frac{4y^3}{27x^2}}Q_{n}^{\nu}(y, t)\,dy\\
& = W_{n, \nu + 1/6}(x, t)\\
& =  t^{-2n}\,\frac{1}{(2t)^{\nu + 1/6 + 1/2}}\,e^{-\frac{x^2}{4t}}
 P_{n, \nu + 1/6}(x, -t)
\end{align*}
\end{theorem}

Since $Q_{n}^{\nu}(y, t)$ is an entire function of order 3/2, the
proof follows by a term for term application of Proposition \ref{Prop12.5}.

This result shows that in a $\nu$-Appell transform theory the functions
$Q_{n}^{\nu}(y, t)$ play the role of the $W_{n, \mu}(x, t)$ in the radial heat
equation. The generalized Hankel translation defined by Bochner \cite{Boch55} or
Delsarte \cite{Dels36} is given by
 \begin{equation}
\begin{aligned}
 f(x \oslash_{\mu} y) & =  \frac{\Gamma(\mu + 1/2)}{\Gamma(\mu)\Gamma(1/2)}\,
 \int_{0}^{\pi}f(\{x^2 + y^2
 - 2xycos\,\theta\}^{1/2})\sin^{2\mu + 1}\,\theta\,d\theta \\
  & =  \int_{0}^{\infty}f(z)D_{\mu}(x, y, z)\,d\mu(z)\,,
\end{aligned}  \label{Eq12.17}
 \end{equation}
where
 \[D_{\mu}(x, y, z) = \frac{2^{3\mu - 5/2}\Gamma(\mu + 1/2)^2}{\Gamma(\mu)\,\pi^{1/2}}\,
 (xyz)^{1 - 2\mu}\Delta(x, y, z)^{2\mu - 2}
\]
 and $\Delta(x, y, z)$ is the area of a triangle with sides $x, y, z$ is g
 there is such a triangle, and otherwise $D(x, y, z) = 0$. The measure is given by
 \[d\mu(x) = \frac{x^{2\mu}}{2^{\mu - 1/2}\Gamma(\mu + 1/2)}\,dx
\]
 In the case $f(z) = z^{2n}$, we get the Bessel binomial
 \begin{align*}
 (x \oslash_{\mu} y)^{2n}
 & =  \int_{0}^{\infty}z^{2n}D_{\mu}(x, y, z)\,d\mu(z) \\
 & =  y^{2n}\,_{2}F_{1}\left[\begin{matrix}
  - n, - n - \mu + 1/2 \\
  \mu + 1/2\end{matrix}\bigg|\frac{x^2}{y^2}\right]\,.
 \end{align*}
 Hirschman developed the Banach algebra for ${\mathcal{L}}^{1}[0, \infty, d\mu]$ with
 translations given by \eqref{Eq12.17}, see \cite{Hirs60}. At the present time a
 Banach algebra associated with $\vartheta_{\nu}$ and the translation $(x \oplus_{\nu}
 y)^{3n}$ is unknown. In this paper the $\nu$-translation $f(x \oplus_{\nu}
 y)$ is defined for a restricted class of functions.

 \begin{theorem}  \label{Thm12.10}
 Let $\nu \geq 0$ and let $\mu = \nu + 1/2$. Then
 \[
 \frac{2^{2n + 4/3}(xy)^{- 2/3}}{\Gamma(1/3)^23^{3n}(1/3)_{n}}\,\int_{0}^{\infty}
 \int_{0}^{\infty}e^{- \frac{4}{27}\,(\frac{w^3}{y^2} + \frac{z^3}{x^2}}
 (z \oplus_{\nu} w)^{3n}\,dz\,dw = (x \oslash_{\mu} y)^{2n}
 \]
 for $x, y > 0$.
 \end{theorem}

The proof follows by interchanging the finite sums in the iterated
 integrals and by Proposition \ref{Prop12.5}.

 The integral relation given by Theorem \ref{Thm12.10} can be extended by a number pf
 classes of functions $f(z)$. For example if $f$ is defined by the Stieltjes integral
 \begin{equation}
 f(z) = \int_{0}^{\infty}{\mathcal{G}}_{\nu}(zx)\,d\beta(x)\,,
 \label{Eq12.22}
 \end{equation}
 where $\beta(x)$ is increasing and bounded with compact support for $d\beta(x)$ in
 $(0, \infty)$. Then
 \[|f(z)| \leq {\mathcal{G}}_{\nu}(a|z|)\int_{0}^{a}d\beta(x) \leq Me^{a|z|}
 \]
 where $a$ is is the least upper bound of the support of $d\beta(x)$. In this case $f$
 is an entire function of growth $\{1, a\}$ and it follows that
\[
 f(z \oplus_{\nu} w) = \int_{0}^{\infty}{\mathcal{G}}_{\nu}(zx){\mathcal{G}}_{\nu}(wx)\,
 d\beta(x)
\]
 with
\[
 |f(z \oplus_{\nu} w)|
 \leq  {\mathcal{G}}_{\nu}(zx){\mathcal{G}}_{\nu}(wx) \int_{0}^{a}d\beta(x)
 \leq  Me^{a|z| + a|w|}\,.
\]
  We define
\[f_{b}(x) = \int_{0}^{\infty}{\bf J}_{\mu}(xt)\,d\beta(t^{2/3})\,.\]
Thus $f_{b}(x)$ is an entire function which is bounded on the real axis. We have
\[
 |f_{b}(x)|  \leq  \int_{0}^{\infty}|{\bf J}_{\mu}(xt)|\,d\beta(t^{2/3})
 \leq  \int_{0}^{\infty}d\beta(t^{2/3}) < \infty
\]
 since $|{\bf J}_{\mu}(xt)| \leq 1$.
 From the Hankel translation theory it follows that
\[f_{b}(x \oslash_{\mu} y) = \int_{0}^{\infty} {\bf J}_{\mu}(tx){\bf J}_{\mu}(ty)\,
 d\beta(t^{2/3})\,.
\]

 \begin{theorem} \label{thm27}
 Let $f(z)$ be defined by \eqref{Eq12.22}. Then for $x, y > 0$,
 \begin{equation}
 \frac{2^{4/3}}{(xy)^{2/3}\Gamma(1/3)^2}\,\int_{0}^{\infty}\int_{0}^{\infty}
 e^{- \frac{4}{27}\,(\frac{w^3}{y^2} + \frac{z^3}{x^2})}\,f(z\oplus_{\nu}w)\,dz\,dw =
 f_{b}(x \oslash_{\mu} y)\,.
 \label{Eq12.26}
 \end{equation}
 \end{theorem}

\begin{proof} The integral in \eqref{Eq12.26} is dominated by
\[
 M\int_{0}^{\infty}\int_{0}^{\infty}
 e^{- \frac{4}{27}\,(\frac{w^3}{y^2} + \frac{z^3}{x^2})}\,dw\,dz < \infty
\]
 and therefore converges absolutely and locally uniformly. Hence we can invert the
 following iterated integrals. We note that by Theorem \ref{Thm12.6}, it follows that
\[
 \int_{0}^{\infty}e^{- \frac{4}{27}\,\frac{z^3}{x^2}}{\mathcal{G}}_{\nu}(tz)\,dz
 = \frac{x^{2/3}\Gamma(1/3)}{2^{2/3}}\,{\bf J}_{\mu}(xt^{3/2})
\]
 We have
 \begin{align*}
&\int_{0}^{\infty}\int_{0}^{\infty}
 e^{- \frac{4}{27}\,(\frac{w^3}{y^2} + \frac{z^3}{x^2})}\int_{0}^{\infty}
 {\mathcal{G}}_{\nu}(zt){\mathcal{G}}_{\nu}(wt)\,d\beta(t)\,dw\,dz \\
& =   \int_{0}^{\infty}d\beta(t)\int_{0}^{\infty}{\mathcal{G}}_{\nu}(wt)
  e^{- \frac{4}{27}\,\frac{w^3}{y^2}}\,dw\int_{0}^{\infty}{\mathcal{G}}_{\nu}(zt)
  e^{- \frac{4}{27}\,\frac{z^3}{x^2}}\,dz \\
& =  \frac{x^{2/3}\Gamma(1/3)}{2^{2/3}}\,\int_{0}^{\infty}{\bf J}_{\mu}(xt^{3/2})\,
  d\beta(t)\,
  \int_{0}^{\infty}{\mathcal{G}}_{\nu}(wt)
  e^{- \frac{4}{27}\,\frac{w^3}{y^2}}\,dw \\
& =  \frac{(xy)^{2/3}\Gamma(1/3)^2}{2^{4/3}}\,\int_{0}^{\infty}
  {\bf J}_{\mu}(xt^{3/2}){\bf J}_{\mu}(yt^{3/2})\,
  d\beta(t) \\
& =  \frac{(xy)^{2/3}\Gamma(1/3)^2}{2^{4/3}}\,f_{b}(x \oslash_{\nu} y),
  \quad\mbox{with the change of variables }t_{*} = t^{3/2}
\end{align*}
 Hence the theorem follows.
 \end{proof}

 The basic translation kernels $(x \oslash_{\nu} y)^{2n}$ and $(x \oplus_{\nu} y)^{3n}$
 can also be related to each other by Beta function integrals.

 \begin{theorem} \label{thm28}
 Let $\nu \geq 0$ and let $\nu_{*} = \nu + 1/6$. Then
 \[(x \oslash_{\nu_{*}} y)^{2n} = \frac{}{}\,\int_{0}^{1}t^{-2/3}(1 - t)^{-2/3}
 (x^{2/3}t^{1/3} \oplus_{\nu} y^{2/3}(1 - t)^{1/3})^{3n}\,dt/B(1/3,n + 1/3)\,.
 \]
 \end{theorem}

\begin{proof} The generalized binomials are related by
 \begin{align*}
\binom{\alpha_{3n}(\nu)}{\alpha_{3k}(\nu)}
& =  \binom{b_{2n}(\nu_{*})}{b_{2k}(\nu_{*})}
 \frac{(1/3)_{n}}{(1/3)_{k}(1/3)_{n - k} }\\
& = \binom{b_{2n}(\nu_{*})}{b_{2k}(\nu_{*})}
 \frac{B(1/3,n + 1/3)}{B(1/3 + k, n - k + 1/3)}\,,
\end{align*}
 where $B(p, q)$ is the Beta function. We have
 \begin{align*}
&\int_{0}^{1}t^{-2/3}(1 - t)^{-2/3}
 (x^{2/3}t^{1/3} \oplus_{\nu} y^{2/3}(1 - t)^{1/3})^{3n}\,dt \\
& =  \sum_{k = 0}^n\binom{\alpha_{3n}(\nu)}{\alpha_{3k}(\nu)}
 x^{2k}y^{2(n - k)}\int_{0}^{1}t^{k - 2/3}(1 - t)^{n - k - 2/3}\,dt \\
& =  \sum_{k = 0}^n\binom{b_{2n}(\nu_{*})}{b_{2k}(\nu_{*})}
 \frac{B(1/3,n + 1/3)}{B(1/3 + k, n - k + 1/3)}\,
 B(1/3 + k, 1/3 + n - k)x^{2k}y^{2(n - k)} \\
& =  B(1/3, n + 1/3)\sum_{k = 0}^n\binom{b_{2n}(\nu_{*})}{b_{2k}(\nu_{*})}
 x^{2k}y^{2(n - k)} \\
& =  B(1/3, n + 1/3)(x \oslash_{\nu_{*}} y)^{2n}
 \end{align*}
 and the theorem follows.
 \end{proof}

 The basis addition formula for Bessel functions is given by the integral
 \[
 {\bf J}_{\nu_{*}}(x \oslash_{\nu_{*}} y)
  =  \int_{0}^{\infty} {\bf J}_{\nu_{*}}(z)D_{\nu_{*}}(x, y, z)\,d\mu_{\nu_{*}}(z)
  =  {\bf J}_{\nu_{*}}(x){\bf J}_{\nu_{*}}(y)\,.
 \]
 Using the previous theorem, a calculation shows that
 \begin{align*}
 {\bf J}_{\nu_{*}}(x \oslash_{\nu} y)
 & =  \frac{1}{B(1/3, 1/3)}\,\int_{0}^{1}
 t^{-2/3}(1 - t)^{-2/3}\\
 &\quad\times{} _{1}F_{2}\left[\begin{matrix} 2/3 \\
 1/3, \nu + 2/3
 \end{matrix}\bigg|\frac{(x^{2/3}t^{1/3} \oplus_{\nu} y^{2/3}(1 - t)^{1/3})^3}
 {4}\right]\,dt \\
 & =  {\bf J}_{\nu_{*}}(x){\bf J}_{\nu_{*}}(y)
 \end{align*}

 \section{Generalized Positive Definite Kernels}

 Bochner \cite{Boch55} obtained a positive definite theory associated with Bessel functions.
 Bochner's main result was extended by Cholewinski, Haimo and Nussbaum \cite{CholHN70}.
 The positive definite results depend on the Banach algebra associated with kernel
 functions, see Hirschman \cite{Hirs60}. In this section we obtain partial results
 concerning positive definite functions associated with the $(x \oplus_{\nu}y)$
 translations. Banach algebra results are not available in this latter case.

 Let $f(x)$ be a function on $0 \leq x < \infty$ for which the $\nu$-translation function
 $f(x \oplus_{\nu}y)$ is well defined. The function $f$ is said to be $\nu$-positive
 definite if
 \begin{equation}
 \label{posdef}
 \sum_{i, j = 1}^na_{i}\bar{a}_{j}f(x_{i} \oplus_{\nu}y_{j}) \geq 0
 \end{equation}
 for all finite sets $0 < x_{1},x_{2},\dots,x_{n}$ and complex $a_{1},a_{2},\dots,
 a_{n}$. We write $f \in PD_{\nu}$. Since the discrete sum \ref{posdef} implies its
 continuous counterpart, see for example Widder \cite{Widd75}, p. 270, we have for
 suitable real valued continuous functions $\phi(x)$ that
 \[
 I(\phi) = \int_{0}^{\infty} \int_{0}^{\infty} \phi(x) \phi(y) f(x_{i} \oplus_{\nu}y_{j})
 \,d\eta_{\nu}(y)\,d\eta_{\nu}(x) \geq 0
 \]
 Let $\phi(x)$ be a function on $0 \leq x < \infty$ such that the integral
 \begin{equation}
 \label{Laplace2}
 \int_{0}^{\infty} |\phi(x)|\,{\mathcal{G}}_{\nu}(xz)\,d\eta_{\nu}(x)
 \end{equation}
 converges locally uniformly for $z$ in $R^{+}$. We define the $LT_{\nu}$ transform of $\phi$ by
 \begin{equation}
 \label{ltrans}
 LT_{\nu}(\phi) = \hat{\phi}(z) = \int_{0}^{\infty} \phi(x)\,{\mathcal{G}}_{\nu}(xz)\,d\eta_{\nu}(x)
 \end{equation}
 This is a generalization of the Laplace transform or of the Hankel transform in the Bessel function
 case. Clearly this transform is linear on the functions for which \ref{Laplace2} converges. If
 $\beta(t)$ is any function on $0 \leq t < \infty$ for which the integral
\[
 \int_{0}^{\infty} {\mathcal{G}}_{\nu}(xt)\,d\beta(t)
\]
 converges locally uniformly for $x$ in $R^{+}$, the $LT_{\nu}$-Stieltjes transform of $\beta(t)$ is
 given by
\[
 \hat{\beta}_{S}(x) = \int_{0}^{\infty} {\mathcal{G}}_{\nu}(xt)\,d\beta(t)
\]

 \subsection*{Examples}
 A. Let $\phi(x) \in C_{00}^{\infty}(R^{+})$, with $\phi(x) = 0$ for $x \geq a$
 then
 \[
 |\hat{\phi}(z)|  \leq  \int_{0}^{\infty} |\phi(x)|
 |{\mathcal{G}}_{\nu}(xz)|\,d\eta_{\nu}(x)
    =  \int_{0}^{a} |\phi(x)||{\mathcal{G}}_{\nu}(xz)|\,d\eta_{\nu}(x)
    \leq  M\,e^{a|z|}\,.
 \]
 Thus $\phi$ is an entire function in $z$ of order 1 and type $a$.

\noindent B. Let $\beta(x)$ be an increasing function on $0 \leq x < \infty$,
which is constant for $x \geq a$,
 then
 \[
 |\hat{\beta}_{S}(x)|  \leq  \int |{\mathcal{G}}_{\nu}(xt)|\,d\beta(t)
 \leq  \int_{0}^{a} |{\mathcal{G}}_{\nu}(xt)|\,d\beta(t)
 \leq  M\,e^{a|x|}\,.
 \]
 Thus $\hat{\beta}_{S}$ has an extension to an entire function in $x$

\noindent C. Let $\phi$ be a continuous function on $0 \leq x < \infty$, such that
 $|\phi(x)| \leq M\exp (-x^{\rho})$ with $\rho > 2$.
 Then $\hat{\phi}(z)$ is given by
 \ref{ltrans} is an entire function in $z$.

 Let $\beta(z)$ be an increasing function as in Example B and $\phi$ be an
 element of $C_{00}^{\infty}(R^{+})$. The function
 \[f(x \oplus _{\nu} y) = \int_{0}^{\infty}{\mathcal{G}}_{\nu}(xz){\mathcal{G}}_{\nu}(yz)\,d\beta(z)\]
 is well-defined and, using Fubini's Theorem, we get
 \begin{equation} \label{eq13.7}
\begin{aligned}
 I(\phi) & =  \int_{0}^{\infty}\int_{0}^{\infty} \phi(x)\bar{\phi}(y)f(x \oplus _{\nu} y)\,
 d\eta_{\nu}(x)d\eta_{\nu}(y)  \\
   & =  \int_{0}^{\infty} d\beta(z) \int_{0}^{\infty} \phi(x){\mathcal{G}}_{\nu}(xz)\,d\eta_{\nu}(x)
   \int_{0}^{\infty} \bar{\phi}(y){\mathcal{G}}_{\nu}(yz)\,d\eta_{\nu}(y)  \\
   & =  \int_{0}^{\infty} |\hat{\phi}(z)|^2\,d\beta(z) > 0
\end{aligned}
 \end{equation}

 \begin{theorem} \label{thm29}
 Let  \[
 f(x) = \int_{0}^{\infty} {\mathcal{G}}(xz)\,d\beta(z)
 \]
 with $\beta$ given by Example B, the $f$ is in $PD_{\nu}$.
 \end{theorem}

\begin{proof} Let $0 <  x_{1},x_{2},\dots,x_{n}$ and
$a_{1},a_{2},\dots,a_{n}$ be
 arbitrary, then
 \[
 \sum_{i,j = 0}^n a_{i}\bar{a}_{j}f(x_{i}\oplus_{\nu}x_{j}) = \int_{0}^{\infty}
 |\sum_{i = 1}^n a_{i}{\mathcal{G}}_{\nu}(x_{i}z)|^2\,d\beta(z) \geq 0\,.
\]
\end{proof}

\subsection*{Example} % 13.2
The source kernel $K_{\nu}(x;t)$ is $\nu$-positive definite. We have
 \[
 K_{\nu}(x;t) = \int_{0}^{\infty} \exp(-ty^3){\mathcal{G}}_{\nu}(xy)
 \,d\eta_{\nu}(y)\]
 and
 \[
 \sum_{i,j = 0}^n a_{i}\bar{a}_{j}K_{\nu}(x_{i}\oplus_{\nu}x_{j};t)
 =  \int_{0}^{\infty} \exp(-ty^3)|
 \sum_{i = 1}^n a_{i}{\mathcal{G}}_{\nu}(x_{i}y)|^2\,d\eta_{nu}(y)
  \geq  0
 \]

 Following S. Bernstein \cite{Bern??} we define a function $f(x)$ on $0 < a < x < b$ to be
 $\nu$-absolutely monotonic
 if it has non-negative $\vartheta_{\nu}$-derivatives of all orders, i.e.,
$\vartheta_{\nu}^{k}f(x) \geq 0$, $0 < a < x < b$; $k = 0,1,2,\dots$.

\subsection*{Examples}% 13.3
The following functions are $\nu$-absolutely monotonic:
 \begin{enumerate}
 \item $f(x) = \sum_{k = 0}^n a_{k}x^{3k}$ with $a_{k} \geq 0$ on $0 \leq x < \infty$.
 \item $f(x) = \sum_{k = 0}^{\infty} a_{k}x^{3k}$ for $0 \leq x \leq \rho$ with $a_{k} \geq 0$.
 \item $f(x) = \int_{0}^{\infty} {\mathcal{G}}_{\nu}(xy)\,d\beta(y)$ for $0 \leq x < \infty$ with
 $\beta(y)$ increasing and constant for  $y \geq a > 0$. Functions given this way are also in
 $PD_{\nu}$ on $0 \leq x < \infty$.
 \end{enumerate}

 \begin{theorem} \label{thm30}
 Let $\phi \in C_{00}^{\infty}(0,\infty)$ be a non-negative function and let
 \[
 u(x\oplus_{\nu}y;t) = \int_{0}^{\infty} \exp(-tz^3)\phi(z){\mathcal{G}}_{\nu}(yz)
 {\mathcal{G}}_{\nu}(xz)\,d\eta_{\nu}(z)\,.\]
Then $u(x\oplus_{\nu}y;t)$ is a $\nu$-positive definite Airy diffusion.
 \end{theorem}

\begin{proof} By Theorem \ref{cont_func}, $u(x\oplus_{\nu}y;t)$ is in
 ${\mathcal{L}}^{1}((0, \infty), d\eta_{\nu}(x))$
 for $t > 0$ and $0 \leq y < \infty$, and clearly $u$ is in
 ${\mathcal{H}}_{\nu}(R^{+},t > 0)$,
 $0 \leq y < \infty$.
 Let $0 < x_{1},x_{2},\dots,x_{n}$ and let $\alpha_{1},\alpha_{2},\dots,\alpha_{n}$ be
 arbitrary complex numbers, then
 \begin{equation} \label{eq13.9}
 \begin{aligned}
 \sum_{i,j = 1}^n \alpha_{i}\bar{\alpha}_{j}u(x_{i}\oplus_{\nu}x_{j};t)
 & =  \int_{0}^{\infty}
 \exp(-ty^3)\phi(y)\sum_{i,j = 1}^n\alpha_{i}\bar{\alpha}_{j}{\mathcal{G}}_{\nu}(x_{i}y)
 {\mathcal{G}}_{\nu}(x_{j}y)\,d\eta_{\nu}(y)  \\
   & =   \int_{0}^{\infty}
 \exp(-ty^3)\phi(y)|\sum_{i = 1}^n\alpha_{i}{\mathcal{G}}_{\nu}(x_{i}y)|^2
 \,d\eta_{\nu}(y)\geq  0
\end{aligned}
 \end{equation}
 since $\phi(y) \geq 0$. Note that Inequality \eqref{eq13.9} is also valid for
 $t = 0$. \end{proof}

 \begin{corollary} \label{cor3}
 The function $u(x\oplus_{\nu}\omega_{6}y;t) = u(z_{\nu};t)$ is a $\nu$-analytic function
 of the umbral variable $z_{\nu}$.
 \end{corollary}

\begin{proof}
We have
\[ u(x\oplus_{\nu}\omega_{6}y;t) = u(z_{\nu};t)
 = \int_{0}^{\infty} \exp(-tw^3)\phi(w)
 {\mathcal{G}}_{\nu}(xw)G_{\nu}(yw)\,d\eta_{\nu}(w)\,.
\]
 Interchanging differentiation and integration, we have
 \[
 \vartheta_{x}\,u(z_{\nu};t) = \int_{0}^{\infty} \exp(-tw^3)\phi(w)(-w)^3
 {\mathcal{G}}_{\nu}(xw)G_{\nu}(yw)\,d\eta_{\nu}(w)
 = -\vartheta_{y}\,u(z_{\nu};t)\,.
 \]
 Thus the $\nu$-Cauchy-Riemann equations hold for $x,y \geq 0$ and therefore $u$ is $\nu$-analytic in the
 ``$z_{\nu}$-umbral plane''.
 \end{proof}

 \begin{corollary} \label{cor4}
 The function
 \[v(x\oplus_{\nu}y;t) = \int_{0}^{\infty} \exp(tw^3\phi(w)
 {\mathcal{G}}_{\nu}(xw){\mathcal{G}}_{\nu}(yw)\,d\eta_{\nu}(w)
\]
 is a $\nu$-positive definite Airy diffusion. Moreover, $v$ is a $\nu$-absolutely
monotonic in the $x$  (or $y$) variable. The function $v(z_{\nu};t)$ is
also $\nu$-analytic.
 \end{corollary}

 The above results are also valid for $\beta(t)$ increasing and constant for $t \geq a > 0$. In this
 case we have
 \begin{gather*}
 u(x\oplus_{\nu}y;t) = \int_{0}^{\infty} \exp(-tw^3
 {\mathcal{G}}_{\nu}(xw){\mathcal{G}}_{\nu}(yw)\,d\beta(t)\,,\\
v(x\oplus_{\nu}y;t) = \int_{0}^{\infty} \exp(tw^3
 {\mathcal{G}}_{\nu}(xw){\mathcal{G}}_{\nu}(yw)\,d\beta(t)\,.
\end{gather*}
 In general functions defined by integrals of the type
 \[f(x) = \int_{0}^{\infty} {\mathcal{G}}_{\nu}(xy)\,d\beta(y) \]
 are absolutely monotonic for suitable increasing functions $\beta(y)$. However, the representation is
 not necessarily unique as the following example demonstrates.

\subsection*{Example}
Let
 \[s_{n} = \int_{0}^{\infty} u^{3n}\,3u^2\exp (-\frac{1}{2}\,u)\sin
 \frac{\sqrt{3}}{2}\,u\,du\,.
\]
 A calculation shows that $s_{n} = 3(3n + 2)!\sin\,\pi(n + 1) = 0$,
 for $n = 0,1,2,\dots\,$. Next
 we have
 \[\mu_{n} = 3\cdot2^{3n + 3}! = \int_{0}^{\infty} u^{3n}\,d\beta(u)
 \]
 where $\beta(u) = \int_{0}^{u} 3\exp (-\frac{1}{2}t)t^2\,dt$.
  Let
 \[\alpha(u) = \int_{0}^{u} 3t^2\exp (-\frac{1}{2}t)\,(1 - \sin
 \frac{\sqrt{3}}{2}t)\,dt\,.
 \]
Then
 \[\int_{0}^{\infty} u^{3n}\,d\alpha(u) = 3\cdot2^{3(n+1)}\,(3n+2)!
 = \int_{0}^{\infty}u^{3n}\,d\beta(u)
\]
 with $\beta(u) \neq \alpha(u)$. Consider
 \[f(x) = \int_{0}^{\infty} {\mathcal{G}}_{\nu}(xu)\,d\beta(u)\,.
\]
 We have
 \[|f(x)| \leq M\int_{0}^{\infty} u^2\exp ((-\frac{1}{2} + x)u)\,du\]
 for $0 \leq x < \frac{1}{2}$. It easily follows that $f$ is absolutely monotonic for
 $0 \leq x < 1/2$. However
 \[
 f(x)  =  \sum_{n = 0}^{\infty} \frac{3\cdot2^{3(n + 1)}\,(3n + 2)!}{3^{3n}\,n!\,
 (\frac{1}{3})_{n}\, (\nu + \frac{2}{3})_{n}}x^{3n}
 =  6^3\,\sum_{n = 0}^{\infty} 2^{3n}\,\frac{(n + \frac{1}{3})(\frac{2}{3})_{n + 1}}{(\nu
   + \frac{2}{3})_{n}}x^{3n}
 \]
 which converges for $|x| < 1/2$, and
 \[f(x) = \int_{0}^{\infty} {\mathcal{G}}_{\nu}(xu)\,d\beta(u) =
 \int_{0}^{\infty} {\mathcal{G}}_{\nu}(xu)\,d\alpha(u)\,.
 \]

 \begin{theorem} \label{thm31}
 Let $\alpha_{c}(t)$ be an increasing function on $0 \leq t < \infty$ with compact
 support in $[0,a]$, $a > 0$, and let
 \[f(z) = \int_{0}^{\infty} {\mathcal{G}}_{\nu}(zt)\,d\alpha_{c}(t)\]
 If $f(x_{0}) = 0$ for some $x_{0} > 0$, then $f(x) \equiv 0$.
 \end{theorem}

 \begin{proof} We have
 \[0 \leq \mu_{n} = \int_{0}^{\infty} t^{3n}\,d\alpha_{c}(t)
 \leq a^{3n}\int_{0}^{a}d\alpha_{c}(t) \leq a^{3n}M
 \]
 and
 \[f(z) = \int_{0}^{\infty} {\mathcal{G}}_{\nu}(zt)\,d\alpha_{c} = \sum_{n = 0}^{\infty}
 M\frac{\mu_{n}\,z^{3n}}{\alpha_{3n}(\nu)}
 \]
 with $f$ an entire function. Since $\mu_{n} \geq 0$, $f(x_{0}) = 0$ implies $\mu_{n} = 0$ for
 $n = 0,1,2,\dots$.
 \end{proof}

 \begin{corollary} \label{cor4a}
 Let
 \begin{equation}  \label{eq13.15}
 u(x,t) = \int_{0}^{\infty} \exp(ty^3)\,{\mathcal{G}}_{\nu}(xy)
 \,d\alpha_{c}(y)\,.
 \end{equation}
 If $u(x_{0},t_{0}) = 0$ for some $x_{0} > 0$ and $t_{0} \geq 0$,
 then $u(x,t) = 0$.
 \end{corollary}

\begin{proof} Clearly $u(x,t)$ is an absolutely monotonic $\nu$-Airy diffusion. The theorem
 implies that $u(x,t_{0}) = 0$ for $x \geq 0$. Hence we obtain
 \[0 \leq \int_{0}^{\infty} \,d\alpha_{c}(y) \leq \int_{0}^{\infty} \exp(t_{0}y^3)\,{\mathcal{G}}_{\nu}
 (xy)\,d\alpha_{c}(y) = 0\]
 and therefore $d\alpha_{c}(y)$ is the zero measure.

 Thus this $\nu$-Airy diffusions given by the representation \ref{eq13.15} are positive functions if
 $\alpha_{c}(y)$ is different from a constant.
 \end{proof}

 \begin{theorem} \label{thm32}
 Let $\alpha_{c}(y)$ be an increasing function with compact support and let
 \begin{equation}
 \label{eq13.16}
 f(x) = \int_{0}^{\infty} {\mathcal{G}}_{\nu}(xy)\,d\alpha_{c}(y)\,.
 \end{equation}
 Then
 \begin{equation}  \label{eq13.17}
 f(x) \geq 0,\quad \left | \begin{matrix}
 f(x) & \vartheta f(x) \\
 \vartheta f(x) & \vartheta'' f(x)
 \end{matrix} \right | \geq 0,\quad
 \left |  \begin{matrix}
 f(x) & \vartheta f(x) & \vartheta^2 f(x)\\
 \vartheta f(x) & \vartheta^2 f(x) & \vartheta^3 f(x) \\
 \vartheta^2 f(x) & \vartheta^3 f(x) &  \vartheta^{4} f(x)
 \end{matrix} \right | \geq 0, \dots
 \end{equation}
 and
 \[f'(x) \geq 0,\quad \left | \begin{matrix}
 \vartheta f(x) & \vartheta^2 f(x) \\
 \vartheta^2 f(x) & \vartheta^3 f(x)
 \end{matrix} \right | \geq 0,\quad
 \left | \begin{matrix}
 \vartheta f(x) & \vartheta^2 f(x) & \vartheta^3 f(x)\\
 \vartheta^2 f(x) & \vartheta^3 f(x) & \vartheta^{4} f(x) \\
 \vartheta^3 f(x) & \vartheta^{4} f(x) &  \vartheta^{5} f(x)
 \end{matrix} \right | \geq 0, \dots
 \]
 for $0 \leq x < \infty$.
 \end{theorem}

\begin{proof} From the integral representation \eqref{eq13.16},
it follows that
 \begin{gather*}
 \vartheta^n f(x) = \int_{0}^{\infty} y^{3n}\,{\mathcal{G}}_{\nu}(xy)
 \,d\alpha_{c}(y)\,,
 \\
[\sum_{i,j = 0}^n \vartheta_{\nu}^{i + j} f(x)a_{i}a_{j}
= \int_{0}^{\infty} {\mathcal{G}}_{\nu}(xy)\left (\sum_{i = 0}^n a_{i}t^{3i}\right )^2
 \,d\alpha_{c}(y) \geq 0
\\
 \sum_{i,j = 0}^n \vartheta_{\nu}^{i + j + m} f(x)a_{i}a_{j}
  = \int_{0}^{\infty} {\mathcal{G}}_{\nu}(xy)y^{3m}
  \left (\sum_{i = 0}^n a_{i}t^{3i}\right )^2\,d\alpha_{c}(y) \geq 0
\end{gather*}
 for $m \geq 1$. Since the quadratic forms are positive definite the associated determinants are
 non-negative.
 Letting $\vartheta^n f(x) = \mu_{n} = \int_{0}^{\infty} y^{3n}\,d\alpha_{c}(y)$, we get that
 the quadratic forms
 \[\sum_{i,j = 0}^n \mu_{i + j}a_{i}a_{j},\quad
 \sum_{i,j = 0}^n \mu_{i + j + 1}a_{i}a_{j}
 \]
 are non-negative. The $\nu$-generating function of the moments $\{\mu_{n}\}_{0}^{\infty}$ is
 \[f(x) = \sum_{n = 0}^{\infty} \frac{\mu_{n}}{\alpha_{3n}(\nu)}x^{3n}
 \]
 an entire function.

 Let
 \begin{equation}
 \label{eq13.21}
 u(x,t) = \int_{0}^{\infty} \exp(ty^3)\,{\mathcal{G}}(xy)\,d\alpha_{c}(y)
 \end{equation}
 then the determinants given by \eqref{eq13.17} with $f(x)$ being replaced by
 $u(x,t)$ for fixed $t \geq 0$ are non-negative. Since $u(x,t)$ is absolutely monotonic in the
 usual Bernstein sense, it follows that the determinants
 \begin{equation}  \label{eq13.22}
 u(x,t) \geq 0,\quad
 \left | \begin{matrix}
 u & D_{t} u \\
 D_{t} u & D_{t}^2 u
 \end{matrix} \right | \geq 0,\quad
 \left | \begin{matrix}
 u & D_{t} u & D_{t}^2 u\\
 D_{t} u & D_{t}^2 u & D_{t}^3 u \\
 D_{t}^2 u & D_{t}^3 & u  D_{t}^{4} u
 \end{matrix} \right | \geq 0, \dots
 \end{equation}
 and
\[ D_{t} u(x,t) \geq 0,\quad \left | \begin{matrix}
 D_{t} u & D_{t}^2 u \\
 D_{t}^2 u & D_{t}^3 u
 \end{matrix} \right | \geq 0,\quad \left |
 \begin{matrix}
 D_{t} u & D_{t}^2 u & D_{t}^3 u\\
 D_{t}^2 u & D_{t}^3 u & D_{t}^{4} u \\
 D_{t}^3 u & D_{t}^{4} u &  D_{t}^{5} u
 \end{matrix} \right | \geq 0, \dots
 \]
 are also non-negative.
 We also obtain that the $\nu$-diffusions given by equation \eqref{eq13.21}
 is logarithmically convex  for
 \[\frac{\partial^2}{\partial t^2}\log u(x,t) =
 \frac{u(x,t)\,\frac{\partial^2}{\partial t^2}u(x,t) - \left(\frac{\partial}{\partial t}
 u(x,t)\right )^2}{u(x,t)^2} \geq 0
 \]
 by \eqref{eq13.22}. From the extension of \eqref{eq13.22} to
 $D_{t}^{m} u(x,t)$, etc., we also get
 \[\frac{u(x,t)}{D_{t} u(x,t)} \geq \frac{D_{t} u(x,t)}{D_{t}^2 u(x,t)} \geq
 \frac{D_{t}^2 u(x,t)}{D_{t}^3 u(x,t)} \geq \dots\,.
 \]
 Thus $D_{t}^{m} u(x,t)$ is also logarithmically convex.
\end{proof}

 \section{$\nu$- Associated Nonlinear Equations}

 Associated with the $\nu$-Airy diffusion equation we obtain a non-linear partial differential
 equation which is linearized by the $\nu$-Airy equation.

 We let $\Delta(\nu) = D_{x}^2 + \frac{2 \nu}{x}\,D_{x} = x^{-2\nu}\,Dx^{2\nu}\,D$ denote the
 Euler-radial operator with $\nu \geq 0$. Further we define a non-linear differential operator by
 \begin{equation} \label{eq15.1}
 \begin{aligned}
 K_{\nu}(\phi)
 & =  x^{-3\nu}\,D_{x}x^{3\nu}\,D_{x} \phi - \frac{3\nu}{x^2}\,\phi -
 \frac{\nu}{x}\,\phi^2 + \frac{1}{9}\,\phi^3 - \phi\phi_{x}  \\
 & =  \frac{\partial^2 \phi}{\partial x^2} + \frac{3\nu}{x}\,\frac{\partial \phi}{\partial x}
  - \frac{3\nu}{x^2}\,\phi - \frac{\nu}{x}\,\phi^2 + \frac{1}{9}\,\phi^3 - \phi\,\frac{\partial \phi}
  {\partial x}
 \end{aligned}
 \end{equation}

 \begin{theorem} \label{thm33}
 Let $\nu \geq 0$, if $u(x,t)$ is a solution of the $\nu$-Airy equation
 \[\vartheta_{\nu} u(x,t) = \frac{\partial u(x,t)}{\partial t}\]
 then $\phi(x,t) = -3\,\frac{\partial}{\partial x}\,u(x,t)/u(x,t)$ is a solution of the non-linear
 partial differential equation
 \begin{equation}
 \label{eq15.2}
 \frac{\partial \phi(x,t)}{\partial t}
  =  \frac{\partial}{\partial x}\,K_{\nu}(\phi)
  =  \vartheta_{\nu} \phi - \phi \Delta(\nu) \phi + (\frac{1}{3}\,\phi^2 - \phi_{x})
  (\phi_{x} - \frac{3\nu}{x^2}) + \frac{6\nu}{x^3}\,\phi
 \end{equation}
 \end{theorem}

\begin{proof} Letting $u_{x} = -\frac{1}{3}\,\phi u$, we get $\frac{{\textstyle u_{xx}}}
 {{\textstyle u}} = \frac{1}{9}\,
 \phi^2 - \frac{1}{3}\,\phi_{x}$. Hence
 \[\frac{\partial u}{\partial t} = D_{x} x^{-3\nu}D_{x} x^{3\nu}u_{x}\]
 implies that
 \begin{align*}
 -3\frac{\partial u}{\partial t}
 & =  D_{x}(\frac{3\nu}{x}\,\phi u + u\phi_{x} + \phi u_{x}) \\
 & =  D_{x}(-\frac{9\nu}{x}\,u_{x} + u\phi_{x} + \phi u_{x}) \\
 & =  u(\frac{9\nu}{x^2}\,\frac{u_{x}}{u} - \frac{\phi u}{x}
 \frac{{\textstyle u_{xx}}}
  u + \phi\,\frac{{\textstyle u_{xx}}}{{\textstyle u}}
  + 2\phi_{x}\,\frac{{\textstyle u_{x}}}{{\textstyle u}} + \phi_{xx})
 \end{align*}
 It follows that
 \[
 -3\frac{\partial \ln u}{\partial t}
  =  \phi _{xx} + \frac{3\nu}{x}\,\phi _{x} - \frac{3\nu}{x^2}
 \,\phi + \frac{1}{9}\,\phi^3 - \frac{\nu}{x}\,\phi^2 - \phi \phi _{x}
  =  K_{\nu}(\phi)
 \]
 and therefore,
 \[\frac{\partial \phi}{\partial t} = \frac{\partial}{\partial x}
 \,K_{\nu}(\phi)
 \]
 \end{proof}

In the case that $\nu = 0$, we have the equation
 \begin{equation}
 \label{eq15.3}
 \frac{\partial \phi}{\partial t} = \frac{\partial ^3\phi}{\partial x^3} - \phi\,
 \frac{\partial ^2 \phi}{\partial x^2} + \frac{1}{3}\,\phi ^3\phi _{x} -
 \left (\frac{\partial
 \phi}{\partial x} \right )^2
 \end{equation}
 which linearizes to the Airy equation
 $\frac{\partial \phi}{\partial t} = \frac{\partial ^3 \phi} {\partial x^2}$.
 The KdV equation
 \[\frac{\partial \phi}{\partial t} + 6\phi\,\frac{\partial \phi}{\partial x} + \frac{\partial ^3 \phi}
 {\partial x^3} = 0\]
 linearizes to the Airy equation $\frac{\partial \phi}{\partial t} + \frac{\partial ^3 \phi}
 {\partial x^3} = 0$. Replacing $t$ by $-t$ gives the form of the Airy
 equation of this paper.

 We note that if $\phi(x,t)$ is a solution of the non-linear equation
 \eqref{eq15.2} then
 formally
 \[
 u(x,t) = B(t)\exp (-\frac{1}{3}\,\int_{0}^{x} \phi(y,t)\,dy)
\]
 is a solution of the $\nu$-Airy equation.

 The differential form of \eqref{eq15.2} suggests a conservation of ``mass'' for
 suitable solutions. If $\phi(x,t)$ is a solution of \eqref{eq15.2} such that
 $\phi$ and all of its derivatives vanish at $\pm \infty$, i.e., $\phi \in C_{0}^{\infty}(R)$, then
 \[
 \int_{-\infty}^{\infty} \frac{\partial \phi(x,t)}{\partial t}\,dx
 = \frac{\partial}{\partial t}\,
 \int_{-\infty}^{\infty} \phi(x,t)\,dx
 = K_{\nu}(\phi)|_{-\infty}^{\infty} = 0
 \]
 which implies the conservation form
 \[\int_{-\infty}^{\infty} \phi(x,t)\,dx = \mbox{constant,}
 \]
 a conservation of mass interpretation.

 Associated with the $\nu$-diffusion polynomials $p_{n}^{\nu}(x,t)$ we obtain an infinite sequence of
 rational
 function solutions of \eqref{eq15.2}. We have
 \[
 p_{n}^{\nu}(x,t) = x^{3n}\,_{3}F_{0}\big[-n,\frac{2}{3} - n,
 \frac{1}{3} - \nu - n\,|\,\big( -\frac{3}{x} \big)^3t\big]\,.
 \]
 Therefore,
 \begin{equation}  \label{eq15.5}
\begin{aligned}
 \omega _{n}^{\nu}(x,t)
 & =  -3\,\frac{\partial}{\partial x}\,p_{n}^{\nu}(x,t)/p_{n}^{\nu}(x,t)  \\
 & = -\frac{9n}{x}\,\frac{_{3}F_{0}\big[1 - n,2/3 - n,1/3 - \nu - n\,|\,
 \big( -\frac{3}{x} \big)^3t\big]}
 {_{3}F_{0}\big[-n,2/3 - n,1/3 - \nu - n\,|\,\big( -\frac{3}{x}
 \big)^3t\big]}
 \end{aligned}
 \end{equation}
 is a solution of the non-linear partial differential equation \eqref{eq15.2}. From this
 equation it follows that
 \[\omega _{n}^{\nu}(x,t) \sim -\frac{9n}{x}
 \]
 as $x \to \infty$.
 The first four $\omega _{n}^{\nu}$'s are given in the table:
\begin{gather*}
\omega _{0}^{\nu}(x,t) = 0\,,\\
\omega _{1}^{\nu}(x,t) = \frac{-9x^2}{x^3 + 9(\nu + 2/3)t}\,\\
\omega _{2}^{\nu}(x,t) = - \frac{18x^{5} + 648(\nu + 5/3)x^2t}
 {x^6 + 72(\nu + 5/3)x^2t + 324(\nu + 2/3)t^2}\,,\\
 \omega _{3}^{\nu}(x,t) = -3\frac{9x^{8} + 1215(\nu + 8/3)x^{5}t + 4536(\nu +
 5/3)_{2}x^2t^2}{x^{9} + \frac{405}{2}\,(\nu + 9/3)x^6t +
 1512(\nu + 5/3)_{2}x^3t^2 + 120\cdot3^6(\nu + 2/3)_{3}t^3}\,.
\end{gather*}
The source solution of the $\nu$-Airy equation also yields a solution of
\eqref{eq15.2}.
 We have
  \[{\mathcal{K}}_{\nu}(x,t) = \frac{2^{2/3}\,\Gamma(\frac{4}{3})}{3^{\nu + 2/3}t^{\nu + 1}}\,\cdot
  J_{-2/3}\left(\frac{2x^{3/2}}{\sqrt{27t}}\right)\]
 and therefore
 \[
 \omega_{\nu}(x,t)
 =  -3\,\frac{\partial}{\partial x}\,{\mathcal{K}}_{\nu}(x,t)/
  {\mathcal{K}}_{\nu}(x,t)
 =  -\left ( \frac{3x}{t}\right)^{1/2}\,\frac{J_{1/3}
 \left(\frac{2x^{3/2}}{\sqrt{27t}}\right)}
  {J_{-2/3}\left(\frac{2x^{3/2}}{\sqrt{27t}}\right)}
 \]
 Let $z = \frac{2x^{3/2}}{\sqrt{27t}}$. The recursion formula for Bessel functions
 $J_{\nu - 1}(z) + J_{\nu + 1}(z) = \frac{2\nu}{z}\,J_{\nu}(z)$ gives a continued fraction representation
 of the solution $\omega_{\nu}(x,t)$. We have
  \[
  \omega_{\nu}(x,t) = -\left(\frac{3x}{t}\right)^{1/2}\left \{{-\frac{4}{3z} -
 \frac{1}{-\frac{{\textstyle 10}}{{\textstyle 3z}} - \frac{{\textstyle 1}}{{\textstyle
 -\frac{{\textstyle 16}}{{\textstyle 3z}} -
 \frac{{\textstyle 1}}{{\textstyle -\frac{{\textstyle 22}}{{\textstyle z}} - \frac{{\textstyle
 J_{-\frac{11}{3}}(z)}}{{\textstyle J_{-\frac{14}{3}}(z)}}}}}}}}\right \}
 \]
Thus the non-linear equation \eqref{eq15.2} has continued fraction solutions.

The associated functions $Q_{n}^{\nu}(x,t)$ also yield solutions of \eqref{eq15.2}. We have
\begin{equation} \label{eq15.7}
\begin{aligned}
W_{n}^{\nu}(x,t) & =  -3\,\frac{\partial}{\partial x}\,Q_{n}^{\nu}(x,t)/Q_{n}^{\nu}(x,t)  \\
 & =  \frac{(n + \nu + 2/3)}{(\nu + 2/3)}\cdot\frac{x^2}{9t}\cdot\frac{
 _{1}F_{2}\left [ \begin{matrix}
 n + \nu + 5/3 \\
 4/3,\nu + 5/3
\end{matrix}\bigg| -\frac{x^3}{27t}\right]}
{_{1}F_{2}\left [\begin{matrix}
 n + \nu + 2/3 \\
 1/3,\nu + 2/3
\end{matrix}\bigg| -\frac{x^3}{27t}\right]}\,.
\end{aligned}
\end{equation}
Using the asymptotic estimates of Marichev \cite[p.~71]{Mari83}, we get
\[W_{n}^{\nu}(x,t) \sim \frac{x^2}{27t}\]
as $x \to \infty$ for $t > 0$.

Next we will show that the non-linear partial differential equation
\eqref{eq15.2} has time periodic solutions. Let
\[
C_{\nu}(x,t) = \exp(ity^3){\mathcal{G}}_{\nu}(\omega_{6}xy)
= \exp(ity^3)E_{\nu}(xy)
\]
which is a complex solution of the $\nu$-Airy equation, where
$\omega_{6} = \exp(i\pi /6)$. Since
\[
\frac{\partial}{\partial x}\,{\mathcal{G}}_{\nu}(x,y) = \frac{x^2y^3}{3(\nu + 2/3)}\,
_{0}F_{2}[4/3,\nu + 5/3\,|\,\left(\frac{xy}{3}\right)^3]
\]
we find that
\begin{align*}
C_{\nu}(x,t) & =  -3\,\frac{\frac{\partial}{\partial x}\,C_{\nu}(x,t)}{C_{\nu}(x,t)} \\
 & =  -i\,\frac{x^2y^3}{(\nu + 2/2)}\cdot \frac{_{0}F_{2}[4/3,\nu + 5/3\,|\,i\left (
 \frac{xy}{3}\right )^3]}{_{0}F_{2}[1/3,\nu + 1/3\,|\,i\left(\frac{xy}{3}\right)^3]}
\end{align*}
a solution that is independent of $t$.

The function $C_{\nu}(x,t)$ is periodic in the $t$ variable, we have
\[
C_{\nu}(x,t + 2\pi k/y^3) = C_{\nu}(x,t)
\]
for $y \neq 0$. The real and imaginary parts of $C_{\nu}(x,t)$ are also
$\nu$-diffusions. We have
\[
\exp(ity^3)E_{\nu}(xy) = (\cos ty^3 + i\sin ty^3)(cos_{\nu}(xy) +
i\sin_{\nu}(xy))\,.
\]
Hence
\[
R_{\nu}(x,t)  =  {\mathcal{R}}e\exp (ity^3)E_{\nu}(xy) \\
  =  \cos(ty^3)\cos _{\nu}(xy) - \sin(ty^3)\sin_{\nu}(xy)
\]
and
\begin{equation}
\label{eq15.10}
I_{\nu}(x,t)  =  dm\exp (ity^3)\,E_{\nu}(xy)  \\
  =  \sin(ty^3)\cos _{\nu}(xy) + \sin_{\nu}(xy)\cos (ty^3)
\end{equation}
are $t$-periodic $\nu$-diffusions. It follows that
\[
{\mathcal{R}}_{\nu}(x,i) = -3\,\frac{\cos(ty^3)\,\frac{\partial}{\partial x}\cos _{\nu}(xy) -
\sin(ty^3)\,\frac{\partial}{\partial x}\sin_{\nu}(xy)}
{\cos(ty^3)\cos _{\nu}(xy) - \sin(ty^3)\sin_{\nu}(xy)}
\]
and
\[
d_{\nu}(x,t) = -3\,\frac{\sin(ty^3)\,\frac{\partial}{\partial x}\cos _{\nu}(xy) +
\cos(ty^3)\,\frac{\partial}{\partial x}\sin_{\nu}(xy)}
{\sin(ty^3)\cos _{\nu}(xy) + \cos(ty^3)\sin_{\nu}(xy)}
\]
are $t$-periodic solutions of the non-linear partial differential equation
\eqref{eq15.2}. Both of these functions can be expressed in
terms of hypergeometric functions.

Actually, there exist a hierarchy of generalized Airy equations and the associated non-linear
partial differential equations. We present a few of the equations and results without proofs.
Let $\nu_{1}$ and $\nu_{2} \geq 0$, then the next simplest higher order Airy equation is given by
\begin{equation} \label{eq15.16}
\begin{aligned}
\frac{\partial u}{\partial t} & =
(x^3\,\frac{\partial ^6}{\partial x^6} + 3(\nu_{1} + \nu_{2} +
3) x^2\,\frac{\partial^{5}}{\partial y^{5}} + 9(2\nu_{1} +
2\nu_{2} + \nu_{1}\nu{2} + 2)x\,
\frac{\partial ^{4}}{\partial x^{4}}   \\
 & \quad + 3(6\nu_{1}\nu_{2} + 3\nu_{1} + 3\nu_{2} + 2)\,\frac{\partial ^3}{\partial x^3} -
 9\,\frac{\nu_{1}\nu_{2}}{x}\,\frac{\partial ^2}{\partial x^2} +
 9\,\frac{\nu_{1}\nu_{2}}{x^2}\,\frac{\partial}{\partial x})\,u
\end{aligned}
\end{equation}
Let $\underline {\nu} = (\nu_{1},\nu_{2})$ and let
\[
p_{n}^{\underline {\nu}}(x,t) = x^{3n}\,_{6}F_{0}(-n,n,2/3 - n,2/3 - n,
1/3 - \nu_{1} - n\,1/3 - \nu_{2} - n\,|\,\frac{3^6}{x^3}t)
\]
then $p_{n}^{\underline {\nu}}(x,t)$ is a polynomial solution of
\ref{eq15.16} for $n = 0,1,2,\dots$. Moreover, we have the
generating function
\begin{align*}
&\exp(tz^3)\,_{0}F_{5}(1,1/3,1/3,\nu_{1} + 2/3,\nu_{2}
+ 2/3\,|\,\big(\frac{xz}{9}\big)^3)\\
& = \sum_{n = 0}^{\infty}\frac{p_{n}^{\underline
{\nu}}(x,t)\,z^{3n}} {3^{6n}\,(n!)^2(1/3)_{n}^2(\nu_{1} +
2/3)_{n}(\nu_{2} + 2/3)_{n}}\,.
\end{align*}

\begin{theorem} \label{thm34}
Let $\nu_{1}$ and $\nu_{2} \geq 0$ and let $u(x,t)$ be a solution
of the higher order Airy diffusion \eqref{eq15.16}. Then $\phi(x,`,t) =
-3\,\frac{\partial}{\partial x}\,u(x,t)/u(x,t)$
is a solution of the non-linear equation
\begin{equation}\label{eq15.18}
\begin{aligned}
\frac{\partial \phi}{\partial t}
& = \frac{\partial}{\partial x}\, \vartheta_{\nu(2)}x^3K_{\nu(1)} -
\frac{1}{3}\,\frac{\partial}{\partial x}(x^3\,
K_{\nu(1)}(\phi)K_{\nu(2)}(\phi)  \\
 & \quad -\frac{2}{3}\,\phi\,\frac{\partial ^2}{\partial x^2}x^3K_{\nu(1)}(\phi) +
 \frac{2}{3}\,\frac{\partial}{\partial x}(\frac{1}{3}\,\phi^2 - \phi _{x})\,\frac{\partial}
 {\partial x}\,(x^3K_{\nu(1)}(\phi))
\end{aligned}
\end{equation}
\end{theorem}
The functions
\begin{align*}
w_{n}^{\underline {\nu}}(x,t)
&=  -3\,\frac{\partial}{\partial x}\,p_{n}^{\underline {\nu}}(x,t)/
p_{n}^{\underline {\nu}}(x,t) \\
 & = \!-\frac{9n}{x}\cdot \frac{_{6}F_{0}(-n,1 - n,2/3 - n,2/3 - n,
1/3 - \nu_{1} - n,\,1/3 - nu_{2} - n\,|\,\frac{3^6}{x^3}t)}
{_{6}F_{0}(-n,-n,2/3 - n, 1/3 - \nu_{1} - n,\,1/3 - nu_{2} -
n\,|\,\frac{3^6}{x^3}t)}
\end{align*}
are rational function solutions of \eqref{eq15.18}, $n = 0,1,2,\dots$.
It can be shown that \eqref{eq15.18} also has $t$-periodic solutions.

 \section{Constant Coefficient $\nu$-Differential Equations}

 In this section, solutions of polynomial operators with argument $\vartheta_{\nu}$ are obtained
 in a number of cases.  First of all we consider solutions of the differential equations
 \begin{equation}
 \label{constcoef}
 (\vartheta_{\nu} - a^3)y(x) = 0
 \end{equation}
 where $a$ is some complex number. For a series of the form $\sum_{n = 0}^{\infty}c_{n}x^{n + \lambda}$,
 we obtain
 \[
 \sum_{n = 0}^{\infty}c_{n}\{(3n + \lambda)(3n + \lambda + 3\nu - 1)(3n + \lambda - 2)x^{3n + \lambda}
 - a^3x^3x^{3n + \lambda}\}
 \]
 with indicial equations $f(\lambda) = \lambda(\lambda + 3\nu - 1)(\lambda - 2)$ and skip number three.
 For $\nu \geq 0$ and $\nu$ different from
 $\frac{1}{3}$, we obtain three linearly independent solutions of \eqref{constcoef}.
 They are, corresponding to the indicial roots 0, 2, and $1 - 3\nu$ respectively
 \begin{gather*}
 G_{\nu}(ax) = \sum_{n = 0}^{\infty} \frac{a^{3n}x^{3n}}{3^{3n}n!(1/3)_{n}(\nu + 2/3)_{n}} =
 _{0}F_{2}[1/3,\nu + 2/3\,|\,\left(\frac{ax}{3}\right)^3]\,,\\
 G_{\nu}^{(2)}(ax) = \sum_{n = 0}^{\infty} \frac{a^{3n}x^{3n + 2}}{3^{3n}n!(5/3)_{n}(\nu + 4/3)_{n}} =
 x^2\,_{0}F_{2}[5/3,\nu + 4/3\,|\,\left(\frac{ax}{3}\right)^3]\,,\\
 \begin{aligned}
 G_{\nu}^{(3)}(ax) &= \sum_{n = 0}^{\infty} \frac{a^{3n}x^{3n + 1 - 3\nu}}{3^{3n}n!(4/3 - \nu)_{n}
 (2/3 - \nu)_{n}n!} \\
 &= x^{1 - 3\nu}\,_{0}F_{2}[4/3 - \nu,2/3 - \nu\,|\,\left(\frac{ax}{3}\right)^3]
\end{aligned}
 \end{gather*}
 In the latter case we must have $\nu$ such that $4/3 - \nu$,
 $2/3 - \nu \neq 0,-1,-2,\dots$.
 The function
 $G_{\nu}(ax)$ is the basic zeta function of this paper associated with $\vartheta_{\nu}$.

 If $p(\vartheta_{\nu} = \sum_{k = 0}^na_{k}\vartheta_{\nu}^{k}$ is a monic polynomial operator,
 we find that $p(\vartheta_{\nu}G_{\nu}(rx)= p(r^3)G_{\nu}(rx)$.
 We let $p(r^3) = \prod_{k = 1}^n(r^3 - r_{k})$. If the $r_{k}$'s are distinct,
 we get the general solution of the equation $p(\vartheta_{\nu})y = 0$ is given by
\[
 y(x) = \sum_{k = 1}^n\{c_{1k}G_{\nu}(r_{k}^{1/3}x) + c_{2k}G_{\nu}^{(2)}(r_{k}^{1/3}x) +
 c_{3k}G_{\nu}^{(3)}(r_{k}^{1/3}x)\}
\]
 We are interested in determining solutions in the repeated root case.
 Thus we need to obtain solutions of the equation $(\vartheta_{\nu} - a^3)^{k}y = 0$. Due to
 the failure of a nice Leibnitz type product formula for
 $\vartheta^{k}(uv)$, an exponential shift type equation is not available for our repeated root equation.
 In the classical case the exponential shift formula
 $(D - a)^{k}(e^{ax}v) = e^{ax}D^{k}v$, leads to solutions in the case of
 repeated roots for constant coefficient equations in $D$. A new method that works
 in a number of cases follows.

 For $m$ a non-negative integer, we define
 \begin{gather*} G_{\nu - m}(ax)
 = \sum_{n = 0}^{\infty} \frac{a^{3n}x^{3n}}{3^{3n}n!(1/3)_{n}(\nu - m + 2/3)_{n}}\,,
 \\
G_{\nu - m}^{(2)}(ax)
 = \sum_{n = 0}^{\infty} \frac{a^{3n}x^{3n + 2}}{3^{3n}n!(5/3)_{n}(\nu
 - m + 4/3)_{n}}
\end{gather*}
 with $\nu$ such that $(\nu - m + 2/3)$ and $(\nu - m + 4/3)$ are not
 zero or a negative integer. The $\nu$-Airy operator
 can be written as
 \[
 \vartheta_{\nu} = \vartheta_{\nu - m} + \frac{3m}{x^3}{\mathcal{N}}({\mathcal{N}} - 2)
 \]
 where ${\mathcal{N}} = xD$ is the numbers operator.

 \begin{theorem} \label{thm35}
 For restricted $\nu$ and $0 \leq k \leq m$
 \begin{equation}
 \label{eq16.9}
 (\vartheta_{\nu} - a^3)^{k}G_{\nu - m}^{(2)}(ax) = \frac{m!}{(m - k)!}\,
 \frac{a^{3k}}{(\nu + 2/3 - m)_{k}}\,G_{\nu - (m - k)}^{(2)}(ax)
 \end{equation}
 \end{theorem}

\begin{proof} We consider the case $k = 1$, the general
 case follows by iteration. By \eqref{eq16.9}, it follows that
 \[(\vartheta_{\nu} - a^3)^{k}G_{\nu - m}^{(2)}(ax) = \frac{3m}{x^3}\,
 {\mathcal{N}}({\mathcal{N}} - 2)G_{\nu - m}(ax)\]
 where ${\mathcal{N}} = xD$ is the number operator. Since
 \[\frac{3m}{x^3}\,{\mathcal{N}}({\mathcal{N}} - 2)x^{3n} = 3m(3n)(3n - 2)x^{3(n - 1)}\]
 we get
 \begin{align*}
 \frac{3m}{x^3}\,{\mathcal{N}}({\mathcal{N}} - 2)G_{\nu - m}(ax)
 & = \sum_{n = 1}^{\infty}\,\frac{m(n - 2/3)x^{3(n - 1)}a^{3n}}{3^{3(n - 1)}(n - 1)!(1/3)_{n}(\nu
 - m + 2/3)_{n}} \\
 & =  \frac{ma^3}{(\nu - m + 2/3)}\,G_{\nu - (m - 1)}(ax)
 \end{align*}
 Since $(\vartheta_{\nu} - a^3)^{m}G_{\nu - (m - 1)}(ax) = \frac{m!a^{3m}}{(\nu + 2/3 - m)_{m}}
 G_{\nu}(ax)$,
  $(\vartheta_{\nu} - a^3)^{m + 1}G_{\nu - m}(ax) = 0$. Hence the
 equation
 \[(\vartheta_{\nu} - a^3)^{m + 1}y(x) = 0\]
 has solutions $G_{\nu}(ax),G_{\nu - 1}(ax),\dots,G_{\nu - m}(ax)$.
\end{proof}

 In the case of repeated roots corresponding to
 $G_{\nu}^{(2)}(ax)$, we have the equations
 \[
 (\vartheta_{\nu} - a^3)^{m}G_{\nu - m}^{(2)}(ax) = \frac{m!}{(m - k)!}\,\frac{a^{3k}}
 {(\nu + 4/3 - m)_{k}}\,G_{\nu - (m - k)}^{(2)}(ax)\,.
 \]
 Therefore, $(\vartheta_{\nu} - a^3)^{m + 1}y(x) = 0$ also has
 $G_{\nu}^{(2)}(ax),G_{\nu - 1}^{(2)}(ax),\dots,G_{\nu - m}^{(2)}(ax)$ as
 solutions.


 Based on the exponential shift equation
 $(D - a)^{k}x^ne^{ax} = \frac{n!}{(n - k)!}x^{n - k}e^{ax}$, we can find solutions
 of the eigenvalue problem for the wave operator $(\frac{\partial}{\partial x} -
 \frac{\partial}{\partial t})$.
 The eigenvalue problem
 \begin{equation}
 \label{eq16.12}
 (\frac{\partial}{\partial x} - \frac{\partial}{\partial t})u(x,t) = \lambda u(x,t)
 \end{equation}
 has polynomial type solutions $u_{n,\lambda}(x,t) = e^{\lambda x}(x + t)^n$.
 We will an analogue of this result for the corresponding $\vartheta_{\nu}$ problem.

 The corresponding eigenvalue problem in the $\nu$-Airy case
 is given by the equation
 \begin{equation}
 \label{eq16.13}
 (\vartheta_{\nu} - \frac{\partial}{\partial t})u(x,t) = \lambda ^3 u(x,t)
 \end{equation}
 Using \eqref{eq15.10}, we find that solutions corresponding
 to $u(x,0) = G_{\nu - n}(\lambda x)$ for suitable $\nu$ are given by
 \begin{equation} \label{eq16.14}
 \begin{aligned}
 C_{n,\lambda}^{\nu}(x,t)
 & =  \exp(t(\vartheta_{\nu} - \lambda ^3))G_{\nu - n}(\lambda x)  \\
 & =  \sum_{k = 0}^{\infty}\frac{t^{k}}{k!}(\vartheta_{\nu} - \lambda ^3)^{k}
  G_{\nu - n}(\lambda x)  \\
 & =  \sum_{k = 0}^{\infty}\,\frac{t^{k}}{k!}\,\frac{n!\lambda ^{3k}G_{\nu
 - (n - k)}(\lambda x)}
  {(n - k)!(\nu + 2/3 - n)_{k}}  \\
 & =  \sum_{k = 0}^{\infty}\binom n k
 \frac{t^{k}\lambda ^{3k}G_{\nu - (n - k)}(\lambda x)}
  {(\nu + 2/3 - n)_{k}}
\end{aligned}
 \end{equation}
 Solutions of the eigenvalue equation \eqref{eq16.13}
 also lead to solutions of the previously encountered non-linear
 partial differential equation
 \[\frac{\partial}{\partial t}\,\phi (x,t) = \frac{\partial}{\partial x}\,
 K_{\nu}(\phi)\,.
 \]
 Let $u_{\lambda}$ be a solution of the eigenvalue equation \eqref{eq16.13} and let
 \[\phi _{\lambda}(x,t) = -3\,\frac{\frac{\partial}{\partial x}\,u_{\lambda}(x,t)}
 {u_{\lambda}(x,t)}\,.
 \]
 Then
 \[
 \frac{\partial u}{\partial t} + \lambda ^3 u
  =  \vartheta u
  =  D(x^{-3\nu}Dx^{3\nu}u_{x}
  =  -\frac{1}{3}\,Dx^{-3\nu}Dx^{3\nu}u\phi\,.
 \]
Therefore, as before we find that
\[
 -\frac{3}{u}\,(\frac{\partial u}{\partial t} + \lambda ^3u) = K_{\nu}(\phi)
\]
 and
\begin{equation} \label{eq16.17}
 \frac{\partial}{\partial t}\phi_{\lambda}(x,t) = DK_{\nu}(\phi)
 \end{equation}
 Thus solutions of \ref{eq16.17} for suitable $\nu$ and
 $n$ a non-negative integer are given by
 \begin{align*}
 \phi_{n,\lambda}^{\nu}(x,t) & =  -3\,\frac{\frac{\partial}{\partial x}\,
 C_{n,\lambda}^{\nu}(x,t)}{C_{n,\lambda}^{\nu}(x,t)} \\
  & =  -3\,\frac{\sum_{k = 0}^n \binom n k
  \frac{t^{k}\lambda^{3k}}{(\nu + 2/3 - n)_{k}}\,\frac{\partial}{\partial x}\,
  G_{\nu - (n - k)}(\lambda x)}{\sum_{k = 0}^n \binom n k
  \frac{t^{k}\lambda^{3k}}{(\nu + 2/3 - n)_{k}}\,
  G_{\nu - (n - k)}(\lambda x)}
 \end{align*}
 which is a quotient of sums of hypergeometric
 functions. That is
 \[
 \phi_{n,\lambda}^{\nu}(x,t) = -x^2\,\frac{\sum_{k = 0}^n\binom n k
\frac{t^{k}\lambda^{3k}}{(\nu + 2/3 - n)_{k}}\,
  _{0}F_{2}[4/3,\nu + 5/3 - n + k\,|\,\left ( \frac{\lambda x}{3} \right )^3]}
  {\sum_{k = 0}^n\binom n k left
\frac{t^{k}\lambda^{3k}}{(\nu + 2/3 - n)_{k}}\,
  _{0}F_{2}[4/3,\nu + 5/3 - n + k\,|\,\left ( \frac{\lambda x}{3} \right )^3]}\]
 for $n = 0,1,2,\dots$.
 In the same manner, we find that
 \[
C_{n,\lambda}^{(2),\nu}(x,t) = \sum_{k = 0}^{\infty}\binom n k
\frac{t^{k}\lambda ^{3k}G_{\nu - (n - k)}^{(2)}(\lambda x)}
  {(\nu + 4/3 - n)_{k}}
\]
 is a solution of the eigenvalue problem \eqref{eq16.12}
for $n = 0,1,2,\dots$. Hence for suitable $\nu$ we obtain
solutions of the non-linear partial differential equation
\eqref{eq16.17} given by
\begin{align*}
\phi_{n,\lambda}^{\nu}(x,t)
& =  -3\,\frac{\frac{\partial}{\partial x}\,
 C_{n,\lambda}^{(2),\nu}(x,t)}{C_{n,\lambda}^{(2),\nu}(x,t)}  \\
& =  -3\,\frac{\sum_{k = 0}^n\binom n k
  \frac{t^{k}\lambda^{3k}}{(\nu + 4/3 - n)_{k}}\,\frac{\partial}{\partial x}\,
  G_{\nu - (n - k)}^{(2)}(\lambda x)}{\sum_{k = 0}^n\binom n k
  \frac{t^{k}\lambda^{3k}}{(\nu + 2/3 - n)_{k}}\,
  G_{\nu - (n - k)}^{(2)}(\lambda x)}  \\
& =  -3\,\frac{\sum_{k = 0}^n\binom n k
\frac{t^{k}\lambda^{3k}}{(\nu + 4/3 - n)_{k}}\,
  _{0}F_{2}[2/3,\nu + 4/3 - n + k\,|\big( \frac{\lambda x}{3} \big)^3]}
  {\sum_{k = 0}^n\binom n k \frac{t^{k}\lambda^{3k}}{(\nu + 2/3 - n)_{k}}\,
  _{0}F_{2}[5/3,\nu + 4/3 - n + k\,|\big( \frac{\lambda x}{3} \big)^3]}
\end{align*}
for $n = 0,1,2,\dots$.

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\end{document}