\documentclass[reqno]{amsart}
\usepackage{hyperref}

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2017 (2017), No. 216, pp. 1--9.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu}
\thanks{\copyright 2017 Texas State University.}
\vspace{8mm}}

\begin{document}
\title[\hfilneg EJDE-2017/216\hfil Convolutions with probability densities]
{Convolutions with probability densities and applications to PDEs}

\author[S. G. Gal \hfil EJDE-2017/216\hfilneg]
{Sorin G. Gal}

\address{Sorin G. Gal \newline
University of Oradea,
Department of Mathematics and Computer Science,
Str. Universit\u{a}\c{t}ii 1,
410087 Oradea, Romania}
\email{galsorin23@gmail.com}

\dedicatory{Communicated by Jerome A Goldstein}

\thanks{Submitted April 4, 2017. Published September 13, 2017.}
\subjclass[2010]{44A35, 35A22, 35C15}
\keywords{Probability density; convolution integral; Fourier transform; 
\hfill\break\indent initial value problem; final value problem}


\begin{abstract}
 In this article we introduce several new convolution operators, generated by
 some known probability densities. By using the inverse Fourier transform
 and taking inverse steps (in the analogues of the classical procedures used
 for example in the heat or Laplace equations), we  deduce the initial and
 final value problems satisfied by the new convolution integrals.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{definition}[theorem]{Definition}
\allowdisplaybreaks

\section{Introduction}

It is well known that the classical Gauss-Weierstrass, Poisson-Cauchy 
and Picard convolution singular integrals are based on convolutions with 
the standard normal density function $e^{-x^2}/\sqrt{\pi}$, standard Cauchy
 density function $\frac{1}{\pi(1+x^2)}$ and Laplace density 
function $e^{-|x|}/2$, respectively. Their approximation properties are studied, 
for example, in \cite{Butzer, Gal1}. Also, by using the Fourier transform 
method, it is known that the solutions of the initial value problems for the 
heat equation and Laplace equation are exactly the Gauss-Weierstrass and 
Poisson-Cauchy convolution singular integrals, respectively, 
see, e.g., \cite[p. 23]{Gold}. On the other hand, in our best knowledge, 
the initial value problem and the partial differential equation corresponding 
to the Picard singular integral, is missing from mathematical literature.
The main aim of the present paper is somehow inverse: 
introducing convolution singular integrals based on some known probability 
densities, we use the inverse Fourier transform in order to find the partial 
differential equations (initial and final value problems) satisfied by 
these integrals, including the Picard singular integral.

\section{Definitions of convolution operators}

In this section we introduce several convolution operators, based on some 
well-known densities of probability.

If $d(t, x)\ge 0$ for $t>0$ and $x\in \mathbb{R}$ is a probability density, 
that is $\int_{-\infty}^{+\infty}d(t, x)d x=1$, then our definitions are based 
on the general known formula
\begin{equation}\label{eq0}
\begin{aligned}
O_{t}(f)(x)
&=d(t, \cdot)*f(\cdot)=\int_{-\infty}^{+\infty}f(u) d(t, x-u) du \\
&=\int_{-\infty}^{+\infty}f(x-v) d(t, v) \,dv.
\end{aligned}
\end{equation}

\begin{definition} \label{def2.1} \rm
(i) For the Maxwell-Boltzmann type probability density 
(see, e.g., \cite[pp. 104, 148-149]{Papou})
$$
d(t, x)=\frac{1}{\sqrt{2 \pi}}  \frac{x^2 e^{-x^2/(2 t^2)}}{t^{3}},\quad
 x\in \mathbb{R},\; t>0
$$
and $f:\mathbb{R}\to \mathbb{R}$, we can formally define the 
Maxwell-Boltzmann convolution operator
\begin{equation}\label{eq17}
S_{t}(f)(x)=\frac{1}{\sqrt{2 \pi}}\int_{-\infty}^{+\infty}f(x-v)
 \frac{v^2 e^{-v^2/(2 t^2)}}{t^{3}}\,dv, \quad t>0,\; x\in \mathbb{R}.
\end{equation}

(ii) For the Laplace type probability density (see, e.g., \cite{Ever})
$$
d(t, x)=\frac{1}{2 t}e^{-|x|/t}, \quad t>0,\; x\in \mathbb{R}
$$
and $f:\mathbb{R}\to \mathbb{R}$, we can formally define the classical 
Picard convolution operator
\begin{equation}\label{eq24}
P_{t}(f)(x)=\frac{1}{2 t}\int_{-\infty}^{+\infty}f(x-v)\cdot e^{-|v|/t}\,dv,\quad
 t>0, \; x\in \mathbb{R}.
\end{equation}

(iii) For the exponential probability density (see, e.g., \cite{Ever,John})
$$
d(t, x)=\frac{t e^{-t|x|}}{2},\quad  x\in \mathbb{R},\; t>0
$$
and $f:\mathbb{R}\to \mathbb{R}$, we can formally define the exponential 
convolution operator
\begin{equation}\label{eq8}
E_{t}(f)(x)=\int_{-\infty}^{+\infty}f(x-v) \frac{t e^{-t|v|}}{2}\,dv, \quad
t>0,\; x\in \mathbb{R}.
\end{equation}

(iv) For any $n\in \mathbb{N}$, $P_{t}(f)(x)$ can be generalized to the so 
called Jackson type generalization of the Picard singular integral defined 
by (see, e.g., \cite{Gal1})
\begin{align*}
P_{n, t}(f)(x)
&=-\frac{1}{2 t}\int_{-\infty}^{+\infty}
\Big(\sum_{k=1}^{n+1}(-1)^{k}\binom{n+1}{k}f(x+k v)e^{-|v|/t}\Big)d v\\
&=\int_{-\infty}^{+\infty}f(x-u)
\Big[\sum_{k=1}^{n+1}(-1)^{k+1}\binom{n+1}{k}( \frac{1}{k})
 \frac{e^{-|u|/(kt)}}{2 t}\Big ]\,du,
\end{align*}
for $t>0$ and $x\in \mathbb{R}$.

(v) Starting from the well known Gauss-Weierstrass operator 
\[
W_{t}(f)(x)=\frac{1}{\sqrt{ \pi t}}\int_{-\infty}^{+\infty}f(x-v)e^{-v^2/t}\,d v,
\]
we can define its Jackson type generalization by (see, e.g., \cite{Gal1})
\begin{align*}
W_{n, t}(f)(x)
&=-\frac{1}{2 C^{*}(t)}\int_{-\infty}^{+\infty}
\Big(\sum_{k=1}^{n+1}(-1)^{k}\binom{n+1}{k}f(x+k v)e^{-v^2/t}\Big)d v \\
&=\int_{-\infty}^{+\infty}f(x-u)
\Big[\sum_{k=1}^{n+1}(-1)^{k+1}\binom{n+1}{k}(\frac{1}{k})
 \frac{e^{-u^2/(k t)}}{2 C^{*}(t)}\Big]\,du,
\end{align*}
for $t>0$ and $x\in \mathbb{R}$,
where $C^{*}(t)=\int_{0}^{\infty}e^{-u^2/t}du=\frac{\sqrt{t \pi}}{2}$.
Therefore, for any $n\in \mathbb{N}$, we can write
$$
W_{n, t}(f)(x)=\int_{-\infty}^{+\infty}f(x-u)
\Big[\sum_{k=1}^{n+1}(-1)^{k+1}\binom{n+1}{k}(\frac{1}{k})
 \frac{e^{-u^2/(k t)}}{\sqrt{\pi t}}\Big]\, du.
$$
\end{definition}

\section{Applications to PDEs}

Concerning the convolution operators defined in Section 2, we can state the 
following applications to PDEs.

\begin{theorem} \label{thm3.1} 
 (i) Suppose that $f, f', f'', f^{\prime \prime \prime}, f^{(4)}:\mathbb{R}\to \mathbb{R}$ are bounded and uniformly continuous on $\mathbb{R}$.
Then the solution of the initial value problem
\begin{gather*}
\frac{\partial u}{\partial t}(x, t)
= t^3 \frac{\partial^4 u}{\partial x^4}(x, t)-t^2 
 \frac{\partial^3 u}{\partial x^2 \partial t}(x, t)
 +3t \frac{\partial^2 u}{\partial x^2}(x, t),\\
\lim_{s\searrow 0}u(x, s)=f(x), \quad t>0,\; x\in \mathbb{R},
\end{gather*}
is $u(x, t):=S_{t}(f)(x)$.

(ii) Suppose that $f, f', f'' :\mathbb{R}\to \mathbb{R}$ are bounded and 
uniformly continuous on $\mathbb{R}$. Then the solution of the initial value problem
\begin{gather*}
\frac{\partial u}{\partial t}(x, t)
=t^2 \frac{\partial^3 u}{\partial x^2 \partial t}(x, t)
+2t \frac{\partial^2 u}{\partial x^2}(x, t),\\
\lim_{s\searrow 0}u(x, s)=f(x), \quad t>0, \; x\in \mathbb{R}
\end{gather*}
is $u(x, t):=P_{t}(f)(x)$.

(iii) Suppose that $f, f', f'' :\mathbb{R}\to \mathbb{R}$ are bounded and 
uniformly continuous on $\mathbb{R}$.
Then the solution of the final value problem
\begin{gather*}
\frac{\partial u}{\partial t}(x, t)
=\frac{1}{t^2} \frac{\partial^{3} u}{\partial x^2 \partial t}(x, t)-
\frac{2}{t^{3}} \frac{\partial^2 u}{\partial x^2}(x, t), \\
\lim_{s\to \infty}u(x, s)=f(x),\quad  t>0, \; x\in \mathbb{R}
\end{gather*}
is $u(x, t):=E_{t}(f)(x)$.

(iv) Suppose that $f, f', f'' :\mathbb{R}\to \mathbb{R}$ are bounded and
 uniformly continuous on $\mathbb{R}$.
Then we have
$$
P_{n, t}(f)(x)=\sum_{k=1}^{n+1}(-1)^{k+1}\binom{n+1}{k} u_{k}(x, t),
$$
where 
\[
u_{k}(x, t)=P_{k t}(f)(x)=\frac{1}{2 k t}
 \int_{-\infty}^{+\infty}f(x-u)e^{-|u|/(k t)}d u, \quad k=1, \dots, n+1\]
 are solutions of the initial value problems (for $t>0$ and $x\in \mathbb{R}$)
$$
\frac{\partial u_{k}}{\partial t}(x, t)
=k^2t^2\frac{\partial^{3} u_{k}}{\partial x^2\partial t}(x, t)+2k^2t
 \frac{\partial^2 u_{k}}{\partial x^2}(x, t), \quad 
 \lim_{s\searrow 0}u_{k}(x, s)=f(x).
$$

(v) Suppose that $f, f', f'' :\mathbb{R}\to \mathbb{R}$ are bounded and uniformly
 continuous on $\mathbb{R}$. Then we have
$$
W_{n, t}(f)(x)=\sum_{k=1}^{n+1}(-1)^{k+1}\binom{n+1}{k}\cdot u_{k}(x, t),
$$
where 
\[
u_{k}(x, t)=\frac{1}{\sqrt{k}}W_{k t}(f)(x)=\frac{1}{k\sqrt{\pi t}}
 \int_{-\infty}^{+\infty}f(x-u)e^{-u^2/(k t)}d u, \quad
k=1, \dots, n+1
\]
 are solutions of the initial value problems
\begin{gather*}
\frac{\partial u_{k}}{\partial t}(x, t)=\frac{k}{4}
 \frac{\partial^2 u_{k}}{\partial x^2}(x, t), \\ 
\lim_{s\searrow 0}u_{k}(x, s)=\frac{1}{\sqrt{k}}f(x),\quad  t>0,\; x\in \mathbb{R}.
\end{gather*}
\end{theorem}

\begin{proof}
Since for the convolution operator given by \eqref{eq0}, in general we have 
$d(t, v)\ge 0$, for all $t>0$ and $v\in \mathbb{R}$, by the standard method 
we easily obtain
\begin{align*}
|O_{t}(f)(x)-f(x)|
&\le \int_{-\infty}^{+\infty}|f(x-v)-f(x)|d(t, v)\,dv \\
&\le \int_{-\infty}^{+\infty}\omega_{1}(f; |v|)_{\mathbb{R}}d(t, v)\,dv\\
&\le 2\omega_{1}(f ; \varphi(t))_{\mathbb{R}},
\end{align*}
where $\omega_{1}(f ; \delta)_{\mathbb{R}}=\sup\{|f(x)-f(y)|;
 x, y \in \mathbb{R}, |x-y|\le \delta\}$ and 
$\varphi(t)=\int_{-\infty}^{+\infty}|v|\cdot d(t, v)d v$.
Evidently  this method is useful only if $\varphi(t)<+\infty$ for all $t>0$.

To deduce the PDEs satisfied by various convolution operators, we 
 need the concepts of Fourier transform of a function $g$. It is defined by
$$
F(g)(\xi)=\hat{g}(\xi)=\frac{1}{\sqrt{2\pi}}
 \int_{-\infty}^{+\infty}g(x)e^{-i \xi x}d x, \quad
\text{if } \int_{-\infty}^{+\infty}|g(x)|d x<+\infty,
$$
and the inverse Fourier transform is defined by
$$
F^{-1}(\hat{g})(x)=g(x)=\frac{1}{\sqrt{2 \pi}}
\int_{-\infty}^{+\infty}\hat{g}(\xi)e^{i \xi x}d \xi.
$$

(i) By making the change of variable $v=\sqrt{2} t s$, we obtain
\begin{align*}
\varphi(t)
&=\frac{1}{t^{3}}\frac{\sqrt{2}}{\sqrt{\pi}}
\int_{0}^{\infty}v^{3}e^{-v^2/(2 t^2)}dv \\
&=\frac{1}{t^{3}}\frac{\sqrt{2}}{\sqrt{\pi}}
 \int_{0}^{\infty}(2\sqrt{2}t^{3} s^{3})e^{-s^2}(\sqrt{2} t )d s \\
& =\frac{4\sqrt{2}}{\sqrt{\pi}} t\int_{0}^{\infty}s^{3}e^{-s^2}d s \\
&=\frac{2\sqrt{2}}{\sqrt{\pi}} t <2 t,
\end{align*}
which  implies
$$
|S_{t}(f)(x)-f(x)|\le 4\omega_{1}(f; t)_{\mathbb{R}},\quad t>0, \; x\in \mathbb{R}.
$$
Taking into account the uniform continuity of $f$, the above inequality 
 implies that $\lim_{t\searrow 0}S_{t}(f)(x)=f(x)$, for all $x\in \mathbb{R}$. 
Therefore we may take, by convention, $S_{0}(f)(x)=f(x)$, for all $x\in \mathbb{R}$.

Now,  to deduce the PDE satisfied by $S_{t}(f)(x)$, we write it in the form
\begin{align*}
S_{t}(f)(x)
&=\frac{1}{\sqrt{2 \pi}}\int_{-\infty}^{+\infty}f(y)
 \frac{(x-y)^2 e^{-(x-y)^2/(2 t^2)}}{t^{3}}\,dy\\
&=\frac{1}{\sqrt{2 \pi}}\int_{-\infty}^{+\infty}f(y)\cdot \hat{g}_{t}(y-x)d y.
\end{align*}
Here, by using standard reasoning and calculation 
(or  WolframAlpha software), we obtain
$$
g_{t}(\xi)=F^{-1}_{w}[w^2e^{-w^2/(2 t^2)}/t^{3}](\xi, t)
=e^{-t^2\xi^2/2}(1-t^2\xi^2),
$$
which implies
\begin{align*}
S_{t}(f)(x)
&=\frac{1}{\sqrt{2 \pi}}\int_{-\infty}^{+\infty}f(y)
\Big[\frac{1}{\sqrt{2 \pi}}\int_{-\infty}^{+\infty}
 e^{-i(y-x)\xi}e^{-t^2\xi^2/2}(1-t^2\xi^2)d \xi\Big]d y \\
 &=\frac{1}{\sqrt{2 \pi}}\int_{-\infty}^{+\infty}
\Big[\frac{1}{\sqrt{2 \pi}}\int_{-\infty}^{+\infty} e^{-i y \xi}f(y)dy\Big]
 e^{i x \xi} e^{-t^2\xi^2/2}(1-t^2\xi^2)d \xi \\
&=\frac{1}{\sqrt{2 \pi}}\int_{-\infty}^{+\infty}e^{i x \xi}\hat{f}(\xi)
 e^{-t^2\xi^2/2}(1-t^2\xi^2)d \xi \\
&=\frac{1}{\sqrt{2 \pi}}\int_{-\infty}^{+\infty}e^{i x \xi}\hat{u}(\xi, t)d \xi
=:u(x, t),
\end{align*}
where
$$
\hat{u}(\xi, t)=\hat{f}(\xi)\cdot e^{-t^2\xi^2/2}(1-t^2\xi^2).
$$
This is equivalent to 
$\hat{u}(\xi, t)\frac{e^{t^2\xi^2/2}}{1-t^2\xi^2}=\hat{f}(\xi)$,
which is equivalent to
$$
\frac{\partial }{\partial t}
\Big[\hat{u}(\xi, t)\frac{e^{t^2\xi^2/2}}{1-t^2\xi^2}\Big]
= \frac{\partial \hat{u}}{\partial t}(\xi, t)
 \frac{e^{t^2 \xi^2/2}}{1-t^2 \xi^2}+\hat{u}(\xi, t)
 \Big(\frac{e^{t^2 \xi^2/2}}{1-t^2 \xi^2}\Big)'_{t}=0.
$$
Note that the above relation can be evidently written under the form
$$
\frac{\partial }{\partial t}
\Big[\hat{u}(\xi, t)\frac{1}{F^{-1}_{w}(d(t, w))(\xi, t)}\Big]=0,
$$
where $d(t, x)$ is the Maxwell-Boltzmann type probability density 
in Definition \ref{def2.1}, (i), entering in the formula for $S_{t}(f)(x)$.

After simple calculations, the above formula is formally equivalent 
to (of course for $1\neq t^2 \xi^2$)
$$
\frac{\partial \hat{u}}{\partial t}(\xi, t)
+t^2\Big(-\xi^2\frac{\partial \hat{u}}{\partial t}(\xi, t)\Big)
-3t[-\xi^2\hat{u}(\xi, t)]-t^3\cdot [\xi^4 \hat{u}(\xi, t)]=0.
$$
Now, taking into account that
$$
\frac{\partial \hat{u}}{\partial t}(\xi, t)
=\widehat{\frac{\partial u}{\partial t}}(\xi, t), \quad
\widehat{\frac{\partial^2 u}{\partial x^2}}(\xi, t) 
= -\xi^2\hat{u}(\xi, t), \quad
 \widehat{\frac{\partial^{4} u}{\partial x^4}}(\xi, t) = \xi^{4}\hat{u}(\xi, t),
$$
and replacing the above, we obtain
$$
F\Big(\frac{\partial u}{\partial t}
+t^2 \frac{\partial^3 u}{\partial x^2 \partial t}
-3t \frac{\partial^2 u}{\partial x^2}-t^3 \frac{\partial^4 u}{\partial x^4}\Big)
(\xi, t)=0,
$$
that is
$$
\frac{\partial u}{\partial t}(x, t)
= t^3 \frac{\partial^4 u}{\partial x^4}(x, t)
-t^2 \frac{\partial^3 u}{\partial x^2 \partial t}(x, t)
+3t \frac{\partial^2 u}{\partial x^2}(x, t).
$$
Finally, following the above steps in inverse order, we arrive at the 
conclusion in the statement.

(ii) By \cite[p. 142, Corollary 3.4.2]{Butzer}, we have
$$
|f(x)-P_{t}(f)(x)|\le C\omega_{2}(f; t)_{\mathbb{R}}.
$$
Therefore, it is immediate that $\lim_{t\searrow 0}P_{t}(f)(x)=f(x)$, 
for all $x\in \mathbb{R}$.

To deduce the PDE satisfied by $P_{t}(f)(x)$, we reason exactly as in case (i). 
Indeed, by standard calculations (or using WolframAlpha software), we obtain
$$
F^{-1}_{w}[e^{-|w|/t}/(2t)](\xi, t)
=\frac{1}{\sqrt{2 \pi}}\frac{1}{1+t^2 \xi^2}
$$
and similar reasoning as in case (i) leads to
$$
P_{t}(f)(x)=\frac{1}{\sqrt{2 \pi}}
\int_{-\infty}^{+\infty}e^{i x \xi}\hat{u}(\xi, t)d \xi:=u(x, t),
$$
where
$$
\hat{u}(\xi, t)=\hat{f}(\xi)\frac{1}{t^2 \xi^2 + 1}.
$$
In fact, directly as in the proof of Theorem \ref{thm3.1}, (i), we can write
$$
\frac{\partial }{\partial t}
\Big[\hat{u}(\xi, t)\frac{1}{F^{-1}_{w}(d(t, w))(\xi, t)}\Big]=0,
$$
where $d(t, x)$ is the Laplace type probability density in 
Definition \ref{def2.1}, (ii), entering in the formula for $P_{t}(f)(x)$.
Therefore,
$$
\frac{\partial }{\partial t}\left [\hat{u}(\xi, t)\cdot (1+t^2\xi^2)\right ]
= \frac{\partial \hat{u}}{\partial t}(\xi, t)
+t^2 \xi^2\frac{\partial \hat{u}}{\partial t}(\xi, t)+2t \xi^2 \hat{u}(\xi, t)=0,
$$
which  leads to
$$
\frac{\partial u}{\partial t}(x, t)
=t^2 \frac{\partial^3 u}{\partial x^2 \partial t}(x, t)
+2t \frac{\partial^2 u}{\partial x^2}(x, t).
$$
Following the above steps, now from the end to the beginning,
 we arrive at the conclusion in the statement.

(iii) Firstly, we observe that $E_{t}(f)(x)=P_{1/t}(f)(x)$, for all 
$t>0$ and $x\in \mathbb{R}$. The, by (ii) we immediately get
$$
|E_{t}(f)(x)-f(x)|=|P_{1/t}(f)(x)-f(x)|\le 2\omega_{1}
\big(f; \frac{1}{t}\big)_{\mathbb{R}}.
$$
Then, again by standard calculations (or  using WolframAlpha), we have 
\[
F^{-1}_{w}(e^{-|w|t})(\xi, t)=\frac{\sqrt{2}}{\sqrt{\pi}}
 \frac{t}{t^2+\xi^2}.
\]
 This implies
$$
F^{-1}(d(t, w))(\xi, t)=\frac{t}{2}
 F^{-1}_{w}(e^{-|w|t})(\xi, t)
=\frac{1}{\sqrt{2 \pi}}\frac{t^2}{t^2+\xi^2}.
$$
It follows that 
\[
\frac{1}{F^{-1}(d(t, w))(\xi, t)}=\sqrt{2 \pi}
\frac{t^2+\xi^2}{t^2}=\sqrt{2 \pi}\big(1+\frac{\xi^2}{t^2}\big).
\]
Therefore, denoting $u(x, t)=E_{t}(f)(x)$, by the  method used at the above 
points, we arrive at the PDE
$$
\frac{\partial}{\partial t}\Big(\hat{u}(\xi, t) \big(1+\frac{\xi^2}{t^2}\big)\Big)
= \frac{\hat{u}}{\partial t}(\xi, t)\big(1+\frac{\xi^2}{t^2}\big)
+\hat{u}(\xi, t)\big(-\frac{2\xi^2}{t^{3}}\big)=0.
$$
This leads to the following PDE, satisfied by $u(x, t)=E_{t}(f)(x)$,
$$
\frac{\partial u}{\partial t}(x, t)
=\frac{1}{t^2} \frac{\partial^{3} u}{\partial x^2 \partial t}(x, t)-
\frac{2}{t^{3}} \frac{\partial^2 u}{\partial x^2}(x, t).
$$
Since $E_{t}(f)(x)=P_{1/t}(f)(x)$, it follows that 
$\lim_{t\nearrow \infty}E_{t}(f)(x)=f(x)$, for all $x\in \mathbb{R}$.

Following the above steps in inverse order, we arrive at the conclusion 
in the statement.

(iv) Concerning the approximation properties of $P_{n, t}(f)(x)$, 
the following estimate was obtained in \cite{Gal1},
$$
|f(x)-P_{n, t}(f)(x)|\le \sum_{k=1}^{n+1} k! 
 \binom{n+1}{k}\cdot \omega_{n+1}(f ; t)_{\mathbb{R}},
$$
where $\omega_{n+1}(f ; \delta)=\sup_{0\le h\le \delta}\{|\Delta_{h}^{n+1}f(x) ; 
x\in \mathbb{R}\}$, with
\[
\Delta_{h}^{n+1}=\sum_{j=0}^{n+1}(-1)^{n+1-j}\binom{n+1}{j}f(x+j h). 
\]
This  implies $\lim_{t\searrow 0}P_{n, t}(f)(x)=f(x)$, for all $x\in \mathbb{R}$.

To deduce the PDE satisfied by $P_{n, t}(f)(x)$, since $F^{-1}_{w}$ 
is linear operator, we use a calculation
(or WolframAlpha software),
$F^{-1}_{w}(e^{-|w|/t})(\xi, t)=\frac{\sqrt{2}}{\sqrt{\pi}}\frac{t}{t^2\xi^2+1}$, 
replacing here $t$ by $k t$, we easily obtain
\begin{align*}
&F^{-1}_{w}\Big[\sum_{k=1}^{n+1}(-1)^{k+1}\binom{n+1}{k}
 \frac{1}{2 t k}e^{-|w|/(k t)}\Big](\xi, t)\\
&=\frac{1}{\sqrt{2 \pi}} \sum_{k=1}^{n+1}(-1)^{k+1}\binom{n+1}{k}
 \frac{1}{k t}\frac{k t}{k^2t^2\xi^2+1} \\
&=\frac{1}{\sqrt{2 \pi}}\cdot \sum_{k=1}^{n+1}(-1)^{k+1}\binom{n+1}{k}
 \frac{1}{k^2t^2\xi^2+1}.
\end{align*}
Therefore, denoting $u(x, t):=P_{n, t}(f)(x)$ we obtain the differential equation
$$
\frac{\partial }{\partial t}\Big[\hat{u}(\xi, t)
 \frac{1}{\sum_{k=1}^{n+1}(-1)^{k+1}\binom{n+1}{k}
 \frac{1}{k^2t^2\xi^2+1}}\Big]=0,
$$
which is equivalent to
$$
\sum_{k=1}^{n+1}(-1)^{k+1}\binom{n+1}{k}
\Big[\frac{\partial \hat{u}}{\partial t}(\xi, t)
 \frac{1}{k^2t^2\xi^2+1}+\hat{u}(\xi, t)
 \frac{2 k^2 t \xi^2}{(k^2t^2\xi^2+1)^2}\Big]=0.
$$
It is worth noting that letting
$$
u_{k}(x, t)=P_{k t}(f)(x)
=\frac{1}{2 k t}\int_{-\infty}^{+\infty}f(x-u)e^{-|u|/(k t)}d u,
$$
we can write
$$
P_{n, t}(f)(x)=\sum_{k=1}^{n+1}(-1)^{k+1}\binom{n+1}{k}\cdot u_{k}(x, t),
$$
where reasoning as above for $P_{t}(f)(x)$, we easily obtain
$$
\frac{\partial \widehat{u_{k}}}{\partial t}(\xi, t)
=k^2t^2\frac{\partial^{3} \widehat{u_{k}}}{\partial x^2\partial t}(\xi, t)
+2k^2t\frac{\partial^2\widehat{u_{k}}}{\partial x^2}(\xi, t)
$$
and which implies
$$
\frac{\partial u_{k}}{\partial t}(x, t)
=k^2t^2\frac{\partial^{3} u_{k}}{\partial x^2\partial t}(x, t)+2k^2t
 \frac{\partial^2 u_{k}}{\partial x^2}(x, t), \quad x\in \mathbb{R},\;
 t>0,\; k=1, \dots, n+1,
$$
with $u_{k}(x, 0)=f(x)$, for all $x\in \mathbb{R}$, $k=1, \dots, n+1$.

(v) Concerning the approximation properties of $W_{n, t}(f)(x)$, reasoning 
as in \cite{Gal1}, we obtain the estimate
$$
|f(x)-W_{n, t}(f)(x)|\le C_{n} \omega_{n+1}(f ; \sqrt{t})_{\mathbb{R}},
$$
where $C_{n}>0$ is a constant independent of $f$, $t$ and $x$. 
This immediately implies that $\lim_{t\searrow 0}W_{n, t}(f)(x)=f(x)$, 
for all $x\in \mathbb{R}$.

To deduce the PDE satisfied by $W_{n, t}(f)(x)$, 
since $F^{-1}_{w}$ is linear operator, using  WolframAlpha software
 we have 
$F^{-1}_{w}(e^{-w^2/t})(\xi, t)=\sqrt{t} e^{-t \xi^2/4}/\sqrt{2}$. 
Replacing  $t$ by $k t$, we easily obtain 
$F^{-1}_{w}(e^{-w^2/(k t)})(\xi, t)
=\sqrt{k t} e^{-k t\xi^2/4}/\sqrt{2}$ and
\begin{align*}
&F^{-1}_{w}\Big[\sum_{k=1}^{n+1}(-1)^{k+1}\binom{n+1}{k}
 \frac{1}{\sqrt{\pi t} k}e^{-w^2/(k t)}\Big](\xi, t) \\
&=\frac{1}{\sqrt{\pi}}\cdot \sum_{k=1}^{n+1}(-1)^{k+1}\binom{n+1}{k}
 \frac{1}{k \sqrt{t}} \frac{\sqrt{k t} e^{-kt\xi^2/4}}{\sqrt{2}} \\
&=\frac{1}{\sqrt{2 \pi}}\cdot \sum_{k=1}^{n+1}(-1)^{k+1}\binom{n+1}{k}
 \frac{1}{\sqrt{k}}\cdot e^{-k t\xi^2/4}.
\end{align*}
Therefore, denoting $u(x, t):=W_{n, t}(f)(x)$ we obtain the differential equation
$$
\frac{\partial }{\partial t}
\Big[\hat{u}(\xi, t)\frac{1}{\sum_{k=1}^{n+1}(-1)^{k+1}
\binom{n+1}{k}\frac{1}{\sqrt{k}} e^{-kt\xi^2/4}}\Big]=0,
$$
which is equivalent to
$$
\sum_{k=1}^{n+1}(-1)^{k+1}\binom{n+1}{k}
\frac{1}{\sqrt{k}}\Big[\frac{\partial \hat{u}}{\partial t}(\xi, t)
 e^{-kt\xi^2/4}+\hat{u}(\xi, t)\frac{k \xi^2}{4} e^{-kt\xi^2/4}\Big]=0.
$$
It is worth noting that letting
$$
u_{k}(x, t)=\frac{1}{\sqrt{k}}W_{k t}(f)(x)
=\frac{1}{k \sqrt{\pi t}}\int_{-\infty}^{+\infty}f(x-u)e^{-u^2/(k t)}d u,
$$
we can write
$$
W_{n, t}(f)(x)=\sum_{k=1}^{n+1}(-1)^{k+1}\binom{n+1}{k}\cdot u_{k}(x, t).
$$
Reasoning as above but for $W_{t}(f)(x)$, we easily obtain
$$
\frac{\partial \widehat{u_{k}}}{\partial t}(\xi, t)
=\frac{k}{4}\frac{\partial^2 \widehat{u_{k}}}{\partial x^2}(\xi, t)
$$
 which implies
$$
\frac{\partial u_{k}}{\partial t}(x, t)
=\frac{k}{4}\frac{\partial^2 u_{k}}{\partial x^2}(x, t), \quad
 x\in \mathbb{R},\; t>0,\; k=1, \dots, n+1,
$$
with $u_{k}(x, 0)=f(x)$, for all $x\in \mathbb{R}$, $k=1, \dots, n+1$. 
\end{proof}

The methods in this paper could be used to make analogous studies for 
the convolutions with other known probability densities, like the Rayleigh 
probability density (abbreviated p.d.), Gumbel p.d., logistic p.d., 
Johnson p.d., Fr\'echet p.d., Gompetz p.d., L\'evy p.d., Lomax p.d. and so on.

It would be also of interest to use the methods in this paper for 
 complex convolutions, based on the ideas and results in the books 
\cite{Gal2,GGG}.


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\end{document}